Engineering Mechanics - Statics
Chapter 1
Problem 1-1 Represent each of the following combinations of units in the cor...

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Engineering Mechanics - Statics

Chapter 1

Problem 1-1 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m/ms (b) μkm (c) ks/mg (d) km⋅ μN Units Used: μN = 10

−6

μkm = 10

N

−6

km

9

Gs = 10 s 3

ks = 10 s mN = 10

−3

−3

ms = 10

N s

Solution: ( a)

m 3m = 1 × 10 ms s m km =1 ms s

( b)

μkm = 1 × 10

−3

m

μkm = 1 mm ( c)

ks 9 s = 1 × 10 mg kg ks Gs =1 mg kg

( d)

−3

km⋅ μN = 1 × 10

mN

km⋅ μN = 1 mm⋅ N

1

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-2 Wood has a density d. What is its density expressed in SI units? Units Used: Mg = 1000 kg Given: d = 4.70

slug ft

3

Solution: 1slug = 14.594 kg d = 2.42

Mg m

3

Problem 1-3 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm (b) mN/μs (c) μm⋅ Mg Solution: 3

( a)

6

Mg 10 kg 10 kg Gg = = = −3 mm m m 10 m Mg Gg = mm m

( b)

mN μs mN μs

( c)

=

10

−3

10 =

N

−6

3

=

s

10 N kN = s s

kN s

(

−6

μm⋅ Mg = 10

)(

3

)

m 10 kg = 10

−3

m⋅ kg

μm⋅ Mg = mm⋅ kg

2

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-4 Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/mm, (c) mN/( kg⋅ μs). Solution: 3

( a)

6

Mg 10 kg 10 kg Gg = = = −3 ms s s 10 s Mg Gg = ms s N = mm

( b)

1N 10

−3

= 10

3N

m

m

=

kN m

N kN = mm m mN

( c)

kg⋅ μs

10

=

mN kg⋅ μs

−3

−6

10 =

N

=

kg⋅ s

kN kg⋅ s

kN kg⋅ s

Problem 1-5 Represent each of the following with SI units having an appropriate prefix: (a) S1, (b) S2, (c) S3. Units Used: kg = 1000 g

−3

ms = 10

s

3

kN = 10 N

Given: S1 = 8653 ms S2 = 8368 N S3 = 0.893 kg Solution: ( a)

S1 = 8.653 s

3

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Engineering Mechanics - Statics

( b)

S2 = 8.368 kN

( c)

S3 = 893 g

Chapter 1

Problem 1-6 Represent each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, and (c) z. Units Used: 6

MN = 10 N μg = 1 × 10

−6

gm

3

kN = 10 N Given: x = 45320 kN

(

y = 568 × 10

5

) mm

z = 0.00563 mg Solution: ( a)

x = 45.3 MN

( b)

y = 56.8 km

( c)

z = 5.63 μg

Problem 1-7 Evaluate ( a⋅ b)/c to three significant figures and express the answer in SI units using an appropriate prefix. Units Used: μm = 10

−6

m

Given: a = ( 204 mm) 4

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

b = ( 0.00457 kg) c = ( 34.6 N) Solution: l =

ab

l = 26.945

c

μm⋅ kg N

Problem 1-8 If a car is traveling at speed v, determine its speed in kilometers per hour and meters per second. Given: v = 55

mi hr

Solution: v = 88.514

km hr

m

v = 24.6

s

Problem 1-9 Convert: (a) S1 to N ⋅ m , (b) S2 to kN/m3, (c) S3 to mm/s. Express the result to three significant figures. Use an appropriate prefix. Units Used: 3

kN = 10 N Given: S1 = 200g lb⋅ ft S2 = 350g

lb ft

S3 = 8

3

ft hr

5

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Solution: ( a)

S1 = 271 N⋅ m

( b)

S2 = 55.0

kN m

S3 = 0.677

( c)

3

mm s

Problem 1-10 What is the weight in newtons of an object that has a mass of: (a) m1, (b) m2, (c) m3? Express the result to three significant figures. Use an appropriate prefix. Units Used: 3

Mg = 10 kg mN = 10

−3

N

3

kN = 10 N Given: m1 = 10 kg m2 = 0.5 gm m3 = 4.50 Mg Solution: ( a)

W = m1 g W = 98.1 N

( b)

W = m2 g W = 4.90 mN

( c)

W = m3 g W = 44.1 kN

6

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-11 If an object has mass m, determine its mass in kilograms. Given: m = 40 slug Solution: m = 584 kg

Problem 1-12 The specific weight (wt./vol.) of brass is ρ. Determine its density (mass/vol.) in SI units. Use an appropriate prefix. Units Used: 3

Mg = 10 kg Given: lb

ρ = 520

ft

3

Solution: Mg

ρ = 8.33

m

3

Problem 1-13 A concrete column has diameter d and length L. If the density (mass/volume) of concrete is ρ, determine the weight of the column in pounds. Units Used: 3

Mg = 10 kg 3

kip = 10 lb Given: d = 350 mm L = 2m

7

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Engineering Mechanics - Statics

ρ = 2.45

Chapter 1

Mg m

3

Solution: 2

d V = π ⎛⎜ ⎟⎞ L ⎝ 2⎠ W = ρ V

V = 192.423 L W = 1.04 kip

Problem 1-14 The density (mass/volume) of aluminum is ρ. Determine its density in SI units. Use an appropriate prefix. Units Used: Mg = 1000 kg Given:

ρ = 5.26

slug ft

3

Solution:

ρ = 2.17

Mg m

3

Problem 1-15 Determine your own mass in kilograms, your weight in newtons, and your height in meters. Solution: Example W = 150 lb m = W

m = 68.039 kg W g = 667.233 N

h = 72 in

h = 1.829 m

8

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-16 Two particles have masses m1 and m2, respectively. If they are a distance d apart, determine the force of gravity acting between them. Compare this result with the weight of each particle. Units Used: G = 66.73 × 10

− 12

m

3 2

kg⋅ s nN = 10

−9

N

Given: m1 = 8 kg m2 = 12 kg d = 800 mm Solution: F =

G m1 m2 d

2

F = 10.0 nN W1 = m1 g

W1 = 78.5 N

W2 = m2 g

W2 = 118 N

W1 F

W2 F

= 7.85 × 10

9

= 1.18 × 10

10

Problem 1-17 Using the base units of the SI system, show that F = G(m1m2)/r2 is a dimensionally homogeneous equation which gives F in newtons. Compute the gravitational force acting between two identical spheres that are touching each other. The mass of each sphere is m1, and the radius is r. Units Used: μN = 10

−6

N

G = 66.73 10

− 12

m

3 2

kg⋅ s

9

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Given: m1 = 150 kg r = 275 mm Solution: F =

G m1 ( 2r)

2

2

F = 4.96 μN Since the force F is measured in Newtons, then the equation is dimensionally homogeneous.

Problem 1-18 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, (c) z. Units Used: 6

MN = 10 N 3

kN = 10 N μm = 10

−6

m

Given: x = ( 200 kN)

2

y = ( 0.005 mm) z = ( 400 m)

2

3

Solution: 2

( a)

x = 0.040 MN

( b)

y = 25.0 μm

( c)

z = 0.0640 km

2

3

10

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-19 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) a 1/b1, (b) a2b2/c2, (c) a3b3. Units Used: μm = 10

−6

6

m

Mm = 10 m

Mg = 10 gm

kg = 10 gm

6

−3

ms = 10

3

s

Given: a1 = 684 μm b1 = 43 ms a2 = 28 ms b2 = 0.0458 Mm c2 = 348 mg a3 = 2.68 mm b3 = 426 Mg Solution: a1

( a)

b1

= 15.9

a2 b2

( b)

c2

mm s

= 3.69 Mm

s kg

a3 b3 = 1.14 km⋅ kg

( c)

Problem 1-20 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) a1/b12 (b) a22b23. Units Used: 6

Mm = 10 m

11

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Given: a1 = 0.631 Mm b1 = 8.60 kg a2 = 35 mm b2 = 48 kg Solution:

( a)

a1 b1

( b)

2

2

= 8.532

km kg

3

2

3

a2 b2 = 135.48 kg ⋅ m

2

12

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-1 Determine the magnitude of the resultant force FR = F1 + F 2 and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 600 N F 2 = 800 N F 3 = 450 N

α = 45 deg β = 60 deg γ = 75 deg

Solution:

ψ = 90 deg − β + α FR =

F1 + F 2 − 2 F1 F 2 cos ( ψ) 2

2

F R = 867 N FR

sin ( ψ)

=

F2

sin ( θ )

⎛ ⎝

θ = asin ⎜F 2 θ = 63.05 deg

sin ( ψ) ⎞ ⎟ FR ⎠

φ = θ+α φ = 108 deg

Problem 2-2 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 80 lb F 2 = 60 lb 13

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Engineering Mechanics - Statics

Chapter 2

θ = 120 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 180 deg − θ ) 2

2

F R = 72.1 lb

⎛ ⎝

β = asin ⎜ F1

sin ( 180 deg − θ ) ⎞ ⎟ FR ⎠

β = 73.9 deg

Problem 2-3 Determine the magnitude of the resultant force F R = F1 + F 2 and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 250 lb F 2 = 375 lb

θ = 30 deg φ = 45 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 90 deg + θ − φ ) 2

2

F R = 178 kg FR

sin ( 90 deg + θ − φ )

⎛ F1

β = asin ⎜

⎝ FR

=

F1

sin ( β )

⎞

sin ( 90 deg + θ − φ )⎟

⎠

β = 37.89 deg 14

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Engineering Mechanics - Statics

Chapter 2

Angle measured ccw from x axis 360 deg − φ + β = 353 deg

Problem 2-4 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive u axis. Given: F 1 = 300 N F 2 = 500 N

α = 30 deg β = 45 deg γ = 70 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 180 deg − β − γ + α ) 2

2

F R = 605 N FR

sin ( 180 deg − β − γ + α )

⎛ ⎝

θ = asin ⎜F 2

=

F2

sin ( θ )

sin ( 180 deg − β − γ + α ) ⎞ ⎟ FR ⎠

θ = 55.40 deg φ = θ+α φ = 85.4 deg Problem 2-5 Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 300 N

α = 30 deg 15

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Engineering Mechanics - Statics

F 2 = 500 N

Chapter 2

β = 45 deg

γ = 70 deg Solution: F1u

=

sin ( γ − α )

F1

sin ( 180 deg − γ ) sin ( γ − α )

F 1u = F1

sin ( 180 deg − γ )

F 1u = 205 N F 1v

sin ( α )

=

F 1v = F 1

F1

sin ( 180 deg − γ ) sin ( α )

sin ( 180 deg − γ )

F 1v = 160 N

Problem 2-6 Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 300 N F 2 = 500 N

α = 30 deg β = 45 deg γ = 70 deg Solution:

⎛ sin ( β ) ⎞ ⎟ ⎝ sin ( γ ) ⎠

F 2u = F2 ⎜

16

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Engineering Mechanics - Statics

Chapter 2

F 2u = 376.2 N

⎡ sin ⎡⎣180 deg − ( β + γ )⎤⎦⎤ ⎥ sin ( γ ) ⎣ ⎦

F 2v = F 2 ⎢

F 2v = 482.2 N

Problem 2-7 Determine the magnitude of the resultant force F R = F 1 + F 2 and its direction measured counterclockwise from the positive u axis. Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg Solution:

α = 180 deg − ( θ 3 + θ 1 ) FR =

F2 + F 1 − 2 F1 F 2 cos ( α ) 2

2

F R = 61.4 lb sin ( θ' ) sin ( α ) = F2 FR

⎛

F2 ⎞

⎝

FR ⎠

θ' = asin ⎜ sin ( α )

⎟

θ' = 51.8 deg θ = θ' − θ 3 θ = 6.8 deg

Problem 2-8 Resolve the force F 1 into components acting along the u and v axes and determine the components. 17

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg Solution: −F u

=

sin ( θ 3 − θ 2 )

Fu =

F1

sin ( θ 2 )

−F1 sin ( θ 3 − θ 2 ) sin ( θ 2 )

F u = −12.9 lb Fv

sin ( 180 deg − θ 3 )

Fv =

=

F1

sin ( θ 2 )

F 1 sin ( 180 deg − θ 3 ) sin ( θ 2 )

F v = 35.4 lb

Problem 2-9 Resolve the force F2 into components acting along the u and v axes and determine the components. Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg 18

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

θ 2 = 30 deg θ 3 = 45 deg Solution: Fu

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

=

F2

sin ( θ 2 )

F2 sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fu =

sin ( θ 2 )

F u = 86.6 lb −Fv

=

sin ( θ 1 ) Fv =

F2

sin ( θ 2 )

−F 2 sin ( θ 1 ) sin ( θ 2 )

F v = −50 lb

Problem 2-10 Determine the components of the F force acting along the u and v axes. Given:

θ 1 = 70 deg θ 2 = 45 deg θ 3 = 60 deg F = 250 N Solution: Fu

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fu =

=

F sin ( θ 2 )

F sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ sin ( θ 2 )

F u = 320 N

19

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Engineering Mechanics - Statics

Fv

sin ( θ 1 )

Fv =

=

Chapter 2

F sin ( θ 2 )

F sin ( θ 1 )

F v = 332 N

sin ( θ 2 )

Problem 2-11 The force F acts on the gear tooth. Resolve this force into two components acting along the lines aa and bb. Given: F = 20 lb

θ 1 = 80 deg θ 2 = 60 deg Solution: F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fa =

Fa

sin ( θ 1 )

F sin ( θ 1 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fb =

=

F sin ( θ 2 )

=

F a = 30.6 lb

Fb

sin ( θ 2 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

F b = 26.9 lb

Problem 2-12 The component of force F acting along line aa is required to be Fa. Determine the magnitude of F and its component along line bb.

20

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F a = 30 lb

θ 1 = 80 deg θ 2 = 60 deg

Solution: Fa

=

sin ( θ 1 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

⎛ sin ( 180 deg − θ 1 − θ 2) ⎞ ⎜ ⎟ sin ( θ 1 ) ⎝ ⎠

F = Fa

Fa

=

sin ( θ 1 ) Fb =

F

F = 19.6 lb

Fb

sin ( θ 2 )

Fa sin ( θ 2 ) sin ( θ 1 )

F b = 26.4 lb

Problem 2-13 A resultant force F is necessary to hold the ballon in place. Resolve this force into components along the tether lines AB and AC, and compute the magnitude of each component.

Given: F = 350 lb

θ 1 = 30 deg θ 2 = 40 deg Solution: F AB

sin ( θ 1 )

=

F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

21

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

sin ( θ 1 ) ⎡ ⎤ ⎢ ⎥ ⎣ sin ⎡⎣180 deg − ( θ 1 + θ 2)⎤⎦⎦

F AB = F

F AB = 186 lb F AC

sin ( θ 2 )

F

=

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ sin ( θ 2 ) ⎡ ⎤ ⎢ ⎥ ⎣sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ ⎦

F AC = F

F AC = 239 lb

Problem 2-14 The post is to be pulled out of the ground using two ropes A and B. Rope A is subjected to force F 1 and is directed at angle θ1 from the horizontal. If the resultant force acting on the post is to be F R, vertically upward, determine the force T in rope B and the corresponding angle θ. Given: F R = 1200 lb F 1 = 600 lb

θ 1 = 60 deg Solution: T =

F 1 + FR − 2 F1 F R cos ( 90 deg − θ 1 ) 2

2

T = 744 lb sin ( 90 − θ 1 ) sin ( θ ) = FR T

⎛ ⎝

θ = asin ⎜ sin ( 90 deg − θ 1 )

F1 ⎞ T

⎟ ⎠

θ = 23.8 deg

22

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-15 Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 250 N F 2 = 150 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 105 deg Solution: F 1v

sin ( θ 1 )

=

F1

sin ( θ 3 )

⎛ sin ( θ 1 ) ⎞ ⎜ ⎟ ⎝ sin ( θ 3 ) ⎠

F 1v = F 1

F 1v = 129 N F 1u

sin ( 180 deg − θ 1 − θ 3 ) F 1u = F1

=

F1

sin ( θ 3 )

⎛ sin ( 180 deg − θ 1 − θ 3 ) ⎞ ⎜ ⎟ sin ( θ 3 ) ⎝ ⎠

F 1u = 183 N

Problem 2-16 Resolve the force F 2 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 250 N 23

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 2 = 150 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 105 deg Solution: F 1v

sin ( θ 1 )

F2

=

F 1v = F 2

sin ( 180 deg − θ 3 ) sin ( θ 1 ) ⎛ ⎞ ⎜ ⎟ ⎝ sin ( 180 deg − θ 3 ) ⎠

F 1v = 77.6 N F2u

sin ( 180 deg − θ 3 ) F 2u = F2

=

F2

sin ( 180 deg − θ 3 )

⎛ sin ( 180 deg − θ 3) ⎞ ⎜ ⎟ ⎝ sin ( 180 deg − θ 3) ⎠

F 2u = 150 N

Problem 2-17 Determine the magnitude and direction of the resultant force F R. Express the result in terms of the magnitudes of the components F 1 and F 2 and the angle φ.

Solution: F R = F 1 + F2 − 2F1 F 2 cos ( 180 deg − φ ) 2

2

2

24

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Since cos ( 180 deg − φ ) = −cos ( φ ), F 1 + F2 + 2 F 1 F2 cos ( φ ) 2

FR =

2

From the figure, tan ( θ ) =

F1 sin ( φ )

F 2 + F 1 cos ( φ )

Problem 2-18 If the tension in the cable is F1, determine the magnitude and direction of the resultant force acting on the pulley. This angle defines the same angle θ of line AB on the tailboard block. Given: F 1 = 400 N

θ 1 = 30 deg

Solution: FR =

F1 + F 1 − 2F 1 F 1 cos ( 90 deg − θ 1 ) 2

2

F R = 400 N sin ( θ 1 ) sin ( 90 deg − θ ) = FR F1

⎛ FR

θ = 90 deg − asin ⎜

⎝ F1

⎞

sin ( θ 1 )⎟

⎠

θ = 60 deg

25

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-19 The riveted bracket supports two forces. Determine the angle θ so that the resultant force is directed along the negative x axis. What is the magnitude of this resultant force? Given: F 1 = 60 lb F 2 = 70 lb

θ 1 = 30 deg Solution: sin ( θ 1 ) sin ( θ ) = F1 F2

⎛

F1 ⎞

⎝

F2 ⎠

θ = asin ⎜ sin ( θ 1 )

⎟

θ = 25.4 deg φ = 180 deg − θ − θ 1 φ = 124.6 deg F1 + F 2 − 2F 1 F 2 cos ( φ ) 2

R =

2

R = 115 lb

Problem 2-20 The plate is subjected to the forces acting on members A and B as shown. Determine the magnitude of the resultant of these forces and its direction measured clockwise from the positive x axis.

Given: F A = 400 lb F B = 500 lb

26

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

θ 1 = 30 deg θ = 60 deg Solution: Cosine law: FR =

FB + FA − 2F B FA cos ( 90 deg − θ + θ 1 ) 2

2

F R = 458 lb Sine law: sin ( 90 deg − θ + θ 1 ) FR

=

sin ( θ − α ) FA

⎛

FA ⎞

⎝

FR ⎠

α = θ − asin ⎜ sin ( 90 deg − θ + θ 1 )

⎟

α = 10.9 deg

Problem 2-21 Determine the angle θ for connecting member B to the plate so that the resultant of FA and FB is directed along the positive x axis. What is the magnitude of the resultant force? Given: F A = 400 lb F B = 500 lb

θ 1 = 30 deg 27

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution: Sine law: sin ( 90 deg − θ 1 ) sin ( θ ) = FA FB

⎛

FA ⎞

⎝

FB ⎠

θ = asin ⎜ sin ( 90 deg − θ 1 )

⎟

θ = 43.9 deg FR

sin ( 90 deg + θ 1 − θ )

FR = FA

=

FA

sin ( θ )

sin ( 90 deg − θ + θ 1 ) sin ( θ )

F R = 561 lb

Problem 2-22 Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by first finding the resultant F' = F1 + F 2 and then forming F R = F' + F 3. Given: F 1 = 30 N F 2 = 20 N F 3 = 50 N

θ = 20 deg c = 3 d = 4

28

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution: c α = atan ⎛⎜ ⎟⎞

⎝ d⎠

F' =

F1 + F 2 − 2F 1 F 2 cos ( 90 deg + θ − α ) 2

2

F' = 30.9 N F'

sin ( ( 90 deg − θ + α ) )

=

⎛ ⎝

F1

sin ( 90 deg − θ − β )

β = 90 deg − θ − asin ⎜F 1

sin ( 90 deg − θ + α ) ⎞ ⎟ F' ⎠

β = 1.5 deg Now add in force F3. FR =

F' + F3 − 2F' F3 cos ( β ) 2

2

F R = 19.2 N FR

sin ( β )

F'

=

sin ( φ )

⎛ ⎝

φ = asin ⎜F'

sin ( β ) ⎞ ⎟ FR ⎠

φ = 2.4 deg

Problem 2-23 Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F' = F 2 + F 3 and then forming F R = F' + F 1. Given: F 1 = 30 N F 2 = 20 N 29

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 3 = 50 N

θ = 20 deg c = 3 d = 4 Solution: F' =

F2 + F 3 − 2F 2 F 3 cos ( ( 90 deg − θ ) ) 2

2

F' = 47.07 N F2

sin ( β )

F'

=

sin ( 90 deg − θ )

⎛ ⎝

β = asin ⎜ F2

sin ( 90 deg − θ ) ⎞ ⎟ F' ⎠

β = 23.53 deg c α = atan ⎛⎜ ⎟⎞

⎝ d⎠

γ = α−β FR =

F' + F1 − 2F' F 1 cos ( γ ) 2

2

F R = 19.2 N FR

sin ( γ )

=

F1

sin ( φ )

⎛ ⎝

φ = asin ⎜F 1

sin ( γ ) ⎞ ⎟ FR ⎠

φ = 21.16 deg ψ = β−φ ψ = 2.37 deg

30

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-24 Resolve the force F into components acting along (a) the x and y axes, and (b) the x and y' axes.

Given: F = 50 lb

α = 65 deg β = 45 deg γ = 30 deg Solution: (a)

F x = F cos ( β ) F x = 35.4 lb F y = F sin ( β ) F y = 35.4 lb

(b)

Fx

sin ( 90 deg − β − γ )

=

F

sin ( 90 deg + γ )

sin ( 90 deg − β − γ )

Fx = F

sin ( 90 deg + γ )

F x = 14.9 lb Fy'

sin ( β )

=

F y' = F

F

sin ( 90 deg + γ ) sin ( β )

sin ( 90 deg + γ )

31

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F y' = 40.8 lb

Problem 2-25 The boat is to be pulled onto the shore using two ropes. Determine the magnitudes of forces T and P acting in each rope in order to develop a resultant force F1, directed along the keel axis aa as shown. Given:

θ = 40 deg θ 1 = 30 deg F 1 = 80 lb

Solution: F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

T = F1

=

T

sin ( θ )

sin ( θ )

sin ( 180 deg − θ − θ 1 )

T = 54.7 lb F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ P = sin ( θ 1 )

=

P

sin ( θ 1 )

F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

P = 42.6 lb

32

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-26 The boat is to be pulled onto the shore using two ropes. If the resultant force is to be F1, directed along the keel aa as shown, determine the magnitudes of forces T and P acting in each rope and the angle θ of P so that the magnitude of P is a minimum. T acts at θ from the keel as shown.

Given:

θ 1 = 30 deg F 1 = 80 lb

Solution: From the figure, P is minimum when

θ + θ 1 = 90 deg θ = 90 deg − θ 1 θ = 60 deg P

sin ( θ 1 )

P =

F1

=

sin ( 90 deg)

F 1 sin ( θ 1 ) sin ( 90 deg)

P = 40 lb T

sin ( θ ) T = F1

=

F1 sin ( 90 deg)

sin ( θ ) sin ( 90 deg)

T = 69.3 lb

33

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-27 The beam is to be hoisted using two chains. Determine the magnitudes of forces FA and FB acting on each chain in order to develop a resultant force T directed along the positive y axis.

Given: T = 600 N

θ 1 = 30 deg θ = 45 deg

Solution: FA

sin ( θ ) FA =

T

=

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ T sin ( θ )

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

F A = 439 N

FB

sin ( θ 1 )

FB = T

=

T

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ sin ( θ 1 )

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

F B = 311 N

34

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-28 The beam is to be hoisted using two chains. If the resultant force is to be F, directed along the positive y axis, determine the magnitudes of forces FA and F B acting on each chain and the orientation θ of F B so that the magnitude of FB is a minimum. Given: F = 600 N

θ 1 = 30 deg Solution: For minimum FB, require

θ = 90 deg − θ 1 θ = 60 deg F A = F cos ( θ 1 ) F A = 520 N F B = F sin ( θ 1 ) F B = 300 N

Problem 2-29 Three chains act on the bracket such that they create a resultant force having magnitude F R. If two of the chains are subjected to known forces, as shown, determine the orientation θ of the third chain,measured clockwise from the positive x axis, so that the magnitude of force F in this chain is a minimum. All forces lie in the x-y plane.What is the magnitude of F ? Hint: First find the resultant of the two known forces. Force F acts in this direction.

35

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F R = 500 lb F 1 = 200 lb F 2 = 300 lb

φ = 30 deg

Solution: Cosine Law: F R1 =

F 1 + F2 − 2 F 1 F2 cos ( 90 deg − φ ) 2

2

F R1 = 264.6 lb Sine Law:

Make F parallel to FR1

sin ( φ + θ ) sin ( 90 deg − φ ) = F1 F R1

⎛

θ = −φ + asin ⎜ sin ( 90 deg − φ )

⎝

⎞ ⎟ FR1 ⎠ F1

θ = 10.9 deg When F is directed along F R1, F will be minimum to create the resultant forces. F = F R − FR1 F = 235 lb

Problem 2-30 Three cables pull on the pipe such that they create a resultant force having magnitude F R. If two of the cables are subjected to known forces, as shown in the figure, determine the direction θ of the third cable so that the magnitude of force F in this cable is a minimum. All forces lie in the 36

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

g x–y plane.What is the magnitude of F ? Hint: First find the resultant of the two known forces.

Given: F R = 900 lb F 1 = 600 lb F 2 = 400 lb

α = 45 deg β = 30 deg

Solution:

F' =

F1 + F 2 − 2F 1 F 2 cos ( 90 deg + α − β ) 2

2

F' = 802.64 lb F = F R − F' F = 97.4 lb sin ( φ ) F1

=

sin ( 90 deg + α − β ) F'

⎛ ⎝

φ = asin ⎜ sin ( 90 deg + α − β )

F1 ⎞ F'

⎟ ⎠

φ = 46.22 deg θ = φ−β θ = 16.2 deg

Problem 2-31 Determine the x and y components of the force F.

37

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F = 800 lb

α = 60 deg β = 40 deg

Solution: F x = F sin ( β ) F y = −F cos ( β ) F x = 514.2 lb F y = −612.8 lb

Problem 2-32 Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis. Given: F 1 = 70 N F 2 = 50 N F 3 = 65 N

θ = 30 deg φ = 45 deg Solution: + →

F Rx = ΣFx;

F RX = F 1 + F 2 cos ( θ ) − F 3 cos ( φ )

38

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑

FR =

F RY = −F2 sin ( θ ) − F 3 sin ( φ )

F Ry = ΣF y; 2

FRX + FRY

Chapter 2

2

⎛ FRY ⎞ ⎟ ⎝ FRX ⎠

θ = atan ⎜

F R = 97.8 N

θ = 46.5 deg

Problem 2-33 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 50 lb F 2 = 35 lb

α = 120 deg β = 25 deg

Solution: + F Rx = ΣF x; →

F Rx = F 1 sin ( α ) − F2 sin ( β ) F Rx = 28.5 lb

+

↑ FRy = ΣFy;

F Ry = −F 1 cos ( α ) − F 2 cos ( β )

39

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F Ry = −6.7 lb FR =

2

FRx + FRy

2

F R = 29.3 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ' = atan ⎜

θ' = 13.3 deg θ = 360 deg − θ' θ = 347 deg

Problem 2-34 Determine the magnitude of the resultant force and its direction , measured counterclockwise from the positive x axis.

Given: F 1 = 850 N F 2 = 625 N F 3 = 750 N

θ = 45 deg φ = 30 deg c = 3 d = 4

Solution: + →

F Rx = SF x;

F RX = F 1

d 2

c +d

2

− F 2 sin ( φ ) − F3 sin ( θ )

40

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑

F Ry = SF y;

F RY = −F1

c 2

2

− F2 cos ( φ ) + F3 cos ( θ )

c +d

F RX = −162.8 N FR =

Chapter 2

2

FRX + FRY

F RY = −520.9 N 2

F R = 546 N

⎛ FRY ⎞ ⎟ ⎝ FRX ⎠

α = atan ⎜

α = 72.64 deg β = α + 180 deg β = 252.6 deg

Problem 2-35 Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the resultant force is directed along the positive x' axis and has a magnitude of FR. Units Used: 3

kN = 10 N Given: F R = 1 kN F 2 = 450 N F 3 = 200 N

α = 45 deg β = 30 deg Solution: + →

F Rx = SF x;

F R cos ( β ) = F 3 + F 2 cos ( α ) + F 1 cos ( θ + β )

+

F Ry = SF y;

−F R sin ( β ) = F2 sin ( α ) − F 1 sin ( θ + β )

↑

41

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 1 cos ( θ + β ) = FR cos ( β ) − F3 − F2 cos ( α ) F 1 sin ( θ + β ) = F 2 sin ( α ) + FR sin ( β )

⎛

F2 sin ( α ) + F R sin ( β )

⎞ ⎟−β ⎝ FR cos ( β ) − F3 − F2 cos ( α ) ⎠

θ = atan ⎜

θ = 37 deg

(FR cos (β ) − F3 − F2 cos (α ))2 + (F2 sin (α ) + FR sin (β ))2

F1 =

F 1 = 889 N

Problem 2-36 Determine the magnitude and direction, measured counterclockwise from the x' axis, of the resultant force of the three forces acting on the bracket.

Given: F 1 = 300 N F 2 = 450 N F 3 = 200 N

α = 45 deg β = 30 deg θ = 20 deg Solution: F Rx = F 1 cos ( θ + β ) + F3 + F2 cos ( α )

F Rx = 711.03 N

F Ry = −F 1 sin ( θ + β ) + F 2 sin ( α )

F Ry = 88.38 N

42

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

FR =

2

FRx + FRy

Chapter 2

2

F R = 717 N

φ (angle from x axis)

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 7.1 deg

φ' (angle from x' axis) φ' = β + φ

φ' = 37.1 deg

Problem 2-37 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

Given: F 1 = 800 N F 2 = 600 N

θ = 40 deg c = 12 d = 5 Solution: + F Rx = ΣF x; →

F Rx = F 1 cos ( θ ) − F 2 ⎜

2

+

F Ry = F 1 sin ( θ ) + F2 ⎜

⎛

d

↑

F Ry = ΣF y;

⎛

c

⎞

2⎟

⎝ c +d ⎠ 2

⎞

2⎟

⎝ c +d ⎠

FR =

2

FRx + FRy

2

F Rx = 59 N

F Ry = 745 N

F R = 747 N

43

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 85.5 deg

Problem 2-38 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Units Used: 3

kN = 10 N Given: F 1 = 30 kN F 2 = 26 kN

θ = 30 deg c = 5 d = 12

Solution: + F Rx = ΣF x; →

F Rx = −F 1 sin ( θ ) −

c ⎛ ⎞ ⎜ 2 2 ⎟ F2 ⎝ c +d ⎠

F Rx = −25 kN

+

F Ry = −F 1 cos ( θ ) +

⎛ d ⎞F ⎜ 2 2⎟ 2 ⎝ c +d ⎠

F Ry = −2 kN

↑ FRy = ΣFy;

FR =

2

FRx + FRy

2

F R = 25.1 kN

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 4.5 deg

β = 180 deg + φ

β = 184.5 deg

44 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-39 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 60 lb F 2 = 70 lb F 3 = 50 lb

θ 1 = 60 deg θ 2 = 45 deg c = 1 d = 1 Solution: d θ 3 = atan ⎛⎜ ⎟⎞

⎝c⎠

F Rx = −F 1 cos ( θ 3 ) − F2 sin ( θ 1 )

F Rx = −103 lb

F Ry = F 1 sin ( θ 3 ) − F 2 cos ( θ 1 ) − F3

F Ry = −42.6 lb

FR =

2

FRx + FRy

2

F R = 111.5 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = 180 deg + atan ⎜

θ = 202 deg

Problem 2-40 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis by summing the rectangular or x, y components of the forces to obtain the resultant force. 45

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 600 N F 2 = 800 N F 3 = 450 N

θ 1 = 60 deg θ 2 = 45 deg θ 3 = 75 deg Solution: + F Rx = ΣF x; → +

↑ FRy = ΣFy;

F Rx = F 1 cos ( θ 2 ) − F2 sin ( θ 1 )

F Rx = −268.556 N

F Ry = F 1 sin ( θ 2 ) + F 2 cos ( θ 1 )

F Ry = 824.264 N

FR =

2

FRx + FRy

2

F R = 867 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = 180 deg − atan ⎜

θ = 108 deg

Problem 2-41 Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by summing the rectangular or x, y components of the forces to obtain the resultant force. Given: F 1 = 30 N F 2 = 20 N

46

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 3 = 50 N

θ = 20 deg c = 3 d = 4

Solution: F Rx = −F 1

F Ry = F 1

FR =

⎛ d ⎞ − F ( sin ( θ ) ) + F 2 3 ⎜ 2 2⎟ ⎝ c +d ⎠

F Rx = 19.2 N

c ⎛ ⎞ ⎜ 2 2 ⎟ − F2 cos ( θ ) ⎝ c +d ⎠ 2

FRx + FRy

F Ry = −0.8 N

2

F R = 19.2 N

⎛ −FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 2.4 deg

Problem 2-42 Determine the magnitude and orientation, measured counterclockwise from the positive y axis, of the resultant force acting on the bracket. Given: F A = 700 N

47

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F B = 600 N

θ = 20 deg φ = 30 deg

Solution: Scalar Notation: Suming the force components algebraically, we have F Rx = ΣF x;

F Rx = F A sin ( φ ) − F B cos ( θ ) F Rx = −213.8 N

F Ry = ΣF y;

F Ry = F A cos ( φ ) + FB sin ( θ ) F Ry = 811.4 N

The magnitude of the resultant force F R is 2

FR =

FRx + FRy

2

F R = 839 N The directional angle θ measured counterclockwise from the positive x axis is

⎛ FRx ⎞ ⎟ ⎝ FRy ⎠

θ = atan ⎜

θ = 14.8 deg

Problem 2-43 Determine the magnitude and direction, measured counterclockwise from the positive x' axis, of the resultant force of the three forces acting on the bracket. Given: F 1 = 300 N

48

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 2 = 200 N F 3 = 180 N

θ = 10 deg θ 1 = 60 deg c = 5 d = 12

Solution: + →

+

↑

F Rx = ΣF x;

F Ry = ΣF y;

F Rx = F 1 sin ( θ 1 + θ ) −

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

F Ry = F 1 cos ( θ 1 + θ ) + F2 +

FR =

2

FRx + FRy

2

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

F Rx = 115.8 N

F Ry = 371.8 N

F R = 389 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 72.7 deg

ψ = φ − ( 90 deg − θ 1 )

ψ = 42.7 deg

Problem 2-44 Determine the x and y components of F 1 and F 2.

49

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 200 N F 2 = 150 N

θ = 45 deg φ = 30 deg Solution: F 1x = F 1 sin ( θ ) F 1x = 141.4 N F 1y = F 1 cos ( θ ) F 1y = 141.4 N F 2x = −F 2 cos ( φ ) F 2x = −129.9 N F 2y = F 2 sin ( φ ) F 2y = 75 N

Problem 2-45 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 200 N F 2 = 150 N

θ = 45 deg φ = 30 deg

50

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Engineering Mechanics - Statics

Chapter 2

Solution: + F Rx = ΣF x; → +

↑ FRy = ΣFy;

F Rx = F 1 sin ( θ ) − F2 cos ( φ )

F Rx = 11.5 N

F Ry = F 1 cos ( θ ) + F 2 sin ( φ )

F Ry = 216.4 N

F =

2

FRx + FRy

2

F = 217 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

β = atan ⎜

β = 87 deg

Problem 2-46 Determine the x and y components of each force acting on the gusset plate of the bridge truss. Given: F 1 = 200 lb

c = 3

F 2 = 400 lb

d = 4

F 3 = 300 lb

e = 3

F 4 = 300 lb

f = 4

Solution: F 1x = −F 1 F 1x = −200 lb F 1y = 0 lb F 2x = F 2 ⎛ ⎜

d 2

⎞

2⎟

⎝ c +d ⎠

F 2x = 320 lb F 2y = −F 2 ⎛

c ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠

F 2y = −240 lb F 3x = F 3 ⎛ ⎜

e 2

⎞

2⎟

⎝ e +f ⎠ 51

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 3x = 180 lb F 3y = F 3 ⎛ ⎜

f 2

⎞

2⎟

⎝ e +f ⎠

F 3y = 240 lb F 4x = −F 4 F 4x = −300 lb F 4y = 0 lb

Problem 2-47 Determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis. Units Used: 3

kN = 10 N

Given: F 1 = 20 kN F 2 = 40 kN F 3 = 50 kN

θ = 60 deg c = 1 d = 1 e = 3 f = 4 Solution: + F Rx = ΣF x; →

F Rx = F 3 ⎛ ⎜

⎞ + F ⎛ d ⎞ − F cos ( θ ) 2⎜ 1 2 2⎟ 2 2⎟ ⎝ e +f ⎠ ⎝ c +d ⎠ f

F Rx = 58.28 kN

52

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Engineering Mechanics - Statics

+

Chapter 2

⎛

c ⎞−F ⎛ ⎞ 2 ⎜ 2 2 ⎟ − F1 sin ( θ ) 2 2⎟ ⎝ e +f ⎠ ⎝ c +d ⎠

↑ FRy = ΣFy;

F Ry = F 3 ⎜

e

F Ry = −15.6 kN F =

2

FRx + FRy

2

F = 60.3 kN

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 15 deg

Problem 2-48 Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the resultant force is directed along the positive x' axis and has magnitude FR. Given: F 2 = 200 N F 3 = 180 N

θ 1 = 60 deg F R = 800 N c = 5 d = 12

Solution: Initial Guesses:

F 1 = 100 N

θ = 10 deg

Given + →

F Rx = ΣF x;

F R sin ( θ 1 ) = F1 sin ( θ 1 + θ ) −

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

53

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Engineering Mechanics - Statics

+

↑

F Ry = ΣF y;

Chapter 2

F R( cos ( θ 1 ) ) = F 1 cos ( θ 1 + θ ) + F2 +

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , θ ) ⎝θ ⎠ F 1 = 869 N

θ = 21.3 deg

Problem 2-49 Determine the magnitude and direction, measured counterclockwise from the positive x' axis, of the resultant force acting on the bracket. Given: F 1 = 300 N F 2 = 200 N F 3 = 180 N

θ 1 = 60 deg θ = 10 deg c = 5 d = 12

Solution: Guesses

F Rx = 100 N

F Ry = 100 N

Given + →

F Rx = ΣF x;

F Rx = F1 sin ( θ 1 + θ ) −

+

F Ry = ΣF y;

F Ry = F1 cos ( θ 1 + θ ) + F 2 +

↑

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠ c ⎛ ⎞ ⎜ 2 2 ⎟ ( F3 ) ⎝ c +d ⎠

54

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Engineering Mechanics - Statics

Chapter 2

⎛ FRx ⎞ ⎜ ⎟ = Find ( FRx , FRy) ⎝ FRy ⎠ FR =

2

FRx + FRy

2

⎛ FRx ⎞ ⎛ 115.8 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FRy ⎠ ⎝ 371.8 ⎠ F R = 389 N

φ = atan ⎜

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = 72.7 deg

φ' = ⎡⎣φ − ( 90 deg − θ 1 )⎤⎦

φ' = 42.7 deg

Problem 2-50 Express each of the three forces acting on the column in Cartesian vector form and compute the magnitude of the resultant force. Given: F 1 = 150 lb

θ = 60 deg

F 2 = 275 lb

c = 4

F 3 = 75 lb

d = 3

Solution: Find the components of each force.

⎛

⎞ ⎟ 2 2 ⎝ c +d ⎠

F 1x = F 1 ⎜

F 1v =

d

⎛ F1x ⎞ ⎜ ⎟ ⎝ F1y ⎠

F 2x = 0 lb

F 2v =

⎛ F2x ⎞ ⎜ ⎟ ⎝ F2y ⎠

⎛

⎞ ⎟ 2 2 ⎝ c +d ⎠

F 1y = F 1 ⎜

F 1v =

−c

⎛ 90 ⎞ ⎜ ⎟ lb ⎝ −120 ⎠

F 2y = −F 2

F 2v =

⎛ 0 ⎞ ⎜ ⎟ lb ⎝ −275 ⎠ 55

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F 3x = −F 3 cos ( θ )

F 3v =

Chapter 2

F 3y = −F 3 sin ( θ )

⎛ F3x ⎞ ⎜ ⎟ ⎝ F3y ⎠

F 3v =

⎛ −37.5 ⎞ ⎜ ⎟ lb ⎝ −65 ⎠

Now find the magnitude of the resultant force. F R = F 1v + F 2v + F 3v

F R = 462.9 lb

Problem 2-51 Determine the magnitude of force F so that the resultant F R of the three forces is as small as possible. What is the minimum magnitude of FR? Units Used: kN = 1000 N Given: F 1 = 5 kN F 2 = 4 kN

θ = 30 deg Solution: Scalar Notation: Suming the force components algebrically, we have + → +

↑

F Rx = ΣF x;

F Rx = F1 − F sin ( θ )

F Ry = ΣF y;

F Ry = F cos ( θ ) − F 2

The magnitude of the resultant force F R is FR =

2

2

F Rx + F Ry =

(F1 − F sin(θ ))2 + (F cos (θ ) − F2)2 56

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F R = F 1 + F2 + F − 2FF1 sin ( θ ) − 2F2 Fcos( θ ) 2

2FR

2

dF R dF

2

= 2F − 2F1 sin ( θ ) − 2F2 cos ( θ )

If F is a minimum, then FR =

2

⎞ ⎛ dFR = 0⎟ ⎜ ⎝ dF ⎠

F = F 1 sin ( θ ) + F2 cos ( θ )

(F1 − F sin(θ ))2 + (F cos (θ ) − F2)2

F = 5.96 kN F R = 2.3 kN

Problem 2-52 Express each of the three forces acting on the bracket in Cartesian vector form with respect to the x and y axes. Determine the magnitude and direction θ of F1 so that the resultant force is directed along the positive x' axis and has magnitude F R. Units Used: kN = 1000 N Given: F R = 600 N F 2 = 350 N F 3 = 100 N

φ = 30 deg Solution: F 2v =

⎛ F2 ⎞ ⎜ ⎟ ⎝0 ⎠

F 3v =

⎛ 0 ⎞ ⎜ ⎟ ⎝ −F 3 ⎠

F 1v =

⎛ F1 cos ( θ ) ⎞ ⎜ ⎟ ⎝ F1 sin ( θ ) ⎠

F 2v =

⎛ 350 ⎞ ⎜ ⎟N ⎝ 0 ⎠

F 3v =

⎛ 0 ⎞ ⎜ ⎟N ⎝ −100 ⎠

57

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 1 = 20 N θ = 10 deg

The initial guesses: Given

⎛ F1 cos ( θ ) ⎞ ⎜ ⎟ + F2v + F3v = ( ) F sin θ 1 ⎝ ⎠ ⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , θ ) ⎝θ ⎠

⎛ FR cos ( φ ) ⎞ ⎜ ⎟ ⎝ FR sin ( φ ) ⎠

F 1 = 434.5 N

θ = 67 deg

Problem 2-53 The three concurrent forces acting on the post produce a resultant force F R = 0. If F 2 = (1/2)F 1, and F 1 is to be 90° from F 2 as shown, determine the required magnitude F 3 expressed in terms of F1 and the angle θ. Solution:

Use the primed coordiates.

ΣFRx = 0

F 3 cos ( θ − 90 deg) − F1 = 0

ΣFRy = 0

−F 3 sin ( θ − 90 deg) + F 2 = 0 tan ( θ − 90 deg) =

F2 F1

=

1 2

1 θ = 90 deg + atan ⎛⎜ ⎟⎞

⎝ 2⎠

θ = 117 deg k =

1

cos ( θ − 90 deg)

k = 1.1

F3 = k F1

58 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-54 Three forces act on the bracket. Determine the magnitude and orientation θ of F2 so that the resultant force is directed along the positive u axis and has magnitude F R. Given: F R = 50 lb F 1 = 80 lb F 3 = 52 lb

φ = 25 deg c = 12 d = 5

Solution: Guesses F 2 = 1 lb

θ = 120 deg

Given F R cos ( φ ) = F1 + F2 cos ( φ + θ ) +

−F R sin ( φ ) = −F 2 sin ( φ + θ ) +

⎛ F2 ⎞ ⎜ ⎟ = Find ( F2 , θ ) ⎝θ ⎠

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

θ = 103.3 deg

F 2 = 88.1 lb

59

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-55 Determine the magnitude and orientation, measured clockwise from the positive x axis, of the resultant force of the three forces acting on the bracket. Given: F 1 = 80 lb F 2 = 150 lb F 3 = 52lb

θ = 55 deg φ = 25 deg c = 12 m d = 5m

Solution:

⎛

⎞ + F cos ( θ + φ ) 2 ⎝ c +d ⎠

F Rx = F 1 + F 3 ⎜

⎛

d

2

F Rx = 126.05 lb

2⎟

⎞ − F sin ( θ + φ ) 2 ⎟ 2 2 ⎝ c +d ⎠ c

F Ry = F 3 ⎜

FR =

2

FRx + FRy

F Ry = −99.7 lb

2

F R = 161 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

β = atan ⎜

β = 38.3 deg

Problem 2-56 Three forces act on the ring. Determine the range of values for the magnitude of P so that the magnitude of the resultant force does not exceed F. Force P is always directed to the right.

60

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Units Used: 3

kN = 10 N Given: F = 2500 N F 1 = 1500 N F 2 = 600 N

θ 1 = 60 deg θ 2 = 45 deg Solution: Initial Guesses:

F Rx = 100 N

F Ry = 100 N

P = 100 N

Given + →

F Rx = ΣF x;

F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 )

+

F Ry = ΣF y;

F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 )

↑

F=

2

F Rx + F Ry

⎛ FRx ⎞ ⎜ ⎟ ⎜ FRy ⎟ = Find ( FRx , FRy , P) ⎜P ⎟ ⎝ max ⎠ Initial Guesses:

2

P max = 1.6 kN

F Rx = −100 N

F Ry = 100 N

P = −2000 N

Given + →

F Rx = ΣF x;

F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 )

+

F Ry = ΣF y;

F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 )

↑

F=

⎛ FRx ⎞ ⎜ ⎟ ⎜ FRy ⎟ = Find ( FRx , FRy , P) ⎜P ⎟ ⎝ min ⎠

2

F Rx + F Ry

2

P min = −1.7 kN

61

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Since P > 0

we conclude that

Chapter 2

0 <= P <= P max = 1.6 kN

Problem 2-57 Determine the magnitude and coordinate direction angles of F1 and F2. Sketch each force on an x, y, z reference. Given:

⎛ 60 ⎞ F 1 = ⎜ −50 ⎟ N ⎜ ⎟ ⎝ 40 ⎠ ⎛ −40 ⎞ F 2 = ⎜ −85 ⎟ N ⎜ ⎟ ⎝ 30 ⎠ Solution: F 1 = 87.7 N

⎛ α1 ⎞ ⎜ ⎟ ⎛ F1 ⎞ ⎜ β 1 ⎟ = acos ⎜ ⎟ ⎝ F1 ⎠ ⎜γ ⎟ ⎝ 1⎠ ⎛ α 1 ⎞ ⎛ 46.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 1 ⎟ = ⎜ 124.7 ⎟ deg ⎜ γ ⎟ ⎝ 62.9 ⎠ ⎝ 1⎠ F 2 = 98.6 N

62

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

⎛ α2 ⎞ ⎜ ⎟ ⎛ F2 ⎞ ⎜ β 2 ⎟ = acos ⎜ ⎟ ⎝ F2 ⎠ ⎜γ ⎟ ⎝ 2⎠ ⎛ α 2 ⎞ ⎛ 113.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 2 ⎟ = ⎜ 149.5 ⎟ deg ⎜ γ ⎟ ⎝ 72.3 ⎠ ⎝ 2⎠

Problem 2-58 Express each force in Cartesian vector form. Units Used: 3

kN = 10 N Given: F 1 = 5 kN F 2 = 2 kN

θ 1 = 60 deg θ 2 = 60 deg θ 3 = 45 deg Solution:

F 1v

⎛ cos ( θ 2) ⎞ ⎜ ⎟ = F1 ⎜ cos ( θ 3 ) ⎟ ⎜ cos θ ⎟ ⎝ ( 1) ⎠

⎛ 2.5 ⎞ F 1v = ⎜ 3.5 ⎟ kN ⎜ ⎟ ⎝ 2.5 ⎠

F 2v

⎛0⎞ = F2 ⎜ −1 ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ F 2v = ⎜ −2 ⎟ kN ⎜ ⎟ ⎝0⎠

63

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-59 Determine the magnitude and coordinate direction angles of the force F acting on the stake. Given: F h = 40 N

θ = 70 deg c = 3 d = 4

Solution:

⎛ c2 + d2 ⎞ ⎟ F = Fh ⎜ ⎝ d ⎠ F = 50 N c ⎛ ⎞ ⎜ 2 2 ⎟F ⎝ c +d ⎠

F x = F h cos ( θ )

F y = F h sin ( θ )

Fz =

F x = 13.7 N

F y = 37.6 N

F z = 30 N

⎛ Fx ⎞ ⎟ ⎝F⎠

⎛ Fy ⎞ ⎟ ⎝F⎠

⎛ Fz ⎞ ⎟ ⎝F⎠

α = acos ⎜

β = acos ⎜

γ = acos ⎜

α = 74.1 deg

β = 41.3 deg

γ = 53.1 deg

Problem 2-60 Express each force in Cartesian vector form.

64

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 400 lb F 2 = 600 lb

θ 1 = 45 deg θ 2 = 60 deg θ 3 = 60 deg θ 4 = 45 deg θ 5 = 30 deg

Solution:

F 1v

⎛ −cos ( θ 2) ⎞ ⎜ ⎟ = F1 ⎜ cos ( θ 3 ) ⎟ ⎜ cos θ ⎟ ( 1) ⎠ ⎝

⎛ −200 ⎞ F 1v = ⎜ 200 ⎟ lb ⎜ ⎟ ⎝ 282.8 ⎠

F 2v

⎛ cos ( θ 5) cos ( θ 4 ) ⎞ ⎜ ⎟ = F2 ⎜ cos ( θ 5 ) sin ( θ 4 ) ⎟ ⎜ −sin θ ( 5) ⎟⎠ ⎝

⎛ 367.4 ⎞ F 2v = ⎜ 367.4 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠

Problem 2-61 The stock S mounted on the lathe is subjected to a force F , which is caused by the die D. Determine the coordinate direction angle β and express the force as a Cartesian vector. Given: F = 60 N

α = 60 deg γ = 30 deg 65

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution: cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

(

2

β = acos 1 − cos ( α ) − cos ( γ ) 2

⎛⎜ −cos ( α ) ⎞⎟ F v = F⎜ −cos ( β ) ⎟ ⎜ −cos ( γ ) ⎟ ⎝ ⎠

2

)

β = 90 deg

⎛ −30 ⎞ Fv = ⎜ 0 ⎟ N ⎜ ⎟ ⎝ −52 ⎠

Problem 2-62 Determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 = 80 lb F 2 = 130 lb

θ = 40 deg φ = 30 deg Solution:

F 1v

⎛⎜ cos ( φ ) cos ( θ ) ⎟⎞ = F1 ⎜ −cos ( φ ) sin ( θ ) ⎟ ⎜ ⎟ sin ( φ ) ⎝ ⎠

⎛ 53.1 ⎞ F 1v = ⎜ −44.5 ⎟ lb ⎜ ⎟ ⎝ 40 ⎠

F 2v

⎛0⎞ = F2 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

⎛ 0 ⎞ F 2v = ⎜ 0 ⎟ lb ⎜ ⎟ ⎝ −130 ⎠

F R = F1v + F2v

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

F R = 113.6 lb

⎛⎜ α ⎞⎟ ⎛ 62.1 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 113.1 ⎟ deg ⎜ γ ⎟ ⎝ 142.4 ⎠ ⎝ ⎠

66

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-63 Specify the coordinate direction angles of F 1 and F 2 and express each force as a cartesian vector. Given: F 1 = 80 lb F 2 = 130 lb

φ = 30 deg θ = 40 deg Solution:

F 1v

⎛⎜ cos ( φ ) cos ( θ ) ⎟⎞ = F1 ⎜ −cos ( φ ) sin ( θ ) ⎟ ⎜ ⎟ sin ( φ ) ⎝ ⎠

⎛ α1 ⎞ ⎜ ⎟ ⎛ F1v ⎞ ⎜ β 1 ⎟ = acos ⎜ ⎟ ⎝ F1 ⎠ ⎜γ ⎟ ⎝ 1⎠

F 2v

⎛0⎞ ⎜ ⎟ = F2 0 ⎜ ⎟ ⎝ −1 ⎠

⎛ α2 ⎞ ⎜ ⎟ ⎛ F2v ⎞ ⎜ β 2 ⎟ = acos ⎜ ⎟ ⎝ F2 ⎠ ⎜γ ⎟ ⎝ 2⎠

⎛ 53.1 ⎞ ⎜ ⎟ F 1v = −44.5 lb ⎜ ⎟ ⎝ 40 ⎠ ⎛ α 1 ⎞ ⎛ 48.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 1 ⎟ = ⎜ 123.8 ⎟ deg ⎜ γ ⎟ ⎝ 60 ⎠ ⎝ 1⎠ ⎛ 0 ⎞ ⎜ 0 ⎟ lb F 2v = ⎜ ⎟ ⎝ −130 ⎠ ⎛ α 2 ⎞ ⎛ 90 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 2 ⎟ = ⎜ 90 ⎟ deg ⎜ γ ⎟ ⎝ 180 ⎠ ⎝ 2⎠

Problem 2-64 The mast is subjected to the three forces shown. Determine the coordinate angles α1, β1, γ1 of F1 so that the resultant force acting on the mast is FRi. Given: F R = 350 N

67

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 1 = 500 N F 2 = 200 N F 3 = 300 N Solution: Guesses

α = 20 deg β = 20 deg γ = 20 deg

Given

⎛⎜ cos ( α ) ⎟⎞ ⎛0⎞ ⎛0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ F 1 ⎜ cos ( β ) ⎟ + F2 0 + F3 −1 = FR⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ cos ( γ ) ⎟ ⎝ −1 ⎠ ⎝0⎠ ⎝0⎠ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎜ β ⎟ = Find ( α , β , γ ) ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 45.6 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 53.1 ⎟ deg ⎜ γ ⎟ ⎝ 66.4 ⎠ ⎝ ⎠

Problem 2-65 The mast is subjected to the three forces shown. Determine the coordinate angles α1, β1, γ1 of F1 so that the resultant force acting on the mast is zero . Given: F 1 = 500 N F 2 = 200 N F 3 = 300 N Solution: Guesses

α = 20 deg β = 20 deg γ = 20 deg

68

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

⎛⎜ cos ( α ) ⎞⎟ ⎛0⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ F 1 ⎜ cos ( β ) ⎟ + F2 0 + F3 −1 = 0 ⎜ ⎟ ⎜ ⎟ ⎜ cos ( γ ) ⎟ ⎝ −1 ⎠ ⎝0⎠ ⎝ ⎠

Given

⎛⎜ α ⎞⎟ ⎜ β ⎟ = Find ( α , β , γ ) ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 90 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 53.1 ⎟ deg ⎜ γ ⎟ ⎝ 66.4 ⎠ ⎝ ⎠

Problem 2-66 The shaft S exerts three force components on the die D. Find the magnitude and direction of the resultant force. Force F2 acts within the octant shown. Given: F 1 = 400 N F 2 = 300 N F 3 = 200 N

α 2 = 60 deg γ 2 = 60 deg c = 3 d = 4 Solution: cos ( α 2 ) + cos ( β 2 ) + cos ( γ 2 ) = 1 2

2

2

Solving for the positive root, 2 2 β 2 = acos ⎛⎝ 1 − cos ( α 2 ) − cos ( γ 2 ) ⎞⎠

F 1v

⎛⎜ F1 ⎞⎟ = ⎜ 0 ⎟ ⎜0 ⎟ ⎝ ⎠

F 2v

β 2 = 45 deg

⎛ cos ( α 2) ⎞ ⎜ ⎟ = F2 ⎜ cos ( β 2 ) ⎟ ⎜ cos γ ⎟ ⎝ ( 2) ⎠

F 3v =

⎛0⎞ ⎜ −d ⎟ ⎟ 2 2⎜ c +d ⎝ c ⎠ F3

69

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Engineering Mechanics - Statics

Chapter 2

F R = F1v + F2v + F3v

F R = 615 N

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 26.6 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 85.1 ⎟ deg ⎜ γ ⎟ ⎝ 64.0 ⎠ ⎝ ⎠

Problem 2-67 The beam is subjected to the two forces shown. Express each force in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 = 630 lb

α = 60 deg

F 2 = 250 lb

β = 135 deg

c = 24

γ = 60 deg

d = 7

Solution:

F 1v =

F 2v

⎛0⎞ ⎜d⎟ 2 2⎜ ⎟ c + d ⎝ −c ⎠ F1

⎛⎜ cos ( α ) ⎟⎞ = F2 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

F R = F1v + F2v

⎛ 0 ⎞ ⎜ ⎟ F 1v = 176.4 lb ⎜ ⎟ ⎝ −604.8 ⎠ ⎛ 125 ⎞ ⎜ ⎟ F 2v = −176.8 lb ⎜ ⎟ ⎝ 125 ⎠ F R = 495.8 lb

70

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Engineering Mechanics - Statics

Chapter 2

⎛ αR ⎞ ⎜ ⎟ ⎛ FR ⎞ ⎜ β R ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ R⎠

⎛ α R ⎞ ⎛ 75.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β R ⎟ = ⎜ 90 ⎟ deg ⎜ γ ⎟ ⎝ 165.4 ⎠ ⎝ R⎠

Problem 2-68 Determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 = 350 N

α = 60 deg

F 2 = 250N

β = 60 deg

c = 3

γ = 45 deg

d = 4

θ = 30 deg

Solution:

F 1v

⎛⎜ cos ( α ) ⎟⎞ = F1 ⎜ cos ( β ) ⎟ ⎜ −cos ( γ ) ⎟ ⎝ ⎠

F 2h = F2 ⎛

d ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠

F 2v

⎛ F2h cos ( θ ) ⎞ ⎜ ⎟ = ⎜ −F 2h sin ( θ ) ⎟ ⎜ F ⎟ 2y ⎝ ⎠

F R = F1v + F2v

⎛ αR ⎞ ⎜ ⎟ ⎛ FR ⎞ ⎜ β R ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ R⎠

⎛ 175 ⎞ F 1v = ⎜ 175 ⎟ N ⎜ ⎟ ⎝ −247.5 ⎠ F 2y = F 2 ⎛

c ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠

⎛ 173.2 ⎞ F 2v = ⎜ −100 ⎟ N ⎜ ⎟ ⎝ 150 ⎠ F R = 369.3 N

⎛ α R ⎞ ⎛ 19.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β R ⎟ = ⎜ 78.3 ⎟ deg ⎜ γ ⎟ ⎝ 105.3 ⎠ ⎝ R⎠

Problem 2-69 Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three 71

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Engineering Mechanics - Statics

Chapter 2

forces acts along the positive y axis and has magnitude F. Given: F = 600 lb F 1 = 180 lb F 2 = 300 lb

α 1 = 30 deg α 2 = 40 deg Solution: Initial guesses:

α = 40 deg β = 50 deg

γ = 50 deg F 3 = 45 lb

Given F Rx = ΣF x;

0 = −F1 + F2 cos ( α 1 ) sin ( α 2 ) + F 3 cos ( α )

F Ry = ΣF y;

F = F2 cos ( α 1 ) cos ( α 2 ) + F3 cos ( β )

F Rz = ΣF z;

0 = −F2 sin ( α 1 ) + F3 cos ( γ ) cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

2

2

F 3 = 428 lb

⎛⎜ α ⎞⎟ ⎛ 88.3 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 20.6 ⎟ deg ⎜ γ ⎟ ⎝ 69.5 ⎠ ⎝ ⎠

Problem 2-70 Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces is zero.

72

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 180 lb

α 1 = 30 deg

F 2 = 300 lb

α 2 = 40 deg

Solution: Initial guesses:

α = 40 deg

γ = 50 deg

β = 50 deg

F 3 = 45 lb

Given F Rx = ΣF x;

0 = −F1 + F2 cos ( α 1 ) sin ( α 2 ) + F 3 cos ( α )

F Ry = ΣF y;

0 = F2 cos ( α 1 ) cos ( α 2 ) + F3 cos ( β )

F Rz = ΣF z;

0 = −F2 sin ( α 1 ) + F3 cos ( γ ) cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

2

2

F 3 = 250 lb

⎛⎜ α ⎞⎟ ⎛ 87.0 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 142.9 ⎟ deg ⎜ γ ⎟ ⎝ 53.1 ⎠ ⎝ ⎠

Problem 2-71 Specify the magnitude F 3 and directions α3, β3, and γ3 of F 3 so that the resultant force of the three forces is FR. Units Used: 3

kN = 10 N Given: F 1 = 12 kN

c = 5

F 2 = 10 kN

d = 12

θ = 30 deg

73

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Engineering Mechanics - Statics

Chapter 2

⎛0⎞ F R = ⎜ 9 ⎟ kN ⎜ ⎟ ⎝0⎠ Solution: F 3x = 1 kN

Initial Guesses:

F 3y = 1 kN

⎛ F3x ⎞ ⎛ 0 ⎞ ⎜ ⎟ F R = ⎜ F3y ⎟ + F 1 ⎜ cos ( θ ) ⎟ + ⎜ ⎟ ⎜F ⎟ ( ) − sin θ ⎝ ⎠ ⎝ 3z ⎠

Given

⎛ F3x ⎞ ⎜ ⎟ ⎜ F3y ⎟ = Find ( F3x , F3y , F3z) ⎜F ⎟ ⎝ 3z ⎠ ⎛ α3 ⎞ ⎜ ⎟ ⎛ F3 ⎞ ⎜ β 3 ⎟ = acos ⎜ ⎟ ⎝ F3 ⎠ ⎜γ ⎟ ⎝ 3⎠

F 3z = 1 kN

⎛ −d ⎞ ⎜0⎟ ⎟ 2 2⎜ c +d ⎝ c ⎠ F2

⎛ F3x ⎞ ⎜ ⎟ F 3 = ⎜ F 3y ⎟ ⎜F ⎟ ⎝ 3z ⎠

⎛ 9.2 ⎞ F 3 = ⎜ −1.4 ⎟ kN ⎜ ⎟ ⎝ 2.2 ⎠

F 3 = 9.6 kN

⎛ α 3 ⎞ ⎛ 15.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 3 ⎟ = ⎜ 98.4 ⎟ deg ⎜ γ ⎟ ⎝ 77.0 ⎠ ⎝ 3⎠

Problem 2-72 The pole is subjected to the force F , which has components acting along the x,y,z axes as shown. Given β and γ, determine the magnitude of the three components of F . Units Used: kN = 1000 N Given: F = 3 kN

β = 30 deg γ = 75 deg Solution: cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

(

2

)

74

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Engineering Mechanics - Statics

α = acos

(

Chapter 2

)

−cos ( β ) − cos ( γ ) + 1 2

2

α = 64.67 deg F x = F cos ( α )

F y = F cos ( β )

F z = F cos ( γ )

F x = 1.28 kN

F y = 2.60 kN

F z = 0.8 kN

Problem 2-73 The pole is subjected to the force F which has components F x and Fz. Determine the magnitudes of F and Fy. Units Used: kN = 1000 N Given: F x = 1.5 kN F z = 1.25 kN

β = 75 deg Solution: cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

2

2

2

⎛ Fx ⎞ 2 ⎛ Fz ⎞ ⎜ ⎟ + cos ( β ) + ⎜ ⎟ = 1 ⎝F⎠ ⎝F⎠ 2

F =

2

Fx + Fz

2 1 − cos ( β )

F y = F cos ( β )

F = 2.02 kN

F y = 0.5 kN

Problem 2-74 The eye bolt is subjected to the cable force F which has a component F x along the x axis, a component F z along the z axis, and a coordinate direction angle β. Determine the magnitude of F .

75

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Engineering Mechanics - Statics

Chapter 2

Given: F x = 60 N F z = −80 N

β = 80 deg Solution: F y = F cos ( β ) F x + F z + Fy cos ( β ) 2

Fy =

2

2

Fy =

2

Fx + Fz

1 − cos ( β ) 2

F =

2

2

2

cos ( β )

2

Fx + Fy + Fz

F y = 17.6 N

F = 102 N

Problem 2-75 Three forces act on the hook. If the resultant force FR has a magnitude and direction as shown, determine the magnitude and the coordinate direction angles of force F3. Given: F R = 120 N F 1 = 80 N F 2 = 110 N c = 3 d = 4

θ = 30 deg φ = 45 deg

76

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Engineering Mechanics - Statics

Chapter 2

Solution:

F 1v

⎛d⎞ ⎛ F1 ⎞ ⎜ ⎟ = ⎜ ⎟ 0 2 2 ⎜ ⎟ c + d ⎝ ⎠⎝ c ⎠

⎛ 64 ⎞ F 1v = ⎜ 0 ⎟ N ⎜ ⎟ ⎝ 48 ⎠

F 2v

⎛0⎞ = F2 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

⎛ 0 ⎞ F 2v = ⎜ 0 ⎟ N ⎜ ⎟ ⎝ −110 ⎠

F Rv

⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟ = F R⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ sin ( φ ) ⎟ ⎝ ⎠

⎛ 42.4 ⎞ F Rv = ⎜ 73.5 ⎟ N ⎜ ⎟ ⎝ 84.9 ⎠

F 3v = FRv − F 1v − F 2v

⎛ −21.6 ⎞ F 3v = ⎜ 73.5 ⎟ N ⎜ ⎟ ⎝ 146.9 ⎠

⎛⎜ α ⎞⎟ ⎛ F3v ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ F3v ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 97.5 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 63.7 ⎟ deg ⎜ γ ⎟ ⎝ 27.5 ⎠ ⎝ ⎠

F 3v = 165.6 N

Problem 2-76 Determine the coordinate direction angles of F 1 and FR. Given: F R = 120 N F 1 = 80 N F 2 = 110 N c = 3 d = 4

θ = 30 deg φ = 45 deg

77

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution:

F 1v

⎛d⎞ ⎛ F1 ⎞ ⎜ ⎟ = ⎜ ⎟ 0 2 2 ⎜ ⎟ c + d ⎝ ⎠⎝ c ⎠

⎛ 64 ⎞ F 1v = ⎜ 0 ⎟ N ⎜ ⎟ ⎝ 48 ⎠

⎛ α1 ⎞ ⎜ ⎟ ⎛ F1v ⎞ ⎜ β 1 ⎟ = acos ⎜ ⎟ ⎝ F1v ⎠ ⎜γ ⎟ ⎝ 1⎠

F Rv

⎛ α 1 ⎞ ⎛ 36.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 1 ⎟ = ⎜ 90 ⎟ deg ⎜ γ ⎟ ⎝ 53.1 ⎠ ⎝ 1⎠

⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟ = F R⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ sin ( φ ) ⎟ ⎝ ⎠

⎛ 42.4 ⎞ F Rv = ⎜ 73.5 ⎟ N ⎜ ⎟ ⎝ 84.9 ⎠

⎛ αR ⎞ ⎜ ⎟ ⎛ FRv ⎞ ⎜ β R ⎟ = acos ⎜ ⎟ ⎝ FRv ⎠ ⎜γ ⎟ ⎝ R⎠

⎛ α R ⎞ ⎛ 69.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β R ⎟ = ⎜ 52.2 ⎟ deg ⎜ γ ⎟ ⎝ 45 ⎠ ⎝ R⎠

Problem 2-77 The pole is subjected to the force F , which has components acting along the x, y, z axes as shown. Given the magnitude of F and the angles α and γ , determine the magnitudes of the components of F . Given: F = 80 N α = 60 deg

γ = 45 deg

Solution:

(

β = acos − 1 − cos ( α ) − cos ( γ ) 2

2

)

β = 120 deg F x = F cos ( α )

F y = F cos ( β )

F z = F cos ( γ )

F x = 40 N

F y = 40 N

F z = 56.6 N

78

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Engineering Mechanics - Statics

Chapter 2

Problem 2-78 Two forces F1 and F 2 act on the bolt. If the resultant force F R has magnitude F R and coordinate direction angles α and β, as shown, determine the magnitude of F2 and its coordinate direction angles. Given: F 1 = 20 lb F R = 50 lb

α = 110 deg β = 80 deg Solution: cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

(

2

γ = acos − 1 − cos ( α ) − cos ( β ) 2

Initial Guesses

Given

F 2x = 1 lb

2

)

γ = 157.44 deg

F 2y = 1 lb

F 2z = 1 lb

⎛⎜ cos ( α ) ⎟⎞ ⎛ 0 ⎞ ⎛⎜ F2x ⎟⎞ ⎜ ⎟ F R⎜ cos ( β ) ⎟ = F 1 0 + ⎜ F2y ⎟ ⎜ ⎟ ⎜ cos ( γ ) ⎟ ⎝ −1 ⎠ ⎜⎝ F2z ⎟⎠ ⎝ ⎠

⎛ F2x ⎞ ⎜ ⎟ F 2 = ⎜ F 2y ⎟ ⎜F ⎟ ⎝ 2z ⎠

⎛ F2x ⎞ ⎜ ⎟ ⎜ F2y ⎟ = Find ( F2x , F2y , F2z) ⎜F ⎟ ⎝ 2z ⎠

⎛ −17.1 ⎞ ⎜ 8.7 ⎟ lb F2 = ⎜ ⎟ ⎝ −26.2 ⎠

F 2 = 32.4 lb

⎛ α2 ⎞ ⎜ ⎟ ⎛ F2 ⎞ ⎜ β 2 ⎟ = acos ⎜ ⎟ ⎝ F2 ⎠ ⎜γ ⎟ ⎝ 2⎠

⎛ α 2 ⎞ ⎛ 121.8 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 2 ⎟ = ⎜ 74.5 ⎟ deg ⎜ γ ⎟ ⎝ 143.8 ⎠ ⎝ 2⎠

Problem 2-79 Given r1, r2, and r3, determine the magnitude and direction of r = 2r1 − r2 + 3r3. 79

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Engineering Mechanics - Statics

Chapter 2

Given:

⎛3⎞ ⎜ ⎟ r1 = −4 m ⎜ ⎟ ⎝3⎠

⎛4⎞ ⎜ ⎟ r2 = 0 m ⎜ ⎟ ⎝ −5 ⎠

⎛3⎞ ⎜ ⎟ r3 = −2 m ⎜ ⎟ ⎝5⎠

Solution: r = 2r1 − r2 + 3r3

⎛ 11 ⎞ ⎜ ⎟ r = −14 m ⎜ ⎟ ⎝ 26 ⎠ ⎛⎜ α ⎞⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠

r = 31.5 m

⎛⎜ α ⎞⎟ ⎛ 69.6 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 116.4 ⎟ deg ⎜ γ ⎟ ⎝ 34.4 ⎠ ⎝ ⎠

Problem 2-80 Represent the position vector r acting from point A(a, b, c) to point B(d, e, f) in Cartesian vector form. Determine its coordinate direction angles and find the distance between points A and B. Given: a = 3m b = 5m c = 6m d = 5m e = −2 m f = 1m Solution:

⎛d − a⎞ ⎜ ⎟ r = e−b ⎜ ⎟ ⎝ f− c⎠

⎛2⎞ ⎜ ⎟ r = −7 m ⎜ ⎟ ⎝ −5 ⎠

r = 8.8 m

80

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Engineering Mechanics - Statics

Chapter 2

α = acos ⎛⎜

d − a⎞

α = 76.9 deg

β = acos ⎛⎜

e − b⎞

β = 142 deg

γ = acos ⎛⎜

f − c⎞

γ = 124 deg

⎟ ⎝ r ⎠

⎟ ⎝ r ⎠

⎟ ⎝ r ⎠

Problem 2-81 A position vector extends from the origin to point A(a, b, c). Determine the angles α, β, γ which the tail of the vector makes with the x, y, z axes, respectively. Given: a = 2m b = 3m

c = 6m

Solution:

⎛a⎞ r = ⎜b⎟ ⎜ ⎟ ⎝c ⎠

⎛2⎞ r = ⎜3⎟ m ⎜ ⎟ ⎝6⎠

⎛⎜ α ⎞⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 73.4 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 64.6 ⎟ deg ⎜ γ ⎟ ⎝ 31.0 ⎠ ⎝ ⎠

Problem 2-82 Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate direction angles. Given: a = 4m

81

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Engineering Mechanics - Statics

Chapter 2

b = 8m c = 3m d = 4m

Solution:

⎛ −c ⎞ r = ⎜ −d − b ⎟ ⎜ ⎟ ⎝ a ⎠ ⎛⎜ α ⎞⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠

⎛ −3 ⎞ r = ⎜ −12 ⎟ m ⎜ ⎟ ⎝ 4 ⎠

r = 13 m

⎛⎜ α ⎞⎟ ⎛ 103.3 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 157.4 ⎟ deg ⎜ γ ⎟ ⎝ 72.1 ⎠ ⎝ ⎠

Problem 2-83 Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate direction angles. Given: a = 8 ft b = 2 ft c = 5 ft

θ = 30 deg φ = 20 deg Solution:

⎛⎜ −c cos ( φ ) sin ( θ ) ⎞⎟ r = ⎜ a − c cos ( φ ) cos ( θ ) ⎟ ⎜ b + c sin ( φ ) ⎟ ⎝ ⎠

⎛ −2.35 ⎞ r = ⎜ 3.93 ⎟ ft ⎜ ⎟ ⎝ 3.71 ⎠

r = 5.89 ft

82

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Engineering Mechanics - Statics

Chapter 2

⎛⎜ α ⎞⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 113.5 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 48.2 ⎟ deg ⎜ γ ⎟ ⎝ 51 ⎠ ⎝ ⎠

Problem 2-84 Determine the length of the connecting rod AB by first formulating a Cartesian position vector from A to B and then determining its magnitude. Given: b = 16 in a = 5 in

α = 30 deg

Solution:

r =

⎛ a sin ( α ) + b ⎞ ⎜ ⎟ ⎝ −a cos ( α ) ⎠

r=

⎛ 18.5 ⎞ ⎜ ⎟ in ⎝ −4.3 ⎠

r = 19 in

Problem 2-85 Determine the length of member AB of the truss by first establishing a Cartesian position vector from A to B and then determining its magnitude. Given: a = 1.2 m b = 0.8 m c = 0.3 m d = 1.5 m

θ = 40 deg 83

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution:

⎛ c + d cot ( θ ) ⎞ ⎜ ⎟ ⎝ d−a ⎠

r =

r=

⎛ 2.09 ⎞ ⎜ ⎟m ⎝ 0.3 ⎠

r = 2.11 m

Problem 2- 86 The positions of point A on the building and point B on the antenna have been measured relative to the electronic distance meter (EDM) at O. Determine the distance between A and B. Hint: Formulate a position vector directed from A to B; then determine its magnitude. Given: a = 460 m b = 653 m

α = 60 deg β = 55 deg θ = 30 deg φ = 40 deg Solution:

rOA

⎛⎜ −a cos ( φ ) sin ( θ ) ⎟⎞ = ⎜ a cos ( φ ) cos ( θ ) ⎟ ⎜ ⎟ a sin ( φ ) ⎝ ⎠

rOB

⎛⎜ −b cos ( β ) sin ( α ) ⎟⎞ = ⎜ −b cos ( β ) cos ( α ) ⎟ ⎜ ⎟ b sin ( β ) ⎝ ⎠

rAB = rOB − rOA

⎛ −148.2 ⎞ ⎜ ⎟ rAB = −492.4 m ⎜ ⎟ ⎝ 239.2 ⎠

rAB = 567.2 m

Problem 2-87 Determine the lengths of cords ACB and CO. The knot at C is located midway between A and B.

84

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Engineering Mechanics - Statics

Chapter 2

Given: a = 3 ft b = 6 ft c = 4 ft

Solution:

rAB

⎛c ⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝ −a ⎠

rAC =

rOA

⎛0⎞ ⎜ ⎟ = 0 ⎜ ⎟ ⎝a⎠

rAB 2

rAB = 7.8 ft rOC = rOA + rAC rOC = 3.91 ft

Problem 2-88 Determine the length of the crankshaft AB by first formulating a Cartesian position vector from A to B and then determining its magnitude.

85

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Engineering Mechanics - Statics

Chapter 2

Given: a = 400 b = 125

θ = 25 deg

Solution:

rAB

⎡a + b sin ( θ ) ⎤ ⎢ ⎥ = −( b cos ( θ ) ) mm ⎢ ⎥ 0 ⎣ ⎦

rAB = 467 mm

Problem 2-89 Determine the length of wires AD, BD, and CD. The ring at D is midway between A and B. Given: a = 0.5 m b = 1.5 m c = 2m d = 2m e = 0.5 m

86

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Engineering Mechanics - Statics

Chapter 2

Solution:

rAD

⎛ −c ⎞ ⎜ 2 ⎟ ⎜ ⎟ d ⎟ ⎜ = ⎜ 2 ⎟ ⎜e b⎟ ⎜ − ⎟ ⎝2 2⎠

⎛ ⎜ ⎜ = ⎜ ⎜ ⎜ ⎜a + ⎝

rCD

rAD

⎛ −1 ⎞ ⎜ 1 ⎟m = ⎜ ⎟ ⎝ −0.5 ⎠

rAD = 1.5 m

rBD = −rAD

rBD = 1.5 m

⎞ ⎟ 2 ⎟ d ⎟ 2 ⎟ b e⎟ − ⎟ 2 2⎠ c

rCD

⎛1⎞ ⎜ ⎟ = 1 m ⎜ ⎟ ⎝1⎠

rCD = 1.7 m

Problem 2-90 Express force F as a Cartesian vector; then determine its coordinate direction angles. Given: F = 600 lb

c = 3 ft

a = 1.5 ft

φ = 60 deg

b = 5 ft

Solution: r = bi + ( a + c sin ( φ ) ) j + ( 0 − c cos ( φ ) ) k r =

b + ( a + c sin ( φ ) ) + ( c cos ( φ ) ) 2

2

2

r=2m

87

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Engineering Mechanics - Statics

Chapter 2

⎡ −( c cos ( φ ) )⎤ ⎥ r ⎣ ⎦

b d = F r

⎛ a + c sin ( φ ) ⎞ e = F⎜ ⎟ r ⎝ ⎠

f = F⎢

d = 452 lb

e = 370 lb

f = −136 lb

F = ( di + ej + fk) lb d α = acos ⎛⎜ ⎞⎟

α = 41.1 deg

e β = acos ⎛⎜ ⎞⎟

β = 51.9 deg

f γ = acos ⎛⎜ ⎞⎟

γ = 103 deg

⎝F⎠

⎝ F⎠

⎝ F⎠

Problem 2-91 Express force F as a Cartesian vector; then determine its coordinate direction angles. Given: a = 1.5 ft b = 5 ft c = 3 ft

θ = 60 deg F = 600 lb

Solution: b ⎛ ⎞ ⎜ ⎟ r = a + c sin ( θ ) ⎜ ⎟ ⎝ −c cos ( θ ) ⎠

F = F

⎛ 452 ⎞ ⎜ ⎟ F = 370 lb ⎜ ⎟ ⎝ −136 ⎠

r r

⎛⎜ α ⎞⎟ ⎛ F ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ F ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 41.1 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 51.9 ⎟ deg ⎜ γ ⎟ ⎝ 103.1 ⎠ ⎝ ⎠

88

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Engineering Mechanics - Statics

Chapter 2

Problem 2-92 Determine the magnitude and coordinate direction angles of the resultant force acting at point A. Given: F 1 = 150 N F 2 = 200 N a = 1.5 m b = 4m c = 3m d = 2m e = 3m

θ = 60 deg Solution: Define the position vectors and then the forces

rAB

⎛ e cos ( θ ) ⎞ = ⎜ a + e sin ( θ ) ⎟ ⎜ ⎟ −b ⎝ ⎠

rAC

⎛ c ⎞ = ⎜a − d⎟ ⎜ ⎟ ⎝ −b ⎠

F 1v = F1

F 2v = F2

⎛ 38 ⎞ F 1v = ⎜ 103.8 ⎟ N ⎜ ⎟ ⎝ −101.4 ⎠

rAB rAB

⎛ 119.4 ⎞ F 2v = ⎜ −19.9 ⎟ N ⎜ ⎟ ⎝ −159.2 ⎠

rAC rAC

Add the forces and find the magnitude of the resultant F R = F1v + F2v

⎛ 157.4 ⎞ F R = ⎜ 83.9 ⎟ N ⎜ ⎟ ⎝ −260.6 ⎠

F R = 316 N

Find the direction cosine angles

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 60.1 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 74.6 ⎟ deg ⎜ γ ⎟ ⎝ 145.6 ⎠ ⎝ ⎠

89

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Engineering Mechanics - Statics

Chapter 2

Problem 2-93 The plate is suspended using the three cables which exert the forces shown. Express each force as a Cartesian vector. Given: F BA = 350 lb F CA = 500 lb F DA = 400 lb a = 3 ft b = 3 ft c = 6 ft d = 14 ft e = 3 ft f = 3 ft g = 2 ft Solution:

rBA

⎛ −e − g ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ d ⎠

rCA

⎛f⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝d⎠

rDA

⎛ −g ⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝d⎠

F BAv = FBA

F CAv = FCA

F DAv = FDA

rBA rBA

rCA rCA

rDA rDA

F BAv

⎛ −109.2 ⎞ ⎜ 131 ⎟ lb = ⎜ ⎟ ⎝ 305.7 ⎠

⎛ 102.5 ⎞ ⎜ ⎟ F CAv = 102.5 lb ⎜ ⎟ ⎝ 478.5 ⎠ ⎛ −52.1 ⎞ ⎜ ⎟ F DAv = −156.2 lb ⎜ ⎟ ⎝ 364.5 ⎠

90

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Engineering Mechanics - Statics

Chapter 2

Problem 2-94 The engine of the lightweight plane is supported by struts that are connected to the space truss that makes up the structure of the plane. The anticipated loading in two of the struts is shown. Express each of these forces as a Cartesian vector. Given: F 1 = 400 lb F 2 = 600 lb a = 0.5 ft b = 0.5 ft c = 3.0 ft d = 2.0 ft e = 0.5 ft f = 3.0 ft Solution:

rCD

⎛c ⎞ ⎜ ⎟ = −b ⎜ ⎟ ⎝a⎠

rAB

⎛ −c ⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝ −e ⎠

F 1v = F1

F 2v = F2

rCD rCD

rAB rAB

⎛ 389.3 ⎞ ⎜ ⎟ F 1v = −64.9 lb ⎜ ⎟ ⎝ 64.9 ⎠ ⎛ −584.0 ⎞ ⎜ 97.3 ⎟ lb F 2v = ⎜ ⎟ ⎝ −97.3 ⎠

Problem 2-95 The window is held open by cable AB. Determine the length of the cable and express the force F acting at A along the cable as a Cartesian vector. 91

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Engineering Mechanics - Statics

Chapter 2

Given: a = 300 mm b = 500 mm c = 150 mm d = 250 mm

θ = 30 deg F = 30 N

Solution:

rAB

⎛ −a cos ( θ ) ⎞ ⎜ c−b ⎟ = ⎜ ⎟ ⎝ d + a sin ( θ ) ⎠

Fv = F

rAB rAB

rAB = 591.6 mm

⎛ −13.2 ⎞ ⎜ ⎟ F v = −17.7 N ⎜ ⎟ ⎝ 20.3 ⎠

Problem 2-96 The force acting on the man, caused by his pulling on the anchor cord, is F. If the length of the cord is L, determine the coordinates A(x, y, z) of the anchor.

92

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Engineering Mechanics - Statics

Chapter 2

Given:

⎛ 40 ⎞ F = ⎜ 20 ⎟ N ⎜ ⎟ ⎝ −50 ⎠ L = 25 m Solution:

r = L

F F

⎛ 14.9 ⎞ r = ⎜ 7.5 ⎟ m ⎜ ⎟ ⎝ −18.6 ⎠

Problem 2-97 Express each of the forces in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 = 80 lb

c = 4 ft

93

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Engineering Mechanics - Statics

Chapter 2

F 2 = 50 lb

d = 2.5 ft

a = 6 ft

e = 12

b = 2 ft

f = 5

Solution:

rAC

⎛⎜ −d ⎞⎟ −c ⎟ = ⎜ ⎜ d⎟ ⎜e f ⎟ ⎝ ⎠

rAB

⎛b⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝ −a ⎠

F R = F1v + F2v

F 1v = F1

F 2v = F2

⎛ −26.2 ⎞ F 1v = ⎜ −41.9 ⎟ lb ⎜ ⎟ ⎝ 62.9 ⎠

rAC rAC

⎛ 6.1 ⎞ F 2v = ⎜ −12.1 ⎟ kg ⎜ ⎟ ⎝ −18.2 ⎠

rAB rAB

⎛ −12.8 ⎞ F R = ⎜ −68.7 ⎟ lb ⎜ ⎟ ⎝ 22.8 ⎠

F R = 73.5 lb

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 1.7 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 2.8 ⎟ ⎜ γ ⎟ ⎝ 1.3 ⎠ ⎝ ⎠

Problem 2-98 The cable attached to the tractor at B exerts force F on the framework. Express this force as a Cartesian vector

94

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Engineering Mechanics - Statics

Chapter 2

Given: F = 350 lb a = 35 ft b = 50 ft

θ = 20 deg

Solution: Find the position vector and then the force

rAB

⎛ b sin ( θ ) ⎞ ⎜ ⎟ = b cos ( θ ) ⎜ ⎟ ⎝ −a ⎠

Fv = F

⎛ 98.1 ⎞ ⎜ ⎟ F v = 269.4 lb ⎜ ⎟ ⎝ −200.7 ⎠

rAB rAB

Problem 2-99 The cable OA exerts force F on point O. If the length of the cable is L, what are the coordinates (x, y, z) of point A? Given:

⎛ 40 ⎞ ⎜ ⎟ F = 60 N ⎜ ⎟ ⎝ 70 ⎠ L = 3m Solution:

r = L

F F

95

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Engineering Mechanics - Statics

Chapter 2

⎛ 1.2 ⎞ r = ⎜ 1.8 ⎟ m ⎜ ⎟ ⎝ 2.1 ⎠

Problem 2-100 Determine the position (x, y, 0) for fixing cable BA so that the resultant of the forces exerted on the pole is directed along its axis, from B toward O, and has magnitude F R. Also, what is the magnitude of force F 3? Given: F 1 = 500 N F 2 = 400 N F R = 1000 N a = 1m b = 2m c = 2m d = 3m

Solution: Initial Guesses F3 = 1 N

x = 1m

y = 1m

Given

⎛0⎞ ⎛ −c ⎞ ⎛a⎞ ⎛x ⎞ F3 F2 ⎛ F1 ⎞ ⎜ ⎟ ⎛ ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟ ⎜ ⎟ FR 0 = ⎜ ⎟ 0 ⎟ + ⎜ 2 2 2 ⎟ ⎜ −b ⎟ + ⎜ 2 2 2 ⎟ ⎜ y ⎟ ⎜ ⎟ 2 2 ⎜ c + d ⎠ ⎝ −d ⎠ ⎝ a + b + d ⎠ ⎝ −d ⎠ ⎝ x + y + d ⎠ ⎝ −d ⎠ ⎝ −1 ⎠ ⎝ ⎛⎜ F3 ⎞⎟ ⎜ x ⎟ = Find ( F3 , x , y) ⎜ y ⎟ ⎝ ⎠

⎛ x ⎞ ⎛ 1.9 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ y ⎠ ⎝ 2.4 ⎠

F 3 = 380 N

96

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Engineering Mechanics - Statics

Chapter 2

Problem 2-101 The cord exerts a force F on the hook. If the cord is length L, determine the location x, y of the point of attachment B, and the height z of the hook. Given:

⎛ 12 ⎞ F = ⎜ 9 ⎟ lb ⎜ ⎟ ⎝ −8 ⎠ L = 8 ft a = 2 ft

Solution: Initial guesses

Given

x = 1 ft

⎛x − a⎞ ⎜ y ⎟=L F ⎜ ⎟ F ⎝ −z ⎠

y = 1 ft

z = 1 ft

⎛x⎞ ⎜ y ⎟ = Find ( x , y , z) ⎜ ⎟ ⎝z⎠

⎛ x ⎞ ⎛ 7.65 ⎞ ⎜ y ⎟ = ⎜ 4.24 ⎟ ft ⎜ ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 3.76 ⎠

Problem 2-102 The cord exerts a force of magnitude F on the hook. If the cord length L, the distance z, and the x component of the force, Fx, are given, determine the location x, y of the point of attachment B of the cord to the ground. Given: F = 30 lb L = 8 ft z = 4 ft F x = 25 lb a = 2 ft

Solution: Guesses x = 1 ft 97

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Engineering Mechanics - Statics

Chapter 2

y = 1 ft

Given Fx =

⎛ x − a ⎞F ⎜ ⎟ ⎝ L ⎠

2

2

2

2

L = ( x − a) + y + z

⎛x⎞ ⎜ ⎟ = Find ( x , y) ⎝ y⎠

⎛ x ⎞ ⎛ 8.67 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ y ⎠ ⎝ 1.89 ⎠

Problem 2-103 Each of the four forces acting at E has magnitude F. Express each force as a Cartesian vector and determine the resultant force. Units used: 3

kN = 10 N Given: F = 28 kN a = 4m b = 6m c = 12 m

Solution: Find the position vectors and then the forces

rEA

⎛b⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝ −c ⎠

F EA = F

rEA rEA

⎛ 12 ⎞ F EA = ⎜ −8 ⎟ kN ⎜ ⎟ ⎝ −24 ⎠ 98

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Engineering Mechanics - Statics

rEB

⎛b⎞ = ⎜a ⎟ ⎜ ⎟ ⎝ −c ⎠

rEC

⎛ −b ⎞ = ⎜ a ⎟ ⎜ ⎟ ⎝ −c ⎠

rED

⎛ −b ⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝ −c ⎠

F EB = F

F EC = F

F ED = F

Chapter 2

rEB rEB

rEC rEC

rED rED

Find the resultant sum F R = FEA + F EB + F EC + FED

⎛ 12 ⎞ F EB = ⎜ 8 ⎟ kN ⎜ ⎟ ⎝ −24 ⎠ ⎛ −12 ⎞ F EC = ⎜ 8 ⎟ kN ⎜ ⎟ ⎝ −24 ⎠ ⎛ −12 ⎞ F ED = ⎜ −8 ⎟ kN ⎜ ⎟ ⎝ −24 ⎠ ⎛ 0 ⎞ F R = ⎜ 0 ⎟ kN ⎜ ⎟ ⎝ −96 ⎠

Problem 2-104 The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles α, β, γ of the resultant force. Units Used: 3

kN = 10 N

Given: x = 20 m

a = 16 m

y = 15 m

b = 18 m

F 1 = 600 N

c = 6m

F 2 = 400 N

d = 4m

F 3 = 800 N

e = 24 m

99

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Engineering Mechanics - Statics

Chapter 2

Solution: Find the position vectors, then the force vectors

rDC

⎛a⎞ = ⎜ −b ⎟ ⎜ ⎟ ⎝ −e ⎠

rDA

⎛x⎞ = ⎜ y ⎟ ⎜ ⎟ ⎝ −e ⎠

rDB

⎛ −c ⎞ = ⎜d ⎟ ⎜ ⎟ ⎝ −e ⎠

F 1v = F1

F 2v = F2

F 3v = F3

⎛ 282.4 ⎞ F 1v = ⎜ −317.6 ⎟ N ⎜ ⎟ ⎝ −423.5 ⎠

rDC rDC

⎛ 230.8 ⎞ F 2v = ⎜ 173.1 ⎟ N ⎜ ⎟ ⎝ −277 ⎠

rDA rDA

⎛ −191.5 ⎞ F 3v = ⎜ 127.7 ⎟ N ⎜ ⎟ ⎝ −766.2 ⎠

rDB rDB

Find the resultant, magnitude, and direction angles

F R = F1v + F2v + F3v

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛ 0.322 ⎞ F R = ⎜ −0.017 ⎟ kN ⎜ ⎟ ⎝ −1.467 ⎠

F R = 1.502 kN

⎛⎜ α ⎞⎟ ⎛ 77.6 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 90.6 ⎟ deg ⎜ γ ⎟ ⎝ 167.6 ⎠ ⎝ ⎠

Problem 2-105 The chandelier is supported by three chains which are concurrent at point O. If the force in each chain has magnitude F, express each force as a Cartesian vector and determine the magnitude and coordinate direction angles of the resultant force.

100

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Engineering Mechanics - Statics

Chapter 2

Given: F = 60 lb a = 6 ft b = 4 ft

θ 1 = 120 deg θ 2 = 120 deg Solution:

θ 3 = 360 deg − θ 1 − θ 2

rOA

⎛ b sin ( θ 1 ) ⎞ ⎜ ⎟ = ⎜ b cos ( θ 1 ) ⎟ ⎜ ⎟ ⎝ −a ⎠

rOB

⎛ b sin ( θ 1 + θ 2 ) ⎞ ⎜ ⎟ = ⎜ b cos ( θ 1 + θ 2 ) ⎟ ⎜ ⎟ −a ⎝ ⎠

rOC

⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝ −a ⎠

FA = F

FB = F

FC = F

rOA rOA

rOB rOB

rOC rOC

FR = FA + FB + FC

⎛ 28.8 ⎞ ⎜ ⎟ F A = −16.6 lb ⎜ ⎟ ⎝ −49.9 ⎠ ⎛ −28.8 ⎞ ⎜ ⎟ F B = −16.6 lb ⎜ ⎟ ⎝ −49.9 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F C = 33.3 lb ⎜ ⎟ ⎝ −49.9 ⎠ F R = 149.8 lb

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 90 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 90 ⎟ deg ⎜ γ ⎟ ⎝ 180 ⎠ ⎝ ⎠

101

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Engineering Mechanics - Statics

Chapter 2

Problem 2-106 The chandelier is supported by three chains which are concurrent at point O. If the resultant force at O has magnitude F R and is directed along the negative z axis, determine the force in each chain assuming F A = F B = F C = F. Given: a = 6 ft b = 4 ft F R = 130 lb Solution: 2

2

a +b FR 3a

F =

F = 52.1 lb

Problem 2-107 Given the three vectors A, B, and D, show that A⋅ ( B + D) = ( A⋅ B) + ( A⋅ D).

Solution: Since the component of (B + D) is equal to the sum of the components of B and D, then A⋅ ( B + D) = A⋅ B + A⋅ D

(QED)

Also, A⋅ ( B + D) = ( A xi + A yj + Azk) ⎡⎣( B x + Dx) i + ( By + Dy) j + ( Bz + Dz) k⎤⎦ =

Ax ( B x + Dx) + Ay ( B y + Dy) + Az( B z + Dz) 102

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Engineering Mechanics - Statics

Chapter 2

=

( Ax Bx + Ay By + Az Bz) + ( Ax Dx + Ay Dy + Az Dz)

=

( A⋅ B) + ( A⋅ D)

(QED)

Problem 2-108 Cable BC exerts force F on the top of the flagpole. Determine the projection of this force along the z axis of the pole. Given: F = 28 N a = 12 m b = 6m c = 4m Solution:

rBC

⎛b⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝ −a ⎠

Fv = F

⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠

rBC rBC

F z = −Fv k

F z = 24 N

Problem 2-109 Determine the angle θ between the tails of the two vectors.

103

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Engineering Mechanics - Statics

Chapter 2

Given: r1 = 9 m r2 = 6 m

α = 60 deg β = 45 deg γ = 120 deg φ = 30 deg ε = 40 deg Solution: Determine the two position vectors and use the dot product to find the angle

r1v

⎛⎜ sin ( ε ) cos ( φ ) ⎟⎞ = r1 ⎜ −sin ( ε ) sin ( φ ) ⎟ ⎜ cos ( ε ) ⎟ ⎝ ⎠ ⎛ r1v⋅ r2v ⎞ ⎟ ⎝ r1v r2v ⎠

θ = acos ⎜

r2v

⎛⎜ cos ( α ) ⎟⎞ = r2 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

θ = 109.4 deg

Problem 2-110 Determine the magnitude of the projected component of r1 along r2, and the projection of r2 along r1. Given: r1 = 9 m r2 = 6 m

α = 60 deg β = 45 deg γ = 120 deg φ = 30 deg ε = 40 deg

104

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Engineering Mechanics - Statics

Chapter 2

Solution: Write the vectors and unit vectors

r1v

⎛⎜ sin ( ε ) cos ( φ ) ⎟⎞ = r1 ⎜ −sin ( ε ) sin ( φ ) ⎟ ⎜ cos ( ε ) ⎟ ⎝ ⎠

⎛ 5.01 ⎞ r1v = ⎜ −2.89 ⎟ m ⎜ ⎟ ⎝ 6.89 ⎠

r2v

⎛⎜ cos ( α ) ⎟⎞ = r2 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

⎛ 3 ⎞ r2v = ⎜ 4.24 ⎟ m ⎜ ⎟ ⎝ −3 ⎠

u1 =

r1v r1v

u2 =

r2v r2v

⎛ 0.557 ⎞ u1 = ⎜ −0.321 ⎟ ⎜ ⎟ ⎝ 0.766 ⎠

⎛ 0.5 ⎞ u2 = ⎜ 0.707 ⎟ ⎜ ⎟ ⎝ −0.5 ⎠

The magnitude of the projection of r1 along r2.

r1v⋅ u2 = 2.99 m

The magnitude of the projection of r2 along r1.

r2v⋅ u1 = 1.99 m

Problem 2-111 Determine the angles θ and φ between the wire segments. Given: a = 0.6 b = 0.8 c = 0.5 d = 0.2 Solution: e = a−d

105

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Engineering Mechanics - Statics

rBA

⎛0⎞ = ⎜ −e ⎟ m ⎜ ⎟ ⎝ −c ⎠

rCB

⎛ −b ⎞ = ⎜ −d ⎟ m ⎜ ⎟ ⎝c ⎠

Chapter 2

rBC

⎛b⎞ = ⎜ d ⎟ ft ⎜ ⎟ ⎝ −c ⎠

θ = acos ⎜

rCD

⎛ −b ⎞ = ⎜ 0 ⎟ ft ⎜ ⎟ ⎝0⎠

φ = acos ⎜

⎛ rBA⋅ rBC ⎞ ⎟ ⎝ rBA rBC ⎠

θ = 74.0 deg

⎛ rCB⋅ rCD ⎞ ⎟ ⎝ rCB rCD ⎠

φ = 33.9 deg

⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠

θ = 64.6 deg

Problem 2-112 Determine the angle θ between the two cords. Given: a = 3m b = 2m c = 6m d = 3m e = 4m

Solution:

rAC

⎛b⎞ = ⎜ a ⎟ ft ⎜ ⎟ ⎝c⎠

rAB

⎛0⎞ = ⎜ −d ⎟ ft ⎜ ⎟ ⎝e ⎠

θ = acos ⎜

Problem 2-113 Determine the angle θ between the two cables. Given: a = 8 ft b = 10 ft 106

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Engineering Mechanics - Statics

Chapter 2

c = 8 ft d = 10 ft e = 4 ft f = 6 ft F AB = 12 lb

Solution:

rAC

⎛a − f⎞ = ⎜ −c ⎟ ft ⎜ ⎟ ⎝ b ⎠

rAB

⎛ −f ⎞ = ⎜ d − c ⎟ ft ⎜ ⎟ ⎝ e ⎠

⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠

θ = acos ⎜

θ = 82.9 deg

Problem 2-114 Determine the projected component of the force F acting in the direction of cable AC. Express the result as a Cartesian vector. Given: F = 12 lb

107

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Engineering Mechanics - Statics

Chapter 2

a = 8 ft b = 10 ft c = 8 ft d = 10 ft e = 4 ft f = 6 ft

Solution:

rAC

⎛a − f⎞ ⎜ −c ⎟ m = ⎜ ⎟ ⎝ b ⎠

rAB

⎛ −f ⎞ ⎜ ⎟ = d−c ⎜ ⎟ ⎝ e ⎠

F AC = ( F AB⋅ uAC) uAC

uAC =

rAC rAC

F AB = F

F AC

rAB rAB

uAC

⎛ 0.2 ⎞ ⎜ ⎟ = −0.6 ⎜ ⎟ ⎝ 0.8 ⎠

⎛ −9.6 ⎞ ⎜ ⎟ F AB = 3.2 lb ⎜ ⎟ ⎝ 6.4 ⎠

⎛ 0.229 ⎞ ⎜ ⎟ = −0.916 lb ⎜ ⎟ ⎝ 1.145 ⎠

Problem 2-115 Determine the components of F that act along rod AC and perpendicular to it. Point B is located at the midpoint of the rod.

108

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Engineering Mechanics - Statics

Chapter 2

Given: F = 600 N c = 4 m a = 4m

d = 3m

b = 6m

e = 4m

Solution: Find the force vector and the unit vector uAC.

rBD

⎛c + d ⎞ ⎜ 2⎟ ⎜ ⎟ e⎟ ⎜ = b− ⎜ 2⎟ ⎜ −a ⎟ ⎜ ⎟ ⎝ 2 ⎠

rAC

⎛ −d ⎞ ⎜ ⎟ = e ⎜ ⎟ ⎝ −a ⎠

⎛ 5.5 ⎞ ⎜ ⎟m rBD = 4 ⎜ ⎟ ⎝ −2 ⎠

rAC

Fv = F

⎛ −3 ⎞ ⎜ ⎟ = 4 m ⎜ ⎟ ⎝ −4 ⎠

uAC =

rBD rBD

rAC rAC

⎛ 465.5 ⎞ ⎜ ⎟ F v = 338.6 N ⎜ ⎟ ⎝ −169.3 ⎠

uAC

⎛ −0.5 ⎞ ⎜ ⎟ = 0.6 ⎜ ⎟ ⎝ −0.6 ⎠

Now find the component parallel to AC. F parallel = Fv ⋅ uAC

F parallel = 99.1 N

The perpendicular component is now found F perpendicular =

2

Fv ⋅ F v − F parallel

F perpendicular = 591.8 N

Problem 2-116 Determine the components of F that act along rod AC and perpendicular to it. Point B is located a distance f along the rod from end C.

109

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Engineering Mechanics - Statics

Chapter 2

Given: F = 600 N

c = 4m

a = 4m

d = 3m

b = 6m

e = 4m

f = 3m Solution: f

r = 2

2

d +e +a

2

Find the force vector and the unit vector uAC.

rBD

⎡c + d( 1 − r) ⎤ ⎢ ⎥ = b − e( 1 − r) ⎢ ⎥ ⎣ −a r ⎦

Fv = F

rAC

⎛ 5.5944 ⎞ ⎜ ⎟ rBD = 3.8741 m ⎜ ⎟ ⎝ −1.8741 ⎠ ⎛ 475.6 ⎞ ⎜ ⎟ F v = 329.3 N ⎜ ⎟ ⎝ −159.3 ⎠

rBD rBD

⎛ −d ⎞ ⎜ ⎟ = e ⎜ ⎟ ⎝ −a ⎠

rAC

⎛ −3 ⎞ ⎜ ⎟ = 4 m ⎜ ⎟ ⎝ −4 ⎠

uAC =

rAC rAC

uAC

⎛ −0.5 ⎞ ⎜ ⎟ = 0.6 ⎜ ⎟ ⎝ −0.6 ⎠

Now find the component parallel to AC. F parallel = Fv ⋅ uAC

F parallel = 82.4 N

The perpendicular component is now found F perpendicular =

2

Fv ⋅ F v − F parallel

F perpendicular = 594.3 N

Problem 2-117 Determine the magnitude of the projected component of the length of cord OA along the Oa axis.

110

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Engineering Mechanics - Statics

Chapter 2

Given: a = 10 ft b = 5 ft c = 15 ft d = 5 ft

θ 1 = 45 deg θ 2 = 60 deg

Solution:

rOA

⎛ cos ( θ 1) cos ( θ 2 ) ⎞ ⎜ ⎟ = d ⎜ cos ( θ 1 ) sin ( θ 2 ) ⎟ ⎜ ⎟ sin ( θ 1 ) ⎝ ⎠

rOa = rOA ⋅ uOa

rOa

⎛c ⎞ = ⎜ a ⎟ ⎜ ⎟ ⎝ −b ⎠

uOa =

rOa rOa

rOa = 2.1 ft

111

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Engineering Mechanics - Statics

Chapter 2

Problem 2-118 Force F acts at the end of the pipe. Determine the magnitudes of the components F1 and F 2 which are directed along the pipe's axis and perpendicular to it. Given:

a = 5 ft

⎛ 0 ⎞ ⎜ 0 ⎟ lb F = ⎜ ⎟ ⎝ −40 ⎠

b = 3 ft c = 3 ft

Solution:

⎛b⎞ ⎜ ⎟ r = a ⎜ ⎟ ⎝ −c ⎠

u =

F1 = F⋅ u F2 =

F⋅ F − F1

r r

F 1 = 18.3 lb 2

F 2 = 35.6 lb

Problem 2-119 Determine the projected component of the force F acting along the axis AB of the pipe.

112

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Engineering Mechanics - Statics

Chapter 2

Given: F = 80 N a = 4m b = 3m c = 12 m d = 2m e = 6m

Solution: Find the force and the unit vector

⎛ −e ⎞ ⎜ ⎟ rA = −a − b ⎜ ⎟ ⎝ d−c ⎠

rAB

⎛ −e ⎞ ⎜ ⎟ = −b ⎜ ⎟ ⎝d⎠

⎛ −6 ⎞ ⎜ ⎟ rA = −7 m ⎜ ⎟ ⎝ −10 ⎠ ⎛ −6 ⎞ ⎜ ⎟ rAB = −3 m ⎜ ⎟ ⎝2⎠

Fv = F

uAB =

rA rA

rAB rAB

⎛ −35.3 ⎞ ⎜ ⎟ F v = −41.2 N ⎜ ⎟ ⎝ −58.8 ⎠ ⎛ −0.9 ⎞ ⎜ ⎟ uAB = −0.4 ⎜ ⎟ ⎝ 0.3 ⎠

Now find the projection using the Dot product. F AB = F v ⋅ uAB

F AB = 31.1 N

Problem 2-120 Determine the angles θ and φ between the axis OA of the pole and each cable, AB and AC. Given: F 1 = 50 N

113

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Engineering Mechanics - Statics

Chapter 2

F 2 = 35 N a = 1m b = 3m c = 2m d = 5m e = 4m f = 6m g = 4m Solution:

rAO

⎛0⎞ ⎜ ⎟ = −g ⎜ ⎟ ⎝−f ⎠

rAB

⎛e⎞ ⎜ ⎟ = a ⎜ ⎟ ⎝−f⎠

rAC

⎛ rAO⋅ rAB ⎞ ⎟ ⎝ rAO rAB ⎠

θ = 52.4 deg

⎛ rAO⋅ rAC ⎞ ⎟ ⎝ rAO rAC ⎠

φ = 68.2 deg

θ = acos ⎜

φ = acos ⎜

⎛ −c ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ −f ⎠

Problem 2-121 The two cables exert the forces shown on the pole. Determine the magnitude of the projected component of each force acting along the axis OA of the pole. Given: F 1 = 50 N F 2 = 35 N a = 1m b = 3m c = 2m 114

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Engineering Mechanics - Statics

Chapter 2

d = 5m e = 4m f = 6m g = 4m Solution:

rAB

⎛e⎞ ⎜ ⎟ = a ⎜ ⎟ ⎝−f⎠

F 1v = F1

F 2v = F2

rAC

rAC rAC rAB rAB

⎛ −c ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ −f ⎠

rAO

⎛0⎞ ⎜ ⎟ = −g ⎜ ⎟ ⎝−f ⎠

F 1AO = F 1v⋅ uAO

F 1AO = 18.5 N

F 2AO = F 2v⋅ uAO

F 2AO = 21.3 N

uAO =

rAO rAO

Problem 2-122 Force F is applied to the handle of the wrench. Determine the angle θ between the tail of the force and the handle AB.

115

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Engineering Mechanics - Statics

Chapter 2

Given: a = 300 mm b = 500 mm F = 80 N

θ 1 = 30 deg θ 2 = 45 deg Solution:

⎛ −cos ( θ 1) sin ( θ 2) ⎞ ⎜ ⎟ F v = F ⎜ cos ( θ 1 ) cos ( θ 2 ) ⎟ ⎜ ⎟ sin ( θ 1 ) ⎝ ⎠ ⎛ Fv⋅ uab ⎞ ⎟ ⎝ F ⎠

θ = acos ⎜

uab

⎛0⎞ ⎜ ⎟ = −1 ⎜ ⎟ ⎝0⎠

θ = 127.8 deg

Problem 2-123 Two cables exert forces on the pipe. Determine the magnitude of the projected component of F 1 along the line of action of F2. Given: F 1 = 30 lb

β = 30 deg

F 2 = 25 lb

γ = 60 deg

α = 30 deg

ε = 60 deg

Solution: We first need to find the third angle ( > 90 deg) that locates force F 2. Initial Guess:

φ = 120 deg

116

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Engineering Mechanics - Statics

Chapter 2

Given cos ( ε ) + cos ( γ ) + cos ( φ ) = 1 2

2

φ = Find ( φ )

2

φ = 135 deg

Find the force F 1v and the unit vector u2.

F 1v

⎛⎜ cos ( α ) sin ( β ) ⎞⎟ = F1 ⎜ cos ( α ) cos ( β ) ⎟ ⎜ −sin ( α ) ⎟ ⎝ ⎠

⎛⎜ cos ( φ ) ⎞⎟ u2 = ⎜ cos ( ε ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠ Now find the projection

⎛ 13 ⎞ F 1v = ⎜ 22.5 ⎟ lb ⎜ ⎟ ⎝ −15 ⎠ ⎛ −0.7 ⎞ u2 = ⎜ 0.5 ⎟ ⎜ ⎟ ⎝ 0.5 ⎠

F 12 = F1v⋅ u2

F 12 = 5.4 lb

Problem 2-124 Determine the angle θ between the two cables attached to the pipe. Given: F 1 = 30 lb

β = 30 deg

F 2 = 25 lb

γ = 60 deg

α = 30 deg

ε = 60 deg

Solution: We first need to find the third angle ( > 90 deg) that locates force F 2. Initial Guesses:

φ = 120 deg

117

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Engineering Mechanics - Statics

Chapter 2

Given cos ( ε ) + cos ( γ ) + cos ( φ ) = 1 2

2

2

φ = Find ( φ )

φ = 135 deg

Find the unit vectors u1 and u2.

⎛⎜ cos ( α ) sin ( β ) ⎞⎟ u1 = ⎜ cos ( α ) cos ( β ) ⎟ ⎜ −sin ( α ) ⎟ ⎝ ⎠

⎛ 0.4 ⎞ u1 = ⎜ 0.8 ⎟ ⎜ ⎟ ⎝ −0.5 ⎠

⎛⎜ cos ( φ ) ⎞⎟ u2 = ⎜ cos ( ε ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

⎛ −0.7 ⎞ u2 = ⎜ 0.5 ⎟ ⎜ ⎟ ⎝ 0.5 ⎠

Find the angle using the dot product

θ = acos ( u1 ⋅ u2 )

θ = 100.4 deg

Problem 2-125 Determine the angle θ between the two cables. Given: a = 7.5 ft b = 2 ft c = 3 ft d = 2 ft e = 3 ft f = 3 ft F 1 = 60 lb F 2 = 30 lb

118

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Engineering Mechanics - Statics

Chapter 2

Solution:

rAC

⎛ −d − b ⎞ = ⎜ −c − f ⎟ ⎜ ⎟ ⎝ a−e ⎠

rAB

⎛ −d = ⎜ −c − ⎜ ⎝ −e

⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠

θ = acos ⎜

⎞ f⎟ ⎟ ⎠

θ = 59.2 deg

Problem 2-126 Determine the projection of the force F 1 along cable AB. Determine the projection of the force F 2 along cable AC. Given: a = 7.5 ft b = 2 ft c = 3 ft d = 2 ft e = 3 ft f = 3 ft F 1 = 60 lb F 2 = 30 lb Solution:

rAC

⎛ −d − b ⎞ = ⎜ −c − f ⎟ ⎜ ⎟ ⎝ a−e ⎠

rAB

⎛ −d = ⎜ −c − ⎜ ⎝ −e

⎞ ⎟ f ⎟ ⎠

uAC =

rAC rAC

F 1v = F1 uAC

F 1AB = F1v⋅ uAB

F 1AB = 30.8 lb

F 2v = F2 uAB

F 2AC = F2v⋅ uAC

F 2AC = 15.4 lb

uAB =

rAB rAB

119

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Engineering Mechanics - Statics

Chapter 2

Problem 2-127 Determine the angle θ between the edges of the sheet-metal bracket. Given: a = 50 mm b = 300 mm c = 250 mm d = 400 mm Solution: Find the unit vectors and use the dot product

⎛d⎞ ⎜ ⎟ r1 = 0 ⎜ ⎟ ⎝c⎠

u1 =

⎛a⎞ ⎜ ⎟ r2 = b ⎜ ⎟ ⎝0⎠

u2 =

r1 r1

r2 r2

⎛ 0.848 ⎞ ⎛ 0.164 ⎞ ⎜ ⎟ ⎜ ⎟ u1 = 0.000 u2 = 0.986 ⎜ ⎟ ⎜ ⎟ ⎝ 0.530 ⎠ ⎝ 0.000 ⎠

θ = acos ( u1 ⋅ u2 )

θ = 82 deg

Problem 2-128 Determine the magnitude of the projected component of the force F acting along the axis BC of the pipe.

120

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Engineering Mechanics - Statics

Chapter 2

Given: F = 100 lb a = 2 ft b = 8 ft c = 6 ft d = 4 ft e = 2 ft

Solution:

rCD

⎛ −c ⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝e⎠

uCD =

F BC = ( FuCD) ⋅ uCB

rCD

rCB

rCD

⎛ −c ⎞ ⎜ ⎟ = −d ⎜ ⎟ ⎝e ⎠

uCB =

rCB rCB

F BC = 10.5 lb

Problem 2-129 Determine the angle θ between pipe segments BA and BC. Given: F = 100 lb a = 3 ft b = 8 ft c = 6 ft d = 4 ft e = 2 ft Solution:

rBC

⎛c⎞ ⎜ ⎟ = d ⎜ ⎟ ⎝ −e ⎠

rBA

⎛ −a ⎞ ⎜ ⎟ = 0 ⎜ ⎟ ⎝0⎠

⎛ rBC⋅ rBA ⎞ ⎟ ⎝ rBC rBA ⎠

θ = acos ⎜

θ = 143.3 deg

121 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-130 Determine the angles θ and φ made between the axes OA of the flag pole and AB and AC, respectively, of each cable. Given: F B = 55 N

c = 2m

F c = 40 N

d = 4m

a = 6m

e = 4m

b = 1.5 m

f = 3m

Solution:

rAO

⎛0⎞ ⎜ ⎟ = −e ⎜ ⎟ ⎝−f⎠

rAB

⎛ b ⎞ ⎜ −e ⎟ = ⎜ ⎟ ⎝a − f⎠

rAC

⎛ −c ⎞ ⎜ −e ⎟ = ⎜ ⎟ ⎝d − f⎠

⎛ rAB⋅ rAO ⎞ ⎟ ⎝ rAB rAO ⎠

θ = 74.4 deg

⎛ rAC⋅ rAO ⎞ ⎟ ⎝ rAC rAO ⎠

φ = 55.4 deg

θ = acos ⎜ φ = acos ⎜

Problem 2-131 Determine the magnitude and coordinate direction angles of F3 so that resultant of the three forces acts along the positive y axis and has magnitude FR.

122

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Engineering Mechanics - Statics

Chapter 2

Given: F R = 600 lb F 1 = 180 lb F 2 = 300 lb

φ = 40 deg θ = 30 deg

Solution: The initial guesses: F 3 = 100 lb

β = 30 deg

α = 10 deg

γ = 60 deg

Given F Rx = ΣF x;

−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0

F Ry = ΣF y;

F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = FR

F Rz = ΣF z;

−F 2 sin ( θ ) + F3 cos ( γ ) = 0 cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

2

Solving:

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

⎛⎜ α ⎞⎟ ⎛ 88.3 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 20.6 ⎟ deg ⎜ γ ⎟ ⎝ 69.5 ⎠ ⎝ ⎠

F 3 = 428.3 lb

123

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Engineering Mechanics - Statics

Chapter 2

Problem 2-132 Determine the magnitude and coordinate direction angles of F3 so that resultant of the three forces is zero. Given: F 1 = 180 lb F 2 = 300 lb

φ = 40 deg θ = 30 deg

Solution: The initial guesses:

α = 10 deg β = 30 deg γ = 60 deg F 3 = 100 lb

Given F Rx = ΣF x;

−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0

F Ry = ΣF y;

F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = 0

F Rz = ΣF z;

−F 2 sin ( θ ) + F3 cos ( γ ) = 0 cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

2

Solving:

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

⎛⎜ α ⎞⎟ ⎛ 87 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 142.9 ⎟ deg ⎜ γ ⎟ ⎝ 53.1 ⎠ ⎝ ⎠

F 3 = 249.6 lb

124

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Engineering Mechanics - Statics

Chapter 2

Problem 2-133 Resolve the force F into two components, one acting parallel and the other acting perpendicular to the u axis. Given: F = 600 lb

θ 1 = 60 deg θ 2 = 20 deg Solution: F perpendicular = F cos ( θ 1 − θ 2 ) F perpendicular = 460 lb F parallel = F sin ( θ 1 − θ 2 ) F parallel = 386 lb

Problem 2-134 The force F has a magnitude F and acts at the midpoint C of the thin rod. Express the force as a Cartesian vector.

125

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Engineering Mechanics - Statics

Chapter 2

Given: F = 80 lb a = 2 ft b = 3 ft c = 6 ft Solution:

rCO

Fv = F

⎛ −b ⎞ ⎜ 2 ⎟ ⎜ ⎟ a ⎟ = ⎜ ⎜ 2 ⎟ ⎜ −c ⎟ ⎜ ⎟ ⎝ 2 ⎠

rCO rCO

⎛ −34.3 ⎞ ⎜ ⎟ F v = 22.9 lb ⎜ ⎟ ⎝ −68.6 ⎠

Problem 2-135 Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by first finding the resultant F' = F1 + F 3 and then forming FR = F ' + F 2. Specify its direction measured counterclockwise from the positive x axis.

126

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 80 N F 2 = 75 N F 3 = 50 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg

Solution:

⎛ −sin ( θ 1 ) ⎞ ⎟ ⎝ cos ( θ 1) ⎠

F 1v = F1 ⎜

⎛ cos ( θ 2 + θ 3 ) ⎞ ⎟ ⎝ sin ( θ 2 + θ 3 ) ⎠

⎛ cos ( θ 3) ⎞ ⎟ ⎝ sin ( θ 3) ⎠

F 2v = F2 ⎜

⎛ −4.6 ⎞ ⎜ ⎟N ⎝ 104.6 ⎠

F' = F1v + F3v

F' =

F R = F' + F2v

FR =

i =

⎛ 14.8 ⎞ ⎜ ⎟N ⎝ 177.1 ⎠

F 3v = F3 ⎜

⎛1⎞ ⎜ ⎟ ⎝0⎠

j =

⎛0⎞ ⎜ ⎟ ⎝1⎠

F R = 177.7 N

⎛ FR j ⎞ ⎟ ⎝ FR i ⎠

θ = atan ⎜

θ = 85.2 deg

Problem 2-136 The leg is held in position by the quadriceps AB, which is attached to the pelvis at A. If the force exerted on this muscle by the pelvis is F , in the direction shown, determine the stabilizing force component acting along the positive y axis and the supporting force component acting along the negative x axis. Given: F = 85 N

127

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Engineering Mechanics - Statics

Chapter 2

θ 1 = 55 deg θ 2 = 45 deg

Solution: F x = F cos ( θ 1 − θ 2 ) F x = 83.7 N F y = F sin ( θ 1 − θ 2 ) F y = 14.8 N

Problem 2-137 Determine the magnitudes of the projected components of the force F in the direction of the cables AB and AC . Given:

⎛ 60 ⎞ ⎜ ⎟ F = 12 N ⎜ ⎟ ⎝ −40 ⎠ a = 3m b = 1.5 m c = 1m d = 0.75 m e = 1m

128

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Engineering Mechanics - Statics

Chapter 2

Solution: Find the unit vectors, then use the dot product

rAB

rAC

⎛ −a ⎞ ⎜ ⎟ = −d ⎜ ⎟ ⎝e ⎠ ⎛ −a ⎞ ⎜ ⎟ = c ⎜ ⎟ ⎝b⎠

⎛ −3 ⎞ ⎜ ⎟ rAB = −0.8 m ⎜ ⎟ ⎝ 1 ⎠ ⎛ −3 ⎞ ⎜ ⎟ rAC = 1 m ⎜ ⎟ ⎝ 1.5 ⎠ ⎛ −78.5 ⎞ ⎜ ⎟ F AB = −19.6 N ⎜ ⎟ ⎝ 26.2 ⎠

F AB = FuAB

uAB =

uAC =

rAB rAB rAC rAC

F AC = FuAC

⎛ −0.9 ⎞ ⎜ ⎟ uAB = −0.2 ⎜ ⎟ ⎝ 0.3 ⎠ ⎛ −0.9 ⎞ ⎜ ⎟ uAC = 0.3 ⎜ ⎟ ⎝ 0.4 ⎠ F AC

⎛ −72.9 ⎞ ⎜ ⎟ = 24.3 N ⎜ ⎟ ⎝ 36.4 ⎠

Problem 2-138 Determine the magnitude and coordinate direction angles of F3 so that resultant of the three forces is zero. Given: F 1 = 180 lb

φ = 40 deg

F 2 = 300 lb

θ = 30 deg

Solution: The initial guesses:

α = 10 deg β = 30 deg γ = 60 deg F 3 = 100 lb

Given F Rx = ΣFx;

−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0

F Ry = ΣF y;

F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = 0 129

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Engineering Mechanics - Statics

Chapter 2

−F 2 sin ( θ ) + F3 cos ( γ ) = 0

F Rz = ΣF z;

cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

2

Solving:

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

⎛⎜ α ⎞⎟ ⎛ 87 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 142.9 ⎟ deg ⎜ γ ⎟ ⎝ 53.1 ⎠ ⎝ ⎠

F 3 = 249.6 lb

Problem 2-139 Determine the angles θ and φ so that the resultant force is directed along the positive x axis and has magnitude FR. . Given: F 1 = 30 lb F 2 = 30 lb F R = 20 lb Solution: Initial Guesses:

θ = 20 deg

φ = 20 deg

Given F1

sin ( φ )

=

F2

sin ( θ )

F R = F 1 + F2 − 2 F 1 F2 cos ( 180 deg − θ − φ ) 2

2

2

130

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Engineering Mechanics - Statics

Chapter 2

⎛θ⎞ ⎜ ⎟ = Find ( θ , φ ) ⎝φ⎠ θ = 70.5 deg φ = 70.5 deg

Problem 2-140 Determine the magnitude of the resultant force and its direction measured counterclockwise from the x axis. Given: F 1 = 300 lb F 2 = 200 lb

θ 1 = 40 deg θ 2 = 100 deg

Solution: F Rx = F 1 cos ( 180 deg − θ 2 ) + F2 cos ( θ 1 )

F Rx = 205.3 lb

F Ry = F 1 sin ( 180 deg − θ 2 ) − F 2 sin ( θ 1 )

F Ry = 166.9 lb

FR =

2

FRx + FRy

2

F R = 265 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 39.1 deg

131

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Engineering Mechanics - Statics

Chapter 3

Problem 3-1 Determine the magnitudes of F1 and F2 so that the particle is in equilibrium. Given: F = 500 N

θ 1 = 45 deg θ 2 = 30deg Solution: Initial Guesses F 1 = 1N

F 2 = 1N

Given + ΣF x = 0; →

F 1 cos ( θ 1 ) + F2 cos ( θ 2 ) − F = 0

+

F 1 sin ( θ 1 ) − F 2 sin ( θ 2 ) = 0

↑ ΣFy = 0;

⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , F2) ⎝ F2 ⎠

⎛ F1 ⎞ ⎛ 259 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ F2 ⎠ ⎝ 366 ⎠

Problem 3-2 Determine the magnitude and direction θ of F so that the particle is in equilibrium. Units Used: 3

kN = 10 N Given: F 1 = 7 kN F 2 = 3 kN c = 4 d = 3 132

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Engineering Mechanics - Statics

Chapter 3

Solution: F = 1kN

The initial guesses:

θ = 30deg

Given Equations of equilibrium: + Σ F x = 0; →

+

↑

Σ F y = 0;

⎛ −d ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ c +d ⎠ c ⎛ ⎞ ⎜ 2 2 ⎟ F1 − F2 − F sin ( θ ) = 0 ⎝ c +d ⎠

⎛F⎞ ⎜ ⎟ = Find ( F , θ ) ⎝θ ⎠

F = 4.94 kN

θ = 31.8 deg

Problem 3-3 Determine the magnitude of F and the orientation θ of the force F 3 so that the particle is in equilibrium. Given: F 1 = 700 N F 2 = 450 N F 3 = 750 N

θ 1 = 15 deg θ 2 = 30 deg

Solution: Initial Guesses:

F = 1N

θ = 10deg

Given + ΣF x = 0; →

F 1 cos ( θ 1 ) − F2 sin ( θ 2 ) − F3 cos ( θ ) = 0 133

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Engineering Mechanics - Statics

+

↑ ΣFy = 0;

Chapter 3

F + F2 cos ( θ 2 ) + F 1 sin ( θ 1 ) − F 3 sin ( θ ) = 0

⎛F⎞ ⎜ ⎟ = Find ( F , θ ) ⎝θ ⎠

F = 28.25 N

θ = 53.02 deg

Problem 3-4 Determine the magnitude and angle θ of F so that the particle is in equilibrium. Units Used: 3

kN = 10 N Given: F 1 = 4.5 kN F 2 = 7.5 kN F 3 = 2.25 kN

α = 60 deg φ = 30 deg

Solution: Guesses: F = 1 kN

θ = 1

Given Equations of Equilibrium: + → +

Σ F x = 0;

↑ Σ F y = 0;

⎛F⎞ ⎜ ⎟ = Find ( F , θ ) ⎝θ ⎠

F cos ( θ ) − F2 sin ( φ ) − F 1 + F 3 cos ( α ) = 0 F sin ( θ ) − F 2 cos ( φ ) − F 3 sin ( α ) = 0 F = 11.05 kN

θ = 49.84 deg

134

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Engineering Mechanics - Statics

Chapter 3

Problem 3-5 The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium. Units Used: 3

kN = 10 N Given: F 1 = 8 kN F 2 = 5 kN

θ 1 = 45 deg θ = 30 deg

Solution: + ΣF x = 0; →

−T cos ( θ ) + F 1 + F 2 sin ( θ 1 ) = 0

T =

F1 + F2 sin ( θ 1 ) cos ( θ )

T = 13.3 kN +

↑ ΣFy = 0;

F − T sin ( θ ) − F 2 cos ( θ 1 ) = 0 F = T sin ( θ ) + F2 cos ( θ 1 ) F = 10.2 kN

Problem 3-6 The gusset plate is subjected to the forces of four members. Determine the force in member B and its proper orientation θ for equilibrium. The forces are concurrent at point O. Units Used: 3

kN = 10 N 135

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Engineering Mechanics - Statics

Chapter 3

Given: F = 12 kN F 1 = 8 kN F 2 = 5 kN

θ 1 = 45 deg Solution: Initial Guesses

T = 1kN

θ = 10deg

Given + ΣF x = 0; →

F 1 − T cos ( θ ) + F 2 sin ( θ 1 ) = 0

+

−T sin ( θ ) − F2 cos ( θ 1 ) + F = 0

↑ ΣFy = 0;

⎛T⎞ ⎜ ⎟ = Find ( T , θ ) ⎝θ⎠

T = 14.31 kN

θ = 36.27 deg

Problem 3-7 Determine the maximum weight of the engine that can be supported without exceeding a tension of T1 in chain AB and T2 in chain AC. Given:

θ = 30 deg T1 = 450 lb T2 = 480 lb Solution: Initial Guesses F AB = T1 F AC = T2 W = 1lb 136

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Engineering Mechanics - Statics

Chapter 3

Given Assuming cable AB reaches the maximum tension F AB = T1. + ΣF x = 0; →

F AC cos ( θ ) − F AB = 0

+

F AC sin ( θ ) − W = 0

↑ ΣFy = 0;

⎛ FAC1 ⎞ ⎜ ⎟ = Find ( FAC , W) ⎝ W1 ⎠ Given

W1 = 259.81 lb

Assuming cable AC reaches the maximum tension FAC = T2.

+ ΣF x = 0; →

F AC cos ( θ ) − F AB = 0

+

F AC sin ( θ ) − W = 0

↑ ΣFy = 0;

⎛ FAB2 ⎞ ⎜ ⎟ = Find ( FAB , W) ⎝ W2 ⎠ W = min ( W1 , W2 )

W2 = 240.00 lb

W = 240.00 lb

Problem 3-8 The engine of mass M is suspended from a vertical chain at A. A second chain is wrapped around the engine and held in position by the spreader bar BC. Determine the compressive force acting along the axis of the bar and the tension forces in segments BA and CA of the chain. Hint: Analyze equilibrium first at A, then at B. Units Used: 3

kN = 10 N Given: M = 200 kg

θ 1 = 55 deg g = 9.81

m 2

s

137

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Engineering Mechanics - Statics

Chapter 3

Solution: Initial guesses: Given

F BA = 1 kN

F CA = 2 kN

Point A

+ ΣF x = 0; →

F CA cos ( θ 1 ) − FBA cos ( θ 1 ) = 0

+

M( g) − FCA sin ( θ 1 ) − FBA sin ( θ 1 ) = 0

↑ ΣFy = 0;

⎛ FBA ⎞ ⎜ ⎟ = Find ( FBA , FCA) ⎝ FCA ⎠

⎛ FBA ⎞ ⎛ 1.20 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ FCA ⎠ ⎝ 1.20 ⎠

At point B: + ΣF x = 0; →

F BA cos ( θ 1 ) − F BC = 0 F BC = FBA cos ( θ 1 )

F BC = 687 N

Problem 3-9 Cords AB and AC can each sustain a maximum tension T. If the drum has weight W, determine the smallest angle θ at which they can be attached to the drum. Given: T = 800 lb W = 900 lb Solution: +

↑ Σ Fy = 0;

W − 2T sin ( θ ) = 0 W⎞ ⎟ ⎝ 2T ⎠

θ = asin ⎛⎜

θ = 34.2 deg

138

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Engineering Mechanics - Statics

Chapter 3

Problem 3-10 The crate of weight W is hoisted using the ropes AB and AC. Each rope can withstand a maximum tension T before it breaks. If AB always remains horizontal, determine the smallest angle θ to which the crate can be hoisted. Given: W = 500 lb T = 2500 lb Solution: TAB = T

Case 1: Assume The initial guess

θ = 30 deg

TAC = 2000 lb

Given + Σ F x = 0; →

TAB − TAC cos ( θ ) = 0

+

TAC sin ( θ ) − W = 0

↑ Σ F y = 0;

⎛ TAC1 ⎞ ⎜ ⎟ = Find ( TAC , θ ) ⎝ θ1 ⎠

θ 1 = 11.31 deg

TAC1 = 2550 lb

TAC = T

Case 1: Assume The initial guess

θ = 30 deg

TAB = 2000 lb

Given + Σ F x = 0; →

TAB − TAC cos ( θ ) = 0

+

TAC sin ( θ ) − W = 0

↑ Σ F y = 0;

⎛ TAB2 ⎞ ⎜ ⎟ = Find ( TAB , θ ) ⎝ θ2 ⎠ θ = max ( θ 1 , θ 2 )

θ 2 = 11.54 deg

TAB2 = 2449 lb

θ = 11.54 deg

139

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Engineering Mechanics - Statics

Chapter 3

Problem 3-11 Two electrically charged pith balls, each having mass M, are suspended from light threads of equal length. Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured distance between them is r. Given: M = 0.2 gm r = 200 mm l = 150 mm d = 50 mm Solution: The initial guesses : T = 200 N

F = 200 N

Given +

↑

Σ F x = 0;

⎛ r − d⎞ = 0 ⎟ ⎝ 2l ⎠

F − T⎜

2 ⎡⎢ 2 r − d ⎞ ⎥⎤ ⎛ ⎢ l − ⎜⎝ 2 ⎟⎠ ⎥ + → Σ Fy = 0; T⎢⎣ ⎥ − Mg = 0 l ⎦

⎛T ⎞ ⎜ ⎟ = Find ( T , F) ⎝F⎠

F = 1.13 × 10

−3

T = 0.00 N

N

Problem 3-12 The towing pendant AB is subjected to the force F which is developed from a tugboat. Determine the force that is in each of the bridles, BC and BD, if the ship is moving forward with constant velocity. Units Used: 3

kN = 10 N

140

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Engineering Mechanics - Statics

Chapter 3

Given: F = 50 kN

θ 1 = 20 deg θ 2 = 30 deg Solution: Initial guesses:

TBC = 1 kN

TBD = 2 kN

Given + ΣF x = 0; →

TBC sin ( θ 2 ) − TBD sin ( θ 1 ) = 0

+

TBC cos ( θ 2 ) + TBD cos ( θ 1 ) − F = 0

↑ ΣFy = 0;

⎛ TBC ⎞ ⎜ ⎟ = Find ( TBC , TBD) ⎝ TBD ⎠

⎛ TBC ⎞ ⎛ 22.32 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ TBD ⎠ ⎝ 32.64 ⎠

Problem 3-13 Determine the stretch in each spring for equilibrium of the block of mass M. The springs are shown in the equilibrium position.

141

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Engineering Mechanics - Statics

Chapter 3

Given: M = 2 kg a = 3m b = 3m c = 4m kAB = 30

N

kAC = 20

N

kAD = 40

N

g = 9.81

m

m

m m 2

s Solution:

The initial guesses: F AB = 1 N F AC = 1 N Given +

→

Σ F x = 0;

+

Σ F y = 0;

↑

F AB⎛

c b ⎛ ⎞ ⎞ ⎜ 2 2 ⎟ − FAC⎜ 2 2 ⎟ = 0 ⎝ a +c ⎠ ⎝ a +b ⎠

F AC⎛

a ⎛ a ⎞ ⎞ ⎜ 2 2 ⎟ + FAB⎜ 2 2 ⎟ − M g = 0 ⎝ a +b ⎠ ⎝ a +c ⎠

⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , FAB) ⎝ FAB ⎠ xAC =

xAB =

F AC kAC FAB kAB

⎛ FAC ⎞ ⎛ 15.86 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FAB ⎠ ⎝ 14.01 ⎠

xAC = 0.79 m

xAB = 0.47 m 142

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Engineering Mechanics - Statics

Chapter 3

Problem 3-14 The unstretched length of spring AB is δ. If the block is held in the equilibrium position shown, determine the mass of the block at D. Given:

δ = 2m a = 3m b = 3m c = 4m kAB = 30

N

kAC = 20

N

kAD = 40

N

g = 9.81

m

m

m m 2

s Solution:

F AB = kAB

(

2

2

a +c −δ

)

The initial guesses: mD = 1 kg

F AC = 1 N

Given + Σ F x = 0; →

+

↑

Σ F y = 0;

F AB⎛

c b ⎛ ⎞ ⎞ ⎜ 2 2 ⎟ − FAC⎜ 2 2 ⎟ = 0 ⎝ a +c ⎠ ⎝ a +b ⎠

F AC⎛

⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , mD) ⎝ mD ⎠

a ⎛ a ⎞ ⎞ ⎜ 2 2 ⎟ + FAB⎜ 2 2 ⎟ − mD g = 0 ⎝ a +b ⎠ ⎝ a +c ⎠ F AC = 101.8 N

mD = 12.8 kg

143

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Engineering Mechanics - Statics

Chapter 3

Problem 3-15 The springs AB and BC have stiffness k and unstretched lengths l/2. Determine the horizontal force F applied to the cord which is attached to the small pulley B so that the displacement of the pulley from the wall is d. Given: l = 6m k = 500

N m

d = 1.5 m

Solution:

⎡

2

⎛ l ⎞ + d2 − ⎜ ⎟ ⎣ ⎝ 2⎠

T = k⎢

l⎤

⎥

T = 177.05 N

2⎦

+ Σ F x = 0; →

d 2

d +

F =

d

( 2T)

( 2T) − F = 0

⎛l⎞ ⎜ ⎟ ⎝ 2⎠

2

F = 158.36 N

2

⎛ l ⎞ + d2 ⎜ ⎟ ⎝ 2⎠

Problem 3-16 The springs AB and BC have stiffness k and an unstretched length of l. Determine the displacement d of the cord from the wall when a force F is applied to the cord.

144

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Engineering Mechanics - Statics

Chapter 3

Given: l = 6m k = 500

N m

F = 175 N Solution: The initial guesses: d = 1m

T = 1N

Given + Σ F x = 0; →

d

−F + ( 2T) 2

d +

⎡

Spring

⎛T⎞ ⎜ ⎟ = Find ( T , d) ⎝d ⎠

T = k⎢ d +

⎣

2

=0

⎛l⎞ ⎜ ⎟ ⎝ 2⎠

2

2 ⎤ ⎛ l ⎞ − l⎥ ⎜ ⎟ 2⎦ ⎝ 2⎠

T = 189.96 N

d = 1.56 m

Problem 3-17 Determine the force in each cable and the force F needed to hold the lamp of mass M in the position shown. Hint: First analyze the equilibrium at B; then, using the result for the force in BC, analyze the equilibrium at C. Given: M = 4 kg

θ 1 = 30 deg θ 2 = 60 deg θ 3 = 30 deg Solution: Initial guesses: 145

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Engineering Mechanics - Statics

TBC = 1 N

Chapter 3

TBA = 2 N

Given At B: + ΣF x = 0; →

TBC cos ( θ 1 ) − TBA cos ( θ 2 ) = 0

+

TBA sin ( θ 2 ) − TBC sin ( θ 1 ) − M g = 0

↑ ΣFy = 0;

⎛ TBC ⎞ ⎜ ⎟ = Find ( TBC , TBA) ⎝ TBA ⎠ At C:

TCD = 1 N

⎛ TBC ⎞ ⎛ 39.24 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ TBA ⎠ ⎝ 67.97 ⎠ F = 2N

Given + ΣF x = 0; →

−TBC cos ( θ 1 ) + TCD cos ( θ 3 ) = 0

+

TBC sin ( θ 1 ) + TCD sin ( θ 3 ) − F = 0

↑ ΣFy = 0;

⎛ TCD ⎞ ⎜ ⎟ = Find ( TCD , F) ⎝ F ⎠

⎛ TCD ⎞ ⎛ 39.24 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ F ⎠ ⎝ 39.24 ⎠

Problem 3-18 The motor at B winds up the cord attached to the crate of weight W with a constant speed. Determine the force in cord CD supporting the pulley and the angle θ for equilibrium. Neglect the

146

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Engineering Mechanics - Statics

Chapter 3

size of the pulley at C. Given: W = 65 lb

c = 12

d = 5

Solution: The initial guesses: θ = 100 deg

F CD = 200 lb

Given Equations of Equilibrium: + Σ F x = 0; →

+

↑Σ Fy = 0;

⎛

⎞=0 2 2⎟ ⎝ c +d ⎠

F CD cos ( θ ) − W⎜

⎛

d

⎞−W=0 ⎟ 2 2 ⎝ c +d ⎠

F CD sin ( θ ) − W⎜

c

⎛ θ ⎞ ⎜ ⎟ = Find ( θ , FCD) ⎝ FCD ⎠ θ = 78.69 deg F CD = 127.5 lb

Problem 3-19 The cords BCA and CD can each support a maximum load T. Determine the maximum weight of the crate that can be hoisted at constant velocity, and the angle θ for equilibrium.

147

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Engineering Mechanics - Statics

Chapter 3

Given: T = 100 lb c = 12 d = 5 The maximum will occur in CD rather than in BCA. Solution : The initial guesses: θ = 100 deg

W = 200 lb

Given Equations of Equilibrium: + Σ F x = 0; →

+

↑

⎛

⎞=0 ⎝ c +d ⎠

T cos ( θ ) − W⎜

⎛ Σ F y = 0; T sin ( θ ) − W⎜

d

2

2⎟

⎞−W=0 ⎝ c +d ⎠ c

2

2⎟

⎛θ⎞ ⎜ ⎟ = Find ( θ , W) ⎝W⎠ θ = 78.69 deg W = 51.0 lb

148

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Engineering Mechanics - Statics

Chapter 3

Problem 3-20 The sack has weight W and is supported by the six cords tied together as shown. Determine the tension in each cord and the angle θ for equilibrium. Cord BC is horizontal. Given: W = 15 lb

θ 1 = 30 deg θ 2 = 45 deg θ 3 = 60 deg Solution: Guesses

TBE = 1 lb

TAB = 1 lb

θ = 20deg

TBC = 1 lb

TAC = 1 lb

TCD = 1 lb

TAH = 1 lb

Given At H: +

↑ ΣFy = 0;

TAH − W = 0

At A: + ΣF x = 0; →

−TAB cos ( θ 2 ) + TAC cos ( θ 3 ) = 0

+

TAB sin ( θ 2 ) + TAC sin ( θ 3 ) − W = 0

↑ ΣFy = 0;

At B: + ΣF x = 0; →

TBC − TBE cos ( θ 1 ) + TAB cos ( θ 2 ) = 0

+

TBE sin ( θ 1 ) − TAB sin ( θ 2 ) = 0

↑ ΣFy = 0;

At C: + ΣF x = 0; →

TCD cos ( θ ) − TBC − TBE cos ( θ 3 ) = 0

+

TCD sin ( θ ) − TAC sin ( θ 3 ) = 0

↑ ΣFy = 0;

149

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Engineering Mechanics - Statics

Chapter 3

⎛ TBE ⎞ ⎜ ⎟ ⎜ TAB ⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TAC ⎟ = Find ( TBE , TAB , TBC , TAC , TCD , TAH , θ ) ⎜T ⎟ ⎜ CD ⎟ ⎜ TAH ⎟ ⎜ ⎟ ⎝ θ ⎠

⎛⎜ TBE ⎞⎟ ⎛⎜ 10.98 ⎞⎟ T ⎜ AB ⎟ 7.76 ⎟ ⎜ ⎟ ⎜ ⎜ TBC ⎟ = ⎜ 4.02 ⎟ lb ⎜ TAC ⎟ ⎜ 10.98 ⎟ ⎟ ⎜ ⎟ ⎜ 13.45 ⎜ ⎟ ⎜ TCD ⎟ ⎜ ⎜ TAH ⎟ ⎝ 15.00 ⎟⎠ ⎝ ⎠ θ = 45.00 deg

Problem 3-21 Each cord can sustain a maximum tension T. Determine the largest weight of the sack that can be supported. Also, determine θ of cord DC for equilibrium. Given: T = 200 lb

θ 1 = 30 deg θ 2 = 45 deg θ 3 = 60 deg

Solution: Solve for W = 1 and then scale the answer at the end.

Guesses

TBE = 1

TAB = 1

TBC = 1

TAC = 1

TCD = 1

TAH = 1

150

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Engineering Mechanics - Statics

Chapter 3

θ = 20 deg Given At H: +

↑ ΣFy = 0;

TAH − W = 0

At A: + ΣF x = 0; →

−TAB cos ( θ 2 ) + TAC cos ( θ 3 ) = 0

+

TAB sin ( θ 2 ) + TAC sin ( θ 3 ) − W = 0

↑ ΣFy = 0;

At B: + ΣF x = 0; →

TBC − TBE cos ( θ 1 ) + TAB cos ( θ 2 ) = 0

+

TBE sin ( θ 1 ) − TAB sin ( θ 2 ) = 0

↑ ΣFy = 0;

At C: + ΣF x = 0; →

TCD cos ( θ ) − TBC − TBE cos ( θ 3 ) = 0

+

TCD sin ( θ ) − TAC sin ( θ 3 ) = 0

↑ ΣFy = 0;

⎛ TBE ⎞ ⎜ ⎟ ⎜ TAB ⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TAC ⎟ = Find ( TBE , TAB , TBC , TAC , TCD , TAH , θ ) ⎜T ⎟ ⎜ CD ⎟ ⎜ TAH ⎟ ⎜ ⎟ ⎝ θ ⎠ W =

T

⎛⎜ TBE ⎞⎟ ⎛⎜ 0.73 ⎞⎟ T ⎜ AB ⎟ 0.52 ⎟ ⎜ ⎟ ⎜ TBC ⎜ ⎟ = ⎜ 0.27 ⎟ ⎜ TAC ⎟ ⎜ 0.73 ⎟ ⎟ ⎜ ⎟ ⎜ 0.90 ⎜ ⎟ ⎜ TCD ⎟ ⎜ ⎜ TAH ⎟ ⎝ 1.00 ⎟⎠ ⎝ ⎠ W = 200.00 lb

max ( TBE , TAB , TBC , TAC , TCD , TAH )

θ = 45.00 deg

151

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Engineering Mechanics - Statics

Chapter 3

Problem 3-22 The block has weight W and is being hoisted at uniform velocity. Determine the angle θ for equilibrium and the required force in each cord. Given: W = 20 lb

φ = 30 deg Solution: The initial guesses:

θ = 10 deg

TAB = 50 lb

Given TAB sin ( φ ) − W sin ( θ ) = 0 TAB cos ( φ ) − W − W cos ( θ ) = 0

⎛ θ ⎞ ⎜ ⎟ = Find ( θ , TAB) ⎝ TAB ⎠ θ = 60.00 deg TAB = 34.6 lb

Problem 3-23 Determine the maximum weight W of the block that can be suspended in the position shown if each cord can support a maximum tension T. Also, what is the angle θ for equilibrium?

152

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Engineering Mechanics - Statics

Chapter 3

Given: T = 80 lb

φ = 30 deg The maximum load will occur in cord AB. Solution: TAB = T The initial guesses:

θ = 100deg W = 200lb Given +

↑Σ Fy = 0;

+ Σ F x = 0; →

TAB cos ( φ ) − W − W cos ( θ ) = 0 TAB sin ( φ ) − W sin ( θ ) = 0

⎛θ⎞ ⎜ ⎟ = Find ( θ , W) ⎝W⎠ W = 46.19 lb

θ = 60.00 deg

Problem 3-24 Two spheres A and B have an equal mass M and are electrostatically charged such that the repulsive force acting between them has magnitude F and is directed along line AB. Determine the angle θ, the tension in cords AC and BC, and the mass M of each sphere.

153

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Engineering Mechanics - Statics

Chapter 3

Unit used: mN = 10

−3

N

Given: F = 20 mN g = 9.81

m 2

s

θ 1 = 30 deg θ 2 = 30 deg Solution: Guesses

TB = 1 mN

M = 1 gm

TA = 1 mN

θ = 30 deg

Given For B: + ΣF x = 0; →

F cos ( θ 2 ) − TB sin ( θ 1 ) = 0

+

F sin ( θ 2 ) + TB cos ( θ 1 ) − M g = 0

↑ ΣFy = 0;

For A: + ΣF x = 0; →

TA sin ( θ ) − F cos ( θ 2 ) = 0

+

TA cos ( θ ) − F sin ( θ 2 ) − M g = 0

↑ ΣFy = 0;

⎛ TA ⎞ ⎜ ⎟ ⎜ TB ⎟ = Find ( T , T , θ , M) A B ⎜θ ⎟ ⎜ ⎟ ⎝M⎠

⎛ TA ⎞ ⎛ 52.92 ⎞ ⎜ ⎟=⎜ ⎟ mN ⎝ TB ⎠ ⎝ 34.64 ⎠

θ = 19.11 deg

M = 4.08 gm

Problem 3-25 Blocks D and F weigh W1 each and block E weighs W2. Determine the sag s for equilibrium. Neglect the size of the pulleys. 154

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Engineering Mechanics - Statics

Chapter 3

Given: W1 = 5 lb W2 = 8 lb a = 4 ft Solution: Sum forces in the y direction Guess

s = 1 ft

Given

2⎜

⎛

⎞W − W = 0 1 2 ⎝ s +a ⎠ s

2

2⎟

s = Find ( s)

s = 5.33 ft

Problem 3-26 If blocks D and F each have weight W1, determine the weight of block E if the sag is s. Neglect the size of the pulleys. Given: W1 = 5 lb s = 3 ft a = 4 ft Solution: Sum forces in the y direction

⎛

⎞W − W = 0 1 ⎝ s +a ⎠

2⎜

s

2

2⎟

155

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Engineering Mechanics - Statics

⎛

Chapter 3

⎞W 1 ⎟ 2 2 + a s ⎝ ⎠ s

W = 2⎜

W = 6.00 lb

Problem 3-27 The block of mass M is supported by two springs having the stiffness shown. Determine the unstretched length of each spring. Units Used: 3

kN = 10 N Given: M = 30 kg l1 = 0.6 m l2 = 0.4 m l3 = 0.5 m kAC = 1.5

kN m

kAB = 1.2

kN m

g = 9.81

m 2

s Solution:

Initial guesses:

F AC = 20 N

F AB = 30 N

Given + ΣF x = 0; →

+

↑

ΣF y = 0;

l2 FAB 2

l2 + l3

− 2

l3 FAB 2

l2 + l3

2

l1 + l3

+ 2

l1 F AC

=0 2

l3 F AC 2

l1 + l3

−Mg=0 2

156

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Engineering Mechanics - Statics

Chapter 3

⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , FAB) ⎝ FAB ⎠ LAB = 0.1 m

Then guess

Given

⎛ FAC ⎞ ⎛ 183.88 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FAB ⎠ ⎝ 226.13 ⎠ LAC = 0.1 m

F AC = kAC ⎛⎝ l1 + l3 − LAC⎞⎠ 2

2

F AB = kAB⎛⎝ l2 + l3 − LAB⎞⎠ 2

2

⎛ LAB ⎞ ⎜ ⎟ = Find ( LAB , LAC) L AC ⎝ ⎠

⎛ LAB ⎞ ⎛ 0.452 ⎞ ⎜ ⎟=⎜ ⎟m 0.658 L ⎝ ⎠ AC ⎝ ⎠

Problem 3-28 Three blocks are supported using the cords and two pulleys. If they have weights of WA = WC = W, WB = kW, determine the angle θ for equilibrium. Given: k = 0.25

Solution: + ΣF x = 0; →

W cos ( φ ) − k W cos ( θ ) = 0

+

W sin ( φ ) + k W sin ( θ ) − W = 0

↑ ΣFy = 0;

cos ( φ ) = k cos ( θ ) sin ( φ ) = 1 − k sin ( θ )

157

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Engineering Mechanics - Statics

Chapter 3

1 = k cos ( θ ) + ( 1 − k sin ( θ ) ) = 1 + k − 2k sin ( θ ) 2

2

k θ = asin ⎛⎜ ⎟⎞

⎝ 2⎠

2

2

θ = 7.18 deg

Problem 3-29 A continuous cable of total length l is wrapped around the small pulleys at A, B, C, and D. If each spring is stretched a distance b, determine the mass M of each block. Neglect the weight of the pulleys and cords. The springs are unstretched when d = l/2. Given: l = 4m k = 500

N m

b = 300 mm g = 9.81

m 2

s

Solution: Fs = k b

F s = 150.00 N

Guesses T = 1N

θ = 10 deg

M = 1 kg

Given 2T sin ( θ ) − F s = 0 −2T cos ( θ ) + M g = 0 b+

l 4

sin ( θ ) =

l 4

⎛T⎞ ⎜ θ ⎟ = Find ( T , θ , M) ⎜ ⎟ ⎝M⎠ 158

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Engineering Mechanics - Statics

Chapter 3

θ = 44.43 deg

T = 107.14 N

M = 15.60 kg

Problem 3-30 Prove Lami's theorem, which states that if three concurrent forces are in equilibrium, each is proportional to the sine of the angle of the other two; that is, P/sin α = Q/sin β = R/sin γ.

Solution: Sine law: R

sin ( 180deg − γ )

=

However, in general R

sin ( γ )

=

Q

sin ( β )

=

Q

sin ( 180deg − β )

=

P

sin ( 180deg − α )

sin ( 180deg − φ ) = sin ( φ ) , hence P

sin ( α )

Q.E.D.

Problem 3-31 A vertical force P is applied to the ends of cord AB of length a and spring AC. If the spring has an unstretched length δ, determine the angle θ for equilibrium. Given: P = 10 lb

δ = 2 ft k = 15

lb ft

a = 2 ft

159

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Engineering Mechanics - Statics

Chapter 3

b = 2 ft Guesses

θ = 10 deg

φ = 10 deg

T = 1 lb

F = 1 lb

x = 1ft Given −T cos ( θ ) + F cos ( φ ) = 0 T sin ( θ ) + F sin ( φ ) − P = 0 F = k( x − δ ) a sin ( θ ) = x sin ( φ ) a cos ( θ ) + x cos ( φ ) = a + b

⎛θ ⎞ ⎜ ⎟ ⎜φ ⎟ ⎜ T ⎟ = Find ( θ , φ , T , F , x) ⎜F⎟ ⎜ ⎟ ⎝x⎠

⎛ T ⎞ ⎛ 10.30 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ F ⎠ ⎝ 9.38 ⎠

θ = 35 deg

Problem 3-32 Determine the unstretched length δ of spring AC if a force P causes the angle θ for equilibrium. Cord AB has length a. Given: P = 80 lb

θ = 60 deg k = 50

lb ft

a = 2 ft 160

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 3

b = 2 ft Guesses

δ = 1ft

φ = 10 deg

T = 1 lb

F = 1 lb

x = 1ft Given −T cos ( θ ) + F cos ( φ ) = 0 T sin ( θ ) + F sin ( φ ) − P = 0 F = k( x − δ ) a sin ( θ ) = x sin ( φ ) a cos ( θ ) + x cos ( φ ) = a + b

⎛δ ⎞ ⎜ ⎟ ⎜φ ⎟ ⎜ T ⎟ = Find ( δ , φ , T , F , x) ⎜F⎟ ⎜ ⎟ ⎝x⎠

⎛ T ⎞ ⎛ 69.28 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ F ⎠ ⎝ 40.00 ⎠

δ = 2.66 ft

Problem 3-33 The flowerpot of mass M is suspended from three wires and supported by the hooks at B and C. Determine the tension in AB and AC for equilibrium. Given: M = 20 kg l1 = 3.5 m l2 = 2 m l3 = 4 m 161

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Engineering Mechanics - Statics

Chapter 3

l4 = 0.5 m g = 9.81

m 2

s Solution: Initial guesses:

TAB = 1 N

TAC = 1 N

θ = 10 deg

φ = 10 deg

Given −TAC cos ( φ ) + TAB cos ( θ ) = 0 TAC sin ( φ ) + TAB sin ( θ ) − M g = 0 l1 cos ( φ ) + l2 cos ( θ ) = l3 l1 sin ( φ ) = l2 sin ( θ ) + l4

⎛ TAB ⎞ ⎜ ⎟ ⎜ TAC ⎟ = Find ( T , T , θ , φ ) AB AC ⎜ θ ⎟ ⎜ ⎟ ⎝ φ ⎠

⎛ θ ⎞ ⎛ 53.13 ⎞ ⎜ ⎟=⎜ ⎟ deg ⎝ φ ⎠ ⎝ 36.87 ⎠

⎛ TAB ⎞ ⎛ 156.96 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ TAC ⎠ ⎝ 117.72 ⎠

Problem 3-34 A car is to be towed using the rope arrangement shown. The towing force required is P. Determine the minimum length l of rope AB so that the tension in either rope AB or AC does not exceed T. Hint: Use the equilibrium condition at point A to determine the required angle θ for attachment, then determine l using trigonometry applied to triangle ABC. Given: P = 600 lb T = 750 lb

φ = 30 deg 162

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Engineering Mechanics - Statics

Chapter 3

d = 4 ft Solution: The initial guesses TAB = T TAC = T

θ = 30 deg l = 2 ft Case 1:

Assume TAC = T

Given + Σ F x = 0; →

TAC cos ( φ ) − TAB cos ( θ ) = 0

+

P − TAC sin ( φ ) − TAB sin ( θ ) = 0

↑

Σ F y = 0;

l

sin ( φ )

=

⎛ TAB ⎞ ⎜ ⎟ ⎜ θ ⎟ = Find ( TAB , θ , l) ⎜ l ⎟ ⎝ 1 ⎠ Case 2:

d

sin ( 180deg − θ − φ )

TAB = 687.39 lb

θ = 19.11 deg

l1 = 2.65 ft

θ = 13.85 deg

l2 = 2.89 ft

Assume TAB = T

Given + Σ F x = 0; →

TAC cos ( φ ) − TAB cos ( θ ) = 0

+

P − TAC sin ( φ ) − TAB sin ( θ ) = 0

↑

Σ F y = 0;

l

sin ( φ )

⎛ TAC ⎞ ⎜ ⎟ ⎜ θ ⎟ = Find ( TAC , θ , l) ⎜ l ⎟ ⎝ 2 ⎠ l = min ( l1 , l2 )

=

d

sin ( 180deg − θ − φ ) TAC = 840.83 lb

l = 2.65 ft

163

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Engineering Mechanics - Statics

Chapter 3

Problem 3-35 Determine the mass of each of the two cylinders if they cause a sag of distance d when suspended from the rings at A and B. Note that s = 0 when the cylinders are removed. Given: d = 0.5 m l1 = 1.5 m l2 = 2 m l3 = 1 m k = 100

N m m

g = 9.81

2

s Solution: TAC = k⎡⎣

( l 1 + d) 2 + l 2 2 −

2

l1 + l2

2⎤

⎦

TAC = 32.84 N

⎛ l1 + d ⎞ ⎟ ⎝ l2 ⎠

θ = atan ⎜

θ = 45 deg

M =

TAC sin ( θ ) g

M = 2.37 kg

164

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Engineering Mechanics - Statics

Chapter 3

Problem 3-36 The sling BAC is used to lift the load W with constant velocity. Determine the force in the sling and plot its value T (ordinate) as a function of its orientation θ , where 0 ≤ θ ≤ 90° .

Solution: W − 2T cos ( θ ) = 0 T=

⎛ W ⎞ ⎜ ⎟ 2 ⎝ cos ( θ ) ⎠ 1

Problem 3-37 The lamp fixture has weight W and is suspended from two springs, each having unstretched length L and stiffness k. Determine the angle θ for equilibrium. Units Used: 3

kN = 10 N Given: W = 10 lb L = 4 ft k = 5

lb ft

a = 4 ft

165

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Engineering Mechanics - Statics

Chapter 3

Solution: The initial guesses: T = 200 lb

θ = 10 deg

Given Spring +

↑Σ Fy = 0;

⎛T⎞ ⎜ ⎟ = Find ( T , θ ) ⎝θ⎠

⎛ a − L⎞ ⎟ ⎝ cos ( θ ) ⎠

T = k⎜

2T sin ( θ ) − W = 0

T = 7.34 lb

θ = 42.97 deg

Problem 3-38 The uniform tank of weight W is suspended by means of a cable, of length l, which is attached to the sides of the tank and passes over the small pulley located at O. If the cable can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cable.What is this tension? Given: W = 200 lb l = 6 ft a = 1 ft b = 2 ft c = b d = 2a Solution: Free Body Diagram: By observation, the force F has to support the entire weight of the tank. Thus, F = W. The tension in

166

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Engineering Mechanics - Statics

Chapter 3

cable is the same throughout the cable.

Equations of Equilibrium W − 2T sin ( θ ) = 0

ΣF y = 0; Attached to CD

θ 1 = acos ⎛⎜

2 a⎞

Attached to AB

θ 2 = acos ⎛⎜

2 b⎞

⎟ ⎝ l ⎠ ⎟ ⎝ l ⎠

θ 1 = 70.53 deg θ 2 = 48.19 deg

We choose the largest angle (which will produce the smallest force)

θ = max ( θ 1 , θ 2 ) T =

1⎛ W

⎜

⎞ ⎟

2 ⎝ sin ( θ ) ⎠

θ = 70.53 deg T = 106 lb

Problem 3-39 A sphere of mass ms rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass mB of block B needed to hold it in the equilibrium position shown. Given: ms = 4 kg a = 0.4 m b = 0.4 m

θ = 60 deg g = 9.81

m 2

s Solution: k =

a 2

b

Geometry: The angle θ1 which the surface make with the horizontal is to be determined first.

167

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Engineering Mechanics - Statics

tan ( θ 1 ) =

dy dx

=2kx

Chapter 3

evaluated at x = a

θ 1 = atan ( 2 k a)

θ 1 = 63.43 deg

Free Body Diagram : The tension in the cord is the same throughout the cord and is equal to the weight of block B, mBg. The initial guesses: mB = 200 kg

F N = 200 N

Given + Σ F x = 0; → +

↑Σ Fy = 0;

mB g cos ( θ ) − FN sin ( θ 1 ) = 0 mB g sin ( θ ) + F N cos ( θ 1 ) − ms g = 0

⎛ mB ⎞ ⎜ ⎟ = Find ( mB , FN) ⎝ FN ⎠

F N = 19.66 N mB = 3.58 kg

Problem 3-40 The pipe of mass M is supported at A by a system of five cords. Determine the force in each cord for equilibrium. Given: M = 30 kg g = 9.81

m

c = 3 d = 4

2

s

θ = 60 deg Solution: Initial guesses: TAB = 1 N

TAE = 1 N

TBC = 1 N

TBD = 1 N

Given TAB sin ( θ ) − M g = 0 TAE − TAB cos ( θ ) = 0 168

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Engineering Mechanics - Statics

Chapter 3

TBD⎛

c ⎞ ⎜ 2 2 ⎟ − TAB sin ( θ ) = 0 ⎝ c +d ⎠

TBD⎛

d ⎞ ⎜ 2 2 ⎟ + TAB cos ( θ ) − TBC = 0 ⎝ c +d ⎠

⎛ TAB ⎞ ⎜ ⎟ ⎜ TAE ⎟ = Find ( T , T , T , T ) AB AE BC BD ⎜ TBC ⎟ ⎜ ⎟ ⎝ TBD ⎠ ⎛ TAB ⎞ ⎛ 339.8 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TAE ⎟ = ⎜ 169.9 ⎟ N ⎜ TBC ⎟ ⎜ 562.3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ TBD ⎠ ⎝ 490.5 ⎠

Problem 3-41 The joint of a space frame is subjected to four forces. Strut OA lies in the x-y plane and strut OB lies in the y-z plane. Determine the forces acting in each of the three struts required for equilibrium. Units Used: 3

kN = 10 N Given: F = 2 kN

θ 1 = 45 deg θ 2 = 40 deg Solution: ΣF x = 0;

−R sin ( θ 1 ) = 0 R = 0

169

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Engineering Mechanics - Statics

Chapter 3

P sin ( θ 2 ) − F = 0

ΣF z = 0;

P =

F

sin ( θ 2 )

P = 3.11 kN Q − P cos ( θ 2 ) = 0

ΣF y = 0;

Q = P cos ( θ 2 ) Q = 2.38 kN

Problem 3-42 Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle. Units Used: 3

kN = 10 N Given: F 4 = 800 N

α = 60 deg β = 30 deg γ = 30 deg c = 3 d = 4

Solution: The initial guesses: F 1 = 100 N

F 2 = 100 N

F 3 = 100 N

Given

⎛ cos ( α ) ⎞ ⎜ 0 ⎟+ F1 ⎜ ⎟ ⎝ sin ( α ) ⎠

⎛ −cos ( γ ) ⎞ ⎛ 0 ⎞ ⎛c ⎞ ⎜ −d ⎟ + F ⎜ −sin ( γ ) ⎟ + F ⎜ sin ( β ) ⎟ = 0 3 4 ⎟ ⎜ ⎟ ⎜ ⎟ 2 2⎜ c +d ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ −cos ( β ) ⎠ F2

170

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Engineering Mechanics - Statics

⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠

Chapter 3

⎛ F1 ⎞ ⎛ 800 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 147 ⎟ N ⎜ F ⎟ ⎝ 564 ⎠ ⎝ 3⎠

Problem 3-43 Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle. Units Used: kN = 1000 N Given: F 4 = 8.5 kN F 5 = 2.8 kN

α = 15 deg β = 30 deg c = 7 d = 24

Solution: Initial Guesses:

F 1 = 1 kN

F 2 = 1 kN

F 3 = 1 kN

Given

⎛ −cos ( β ) ⎞ ⎜ 0 ⎟+ F1 ⎜ ⎟ ⎝ sin ( β ) ⎠

⎛ −sin ( α ) ⎞ ⎛ −c ⎞ ⎛1⎞ ⎛0⎞ ⎜ −d ⎟ + F ⎜ 0 ⎟ + F ⎜ cos ( α ) ⎟ + F ⎜ 0 ⎟ = 0 3 4 5 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 2⎜ c +d ⎝ 0 ⎠ ⎝0⎠ ⎝ −1 ⎠ ⎝ 0 ⎠ F2

171

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Engineering Mechanics - Statics

⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠

Chapter 3

⎛ F1 ⎞ ⎛ 5.60 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 8.55 ⎟ kN ⎜ F ⎟ ⎝ 9.44 ⎠ ⎝ 3⎠

Problem 3-44 Determine the magnitudes of F1, F 2 and F3 for equilibrium of the particle F = {- 9i - 8j - 5k}. Units Used: 3

kN = 10 N

Given:

⎛ −9 ⎞ F = ⎜ −8 ⎟ kN ⎜ ⎟ ⎝ −5 ⎠ a = 4m b = 2m c = 4m

θ 1 = 30 deg θ 2 = 60 deg θ 3 = 135 deg θ 4 = 60 deg θ 5 = 60 deg Solution: Initial guesses:

F 1 = 8 kN F 2 = 3 kN F 3 = 12 kN

172

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Engineering Mechanics - Statics

Chapter 3

Given

⎛ cos ( θ 2 ) cos ( θ 1) ⎞ ⎛ cos ( θ 3 ) ⎞ ⎜ ⎟ ⎜ ⎟ F 1 ⎜ −cos ( θ 2 ) sin ( θ 1 ) ⎟ + F 2 ⎜ cos ( θ 5 ) ⎟ + ⎜ ⎟ ⎜ cos θ ⎟ sin ( θ 2 ) ⎝ ⎠ ⎝ ( 4) ⎠ ⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠

⎛a⎞ ⎜ c ⎟+F=0 ⎟ 2 2 2⎜ a + b + c ⎝ −b ⎠ F3

⎛ F1 ⎞ ⎛ 8.26 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 3.84 ⎟ kN ⎜ F ⎟ ⎝ 12.21 ⎠ ⎝ 3⎠

Problem 3-45 The three cables are used to support the lamp of weight W. Determine the force developed in each cable for equilibrium. Units Used: 3

kN = 10 N Given: W = 800 N

b = 4m

a = 4m

c = 2m

Solution: Initial Guesses: F AB = 1 N

F AC = 1 N

F AD = 1 N

173

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Engineering Mechanics - Statics

Chapter 3

Given

⎛0⎞ ⎛1⎞ ⎜ ⎟ F AB 1 + FAC ⎜ 0 ⎟ + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠

⎛ −c ⎞ ⎛0⎞ ⎜ −b ⎟ + W⎜ 0 ⎟ = 0 ⎟ ⎜ ⎟ 2 2 2⎜ a +b +c ⎝ a ⎠ ⎝ −1 ⎠ F AD

⎛ FAB ⎞ ⎜ ⎟ F ⎜ AC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

⎛ FAB ⎞ ⎛ 800 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ 400 ⎟ N ⎜ F ⎟ ⎝ 1200 ⎠ ⎝ AD ⎠

Problem 3-46 Determine the force in each cable needed to support the load W. Given: a = 8 ft b = 6 ft c = 2 ft d = 2 ft e = 6 ft W = 500 lb

Solution: Initial guesses: F CD = 600 lb

F CA = 195 lb

F CB = 195 lb

174

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Engineering Mechanics - Statics

Chapter 3

Given

⎛c⎞ ⎜ −e ⎟ + 2 2⎜ ⎟ c +e ⎝0 ⎠ FCA

⎛ −d ⎞ ⎜ −e ⎟ + ⎟ 2 2⎜ d +e ⎝ 0 ⎠ F CB

⎛0⎞ ⎛0⎞ ⎜ b ⎟ + W⎜ 0 ⎟ = 0 ⎜ ⎟ 2 2⎜ ⎟ a + b ⎝a⎠ ⎝ −1 ⎠ F CD

⎛ FCD ⎞ ⎜ ⎟ F ⎜ CA ⎟ = Find ( FCD , FCA , FCB) ⎜F ⎟ ⎝ CB ⎠

⎛ FCD ⎞ ⎛ 625 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ CA ⎟ = ⎜ 198 ⎟ lb ⎜ F ⎟ ⎝ 198 ⎠ ⎝ CB ⎠

Problem 3-47 Determine the stretch in each of the two springs required to hold the crate of mass mc in the equilibrium position shown. Each spring has an unstretched length δ and a stiffness k. Given: mc = 20 kg

δ = 2m k = 300

N m

a = 4m b = 6m c = 12 m Solution: Initial Guesses F OA = 1 N F OB = 1 N F OC = 1 N Given

⎛0⎞ ⎛ −1 ⎞ ⎜ ⎟ F OA −1 + F OB⎜ 0 ⎟ + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠

⎛b⎞ ⎛0⎞ ⎜ a ⎟ + m g⎜ 0 ⎟ = 0 c ⎜ ⎟ 2 2 2⎜ ⎟ a + b + c ⎝c⎠ ⎝ −1 ⎠ F OC

175

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Engineering Mechanics - Statics

Chapter 3

⎛ FOA ⎞ ⎜ ⎟ ⎜ FOB ⎟ = Find ( FOA , FOB , FOC) ⎜F ⎟ ⎝ OC ⎠ δ OA =

δ OB =

F OA k F OB k

⎛ FOA ⎞ ⎛ 65.40 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FOB ⎟ = ⎜ 98.10 ⎟ N ⎜ F ⎟ ⎝ 228.90 ⎠ ⎝ OC ⎠

δ OA = 218 mm

δ OB = 327 mm

Problem 3-48 If the bucket and its contents have total weight W, determine the force in the supporting cables DA, DB, and DC. Given: W = 20 lb a = 3 ft b = 4.5 ft c = 2.5 ft d = 3 ft e = 1.5 ft f = 1.5 ft

176

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Engineering Mechanics - Statics

Chapter 3

Solution: The initial guesses: F DA = 40 lb

F DB = 20 lb

F DC = 30 lb

Given Σ F x = 0;

b−e e ⎡ ⎤F − ⎡ ⎤F = 0 DA ⎢ DC ⎢ ⎥ 2 2 2 2 2 2⎥ ⎣ ( b − e) + f + a ⎦ ⎣ e + d + ( c − f) ⎦

Σ F y = 0;

−f c− f ⎡ ⎤F + ⎡ ⎤F − F = 0 DA ⎢ DC DB ⎢ ⎥ 2 2 2 2 2 2⎥ + f + a + d + ( c − f ) ( b − e ) e ⎣ ⎦ ⎣ ⎦

Σ F z = 0;

a d ⎡ ⎤F + ⎡ ⎤F − W = 0 DA ⎢ DC ⎢ ⎥ 2 2 2 2 2 2⎥ ⎣ ( b − e) + f + a ⎦ ⎣ e + d + ( c − f) ⎦

⎛ FDA ⎞ ⎜ ⎟ ⎜ FDB ⎟ = Find ( FDA , FDB , FDC) ⎜F ⎟ ⎝ DC ⎠

⎛ FDA ⎞ ⎛ 10.00 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDB ⎟ = ⎜ 1.11 ⎟ lb ⎜ F ⎟ ⎝ 15.56 ⎠ ⎝ DC ⎠

Problem 3-49

The crate which of weight F is to be hoisted with constant velocity from the hold of a ship using the cable arrangement shown. Determine the tension in each of the three cables for equilibrium. 177

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Engineering Mechanics - Statics

Chapter 3

Units Used: 3

kN = 10 N Given: F = 2.5 kN a = 3m b = 1m c = 0.75 m d = 1m e = 1.5 m f = 3m Solution: The initial guesses F AD = 3 kN

F AC = 3 kN −c

Given 2

2

c +b +a

2

b 2

2

c +b +a

2

F AD +

2

c +b +a

d 2

2

d +e +a

F AD +

2

F AD +

⎛ FAD ⎞ ⎜ ⎟ F ⎜ AC ⎟ = Find ( FAD , FAC , FAB) ⎜F ⎟ ⎝ AB ⎠

2

e 2

2

d +e +a

−a 2

F AB = 3 kN

2

F AC +

2

d +e +a

2

2

d + f +a

F AC +

2

F AC +

2

−f 2

2

d + f +a

−a 2

d

2

F AB = 0

F AB = 0

−a 2

2

d + f +a

2

F AB + F = 0

⎛ FAD ⎞ ⎛ 1.55 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ 0.46 ⎟ kN ⎜ F ⎟ ⎝ 0.98 ⎠ ⎝ AB ⎠

178

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Engineering Mechanics - Statics

Chapter 3

Problem 3-50 The lamp has mass ml and is supported by pole AO and cables AB and AC. If the force in the pole acts along its axis, determine the forces in AO, AB, and AC for equilibrium. Given: ml = 15 kg

d = 1.5 m

a = 6m

e = 4m

b = 1.5 m

f = 1.5 m

c = 2m

g = 9.81

m 2

s Solution: The initial guesses: F AO = 100 N F AB = 200 N F AC = 300 N Given Equilibrium equations: c 2

2

c +b +a

2

F AO −

2

2

c +b +a a 2

2

c +b +a

2

2

2

( c + e) + ( b + f) + a

b

−

c+e

2

2

2

( c + e) + ( b + f) + a a 2

2

( c + e) + ( b + f) + a

⎛ FAO ⎞ ⎜ ⎟ ⎜ FAB ⎟ = Find ( FAO , FAB , FAC) ⎜F ⎟ ⎝ AC ⎠

2

c 2

2

c + ( b + d) + a

b+ f

F AO +

F AO −

2

F AB −

2

F AC = 0

b+d

F AB +

F AB −

2

2

2

c + ( b + d) + a a 2

2

c + ( b + d) + a

2

2

F AC = 0

F AC − ml g = 0

⎛ FAO ⎞ ⎛ 318.82 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ 110.36 ⎟ N ⎜ F ⎟ ⎝ 85.84 ⎠ ⎝ AC ⎠

179

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Engineering Mechanics - Statics

Chapter 3

Problem 3-51 Cables AB and AC can sustain a maximum tension Tmax, and the pole can support a maximum compression Pmax. Determine the maximum weight of the lamp that can be supported in the position shown. The force in the pole acts along the axis of the pole. Given: Tmax = 500 N

c = 2m

Pmax = 300 N

d = 1.5 m

a = 6m

e = 4m

b = 1.5 m

f = 1.5 m

Solution: Lengths 2

2

2

AO =

a +b +c

AB =

a + ( c + e) + ( b + d)

AC =

a + c + ( b + d)

2

2

2

2

2

2

The initial guesses: F AO = Pmax

F AB = Tmax

F AC = Tmax

W = 300N

Case 1 Assume the pole reaches maximum compression Given

⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞ ⎛0⎞ AB ⎜ AC ⎜ ⎟ ⎟ ⎟ −b + b+ f + b + d + W⎜ 0 ⎟ = 0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AO AB AC ⎝a⎠ ⎝ −a ⎠ ⎝ −a ⎠ ⎝ −1 ⎠

F AO ⎜

⎛ W1 ⎞ ⎜ ⎟ ⎜ FAB1 ⎟ = Find ( W , FAB , FAC) ⎜F ⎟ ⎝ AC1 ⎠

⎛ W1 ⎞ ⎛ 138.46 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB1 ⎟ = ⎜ 103.85 ⎟ N ⎜F ⎟ ⎝ AC1 ⎠ ⎝ 80.77 ⎠

180

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Engineering Mechanics - Statics

Chapter 3

Case 2 Assume that cable AB reaches maximum tension Given

⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞ ⎛0⎞ AB ⎜ AC ⎜ ⎟ ⎟ ⎟ −b + b+ f + b + d + W⎜ 0 ⎟ = 0 ⎟ AC ⎜ ⎟ ⎜ ⎟ AO ⎜ ⎟ AB ⎜ ⎝a⎠ ⎝ −a ⎠ ⎝ −a ⎠ ⎝ −1 ⎠

F AO ⎜

⎛ W2 ⎞ ⎜ ⎟ ⎜ FAO2 ⎟ = Find ( W , FAO , FAC) ⎜F ⎟ ⎝ AC2 ⎠

⎛ W2 ⎞ ⎛ 666.67 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAO2 ⎟ = ⎜ 1444.44 ⎟ N ⎜F ⎟ ⎝ AC2 ⎠ ⎝ 388.89 ⎠

Case 3 Assume that cable AC reaches maximum tension Given

⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞ ⎛0⎞ AB ⎜ AC ⎜ ⎟ ⎟ ⎟ −b + b+ f + b + d + W⎜ 0 ⎟ = 0 ⎟ AC ⎜ ⎟ ⎜ ⎟ AO ⎜ ⎟ AB ⎜ ⎝a⎠ ⎝ −a ⎠ ⎝ −a ⎠ ⎝ −1 ⎠

F AO ⎜

⎛ W3 ⎞ ⎜ ⎟ ⎜ FAO3 ⎟ = Find ( W , FAO , FAB) ⎜F ⎟ ⎝ AB3 ⎠ Final Answer

⎛ W3 ⎞ ⎛ 857.14 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAO3 ⎟ = ⎜ 1857.14 ⎟ N ⎜F ⎟ ⎝ AB3 ⎠ ⎝ 642.86 ⎠

W = min ( W1 , W2 , W3 )

W = 138.46 N

Problem 3-52 Determine the tension in cables AB, AC, and AD, required to hold the crate of weight W in equilibrium. Given: W = 60 lb a = 6 ft b = 12 ft c = 8 ft 181

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Engineering Mechanics - Statics

Chapter 3

d = 9 ft e = 4 ft f = 6 ft

Solution: The initial guesses: TB = 100 lb TC = 100 lb TD = 100 lb Given Σ F x = 0;

b

TB −

2

2

b +c +d Σ F y = 0;

d 2

2

b +c +d Σ F z = 0;

2

2

TC −

2

2

2

2

2

2

b +e + f e 2

b +e + f

c

−W +

b

TC −

b +c +d

2

TC +

2

TD = 0

TD = 0

f 2

2

b +e + f

2

TD = 0

Solving

⎛ TB ⎞ ⎜ ⎟ ⎜ TC ⎟ = Find ( TB , TC , TD) ⎜T ⎟ ⎝ D⎠

⎛ TB ⎞ ⎛ 108.84 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TC ⎟ = ⎜ 47.44 ⎟ lb ⎜ T ⎟ ⎝ 87.91 ⎠ ⎝ D⎠

Problem 3-53 The bucket has weight W. Determine the tension developed in each cord for equilibrium. Given: W = 20 lb

182

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Engineering Mechanics - Statics

Chapter 3

a = 2 ft b = 2 ft c = 8 ft d = 7 ft e = 3 ft f = a Solution: F DA = 20 lb

Initial Guesses:

F DB = 10 lb

F DC = 15 lb

Given c−e

ΣF x = 0;

2

2

( c − e) + b + a

2

−b

ΣF x = 0;

2

2

( c − e) + b + a

2

a

ΣF y = 0;

2

2

( c − e) + b + a

2

F DA +

−e 2

2

e + ( d − b) + f

F DA +

d−b 2

2

e + ( d − b) + f F DA +

2

2

f 2

2

e + ( d − b) + f

2

F DC +

−e 2

2

e +b + f

F DC +

−b 2

2

2

2

2

e +b + f F DC +

2

f 2

e +b + f

F DB = 0

F DB = 0

F DB − W = 0

⎛ FDA ⎞ ⎜ ⎟ ⎜ FDB ⎟ = Find ( FDA , FDB , FDC) ⎜F ⎟ ⎝ DC ⎠ ⎛ FDA ⎞ ⎛ 21.54 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDB ⎟ = ⎜ 13.99 ⎟ lb ⎜ F ⎟ ⎝ 17.61 ⎠ ⎝ DC ⎠

Problem 3-54 The mast OA is supported by three cables. If cable AB is subjected to tension T, determine the tension in cables AC and AD and the vertical force F which the mast exerts along its axis on the collar at A. Given: T = 500 N 183

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Engineering Mechanics - Statics

Chapter 3

a = 6m b = 3m c = 6m d = 3m e = 2m f = 1.5 m g = 2m Solution: Initial Guesses:

F AC = 90 N

F AD = 350 N

F = 750 N

Given Σ F x = 0;

e 2

2

e +d +a Σ F y = 0;

2

2

−a 2

2

2

e +d +a

2

f +g +a

2

2

f +g +a

⎛ FAC ⎞ ⎜ ⎟ F ⎜ AD ⎟ = Find ( FAC , FAD , F) ⎜ F ⎟ ⎝ ⎠

2

a

T− 2

2

g

T+

e +d +a Σ F z = 0;

2

d 2

f

T−

2

2

f +g +a

2

F AC −

b 2

2

2

2

2

2

2

b +c +a F AC −

c 2

b +c +a F AC −

a 2

b +c +a

F AD = 0

F AD = 0

F AD + F = 0

⎛ FAC ⎞ ⎛ 92.9 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AD ⎟ = ⎜ 364.3 ⎟ N ⎜ F ⎟ ⎝ 757.1 ⎠ ⎝ ⎠

Problem 3-55 The ends of the three cables are attached to a ring at A and to the edge of the uniform plate of mass M. Determine the tension in each of the cables for equilibrium. Given: M = 150 kg

e = 4m 184

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Engineering Mechanics - Statics

Chapter 3

a = 2m

f = 6m

b = 10 m

g = 6m

c = 12 m

h = 6m

d = 2m

i = 2m

gravity = 9.81

m 2

s Solution: The initial guesses: F B = 15 N F C = 16 N F D = 16 N Given Σ F x = 0;

F B( f − i ) 2

2

+ 2

2

( f − i) + h + c

Σ F y = 0;

F B ( − h) 2

2

F B( −c) 2

2

( f − i) + h + c

⎛ FB ⎞ ⎜ ⎟ ⎜ FC ⎟ = Find ( FB , FC , FD) ⎜F ⎟ ⎝ D⎠

2

2

+ 2

2

FC( −c) 2

2

( d + e) + ( h − a) + c

2

=0 2

FD( g) 2

2

=0 2

e +g +c

+ 2

FD( −e) e +g +c

( d + e) + ( h − a) + c

+ 2

2

F C [ − ( h − a) ] 2

( f − i) + h + c

Σ F z = 0;

+

( d + e) + ( h − a) + c

+ 2

F C( −d − e)

FD( −c) 2

2

+ M gravity = 0 2

e +g +c

⎛ FB ⎞ ⎛ 858 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FC ⎟ = ⎜ 0 ⎟ N ⎜ F ⎟ ⎝ 858 ⎠ ⎝ D⎠

Problem 3-56 The ends of the three cables are attached to a ring at A and to the edge of the uniform plate. Determine the largest mass the plate can have if each cable can support a maximum tension of T.

185

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Engineering Mechanics - Statics

Chapter 3

3

kN = 10 N Given: T = 15 kN

e = 4m

a = 2m

f = 6m

b = 10 m

g = 6m

c = 12 m

h = 6m

d = 2m

i = 2m

gravity = 9.81

m 2

s

Solution: The initial guesses:

FB = T

FC = T

FD = T

M = 1 kg

Case 1: Assume that cable B reaches maximum tension Given Σ F x = 0;

F B( f − i ) 2

2

+ 2

2

( f − i) + h + c

Σ F y = 0;

F B ( − h) 2

2

F B( −c) 2

2

( f − i) + h + c

⎛ M1 ⎞ ⎜ ⎟ ⎜ FC1 ⎟ = Find ( M , FC , FD) ⎜F ⎟ ⎝ D1 ⎠

2

2

FC( −c) 2

2

( d + e) + ( h − a) + c

⎛ FC1 ⎞ ⎛ −0.00 ⎞ ⎜ ⎟=⎜ ⎟ kN 15.00 F ⎝ ⎠ D1 ⎝ ⎠

=0 2

FD( g) 2

2

=0 2

e +g +c

+ 2

2

e +g +c

+ 2

FD( −e) 2

( d + e) + ( h − a) + c

+ 2

2

F C [ − ( h − a) ] 2

( f − i) + h + c

Σ F z = 0;

+

( d + e) + ( h − a) + c

+ 2

F C( −d − e)

FD( −c) 2

2

+ M gravity = 0 2

e +g +c

M1 = 2621.23 kg

186

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Engineering Mechanics - Statics

Chapter 3

Case 2: Assume that cable D reaches maximum tension Given F B( f − i )

Σ F x = 0;

2

2

2

2

( f − i) + h + c F B ( − h)

Σ F y = 0;

2

2

F B( −c) 2

2

2

( f − i) + h + c

⎛ M2 ⎞ ⎜ ⎟ ⎜ FB2 ⎟ = Find ( M , FB , FC) ⎜F ⎟ ⎝ C2 ⎠

2

2

FC( −c) 2

2

( d + e) + ( h − a) + c

⎛ FB2 ⎞ ⎛ 15.00 ⎞ ⎜ ⎟=⎜ ⎟ kN 0.00 F ⎝ ⎠ C2 ⎝ ⎠

=0 2

FD( g) 2

2

=0 2

e +g +c

+ 2

2

e +g +c

+ 2

FD( −e) 2

( d + e) + ( h − a) + c

+ 2

2

F C [ − ( h − a) ]

+ 2

+

( d + e) + ( h − a) + c

( f − i) + h + c

Σ F z = 0;

F C( −d − e)

+

FD( −c) 2

2

+ M gravity = 0 2

e +g +c

M2 = 2621.23 kg

For this set of number FC = 0 for any mass that is applied. For a different set of numbers it would be necessary to also check case 3: Assume that the cable C reaches a maximum. M = min ( M1 , M2 )

M = 2621.23 kg

Problem 3-57 The crate of weight W is suspended from the cable system shown. Determine the force in each segment of the cable, i.e., AB, AC, CD, CE, and CF. Hint: First analyze the equilibrium of point A, then using the result for AC, analyze the equilibrium of point C. Units Used: kip = 1000 lb Given: W = 500 lb a = 10 ft b = 24 ft c = 24 ft

187

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Engineering Mechanics - Statics

Chapter 3

d = 7 ft e = 7 ft

θ 1 = 20 deg θ 2 = 35 deg

Solution:

At A:

Initial guesses:

F AC = 570 lb

F AB = 500 lb

Given + ΣF x = 0; →

F AB cos ( θ 1 ) − F AC cos ( θ 2 ) = 0

+

F AB sin ( θ 1 ) + FAC sin ( θ 2 ) − W = 0

↑ ΣFy = 0;

⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , FAB) ⎝ FAB ⎠ At C:

⎛ FAC ⎞ ⎛ 574 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FAB ⎠ ⎝ 500 ⎠

Initial Guesses

F CD = 1 lb

F CE = 1 lb

F CF = 1 lb

Given

⎛ cos ( θ 2 ) ⎞ ⎜ ⎟ F AC⎜ 0 ⎟+ ⎜ −sin ( θ ) ⎟ 2 ⎠ ⎝

⎛ −a ⎞ ⎜ 0⎟+ ⎟ 2 2⎜ a +b ⎝ b ⎠ FCD

⎛0⎞ ⎜d⎟+ 2 2⎜ ⎟ c + d ⎝ −c ⎠ F CE

⎛0⎞ ⎜ −e ⎟ = 0 2 2⎜ ⎟ c + e ⎝ −c ⎠ F CF

188

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Engineering Mechanics - Statics

Chapter 3

⎛ FCD ⎞ ⎜ ⎟ F ⎜ CE ⎟ = Find ( FCD , FCE , FCF ) ⎜F ⎟ ⎝ CF ⎠

F CD = 1.22 kip

⎛ FCE ⎞ ⎛ 416 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FCF ⎠ ⎝ 416 ⎠

Problem 3-58 The chandelier of weight W is supported by three wires as shown. Determine the force in each wire for equilibrium. Given: W = 80 lb r = 1 ft h = 2.4 ft

Solution: The initial guesses: F AB = 40 lb F AC = 30 lb F AD = 30 lb

Given r

Σ F x = 0;

2

2

FAC −

r +h Σ F y = 0;

−r 2

2

r +h

FAD +

r cos ( 45 deg) 2

r +h

2

r cos ( 45 deg) 2

r +h

2

FAB = 0

FAB = 0

189

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Engineering Mechanics - Statics

Chapter 3

h

Σ F z = 0;

2

2

FAC +

r +h

h 2

r +h

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

2

F AD +

h 2

2

FAB − W = 0

r +h

⎛ FAB ⎞ ⎛ 35.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC ⎟ = ⎜ 25.4 ⎟ lb ⎜ F ⎟ ⎝ 25.4 ⎠ ⎝ AD ⎠

Problem 3-59 If each wire can sustain a maximum tension Tmax before it fails, determine the greatest weight of the chandelier the wires will support in the position shown. Given: Tmax = 120 lb r = 1 ft h = 2.4 ft

Solution: The initial guesses: F AB = Tmax F AC = Tmax F AD = Tmax W = Tmax Case 1 Assume that cable AB has maximum tension Given r

Σ F x = 0;

2

2

FAC −

r +h Σ F y = 0;

−r 2

2

r +h

FAD +

r cos ( 45deg) 2

r +h

2

r cos ( 45deg) 2

r +h

2

F AB = 0

F AB = 0

190

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Engineering Mechanics - Statics

Chapter 3

h

Σ Fz = 0;

2

2

FAC +

r +h

h 2

r +h

⎛ W1 ⎞ ⎜ ⎟ ⎜ FAC1 ⎟ = Find ( W , FAC , FAD) ⎜F ⎟ ⎝ AD1 ⎠

2

F AD +

h 2

2

FAB − W = 0

r +h

⎛ W1 ⎞ ⎛ 267.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC1 ⎟ = ⎜ 84.9 ⎟ lb ⎜F ⎟ ⎝ AD1 ⎠ ⎝ 84.9 ⎠

Case 2 Assume that cable AC has maximum tension Given r

Σ F x = 0;

2

2

FAC −

r cos ( 45 deg) 2

r +h Σ F y = 0;

−r 2

r +h

FAD +

2

r cos ( 45 deg) 2

r +h

r +h

h

Σ F z = 0;

2

2

2

FAC +

r +h

2

h 2

r +h

⎛ W2 ⎞ ⎜ ⎟ ⎜ FAB2 ⎟ = Find ( W , FAB , FAD) ⎜F ⎟ ⎝ AD2 ⎠

2

FAB = 0

FAB = 0

F AD +

h 2

2

FAB − W = 0

r +h

⎛ W2 ⎞ ⎛ 378.2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB2 ⎟ = ⎜ 169.7 ⎟ lb ⎜F ⎟ ⎝ AD2 ⎠ ⎝ 120 ⎠

Case 3 Assume that cable AD has maximum tension Given r

Σ F x = 0;

2

2

FAC −

r +h Σ F y = 0;

−r 2

2

2

r +h

h 2

r +h

2

r cos ( 45deg)

r +h Σ F z = 0;

2

r +h

FAD +

2

r cos ( 45deg)

FAC +

2

h 2

r +h

2

F AB = 0

F AB = 0

F AD +

h 2

2

FAB − W = 0

r +h

191

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Engineering Mechanics - Statics

Chapter 3

⎛ W3 ⎞ ⎜ ⎟ ⎜ FAB3 ⎟ = Find ( W , FAB , FAC) ⎜F ⎟ ⎝ AC3 ⎠ W = min ( W1 , W2 , W3 )

⎛ W3 ⎞ ⎛ 378.2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB3 ⎟ = ⎜ 169.7 ⎟ lb ⎜F ⎟ ⎝ AC3 ⎠ ⎝ 120 ⎠ W = 267.42 lb

Problem 3-60 Determine the force in each cable used to lift the surge arrester of mass M at constant velocity. Units Used: 3

kN = 10 N 3

Mg = 10 kg Given: M = 9.50 Mg a = 2m b = 0.5 m

θ = 45 deg

Solution: Initial guesses:

F B = 50 kN

F C = 30 kN

Given b

ΣF x = 0;

FB

ΣF y = 0;

FC

ΣF z = 0;

M g − FB

2

b +a

− FC

2

2

=0

2

b +a

−b sin ( θ ) 2

b cos ( θ )

F D = 10 kN

+ FD

2

b +a

b 2

=0

b +a a

2

2

b +a

− FC

2

a 2

b +a

2

− FD

a 2

=0 2

b +a

192

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Engineering Mechanics - Statics

Chapter 3

⎛ FB ⎞ ⎜ ⎟ ⎜ FC ⎟ = Find ( FB , FC , FD) ⎜F ⎟ ⎝ D⎠

⎛ FB ⎞ ⎛ 28.13 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FC ⎟ = ⎜ 39.78 ⎟ kN ⎜ F ⎟ ⎝ 28.13 ⎠ ⎝ D⎠

Problem 3-61 The cylinder of weight W is supported by three chains as shown. Determine the force in each chain for equilibrium.

Given: W = 800 lb r = 1 ft d = 1 ft Solution: The initial guesses: F AB = 1 lb F AC = 1 lb F AD = 1 lb Given Σ F x = 0;

r 2

2

FAC −

r cos ( 45 deg) 2

r +d Σ F y = 0;

−r 2

2

r +d FAD +

r +d d Σ F z = 0;

2

2

2

r sin ( 45 deg) 2

2

FAB = 0

F AB = 0

r +d FAD +

r +d

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

d 2

r +d

2

F AC +

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ ⎜F ⎟ = ⎝ AD ⎠

d 2

2

FAB − W = 0

r +d

⎛ 469 ⎞ ⎜ 331 ⎟ ⎜ ⎟ ⎝ 331 ⎠ lb

193

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Engineering Mechanics - Statics

Chapter 3

Problem 3-62 The triangular frame ABC can be adjusted vertically between the three equal-length cords. If it remains in a horizontal plane, determine the required distance s so that the tension in each of the cords, OA, OB, and OC, equals F . The lamp has a mass M. Given: F = 20 N M = 5 kg m

g = 9.81

2

s d = 0.5 m Solution:

3F cos ( γ ) = M g

ΣF z = 0;

γ = acos ⎛⎜

M g⎞

⎟ ⎝ 3F ⎠

2d cos ( 30 deg)

Geometry

3 s =

γ = 35.16 deg = s tan ( γ )

2d cos ( 30 deg) 3 tan ( γ )

s = 410 mm

Problem 3-63 Determine the force in each cable needed to support the platform of weight W. Units Used: 3

kip = 10 lb

194

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Engineering Mechanics - Statics

Chapter 3

Given: W = 3500 lb

d = 4 ft

a = 2 ft

e = 3 ft f = 3 ft

b = 4 ft c = 4 ft

g = 10 ft

Solution: The initial guesses: F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

Given −b 2

2

2

c−d

FAD +

2

g + ( e − a) + b e−a 2

2

2

2

2

2

g + ( e − a) + b

2

( c − d) + e + g e

FAD +

2

g + ( e − a) + b −g

2

2

2

g 2

2

2

e + ( c − d) + g

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

2

2

2

2

2

c + f +g f

FAC −

e + ( c − d) + g

FAD −

c

FAC +

FAC −

2

c + f +g

g 2

2

2

F AB = 0

F AB = 0

F AB + W = 0

g + f +c

⎛ FAB ⎞ ⎛ 1.467 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC ⎟ = ⎜ 0.914 ⎟ kip ⎜ F ⎟ ⎝ 1.42 ⎠ ⎝ AD ⎠

195

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Engineering Mechanics - Statics

Chapter 3

Problem 3-64 A flowerpot of mass M is supported at A by the three cords. Determine the force acting in each cord for equilibrium. Given: M = 25 kg g = 9.81

m 2

s

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 60 deg θ 4 = 45 deg Solution: Initial guesses: F AB = 1 N F AD = 1 N F AC = 1 N Given ΣF x = 0;

F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0

ΣF y = 0;

−F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0

ΣF z = 0;

F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − M g = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

⎛ FAB ⎞ ⎛ 219.89 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC ⎟ = ⎜ 103.65 ⎟ N ⎜ F ⎟ ⎝ 103.65 ⎠ ⎝ AD ⎠

Problem 3-65 If each cord can sustain a maximum tension of T before it fails, determine the greatest weight of the flowerpot the cords can support.

196

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Engineering Mechanics - Statics

Chapter 3

Given: T = 50 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 60 deg θ 4 = 45 deg Solution: Initial guesses: F AB = T F AD = T F AC = T W = T Case 1

Assume that AB reaches maximum tension

Given ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0 ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0 ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0

⎛ W1 ⎞ ⎜ ⎟ ⎜ FAC1 ⎟ = Find ( W , FAC , FAD) ⎜F ⎟ ⎝ AD1 ⎠ Case 2

⎛ W1 ⎞ ⎛ 55.77 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC1 ⎟ = ⎜ 23.57 ⎟ N ⎜F ⎟ ⎝ AD1 ⎠ ⎝ 23.57 ⎠

Assume that AC reaches maximum tension

197

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Engineering Mechanics - Statics

Chapter 3

Given ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0 ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0 ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0

⎛ W2 ⎞ ⎜ ⎟ ⎜ FAB2 ⎟ = Find ( W , FAB , FAD) ⎜F ⎟ ⎝ AD2 ⎠ Case 3

⎛ W2 ⎞ ⎛ 118.30 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB2 ⎟ = ⎜ 106.07 ⎟ N ⎜F ⎟ ⎝ AD2 ⎠ ⎝ 50.00 ⎠

Assume that AD reaches maximum tension

Given ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0 ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0 ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0

⎛ W3 ⎞ ⎜ ⎟ ⎜ FAB3 ⎟ = Find ( W , FAB , FAC) ⎜F ⎟ ⎝ AC3 ⎠ W = min ( W1 , W2 , W3 )

⎛ W3 ⎞ ⎛ 118.30 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB3 ⎟ = ⎜ 106.07 ⎟ N ⎜F ⎟ ⎝ AC3 ⎠ ⎝ 50.00 ⎠ W = 55.77 N

198

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Engineering Mechanics - Statics

Chapter 3

Problem 3-66 The pipe is held in place by the vice. If the bolt exerts force P on the pipe in the direction shown, determine the forces F A and F B that the smooth contacts at A and B exerton the pipe. Given: P = 50 lb

θ = 30 deg c = 3 d = 4

Solution: Initial Guesses

F A = 1 lb

F B = 1 lb

Given + Σ F x = 0; → +

↑Σ Fy = 0;

⎛ FA ⎞ ⎜ ⎟ = Find ( FA , FB) ⎝ FB ⎠

⎛

⎞=0 ⎟ 2 2 ⎝ c +d ⎠

F B − FA sin ( θ ) − P⎜

⎛

d

⎞=0 ⎟ 2 2 ⎝ c +d ⎠

−F A cos ( θ ) + P⎜

c

⎛ FA ⎞ ⎛ 34.6 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FB ⎠ ⎝ 57.3 ⎠

Problem 3-67 When y is zero, the springs sustain force F0. Determine the magnitude of the applied vertical forces F and -F required to pull point A away from point B a distance y1. The ends of cords CAD and CBD are attached to rings at C and D.

199

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Engineering Mechanics - Statics

Chapter 3

Given: F 0 = 60 lb k = 40

lb ft

d = 2 ft y1 = 2 ft Solution: Initial spring stretch: s1 =

F0

s1 = 1.50 ft

k

Initial guesses: F s = 1 lb

T = 1 lb

F = 1 lb

Given 2

⎛ y1 ⎞ d −⎜ ⎟ ⎝ 2⎠ T−F =0 s 2

F−

y1 2d

T=0

⎛⎜ Fs ⎞⎟ ⎜ T ⎟ = Find ( Fs , T , F) ⎜F⎟ ⎝ ⎠

d

⎛ Fs ⎞ ⎛ 70.72 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ T ⎠ ⎝ 81.66 ⎠

2 ⎡ ⎤ 2 ⎛ y1 ⎞ ⎢ F s = k d − d − ⎜ ⎟ + s1⎥ ⎣ ⎝2⎠ ⎦

F = 40.83 lb

Problem 3-68 When y is zero, the springs are each stretched a distance δ. Determine the distance y if a force F is applied to points A and B as shown. The ends of cords CAD and CBD are attached to

200

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Engineering Mechanics - Statics

Chapter 3

pp p rings at C and D. Given:

δ = 1.5 ft k = 40

lb ft

d = 2 ft F = 60 lb Solution: Initial guesses: F s = 1 lb

T = 1 lb

y = 1 ft

Given F−

y T=0 2d 2

y d − ⎛⎜ ⎟⎞ ⎝ 2⎠ T − F = 0 s 2

d

⎛⎜ Fs ⎞⎟ ⎜ T ⎟ = Find ( Fs , T , y) ⎜y⎟ ⎝ ⎠

⎡

F s = k⎢d −

⎣

2

d −

2 ⎤ ⎛ y ⎞ + δ⎥ ⎜ ⎟ ⎝ 2⎠ ⎦

⎛ Fs ⎞ ⎛ 76.92 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ T ⎠ ⎝ 97.55 ⎠

y = 2.46 ft

Problem 3-69 Cord AB of length a is attached to the end B of a spring having an unstretched length b. The other end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If a weight W is suspended from B, determine the angle θ of cord AB for equilibrium.

201

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Engineering Mechanics - Statics

Chapter 3

Given: a = 5 ft b = 5 ft k = 10

lb ft

W = 10 lb Solution: Initial Guesses F BA = 1 lb F sp = 1 lb

θ = 30 deg Given F sp − FBA cos ( θ ) = 0 F BA sin ( θ ) − W = 0 F sp = k( a − a cos ( θ ) )

⎛ Fsp ⎞ ⎜ ⎟ ⎜ FBA ⎟ = Find ( Fsp , FBA , θ ) ⎜ ⎟ ⎝ θ ⎠

⎛ Fsp ⎞ ⎛ 11.82 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FBA ⎠ ⎝ 15.49 ⎠

θ = 40.22 deg

Problem 3-70 The uniform crate of mass M is suspended by using a cord of length l that is attached to the sides of the crate and passes over the small pulley at O. If the cord can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cord and specify the cord tension in this case.

202

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Engineering Mechanics - Statics

Chapter 3

Given: M = 50 kg g = 9.81

m 2

s a = 0.6 m b = 1.5 m l = 2m c =

a

d =

b

2

2

Solution: Case 1 Attached at A and B

Given

⎡⎢ l 2 2 ⎥⎤ ⎛ ⎞ −d ⎢ ⎜⎝ 2 ⎟⎠ ⎥ Mg − ⎢ ⎥ 2T = 0 l ⎢ ⎥ 2 ⎣ ⎦

Case 2 Attached at C and D

Given

Guess

Guess

⎡⎢ l 2 2 ⎥⎤ ⎛ ⎞ −c ⎢ ⎜⎝ 2 ⎟⎠ ⎥ Mg − ⎢ ⎥ 2T = 0 l ⎢ ⎥ 2 ⎣ ⎦

Choose the arrangement that gives the smallest tension.

T = 1N

T1 = Find ( T)

T1 = 370.78 N

T = 1N

T2 = Find ( T)

T = min ( T1 , T2 )

T2 = 257.09 N

T = 257.09 N

Problem 3-71 The man attempts to pull the log at C by using the three ropes. Determines the direction θ in which he should pull on his rope with a force P, so that he exerts a maximum 203

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Engineering Mechanics - Statics

Chapter 3

p p force on the log. What is the force on the log for this case ? Also, determine the direction in which he should pull in order to maximize the force in the rope attached to B.What is this maximum force? Given: P = 80 lb

φ = 150 deg

Solution: + Σ F x = 0; → +

↑Σ Fy = 0;

F AC =

F AB + P cos ( θ ) − F AC sin ( φ − 90 deg) = 0 P sin ( θ ) − F AC cos ( φ − 90 deg) = 0

P sin ( θ )

In order to maximize FAC we choose

cos ( φ − 90deg)

Thus

θ = 90 deg

F AC =

P sin ( θ )

cos ( φ − 90deg)

sin ( θ ) = 1.

F AC = 160.00 lb

Now let's find the force in the rope AB. F AB = −P cos ( θ ) + F AC sin ( φ − 90 deg) F AB = −P cos ( θ ) +

F AB = P

P sin ( θ ) sin ( φ − 90 deg) cos ( φ − 90 deg)

sin ( θ ) sin ( φ − 90 deg) − cos ( θ ) cos ( φ − 90 deg) cos ( φ − 90 deg)

= −P

cos ( θ + φ − 90deg) cos ( φ − 90 deg)

204

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Engineering Mechanics - Statics

Chapter 3

cos ( θ + φ − 90 deg) = −1

In order to maximize the force we set

θ + φ − 90 deg = 180 deg θ = 270 deg − φ

F AB = −P

θ = 120.00 deg

cos ( θ + φ − 90 deg)

F AB = 160.00 lb

cos ( φ − 90 deg)

Problem 3-72 The "scale" consists of a known weight W which is suspended at A from a cord of total length L. Determine the weight w at B if A is at a distance y for equilibrium. Neglect the sizes and weights of the pulleys.

Solution: +

↑ ΣFy = 0;

Geometry

2W sin ( θ ) − w = 0

h=

⎛L − ⎜ ⎝ 2

y⎞

⎟ ⎠

2

2

−

⎛ d ⎞ = 1 ( L − y) 2 − d2 ⎜ ⎟ 2 ⎝ 2⎠

205

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Engineering Mechanics - Statics

Chapter 3

⎡ 1 ( L − y) 2 + d2 ⎤ ⎢2 ⎥ w = 2W⎢ ⎥ L− y ⎢ ⎥ 2 ⎣ ⎦

w=

2W L−y

2

2

( L − y) − d

Problem 3-73 Determine the maximum weight W that can be supported in the position shown if each cable AC and AB can support a maximum tension of F before it fails. Given:

θ = 30 deg F = 600 lb c = 12 d = 5

Solution: Initial Guesses

F AB = F

F AC = F

W = F

Case 1 Assume that cable AC reaches maximum tension

Given

F AC sin ( θ ) −

d 2

FAB = 0

2

c +d F AC cos ( θ ) +

c 2

2

FAB − W = 0

c +d

⎛ W1 ⎞ ⎜ ⎟ = Find ( W , FAB) ⎝ FAB1 ⎠

⎛ W1 ⎞ ⎛ 1239.62 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FAB1 ⎠ ⎝ 780.00 ⎠

Case 2 Assume that cable AB reaches maximum tension

206

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Engineering Mechanics - Statics

F AC sin ( θ ) −

Given

Chapter 3

d 2

FAB = 0

2

c +d F AC cos ( θ ) +

c 2

2

FAB − W = 0

c +d

⎛ W2 ⎞ ⎜ ⎟ = Find ( W , FAC) ⎝ FAC2 ⎠ W = min ( W1 , W2 )

⎛ W2 ⎞ ⎛ 953.55 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FAC2 ⎠ ⎝ 461.54 ⎠ W = 953.6 lb

Problem 3-74 If the spring on rope OB has been stretched a distance δ. and fixed in place as shown, determine the tension developed in each of the other three ropes in order to hold the weight W in equilibrium. Rope OD lies in the x-y plane. Given: a = 2 ft b = 4 ft c = 3 ft d = 4 ft e = 4 ft f = 4 ft xB = −2 ft yB = −3 ft zB = 3 ft

θ = 30 deg k = 20

lb in

δ = 2 in W = 225 lb

207

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Engineering Mechanics - Statics

Chapter 3

Solution: F OA = 10 lb

Initial Guesses

F OC = 10 lb

F OD = 10 lb

Given c 2

2

2

a +b +c −b 2

2

2

2

2

2

a +b +c

2

2

2

yB

F OA + kδ

2

2

2

⎛ FOA ⎞ ⎜ ⎟ ⎜ FOC ⎟ = Find ( FOA , FOC , FOD) ⎜F ⎟ ⎝ OD ⎠

2

2

2

2

2

f 2

d +e + f e

+

xB + yB + zB

2

d +e + f

2

zB 2

2

+

xB + yB + zB F OA + kδ

−d

+

xB + yB + zB

a +b +c a

xB

F OA + kδ

2

2

d +e + f

FOC + F OD sin ( θ ) = 0

FOC + F OD cos ( θ ) = 0

FOC − W = 0

⎛ FOA ⎞ ⎛ 201.6 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FOC ⎟ = ⎜ 215.7 ⎟ lb ⎜ F ⎟ ⎝ 58.6 ⎠ ⎝ OD ⎠

Problem 3-75 The joint of a space frame is subjected to four member forces. Member OA lies in the x - y plane and member OB lies in the y - z plane. Determine the forces acting in each of the members required for equilibrium of the joint. Given: F 4 = 200 lb

θ = 40 deg φ = 45 deg Solution: The initial guesses : F 1 = 200 lb

F 2 = 200 lb

F 3 = 200 lb

208

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Engineering Mechanics - Statics

Chapter 3

Given Σ F y = 0;

F 3 + F 1 cos ( φ ) − F 2 cos ( θ ) = 0

Σ F x = 0;

−F 1 sin ( φ ) = 0

Σ F z = 0;

F 2 sin ( θ ) − F4 = 0

⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠ ⎛ F1 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 311.1 ⎟ lb ⎜ F ⎟ ⎝ 238.4 ⎠ ⎝ 3⎠

209

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Engineering Mechanics - Statics

Chapter 4

Problem 4-1 If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e., A × ( B + D) = ( A × B) + ( A × D). Solution: Consider the three vectors; with A vertical. Note triangle obd is perpendicular to A. od = A × ( B + D) = A

(

B + D ) sin ( θ 3 )

ob = A × B = A B sin ( θ 1 ) bd = A × D = A B sin ( θ 2 ) Also, these three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product is proportional to the length of each side of the triangle. The three vector cross - products also form a closed triangle o'b'd' which is similar to triangle obd. Thus from the figure, A × ( B + D) = A × B + A × D

(QED)

Note also, A = Axi + Ayj + A zk B = Bxi + B yj + B zK D = Dxi + Dyj + Dzk j k ⎞ ⎛ i ⎜ ⎟ Ay Az ⎟ A × ( B + D) = ⎜ Ax ⎜B + D B + D B + D ⎟ x y y z z⎠ ⎝ x =

=

⎡⎣Ay( Bz + Dz) − Az( By + Dy)⎤⎦ i − ⎡⎣Ax( Bz + Dz) − Az( Bx + Dx)⎤⎦ j + ⎡⎣Ax( By + Dy) − Ay( Bx − Dx)⎤⎦ k ⎡⎣( Ay Bz − Az By) i − ( Ax Bz − Az Bx) j + ( Ax By − Ay Bx) k⎤⎦ ... + ⎡⎣( Ay Dz − A z Dy) i − ( Ax Dz − A z Dx) j + ( A x Dy − A y Dx) k⎤⎦ 210

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Engineering Mechanics - Statics

Chapter 4

⎛ i j k⎞ ⎛ i j k⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ Ax Ay Az ⎟ + ⎜ Ax Ay Az ⎟ ⎜B B B ⎟ ⎜D D D ⎟ ⎝ x y z⎠ ⎝ x y z⎠

( A × B) + ( A × D)

=

(QED)

Problem 4-2 Prove the triple scalar product identity A⋅ ( B × C) = ( A × B) ⋅ C. Solution: As shown in the figure Area = B ( C sin ( θ ) ) = B × C Thus, Volume of parallelopiped is B × C h But, h = A⋅ u

B× C

⎛ B× C ⎞ ⎟ ⎝ B× C ⎠

= A⋅ ⎜

Thus, Volume = A⋅ ( B × C) Since A × B⋅ C represents this same volume then A⋅ ( B × C) = ( A × B) ⋅ C

(QED)

Also, LHS = A⋅ ( B × C)

⎛ i j k⎞ ⎜ ⎟ = ( A xi + A yj + Azk) ⎜ Bx By Bz ⎟ ⎜C C C ⎟ ⎝ x y z⎠ = Ax( By Cz − B z Cy) − Ay( Bx Cz − B z Cx) + Az( B x Cy − B y Cx) = Ax B y Cz − A x Bz Cy − Ay B x Cz + A y Bz Cx + Az Bx Cy − A z B y Cx RHS = ( A × B) ⋅ C 211

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Engineering Mechanics - Statics

Chapter 4

⎛ i j k⎞ ⎜ ⎟ A A A x y z ⎜ ⎟ ( Cxi + Cyj + Czk) ⎜B B B ⎟ ⎝ x y z⎠

=

= Cx( Ay B z − Az By) − Cy( A x Bz − A z B x) + Cz( A x By − Ay B x) =

Ax B y Cz − A x Bz Cy − Ay B x Cz + A y Bz Cx + Az Bx Cy − A z B y Cx

Thus,

LHS = RHS

A⋅ B × C = A × B⋅ C

(QED)

Problem 4-3 Given the three nonzero vectors A, B, and C, show that if A⋅ ( B × C) = 0, the three vectors must lie in the same plane. Solution: Consider, A⋅ ( B × C) = A B × C cos ( θ ) =

(

=

A cos ( θ ) ) B × C h

B× C

= BC h sin ( φ ) = volume of parallelepiped. If A⋅ ( B × C) = 0, then the volume equals zero, so that A, B, and C are coplanar.

Problem 4-4 Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point O. Given: F 1 = 40 lb F 2 = 60 lb

212

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Engineering Mechanics - Statics

Chapter 4

θ 1 = 30 deg θ 2 = 45 deg a = 5 in b = 13 in c = 3 in d = 6 in e = 3 in f = 6 in Solution: MRO =ΣMO; MRO = F 1 cos ( θ 2 ) e − F1 sin ( θ 2 ) f − F2 cos ( θ 1 ) b − a − F2 sin ( θ 1 ) a 2

MRO = −858 lb⋅ in

2

MRO = 858 lb⋅ in

Problem 4-5 Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point P. Units Used: kip = 1000 lb Given: F 1 = 40 lb

b = 13 in

F 2 = 60 lb

c = 3 in

θ 1 = 30 deg

d = 6 in

θ 2 = 45 deg

e = 3 in

a = 5 in

f = 6 in

Solution: MRP = ΣMP; MRP = F1 cos ( θ 2 ) ( e + c) − F 1 sin ( θ 2 ) ( d + f) − F2 cos ( θ 1 ) + −F2 sin ( θ 1 ) ( a − c)

(

2

2

)

b − a + d ...

213

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Engineering Mechanics - Statics

Chapter 4

MRP = −1165 lb⋅ in

MRP = 1.17 kip⋅ in

Problem 4-6 Determine the magnitude of the force F that should be applied at the end of the lever such that this force creates a clockwise moment M about point O. Given: M = 15 N m

φ = 60 deg θ = 30 deg a = 50 mm b = 300 mm Solution: M = F cos ( θ ) ( a + b sin ( φ ) ) − F sin ( θ ) ( b cos ( φ ) )

F =

M

cos ( θ ) ( a + b sin ( φ ) ) − sin ( θ ) ( b cos ( φ ) )

F = 77.6 N

Problem 4-7 Determine the angle θ (0 <= θ <= 90 deg) so that the force F develops a clockwise moment M about point O. Given: F = 100 N

φ = 60 deg

214

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Engineering Mechanics - Statics

Chapter 4

M = 20 N⋅ m a = 50 mm

θ = 30 deg

b = 300 mm

Solution:

θ = 30 deg

Initial Guess

Given M = F cos ( θ ) ( a + b sin ( φ ) ) − F sin ( θ ) ( b cos ( φ ) )

θ = Find ( θ )

θ = 28.6 deg

Problem 4-8 Determine the magnitude and directional sense of the moment of the forces about point O. Units Used: 3

kN = 10 N Given: F B = 260 N

e = 2m

a = 4m

f = 12

b = 3m

g = 5

c = 5m

θ = 30 deg

d = 2m

F A = 400 N

Solution: Mo = F A sin ( θ ) d + FA cos ( θ ) c + F B

f 2

( a + e) 2

f +g Mo = 3.57 kN⋅ m

(positive means counterclockwise)

215

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Engineering Mechanics - Statics

Chapter 4

Problem 4-9 Determine the magnitude and directional sense of the moment of the forces about point P.

Units Used: 3

kN = 10 N Given: F B = 260 N

e = 2m

a = 4m

f = 12

b = 3m

g = 5

c = 5m

θ = 30 deg

d = 2m F A = 400 N Solution: Mp = F B

g 2

f +g

2

b + FB

Mp = 3.15 kN⋅ m

f 2

2

e − FA sin ( θ ) ( a − d) + FA cos ( θ ) ( b + c)

f +g

(positive means counterclockwise)

Problem 4-10 A force F is applied to the wrench. Determine the moment of this force about point O. Solve the problem using both a scalar analysis and a vector analysis.

216

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Engineering Mechanics - Statics

Chapter 4

Given: F = 40 N

θ = 20 deg a = 30 mm b = 200 mm Scalar Solution MO = −F cos ( θ ) b + F sin ( θ ) a MO = −7.11 N⋅ m

MO = 7.11 N⋅ m

Vector Solution

⎛ b ⎞ ⎛ −F sin ( θ ) ⎞ ⎜ ⎟ ⎜ ⎟ MO = a × −F cos ( θ ) ⎜ ⎟ ⎜ ⎟ 0 ⎝0⎠ ⎝ ⎠

⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −7.11 ⎠

MO = 7.107 N⋅ m

Problem 4-11 Determine the magnitude and directional sense of the resultant moment of the forces about point O. Units Used: 3

kip = 10 lb Given: F 1 = 300 lb

e = 10 ft

F 2 = 250 lb

f = 4

a = 6 ft

g = 3

b = 3 ft

θ = 30 deg

c = 4 ft

φ = 30 deg

d = 4 ft

217

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Engineering Mechanics - Statics

Chapter 4

Solution: Mo = F 2

f 2

f +g Mo = 2.42 kip⋅ ft

2

e sin ( φ ) + F2

g 2

2

e cos ( φ ) + F1 sin ( θ ) a − F1 cos ( θ ) b

f +g

positive means clockwise

Problem 4-12 To correct a birth defect, the tibia of the leg is straightened using three wires that are attached through holes made in the bone and then to an external brace that is worn by the patient. Determine the moment of each wire force about joint A. Given: F1 = 4 N

d = 0.15 m

F2 = 8 N

e = 20 mm

F3 = 6 N

f = 35 mm

a = 0.2 m

g = 15 mm

b = 0.35 m

θ 1 = 30 deg

c = 0.25 m

θ 2 = 15 deg

Solution:

Positive means counterclockwise

MA1 = F 1 cos ( θ 2 ) d + F1 sin ( θ 2 ) e

MA1 = 0.6 N⋅ m

MA2 = F 2 ( c + d)

MA2 = 3.2 N⋅ m

MA3 = F 3 cos ( θ 1 ) ( b + c + d) − F 3 sin ( θ 1 ) g MA3 = 3.852 N⋅ m

Problem 4-13 To correct a birth defect, the tibia of the leg is straightened using three wires that are attached through holes made in the bone and then to an external brace that is worn by the patient. Determine the moment of each wire force about joint B. Given: F1 = 4 N

d = 0.15 m

218

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Engineering Mechanics - Statics

F2 = 8 N

e = 20 mm

F3 = 6 N

f = 35 mm

a = 0.2 m

g = 15 mm

b = 0.35 m

θ 1 = 30 deg

c = 0.25 m

θ 2 = 15 deg

Chapter 4

Solution: Positive means clockwise MB1 = F 1 cos ( θ 2 ) ( a + b + c) − F 1 sin ( θ 2 ) e MB1 = 3.07 N⋅ m MB2 = F 2 ( a + b)

MB2 = 4.4 N⋅ m

MB3 = F 3 cos ( θ 1 ) a + F 3 sin ( θ 1 ) g

MB3 = 1.084 N⋅ m

Problem 4-14 Determine the moment of each force about the bolt located at A. Given: F B = 40 lb

a = 2.5 ft

α = 20 deg

F C = 50 lb

b = 0.75 ft

β = 25 deg

γ = 30 deg

219

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Engineering Mechanics - Statics

Chapter 4

Solution: MB = FB cos ( β ) a MB = 90.6 lb⋅ ft MC = F C cos ( γ ) ( a + b) MC = 141 lb⋅ ft

Problem 4-15 Determine the resultant moment about the bolt located at A. Given: F B = 30 lb F C = 45 lb a = 2.5 ft b = 0.75 ft

α = 20 deg β = 25 deg γ = 30 deg Solution: MA = FB cos ( β ) a + FC cos ( γ ) ( a + b)

MA = 195 lb⋅ ft

Problem 4-16 The elbow joint is flexed using the biceps brachii muscle, which remains essentially vertical as the arm moves in the vertical plane. If this muscle is located a distance a from the pivot point A on the humerus, determine the variation of the moment capacity about A if the constant force developed by the muscle is F. Plot these results of M vs. θ for −60 ≤ θ ≤ 80.

220

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N

Given: a = 16 mm F = 2.30 kN

θ = ( −60 .. 80) Solution: MA ( θ ) = F ( a) cos ( θ deg)

N.m

50

MA( θ )

0

50

0

50

100

θ

Problem 4-17 The Snorkel Co.produces the articulating boom platform that can support weight W. If the boom is in the position shown, determine the moment of this force about points A, B, and C. Units Used: 3

kip = 10 lb

221

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Engineering Mechanics - Statics

Chapter 4

Given: a = 3 ft b = 16 ft c = 15 ft

θ 1 = 30 deg θ 2 = 70 deg W = 550 lb

Solution: MA = W a

MA = 1.65 kip⋅ ft

MB = W( a + b cos ( θ 1 ) )

MB = 9.27 kip⋅ ft

MC = W( a + b cos ( θ 1 ) − c cos ( θ 2 ) )

MC = 6.45 kip⋅ ft

Problem 4-18 Determine the direction θ ( 0° ≤ θ ≤ 180°) of the force F so that it produces (a) the maximum moment about point A and (b) the minimum moment about point A. Compute the moment in each case. Given: F = 40 lb a = 8 ft b = 2 ft

222

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Engineering Mechanics - Statics

Chapter 4

Solution: The maximum occurs when the force is perpendicular to the line between A and the point of application of the force. The minimum occurs when the force is parallel to this line.

( a)

MAmax = F

2

a +b

2

b φ a = atan ⎛⎜ ⎟⎞

φ a = 14.04 deg

θ a = 90 deg − φ a

θ a = 76.0 deg

MAmin = 0 lb⋅ ft

MAmin = 0 lb⋅ ft

b φ b = atan ⎛⎜ ⎟⎞

φ b = 14.04 deg

θ b = 180 deg − φ b

θ b = 166 deg

⎝ a⎠

( b)

MAmax = 329.848 lb⋅ ft

⎝ a⎠

Problem 4-19 The rod on the power control mechanism for a business jet is subjected to force F. Determine the moment of this force about the bearing at A. Given: F = 80 N

θ 1 = 20 deg

a = 150 mm θ 2 = 60 deg

Solution: MA = F cos ( θ 1 ) ( a) sin ( θ 2 ) − F sin ( θ 1 ) ( a) cos ( θ 2 )

MA = 7.71 N⋅ m

Problem 4-20 The boom has length L, weight Wb, and mass center at G. If the maximum moment that can be developed by the motor at A is M, determine the maximum load W, having a mass center at G', that can be lifted. 223

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Engineering Mechanics - Statics

Chapter 4

Given: L = 30 ft Wb = 800 lb a = 14 ft b = 2 ft

θ = 30 deg 3

M = 20 × 10 lb⋅ ft Solution: M = Wb ( L − a) cos ( θ ) + W ( L cos ( θ ) + b) W =

M − Wb ( L − a) cos ( θ )

W = 319 lb

L cos ( θ ) + b

Problem 4-21 The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench. If a force P is applied to the wrench at B in the direction shown, determine the moment it creates about the nut at C. What is the magnitude of force F at A so that it creates the opposite moment about C ? Given: P = 50 N

θ = 60 deg a = 400 mm

b = 300 mm c = 5 d = 12

Solution: (a)

MA = P sin ( θ ) b MA = 13.0 N⋅ m

(b)

MA − F

d 2

c +d

a=0 2

224

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Engineering Mechanics - Statics

Chapter 4

⎛ c2 + d2 ⎞ ⎟ F = MA ⎜ ⎝ da ⎠ F = 35.2 N

Problem 4-22 Determine the clockwise direction θ ( 0 deg ≤ θ ≤ 180 deg) of the force F so that it produces (a) the maximum moment about point A and (b) no moment about point A. Compute the moment in each case. Given: F = 80 lb a = 4 ft b = 1 ft

Solution: (a)

2

MAmax = 330 lb⋅ ft

b φ = atan ⎛⎜ ⎟⎞

φ = 14.0 deg

θ a = 90 deg + φ

θ a = 104 deg

⎝ a⎠

( b)

2

MAmax = F a + b

MAmin = 0 b θ b = atan ⎛⎜ ⎟⎞

⎝ a⎠

θ b = 14.04 deg

Problem 4-23 The Y-type structure is used to support the high voltage transmission cables. If the supporting cables each exert a force F on the structure at B, determine the moment of each force about point A. Also, by the principle of transmissibility, locate the forces at points C and D and determine the moments.

225

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Engineering Mechanics - Statics

Chapter 4

Units Used: kip = 1000 lb Given: F = 275 lb a = 85 ft

θ = 30 deg Solution: MA1 = F sin ( θ ) a

MA1 = 11.7 kip⋅ ft

MA2 = F sin ( θ ) a

MA2 = 11.7 kip⋅ ft

Also

b = ( a)tan ( θ )

MA1 = F cos ( θ ) b

MA1 = 11.7 kip⋅ ft

MA2 = F cos ( θ ) b

MA2 = 11.7 kip⋅ ft

Problem 4-24 The force F acts on the end of the pipe at B. Determine (a) the moment of this force about point A, and (b) the magnitude and direction of a horizantal force, applied at C, which produces the same moment. Given: F = 70 N a = 0.9 m b = 0.3 m c = 0.7 m

θ = 60 deg Solution:

(a)

MA = F sin ( θ ) c + F cos ( θ ) a

MA = 73.9 N⋅ m 226

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Engineering Mechanics - Statics

Chapter 4

F C ( a) = MA

(b)

FC =

MA

F C = 82.2 N

a

Problem 4-25 The force F acts on the end of the pipe at B. Determine the angles θ ( 0° ≤ θ ≤ 180° ) of the force that will produce maximum and minimum moments about point A. What are the magnitudes of these moments? Given: F = 70 N a = 0.9 m b = 0.3 m c = 0.7 m Solution: MA = F sin ( θ ) c + F cos ( θ ) a

For maximum moment

d dθ

MA = c F cos ( θ ) − a F sin ( θ ) = 0

c θ max = atan ⎛⎜ ⎟⎞

θ max = 37.9 deg

MAmax = F sin ( θ max) c + F cos ( θ max) a

MAmax = 79.812 N⋅ m

⎝ a⎠

For minimum moment

MA = F sin ( θ ) c + F cos ( θ ) a = 0 −a ⎞ ⎟ ⎝c⎠

θ min = 180 deg + atan ⎛⎜

θ min = 128 deg

MAmin = F c sin ( θ min ) + F ( a) cos ( θ min )

MAmin = 0 N⋅ m

Problem 4-26 The towline exerts force P at the end of the crane boom of length L. Determine the placement 227

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Engineering Mechanics - Statics

Chapter 4

g p x of the hook at A so that this force creates a maximum moment about point O. What is this moment? Unit Used: 3

kN = 10 N Given: P = 4 kN L = 20 m

θ = 30 deg a = 1.5 m Solution: Maximum moment, OB ⊥ BA Guesses

x = 1m

d = 1 m (Length of the cable from B to A)

Given

L cos ( θ ) + d sin ( θ ) = x a + L sin ( θ ) = d cos ( θ )

⎛x⎞ ⎜ ⎟ = Find ( x , d) ⎝d⎠

x = 23.96 m

Mmax = P L

Mmax = 80 kN⋅ m

Problem 4-27 The towline exerts force P at the end of the crane boom of length L. Determine the position θ of the boom so that this force creates a maximum moment about point O. What is this moment? Units Used: 3

kN = 10 N

228

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Engineering Mechanics - Statics

Chapter 4

Given: P = 4 kN x = 25 m L = 20 m a = 1.5 m Solution: Maximum moment, OB ⊥ BA Guesses

θ = 30 deg

Given

L cos ( θ ) + d sin ( θ ) = x

d = 1m

(length of cable from B to A)

a + L sin ( θ ) = d cos ( θ )

⎛θ⎞ ⎜ ⎟ = Find ( θ , d) ⎝d ⎠

θ = 33.573 deg

Mmax = P L

Mmax = 80 kN⋅ m

Problem 4-28 Determine the resultant moment of the forces about point A. Solve the problem first by considering each force as a whole, and then by using the principle of moments.

229

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: F 1 = 250 N

a = 2m

F 2 = 300 N

b = 3m

F 3 = 500 N

c = 4m

θ 1 = 60 deg

d = 3

θ 2 = 30 deg

e = 4

Solution Using Whole Forces:

Geometry

d α = atan ⎛⎜ ⎟⎞

⎝e⎠

L = ⎛⎜ a + b −

d

⎝

e

c⎟⎞

e

⎠ e2 + d2

MA = −F1 ⎡⎣( a)cos ( θ 2 )⎤⎦ − F2 ( a + b) sin ( θ 1 ) − F3 L

MA = −2.532 kN⋅ m

Solution Using Principle of Moments: MA = −F1 cos ( θ 2 ) a − F2 sin ( θ 1 ) ( a + b) + F3

d 2

2

d +e

c − F3

e 2

( a + b) 2

d +e

3

MA = −2.532 × 10 N⋅ m

Problem 4-29 If the resultant moment about point A is M clockwise, determine the magnitude of F 3.

230

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: M = 4.8 kN⋅ m a = 2 m F 1 = 300 N

b = 3m

F 2 = 400 N

c = 4m

θ 1 = 60 deg

d = 3

θ 2 = 30 deg

e = 4

Solution: Initial Guess

F3 = 1 N

Given −M = −F 1 cos ( θ 2 ) a − F 2 sin ( θ 1 ) ( a + b) + F 3 F 3 = Find ( F3 )

e ⎛ d ⎞c − F ⎛ ⎞ 3 ⎜ 2 2⎟ ⎜ 2 2 ⎟ ( a + b) ⎝ d +e ⎠ ⎝ d +e ⎠

F 3 = 1.593 kN

Problem 4-30 The flat-belt tensioner is manufactured by the Daton Co. and is used with V-belt drives on poultry and livestock fans. If the tension in the belt is F, when the pulley is not turning, determine the moment of each of these forces about the pin at A.

231

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Engineering Mechanics - Statics

Chapter 4

Given: F = 52 lb a = 8 in b = 5 in c = 6 in

θ 1 = 30 deg θ 2 = 20 deg

Solution: MA1 = F cos ( θ 1 ) ( a + c cos ( θ 1 ) ) − F sin ( θ 1 ) ( b − c sin ( θ 1 ) ) MA1 = 542 lb⋅ in MA2 = F cos ( θ 2 ) ( a − c cos ( θ 2 ) ) − F ( sin ( θ 2 ) ) ( b + c sin ( θ 2 ) ) MA2 = −10.01 lb⋅ in

Problem 4-31 The worker is using the bar to pull two pipes together in order to complete the connection. If he applies a horizantal force F to the handle of the lever, determine the moment of this force about the end A. What would be the tension T in the cable needed to cause the opposite moment about point A. Given: F = 80 lb

θ 1 = 40 deg

θ 2 = 20 deg

a = 0.5 ft

b = 4.5 ft

232

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Engineering Mechanics - Statics

Chapter 4

Solution: MA = F( a + b) cos ( θ 1 ) MA = 306 lb⋅ ft Require

MA = T cos ( θ 2 ) ( a) cos ( θ 1 ) + T sin ( θ 2 ) ( a) sin ( θ 1 ) T =

MA

( a) ( cos ( θ 2 ) cos ( θ 1 ) + sin ( θ 2 ) sin ( θ 1 ) )

T = 652 lb

Problem 4-32 If it takes a force F to pull the nail out, determine the smallest vertical force P that must be applied to the handle of the crowbar. Hint: This requires the moment of F about point A to be equal to the moment of P about A. Why? Given: F = 125 lb a = 14 in b = 3 in c = 1.5 in

θ 1 = 20 deg θ 2 = 60 deg

233

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Engineering Mechanics - Statics

Chapter 4

Solution: MF = F sin ( θ 2 ) ( b)

MF = 325 lb⋅ in

P ⎡⎣( a)cos ( θ 1 ) + ( c)sin ( θ 1 )⎤⎦ = MF

P =

MF

( a) cos ( θ 1 ) + ( c) sin ( θ 1 )

P = 23.8 lb

Problem 4-33 The pipe wrench is activated by pulling on the cable segment with a horizantal force F . Determine the moment MA produced by the wrench on the pipe at θ. Neglect the size of the pulley. Given: F = 500 N a = 0.2 m b = 0.5 m c = 0.4 m

θ = 20 deg

Solution: Initial Guesses

φ = 20 deg MA = 1 N⋅ m Given b − c sin ( θ )

c cos ( θ ) − a

= tan ( φ − θ )

MA = F c sin ( φ )

⎛ φ ⎞ ⎜ ⎟ = Find ( φ , MA) ⎝ MA ⎠

φ = 84.161 deg

MA = 199 N⋅ m

234

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Engineering Mechanics - Statics

Chapter 4

Problem 4-34 Determine the moment of the force at A about point O. Express the result as a Cartesian vector. Given:

⎛ 60 ⎞ ⎜ ⎟ F = −30 N ⎜ ⎟ ⎝ −20 ⎠ a = 4m

d = 4m

b = 7m

e = 6m

c = 3m

f = 2m

Solution:

rOA

⎛ −c ⎞ ⎜ ⎟ = −b ⎜ ⎟ ⎝a⎠

MO = rOA × F

⎛ 260 ⎞ ⎜ ⎟ MO = 180 N⋅ m ⎜ ⎟ ⎝ 510 ⎠

Problem 4-35 Determine the moment of the force at A about point P. Express the result as a Cartesian vector. Given: a = 4m

b = 7m

c = 3m

d = 4m

e = 6m

f = 2m

⎛ 60 ⎞ ⎜ ⎟ F = −30 N ⎜ ⎟ ⎝ −20 ⎠ Solution:

rPA

⎛ −c − d ⎞ ⎜ ⎟ = −b − e ⎜ ⎟ ⎝ a+ f ⎠

MP = rPA × F

⎛ 440 ⎞ ⎜ ⎟ MP = 220 N⋅ m ⎜ ⎟ ⎝ 990 ⎠

235

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Engineering Mechanics - Statics

Chapter 4

Problem 4-36 Determine the moment of the force F at A about point O. Express the result as a cartesian vector. Units Used: 3

kN = 10 N Given: F = 13 kN a = 6m b = 2.5 m c = 3m d = 3m e = 8m f = 6m g = 4m h = 8m Solution:

rAB

⎛b − g⎞ ⎜ ⎟ = c+d ⎜ ⎟ ⎝h − a⎠

MO = rOA × F1

rOA

⎛ −b ⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝a⎠

F1 = F

rAB rAB

⎛ −84 ⎞ ⎜ ⎟ MO = −8 kN⋅ m ⎜ ⎟ ⎝ −39 ⎠

Problem 4-37 Determine the moment of the force F at A about point P. Express the result as a Cartesian vector. Units Used:

3

kN = 10 N

236

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Engineering Mechanics - Statics

Chapter 4

Given: F = 13 kN a = 6m b = 2.5 m c = 3m d = 3m e = 8m f = 6m g = 4m h = 8m Solution:

rAB

⎛b − g⎞ ⎜ ⎟ = c+d ⎜ ⎟ ⎝h − a⎠

MO = rPA × F 1

rPA

⎛ −b − f ⎞ ⎜ ⎟ = −c − e ⎜ ⎟ ⎝ a ⎠

F1 = F

rAB rAB

⎛ −116 ⎞ ⎜ 16 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −135 ⎠

Problem 4-38 The curved rod lies in the x-y plane and has radius r. If a force F acts at its end as shown, determine the moment of this force about point O. Given: r = 3m

a = 1m

θ = 45 deg

F = 80 N b = 2 m

237

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Engineering Mechanics - Statics

Chapter 4

Solution:

rAC

⎛a⎞ ⎜ ⎟ = −r ⎜ ⎟ ⎝ −b ⎠

Fv = F

rOA

rAC

⎛1⎞ ⎜ ⎟ = −3 m ⎜ ⎟ ⎝ −2 ⎠

⎛ 21.381 ⎞ ⎜ ⎟ F v = −64.143 N ⎜ ⎟ ⎝ −42.762 ⎠

rAC rAC

⎛r⎞ ⎜ ⎟ = r ⎜ ⎟ ⎝0⎠

rOA

MO = rOA × Fv

⎛3⎞ ⎜ ⎟ = 3 m ⎜ ⎟ ⎝0⎠

⎛ −128.285 ⎞ ⎜ ⎟ MO = 128.285 N⋅ m ⎜ ⎟ ⎝ −256.571 ⎠

Problem 4-39 The curved rod lies in the x-y plane and has a radius r. If a force F acts at its end as shown, determine the moment of this force about point B. Given: F = 80 N

c = 3m

a = 1m

r = 3m

238

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Engineering Mechanics - Statics

Chapter 4

θ = 45 deg

b = 2m

Solution:

rAC

⎛a⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝ −b ⎠

Fv = F

rAC rBA

rAC

MB = rBA × F v

⎛ r cos ( θ ) ⎞ ⎜ ⎟ = r − r sin ( θ ) ⎜ ⎟ 0 ⎠ ⎝

⎛ −37.6 ⎞ ⎜ 90.7 ⎟ N⋅ m MB = ⎜ ⎟ ⎝ −154.9 ⎠

Problem 4-40 The force F acts at the end of the beam. Determine the moment of the force about point A. Given:

⎛ 600 ⎞ ⎜ ⎟ F = 300 N ⎜ ⎟ ⎝ −600 ⎠ a = 1.2 m b = 0.2 m c = 0.4 m Solution:

rAB

⎛b⎞ ⎜ ⎟ = a ⎜ ⎟ ⎝0⎠

MA = rAB × F

⎛ −720 ⎞ ⎜ ⎟ MA = 120 N⋅ m ⎜ ⎟ ⎝ −660 ⎠

239

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Engineering Mechanics - Statics

Chapter 4

Problem 4-41 The pole supports a traffic light of weight W. Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A. Given: W = 22 lb

a = 12 ft

θ = 30 deg

Solution:

⎡ ( a)sin ( θ ) ⎤ ⎢ ⎥ r = ( a)cos ( θ ) ⎢ ⎥ ⎣ 0 ⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ F = ⎜ ⎟ ⎝ −W ⎠ MA = r × F

⎛ −229 ⎞ ⎜ ⎟ MA = 132 lb⋅ ft ⎜ ⎟ ⎝ 0 ⎠

Problem 4-42 The man pulls on the rope with a force F. Determine the moment that this force exerts about the base of the pole at O. Solve the problem two ways, i.e., by using a position vector from O to A, then O to B. Given: F = 20 N a = 3m

240

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Engineering Mechanics - Statics

Chapter 4

b = 4m c = 1.5 m d = 10.5 m

Solution:

rAB

rOB

⎛ b ⎞ ⎜ −a ⎟ = ⎜ ⎟ ⎝c − d⎠ ⎛b⎞ ⎜ ⎟ = −a ⎜ ⎟ ⎝c ⎠

MO1 = rOA × F v

MO2 = rOB × Fv

rOA

⎛0⎞ ⎜ ⎟ = 0 ⎜ ⎟ ⎝d⎠

Fv = F

rAB rAB

MO1

⎛ 61.2 ⎞ ⎜ ⎟ = 81.6 N⋅ m ⎜ ⎟ ⎝ 0 ⎠

MO2

⎛ 61.2 ⎞ ⎜ ⎟ = 81.6 N⋅ m ⎜ ⎟ ⎝ −0 ⎠

Problem 4-43 Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has radius r, to fail at the support C. This requires a moment to be developed at C of magnitude M.

241

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Engineering Mechanics - Statics

Chapter 4

Given: r = 5 ft M = 80 lb⋅ ft

θ = 60 deg a = 7 ft b = 6 ft

Solution:

rAB

b ⎞ ⎛ ⎜ = a − r sin ( θ ) ⎟ ⎜ ⎟ ⎝ −r cos ( θ ) ⎠

Guess Given

uAB =

rAB

rCB

rAB

⎛b⎞ = ⎜a⎟ ⎜ ⎟ ⎝ −r ⎠

F = 1 lb rCB × ( F uAB) = M

F = Find ( F)

F = 18.6 lb

Problem 4-44 The pipe assembly is subjected to the force F . Determine the moment of this force about point A.

242

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Engineering Mechanics - Statics

Chapter 4

Given: F = 80 N a = 400 mm b = 300 mm c = 200 mm d = 250 mm

θ = 40 deg φ = 30 deg Solution:

rAC

⎛b + d⎞ ⎜ a ⎟ = ⎜ ⎟ ⎝ −c ⎠

MA = rAC × Fv

⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟ F v = F ⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ −sin ( φ ) ⎟ ⎝ ⎠ ⎛ −5.385 ⎞ ⎜ ⎟ MA = 13.093 N⋅ m ⎜ ⎟ ⎝ 11.377 ⎠

Problem 4-45 The pipe assembly is subjected to the force F . Determine the moment of this force about point B.

243

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Engineering Mechanics - Statics

Chapter 4

Given: F = 80 N a = 400 mm b = 300 mm c = 200 mm d = 250 mm

θ = 40 deg φ = 30 deg Solution:

rBC

⎛b + d⎞ = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎛ 550 ⎞ rBC = ⎜ 0 ⎟ mm ⎜ ⎟ ⎝ −200 ⎠

⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟ F v = F ⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ −sin ( φ ) ⎟ ⎝ ⎠ MB = rBC × F v

⎛ 44.534 ⎞ F v = ⎜ 53.073 ⎟ N ⎜ ⎟ ⎝ −40 ⎠ ⎛ 10.615 ⎞ MB = ⎜ 13.093 ⎟ N⋅ m ⎜ ⎟ ⎝ 29.19 ⎠

Problem 4-46 The x-ray machine is used for medical diagnosis. If the camera and housing at C have mass M and a mass center at G, determine the moment of its weight about point O when it is in the position shown. Units Used: 3

kN = 10 N

244

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Engineering Mechanics - Statics

Chapter 4

Given: M = 150 kg a = 1.2 m b = 1.5 m

θ = 60 deg g = 9.81

m 2

s Solution:

⎡ −( b) cos ( θ ) ⎤ ⎛ 0 ⎞ ⎢ ⎥×⎜ 0 ⎟ MO = a ⎢ ⎥ ⎜ ⎟ ⎣ ( b)sin ( θ ) ⎦ ⎝ −M g ⎠

⎛ −1.77 ⎞ ⎜ ⎟ MO = −1.1 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠

Problem 4-47 Using Cartesian vector analysis, determine the resultant moment of the three forces about the base of the column at A. Units Used : 3

kN = 10 N Given:

⎛ 400 ⎞ ⎜ ⎟ F 1 = 300 N ⎜ ⎟ ⎝ 120 ⎠ ⎛ 100 ⎞ ⎜ ⎟ F 2 = −100 N ⎜ ⎟ ⎝ −60 ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟N F3 = ⎜ ⎟ ⎝ −500 ⎠ a = 4m b = 8m 245

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Engineering Mechanics - Statics

Chapter 4

c = 1m Solution:

rAB

⎛ 0 ⎞ ⎜ 0 ⎟ = ⎜ ⎟ ⎝a + b⎠

rA3

⎛0⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝b⎠

The individual moments MA1 = rAB × F1

MA2 = rAB × F2

MA3 = rA3 × F3

⎛ −3.6 ⎞ ⎜ ⎟ MA1 = 4.8 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠

⎛ 1.2 ⎞ ⎜ ⎟ MA2 = 1.2 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠

⎛ 0.5 ⎞ ⎜ ⎟ MA3 = 0 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠

The total moment MA = MA1 + MA2 + MA3

⎛ −1.9 ⎞ ⎜ 6 ⎟ kN⋅ m MA = ⎜ ⎟ ⎝ 0 ⎠

Problem 4-48 A force F produces a moment MO about the origin of coordinates, point O. If the force acts at a point having the given x coordinate, determine the y and z coordinates. Units Used :

3

kN = 10 N

Given:

⎛6⎞ ⎜ ⎟ F = −2 kN ⎜ ⎟ ⎝1⎠ ⎛ 4 ⎞ ⎜ 5 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −14 ⎠ x = 1m Solution: The initial guesses: y = 1m

z = 1m 246

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Engineering Mechanics - Statics

Chapter 4

Given

⎛x⎞ ⎜ y⎟ × F = M O ⎜ ⎟ ⎝z⎠

⎛ y⎞ ⎜ ⎟ = Find ( y , z) ⎝z⎠

⎛ y⎞ ⎛2⎞ ⎜ ⎟=⎜ ⎟m ⎝ z⎠ ⎝1⎠

Problem 4-49 The force F creates a moment about point O of MO. If the force passes through a point having the given x coordinate, determine the y and z coordinates of the point. Also, realizing that MO = Fd, determine the perpendicular distance d from point O to the line of action of F . Given:

⎛6⎞ F = ⎜8 ⎟ N ⎜ ⎟ ⎝ 10 ⎠ ⎛ −14 ⎞ MO = ⎜ 8 ⎟ N⋅ m ⎜ ⎟ ⎝ 2 ⎠ x = 1m Solution: The initial guesses:

y = 1m

z = 1m

Given

⎛x⎞ ⎜ y⎟ × F = M O ⎜ ⎟ ⎝z⎠ d =

MO F

⎛ y⎞ ⎜ ⎟ = Find ( y , z) ⎝z⎠

⎛ y⎞ ⎛1⎞ ⎜ ⎟=⎜ ⎟m ⎝ z⎠ ⎝3⎠

d = 1.149 m

Problem 4-50 The force F produces a moment MO about the origin of coordinates, point O. If the force acts at a point having the given x-coordinate, determine the y and z coordinates.

247

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: x = 1m

⎛6⎞ ⎜ ⎟ F = −2 kN ⎜ ⎟ ⎝1⎠ ⎛ 4 ⎞ ⎜ 5 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −14 ⎠ Solution: y = 1m

Initial Guesses:

Given

⎛x ⎞ ⎜ ⎟ MO = y × F ⎜ ⎟ ⎝z⎠

z = 1m

⎛ y⎞ ⎜ ⎟ = Find ( y , z) ⎝z⎠

⎛ y⎞ ⎛2⎞ ⎜ ⎟=⎜ ⎟m ⎝ z⎠ ⎝1⎠

Problem 4-51 Determine the moment of the force F about the Oa axis. Express the result as a Cartesian vector. Given:

⎛ 50 ⎞ ⎜ ⎟ F = −20 N ⎜ ⎟ ⎝ 20 ⎠ a = 6m b = 2m c = 1m d = 3m e = 4m

248

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Engineering Mechanics - Statics

Chapter 4

Solution:

rOF

⎛c ⎞ = ⎜ −b ⎟ ⎜ ⎟ ⎝a⎠

MOa =

rOa

⎛0⎞ = ⎜e⎟ ⎜ ⎟ ⎝d⎠

⎡⎣( rOF × F) ⋅ uOa⎤⎦ uOa

uOa =

MOa

rOa rOa

⎛ 0 ⎞ = ⎜ 217.6 ⎟ N⋅ m ⎜ ⎟ ⎝ 163.2 ⎠

Problem 4-52 Determine the moment of the force F about the aa axis. Express the result as a Cartesian vector. Given: F = 600 lb a = 6 ft b = 3 ft c = 2 ft d = 4 ft e = 4 ft f = 2 ft

Solution:

Fv =

Maa =

⎛ −d ⎞ ⎜e ⎟ ⎟ 2 2 2⎜ c +d +e ⎝ c ⎠

⎛d⎞ r = ⎜0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎡⎣( r × Fv) ⋅ uaa⎤⎦ uaa

⎛ −441 ⎞ Maa = ⎜ −294 ⎟ lb⋅ ft ⎜ ⎟ ⎝ 882 ⎠

F

uaa =

⎛ −b ⎞ ⎜−f ⎟ ⎟ 2 2 2⎜ a +b + f ⎝ a ⎠ 1

249

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Engineering Mechanics - Statics

Chapter 4

Problem 4-53 Determine the resultant moment of the two forces about the Oa axis. Express the result as a Cartesian vector. Given: F 1 = 80 lb F 2 = 50 lb

α = 120 deg β = 60 deg γ = 45 deg a = 5 ft b = 4 ft c = 6 ft

θ = 30 deg φ = 30 deg Solution:

F 1v

⎛⎜ cos ( α ) ⎟⎞ = F1 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

⎡ ( b)sin ( θ ) ⎤ ⎢ ⎥ r1 = ( b)cos ( θ ) ⎢ ⎥ ⎣ c ⎦

uaa

⎛ cos ( φ ) ⎞ ⎜ ⎟ = −sin ( φ ) ⎜ ⎟ ⎝ 0 ⎠

F 2v

⎛⎜ 0 ⎞⎟ = ⎜ 0 ⎟ ⎜ F2 ⎟ ⎝ ⎠

0 ⎤ ⎡ ⎢ r2 = −( a) sin ( φ ) ⎥ ⎢ ⎥ 0 ⎣ ⎦

Maa =

⎡⎣( r1 × F1v + r2 × F2v) uaa⎤⎦ uaa

⎛ 26.132 ⎞ ⎜ ⎟ Maa = −15.087 lb⋅ ft ⎜ ⎟ ⎝ 0 ⎠

250

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Engineering Mechanics - Statics

Chapter 4

Problem 4-54 The force F is applied to the handle of the box wrench. Determine the component of the moment of this force about the z axis which is effective in loosening the bolt. Given: a = 3 in b = 8 in c = 2 in

⎛8⎞ F = ⎜ −1 ⎟ lb ⎜ ⎟ ⎝1⎠ Solution:

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠

⎛c ⎞ r = ⎜ −b ⎟ ⎜ ⎟ ⎝a⎠

Mz = ( r × F ) ⋅ k

Mz = 62 lb⋅ in

Problem 4-55 The force F acts on the gear in the direction shown. Determine the moment of this force about the y axis. Given: F = 50 lb a = 3 in

θ 1 = 60 deg θ 2 = 45 deg θ 3 = 120 deg

251

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛0⎞ j = ⎜1⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ r = ⎜0⎟ ⎜ ⎟ ⎝a⎠

⎛ −cos ( θ 3) ⎞ ⎜ ⎟ F v = F ⎜ −cos ( θ 2 ) ⎟ My = ( r × Fv ) ⋅ j ⎜ −cos θ ⎟ ( 1) ⎠ ⎝

My = 75 lb⋅ in

Problem 4-56 The RollerBall skate is an in-line tandem skate that uses two large spherical wheels on each skate, rather than traditional wafer-shape wheels. During skating the two forces acting on the wheel of one skate consist of a normal force F 2 and a friction force F1. Determine the moment of both of these forces about the axle AB of the wheel. Given:

θ = 30 deg F 1 = 13 lb F 2 = 78 lb a = 1.25 in Solution:

⎛ F1 ⎞ ⎛0⎞ ⎜ ⎟ F = ⎜ F2 ⎟ r = ⎜ −a ⎟ ⎜ ⎟ ⎜0 ⎟ ⎝0⎠ ⎝ ⎠ ⎛ cos ( θ ) ⎞ ab = ⎜ −sin ( θ ) ⎟ ⎜ ⎟ ⎝ 0 ⎠

Mab = ( r × F ) ⋅ ab

Mab = 0 lb⋅ in

Problem 4-57 The cutting tool on the lathe exerts a force F on the shaft in the direction shown. Determine the moment of this force about the y axis of the shaft.

252

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given:

⎛6⎞ ⎜ ⎟ F = −4 kN ⎜ ⎟ ⎝ −7 ⎠ a = 30 mm

θ = 40 deg Solution:

⎛ cos ( θ ) ⎞ ⎜ 0 ⎟ r = a ⎜ ⎟ ⎝ sin ( θ ) ⎠

⎛0⎞ ⎜ ⎟ j = 1 ⎜ ⎟ ⎝0⎠

My = ( r × F) ⋅ j

My = 0.277 kN⋅ m

Problem 4-58 The hood of the automobile is supported by the strut AB, which exerts a force F on the hood. Determine the moment of this force about the hinged axis y. Given: F = 24 lb

a = 2 ft

b = 4 ft

c = 2 ft

d = 4 ft

253

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛b⎞ ⎜ ⎟ rA = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ j = 1 ⎜ ⎟ ⎝0⎠

rAB

⎛ −b + c ⎞ ⎜ a ⎟ = ⎜ ⎟ ⎝ d ⎠

Fv = F

My = ( rA × F v ) ⋅ j

rAB rAB

⎛ −9.798 ⎞ ⎜ ⎟ F v = 9.798 lb ⎜ ⎟ ⎝ 19.596 ⎠

My = −78.384 lb⋅ ft

Problem 4-59 The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F at A. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. If the force F can be applied at A in any other direction, will it be possible to turn the nut?

254

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Engineering Mechanics - Statics

Chapter 4

Given: F = 30 N M = 14 N⋅ m a = 0.25 m b = 0.3 m c = 0.5 m d = 0.1 m Solution: 2

Mx = F c − b

2

Mx = 12 N⋅ m Mx < M

No

For Mxmax, apply force perpendicular to the handle and the x-axis. Mxmax = F c Mxmax = 15 N⋅ m Mxmax > M

Yes

Problem 4-60 The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F . Assume that the cheater pipe AB is slipped over the handle of the wrench and the F force can be applied at any point and in any direction on the assembly. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. Given: F 1 = 30 N

M = 14 N⋅ m

a = 0.25 m

b = 0.3 m

c = 0.5 m

d = 0.1 m

255

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Engineering Mechanics - Statics

Chapter 4

Solution: Mx = F1

a+c c

2

2

c −b

Mx = 18 N⋅ m Mx > M

Yes

Mxmax occurs when force is applied perpendicular to both the handle and the x-axis. Mxmax = F1 ( a + c) Mxmax = 22.5 N⋅ m Mxmax > M

Yes

Problem 4-61 The bevel gear is subjected to the force F which is caused from contact with another gear. Determine the moment of this force about the y axis of the gear shaft. Given: a = 30 mm b = 40 mm

⎛ 20 ⎞ ⎜ 8 ⎟N F = ⎜ ⎟ ⎝ −15 ⎠

256

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛ −b ⎞ ⎜ ⎟ r = 0 ⎜ ⎟ ⎝a⎠

⎛0⎞ ⎜ ⎟ j = 1 ⎜ ⎟ ⎝0⎠

My = ( r × F) ⋅ j

My = 0 N⋅ m

Problem 4-62 The wooden shaft is held in a lathe. The cutting tool exerts force F on the shaft in the direction shown. Determine the moment of this force about the x axis of the shaft. Express the result as a Cartesian vector. The distance OA is a. Given: a = 25 mm

θ = 30 deg

⎛ −5 ⎞ ⎜ ⎟ F = −3 N ⎜ ⎟ ⎝8⎠ Solution:

⎡ 0 ⎤ ⎢ ⎥ r = ( a)cos ( θ ) ⎢ ⎥ ⎣ ( a)sin ( θ ) ⎦

⎛1⎞ ⎜ ⎟ i = 0 ⎜ ⎟ ⎝0⎠

Mx =

⎡⎣( r × F) ⋅ i⎤⎦ i

⎛ 0.211 ⎞ ⎜ 0 ⎟ N⋅ m Mx = ⎜ ⎟ ⎝ 0 ⎠

Problem 4-63 Determine the magnitude of the moment of the force F about the base line CA of the tripod. Given:

⎛ 50 ⎞ ⎜ ⎟ F = −20 N ⎜ ⎟ ⎝ −80 ⎠

257

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Engineering Mechanics - Statics

Chapter 4

a = 4m b = 2.5 m c = 1m d = 0.5 m e = 2m f = 1.5 m g = 2m

Solution:

rCA

⎛ −g ⎞ = ⎜ e ⎟ ⎜ ⎟ ⎝0⎠

uCA =

rCA rCA

rCD

⎛b − g⎞ = ⎜ e ⎟ ⎜ ⎟ ⎝ a ⎠

MCA = ( rCD × F) ⋅ uCA

MCA = 226 N⋅ m

Problem 4-64 The flex-headed ratchet wrench is subjected to force P, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A. Given: P = 16 lb

a = 10 in

θ = 60 deg b = 0.75 in Solution: M = P ⎡⎣b + ( a)sin ( θ )⎤⎦

M = 150.564 lb⋅ in

258

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Engineering Mechanics - Statics

Chapter 4

Problem 4-65 If a torque or moment M is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench. Given: M = 80 lb⋅ in

θ = 60 deg a = 10 in b = 0.75 in Solution: M = P ⎡⎣b + ( a)sin ( θ )⎤⎦

P =

M

b + ( a)sin ( θ )

P = 8.50 lb

Problem 4-66 The A-frame is being hoisted into an upright position by the vertical force F . Determine the moment of this force about the y axis when the frame is in the position shown. Given: F = 80 lb a = 6 ft b = 6 ft

θ = 30 deg φ = 15 deg Solution: Using the primed coordinates we have

259

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Engineering Mechanics - Statics

⎛ −sin ( θ ) ⎞ ⎜ ⎟ j = cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

Chapter 4

⎛0⎞ ⎜ ⎟ Fv = F 0 ⎜ ⎟ ⎝1⎠

My = ( rAC × F v ) ⋅ j

rAC

⎛ −b cos ( φ ) ⎞ ⎜ ⎟ a ⎟ = ⎜ 2 ⎜ ⎟ ⎜ b sin ( φ ) ⎟ ⎝ ⎠

My = 281.528 lb⋅ ft

Problem 4-67 Determine the moment of each force acting on the handle of the wrench about the a axis. Given:

⎛ −2 ⎞ ⎜ ⎟ F 1 = 4 lb ⎜ ⎟ ⎝ −8 ⎠

⎛3⎞ ⎜ ⎟ F 2 = 2 lb ⎜ ⎟ ⎝ −6 ⎠

b = 6 in c = 4 in d = 3.5 in

θ = 45 deg

Solution:

⎛ cos ( θ ) ⎞ ⎜ 0 ⎟ ua = ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎜ 0 ⎟ r1 = b 0 + ( c + d) ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠ M1a = ( r1 × F1 ) ⋅ ua

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ + c⎜ 0 ⎟ r2 = b ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

M1a = 30 lb⋅ in

260

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Engineering Mechanics - Statics

M2a = ( r2 × F2 ) ⋅ ua

Chapter 4

M2a = 8 lb⋅ in

Problem 4-68 Determine the moment of each force acting on the handle of the wrench about the z axis. Given:

⎛ −2 ⎞ ⎜ ⎟ F 1 = 4 lb ⎜ ⎟ ⎝ −8 ⎠

⎛3⎞ ⎜ ⎟ F 2 = 2 lb ⎜ ⎟ ⎝ −6 ⎠

b = 6 in c = 4 in d = 3.5 in

θ = 45 deg

Solution:

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎜ 0 ⎟ r1 = b 0 + ( c + d) ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ + c⎜ 0 ⎟ r2 = b ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

M1z = ( r1 × F 1 ) ⋅ k

M1z = 38.2 lb⋅ in

M2z = ( r2 × F 2 ) ⋅ k

M2z = 14.1 lb⋅ in

⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠

Problem 4-69 Determine the magnitude and sense of the couple moment. Units Used: 3

kN = 10 N 261

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Engineering Mechanics - Statics

Chapter 4

Given: F = 5 kN

θ = 30 deg a = 0.5 m b = 4m c = 2m d = 1m Solution: MC = F cos ( θ ) ( a + c) + F sin ( θ ) ( b − d) MC = 18.325 kN⋅ m

Problem 4-70 Determine the magnitude and sense of the couple moment. Each force has a magnitude F. Given: F = 65 lb a = 2 ft b = 1.5 ft c = 4 ft d = 6 ft e = 3 ft

Solution: Mc = ΣMB;

⎡ ⎛

⎞ ( d + a)⎤ + ⎡F⎛ e ⎞ ( c + a)⎤ ⎥ ⎢ ⎜ 2 2⎟ ⎥ 2 2⎟ ⎣ ⎝ c +e ⎠ ⎦ ⎣ ⎝ c +e ⎠ ⎦

MC = ⎢F⎜

c

MC = 650 lb⋅ ft

(Counterclockwise)

Problem 4-71 Determine the magnitude and sense of the couple moment. 262

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kip = 10 lb Given: F = 150 lb a = 8 ft b = 6 ft c = 8 ft d = 6 ft e = 6 ft f = 8 ft Solution: MC = ΣMA;

MC = F

d 2

d + f MC = 3120 lb⋅ ft

f

( a + f) + F 2

2

d + f

( c + d) 2

MC = 3.120 kip⋅ ft

Problem 4-72 If the couple moment has magnitude M, determine the magnitude F of the couple forces. Given: M = 300 lb⋅ ft a = 6 ft b = 12 ft c = 1 ft d = 2 ft e = 12 ft f = 7 ft

263

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Engineering Mechanics - Statics

Chapter 4

Solution:

M=F

( f − d) ( b + e) ⎤ ⎡ e( f + a) − ⎢ 2 2 2 2⎥ ( f − d) + e ⎦ ⎣ ( f − d) + e

M

F =

e( f + a) 2

−

F = 108 lb

( f − d) ( b + e)

2

2

( f − d) + e

2

( f − d) + e

Problem 4-73 A clockwise couple M is resisted by the shaft of the electric motor. Determine the magnitude of the reactive forces −R and R which act at supports A and B so that the resultant of the two couples is zero. Given: a = 150 mm

θ = 60 deg M = 5 N⋅ m

Solution: MC = −M +

2R a

tan ( θ )

=0

R =

M tan ( θ ) 2

a

R = 28.9 N

Problem 4-74 The resultant couple moment created by the two couples acting on the disk is MR. Determine the magnitude of force T.

264

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kip = 10 lb Given:

⎛0⎞ ⎜ ⎟ MR = 0 kip⋅ in ⎜ ⎟ ⎝ 10 ⎠ a = 4 in b = 2 in c = 3 in Solution: Initial Guess

Given

T = 1 kip

⎛ a ⎞ ⎛ 0 ⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛ −b − c ⎞ ⎛ 0 ⎞ ⎜ 0 ⎟ × ⎜ T ⎟ + ⎜ 0 ⎟ × ⎜ −T ⎟ + ⎜ 0 ⎟ × ⎜ −T ⎟ = M R ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠

T = Find ( T)

T = 0.909 kip

Problem 4-75 Three couple moments act on the pipe assembly. Determine the magnitude of M3 and the bend angle θ so that the resultant couple moment is zero.

Given:

θ 1 = 45 deg M1 = 900 N⋅ m M2 = 500 N⋅ m

265

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Engineering Mechanics - Statics

Chapter 4

Solution: Initial guesses:

θ = 10 deg

M3 = 10 N⋅ m

Given + ΣMx = 0; →

M1 − M3 cos ( θ ) − M2 cos ( θ 1 ) = 0

+

M3 sin ( θ ) − M2 sin ( θ 1 ) = 0

↑ ΣMy = 0;

⎛ θ ⎞ ⎜ ⎟ = Find ( θ , M3) ⎝ M3 ⎠

θ = 32.9 deg

M3 = 651 N⋅ m

Problem 4-76 The floor causes couple moments MA and MB on the brushes of the polishing machine. Determine the magnitude of the couple forces that must be developed by the operator on the handles so that the resultant couple moment on the polisher is zero. What is the magnitude of these forces if the brush at B suddenly stops so that MB = 0? Given: a = 0.3 m MA = 40 N⋅ m MB = 30 N⋅ m Solution:

MA − MB − F 1 a = 0

F1 =

MA − F 2 a = 0

F2 =

MA − MB a MA

F 1 = 33.3 N

F 2 = 133 N

a

266

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Engineering Mechanics - Statics

Chapter 4

Problem 4-77 The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is M clockwise. Given: F 1 = 600 N F 2 = 250 N a = 1m

θ = 40 deg M = 400 N⋅ m Solution: Initial Guess

Given

F1

F = 1N

⎛ a ⎞ − F a − F ⎛ a ⎞ = −M ⎜ ⎟ 2 ⎜ ⎟ ⎝ 2 cos ( θ ) ⎠ ⎝ 2 cos ( θ ) ⎠

F = Find ( F)

F = 830 N

Problem 4-78 Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is M counterclockwise. Where on the beam does the resultant couple moment act? Given: M = 450 lb⋅ ft P = 200 lb a = 1.5 ft b = 1.25 ft c = 2 ft

θ = 30 deg

267

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Engineering Mechanics - Statics

Chapter 4

Solution: MR = Σ M

M = F b cos ( θ ) + P a

F =

M − Pa

b cos ( θ )

F = 139 lb

The resultant couple moment is a free vector. It can act at any point on the beam.

Problem 4-79 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O. Given:

⎛0⎞ F = ⎜0 ⎟ N ⎜ ⎟ ⎝ 25 ⎠ a = 300 mm b = 150 mm c = 400 mm d = 200 mm e = 200 mm

Solution:

( a)

( b)

rAB

⎛ −e − b ⎞ = ⎜ −c + d ⎟ ⎜ ⎟ ⎝ 0 ⎠

M = rAB × F

rOB

⎛a⎞ = ⎜d⎟ ⎜ ⎟ ⎝0⎠

⎛a + b + e⎞ ⎟ c = ⎜ ⎜ ⎟ ⎝ 0 ⎠

rOA

⎛ −5 ⎞ M = ⎜ 8.75 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

⎛ −5 ⎞ M = ⎜ 8.75 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = rOB × F + rOA × ( −F)

268

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Engineering Mechanics - Statics

Chapter 4

Problem 4-80 If the couple moment acting on the pipe has magnitude M, determine the magnitude F of the vertical force applied to each wrench. Given: M = 400 N⋅ m a = 300 mm b = 150 mm c = 400 mm d = 200 mm e = 200 mm

Solution:

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠

rAB

⎛ −e − b ⎞ = ⎜ −c + d ⎟ ⎜ ⎟ ⎝ 0 ⎠

Guesss

F = 1N rAB × ( Fk) = M

Given

F = Find ( F)

F = 992.278 N

Problem 4-81 Determine the resultant couple moment acting on the beam. Solve the problem two ways: (a) sum moments about point O; and (b) sum moments about point A. Units Used: 3

kN = 10 N

269

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 2 kN

θ 1 = 30 deg

F 2 = 8 kN

θ 2 = 45 deg

a = 0.3 m b = 1.5 m c = 1.8 m Solution: MR = ΣMO;

( a)

MRa = ( F1 cos ( θ 1 ) + F 2 cos ( θ 2 ) ) c + ( F2 cos ( θ 2 ) − F 1 sin ( θ 1 ) ) a ... + −( F2 cos ( θ 2 ) + F 1 cos ( θ 1 ) ) ( b + c) MRa = −9.69 kN⋅ m MR = ΣMA;

( b)

MRb = ( F2 sin ( θ 2 ) − F1 sin ( θ 1 ) ) a − ( F 2 cos ( θ 2 ) + F1 cos ( θ 1 ) ) b MRb = −9.69 kN⋅ m

Problem 4-82 Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is M counterclockwise. Where on the beam does the resultant couple act? Given: M = 300 lb⋅ ft a = 4 ft b = 1.5 ft P = 200 lb c = 3 d = 4

270

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Engineering Mechanics - Statics

Chapter 4

Solution: c

M=

Fa +

2

2

2

2⎛ M + P b ⎞

c +d F =

d

c +d

F = 167 lb

2

c +d

Fb − Pb 2

⎜ ⎟ ⎝ ca + d b⎠

Resultant couple can act anywhere.

Problem 4-83 Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the couple forces F 1. Given: F 1 = 80 lb F 2 = 50 lb a = 1 ft b = 3 ft c = 2 ft e = 3 ft f = 3 g = 4

θ = 30 deg Solution: g ⎡−F cos ( θ ) e + ⎛ ⎞ ⎤ ⎢ 2 ⎜ 2 2 ⎟ F1 d⎥ = 0 ⎣ ⎝ g +f ⎠ ⎦

d =

F2 F1

⎛ g2 + f 2 ⎞ ⎟ g ⎝ ⎠

cos ( θ ) e ⎜

d = 2.03 ft

271

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Engineering Mechanics - Statics

Chapter 4

Problem 4-84 Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4-13) and (b) summing the moments of all the force components about point A. Given: F 1 = 80 lb

c = 2 ft

g = 4

F 2 = 50 lb

d = 4 ft

θ = 30 deg

a = 1 ft

e = 3 ft

b = 3 ft

f = 3

Solution: ( a)

M = Σ ( r × F)

⎛ e ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −g ⎞⎤ 1 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ M = 0 × F 2 −cos ( θ ) + d × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦

⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft M= ⎜ ⎟ ⎝ 126.096 ⎠

( b)

Summing the moments of all force components aboout point A.

M1 =

g ⎛ −g ⎞ F b + ⎛ ⎞ 1 ⎜ 2 2⎟ ⎜ 2 2 ⎟ F 1 ( b + d) ⎝ f +g ⎠ ⎝ f +g ⎠

M2 = F 2 cos ( θ ) c − F 2 sin ( θ ) ( a + b + d) − F 2 cos ( θ ) ( c + e) + F 2 sin ( θ ) ( a + b + d) M = M1 + M2

M = 126.096 lb⋅ ft

Problem 4-85 Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4 -13) and (b) summing the moments of all the force components about point B.

272

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 80 lb

d = 4ft

F 2 = 50 lb

e = 3 ft

a = 1 ft

f = 3 ft

b = 3 ft

g = 4 ft

c = 2 ft

θ = 30 deg

Solution: ( a)

M = Σ ( r × F)

⎛ e ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −g ⎞⎤ 1 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ M = 0 × F 2 −cos ( θ ) + d × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ( b)

⎛ 0 ⎞ ⎟ lb⋅ ft M=⎜ 0 ⎜ ⎟ ⎝ 126.096 ⎠

Summing the moments of all force components about point B. M1 =

g g ⎛ ⎛ ⎞ ⎞ ⎜ 2 2 ⎟ F1( a + d) − ⎜ 2 2 ⎟ F1 a ⎝ f +g ⎠ ⎝ f +g ⎠

M2 = F 2 cos ( θ ) c − F 2 cos ( θ ) ( c + e) M = M1 + M2

M = 126.096 lb⋅ ft

Problem 4-86 Determine the couple moment. Express the result as a Cartesian vector.

273

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Engineering Mechanics - Statics

Chapter 4

Given:

⎛8⎞ F = ⎜ −4 ⎟ N ⎜ ⎟ ⎝ 10 ⎠ a = 5m b = 3m c = 4m d = 2m e = 3m Solution:

⎛ −b − e ⎞ r = ⎜ c+d ⎟ ⎜ ⎟ ⎝ −a ⎠

M = r× F

⎛ 40 ⎞ M = ⎜ 20 ⎟ N⋅ m ⎜ ⎟ ⎝ −24 ⎠

Problem 4-87 Determine the couple moment. Express the result as a Cartesian vector. Given: F = 80 N a = 6m b = 10 m c = 10 m d = 5m e = 4m f = 4m

274

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Engineering Mechanics - Statics

Chapter 4

Solution: u =

⎛ b ⎞ ⎜c + d⎟ ⎟ 2 2 2⎜ a + b + ( c + d) ⎝ − a ⎠ 1

F v = Fu

⎛−f − b⎞ r = ⎜ −d − c ⎟ ⎜ ⎟ ⎝ e+a ⎠

⎛ −252.6 ⎞ M = ⎜ 67.4 ⎟ N⋅ m ⎜ ⎟ ⎝ −252.6 ⎠

M = r × Fv

Problem 4-88 If the resultant couple of the two couples acting on the fire hydrant is MR = { −15i + 30j} N ⋅ m, determine the force magnitude P. Given: a = 0.2 m b = 0.150 m

⎛ −15 ⎞ M = ⎜ 30 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠ F = 75 N Solution: Initial guess

P = 1N

Given

⎛ −F a ⎞ M = ⎜ Pb ⎟ ⎜ ⎟ ⎝ 0 ⎠

P = Find ( P)

P = 200 N

Problem 4-89 If the resultant couple of the three couples acting on the triangular block is to be zero, determine the magnitude of forces F and P . 275

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 150 N a = 300 mm b = 400 mm d = 600 mm

Solution:

Initial guesses:

Given

F = 1N

P = 1N

⎛ d ⎞ ⎛ 0 ⎞ ⎛ d ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ −F1 ⎟⎞ ⎛ 0 ⎞ ⎛⎜ F1 ⎞⎟ ⎜ 0 ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ −P ⎟ + ⎜ b ⎟ × ⎜ ⎟ + 0 × =0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜0 ⎟ ⎝ a ⎠ ⎝ F ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ a ⎠ ⎜⎝ 0 ⎟⎠

⎛F⎞ ⎜ ⎟ = Find ( F , P) ⎝P⎠

⎛ F ⎞ ⎛ 75 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ P ⎠ ⎝ 100 ⎠

Problem 4-90 Determine the couple moment that acts on the assembly. Express the result as a Cartesian vector. Member BA lies in the x-y plane.

276

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Engineering Mechanics - Statics

Chapter 4

Given:

⎛ 0 ⎞ ⎜ 0 ⎟N F = ⎜ ⎟ ⎝ 100 ⎠ a = 300 mm b = 150 mm c = 200 mm d = 200 mm

θ = 60 deg

Solution:

⎡−( c + d) sin ( θ ) − b cos ( θ ) ⎤ ⎢ ⎥ r = −b sin ( θ ) + ( c + d)cos ( θ ) ⎢ ⎥ 0 ⎦ ⎣

M = r× F

⎛ 7.01 ⎞ ⎜ ⎟ M = 42.14 N⋅ m ⎜ ⎟ ⎝ 0.00 ⎠

Problem 4-91 If the magnitude of the resultant couple moment is M, determine the magnitude F of the forces applied to the wrenches. Given: M = 15 N⋅ m

c = 200 mm

a = 300 mm

d = 200 mm

b = 150 mm

θ = 60 deg

277

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎡−( c + d) sin ( θ ) − b cos ( θ ) ⎤ ⎢ ⎥ r = −b sin ( θ ) + ( c + d)cos ( θ ) ⎢ ⎥ 0 ⎦ ⎣ Guess Given

⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠

F = 1N r × ( Fk) = M

F = Find ( F)

F = 35.112 N

Problem 4-92 The gears are subjected to the couple moments shown. Determine the magnitude and coordinate direction angles of the resultant couple moment. Given: M1 = 40 lb⋅ ft M2 = 30 lb⋅ ft

θ 1 = 20 deg θ 2 = 15 deg θ 3 = 30 deg

Solution:

⎛ M1 cos ( θ 1) sin ( θ 2) ⎞ ⎜ ⎟ M1 = ⎜ M1 cos ( θ 1 ) cos ( θ 2 ) ⎟ ⎜ −M sin θ ( 1) ⎟⎠ 1 ⎝ MR = M1 + M2

⎛⎜ α ⎞⎟ ⎛ MR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ MR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛ −M2 sin ( θ 3) ⎞ ⎜ ⎟ M2 = ⎜ M2 cos ( θ 3 ) ⎟ ⎜ ⎟ 0 ⎝ ⎠ MR = 64.0 lb⋅ ft

⎛⎜ α ⎞⎟ ⎛ 94.7 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 13.2 ⎟ deg ⎜ γ ⎟ ⎝ 102.3 ⎠ ⎝ ⎠

278

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Engineering Mechanics - Statics

Chapter 4

Problem 4-93 Express the moment of the couple acting on the rod in Cartesian vector form. What is the magnitude of the couple moment? Given:

⎛ 14 ⎞ ⎜ ⎟ F = −8 N ⎜ ⎟ ⎝ −6 ⎠ a = 1.5 m b = 0.5 m c = 0.5 m d = 0.8 m Solution:

⎛d⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ M = a × F + 0 × ( −F) ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝b⎠

⎛ −17 ⎞ ⎜ ⎟ M = −9.2 N⋅ m ⎜ ⎟ ⎝ −27.4 ⎠

M = 33.532 N⋅ m

Problem 4-94 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O.

279

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Engineering Mechanics - Statics

Chapter 4

Given: a = 0.3 m b = 0.4 m c = 0.6 m

⎛ −6 ⎞ F = ⎜ 2 ⎟ N ⎜ ⎟ ⎝3⎠ Solution:

⎛0⎞ M = ⎜ c ⎟×F ⎜ ⎟ ⎝ −b ⎠

(a)

⎛ 2.6 ⎞ M = ⎜ 2.4 ⎟ N⋅ m ⎜ ⎟ ⎝ 3.6 ⎠

⎛0⎞ ⎛ 0 ⎞ ⎜ ⎟ M = 0 × ( −F ) + ⎜ c ⎟ × F ⎜ ⎟ ⎜ ⎟ ⎝ −a ⎠ ⎝ −a − b ⎠

(b)

⎛ 2.6 ⎞ M = ⎜ 2.4 ⎟ N⋅ m ⎜ ⎟ ⎝ 3.6 ⎠

Problem 4-95 A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment. Given: F 1 = 35 N θ = 60 deg F 2 = 25 N r1 = 175 mm r2 = 175 mm Solution:

⎛⎜ −F1 2 r1 ⎞⎟ ⎛⎜ −F2 2r2 cos ( θ ) ⎞⎟ M = ⎜ 0 ⎟ + ⎜ −F2 2 r2 sin ( θ ) ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ 0 ⎠

⎛ −16.63 ⎞ M = ⎜ −7.58 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = 18.3 N⋅ m

280

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Engineering Mechanics - Statics

Chapter 4

⎛⎜ α ⎞⎟ ⎛ M ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ M ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 155.496 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 114.504 ⎟ deg ⎜ γ ⎟ ⎝ 90 ⎠ ⎝ ⎠

Problem 4-96 Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment? Given: F = 125 N a = 150 mm b = 150 mm c = 200 mm d = 600 mm

Solution:

⎛ c ⎞ ⎛0⎞ M = ⎜a + b⎟ × ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝F⎠

⎛ 37.5 ⎞ M = ⎜ −25 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = 45.1 N⋅ m

Problem 4-97 If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the forces applied to the wrenches.

281

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Engineering Mechanics - Statics

Chapter 4

Given: M = 300 N⋅ m a = 150 mm b = 150 mm c = 200 mm d = 600 mm Solution: F = 1N

Initial guess:

Given

⎛ c ⎞ ⎛0⎞ ⎜a + b⎟ × ⎜ 0 ⎟ = M ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝F⎠

F = Find ( F)

F = 832.1 N

Problem 4-98 Replace the force at A by an equivalent force and couple moment at point O. Given: F = 375 N a = 2m b = 4m c = 2m d = 1m

θ = 30 deg

282

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛ sin ( θ ) ⎞ ⎜ ⎟ F v = F −cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

⎛ 187.5 ⎞ ⎜ ⎟ F v = −324.76 N ⎜ ⎟ ⎝ 0 ⎠

⎛ −a ⎞ ⎜ ⎟ MO = b × Fv ⎜ ⎟ ⎝0⎠

⎛ 0 ⎞ ⎜ ⎟ N⋅ m 0 MO = ⎜ ⎟ ⎝ −100.481 ⎠

Problem 4-99 Replace the force at A by an equivalent force and couple moment at point P . Given: F = 375 N a = 2m b = 4m c = 2m d = 1m

θ = 30 deg Solution:

⎛ sin ( θ ) ⎞ ⎜ ⎟ F v = F −cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

⎛ 187.5 ⎞ ⎜ ⎟ F v = −324.76 N ⎜ ⎟ ⎝ 0 ⎠

⎛ −a − c ⎞ ⎜ ⎟ MP = b − d × Fv ⎜ ⎟ ⎝ 0 ⎠

⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MP = ⎜ ⎟ ⎝ 736.538 ⎠

Problem 4-100 Replace the force system by an equivalent resultant force and couple moment at point O.

283

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 60 lb

a = 2 ft

F 2 = 85 lb

b = 3 ft

F 3 = 25 lb

c = 6 ft

θ = 45 deg

d = 4 ft

e = 3 f = 4 Solution:

⎛⎜ 0 ⎟⎞ F = ⎜ −F1 ⎟ + ⎜ 0 ⎟ ⎝ ⎠

⎛ cos ( θ ) ⎞ ⎛ −e ⎞ ⎜ − f ⎟ + F ⎜ sin ( θ ) ⎟ 3 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝ 0 ⎠ F2

⎛ −33.322 ⎞ F = ⎜ −110.322 ⎟ lb ⎜ ⎟ 0 ⎝ ⎠

F = 115.245 lb

⎛ −c ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0 ⎞ ⎡ F ⎛ −e ⎞⎤ ⎛ d ⎞ ⎡ ⎛ cos ( θ ) ⎞⎤ 2 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ + ⎜ b ⎟ × ⎢F ⎜ sin ( θ ) ⎟⎥ MO = 0 × ⎜ −F1 ⎟ + a × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 3⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ MO = ⎜ 0 ⎟ lb⋅ ft ⎜ ⎟ ⎝ 480 ⎠

MO = 480 lb⋅ ft

Problem 4-101 Replace the force system by an equivalent resultant force and couple moment at point P.

284

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 60 lb

a = 2 ft

F 2 = 85 lb

b = 3 ft

F 3 = 25 lb

c = 6 ft

θ = 45 deg

d = 4 ft

e = 3

f = 4

Solution:

⎛⎜ 0 ⎟⎞ F = ⎜ −F1 ⎟ + ⎜ 0 ⎟ ⎝ ⎠

⎛ cos ( θ ) ⎞ ⎛ −e ⎞ ⎜ − f ⎟ + F ⎜ sin ( θ ) ⎟ 3 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝ 0 ⎠ F2

⎛ −33.322 ⎞ ⎜ ⎟ F = −110.322 lb ⎜ ⎟ 0 ⎝ ⎠

F = 115.245 lb

⎛ −c − d ⎞ ⎛⎜ 0 ⎟⎞ ⎛ −d ⎞ ⎡ F ⎛ −e ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ cos ( θ ) ⎞⎤ 2 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ + ⎜ b ⎟ × ⎢F ⎜ sin ( θ ) ⎟⎥ 0 MP = × ⎜ −F 1 ⎟ + a × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 3 ⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MP = ⎜ ⎟ ⎝ 921 ⎠

MP = 921 lb⋅ ft

Problem 4-102 Replace the force system by an equivalent force and couple moment at point O. Units Used: 3

kip = 10 lb Given: F 1 = 430 lb a = 2 ft

F 2 = 260 lb e = 5 ft 285

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 4

b = 8 ft

f = 12

c = 3 ft

g = 5

d = a

θ = 60 deg

Solution:

⎛ −sin ( θ ) ⎞ ⎜ ⎟ F R = F1 −cos ( θ ) + ⎜ ⎟ ⎝ 0 ⎠

⎛g⎞ ⎜f⎟ 2 2⎜ ⎟ g + f ⎝0⎠

⎛ −272 ⎞ ⎜ 25 ⎟ lb FR = ⎜ ⎟ ⎝ 0 ⎠

F2

⎛ −d ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ e ⎞ ⎡ F ⎛ g ⎞⎤ 2 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ f ⎟⎥ MO = b × F 1 −cos ( θ ) + 0 × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + f ⎝ 0 ⎠⎦

F R = 274 lb

⎛ 0 ⎞ ⎜ 0 ⎟ kip⋅ ft MO = ⎜ ⎟ ⎝ 4.609 ⎠

Problem 4-103 Replace the force system by an equivalent force and couple moment at point P. Units Used: 3

kip = 10 lb Given: F 1 = 430 lb

F 2 = 260 lb

a = 2 ft

e = 5 ft

b = 8 ft

f = 12

c = 3 ft

g = 5

d = a

θ = 60 deg

Solution:

⎛ −sin ( θ ) ⎞ ⎜ ⎟ F R = F1 −cos ( θ ) + ⎜ ⎟ ⎝ 0 ⎠

⎛g⎞ ⎜f⎟ 2 2⎜ ⎟ g + f ⎝0⎠

⎛ −272 ⎞ ⎜ 25 ⎟ lb FR = ⎜ ⎟ ⎝ 0 ⎠

F2

F R = 274 lb

286

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Engineering Mechanics - Statics

Chapter 4

⎛ 0 ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ d + e ⎞ ⎡ F ⎛ g ⎞⎤ 2 ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ c ⎟ × ⎢ ⎜ f ⎟⎥ MP = b + c × F 1 −cos ( θ ) + ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + f ⎝ 0 ⎠⎦

⎛ 0 ⎞ ⎜ 0 ⎟ kip⋅ ft MP = ⎜ ⎟ ⎝ 5.476 ⎠

Problem 4-104 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point O. Given: F 1 = 30 lb

a = 1 ft

d = 3

F 2 = 40 lb

b = 3 ft

e = 4

F 3 = 60 lb

c = 2 ft

Solution:

⎛0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 −1 + F2 0 + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎛ 4 ⎞ ⎜ ⎟ F R = −78 lb ⎜ ⎟ ⎝ 0 ⎠

⎛ −d ⎞ ⎜ −e ⎟ ⎟ 2 2⎜ d +e ⎝ 0 ⎠ F3

F R = 78.1 lb

⎛ 0 ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ 1 ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −d ⎞⎤ 3 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ MO = a + b + c × F1 −1 + c × F 2 0 + b + c × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ d + e ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MO = ⎜ ⎟ ⎝ 100 ⎠

Problem 4-105 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point P.

287

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 30 lb F 2 = 40 lb F 3 = 60 lb a = 1 ft b = 3 ft c = 2 ft d = 3 e = 4 Solution:

⎛0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 −1 + F2 0 + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎛ 4 ⎞ ⎜ ⎟ F R = −78 lb ⎜ ⎟ ⎝ 0 ⎠

⎛ −d ⎞ ⎜ −e ⎟ ⎟ 2 2⎜ d +e ⎝ 0 ⎠ F3

F R = 78.1 lb

⎡⎛ 0 ⎞ ⎤ ⎡ ⎛ 0 ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ 1 ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −d ⎞⎤ 3 ⎢ ⎜ ⎟ ⎥ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ MP = 0 ft × F1 −1 + −a − b × F2 0 + −a × ⎢⎜ ⎟ ⎥ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎣⎝ 0 ⎠ ⎦ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ d + e ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MP = ⎜ ⎟ ⎝ 124 ⎠

Problem 4-106 Replace the force and couple system by an equivalent force and couple moment at point O. Units Used: 3

kN = 10 N 288

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 4

Given: M = 8 kN m

θ = 60 deg

a = 3m

f = 12

b = 3m

g = 5

c = 4m

F 1 = 6 kN

d = 4m

F 2 = 4 kN

e = 5m Solution:

FR =

⎛ −cos ( θ ) ⎞ ⎛g⎞ ⎜ f ⎟ + F ⎜ −sin ( θ ) ⎟ 2 ⎜ ⎟ 2 2⎜ ⎟ f + g ⎝0⎠ ⎝ 0 ⎠ F1

⎛ 0.308 ⎞ ⎜ ⎟ F R = 2.074 kN ⎜ ⎟ ⎝ 0 ⎠

F R = 2.097 kN

⎛ 0 ⎞ ⎛ −c ⎞ ⎡ F ⎛ g ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ −cos ( θ ) ⎞⎤ 1 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ f ⎟⎥ + ⎜ −d ⎟ × ⎢F ⎜ −sin ( θ ) ⎟⎥ MO = 0 + −e × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 ⎜ ⎟⎥ ⎝ M ⎠ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −10.615 ⎠

Problem 4-107 Replace the force and couple system by an equivalent force and couple moment at point P. Units Used: 3

kN = 10 N

289

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Engineering Mechanics - Statics

Chapter 4

Given: M = 8 kN⋅ m

θ = 60 deg

a = 3m

f = 12

b = 3m

g = 5

c = 4m

F 1 = 6 kN

d = 4m

F 2 = 4 kN

e = 5m Solution:

FR =

⎛ −cos ( θ ) ⎞ ⎛g⎞ ⎜ f ⎟ + F ⎜ −sin ( θ ) ⎟ 2 ⎜ ⎟ 2 2⎜ ⎟ f + g ⎝0⎠ ⎝ 0 ⎠ F1

⎛ 0.308 ⎞ F R = ⎜ 2.074 ⎟ kN ⎜ ⎟ ⎝ 0 ⎠

F R = 2.097 kN

⎛ 0 ⎞ ⎛ −c − b ⎞ ⎡ F ⎛ g ⎞⎤ ⎛ −b ⎞ ⎡ ⎛ −cos ( θ ) ⎞⎤ 1 ⎜ f ⎟⎥ + ⎜ −d ⎟ × ⎢F ⎜ −sin ( θ ) ⎟⎥ MP = ⎜ 0 ⎟ + ⎜ −e ⎟ × ⎢ ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2⎜ ⎟⎥ ⎝ M ⎠ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎟ kN⋅ m 0 MP = ⎜ ⎜ ⎟ ⎝ −16.838 ⎠

Problem 4-108 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point O. Given: F 1 = 125 lb F 2 = 350 lb F 3 = 850 lb

290

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Engineering Mechanics - Statics

Chapter 4

a = 2 ft b = 6 ft c = 3 ft d = 4 ft

Solution: F Ry = F 3 − F 2 − F 1

F Ry = 375 lb

F Ry x = F3 ( b + c) − F 2 ( b) + F1 ( a) x =

F3 ( b + c) − F 2 ( b) + F1 a FRy

x = 15.5 ft

Problem 4-109 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point P. Given: F 1 = 125 lb

a = 2 ft

F 2 = 350 lb

b = 6 ft

F 3 = 850 lb

c = 3 ft d = 4 ft

Solution: F Ry = F 3 − F 2 − F 1

F Ry = 375 lb

F Ry x = F2 ( d + c) − F 3 ( d) + F1 ( a + b + c + d) x =

F2 d + F2 c − F3 d + F1 a + F1 b + F1 c + F1 d

x = 2.47 ft

F Ry (to the right of P)

291

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Engineering Mechanics - Statics

Chapter 4

Problem 4-110 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment acting at point O. Express the results in Cartesian vector form.

Given: a = 120 mm b = 800 mm

Solution:

⎛ −50 ⎞ F t = ⎜ 80 ⎟ N ⎜ ⎟ ⎝ −158 ⎠

⎛ −20 ⎞ F h = ⎜ 60 ⎟ N ⎜ ⎟ ⎝ −250 ⎠

⎛ −6 ⎞ Mt = ⎜ 4 ⎟ N⋅ m ⎜ ⎟ ⎝2⎠

⎛ −20 ⎞ Mh = ⎜ 8 ⎟ N⋅ m ⎜ ⎟ ⎝ 3 ⎠

FR = Ft + Fh

⎛ −70 ⎞ F R = ⎜ 140 ⎟ N ⎜ ⎟ ⎝ −408 ⎠ r0Ft

⎛a⎞ = ⎜0⎟ ⎜ ⎟ ⎝0⎠

MRP = ( r0Ft × F t) + Mt + Mh

⎛ −26 ⎞ MRP = ⎜ 31 ⎟ N⋅ m ⎜ ⎟ ⎝ 14.6 ⎠

Problem 4-111 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment 292

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Engineering Mechanics - Statics

Chapter 4

acting at point P. Express the results in Cartesian vector form. Given: a = 120 mm b = 800 mm

⎛ −50 ⎞ ⎜ 80 ⎟ N Ft = ⎜ ⎟ ⎝ −158 ⎠ ⎛ −6 ⎞ ⎜ ⎟ Mt = 4 N⋅ m ⎜ ⎟ ⎝2⎠ ⎛ −20 ⎞ ⎜ 60 ⎟ N Fh = ⎜ ⎟ ⎝ −250 ⎠ ⎛ −20 ⎞ ⎜ 8 ⎟ N⋅ m Mh = ⎜ ⎟ ⎝ 3 ⎠ Solution: FR = Ft + Fh

⎛ −70 ⎞ ⎜ ⎟ F R = 140 N ⎜ ⎟ ⎝ −408 ⎠

⎛b⎞ ⎛a + b⎞ ⎜ ⎟ ⎜ 0 ⎟×F MP = Mt + Mh + 0 × F h + ⎜ ⎟ ⎜ ⎟ t ⎝0⎠ ⎝ 0 ⎠

⎛ −26 ⎞ ⎜ ⎟ MP = 357.4 N⋅ m ⎜ ⎟ ⎝ 126.6 ⎠

Problem 4-112 Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B.

293

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 500 lb F 2 = 200 lb F 3 = 260 lb a = 5 ft

e = 3

b = 3 ft

f = 4

c = 2 ft

g = 12

d = 4 ft

h = 5

Solution: FR =

⎛−f⎞ ⎛0⎞ ⎜ −e ⎟ + F ⎜ −1 ⎟ + 2 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝0⎠ F1

Initial guess:

⎛h⎞ ⎜ −g ⎟ ⎟ 2 2⎜ g +h ⎝ 0 ⎠ F3

⎛ −300 ⎞ F R = ⎜ −740 ⎟ lb ⎜ ⎟ ⎝ 0 ⎠

F R = 798 lb

x = 1 ft

Given

⎛a⎞ ⎡ F ⎛ − f ⎞⎤ ⎛ a + b ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ a + b + c ⎞ ⎡ F ⎛ h ⎞⎤ ⎛ −x ⎞ 1 3 ⎜0⎟ × ⎢ ⎜ −e ⎟⎥ + ⎜ 0 ⎟ × ⎢F ⎜ −1 ⎟⎥ + ⎜ 0 ⎟ × ⎢ ⎜ −g ⎟⎥ = ⎜ 0 ⎟ × F ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ R ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + h ⎝ 0 ⎠⎦ ⎝ 0 ⎠ x = Find ( x)

x = −7.432 ft

Problem 4-113 Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B. Given: F 1 = 500 lb F 2 = 200 lb F 3 = 260 lb a = 5 ft

e = 3

294

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Engineering Mechanics - Statics

b = 3 ft

f = 4

c = 2 ft

g = 12

d = 4 ft

h = 5

Chapter 4

Solution: FR =

⎛−f⎞ ⎛0⎞ ⎜ −e ⎟ + F ⎜ −1 ⎟ + 2 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝0⎠ F1

Initial guess:

⎛h⎞ ⎜ −g ⎟ ⎟ 2 2⎜ g +h ⎝ 0 ⎠ F3

⎛ −300 ⎞ F R = ⎜ −740 ⎟ lb ⎜ ⎟ ⎝ 0 ⎠

F R = 798 lb

x = 1ft

Given

⎛ −b − c − d ⎞ ⎡ F ⎛ − f ⎞⎤ ⎛ −c − d ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ −d ⎞ ⎡ F ⎛ h ⎞⎤ ⎛ −x ⎞ 1 3 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −g ⎟⎥ = ⎜ 0 ⎟ × F 0 −e + × 0 × F 2 −1 + 0 × ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ 0 ⎝ ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + h ⎝ 0 ⎠⎦ ⎝ 0 ⎠ x = Find ( x)

x = 6.568 ft

measured to the left of B

Problem 4-114 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. Given: F 1 = 300 lb F 2 = 200 lb F 3 = 400 lb F 4 = 200 lb

M = 600 lb⋅ ft a = 3 ft b = 4 ft c = 2 ft d = 7 ft

Solution: F Rx = −F 4

F Rx = −200 lb

F Ry = −F 1 − F 2 − F 3

F Ry = −900 lb

295

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Engineering Mechanics - Statics

2

F =

FRx + FRy

Chapter 4

2

F = 922 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 77.5 deg

F Ry x = −F2 a − F 3 ( a + b) − F 4 c + M x = −

F2 ( a) + F 3 ( a + b) + F 4 c − M

x = 3.556 ft

F Ry

Problem 4-115 Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end A. Given: F 1 = 450 N

a = 2m

F 2 = 300 N

b = 4m

F 3 = 700 N

c = 3m

θ = 60 deg

M = 1500 N⋅ m

φ = 30 deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ )

F Rx = −125 N

F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2

F Ry = −1.296 × 10 N

F =

2

FRx + FRy

3

3

2

F = 1.302 × 10 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 84.5 deg

F Ry( x) = −F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M x =

−F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M F Ry

x = 7.36 m

Problem 4-116 Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end B. 296

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 450 N

a = 2m

F 2 = 300 N

b = 4m

F 3 = 700 N

c = 3m

θ = 60 deg

M = 1500 N⋅ m

φ = 30deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ )

F Rx = −125 N

F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2

F Ry = −1.296 × 10 N

F =

2

FRx + FRy

3

3

2

F = 1.302 × 10 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 84.5 deg

F Ry x = F1 sin ( θ ) b − F3 cos ( φ ) c − M x =

F1 sin ( θ ) b − F3 cos ( φ ) c − M

x = 1.36 m

F Ry

(to the right)

Problem 4-117 Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. Given: F 1 = 200 N F 2 = 450 N M = 200 N⋅ m a = 0.2 m b = 1.5 m c = 2m d = 1.5 m

297

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Engineering Mechanics - Statics

Chapter 4

θ = 30 deg Solution:

⎛ −sin ( θ ) ⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 1 + F2 −cos ( θ ) ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝ 0 ⎠ ⎛ −225 ⎞ ⎜ ⎟ F R = −190 N ⎜ ⎟ ⎝ 0 ⎠

F R = 294 N

⎛ b + c ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ b ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛0⎞ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ a MO = × F 1 + a × F −cos ( θ ) + M 0 ⎜ ⎟ ⎢ 1⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 ⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ −1 ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −39.6 ⎠

Problem 4-118 Determine the magnitude and direction θ of force F and its placement d on the beam so that the loading system is equivalent to a resultant force FR acting vertically downward at point A and a clockwise couple moment M. Units Used: 3

kN = 10 N Given: F 1 = 5 kN

a = 3m

F 2 = 3 kN

b = 4m

F R = 12 kN

c = 6m

M = 50 kN⋅ m

e = 7

f = 24

Solution: Initial guesses:

F = 1 kN

θ = 30 deg

d = 2m

298

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Engineering Mechanics - Statics

Given

Chapter 4

⎛ −e ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ e +f ⎠ ⎛ − f ⎞ F − F sin ( θ ) − F = −F 2 R ⎜ 2 2⎟ 1 ⎝ e +f ⎠ f ⎛ ⎞ ⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M ⎝ e +f ⎠

⎛F⎞ ⎜ θ ⎟ = Find ( F , θ , d) ⎜ ⎟ ⎝d⎠

θ = 71.565 deg

F = 4.427 kN

d = 3.524 m

Problem 4-119 Determine the magnitude and direction θ of force F and its placement d on the beam so that the loading system is equivalent to a resultant force F R acting vertically downward at point A and a clockwise couple moment M. Units Used: 3

kN = 10 N Given: F 1 = 5 kN

a = 3m

F 2 = 3 kN

b = 4m

F R = 10 kN

c = 6m

M = 45 kN⋅ m e = 7 f = 24 Solution: Initial guesses:

Given

F = 1 kN

θ = 30 deg

d = 1m

⎛ −e ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ e +f ⎠ ⎛ − f ⎞ F − F sin ( θ ) − F = −F 2 R ⎜ 2 2⎟ 1 ⎝ e +f ⎠ 299

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Engineering Mechanics - Statics

Chapter 4

f ⎛ ⎞ ⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M ⎝ e +f ⎠

⎛F⎞ ⎜ θ ⎟ = Find ( F , θ , d) ⎜ ⎟ ⎝d⎠

θ = 57.529 deg

F = 2.608 kN

d = 2.636 m

Problem 4-120 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. Given: F 1 = 500 N

a = 3m b = 2m

F 2 = 300 N

c = 1m

F 3 = 250 N

d = 2m

M = 400 N⋅ m

e = 3m

θ = 60 deg

f = 3 g = 4

Solution:

⎛

⎞ − F ( cos ( θ ) ) 1 ⎟ 2 2 ⎝ g +f ⎠ g

F Rx = −F 3 ⎜

F Rx = −450 N

⎛

f

F Ry = −883.0127 N

FRx + FRy

2

F R = 991.066 N

⎞ − F sin ( θ ) 1 ⎟ 2 2 ⎝ f +g ⎠

F Ry = −F 2 − F 3 ⎜

FR =

2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 62.996 deg

300

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Engineering Mechanics - Statics

Chapter 4

−F Rx( y) = M + F 1 cos ( θ ) a + F3

g 2

g + f M + F 1 cos ( θ ) a + F3

g 2

g + f

y =

2

2

( b + a) − F2 ( d) − F 3 ⎛ ⎜

f 2

⎞ ( d + e)

2⎟

⎝ g +f ⎠

( b + a) − F2 ( d) − F 3 ⎛

f ⎞ ⎜ 2 2 ⎟ ( d + e) ⎝ g +f ⎠

−FRx y = 1.78 m

Problem 4-121 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C. Given: F 1 = 500 N

a = 3m b = 2m

F 2 = 300 N

c = 1m

F 3 = 250 N

d = 2m

M = 400 N⋅ m

e = 3m

θ = 60 deg

f = 3 g = 4

Solution: F Rx = −F 3 ⎛ ⎜

⎞ − F ( cos ( θ ) ) 1 ⎝ g +f ⎠ g

2

F Ry = −F 2 − F 3 ⎛ ⎜

F Rx = −450 N

2⎟

⎞ − F sin ( θ ) 1 ⎝ f +g ⎠

FR =

2

f

2

FRx + FRy

F Ry = −883.0127 N

2⎟

2

F R = 991.066 N

301

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Engineering Mechanics - Statics

Chapter 4

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 62.996 deg

⎛

⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ ) 1 1 ⎝ g +f ⎠

F Ry( x) = M − F2 ( d + c) − F 3 ⎜

⎛

f

2

2⎟

⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ ) 1 1 ⎝ g +f ⎠

M − F2 ( d + c) − F 3 ⎜ x =

f

2

2⎟

F Ry x = 2.64 m

Problem 4-122 Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member AB, measured from point A. Given: F 1 = 35 lb a = 2 ft F 2 = 20 lb b = 4 ft F 3 = 25 lb c = 3 ft

θ = 30 deg d = 2 ft Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = −F 1 cos ( θ ) − F 2

F Ry = −50.311 lb

FR =

2

FRx + FRy

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

2

F R = 65.9 lb

θ 1 = −49.8 deg

F Ry x = −F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

302

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Engineering Mechanics - Statics

x =

Chapter 4

−F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

x = 2.099 ft

FRy

Problem 4-123 Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member BC, measured from point B. Given: F 1 = 35 lb F 2 = 20 lb F 3 = 25 lb

θ = 30 deg a = 2 ft b = 4 ft c = 3 ft d = 2 ft

Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = −F 1 cos ( θ ) − F 2

F Ry = −50.311 lb

FR =

2

FRx + FRy

2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

F R = 65.9 lb

θ 1 = −49.8 deg

F Rx y = F 1 cos ( θ ) b + F3 ( c)

303

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Engineering Mechanics - Statics

y =

F 1 cos ( θ ) b + F3 ( c) FRx

Chapter 4

y = 4.617 ft

(Below point B)

Problem 4-124 Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A. Given: F 1 = 35 lb

a = 2 ft

F 2 = 20 lb

b = 4 ft

F 3 = 25 lb

c = 3 ft

θ = 30 deg

d = 2 ft

Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = F 1 cos ( θ ) + F 2

F Ry = 50.311 lb

FR =

2

FRx + FRy

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

2

F R = 65.9 lb

θ 1 = 49.8 deg

MRA = −F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

MRA = −106 lb⋅ ft

Problem 4-125 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point O. Express the results in Cartesian vector form.

304

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠

a = 3m b = 3m

⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠

c = 4m

e = 5m f = 6m g = 5m

d = 6m

Solution:

⎛−f⎞ ⎜ ⎟ MR = M + e × F ⎜ ⎟ ⎝g⎠

FR = F

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −10 ⎞ ⎜ ⎟ MR = 18 kN⋅ m ⎜ ⎟ ⎝ −56 ⎠

Problem 4-126 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠ ⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠ a = 3m b = 3m

e = 5m

c = 4m

f = 6m 305

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Engineering Mechanics - Statics

d = 6m

Chapter 4

g = 5m

Solution:

⎛ −f ⎞ ⎜ e ⎟×F MR = M + ⎜ ⎟ ⎝d + g⎠

FR = F

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −46 ⎞ ⎜ ⎟ MR = 66 kN⋅ m ⎜ ⎟ ⎝ −56 ⎠

Problem 4-127 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point Q. Express the results in Cartesian vector form. Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠ ⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠ a = 3m b = 3m

e = 5m

c = 4m

f = 6m

d = 6m

g = 5m

Solution:

FR = F

⎛0⎞ ⎜ ⎟ MR = M + e × F ⎜ ⎟ ⎝g⎠

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −10 ⎞ ⎜ ⎟ MR = −30 kN⋅ m ⎜ ⎟ ⎝ −20 ⎠

Problem 4-128 The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the −k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in 306

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Engineering Mechanics - Statics

Chapter 4

p y Cartesian vector form.

q

p

p

p

Given: F 1 = 40 N

r = 80 mm

F 2 = 40 N

a = 300 mm

θ = 0 deg Solution:

F 1v

F 2v

⎛0⎞ = F1 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

⎛ 0 ⎞ = F2 ⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ −sin ( θ ) ⎠

⎛ −a ⎞ r2 = ⎜ −r sin ( θ ) ⎟ ⎜ ⎟ ⎝ r cos ( θ ) ⎠

⎛ −a ⎞ r1 = ⎜ r ⎟ ⎜ ⎟ ⎝0⎠

F R = F1v + F2v

MA = r1 × F1v + r2 × F2v

⎛ 0 ⎞ F R = ⎜ −40 ⎟ N ⎜ ⎟ ⎝ −40 ⎠

⎛ 0 ⎞ MA = ⎜ −12 ⎟ N⋅ m ⎜ ⎟ ⎝ 12 ⎠

Problem 4-129 The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the −k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form.

307

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 40 N F 2 = 40 N

θ = 0 deg r = 80 mm a = 300 mm

θ = 45 deg

Solution:

F 1v

⎛0⎞ = F1 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

F R = F1v + F2v

⎛ 0 ⎞ F R = ⎜ −28.28 ⎟ N ⎜ ⎟ ⎝ −68.28 ⎠

F 2v

⎛ 0 ⎞ = F2 ⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ −sin ( θ ) ⎠

⎛ −a ⎞ r1 = ⎜ r ⎟ ⎜ ⎟ ⎝0⎠

⎛ −a ⎞ r2 = ⎜ −r sin ( θ ) ⎟ ⎜ ⎟ ⎝ r cos ( θ ) ⎠

MA = r1 × F1v + r2 × F2v

⎛ 0 ⎞ MA = ⎜ −20.49 ⎟ N⋅ m ⎜ ⎟ ⎝ 8.49 ⎠

Problem 4-130 Replace this system by an equivalent resultant force and couple moment acting at O. Express the results in Cartesian vector form. Given: F 1 = 50 N F 2 = 80 N

308

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Engineering Mechanics - Statics

Chapter 4

F 3 = 180 N a = 1.25 m b = 0.5 m c = 0.75 m

Solution:

⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎟⎞ ⎛⎜ 0 ⎟⎞ FR = ⎜ 0 ⎟ + ⎜ 0 ⎟ + ⎜ 0 ⎟ ⎜ F 1 ⎟ ⎜ −F2 ⎟ ⎜ −F 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟N FR = ⎜ ⎟ ⎝ −210 ⎠ ⎛ a + c ⎞ ⎛⎜ 0 ⎞⎟ ⎛ a ⎞ ⎛⎜ 0 ⎟⎞ ⎛ a ⎞ ⎛⎜ 0 ⎟⎞ ⎜ b ⎟ × 0 + ⎜b⎟ × 0 + ⎜0⎟ × 0 MO = ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 0 ⎠ ⎝ F1 ⎠ ⎝ 0 ⎠ ⎝ −F2 ⎠ ⎝ 0 ⎠ ⎝ −F3 ⎟⎠

⎛ −15 ⎞ ⎜ ⎟ MO = 225 N⋅ m ⎜ ⎟ ⎝ 0 ⎠

Problem 4-131 Handle forces F 1 and F 2 are applied to the electric drill. Replace this system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form. Given: a = 0.15 m b = 0.25 m c = 0.3 m

⎛ 6 ⎞ ⎜ ⎟ F 1 = −3 N ⎜ ⎟ ⎝ −10 ⎠ ⎛0⎞ ⎜ ⎟ F2 = 2 N ⎜ ⎟ ⎝ −4 ⎠ 309

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Engineering Mechanics - Statics

Chapter 4

Solution: FR = F1 + F2

⎛ 6 ⎞ ⎜ ⎟ F R = −1 N ⎜ ⎟ ⎝ −14 ⎠

⎛a⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ MO = 0 × F1 + −b × F 2 ⎜ ⎟ ⎜ ⎟ ⎝c⎠ ⎝c ⎠

⎛ 1.3 ⎞ ⎜ 3.3 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −0.45 ⎠

Problem 4-132 A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of F R for the rectus, FO for the oblique, FL for the lumbar latissimus dorsi, and F E for the erector spinae. These loadings are symmetric with respect to the y - z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form. Given: F R = 35 N

a = 75 mm

F O = 45 N

b = 45 mm

F L = 23 N

c = 15 mm

F E = 32 N

d = 50 mm

e = 40 mm

f = 30 mm

Solution: F Res = ΣFi;

F Res = 2( FR + F O + F L + F E)

F Res = 270 N

MROx = ΣMOx;

MRO = −2FR a + 2F E c + 2F L b

MRO = −2.22 N⋅ m

Problem 4-133 The building slab is subjected to four parallel column loadings.Determine the equivalent resultant force and specify its location (x, y) on the slab.

310

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: F 1 = 30 kN

a = 3m

F 2 = 40 kN

b = 8m

F 3 = 20 kN

c = 2m

F 4 = 50 kN

d = 6m e = 4m

Solution: +

↑ FR = ΣFx;

FR = F1 + F2 + F3 + F4 F R = 140 kN

MRx = ΣMx;

−F R( y) = −( F4 ) ( a) −

y =

⎡⎣( F1 ) ( a + b)⎤⎦ − ⎡⎣( F2 ) ( a + b + c)⎤⎦

F4 a + F1 a + F1 b + F2 a + F2 b + F2 c FR

y = 7.14 m MRy = ΣMy;

(FR)x = (F4)( e) + (F3)( d + e) + (F2)( b + c) x =

F4 e + F3 d + F3 e + F2 b + F2 c FR

x = 5.71 m

Problem 4-134 The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab.

311

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: F 1 = 20 kN

a = 3m

F 2 = 50 kN

b = 8m

F 3 = 20 kN

c = 2m

F 4 = 50 kN

d = 6m e = 4m

Solution: FR = F1 + F2 + F3 + F4

F R = 140 kN

F R x = F2 e + F1 ( d + e) + F 2 ( d + e) x =

2 F2 e + F1 d + F1 e + F2 d

x = 6.43 m

FR

−F R y = −F 2 a − F3 ( a + b) − F2 ( a + b + c)

y =

2 F2 a + F3 a + F3 b + F2 b + F2 c FR

y = 7.29 m

Problem 4-135 The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the magnitude F of the couple forces so that the system can be simplified to a wrench acting at point C. Given: a = 0.6 m

312

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Engineering Mechanics - Statics

Chapter 4

b = 0.8 m c = 0.25 m d = 0.7 m e = 0.3 m f = 0.3 m g = 0.5 m h = 0.25 m P = 60 N Q = 40 N Solution: Initial Guess

F = 1N

MC = 1 N⋅ m

Given

⎛⎜ −MC ⎞⎟ ⎡ −P( c + h) ⎤ ⎡ 0 ⎢ ⎥+⎢ 0 ⎜ 0 ⎟=⎢ 0 ⎥ ⎢ ⎜ 0 ⎟ ⎣ 0 ⎦ ⎣ −F ( e + ⎝ ⎠ ⎛ F ⎞ ⎜ ⎟ = Find ( F , MC) ⎝ MC ⎠

⎤ ⎛ a ⎞ ⎛ −Q ⎞ ⎥ + ⎜b⎟ × ⎜ 0 ⎟ ⎥ ⎜ ⎟ ⎜ ⎟ f) ⎦ ⎝ 0 ⎠ ⎝ 0 ⎠

MC = 30 N⋅ m

F = 53.3 N

Problem 4-136 The three forces acting on the block each have a magnitude F 1 = F2 = F3. Replace this system by a wrench and specify the point where the wrench intersects the z axis, measured from point O. Given: F 1 = 10 lb

a = 6 ft

F2 = F1

b = 6 ft

F3 = F1

c = 2 ft

Solution: The vectors

313

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Engineering Mechanics - Statics

F 1v =

⎛b⎞ ⎜ −a ⎟ ⎟ 2 2⎜ b +a ⎝ 0 ⎠ F1

Chapter 4

F 2v

⎛⎜ 0 ⎞⎟ = ⎜ −F 2 ⎟ ⎜ 0 ⎟ ⎝ ⎠

F 3v =

M = 1 lb⋅ ft

R x = 1 lb

⎛ −b ⎞ ⎜a⎟ ⎟ 2 2⎜ b +a ⎝ 0 ⎠ F3

Place the wrench in the x - z plane. x = 1ft z = 1ft

Guesses

R y = 1 lb

R z = 1 lb

⎛ Rx ⎞ ⎜ ⎟ Given ⎜ Ry ⎟ = F1v + F2v + F3v ⎜R ⎟ ⎝ z⎠ ⎛ Rx ⎞ ⎛ 0 ⎞ ⎛ x ⎞ ⎛⎜ Rx ⎟⎞ ⎛0⎞ ⎛b⎞ ⎜ ⎟ M ⎜0⎟ × R + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ry ⎟ = ⎜ a ⎟ × F2v + ⎜ a ⎟ × F1v + ⎜ 0 ⎟ × F3v ⎜ ⎟ ⎜ y⎟ 2 2 2 Rx + Ry + Rz ⎜ R ⎟ ⎝ c ⎠ ⎝ z ⎠ ⎜⎝ Rz ⎟⎠ ⎝0⎠ ⎝c ⎠ ⎝ z⎠ ⎛x ⎞ ⎜ ⎟ ⎜ z ⎟ ⎛ Rx ⎞ ⎜ ⎟ ⎜M ⎟ M Mv = ⎜ R ⎟ = Find ( x , z , M , Rx , Ry , Rz) ⎜ Ry ⎟ 2 2 2 ⎜ x⎟ Rx + Ry + Rz ⎜ R ⎟ ⎝ z⎠ R ⎜ y⎟ ⎜ ⎟ ⎝ Rz ⎠ ⎛ Rx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎛x⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎟ Mv = −14.142 lb⋅ ft ⎜ ⎟=⎜ ⎟ ft ⎜ Ry ⎟ = ⎜ −10 ⎟ lb ⎜ ⎟ ⎝ z ⎠ ⎝ 0.586 ⎠ ⎜R ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ z⎠ Problem 4-137 Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate. Units Used: 3

kN = 10 N

314

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Engineering Mechanics - Statics

Chapter 4

Given: F A = 500 N F B = 800 N F C = 300 N a = 4m b = 6m

Solution:

⎛ FA ⎞ ⎜ ⎟ FR = ⎜ FC ⎟ ⎜F ⎟ ⎝ B⎠ Guesses

F R = 0.9899 kN

x = 1m

Given

M

FR FR

y = 1m

M = 100 N⋅ m

⎛x⎞ + ⎜ y ⎟ × FR = ⎜ ⎟ ⎝0⎠

⎛M⎞ ⎜ x ⎟ = Find ( M , x , y) ⎜ ⎟ ⎝y⎠

⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0 ⎞ ⎛⎜ 0 ⎞⎟ ⎜a⎟ × F + ⎜a⎟ × 0 ⎜ ⎟ ⎜ C⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ FB ⎟⎠ ⎛ x ⎞ ⎛ 1.163 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ y ⎠ ⎝ 2.061 ⎠

M = 3.07 kN⋅ m

Problem 4-138 Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(y, z) where its line of action intersects the plate. Given: F A = 80 lb

a = 12 ft

F B = 60 lb

b = 12 ft

F C = 40 lb

315

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛ −F C ⎞ ⎜ ⎟ F R = ⎜ −F B ⎟ ⎜ −F ⎟ ⎝ A⎠

F R = 108 lb

y = 1 ft

Guesses

Given

M

FR FR

z = 1 ft

⎛0⎞ + ⎜ y ⎟ × FR = ⎜ ⎟ ⎝z⎠

⎛M⎞ ⎜ y ⎟ = Find ( M , y , z) ⎜ ⎟ ⎝z⎠

M = 1 lb⋅ ft

⎛ 0 ⎞ ⎛⎜ −FC ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜a⎟ × ⎜ ⎟ ⎜ 0 ⎟ + a × ⎜ −FB ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ b ⎠ ⎜⎝ 0 ⎟⎠ ⎛ y ⎞ ⎛ 0.414 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ z ⎠ ⎝ 8.69 ⎠

M = −624 lb⋅ ft

Problem 4-139 The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. Given: w1 = 2

lb ft

w2 = 3.5

lb ft

a = 2.75 ft b = 4 ft c = 1.5 ft R = 1 lb

Solution: Guesses Given

d = 1ft

w1 b + w2 c = R w1 b⎛⎜ a −

⎝

⎛R⎞ ⎜ ⎟ = Find ( R , d) ⎝d⎠

b⎞

⎞ ⎛c ⎟ − w2 c⎜ + b − a⎟ = −d R 2⎠ ⎝2 ⎠ R = 13.25 lb

d = 0.34 ft

316

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Engineering Mechanics - Statics

Chapter 4

Problem 4-140 Replace the loading by an equivalent resultant force and couple moment acting at point A. Units Used: 3

kN = 10 N Given: w1 = 600

N

w2 = 600

N

m

m

a = 2.5 m b = 2.5 m Solution: F R = w1 a − w2 b MRA = w1 a⎛⎜

a + b⎞

⎟ ⎝ 2 ⎠

FR = 0 N MRA = 3.75 kN⋅ m

Problem 4-141 Replace the loading by an equivalent force and couple moment acting at point O. Units Used: 3

kN = 10 N Given: w = 6

kN m

F = 15 kN M = 500 kN⋅ m a = 7.5 m b = 4.5 m

317

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Engineering Mechanics - Statics

Chapter 4

Solution: FR =

1 2

w( a + b) + F

MR = −M −

F R = 51.0 kN

⎛ 1 w a⎞ ⎛ 2 a⎞ − ⎛ 1 w b⎞ ⎛a + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝

b⎞

⎟ − F( a + b)

3⎠

MR = −914 kN⋅ m

Problem 4-142 Replace the loading by a single resultant force, and specify the location of the force on the beam measured from point O. Units Used: 3

kN = 10 N Given: w = 6

kN m

F = 15 kN M = 500 kN⋅ m a = 7.5 m b = 4.5 m Solution: Initial Guesses:

F R = 1 kN

d = 1m

Given FR =

1 2

w( a + b) + F

−F R d = −M −

⎛ 1 w a⎞ ⎛ 2 a⎞ − ⎛ 1 w b⎞ ⎛ a + b ⎞ − F( a + b) ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝ 3 ⎠

⎛ FR ⎞ ⎜ ⎟ = Find ( FR , d) ⎝ d ⎠

F R = 51 kN

d = 17.922 m

Problem 4-143 The column is used to support the floor which exerts a force P on the top of the column. The effect of soil pressure along its side is distributed as shown. Replace this loading by an 318

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Engineering Mechanics - Statics

Chapter 4

p g p g y equivalent resultant force and specify where it acts along the column, measured from its base A. 3

kip = 10 lb

Units Used: Given:

P = 3000 lb w1 = 80

lb ft

w2 = 200

lb ft

h = 9 ft Solution: F Rx = w1 h +

1 (w2 − w1)h 2

F Rx = 1260 lb

F Ry = P

2

FR =

FRx + P

2

⎞ θ = atan ⎛⎜ ⎟ F ⎝ Rx ⎠ P

θ = 67.2 deg

1 h h w2 − w1 ) h + w1 h ( 2 3 2

F Rx y =

y =

F R = 3.25 kip

1 2 w2 + 2 w1 h FRx 6

y = 3.86 ft

Problem 4-144 Replace the loading by an equivalent force and couple moment at point O. Units Used: 3

kN = 10 N Given: kN m

w1 = 15

w2 = 5

kN m 319

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Engineering Mechanics - Statics

Chapter 4

d = 9m Solution: FR =

1 (w1 + w2)d 2

MRO = w2 d

F R = 90 kN

d 1 d + ( w1 − w2 ) d 2 2 3

MRO = 338 kN⋅ m

Problem 4-145 Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C. Units Used: 3

kip = 10 lb Given: w = 800

lb ft

a = 15 ft b = 15 ft

θ = 30 deg Solution: FR = w a + FR x = w a

wa x =

a 2

wb 2

F R = 18 kip

a wb⎛ b⎞ + ⎜a + ⎟ 2 ⎝ 2 3⎠ +

b⎞ ⎜a + ⎟ 2 ⎝ 3⎠

wb⎛ FR

x = 11.7 ft

Problem 4-146 The beam supports the distributed load caused by the sandbags. Determine the resultant force on the beam and specify its location measured from point A.

320

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Engineering Mechanics - Statics

Units Used:

Chapter 4

3

kN = 10 N

Given: w1 = 1.5 w2 = 1

kN

a = 3m

m

kN

b = 3m

m

w3 = 2.5

kN

c = 1.5 m

m

Solution: F R = w1 a + w2 b + w3 c MA = w1 a

a 2

+ w2 b⎛⎜ a +

⎝

F R = 11.25 kN b⎞

c⎞ ⎛ ⎟ + w3 c⎜ a + b + ⎟ 2⎠ 2⎠ ⎝

MA = 45.563 kN⋅ m

d =

MA

d = 4.05 m

FR

Problem 4-147 Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is M clockwise. Units Used: 3

kN = 10 N Given: w1 = 4

kN

w2 = 2.5

m

M = 8 kN⋅ m

kN m

c = 9m

Solution: Initial Guesses: Given

−1 2

a = 1m w1 b +

1 2

b = 1m

w2 c = 0

321

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Engineering Mechanics - Statics

1 2

w1 b⎛⎜ a +

⎝

⎛a⎞ ⎜ ⎟ = Find ( a , b) ⎝b⎠

Chapter 4

2b ⎞

1

2c

⎟ − w2 c = −M 3⎠ 2 3 ⎛ a ⎞ ⎛ 1.539 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ b ⎠ ⎝ 5.625 ⎠

Problem 4-148 Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A. Units Used: 3

kN = 10 N Given: w1 = 800

N

w2 = 200

N

m

m

a = 2m b = 3m Solution: F R = w2 b + w1 a + x F R = w1 a

w1 a x =

a 2

+

1 2

(w1 − w2)b

F R = 3.10 kN

b b w1 − w2 ) b⎛⎜ a + ⎞⎟ + w2 b⎛⎜ a + ⎞⎟ ( 2 3 2 1

⎝

⎠

⎝

a 1 ⎛ b⎞ ⎛ b⎞ + ( w1 − w2 ) b⎜ a + ⎟ + w2 b⎜ a + ⎟ 2 2 ⎝ 3⎠ ⎝ 2⎠ FR

⎠

x = 2.06 m

Problem 4-149 The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O. Units Used: 322

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Engineering Mechanics - Statics

Chapter 4

3

kip = 10 lb Given: w1 = 50

lb ft

w2 = 300

lb

w3 = 100

lb

ft

ft

a = 12 ft b = 9 ft Solution :

(w2 − w1)a + 2 (w2 − w3)b + w3 b 2 1

F R = w1 a + F R d = w1 a

a 2

1

+

(w2 − w1)a 2 1

2

d =

2a 3

+

F R = 3.9 kip

b b w2 − w3 ) b⎛⎜ a + ⎞⎟ + w3 b⎛⎜ a + ⎞⎟ ( 2 3 2 1

2

⎝

2

⎠

3 w3 b a + 2 w3 b + w1 a + 2 a w2 + 3 b w2 a + w2 b 6FR

⎝

⎠

2

d = 11.3 ft

Problem 4-150 The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero. Given: w1 = 40

lb

w2 = 60

lb

ft

c = 10ft d = 6 ft

ft

Solution: Initial Guesses:

a = 1 ft

b = 1ft

323

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Engineering Mechanics - Statics

Chapter 4

Given 1 2

1

w2 d − w1 b = 0

⎛a⎞ ⎜ ⎟ = Find ( a , b) ⎝b⎠

2

w2 d⎛⎜ c +

⎝

d⎞

⎛ ⎟ − w1 b⎜ a + 3⎠ ⎝

b⎞

⎟=0

2⎠

⎛ a ⎞ ⎛ 9.75 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ b ⎠ ⎝ 4.5 ⎠

Problem 4-151 Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point B. Units Used: 3

kip = 10 lb Given: w1 = 800

lb

w2 = 500

lb

ft

ft

a = 12 ft b = 9 ft Solution: FR =

1 2

a w1 +

1 2

(w1 − w2)b + w2 b

F R = 10.65 kip

a 1 1 b b F R x = − a w1 + ( w1 − w2 ) b + w2 b 3 2 2 3 2

x =

a 1 b b 1 − a w1 + ( w1 − w2 ) b + w2 b 3 2 3 2 2 FR

x = 0.479 ft ( to the right of B )

Problem 4-152 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A.

324

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Engineering Mechanics - Statics

Chapter 4

Given: w1 = 200

N m

w2 = 100

N m

w3 = 200

N m

a = 5m b = 6m Solution: F Rx = −w3 a F Ry =

F Rx = −1000 N

−1 (w1 + w2)b 2

− y FRx = w3 a

w3 a y =

a 2

F Ry = −900 N

a b 1 b − w2 b − ( w1 − w2 ) b 2 2 2 3

− w2 b

b

−

(w1 − w2)b 3 2 1

b

2 −FRx

y = 0.1 m

Problem 4-153 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C. Units Used: 3

kN = 10 N

325

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Engineering Mechanics - Statics

Chapter 4

Given: w1 = 200

N

w2 = 100

N

w3 = 200

N

m

m

m

a = 5m b = 6m Solution : F Rx = −w3 a F Ry =

−1 2

(w1 + w2)b

−x F Ry = −w3 a

−w3 a x =

F Rx = −1000 N

a 2

+ w2 b

F Ry = −900 N b 2

+

(w1 − w2)b 2 1

2b 3

a b 1 2b + w2 b + ( w1 − w2 ) b 2 2 2 3

x = 0.556 m

−FRy

⎛ FRx ⎞ ⎜ ⎟ = 1.345 kN ⎝ FRy ⎠ Problem 4-154 Replace the loading by an equivalent resultant force and couple moment acting at point O. Units Used: 3

kN = 10 N Given: w1 = 7.5

kN m

326

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Engineering Mechanics - Statics

w2 = 20

Chapter 4

kN m

a = 3m b = 3m c = 4.5 m

Solution:

(w2 − w1)c + w1 c + w1 b + 2 1

FR =

1 2

w1 a

F R = 95.6 kN MRo = −

c c b w2 − w1 ) c − w1 c − w1 b⎛⎜ c + ⎞⎟ − ( 2 3 2 2 1

⎝

⎠

1 2

w1 a⎛⎜ b + c +

⎝

a⎞

⎟

3⎠

MRo = −349 kN⋅ m

Problem 4-155 Determine the equivalent resultant force and couple moment at point O. Units Used: 3

kN = 10 N Given: a = 3m wO = 3

kN m

x w ( x) = wO ⎛⎜ ⎟⎞ ⎝ a⎠

2

Solution: ⌠a F R = ⎮ w ( x) dx ⌡0

F R = 3 kN

⌠a MO = ⎮ w ( x) ( a − x) dx ⌡0

MO = 2.25 kN⋅ m

327

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Engineering Mechanics - Statics

Chapter 4

Problem 4-156 Wind has blown sand over a platform such that the intensity of the load can be approximated by 3

⎛ x ⎞ . Simplify this distributed loading to an equivalent resultant force and ⎟ ⎝d⎠

the function w = w0 ⎜

specify the magnitude and location of the force, measured from A.

Units Used: 3

kN = 10 N Given: w0 = 500

N m

d = 10 m

⎛x⎞ ⎟ ⎝ d⎠

3

w ( x) = w0 ⎜ Solution: d

⌠ F R = ⎮ w ( x) dx ⌡0

d =

⌠d ⎮ x w ( x) dx ⌡0 FR

F R = 1.25 kN

d=8m

Problem 4-157 Determine the equivalent resultant force and its location, measured from point O.

328

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Engineering Mechanics - Statics

Chapter 4

Solution: L

⌠ 2w0 L ⎛ πx⎞ F R = ⎮ w0 sin ⎜ ⎟ dx = ⎮ π ⎝L⎠ ⌡ 0

L

d=

⌠ ⎮ x w sin ⎛ π x ⎞ dx ⎜ ⎟ 0 ⎮ ⎝L⎠ ⌡ 0

FR

=

L 2

Problem 4-158 Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and specify where it acts, measured from point A. Given: a = 3 ft k = 86

lb ft

3 2

w ( x) = k x Solution: a

⌠ F R = ⎮ w ( x) dx ⌡0

x =

⌠a ⎮ x w ( x) dx ⌡0 FR

F R = 774 lb

x = 2.25 ft

Problem 4-159 Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To 329

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Engineering Mechanics - Statics

Chapter 4

y g y p j y alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A. Given: a = 0.5 ft w0 = 12

lb ft

lb

k = 24

ft

3

2

w ( x) = w0 + kx

Solution: ⌠a F R = ⎮ w ( x) dx ⌡0

F R = 7 lb

a

x =

⌠ ⎮ x w ( x) dx ⌡0 FR

x = 0.268 ft

Problem 4-160 Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integrals using Simpson's rule.

Units Used: 3

kN = 10 N Given: c1 = 5 c2 = 16 a = 3

330

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Engineering Mechanics - Statics

Chapter 4

b = 1 Solution: a+ b

⌠ FR = ⎮ ⌡0

2

c1 x +

c2 + x dx

x c1 x +

c2 + x dx

F R = 14.9

a+ b

d =

⌠ ⎮ ⌡0

2

d = 2.27

FR

Problem 4-161 Determine the coordinate direction angles of F, which is applied to the end A of the pipe assembly, so that the moment of F about O is zero. Given: F = 20 lb a = 8 in b = 6 in c = 6 in d = 10 in Solution: Require Mo = 0. This happens when force F is directed either towards or away from point O.

⎛ c ⎞ r = ⎜a + b⎟ ⎜ ⎟ ⎝ d ⎠

u =

r r

⎛ 0.329 ⎞ u = ⎜ 0.768 ⎟ ⎜ ⎟ ⎝ 0.549 ⎠

If the force points away from O, then

⎛⎜ α ⎞⎟ ⎜ β ⎟ = acos ( u) ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 70.774 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 39.794 ⎟ deg ⎜ γ ⎟ ⎝ 56.714 ⎠ ⎝ ⎠

If the force points towards O, then 331

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Engineering Mechanics - Statics

⎛⎜ α ⎞⎟ ⎜ β ⎟ = acos ( −u) ⎜γ ⎟ ⎝ ⎠

Chapter 4

⎛⎜ α ⎞⎟ ⎛ 109.226 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 140.206 ⎟ deg ⎜ γ ⎟ ⎝ 123.286 ⎠ ⎝ ⎠

Problem 4-162 Determine the moment of the force F about point O. The force has coordinate direction angles α, β, γ. Express the result as a Cartesian vector. Given: F = 20 lb

a = 8 in

α = 60 deg

b = 6 in

β = 120 deg

c = 6 in

γ = 45 deg

d = 10 in

Solution:

⎛ c ⎞ r = ⎜a + b⎟ ⎜ ⎟ ⎝ d ⎠

⎛⎜ cos ( α ) ⎟⎞ F v = F⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

M = r × Fv

⎛ 297.99 ⎞ M = ⎜ 15.147 ⎟ lb⋅ in ⎜ ⎟ ⎝ −200 ⎠

Problem 4-163 Replace the force at A by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used :

332

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Engineering Mechanics - Statics

Chapter 4

3

kN = 10 N Given: a = 4m b = 6m c = 8m d = 4m

⎛ −300 ⎞ ⎜ ⎟ F = 200 N ⎜ ⎟ ⎝ −500 ⎠ Solution : FR = F

⎛ −300 ⎞ ⎜ ⎟ F R = 200 N ⎜ ⎟ ⎝ −500 ⎠

⎛ −a − c ⎞ ⎜ b ⎟×F MP = ⎜ ⎟ ⎝ d ⎠

⎛ −3.8 ⎞ ⎜ ⎟ MP = −7.2 kN⋅ m ⎜ ⎟ ⎝ −0.6 ⎠

Problem 4-164 Determine the moment of the force FC about the door hinge at A. Express the result as a Cartesian vector.

333

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Engineering Mechanics - Statics

Chapter 4

Given: F = 250 N b = 1m c = 2.5 m d = 1.5 m e = 0.5 m

θ = 30 deg

Solution:

rCB

⎛ c−e ⎞ ⎜ ⎟ = b + d cos ( θ ) ⎜ ⎟ ⎝ −d sin ( θ ) ⎠

MA = rAB × F v

rAB

⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝0⎠

Fv = F

rCB rCB

⎛ −59.7 ⎞ ⎜ 0.0 ⎟ N⋅ m MA = ⎜ ⎟ ⎝ −159.3 ⎠

Problem 4-165 Determine the magnitude of the moment of the force FC about the hinged axis aa of the door.

334

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Engineering Mechanics - Statics

Chapter 4

Given: F = 250 N b = 1m c = 2.5 m d = 1.5 m e = 0.5 m

θ = 30 deg

Solution:

rCB

⎛ c−e ⎞ ⎜ ⎟ = b + d cos ( θ ) ⎜ ⎟ ⎝ −d sin ( θ ) ⎠

Maa = ( rAB × Fv) ⋅ ua

rAB

⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝0⎠

Fv = F

rCB rCB

⎛1⎞ ⎜ ⎟ ua = 0 ⎜ ⎟ ⎝0⎠

Maa = −59.7 N⋅ m

Problem 4-166 A force F1 acts vertically downward on the Z-bracket. Determine the moment of this force about the bolt axis (z axis), which is directed at angle θ from the vertical.

335

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 80 N a = 100 mm b = 300 mm c = 200 mm

θ = 15 deg Solution:

⎛ −b ⎞ ⎜ ⎟ r = a+c ⎜ ⎟ ⎝ 0 ⎠ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ F = F1 ⎜ ⎟ ⎝ −cos ( θ ) ⎠ ⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠ Mz = ( r × F ) k Mz = −6.212 N⋅ m

Problem 4-167 Replace the force F having acting at point A by an equivalent force and couple moment at point C. Units Used:

3

kip = 10 lb

Given: F = 50 lb a = 10 ft b = 20 ft c = 15 ft

336

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Engineering Mechanics - Statics

Chapter 4

d = 10 ft e = 30 ft

Solution :

rAB

⎛d⎞ = ⎜ c ⎟ ⎜ ⎟ ⎝ −e ⎠

Fv = F

rCA

rAB rAB

⎛ 0 ⎞ = ⎜a + b⎟ ⎜ ⎟ ⎝ e ⎠

FR = Fv

⎛ 14.286 ⎞ F R = ⎜ 21.429 ⎟ lb ⎜ ⎟ ⎝ −42.857 ⎠

MR = rCA × Fv

⎛ −1.929 ⎞ MR = ⎜ 0.429 ⎟ kip⋅ ft ⎜ ⎟ ⎝ −0.429 ⎠

Problem 4-168 The horizontal force F acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis? Given: F = 30 N a = 50 mm b = 200 mm c = 10 mm

θ = 45 deg Solution:

⎛ sin ( θ ) ⎞ F v = F⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ 0 ⎠

rOA

⎛ −c ⎞ = ⎜b ⎟ ⎜ ⎟ ⎝a⎠

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠ 337

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Engineering Mechanics - Statics

Mz = ( rOA × F v ) k

Chapter 4

Mz = −4.03 N⋅ m

Problem 4-169 The horizontal force F acts on the handle of the wrench. Determine the moment of this force about point O. Specify the coordinate direction angles α, β, γ of the moment axis. Given: F = 30 N

c = 10 mm

a = 50 mm θ = 45 deg b = 200 mm Solution:

⎛ sin ( θ ) ⎞ F v = F⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ 0 ⎠ MO = rOA × Fv

⎛⎜ α ⎞⎟ ⎛ MO ⎞ ⎜ β ⎟ = acos ⎜ ⎟ MO ⎠ ⎝ ⎜γ ⎟ ⎝ ⎠

rOA

⎛ −c ⎞ = ⎜b ⎟ ⎜ ⎟ ⎝a⎠

⎛ 1.06 ⎞ MO = ⎜ 1.06 ⎟ N⋅ m ⎜ ⎟ ⎝ −4.03 ⎠ ⎛⎜ α ⎞⎟ ⎛ 75.7 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 75.7 ⎟ deg ⎜ γ ⎟ ⎝ 159.6 ⎠ ⎝ ⎠

Problem 4-170 If the resultant couple moment of the three couples acting on the triangular block is to be zero, determine the magnitudes of forces F and P.

338

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 10 lb a = 3 in b = 4 in c = 6 in d = 3 in

θ = 30 deg

Solution: Initial Guesses:

F = 1 lb

P = 1 lb

Given

⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ −F c ⎟ + ⎜ 0 ⎟ + ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ −P c ⎠

⎛0⎞ ⎜a⎟ = 0 2 2⎜ ⎟ a + b ⎝b⎠

⎛F⎞ ⎜ ⎟ = Find ( F , P) ⎝P⎠

F1 d

⎛F⎞ ⎛3⎞ ⎜ ⎟ = ⎜ ⎟ lb ⎝P⎠ ⎝4⎠

339

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Engineering Mechanics - Statics

Chapter 5

Problem 5-1 Draw the free-body diagram of the sphere of weight W resting between the smooth inclined planes. Explain the significance of each force on the diagram.

Given: W = 10 lb

θ 1 = 105 deg θ 2 = 45 deg

Solution:

NA, NB force of plane on sphere. W force of gravity on sphere.

Problem 5-2 Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B. Given: F = 8 lb a = 1.5 ft b = 0.2 ft c = 2 ft

340

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Engineering Mechanics - Statics

Chapter 5

Solution:

Problem 5-3 Draw the free-body diagram of the beam supported at A by a fixed support and at B by a roller. Explain the significance of each force on the diagram. Given: w = 40

lb ft

a = 3 ft b = 4 ft

θ = 30 deg

Solution:

A x, A y, MA effect of wall on beam. NB force of roller on beam. wa 2

resultant force of distributed load on beam.

Problem 5-4 Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by member (link) BC.

341

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: F = 8 kN a = 3m b = 4m c = 0.4 m d = 3 e = 4

Solution:

Problem 5-5 Draw the free-body diagram of the C-bracket supported at A, B, and C by rollers. Explain the significance of each forcce on the diagram. Given: a = 3 ft b = 4 ft

θ 1 = 30 deg θ 2 = 20 deg F = 200 lb 342

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Engineering Mechanics - Statics

Chapter 5

Solution: NA , NB , NC force of rollers on beam.

Problem 5-6 Draw the free-body diagram of the smooth rod of mass M which rests inside the glass. Explain the significance of each force on the diagram. Given: M = 20 gm a = 75 mm b = 200 mm

θ = 40 deg

Solution: A x , A y , NB force of glass on rod. M(g) N force of gravity on rod.

Problem 5-7 Draw the free-body diagram of the “spanner wrench” subjected to the force F. The support at A can be considered a pin, and the surface of contact at B is smooth. Explain the significance of 343

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Engineering Mechanics - Statics

Chapter 5

p each force on the diagram.

p

g

Given: F = 20 lb a = 1 in b = 6 in

Solution: A x, A y, NB force of cylinder on wrench.

Problem 5-8 Draw the free-body diagram of the automobile, which is being towed at constant velocity up the incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are free to roll. Explain the significance of each force on the diagram. Units Used: 3

Mg = 10 kg Given: M = 5 Mg

d = 1.50 m

a = 0.3 m

e = 0.6 m

b = 0.75 m

θ 1 = 20 deg

c = 1m

θ 2 = 30 deg

g = 9.81

m 2

s

344

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Engineering Mechanics - Statics

Chapter 5

Solution: NA, NB force of road on car. F force of cable on car. Mg force of gravity on car.

Problem 5-9 Draw the free-body diagram of the uniform bar, which has mass M and center of mass at G. The supports A, B, and C are smooth. Given: M = 100 kg a = 1.75 m b = 1.25 m c = 0.5 m d = 0.2 m g = 9.81

m 2

s Solution:

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Engineering Mechanics - Statics

Chapter 5

Problem 5-10 Draw the free-body diagram of the beam, which is pin-connected at A and rocker-supported at B. Given: F = 500 N M = 800 N⋅ m a = 8m b = 4m c = 5m

Solution:

Problem 5-11 The sphere of weight W rests between the smooth

inclined planes. Determine the reaactions at the supports. Given: W = 10 lb

θ 1 = 105 deg θ 2 = 45 deg Solution: Initial guesses 346

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Engineering Mechanics - Statics

NA = 1 lb

Chapter 5

NB = 1 lb

Given NB cos ( θ 1 − 90 deg) − NA cos ( θ 2 ) = 0 NA sin ( θ 2 ) − NB sin ( θ 1 − 90 deg) − W = 0

⎛ NA ⎞ ⎜ ⎟ = Find ( NA , NB) ⎝ NB ⎠

⎛ NA ⎞ ⎛ 19.3 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ NB ⎠ ⎝ 14.1 ⎠

Problem 5-12 Determine the magnitude of the resultant force acting at pin A of the handpunch. Given: F = 8 lb a = 1.5 ft b = 0.2 ft c = 2 ft

Solution:

Σ F x = 0; Σ M = 0;

Ax − F = 0

Ax = F

F a − Ay c = 0

Ay = F

Ax = 8 lb a c

Ay = 6 lb

347

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Engineering Mechanics - Statics

FA =

Chapter 5

2

Ax + Ay

2

F A = 10 lb

Problem 5-13 The C-bracket is supported at A, B, and C by rollers. Determine the reactions at the supports. Given: a = 3 ft b = 4 ft

θ 1 = 30 deg θ 2 = 20 deg F = 200 lb Solution: Initial Guesses: NA = 1 lb NB = 1 lb NC = 1 lb Given NA a − F b = 0 NB sin ( θ 1 ) − NC sin ( θ 2 ) = 0 NB cos ( θ 1 ) + NC cos ( θ 2 ) − NA − F = 0

⎛ NA ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NA , NB , NC) ⎜N ⎟ ⎝ C⎠

⎛ NA ⎞ ⎛ 266.7 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NB ⎟ = ⎜ 208.4 ⎟ lb ⎜ N ⎟ ⎝ 304.6 ⎠ ⎝ C⎠

348

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Engineering Mechanics - Statics

Chapter 5

Problem 5-14 The smooth rod of mass M rests inside the glass. Determine the reactions on the rod. Given: M = 20 gm a = 75 mm b = 200 mm

θ = 40 deg g = 9.81

m 2

s Solution: Initial Guesses: Ax = 1 N

Ay = 1 N

NB = 1 N

Given Ax − NB sin ( θ ) = 0 Ay − M g + NB cos ( θ ) = 0 −M g

a+b cos ( θ ) + NB b = 0 2

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , NB) ⎜N ⎟ ⎝ B⎠

⎛ Ax ⎞ ⎛ 0.066 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0.117 ⎟ N ⎜ N ⎟ ⎝ 0.103 ⎠ ⎝ B⎠

Problem 5-15 The “spanner wrench” is subjected to the force F. The support at A can be considered a pin, and the surface of contact at B is smooth. Determine the reactions on the spanner wrench.

349

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Engineering Mechanics - Statics

Chapter 5

Given: F = 20 lb a = 1 in b = 6 in Solution: Initial Guesses: Ax = 1 lb Ay = 1 lb NB = 1 lb Given − Ax + NB = 0

Ay − F = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , NB) ⎜N ⎟ ⎝ B⎠

−F ( a + b) + Ax a = 0

⎛ Ax ⎞ ⎛ 140 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 20 ⎟ lb ⎜ N ⎟ ⎝ 140 ⎠ ⎝ B⎠

Problem 5-16 The automobile is being towed at constant velocity up the incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are free to roll. Determine the reactions on both wheels at A and B and the tension in the cable at C. Units Used: 3

Mg = 10 kg

3

kN = 10 N

350

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Engineering Mechanics - Statics

Chapter 5

Given: M = 5 Mg

d = 1.50 m

a = 0.3 m

e = 0.6 m

b = 0.75 m

θ 1 = 20 deg

c = 1m

θ 2 = 30 deg

g = 9.81

m 2

s Solution: Guesses

F = 1 kN

NA = 1 kN

NB = 1 kN

Given NA + NB + F sin ( θ 2 ) − M g cos ( θ 1 ) = 0 −F cos ( θ 2 ) + M g sin ( θ 1 ) = 0 F cos ( θ 2 ) a − F sin ( θ 2 ) b − M g cos ( θ 1 ) c − M g sin ( θ 1 ) e + NB( c + d) = 0

⎛F ⎞ ⎜ ⎟ ⎜ NA ⎟ = Find ( F , NA , NB) ⎜ NB ⎟ ⎝ ⎠

⎛ F ⎞ ⎛ 19.37 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NA ⎟ = ⎜ 13.05 ⎟ kN ⎜ NB ⎟ ⎝ 23.36 ⎠ ⎝ ⎠

Problem 5-17 The uniform bar has mass M and center of mass at G. The supports A, B, and C are smooth. Determine the reactions at the points of contact at A, B, and C. Given: M = 100 kg a = 1.75 m b = 1.25 m

351

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Engineering Mechanics - Statics

Chapter 5

c = 0.5 m d = 0.2 m

θ = 30 deg g = 9.81

m 2

s Solution:

The initial guesses: NA = 20 N NB = 30 N NC = 40 N Given

ΣMA = 0; +

↑Σ Fy = 0; +

↑Σ Fy = 0;

−M g cos ( θ ) a − M g sin ( θ )

d + NB sin ( θ ) d + NC( a + b) = 0 2

NB − M g + NC cos ( θ ) = 0 NA − NC sin ( θ ) = 0

⎛ NC ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NC , NB , NA) ⎜N ⎟ ⎝ A⎠

⎛ NC ⎞ ⎛ 493 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NB ⎟ = ⎜ 554 ⎟ N ⎜ N ⎟ ⎝ 247 ⎠ ⎝ A⎠

Problem 5-18 The beam is pin-connected at A and rocker-supported at B. Determine the reactions at the pin A and at the roller at B. Given: F = 500 N M = 800 N⋅ m a = 8m

352

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Engineering Mechanics - Statics

Chapter 5

b = 4m c = 5m Solution:

⎞ ⎟ ⎝ a + b⎠

α = atan ⎛⎜

c

ΣMA = 0;

−F

a

cos ( α )

By =

− M + By a = 0

F a + M cos ( α ) cos ( α ) a

B y = 642 N + Σ F x = 0; → +

↑Σ Fy = 0;

− Ax + F sin ( α ) = 0

Ax = F sin ( α )

Ax = 192 N

− Ay − F cos ( α ) + B y = 0

Ay = −F cos ( α ) + By

Ay = 180 N

Problem 5-19 Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam. Given: F 1 = 600 N F 2 = 400 N

θ = 15 deg a = 4m b = 8m c = 3 d = 4 Solution:

ΣMA = 0;

B y( a + b) − F2 cos ( θ ) ( a + b) − F1 a = 0 353

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Engineering Mechanics - Statics

By = + Σ F x = 0; →

Chapter 5

F 2 cos ( θ ) ( a + b) + F 1 a a+b

Ax − F 2 sin ( θ ) = 0 Ax = F2 sin ( θ )

+

↑Σ Fy = 0;

B y = 586 N

Ax = 104 N

Ay − F 2 cos ( θ ) + B y − F1 = 0 Ay = F2 cos ( θ ) − B y + F1

FA =

2

Ax + Ay

2

Ay = 400 N

F A = 413 N

Problem 5-20 Determine the reactions at the supports. Given: w = 250

lb ft

a = 6 ft b = 6 ft c = 6 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

B y = 1 lb

Given Ax = 0

wa

Ay + B y − w( a + b) −

⎛ a ⎞ − w b⎛ b ⎞ − 1 w c⎛b + ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 2⎠ ⎝ 2⎠ 2 ⎝

1 2

wc = 0

c⎞

⎟ + By b = 0

3⎠

354

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Engineering Mechanics - Statics

Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , By) ⎜B ⎟ ⎝ y⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 2750 ⎟ lb ⎜ B ⎟ ⎝ 1000 ⎠ ⎝ y⎠

Problem 5-21 When holding the stone of weight W in equilibrium, the humerus H, assumed to be smooth, exerts normal forces FC and F A on the radius C and ulna A as shown. Determine these forces and the force F B that the biceps B exerts on the radius for equilibrium. The stone has a center of mass at G. Neglect the weight of the arm. Given: W = 5 lb

θ = 75 deg a = 2 in b = 0.8 in c = 14 in

Solution:

ΣMB = 0;

− W ( c − a) + F A a = 0 FA = W

⎛ c − a⎞ ⎜ ⎟ ⎝ a ⎠

F A = 30 lb +

↑Σ Fy = 0;

F B sin ( θ ) − W − FA = 0

FB =

W + FA sin ( θ )

F B = 36.2 lb + Σ F x = 0; →

F C − F B cos ( θ ) = 0 355

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Engineering Mechanics - Statics

Chapter 5

F C = FB cos ( θ ) F C = 9.378 lb

Problem 5-22 The uniform door has a weight W and a center of gravity at G. Determine the reactions at the hinges if the hinge at A supports only a horizontal reaction on the door, whereas the hinge at B exerts both horizontal and vertical reactions. Given: W = 100 lb a = 3 ft b = 3 ft c = 0.5 ft d = 2 ft

Solution: ΣMB = 0;

W d − A x ( a + b) = 0 Ax = W

⎛ d ⎞ ⎜ ⎟ ⎝ a + b⎠

Ax = 33.3 lb

ΣF x = 0;

Bx = Ax

B x = 33.3 lb

ΣF y = 0;

By = W

B y = 100 lb

Problem 5-23 The ramp of a ship has weight W and center of gravity at G. Determine the cable force in CD needed to just start lifting the ramp, (i.e., so the reaction at B becomes zero). Also, determine the horizontal and vertical components of force at the hinge (pin) at A.

356

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Engineering Mechanics - Statics

Chapter 5

Given: W = 200 lb

a = 4 ft

θ = 30 deg

b = 3 ft

φ = 20 deg

c = 6 ft

Solution:

ΣMA = 0; −F CD cos ( θ ) ( b + c) cos ( φ ) + FCD sin ( θ ) ( b + c) sin ( φ ) + W c cos ( φ ) = 0

F CD =

W c cos ( φ )

( b + c) ( cos ( θ ) cos ( φ ) − sin ( θ ) sin ( φ ) )

+ Σ F x = 0; →

F CD sin ( θ ) − A x = 0 Ax = FCD sin ( θ )

+

↑Σ Fy = 0;

F CD = 195 lb

Ax = 97.5 lb

Ay − W + F CD cos ( θ ) = 0 Ay = W − FCD cos ( θ )

Ay = 31.2 lb

Problem 5-24 The drainpipe of mass M is held in the tines of the fork lift. Determine the normal forces at A and B as functions of the blade angle θ and plot the results of force (ordinate) versus θ (abscissa) for 0 ≤ θ ≤ 90 deg.

357

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Engineering Mechanics - Statics

Chapter 5

Units used: 3

Mg = 10 kg Given: M = 1.4 Mg a = 0.4 m g = 9.81

m 2

s Solution:

θ = 0 .. 90 NA ( θ ) =

M g sin ( θ deg) 3

10 NB ( θ ) =

M g cos ( θ deg) 3

10

Force in kN

15 NA( θ )10 NB( θ )

5

0

0

20

40

60

80

100

θ

Angle in Degrees

Problem 5-25 While slowly walking, a man having a total mass M places all his weight on one foot. Assuming that the normal force NC of the ground acts on his foot at C, determine the resultant vertical compressive force F B which the tibia T exerts on the astragalus B, and the vertical tension FA in the achilles tendon A at the instant shown.

358

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: M = 80 kg a = 15 mm b = 5 mm c = 20 mm d = 100 mm

Solution: NC = M g NC = 785 N ΣMA = 0;

−F B c + NC ( c + d) = 0 F B = NC

⎛ c + d⎞ ⎜ ⎟ ⎝ c ⎠

F B = 4.71 kN ΣF y = 0;

F A − FB + NC = 0 F A = F B − NC F A = 3.92 kN

Problem 5-26 Determine the reactions at the roller A and pin B.

359

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Engineering Mechanics - Statics

Chapter 5

Given: M = 800 lb ft

c = 3 ft

F = 390 lb

d = 5

a = 8 ft

e = 12

b = 4 ft

θ = 30 deg

Solution: Guesses

R A = 1 lb B x = 1 lb

B y = 1 lb

Given R A sin ( θ ) + B x −

⎛ d ⎞F = 0 ⎜ 2 2⎟ ⎝ e +d ⎠

R A cos ( θ ) + By −

e ⎛ ⎞ ⎜ 2 2 ⎟F = 0 ⎝ e +d ⎠

M − R A cos ( θ ) ( a + b) + Bx c = 0

⎛ RA ⎞ ⎜ ⎟ ⎜ Bx ⎟ = Find ( RA , Bx , By) ⎜B ⎟ ⎝ y⎠

R A = 105.1 lb

⎛ Bx ⎞ ⎛ 97.4 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ By ⎠ ⎝ 269 ⎠

Problem 5-27 The platform assembly has weight W1 and center of gravity at G1. If it is intended to support a maximum load W2 placed at point G2,,determine the smallest counterweight W that should be placed at B in order to prevent the platform from tipping over. Given: W1 = 250 lb

a = 1 ft

c = 1 ft

e = 6 ft

W2 = 400 lb

b = 6 ft

d = 8 ft

f = 2 ft

360

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Engineering Mechanics - Statics

Chapter 5

Solution: When tipping occurs, R c = 0

ΣMD = 0;

−W2 f + W1 c + WB ( b + c) = 0

WB =

W2 f − W1 c b+c

WB = 78.6 lb

Problem 5-28 The articulated crane boom has a weight W and mass center at G. If it supports a load L, determine the force acting at the pin A and the compression in the hydraulic cylinder BC when the boom is in the position shown. Units Used: 3

kip = 10 lb Given: W = 125 lb

361

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Engineering Mechanics - Statics

Chapter 5

L = 600 lb a = 4 ft b = 1 ft c = 1 ft d = 8 ft

θ = 40 deg

Solution: Guesses

Given

Ax = 1 lb

Ay = 1 lb

F B = 1 lb

− Ax + F B cos ( θ ) = 0

− Ay + F B sin ( θ ) − W − L = 0

F B cos ( θ ) b + F B sin ( θ ) c − W a − L( d + c) = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , FB) ⎜F ⎟ ⎝ B⎠

F B = 4.19 kip

⎛ Ax ⎞ ⎛ 3.208 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ Ay ⎠ ⎝ 1.967 ⎠

Problem 5-29 The device is used to hold an elevator door open. If the spring has stiffness k and it is compressed a distnace δ, determine the horizontal and vertical components of reaction at the pin A and the resultant force at the wheel bearing B. Given: N m

b = 125 mm

δ = 0.2 m

c = 100 mm

a = 150 mm

θ = 30 deg

k = 40

362

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Engineering Mechanics - Statics

Solution:

ΣMA = 0;

Chapter 5

F s = kδ −F s a + F B cos ( θ ) ( a + b) − F B sin ( θ ) c = 0 FB = Fs

a

cos ( θ ) ( a + b) − sin ( θ ) c

F B = 6.378 N + Σ F x = 0; →

Ax − F B sin ( θ ) = 0 Ax = FB sin ( θ ) Ax = 3.189 N

+

↑Σ Fy = 0;

Ay − F s + F B cos ( θ ) = 0 Ay = Fs − FB cos ( θ ) Ay = 2.477 N

Problem 5-30 Determine the reactions on the bent rod which is supported by a smooth surface at B and by a collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod. Given: F = 100 lb M = 200 lb ft a = 3 ft b = 3 ft c = 2ft

363

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Engineering Mechanics - Statics

Chapter 5

d = 3 e = 4 f = 12 g = 5 Solution: Initial Guesses: NA = 20 lb

NB = 10 lb

MA = 30 lb ft

Given f g ⎞ ⎞ ⎛ ⎛ ⎜ 2 2 ⎟ ( a + b) − NB ⎜ 2 2 ⎟ c = 0 ⎝ f +g ⎠ ⎝ f +g ⎠

ΣMA = 0;

MA − F a − M + NB

ΣF x = 0;

NA

e g ⎞ ⎞ ⎛ ⎛ ⎜ 2 2 ⎟ − NB ⎜ 2 2 ⎟ = 0 ⎝ e +d ⎠ ⎝ f +g ⎠

ΣF y = 0;

NA

f ⎞−F=0 ⎛ d ⎞+N ⎛ B⎜ ⎜ 2 2⎟ ⎟ 2 2 ⎝ e +d ⎠ ⎝ f +g ⎠

⎛ NA ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NA , NB , MA) ⎜M ⎟ ⎝ A⎠

⎛ NA ⎞ ⎛ 39.7 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ NB ⎠ ⎝ 82.5 ⎠

MA = 106 lb⋅ ft

Problem 5-31 The cantilevered jib crane is used to support the load F. If the trolley T can be placed anywhere in the range x1 ≤ x ≤ x2, determine the maximum magnitude of reaction at the supports A and B. Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, whereas the one at A does not. Units Used: kip = 1000 lb Given: F = 780 lb

364

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Engineering Mechanics - Statics

Chapter 5

a = 4 ft b = 8 ft x1 = 1.5 ft x2 = 7.5 ft Solution: The maximum occurs when x = x2

ΣMA = 0;

−F x2 + B x a = 0

Bx = F

x2 a 3

B x = 1.462 × 10 lb + Σ F x = 0; → +

↑Σ Fy = 0; FB =

Ax − Bx = 0 By − F = 0 2

Bx + By

Ax = Bx By = F

2

3

Ax = 1.462 × 10 lb B y = 780 lb

F B = 1.657 kip

Problem 5-32 The uniform rod AB has weight W. Determine the force in the cable when the rod is in the position shown. Given: W = 15 lb L = 5 ft

365

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Engineering Mechanics - Statics

Chapter 5

θ 1 = 30 deg θ 2 = 10 deg

Solution:

ΣMA = 0;

NB L sin ( θ 1 + θ 2 ) − W

NB =

⎛ L ⎞ cos ( θ + θ ) = 0 ⎜ ⎟ 1 2 ⎝2⎠

W cos ( θ 1 + θ 2 ) 2 sin ( θ 1 + θ 2 )

NB = 8.938 lb ΣF x = 0;

T cos ( θ 2 ) − NB

T =

NB

cos ( θ 2 )

T = 9.08 lb

Problem 5-33 The power pole supports the three lines, each line exerting a vertical force on the pole due to its weight as shown. Determine the reactions at the fixed support D. If it is possible for wind or ice to snap the lines, determine which line(s) when removed create(s) a condition for the greatest moment reaction at D. Units Used: 3

kip = 10 lb

366

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Engineering Mechanics - Statics

Chapter 5

Given: W1 = 800 lb W2 = 450 lb W3 = 400 lb a = 2ft b = 4 ft c = 3 ft Solution: + Σ F x = 0; → +

↑Σ Fy = 0;

Dx = 0 Dy − ( W1 + W2 + W3 ) = 0 Dy = W1 + W2 + W3

ΣMD = 0;

Dy = 1.65 kip

−W2 b − W3 c + W1 a + MD = 0 MD = W2 b + W3 c − W1 a

MD = 1.4 kip⋅ ft

Examine all cases. For these numbers we require line 1 to snap. MDmax = W2 b + W3 c

MDmax = 3 kip⋅ ft

Problem 5-34 The picnic table has a weight WT and a center of gravity at GT . If a man weighing WM has a center of gravity at GM and sits down in the centered position shown, determine the vertical reaction at each of the two legs at B.Neglect the thickness of the legs. What can you conclude from the results? Given: WT = 50 lb

367

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Engineering Mechanics - Statics

Chapter 5

WM = 225 lb a = 6 in b = 20 in c = 20 in

Solution: ΣMA = 0;

2 NB( b + c) + WM a − WT b = 0

NB =

WT b − WM a 2 ( b + c)

NB = −4.37 lb Since NB has a negative sign, the table will tip over.

Problem 5-35 If the wheelbarrow and its contents have a mass of M and center of mass at G, determine the magnitude of the resultant force which the man must exert on each of the two handles in order to hold the wheelbarrow in equilibrium. Given: M = 60 kg a = 0.6 m b = 0.5 m c = 0.9 m d = 0.5 m g = 9.81

m 2

s

368

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Engineering Mechanics - Statics

Chapter 5

Solution:

ΣMB = 0;

− Ay ( b + c) + M g c = 0 Ay =

Mgc b+c

Ay = 378.386 N + Σ F x = 0; →

B x = 0N

+

Ay − M g + 2 By = 0

↑Σ Fy = 0;

By =

Bx = 0

M g − Ay 2

B y = 105.107 N

Problem 5-36 The man has weight W and stands at the center of the plank. If the planes at A and B are smooth, determine the tension in the cord in terms of W and θ.

Solution: ΣMB = 0;

W

L cos ( φ ) − NA L cos ( φ ) = 0 2

NA = ΣF x = 0;

W 2

T cos ( θ ) − NB sin ( θ ) = 0

(1)

369

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Engineering Mechanics - Statics

Chapter 5

T sin ( θ ) + NB cos ( θ ) + NA − W = 0

ΣF y = 0;

(2)

Solving Eqs. (1) and (2) yields:

T=

W sin ( θ ) 2

NB =

W cos ( θ ) 2

Problem 5-37 When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is F , determine the unstretched length of the spring if the stiffness of the spring is k. Given: F = 3N k = 80

N m

a = 100 mm b = 50 mm c = 40 mm d = 10 mm

θ = 30 deg Solution: ΣMD = 0; F s b − F cos ( θ ) c − F sin ( θ ) d = 0 Fs = F

cos ( θ ) c + sin ( θ ) d b

Fs = k x

x =

L0 = a − x

Fs k

F s = 2.378 N

x = 29.73 mm L0 = 70.3 mm

370 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 5

Problem 5-38 The telephone pole of negligible thickness is subjected to the force F directed as shown. It is supported by the cable BCD and can be assumed pinned at its base A. In order to provide clearance for a sidewalk right of way, where D is located, the strut CE is attached at C, as shown by the dashed lines (cable segment CD is removed). If the tension in CD' is to be twice the tension in BCD, determine the height h for placement of the strut CE. Given: F = 80 lb

θ = 30 deg a = 30 ft b = 10 ft

Solution:

+ ΣMA = 0;

−F cos ( θ ) a +

b ⎛ ⎞ ⎜ 2 2 ⎟ TBCD a = 0 ⎝ a +b ⎠

TBCD = F cos ( θ )

2

2

a +b b

TBCD = 219.089 lb

Require TCD' = 2 TBCD + ΣMA = 0;

TCD' = 438.178 lb

TCD' d − F cos ( θ ) a = 0

d = Fa

⎛ cos ( θ ) ⎞ ⎜T ⎟ ⎝ CD' ⎠

d = 4.7434 ft

Geometry: a−h a = d b

h = a−a

⎛ d⎞ ⎜ ⎟ ⎝ b⎠

h = 15.8 ft

371

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Engineering Mechanics - Statics

Chapter 5

Problem 5-39 The worker uses the hand truck to move material down the ramp. If the truck and its contents are held in the position shown and have weight W with center of gravity at G, determine the resultant normal force of both wheels on the ground A and the magnitude of the force required at the grip B. Given: W = 100 lb

e = 1.5 ft

a = 1 ft

f = 0.5 ft

b = 1.5 ft

θ = 60 deg

c = 2 ft

φ = 30 deg

d = 1.75 ft

Solution: ΣMB = 0; NA cos ( θ − φ ) ( b + c + d) + NA sin ( θ − φ ) ( a − f) − W cos ( θ ) ( b + c) − W sin ( θ ) ( e + a) = 0

NA =

W cos ( θ ) ( b + c) + W sin ( θ ) ( e + a)

cos ( θ − φ ) ( b + c + d) + sin ( θ − φ ) ( a − f)

NA = 81.621 lb

ΣF x = 0;

−B x + NA sin ( φ ) = 0

B x = NA sin ( φ )

B x = 40.811 lb

ΣF y = 0;

B y + NA ( cos ( φ ) − W = 0)

B y = W − NA cos ( φ )

B y = 29.314 lb

FB =

2

Bx + By

2

F B = 50.2 lb

372

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Engineering Mechanics - Statics

Chapter 5

T sin ( θ ) + NB cos ( θ ) + NA − W = 0

ΣF y = 0;

(2)

Solving Eqs. (1) and (2) yields:

T=

W sin ( θ ) 2

NB =

W cos ( θ ) 2

Problem 5-37 When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is F , determine the unstretched length of the spring if the stiffness of the spring is k. Given: F = 3N k = 80

N m

a = 100 mm b = 50 mm c = 40 mm d = 10 mm

θ = 30 deg Solution: ΣMD = 0; F s b − F cos ( θ ) c − F sin ( θ ) d = 0 Fs = F

cos ( θ ) c + sin ( θ ) d b

Fs = k x

x =

L0 = a − x

Fs k

F s = 2.378 N

x = 29.73 mm L0 = 70.3 mm 373

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 5

The shelf supports the electric motor which has mass m1 and mass center at Gm. The platform upon which it rests has mass m2 and mass center at Gp. Assuming that a single bolt B holds the shelf up and the bracket bears against the smooth wall at A, determine this normal force at A and the horizontal and vertical components of reaction of the bolt B on the bracket. Given: m1 = 15 kg

c = 50 mm

m2 = 4 kg

d = 200 mm

a = 60 mm

e = 150 mm

b = 40 mm g = 9.81

m 2

s Solution: ΣMA = 0;

B x a − m2 g d − m1 g( d + e) = 0

Bx = g + Σ F x = 0; →

m2 d + m1 ( d + e)

B x = 989 N

a

Ax − Bx = 0 Ax = Bx

+

↑Σ Fy = 0;

Ax = 989 N

B y − m2 g − m1 g = 0 B y = m2 g + m1 g

B y = 186 N

Problem 5-42 A cantilever beam, having an extended length L, is subjected to a vertical force F. Assuming that the wall resists this load with linearly varying distributed loads over the length a of the beam portion inside the wall, determine the intensities w1 and w2 for equilibrium.

374

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: F = 500 N a = 0.15 m L = 3m Solution: The initial guesses w1 = 1

kN m

w2 = 1

kN m

Given +

↑Σ Fy = 0; ΣMA = 0;

⎛ w1 ⎞ ⎜ ⎟ = Find ( w1 , w2) ⎝ w2 ⎠

1

1 w1 a − w2 a − F = 0 2 2 −F L −

1 2

1 ⎛ 2 a⎞ ⎟ + w2 a⎜ ⎟ = 0 ⎝ 3⎠ 2 ⎝ 3 ⎠

w1 a⎛⎜

a⎞

⎛ w1 ⎞ ⎛ 413 ⎞ kN ⎜ ⎟=⎜ ⎟ ⎝ w2 ⎠ ⎝ 407 ⎠ m

Problem 5-43 The upper portion of the crane boom consists of the jib AB, which is supported by the pin at A, the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If the load F is supported by the hoist line, which passes over the pulley at B, determine the magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has a radius of r.

375

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: F = 5 kN r = 0.1 m a = r b = 1.5 m c = 5m Solution: From pulley, tension in the hoist line is ΣMB = 0;

T( a) − F( r) = 0

T = F

r T = 5 kN

a

From the jib, ΣMA = 0;

b+a

−F ( c) + TBC

c=0

2

c + ( b + a) 2

c + ( b + a)

TBC = F

+

↑Σ Fy = 0;

2

2

TBC = 16.406 kN

b+a b+a ⎤−F=0 ⎢ 2 2⎥ ⎣ c + ( b + a) ⎦

− Ay + TBC ⎡

b+a ⎤−F ⎢ 2 2⎥ ⎣ c + ( b + a) ⎦

Ay = TBC⎡

+ Σ F x = 0; →

Ay = 0 kN

⎤−F=0 ⎣ c + ( b + a) ⎦

Ax − TBC ⎡ ⎢

c

2

2⎥

376

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Engineering Mechanics - Statics

Chapter 5

c

Ax = TBC

2

c + ( b + a) FA =

2

Ax + Ay

+F 2

2

Ax = 20.6 kN

F A = 20.6 kN

Problem 5-44 The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center of gravity at G2. Determine the smallest angle of tilt θ of the boom, without causing the crane to overturn if the suspended load has weight W. Neglect the thickness of the tracks at A and B. Given: W1 = 120000 lb W2 = 30000 lb W = 40000 lb a = 4 ft b = 6 ft c = 3 ft d = 12 ft e = 15 ft Solution: When tipping occurs, R A = 0 ΣMB = 0;

−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ + W1 ( b + c) = 0

⎡W2 c + W c + W1 ( b + c) ⎤ ⎥ ⎣ W2 d + W ( d + e) ⎦

θ = acos ⎢

θ = 26.4 deg

377

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Engineering Mechanics - Statics

Chapter 5

Problem 5-45 The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center of gravity at G2. If the suspended load has weight W determine the normal reactions at the tracks A and B. For the calculation, neglect the thickness of the tracks . Units Used: 3

kip = 10 lb Given: W1 = 120000 lb

a = 4 ft

W2 = 30000 lb

b = 6 ft

W = 16000 lb

c = 3 ft

θ = 30 deg

d = 12 ft e = 15 ft

Solution: ΣMB = 0; −W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ − RA( a + b + c) + W1 ( b + c) = 0

RA = +

↑

−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ + W1 ( b + c)

Σ F y = 0;

a+b+c

R A = 40.9 kip

R A + RB − W1 − W2 − W = 0 R B = −R A + W1 + W2 + W

R B = 125 kip

Problem 5-46 The man attempts to support the load of boards having a weight W and a center of gravity at G. If he is standing on a smooth floor, determine the smallest angle θ at which he can hold them up in the position shown. Neglect his weight .

378

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Engineering Mechanics - Statics

Chapter 5

Given: a = 0.5 ft b = 3 ft c = 4 ft d = 4 ft

Solution: ΣMB = 0;

−NA( a + b) + W( b − c cos ( θ ) ) = 0

As θ becomes smaller, NA goes to 0 so that, cos ( θ ) =

b c

b θ = acos ⎛⎜ ⎟⎞

⎝c⎠

θ = 41.4 deg

Problem 5-47 The motor has a weight W. Determine the force that each of the chains exerts on the supporting hooks at A, B, and C. Neglect the size of the hooks and the thickness of the beam.

379

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Engineering Mechanics - Statics

Chapter 5

Given: W = 850 lb a = 0.5 ft b = 1 ft c = 1.5 ft

θ 1 = 10 deg θ 2 = 30 deg θ 3 = 10 deg

Solution: Guesses F A = 1 lb

F B = 1 lb

F C = 1 lb

Given ΣMB = 0;

F A cos ( θ 3 ) b + W a − FC cos ( θ 1 ) ( a + c) = 0

ΣF x = 0;

F C sin ( θ 1 ) − F B sin ( θ 2 ) − FA sin ( θ 3 ) = 0

ΣF y = 0;

W − F A cos ( θ 3 ) − F B cos ( θ 2 ) − F C cos ( θ 1 ) = 0

⎛ FA ⎞ ⎜ ⎟ ⎜ FB ⎟ = Find ( FA , FB , FC) ⎜F ⎟ ⎝ C⎠ ⎛ FA ⎞ ⎛ 432 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FB ⎟ = ⎜ −0 ⎟ lb ⎜ F ⎟ ⎝ 432 ⎠ ⎝ C⎠

380

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Engineering Mechanics - Statics

Chapter 5

Problem 5-48 The boom supports the two vertical loads. Neglect the size of the collars at D and B and the thickness of the boom, and compute the horizontal and vertical components of force at the pin A and the force in cable CB. Given: F 1 = 800 N F 2 = 350 N a = 1.5 m b = 1m c = 3 d = 4

θ = 30 deg Solution: ΣMA = 0; −F 1 a cos ( θ ) − F2 ( a + b) cos ( θ ) +

d 2

2

FCB ( a + b) sin ( θ ) +

c +d

F CB

d 2

c +d Ax =

2

FCB ( a + b) cos ( θ ) = 0

F CB = 0

d 2

2

F CB = 782 N

d sin ( θ ) ( a + b) + c cos ( θ ) ( a + b) Ax −

2

c +d

2 2 ⎡⎣F1 a + F2 ( a + b)⎤⎦ cos ( θ ) c + d =

+ Σ F x = 0; →

c

2

Ax = 625 N

F CB

c +d +

↑

Σ F y = 0;

Ay − F1 − F2 +

c 2

c +d Ay = F1 + F2 −

2

F CB = 0

c 2

2

FCB

Ay = 681 N

c +d

Problem 5-49 The boom is intended to support two vertical loads F1 and F 2. If the cable CB can sustain a 381

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Engineering Mechanics - Statics

Chapter 5

maximum load Tmax before it fails, determine the critical loads if F1 = 2F2. Also, what is the magnitude of the maximum reaction at pin A? Units Used: 3

kN = 10 N Given: Tmax = 1500 N a = 1.5 m b = 1m c = 3 d = 4

θ = 30deg Solution: ΣMA = 0;

F1 = 2 F2

−2 F 2 a cos ( θ ) − F2 ( a + b) cos ( θ ) +

d 2

2

Tmax( a + b) sin ( θ ) +

c +d

F2 =

( a + b)Tmax( d sin ( θ ) + c cos ( θ ) ) 2

F1 = 2 F2 + ΣF x = 0; →

Ax −

d 2

c +d Ax =

2

2

Tmax( a + b) cos ( θ ) = 0

F 1 = 1.448 kN

Tmax = 0

d 2

2

c +d

F 2 = 724 N

c + d cos ( θ ) ( 3 a + b) 2

c

2

Ax = 1.20 kN

Tmax

c +d

+

↑Σ Fy = 0;

Ay − F2 − F1 +

c 2

c +d

2

Tmax = 0

382

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Engineering Mechanics - Statics

Chapter 5

c

Ay = F2 + F1 −

2

2

Ay = 1.27 kN

Tmax

c +d

2

FA =

Ax + Ay

2

F A = 1.749 kN

Problem 5-50 The uniform rod of length L and weight W is supported on the smooth planes. Determine its position θ for equilibrium. Neglect the thickness of the rod.

Solution: L

cos ( θ ) + NA cos ( φ − θ ) L = 0

ΣMB = 0;

−W

ΣMA = 0;

W

ΣF x = 0;

NB sin ( ψ) − NA sin ( φ ) = 0

W cos ( θ )

2 cos ( ψ + θ )

L 2

2

NA =

cos ( θ ) − NB cos ( ψ + θ ) L = 0

sin ( ψ) −

W cos ( θ )

2 cos ( φ − θ )

NB =

W cos ( θ )

2 cos ( φ − θ ) W cos ( θ )

2 cos ( ψ + θ )

sin ( φ ) = 0

sin ( ψ) cos ( φ − θ ) − sin ( φ ) cos ( ψ + θ ) = 0 sin ( ψ) ( cos ( φ ) cos ( θ ) + sin ( φ ) sin ( θ ) ) − sin ( φ ) ( cos ( ψ) cos ( θ ) − sin ( ψ) sin ( θ ) ) = 0 2 sin ( ψ) sin ( φ ) sin ( θ ) = ( sin ( φ ) cos ( ψ) − sin ( ψ) cos ( φ ) ) cos ( θ ) tan ( θ ) =

sin ( φ ) cos ( ψ) − sin ( ψ) cos ( φ ) 2 sin ( ψ) sin ( φ )

=

cot ( ψ) − cot ( φ ) 2

⎛ cot ( ψ) − cot ( φ ) ⎞ ⎟ 2 ⎝ ⎠

θ = atan ⎜

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Engineering Mechanics - Statics

Chapter 5

Problem 5-51 The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in place by the spring which has unstretched length δ. Determine the magnitude of the resultant force at A and the normal force on the peg at B when the lever is in the position shown. Given:

δ = 200 mm k = 5

N m

a = 100 mm b = 300 mm c = 300 mm

θ = 30 deg

Solution: Using the law of cosines and the law of sines c + ( a + b) − 2 c( a + b) cos ( 180 deg − θ ) 2

l =

sin ( φ ) c

=

2

sin ( 180 deg − θ ) l

Fs = k s = k (l − δ )

ΣMA = 0;

−F s sin ( φ ) ( a + b) + NB a = 0

⎛ sin (180 deg − θ ) ⎞ ⎟ l ⎝ ⎠

φ = asin ⎜c

φ = 12.808 deg

F s = k( l − δ )

F s = 2.3832 N

NB = F s sin ( φ )

a+b a

NB = 2.11 N

ΣF x = 0;

Ax − F s cos ( φ ) = 0

Ax = Fs cos ( φ )

Ax = 2.3239 N

ΣF y = 0;

Ay + NB − Fs sin ( φ ) = 0

Ay = Fs sin ( φ ) − NB

Ay = −1.5850 N

FA =

2

Ax + Ay

2

F A = 2.813 N

384 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 5

Problem 5-52 The rigid beam of negligible weight is supported horizontally by two springs and a pin. If the springs are uncompressed when the load is removed, determine the force in each spring when the load P is applied. Also, compute the vertical deflection of end C. Assume the spring stiffness k is large enough so that only small deflections occur. Hint: The beam rotates about A so the deflections in the springs can be related. Solution: ΣMA = 0; 3 F B L + FC2 L − P L = 0 2 F B + 2 FC = 1.5 P ΔC = 2 ΔB FC k

=

2 FB k

F C = 2FB 5 F B = 1.5 P F B = 0.3 P F c = 0.6 P ΔC =

0.6 P k

Problem 5-53 The rod supports a weight W and is pinned at its end A. If it is also subjected to a couple moment of M, determine the angle θ for equilibrium.The spring has an unstretched length δ and a stiffness k.

385

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Engineering Mechanics - Statics

Chapter 5

Given: W = 200 lb M = 100 lb ft

δ = 2 ft k = 50

lb ft

a = 3 ft b = 3 ft c = 2 ft Solution: Initial Guess:

θ = 10 deg

Given k⎡⎣( a + b)sin ( θ ) + c − δ⎤⎦ ( a + b) cos ( θ ) − W a cos ( θ ) − M = 0

θ = Find ( θ )

θ = 23.2 deg

Problem 5-54 The smooth pipe rests against the wall at the points of contact A, B, and C. Determine the reactions at these points needed to support the vertical force F . Neglect the pipe's thickness in the calculation. Given: F = 45 lb

386

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Engineering Mechanics - Statics

Chapter 5

θ = 30 deg a = 16 in b = 20 in c = 8 in

Solution: R A = 1 lb

Initial Guesses:

R B = 1 lb

R C = 1 lb

Given F cos ( θ ) ( a + b) − F sin ( θ ) c − RC b + R B c tan ( θ ) = 0

ΣMA = 0;

R C cos ( θ ) − R B cos ( θ ) − F = 0

+

↑Σ Fy = 0;

+ Σ F x = 0; →

R A + RB sin ( θ ) − RC sin ( θ ) = 0

⎛ RA ⎞ ⎜ ⎟ ⎜ RB ⎟ = Find ( RA , RB , RC) ⎜R ⎟ ⎝ C⎠

⎛ RA ⎞ ⎛ 25.981 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ RB ⎟ = ⎜ 11.945 ⎟ lb ⎜ R ⎟ ⎝ 63.907 ⎠ ⎝ C⎠

Problem 5-55 The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the stiffness of the springs at A and B is k, and the strip is originally horizontal when the springs are unstretched, determine the smallest force needed to close the contact gap at C. Units Used: mN = 10

−3

N

387

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Engineering Mechanics - Statics

Chapter 5

Given: a = 50 mm b = 50 mm c = 10 mm k = 5

N m

Solution: Initial Guesses:

F = 0.5 N

yA = 1 mm

yB = 1 mm

Given c − yA a+b

=

yB − yA a

⎛ yA ⎞ ⎜ ⎟ ⎜ yB ⎟ = Find ( yA , yB , F) ⎜F⎟ ⎝ ⎠

k yA + k yB − F = 0

⎛ yA ⎞ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ ⎟ mm ⎝ yB ⎠ ⎝ 4 ⎠

k yB a − F( a + b) = 0

F = 10 mN

Problem 5-56 The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine the maximum stiffness k of the springs at A and B so that the contact at C closes when the vertical force developed there is F . Originally the strip is horizontal as shown. Units Used: mN = 10

−3

N

Given: a = 50 mm b = 50 mm c = 10 mm F = 0.5 N

388

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Engineering Mechanics - Statics

Chapter 5

Solution:

Initial Guesses: k = 1

N

yA = 1 mm

m

yB = 1 mm

Given c − yA a+b

=

yB − yA

k yA + k yB − F = 0

a

⎛ yA ⎞ ⎜ ⎟ ⎜ yB ⎟ = Find ( yA , yB , k) ⎜k ⎟ ⎝ ⎠

⎛ yA ⎞ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ ⎟ mm ⎝ yB ⎠ ⎝ 4 ⎠

k yB a − F( a + b) = 0

k = 250

N m

Problem 5-57 Determine the distance d for placement of the load P for equilibrium of the smooth bar in the position θ as shown. Neglect the weight of the bar.

Solution: +

↑Σ Fy = 0; Σ MA = 0;

R cos ( θ ) − P = 0 −P d cos ( θ ) + R R d cos ( θ ) = R 2

d=

a

cos ( θ )

=0

a

cos ( θ )

a cos ( θ )

3

Problem 5-58 The wheelbarrow and its contents have mass m and center of mass at G. Determine the greatest 389

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Engineering Mechanics - Statics

Chapter 5

angle of tilt θ without causing the wheelbarrow to tip over.

g

Solution: Require point G to be over the wheel axle for tipping. Thus b cos ( θ ) = a sin ( θ ) − 1⎛ b ⎞

θ = tan

⎜ ⎟ ⎝ a⎠

Problem 5-59 Determine the force P needed to pull the roller of mass M over the smooth step. Given: M = 50 kg a = 0.6 m b = 0.1 m

θ = 60 deg θ 1 = 20 deg g = 9.81

m 2

s

390

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Engineering Mechanics - Statics

Chapter 5

Solution:

φ = acos ⎛⎜

a − b⎞

⎟ ⎝ a ⎠

φ = 33.56 deg ΣMB = 0,

M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) ... = 0 + P cos ( θ ) ( a − b) − P sin ( θ ) a sin ( φ )

⎡sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤ ⎥ ⎣ cos ( θ ) ( a − b) + sin ( θ ) a sin ( φ ) ⎦

P = M g⎢

P = 441 N

Problem 5-60 Determine the magnitude and direction θ of the minimum force P needed to pull the roller of mass M over the smooth step. Given: a = 0.6 m b = 0.1 m

θ 1 = 20 deg M = 50 kg g = 9.81

m 2

s Solution:

For Pmin, NA tends to 0

φ = acos ⎛⎜

a − b⎞

⎟ ⎝ a ⎠

φ = 33.56 deg ΣMB = 0 M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) ... = 0 + ⎡⎣−P cos ( θ ) ( a − b)⎤⎦ − P sin ( θ ) a sin ( φ )

391

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Engineering Mechanics - Statics

P=

Chapter 5

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦ cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )

For Pmin : dP dθ

=

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦

⎡⎣cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )⎤⎦ θ = atan ⎛⎜ sin ( φ ) ⋅

which gives,

P =

⎝

a

2

⎡⎣a sin ( φ ) cos ( θ ) − ( a − b)sin ( θ )⎤⎦ = 0

⎞ ⎟

θ = 33.6 deg

a − b⎠

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦ cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )

P = 395 N

Problem 5-61 A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a radius r. Determine the angle of inclination θ for equilibrium.

Solution: By Observation φ = θ. Equilibirium : ΣMA = 0; NB2 r cos ( θ ) − W NB =

L 2

cos ( θ ) = 0

WL 4r

ΣF x = 0; NA cos ( θ ) − W sin ( θ ) = 0

NA = W tan ( θ ) 392

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Engineering Mechanics - Statics

Chapter 5

ΣF y = 0; W tan ( θ ) sin ( θ ) +

WL 4r

− W cos ( θ ) = 0

sin ( θ ) − cos ( θ ) = 1 − 2 cos ( θ ) = 2

2

2 cos ( θ ) − 2

cos ( θ ) =

L 4r

L+

2

−L 4r

cos ( θ )

cos ( θ ) − 1 = 0 2

⎛ L + L2 + 128 r2 ⎞ ⎟ 16 r ⎝ ⎠

2

L + 128 r

θ = acos ⎜

16 r

Problem 5-62 The disk has mass M and is supported on the smooth cylindrical surface by a spring having stiffness k and unstretched length l0. The spring remains in the horizontal position since its end A is attached to the small roller guide which has negligible weight. Determine the angle θ to the nearest degree for equilibrium of the roller. Given: M = 20 kg k = 400

N m

l0 = 1 m r = 2m g = 9.81

m 2

s a = 0.2 m Guesses

F = 10 N

Solution:

R = 10 N θ = 30 deg

Given

+ Σ F y = 0; →

R sin ( θ ) − M g = 0

+

R cos ( θ ) − F = 0

↑Σ Fx = 0;

Spring

F = k⎡⎣( r + a)cos ( θ ) − l0⎤⎦ 393

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Engineering Mechanics - Statics

⎛F⎞ ⎜ R ⎟ = Find ( F , R , θ ) ⎜ ⎟ ⎝θ ⎠

Chapter 5

⎛ F ⎞ ⎛ 163.633 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ R ⎠ ⎝ 255.481 ⎠

θ = 50.171 deg

There is also another answer that we can find by choosing different starting guesses. Guesses

F = 200 N R = 200 N θ = 20 deg

Solution: + Σ F y = 0; → +

↑Σ Fx = 0;

Spring

Given R sin ( θ ) − M g = 0 R cos ( θ ) − F = 0 F = k⎡⎣( r + a)cos ( θ ) − l0⎤⎦

⎛F⎞ ⎜ R ⎟ = Find ( F , R , θ ) ⎜ ⎟ ⎝θ ⎠

⎛ F ⎞ ⎛ 383.372 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ R ⎠ ⎝ 430.66 ⎠

θ = 27.102 deg

Problem 5-63 Determine the x, y, z components of reaction at the fixed wall A.The force F2 is parallel to the z axis and the force F1 is parallel to the y axis. Given: a = 2m

d = 2m

b = 1m

F 1 = 200 N

c = 2.5 m

F 2 = 150 N

394

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Engineering Mechanics - Statics

Chapter 5

Solution: ΣF x = 0;

Ax = 0

ΣF x = 0;

Ay = −F1 Ay = −200 N

ΣF x = 0;

Az = F2 Az = 150 N

ΣΜx = 0;

MAx = −F2 a + F 1 d MAx = 100 N⋅ m

ΣΜy = 0;

MAy = 0

ΣMz = 0;

MAz = F 1 c MAz = 500 N⋅ m

Problem 5-64 The wing of the jet aircraft is subjected to thrust T from its engine and the resultant lift force L. If the mass of the wing is M and the mass center is at G, determine the x, y, z components of reaction where the wing is fixed to the fuselage at A. Units Used: 3

Mg = 10 kg 3

kN = 10 N g = 9.81

m 2

s

395

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Engineering Mechanics - Statics

Chapter 5

Given: T = 8 kN L = 45 kN M = 2.1 Mg a = 2.5 m b = 5m 395

c = 3m d = 7m Solution: ΣF x = 0;

− Ax + T = 0 Ax = T

Ax = 8 kN

ΣF y = 0;

Ay = 0

Ay = 0

ΣF z = 0;

− Az − M g + L = 0 Az = L − M g

ΣMy = 0;

Az = 24.4 kN

M y − T ( a) = 0 My = T a

ΣMx = 0;

My = 20.0 kN⋅ m

L( b + c + d) − M g b − Mx = 0 M x = L ( b + c + d) − M g b

ΣMz = 0;

Mx = 572 kN⋅ m

Mz − T( b + c) = 0 Mz = T( b + c)

Mz = 64.0 kN⋅ m

Problem 5-65 The uniform concrete slab has weight W. Determine the tension in each of the three parallel supporting cables when the slab is held in the horizontal plane as shown.

396

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kip = 10 lb Given: W = 5500 lb a = 6 ft b = 3 ft c = 3 ft d = 6 ft

Solution: Equations of Equilibrium : The cable tension TB can be obtained directly by summing moments about the y axis.

ΣMy = 0;

ΣMx = 0;

W

d 2

− TB d = 0

TC a + TB( a + b) − W⎛⎜

⎝

TC =

ΣF z = 0;

TB =

W 2

TB = 2.75 kip

a + b + c⎞

⎟=0 ⎠

2

1⎡ a + b + c − TB( a + b)⎥⎤ ⎢W a⎣ 2 ⎦

TA + TB + TC − W = 0

TA = − TB − TC + W

TC = 1.375 kip

TA = 1.375 kip

397

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Engineering Mechanics - Statics

Chapter 5

Problem 5-66 The air-conditioning unit is hoisted to the roof of a building using the three cables. If the tensions in the cables are TA, TB and TC, determine the weight of the unit and the location (x, y) of its center of gravity G. Given: TA = 250 lb TB = 300 lb TC = 200 lb a = 5 ft b = 4 ft c = 3 ft d = 7 ft e = 6 ft Solution: ΣF z = 0; TA + TB + TC − W = 0 W = TA + TB + TC ΣMy = 0;

W x − T A ( c + d) − T C d = 0 x =

ΣMx = 0;

W = 750 lb

TA( c + d) + TC d W

x = 5.2 ft

TA a + TB( a + b − e) + TC( a + b) − W y = 0

y =

TA a + TB( a + b − e) + TC( a + b) W

y = 5.267 ft

Problem 5-67 The platform truck supports the three loadings shown. Determine the normal reactions on each of its three wheels. 398

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Engineering Mechanics - Statics

Chapter 5

Given: F 1 = 380 lb F 2 = 500 lb F 3 = 800 lb a = 8 in

b = 12 in c = 10 in d = 5 in e = 12 in f = 12 in

Solution: The initail guesses are

F A = 1 lb

F B = 1 lb

F C = 1 lb

Given ΣMx = 0;

F 1 ( c + d) + F2 ( b + c + d) + F3 d − F A( a + b + c + d) = 0

ΣMy = 0;

F1 e − FB e − F2 f + FC f = 0

ΣF y = 0;

FB + FC − F2 + FA − F1 − F3 = 0

⎛ FA ⎞ ⎜ ⎟ ⎜ FB ⎟ = Find ( FA , FB , FC) ⎜F ⎟ ⎝ C⎠

⎛ FA ⎞ ⎛ 663 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FB ⎟ = ⎜ 449 ⎟ lb ⎜ F ⎟ ⎝ 569 ⎠ ⎝ C⎠

Problem 5-68 Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights WA, WB and WC, determine the normal reactions of the wheels D, E, and F on the ground.

399

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kip = 10 lb Given: WA = 45000 lb WB = 8000 lb WC = 6000 lb a = 8 ft

e = 20 ft

b = 6 ft

f = 4 ft

c = 8 ft

g = 3 ft

d = 6 ft Solution: Initial guesses: R D = 1 kip

R E = 1 kip

R F = 1 kip

Given ΣMx = 0;

WB b − RD( a + b) − WC c + R E( c + d) = 0

ΣMy = 0;

W B f + W A( g + f ) + W C f − R F( e + g + f ) = 0

ΣF z = 0;

R D + R E + RF − WA − WB − WC = 0

⎛ RD ⎞ ⎜ ⎟ ⎜ RE ⎟ = Find ( RD , RE , RF) ⎜R ⎟ ⎝ F⎠

⎛ RD ⎞ ⎛ 22.6 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ RE ⎟ = ⎜ 22.6 ⎟ kip ⎜ R ⎟ ⎝ 13.7 ⎠ ⎝ F⎠

Problem 5-69 If the cable can be subjected to a maximum tension T, determine the maximum force F which may be applied to the plate. Compute the x, y, z components of reaction at the hinge A for this loading.

400

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Engineering Mechanics - Statics

Chapter 5

Given: a = 3 ft b = 2 ft c = 1 ft d = 3 ft e = 9 ft T = 300 lb Solution: Initial guesses: F = 10 lb

MAx = 10 lb ft

MAz = 10 lb ft

Ax = 10 lb

Ay = 10 lb

Az = 10 lb

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ = 0 ⎜ A ⎟ ⎝T − F⎠ ⎝ z⎠

⎛ MAx ⎞ ⎛ a ⎞ ⎛ 0 ⎞ ⎛ e ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ −c ⎟ × ⎜ 0 ⎟ + ⎜ −b − c ⎟ × ⎜ 0 ⎟ = 0 ⎜ MAz ⎟ ⎝ 0 ⎠ ⎝ −F ⎠ ⎝ 0 ⎠ ⎝ T ⎠ ⎝ ⎠

⎛ F ⎞ ⎜ A ⎟ ⎜ x ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( F , Ax , Ay , Az , MAx , MAz) Az ⎜ ⎟ ⎜ MAx ⎟ ⎟ ⎜ ⎝ MAz ⎠

⎛ MAx ⎞ ⎛ 0 ⎞ ⎜ ⎟ = ⎜ ⎟ lb ft ⎝ MAz ⎠ ⎝ 0 ⎠ ⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ lb ⎜ A ⎟ ⎝ 600 ⎠ ⎝ z⎠

F = 900 lb

Problem 5-70 The boom AB is held in equilibrium by a ball-and-socket joint A and a pulley and cord system as shown. Determine the x, y, z components of reaction at A and the tension in cable DEC.

401

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Engineering Mechanics - Statics

Chapter 5

Given:

F

⎛ 0 ⎞ = ⎜ 0 ⎟ lb ⎜ ⎟ ⎝ −1500 ⎠

a = 5 ft b = 4 ft c = b d = 5 ft e = 5 ft f = 2 ft Solution:

α = atan ⎛⎜

a

⎞ ⎟

⎝ d + e⎠

L =

2

a + ( d + e)

β = atan ⎛⎜

2

⎞ ⎟ fL ⎜L − ⎟ d+ e⎠ ⎝

Guesses

b

TBE = 1 lb TDEC = 1 lb

Ax = 1 lb Ay = 1 lb Given

Az = 1 lb

2 TDEC cos ( β ) = TBE

0 ⎞ ⎛0⎞ ⎛ 0 ⎞ ⎛⎜ ⎜ d ⎟ × F + ⎜ d + e ⎟ × −TBE cos ( α ) ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝0⎠ ⎝ 0 ⎠ ⎝ TBE sin ( α ) ⎟⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + F + TBE⎜ −cos ( α ) ⎟ = 0 ⎜A ⎟ ⎝ sin ( α ) ⎠ ⎝ z⎠

402

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Engineering Mechanics - Statics

Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ Az ⎟ = Find ( A , A , A , T , T x y z BE DEC) TDEC = 919 lb ⎜ ⎟ ⎜ TBE ⎟ ⎜T ⎟ ⎝ DEC ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 1.5 × 103 ⎟ lb ⎜A ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 750 ⎠

Problem 5-71 The cable CED can sustain a maximum tension Tmax before it fails. Determine the greatest vertical force F that can be applied to the boom. Also, what are the x, y, z components of reaction at the ball-and-socket joint A? Given: Tmax = 800 lb a = 5 ft b = 4 ft c = b d = 5 ft e = 5 ft f = 2 ft Solution: a ⎞ ⎟ ⎝ d + e⎠

α = atan ⎛⎜ L =

2

a + ( d + e)

β = atan ⎛⎜

2

⎞ fL ⎟ ⎜L − ⎟ d+ e⎠ ⎝

Guesses

b

TBE = 1 lb

Ax = 1 lb Ay = 1 lb Given

TDEC = Tmax

F = 1 lb Az = 1 lb

2 TDEC cos ( β ) = TBE

403

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Engineering Mechanics - Statics

Chapter 5

0 ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ ⎜ d ⎟ × ⎜ 0 ⎟ + ⎜ d + e ⎟ × −TBE cos ( α ) ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝ 0 ⎠ ⎝ −F ⎠ ⎝ 0 ⎠ ⎝ TBE sin ( α ) ⎟⎠

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ = Find ( A , A , A , T , F) x y z BE ⎜ z ⎟ ⎜ TBE ⎟ ⎜ ⎟ ⎝ F ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ + TBE⎜ −cos ( α ) ⎟ = 0 ⎜ A ⎟ ⎝ −F ⎠ ⎝ sin ( α ) ⎠ ⎝ z⎠

F = 1306 lb

0 ⎛ Ax ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 1.306 × 103 ⎟ lb ⎜A ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 653.197 ⎠

Problem 5-72 The uniform table has a weight W and is supported by the framework shown. Determine the smallest vertical force P that can be applied to its surface that will cause it to tip over. Where should this force be applied? Given: W = 20 lb a = 3.5 ft b = 2.5 ft c = 3 ft e = 1.5 ft f = 1 ft

Solution: f θ = atan ⎛⎜ ⎟⎞

θ = 33.69 deg

d = e sin ( θ )

d = 0.832 ft

⎛ a −e⎞ ⎜ 2 ⎟ φ = atan ⎜ ⎟ ⎜ b ⎟ ⎝ 2 ⎠

φ = 11.31 deg

⎝ e⎠

404

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Engineering Mechanics - Statics

2

d' =

⎛ a − e⎞ + ⎛ b ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠

Chapter 5

2

d' = 1.275 ft

Tipping will occur about the g - g axis. Require P to be applied at the corner of the table for Pmin. W d = P d' sin ( 90 deg − φ + θ ) P = W

d

P = 14.1 lb

d' sin ( 90 deg − φ + θ )

Problem 5-73 The windlass is subjected to load W. Determine the horizontal force P needed to hold the handle in the position shown, and the components of reaction at the ball-and-socket joint A and the smooth journal bearing B. The bearing at B is in proper alignment and exerts only force reactions perpendicular to the shaft on the windlass. Given: W = 150 lb a = 2 ft b = 2 ft c = 1 ft d = 1 ft e = 1 ft f = 0.5 ft Solution: ΣMy = 0;

W f−P d=0 P =

W f d

ΣF y = 0;

Ay = 0 lb

ΣMx = 0;

−W a + B z( a + b) = 0

P = 75 lb Ay = 0

405

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Engineering Mechanics - Statics

Bz = ΣF z = 0;

Chapter 5

Wa

B z = 75 lb

a+b

Az + Bz − W = 0 Az = W − Bz

ΣMz = 0;

B x( a + b) − ( a + b + c + e)P = 0 Bx =

ΣF x = 0;

Az = 75 lb

P ( a + b + c + e) a+b

B x = 112 lb

Ax − Bx + P = 0 Ax = Bx − P

Ax = 37.5 lb

Problem 5-74 A ball of mass M rests between the grooves A and B of the incline and against a vertical wall at C. If all three surfaces of contact are smooth, determine the reactions of the surfaces on the ball. Hint: Use the x, y, z axes, with origin at the center of the ball, and the z axis inclined as shown. Given: M = 2 kg

θ 1 = 10 deg θ 2 = 45 deg

Solution: ΣF x = 0; F c cos ( θ 1 ) − M g sin ( θ 1 ) = 0 F c = M g⋅ tan ( θ 1 ) F c = 0.32 kg⋅ m

406

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Engineering Mechanics - Statics

Chapter 5

ΣF y = 0; NA cos ( θ 2 ) − NB cos ( θ 2 ) = 0 NA = NB ΣF z = 0; 2 NA sin ( θ 2 ) − M g cos ( θ 1 ) − F c sin ( θ 1 ) = 0

NA =

1 M g⋅ cos ( θ 1 ) + F c⋅ sin ( θ 1 ) ⋅ 2 sin ( θ 2 )

NA = 1.3 kg⋅ m N = NA = NB

Problem 5-75 Member AB is supported by cable BC and at A by a square rod which fits loosely through the square hole at the end joint of the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the cylinder of weight W in equilibrium. Units Used: 3

kip = 10 lb Given: W = 800 lb a = 2 ft b = 6 ft c = 3 ft Solution:

⎛

⎞=0 ⎝ c +b +a ⎠

ΣF x = 0

F BC ⎜

ΣF y = 0

Ay = 0

c

2

2

2⎟

Ay = 0lb

F BC = 0 lb

Ay = 0 lb

407

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Engineering Mechanics - Statics

Chapter 5

ΣF z = 0

Az − W = 0

ΣMx = 0

MAx − W b = 0

ΣMy = 0 ΣMz = 0

Az = W

Az = 800 lb

MAx = W b

MAx = 4.80 kip⋅ ft

MAy = 0 lb ft

MAy = 0 lb⋅ ft

MAz = 0 lb ft

MAz = 0 lb⋅ ft

Problem 5-76 The pipe assembly supports the vertical loads shown. Determine the components of reaction at the ball-and-socket joint A and the tension in the supporting cables BC and BD. Units Used: 3

kN = 10 N Given: F 1 = 3 kN d = 2 m F 2 = 4 kN e = 1.5 m a = 1m

g = 1m

b = 1.5 m

h = 3m

c = 3m

i = 2m

f = c−e

j = 2m

Solution: The initial guesses are: TBD = 1 kN

TBC = 1 kN

Ax = 1 kN

Ay = 1 kN

Az = 1 kN

The vectors

408

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Engineering Mechanics - Statics

⎛ 0 r1 = ⎜ a + b + ⎜ ⎝ d

rBC

⎞ f⎟ ⎟ ⎠

⎛ i ⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝h − g⎠

uBC =

Given

rBC rBC

Chapter 5

⎛ 0 ⎞ r2 = ⎜ a + b + c ⎟ ⎜ ⎟ ⎝ d ⎠

rBD

⎛ −j ⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝h − g⎠

uBD =

rBD rBD

rAB

⎛0⎞ = ⎜a⎟ ⎜ ⎟ ⎝g⎠

⎛1⎞ i = ⎜0⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ j = ⎜1⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠

Ax i + Ay j + Az k − F1 k − F2 k + TBDuBD + TBC uBC = 0 rAB × ( TBDuBD + TBCuBC) + r1 × ( −F 1 k) + r2 × ( −F 2 k) = 0

⎛ TBD ⎞ ⎜ ⎟ ⎜ TBC ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BD BC x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBD ⎞ ⎛ 17 ⎞ ⎜ ⎟ = ⎜ ⎟ kN ⎝ TBC ⎠ ⎝ 17 ⎠ ⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 11.333 ⎟ kN ⎜ A ⎟ ⎝ −15.667 ⎠ ⎝ z⎠

Problem 5-77 The hatch door has a weight W and center of gravity at G. If the force F applied to the handle at C has coordinate direction angles of α, β and γ, determine the magnitude of F needed to hold the door slightly open as shown. The hinges are in proper alignment and exert only force reactions on the door. Determine the components of these reactions if A exerts only x and z components of force and B exerts x, y, z force components. Given: W = 80 lb

α = 60 deg

409

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Engineering Mechanics - Statics

Chapter 5

β = 45 deg γ = 60 deg a = 3 ft b = 2 ft c = 4 ft d = 3 ft Solution: Initial Guesses: Ax = 1 lb

Az = 1 lb

F = 1 lb

B x = 1 lb

B y = 1 lb

B z = 1 lb

Given

⎛ Ax ⎞ ⎛ Bx ⎞ ⎛⎜ cos ( α ) ⎟⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ By ⎟ + F⎜ cos ( β ) ⎟ + ⎜ 0 ⎟ = 0 ⎜ Az ⎟ ⎜ B ⎟ ⎜ cos ( γ ) ⎟ ⎝ −W ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ z⎠ ⎛ a + b ⎞ ⎢⎡ ⎛⎜ cos ( α ) ⎟⎞⎥⎤ ⎛ a ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎜ 0 ⎟ × F cos ( β ) + ⎜ c ⎟ × ⎜ 0 ⎟ + ⎜ c + d ⎟ × B = 0 ⎟⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎢ ⎜ ⎟ ⎜ y⎟ ⎢ ⎜ ⎟ ⎥ ⎝ 0 ⎠ ⎣ ⎝ cos ( γ ) ⎠⎦ ⎝ 0 ⎠ ⎝ −W ⎠ ⎝ 0 ⎠ ⎜⎝ Bz ⎟⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Az ⎟ ⎜ Bx ⎟ ⎜ ⎟ = Find ( Ax , Az , Bx , By , Bz , F) ⎜ By ⎟ ⎜ Bz ⎟ ⎜ ⎟ ⎝F ⎠

⎛ Ax ⎞ ⎛ −96.5 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Az ⎠ ⎝ −13.7 ⎠ ⎛ Bx ⎞ ⎛ 48.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ By ⎟ = ⎜ −67.9 ⎟ lb ⎜ B ⎟ ⎝ 45.7 ⎠ ⎝ z⎠ F = 96 lb

410

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Engineering Mechanics - Statics

Chapter 5

Problem 5-78 The hatch door has a weight W and center of gravity at G. If the force F applied to the handle at C has coordinate direction angles α, β, γ determine the magnitude of F needed to hold the door slightly open as shown. If the hinge at A becomes loose from its attachment and is ineffective, what are the x, y, z components of reaction at hinge B? Given: W = 80 lb

α = 60 deg β = 45 deg γ = 60 deg a = 3 ft b = 2 ft c = 4 ft d = 3 ft Solution: a

ΣMy = 0;

F = W

ΣF x = 0;

B x + F cos ( α ) = 0

cos ( γ ) ( a + b)

B x = −F cos ( α ) ΣF y = 0;

B y = −67.9 lb

B z − W + F cos ( γ ) = 0 B z = W − F cos ( γ )

ΣMx = 0;

B x = −48 lb

B y + F cos ( β ) = 0 B y = −F cos ( β )

ΣF z = 0;

F = 96 lb

B z = 32 lb

MBx + W d − F cos ( γ ) ( c + d) = 0 MBx = −W d + F cos ( γ ) ( c + d)

ΣMz = 0;

MBx = 96 lb⋅ ft

MBz + F cos ( α ) ( c + d) + F cos ( β ) ( a + b) = 0 MBz = −F cos ( α ) ( c + d) − F cos ( β ) ( a + b)

MBz = −675 lb⋅ ft

411

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Engineering Mechanics - Statics

Chapter 5

Problem 5-79 The bent rod is supported at A, B, and C by smooth journal bearings. Compute the x, y, z components of reaction at the bearings if the rod is subjected to forces F 1 and F 2. F1 lies in the y-z plane. The bearings are in proper alignment and exert only force reactions on the rod. Given: F 1 = 300 lb

d = 3 ft

F 2 = 250 lb

e = 5 ft

a = 1 ft

α = 30 deg

b = 4 ft

β = 45 deg

c = 2 ft

θ = 45 deg

Solution: The initial guesses: Ax = 100 lb

Ay = 200 lb

B x = 300 lb

B z = 400 lb

Cy = 500 lb

Cz = 600 lb

Given Ax + B x + F2 cos ( β ) sin ( α ) = 0 Ay + Cy − F1 cos ( θ ) + F2 cos ( β ) cos ( α ) = 0 B z + Cz − F 1 sin ( θ ) − F2 sin ( β ) = 0 F 1 cos ( θ ) ( a + b) + F 1 sin ( θ ) ( c + d) − B z d − A y b = 0 Ax b + Cz e = 0 Ax( c + d) + Bx d − Cy e = 0

412

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Engineering Mechanics - Statics

Chapter 5

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) x y x z y z ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cy ⎟ ⎜ Cz ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 632.883 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ −141.081 ⎠ ⎛ Bx ⎞ ⎛ −721.271 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Bz ⎠ ⎝ 895.215 ⎠ ⎛ Cy ⎞ ⎛ 200.12 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Cz ⎠ ⎝ −506.306 ⎠

Problem 5-80 The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F2 which will cause the reaction Cy at the bearing C to be equal to zero. The bearings are in proper alignment and exert only force reactions on the rod. Given: F 1 = 300 lb

d = 3 ft

Cy = 0 lb

e = 5 ft

a = 1 ft

α = 30 deg

b = 4 ft

β = 45 deg

c = 2 ft

θ = 45 deg

Solution: The initial guesses: Ax = 100 lb

Ay = 200 lb

B x = 300 lb

B z = 400 lb

F 2 = 500 lb

Cz = 600 lb

Given Ax + B x + F2 cos ( β ) sin ( α ) = 0

413

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Engineering Mechanics - Statics

Chapter 5

Ay + Cy − F1 cos ( θ ) + F2 cos ( β ) cos ( α ) = 0 B z + Cz − F 1 sin ( θ ) − F2 sin ( β ) = 0 F 1 cos ( θ ) ( a + b) + F 1 sin ( θ ) ( c + d) − B z d − A y b = 0 Ax b + Cz e = 0 Ax( c + d) + Bx d − Cy e = 0

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , F ) x y x z z 2 ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cz ⎟ ⎜ F2 ⎟ ⎝ ⎠

F 2 = 673.704 lb

Problem 5-81 Determine the tension in cables BD and CD and the x, y, z components of reaction at the ball-and-socket joint at A. Given: F = 300 N a = 3m b = 1m c = 0.5 m d = 1.5 m Solution:

rBD

⎛ −b ⎞ ⎜ ⎟ = d ⎜ ⎟ ⎝a⎠

414

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Engineering Mechanics - Statics

rCD

Chapter 5

⎛ −b ⎞ = ⎜ −d ⎟ ⎜ ⎟ ⎝a⎠

Initial Guesses:

TBD = 1 N

TCD = 1 N

Ax = 1 N

Ay = 1 N

Az = 1 N

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ rBD rCD + TCD +⎜ 0 ⎟ =0 ⎜ Ay ⎟ + TBD ⎜ ⎟ rBD rCD ⎜A ⎟ ⎝ −F ⎠ ⎝ z⎠ ⎛d⎞ ⎛d⎞ ⎛d − c⎞ ⎛ 0 ⎞ rBD ⎞ ⎜ ⎟ ⎛ rCD ⎞ ⎜ ⎜ −d ⎟ × ⎛ T + d × ⎜ TCD + 0 ⎟×⎜ 0 ⎟ = 0 ⎜ ⎟ ⎟ BD ⎜ ⎟ ⎝ ⎟ ⎜ ⎟ rBD ⎠ ⎜ ⎟ ⎝ rCD ⎠ ⎜ ⎝0⎠ ⎝0⎠ ⎝ 0 ⎠ ⎝ −F ⎠ ⎛ TBD ⎞ ⎜ ⎟ ⎜ TCD ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BD CD x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBD ⎞ ⎛ 116.7 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ TCD ⎠ ⎝ 116.7 ⎠ ⎛ Ax ⎞ ⎛ 66.7 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ N ⎜ A ⎟ ⎝ 100 ⎠ ⎝ z⎠

Problem 5-82 Determine the tensions in the cables and the components of reaction acting on the smooth collar at A necessary to hold the sign of weight W in equilibrium. The center of gravity for the sign is at G. Given: W = 50 lb

f = 2.5 ft

a = 4 ft

g = 1 ft

b = 3 ft

h = 1 ft

415

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Engineering Mechanics - Statics

Chapter 5

c = 2 ft

i = 2 ft

d = 2 ft

j = 2 ft

e = 2.5 ft

k = 3 ft

Solution: The initial guesses are: TBC = 10 lb

Ax = 10 lb

MAx = 10 lb⋅ ft

TDE = 10 lb

Ay = 10 lb

MAy = 10 lb⋅ ft

Given TDE

( a − k)

2

2

( a − k) + i + j −i

TDE 2

( a − k) + i + j

j

TBC 2

2

2

2

TBC

2

2

2

+ Ax = 0

+ Ay = 0

−W=0 2

2

( d − b) + i + c

i 2

2

( a − k) + i + j

MAy − TDE k

2

( d − b) + i + c

+c

( a − k) + i + j

MAx + TDE j

2

( d − b) + i + c

2

TDE 2

2

−i 2

TBC

+ ( −b + d)

2

j 2

2

( a − k) + i + j

2

i

+ c TBC

2

− Wi = 0 2

2

2

2

( d − b) + i + c

d

+ TBC c

2

+ W ( k − f) = 0

( d − b) + i + c

416

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Engineering Mechanics - Statics

Chapter 5

i

−TDE a

2

2

( a − k) + i + j

2

+ TBC b

i 2

=0 2

2

( d − b) + i + c

⎛⎜ MAx ⎞⎟ ⎜ MAy ⎟ ⎜ ⎟ ⎜ TBC ⎟ = Find ( M , M , T , T , A , A ) Ax Ay BC DE x y ⎜ TDE ⎟ ⎜ ⎟ ⎜ Ax ⎟ ⎜ Ay ⎟ ⎝ ⎠

⎛ TBC ⎞ ⎛ 42.857 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ TDE ⎠ ⎝ 32.143 ⎠ ⎛ Ax ⎞ ⎛ 3.571 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ 50 ⎠ ⎛ MAx ⎞ ⎛ 2.698 × 10− 13 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAy ⎠ ⎝ −17.857 ⎠

Problem 5-83 The member is supported by a pin at A and a cable BC. If the load at D is W, determine the x, y, z components of reaction at these supports. Units Used: 3

kip = 10 lb Given: W = 300 lb a = 1 ft b = 2 ft c = 6 ft d = 2 ft e = 2 ft f = 2 ft

417

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Engineering Mechanics - Statics

Chapter 5

Solution: Initial Guesses: TBC = 1 lb

Ax = 1 lb

Ay = 1 lb

Az = 1 lb

MAy = 1 lb ft

MAz = 1 lb ft

Given

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ + ⎜A ⎟ ⎝ z⎠

⎛e + f − a⎞ ⎛ 0 ⎞ ⎜ −c ⎟ + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎟ 2 2 2⎜ b + c + ( e + f − a) ⎝ b ⎠ ⎝ −W ⎠ TBC

⎛ 0 ⎞ ⎛ −e ⎞ ⎛ 0 ⎞ ⎛ −a ⎞ ⎡ ⎛ e + f − a ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ TBC ⎜ −c ⎟⎥ = 0 ⎜ MAy ⎟ + ⎜ c ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎢ ⎜ ⎟⎥ ⎜ MAz ⎟ ⎝ 0 ⎠ ⎝ −W ⎠ ⎝ b ⎠ ⎣ b2 + c2 + ( e + f − a) 2 ⎝ b ⎠⎦ ⎝ ⎠ ⎛⎜ TBC ⎞⎟ ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ = Find ( T , A , A , A , M , M ) BC x y z Ay Az ⎜ Az ⎟ ⎜ ⎟ ⎜ MAy ⎟ ⎜ MAz ⎟ ⎝ ⎠

TBC = 1.05 kip

⎛ Ax ⎞ ⎛ −450 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 900 ⎟ lb ⎜A ⎟ ⎝ 0 ⎠ ⎝ z⎠ ⎛ MAy ⎞ ⎛ −600 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAz ⎠ ⎝ −900 ⎠

Problem 5-84 Determine the x, y, z components of reaction at the pin A and the tension in the cable BC necessary for equilibrium of the rod.

418

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Engineering Mechanics - Statics

Chapter 5

Given: F = 350 lb

e = 12 ft

a = 4 ft

f = 4 ft

b = 5 ft

g = 10 ft

c = 4 ft

h = 4 ft

d = 2 ft

i = 10 ft

Solution: Initial Guesses: F BC = 1 lb Ay = 1 lb MAy = 1 lb⋅ ft Ax = 1 lb Az = 1 lb MAz = 1 lb⋅ ft Given

⎛ 0 ⎞ ⎛d⎞ ⎡ ⎛ g − d ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎢ F ⎜ e − c ⎟⎥ ... = 0 ⎜ MAy ⎟ + ⎜ c ⎟ × ⎢ ⎟⎥ 2 2 2⎜ ⎜ MAz ⎟ ⎝ 0 ⎠ ⎣ ( g − d) + ( e − c) + f ⎝ − f ⎠⎦ ⎝ ⎠ ⎛ −a ⎞ ⎡ ⎛ a − h ⎞⎤ F BC ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ + 0 × ⎜ ⎟ ⎢ ⎟⎥ 2 2 2⎜ ⎝ b ⎠ ⎣ ( a − h) + e + b ⎝ b ⎠⎦ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ + ⎜A ⎟ ⎝ z⎠

⎛g − d⎞ ⎜e− c⎟ + ⎟ 2 2 2⎜ ( g − d) + ( e − c) + f ⎝ − f ⎠ F

⎛a − h⎞ ⎜ −e ⎟ = 0 ⎟ 2 2 2⎜ ( a − h) + e + b ⎝ b ⎠ F BC

419

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Engineering Mechanics - Statics

Chapter 5

⎛⎜ FBC ⎞⎟ ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ = Find ( F , A , A , A , M , M ) BC x y z Ay Az ⎜ Az ⎟ ⎜ ⎟ ⎜ MAy ⎟ ⎜ MAz ⎟ ⎝ ⎠

F BC = 101 lb

⎛ Ax ⎞ ⎛ −233.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ −140 ⎟ lb ⎜ A ⎟ ⎝ 77.8 ⎠ ⎝ z⎠ ⎛ MAy ⎞ ⎛ −388.9 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAz ⎠ ⎝ 93.3 ⎠

Problem 5-85 Rod AB is supported by a ball-and-socket joint at A and a cable at B. Determine the x, y, z components of reaction at these supports if the rod is subjected to a vertical force F as shown. Given: F = 50 lb a = 2 ft

c = 2 ft

b = 4 ft

d = 2 ft

Solution: TB = 10 lb

Ax = 10 lb

Ay = 10 lb

Az = 10 lb

B y = 10 lb Given ΣF x = 0;

− TB + A x = 0

ΣF y = 0;

Ay + By = 0

420

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Engineering Mechanics - Statics

ΣF z = 0;

Chapter 5

−F + A z = 0

ΣMAx = 0;

F ( c) − By( b) = 0

ΣMAy = 0;

F ( a) − TB( b) = 0

ΣMAz = 0;

B y( a) − TB( c) = 0

Solving,

⎛ TB ⎞ ⎜ ⎟ ⎜ Ax ⎟ ⎜ Ay ⎟ = Find ( T , A , A , A , B ) B x y z y ⎜ ⎟ ⎜ Az ⎟ ⎜B ⎟ ⎝ y⎠

⎛ TB ⎞ ⎛ 25 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ax ⎟ ⎜ 25 ⎟ ⎜ Ay ⎟ = ⎜ −25 ⎟ lb ⎜ ⎟ ⎜ ⎟ ⎜ Az ⎟ ⎜ 50 ⎟ ⎜ B ⎟ ⎝ 25 ⎠ ⎝ y⎠

Problem 5-86 The member is supported by a square rod which fits loosely through a smooth square hole of the attached collar at A and by a roller at B. Determine the x, y, z components of reaction at these supports when the member is subjected to the loading shown. Given: M = 50 lb⋅ ft

⎛ 20 ⎞ ⎜ ⎟ F = −40 lb ⎜ ⎟ ⎝ −30 ⎠ a = 2 ft b = 1 ft c = 2 ft Solution: Initial Guesses Ax = 1 lb

Ay = 1 lb

421

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Engineering Mechanics - Statics

Chapter 5

MAx = 1 lb ft

MAy = 1 lb ft

MAz = 1 lb ft

B z = 1 lb

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ + F = 0 ⎜ 0 ⎟ ⎜ Bz ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ MAx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎡⎛ 0 ⎞ ⎛ 0 ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎜ MAy ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟ + ⎢⎜ a + b ⎟ × F + ⎜ 0 ⎟⎥ = 0 ⎜ M ⎟ ⎝ 0 ⎠ ⎜ Bz ⎟ ⎣⎝ −c ⎠ ⎝ −M ⎠⎦ ⎝ ⎠ ⎝ Az ⎠ ⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ MAx ⎟ = Find ( A , A , M , M , M , B ) x y Ax Ay Az z ⎜ MAy ⎟ ⎜ ⎟ ⎜ MAz ⎟ ⎜ Bz ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ −20 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ 40 ⎠ B z = 30 lb

⎛ MAx ⎞ ⎛ 110 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ MAy ⎟ = ⎜ 40 ⎟ lb⋅ ft ⎜ M ⎟ ⎝ 110 ⎠ ⎝ Az ⎠

Problem 5-87 The platform has mass M and center of mass located at G. If it is lifted using the three cables, determine the force in each of these cables. Units Used: 3

Mg = 10 kg

3

kN = 10 N

g = 9.81

m 2

s

422

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Engineering Mechanics - Statics

Chapter 5

Given: M = 3 Mg a = 4m b = 3m c = 3m d = 4m e = 2m Solution: The initial guesses are: F AC = 10 N

F BC = 10 N

F DE = 10 N

Given b ( F AC) 2

2

a +b

−

c ( F BC) 2

=0

2

a +c

M g e − ( F AC) a

d+e 2

2

− FBC

a +b a 2

2

a( d + e) 2

=0

2

a +c

F BC( b + c) − M g b + FDE b = 0

a +c

⎛ FAC ⎞ ⎜ ⎟ F ⎜ BC ⎟ = Find ( FAC , FBC , FDE) ⎜F ⎟ ⎝ DE ⎠

⎛ FAC ⎞ ⎜ ⎟ F ⎜ BC ⎟ = kN ⎜F ⎟ ⎝ DE ⎠

423

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Engineering Mechanics - Statics

Chapter 5

Problem 5-88 The platform has a mass of M and center of mass located at G. If it is lifted using the three cables, determine the force in each of the cables. Solve for each force by using a single moment equation of equilibrium. Units Used: Mg = 1000 kg 3

kN = 10 N g = 9.81

m 2

s Given: M = 2 Mg

c = 3m

a = 4m

d = 4m

b = 3m

e = 2m

Solution:

rBC

⎛0⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝a⎠

rAD

⎛ −e − d ⎞ = ⎜ b ⎟ ⎜ ⎟ ⎝ 0 ⎠

rAC

⎛0⎞ = ⎜b⎟ ⎜ ⎟ ⎝a⎠

rBD

⎛ −d − e ⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝ 0 ⎠

First find FDE. ΣMy' = 0;

F DE( d + e) − M g d = 0

Next find FBC.

Guess

⎡⎛ e ⎞ ⎛ 0 Given ⎢⎜ 0 ⎟ × ⎜ 0 ⎢⎜ ⎟ ⎜ ⎣⎝ 0 ⎠ ⎝ −M

F DE =

Mgd d+e

2

F DE = 4.1 s kN

F BC = 1 kN

⎞ ⎛e + d⎞ rBC ⎟ + ⎜ c ⎟ × ⎛F ⎜ BC ⎟ ⎜ ⎟ ⎝ rBC g⎠ ⎝ 0 ⎠

⎤ ⎞⎥ ⎟⎥ rAD = 0 ⎠ ⎦

F BC = Find ( FBC )

F BC = kN Now find FAC.

Guess

F AC = 1 kN

424

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Engineering Mechanics - Statics

⎡⎛ e ⎞ ⎛ 0 Given ⎢⎜ 0 ⎟ × ⎜ 0 ⎢⎜ ⎟ ⎜ ⎣⎝ 0 ⎠ ⎝ −M

Chapter 5

⎞ ⎛e + d⎞ rAC ⎟ + ⎜ −b ⎟ × ⎛ F ⎜ AC ⎟ ⎜ ⎟ ⎝ rAC g⎠ ⎝ 0 ⎠

⎤ ⎞⎥ ⎟⎥ rBD = 0 ⎠ ⎦

F AC = Find ( FAC )

F AC = kN

Problem 5-89 The cables exert the forces shown on the pole. Assuming the pole is supported by a ball-and-socket joint at its base, determine the components of reaction at A. The forces F 1 and F2 lie in a horizontal plane. Given: F 1 = 140 lb F 2 = 75 lb

θ = 30 deg a = 5 ft b = 10 ft c = 15 ft

Solution: The initial guesses are TBC = 100 lb

TBD = 100 lb

Ax = 100 lb

Ay = 100 lb

Az = 100 lb

Given c ⎞ − T a⎛ ⎞=0 BD ⎜ ⎟ ⎟ 2 2 2 2 2 ⎝ a +b +c ⎠ ⎝ a +c ⎠

(F1 cos (θ ) + F2)c − TBC a⎛⎜

c

⎞=0 ⎝ a +b +c ⎠

F 1 sin ( θ ) c − b TBC ⎛ ⎜

c

2

2

2⎟

425

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Engineering Mechanics - Statics

⎛

Chapter 5

⎞ ⎟=0 2 2 2 a + b + c ⎝ ⎠

Ax + F 1 sin ( θ ) − b⎜

TBC

TBC ⎞ ⎞ + a⎛ =0 ⎜ ⎟ ⎟ 2 2 2 2 2 + c a + b + c a ⎝ ⎠ ⎝ ⎠

⎛

Ay − F 1 cos ( θ ) − F 2 + TBD⎜

a

TBC ⎛ TBD ⎞ ⎛ ⎞ − c =0 ⎟ ⎜ ⎟ 2 2 2 2 2 a + c a + b + c ⎝ ⎠ ⎝ ⎠

Az − c⎜

⎛ TBC ⎞ ⎜ ⎟ ⎜ TBD ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BC BD x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBC ⎞ ⎛ 131.0 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ TBD ⎠ ⎝ 509.9 ⎠ ⎛ Ax ⎞ ⎛ −0.0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0.0 ⎟ lb ⎜ A ⎟ ⎝ 588.7 ⎠ ⎝ z⎠

Problem 5-90 The silo has a weight W, a center of gravity at G and a radius r. Determine the vertical component of force that each of the three struts at A, B, and C exerts on the silo if it is subjected to a resultant wind loading of F which acts in the direction shown. Given: W = 3500 lb F = 250 lb

θ 1 = 30 deg θ 2 = 120 deg θ 3 = 30 deg r = 5 ft b = 12 ft c = 15 ft

426

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Engineering Mechanics - Statics

Chapter 5

Solution: Initial Guesses:

Az = 1 lb

B z = 2 lb

Cz = 31 lb

Given ΣMy = 0; B z r cos ( θ 1 ) − Cz r cos ( θ 1 ) − F sin ( θ 3 ) c = 0

[1]

ΣMx = 0; −B z r sin ( θ 1 ) − Cz r sin ( θ 1 ) + Az r − F cos ( θ 3 ) c = 0

[2]

ΣF z = 0; Az + Bz + Cz = W

[3]

Solving Eqs.[1], [2] and [3] yields:

⎛ Az ⎞ ⎜ ⎟ ⎜ Bz ⎟ = Find ( Az , Bz , Cz) ⎜C ⎟ ⎝ z⎠

⎛ Az ⎞ ⎛ 1600 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Bz ⎟ = ⎜ 1167 ⎟ lb ⎜ C ⎟ ⎝ 734 ⎠ ⎝ z⎠

Problem 5-91 The shaft assembly is supported by two smooth journal bearings A and B and a short link DC. If a couple moment is applied to the shaft as shown, determine the components of force reaction at the bearings and the force in the link. The link lies in a plane parallel to the y-z plane and the bearings are properly aligned on the shaft. Units Used: 3

kN = 10 N Given: M = 250 N⋅ m a = 400 mm b = 300 mm c = 250 mm d = 120 mm

θ = 30 deg 427

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 5

φ = 20 deg Solution: Initial Guesses: Ay = 1 kN

Az = 1 kN

B z = 1 kN

F CD = 1 kN

B y = 1 kN

Given 0 ⎞ ⎛0 ⎞ ⎛0⎞ ⎛ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ By ⎟ + ⎜ −FCD cos ( φ ) ⎟ = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ F sin ( φ ) ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ CD ⎠ 0 ⎞ ⎛ −a − b ⎞ ⎛ 0 ⎞ ⎛ −M ⎞ ⎛ d − a ⎞ ⎛⎜ ⎜ c sin ( θ ) ⎟ × −FCD cos ( φ ) ⎟ + ⎜ 0 ⎟ × ⎜ By ⎟ + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ c cos ( θ ) ⎠ ⎝ FCD sin ( φ ) ⎠ ⎝ 0 ⎠ ⎜⎝ Bz ⎟⎠ ⎝ 0 ⎠

⎛ Ay ⎞ ⎜ ⎟ A z ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , B , F ) y z y z CD ⎜ ⎟ ⎜ Bz ⎟ ⎜F ⎟ ⎝ CD ⎠

⎛ Ay ⎞ ⎛ 573 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ Az ⎠ ⎝ −208 ⎠ ⎛ By ⎞ ⎛ 382 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ Bz ⎠ ⎝ −139 ⎠ F CD = 1.015 kN

Problem 5-92 If neither the pin at A nor the roller at B can support a load no greater than F max, determine the maximum intensity of the distributed load w, so that failure of a support does not occur.

428

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: F max = 6 kN a = 3m b = 3m

Solution: The greatest reaction is at A. Require ΣMB = 0;

−F max( a + b) + w a

w =

F max( a + b) a⎛⎜

a

⎝2

+ b⎞⎟ +

⎠

2

⎛ a + b⎞ + 1 w b 2 b = 0 ⎜ ⎟ 3 ⎝2 ⎠ 2 w = 2.18

b

kN m

3

Problem 5-93 If the maximum intensity of the distributed load acting on the beam is w, determine the reactions at the pin A and roller B. Units Used: 3

kN = 10 N Given: F = 6 kN a = 3m b = 3m

429

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Engineering Mechanics - Statics

w = 4

Chapter 5

kN m

Solution: ΣF x = 0;

Ax = 0

ΣMA = 0;

−w a

By =

a 2 1 6

−

1 2

w b⎛⎜ a +

⎝

⎟ + By( a + b) = 0

3⎠

2

w

3a + 3 a b + b

Ay + By − w a −

ΣF y = 0;

b⎞

2

B y = 7 kN

a+b 1 2

Ay = −By + w a +

wb=0 1 2

Ay = 11 kN

wb

Problem 5-94 Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member. Units Used: 3

kN = 10 N Given: F 1 = 10 kN F 2 = 6 kN a = 0.6 m b = 0.6 m c = 0.8 m d = 0.4 m

θ = 60 deg

430

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Engineering Mechanics - Statics

Chapter 5

Solution: Initial Guesses: NA = 1 kN B x = 1 kN B y = 1 kN Given B x − F2 sin ( θ ) = 0 B y + NA − F 1 − F 2 cos ( θ ) = 0 F 2 d + F1 ⎡⎣b + ( c + d)cos ( θ )⎤⎦ − NA⎡⎣a + b + ( c + d)cos ( θ )⎤⎦ = 0

⎛ NA ⎞ ⎜ ⎟ ⎜ Bx ⎟ = Find ( NA , Bx , By) ⎜B ⎟ ⎝ y⎠

⎛ NA ⎞ ⎛ 8 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 5.196 ⎟ kN ⎜B ⎟ ⎝ 5 ⎠ ⎝ y⎠

Problem 5-95 The symmetrical shelf is subjected to uniform pressure P. Support is provided by a bolt (or pin) located at each end A and A' and by the symmetrical brace arms, which bear against the smooth wall on both sides at B and B'. Determine the force resisted by each bolt at the wall and the normal force at B for equilibrium. Units Used: 3

kPa = 10 Pa Given: P = 4 kPa a = 0.15 m b = 0.2 m c = 1.5 m Solution: ΣMA = 0;

431

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Engineering Mechanics - Statics

NB a − P ⎛⎜ b

Chapter 5

c⎞ b

⎟ =0 ⎝ 2⎠ 2 2

NB = P

b c

NB = 400 N

4a

ΣF x = 0; Ax = NB

Ax = 400 N

ΣF y = 0; Ay = P b

FA =

c

Ay = 600 N

2 2

Ax + Ay

2

F A = 721 N

Problem 5-96 A uniform beam having a weight W supports a vertical load F . If the ground pressure varies linearly as shown, determine the load intensities w1 and w2 measured in lb/ft, necessary for equilibrium. Given: W = 200 lb F = 800 lb a = 7 ft b = 6 ft Solution: Initial Guesses: w1 = 1

lb ft

w2 = 1

lb ft

432

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Engineering Mechanics - Statics

Chapter 5

Given 1

w1 ( a + b) +

w1 ( a + b)

2

(w2 − w1)( a + b) − F − W = 0

a+b 2

+

(w2 − w1)( a + b) 3 ( a + b) − W 2 1

⎛ w1 ⎞ ⎜ ⎟ = Find ( w1 , w2) ⎝ w2 ⎠

2

a+b 2

− Fa = 0

⎛ w1 ⎞ ⎛ 62.7 ⎞ lb ⎜ ⎟=⎜ ⎟ ⎝ w2 ⎠ ⎝ 91.1 ⎠ ft

Problem 5-97 The uniform ladder rests along the wall of a building at A and on the roof at B. If the ladder has a weight W and the surfaces at A and B are assumed smooth, determine the angle θ for equilibrium. Given: a = 18 ft W = 25 lb

θ 1 = 40 deg Solution: Initial guesses: R A = 10 lb R B = 10 lb

θ = 10 deg Given ΣMB = 0;

−R A a sin ( θ ) + W

ΣF x = 0;

R A − RB sin ( θ 1 ) = 0

ΣF y = 0;

R B cos ( θ 1 ) − W = 0

a 2

cos ( θ ) = 0

Solving,

433

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Engineering Mechanics - Statics

Chapter 5

⎛ RB ⎞ ⎜ ⎟ ⎜ RA ⎟ = Find ( RB , RA , θ ) ⎜ ⎟ ⎝θ ⎠

⎛ RA ⎞ ⎛ 21 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ RB ⎠ ⎝ 32.6 ⎠

θ = 30.8 deg

Problem 5-98 Determine the x, y, z components of reaction at the ball supports B and C and the ball-and-socket A (not shown) for the uniformly loaded plate. Given: P = 2

lb ft

2

a = 4 ft b = 1 ft c = 2 ft d = 2 ft

Solution: The initial guesses are

Ax = 1 lb

Ay = 1 lb

Az = 1 lb

B z = 1 lb

Cz = 1 lb

Given ΣF x = 0;

Ax = 0

ΣF y = 0;

Ay = 0

ΣF z = 0; Az + Bz + Cz − P a c = 0 c⎞ ⎟ + Cz b = 0 ⎝ 2⎠

ΣMx = 0; c B z − P a c⎛⎜

a⎞ ⎟ − Cz a = 0 ⎝ 2⎠

ΣMy = 0; −B z ( a − d) + P a c⎛⎜

434

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Engineering Mechanics - Statics

Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ Az ⎟ = Find ( A , A , A , B , C ) x y z z z ⎜ ⎟ ⎜ Bz ⎟ ⎜C ⎟ ⎝ z⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ lb ⎜ A ⎟ ⎝ 5.333 ⎠ ⎝ z⎠

⎛ Bz ⎞ ⎛ 5.333 ⎞ ⎜ ⎟=⎜ ⎟ lb 5.333 C ⎝ ⎠ z ⎝ ⎠

Problem 5-99 A vertical force F acts on the crankshaft. Determine the horizontal equilibrium force P that must be applied to the handle and the x, y, z components of force at the smooth journal bearing A and the thrust bearing B. The bearings are properly aligned and exert the force reactions on the shaft. Given: F = 80 lb a = 10 in b = 14 in c = 14 in d = 8 in e = 6 in f = 4 in

Solution: ΣMy = 0;

P d−F a= 0 P = F

ΣMx = 0;

P = 100 lb

B z( b + c) − F c = 0 Bz = F

ΣMz = 0;

⎛ a⎞ ⎜ ⎟ ⎝ d⎠ ⎛ c ⎞ ⎜ ⎟ ⎝ b + c⎠

B z = 40 lb

−B x( b + c) − P ( e + f) = 0

435

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Engineering Mechanics - Statics

B x = −P ΣF x = 0;

⎛ e + f ⎞ B = −35.7 lb ⎜ ⎟ x ⎝ b + c⎠

Ax + Bx − P = 0 Ax = −Bx + P

Ax = 135.7 lb By = 0

ΣF y = 0; ΣF z = 0;

Chapter 5

Az + Bz − F = 0 Az = −B z + F

Az = 40 lb

Problem 5-100 The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is kA , determine the required stiffness of the spring at B so that if the beam is loaded with the force F , it remains in the horizontal position both before and after loading. Units Used: 3

kN = 10 N Given: kA = 5

kN m

F = 800 N

a = 1m b = 2m

Solution: Equilibrium: ΣMA = 0;

F B ( a + b) − F a = 0 FB = F

⎛ a ⎞ ⎜ ⎟ ⎝ a + b⎠

F B = 266.667 N ΣMB = 0;

F b − FA( a + b) = 0 436

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Engineering Mechanics - Statics

Chapter 5

FA = F

⎛ b ⎞ ⎜ ⎟ ⎝ a + b⎠

F A = 533.333 N Spring force formula: xA = xB

FA kA

=

FB kB

kB =

FB FA

kA

kB = 2.5

kN m

437

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Problem 6-1 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 7 kN P 2 = 7 kN Solution:

θ = 45 deg Initial Guesses: F AB = 1 kN F DC = 1 kN

F AD = 1 kN

F DB = 1 kN

F CB = 1 kN

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint D:

F DB cos ( θ ) − F AD cos ( θ ) + F DC cos ( θ ) = 0

(FAD + FDB − FDC)sin(θ ) − P2 = 0 Joint C:

F CB + FDC sin ( θ ) = 0

⎛⎜ FAB ⎞⎟ ⎜ FAD ⎟ ⎜ ⎟ ⎜ FDB ⎟ = Find ( FAB , FAD , FDB , FDC , FCB) ⎜F ⎟ ⎜ DC ⎟ ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ 7 ⎞ ⎜ FAD ⎟ ⎜ −9.9 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDB ⎟ = ⎜ 4.95 ⎟ kN ⎜ F ⎟ ⎜ −14.85 ⎟ ⎜ DC ⎟ ⎜ ⎟ 10.5 ⎝ ⎠ ⎜ FCB ⎟ ⎝ ⎠

Positive means Tension, Negative means Compression

438

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Engineering Mechanics - Statics

Chapter 6

Problem 6-2 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 8 kN P 2 = 10 kN Solution:

θ = 45 deg Initial Guesses: F AB = 1 kN F DC = 1 kN

F AD = 1 kN

F DB = 1 kN

F CB = 1 kN

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint D:

F DB cos ( θ ) − F AD cos ( θ ) + F DC cos ( θ ) = 0

(FAD + FDB − FDC)sin(θ ) − P2 = 0 Joint C:

F CB + FDC sin ( θ ) = 0

⎛⎜ FAB ⎞⎟ ⎜ FAD ⎟ ⎜ ⎟ F ⎜ DB ⎟ = Find ( FAB , FAD , FDB , FDC , FCB) ⎜F ⎟ ⎜ DC ⎟ ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ 8 ⎞ ⎜ FAD ⎟ ⎜ −11.31 ⎟ ⎟ ⎜ ⎟ ⎜ F 7.07 = ⎟ kN ⎜ DB ⎟ ⎜ ⎜ F ⎟ ⎜ −18.38 ⎟ ⎜ DC ⎟ ⎜ ⎟ ⎜ FCB ⎟ ⎝ 13 ⎠ ⎝ ⎠

Positive means Tension, Negative means Compression

439

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Engineering Mechanics - Statics

Chapter 6

Problem 6-3 The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Units Used: 3

kip = 10 lb Given: P 1 = 600 lb P 2 = 400 lb a = 4 ft

θ = 45 deg Solution: Initial Guesses F AB = 1 lb

F AD = 1 lb

F DC = 1 lb

F BC = 1 lb

F BD = 1 lb

F DE = 1 lb

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint B:

F BC − F AB = 0 −P 2 − F BD = 0

Joint D:

(FDC − FAD)cos (θ ) + FDE = 0 (FDC + FAD)sin(θ ) + FBD = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AD BC BD DC DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDC ⎟ ⎜ ⎟ ⎝ FDE ⎠

440

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Engineering Mechanics - Statics

⎛ FAB ⎞ ⎜ ⎟ ⎛ 600 ⎞ F AD ⎜ ⎟ ⎜ −849 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BC ⎟ = 600 ⎟ lb ⎜ FBD ⎟ ⎜ −400 ⎟ ⎟ ⎜ ⎟ ⎜ 1414 ⎜ ⎟ ⎜ FDC ⎟ ⎜ ⎜ ⎟ ⎝ −1600 ⎟⎠ F ⎝ DE ⎠

Chapter 6

Positive means Tension, Negative means Compression

Problem 6-4 The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Units Used: 3

kip = 10 lb Given: P 1 = 800 lb P 2 = 0 lb a = 4 ft

θ = 45 deg Solution: Initial Guesses F AB = 1 lb

F AD = 1 lb

F DC = 1 lb

F BC = 1 lb

F BD = 1 lb

F DE = 1 lb

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint B:

F BC − F AB = 0 −P 2 − F BD = 0

441

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Joint D:

Chapter 6

(FDC − FAD)cos (θ ) + FDE = 0 (FDC + FAD)sin(θ ) + FBD = 0

⎛ FAB ⎞ ⎜ ⎟ F ⎜ AD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AD BC BD DC DE) ⎜ FBD ⎟ ⎜ ⎟ F ⎜ DC ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 800 ⎞ ⎜ FAD ⎟ ⎜ −1131 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BC ⎟ = 800 ⎟ lb ⎜ FBD ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDC ⎟ ⎜ 1131 ⎟ ⎜ ⎟ ⎜⎝ −1600 ⎟⎠ F ⎝ DE ⎠

Positive means Tension, Negative means Compression

Problem 6-5 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 20 kN P 2 = 10 kN a = 1.5 m e = 2m Solution:

e θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

442

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Engineering Mechanics - Statics

Chapter 6

Initial Guesses: F AB = 1 kN F AG = 1 kN F CF = 1 kN F BC = 1 kN F BG = 1 kN F DE = 1 kN F CG = 1 kN F FG = 1 kN F EF = 1 kN F CD = 1 kN F DF = 1 kN Given Joint B

F BC − F AB cos ( θ ) = 0 −F BG − F AB sin ( θ ) = 0

Joint G

F FG + F CG cos ( θ ) − F AG = 0 F CG sin ( θ ) + FBG − P1 = 0

Joint C

(

)

−F BC + F CD + FCF − F CG cos ( θ ) = 0

(

)

− FCG + FCF sin ( θ ) = 0 Joint D

−F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0

Joint F

F EF − F FG − F CF cos ( θ ) = 0 F DF + F CF sin ( θ ) − P 2 = 0

Joint E

−F DE cos ( θ ) − F EF = 0

443

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCG ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FAG ⎟ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBC , FCG , FCD , FAG , FBG , FFG , FDF , FCF , FDE , FEF) ⎜F ⎟ ⎜ FG ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛⎜ −21.88 ⎟⎞ ⎜ FBC ⎟ ⎜ −13.13 ⎟ kN ⎜ ⎟=⎜ ⎟ F 3.13 CG ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ −9.37 ⎠ ⎝ CD ⎠

⎛ FAG ⎞ ⎜ ⎟ ⎛⎜ 13.13 ⎞⎟ ⎜ FBG ⎟ ⎜ 17.5 ⎟ kN ⎜ ⎟=⎜ ⎟ F 11.25 FG ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ 12.5 ⎠ ⎝ DF ⎠

⎛ FCF ⎞ ⎛ −3.13 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ DE ⎟ = ⎜ −15.62 ⎟ kN ⎜ F ⎟ ⎝ 9.37 ⎠ ⎝ EF ⎠

Positive means Tension, Negative means Compression

Problem 6-6 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 40 kN P 2 = 20 kN a = 1.5 m e = 2m

444

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

e θ = atan ⎛⎜ ⎞⎟

Solution:

⎝ a⎠

Initial Guesses: F AB = 1 kN F AG = 1 kN F CF = 1 kN F BC = 1 kN F BG = 1 kN F DE = 1 kN F CG = 1 kN F FG = 1 kN F EF = 1 kN F CD = 1 kN F DF = 1 kN Given Joint B

F BC − F AB cos ( θ ) = 0 −F BG − F AB sin ( θ ) = 0

Joint G

F FG + F CG cos ( θ ) − F AG = 0 F CG sin ( θ ) + FBG − P1 = 0

Joint C

(

)

−F BC + F CD + FCF − F CG cos ( θ ) = 0

(

)

− FCG + FCF sin ( θ ) = 0 Joint D

−F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0

Joint F

F EF − F FG − F CF cos ( θ ) = 0 F DF + F CF sin ( θ ) − P 2 = 0

Joint E

−F DE cos ( θ ) − F EF = 0

445

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCG ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FAG ⎟ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBC , FCG , FCD , FAG , FBG , FFG , FDF , FCF , FDE , FEF) ⎜F ⎟ ⎜ FG ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛⎜ −43.75 ⎟⎞ F ⎜ BC ⎟ ⎜ −26.25 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCG ⎟ ⎜ 6.25 ⎟ ⎜ F ⎟ ⎝ −18.75 ⎠ ⎝ CD ⎠

⎛ FAG ⎞ ⎜ ⎟ ⎛⎜ 26.25 ⎞⎟ F ⎜ BG ⎟ ⎜ 35 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FFG ⎟ ⎜ 22.5 ⎟ ⎜ F ⎟ ⎝ 25 ⎠ ⎝ DF ⎠

⎛ FCF ⎞ ⎛ −6.25 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −31.25 ⎟ kN ⎜ F ⎟ ⎝ 18.75 ⎠ ⎝ EF ⎠

Positive means Tension, Negative means Compression

Problem 6-7 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 3 kN F 2 = 8 kN F 3 = 4 kN F 4 = 10 kN

446

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Engineering Mechanics - Statics

Chapter 6

a = 2m b = 1.5 m b θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ a⎠

Initial Guesses F BA = 1 kN

F BC = 1 kN

F AF = 1 kN

F CD = 1 kN

F DF = 1 kN

F ED = 1 kN

F AC = 1 kN F CF = 1 kN F EF = 1 kN

Given Joint B

F 1 + F BC = 0 −F 2 − F BA = 0

Joint C

F CD − F BC − F AC cos ( θ ) = 0 −F 3 − F AC sin ( θ ) − FCF = 0

Joint E

−F EF = 0

Joint D

−F CD − F DF cos ( θ ) = 0 −F 4 − F DF sin ( θ ) − FED = 0

Joint F

−F AF + F EF + F DF cos ( θ ) = 0 F CF + FDF sin ( θ ) = 0

⎛ FBA ⎞ ⎜ ⎟ ⎜ FAF ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( FBA , FAF , FDF , FBC , FCD , FED , FAC , FCF , FEF) ⎜F ⎟ ⎜ ED ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎟ ⎜ ⎝ FEF ⎠

447

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ −8 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 4.167 ⎟ ⎜ DF ⎟ ⎜ 5.208 ⎟ ⎜ FBC ⎟ ⎜ −3 ⎟ ⎟ ⎜ ⎟ ⎜ F − 4.167 = ⎟ kN ⎜ CD ⎟ ⎜ ⎜ F ⎟ ⎜ −13.125 ⎟ ⎟ ⎜ ED ⎟ ⎜ ⎜ FAC ⎟ ⎜ −1.458 ⎟ ⎜ ⎟ ⎜ −3.125 ⎟ F ⎜ CF ⎟ ⎜ ⎟ ⎟ ⎝ 0 ⎠ ⎜ ⎝ FEF ⎠

Positive means tension, Negative means compression.

Problem 6-8 Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

Solution: ΣMA = 0;

−P a + Cy2a − P a = 0 Cy = P

Joint C: ΣF x = 0;

1

ΣF y = 0;

P+

2

F BC − 1 17

4 17

F CD −

FCD = 0 1 2

F BC = 0

448

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

F BC =

4 2P

F CD =

17 P

1

FCD +

3

3

= 1.886 P (C) = 1.374 P (T)

Joint B: ΣF x = 0;

P−

ΣFy = 0;

1

2

2

F CD +

2

2P

F AB = F BD =

1

3 5P 3

1 2

FAB = 0

F AB − F BD = 0

= 0.471P

= 1.667P

( C)

( T)

Joint D: ΣF x = 0;

F DA = F CD = 1.374P

(T)

Problem 6-9 The maximum allowable tensile force in the members of the truss is Tmax, and the maximum allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads that can be applied to the truss. Given: Tmax = 1500 lb Cmax = 800 lb Solution: Set

P = 1 lb

Initial Guesses F AB = 1 lb

F AD = 1 lb

F BC = 1 lb

F CD = 1 lb

F BD = 1 lb

449

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given Joint B

(FBC − FAB)

1

+P=0

2

(

−F BD − FAB + FBC Joint D

(FCD − FAD)

4

)

1

=0

2

=0

17

(

)

F BD − P − F AD + F CD Joint C

−F BC

1 2

− F CD

4

1

=0

17

=0

17

⎛⎜ FAB ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FAD ⎟ = Find ( FAB , FBC , FAD , FCD , FBD) ⎜F ⎟ ⎜ CD ⎟ ⎜ FBD ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ −0.471 ⎞ ⎜ FBC ⎟ ⎜ −1.886 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAD ⎟ = ⎜ 1.374 ⎟ lb ⎜ F ⎟ ⎜ 1.374 ⎟ ⎜ CD ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎝ 1.667 ⎠ ⎝ ⎠

Now find the critical load P1 = P

Tmax

(

max F AB , F BC , F AD , F CD , F BD

P2 = P

)

P 1 = 900 lb

)

P 2 = 424.264 lb

Cmax

(

min F AB , F BC , F AD , F CD , F BD

(

)

P = min P 1 , P 2

P = 424.3 lb

Problem 6-10 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 0 lb

450

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

P 2 = 1000 lb a = 10 ft b = 10 ft Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AG = 1 lb

F BG = 1 lb

F BC = 1 lb

F DC = 1 lb

F DE = 1 lb

F EG = 1 lb

F EC = 1 lb

F CG = 1 lb

Given Joint B

F BC − F AB = 0 F BG − P 1 = 0

Joint G

(FCG − FAG)cos (θ ) + FEG = 0 (

)

− FCG + FAG sin ( θ ) − FBG = 0 Joint C

F DC − FBC − FCG cos ( θ ) = 0 F EC + F CG sin ( θ ) − P2 = 0

Joint E

F DE cos ( θ ) − F EG = 0 −F EC − F DE sin ( θ ) = 0

Joint D

−F DE cos ( θ ) − F DC = 0

451

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ FAG ⎟ ⎜ ⎟ F ⎜ DC ⎟ = Find ( FAB , FBC , FEG , FAG , FDC , FEC , FBG , FDE , FCG) ⎜F ⎟ ⎜ EC ⎟ ⎜ FBG ⎟ ⎜ ⎟ F ⎜ DE ⎟ ⎟ ⎜ ⎝ FCG ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 333 ⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 333 ⎟ ⎜ EG ⎟ ⎜ −667 ⎟ ⎜ FAG ⎟ ⎜ −471 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDC ⎟ = ⎜ 667 ⎟ lb ⎜ F ⎟ ⎜ 667 ⎟ ⎟ ⎜ EC ⎟ ⎜ ⎜ FBG ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ −943 ⎟ ⎜ FDE ⎟ ⎜ ⎟ ⎟ ⎝ 471 ⎠ ⎜ ⎝ FCG ⎠

Positive means tension, Negative means compression.

Problem 6-11 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 500 lb P 2 = 1500 lb

452

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Engineering Mechanics - Statics

Chapter 6

a = 10 ft b = 10 ft Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AG = 1 lb

F BG = 1 lb

F BC = 1 lb

F DC = 1 lb

F DE = 1 lb

F EG = 1 lb

F EC = 1 lb

F CG = 1 lb

Given Joint B

F BC − F AB = 0 F BG − P 1 = 0

Joint G

(FCG − FAG)cos (θ ) + FEG = 0 (

)

− FCG + FAG sin ( θ ) − FBG = 0 Joint C

F DC − FBC − FCG cos ( θ ) = 0 F EC + F CG sin ( θ ) − P2 = 0

Joint E

F DE cos ( θ ) − F EG = 0 −F EC − F DE sin ( θ ) = 0

Joint D

−F DE cos ( θ ) − F DC = 0

453

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ FAG ⎟ ⎜ ⎟ F ⎜ DC ⎟ = Find ( FAB , FBC , FEG , FAG , FDC , FEC , FBG , FDE , FCG) ⎜F ⎟ ⎜ EC ⎟ ⎜ FBG ⎟ ⎜ ⎟ F ⎜ DE ⎟ ⎟ ⎜ ⎝ FCG ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 833 ⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 833 ⎟ ⎜ EG ⎟ ⎜ −1167 ⎟ ⎜ FAG ⎟ ⎜ −1179 ⎟ ⎟ ⎜ ⎟ ⎜ F 1167 = ⎟ lb ⎜ DC ⎟ ⎜ ⎜ F ⎟ ⎜ 1167 ⎟ ⎟ ⎜ EC ⎟ ⎜ ⎜ FBG ⎟ ⎜ 500 ⎟ ⎜ ⎟ ⎜ −1650 ⎟ F ⎜ DE ⎟ ⎜ ⎟ ⎟ ⎝ 471 ⎠ ⎜ ⎝ FCG ⎠

Positive means tension, Negative means compression.

Problem 6-12 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 10 kN P 2 = 15 kN

454

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Engineering Mechanics - Statics

Chapter 6

a = 2m b = 4m c = 4m Solution:

c α = atan ⎛⎜ ⎟⎞

⎝ a⎠

c β = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses: F AB = 1 kN

F AF = 1 kN

F GB = 1 kN

F BF = 1 kN

F FC = 1 kN

F FE = 1 kN

F BC = 1 kN

F EC = 1 kN

F CD = 1 kN

F ED = 1 kN Given Joint B

−F GB + F BC − F AB cos ( α ) = 0

Joint F

−F AB sin ( α ) − FBF = 0 −F AF + F FE + F FC cos ( β ) = 0 F BF + F FC sin ( β ) − P 1 = 0

Joint C

−F BC − F FC cos ( β ) + FCD cos ( α ) = 0 −F FC sin ( β ) − F CD sin ( α ) − FEC = 0

Joint E

−F FE + F ED = 0 F EC − P 2 = 0

Joint D

−F CD cos ( α ) − F ED = 0 F CD sin ( α ) = 0

455

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎜ ⎟ = Find ( FAB , FBF , FBC , FED , FAF , FFC , FEC , FGB , FFE , FCD) F FC ⎜ ⎟ ⎜F ⎟ ⎜ EC ⎟ ⎜ FGB ⎟ ⎜ ⎟ F ⎜ FE ⎟ ⎜F ⎟ ⎝ CD ⎠ ⎛⎜ FAB ⎞⎟ ⎛ −27.951 ⎞ ⎜ FBF ⎟ ⎜ 25 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FBC ⎟ = ⎜ 15 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ ED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎝ −15 ⎠ ⎝ ⎠

⎛⎜ FFC ⎞⎟ ⎛ −21.213 ⎞ ⎜ FEC ⎟ ⎜ 15 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGB ⎟ = ⎜ 27.5 ⎟ kN Positive means Tension, Negative means Compression ⎜F ⎟ ⎜ 0 ⎟ FE ⎜ ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎝ 0 ⎠ ⎝ ⎠

Problem 6-13 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 0 kN P 2 = 20 kN a = 2m b = 4m c = 4m

456

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Engineering Mechanics - Statics

Solution:

c α = atan ⎛⎜ ⎞⎟

⎝ a⎠

Chapter 6

c β = atan ⎛⎜ ⎞⎟

⎝ b⎠

Initial Guesses: F AB = 1 kN

F AF = 1 kN

F GB = 1 kN

F BF = 1 kN

F FC = 1 kN

F FE = 1 kN

F BC = 1 kN

F EC = 1 kN

F CD = 1 kN

F ED = 1 kN Given Joint B

−F GB + F BC − F AB cos ( α ) = 0

Joint F

−F AB sin ( α ) − FBF = 0 −F AF + F FE + F FC cos ( β ) = 0 F BF + F FC sin ( β ) − P 1 = 0

Joint C

−F BC − F FC cos ( β ) + FCD cos ( α ) = 0 −F FC sin ( β ) − F CD sin ( α ) − FEC = 0

Joint E

−F FE + F ED = 0 F EC − P 2 = 0

Joint D

−F CD cos ( α ) − F ED = 0 F CD sin ( α ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎜ ⎟ = Find ( FAB , FBF , FBC , FED , FAF , FFC , FEC , FGB , FFE , FCD) FFC ⎜ ⎟ ⎜F ⎟ ⎜ EC ⎟ ⎜ FGB ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎜F ⎟ ⎝ CD ⎠

457

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ FAB ⎞⎟ ⎛ −22.361 ⎞ ⎜ FBF ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FBC ⎟ = ⎜ 20 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ ED ⎟ ⎜ ⎟ − 20 ⎝ ⎠ ⎜ FAF ⎟ ⎝ ⎠

⎛⎜ FFC ⎞⎟ ⎛ −28.284 ⎞ ⎜ FEC ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGB ⎟ = ⎜ 30 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ FE ⎟ ⎜ ⎟ 0 ⎝ ⎠ ⎜ FCD ⎟ ⎝ ⎠

Positive means Tension, Negative means Compression

Problem 6-14 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 100 lb P 2 = 200 lb P 3 = 300 lb a = 10 ft b = 10 ft

θ = 30 deg Solution:

b φ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AF = 1 lb

F BC = 1 lb

F BF = 1 lb

F FC = 1 lb

F FE = 1 lb

F ED = 1 lb

F EC = 1 lb

F CD = 1 lb

458

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given Joint B

F BC − F AB cos ( φ ) = 0 −F BF − F AB sin ( φ ) = 0

Joint F

−F AF + F FE + F FC cos ( φ ) = 0 −P 2 + F BF + F FC sin ( φ ) = 0

Joint C

−F BC + F CD cos ( φ ) − F FC cos ( φ ) = 0 −F EC − F CD sin ( φ ) − FFC sin ( φ ) = 0

Joint E

−F FE + F ED = 0 F EC − P 3 = 0

Joint D

−F ED cos ( θ ) − F CD cos ( φ + θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ F AF ⎜ ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBF ⎟ ⎜ ⎟ F ⎜ FC ⎟ = Find ( FAB , FAF , FBC , FBF , FFC , FFE , FED , FEC , FCD) ⎜F ⎟ ⎜ FE ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FEC ⎟ ⎜ ⎟ ⎝ FCD ⎠

459

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ −330.0 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 79.4 ⎟ ⎜ BC ⎟ ⎜ −233.3 ⎟ ⎜ FBF ⎟ ⎜ 233.3 ⎟ ⎟ ⎜ ⎟ ⎜ F − 47.1 = ⎟ lb ⎜ FC ⎟ ⎜ ⎜ F ⎟ ⎜ 112.7 ⎟ ⎟ ⎜ FE ⎟ ⎜ ⎜ FED ⎟ ⎜ 112.7 ⎟ ⎜ ⎟ ⎜ 300.0 ⎟ F ⎜ EC ⎟ ⎜ ⎟ ⎟ ⎝ −377.1 ⎠ ⎜ ⎝ FCD ⎠

Positive means Tension, Negative means Compression

Problem 6-15 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 400 lb P 2 = 400 lb P 3 = 0 lb a = 10 ft b = 10 ft

θ = 30 deg Solution:

b φ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AF = 1 lb

F BC = 1 lb

F BF = 1 lb

F FC = 1 lb

F FE = 1lb

F ED = 1 lb

F EC = 1 lb

F CD = 1 lb

460

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Engineering Mechanics - Statics

Chapter 6

Given Joint B

F BC − F AB cos ( φ ) = 0 −F BF − F AB sin ( φ ) = 0

Joint F

−F AF + F FE + F FC cos ( φ ) = 0 −P 2 + F BF + F FC sin ( φ ) = 0

Joint C

−F BC + F CD cos ( φ ) − F FC cos ( φ ) = 0 −F EC − F CD sin ( φ ) − FFC sin ( φ ) = 0

Joint E

−F FE + F ED = 0 F EC − P 3 = 0

Joint D

−F ED cos ( θ ) − F CD cos ( φ + θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBF ⎟ ⎜ ⎟ ⎜ FFC ⎟ = Find ( FAB , FAF , FBC , FBF , FFC , FFE , FED , FEC , FCD) ⎜F ⎟ ⎜ FE ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FEC ⎟ ⎜ ⎟ ⎝ FCD ⎠

461

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Engineering Mechanics - Statics

⎛ FAB ⎞ ⎜ ⎟ ⎛ −377.1 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 189.7 ⎟ ⎜ BC ⎟ ⎜ −266.7 ⎟ ⎜ FBF ⎟ ⎜ 266.7 ⎟ ⎟ ⎜ ⎟ ⎜ F 188.6 = ⎟ lb ⎜ FC ⎟ ⎜ ⎜ F ⎟ ⎜ 56.4 ⎟ ⎟ ⎜ FE ⎟ ⎜ ⎜ FED ⎟ ⎜ 56.4 ⎟ ⎜ ⎟ ⎜ 0.0 ⎟ F ⎜ EC ⎟ ⎜ ⎟ ⎟ ⎝ −188.6 ⎠ ⎜ ⎝ FCD ⎠

Chapter 6

Positive means Tension, Negative means Compression

Problem 6-16 Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

Solution: Support reactions: ΣME = 0;

3 Ax d − P d = 0 2

Ax =

2P 3

462

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Engineering Mechanics - Statics

Chapter 6

2P

ΣF x = 0;

Ax − Ex = 0

Ex =

ΣF y = 0;

Ey − P = 0

Ey = P

3

Joint E: 2

ΣF x = 0;

F EC

ΣF y = 0;

P − FED − FEC

13

− Ex = 0

13

F EC =

3

=0

13

3

P = 1.20P (T)

F ED = 0

Joint A: 1

− F AD

1

2

=0

ΣF y = 0;

F AB

ΣF x = 0;

Ax − 2FAB

5

=0

5

F AB = F AD

F AB = F AD =

5

Joint D: ΣF x = 0;

F AD

ΣF y = 0;

2FAD

2 5

− F DC

1 5

2

=0

5

− FDB = 0

F DC = F DB =

5 6 P

5 6

P = 0.373P (C)

P = 0.373P (C)

(T)

3

Joint B: ΣF x = 0; F AB

1 5

− F BC

1 5

=0

F BC =

5 6

P = 0.373P (C)

463

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Problem 6-17 The maximum allowable tensile force in the members of the truss is Tmax and the maximum allowable compressive force is Cmax. Determine the maximum magnitude of the load P that can be applied to the truss. Units Used: 3

kN = 10 N Given: Tmax = 5 kN Cmax = 3 kN d = 2m Solution: P = 1 kN

Set

Initial Guesses: F AD = 1 kN

F AB = 1 kN

F BC = 1 kN

F BD = 1 kN

F CD = 1 kN

F CE = 1 kN

F DE = 1 kN

Given Joint A

Joint B

Joint D

F AD F BC

1 5 2 5

− F AB

1

− F AB

2

=0

5 =0

5

(FBC + FAB)

1

(FCD − FAD)

2

5

+ F BD = 0 =0

5

(

)

1

F DE − F BD − FAD + FCD Joint C

− FCD + FBC

(

)

(FCD − FBC )

1

2 5

5

2

− F CE

+ F CE

=0

5 =0

13 3

−P=0

13

464

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Engineering Mechanics - Statics

Chapter 6

⎛ FAD ⎞ ⎜ ⎟ ⎜ FAB ⎟ ⎜ ⎟ FBC ⎜ ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F ( AD AB BC BD CD CE DE) ⎜ BD ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ FCE ⎟ ⎜F ⎟ ⎝ DE ⎠ ⎛ FAD ⎞ ⎜ ⎟ ⎛ −0.373 ⎞ ⎟ ⎜ FAB ⎟ ⎜ ⎜ ⎟ ⎜ −0.373 ⎟ ⎜ FBC ⎟ ⎜ −0.373 ⎟ ⎜ F ⎟ = ⎜ 0.333 ⎟ kN ⎟ ⎜ BD ⎟ ⎜ ⎜ FCD ⎟ ⎜ −0.373 ⎟ ⎜ ⎟ ⎜ 1.202 ⎟ ⎟ ⎜ FCE ⎟ ⎜ 0 ⎝ ⎠ ⎜F ⎟ ⎝ DE ⎠ Now Scale the answer P1 = P

P2 = P

Tmax

(

max F AD , F AB , F BC , F BD , F CD , F CE , FDE

)

Cmax min F AD , F AB , F BC , F BD , F CD , F CE , FDE

(

(

)

P = min P 1 , P 2

)

P = 4.16 kN

Problem 6-18 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The horizontal force component at A must be zero. Why?

465

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Engineering Mechanics - Statics

Chapter 6

Units Used: 3

kip = 10 lb Given: F 1 = 600 lb F 2 = 800 lb a = 4 ft b = 3 ft

θ = 60 deg Solution: Initial Guesses F BA = 1 lb

F BD = 1 lb

F CB = 1 lb

F CD = 1 lb

Given Joint C

Joint B

−F CB − F2 cos ( θ ) = 0 F CB + FBD

b 2

a +b

−F CD − F 2 sin ( θ ) = 0 =0

2

⎛ FBA ⎞ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ = Find ( FBA , FBD , FCB , FCD) ⎜ FCB ⎟ ⎜F ⎟ ⎝ CD ⎠

−F BA − F BD

a 2

2

a +b

− F1 = 0

⎛ FBA ⎞ ⎛ 3 ⎜ ⎟ −1.133 × 10 ⎞ ⎜ ⎟ Positive means Tension ⎜ FBD ⎟ 666.667 ⎟ lb Negative means ⎜ ⎟=⎜ ⎜ ⎟ Compression F −400 ⎜ CB ⎟ ⎜ ⎜ F ⎟ ⎝ −692.82 ⎟⎠ ⎝ CD ⎠

Problem 6-19 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The resultant force at the pin E acts along member ED. Why? Units Used: 3

kN = 10 N

466

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 3 kN F 2 = 2 kN a = 3m b = 4m Solution: Initial Guesses: F CB = 1 kN

F CD = 1 kN

F BA = 1 kN

F BD = 1 kN

F DA = 1 kN

F DE = 1 kN

Given Joint C

−F 2 − F CD

Joint B

2a

−F CB − FCD

=0

2

( 2 a) + b b 2

2

=0 2

( 2 a) + b

−F BA + F CB = 0 −F 1 − F BD = 0

Joint D

(FCD − FDA − FDE) (

2a 2

=0 2

( 2 a) + b

F BD + FCD + FDA − FDE

)

b 2

=0 2

( 2 a) + b

467

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FCB ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ BA ⎟ = Find F , F , F , F , F , F ( CB CD BA BD DA DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FCB ⎞ ⎜ ⎟ ⎛ 3 ⎞ F ⎜ CD ⎟ ⎜ −3.606 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ BA 3 ⎜ ⎟= ⎜ ⎟ kN ⎜ FBD ⎟ −3 ⎟ ⎜ ⎜ ⎟ ⎜ FDA ⎟ ⎜ 2.704 ⎟ ⎜ ⎟ ⎜⎝ −6.31 ⎟⎠ F ⎝ DE ⎠

Positive means Tension, Negative means Compression

Problem 6-20 Each member of the truss is uniform and has a mass density ρ. Determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member. Given:

ρ = 8

kg m

g = 9.81

m 2

s F1 = 0 N F2 = 0 N a = 3m b = 4m

468

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Solution: Initial Guesses: F CB = 1 N

F CD = 1 N

F BA = 1 N

F BD = 1 N

F DA = 1 N

F DE = 1 N

Given Joint C

−F CB − FCD

−F 2 − F CD

Joint B

2a 2

( 2 a) + b

=0 2

−F BA + F CB = 0

⎛ b⎞ −F 1 − F BD − ρ g⎜ a + ⎟ = 0 ⎝ 4⎠ Joint D

2 2⎤ ⎛ a⎞ + ⎛ b⎞ ⎥ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 4⎠ ⎦

⎡a − ρ g⎢ + 2 2 ⎣2 ( 2 a) + b b

(

F BD + FCD + FDA − FDE

(FCD − FDA − FDE)

⎡b − ρ g⎢ + 3 2 2 ⎣4 ( 2 a) + b

)

b

2a 2

2 2⎤ ⎛ a⎞ + ⎛ b⎞ ⎥ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 4⎠ ⎦

=0 2

( 2 a) + b

⎛ FCB ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ BA ⎟ = Find F , F , F , F , F , F ( CB CD BA BD DA DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎟ ⎝ FDE ⎠

469

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Engineering Mechanics - Statics

⎛ FCB ⎞ ⎜ ⎟ ⎛ 389 ⎞ F CD ⎜ ⎟ ⎜ −467 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BA ⎟ = 389 ⎟ N ⎜ FBD ⎟ ⎜ −314 ⎟ ⎟ ⎜ ⎟ ⎜ 736 ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎜ ⎟ ⎝ −1204 ⎟⎠ F ⎝ DE ⎠

Chapter 6

Positive means Tension, Negative means Compression

Problem 6-21 Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

Solution: Joint B: +

↑Σ Fy = 0;

F BA sin ( 2 θ ) − P = 0 F BA = P csc ( 2 θ )

+ Σ F x = 0; →

( C)

F BAcos( 2 θ ) − FBC = 0 F BC = Pcot( 2 θ )

( C)

Joint C: + Σ F x = 0; →

P cot ( 2 θ ) + P + F CD cos ( 2 θ ) − F CA cos ( θ ) = 0

+

F CD sin ( 2 θ ) − FCA sin ( θ ) = 0

↑Σ Fy = 0;

470

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Engineering Mechanics - Statics

F CA =

Chapter 6

cot ( 2 θ ) + 1

cos ( θ ) − sin ( θ ) cot( 2 θ )

P

F CA = ( cot ( θ ) csc ( θ ) − sin ( θ ) + 2 cos ( θ ) ) P

( T)

F CD = ( cot ( 2 θ ) + 1) P

( C)

Joint D: + Σ F x = 0; →

F DA − ⎡⎣cot( 2 θ ) + 1⎤⎦ ⎡⎣cos( 2 θ )⎤⎦ P = 0 F DA = ⎡⎣cot( 2 θ ) + 1⎤⎦ ⎡⎣cos( 2 θ )⎤⎦ P

( C)

Problem 6-22 The maximum allowable tensile force in the members of the truss is Tmax, and the maximum allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads that can be applied to the truss. Units Used: 3

kN = 10 N Given: Tmax = 2 kN Cmax = 1.2 kN L = 2m

θ = 30 deg Solution: Initial guesses (assume all bars are in tension). Use a unit load for P and then scale the answer later. F BA = 1 kN

F BC = 1 kN

F CA = 1 kN

F CD = 1 kN

F DA = 1 kN

P = 1 kN

471

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Engineering Mechanics - Statics

Chapter 6

Given Joint B +

↑ Σ Fy = 0; + Σ F x = 0; →

−F BA sin ( 2 θ ) − P = 0 −F BA cos ( 2 θ ) + F BC = 0

Joint C +

↑Σ Fy = 0;

−F CA sin ( θ ) − F CD sin ( 2 θ ) = 0

+ Σ F x = 0; →

Joint D

−F BC + P − FCD cos ( 2 θ ) − FCA cos ( θ ) = 0

+ Σ F x = 0; →

−F DA + F CD cos ( 2 θ ) = 0

⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCA ⎟ = Find ( FBA , FBC , FCA , FCD , FDA) ⎜F ⎟ ⎜ CD ⎟ ⎜ FDA ⎟ ⎝ ⎠

⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ans = ⎜ FCA ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FDA ⎟ ⎝ ⎠

⎛ −1.155 ⎞ ⎜ −0.577 ⎟ ⎜ ⎟ ans = ⎜ 2.732 ⎟ kN ⎜ −1.577 ⎟ ⎜ ⎟ ⎝ −0.789 ⎠

Now find the biggest tension and the biggest compression. T = max ( ans)

T = 2.732 kN

C = min ( ans)

C = −1.577 kN

Decide which is more important and scale the answer

⎡⎛ Tmax ⎞ ⎤ ⎢⎜ ⎟ ⎥ T ⎢ ⎜ ⎟ P⎥ P = min ⎢⎜ −Cmax ⎟ ⎥ ⎢⎜ ⎟ ⎥ ⎣⎝ C ⎠ ⎦

P = 732.051 N

Problem 6-23

472

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Engineering Mechanics - Statics

Chapter 6

The Fink truss supports the loads shown. Determine the force in each member and state if the members are in tension or compression. Approximate each joint as a pin. Units Used: 3

kip = 10 lb Given: F 1 = 500 lb

a = 2.5 ft

F 2 = 1 kip

θ = 30 deg

F 3 = 1 kip Solution: Entire truss:

(

)

ΣF x = 0;

E x = F1 + F2 + F3 + F2 + F1 sin ( θ )

ΣME = 0;

− Ay 4a cos ( θ ) + F 14a + F 23a + F 32a + F 2 a = 0 Ay =

ΣF y = 0;

2 F1 + 2 F2 + F3

E x = 2000 lb

Ay = 2309.4 lb

2 cos ( θ )

E y = − Ay + 2 cos ( θ ) F1 + 2 cos ( θ ) F 2 + cos ( θ ) F3

E y = 1154.7 lb

Joint A: ΣF y = 0;

F AB =

−cos ( θ ) F 1 + Ay sin ( θ )

F AB = 3.75 kip ΣF x = 0;

(C)

F AH = −sin ( θ ) F 1 + F AB cos ( θ ) F AH = 3 kip

(T)

473

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Engineering Mechanics - Statics

Chapter 6

Joint B: ΣF x = 0;

F BC = FAB

F BC = 3.75 kip

ΣF y = 0;

F BH = F 2

F BH = 1 kip

(C)

(C)

Joint H: Σ F y = 0; ΣF x = 0;

F HC = F 2

F HC = 1 kip

(T)

F GH = −F2 cos ( −90 deg + θ ) − FHC cos ( −90 deg + θ ) + F AH F GH = 2 kip

(T)

Joint E: ΣF y = 0;

F EF =

(

− F 1 − E x sin ( θ ) − Ey cos ( θ )

F EF = 3 kip ΣF x = 0;

sin ( θ )

)

(T)

F ED = −Ey sin ( θ ) + Ex cos ( θ ) + F EF cos ( θ ) F ED = 3.75 kip

(C)

Joint D: ΣF x = 0;

F DC = F ED F DC = 3.75 kip

ΣF y = 0;

(C)

F DF = F2 F DF = 1 kip

(C)

Joint C: ΣF x = 0;

F CF = F HC F CF = 1 kip

ΣF y = 0;

(T)

F CG = F3 + FHC cos ( 90 deg − θ ) ( 2)

F CG = 2 kip

(C)

F FG = FEF − FCF cos ( 90 deg − θ ) ( 2)

F FG = 2 kip

(T)

Joint F: ΣF x = 0;

474

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Engineering Mechanics - Statics

Chapter 6

Problem 6-24 Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression.

Solution: ΣΜ A = 0; +

↑ ΣF y = 0;

P

L 2L +P − Dy L = 0 3 3

Ay + Dy − 2 P = 0

Joint F: +

1

↑ ΣF y = 0;

F FB

+ ΣF x = 0; →

F FD − F FE − F FB

−P=0

2 1

=0

2

475

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Engineering Mechanics - Statics

Chapter 6

Joint E: +

↑ ΣF y = 0;

+ ΣF x = 0; →

1

F EC

−P=0

2 1

F EF − F EA + F EC

=0

2

Joint B: +

↑ ΣF y = 0;

F BA

+ ΣF x = 0; →

F BA

1 2 1 2

+ F BD

1

+ F FB

1

1

− F FB

5

=0

2

⎛ 2 ⎞=0 ⎟ ⎝ 5⎠

− F BD ⎜

2

Joint C: +

↑ ΣF y = 0;

F CA

+ ΣF x = 0; →

F CA

1 5 2 5

+ FCD

1

− FEC

1

2

2

− FEC

1

− FCD

1

=0

2 =0

2

476

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Engineering Mechanics - Statics

Chapter 6

Joint A: + ΣF x = 0; →

F AE − F BA

1 2

− F CA

2

=0

5

Solving we find F EF = 0.667 P ( T ) F FD = 1.67 P ( T) F AB = 0.471 P ( C) F AE = 1.67 P ( T) F AC = 1.49 P ( C) F BF = 1.41 P ( T) F BD = 1.49 P ( C) F EC = 1.41 P ( T) F CD = 0.471 P ( C)

Problem 6-25 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The vertical component of force at C must equal zero. Why? Units Used: 3

kN = 10 N Given: F 1 = 6 kN

477

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Engineering Mechanics - Statics

Chapter 6

F 2 = 8 kN a = 1.5 m b = 2m c = 2m

Solution: Initial Guesses: F AB = 1 kN

F AE = 1 kN

F EB = 1 kN

F BC = 1 kN

F BD = 1 kN

F ED = 1 kN

Given Joint A

F AB

F AB

Joint E

a 2

2

a +c c 2

2

a +c

+ F AE = 0 − F1 = 0

F ED − F AE = 0 F EB − F 2 = 0

Joint B

F BC + F BD −F EB − F BD

b 2

2

b +c c 2

2

b +c

− F AB − F AB

a 2

=0 2

a +c c 2

=0 2

a +c

⎛ FAB ⎞ ⎜ ⎟ F AE ⎜ ⎟ ⎜F ⎟ ⎜ EB ⎟ = Find F , F , F , F , F , F ( AB AE EB BC BD ED) ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎝ FED ⎠

478

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ 7.5 ⎞ F AE ⎜ ⎟ ⎜ −4.5 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ 8 ⎜ EB ⎟ = kN ⎜ FBC ⎟ ⎜ 18.5 ⎟ ⎟ ⎜ ⎟ ⎜ − 19.799 ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎜ ⎟ ⎝ −4.5 ⎟⎠ F ⎝ ED ⎠

Positive means Tension, Negative means Compresson.

Problem 6-26 Each member of the truss is uniform and has a mass density ρ. Remove the external loads F1 and F 2 and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member. Given: F1 = 0 F2 = 0

ρ = 8

kg m

a = 1.5 m b = 2m c = 2m g = 9.81

m 2

s Solution:

Find the weights of each bar. 2

2

WAB = ρ g a + c

WBC = ρ g b

WBE = ρ g c

WAE = ρ g a

WBD = ρ g b + c

2

2

WDE = ρ g b

479

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Engineering Mechanics - Statics

Guesses

Chapter 6

F AB = 1 N

F AE = 1 N

F BC = 1 N

F BE = 1 N

F BD = 1 N

F DE = 1 N

Given Joint A

a

F AE +

2

a +c

c 2

2

a +c Joint E

F AB −

F AB = 0 WAB + WAE 2

=0

F DE − F AE = 0 F BE −

Joint B

2

F BC +

WAE + WBE + WDE 2 b 2

2

b +c

−c

F BD −

F AB − F BE − 2 2 a +c

=0

a 2

2

a +c c 2

2

b +c

F AB = 0

F BD −

WAB + WBE + WBD + WBC 2

=0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAE ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AE BC BD BE DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 196 ⎞ F ⎜ AE ⎟ ⎜ −118 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ BC 857 ⎜ ⎟= ⎜ ⎟N ⎜ FBD ⎟ −1045 ⎜ ⎟ ⎜ ⎟ 216 ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ ⎜⎝ −118 ⎟⎠ FDE ⎝ ⎠

Positive means tension, Negative means Compression.

480

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Engineering Mechanics - Statics

Chapter 6

Problem 6-27 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 4 kN P 2 = 0 kN a = 2m

θ = 15 deg Solution: Take advantage of the symetry. Initial Guesses: F BD = 1 kN

F CD = 1 kN

F CA = 1 kN

F BC = 1 kN

Given Joint D Joint B

−P 1 2

F AB = 1 kN

− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0

−P 2 cos ( 2 θ ) − F BC = 0 F BD − F AB − P 2 sin ( 2 θ ) = 0

Joint C

F CD cos ( θ ) − F CA cos ( θ ) = 0

(FCD + FCA)sin(θ ) + FBC = 0 ⎛⎜ FBD ⎞⎟ ⎜ FCD ⎟ ⎜ ⎟ F ⎜ AB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC ) ⎜F ⎟ ⎜ CA ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛⎜ FBD ⎞⎟ ⎜ FED ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ ⎟ F F = ⎜ GF ⎟ ⎜ AB ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ CA ⎟ ⎜ FFE ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎝ ⎠

481

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Engineering Mechanics - Statics

⎛⎜ FBD ⎞⎟ ⎛ −4 ⎞ ⎜ FCD ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ −4 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ CA ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ 0 ⎠ ⎝ ⎠

Chapter 6

⎛⎜ FFD ⎞⎟ ⎛ −4 ⎞ ⎜ FED ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ −4 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ EG ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎝ 0 ⎠ ⎝ ⎠

Positvive means Tension, Negative means Compression

Problem 6-28 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 2 kN P 2 = 4 kN a = 2m

θ = 15 deg Solution: Take advantage of the symmetry. Initial Guesses: F BD = 1 kN

F CD = 1 kN

F CA = 1 kN

F BC = 1 kN

Given Joint D Joint B

−P 1 2

F AB = 1 kN

− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0

−P 2 cos ( 2 θ ) − F BC = 0 F BD − F AB − P 2 sin ( 2θ ) = 0

Joint C

F CD cos ( θ ) − F CA cos ( θ ) = 0

(FCD + FCA)sin(θ ) + FBC = 0 482

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛⎜ FBD ⎞⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ FAB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC ) ⎜F ⎟ ⎜ CA ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎛⎜ FBD ⎞⎟ ⎛ −11.46 ⎞ ⎜ FCD ⎟ ⎜ 6.69 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ −13.46 ⎟ kN ⎜ F ⎟ ⎜ 6.69 ⎟ ⎜ CA ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ −3.46 ⎠ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛⎜ FBD ⎞⎟ ⎜ FED ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ FAB ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ CA ⎟ ⎜ FFE ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛ −11.46 ⎞ ⎜ FED ⎟ ⎜ 6.69 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ −13.46 ⎟ kN ⎜ F ⎟ ⎜ 6.69 ⎟ ⎜ EG ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎝ −3.46 ⎠ ⎝ ⎠

Positvive means Tension, Negative means Compression

Problem 6-29 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 2 kip F 2 = 1.5 kip F 3 = 3 kip F 4 = 3 kip a = 4 ft b = 10 ft Solution:

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses

483

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Engineering Mechanics - Statics

Chapter 6

F AB = 1 lb

F BC = 1 lb

F CD = 1 lb

F DE = 1 lb

F AI = 1 lb

F BI = 1 lb

F CI = 1 lb

F CG = 1 lb

F CF = 1 lb

F DF = 1 lb

F EF = 1 lb

F HI = 1 lb

F GI = 1 lb

F GH = 1 lb

F FG = 1 lb

Given Joint A

F AI cos ( θ ) + FAB = 0

Joint B

F BC − F AB = 0 F BI = 0

Joint C

(

)

F CD − F BC + FCF − F CI cos ( θ ) = 0

(

)

F CG + FCF + F CI sin ( θ ) = 0 Joint D

F DE − F CD = 0 F DF = 0

Joint I

(

)

F 2 + FGI + F CI − F AI cos ( θ ) = 0

(

)

F HI − FBI + FGI − F AI − FCI sin ( θ ) = 0 Joint H

F GH cos ( θ ) + F 1 = 0 −F GH sin ( θ ) − FHI = 0

Joint G

Joint F

(FFG − FGH − FGI )cos (θ ) = 0 −F 3 − F CG + ( FGH − FFG − FGI ) sin ( θ ) = 0 (FEF − FFG − FCF)cos (θ ) = 0 (FFG − FCF − FEF)sin(θ ) − F4 − FDF = 0

484

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜F ⎟ ⎜ DE ⎟ ⎜ FAI ⎟ ⎜ ⎟ ⎜ FBI ⎟ ⎜ ⎟ ⎜ FCI ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F , F , F , F , F , F , F , F , F ( AB BC CD DE AI BI CI CG CF DF EF HI GI GH ⎜ CG ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎜F ⎟ ⎜ EF ⎟ ⎜ FHI ⎟ ⎜ ⎟ ⎜ FGI ⎟ ⎜F ⎟ ⎜ GH ⎟ ⎜ FFG ⎟ ⎝ ⎠ ⎛⎜ FAB ⎞⎟ ⎛ 3.75 ⎞ ⎜ FBC ⎟ ⎜ 3.75 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCD ⎟ = ⎜ 7.75 ⎟ kip ⎜ F ⎟ ⎜ 7.75 ⎟ ⎜ DE ⎟ ⎜ ⎟ ⎜ FAI ⎟ ⎝ −4.04 ⎠ ⎝ ⎠

⎛⎜ FBI ⎞⎟ ⎛ 0 ⎞ ⎜ FCI ⎟ ⎜ 0.27 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCG ⎟ = ⎜ 1.4 ⎟ kip ⎜ F ⎟ ⎜ −4.04 ⎟ ⎜ CF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎝ 0 ⎠ ⎝ ⎠

⎛⎜ FEF ⎟⎞ ⎛ −12.12 ⎞ ⎜ FHI ⎟ ⎜ 0.8 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGI ⎟ = ⎜ −5.92 ⎟ kip ⎜ F ⎟ ⎜ −2.15 ⎟ ⎜ GH ⎟ ⎜ ⎟ ⎜ FFG ⎟ ⎝ −8.08 ⎠ ⎝ ⎠

Positive means Tension, Negative means Compression

Problem 6-30 The Howe bridge truss is subjected to the loading shown. Determine the force in members DE, EH, and HG, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

485

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 30 kN F 2 = 20 kN F 3 = 20 kN F 4 = 40 kN a = 4m b = 4m Solution: −F 2 a − F 3( 2a) − F 4( 3a) + Gy( 4a) = 0 Gy =

F2 + 2F 3 + 3F4 4

Gy = 45 kN Guesses

F DE = 1 kN

F EH = 1 kN

F HG = 1 kN

Given −F DE − F HG = 0

Gy − F 4 − F EH = 0

F DE b + Gy a = 0

⎛ FDE ⎞ ⎜ ⎟ F ⎜ EH ⎟ = Find ( FDE , FEH , FHG) ⎜F ⎟ ⎝ HG ⎠

⎛ FDE ⎞ ⎛ −45 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ EH ⎟ = ⎜ 5 ⎟ kN ⎜ F ⎟ ⎝ 45 ⎠ ⎝ HG ⎠

Positive (T) Negative (C)

Problem 6-31 The Pratt bridge truss is subjected to the loading shown. Determine the force in members LD, LK, CD, and KD, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

486

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 50 kN F 2 = 50 kN F 3 = 50 kN a = 4m b = 3m Solution: Ax = 0 Ay =

3F 3 + 4F2 + 5F 1 6

Guesses F LD = 1 kN

F LK = 1 kN

F CD = 1 kN

F KD = 1 kN

Given F 2 b + F 1( 2b) − Ay( 3b) − F LK a = 0 F CD a + F1 b − Ay( 2b) = 0 Ay − F1 − F2 −

a ⎞ ⎛ ⎜ 2 2 ⎟ FLD = 0 ⎝ a +b ⎠

−F 3 − F KD = 0

⎛ FLD ⎞ ⎜ ⎟ ⎜ FLK ⎟ ⎜ ⎟ = Find ( FLD , FLK , FCD , FKD) ⎜ FCD ⎟ ⎜F ⎟ ⎝ KD ⎠

⎛ FLD ⎞ ⎜ ⎟ ⎛⎜ 0 ⎟⎞ F ⎜ LK ⎟ ⎜ −112.5 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCD ⎟ ⎜ 112.5 ⎟ ⎜ F ⎟ ⎝ −50 ⎠ ⎝ KD ⎠

Positive (T) Negative (C)

487

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Engineering Mechanics - Statics

Chapter 6

Problem 6-32 The Pratt bridge truss is subjected to the loading shown. Determine the force in members JI, JE, and DE, and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 50 kN F 2 = 50 kN F 3 = 50 kN a = 4m b = 3m Solution: Initial Guesses Gy = 1 kN F JI = 1 kN F JE = 1 kN F DE = 1 kN Given Entire Truss −F 1 b − F2( 2b) − F3( 3b) + Gy( 6b) = 0 Section −F DE − F JI = 0

F JE + Gy = 0

Gy( 2b) − F DE a = 0 488

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Engineering Mechanics - Statics

Chapter 6

⎛ Gy ⎞ ⎜ ⎟ ⎜ FJI ⎟ ⎜ ⎟ = Find ( Gy , FJI , FJE , FDE) Gy = 50 kN F JE ⎜ ⎟ ⎜F ⎟ ⎝ DE ⎠

⎛ FJI ⎞ ⎛ −75 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JE ⎟ = ⎜ −50 ⎟ kN ⎜ F ⎟ ⎝ 75 ⎠ ⎝ DE ⎠ Positive means Tension, Negative means Compression

Problem 6-33 The roof truss supports the vertical loading shown. Determine the force in members BC, CK, and KJ and state if these members are in tension or compression.

Units Used: 3

kN = 10 N Given: F 1 = 4 kN F 2 = 8 kN a = 2m b = 3m

489

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Engineering Mechanics - Statics

Chapter 6

Solution: Initial Guesses Ax = 1 kN

Ay = 1 kN

F BC = 1 kN

F CK = 1 kN

F KJ = 1 kN Given Ax = 0 F 2( 3a) + F 1( 4a) − Ay( 6a) = 0

⎛ 2b ⎞ ⎟ + Ax⎜ ⎟ − Ay( 2a) = 0 ⎝3⎠ ⎝3⎠

F KJ ⎛⎜

2b ⎞

F KJ + A x +

3a ⎞F = 0 ⎛ BC ⎜ 2 ⎟ 2 ⎝ b + 9a ⎠

F CK + A y +

b ⎞F = 0 ⎛ BC ⎜ 2 2⎟ ⎝ b + 9a ⎠

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ F ⎜ KJ ⎟ = Find ( Ax , Ay , FKJ , FCK , FBC ) ⎜F ⎟ ⎜ CK ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 6.667 ⎟ ⎜ ⎟ ⎜ ⎟ F 13.333 = ⎜ KJ ⎟ ⎜ ⎟ kN Positive (T) Negative (C) ⎜F ⎟ ⎜ 0 ⎟ CK ⎜ ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ −14.907 ⎠ ⎝ ⎠

Problem 6-34 Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the deck of a bridge. State if these members are in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 4000 lb F 2 = 8000 lb

490

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Engineering Mechanics - Statics

Chapter 6

F 3 = 5000 lb a = 9 ft b = 12 ft

Solution Initial Guesses:

F DJ = 1 kip

Ay = 1 kip

F CD = 1 kip

F CJ = 1 kip

F KJ = 1 kip

Given F 3 a + F 2( 4a) + F 1( 5a) − Ay( 6a) = 0 − Ay( 2a) + F 1 a + FKJ b = 0 F CD + F KJ +

a ⎛ ⎞ ⎜ 2 2 ⎟ FCJ = 0 ⎝ a +b ⎠

Ay − F1 − F2 −

b ⎛ ⎞ ⎜ 2 2 ⎟ FCJ = 0 ⎝ a +b ⎠

−F DJ = 0

⎛⎜ Ay ⎞⎟ ⎜ FKJ ⎟ ⎜ ⎟ ⎜ FCJ ⎟ = Find ( Ay , FKJ , FCJ , FDJ , FCD) ⎜F ⎟ ⎜ DJ ⎟ ⎜ FCD ⎟ ⎝ ⎠

⎛⎜ Ay ⎞⎟ ⎛ 9.5 ⎞ ⎜ FKJ ⎟ ⎜ 11.25 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCJ ⎟ = ⎜ −3.125 ⎟ kip ⎜F ⎟ ⎜ 0 ⎟ ⎜ DJ ⎟ ⎜ ⎟ − 9.375 ⎝ ⎠ ⎜ FCD ⎟ ⎝ ⎠

Positive (T) Negative (C)

491

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Engineering Mechanics - Statics

Chapter 6

Problem 6-35 Determine the force in members EI and JI of the truss which serves to support the deck of a bridge. State if these members are in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 4000 lb F 2 = 8000 lb F 3 = 5000 lb a = 9 ft b = 12 ft Solution: Initial Guesses: Gy = 1 kip

F EI = 1 kip

F JI = 1 kip

Given −F 1 a − F22a − F35a + Gy6a = 0 Gy2a − F 3 a − FJI b = 0 F EI − F3 + Gy = 0

⎛ Gy ⎞ ⎜ ⎟ F ⎜ JI ⎟ = Find ( Gy , FJI , FEI ) ⎜F ⎟ ⎝ EI ⎠

⎛ Gy ⎞ ⎛ 7.5 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JI ⎟ = ⎜ 7.5 ⎟ kip ⎜ F ⎟ ⎝ −2.5 ⎠ ⎝ EI ⎠

Positive (T) Negative (C)

Problem 6-36 Determine the force in members BE, EF, and CB, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

492

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 5 kN

F 4 = 10 kN

F 2 = 10 kN

a = 4m

F 3 = 5 kN

b = 4m

a θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ b⎠

Inital Guesses F CB = 1 kN F BE = 1 kN

F EF = 1 kN

Given F 1 + F 2 − F BE cos ( θ ) = 0 −F CB − FEF − FBE sin ( θ ) − F3 = 0 −F 1 a + F CB b = 0

⎛ FCB ⎞ ⎜ ⎟ F ⎜ BE ⎟ = Find ( FCB , FBE , FEF) ⎜F ⎟ ⎝ EF ⎠ ⎛ FCB ⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FBE ⎟ = ⎜ 21.2 ⎟ kN ⎜ F ⎟ ⎝ −25 ⎠ ⎝ EF ⎠

Positive (T) Negative (C)

Problem 6-37 Determine the force in members BF, BG, and AB, and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 5 kN

F 4 = 10 kN

493

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Engineering Mechanics - Statics

F 2 = 10 kN

a = 4m

F 3 = 5 kN

b = 4m

Solution:

Chapter 6

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Inital Guesses F AB = 1 kN F BG = 1 kN

F BF = 1 kN

Given F 1 + F 2 + F 4 + F BG cos ( θ ) = 0 −F 1 3a − F 22a − F 4 a + FAB b = 0 −F BF = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBG , FBF) ⎜F ⎟ ⎝ BF ⎠ ⎛ FAB ⎞ ⎛ 45 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ BG ⎟ = ⎜ −35.4 ⎟ kN Positive (T) Negative (C) ⎜F ⎟ ⎝ 0 ⎠ BF ⎝ ⎠

Problem 6-38 Determine the force developed in members GB and GF of the bridge truss and state if these members are in tension or compression. Given: F 1 = 600 lb

494

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Engineering Mechanics - Statics

Chapter 6

F 2 = 800 lb a = 10 ft b = 10 ft c = 4 ft Solution: Initial Guesses Ax = 1 lb

Ay = 1 lb

F GB = 1 lb

F GF = 1 lb

Given F 2 b + F1( b + 2c) − A y2( b + c) = 0 Ax = 0 Ay − F GB = 0 − Ay b − FGF a = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , FGB , FGF) ⎜ FGB ⎟ ⎜F ⎟ ⎝ GF ⎠ ⎛ Ax ⎞ ⎞ ⎜ ⎟ ⎛⎜ 0 ⎟ ⎜ Ay ⎟ ⎜ 671.429 ⎟ lb Positive (T) ⎜ ⎟=⎜ FGB 671.429 ⎟ Negative (C) ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ −671.429 ⎠ ⎝ GF ⎠

Problem 6-39 Determine the force members BC, FC, and FE, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

495

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 6 kN F 2 = 6 kN a = 3m b = 3m Solution:

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses Dy = 1 kN

F BC = 1 kN

F FC = 1 kN

F FE = 1 kN

Given −F 1 b − F2( 2b) + Dy( 3b) = 0 Dy b − FFE cos ( θ ) a = 0

(

)

−F FC − FBC + FFE cos ( θ ) = 0

(

)

−F 2 + Dy + FFE + FBC sin ( θ ) = 0

⎛ Dy ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( Dy , FBC , FFC , FFE) ⎜ FFC ⎟ ⎜F ⎟ ⎝ FE ⎠ ⎛ Dy ⎞ ⎜ ⎟ ⎛⎜ 6 ⎟⎞ F ⎜ BC ⎟ ⎜ −8.49 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FFC ⎟ ⎜ 0 ⎟ ⎜ F ⎟ ⎝ 8.49 ⎠ ⎝ FE ⎠

Positive (T) Negative (C)

496

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Engineering Mechanics - Statics

Chapter 6

Problem 6-40 Determine the force in members IC and CG of the truss and state if these members are in tension or compression. Also, indicate all zero-force members. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN a = 1.5 m b = 2m Solution: By inspection of joints B, D, H and I. AB, BC, CD, DE, HI, and GI are all zero-force members. Guesses

Ay = 1 kN

Given

− Ay( 4a) + F 1( 2a) + F 2 a = 0 − Ay( 2a) − −a 2

2

a +b −b 2

2

a +b

F IC = 1 kN

b 2

2

a +b

FIC +

FIC −

FIC a −

a 2

a +b

2

b 2

a +b

2

⎛ Ay ⎞ ⎜ ⎟ ⎜ FIC ⎟ ⎜ ⎟ = Find ( Ay , FIC , FCG , FCJ ) ⎜ FCG ⎟ ⎜F ⎟ ⎝ CJ ⎠

F CG = 1 kN

a 2

2

a +b

F CJ = 1 kN

FIC b = 0

F CJ = 0

F CJ − FCG = 0

⎛ Ay ⎞ ⎜ ⎟ ⎛⎜ 4.5 ⎟⎞ F ⎜ IC ⎟ ⎜ −5.625 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCG ⎟ ⎜ 9 ⎟ ⎜ F ⎟ ⎝ −5.625 ⎠ ⎝ CJ ⎠

Positive (T) Negative (C)

Problem 6-41 Determine the force in members JE and GF of the truss and state if these members are in tension or compression. Also, indicate all zero-force members. 497

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Engineering Mechanics - Statics

Chapter 6

Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN a = 1.5 m b = 2m Solution: By inspection of joints B, D, H and I. AB, BC, CD, DE, HI, and GI are all zero-force members. E y = 1 kN

Guesses Given

F JE = 1 kN

F GF = 1 kN

−F 1( 2a) − F 2( 3a) + E y( 4a) = 0 Ey +

b 2

a +b

−a 2

2

a +b

2

F JE = 0

FJE − FGF = 0

⎛ Ey ⎞ ⎜ ⎟ F ⎜ JE ⎟ = Find ( Ey , FJE , FGF) ⎜F ⎟ ⎝ GF ⎠

⎛ Ey ⎞ ⎛ 7.5 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JE ⎟ = ⎜ −9.375 ⎟ kN Positive (T) Negative (C) ⎜ F ⎟ ⎝ 5.625 ⎠ ⎝ GF ⎠

Problem 6-42 Determine the force in members BC, HC, and HG. After the truss is sectioned use a single equation of equilibrium for the calculation of each force. State if these members are in tension or compression. Units Used: 3

kN = 10 N

498

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 2 kN F 4 = 5 kN a = 5 m F 2 = 4 kN F 5 = 3 kN b = 2 m F 3 = 4 kN

c = 3m

Solution: Guesses Ax = 1 kN

Ay = 1 kN

F BC = 1 kN

F HC = 1 kN

F HG = 1 kN

d = 1m

Given c b = a+d a

− Ax = 0

(F1 − Ay)( 4a) + F2( 3a) + F3( 2a) + F4( a) = 0 (F1 − Ay)( a) + Ax( c) − FBC ( c) = 0 (F1 − Ay)( 2a) + F2( a) + ( A y − F 1) ( d) − F 2( a + d ) +

a 2

a +b

2

F HG( c) +

c 2

2

a +c

b 2

a +b

FHC ( a + d) +

⎛ Ay ⎞ ⎜ ⎟ ⎜ Ax ⎟ ⎜ ⎟ FBC ⎜ ⎟ = Find A , A , F , F , F , d ( y x BC HC HG ) ⎜F ⎟ HC ⎜ ⎟ ⎜ FHG ⎟ ⎜ ⎟ ⎝ d ⎠

2

F HG( a) = 0 a 2

2

a +c

F HC ( c) = 0

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ = kN ⎜ Ay ⎟ ⎜⎝ 8.25 ⎟⎠ ⎝ ⎠ d = 2.5 m

⎛ FBC ⎞ ⎛ −10.417 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FHC ⎟ = ⎜ 2.235 ⎟ kN ⎜ F ⎟ ⎝ 9.155 ⎠ ⎝ HG ⎠ Positive (T) Negative (C)

Problem 6-43 Determine the force in members CD, CF, and CG and state if these members are in tension or 499

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Engineering Mechanics - Statics

Chapter 6

compression. Units Used: 3

kN = 10 N Given: F 1 = 2 kN F 4 = 5 kN a = 5 m F 2 = 4 kN F 5 = 3 kN b = 2 m F 3 = 4 kN

c = 3m

Solution: Guesses E y = 1 kN

F CD = 1 kN

F CF = 1 kN

F CG = 1 kN

F FG = 1 kN

F GH = 1 kN

Given

(

)

−F 2( a) − F3( 2a) − F4( 3a) + E y − F5 ( 4a) = 0

(

)

(

)

F CD( c) + Ey − F 5 ( a) = 0

a

−F 4( a) − F 5 − E y ( 2a) −

a 2

2

a +b b 2

2

a +b

(F5 − Ey)

FFG −

a 2

a +b

2

2

2

a +b

FFG( b + c) = 0

F GH = 0

(FFG + FGH ) + FCG = 0 a( c − b) b

+ F4⎡⎢a +

⎣

a( c − b) ⎤ b

⎥− ⎦

c 2

2

a +c

F CF⎡⎢2 a +

⎣

a( c − b) ⎤ b

⎥=0 ⎦

500

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Engineering Mechanics - Statics

Chapter 6

⎛ Ey ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ CF ⎟ = Find E , F , F , F , F , F ( y CD CF CG FG GH ) ⎜ FCG ⎟ ⎜ ⎟ ⎜ FFG ⎟ ⎜ ⎟ ⎝ FGH ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛ 9.75 ⎞ F CD ⎜ ⎟ ⎜ −11.25 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ CF ⎟ = 3.207 ⎟ kN Positive (T) ⎜ FCG ⎟ ⎜ −6.8 ⎟ Negative (C) ⎟ ⎜ ⎟ ⎜ ⎜ FFG ⎟ ⎜ 9.155 ⎟ ⎜ ⎟ ⎜⎝ 9.155 ⎟⎠ F ⎝ GH ⎠

Problem 6-44 Determine the force in members OE, LE, and LK of the Baltimore truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 2 kN

a = 2m

F 2 = 2 kN

b = 2m

F 3 = 5 kN F 4 = 3 kN Solution: Ax = 0 kN Initial Guesses Ay = 1 kN

F OE = 1 kN

F DE = 1 kN F LK = 1 kN F LE = 1 kN Given F LE = 0 F 4( 3b) + F 3( 4b) + F 2( 5b) + F 1( 6b) − Ay( 8b) = 0 F LK + FDE + FOE

b 2

a +b

=0 2

501

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Engineering Mechanics - Statics

Ay − F 1 − F 2 − F OE

Chapter 6

a 2

=0 2

a +b

−F LK ( 2a) + F2( b) + F 1( 2b) − Ay( 4b) = 0

⎛⎜ Ay ⎞⎟ ⎜ FOE ⎟ ⎜ ⎟ ⎜ FDE ⎟ = Find ( Ay , FOE , FDE , FLK , FLE) ⎜F ⎟ ⎜ LK ⎟ ⎜ FLE ⎟ ⎝ ⎠ ⎛⎜ Ay ⎞⎟ ⎛ 6.375 ⎞ = kN ⎜ FDE ⎟ ⎜⎝ 7.375 ⎟⎠ ⎝ ⎠

⎛ FOE ⎞ ⎛ 3.36 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FLE ⎟ = ⎜ 0 ⎟ kN ⎜ F ⎟ ⎝ −9.75 ⎠ ⎝ LK ⎠

Positive (T) Negative (C)

Problem 6-45 Determine the force in member GJ of the truss and state if this member is in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 1000 lb F 2 = 1000 lb F 3 = 1000 lb F 4 = 1000 lb a = 10 ft

θ = 30 deg

502

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= Chapter 6

Engineering Mechanics - Statics

Solution: Guess

E y = 1 lb

F GJ = 1 lb

Given −F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = 0 −F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = 0

⎛⎜ Ey ⎟⎞ = Find ( Ey , F GJ ) ⎜ FGJ ⎟ ⎝ ⎠

⎛⎜ Ey ⎟⎞ ⎛ 1.5 ⎞ = kip ⎜ FGJ ⎟ ⎜⎝ −2 ⎟⎠ ⎝ ⎠

Positive (T) Negative (C)

Problem 6-46 Determine the force in member GC of the truss and state if this member is in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 1000 lb F 2 = 1000 lb F 3 = 1000 lb F 4 = 1000 lb a = 10 ft

θ = 30 deg Solution: Guess

E y = 1 lb

F GJ = 1 lb

F HG = 1 lb

F GC = 1 lb

Given −F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = 0 −F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = 0 503

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Engineering Mechanics - Statics

Chapter 6

−F HG cos ( θ ) + FGJ cos ( θ ) = 0

(

)

−F 3 − F GC − F HG + FGJ sin ( θ ) = 0

⎛ Ey ⎞ ⎜ ⎟ ⎜ FGJ ⎟ ⎜ ⎟ = Find ( Ey , FGJ , FGC , FHG) F GC ⎜ ⎟ ⎜F ⎟ ⎝ HG ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛⎜ 1.5 ⎟⎞ ⎜ FGJ ⎟ ⎜ −2 ⎟ ⎜ ⎟ = ⎜ ⎟ kip F GC ⎜ ⎟ ⎜ 1 ⎟ ⎜ F ⎟ ⎝ −2 ⎠ ⎝ HG ⎠

Positive (T) Negative (C)

Problem 6-47 Determine the force in members KJ, JN, and CD, and state if the members are in tension or compression. Also indicate all zero-force members. Units Used: 3

kip = 10 lb Given: F = 3 kip a = 20 ft b = 30 ft c = 20 ft Solution:

Ax = 0 2c ⎞ ⎟ ⎝ 2a + b ⎠

θ = atan ⎛⎜

2c ⎞ ⎟ ⎝b⎠

φ = atan ⎛⎜

Initial Guesses: Ay = 1 lb

F CD = 1 lb

F KJ = 1 lb

F JN = 1 lb

504

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Engineering Mechanics - Statics

Chapter 6

Given

⎛ ⎝

F⎜a +

b⎞ ⎟ − Ay( 2a + b) = 0 2⎠

⎛ b⎞ F CD c − Ay⎜ a + ⎟ = 0 ⎝ 2⎠ F CD + F JN cos ( φ ) + F KJ cos ( θ ) = 0 Ay + F JN sin ( φ ) + FKJ sin ( θ ) = 0

⎛ Ay ⎞ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ = Find ( Ay , FCD , FJN , FKJ ) ⎜ FJN ⎟ ⎜F ⎟ ⎝ KJ ⎠

Ay = 1.5 kip

⎛ FCD ⎞ ⎛ 2.625 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FJN ⎟ = ⎜ 0 ⎟ kip ⎜ F ⎟ ⎝ −3.023 ⎠ ⎝ KJ ⎠ Positive (T), Negative (C)

Problem 6-48 Determine the force in members BG, HG, and BC of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 7 kN F 3 = 4 kN a = 3m b = 3m c = 4.5 m

505

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Engineering Mechanics - Statics

Chapter 6

Solution:

Initial Guesses

F BG = 1 kN

Ax = 1 kN

F HG = 1 kN

Ay = 1 kN

F BC = 1 kN

Given − Ax = 0 − Ay( a) −

a ⎡ ⎤ F ( b) = 0 HG ⎢ 2 2⎥ ⎣ ( c − b) + a ⎦

F 3( a) + F2( 2a) + F1( 3a) − A y( 4a) = 0 F BC +

a ⎡ ⎤F + ⎛ a ⎞F − A = 0 HG ⎜ BG x ⎢ 2 2⎥ 2 2⎟ ⎣ ( c − b) + a ⎦ ⎝ a +c ⎠

Ay − F1 +

c−b c ⎡ ⎤F + ⎛ ⎞F = 0 HG ⎜ BG ⎢ ⎥ ⎟ 2 2 2 2 ⎣ ( c − b) + a ⎦ ⎝ a +c ⎠ 506

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ FHG ⎟ = Find ( Ax , Ay , FHG , FBG , FBC ) ⎜F ⎟ ⎜ BG ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 9 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FHG ⎟ = ⎜ −10.062 ⎟ kN ⎜ F ⎟ ⎜ 1.803 ⎟ ⎜ BG ⎟ ⎜ ⎟ 8 ⎝ ⎠ ⎜ FBC ⎟ ⎝ ⎠

Positive (T) Negative (C)

Problem 6-49 The skewed truss carries the load shown. Determine the force in members CB, BE, and EF and state if these members are in tension or compression. Assume that all joints are pinned.

Solution: ΣMB = 0;

−P d − F EF d = 0

ΣME = 0;

−P d +

+ Σ F x = 0; →

2

F CB d = 0

F CB =

FCB − F BE = 0

P BE =

5 P−

1 5

F EF = −P 5 P 2 P 2

F EF = P

( C)

F CB = 1.12P

( T)

F BE = 0.5P

( T)

Problem 6-50 The skewed truss carries the load shown. Determine the force in members AB, BF, and EF and state if these members are in tension or compression. Assume that all joints are pinned. 507

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Engineering Mechanics - Statics

Chapter 6

Solution: ΣMF = 0;

−P 2d + P d + FAB d = 0

F AB = P

F AB = P

( T)

ΣMB = 0;

−P d − F EF d = 0

F EF = −P

F EF = P

( C)

F BE = − 2 P

F BF = 1.41P

( C)

+ Σ F x = 0; →

P + FBF

1

=0

2

Problem 6-51 Determine the force developed in members BC and CH of the roof truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 1.5 kN F 2 = 2 kN

508

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Engineering Mechanics - Statics

Chapter 6

a = 1.5 m b = 1m c = 2m d = 0.8 m Solution: a θ = atan ⎛⎜ ⎟⎞

⎝c⎠

a ⎞ ⎟ ⎝ c − b⎠

φ = atan ⎛⎜

Initial Guesses: E y = 1 kN

F BC = 1 kN

F CH = 1 kN

Given −F 2( d) − F1( c) + Ey( 2c) = 0 F BC sin ( θ ) ( c) + FCH sin ( φ ) ( c − b) + E y( c) = 0 −F BC sin ( θ ) − FCH sin ( φ ) − F1 + Ey = 0

⎛ Ey ⎞ ⎜ ⎟ ⎜ FBC ⎟ = Find ( Ey , FBC , FCH ) Ey = 1.15 kN ⎜F ⎟ ⎝ CH ⎠

⎛⎜ FBC ⎞⎟ ⎛ −3.25 ⎞ = kN ⎜ FCH ⎟ ⎜⎝ 1.923 ⎟⎠ ⎝ ⎠

Positive (T) Negative (C)

Problem 6-52 Determine the force in members CD and GF of the truss and state if the members are in tension or compression. Also indicate all zero-force members. Units Used: 3

kN = 10 N Given: F 1 = 1.5 kN F 2 = 2 kN 509

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Engineering Mechanics - Statics

Chapter 6

a = 1.5 m b = 1m c = 2m d = 0.8 m Solution: a θ = atan ⎛⎜ ⎟⎞

⎝c⎠

⎞ ⎟ ⎝ c − b⎠

φ = atan ⎛⎜

a

Initial Guesses: E y = 1 kN

F CD = 1 kN

F GF = 1 kN

Given −F 2( d) − F1( c) + Ey( 2c) = 0 E y( b) + F CD sin ( θ ) ( b) = 0 E y( c) − FGF( a) = 0

⎛ Ey ⎞ ⎜ ⎟ F ⎜ CD ⎟ = Find ( Ey , FCD , FGF) Ey = 1.15 kN ⎜F ⎟ ⎝ GF ⎠

⎛⎜ FCD ⎞⎟ ⎛ −1.917 ⎞ Positive (T) = kN Negative (C) ⎜ FGF ⎟ ⎜⎝ 1.533 ⎟⎠ ⎝ ⎠ DF and CF are zero force members.

Problem 6-53 Determine the force in members DE, DL, and ML of the roof truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 12 kN F 3 = 12 kN

510

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Engineering Mechanics - Statics

Chapter 6

F 4 = 12 kN a = 4m b = 3m c = 6m Solution:

θ = atan ⎛⎜

c − b⎞

⎟ ⎝ 3a ⎠

⎡b + ⎢ φ = atan ⎢ ⎣

2 ⎤ ( c − b) ⎥ 3 a

⎥ ⎦

Initial Guesses: Ay = 1 kN

F ML = 1 kN

F DL = 1 kN

F DE = 1 kN

Given F 2( a) + F3( 2a) + F4( 3a) + F3( 4a) + F2( 5a) + F1( 6a) − A y( 6a) = 0 2 F 1( 2a) + F 2( a) − A y( 2a) + FML⎡⎢b + ( c − b)⎤⎥ = 0 ⎣ 3 ⎦ Ay − F 1 − F 2 − F 3 + F DE sin ( θ ) − FDL sin ( φ ) = 0 F ML + F DL cos ( φ ) + FDE cos ( θ ) = 0

⎛ Ay ⎞ ⎜ ⎟ ⎜ FML ⎟ ⎜ ⎟ = Find ( Ay , FML , FDE , FDL) ⎜ FDE ⎟ ⎜F ⎟ ⎝ DL ⎠

Ay = 36 kN

⎛ FML ⎞ ⎛ 38.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −37.1 ⎟ kN ⎜ F ⎟ ⎝ −3.8 ⎠ ⎝ DL ⎠ Positive (T), Negative (C)

Problem 6-54 Determine the force in members EF and EL of the roof truss and state if the members are in 511

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 12 kN F 3 = 12 kN F 4 = 12 kN a = 4m b = 3m c = 6m Solution:

θ = atan ⎛⎜

c − b⎞

⎟ ⎝ 3a ⎠

Initial Guesses: Iy = 1 kN

F EF = 1 kN

F EL = 1 kN Given −F 2( a) − F3( 2a) − F4( 3a) − F3( 4a) − F2( 5a) − F1( 6a) + Iy( 6a) = 0 −F 3( a) − F2( 2a) − F1( 3a) + Iy( 3a) + F EF cos ( θ ) ( c) = 0 −F 4 − F EL − 2F EF sin ( θ ) = 0

⎛ Iy ⎞ ⎜ ⎟ ⎜ FEF ⎟ = Find ( Iy , FEF , FEL) ⎜F ⎟ ⎝ EL ⎠

Iy = 36 kN

⎛⎜ FEF ⎞⎟ ⎛ −37.108 ⎞ = kN ⎜ FEL ⎟ ⎜⎝ 6 ⎟⎠ ⎝ ⎠

Positive (T) Negative (C)

Problem 6-55 Two space trusses are used to equally support the uniform sign of mass M. Determine the force developed in members AB, AC, and BC of truss ABCD and state if the members are in tension or compression. Horizontal short links support the truss at joints B and D and there is a ball-and512

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Engineering Mechanics - Statics

Chapter 6

p socket joint at C.

pp

j

Given: M = 50 kg

g = 9.81

m 2

s

a = 0.25 m b = 0.5 m c = 2m Solution:

h =

2

2

b −a

⎛ −a ⎞ AB = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠

⎛a⎞ AD = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠

⎛ 2a ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛a⎞ BC = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −h ⎠

⎛0⎞ AC = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 N

F AD = 1 N

F AC = 1 N

F BC = 1 N

F BD = 1 N

By = 1 N

Given

⎛⎜ 0 ⎟⎞ 0 ⎟ AB AD AC F AB + FAD + FAC +⎜ =0 ⎜ −M g ⎟ AB AD AC ⎜ 2 ⎟ ⎝ ⎠ ⎛⎜ 0 ⎟⎞ F AB + FBD + F BC + ⎜ −B y ⎟ = 0 AB BD BC ⎜ 0 ⎟ ⎝ ⎠ −AB

BD

BC

513

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ F AD ⎜ ⎟ ⎜F ⎟ ⎜ AC ⎟ = Find F , F , F , F , F , B ( AB AD AC BC BD y) ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎝ By ⎠

⎛ By ⎞ ⎛ 566 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AD ⎟ = ⎜ 584 ⎟ N ⎜F ⎟ ⎝ 0 ⎠ ⎝ BD ⎠ ⎛ FAB ⎞ ⎛ 584 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ −1133 ⎟ N ⎜ F ⎟ ⎝ −142 ⎠ ⎝ BC ⎠ Positive (T), Negative (C)

Problem 6-56 Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at B, C, and D. Given: F = 600 N a = 3m b = 1m c = 1.5 m Solution:

⎛b⎞ ⎜ ⎟ AB = −c ⎜ ⎟ ⎝ −a ⎠ ⎛0⎞ ⎜ ⎟ AC = c ⎜ ⎟ ⎝ −a ⎠ ⎛ −b ⎞ ⎜ ⎟ AD = −c ⎜ ⎟ ⎝ −a ⎠

⎛ −b ⎞ ⎜ ⎟ CD = −2c ⎜ ⎟ ⎝ 0 ⎠

514

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Engineering Mechanics - Statics

⎛ b ⎞ CB = ⎜ −2c ⎟ ⎜ ⎟ ⎝ 0 ⎠

Chapter 6

⎛ −2b ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝ 0 ⎠

Guesses F BA = 1 N

F BC = 1 N

F CA = 1 N

F DA = 1 N

F BD = 1 N

F DC = 1 N

By = 1 N

Bz = 1 N

Cz = 1 N

Given

⎛ 0 ⎞ F BA + FCA + F DA +⎜ 0 ⎟ =0 ⎜ ⎟ AB AC AD ⎝ −F ⎠ AB

AC

AD

⎛⎜ 0 ⎟⎞ F CA + FDC + F BC +⎜0 ⎟ =0 AC CD CB ⎜ Cz ⎟ ⎝ ⎠ −AC

CD

CB

⎛0⎞ ⎜ ⎟ F BC + FBD + F BA + ⎜ By ⎟ = 0 CB BD AB ⎜B ⎟ ⎝ z⎠ −CB

BD

−AB

⎛ FBA ⎞ ⎜ ⎟ F BC ⎜ ⎟ ⎜F ⎟ ⎜ CA ⎟ ⎜ FDA ⎟ ⎜ ⎟ F ⎜ BD ⎟ = Find ( FBA , FBC , FCA , FDA , FBD , FDC , By , Bz , Cz) ⎜F ⎟ ⎜ DC ⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Bz ⎟ ⎟ ⎜ ⎝ Cz ⎠

515

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Engineering Mechanics - Statics

⎛ By ⎞ ⎛ 1.421 × 10− 14 ⎞ ⎜ ⎟ ⎜ ⎟ B = ⎜ z⎟ ⎜ ⎟N 150 ⎜C ⎟ ⎜ ⎟ 300 ⎠ ⎝ z⎠ ⎝

Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ −175 ⎞ F BC ⎜ ⎟ ⎜ 79.1 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ CA ⎟ = −335.4 ⎟ N ⎜ FDA ⎟ ⎜ −175 ⎟ ⎟ ⎜ ⎟ ⎜ 25 ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎜ ⎟ ⎝ 79.1 ⎟⎠ F ⎝ DC ⎠

Positive (T), Negative (C)

Problem 6-57 Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at A, B, and C. Given: a = 4 ft b = 2 ft c = 3 ft d = 2 ft e = 8 ft

⎛ 0 ⎞ ⎜ ⎟ F = 500 lb ⎜ ⎟ ⎝ 0 ⎠ Solution:

⎛ −c ⎞ ⎜ ⎟ AD = a ⎜ ⎟ ⎝e⎠

⎛ −c ⎞ ⎜ ⎟ BD = −b ⎜ ⎟ ⎝e ⎠

⎛d⎞ ⎜ ⎟ CD = −b ⎜ ⎟ ⎝e ⎠

⎛ 0 ⎞ ⎜ ⎟ AB = a + b ⎜ ⎟ ⎝ 0 ⎠

⎛ −c − d ⎞ ⎛ −c − d ⎞ ⎜ ⎟ ⎜ 0 ⎟ AC = a + b BC = ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ Guesses

516

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Engineering Mechanics - Statics

Chapter 6

F BA = 1 lb

F BC = 1 lb

F BD = 1 lb

F AD = 1 lb

F AC = 1 lb

F CD = 1 lb

Ay = 1 lb

Az = 1 lb

B x = 1 lb

B z = 1 lb

Cy = 1 lb

Cz = 1 lb

Given F + FAD

−AD AD

+ FBD

−BD BD

+ F CD

−CD CD

=0

⎛0 ⎞ ⎜ ⎟ F AD + F BA + FAC + ⎜ Ay ⎟ = 0 AD AB AC ⎜A ⎟ ⎝ z⎠ AD

AB

AC

⎛ Bx ⎞ ⎜ ⎟ −AB BC BD F BA + FBC + F BD +⎜ 0 ⎟ =0 AB BC BD ⎜B ⎟ ⎝ z⎠ ⎛0⎞ CD −AC −BC ⎜ C ⎟ F CD + F AC + F BC + ⎜ y⎟ = 0 CD AC BC ⎜C ⎟ ⎝ z⎠ ⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜F ⎟ ⎜ AD ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ = Find ( FBA , FBC , FBD , FAD , FAC , FCD , Ay , Az , Bx , Bz , Cy , Cz) ⎜ Ay ⎟ ⎜ A ⎟ ⎜ z ⎟ ⎜ Bx ⎟ ⎜ ⎟ ⎜ Bz ⎟ ⎜ C ⎟ ⎜ y ⎟ ⎜ Cz ⎟ ⎝ ⎠

517

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Engineering Mechanics - Statics

⎛ Ay ⎞ ⎜ ⎟ ⎛ −200 ⎞ ⎜ Az ⎟ ⎜ −667 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ x ⎟ = 0 ⎟ lb ⎜ Bz ⎟ ⎜ 667 ⎟ ⎟ ⎜ ⎟ ⎜ − 300 ⎜ ⎟ ⎜ Cy ⎟ ⎜ ⎜ ⎟ ⎝ 0 ⎟⎠ ⎝ Cz ⎠

Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ 167 ⎞ F BC ⎜ ⎟ ⎜ 250 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BD ⎟ = −731 ⎟ lb ⎜ FAD ⎟ ⎜ 786 ⎟ ⎟ ⎜ ⎟ ⎜ − 391 ⎜ ⎟ ⎜ FAC ⎟ ⎜ ⎜ ⎟ ⎝ 0 ⎟⎠ F ⎝ CD ⎠

Positive (T) Negative (C)

Problem 6-58 The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Given:

⎛ 0 ⎞ ⎜ 0 ⎟ lb F1 = ⎜ ⎟ ⎝ −500 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F 2 = 400 lb ⎜ ⎟ ⎝ 0 ⎠ a = 4 ft b = 3 ft c = 3 ft Solution: Find the external reactions Guesses E y = 1 lb

Cy = 1 lb

Cz = 1 lb

Dx = 1 lb

Dy = 1 lb

Dz = 1 lb

518

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given

⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0 ⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ D ⎛0⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛⎜ x ⎞⎟ ⎛ −b ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + ⎜ 0 ⎟ × Cy = 0 ⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎝ c ⎠ ⎜⎝ Dz ⎟⎠ ⎝ c ⎠ ⎜⎝ Cz ⎟⎠

⎛ Ey ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜C ⎟ ⎜ z ⎟ = Find E , C , C , D , D , D ( y y z x y z) ⎜ Dx ⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎝ Dz ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛ 266.667 ⎞ ⎜ Cy ⎟ ⎜ −400 ⎟ ⎟ ⎜C ⎟ ⎜ ⎜ ⎟ 0 ⎜ z⎟ = ⎟ lb ⎜ Dx ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ − 266.667 ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎜ ⎟ ⎝ 500 ⎟⎠ ⎝ Dz ⎠

Now find the force in each member.

⎛ −b ⎞ ⎛ −b ⎞ ⎜ ⎟ AB = 0 AC = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝c ⎠

⎛0⎞ AD = ⎜ −a ⎟ ⎜ ⎟ ⎝c ⎠

⎛0⎞ AE = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛b⎞ ⎜ ⎟ BC = −a BE = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ c ⎝ ⎠ ⎝0⎠

⎛0⎞ BF = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛b⎞ CD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛0⎞ ⎜ ⎟ CF = 0 DE = ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝ −c ⎠

⎛ −b ⎞ DF = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎛ −b ⎞ EF = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F BC = 1 lb

F BE = 1 lb

F BF = 1 lb

F CD = 1 lb

519

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F CF = 1 lb

Chapter 6

F DE = 1lb

F DF = 1 lb

F EF = 1 lb

Given F 1 + F AB

AB

F 2 + F BC

BC

AB

BC

+ FAC

AC

+ FBF

BF

AC

BF

+ FAD

AD

+ FBE

BE

AD

BE

+ FAE + F AB

AE AE −AB AB

=0

=0

⎛⎜ 0 ⎟⎞ −AE −BE EF −DE + FBE + F EF + F DE =0 ⎜ Ey ⎟ + FAE AE BE EF DE ⎜0⎟ ⎝ ⎠ F BF

−BF BF

+ F CF

−CF CF

+ F DF

−DF DF

+ FEF

−EF EF

=0

⎛0⎞ ⎜ ⎟ −BC −AC CD CF + FAC + FCD + FCF =0 ⎜ Cy ⎟ + FBC BC AC CD CF ⎜C ⎟ ⎝ z⎠ ⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF) ⎜ FBF ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

520

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Engineering Mechanics - Statics

⎛ FAB ⎞ ⎜ ⎟ ⎛ −300 ⎞ F AC ⎜ ⎟ ⎜ 583.095 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ AD ⎟ = 333.333 ⎟ lb ⎜ FAE ⎟ ⎜ −666.667 ⎟ ⎟ ⎜ ⎟ ⎜ 0 ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎜ ⎟ ⎝ 500 ⎟⎠ F ⎝ BE ⎠

Chapter 6

⎛ FBF ⎞ ⎜ ⎟ ⎛ 0 ⎞ F CD ⎜ ⎟ ⎜ −300 ⎟ ⎟ ⎜F ⎟ ⎜ Positive (T) ⎜ ⎜ CF ⎟ = −300 ⎟ lb Negative (C) ⎜ FDE ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ 424.264 ⎜ ⎟ ⎜ FDF ⎟ ⎜ ⎜ ⎟ ⎝ −300 ⎟⎠ F ⎝ EF ⎠

Problem 6-59 The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Given:

⎛ 200 ⎞ ⎜ ⎟ F 1 = 300 lb ⎜ ⎟ ⎝ −500 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F 2 = 400 lb ⎜ ⎟ ⎝ 0 ⎠ a = 4 ft b = 3 ft c = 3 ft Solution: Find the external reactions Guesses E y = 1 lb

Cy = 1 lb

Cz = 1 lb

Dx = 1 lb

Dy = 1 lb

Dz = 1 lb

521

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Engineering Mechanics - Statics

Chapter 6

Given

⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0 ⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ D ⎛0⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛⎜ x ⎞⎟ ⎛ −b ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + ⎜ 0 ⎟ × Cy = 0 ⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎝ c ⎠ ⎜⎝ Dz ⎟⎠ ⎝ c ⎠ ⎜⎝ Cz ⎟⎠

⎛ Ey ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜C ⎟ ⎜ z ⎟ = Find E , C , C , D , D , D ( y y z x y z) ⎜ Dx ⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎝ Dz ⎠

⎛ Ey ⎞ ⎞ ⎜ ⎟ ⎛ −33.333 ⎜ ⎟ C ⎜ y⎟ −666.667 ⎟ ⎜C ⎟ ⎜ 200 ⎜ ⎟ z ⎜ ⎟= ⎜ ⎟ lb −200 ⎜ Dx ⎟ ⎜ ⎟ ⎜ ⎟ − 13 ⎟ ⎜ −1.253 × 10 ⎜ Dy ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 300 ⎠ Dz ⎝ ⎠

Now find the force in each member.

⎛ −b ⎞ ⎛ −b ⎞ ⎜ ⎟ AB = 0 AC = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝c ⎠

⎛0⎞ AD = ⎜ −a ⎟ ⎜ ⎟ ⎝c ⎠

⎛0⎞ AE = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛b⎞ ⎜ ⎟ BC = −a BE = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝c ⎠ ⎝0⎠

⎛0⎞ BF = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛b⎞ CD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛0⎞ ⎜ ⎟ CF = 0 DE = ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝ −c ⎠

⎛ −b ⎞ DF = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎛ −b ⎞ EF = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F BC = 1 lb

F BE = 1 lb

F BF = 1 lb

F CD = 1 lb

F CF = 1 lb

F DE = 1lb

F DF = 1 lb

F EF = 1 lb

Given 522

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F 1 + F AB

AB

F 2 + F BC

BC

AB

BC

Chapter 6

+ FAC

AC

+ FBF

BF

AC

BF

+ FAD

AD

+ FBE

BE

AD

BE

+ FAE + F AB

AE AE −AB AB

=0

=0

⎛⎜ 0 ⎟⎞ −AE −BE EF −DE + FBE + F EF + F DE =0 ⎜ Ey ⎟ + FAE AE BE EF DE ⎜0⎟ ⎝ ⎠ F BF

−BF BF

+ F CF

−CF CF

+ F DF

−DF DF

+ FEF

−EF EF

=0

⎛0⎞ ⎜ ⎟ −BC −AC CD CF + FAC + FCD + FCF =0 ⎜ Cy ⎟ + FBC BC AC CD CF ⎜C ⎟ ⎝ z⎠ ⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF) F BF ⎜ ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

523

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎛ FBF ⎞ −300 ⎜ ⎟ ⎛ ⎜⎞ ⎟ ⎛ 0 ⎞ ⎜ ⎟ F F AC CD ⎜ ⎟ ⎜ ⎟ ⎜ −500 ⎟ 971.825 ⎜ ⎟ ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ Positive (T) − 11 ⎜ ⎟ ⎜ ⎜ AD ⎟ = 1.121 × 10 ⎜ lbCF ⎟ = −300 ⎟ lb Negative (C) ⎜ FAE ⎟ ⎜ −366.667 ⎜⎟ FDE ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜⎟ ⎟ ⎜ 424.264 ⎜ ⎟ ⎜ ⎟ 0 ⎜ FBC ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎝ ⎜⎠ ⎟ ⎝ −300 ⎟⎠ 500 F F ⎝ BE ⎠ ⎝ EF ⎠

Problem 6-60 Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by a ball-and-socket joints at A, B, and E. Hint: The support reaction at E acts along member EC. Why? Given:

⎛ −200 ⎞ ⎜ ⎟ F = 400 N ⎜ ⎟ ⎝ 0 ⎠

a = 2m b = 1.5 m c = 5m d = 1m e = 2m

Solution:

⎛ 0 ⎞ ⎜ ⎟ AC = a + b ⎜ ⎟ ⎝ 0 ⎠

⎛d⎞ ⎜ ⎟ AD = a ⎜ ⎟ ⎝e ⎠

⎛ −c − d ⎞ ⎜ 0 ⎟ BC = ⎜ ⎟ ⎝ 0 ⎠

⎛ −c ⎞ ⎜ ⎟ BD = −b ⎜ ⎟ ⎝e ⎠

Guesses

Given

⎛d⎞ ⎜ ⎟ CD = −b ⎜ ⎟ ⎝e ⎠

F AC = 1 N

F AD = 1 N

F BC = 1 N

F CD = 1 N

F EC = 1 N

F BD = 1 N

F + FAD

−AD AD

+ FBD

−BD BD

+ F CD

−CD CD

=0

524

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛⎜ 0 ⎞⎟ F CD + F BC + FAC +⎜ 0 ⎟ =0 CD BC AC ⎜ −FEC ⎟ ⎝ ⎠ CD

−BC

−AC

⎛ FAC ⎞ ⎜ ⎟ F ⎜ AD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AC AD BC BD CD EC ) ⎜ FBD ⎟ ⎜ ⎟ F ⎜ CD ⎟ ⎜ ⎟ ⎝ FEC ⎠

⎛ FAC ⎞ ⎜ ⎟ ⎛ 221 ⎞ F ⎜ AD ⎟ ⎜ 343 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ BC ⎟ = ⎜ 148 ⎟ N ⎜ FBD ⎟ ⎜ 186 ⎟ ⎟ ⎜ ⎟ ⎜ − 397 ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎜ ⎟ ⎝ −295 ⎟⎠ F ⎝ EC ⎠

Positive (T) Negative (C)

Problem 6-61 Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at C, D, E, and G. Units Used: 3

kN = 10 N Given: F = 3 kN a = 2m b = 1.5 m c = 2m d = 1m e = 1m Solution: Fv =

⎛0⎞ ⎜b⎟ 2 2⎜ ⎟ b + c ⎝ −c ⎠ F

525

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Engineering Mechanics - Statics

Chapter 6

uAG =

⎛ −e ⎞ ⎜ −a ⎟ ⎟ 2 2⎜ a +e ⎝ 0 ⎠

uAE =

⎛d⎞ ⎜ −a ⎟ ⎟ 2 2⎜ a +d ⎝ 0 ⎠

1

1

⎛0⎞ ⎜ ⎟ uAB = 0 ⎜ ⎟ ⎝ −1 ⎠

uBC = uAE uBD = uAG

⎛d⎞ ⎜ −a ⎟ ⎟ 2 2 2⎜ a +c +d ⎝ c ⎠ 1

uBE =

⎛ −e ⎞ ⎜ −a ⎟ ⎟ 2 2 2⎜ a +e +c ⎝ c ⎠ 1

uBG =

Guesses F AB = 1 kN

F AE = 1 kN

F BC = 1 kN

F BD = 1 kN

F BE = 1 kN

F BG = 1 kN

F AG = 1 kN

Given −c 2

c +b

2

e 2

a

F ( a) −

2

a +e

2

2

a +d F BD( a) −

FBC ( c) −

d 2

a +d

2

a 2

2

a +e

FBD( c) = 0

F BC ( a) = 0

F v + F AEuAE + FAGuAG + F ABuAB = 0 −F AB uAB + FBGuBG + FBEuBE + F BC uBC + FBDuBD = 0

526

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAE ⎟ ⎜ ⎟ FAG ⎜ ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F ( AB AE AG BC BD BE BG) ⎜ BC ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜F ⎟ ⎝ BG ⎠

⎛ FAB ⎞ ⎜ ⎟ ⎛ −2.4 ⎞ ⎟ ⎜ FAE ⎟ ⎜ ⎜ ⎟ ⎜ 1.006 ⎟ ⎜ FAG ⎟ ⎜ 1.006 ⎟ ⎜ F ⎟ = ⎜ −1.342 ⎟ kN ⎟ ⎜ BC ⎟ ⎜ ⎜ FBD ⎟ ⎜ −1.342 ⎟ ⎜ ⎟ ⎜ 1.8 ⎟ ⎟ ⎜ FBE ⎟ ⎜ 1.8 ⎝ ⎠ ⎜F ⎟ ⎝ BG ⎠ Positive (T) Negative (C)

Problem 6-62 Determine the force in members BD, AD, and AF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F. Given:

⎛ 0 ⎞ ⎜ ⎟ F = 250 lb ⎜ ⎟ ⎝ −250 ⎠ a = 6 ft b = 6 ft

θ = 60 deg Solution: Find the external reactions h = b sin ( θ ) Guesses Ax = 1 lb

B y = 1 lb

B z = 1 lb

Dy = 1 lb

F y = 1 lb

F z = 1 lb

Given

⎛

⎞

⎛

⎞ 527

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛⎜ Ax ⎞⎟ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎞⎟ F + ⎜ 0 ⎟ + ⎜ By ⎟ + ⎜ Dy ⎟ + ⎜ F y ⎟ = 0 ⎜ 0 ⎟ ⎜B ⎟ ⎜ 0 ⎟ ⎜F ⎟ ⎝ ⎠ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ ⎛ 0.5b ⎞ ⎛ b ⎞ ⎛⎜ Ax ⎞⎟ ⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0.5b ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜a⎟ × ⎜ ⎟ B ⎜ 0 ⎟ × D + ⎜ a ⎟ × Fy = 0 ⎜ 0 ⎟ + 0 × ⎜ y⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ h ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ h ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ Fz ⎟⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜B ⎟ ⎜ z ⎟ = Find A , B , B , D , F , F ( x y z y y z) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ ⎝ Fz ⎠

⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ By ⎟ ⎜ 72 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ z ⎟ = 125 ⎟ lb ⎜ Dy ⎟ ⎜ −394 ⎟ ⎟ ⎜ ⎟ ⎜ 72 ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎜ ⎟ ⎝ 125 ⎟⎠ ⎝ Fz ⎠

Now find the forces in the members

⎛0⎞ ⎜ ⎟ AB = −a ⎜ ⎟ ⎝0⎠

⎛ −b ⎞ ⎜ ⎟ AC = −a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ −a ⎟ AD = ⎜ ⎟ ⎝ h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ AE = ⎜ ⎟ ⎝ h ⎠ ⎛ −0.5b ⎞ ⎜ 0 ⎟ BD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ AF = 0 ⎜ ⎟ ⎝0⎠ ⎛ 0.5b ⎞ ⎜ 0 ⎟ CD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ BC = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ CF = a ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎜ ⎟ DE = a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ a ⎟ DF = ⎜ ⎟ ⎝ −h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ EF = ⎜ ⎟ ⎝ −h ⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F AF = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

F CF = 1 lb

528

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F DE = 1 lb

F DF = 1 lb

Chapter 6

F EF = 1 lb

Given −DE −AE EF F + FDE + F AE + FEF =0 DE AE EF F CF

CD −BC −AC + F CD + F BC + FAC =0 CF CD BC AC CF

⎛⎜ 0 ⎞⎟ F DE + FDF + F AD + F BD + FCD + ⎜ Dy ⎟ = 0 DE DF AD BD CD ⎜0 ⎟ ⎝ ⎠ ⎛0⎞ ⎜ ⎟ −AB BC BD F AB + FBC + F BD + ⎜ By ⎟ = 0 AB BC BD ⎜B ⎟ ⎝ z⎠ DE

DF

−AD

−BD

−CD

⎛⎜ Ax ⎞⎟ F AB + FAC + FAF + F AD + F AE +⎜ 0 ⎟ =0 AB AC AF AD AE ⎜0 ⎟ ⎝ ⎠ AB

AC

AF

AD

AE

⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FAF ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FAF , FBC , FBD , FCD , FCF , FDE , FDF , FEF) ⎜ FBD ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

529

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Engineering Mechanics - Statics

⎛ FBD ⎞ ⎛ −144.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAD ⎟ = ⎜ 204.1 ⎟ lb ⎜ F ⎟ ⎝ 72.2 ⎠ ⎝ AF ⎠

Chapter 6

Positive (T) Negative (C)

Problem 6-63 Determine the force in members CF, EF, and DF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F. Given:

⎛ 0 ⎞ ⎜ ⎟ F = 250 lb ⎜ ⎟ ⎝ −250 ⎠ a = 6 ft b = 6 ft

θ = 60 deg Solution: Find the external reactions h = b sin ( θ ) Guesses Ax = 1 lb

B y = 1 lb

B z = 1 lb

Dy = 1 lb

F y = 1 lb

F z = 1 lb

Given

⎛⎜ Ax ⎞⎟ ⎛⎜ 0 ⎟⎞ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎟⎞ F + ⎜ 0 ⎟ + ⎜ By ⎟ + ⎜ Dy ⎟ + ⎜ F y ⎟ = 0 ⎜ 0 ⎟ ⎜B ⎟ ⎜ 0 ⎟ ⎜F ⎟ ⎝ ⎠ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ ⎛ 0.5b ⎞ ⎛ b ⎞ ⎛⎜ Ax ⎞⎟ ⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0.5b ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜a⎟ × ⎜ ⎟ ⎜ 0 ⎟ × D + ⎜ a ⎟ × Fy = 0 + 0 × By + ⎜ ⎟ ⎜ ⎟ ⎜0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ h ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ h ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ Fz ⎟⎠

530

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜B ⎟ ⎜ z ⎟ = Find A , B , B , D , F , F ( x y z y y z) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ ⎝ Fz ⎠

⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ By ⎟ ⎜ 72 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ z ⎟ = 125 ⎟ lb ⎜ Dy ⎟ ⎜ −394 ⎟ ⎟ ⎜ ⎟ ⎜ 72 ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎜ ⎟ ⎝ 125 ⎟⎠ ⎝ Fz ⎠

Now find the forces in the members

⎛0⎞ ⎜ ⎟ AB = −a ⎜ ⎟ ⎝0⎠

⎛ −b ⎞ ⎜ ⎟ AC = −a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ −a ⎟ AD = ⎜ ⎟ ⎝ h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ AE = ⎜ ⎟ ⎝ h ⎠ ⎛ −0.5b ⎞ ⎜ 0 ⎟ BD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ AF = 0 ⎜ ⎟ ⎝0⎠ ⎛ 0.5b ⎞ ⎜ 0 ⎟ CD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ BC = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ CF = a ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎜ ⎟ DE = a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ a ⎟ DF = ⎜ ⎟ ⎝ −h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ EF = ⎜ ⎟ ⎝ −h ⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F AF = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

F CF = 1 lb

F DE = 1 lb

F DF = 1 lb

F EF = 1 lb

Given F + FDE

F CF

CF CF

−DE DE

+ F AE

+ F CD

−AE

CD CD

AE

+ FEF

+ F BC

EF EF

−BC BC

=0

+ FAC

−AC AC

=0

531

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ 0 ⎞⎟ F DE + FDF + F AD + F BD + FCD + ⎜ Dy ⎟ = 0 DE DF AD BD CD ⎜0 ⎟ ⎝ ⎠ DE

DF

−AD

−BD

−CD

⎛0⎞ ⎜ ⎟ F AB + FBC + F BD + ⎜ By ⎟ = 0 AB BC BD ⎜B ⎟ ⎝ z⎠ −AB

BC

BD

⎛⎜ Ax ⎞⎟ F AB + FAC + FAF + F AD + F AE +⎜ 0 ⎟ =0 AB AC AF AD AE ⎜0 ⎟ ⎝ ⎠ AB

AC

AF

AD

AE

⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FAF ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FAF , FBC , FBD , FCD , FCF , FDE , FDF , FEF) ⎜ FBD ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

⎛ FCF ⎞ ⎛ 72.2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FEF ⎟ = ⎜ −144.3 ⎟ lb ⎜F ⎟ ⎝ 0 ⎠ ⎝ DF ⎠

Positive (T) Negative (C)

532

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Engineering Mechanics - Statics

Chapter 6

Problem 6-64 Determine the force developed in each member of the space truss and state if the members are in tension or compression. The crate has weight W. Given: W = 150 lb a = 6 ft b = 6 ft c = 6 ft

Solution: h =

2

Unit Vectors

c −

⎛ a⎞ ⎜ ⎟ ⎝ 2⎠

2

uAD =

⎛⎜ −a ⎟⎞ 1 ⎜ 2 ⎟ 2⎜ 0 ⎟ 2 ⎛ a⎞ ⎜ ⎟ h +⎜ ⎟ ⎝ 2⎠ ⎝ h ⎠

uBD =

⎛⎜ a ⎟⎞ 1 ⎜2⎟ 2⎜ 0 ⎟ a 2 h + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

uAC =

⎛⎜ − a ⎟⎞ 1 ⎜ 2⎟ 2⎜ b ⎟ a 2 2 h + b + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

uBC =

⎛⎜ a ⎟⎞ 1 ⎜2⎟ 2⎜ b ⎟ a 2 2 h + b + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

Guesses F AB = 1 lb

B y = 1 lb

Ax = 1 lb

F AC = 1 lb

Ay = 1 lb

F AD = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

533

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given

⎛⎜ 0 ⎟⎞ ⎜ −FCD ⎟ − FAC uAC − FBC uBC = 0 ⎜ −W ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎜ ⎟ B ⎜ y ⎟ + FBC uBC + FBDuBD = 0 ⎜ ⎟ ⎝ 0 ⎠ ⎛ Ax − FAB ⎞ ⎜ ⎟ A ⎜ ⎟ + FAC uAC + FADuAD = 0 y ⎜ ⎟ 0 ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ B ⎟ ⎜ y ⎟ ⎜ FAB ⎟ ⎜ ⎟ ⎜ FAC ⎟ = Find ( Ax , Ay , By , FAB , FAC , FAD , FBC , FBD , FCD) ⎜F ⎟ ⎜ AD ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎟ ⎛ FAB ⎞ ⎜ ⎜ ⎟ ⎛ 0.0 ⎞ ⎝ FCD ⎠ F ⎜ AC ⎟ ⎜ −122.5 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ AD 86.6 Positive (T) ⎜ ⎟= lb ⎜ ⎟ Negative (C) ⎜ FBC ⎟ −122.5 ⎜ ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ 86.6 ⎟ ⎜ ⎟ ⎜⎝ 173.2 ⎟⎠ FCD ⎝ ⎠

Problem 6-65 The space truss is used to support vertical forces at joints B, C, and D. Determine the force in each member and state if the members are in tension or compression.

534

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Engineering Mechanics - Statics

Chapter 6

Units Used: 3

kN = 10 N Given: F 1 = 6 kN

a = 0.75 m

F 2 = 8 kN

b = 1.00 m

F 3 = 9 kN

c = 1.5 m

Solution: Assume that the connections at A, E, and F are rollers Guesses F BC = 1 kN

F CF = 1 kN

F CD = 1 kN

F AD = 1 kN

F DF = 1 kN

F DE = 1 kN

F BD = 1 kN

F BA = 1 kN

F EF = 1 kN

F AE = 1 kN

F AF = 1 kN Given Joint C F BC = 0

F CD = 0

−F 2 − F CF = 0 Joint D a 2

2

a +b

FBD +

a 2

2

2

a +b +c

FAD = 0

535

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Engineering Mechanics - Statics

b

−F CD −

2

−b

+ 2

a +b

2

2

2

2

2

b +c

−c 2

F AD −

c

−F 3 − F DE − +

FBD ...

2

a +b +c

2

a +b +c

Chapter 6

=0 b 2

2

b +c

F DF

F DF ... = 0

F AD

Joint B a

−F BC −

2

2

a +b

FBD = 0

b 2

2

a +b

FBD = 0

−F 1 − F BA = 0

Joint E a 2

2

a +b

FAE = 0

−F EF −

b 2

2

a +b

FAE = 0

⎛ FBC ⎞ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜F ⎟ ⎜ AD ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FDE ⎟ = Find ( FBC , FCF , FCD , FAD , FDF , FDE , FBD , FBA , FEF , FAE , FAF) ⎜F ⎟ ⎜ BD ⎟ ⎜ FBA ⎟ ⎜ ⎟ ⎜ FEF ⎟ ⎜ ⎟ ⎜ FAE ⎟ ⎜F ⎟ ⎝ AF ⎠

536

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Engineering Mechanics - Statics

⎛ FBC ⎞ ⎜ ⎟ ⎜ FCF ⎟ ⎛⎜ 0.00 ⎟⎞ ⎜ ⎟ ⎜ −8.00 ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 0.00 ⎟ AD ⎜ ⎟ ⎜ 0.00 ⎟ ⎜ FDF ⎟ ⎜ 0.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −9.00 ⎟ kN ⎜ F ⎟ ⎜ 0.00 ⎟ ⎜ BD ⎟ ⎜ ⎟ ⎜ FBA ⎟ ⎜ −6.00 ⎟ ⎜ ⎟ ⎜ 0.00 ⎟ ⎜ FEF ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0.00 ⎟ F AE ⎜ ⎟ ⎝ 0.00 ⎠ ⎜F ⎟ ⎝ AF ⎠

Chapter 6

Positive (T) Negative (C)

Problem 6-66 A force P is applied to the handles of the pliers. Determine the force developed on the smooth bolt B and the reaction that pin A exerts on its attached members. Given: P = 8 lb a = 1.25 in b = 5 in c = 1.5 in Solution: ΣMA = 0;

−R B c + P b = 0 RB = P

b c

R B = 26.7 lb ΣF x = 0;

Ax = 0

ΣF y = 0;

Ay − P − RB = 0 537

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Ay = P + RB Ay = 34.7 lb

Problem 6-67 The eye hook has a positive locking latch when it supports the load because its two parts are pin-connected at A and they bear against one another along the smooth surface at B. Determine the resultant force at the pin and the normal force at B when the eye hook supports load F . Given: F = 800 lb a = 0.25 in b = 3 in c = 2 in

θ = 30 deg

Solution: Σ MA = 0;

−F B cos ( 90 deg − θ ) ( b) − F B sin ( 90 deg − θ ) ( c) + F a = 0 FB = F

+

↑Σ Fy = 0;

+ Σ F x = 0; →

a

cos ( 90 deg − θ ) b + sin ( 90 deg − θ ) c

F B = 61.9 lb

−F − FB sin ( 90 deg − θ ) + A y = 0 Ay = F + F B sin ( 90 deg − θ )

Ay = 854 lb

Ax − F B cos ( 90 deg − θ ) = 0 Ax = FB cos ( 90 deg − θ ) FA =

2

2

Ax + Ay

Ax = 30.9 lb F A = 854 lb

Problem 6-68 Determine the force P needed to hold the block of mass F in equilibrium. 538

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given: F = 20 lb Solution: Pulley B:

ΣF y = 0;

2P − T = 0

Pulley A:

ΣF y = 0;

2T − F = 0

T =

1 F 2

2P = T

T = 10 lb P =

1 T 2

P = 5 lb

Problem 6-69 The link is used to hold the rod in place. Determine the required axial force on the screw at E if the largest force to be exerted on the rod at B, C or D is to be F max. Also, find the magnitude of the force reaction at pin A. Assume all surfaces of contact are smooth. Given: F max = 100 lb a = 100 mm b = 80 mm c = 50 mm

θ = 45 deg

539

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Engineering Mechanics - Statics

Chapter 6

Solution: Assign an initial value for R E. This will be scaled at the end of the problem. Guesses

Given

Ax = 1 lb

Ay = 1 lb

R B = 1 lb

R C = 1 lb

R D = 1 lb

R E = 1 lb

− Ax + R E − R B cos ( θ ) = 0

Ay − R B sin ( θ ) = 0

R E a − RB cos ( θ ) ( a + b + c cos ( θ ) ) − RB sin ( θ ) c sin ( θ ) = 0 R B cos ( θ ) − R D = 0

R B sin ( θ ) − RC = 0

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ RB ⎟ = Find ( Ax , Ay , RB , RC , RD) ⎜R ⎟ ⎜ C⎟ ⎜ RD ⎟ ⎝ ⎠

⎛⎜ Ax ⎟⎞ ⎛ 0.601 ⎞ ⎜ Ay ⎟ ⎜ 0.399 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ RB ⎟ = ⎜ 0.564 ⎟ lb ⎜ R ⎟ ⎜ 0.399 ⎟ ⎜ C⎟ ⎜ ⎟ 0.399 ⎝ ⎠ ⎜ RD ⎟ ⎝ ⎠

Now find the critical load and scale the problem

⎛ RB ⎞ ⎜ ⎟ ans = ⎜ R C ⎟ ⎜R ⎟ ⎝ D⎠

F scale =

Fmax

R E = Fscale RE

max ( ans )

F A = Fscale

2

2

Ax + Ay

R E = 177.3 lb

F A = 127.9 lb

Problem 6-70 The man of weight W1 attempts to lift himself and the seat of weight W2 using the rope and pulley system shown. Determine the force at A needed to do so, and also find his reaction on the seat.

540

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Engineering Mechanics - Statics

Chapter 6

Given: W1 = 150 lb W2 = 10 lb Solution: Pulley C: ΣF y = 0;

3T − R = 0

Pulley B: ΣF y = 0;

3R − P = 0

Thus,

P = 9T

Man and seat: ΣF y = 0;

T + P − W1 − W2 = 0 10T = W1 + W2 T =

W1 + W2 10

P = 9T

T = 16 lb P = 144 lb

Seat: ΣF y = 0;

P − N − W2 = 0 N = P − W2

N = 134 lb

Problem 6-71 Determine the horizontal and vertical components of force that pins A and C exert on the frame. Given: F = 500 N a = 0.8 m

d = 0.4 m

541

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Engineering Mechanics - Statics

Chapter 6

b = 0.9 m

e = 1.2 m

c = 0.5 m

θ = 45deg

Solution: BC is a two-force member Member AB : ΣMA = 0;

e

−F c + FBC

2

2

a +e

b + F BC

2

a 2

( c + d) = 0 2

a +e

2

a +e F BC = F c eb+a c+a d

F BC = 200.3 N

Thus, Cx = F BC

Cy = F BC

ΣF x = 0;

ΣF y = 0;

Ax − F BC

e 2

Cx = 167 N

2

a +e a 2

Cy = 111 N

2

a +e e 2

=0

Ax = FBC

2

a +e

Ay − F + FBC

a 2

=0 2

a +e

e 2

Ax = 167 N

2

a +e

Ay = F − F BC

a 2

2

a +e

Ay = 389 N

Problem 6-72 Determine the horizontal and vertical components of force that pins A and C exert on the frame. Units Used: 3

kN = 10 N Given: F 1 = 1 kN

542

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Engineering Mechanics - Statics

Chapter 6

F 2 = 500 N

θ = 45 deg a = 0.2 m b = 0.2 m c = 0.4 m d = 0.4 m Solution: Guesses Ax = 1 N

Ay = 1 N

Cx = 1 N

Cy = 1 N

Given Ax − Cx = 0

Ay + Cy − F1 − F2 = 0

F1 a − Ay 2 a + Ax d = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠

−F 2 b + Cy( b + c) − Cx d = 0

⎛ Ax ⎞ ⎜ ⎟ ⎛⎜ 500 ⎞⎟ ⎜ Ay ⎟ ⎜ 1000 ⎟ N ⎜ ⎟=⎜ ⎟ C 500 x ⎜ ⎟ ⎜ ⎟ ⎜ C ⎟ ⎝ 500 ⎠ ⎝ y⎠

Problem 6-73 The truck exerts the three forces shown on the girders of the bridge. Determine the reactions at the supports when the truck is in the position shown. The girders are connected together by a short vertical link DC. Units Used: 3

kip = 10 lb

543

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Engineering Mechanics - Statics

Chapter 6

Given: a = 55 ft

f = 12 ft

b = 10 ft

F 1 = 5 kip

c = 48 ft

F 2 = 4 kip

d = 5 ft

F 3 = 2 kip

e = 2 ft Solution:

Member CE: ΣMC = 0;

−F 3 e + E y( e + c) = 0

e Ey = F3 e+c

E y = 80 lb

ΣF y = 0;

Cy − F3 + Ey = 0

Cy = F 3 − E y

Cy = 1920 lb

Member ABD: ΣMA = 0;

−F 1 a − F2( d + a) − Cy( a + d + b) + By( a + d) = 0 By =

ΣF y = 0;

F 1 a + F2( d + a) + Cy( a + d + b) d+a

B y = 10.8 kip

Ay − F 1 + B y − F2 − Cy = 0 Ay = Cy + F 1 − B y + F2

Ay = 96.7 lb

Problem 6-74 Determine the greatest force P that can be applied to the frame if the largest force resultant acting at A can have a magnitude F max. Units Used: 3

kN = 10 N

544

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given: F max = 2 kN a = 0.75 m b = 0.75 m c = 0.5 m d = 0.1 m Solution: Σ MA = 0;

T( c + d) − P( a + b) = 0

+ Σ F x = 0; →

Ax − T = 0

+

↑Σ Fy = 0;

Ay − P = 0

Thus,

T=

a+b c+d

P

Ay = P

Ax =

a+b c+d

P

Require, F max = P =

2

2

Ax + Ay F max 2

⎛ a + b⎞ + 1 ⎜ ⎟ ⎝c + d⎠ P = 743 N

Problem 6-75 The compound beam is pin supported at B and supported by rockers at A and C. There is a hinge (pin) at D. Determine the reactions at the supports. Units Used: 3

kN = 10 N

545

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 7 kN

a = 4m

F 2 = 6 kN

b = 2m

F 3 = 16 kN

c = 3m

θ = 60 deg

d = 4m

Solution: Member DC : ΣMD = 0;

−F 1 sin ( θ ) ( a − c) + Cy a = 0 Cy = F 1 sin ( θ )

ΣF y = 0;

a−c

Cy = 1.52 kN

a

Dy − F 1 sin ( θ ) + Cy = 0 Dy = F1 sin ( θ ) − Cy

ΣF x = 0;

Dy = 4.55 kN

Dx − F 1 cos ( θ ) = 0 Dx = F1 cos ( θ )

Dx = 3.5 kN

Member ABD : ΣMA = 0;

−F 3 a − F2( 2 a + b) − Dy( 3 a + b) + By2 a = 0 By =

ΣF y = 0;

F 3 a + F2( 2a + b) + Dy ( 3a + b) 2a

Ay − F 3 + B y − F2 − Dy = 0 Ay = Dy + F3 − By + F 2

ΣF x = 0;

B y = 23.5 kN

Ay = 3.09 kN

B x − F1 cos ( θ ) = 0 B x = F 1 cos ( θ )

B x = 3.5 kN

Problem 6-76 The compound beam is fixed supported at A and supported by rockers at B and C. If there are hinges (pins) at D and E, determine the reactions at the supports A, B, and C.

546

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Engineering Mechanics - Statics

Chapter 6

Units Used: 3

kN = 10 N Given: a = 2 m M = 48 kN⋅ m b = 4m

kN w1 = 8 m

c = 2m

kN w2 = 6 d = 6m m e = 3m

Solution: Guesses Ax = 1 N

Ay = 1 N

MA = 1 N m

Dx = 1 N

Dy = 1 N

By = 1 N

Ey = 1 N

Ex = 1 N

Cy = 1 N

Given Ay − w2 a − Dy = 0 a MA − w2 a − Dy a = 0 2 −w1

( b + c) 2

E y − w1 −w1⎛⎜

2

d+e 2

+ B y b − E y( b + c) = 0 + Cy = 0

− Ax − Dx = 0 Dy − w1( b + c) + By − E y = 0 Dx + E x = 0 −E x = 0

d + e⎞⎛ d + e⎞

⎟⎜ ⎟ + Cy d − M = 0 ⎝ 2 ⎠⎝ 3 ⎠

547

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Engineering Mechanics - Statics

Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜M ⎟ ⎜ A⎟ ⎜ Dx ⎟ ⎜ ⎟ D ⎜ y ⎟ = Find ( Ax , Ay , MA , Dx , Dy , By , Ey , Ex , Cy) ⎜B ⎟ ⎜ y⎟ ⎜ Ey ⎟ ⎜ ⎟ E ⎜ x⎟ ⎟ ⎜ ⎝ Cy ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ = kN ⎜ Ay ⎟ ⎜⎝ 19 ⎟⎠ ⎝ ⎠ MA = 26 kN m B y = 51 kN Cy = 26 kN

Problem 6-77 Determine the reactions at supports A and B. Units Used: 3

kip = 10 lb Given: lb w1 = 500 ft lb w2 = 700 ft a = 6 ft b = 8 ft c = 9 ft d = 6 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

B x = 1 lb

B y = 1 lb

F CD = 1 lb MA = 1 lb⋅ ft

548

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given Ay − w1 a −

MA − w1 a a −

d 2

b +d

2

F CD = 0

d 2

d +b

2

Ax −

b 2

b +d

b

F CD2 a = 0

2

2

b +d

d

2

FCD = 0

FCD − Bx = 0

c c B y c − w2 =0 2 3

c FCD − w2 + B y = 0 2 2 2 b +d

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ B ⎟ ⎜ x ⎟ = Find A , A , B , B , F , M ( x y x y CD A) ⎜ By ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎝ MA ⎠

⎛⎜ Ax ⎞⎟ ⎛ 2.8 ⎞ = kip ⎜ Ay ⎟ ⎜⎝ 5.1 ⎟⎠ ⎝ ⎠ MA = 43.2 kip⋅ ft

⎛⎜ Bx ⎟⎞ ⎛ 2.8 ⎞ = kip ⎜ By ⎟ ⎜⎝ 1.05 ⎟⎠ ⎝ ⎠

Problem 6-78 Determine the horizontal and vertical components of force at C which member ABC exerts on member CEF. Given: F = 300 lb a = 4 ft b = 6 ft c = 3 ft r = 1 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

F y = 1 lb

549

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Engineering Mechanics - Statics

Cx = 1 lb

Chapter 6

Cy = 1 lb

Given b b Ax a − A y − Cx a − Cy = 0 2 2 Cx a − F r = 0 Ax = 0 Ay + Fy − F = 0 F y b − F ( b + c + r) = 0

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Fy ⎟ = Find ( Ax , Ay , Fy , Cx , Cy) ⎜C ⎟ ⎜ x⎟ ⎜ Cy ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ −200 ⎟ lb ⎜ F ⎟ ⎝ 500 ⎠ ⎝ y⎠

⎛⎜ Cx ⎟⎞ ⎛ 75 ⎞ = lb ⎜ Cy ⎟ ⎜⎝ 100 ⎟⎠ ⎝ ⎠

Problem 6-79 Determine the horizontal and vertical components of force that the pins at A, B, and C exert on their connecting members. Units Used: 3

kN = 10 N Given: F = 800 N a = 1m r = 50 mm b = 0.2 m Solution: −F ( a + r) + A x b = 0 550

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Engineering Mechanics - Statics

Ax = F

a+r b

Chapter 6

Ax = 4.2 kN

− Ax + Bx = 0 Bx = Ax

B x = 4.2 kN

−F r − Ay b + A x b = 0 Ay =

−F r + A x b b

Ay = 4 kN

Ay − By − F = 0 By = Ay − F

B y = 3.2 kN

− Ax + F + Cx = 0 Cx = Ax − F

Cx = 3.4 kN

Ay − Cy = 0 Cy = Ay

Cy = 4 kN

Problem 6-80 Operation of exhaust and intake valves in an automobile engine consists of the cam C, push rod DE, rocker arm EFG which is pinned at F, and a spring and valve, V. If the spring is compressed a distance δ when the valve is open as shown, determine the normal force acting on the cam lobe at C. Assume the cam and bearings at H, I, and J are smooth.The spring has a stiffness k. Given: a = 25 mm b = 40 mm

δ = 20 mm k = 300

N m

551

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Engineering Mechanics - Statics

Chapter 6

Solution: F s = kδ

Fs = 6 N

ΣF y = 0; −F G + Fs = 0 FG = Fs

FG = 6 N

ΣMF = 0; FG b + T a = 0 b T = FG a

T = 9.60 N

552

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Engineering Mechanics - Statics

Chapter 6

Problem 6-81 Determine the force P on the cord, and the angle θ that the pulley-supporting link AB makes with the vertical. Neglect the mass of the pulleys and the link. The block has weight W and the cord is attached to the pin at B. The pulleys have radii of r1 and r2. Given: W = 200 lb r1 = 2 in r2 = 1 in

φ = 45 deg Solution: The initial guesses are

θ = 30 deg +

↑Σ Fy = 0;

F AB = 30 lb 2P − W = 0 P =

1 2

W

P = 100 lb

Given + Σ F x = 0; →

P cos ( φ ) − FAB sin ( θ ) = 0

+

F AB cos ( θ ) − P − P − P sin ( φ ) = 0

↑Σ Fy = 0;

⎛ θ ⎞ ⎜ F ⎟ = Find ( θ , FAB) ⎝ AB ⎠

θ = 14.6 deg

F AB = 280 lb

Problem 6-82 The nail cutter consists of the handle and the two cutting blades. Assuming the blades are pin connected at B and the surface at D is smooth, determine the normal force on the fingernail when a force F is applied to the handles as shown.The pin AC slides through a smooth hole at A and is attached to the bottom member at C.

553

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Engineering Mechanics - Statics

Chapter 6

Given: F = 1 lb a = 0.25 in b = 1.5 in Solution: Handle : ΣMD = 0;

FA a − F b = 0 FA = F

ΣF y = 0;

⎛ b⎞ ⎜ ⎟ ⎝ a⎠

F A = 6 lb

ND − FA − F = 0 ND = F A + F

ND = 7 lb

Top blade : ΣMB = 0;

ND b − F N( 2 a + b) = 0 b ⎞ F N = ND ⎛⎜ ⎟ ⎝ 2 a + b⎠

F N = 5.25 lb

Problem 6-83 The wall crane supports load F. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W? Units Used:

3

kip = 10 lb

Given: F = 700 lb

554

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Engineering Mechanics - Statics

Chapter 6

a = 4 ft b = 4 ft c = 4 ft

θ = 60 deg Solution: Pulley E: +

↑Σ Fy = 0; T =

1 2

2T − F = 0 T = 350 lb

F

This is the force in the cable at the winch W a φ = atan ⎛⎜ ⎟⎞

Member ABC:

⎝ b⎠

Σ MA = 0;

(

)

−F ( b + c) + TBD sin ( φ ) − T sin ( θ ) b = 0

⎛ b + c ⎞ + T sin ( θ ) ⎟ ⎝ b ⎠

F⎜ TBD =

sin ( φ ) 3

TBD = 2.409 × 10 lb +

↑Σ Fy = 0; − Ay + TBD sin ( φ ) − T sin ( θ ) − F = 0 Ay = TBD sin ( φ ) − T sin ( θ ) − F

Ay = 700 lb

+ Σ F x = 0; →

Ax − TBD cos ( φ ) − T cos ( θ ) = 0 Ax = TBD cos ( φ ) + T cos ( θ )

Ax = 1.878 kip

At D: 555

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Engineering Mechanics - Statics

Chapter 6

Dx = TBD cos ( φ )

Dx = 1.703 kip

Dy = TBD sin ( φ )

Dy = 1.703 kip

Problem 6-84 Determine the force that the smooth roller C exerts on beam AB. Also, what are the horizontal and vertical components of reaction at pin A? Neglect the weight of the frame and roller. Given: M = 60 lb⋅ ft a = 3 ft b = 4 ft c = 0.5 ft Solution: Σ MA = 0;

−M + Dx c = 0 M Dx = c Dx = 120 lb

+ Σ F x = 0; →

+

↑Σ Fy = 0; Σ MB = 0;

Ax − Dx = 0 Ax = Dx

Ax = 120 lb

Ay = 0

Ay = 0 lb

−Nc b + Dx c = 0 c Nc = Dx b

Nc = 15.0 lb

Problem 6-85 Determine the horizontal and vertical components of force which the pins exert on member ABC.

556

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Engineering Mechanics - Statics

Chapter 6

Given: W = 80 lb a = 6 ft b = 9 ft c = 3 ft r = 0.5 ft Solution: + Σ F x = 0; →

− Ax + W = 0 Ax = W

+

↑Σ Fy = 0;

Ax = 80 lb

Ay − W = 0 Ay = W

Σ MC = 0;

Ay = 80 lb

Ay( a + b) − B y b = 0 By = Ay

a+b b

B y = 133 lb Σ MD = 0;

−W( c − r) + By b − Bx c = 0 Bx =

+ Σ F x = 0; →

B y b − W ( c − r)

B x = 333 lb

c

Ax + B x − Cx = 0 Cx = Ax + B x

+

↑Σ Fy = 0;

Cx = 413 lb

− Ay + B y − Cy = 0 Cy = B y − A y

Cy = 53.3 lb

Problem 6-86 The floor beams AB and BC are stiffened using the two tie rods CD and AD. Determine the force along each rod when the floor beams are subjected to a uniform load w. Assume the three contacting members at B are smooth and the joints at A, C, and D are pins. Hint: 557

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Engineering Mechanics - Statics

Chapter 6

g j Members AD, CD, and BD are two-force members

p

3

kip = 10 lb

Units Used: Given: w = 80

lb ft

b = 5 ft a = 12 ft Solution: Due to summetry: Cy =

w( 2a) 2

Cy = 960 lb

Member BC : ΣMB = 0;

a Cy( a) − w a⎛⎜ ⎟⎞ − T⎛ ⎜ 2

⎝ ⎠

a T = ⎛⎜ Cy − w ⎟⎞ 2⎠ ⎝

⎞a = 0 2 2⎟ ⎝ a +b ⎠

2

b

2

a +b

T = 1.248 kip

b

Problem 6-87 Determine the horizontal and vertical components of force at pins B and C. Given: F = 50 lb

c = 6 ft

a = 4 ft

d = 1.5 ft

558

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Engineering Mechanics - Statics

b = 4 ft Solution:

Chapter 6

r = 0.5 ft Guesses

Cx = 1 lb

Cy = 1 lb

B x = 1 lb

B y = 1 lb

Given F ( a − r) + Cx c − Cy( a + b) = 0 −F ( a − r) − F( d + r) + Cy( a + b) = 0 −B x + F + Cx = 0 B y − F + Cy = 0

⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜ ⎟ = Find ( Bx , By , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠ ⎛ Bx ⎞ ⎜ ⎟ ⎛⎜ 66.667 ⎞⎟ ⎜ By ⎟ ⎜ 15.625 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Cx ⎟ ⎜ 16.667 ⎟ ⎜ C ⎟ ⎝ 34.375 ⎠ ⎝ y⎠

559

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Engineering Mechanics - Statics

Chapter 6

Problem 6-88 The skid steer loader has a mass M1, and in the position shown the center of mass is at G1. If there is a stone of mass M2 in the bucket, with center of mass at G2 determine the reactions of each pair of wheels A and B on the ground and the force in the hydraulic cylinder CD and at the pin E. There is a similar linkage on each side of the loader. Units Used: 3

Mg = 10 kg 3

kN = 10 N Given: M1 = 1.18 Mg M2 = 300 kg a = 1.25 m

d = 0.15 m

b = 1.5 m

e = 0.5 m

c = 0.75 m

θ = 30 deg

Solution:

Entire System:

ΣMA = 0;

M2 g b − M1 g( c − d) + NB c = 0 NB =

ΣF y = 0;

M1 g( c − d) − M2 g b c

NB = 3.37 kN

(Both wheels)

NA = 11.1 kN

(Both wheels)

NB − M2 g − M1 g + NA = 0 NA = −NB + M2 g + M1 g

Upper member: ΣME = 0;

M2 g( a + b) − 2 F CD sin ( θ ) a = 0

560

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Engineering Mechanics - Statics

F CD = ΣF x = 0; ΣF y = 0;

Chapter 6

M 2 g( a + b)

F CD = 6.5 kN

2 sin ( θ ) a

E x = F CD( cos ( θ ) ) M2 g Ey − + FCD sin ( θ ) = 0 2 Ey = FR =

M2 g 2

E x = 5607 N

− F CD sin ( θ )

2

E y = −1766 N

2

Ex + Ey

F R = 5.879 kN

Problem 6-89 Determine the horizontal and vertical components of force at each pin. The suspended cylinder has a weight W. Given: W = 80 lb

d = 6 ft

a = 3 ft

e = 2 ft

b = 4 ft

r = 1 ft

c = 4 ft Solution: Guesses Ax = 1 lb

B x = 1 lb

B y = 1 lb

F CD = 1 lb

E x = 1 lb

E y = 1 lb

Given Ex − Bx + W = 0 −E y + By = 0

561

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Engineering Mechanics - Statics

Chapter 6

−W r + Ey a − Ex d = 0 −B y +

−B y a +

c 2

FCD − W = 0

2

c +d c 2

c +d

− Ax + Bx +

2

F CD d − W( d + e − r) = 0

d 2

2

c +d

FCD − W = 0

⎛ Ax ⎞ ⎜ ⎟ B x ⎜ ⎟ ⎜ B ⎟ ⎜ y ⎟ = Find A , B , B , F , E , E ( x x y CD x y) ⎜ FCD ⎟ ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎟ ⎝ Ey ⎠ Cx = F CD

d 2

c +d

2

Cy = F CD

c 2

c +d

2

Dx = −Cx

Dy = −Cy

⎛ Ax ⎞ ⎜ ⎟ ⎛ 160 ⎞ ⎜ Bx ⎟ ⎜ ⎟ 80 ⎜B ⎟ ⎜ ⎟ 26.667 ⎜ y⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ 160 ⎜ ⎟ ⎜ ⎟ 106.667 C = ⎜ y⎟ ⎜ ⎟ lb ⎜D ⎟ ⎜ ⎟ −160 ⎜ x⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ −106.667 ⎟ ⎜ ⎟ ⎜ −8.694 × 10− 13 ⎟ ⎜ Ex ⎟ ⎜ ⎟ 26.667 ⎠ ⎜ ⎟ ⎝ Ey ⎝ ⎠

562

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Engineering Mechanics - Statics

Chapter 6

Problem 6-90 The two-member frame is pin connected at C, D, and E. The cable is attached to A, passes over the smooth peg at B, and is attached to a load W. Determine the horizontal and vertical reactions at each pin. Given: a = 2 ft b = 1 ft c = 0.75 ft W = 100 lb Solution: d =

c

( a + 2 b) b Initial guesses: Cx = 1 lb

Cy = 1 lb

Dx = 1 lb

Dy = 1 lb

E x = 1 lb

E y = 1 lb

Given −Dx + Cx − W = 0 −Dy + Cy − W = 0 −Cx c + Cy b + W d − W( a + 2 b) = 0 W − Cx + Ex = 0 E y − Cy = 0 −W d + Cx c + Cy b = 0

562

563

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Engineering Mechanics - Statics

Chapter 6

⎛ Cx ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜D ⎟ ⎜ x ⎟ = Find C , C , D , D , E , E ( x y x y x y) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎟ ⎝ Ey ⎠

⎛ Cx ⎞ ⎜ ⎟ ⎛ 133 ⎞ ⎜ Cy ⎟ ⎜ 200 ⎟ ⎟ ⎜D ⎟ ⎜ ⎜ ⎜ x ⎟ = 33 ⎟ lb ⎜ Dy ⎟ ⎜ 100 ⎟ ⎟ ⎜ ⎟ ⎜ 33 ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎜ ⎟ ⎝ 200 ⎟⎠ ⎝ Ey ⎠

Problem 6-91 Determine the horizontal and vertical components of force which the pins at A, B, and C exert on member ABC of the frame. Given: F 1 = 400 N F 2 = 300 N F 3 = 300 N a = 1.5 m b = 2m c = 1.5 m d = 2.5 m f = 1.5 m g = 2m e = a+b+c−d Solution: Guesses Ay = 1 N F BD = 1 N

Cx = 1 N

Cy = 1 N

F BE = 1 N

564

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Engineering Mechanics - Statics

Chapter 6

Given F 1 g + F2( a + b) + F3 a − Ay( f + g) = 0 F 1 g − Cy( f + g) = 0 Cx e = 0 −Cx −

f+g 2

e + ( f + g)

Ay − Cy −

2

FBD +

e 2

e + ( f + g)

2

f+g 2

d + ( f + g)

FBD −

2

F BE = 0

d 2

d + ( f + g)

2

F BE = 0

⎛⎜ Ay ⎞⎟ ⎜ Cx ⎟ ⎜ ⎟ C ⎜ y ⎟ = Find ( Ay , Cx , Cy , FBD , FBE) ⎜F ⎟ ⎜ BD ⎟ ⎜ FBE ⎟ ⎝ ⎠ Bx = −

By =

f+g 2

e + ( f + g) e 2

e + ( f + g)

2

2

F BD +

F BD +

f+g 2

d + ( f + g) d 2

d + ( f + g)

2

2

FBE

FBE

Ay = 657 N

⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ = N ⎜ By ⎟ ⎜⎝ 429 ⎟⎠ ⎝ ⎠ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎞ = N ⎜ Cy ⎟ ⎜⎝ 229 ⎟⎠ ⎝ ⎠

565

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Engineering Mechanics - Statics

Chapter 6

Problem 6-92 The derrick is pin-connected to the pivot at A. Determine the largest mass that can be supported by the derrick if the maximum force that can be sustained by the pin at A is Fmax. Units Used: 3

kN = 10 N m

g = 9.81

2

s 3

Mg = 10 kg Given: F max = 18 kN L = 5m

θ = 60 deg Solution: AB is a two-force member. Require F AB = Fmax +

↑

Σ F y = 0;

F AB sin ( θ ) − M = 2

Mg

⎛ FAB ⎞ ⎜ ⎟ ⎝ g ⎠

2

sin ( θ ) − W = 0

⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎝ sin ( θ ) + 2 ⎠

M = 5.439

1 2

Mg

s

Problem 6-93 Determine the required mass of the suspended cylinder if the tension in the chain wrapped around the freely turning gear is T. Also, what is the magnitude of the resultant force on pin A? Units Used: 3

kN = 10 N g = 9.8

m 2

s

566

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Engineering Mechanics - Statics

Chapter 6

Given: T = 2 kN L = 2 ft

θ = 30 deg φ = 45 deg

Solution: Σ MA = 0;

−2 T L cos ( θ ) + M g cos ( φ ) L cos ( θ ) + M g sin ( φ ) L sin ( θ ) = 0 M =

2 T cos ( θ )

(cos ( φ ) cos ( θ ) + sin (φ ) sin ( θ )) g

M = 1793 +

→ Σ Fx = 0;

1 2

kg

s

2 T − M g cos ( φ ) − A x = 0 Ax = 2 T − M g cos ( φ )

+

↑Σ Fy = 0;

M g sin ( φ ) − Ay = 0 Ay = M g sin ( φ ) FA =

2

2

Ax + Ay

F A = 2.928 kN

567

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Problem 6-94 The tongs consist of two jaws pinned to links at A, B, C, and D. Determine the horizontal and vertical components of force exerted on the stone of weight W at F and G in order to lift it. Given: a = 1 ft b = 2 ft c = 1.5 ft d = 1 ft W = 500 lb Solution: Guesses F x = 1 lb F y = 1 lb F AD = 1 lb F BE = 1 lb Given

2 Fy − W = 0 F AD − F x −

F AD b − Fx( b + c) = 0 a ⎞ ⎛ ⎜ 2 2 ⎟ FBE = 0 ⎝ a +d ⎠

⎛ Fx ⎞ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ = Find ( Fx , Fy , FAD , FBE) ⎜ FAD ⎟ ⎜F ⎟ ⎝ BE ⎠

−F y +

d ⎞ ⎛ ⎜ 2 2 ⎟ FBE = 0 ⎝ a +d ⎠

⎛⎜ Gx ⎞⎟ ⎛⎜ Fx ⎟⎞ = ⎜ Gy ⎟ ⎜ Fy ⎟ ⎝ ⎠ ⎝ ⎠

⎛ Fx ⎞ ⎜ ⎟ ⎛⎜ 333 ⎟⎞ ⎜ Fy ⎟ ⎜ 250 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Gx ⎟ ⎜ 333 ⎟ ⎜ G ⎟ ⎝ 250 ⎠ ⎝ y⎠

Problem 6-95 Determine the force P on the cable if the spring is compressed a distance δ when the mechanism is in the position shown. The spring has a stiffness k.

568

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given:

δ = 0.5 in

c = 6 in

lb

d = 6 in

k = 800

ft

a = 24 in

e = 4 in

b = 6 in

θ = 30 deg

Solution: F E = kδ

F E = 33.333 lb

The initial guesses are P = 20 lb

B x = 11 lb

B y = 34 lb

F CD = 34 lb

Given ΣMA = 0;

B x b + B y c − F E ( a + b) = 0

ΣMD = 0;

By d − P e = 0

+ Σ F x = 0; →

ΣMB = 0;

−B x + FCD cos ( θ ) = 0 F CD sin ( θ ) d − P ( d + e) = 0

⎛⎜ FCD ⎞⎟ ⎜ Bx ⎟ ⎜ ⎟ = Find ( FCD , Bx , By , P) B ⎜ y ⎟ ⎜ P ⎟ ⎝ ⎠ B x = 135.398 lb

B y = 31.269 lb

F CD = 156.344 lb

P = 46.903 lb

569

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Problem 6-96 The scale consists of five pin-connected members. Determine the load W on the pan EG if a weight F is suspended from the hook at A. Given: F = 3 lb

b = 3 in

a = 5 in

c = 4 in

d = 6 in

f = 2 in

e = 8 in Solution: Guesses

TC = 10 lb

TD = 10 lb

TG = 10 lb

W = 10 lb

Given Member ABCD:

ΣMB = 0;

F a − TC b − TD( b + e − d) = 0 Member EG: ΣMG = 0;

− TC e + W c = 0

ΣF y = 0;

TG − W + TC = 0

Member FH: ΣMH = 0;

−TD( d + f) + TG f = 0

⎛⎜ TC ⎟⎞ ⎜ TD ⎟ ⎜ ⎟ = Find ( TC , TD , TG , W) ⎜ TG ⎟ ⎜W⎟ ⎝ ⎠ W = 7.06 lb

Problem 6-97 The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF, 570

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

g p gg which are operated by the hydraulic cylinder H. The toggles push the movable bar G forward, pressing the plate p into the cavity. If the force which the plate exerts on the head is P, determine the force F in the hydraulic cylinder for the given angle θ. Units Used: 3

kN = 10 N Given: P = 12 kN a = 200 mm

θ = 30 deg Solution: Member EF: ΣME = 0;

−F y a cos ( θ ) + Fy =

ΣF x = 0;

Ex − Ex =

ΣF y = 0;

2

2

2

a sin ( θ ) = 0

tan ( θ )

P

P

P

F y = 3.464 kN

=0

P

E x = 6 kN

2

Ey − Fy = 0 Ey = Fy

E y = 3.464 kN

Joint E: ΣF x = 0;

−F DE cos ( θ ) − E x = 0 F DE =

−E x

cos ( θ )

F DE = −6.928 kN

F − Ey + F DE sin ( θ ) = 0

ΣF y = 0;

F = E y − FDE sin ( θ )

F = 6.93 kN

Problem 6-98 Determine the horizontal and vertical components of force at pins A and C of the two-member frame.

571

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given: N w1 = 500 m N w2 = 400 m N w3 = 600 m a = 3m b = 3m

Solution: Guesses Ax = 1 N

Ay = 1 N Cx = 1 N

Cy = 1 N

Given 1 Ay + Cy − w1 a − w2 a = 0 2

Ax a −

1 − Ax + Cx + w3 b = 0 2

− Ay a +

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠

1 2a 1 b a w1 a − w3 b − w2 a = 0 2 3 2 3 2 1 a w1 a = 0 2 3

⎛ Ax ⎞ ⎜ ⎟ ⎛⎜ 1400 ⎞⎟ ⎜ Ay ⎟ ⎜ 250 ⎟ N ⎜ ⎟=⎜ ⎟ C 500 x ⎜ ⎟ ⎜ ⎟ ⎜ C ⎟ ⎝ 1700 ⎠ ⎝ y⎠

572

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Engineering Mechanics - Statics

Chapter 6

Problem 6-99 The truck rests on the scale, which consists of a series of compound levers. If a mass M1 is placed on the pan P and it is required that the weight is located at a distance x to balance the “beam” ABC, determine the mass of the truck. There are pins at all lettered points. Is it necessary for the truck to be symmetrically placed on the scale? Explain. Units Used: 3

Mg = 10 kg

g = 9.81

m 2

s Given: M1 = 15 kg

FD = 3 m

x = 0.480 m

EF = 0.2 m

a = 0.2 m HI = 0.1 m

GH = 2.5 m

KJ = HI

KG = GH

Solution: Member ABC : ΣMB = 0;

−M1 g x + F AD a = 0

x F AD = M1 g a Member EFD : ΣME = 0;

2

F AD = 72 s N

−F y EF + F AD( FD + EF) = 0

⎛ FD + EF ⎞ ⎟ ⎝ EF ⎠

F y = F AD⎜

2

F y = 1152 s N

573

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Engineering Mechanics - Statics

Chapter 6

Member GHI : ΣMI = 0;

Hy HI − GY( GH + HI ) = 0

Member JKG : ΣMJ = 0;

(Fy − Gy)( KJ + GH ) − Ky( KJ ) = 0 Ky + Hy = Fy

KJ + KG HI

Scale Platform : ΣF y = 0;

Ky + Hy = W W = Fy⎛⎜

⎝

M =

W g

KJ + KG ⎞ HI

⎟ ⎠

M = 14.98 Mg

Because KJ = HI and KG = GH it doesn't matter where the truck is on the scale.

Problem 6-100 By squeezing on the hand brake of the bicycle, the rider subjects the brake cable to a tension T If the caliper mechanism is pin-connected to the bicycle frame at B, determine the normal force each brake pad exerts on the rim of the wheel. Is this the force that stops the wheel from turning? Explain. Given: T = 50 lb a = 2.5 in b = 3 in

574

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Engineering Mechanics - Statics

Chapter 6

Solution: Σ MB = 0;

−N b + T a = 0

N = T

a b

N = 41.7 lb

This normal force does not stop the wheel from turning. A frictional force (see Chapter 8), which acts along the wheel's rim stops the wheel.

Problem 6-101 If a force of magnitude P is applied perpendicular to the handle of the mechanism, determine the magnitude of force F for equilibrium. The members are pin-connected at A, B, C, and D. Given: P = 6 lb a = 25 in b = 4 in c = 5 in d = 4 in e = 5 in f = 5 in g = 30 in Solution: Σ MA = 0;

F BC b − P a = 0 F BC =

Pa b

575

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Engineering Mechanics - Statics

Chapter 6

F BC = 37.5 lb + Σ F x = 0; →

− Ax + P = 0 Ax = P

+

↑Σ Fy = 0;

Ax = 6 lb

− Ay + F BC = 0 Ay = FBC

Σ MD = 0;

Ay = 37.5 lb

−e Ax − Ay( b + c) + ( g + b + c)F = 0 F =

e Ax + Ay( b + c)

F = 9.423 lb

g+b+c

Problem 6-102 The pillar crane is subjected to the load having a mass M. Determine the force developed in the tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied in the position shown. Units Used: 3

kN = 10 N Given: M = 500 kg a = 1.8 m b = 2.4 m

θ 1 = 10 deg θ 2 = 20 deg g = 9.81

m 2

s Solution:

initial guesses: F CB = 10 kN

F AB = 10 kN

576

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Engineering Mechanics - Statics

Chapter 6

Given −M

( )

( )

( )

( )

g cos θ 1 − F AB cos θ 2 + FCB 2

−M

g sin θ 1 − FAB sin θ 2 + FCB 2

⎛ FAB ⎞

⎜ ⎟ = Find F , F ( AB CB) ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ Cx ⎟⎞ ⎜ Cy ⎟ ⎝ ⎠

b

=0

2

2

a +b a 2

− Mg = 0 2

a +b

=

⎛b⎞ ⎜ ⎟ 2 2 a a +b ⎝ ⎠ FCB

⎛ FAB ⎞ ⎛ 9.7 ⎞ ⎜ ⎟ ⎜ ⎟ C ⎜ x ⎟ = ⎜ 11.53 ⎟ kN ⎜ C ⎟ ⎝ 8.65 ⎠ ⎝ y ⎠

Problem 6-103 The tower truss has a weight W and a center of gravity at G. The rope system is used to hoist it into the vertical position. If rope CB is attached to the top of the shear leg AC and a second rope CD is attached to the truss, determine the required tension in BC to hold the truss in the position shown. The base of the truss and the shear leg bears against the stake at A, which can be considered as a pin. Also, compute the compressive force acting along the shear leg. Given: W = 575 lb

θ = 40 deg a = 5 ft b = 3 ft c = 10 ft d = 4 ft e = 8 ft Solution: Entire system:

ΣMA = 0;

TBC cos ( θ ) ( d + e) − TBC sin ( θ ) a − W( a + b) = 0 TBC =

W( a + b)

cos ( θ ) ( d + e) − sin ( θ ) a

TBC = 769 lb 577

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Engineering Mechanics - Statics

Chapter 6

CA is a two-force member.

At C:

e ⎞ ⎟ ⎝ b + c⎠

φ = atan ⎛⎜ initial guesses:

F CA = 500 lb

TCD = 300 lb

Given ΣF x = 0;

ΣF y = 0;

F CA

F CA

a 2

a + ( d + e)

2

d+e 2

a + ( d + e)

⎛⎜ FCA ⎟⎞ = Find ( FCA , TCD) ⎜ TCD ⎟ ⎝ ⎠

2

+ TCD cos ( φ ) − TBC cos ( θ ) = 0

− TCD sin ( φ ) − TBC sin ( θ ) = 0

TCD = 358 lb

F CA = 739 lb

Problem 6-104 The constant moment M is applied to the crank shaft. Determine the compressive force P that is exerted on the piston for equilibrium as a function of θ. Plot the results of P (ordinate) versus θ (abscissa) for 0 deg ≤ θ ≤ 90 deg. Given: a = 0.2 m b = 0.45 m M = 50 N⋅ m Solution: a cos ( θ ) = b sin ( φ )

φ = asin ⎛⎜ cos ( θ )⎟⎞ a ⎝b

⎠

−M + F BC cos ( θ − φ ) a = 0

578

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Engineering Mechanics - Statics

F BC =

Chapter 6

M

a cos ( θ − φ )

P = FBC cos ( φ ) =

M cos ( φ )

a cos ( θ − φ )

This function goes to infinity at θ = 90 deg, so we will only plot it to θ = 80 deg.

θ = 0 , 0.1 .. 80 φ ( θ ) = asin ⎛⎜ cos ( θ deg)⎟⎞

P (θ) =

a

⎝b

⎠

M cos ( φ ( θ ) )

a cos ( θ deg − φ ( θ ) )

Newtons

1500

P( θ )

1000 500 0

0

20

40

60

80

θ

Degrees

Problem 6-105 Five coins are stacked in the smooth plastic container shown. If each coin has weight W, determine the normal reactions of the bottom coin on the container at points A and B. Given: W = 0.0235 lb a = 3 b = 4

579

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Engineering Mechanics - Statics

Chapter 6

Solution: All coins : ΣF y = 0;

NB = 5W NB = 0.1175 lb

Bottom coin : ΣF y = 0;

⎛

⎞=0 ⎝ a +b ⎠ b

NB − W − N⎜

(

N = NB − W

2⎟

2

⎛ a2 + b2 ⎞ )⎜⎝ b ⎟⎠

N = 0.1175 lb ΣF x = 0;

⎛

NA = N⎜

a 2

⎞

2⎟

⎝ a +b ⎠

NA = 0.0705 lb

Problem 6-106 Determine the horizontal and vertical components of force at pin B and the normal force the pin at C exerts on the smooth slot. Also, determine the moment and horizontal and vertical reactions of force at A. There is a pulley at E. Given: F = 50 lb a = 4 ft b = 3 ft Solution: Guesses B x = 1 lb B y = 1 lb NC = 1 lb 580

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Engineering Mechanics - Statics

Ax = 1 lb

Ay = 1 lb

Chapter 6

MA = 1 lb⋅ ft

Given Bx +

a ⎛ ⎞ ⎜ 2 2 ⎟ NC − F = 0 ⎝ a +b ⎠

By −

b ⎛ ⎞ ⎜ 2 2 ⎟ NC − F = 0 ⎝ a +b ⎠

(F + Bx)a − (F + By)b = 0 F−

a ⎛ ⎞ ⎜ 2 2 ⎟ NC − Ax = 0 ⎝ a +b ⎠

b ⎛ ⎞ ⎜ 2 2 ⎟ NC − Ay = 0 ⎝ a +b ⎠ −F 2 a +

a ⎛ ⎞ ⎜ 2 2 ⎟ NC a + MA = 0 ⎝ a +b ⎠

⎛ Bx ⎞ ⎜ ⎟ B y ⎜ ⎟ ⎜N ⎟ ⎜ C ⎟ = Find B , B , N , A , A , M ( x y C x y A) ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎝ MA ⎠

NC = 20 lb

⎛⎜ Bx ⎟⎞ ⎛ 34 ⎞ = lb ⎜ By ⎟ ⎜⎝ 62 ⎟⎠ ⎝ ⎠ ⎛⎜ Ax ⎞⎟ ⎛ 34 ⎞ = lb ⎜ Ay ⎟ ⎜⎝ 12 ⎟⎠ ⎝ ⎠ MA = 336 lb⋅ ft

Problem 6-107 A force F is applied to the handles of the vise grip. Determine the compressive force developed on the smooth bolt shank A at the jaws. 581

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Engineering Mechanics - Statics

Chapter 6

Given: F = 5 lb

b = 1 in

a = 1.5 in

c = 3 in

d = 0.75 in

e = 1 in

θ = 20 deg Solution: From FBD (a) ΣME = 0;

⎡

⎤b = 0 ⎣ c + ( d + e) ⎦

F ( b + c) − F CD⎢

d+e

2

2⎥

⎡ c2 + ( d + e) 2⎤ ⎥ ⎣ b( d + e) ⎦

F CD = F( b + c) ⎢ ΣF x = 0;

⎡

F CD = 39.693 lb

⎤ ⎥ 2 2 ⎣ c + ( d + e) ⎦

E x = F CD⎢

c

E x = 34.286 lb From FBD (b) ΣMB = 0;

NA sin ( θ ) d + NA cos ( θ ) a − Ex( d + e) = 0

NA = Ex

d+e ⎛ ⎜ ( ) ⎝ sin θ d + cos ( θ )

⎞ ⎟

a⎠

NA = 36.0 lb

Problem 6-108 If a force of magnitude P is applied to the grip of the clamp, determine the compressive force F that the wood block exerts on the clamp.

582

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Engineering Mechanics - Statics

Chapter 6

Given: P = 10 lb a = 2 in b = 2 in c = 0.5 in d = 0.75 in e = 1.5 in

Solution: Define

b φ = atan ⎛⎜ ⎟⎞

φ = 69.444 deg

⎝ d⎠

From FBD (a), ΣMB = 0;

F CD cos ( φ ) c − P ( a + b + c) = 0 F CD =

+

↑Σ Fy = 0;

P( a + b + c)

F CD = 256.32 lb

cos ( φ ) ( c)

F CD sin ( φ ) − By = 0 B y = F CD sin ( φ )

B y = 240 lb

From FBD (b), ΣMA = 0;

By d − F e = 0 F =

By d

F = 120 lb

e

583

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Engineering Mechanics - Statics

Chapter 6

Problem 6-109 The hoist supports the engine of mass M. Determine the force in member DB and in the hydraulic cylinder H of member FB. Units Used: 3

kN = 10 N Given: M = 125 kg

d = 1m

a = 1m

e = 1m

b = 2m

f = 2m

c = 2m

g = 9.81

m 2

s Solution: Member GFE: ΣME = 0; −F FB⎡ ⎢

⎤ b + M g ( a + b) = 0 2 2⎥ ⎣ ( c + d) + ( b − e) ⎦

F FB = M g

c+d

⎡ a + b ⎤ ( c + d) 2 + ( b − e) 2 ⎢ ⎥ ⎣ b( c + d) ⎦

F FB = 1.94 kN ΣF x = 0;

⎡

⎤=0 ⎣ ( c + d) + ( b − e) ⎦ b−e

E x − FFB⎢

2⎥

2

⎡

⎤

b−e

E x = F FB⎢

2⎥

2

⎣ ( c + d) + ( b − e) ⎦

Member EDC: ΣΜc = 0;

⎛

⎞d = 0 ⎟ 2 2 ⎝ e +d ⎠

E x( c + d) − F DB⎜

F DB = Ex

⎛ c + d⎞ ⎜ ⎟ ⎝ ed ⎠

e

2

2

e +d

F DB = 2.601 kN

584

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Engineering Mechanics - Statics

Chapter 6

Problem 6-110 The flat-bed trailer has weight W1 and center of gravity at GT . It is pin-connected to the cab at D. The cab has a weight W2 and center of gravity at GC. Determine the range of values x for the position of the load L of weight W3 so that no axle is subjected to a force greater than FMax. The load has a center of gravity at GL . Given: W1 = 7000 lb

a = 4 ft

W2 = 6000 lb

b = 6 ft

W3 = 2000 lb

c = 3 ft

F max = 5500 lb d = 10 ft e = 12 ft Solution: Case 1:

Assume

Guesses

Ay = Fmax x = 1 ft

Given

Ay = Fmax B y = F max

Cy = F max

Dy = Fmax

Ay + B y − W2 − Dy = 0 −W2 a − Dy( a + b) + B y( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3 x + W1 e − Dy( c + d + e) = 0

⎛⎜ By ⎞⎟ ⎜ Cy ⎟ ⎜ ⎟ = Find ( By , Cy , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠ ⎛ Ay ⎞ ⎛⎜ 5.5 × 103 ⎟⎞ ⎜ ⎟ x1 = 30.917 ft ⎜ By ⎟ = ⎜ 6.333 × 103 ⎟ lb x1 = x ⎟ ⎜C ⎟ ⎜ Since By > Fmax then this solution is no good. ⎝ y ⎠ ⎜⎝ 3.167 × 103 ⎟⎠ Case 2:

Assume

Guesses

Ay = Fmax

B y = F max B y = F max

Cy = F max 585

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Engineering Mechanics - Statics

x = 1 ft Given

Chapter 6

Dy = Fmax

Ay + B y − W2 − Dy = 0 −W2 a − Dy( a + b) + B y( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3 x + W1 e − Dy( c + d + e) = 0

⎛⎜ Ay ⎞⎟ ⎜ Cy ⎟ ⎜ ⎟ = Find ( Ay , Cy , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠

⎛ Ay ⎞ ⎛⎜ 5.25 × 103 ⎟⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 5.5 × 103 ⎟ lb x2 = x ⎟ ⎜C ⎟ ⎜ 3 ⎜ ⎝ y ⎠ ⎝ 4.25 × 10 ⎟⎠

x2 = 17.375 ft

Since Ay < Fmax and Cy < F max then this solution is good. Cy = F max

Case 3:

Assume

Guesses

Ay = Fmax

B y = F max

x = 1 ft

Dy = Fmax

Given

Cy = F max

Ay + B y − W2 − Dy = 0 −W2 a − Dy( a + b) + B y( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3 x + W1 e − Dy( c + d + e) = 0

⎛⎜ Ay ⎞⎟ ⎜ By ⎟ ⎜ ⎟ = Find ( Ay , By , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠

⎛ Ay ⎞ ⎛⎜ 4.962 × 103 ⎟⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 4.538 × 103 ⎟ lb x3 = x ⎟ ⎜C ⎟ ⎜ 3 ⎜ ⎝ y ⎠ ⎝ 5.5 × 10 ⎟⎠

x3 = 1.75 ft

Since Ay < Fmax and B y < F max then this solution is good. We conclude that x3 = 1.75 ft < x < x2 = 17.375 ft

586

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Engineering Mechanics - Statics

Chapter 6

Problem 6-111 Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovel in equilibrium. The shovel load has a mass W and a center of gravity at G. All joints are pin connected. Units Used: 3

Mg = 10 kg 3

kN = 10 N Given: a = 0.25 m θ 1 = 30 deg b = 0.25 m θ 2 = 10 deg c = 1.5 m

θ 3 = 60 deg

d = 2m

W = 1.25 Mg

e = 0.5 m Solution: Assembly FHG :

(

( )) = 0

ΣMH = 0; −[ W g( e) ] + FEF c sin θ 1 F EF = W g

⎛ e ⎞ F = 8.175 kN (T) ⎜ c sin θ ⎟ EF ( ) 1 ⎝ ⎠

Assembly CEFHG:

(

)

( )

ΣMC = 0; F AD cos θ 1 + θ 2 b − W g⎡( a + b + c)cos θ 2 + e⎤ = 0 ⎣ ⎦ F AD = W g

⎛ cos ( θ 2) a + cos ( θ 2) b + cos ( θ 2) c + e ⎞ ⎜ ⎟ cos ( θ 1 + θ 2) b ⎝ ⎠

F AD = 158 kN (C)

587

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Engineering Mechanics - Statics

Chapter 6

Problem 6-112 The aircraft-hangar door opens and closes slowly by means of a motor which draws in the cable AB. If the door is made in two sections (bifold) and each section has a uniform weight W and length L, determine the force in the cable as a function of the door's position θ. The sections are pin-connected at C and D and the bottom is attached to a roller that travels along the vertical track.

587

Solution: L

⎛ θ ⎞ − 2L sin ⎛ θ ⎞ N = 0 ⎟ ⎜ ⎟ A ⎝2⎠ ⎝2⎠

cos ⎜

ΣMD = 0;

2W

ΣΜC = 0;

T L cos ⎜

2

⎛ θ ⎞ − N L sin ⎛ θ ⎞ − W L cos ⎛ θ ⎞ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ A 2 ⎝2⎠ ⎝2⎠ ⎝2⎠

W ⎛θ⎞ NA = cot ⎜ ⎟ 2 ⎝2⎠ T=W

Problem 6-113 A man having weight W attempts to lift himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight of the platform. Given: W = 175 lb

588

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Solution:

(a) Bar: +

↑Σ Fy = 0;

2

⎛ F⎞ − 2 ⎛ W⎞ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝2⎠

F = W

F = 175 lb

Man: +

↑Σ Fy = 0;

NC − W − 2

⎛ F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W + F

NC = 350 lb

( b) Bar: +

↑

Σ F y = 0;

2

⎛ W⎞ − 2 F = 0 ⎜ ⎟ 2 ⎝4⎠

F =

W 2

Man: +

↑Σ Fy = 0;

NC − W + 2

F = 87.5 lb

⎛ F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W − F

NC = 87.5 lb

589

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Problem 6-114 A man having weight W1 attempts to lift himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has weight W2.

Given: W1 = 175 lb W2 = 30 lb Solution: (a) Bar: +

↑Σ Fy = 0;

2

F 2

(

)

− W1 + W2 = 0

F = W1 + W2 F = 205 lb Man: +

↑

Σ F y = 0;

F NC − W1 − 2 =0 2 NC = F + W1 NC = 380 lb

( b) Bar: +

↑

Σ F y = 0;

−2

⎛ F ⎞ + 2 ⎛⎜ W1 + W2 ⎞⎟ = 0 ⎜ ⎟ ⎝ 2⎠ ⎝ 4 ⎠

590

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F =

Chapter 6

W1 + W2 2

F = 102 lb Man: +

↑

Σ F y = 0;

NC − W1 + 2

⎛F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W1 − F NC = 72.5 lb

Problem 6-115 The piston C moves vertically between the two smooth walls. If the spring has stiffness k and is unstretched when θ = 0, determine the couple M that must be applied to AB to hold the mechanism in equilibrium. Given: k = 15

lb in

θ = 30 deg a = 8 in b = 12 in

Solution: Geometry: b sin ( ψ) = a sin ( θ )

ψ = asin ⎛⎜ sin ( θ )

⎝

a⎞ ⎟ b⎠

φ = 180 deg − ψ − θ

ψ = 19.471 deg φ = 130.529 deg

591

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

rAC

sin ( φ )

=

b

Chapter 6

rAC = b

sin ( θ )

⎛ sin ( φ ) ⎞ ⎜ ⎟ ⎝ sin ( θ ) ⎠

rAC = 18.242 in

Free Body Diagram: The solution for this problem will be simplified if one realizes that member CB is a two force member. Since the spring stretches x = ( a + b) − rAC

x = 1.758 in F sp = k x

the spring force is

F sp = 26.371 lb

Equations of Equilibrium: Using the method of joints +

↑

F CB cos ( ψ) − Fsp = 0

Σ F y = 0;

F CB =

Fsp

cos ( ψ)

F CB = 27.971 lb

From FBD of bar AB F CB sin ( φ ) a − M = 0

+ ΣMA = 0;

M = FCB sin ( φ ) a

M = 14.2 lb⋅ ft

Problem 6-116 The compound shears are used to cut metal parts. Determine the vertical cutting force exerted on the rod R if a force F is applied at the grip G. The lobe CDE is in smooth contact with the head of the shear blade at E. Given: F = 20 lb

e = 0.5 ft

a = 1.4 ft

f = 0.5 ft

b = 0.2 ft

g = 0.5 ft

c = 2 ft

h = 2.5 ft

d = 0.75 ft

θ = 60 deg

Solution: Member AG: ΣMA = 0;

F ( a + b) − FBC b sin ( θ ) = 0

592

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Engineering Mechanics - Statics

F BC = F

⎛ a+b ⎞ ⎜ ⎟ ⎝ b sin ( θ ) ⎠

Chapter 6

F BC = 184.75 lb

Lobe: ΣMD = 0;

F BC f − NE e = 0 f NE = FBC e

NE = 184.75 lb

Head: ΣMF = 0;

−NE( h + g) + h NR = 0 NR = NE

⎛ h + g⎞ ⎜ ⎟ ⎝ h ⎠

NR = 222 lb

Problem 6-117 The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear C. Note that AB is pinned at its ends to gear C and the underside of the table EF, which is allowed to move vertically due to the smooth guides at E and F. If the gears exert tangential forces between them, determine the compressive force developed on the cylinder S when a vertical force F is applied to the handle of the press. Given: F = 40 N a = 0.5 m b = 0.2 m c = 1.2 m d = 0.35 m e = 0.65 m Solution: Member GD: ΣMG = 0;

−F a + F CG b = 0

593

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Engineering Mechanics - Statics

F CG = F

Chapter 6

a

F CG = 100 N

b

Sector gear :

⎞d = 0 2 2⎟ ⎝ c +d ⎠

ΣMH = 0; F CG( d + e) − FAB⎛ ⎜

c

⎛ c2 + d2 ⎞ ⎟ F = 297.62 N F AB = FCG ( d + e) ⎜ ⎝ c d ⎠ AB Table: ΣF y = 0;

c ⎞ ⎜ 2 2 ⎟ − Fs = 0 ⎝ c +d ⎠

F AB⎛

F s = FAB

c ⎞ ⎛ ⎜ 2 2⎟ ⎝ c +d ⎠

F s = 286 N

Problem 6-118 The mechanism is used to hide kitchen appliances under a cabinet by allowing the shelf to rotate downward. If the mixer has weight W, is centered on the shelf, and has a mass center at G, determine the stretch in the spring necessary to hold the shelf in the equilibrium position shown. There is a similar mechanism on each side of the shelf, so that each mechanism supports half of the load W. The springs each have stiffness k. Given: W = 10 lb k = 4

a = 2 in

lb

b = 4 in

in

φ = 30 deg

c = 15 in

θ = 30 deg

d = 6 in

Solution: ΣMF = 0;

W 2

b − a F ED cos ( φ ) = 0

594

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Engineering Mechanics - Statics

F ED =

Wb

2 a cos ( φ )

+ Σ F x = 0; →

↑Σ Fy = 0;

Fy =

W 2

F ED = 11.547 lb

−F x + FED cos ( φ ) = 0

F x = F ED cos ( φ ) +

Chapter 6

−W 2

F x = 10 lb + F y − FED sin ( φ ) = 0

+ F ED⋅ sin ( φ )

F y = 10.774 lb

Member FBA: ΣMA = 0;

F s = k s;

F y( c + d) cos ( φ ) − F x( c + d) sin ( φ ) − Fs sin ( θ + φ ) d = 0 Fy( c + d) cos ( φ ) − Fx( c + d) sin ( φ ) Fs = F s = 17.5 lb d sin ( θ + φ ) Fs = k x

x =

Fs

x = 4.375 in

k

Problem 6-119 If each of the three links of the mechanism has a weight W, determine the angle θ for equilibrium.The spring, which always remains horizontal, is unstretched when θ = 0°.

595

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Engineering Mechanics - Statics

Chapter 6

Given: W = 25 lb k = 60

lb ft

a = 4 ft b = 4 ft Solution: Guesses

θ = 30 deg

B x = 10 lb

B y = 10 lb

Cx = 10 lb

Cy = 10 lb

Given

⎛ a ⎞ sin ( θ ) − C a sin ( θ ) + C a cos ( θ ) = 0 ⎟ y x ⎝ 2⎠

−W⎜

⎛ b⎞ + C b = 0 ⎟ y ⎝ 2⎠

−W⎜

B y + Cy − W = 0 −B x + Cx = 0

⎛ a⎞ ⎛ a⎞ ⎛ a⎞ −B x a cos ( θ ) − B y a sin ( θ ) − W⎜ ⎟ sin ( θ ) + k⎜ ⎟ sin ( θ ) ⎜ ⎟ cos ( θ ) = 0 ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜ C ⎟ = Find B , B , C , C , θ (x y x y ) ⎜ x⎟ ⎜ Cy ⎟ ⎜ ⎟ ⎝θ ⎠

⎛ Bx ⎞ ⎜ ⎟ ⎛⎜ 16.583 ⎞⎟ ⎜ By ⎟ ⎜ 12.5 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Cx ⎟ ⎜ 16.583 ⎟ ⎜ C ⎟ ⎝ 12.5 ⎠ ⎝ y⎠

θ = 33.6 deg

Problem 6-120 Determine the required force P that must be applied at the blade of the pruning shears so that the blade exerts a normal force F on the twig at E.

596

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Engineering Mechanics - Statics

Chapter 6

Given: F = 20 lb a = 0.5 in b = 4 in c = 0.75 in d = 0.75 in e = 1 in Solution: initial guesses: Ax = 1 lb

Ay = 1 lb

Dy = 10 lb P = 20 lb

Dx = 10 lb F CB = 20 lb

Given − P ( b + c + d) − A x a + F e = 0 Dy − P − A y − F = 0 Dx − Ax = 0 − Ay( d) − A x( a) + ( b + c)P = 0

⎞=0 2 2⎟ ⎝ c +a ⎠

Ax − F CB⎛ ⎜

c

⎞=0 ⎟ 2 2 ⎝ a +c ⎠

Ay + P − FCB⎛ ⎜

a

597

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Engineering Mechanics - Statics

Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ Dx ⎜ ⎟ = Find A , A , D , D , P , F ( x y x y CB) ⎜ D ⎟ y ⎜ ⎟ ⎜ P ⎟ ⎜F ⎟ ⎝ CB ⎠

⎛⎜ Ax ⎟⎞ ⎛ 13.333 ⎞ ⎜ Ay ⎟ ⎜ 6.465 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dx ⎟ = ⎜ 13.333 ⎟ lb ⎜ D ⎟ ⎜ 28.889 ⎟ ⎜ y ⎟ ⎜ ⎟ 16.025 ⎝ ⎠ ⎜ FCB ⎟ ⎝ ⎠

P = 2.424 lb

Problem 6-121 The three power lines exert the forces shown on the truss joints, which in turn are pin-connected to the poles AH and EG. Determine the force in the guy cable AI and the pin reaction at the support H. Units Used: 3

kip = 10 lb Given: F 1 = 800 lb

d = 125 ft

F 2 = 800 lb

e = 50 ft

a = 40 ft

f = 30 ft

b = 20 ft

g = 30 ft

c = 20 ft Solution: AH is a two-force member. c θ = atan ⎛⎜ ⎟⎞

φ = atan ⎛⎜

d β = atan ⎛⎜ ⎟⎞

e γ = atan ⎛⎜ ⎟⎞

⎝ b⎠

⎝ f⎠

⎞ ⎟ ⎝ a + b⎠ c

⎝ d⎠

α = 90 deg − β + γ

Guesses F AB = 1 lb

F BC = 1 lb

F CA = 1 lb

F AI = 1 lb

F H = 1 lb

598

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Engineering Mechanics - Statics

Chapter 6

Given F BC − F AB cos ( θ ) = 0 F AB sin ( θ ) − F1 = 0 2 F CA sin ( φ ) − F 2 = 0 −F AI sin ( α ) + F AB sin ( β − θ ) + F CA sin ( β − φ ) = 0 −F AI cos ( α ) − FAB cos ( β − θ ) − F CA cos ( β − φ ) + F H = 0

⎛⎜ FAB ⎞⎟ ⎜ FCA ⎟ ⎜ ⎟ F ⎜ BC ⎟ = Find ( FAB , FCA , FBC , FAI , FH) ⎜F ⎟ ⎜ AI ⎟ ⎜ FH ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎛ 1.131 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FCA ⎟ = ⎜ 1.265 ⎟ kip ⎜ F ⎟ ⎝ 0.8 ⎠ ⎝ BC ⎠

⎛⎜ FAI ⎞⎟ ⎛ 2.881 ⎞ = kip ⎜ FH ⎟ ⎜⎝ 3.985 ⎟⎠ ⎝ ⎠

Problem 6-122 The hydraulic crane is used to lift the load of weight W. Determine the force in the hydraulic cylinder AB and the force in links AC and AD when the load is held in the position shown. Units Used: 3

kip = 10 lb Given:

W = 1400 lb a = 8 ft

c = 1 ft

b = 7 ft

γ = 70 deg

599

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Engineering Mechanics - Statics

Chapter 6

Solution: ΣMD = 0; F CA sin ( 60 deg)c − W a = 0 F CA =

Wa sin ( 60 deg) c

+

↑Σ Fy = 0;

F CA sin ( 60 deg) − FAB sin ( γ ) = 0

F AB = FCA + Σ F x = 0; →

F CA = 12.9 kip

sin ( 60 deg) sin ( γ )

F AB = 11.9 kip

−F AB cos ( γ ) + FCA cos ( 60 deg) − FAD = 0

F AD = −FAB cos ( γ ) + F CA cos ( 60 deg)

F AD = 2.39 kip

Problem 6-123 The kinetic sculpture requires that each of the three pinned beams be in perfect balance at all times during its slow motion. If each member has a uniform weight density γ and length L, determine the necessary counterweights W1, W2 and W3 which must be added to the ends of each member to keep the system in balance for any position. Neglect the size of the counterweights. Given:

γ = 2

lb ft

L = 3 ft a = 1 ft

Solution: ΣMA = 0;

W1 a cos ( θ ) − γ L cos ( θ ) ⎛⎜

L

⎝2

− a⎟⎞ = 0

⎠

600

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Engineering Mechanics - Statics

Chapter 6

γ L⎛⎜

⎝2

W1 = +

↑Σ Fy = 0;

L

− a⎞⎟

⎠

a

W1 = 3 lb

R A − W1 − γ L = 0 R A = W1 + γ L

R A = 9 lb

ΣMB = 0;

⎛ L − a⎞ − R ( L − a) cos ( φ ) = 0 ⎟ A ⎝2 ⎠

W2 a cos ( φ ) − γ L cos ( φ ) ⎜

γ L⎛⎜ W2 =

L

⎝2

+

↑Σ Fy = 0;

− a⎟⎞ + R A( L − a)

⎠

a

W2 = 21 lb

R B − W2 − RA − γ L = 0 R B = W2 + R A + γ L

R B = 36 lb

L ΣMC = 0; R B( L − a) cos ( φ ) + γ L⎛⎜ − a⎟⎞ cos ( φ ) − W3 a cos ( φ ) = 0 2

⎝ ⎠ L R B( L − a) + γL⎛⎜ − a⎟⎞ ⎝2 ⎠ W3 = a

W3 = 75 lb

Problem 6-124 The three-member frame is connected at its ends using ball-and-socket joints. Determine the x, y, z components of reaction at B and the tension in member ED. The force acting at D is F. Given:

⎛ 135 ⎞ ⎜ ⎟ F = 200 lb ⎜ ⎟ ⎝ −180 ⎠ a = 6 ft

e = 3 ft

b = 4 ft

f = 1 ft

601

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Engineering Mechanics - Statics

d = 6 ft

Chapter 6

g = 2 ft

c = g+ f Solution: AC and DE are two-force members. Define some vectors

⎛ −e ⎞ ⎜ ⎟ rDE = −b − g ⎜ ⎟ ⎝ a ⎠

uDE =

⎛ −d − e ⎞ ⎜ −b ⎟ rAC = ⎜ ⎟ ⎝ 0 ⎠

uAC =

⎛e⎞ ⎜ ⎟ rBD = − f ⎜ ⎟ ⎝0⎠

⎛e + d⎞ ⎜ −c ⎟ rBA = ⎜ ⎟ ⎝ 0 ⎠

rDE rDE

rAC rAC

Guesses B x = 1 lb

B y = 1 lb

B z = 1 lb

F DE = 1 lb

F AC = 1 lb

Given

⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ + FDE uDE + FAC uAC + F = ⎜B ⎟ ⎝ z⎠

0

⎛⎜ Bx ⎞⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Bz ⎟ = Find ( Bx , By , Bz , FDE , FAC ) ⎜F ⎟ ⎜ DE ⎟ ⎜ FAC ⎟ ⎝ ⎠

(

)

(

)

rBD × FDEuDE + F + rBA × FAC uAC = 0

⎛ Bx ⎞ ⎛ −30 ⎞ F ⎜ ⎟ ⎜ ⎟ ⎛⎜ DE ⎞⎟ ⎛ 270 ⎞ = lb ⎜ By ⎟ = ⎜ −13.333 ⎟ lb⎜ ⎟ ⎜ 16.415 ⎟⎠ F ⎜B ⎟ ⎜ − 12 ⎟ ⎝ AC ⎠ ⎝ ⎝ z ⎠ ⎝ 3.039 × 10 ⎠

602

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Engineering Mechanics - Statics

Chapter 6

Problem 6-125 The four-member "A" frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is F max, determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pon at G only exert force components on the frame. Given: F max = 800 N a = 300 mm b = 600 mm c = 600 mm Solution: ΣMx = 0; b

−P 2 c +

2

2

b +c

Fmax c = 0

F max b

P = 2

2

P = 282.843 N 2

b +c

c

B z + Dz − Fmax

2

b +c

F max c

Bz = 2

2

=0 2

Dz = B z

2

b +c

Dz = Bz

B z = 283 N Dz = 283 N

b

B y + Dy − Fmax

2

b +c

F max b

By = 2

2

=0 2

2

b +c

Dy = By

Dy = B y

B y = 283 N

Dy = 283 N

B x = Dx = 0

Problem 6-126 The structure is subjected to the loading shown. Member AD is supported by a cable AB and a roller at C and fits through a smooth circular hole at D. Member ED is supported by a roller at 603

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

g pp y D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of reaction at E and the tension in cable AB. Units Used: 3

kN = 10 N Given:

⎛ 0 ⎞ F = ⎜ 0 ⎟ kN ⎜ ⎟ ⎝ −2.5 ⎠ a = 0.5 m

d = 0.3 m

b = 0.4 m

e = 0.8 m

c = 0.3 m Solution: Guesses

⎛ −c − d ⎞ AB = ⎜ 0 ⎟ ⎜ ⎟ ⎝ e ⎠

F AB = 1 kN

Dx = 1 kN

Dz = 1 kN

Dz2 = 1 kN

E x = 1 kN

E y = 1 kN

MDx = 1 kN⋅ m

MDz = 1 kN⋅ m

Cx = 1 kN

MEx = 1 kN⋅ m

MEy = 1⋅ kN m

Given

⎛⎜ Cx ⎟⎞ ⎛⎜ Dx ⎞⎟ F + FAB +⎜ 0 ⎟+⎜ 0 ⎟ =0 AB ⎜ 0 ⎟ ⎜D ⎟ ⎝ ⎠ ⎝ z⎠ AB

⎛ MDx ⎞ ⎛ 0 ⎞ ⎛ Cx ⎞ ⎛ d ⎞ ⎛c + d⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB ⎞ ⎜ ⎟ ⎛ ⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × F + ⎜ b ⎟ × ⎜ FAB ⎟=0 AB ⎠ ⎝ ⎜M ⎟ ⎝0⎠ ⎜ 0 ⎟ ⎝0⎠ ⎝ 0 ⎠ ⎝ ⎠ ⎝ Dz ⎠

⎛ −Dx ⎞ ⎛ Ex ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ Ey ⎟ = 0 ⎜D − D ⎟ ⎜ ⎟ z⎠ ⎝ 0 ⎠ ⎝ z2

⎛ −MDx ⎞ ⎛ MEx ⎞ ⎛ 0 ⎞ ⎛ Dx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ MEy ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟ = 0 ⎜ −M ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ Dz ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Dz − Dz2 ⎠ 604

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ Cx ⎞ ⎜ ⎟ ⎜ Dx ⎟ ⎜ ⎟ D z ⎜ ⎟ ⎜D ⎟ ⎜ z2 ⎟ ⎜ Ex ⎟ ⎜ ⎟ E ⎜ y ⎟ = Find ( Cx , Dx , Dz , Dz2 , Ex , Ey , FAB , MDx , MDz , MEx , MEy) ⎜F ⎟ ⎜ AB ⎟ ⎜ MDx ⎟ ⎜ ⎟ ⎜ MDz ⎟ ⎜ ⎟ ⎜ MEx ⎟ ⎜M ⎟ ⎝ Ey ⎠ ⎛ Cx ⎞ ⎜ ⎟ ⎛⎜ 0.937 ⎞⎟ D ⎜ x⎟ ⎜ 0 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ Dz ⎟ ⎜ 1.25 ⎟ ⎜ D ⎟ ⎝ 1.25 ⎠ ⎝ z2 ⎠ ⎛⎜ Ex ⎟⎞ ⎛ 0 ⎞ = kN ⎜ Ey ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

⎛⎜ MDx ⎞⎟ ⎛ 0.5 ⎞ = kN⋅ m ⎜ MDz ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

⎛⎜ MEx ⎟⎞ ⎛ 0.5 ⎞ = kN⋅ m ⎜ MEy ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

F AB = 1.562 kN

Problem 6-127 The structure is subjected to the loadings shown.Member AB is supported by a ball-and-socket at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C.

605

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Engineering Mechanics - Statics

Chapter 6

Given: a = 2m

M = 800 N⋅ m

b = 1.5 m

F = 250 N

c = 3m

θ 1 = 60 deg

d = 4m

θ 2 = 45 deg

Solution:

θ 3 = 60 deg

Guesses Bx = 1 N

By = 1 N

Ax = 1 N

Ay = 1 N

Az = 1 N

Cx = 1 N

Cy = 1 N

Cz = 1 N

MBx = 1 N⋅ m

MBy = 1 N⋅ m

MCy = 1 N⋅ m

MCz = 1 N⋅ m

Given

⎛ Ax ⎞ ⎛ Bx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ By ⎟ = 0 ⎜A ⎟ ⎜ ⎟ ⎝ z⎠ ⎝ 0 ⎠ ⎛ c ⎞ ⎛⎜ Bx ⎟⎞ ⎛ M ⎞ ⎛⎜ −MBx ⎞⎟ ⎜ a ⎟ × B + ⎜ 0 ⎟ + −M =0 ⎜ ⎟ ⎜ y ⎟ ⎜ ⎟ ⎜ By ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎛ cos ( θ 1) ⎞ ⎛ −Bx ⎞ ⎛ Cx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ F ⎜ cos ( θ 2) ⎟ + ⎜ −B y ⎟ + ⎜ Cy ⎟ = 0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ cos ( θ 3) ⎠ ⎝ 0 ⎠ ⎝ Cz ⎠ ⎡ ⎛ cos θ ⎞⎤ ⎛ 0 ⎞ ⎢ ⎜ ( 1) ⎟⎥ ⎛ 0 ⎞ ⎛⎜ −Bx ⎞⎟ ⎛⎜ MBx ⎟⎞ ⎛⎜ 0 ⎞⎟ ⎜ 0 ⎟ × F cos ( θ ) + ⎜ 0 ⎟ × −B + M + MCy ⎟ = 0 2 ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎜ ⎟ ⎜ y ⎟ ⎜ By ⎟ ⎜ ⎝ b + d ⎠ ⎢ ⎜ cos ( θ 3) ⎟⎥ ⎝ b ⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ MCz ⎟⎠ ⎣ ⎝ ⎠⎦

606

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Az ⎟ ⎜ C ⎟ ⎜ x ⎟ ⎜ Cy ⎟ ⎜ ⎟ ⎜ Cz ⎟ ⎜ ⎟ = Find ( Ax , Ay , Az , Cx , Cy , Cz , Bx , By , MBx , MBy , MCy , MCz ) ⎜ Bx ⎟ ⎜ B ⎟ ⎜ y ⎟ ⎜ MBx ⎟ ⎜ ⎟ ⎜ MBy ⎟ ⎜M ⎟ ⎜ Cy ⎟ ⎜ MCz ⎟ ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎛ −172.3 ⎞ ⎜ Ay ⎟ ⎜ −114.8 ⎟ ⎟ ⎜A ⎟ ⎜ ⎜ ⎟ z 0 ⎜ ⎟= ⎜ ⎟N ⎜ Cx ⎟ 47.3 ⎜ ⎟ ⎜ ⎟ − 61.9 ⎜ ⎟ ⎜ Cy ⎟ ⎜ ⎜ ⎟ ⎝ −125 ⎟⎠ ⎝ Cz ⎠

⎛⎜ MCy ⎞⎟ ⎛ −429 ⎞ = N⋅ m ⎜ MCz ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

Problem 6-128 Determine the resultant forces at pins B and C on member ABC of the four-member frame. Given: w = 150

lb ft

a = 5 ft b = 2 ft

607

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Engineering Mechanics - Statics

Chapter 6

c = 2 ft e = 4 ft d = a+b−c

Solution: The initial guesses are F CD = 20 lb

F BE = 40 lb

Given F CD( c + d) − FBE

⎛ a + b⎞ + ⎟ ⎝ 2 ⎠

−w( a + b) ⎜

ec

=0 2

2

( d − b) + e F BE e 2

2

( d − b) + e

⎛⎜ FCD ⎞⎟ = Find ( F CD , F BE) ⎜ FBE ⎟ ⎝ ⎠

a − FCD( a + b) = 0

⎛⎜ FCD ⎞⎟ ⎛ 350 ⎞ = lb ⎜ FBE ⎟ ⎜⎝ 1531 ⎟⎠ ⎝ ⎠

Problem 6-129 The mechanism consists of identical meshed gears A and B and arms which are fixed to the gears. The spring attached to the ends of the arms has an unstretched length δ and a stiffness k. If a torque M is applied to gear A, determine the angle θ through which each arm rotates. The gears are each pinned to fixed supports at their centers.

608

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Engineering Mechanics - Statics

Chapter 6

Given:

δ = 100 mm k = 250

N m

M = 6 N⋅ m r =

δ 2

a = 150 mm Solution: ΣMA = 0;

−F r − P a cos ( θ ) + M = 0

ΣMB = 0;

P a cos ( θ ) − F r = 0 2P a cos ( θ ) = M 2k( 2a) sin ( θ ) a cos ( θ ) = M 2k a sin ( 2θ ) = M 2

θ =

1 2

⎞ ⎟ 2 ⎝ 2ka ⎠

asin ⎛⎜

M

θ = 16.1 deg

Problem 6-130 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: kN = 1000 N

609

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 20 kN F 2 = 10 kN a = 1.5 m b = 2m Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Guesses

F AG = 1 kN

F BG = 1 kN

F GC = 1 kN

F GF = 1 kN

F AB = 1 kN

F BC = 1 kN

F CD = 1 kN

F CF = 1 kN

F DF = 1 kN

F DE = 1 kN

F EF = 1 kN

Given −F AB cos ( θ ) + F BC = 0 −F AB sin ( θ ) − FBG = 0 F GC cos ( θ ) + FGF − FAG = 0 F GC sin ( θ ) + F BG − F 1 = 0 −F BC + F CD − F GC cos ( θ ) + FCF cos ( θ ) = 0 −F GC sin ( θ ) − F CF sin ( θ ) = 0 −F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0 −F GF − F CF cos ( θ ) + FEF = 0

610

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Engineering Mechanics - Statics

Chapter 6

F DF + F CF sin ( θ ) − F 2 = 0 −F DE cos ( θ ) − F EF = 0

611

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Engineering Mechanics - Statics

Chapter 6

⎛ FAG ⎞ ⎜ ⎟ ⎜ FBG ⎟ ⎜ ⎟ ⎜ FGC ⎟ ⎜F ⎟ ⎜ GF ⎟ ⎜ FAB ⎟ ⎜ ⎟ ⎜ FBC ⎟ = Find ( FAG , FBG , FGC , FGF , FAB , FBC , FCD , FCF , FDF , FDE , FEF) ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAG ⎞ ⎜ ⎟ ⎜ FBG ⎟ ⎛⎜ 13.13 ⎟⎞ ⎜ ⎟ ⎜ 17.50 ⎟ F GC ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 3.13 ⎟ ⎜ GF ⎟ ⎜ 11.25 ⎟ ⎜ FAB ⎟ ⎜ −21.88 ⎟ ⎜ ⎟ ⎜ ⎟ F ⎜ BC ⎟ = ⎜ −13.13 ⎟ kN ⎜ F ⎟ ⎜ −9.37 ⎟ ⎜ CD ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ −3.13 ⎟ ⎜ ⎟ ⎜ 12.50 ⎟ F ⎜ DF ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −15.62 ⎟ ⎜ FDE ⎟ ⎝ 9.37 ⎠ ⎜F ⎟ ⎝ EF ⎠

Positive (T) Negative (C)

Problem 6-131 The spring has an unstretched length δ. Determine the angle θ for equilibrium if the uniform links each have a mass mlink.

612

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Engineering Mechanics - Statics

Chapter 6

Given: mlink = 5 kg

δ = 0.3 m N

k = 400

m

a = 0.1 m b = 0.6 m g = 9.81

m 2

s Solution: Guesses

θ = 10 deg F BD = 1 N Ex = 1 N Given mlink g

a+b

cos ( θ ) − FBD b cos ( θ ) + E x b sin ( θ ) = 0

2

−2 mlink g

a+b 2

cos ( θ ) + Ex 2 b sin ( θ ) = 0

F BD = k( 2 b sin ( θ ) − δ )

⎛ FBD ⎞ ⎜ ⎟ ⎜ Ex ⎟ = Find ( FBD , Ex , θ ) ⎜ ⎟ ⎝ θ ⎠

θ = 21.7 deg

Problem 6-132 The spring has an unstretched length δ. Determine the mass mlink of each uniform link if the angle for equilibrium is θ. Given:

δ = 0.3 m

613

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Engineering Mechanics - Statics

Chapter 6

θ = 20 deg k = 400

N m

a = 0.1 m b = 0.6 m g = 9.81

m 2

s Solution: Guesses

Ey = 1 N

mlink = 1 kg

Fs = 1 N Given F s = ( 2 b sin ( θ ) − δ ) k mlink g

a+b cos ( θ ) − Fs b cos ( θ ) + E y b sin ( θ ) = 0 2

−2mlink g

a+b cos ( θ ) + E y2b sin ( θ ) = 0 2

⎛ mlink ⎞ ⎜ ⎟ ⎜ Fs ⎟ = Find ( mlink , Fs , Ey) ⎜ E ⎟ ⎝ y ⎠ mlink = 3.859 kg y = sin ( θ ) 2( b) y = 2 b sin ( θ ) F s = ( y − δ ) ( k) F s = 44.17 N

614

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Engineering Mechanics - Statics

ΣMA = 0;

Chapter 6

a+b

( cos (θ ) ) = 0 ⎡⎣Ey ( 2) ( a + b) sin ( θ ) − 2( M) ( g)⎤⎦ 2 ⎡E ( 2) sin ( θ ) − 2 ( m) ⎢ y ⎣

g⎤

⎥ ( cos ( θ ) = 0

2⎦

⇒ E y ( 2 sin ( θ ) ) = m ( g) ( cos ( θ ) ) Ey = ΣMC = 0;

m( g) ( cos ( θ ) ) 2 sin ( θ )

m( g) ( cos ( θ ) ) 2 sin ( θ )

( a + b) sin ( θ ) + m( g) ⎛⎜

a + b⎞

⎟ cos ( θ ) − Fs( b cos ( θ ) ) = 0 ⎝ 2 ⎠

b m = Fs g ( a + b) m = 3.859 kg

Problem 6-133 Determine the horizontal and vertical components of force that the pins A and B exert on the two-member frame. Given: w = 400

N m

a = 1.5 m b = 1m c = 1m F = 0N

θ = 60 deg Solution: Guesses Ax = 1 N

Ay = 1 N

Bx = 1 N

615

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Engineering Mechanics - Statics

By = 1 N

Cx = 1 N

Chapter 6

Cy = 1 N

Given −w a

a + Cx a sin ( θ ) − Cy a cos ( θ ) = 0 2

F c − Cx c − Cy b = 0

Ay − Cy − w a cos ( θ ) = 0

− Ax − Cx + w a sin ( θ ) = 0

Cx − Bx − F = 0 Cy + By = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜B ⎟ ⎜ x ⎟ = Find A , A , B , B , C , C ( x y x y x y) ⎜ By ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ ⎝ Cy ⎠

⎛⎜ Ax ⎞⎟ ⎛ 300.0 ⎞ = N ⎜ Ay ⎟ ⎜⎝ 80.4 ⎟⎠ ⎝ ⎠

⎛⎜ Bx ⎟⎞ ⎛ 220 ⎞ = N ⎜ By ⎟ ⎜⎝ 220 ⎟⎠ ⎝ ⎠

Problem 6-134 Determine the horizontal and vertical components of force that the pins A and B exert on the two-member frame. Given: w = 400

N m

a = 1.5 m b = 1m c = 1m F = 500 N

θ = 60 deg

616

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Engineering Mechanics - Statics

Chapter 6

Solution: Guesses Ax = 1 N

Ay = 1 N

Bx = 1 N

By = 1 N

Cx = 1 N

Cy = 1 N

Given −w a

a 2

+ Cx a sin ( θ ) − Cy a cos ( θ ) = 0

Ay − Cy − w a cos ( θ ) = 0 Cx + Bx − F = 0

F c − Cx c − Cy b = 0 − Ax − Cx + w a sin ( θ ) = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜B ⎟ ⎜ x ⎟ = Find A , A , B , B , C , C ( x y x y x y) ⎜ By ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ ⎝ Cy ⎠

Cy − By = 0

⎛⎜ Ax ⎞⎟ ⎛ 117.0 ⎞ = N ⎜ Ay ⎟ ⎜⎝ 397.4 ⎟⎠ ⎝ ⎠

⎛⎜ Bx ⎟⎞ ⎛ 97.4 ⎞ = N ⎜ By ⎟ ⎜⎝ 97.4 ⎟⎠ ⎝ ⎠

Problem 6-135 Determine the force in each member of the truss and indicate whether the members are in tension or compression. Units Used: kip = 1000 lb Given: F 1 = 1000 lb

b = 8 ft

F 2 = 500 lb

c = 4 ft

a = 4 ft

Solution: Joint B: 617

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Engineering Mechanics - Statics

Chapter 6

Initial Guesses: Given F BC = 100 lb

F BA = 150 lb

ΣF x = 0;

b c F 1 − F BC cos ⎛⎜ atan ⎛⎜ ⎟⎞ ⎞⎟ − FBA cos ⎜⎛ atan ⎛⎜ ⎟⎞ ⎞⎟ = 0 a a

ΣF y = 0;

−F BC sin ⎜⎛ atan ⎛⎜

⎝

⎝ ⎠⎠

⎝

⎝ ⎠⎠

⎛ ⎛ c ⎞⎞ ⎟ ⎟ + FBA sin ⎜atan ⎜ ⎟ ⎟ − F2 = 0 ⎝ a ⎠⎠ ⎝ ⎝ a ⎠⎠

⎝

b ⎞⎞

⎛⎜ FBC ⎞⎟ = Find ( F BC , F BA) ⎜ FBA ⎟ ⎝ ⎠ F BC = 373 lb(C) F BA = 1178.51 lb F BA = 1.179 kip (C) Joint A: ΣF y = 0;

c F AC − FBA sin ⎜⎛ atan ⎛⎜ ⎟⎞ ⎞⎟ = 0 a

(

)

⎝

c

F AC = FBA

⎝ ⎠⎠

2

a

2

a +c 2

a F AC = 833 lb (T)

Problem 6-136 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kip = 10 lb Given: F = 1000 lb a = 10 ft b = 10 ft

618

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Engineering Mechanics - Statics

Solution:

Chapter 6

b θ = atan ⎛⎜ ⎞⎟

⎝ a⎠

Guesses F AB = 1 lb

F AG = 1 lb

F BC = 1 lb

F BG = 1 lb

F CD = 1 lb

F CE = 1 lb

F CG = 1 lb

F DE = 1 lb

F EG = 1 lb Given F AB + F AG cos ( θ ) = 0 −F AB + F BC = 0 F BG = 0 F CD − F BC − F CG cos ( θ ) = 0 F CE − F + F CG sin ( θ ) = 0 −F CD − F DE cos ( θ ) = 0 F DE cos ( θ ) − F EG = 0 −F CE − FDE sin ( θ ) = 0 F EG − F AG cos ( θ ) + F CG cos ( θ ) = 0 −F AG sin ( θ ) − FBG − FCG sin ( θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAG ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBG ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( FAB , FAG , FBC , FBG , FCD , FCE , FCG , FDE , FEG) ⎜F ⎟ ⎜ CE ⎟ ⎜ FCG ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎟ ⎜ ⎝ FEG ⎠

619

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ 333 ⎞ ⎜ FAG ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ −471 ⎟ ⎜ BC ⎟ ⎜ 333 ⎟ ⎜ FBG ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ F 667 = ⎟ lb ⎜ CD ⎟ ⎜ ⎜ F ⎟ ⎜ 667 ⎟ ⎟ ⎜ CE ⎟ ⎜ ⎜ FCG ⎟ ⎜ 471 ⎟ ⎜ ⎟ ⎜ −943 ⎟ F ⎜ DE ⎟ ⎜ ⎟ ⎟ ⎝ −667 ⎠ ⎜ ⎝ FEG ⎠

Positive (T), Negative (C)

Problem 6-137 Determine the force in members AB, AD, and AC of the space truss and state if the members are in tension or compression. The force F is vertical. Units Used: 3

kip = 10 lb Given: F = 600 lb a = 1.5 ft b = 2 ft c = 8 ft Solution:

⎛a⎞ AB = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

⎛ −a ⎞ AC = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ AD = ⎜ −c ⎟ ⎜ ⎟ ⎝b⎠ Guesses

F AB = 1 lb

F AC = 1 lb

F AD = 1 lb 620

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Given

Chapter 6

⎛ 0 ⎞ F AB + FAC + FAD +⎜ 0 ⎟ =0 ⎜ ⎟ AB AC AD ⎝ −F ⎠ AB

AC

AD

⎛ FAB ⎞ ⎜ ⎟ F ⎜ AC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠ ⎛ FAB ⎞ ⎛ −1.221 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ −1.221 ⎟ kip ⎜ F ⎟ ⎝ 2.474 ⎠ ⎝ AD ⎠

Positive (T) Negative (C)

621

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Engineering Mechanics - Statics

Chapter 7

Problem 7-1 The column is fixed to the floor and is subjected to the loads shown. Determine the internal normal force, shear force, and moment at points A and B. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN F 3 = 8 kN a = 150 mm b = 150 mm c = 150 mm Solution: Free body Diagram: The support reaction need not be computed in this case. Internal Forces: Applying equations of equillibrium to the top segment sectioned through point A, we have + Σ F x = 0; →

VA = 0

+

NA − F1 − F2 = 0

NA = F 1 + F 2

NA = 12.0 kN

F 1 a − F2 b − MA = 0

MA = F1 a − F 2 b

MA = 0 kN⋅ m

↑Σ Fy = 0; ΣMA = 0;

VA = 0

Applying equations of equillibrium to the top segment sectioned through point B, we have + Σ F x = 0; →

VB = 0

+

NB − F1 − F2 − F3 = 0

NB = F 1 + F 2 + F 3

NB = 20.0 kN

F 1 a − F 2 b − F3 c + MB = 0

MB = −F1 a + F 2 b + F3 c

MB = 1.20 kN⋅ m

↑Σ Fy = 0;

+ΣMB = 0;

VB = 0

Problem 7-2 The axial forces act on the shaft as shown. Determine the internal normal forces at points A and B. 622

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given: F 1 = 20 lb F 2 = 50 lb F 3 = 10 lb

Solution: Section A: ΣF z = 0;

F 2 − 2 F1 − NA = 0 NA = F 2 − 2 F 1 NA = 10.00 lb

Section B: ΣF z = 0;

F 2 − 2 F1 − NA + NB = 0 NB = −F 2 + 2 F1 + NA NB = 0.00 lb

Problem 7-3 The shaft is supported by smooth bearings at A and B and subjected to the torques shown. Determine the internal torque at points C, D, and E. Given: M1 = 400 N⋅ m M2 = 150 N⋅ m M3 = 550 N⋅ m

623

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Engineering Mechanics - Statics

Chapter 7

Solution: Section C: ΣMx = 0;

TC = 0

Section D: ΣMx = 0;

TD − M 1 = 0 TD = M 1 TD = 400.00 N⋅ m

Section E: ΣMx = 0;

M 1 + M 2 − TE = 0 TE = M 1 + M 2 TE = 550.00 N⋅ m

Problem 7-4 Three torques act on the shaft. Determine the internal torque at points A, B, C, and D. Given: M1 = 300 N⋅ m M2 = 400 N⋅ m M3 = 200 N⋅ m Solution: Section A: ΣΜx = 0;

− TA + M 1 − M 2 + M 3 = 0 TA = M 1 − M 2 + M 3 TA = 100.00 N⋅ m

Section B: ΣMx = 0;

TB + M 3 − M 2 = 0

624

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

TB = − M 3 + M 2 TB = 200.00 N⋅ m Section C: ΣΜx = 0;

− TC + M 3 = 0 TC = M 3 TC = 200.00 N⋅ m

Section D: ΣΜx = 0;

TD = 0

Problem 7-5 The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the normal force, shear force, and moment at a section passing through (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the force F 2. Units Used: 3

kip = 10 lb Given: F 1 = 2.5 kip F 2 = 3 kip w = 75

lb ft

a = 6 ft

b = 12 ft c = 2 ft

Solution: Σ MB = 0;

625

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

b − Ay( b + c) + F1 ( a + b + c) + w b⎛⎜ c + ⎞⎟ + F2 c = 0 ⎝ 2⎠ F1 ( a + b + c) + w b⎛⎜ c +

⎝

Ay =

b⎞

⎟ + F2 c

2⎠

b+c

Ay = 4514 lb + Σ F x = 0; → +

↑Σ Fy = 0;

B x = 0 lb Ay − F1 − w b − F2 + By = 0 By = − Ay + F1 + w b + F2

Σ MC = 0; + Σ F x = 0; → +

↑

Σ F y = 0; Σ MD = 0;

+ → +

↑

Σ F x = 0; Σ F y = 0;

F 1 a + Mc = 0

B y = 1886 lb

MC = −F 1 a

NC = 0 lb

MC = −15.0 kip⋅ ft NC = 0.00 lb

−F 1 + Ay − V C = 0 −MD + By c = 0

VC = Ay − F1

V C = 2.01 kip

MD = B y c

MD = 3.77 kip⋅ ft

ND = 0 lb

ND = 0.00 lb

VD − F2 + By = 0

VD = F2 − By

V D = 1.11 kip

Problem 7-6 Determine the internal normal force, shear force, and moment at point C. Given: M = 400 lb⋅ ft a = 4 ft b = 12 ft

626

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Engineering Mechanics - Statics

Chapter 7

Solution: Beam: ΣMB = 0;

M − Ay( a + b) = 0 Ay =

M a+b

Ay = 25.00 lb

Segment AC: ΣF x = 0;

NC = 0

ΣF y = 0;

Ay − VC = 0 VC = Ay V C = 25.00 lb

ΣMC = 0;

− Ay a + MC = 0 MC = Ay a MC = 100.00 lb⋅ ft

Problem 7-7 Determine the internal normal force, shear force, and moment at point C. Units Used: kN = 103 N Given: F 1 = 30 kN F 2 = 50 kN F 3 = 25 kN a = 1.5 m b = 3m

θ = 30 deg Solution: ΣF x = 0;

−NC + F 3 cos ( θ ) = 0 627

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

NC = F3 cos ( θ ) NC = 21.7 kN V C − F 2 − F 3 sin ( θ ) = 0

ΣF y = 0;

V C = F2 + F3 sin ( θ ) V C = 62.50 kN ΣMC = 0; −MC − F2 b − F 3 sin ( θ ) 2 b = 0 MC = −F 2 b − F3 sin ( θ ) 2b MC = −225.00 kN⋅ m

Problem 7-8 Determine the normal force, shear force, and moment at a section passing through point C. Assume the support at A can be approximated by a pin and B as a roller. Units used: 3

kip = 10 lb Given: F 1 = 10 kip

a = 6 ft

F 2 = 8 kip

b = 12 ft

kip ft

c = 12 ft

w = 0.8

d = 6 ft

Solution: ΣMA = 0;

⎛ b + c ⎞ − F ( b + c + d) + B ( b + c) + F a = ⎟ 2 y 1 ⎝ 2 ⎠

−w( b + c) ⎜

w

( b + c)

By = + Σ F x = 0; →

2

0

2

+ F 2 ( b + c + d) − F 1 a B y = 17.1 kip

b+c NC = 0 628

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑Σ Fy = 0;

Chapter 7

VC − w c + By − F2 = 0 VC = w c − By + F2

V C = 0.5 kip

⎛ c ⎞ + B c − F ( c + d) = ⎟ y 2 ⎝ 2⎠

−MC − w c⎜

ΣMC = 0;

⎛ c2 ⎞ ⎟ + B y c − F 2 ( c + d) ⎝2⎠

MC = −w⎜

0

MC = 3.6 kip⋅ ft

Problem 7-9 The beam AB will fail if the maximum internal moment at D reaches Mmax or the normal force in member BC becomes Pmax. Determine the largest load w it can support. Given: Mmax = 800 N⋅ m P max = 1500 N a = 4m b = 4m c = 4m d = 3m Solution:

⎛ a + b ⎞ − A ( a + b) = ⎟ y ⎝ 2 ⎠

w ( a + b) ⎜ Ay =

w( a + b) 2

⎛ a⎞ − A a + M = ⎟ y D ⎝2⎠

w a⎜

MD = w

0

⎛ a b⎞ ⎜2 ⎟ ⎝ ⎠

Ay − w( a + b) +

T=

0

w( a + b)

⎛ d ⎞T = ⎜ 2 2⎟ ⎝ c +d ⎠ 2

c +d

0

2

2d

Assume the maximum moment has been reached 629

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Engineering Mechanics - Statics

Chapter 7

MD = Mmax

2 MD

w1 =

ab

w1 = 100

N m

w2 = 225

N m

Assume that the maximum normal force in BC has been reached T = Pmax

T2 d

w2 =

2

( a + b) c + d

2

w = min ( w1 , w2 )

Now choose the critical load

w = 100

N m

Problem 7-10 Determine the shear force and moment acting at a section passing through point C in the beam. Units Used: 3

kip = 10 lb Given: w = 3

kip ft

a = 6 ft b = 18 ft Solution: ΣMB = 0;

− Ay b + Ay =

ΣMC = 0;

1 6

− Ay a +

1 2

wb

⎛ b⎞ = ⎜3⎟ ⎝ ⎠

0

Ay = 9 kip

wb

a⎞ ⎛ a⎞ w ⎟ a ⎜ ⎟ + MC = ⎜ 2 ⎝ b⎠ ⎝ 3 ⎠ 1⎛

0

3

wa MC = Ay a − 6b MC = 48 kip⋅ ft +

↑

Σ F y = 0;

Ay −

1 2

⎛w a ⎞ a − V = ⎜ ⎟ C ⎝ b⎠

0

2

wa VC = Ay − 2b

630

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Engineering Mechanics - Statics

Chapter 7

V C = 6 kip

Problem 7-11 Determine the internal normal force, shear force, and moment at points E and D of the compound beam. Given: M = 200 N⋅ m

c = 4m

F = 800 N

d = 2m

a = 2m

e = 2m

b = 2m Solution: Segment BC : M d+e

−M + Cy( d + e) = 0

Cy =

−B y + Cy = 0

B y = Cy

Segment EC : −NE = 0

NE = 0 N

V E + Cy = 0

NE = 0.00

V E = −Cy

−ME − M + Cy e = 0

V E = −50.00 N

ME = Cy e − M

ME = −100.00 N⋅ m

Segment DB : −ND = 0

ND = 0 N

ND = 0.00

VD − F + By = 0

VD = F − By

V D = 750.00 N

−MD − F b + By( b + c) = 0 MD = −F b + B y( b + c)

MD = −1300 N⋅ m

631

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Engineering Mechanics - Statics

Chapter 7

Problem 7-12 The boom DF of the jib crane and the column DE have a uniform weight density γ. If the hoist and load have weight W, determine the normal force, shear force, and moment in the crane at sections passing through points A, B, and C. Treat the boom tip, beyond the hoist, as weightless. Given: W = 300 lb

γ = 50

lb ft

a = 7 ft b = 5 ft c = 2 ft d = 8 ft e = 3 ft Solution: + ΣF x = 0; →

−NA = 0

+

VA − W − γ e = 0

↑ ΣFy = 0;

NA = 0 lb

VA = W + γ e ΣΜΑ = 0;

+ → +

ΣF x = 0;

↑ ΣFy = 0; ΣΜΒ = 0;

NA = 0.00 lb

V A = 450 lb

⎛e⎞ − We = ⎟ ⎝ 2⎠ ⎛ e2 ⎞ MA = γ ⎜ ⎟ + W e ⎝2⎠ MA − γ e⎜

0

−NB = 0

MA = 1125.00 lb⋅ ft NB = 0 lb

V B − γ ( d + e) − W = 0

V B = γ ( d + e) + W

⎛ d + e ⎞ − W( d + e) = ⎟ ⎝ 2 ⎠

MB − γ ( d + e) ⎜ MB =

1 2

NB = 0.00 lb

2

γ ( d + e) + W( d + e)

V B = 850 lb

0

MB = 6325.00 lb⋅ ft

632

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Engineering Mechanics - Statics

Chapter 7

+ →

ΣF x = 0;

VC = 0

+

ΣF y = 0;

NC − ( c + d + e) γ − W − γ ( b) = 0

↑

V C = 0 lb

V C = 0.00 lb

NC = γ ( c + d + e + b) + W ΣΜC = 0;

NC = 1200.00 lb

MC − ( c + d + e) γ

⎛ c + d + e ⎞ − W( c + d + e) = ⎜ 2 ⎟ ⎝ ⎠

MC = ( c + d + e) γ

⎛ c + d + e ⎞ + W( c + d + e) ⎜ 2 ⎟ ⎝ ⎠

0

MC = 8125.00 lb⋅ ft

Problem 7-13 Determine the internal normal force, shear force, and moment at point C. Units Used: 3

kip = 10 lb Given: a = 0.5 ft

d = 8 ft

b = 2 ft

e = 4 ft

c = 3 ft

w = 150

lb ft

Solution: Entire beam: ΣMA = 0;

⎛ d + e ⎞ + T ( a + b) = ⎟ ⎝ 2 ⎠

−w( d + e) ⎜

w ( d + e) T = 2 ( a + b) ΣF x = 0;

2

T = 4.32 kip

Ax − T = 0 Ax = T

ΣF y = 0;

0

Ax = 4.32 kip

Ay − w( d + e) = 0 Ay = w( d + e)

Ay = 1.80 kip

Segment AC: 633

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Engineering Mechanics - Statics

Chapter 7

ΣF x = 0;

Ax + NC = 0

NC = − A x

NC = −4.32 kip

ΣF y = 0;

Ay − w c − VC = 0

VC = Ay − w c

V C = 1.35 kip

ΣMC = 0;

− Ay c + w c⎜

⎛c⎞ + M = ⎟ C ⎝2⎠

MC = Ay c − w

0

⎛ c2 ⎞ ⎜ ⎟ ⎝2⎠

MC = 4.72 kip⋅ ft

Problem 7-14 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: 3

kN = 10 N Given: w = 400

N m

a = 2.5 m b = 3m c = 6m

Solution: ΣMA = 0;

−1 2

⎛ 2 c⎞ + F ⎛ a ⎞ c = ⎜3 ⎟ BC ⎜ ⎝ ⎠ 2 2⎟ ⎝ a +c ⎠

wc

F BC = +

→ Σ Fx = 0;

↑Σ Fy = 0;

3

wc

⎛ a2 + c2 ⎞ ⎜ ⎟ ⎝ a ⎠

⎛ c ⎞F − A = x ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ Ax =

+

1

Ay −

⎛ c ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ 1 2

wc +

0

F BC = 2080 N

0

Ax = 1920 N

⎛ a ⎞F = ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

0

634

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Engineering Mechanics - Statics

Ay = + Σ F x = 0; → +

↑Σ Fy = 0;

Chapter 7

1 2

wc −

⎛ a ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

ND − Ax = 0 Ay −

Ay = 400 N

ND = A x

b⎞ ⎜w ⎟ b − VD = c⎠

1⎛ 2⎝

0

2⎞

1 ⎛b V D = A y − w⎜ 2 ⎝ c

ΣF y = 0;

− Ay b +

⎟ ⎠

V D = 100 N

b b w ⎞⎟ b⎛⎜ ⎟⎞ + MD = ⎜ 2⎝ c⎠ ⎝3⎠ 1⎛

MD = Ay b −

1 6

ND = 1.920 kN

0

⎛ b3 ⎞ w⎜ ⎟ ⎝c⎠

MD = 900 N m

Problem 7-15 The beam has weight density γ. Determine the internal normal force, shear force, and moment at point C. Units Used: kip = 103 lb Given:

γ = 280

lb ft

a = 3 ft b = 7 ft c = 8 ft d = 6 ft

635

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Solution: c θ = atan ⎛⎜ ⎟⎞

W = γ ( a + b)

⎝ d⎠

Guesses Ax = 1 lb

Ay = 1 lb

B x = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb ft

Given Entire beam: Ax − Bx = 0

Ay − W = 0

Bx c − W

⎛ d⎞ = ⎜ 2⎟ ⎝ ⎠

0

Bottom Section Ax − NC cos ( θ ) + V C sin ( θ ) = 0 Ay − W

⎛ a ⎞ − N sin ( θ ) − V cos ( θ ) = ⎜ ⎟ C C ⎝ a + b⎠

MC − VC a − W

⎛ a ⎞ ⎜ ⎟ ⎝ a + b⎠

⎛ a ⎞ cos ( θ ) = ⎜ 2⎟ ⎝ ⎠

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , N , V , M ) x y x C C C ⎜ NC ⎟ ⎜ ⎟ ⎜ VC ⎟ ⎜ MC ⎟ ⎝ ⎠

0

0

⎛ Ax ⎞ ⎛ 1.05 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 2.80 ⎟ kip ⎜ B ⎟ ⎜⎝ 1.05 ⎟⎠ ⎝ x⎠

⎛ NC ⎞ ⎛ 2.20 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ VC ⎠ ⎝ 0.34 ⎠ MC = 1.76 kip⋅ ft

Problem 7-16 Determine the internal normal force, shear force, and moment at points C and D of the beam.

636

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Engineering Mechanics - Statics

Chapter 7

Units Used: 3

kip = 10 lb Given: w1 = 60

lb ft

a = 12 ft b = 15 ft

lb w2 = 40 ft

c = 10 ft d = 5 ft

F = 690 lb e = 12

f = 5 e θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ f⎠

Guesses B y = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb⋅ ft

ND = 1 lb

V D = 1 lb

MD = 1 lb⋅ ft Given

⎛ b ⎞ ... = ⎟ ⎝ 2⎠

B y b − F sin ( θ ) ( b + c) − w2 b⎜ +

−1 2

(w1 − w2)b⎛⎜ 3 ⎟⎞

0

b

⎝ ⎠

−NC − F cos ( θ ) = 0 b − a⎞ ⎟ ( b − a) ... = 0 ⎝ b ⎠ + B y − w2 ( b − a) − F sin ( θ ) VC −

(w1 − w2) ⎛⎜ 2 1

−ND − F cos ( θ ) = 0 V D − F sin ( θ ) = 0 −MD − F sin ( θ ) d = 0

⎛ b − a⎞ − ⎟ ⎝ 2 ⎠

−MC − w2 ( b − a) ⎜

b − a⎞ ⎛ b − a⎞ ⎟ ( b − a) ⎜ 3 ⎟ ... = 0 ⎝ ⎠ ⎝ b ⎠

⎛ w1 − w2 ) ⎜ ( 2 1

+ B y ( b − a) − F sin ( θ ) ( c + b − a)

637

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Engineering Mechanics - Statics

Chapter 7

⎛ By ⎞ ⎜ ⎟ ⎜ NC ⎟ ⎜ VC ⎟ ⎜ ⎟ M ⎜ C ⎟ = Find ( By , NC , VC , MC , ND , VD , MD) ⎜N ⎟ ⎜ D⎟ ⎜ VD ⎟ ⎜ ⎟ ⎝ MD ⎠

⎛ NC ⎞ ⎛ −265 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ VC ⎟ = ⎜ −649 ⎟ lb ⎜ ND ⎟ ⎜ −265 ⎟ ⎜ ⎟ ⎜ 637 ⎟ ⎠ ⎝ VD ⎠ ⎝ ⎛ MC ⎞ ⎛ −4.23 ⎞ ⎜ ⎟=⎜ ⎟ kip⋅ ft ⎝ MD ⎠ ⎝ −3.18 ⎠

Problem 7-17 Determine the normal force, shear force, and moment acting at a section passing through point C. 3

kip = 10 lb

Units Used: Given:

F 1 = 800 lb F 2 = 700 lb F 3 = 600 lb

θ = 30 deg a = 1.5 ft b = 1.5 ft c = 3 ft d = 2 ft e = 1 ft f = a+b+c−d−e Solution: Guesses

B y = 1 lb

Ax = 1 lb

Ay = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb⋅ ft

638

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given − Ax + V C sin ( θ ) + NC cos ( θ ) = 0 Ay − V C cos ( θ ) + NC sin ( θ ) = 0 MC − A x ( a) sin ( θ ) − Ay ( a) cos ( θ ) = 0 − Ax + F 1 sin ( θ ) − F3 sin ( θ ) = 0 Ay + B y − F2 − F1 cos ( θ ) − F3 cos ( θ ) = 0 −F 1 ( a + b) − F 2 ( a + b + c) cos ( θ ) − F 3 cos ( θ ) ( a + b + c + d + e) cos ( θ ) ... = 0 + F 3 sin ( θ ) f sin ( θ ) + B y2 ( a + b + c) cos ( θ )

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , N , V , M ) x y y C C C ⎜ NC ⎟ ⎜ ⎟ ⎜ VC ⎟ ⎜ MC ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 100 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 985 ⎟ lb ⎜ B ⎟ ⎜⎝ 927 ⎟⎠ ⎝ y⎠ ⎛ NC ⎞ ⎛ −406 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ VC ⎠ ⎝ 903 ⎠ MC = 1.355 kip⋅ ft

Problem 7-18 Determine the normal force, shear force, and moment acting at a section passing through point D. Units Used:

kip = 103 lb

639

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given: F 1 = 800 lb F 2 = 700 lb F 3 = 600 lb

θ = 30 deg a = 1.5 ft b = 1.5 ft c = 3 ft d = 2 ft e = 1 ft f = a+b+c−d−e Solution: Guesses

B y = 1 lb

Ax = 1 lb

Ay = 1 lb

ND = 1 lb

V D = 1 lb

MD = 1 lb⋅ ft

Given V D sin ( θ ) − ND cos ( θ ) − F3 sin ( θ ) = 0 B y + VD cos ( θ ) + ND sin ( θ ) − F 3 cos ( θ ) = 0 −MD − F3 e + By( e + f) cos ( θ ) = 0 Ay + B y − F2 − F1 cos ( θ ) − F3 cos ( θ ) = 0 − Ax + F 1 sin ( θ ) − F3 sin ( θ ) = 0 −F 1 ( a + b) − F 2 ( a + b + c) cos ( θ ) − F 3 cos ( θ ) ( a + b + c + d + e) cos ( θ ) ... = 0 + F 3 sin ( θ ) f sin ( θ ) + B y2 ( a + b + c) cos ( θ )

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , N , V , M ) x y y D D D ⎜ ND ⎟ ⎜ ⎟ ⎜ VD ⎟ ⎜ MD ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 100 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 985 ⎟ lb ⎜ B ⎟ ⎜⎝ 927 ⎟⎠ ⎝ y⎠ ⎛ ND ⎞ ⎛ −464 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ VD ⎠ ⎝ −203 ⎠ MD = 2.61 kip⋅ ft

640

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-19 Determine the normal force, shear force, and moment at a section passing through point C. Units Used: 3

kN = 10 N Given: P = 8 kN

c = 0.75 m

a = 0.75m

d = 0.5 m

b = 0.75m

r = 0.1 m

Solution: ΣMA = 0;

−T ( d + r) + P( a + b + c) = 0 T = P

⎛ a + b + c⎞ T = ⎜ ⎟ ⎝ d+r ⎠

30 kN

+ Σ F x = 0; →

Ax = T

Ax = 30 kN

+

Ay = P

Ay = 8 kN

↑Σ Fy = 0;

+ Σ F x = 0; →

−NC − T = 0 NC = −T

+

↑Σ Fy = 0;

VC + P = 0 V C = −P

ΣMC = 0;

NC = −30 kN

V C = −8 kN

−MC + P c = 0 MC = P c

MC = 6 kN⋅ m

Problem 7-20 The cable will fail when subjected to a tension Tmax. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading. Units Used: 3

kN = 10 N 641

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given: Tmax = 2 kN a = 0.75 m b = 0.75 m c = 0.75 m d = 0.5 m r = 0.1 m Solution: ΣMA = 0;

−Tmax( r + d) + P ( a + b + c) = 0

⎛ d+r ⎞ ⎟ ⎝ a + b + c⎠

P = Tmax⎜

P = 0.533 kN

+ Σ F x = 0; →

Tmax − Ax = 0

Ax = Tmax

Ax = 2 kN

+

Ay − P = 0

Ay = P

Ay = 0.533 kN

+ Σ F x = 0; →

−NC − Ax = 0

NC = − A x

NC = −2 kN

VC = Ay

V C = 0.533 kN

MC = Ay c

MC = 0.400 kN⋅ m

↑Σ Fy = 0;

+

↑Σ Fy = 0; ΣMC = 0;

−V C + Ay = 0 −MC + A y c = 0

Problem 7-21 Determine the internal shear force and moment acting at point C of the beam. Units Used: 3

kip = 10 lb Given: w = 2

kip ft

a = 9 ft

642

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Solution: ΣF x = 0;

NC = 0

ΣF y = 0;

wa

ΣMC = 0;

MC −

2

NC = 0

wa

−

− VC = 0

2

⎛ w a⎞a + ⎛ w a ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠

VC = 0

⎛ a⎞ = ⎜3⎟ ⎝ ⎠

0

2

MC =

wa

MC = 54.00 kip⋅ ft

3

Problem 7-22 Determine the internal shear force and moment acting at point D of the beam. Units Used: kip = 103 lb Given: w = 2

kip ft

a = 6 ft b = 9 ft Solution: ΣF x = 0; ΣF y = 0;

ND = 0 wb 2

− w⎛⎜

VD =

a⎞

⎛ a⎞ ⎟ ⎜ 2 ⎟ − VD = ⎝ b⎠ ⎝ ⎠

wb 2

− w⎛⎜

0

a⎞

⎛ a⎞ ⎟ ⎜ 2⎟ ⎝ b⎠ ⎝ ⎠

V D = 5.00 kip ΣMD = 0;

⎛ w b ⎞a + ⎛ w a ⎞⎛ a ⎞ ⎜ 2 ⎟ ⎜ ⎟⎜ 2 ⎟ ⎝ ⎠ ⎝ b ⎠⎝ ⎠ ⎛ w a3 ⎞ wb⎞ ⎟ MD = ⎛⎜ a−⎜ ⎟ ⎝ 2 ⎠ ⎝ 6b ⎠ MD −

⎛ a⎞ = ⎜ 3⎟ ⎝ ⎠

0

MD = 46.00 kip⋅ ft

643

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-23 The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the internal normal force, shear force, and moment at (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the F2 force. Units Used: kip = 103 lb Given: F 1 = 2500 lb

a = 6 ft

F 2 = 3000 lb

b = 12 ft

w = 75

lb ft

c = 2 ft

Solution: ΣMB = 0;

− Ay( b + c) + F1 ( a + b + c) + w b

Ay =

(2

2

b+c

B x = 0 lb

ΣF y = 0;

Ay − F1 − w b − F2 + By = 0 By = − Ay + F1 + w b + F2

ΣMC = 0;

ΣF x = 0; ΣF y = 0;

)

1 2 F 1 ( a + b + c) + w b + 2 c b + 2 F 2 c

ΣF x = 0;

Segment AC :

⎛ b + c⎞ + F c = ⎜2 ⎟ 2 ⎝ ⎠

0

Ay = 4514 lb

B y = 1886 lb

F 1 a + MC = 0 MC = −F 1 a

MC = −15 kip⋅ ft

NC = 0

NC = 0

−F 1 + Ay − V C = 0 VC = Ay − F1

V C = 2.01 kip

Segment BD: ΣMD = 0;

−MD + By c = 0 MD = B y c

MD = 3.77 kip⋅ ft

ΣF x = 0;

ND = 0

ND = 0

ΣF y = 0;

VD − F2 + By = 0 VD = F2 − By

V D = 1.11 kip

644

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-24 The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam. Neglect the weight of the beams. Units Used: kip = 103 lb Given: P = 5000 lb a = 2 ft b = 10 ft Solution: Segment: ΣMC = 0; MC +

⎛ P ⎞b = ⎜2⎟ ⎝ ⎠

MC = −

P 2

0

b

MC = −25.00 kip⋅ ft

Problem 7-25 The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam. Assume that each beam has a uniform weight density γ. Units Used: kip = 103 lb Given: P = 5000 lb 645

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

γ = 150

Chapter 7

lb ft

a = 2 ft b = 10 ft Solution: Beam: +

↑Σ Fy = 0;

P − 2 γ ( a + b) − 2 R = 0 R =

P 2

− γ ( a + b)

R = 700 lb Segment: ΣMC = 0;

M C + R b + γ ( a + b) MC = −R b − γ

⎛ a + b⎞ = ⎜ 2 ⎟ ⎝ ⎠

( a + b)

0

2

2

MC = −17.8 kip⋅ ft

Problem 7-26 Determine the normal force, shear force, and moment in the beam at sections passing through points D and E. Point E is just to the right of the F load. Units Used: kip = 103 lb Given: w = 1.5

a = 6 ft kip ft

F = 3 kip

b = 6 ft c = 4 ft d = 4 ft

646

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Engineering Mechanics - Statics

Chapter 7

Solution: Σ MB = 0;

1 2

⎛ a + b ⎞ − A ( a + b) = ⎟ y ⎝ 3 ⎠

w( a + b) ⎜ 1 2

Ay =

w( a + b)

0

⎛ a + b⎞ ⎜ 3 ⎟ ⎝ ⎠

a+b

Ay = 3 kip + Σ F x = 0; →

Bx = 0

+

By + Ay −

↑

Σ F y = 0;

1

w( a + b) = 0

2

1

By = − Ay +

2

w( a + b)

B y = 6 kip + Σ F x = 0; →

ND = 0

+

Ay −

↑

Σ F y = 0;

⎞a − V = ⎜ ⎟ D 2 ⎝ a + b⎠ 1⎛ aw

VD = Ay −

Σ MD = 0;

MD +

1⎛ aw

⎜

⎞a ⎟

V D = 0.75 kip

2 ⎝ a + b⎠

⎞a ⎛ a ⎞ − A a = ⎜ ⎟ ⎜ ⎟ y 2 ⎝ a + b⎠ ⎝ 3⎠ 1⎛ aw

MD =

−1 ⎛ a w ⎞ ⎛ a ⎞ ⎜ ⎟ a ⎜ ⎟ + Ay a 2 ⎝ a + b⎠ ⎝ 3 ⎠

+ Σ F x = 0; →

NE = 0

+

−V E − F − B y = 0

↑Σ Fy = 0;

0

V E = −F − By ΣME = 0;

0

MD = 13.5 kip⋅ ft

V E = −9 kip

ME + B y c = 0 ME = −By c

ME = −24.0 kip⋅ ft

647

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-27 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: kN = 103 N Given: w1 = 200

N

w2 = 400

N

m

m

a = 2.5 m b = 3m c = 6m Solution: Σ MA = 0;

F BC + Σ F x = 0; → +

↑Σ Fy = 0;

−w1 c⎛⎜

c⎞

→ Σ Fx = 0; +

↑

Σ F y = 0;

2

(w2 − w1)c ⎛⎜ 3 ⎟⎞ + 2c

Ax =

2

1 2

2

2

(FBC c) = 0

a +c

2

a +c

F BC = 2600 N

ac

⎛ c ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

Ay − w1 c −

a

⎝ ⎠

2 ⎡ c2 c⎤ ⎢ = w1 + ( w2 − w1 ) ⎥ 3⎦ ⎣ 2

Ay = w1 c + +

1

⎟− ⎝ 2⎠

Ax = 2400 N

(w2 − w1)c + ⎛⎜

⎞F = BC ⎟ 2 2 + c a ⎝ ⎠

(w2 − w1)c − ⎛⎜ 2 1

a

⎞F

a

2⎟

2

⎝ a +c ⎠

− Ax + ND = 0

0

BC

ND = A x

ND = 2.40 kN

1 b Ay − w1 b − ( w2 − w1 ) ⎛⎜ ⎟⎞ b − V D = 0 2 ⎝c⎠ 2⎞

⎛b 1 V D = A y − w1 b − ( w2 − w1 ) ⎜ 2 ⎝c Σ MD = 0;

Ay = 800 N

− Ay b + w1 b⎛⎜

b⎞

⎟+ ⎝2⎠

⎟ ⎠

V D = 50 N

(w2 − w1)⎛⎜ c ⎟⎞ b ⎛⎜ 3 ⎟⎞ + MD = 0 2 1

⎛ b2 ⎞ MD = Ay( b) − w1 ⎜ ⎟ − ⎝2⎠

b

b

⎝ ⎠

⎝ ⎠

⎛ b3 ⎞ (w2 − w1)⎜ 3c ⎟ 2 ⎝ ⎠ 1

648

MD = 1.35 kN⋅ m

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-28 Determine the normal force, shear force, and moment at sections passing through points E and F. Member BC is pinned at B and there is a smooth slot in it at C. The pin at C is fixed to member CD. Units Used: 3

kip = 10 lb Given: M = 350 lb⋅ ft w = 80

lb ft

c = 2 ft

F = 500 lb

d = 3 ft

θ = 60 deg

e = 2 ft

a = 2 ft

f = 4 ft

b = 1 ft

g = 2 ft

Solution: Σ MB = 0; −1 2

wd

⎛ 2d ⎞ − F sin ( θ ) d + C ( d + e) = ⎜3⎟ y ⎝ ⎠

⎛ w d2 ⎞ ⎜ ⎟ + F sin ( θ ) d 3 ⎠ ⎝ Cy =

Cy = 307.8 lb

d+e

+ Σ F x = 0; →

B x − F cos ( θ ) = 0

+

By −

↑

Σ F y = 0;

By = + Σ F x = 0; →

0

1 2

B x = F cos ( θ )

B x = 250 lb

w d − F sin ( θ ) + Cy = 0

1 2

w d + F sin ( θ ) − Cy

−NE − Bx = 0

NE = −B x

B y = 245.2 lb NE = −250 lb

649

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑Σ Fy = 0;

Chapter 7

VE − By = 0

VE = By

V E = 245 lb

−ME − B y c = 0

ME = −By c

ME = −490 lb⋅ ft

+ Σ F x = 0; →

NF = 0

NF = 0 lb

NF = 0.00 lb

+

−Cy − VF = 0

V F = −Cy

V F = −308 lb

Cy( f) + MF = 0

MF = − f Cy

MF = −1.23 kip⋅ ft

Σ ME = 0

↑Σ Fy = 0; Σ MF = 0;

Problem 7-29 The bolt shank is subjected to a tension F. Determine the internal normal force, shear force, and moment at point C. Given: F = 80 lb a = 6 in

Solution: ΣF x = 0;

NC + F = 0 NC = −F NC = −80.00 lb

ΣF y = 0;

VC = 0

ΣMC = 0;

MC + F a = 0 MC = −F a MC = −480.00 lb⋅ in

Problem 7-30 Determine the normal force, shear force, and moment acting at sections passing through points B and C on the curved rod.

650

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Engineering Mechanics - Statics

Chapter 7

Units Used: kip = 103 lb Given: F 1 = 300 lb F 2 = 400 lb

θ = 30 deg

r = 2 ft

φ = 45 deg

Solution: Σ F x = 0;

F 2 sin ( θ ) − F1 cos ( θ ) + NB = 0 NB = −F 2 sin ( θ ) + F1 cos ( θ ) NB = 59.8 lb

Σ F y = 0;

V B + F2 cos ( θ ) + F1 sin ( θ ) = 0 V B = −F 2 cos ( θ ) − F 1 sin ( θ ) V B = −496 lb

Σ MB = 0;

MB + F 2 r sin ( θ ) + F 1 ( r − r cos ( θ ) ) = 0 MB = −F2 r sin ( θ ) − F1 r( 1 − cos ( θ ) ) MB = −480 lb⋅ ft

+ Σ F x = 0; →

F2 − Ax = 0

Ax = F2

Ax = 400 lb

+

Ay − F1 = 0

Ay = F1

Ay = 300 lb

↑Σ Fy = 0; Σ MA = 0;

−MA + F 1 2 r = 0

MA = 2 F 1 r

Σ F x = 0;

NC + Ax sin ( φ ) + A y cos ( φ ) = 0

MA = 1200 lb⋅ ft

NC = − A x sin ( φ ) − Ay cos ( φ ) Σ F y = 0;

V C − Ax cos ( φ ) + Ay sin ( φ ) = 0 V C = A x cos ( φ ) − A y sin ( φ )

Σ MC = 0;

NC = −495 lb

V C = 70.7 lb

−MC − MA − A x r sin ( φ ) + A y( r − r cos ( φ ) ) = 0

651

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

MC = −MA − Ax r sin ( φ ) + Ay r( 1 − cos ( φ ) )

MC = −1.59 kip⋅ ft

Problem 7-31 The cantilevered rack is used to support each end of a smooth pipe that has total weight W. Determine the normal force, shear force, and moment that act in the arm at its fixed support A along a vertical section. Units Used: 3

kip = 10 lb Given: W = 300 lb r = 6 in

θ = 30 deg

Solution: Pipe: +

↑

Σ F y = 0;

NB cos ( θ ) −

NB =

1 2

W 2

=0

⎛ W ⎞ ⎜ ( )⎟ ⎝ cos θ ⎠

NB = 173.205 lb

Rack: + Σ F x = 0; →

−NA + NB sin ( θ ) = 0 NA = NB sin ( θ )

+

↑Σ Fy = 0;

NA = 86.6 lb

V A − NB cos ( θ ) = 0 V A = NB cos ( θ )

Σ MA = 0;

MA − NB

⎛ r + r sin ( θ ) ⎞ = ⎜ ⎟ ⎝ cos ( θ ) ⎠

MA = NB

⎛ r + r sin ( θ ) ⎞ ⎜ ⎟ ⎝ cos ( θ ) ⎠

V A = 150 lb 0

MA = 1.800 kip⋅ in

652

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-32 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: 3

kN = 10 N

Given: kN m

w = 0.75 F = 4 kN a = 1.5 m

d = 1.5 m

b = 1.5 m

e = 3

c = 2.5 m

f = 4

Solution: Σ MC = 0; −B x( c + d) +

f ⎛ ⎞ ⎜ 2 2 ⎟F d = ⎝ e +f ⎠ fdF

Bx =

2

2

0

B x = 1.2 kN

e + f ( c + d) Σ MA = 0;

⎛ c + d ⎞ + B ( a + b) + B ( c + d) = ⎟ y x ⎝ 2 ⎠

−w( c + d) ⎜

⎡ ( c + d) 2⎤ ⎥ − B x ( c + d) w⎢ 2 ⎣ ⎦ By = a+b

0

B y = 0.40 kN

+ Σ F x = 0; →

−ND − B x = 0

ND = −Bx

ND = −1.2 kN

+

VD + By = 0

V D = −By

V D = −0.4 kN

↑Σ Fy = 0;

653

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Σ MD = 0;

−MD + By b = 0

Chapter 7

MD = B y b

MD = 0.6 kN⋅ m

Problem 7-33 Determine the internal normal force, shear force, and moment acting at point A of the smooth hook. Given:

θ = 45 deg a = 2 in F = 20 lb Solution: ΣF x = 0;

NA − F cos ( θ ) = 0 NA = F cos ( θ )

ΣF y = 0;

V A − F sin ( θ ) = 0 V A = F sin ( θ )

ΣMB = 0;

NA = 14.1 lb

V A = 14.1 lb

MA − NA a = 0 MA = NA a

MA = 28.3 lb⋅ in

Problem 7-34 Determine the internal normal force, shear force, and moment acting at points B and C on the curved rod.

654

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Units Used: kip = 103 lb Given:

θ 2 = 30 deg

F = 500 lb r = 2 ft

a = 3

θ 1 = 45 deg

b = 4

Solution: ΣF N = 0;

⎛ F b ⎞ sin ( θ ) − ⎛ F a ⎞ cos ( θ ) + N = 2 2 B ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ a +b ⎠ ⎝ a +b ⎠ ⎡( a)cos ( θ 2 ) − b sin ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

NB = F ⎢

ΣF V = 0;

VB +

ΣMB = 0;

NB = 59.8 lb

⎛ F b ⎞ cos ( θ ) + ⎛ F a ⎞ sin ( θ ) = 2 2 ⎜ 2 2⎟ ⎜ 2 2⎟ + b + b a a ⎝ ⎠ ⎝ ⎠

V B = −F

MB +

0

⎡⎢ b cos ( θ 2) + ( a)sin ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

0

V B = −496 lb

⎛ F b ⎞ r sin ( θ ) + F⎛ a ⎞ ( r − r cos ( θ ) ) = 2 2 ⎜ 2 2⎟ ⎜ 2 2⎟ + b + b a a ⎝ ⎠ ⎝ ⎠

MB = F r

⎡⎢−b sin ( θ 2) − a + ( a)cos ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

0

MB = −480 lb⋅ ft

Also, ΣF x = 0;

Fb

− Ax +

=0

2

a +b Ax = F

ΣF y = 0;

Ay −

2

⎛ b ⎞ ⎜ 2 2⎟ ⎝ a +b ⎠ Fa 2

a +b

Ax = 400.00 lb

=0 2

655

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Ay = F

ΣMA = 0;

−MA +

MA =

Chapter 7

⎛ a ⎞ ⎜ 2 2⎟ ⎝ a +b ⎠

⎛ F a ⎞ 2r = ⎜ 2 2⎟ ⎝ a +b ⎠ 2F r a 2

2

Ay = 300.00 lb

0

MA = 1200 lb⋅ ft

a +b ΣF x = 0;

NC + Ax sin ( θ 1 ) + Ay cos ( θ 1 ) = 0 NC = − A x sin ( θ 1 ) − A y cos ( θ 1 )

ΣF y = 0;

NC = −495 lb

V C − Ax cos ( θ 1 ) + A y sin ( θ 1 ) = 0 V C = A x cos ( θ 1 ) − Ay sin ( θ 1 )

ΣMC = 0;

V C = 70.7 lb

−MC − MA + A y( r − r cos ( θ 1 ) ) − Ax r sin ( θ 1 ) = 0 MC = −MA + Ay( r − r cos ( θ 1 ) ) − A x r sin ( θ 1 )

MC = −1.59 kip⋅ ft

Problem 7-35 Determine the ratio a/b for which the shear force will be zero at the midpoint C of the beam.

Solution: Find Ay:

ΣMB = 0;

656

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

( 2 a + b) w ⎡⎢ ( b − a)⎤⎥ − Ay b = 0

1

1

⎣3

2

Ay =

⎦

w ( 2 a + b) ( b − a) 6b

This problem requires

V C = 0.

Summing forces vertically for the section, we have +

↑Σ Fy = 0; w 1 ( 2a + b) ( b − a) − 2 6b

⎛a + ⎜ ⎝

b⎞ w

⎟

2⎠ 2

=0

w w ( 2a + b) ( b − a) = ( 2a + b) 8 6b 4 ( b − a) = 3 b

b = 4a a 1 = 4 b

Problem 7-36 The semicircular arch is subjected to a uniform distributed load along its axis of w0 per unit length. Determine the internal normal force, shear force, and moment in the arch at angle θ. Given:

θ = 45 deg

Solution: Resultants of distributed load:

657

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Engineering Mechanics - Statics

Chapter 7

θ ⌠ F Rx = ⎮ w0 r dθ sin ( θ ) = r w0 ( 1 − cos ( θ ) ) ⌡0

F Rx = r w0 ( 1 − cos ( θ ) ) θ ⌠ F Ry = ⎮ w0 r dθ cos ( θ ) = r w0 sin ( θ ) ⌡0

F Rx = r w0 ( sin ( θ ) ) θ ⌠ 2 MRo = ⎮ w0 r dθ r = r w0 θ ⌡0

Σ F x = 0;

−V + FRx cos ( θ ) − F Ry sin ( θ ) = 0

V = ⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ cos ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ sin ( θ ) V = w0 r( cos ( θ ) − 1 ) a = cos ( θ ) − 1 a = −0.293 Σ F y = 0;

V = a r w0

N + FRy cos ( θ ) + F Rx sin ( θ ) = 0 N = −⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ sin ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ cos ( θ ) N = −w0 r sin ( θ ) b = −sin ( θ ) b = −0.707

ΣMo = 0;

N = w0 r b

−M + r w0 ( θ ) + b r w0 r = 0 2

M = w0 r ( θ + b) 2

c = θ+b c = 0.0783

2

M = c r w0

658

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-37 The semicircular arch is subjected to a uniform distributed load along its axis of w0 per unit length. Determine the internal normal force, shear force, and moment in the arch at angle θ. Given:

θ = 120 deg Solution: Resultants of distributed load: θ ⌠ F Rx = ⎮ w0 r dθ sin ( θ ) = r w0 ( 1 − cos ( θ ) ) ⌡0

F Rx = r w0 ( 1 − cos ( θ ) ) θ ⌠ F Ry = ⎮ w0 r dθ cos ( θ ) = r w0 sin ( θ ) ⌡0

F Rx = r w0 ( sin ( θ ) ) θ ⌠ 2 MRo = ⎮ w0 r dθ r = r w0 θ ⌡0

Σ F x = 0;

−V + FRx cos ( θ ) − F Ry sin ( θ ) = 0 V = ⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ cos ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ sin ( θ ) V = w0 r( cos ( θ ) − 1 ) a = cos ( θ ) − 1 a = −1.500

Σ F y = 0;

V = a r w0

N + FRy cos ( θ ) + F Rx sin ( θ ) = 0 N = −⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ sin ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ cos ( θ ) N = −w0 r sin ( θ ) b = −sin ( θ ) 659

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

b = −0.866 Σ Mo = 0;

N = w0 ⋅ r⋅ b

−M + r w0 ( θ ) + b r w0 r = 0 2

M = w0 r ( θ + b) 2

c = θ+b c = 1.2284

2

M = c r w0

Problem 7-38 Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe. Units Used: kip = 103 lb Given:

⎛ 0 ⎞ F 1 = ⎜ 350 ⎟ lb ⎜ ⎟ ⎝ −400 ⎠ ⎛ 150 ⎞ F 2 = ⎜ 0 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠ a = 1.5 ft

b = 2 ft

c = 3 ft

Solution:

⎛c⎞ r1 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ r2 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

F C = −F1 − F2

⎛ −150.00 ⎞ F C = ⎜ −350.00 ⎟ lb ⎜ ⎟ ⎝ 700.00 ⎠

660

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

MC = −r1 × F1 − r2 × F2

Chapter 7

⎛ 1400.00 ⎞ MC = ⎜ −1200.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −750.00 ⎠

Problem 7-39 Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe. Units Used: kip = 103 lb Given:

⎛ −80 ⎞ F 1 = ⎜ 200 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠ ⎛ 250 ⎞ F 2 = ⎜ −150 ⎟ lb ⎜ ⎟ ⎝ −200 ⎠ a = 1.5 ft

b = 2 ft

c = 3 ft

Solution:

⎛c⎞ r1 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ r2 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

F C = −F1 − F2

⎛ −170.00 ⎞ F C = ⎜ −50.00 ⎟ lb ⎜ ⎟ ⎝ 500.00 ⎠

MC = −r1 × F1 − r2 × F2

⎛ 1000.00 ⎞ MC = ⎜ −900.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −260.00 ⎠

Problem 7-40 Determine the x, y, z components of internal loading in the rod at point D.

661

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Engineering Mechanics - Statics

Chapter 7

Units Used: kN = 103 N Given: M = 3 kN⋅ m

⎛ 7 ⎞ F = ⎜ −12 ⎟ kN ⎜ ⎟ ⎝ −5 ⎠ a = 0.75 m b = 0.2 m c = 0.2 m d = 0.6 m e = 1m Solution: Guesses Cx = 1 N

Cy = 1 N

Bx = 1 N

Bz = 1 N

Ay = 1 N

Az = 1 N

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ Bx ⎟⎞ ⎛⎜ Cx ⎟⎞ ⎜ Ay ⎟ + ⎜ 0 ⎟ + ⎜ Cy ⎟ + F = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ −e ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎛ 0 ⎞ ⎞ ⎜ b + c + d ⎟ × ⎜ Ay ⎟ + ⎜ b + c ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎜ C ⎟ + ⎜ b + c + d ⎟ × F + ⎜ 0 ⎟ = 0 y ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ 0 a ⎠ ⎝ Bz ⎠ ⎝ ⎠ ⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎝ −M ⎠ ⎝ 0 ⎠ ⎝ Az ⎠ ⎝

662

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Engineering Mechanics - Statics

⎛⎜ Ay ⎞⎟ ⎜ Az ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) y z x z x y ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ Cy ⎟ ⎝ ⎠

Chapter 7

⎛⎜ Ay ⎞⎟ ⎛ −53.60 ⎞ ⎜ Az ⎟ ⎜ 87.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 109.00 ⎟ kN ⎜ Bz ⎟ ⎜ −82.00 ⎟ ⎜ ⎟ ⎜ −116.00 ⎟ ⎜ Cx ⎟ ⎜ ⎟ 65.60 ⎝ ⎠ ⎜ Cy ⎟ ⎝ ⎠

Guesses V Dx = 1 N

NDy = 1 N

V Dz = 1 N

MDx = 1 N⋅ m

MDy = 1 N⋅ m

MDz = 1 N⋅ m

Given

⎛⎜ Cx ⎟⎞ ⎛⎜ VDx ⎟⎞ ⎜ Cy ⎟ + ⎜ NDy ⎟ = 0 ⎜ 0 ⎟ ⎜V ⎟ ⎝ ⎠ ⎝ Dz ⎠ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛⎜ MDx ⎟⎞ ⎛ 0 ⎞ ⎜ −b ⎟ × ⎜ C ⎟ + ⎜ MDy ⎟ + ⎜ 0 ⎟ = 0 ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎝ a ⎠ ⎝ 0 ⎠ ⎝ MDz ⎟⎠ ⎝ −M ⎠ ⎛⎜ VDx ⎟⎞ ⎜ NDy ⎟ ⎜ ⎟ ⎜ VDz ⎟ = Find ( V , N , V , M , M , M ) Dx Dy Dz Dx Dy Dz ⎜ MDx ⎟ ⎜ ⎟ ⎜ MDy ⎟ ⎜ MDz ⎟ ⎝ ⎠

⎛ VDx ⎞ ⎛ 116.00 ⎞ ⎜ ⎟ ⎜ NDy ⎟ = ⎜ −65.60 ⎟ kN ⎜ V ⎟ ⎜⎝ 0.00 ⎟⎠ ⎝ Dz ⎠ ⎛ MDx ⎞ ⎛ 49.20 ⎞ ⎜ ⎟ ⎜ MDy ⎟ = ⎜ 87.00 ⎟ kN⋅ m ⎜ M ⎟ ⎜⎝ 26.20 ⎟⎠ ⎝ Dz ⎠

663

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Engineering Mechanics - Statics

Chapter 7

Problem 7-41 Determine the x, y, z components of internal loading in the rod at point E. Units Used: 3

kN = 10 N Given: M = 3 kN⋅ m

⎛ 7 ⎞ F = ⎜ −12 ⎟ kN ⎜ ⎟ ⎝ −5 ⎠ a = 0.75 m b = 0.4 m c = 0.6 m d = 0.5 m e = 0.5 m Solution: Guesses Cx = 1 N

Cy = 1 N

Bx = 1 N

Bz = 1 N

Ay = 1 N

Az = 1 N

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ Bx ⎟⎞ ⎛⎜ Cx ⎟⎞ ⎜ Ay ⎟ + ⎜ 0 ⎟ + ⎜ Cy ⎟ + F = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ −d − e ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ b + c ⎟ × ⎜ Ay ⎟ + ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎜ C ⎟ + ⎜ b + c ⎟ × F + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ 0 ⎝ −M ⎠ ⎝ ⎠ ⎝ Az ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ a ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠

664

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Engineering Mechanics - Statics

⎛⎜ Ay ⎞⎟ ⎜ Az ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) y z x z x y ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ Cy ⎟ ⎝ ⎠

Chapter 7

⎛⎜ Ay ⎞⎟ ⎛ −53.60 ⎞ ⎜ Az ⎟ ⎜ 87.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 109.00 ⎟ kN ⎜ Bz ⎟ ⎜ −82.00 ⎟ ⎜ ⎟ ⎜ −116.00 ⎟ ⎜ Cx ⎟ ⎜ ⎟ 65.60 ⎝ ⎠ ⎜ Cy ⎟ ⎝ ⎠

Guesses NEx = 1 N

V Ey = 1 N

V Ez = 1 N

MEx = 1 N⋅ m

MEy = 1 N⋅ m

MEz = 1 N⋅ m

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ NEx ⎟⎞ ⎜ Ay ⎟ + ⎜ VEy ⎟ = 0 ⎜ Az ⎟ ⎜ V ⎟ ⎝ ⎠ ⎝ Ez ⎠ ⎛ −e ⎞ ⎛⎜ 0 ⎞⎟ ⎛⎜ MEx ⎞⎟ ⎜ 0 ⎟ × Ay + M =0 ⎜ ⎟ ⎜ ⎟ ⎜ Ey ⎟ ⎝ 0 ⎠ ⎜⎝ Az ⎟⎠ ⎜⎝ MEz ⎟⎠ ⎛⎜ NEx ⎞⎟ ⎜ VEy ⎟ ⎜ ⎟ ⎜ VEz ⎟ = Find ( N , V , V , M , M , M ) Ex Ey Ez Ex Ey Ez ⎜ MEx ⎟ ⎜ ⎟ ⎜ MEy ⎟ ⎜ MEz ⎟ ⎝ ⎠

⎛ NEx ⎞ ⎛ 0.00 ⎞ ⎜ ⎟ ⎜ VEy ⎟ = ⎜ 53.60 ⎟ kN ⎜ V ⎟ ⎜⎝ −87.00 ⎟⎠ ⎝ Ez ⎠ ⎛ MEx ⎞ ⎛ 0.00 ⎞ ⎜ ⎟ ⎜ MEy ⎟ = ⎜ −43.50 ⎟ kN⋅ m ⎜ M ⎟ ⎜⎝ −26.80 ⎟⎠ ⎝ Ez ⎠

Problem 7-42 Draw the shear and moment diagrams for the shaft in terms of the parameters shown; There is 665

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

g a thrust bearing at A and a journal bearing at B.

p

Units Used: 3

kN = 10 N Given: P = 9 kN a = 2m L = 6m Solution: P ( L − a) − Ay L = 0

Ay = P

L−a L

x1 = 0 , 0.01a .. a Ay − V 1 ( x) = 0

V 1 ( x) =

M1 ( x) − Ay x = 0

M1 ( x) =

Ay kN

Ay x kN⋅ m

x2 = a , 1.01a .. L Ay − P − V2 ( x) = 0

V 2 ( x) =

Ay − P kN

M2 ( x) − Ay x + P ( x − a) = 0

M2 ( x) =

A y x − P ( x − a) kN⋅ m

666

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Engineering Mechanics - Statics

Chapter 7

Force (kN)

10

V1( x1) 5 V2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Moment (kN-m)

15 10

M1( x1) M2( x2)

5 0 5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Problem 7-43 Draw the shear and moment diagrams for the beam in terms of the parameters shown. Given: P = 800 lb

a = 5 ft

L = 12 ft

Solution:

667

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Engineering Mechanics - Statics

x1 = 0 , 0.01a .. a

Chapter 7

x2 = a , 1.01a .. L − a

x3 = L − a , 1.01( L − a) .. L

V 2 ( x) = 0

V 3 ( x) =

P

V 1 ( x) =

lb Px

M1 ( x) =

M2 ( x) =

lb⋅ ft

Pa

M3 ( x) =

lb⋅ ft

−P lb P( L − x) lb⋅ ft

Force (lb)

1000

V1( x1)

500

V2( x2)

0

V3( x3)

500

1000

0

2

4

6

8

10

8

10

12

x1 x2 x3 , , ft ft ft

Distance (ft)

Moment (lb-ft)

4000

M1( x1) M2( x2)

M3( x3)2000

0

0

2

4

6

12

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-44 Draw the shear and moment diagrams for the beam (a) in terms of the parameters 668

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

g shown; (b) set M0 and L as given.

( )

p

Given: M0 = 500 N⋅ m L = 8m Solution: For

0 ≤ x≤

L 3

+

↑Σ Fy = 0;

V1 = 0

ΣMx = 0;

For

L 3

≤x≤

M1 = 0

2L 3

+

↑Σ Fy = 0; ΣMx = 0;

For

2L 3

M2 = M0

≤x≤L

+

↑Σ Fy = 0; ΣMx = 0;

( b)

V2 = 0

V3 = 0 M3 = 0

x1 = 0 , 0.01L ..

L 3

x2 =

L L 2L , 1.01 .. 3

3

3

x3 =

2L 2L 3

,

3

V 1 ( x1 ) = 0

V 2 ( x2 ) = 0

V 3 ( x3 ) = 0

M1 ( x1 ) = 0

M2 ( x2 ) = M0

M3 ( x3 ) = 0

1.01 .. L

669

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

1

Shear force in N

0.5

V1( x1) V2( x2)

0

V3( x3)

0.5

1

0

1

2

3

4

5

6

7

8

9

x1 , x2 , x3

Distance in m

600

Moment in N - m

400

M1( x1) M2( x2)

M3( x3)200

0

0

1

2

3

4

5

6

7

8

9

x1 , x2 , x3

Distance in m

Problem 7-45 The beam will fail when the maximum shear force is V max or the maximum bending moment is Mmax. Determine the magnitude M0 of the largest couple moments it will support.

670

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Units Used: 3

kN = 10 N Given: L = 9m V max = 5 kN Mmax = 2 kN⋅ m

Solution: The shear force is zero everywhere in the beam. The moment is zero in the first third and the last third of the bam. In the middle section of the beam the moment is

M = M0

M0 = Mmax

Thus the beam will fail when

M0 = 2.00 kN⋅ m

Problem 7-46 The shaft is supported by a thrust bearing at A and a journal bearing at B. Draw the shear and moment diagrams for the shaft in terms of the parameters shown. Given: w = 500

lb ft

L = 10 ft Solution:

For

ΣF y = 0;

wL

ΣM = 0;

2

0 ≤ x< L

− wx − V = 0

−w L 2

⎛x⎞ + M = ⎟ ⎝ 2⎠

x + w x⎜

V ( x) =

0

w 2

( L − 2 x)

M ( x) =

1

lb

1 2 ( Lx − x ) 2 lb⋅ ft

w

671

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (lb)

5000

V( x )

0

5000

0

2

4

6

8

10

x ft

Distance (ft) 1 .10

Moment (lb-ft)

4

M( x)

0

1 .10

4

0

2

4

6

8

10

x ft

Distance (ft)

Problem 7-47 The shaft is supported by a thrust bearing at A and a journal bearing at B. The shaft will fail when the maximum moment is Mmax. Determine the largest uniformly distributed load w the shaft will support. Units Used: 3

kip = 10 lb Given: L = 10 ft Mmax = 5 kip⋅ ft

672

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Engineering Mechanics - Statics

Chapter 7

Solution: wL 2

−

− wx − V = 0

wL 2

V = −w x +

⎛x⎞ x + wx ⎜ ⎟ + M = ⎝2⎠

wL 2

2 wL⎞ wx ⎛ M=⎜ ⎟x − 2 ⎝ 2 ⎠

0

From the Moment Diagram, wL

Mmax =

2

w =

8

8 Mmax

L

2

w = 400.00

lb ft

Problem 7-48 Draw the shear and moment diagrams for the beam. 3

kN = 10 N

Units Used: Given: w = 2

kN m

L = 5m MB = 5 kN⋅ m Solution: ΣF y = 0; −V ( x) + wL − wx = 0

ΣM = 0;

V ( x) = ( w L − w x)

⎛ L ⎞ + M − w L x + w x⎛ x ⎞ = ⎟ ⎜ 2⎟ B ⎝2⎠ ⎝ ⎠

M ( x) + w L⎜

⎡ ⎣

1

kN

0

⎛ x ⎞ − w L⎛ L ⎞ − M ⎤ 1 ⎟ ⎜2⎟ B⎥ ⎝2⎠ ⎝ ⎠ ⎦ kN⋅ m

M ( x) = ⎢w L x − w x⎜

673

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (kN)

10

V( x ) 5

0

0

1

2

3

4

5

4

5

x

Distance (m)

Moment (kN-m)

0

M( x) 20

40

0

1

2

3

x

Distance (m)

Problem 7-49 Draw the shear and moment diagrams for the beam. 3

kN = 10 N

Units Used: Given: w = 3

kN m

F = 10 kN

L = 6m Solution: V ( x) − w( L − x) − F = 0 V ( x) = [ w( L − x) + F ]

1

kN

⎛ L − x ⎞ − F( L − x) = ⎟ ⎝ 2 ⎠

−M ( x) − w( L − x) ⎜

0

674

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

⎡

( L − x)

⎣

2

M ( x) = ⎢−w

2

Chapter 7

⎤

− F ( L − x)⎥

1

⎦ kN⋅ m

Force (kN)

40

V( x) 20

0

0

1

2

3

4

5

6

x

Distance (m) Moment (kN-m)

0 50

M( x) 100 150

0

1

2

3

4

5

6

x

Distance (m)

Problem 7-50 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: a = 2m

b = 4m

w = 1.5

kN m

Solution:

⎛ b − a⎞ − A b = ⎟ y ⎝ 2 ⎠

w( b − a) ⎜

0

Ay =

w ( b − a) 2b

2

Ay = 0.75 kN

x1 = 0 , 0.01a .. a Ay − V 1 ( x) = 0

V 1 ( x) = Ay

1

kN 675

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

− Ay x + M1 ( x) = 0

M1 ( x) = A y x

1

kN⋅ m

x2 = b − a , 1.01( b − a) .. b 1

Ay − w( x − a) − V 2 ( x) = 0

V 2 ( x) = ⎡⎣A y − w( x − a)⎤⎦

⎛ x − a ⎞ + M ( x) = − Ay x + w( x − a) ⎜ ⎟ 2 ⎝ 2 ⎠

2 ⎡ ( x − a) ⎤ ⎢ ⎥ 1 M2 ( x) = Ay x − w 2 ⎣ ⎦ kN⋅ m

0

kN

Force (kN)

2

V1( x1) 0 V2( x2)

2

4

0

0.5

1

1.5

2

2.5

3

2.5

3

3.5

4

x1 , x2

Distance (m)

Moment (kN-m)

2

M1( x1) 1 M2( x2) 0

1

0

0.5

1

1.5

2

3.5

4

x1 , x2

Distance (m)

676

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-51 Draw the shear and moment diagrams for the beam. Given: L = 20 ft

w = 250

lb ft

M1 = 150 lb⋅ ft

Solution:

⎛ L⎞ − M − A l = ⎟ 2 y ⎝2⎠

M 1 + w L⎜

V ( x) = ( Ay − w x)

M2 = 150 lb⋅ ft

M1 − M2 + w Ay =

0

⎛ L2 ⎞ ⎜ ⎟ ⎝2⎠

L

Ay = 2500 lb

⎡ ⎛ x2 ⎞ ⎤ 1 ⎢ M ( x) = A y x − w⎜ ⎟ − M1⎥ ⎣ ⎝2⎠ ⎦ lb⋅ ft

1

lb

Force in lb

2000

V( x )

0

2000 0

5

10

15

20

x ft

Distance in ft

677

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

1.5 .10

Moment in lb ft

4

1 .10

4

M( x) 5000

0 0

5

10

15

20

x ft

Distance in ft

Problem 7-52 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: w = 40

kN m

F = 20 kN

M = 150 kN⋅ m

Solution: − Ay a + w a

⎛ a⎞ − F b − M = ⎜ 2⎟ ⎝ ⎠

0

x1 = 0 , 0.01a .. a 1

kN

⎛x

M1 ( x) = ⎢Ay x − w⎜

⎣

b = 3m

⎡ ⎛ a2 ⎞ ⎤ ⎟ − F b − M⎥ ⎛⎜ 1 ⎟⎞ ⎣ ⎝2⎠ ⎦⎝ a ⎠

Ay = ⎢w⎜

Ay = 133.75 kN

x2 = a , 1.01a .. a + b

V 1 ( x) = ( A y − w x)

⎡

a = 8m

2 ⎞⎤

⎟⎥ 1 ⎝ 2 ⎠⎦ kN⋅ m

V 2 ( x) = F

1

kN

M2 ( x) = [ −F( a + b − x) − M]

1

kN⋅ m

678

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force in kN

200

V1( x1) V2( x2)

0

200

0

2

4

6

8

10

x1 , x2

Distance in m

Moment (kN-m)

400

M1( x1)

200

M2( x2)

0 200 400

0

2

4

6

8

10

x1 , x2

Distance (m)

Problem 7-53 Draw the shear and moment diagrams for the beam.

679

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Solution: 0 ≤ x< a ΣF y = 0; −V − w x = 0

V = −w x ΣM = 0;

M + w x⎛⎜

x⎞

M = −w

x

⎟= ⎝2⎠

0

2

2

a < x ≤ 2a ΣF y = 0; −V + 2 w a − w x = 0

V = w( 2 a − x) ΣM = 0;

M + w x⎛⎜

x⎞

⎟ − 2 w a( x − a) = ⎝2⎠

0

2

M = 2 w a( x − a) −

wx 2

Problem 7-54 Draw the shear and bending-moment diagrams for beam ABC. Note that there is a pin at B.

680

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Engineering Mechanics - Statics

Chapter 7

Solution: Support Reactions: From FBD (b), wL ⎛ L⎞

⎛ L⎞ ⎜ 4 ⎟ − B y⎜ 2 ⎟ = ⎝ ⎠ ⎝ ⎠

2

wL

By =

4

Ay −

From FBD (a),

wL 2

−

B y

=0

3w L

Ay =

−B y

0

4

L 2

− w⎛⎜

MA = w

⎛ L⎞ + M = ⎟ ⎜ ⎟ A ⎝2⎠ ⎝4⎠ L⎞

0

⎛ L2 ⎞ ⎜ ⎟ ⎝4⎠

Shear and Moment Functions: From FBD (c) For 0 ≤ x≤ L

V=

w 4

Ay − w x − V = 0

( 3 L − 4 x)

x MA − Ay x + w x ⎛⎜ ⎟⎞ + M = 0 ⎝ 2⎠

M=

w 4

(3L x − 2x2 − L2)

681

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Engineering Mechanics - Statics

Chapter 7

Problem 7-55 The beam has depth a and is subjected to a uniform distributed loading w which acts at an angle θ from the vertical as shown. Determine the internal normal force, shear force, and moment in the beam as a function of x. Hint:The moment loading is to be determined from a point along the centerline of the beam (x axis). Given: a = 2 ft L = 10 ft

θ = 30 deg w = 50

lb ft

Solution: 0 ≤ x≤ L

N + w sin ( θ ) x = 0

ΣF x = 0;

N ( x) = −w sin ( θ ) x −V − w cos ( θ ) x = 0

ΣF y = 0;

V = −w cos ( θ ) x

⎛ x ⎞ − w sin ( θ ) x⎛ a ⎞ + M = ⎟ ⎜ 2⎟ ⎝2⎠ ⎝ ⎠

w cos ( θ ) x⎜

ΣM = 0;

M ( x) = −w cos ( θ )

0

⎛ x2 ⎞ ⎜ ⎟ + w sin ( θ ) ⎛⎜ x a ⎞⎟ ⎝2⎠ ⎝2⎠

Problem 7-56 Draw the shear and moment diagrams for the beam. Given: w = 250

lb ft

L = 12 ft

682

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Engineering Mechanics - Statics

Chapter 7

Solution: 1

2

wx =0 L

ΣF y = 0;

−V −

ΣM = 0;

x ⎛ w x⎞ 1 M + ⎜ ⎟ ⎛⎜ x⎟⎞ = 0 2 ⎝ L ⎠⎝ 3 ⎠

2

x

V ( x) = −

wx 1 2 L lb 3

−w x M ( x) = 6L

1

lb⋅ ft

Force (lb)

0

V( x) 1000

2000

0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (lb-ft)

0

M( x)

5000

1 .10

4

0

2

4

6

8

10

12

x ft

Distance (ft)

Problem 7-57 The beam will fail when the maximum shear force is V max or the maximum moment is Mmax. Determine the largest intensity w of the distributed loading it will support.

683

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Engineering Mechanics - Statics

Chapter 7

Given: L = 18 ft V max = 800 lb Mmax = 1200 lb⋅ ft Solution: For 0 ≤ x ≤ L 2

V=

M=

2L

w1 = 2

6L

2

⎛ Vmax ⎞ ⎜ ⎟ ⎝ L ⎠

Mmax =

−w x

wL

V max =

w2 = 6

3

−w x

w2 L

w1 = 88.9

lb ft

2

6

⎛ Mmax ⎞ ⎜ 2 ⎟ ⎝ L ⎠

w2 = 22.2

Now choose the critical case

lb ft

w = min ( w1 , w2 )

w = 22.22

lb ft

Problem 7-58 The beam will fail when the maximum internal moment is Mmax. Determine the position x of the concentrated force P and its smallest magnitude that will cause failure.

684

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Engineering Mechanics - Statics

Chapter 7

Solution: For ξ < x,

M1 =

For ξ > x,

M2 =

Pξ ( L − x) L P x( L − ξ ) L

Note that M1 = M2 when x = ξ Mmax = M1 = M2 =

d dx

P x( L − x) L

(L x − x2) = L − 2x

Thus,

Mmax =

x=

=

P L

(L x − x2)

L 2

P ⎛ L ⎞⎛

L⎞ P ⎛ L⎞ ⎜ 2 ⎟ ⎜L − 2 ⎟ = 2 ⎜ 2 ⎟ L ⎝ ⎠⎝ ⎠ ⎝ ⎠

P=

4 Mmax

L

Problem 7-59 Draw the shear and moment diagrams for the beam. Given: w = 30

lb

MC = 180 lb⋅ ft

ft

a = 9 ft

b = 4.5 ft

Solution: − Ay a +

1 2

wa

⎛ a⎞ − M = ⎜ 3⎟ C ⎝ ⎠

0

685

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Ay =

wa 6

Ay + By −

By =

wa 2

MC

− 1 2

Chapter 7

a wa = 0

− Ay

x1 = 0 , 0.01a .. a Ay −

1 2

⎛x⎞ w ⎜ ⎟ x − V1 ( x) = ⎝ a⎠

2 ⎛ wx ⎞ 1 ⎜ ⎟ V 1 ( x) = A y − 2 a ⎠ lb ⎝

0

⎛x⎞ ⎛x⎞ − Ay x + w⎜ ⎟ x⎜ ⎟ + M1 ( x) = 2 ⎝ a⎠ ⎝ 3 ⎠ 1

3 ⎛ wx ⎞ 1 ⎜ ⎟ M1 ( x) = Ay x − 6 a ⎠ lb⋅ ft ⎝

0

x2 = a , 1.01a .. a + b Ay −

1 2

w a + By − V 2 ( x) = 0

− Ay x +

1 2

⎛ ⎝

w a ⎜x −

V 2 ( x) =

2a ⎞ 3

⎟ − By( x − a) + M2 ( x) = ⎠

⎛A + B − wa⎞ 1 ⎜ y y 2 ⎟ ⎝ ⎠ lb

0

⎡ ⎣

M2 ( x) = ⎢Ay x + B y( x − a) −

wa 2

⎛ ⎝

⋅ ⎜x −

2 a ⎞⎤ 3

1

⎟⎥ ⎠⎦ lb⋅ ft

Force (lb)

100

V1( x1) V2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Distance (ft)

686

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Engineering Mechanics - Statics

Chapter 7

Moment (lb-ft)

100

M1( x1) M2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Diastance (ft)

Problem 7-60 The cantilevered beam is made of material having a specific weight γ. Determine the shear and moment in the beam as a function of x. Solution: By similar triangles y h = x d

y=

h x d

W = γV = γ

⎛⎜ 1 y x t⎟⎞ = γ ⎡⎢ 1 ⎛⎜ h x⎟⎞ x t⎤⎥ = ⎛⎜ γ h t ⎟⎞ x2 ⎝2 ⎠ ⎣2 ⎝ d ⎠ ⎦ ⎝ 2d ⎠

ΣF y = 0;

V−

⎛ γ h t ⎞ x2 = ⎜ ⎟ ⎝ 2d ⎠

V=

ΣM = 0;

−M −

0

⎛ γ h t ⎞ x2 ⎜ ⎟ ⎝ 2d ⎠ ⎛ γ h t ⎞ x2 ⎛ x ⎞ = ⎜ ⎟ ⎜3⎟ ⎝ 2d ⎠ ⎝ ⎠

0

687

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Engineering Mechanics - Statics

M=−

Chapter 7

γh t 3 6d

x

Problem 7-61 Draw the shear and moment diagrams for the beam. 3

kip = 10 lb

Given:

w1 = 30

lb ft

w2 = 120

lb ft

L = 12 ft Solution:

w1 L

⎛ L⎞ + ⎜2⎟ ⎝ ⎠

(w2 − w1)L ⎛⎜ 3 ⎞⎟ − Ay L = 0 2 L

1

⎝ ⎠

⎛ L ⎞ + ⎛⎜ w2 − w1 ⎞⎟ ⎛ L ⎞ ⎟ ⎜ ⎟ ⎝2⎠ ⎝ 2 ⎠ ⎝3⎠

Ay = w1 ⎜

Ay − w1 x −

(w2 − w1)⎛⎜ L ⎞⎟ x − V ( x) 2 x

1

⎝ ⎠

⎡ V ( x) = ⎢Ay − w1 x − ⎣ ⎛x⎞ + ⎟ ⎝ 2⎠

− Ay x + w1 x⎜

=0

⎛ x2 ⎞⎤ 1 (w2 − w1)⎜ L ⎟⎥ lb 2 ⎝ ⎠⎦ 1

(w2 − w1)⎛⎜ L ⎞⎟ x ⎛⎜ 3 ⎟⎞ + M ( x) 2 1

x

x

⎝ ⎠

⎝ ⎠

=0

⎡ ⎛ x2 ⎞ ⎛ x3 ⎞⎤ 1 ⎜ ⎟ M ( x) = ⎢A y x − w1 − ( w2 − w1 ) ⎜ ⎟⎥ ⎝2⎠ ⎣ ⎝ 6 L ⎠⎦ kip⋅ ft

688

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (lb)

500

0

V( x )

500 0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (kip-ft)

2

M( x) 1

0

0

2

4

6

8

10

12

x ft

Distance (ft)

Problem 7-62 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb Given: F 1 = 20 kip

F 2 = 20 kip

w = 4

kip ft

a = 15 ft

b = 30 ft

c = 15 ft

Solution:

⎛ b⎞ − F c − A b = ⎟ 2 y ⎝ 2⎠

F 1 ( a + b) + w b⎜

⎛ a + b⎞ + wb − F ⎛ c ⎞ ⎟ 2⎜ ⎟ ⎝ b ⎠ 2 ⎝ b⎠

Ay = F1 ⎜

0

Ay + By − F1 − F2 − w b = 0

By = F1 + F2 + w b − Ay 689

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

x1 = 0 , 0.01a .. a 1

−F 1 − V 1 ( x) = 0

V 1 ( x) = −F 1

F 1 x + M1 ( x) = 0

M1 ( x) = −F1 x

kip 1

kip⋅ ft

x2 = a , 1.01a .. a + b 1

V 2 ( x) = ⎡⎣−F1 − w( x − a) + A y⎤⎦

−F 1 − w( x − a) + Ay − V 2 ( x) = 0

⎛ x − a ⎞ + M ( x) = ⎟ 2 ⎝ 2 ⎠

F 1 x − Ay( x − a) + w( x − a) ⎜

kip

0

⎡

( x − a)

⎣

2

M2 ( x) = ⎢−F 1 x + Ay( x − a) − w

2⎤

⎥ 1 ⎦ kip⋅ ft

x3 = a + b , 1.01( a + b) .. a + b + c 1

V 3 ( x) − F2 = 0

V 3 ( x) = F 2

−M3 ( x) − F 2 ( a + b + c − x) = 0

M3 ( x) = −F2 ( a + b + c − x)

kip 1

kip⋅ ft

Force (kip)

100

V1( x1) V2( x2) V3( x3)

50

0

50

100

0

10

20

30

40

50

60

x1 x2 x3 , , ft ft ft

Distance (ft)

690

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

200

Moment (kip-ft)

M1( x1)

0

M2( x2) M3( x3)

200

400

0

10

20

30

40

50

60

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-63 Express the x, y, z components of internal loading in the rod at the specific value for y, where 0 < y< a Units Used: 3

kip = 10 lb Given:

y = 2.5 ft

w = 800

lb F = 1500 lb ft

a = 4 ft

b = 2 ft

Solution:

In general we have

⎡ F ⎤ ⎥ V = ⎢ 0 ⎢ ⎥ ⎣w( a − y) ⎦

691

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.

Engineering Mechanics - Statics

⎡⎢ w( a − y) ⎛ a − ⎜ 2 ⎝ ⎢ M = ⎢ −F b ⎢ −F( a − y) ⎣

Chapter 7

y⎞⎤

⎟⎥ ⎠⎥ ⎥ ⎥ ⎦

For these values we have

⎛ 1500.00 ⎞ V = ⎜ 0.00 ⎟ lb ⎜ ⎟ ⎝ 1200.00 ⎠

⎛ 900.00 ⎞ M = ⎜ −3000.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −2250.00 ⎠

Problem 7-64 Determine the normal force, shear force, and moment in the curved rod as a function of θ. Given: c = 3 d = 4

Solution: For

0 ≤ θ ≤ π

ΣF x = 0;

ΣF y = 0;

N−

⎛ d ⎞ P cos ( θ ) − ⎛ c ⎞ P sin ( θ ) = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠

N=

⎛ P ⎞ ( d cos ( θ ) + c sin ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

V−

⎛ d ⎞ P sin ( θ ) + ⎛ c ⎞ P cos ( θ ) = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠

V=

⎛ P ⎞ ( d sin ( θ ) − c cos ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

0

0

692

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

ΣM = 0;

Chapter 7

⎛ −d ⎞ P( r − r cos ( θ ) ) + ⎛ c ⎞ P r sin ( θ ) + M = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠ M=

0

⎛ P r ⎞ ( d − d cos ( θ ) − c sin ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

Problem 7-65 The quarter circular rod lies in the horizontal plane and supports a vertical force P at its end. Determine the magnitudes of the components of the internal shear force, moment, and torque acting in the rod as a function of the angle θ.

Solution: ΣF z = 0;

V= P

ΣMx = 0;

M + P r cos ( θ ) = 0 M = −P r cos ( θ ) M = P r cos ( θ )

ΣMy = 0;

T + P r( l − sin ( θ ) ) = 0 T = −P r ( l − sin ( θ ) T = P r ( 1 − sin ( θ )

693

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-66 Draw the shear and moment diagrams for the beam. Given: MB = 800 lb⋅ ft a = 5 ft F = 100 lb

b = 5 ft

Solution: V 1 ( x) = F

x2 = a , 1.01a .. a + b

V 2 ( x) = F

Force (lb)

x1 = 0 , 0.01a .. a

1

M1 ( x) = ⎡⎣−F ( a + b − x) − MB⎤⎦

lb 1

M2 ( x) = −F( a + b − x)

lb

1

lb⋅ ft

1

lb⋅ ft

V1( x1) 100 V2( x2)

99.9

99.8

0

2

4

6

8

10

x1 x2 , ft ft

Distance (ft)

Moment (lb-ft)

0

M1( x1) M2( x2)

1000

2000

0

2

4

6

8

10

x1 x2 , ft ft

Distane (ft)

694

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-67 Draw the shear and moment diagrams for the beam. kN = 103 N

Units Used: Given: w = 3

kN

F = 10 kN

m

L = 6m Solution: V ( x) = [ w( L − x) + F ]

⎡

( L − x)

⎣

2

M ( x) = ⎢−w

2

1

kN

⎤

− F ( L − x)⎥

1

⎦ kN⋅ m

Force (kN)

40

V( x) 20

0

0

1

2

3

4

5

6

x

Distance (m)

Moment (kN-m)

0 50

M( x) 100 150

0

1

2

3

4

5

6

x

Distance (m)

695

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-68 Draw the shear and moment diagrams for the beam. 3

kN = 10 N

Units Used: Given:

F = 7 kN

Solution: Given

M = 12 kN⋅ m A = 1N

Guesses A+B−F=0

a = 2m

b = 2m

c = 4m

B = 1N

⎛A⎞ ⎜ ⎟ = Find ( A , B) ⎝B ⎠

−F a − M + B ( a + b + c) = 0

x1 = 0 , 0.01a .. a

x2 = a , 1.01a .. a + b

x3 = a + b , 1.01( a + b) .. a + b + c

V 1 ( x1 ) = A

V 2 ( x2 ) = ( A − F)

V 3 ( x3 ) = −B

M1 ( x1 ) =

1

kN A x1

kN⋅ m

M2 ( x2 ) =

1

kN

A x2 − F( x2 − a) kN⋅ m

1

kN

M3 ( x3 ) = B ( a + b + c − x3 )

1

kN⋅ m

Force (kN)

5

V1( x1) V2( x2) V3( x3)

0

5

0

1

2

3

4

5

6

7

x1 , x2 , x3

Distance (m)

696

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8

Engineering Mechanics - Statics

Chapter 7

Moment (kN-m)

15

M1( x1)

10

M2( x2)

5

M3( x3)

0

5

0

1

2

3

4

5

6

7

8

x1 , x2 , x3

Distance (m)

Problem 7-69 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: a = 2m

b = 4m

w = 1.5

kN m

Solution:

⎛ b − a⎞ − A b = ⎟ y ⎝ 2 ⎠

w( b − a) ⎜

x1 = 0 , 0.01a .. a

V 1 ( x) = Ay

x2 = b − a , 1.01( b − a) .. b

Ay =

0 1

kN

w ( b − a) 2b

2

Ay = 0.75 kN

M1 ( x) = A y x

V 2 ( x) = ⎡⎣A y − w( x − a)⎤⎦

1

kN⋅ m

2 ⎡ ( x − a) ⎤ 1 ⎢ ⎥ kN M2 ( x) = Ay x − w 2 ⎣ ⎦ kN⋅ m

1

697

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (kN)

2

V1( x1) 0 V2( x2)

2

4

0

0.5

1

1.5

2

2.5

3

2.5

3

3.5

4

x1 , x2

Distance (m)

Moment (kN-m)

2

M1( x1) 1 M2( x2) 0

1

0

0.5

1

1.5

2

3.5

4

x1 , x2

Distance (m)

Problem 7-70 Draw the shear and moment diagrams for the beam. Given: lb ft

w = 30

MC = 180 lb⋅ ft

a = 9 ft

b = 4.5 ft

Solution: − Ay a +

Ay =

1 2

wa 6

⎛ a⎞ − M = ⎟ C ⎝ 3⎠

w a⎜

−

0

MC a 698

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Ay + By − By =

wa 2

1

Chapter 7

wa = 0

2

− Ay

x1 = 0 , 0.01a .. a 2 ⎛ wx ⎞ 1 ⎜ Ay − ⎟ 2 a ⎠ lb ⎝

V 1 ( x) =

M1 ( x) =

3 ⎛ wx ⎞ 1 ⎜ Ay x − ⎟ 6 a ⎠ lb⋅ ft ⎝

x2 = a , 1.01a .. a + b

⎛A + B − wa⎞ 1 ⎜ y y 2 ⎟ ⎝ ⎠ lb

V 2 ( x) =

⎡ ⎣

M2 ( x) = ⎢Ay x + B y( x − a) −

wa⎛ 2

⎜x − ⎝

2 a ⎞⎤ 1 3

⎟⎥ ⎠⎦ lb⋅ ft

Force (lb)

100

V1( x1) V2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Distance (ft)

699

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (lb-ft)

100

M1( x1) M2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Diastance (ft)

Problem 7-71 Draw the shear and moment diagrams for the beam. 3

kip = 10 lb

Given:

w1 = 30

lb ft

w2 = 120

lb ft

L = 12 ft Solution:

⎛ L⎞ + ⎟ ⎝2⎠

w1 L⎜

(w2 − w1)L⎛⎜ 3 ⎞⎟ − Ay L = 0 2 L

1

⎝ ⎠

⎛ L ⎞ + ⎛⎜ w2 − w1 ⎞⎟ ⎛ L ⎞ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 3 ⎠

Ay = w1 ⎜

⎡ ⎣

⎦

⎡

x

⎣

2

M ( x) = ⎢A y x − w1

2⎤

x 1 w2 − w1 ) ⎥ ( 2 L lb 1

V ( x) = ⎢Ay − w1 x −

2

− ( w2 − w1 )

x ⎤ 1 ⎥ 6 L⎦ kip⋅ ft 3

700

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (lb)

500

0

V( x )

500 0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (kip-ft)

2

M( x) 1

0

0

2

4

6

8

10

12

x ft

Distance (ft)

Problem 7-72 Draw the shear and moment diagrams for the shaft. The support at A is a journal bearing and at B it is a thrust bearing. Given: F 1 = 400 lb w = 100

Solution:

lb in

F 2 = 800 lb a = 4 in

b = 12 in

c = 4 in

⎛ b ⎞ − F b + B( b + c) = ⎟ 2 ⎝2⎠

F 1 a − w b⎜

⎛ b2 ⎞ ⎟ + F2 b − F1 a ⎝2⎠ B=

w⎜ 0

B =

b+c

950.00 lb

701

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

x1 = 0 , 0.01a .. a

Chapter 7

V 1 ( x) = −F 1

x2 = a , 1.01a .. a + b M2 ( x2 )

1

M1 ( x) = −F1 x

lb

V 2 ( x2 ) = ⎡⎣−B + F2 + w( a + b − x2 )⎤⎦

1

lb⋅ in

1

lb

2 ⎡ a + b − x2 ) ⎤ 1 ( ⎥ = ⎢B( a + b + c − x2 ) − F2 ( a + b − x2 ) − w⋅ 2 ⎣ ⎦ lb⋅ in

V 3 ( x3 ) =

x3 = a + b , 1.01( a + b) .. a + b + c

−B lb

M3 ( x3 ) = B ( a + b + c − x3 )

1

lb⋅ in

Force (lb)

2000

V1( x1) V2( x2)

1000

V3( x3)

0

1000

0

5

10

15

20

x1 x2 x3 , , in in in

Distance (in)

Moment (lb-in)

4000

M1( x1) M2( x2) M3( x3)

2000

0

2000

0

5

10

15

x1 x2 x3 , , in in in

Distance (ini)

702

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

20

Engineering Mechanics - Statics

Chapter 7

Problem 7-73 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: F 1 = 10 kN

F 2 = 10 kN

M0 = 12 kN⋅ m

a = 2m

b = 2m

c = 2m

d = 2m

Solution: F 1 ( b + c + d) − M0 + F2 d − Ay( a + b + c + d) = 0

Ay =

F 1 ( b + c + d) − M 0 + F 2 d

Ay = 8.50 kN

a+b+c+d

Ay + By − F1 − F2 = 0

By = F1 + F2 − Ay

B y = 11.50 kN

x1 = 0 , 0.01a .. a V 1 ( x) = Ay

1

M1 ( x) = A y x

kN

1

kN⋅ m

x2 = a , 1.01a .. a + b V 2 ( x) = ( A y − F1 )

1

kN

M2 ( x) = ⎡⎣Ay x − F 1 ( x − a)⎤⎦

1

kN⋅ m

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = ( A y − F1 )

1

kN

M3 ( x) = ⎡⎣Ay x − F 1 ( x − a) + M0⎤⎦

1

kN⋅ m

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x) = −B y

1

kN

M4 ( x) = By( a + b + c + d − x)

1

kN⋅ m

703

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

10

Force (kN)

V1( x1)

5

V2( x2) 0 V3( x3) 5

V4( x4)

10 15

0

1

2

3

4

5

6

7

5

6

7

8

x1 , x2 , x3 , x4

Distance (m)

Moment (kN-m)

30

M1( x1) M2( x2)

20

M3( x3) M4( x4)10

0

0

1

2

3

4

8

x1 , x2 , x3 , x4

Distance (m)

Problem 7-74 Draw the shear and moment diagrams for the shaft. The support at A is a journal bearing and at B it is a thrust bearing.

704

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given: F = 200 lb

w = 100

lb

M = 300 lb⋅ ft

ft

Solution: F ( a + b) − A b + w b⎛⎜

F( a + b) +

b⎞

⎟−M= ⎝ 2⎠

x1 = 0 , 0.01a .. a

a = 1 ft

A =

0

V 1 ( x1 ) = −F

⎛ w b2 ⎞ ⎜ ⎟−M ⎝ 2 ⎠

M1 ( x1 ) = −F x1

1

lb

M2 ( x2 )

c = 1 ft

A = 375.00 lb

b

V 2 ( x2 ) = ⎡⎣−F + A − w( x2 − a)⎤⎦

x2 = a , 1.01a .. a + b

b = 4 ft

1

lb⋅ ft

1

lb

⎡ (x2 − a)2⎥⎤ 1 ⎢ = −F x2 + A ( x2 − a) − w 2 ⎣ ⎦ lb⋅ ft V 3 ( x3 ) = 0

x3 = a + b , 1.01( a + b) .. a + b + c

1

lb

M3 ( x3 ) = −M

1

lb⋅ ft

Force (lb)

200

V1( x1) V2( x2) V3( x3)

0

200

400

0

1

2

3

4

5

x1 x2 x3 , , ft ft ft

Distance (ft)

705

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6

Engineering Mechanics - Statics

Chapter 7

0

Moment (lb-ft)

M1( x1)

100

M2( x2) M3( x3)

200

300

0

1

2

3

4

5

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-75 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given:

F = 8 kN c = 2m

M = 20 kN⋅ m

w = 15

kN m

a = 2m

b = 1m

d = 3m

Solution:

⎛ d⎞ = ⎟ ⎝2⎠

− A( a + b + c) − M + F c − w d⎜

A =

0

x1 = 0 , 0.01a .. a

V 1 ( x1 ) = A

x2 = a , 1.01a .. a + b

V 2 ( x2 ) = A

x3 = a + b , 1.01( a + b) .. a + b + c

⎛ d2 ⎞ ⎟ ⎝2⎠

F c − M − w⎜

1

kN 1

kN

A = −14.30 kN

a+b+c M1 ( x1 ) = A x1

1

kN⋅ m

M2 ( x2 ) = ( A x2 + M)

V 3 ( x3 ) = ( A − F)

1

kN⋅ m

1

kN

M3 ( x3 ) = ⎡⎣A x3 + M − F ( x3 − a − b)⎤⎦

1

kN⋅ m

706

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6

Engineering Mechanics - Statics

Chapter 7

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x4 ) = w( a + b + c + d − x4 )

M4 ( x4 ) = −w

1

kN

(a + b + c + d − x4)2

1

2

kN⋅ m

60

Force (kN)

V1( x1)

40

V2( x2) 20 V3( x3) 0

V4( x4)

20 40

0

1

2

3

4

5

6

7

8

5

6

7

8

x1 , x2 , x3 , x4

Distance (m)

Moment (kN-m)

0

M1( x1) 20 M2( x2) M3( x3) 40 M4( x4) 60

80

0

1

2

3

4

x1 , x2 , x3 , x4

Distance (m)

707

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-76 Draw the shear and moment diagrams for the shaft. The support at A is a thrust bearing and at B it is a journal bearing. Units Used: 3

kN = 10 N Given:

w = 2

kN m

F = 4 kN

⎛ a⎞ = ⎟ ⎝2⎠

B ( a + b ) − F a − w a⎜

Solution:

a = 0.8 m

x2 = a , 1.01a .. a + b

B =

0

a+b

A = wa + F − B

V 1 ( x1 ) = ( A − w x1 ) V 2 ( x2 ) = −B

⎛ a2 ⎞ ⎟ ⎝2⎠

F a + w⎜

A + B − wa − F = 0 x1 = 0 , 0.01a .. a

b = 0.2 m

M1 ( x1 )

1

kN

B = 3.84 kN A = 1.76 kN

⎡ ⎛ x1 2 ⎞⎤ 1 ⎢ ⎟⎥ = A x1 − w⎜ ⎣ ⎝ 2 ⎠⎦ kN⋅ m

M2 ( x2 ) = B ( a + b − x2 )

1

kN

1

kN⋅ m

Force (kN)

5

V1( x1) V2( x2)

0

5

0

0.2

0.4

0.6

0.8

x1 , x2

Distance (m)

Moment (kN-m)

1

M1( x1) 0.5 M2( x2)

0

0.5

0

0.2

0.4

0.6

0.8

x1 , x2

Distance (m)

708 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-77 Draw the shear and moment diagrams for the beam. Given:

w = 20

lb ft

MB = 160 lb⋅ ft

a = 20 ft

b = 20 ft

Solution:

⎛ a⎞ −w a⎜ ⎟ − MB + Cy( a + b) = ⎝ 2⎠

0

Ay − w a + Cy = 0

⎛ a2 ⎞ 1 ⎞ Cy = ⎜ w + MB⎟ ⎛⎜ ⎟ ⎝ 2 ⎠⎝ a + b ⎠

Cy = 104.00 lb

Ay = w a − Cy

Ay = 296.00 lb

x1 = 0 , 0.01a .. a V 1 ( x) = ( A y − w x)

2 ⎛ x ⎞ 1 ⎜ M1 ( x) = Ay x − w ⎟ 2 ⎠ lb⋅ ft ⎝

1

lb

x2 = a , 1.01a .. a + b V 2 ( x) = −Cy

1

M2 ( x) = Cy( a + b − x)

lb

1

lb⋅ ft

Force (lb)

400

V1( x1) 200 V2( x2)

0

200

0

5

10

15

20

25

30

35

40

x1 x2 , ft ft

Distance (ft)

709

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (lb-ft)

3000

M1( x1)2000 M2( x2) 1000

0

0

10

20

30

40

x1 x2 , ft ft

Distance (ft)

Problem 7-78 The beam will fail when the maximum moment is Mmax or the maximum shear is Vmax. Determine the largest distributed load w the beam will support. 3

kip = 10 lb

Units used: Given:

Mmax = 30 kip⋅ ft

Solution:

−A b + w

kip ft

w = 1

Set

V max = 8 kip

b ⎞ + b⎟ + w b = ⎜ 2⎝3 2 ⎠ a 2

b = 6 ft

and then scale the answer at the end

a⎛a

A + B − wb − w

a = 6 ft

w A =

0

b ⎞ + b⎟ + w ⎜ 2⎝3 2 ⎠ b

⎛ ⎝

B = w⎜ b +

=0

2

a⎛a

a⎞

⎟−A

2⎠

A = 7.00 kip

B = 2.00 kip

Shear limit - check critical points to the left and right of A and at B

⎛ ⎝

V big = max ⎜ B , w

wshear =

⎛ Vmax ⎞ ⎜ ⎟w ⎝ Vbig ⎠

a 2

, w

a 2

−A

⎞ ⎟ ⎠

wshear = 2.00

V big = 4.00 kip

kip ft

710

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment limit - check critical points at A and betwen A and B B ⎛ a⎞⎛ a ⎞ MA = −w⎜ ⎟ ⎜ ⎟ x = w ⎝ 2 ⎠⎝ 3 ⎠ Mbig = max ( MA , MAB wmoment =

MAB

)

⎛ x2 ⎞ = B x − w⎜ ⎟ ⎝2⎠

⎛ MA ⎞ ⎛ −6.00 ⎞ ⎜ ⎟=⎜ ⎟ kip⋅ ft ⎝ MAB ⎠ ⎝ 2.00 ⎠

Mbig = 6.00 kip⋅ ft

⎛ Mmax ⎞ ⎜ ⎟w ⎝ Mbig ⎠

wmoment = 5.00

kip ft

wans = min ( wshear , wmoment)

Choose the critical case

wans = 2.00

kip ft

Problem 7-79 The beam consists of two segments pin connected at B. Draw the shear and moment diagrams for the beam.

Given: F = 700 lb

w = 150

lb ft

Solution:

⎛ c ⎞ − M + Cc = ⎟ ⎝2⎠

−w c⎜

B + C − wc = 0 x1 = 0 , 0.01a .. a x2 = a , 1.01a .. a + b

M = 800 lb ft

⎛ c2 ⎞ ⎟+M ⎝2⎠

a = 8 ft

b = 4 ft

c = 6 ft

w⎜ 0

C =

C = 583.33 lb

c

B = wc − C V 1 ( x1 ) = ( B + F ) V 2 ( x2 ) = B

x3 = a + b , 1.01( a + b) .. a + b + c M3 ( x3 )

B = 316.67 lb M1 ( x1 ) = ⎡⎣−F( a − x1 ) − B ( a + b − x1 )⎤⎦

1

lb

M2 ( x2 ) = −B ( a + b − x2 )

1

lb

V 3 ( x3 ) = ⎡⎣−C + w( a + b + c − x3 )⎤⎦

1

lb⋅ ft 1

lb

2 ⎡ ⎤ 1 a + b + c − x3 ) ( = ⎢C( a + b + c − x3 ) − w − M⎥ 2 ⎣ ⎦ lb⋅ ft

711

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

lb⋅ ft

Engineering Mechanics - Statics

Chapter 7

Force (lb)

1500

V1( x1)

1000

V2( x2)

500

V3( x3)

0 500 1000

0

2

4

6

8

10

12

14

16

18

20

x1 x2 x3 , , ft ft ft

Distance (ft)

Moment (lb-ft)

5000

M1( x1)

0

M2( x2) M3( x3)

5000

1 .10

4

0

2

4

6

8

10

12

14

16

18

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-80 The beam consists of three segments pin connected at B and E. Draw the shear and moment diagrams for the beam. 3

Units Used:

kN = 10 N

Given:

MA = 8 kN⋅ m

F = 15 kN

w = 3

c = 2m

d = 2m

e = 2m

kN m

a = 3m

b = 2m

f = 4m

712

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

20

Engineering Mechanics - Statics

Guesses

Chapter 7

Ay = 1 N

By = 1 N

Cy = 1 N

Dy = 1 N

Ey = 1 N

Fy = 1 N

Given Ay + Cy + Dy + Fy − F − w f = 0

F b − MA − Ay( a + b) = 0

−w f⎛⎜

M A + F a + B y ( a + b) = 0

f⎞

⎟ + Fy f = ⎝2⎠

0

B y + Cy + Dy + E y = 0

−B y c + Dy d + E y( d + e) = 0

⎛⎜ Ay ⎞⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Cy ⎟ = Find ( A , B , C , D , E , F ) y y y y y y ⎜ Dy ⎟ ⎜ ⎟ ⎜ Ey ⎟ ⎜ Fy ⎟ ⎝ ⎠

⎛⎜ Ay ⎞⎟ ⎛ 4.40 ⎞ ⎜ By ⎟ ⎜ −10.60 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ Cy ⎟ = ⎜ 15.20 ⎟ kN ⎜ Dy ⎟ ⎜ 1.40 ⎟ ⎜ ⎟ ⎜ −6.00 ⎟ ⎟ ⎜ Ey ⎟ ⎜ ⎜ Fy ⎟ ⎝ 6.00 ⎠ ⎝ ⎠

x1 = 0 , 0.01a .. a M1 ( x) = ( Ay x + MA)

1

V 1 ( x) = Ay

kN x2 = a , 1.01a .. a + b V 2 ( x) = ( A y − F)

1

kN

1

kN⋅ m

M2 ( x) = ⎡⎣Ay x + MA − F ( x − a)⎤⎦

1

kN⋅ m

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = B y

1

M3 ( x) = By( x − a − b)

kN

1

kN⋅ m

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x) = ( By + Cy)

1

kN

M4 ( x) = ⎡⎣B y( x − a − b) + Cy( x − a − b − c)⎤⎦

1

kN⋅ m

x5 = a + b + c + d , 1.01( a + b + c + d) .. a + b + c + d + e V 5 ( x) = −E y

1

kN

M5 ( x) = Ey( a + b + c + d + e − x)

1

kN⋅ m

x6 = a + b + c + d + e , 1.01( a + b + c + d + e) .. a + b + c + d + e + f 713

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

V 6 ( x) = ⎡⎣−Fy + w( a + b + c + d + e + f − x)⎤⎦

1

kN

2 ⎡ ( a + b + c + d + e + f − x) ⎤ 1 ⎢ ⎥ M6 ( x) = F y( a + b + c + d + e + f − x) − w 2 ⎣ ⎦ kN⋅ m

10

V1( x1) 5

Force (kN)

V2( x2) V3( x3) 0 V4( x4) V5( x5) 5 V6( x6) 10

15

0

2

4

6

8

10

12

14

x1 , x2 , x3 , x4 , x5 , x6

Distance (m)

714

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

30

M1( x1) 20

Moment (kN-m)

M2( x2) M3( x3)

10

M4( x4) 0 M5( x5) M6( x6)

10

20

30

0

2

4

6

8

10

12

14

x1 , x2 , x3 , x4 , x5 , x6

Distance (m)

Problem 7-81 Draw the shear and moment diagrams for the beam.

Solutions: Support Reactions: ΣMx = 0;

⎛ L ⎞ − w0 L ⎛ 4 L ⎞ = ⎟ ⎜ ⎟ 2 ⎝ 3 ⎠ ⎝2⎠

B y L − w0 L⎜

0

By =

7 w0 L 6

715

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑

Σ F y = 0;

Ay +

Chapter 7

⎛ 7w0 L ⎞ ⎛ w0 L ⎞ ⎜ ⎟ − w0 L − ⎜ ⎟= ⎝ 6 ⎠ ⎝ 2 ⎠

0

Ay =

w0 L 3

716

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-82 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb Given: F = 2000 lb

a = 9 ft

lb ft

b = 9 ft

w = 500 Solution:

⎛ 2b ⎞ ⎛ a⎞ w b⎜ ⎟ + F b + w a⎜ b + ⎟ − A y ( a + b ) = 2 ⎝3⎠ ⎝ 2⎠ 1

Ay + By − F − w a − Ay = 5.13 kip

1 2

0

wb = 0

⎛ w b2 ⎞ ⎜ ⎟ + F b + w a⎛⎜b + 3 ⎠ ⎝ ⎝ Ay =

a⎞

a+b

By = w a +

1 2

⎟

2⎠

w b − Ay + F

B y = 3.63 kip

x1 = 0 , 0.01a .. a

717

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

V 1 ( x) = ( A y − w x)

Chapter 7

x M1 ( x) = ⎡⎢Ay x − w x⎛⎜ ⎞⎟⎥⎤ 2

1

⎣

kip

1

⎝ ⎠⎦ kip⋅ ft

x2 = a , 1.01a .. a + b V 2 ( x) = ⎡⎢−By +

⎣

1 2

w⎛⎜

⎝

a + b − x⎞ b

M2 ( x) = ⎡⎢B y( a + b − x) −

⎣

⎤ 1 ⎟ ( a + b − x)⎥ ⎠ ⎦ kip

1 2

w⎛⎜

⎝

a + b − x⎞ b

⎛ a + b − x ⎞⎤ 1 ⎟ ( a + b − x) ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kip⋅ ft ⎠

Force (kip)

10

V1( x1) 5 V2( x2) 0

5

0

5

10

15

x1 x2 , ft ft

Distance (ft)

Moment (kip-ft)

30

M1( x1)

20

M2( x2)

10 0 10

0

5

10

15

x1 x2 , ft ft

Distance (ft)

718

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-83 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: w = 3

kN m

a = 3m

b = 3m

Solution:

wb

− Ay( a + b) + w

Ay + By −

2

⎛ w b ⎞ ⎛ 2b ⎞ ⎜ ⎝

⎟⎜ ⎠⎝

2

3

⎛ w a ⎞⎛

⎟+⎜ ⎠ ⎝

2

a⎞

⎟ ⎜b + 3 ⎟ = ⎠⎝ ⎠

0

( a + b) = 0

+

3

Ay = By =

2

⎛ w a ⎞ ⎛b + a ⎞ ⎜ 2 ⎟⎜ 3 ⎟ ⎝ ⎠⎝ ⎠ a+b

w 2

( a + b) − A y

x1 = 0 , 0.01a .. a

⎡ ⎣

V 1 ( x) = ⎢A y −

1 2

⎛ x ⎞ x⎤ 1 ⎟⎥ ⎝ a ⎠ ⎦ kN

⎡ ⎣

w⎜

M1 ( x) = ⎢Ay x −

1 2

⎛ x ⎞ x⎛ x ⎞⎤ 1 ⎟ ⎜ ⎟⎥ ⎝ a ⎠ ⎝ 3 ⎠⎦ kN⋅ m

w⎜

x2 = a , 1.01a .. a + b

⎡ ⎣

V 2 ( x) = ⎢−By +

1 2

⎛ a + b − x ⎞ ( a + b − x)⎤ 1 ⎟ ⎥ ⎝ b ⎠ ⎦ kN

w⎜

⎡ ⎣

M2 ( x) = ⎢B y( a + b − x) −

1 2

⎛ a + b − x ⎞ ( a + b − x) ⎛ a + b − x ⎞⎤ 1 ⎟ ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kN⋅ m ⎝ b ⎠

w⎜

Force (kN)

5

V1( x1) V2( x2)

0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

719

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (kN-m)

10

M1( x1) 5 M2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Problem 7-84 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb Given: w = 100 Guesses: Given

lb ft

M0 = 9 kip⋅ ft

Ay = 1 lb

1 2

⎛ Ay ⎞ ⎜ ⎟ = Find ( Ay , By) ⎝ By ⎠

b = 6 ft

c = 4 ft

B y = 1 lb

−M0 + By( a + b) − Ay + By −

a = 6 ft

1 2

⎛ 2 a ⎞ − 1 w b⎛ a + ⎟ ⎜ ⎝3⎠ 2 ⎝

w a⎜

b⎞

⎟=

0

M1 ( x) = ⎢Ay x −

1

3⎠

w( a + b) = 0

⎛ Ay ⎞ ⎛ −0.45 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ By ⎠ ⎝ 1.05 ⎠

x1 = 0 , 0.01a .. a

⎡ ⎣

V 1 ( x) = ⎢A y −

1 2

⎛ x ⎞ x⎤ 1 ⎟⎥ ⎝ a ⎠ ⎦ lb

w⎜

⎡ ⎣

2

⎛ x ⎞ x⎛ x ⎞⎤ 1 ⎟ ⎜ ⎟⎥ ⎝ a ⎠ ⎝ 3 ⎠⎦ kip⋅ ft

w⎜

x2 = a , 1.01a .. a + b

720

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

V 2 ( x) = ⎡⎢−By + w⎛⎜ 2 ⎝ ⎣ 1

Chapter 7

a + b − x⎞ b

⎤1 ⎟ ( a + b − x)⎥ ⎠ ⎦ lb

M2 ( x) = ⎡⎢B y( a + b − x) − M0 −

⎣

1 2

w⎛⎜

⎝

a + b − x⎞

⎛ a + b − x ⎞⎤ 1 ⎟ ( a + b − x) ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kip⋅ ft ⎠

b

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = 0

M3 ( x) = −M0

1

kip⋅ ft

V1( x1) V2( x2)

500

V3( x3) 1000

1500

0

2

4

6

8

10

12

14

16

x1 x2 x3 , , ft ft ft

Distance (ft) 0

Moment (kip-ft)

Force (lb)

0

M1( x1) M2( x2) M3( x3)

5

10

0

5

10

15

x1 x2 x3 , , ft ft ft

Distance (ft)

721

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-85 Draw the shear and moment diagrams for the beam.

Solution:

722

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-86 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given:

kN m

w = 2

a = 3m

b = 3m

Solution: x1 = 0 , 0.01a .. a V 1 ( x) =

⎡ 1 w b + 1 w⎛ a − x ⎞ ( a − x)⎤ 1 ⎢2 ⎜ ⎟ ⎥ 2 ⎝ a ⎠ ⎣ ⎦ kN

M1 ( x) =

⎡−1 w b⎛ 2 b + a − x⎞ − 1 w⎛ a − x ⎞ ( a − x) ⎛ a − x ⎞⎤ 1 ⎢2 ⎜3 ⎟ 2 ⎜ ⎟ ⎜ 3 ⎟⎥ ⎝ ⎠ ⎝ ⎠⎦ kN⋅ m ⎣ ⎝ a ⎠

x2 = a , 1.01a .. a + b V 2 ( x) =

⎡ 1 w b − 1 w⎛ x − a ⎞ ( x − a)⎤ 1 ⎢2 ⎜ ⎟ ⎥ 2 ⎝ a ⎠ ⎣ ⎦ kN

M2 ( x) =

⎡−1 w b⎛a + 2b − x⎞ − 1 w⎛ x − a ⎞ ( x − a) ⎛ x − a ⎞⎤ 1 ⎢2 ⎜ ⎟ 2 ⎜ ⎟ ⎜ 3 ⎟⎥ 3 ⎝ ⎠ ⎝ ⎠⎦ kN⋅ m ⎣ ⎝ b ⎠

Force (kN)

10

V1( x1) 5 V2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

723

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (kN-m)

0

M1( x1) M2( x2)

10

20

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Problem 7-87 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb

Given: w = 5

kip ft

M1 = 15 kip⋅ ft

M2 = 15 kip⋅ ft

a = 6 ft

b = 10 ft

c = 6 ft

Solution:

⎛ a ⎞ ⎛b + ⎟⎜ ⎝ 2 ⎠⎝

M1 − A b − M2 + w⎜

a⎞

⎛ b⎞ ⎛ c ⎞⎛ c ⎞ + w b⎜ ⎟ − w⎜ ⎟ ⎜ ⎟ = ⎟ 3⎠ ⎝2⎠ ⎝ 2 ⎠⎝ 3 ⎠

0

⎛ a + c⎞ = ⎟ ⎝ 2 ⎠

A + B − w b − w⎜

2 2 ⎛ a ⎞ ⎛b + a ⎞ + w⎛⎜ b ⎟⎞ − w⎛⎜ c ⎞⎟ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 3 ⎠ ⎝2⎠ ⎝6⎠

M1 − M2 + w⎜ A =

b

⎛ ⎝

B = w⎜ b +

a + c⎞ 2

x1 = 0 , 0.01a .. a

⎟−A ⎠

⎛ A ⎞ ⎛ 40.00 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ B ⎠ ⎝ 40.00 ⎠ ⎛ ⎝

V 1 ( x) = ⎜ −w

M1p ( x) =

x⎞ x 1 ⎟ a ⎠ 2 kip

⎡⎛ −w x ⎞ x x − M ⎤ 1 ⎢⎜ ⎟ 1⎥ ⎣⎝ a ⎠ 2 3 ⎦ kip⋅ ft 724

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

0

Engineering Mechanics - Statics

Chapter 7

x2 = a , 1.01a .. a + b

⎡ ⎣

V 2 ( x) = ⎢A − w

⎡ ⎣

a 2

⎤ 1 ⎦ kip

− w( x − a)⎥

M2p ( x) = ⎢−M1 − w

a⎛

⎜x − 2⎝

2a ⎞ 3

⎛ x − a ⎞⎤ 1 ⎟ + A( x − a) − w( x − a) ⎜ 2 ⎟⎥ ⎠ ⎝ ⎠⎦ kip⋅ ft

x3 = a + b , 1.01( a + b) .. a + b + c

⎛ a + b + c − x⎞⎛ a + b + c − x⎞ 1 ⎟⎜ ⎟ 2 c ⎠ kip ⎝ ⎠⎝

V 3 ( x) = w⎜

⎡ ⎛ a + b + c − x⎞⎛ a + b + c − x⎞⎛ a + b + c − x⎞ − M ⎤ 1 ⎟⎜ ⎟⎜ ⎟ 2⎥ 2 3 c ⎠⎝ ⎠ ⎣ ⎝ ⎠⎝ ⎦ kip⋅ ft

M3p ( x) = ⎢−w⎜

Force (lb)

40

V1( x1) V2( x2) V3( x3)

20

0

20

40

0

5

10

15

20

x1 x2 x3 , , ft ft ft

Distance (ft)

725

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

20

Moment (kip-ft)

M1p( x1) M2p( x2) M3p( x3)

0

20

40

60

0

5

10

15

20

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-88 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb

Given: w1 = 2

kip ft

w2 = 1

kip ft

a = 15 ft

Solution: x = 0 , 0.01a .. a

⎡ ⎣

⎛ ⎝

V ( x) = ⎢w2 x − ⎜ w1

x ⎞ x⎤ 1 ⎟ ⎥ a ⎠ 2⎦ kip

⎡ ⎣

M ( x) = ⎢w2 x

x 2

⎛ ⎝

− ⎜ w1

x ⎞ x x⎤ 1 ⎟ ⎥ a ⎠ 2 3⎦ kip⋅ ft

726

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (kip)

4

V( x ) 2

0

0

2

4

6

8

10

12

14

x ft

Distance (ft) Moment (kip-ft)

40

M( x) 20

0

0

2

4

6

8

10

12

14

x ft

Distance (ft)

Problem 7-89 Determine the force P needed to hold the cable in the position shown, i.e., so segment BC remains horizontal. Also, compute the sag yB and the maximum tension in the cable. Units Used: 3

kN = 10 N

727

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given: a = 4m

F 1 = 4 kN

b = 6m

F 2 = 6 kN

c = 3m d = 2m e = 3m Solution: Initial guesses: yB = 1 m P = 1 kN

TAB = 1 kN TBC = 1 kN TCD = 1 kN TDE = 1 kN

Given

⎛ −a ⎞ T + T = AB BC ⎜ 2 2⎟ + y a B ⎠ ⎝ yB ⎛⎜ ⎞⎟ TAB − F1 = ⎜ a2 + yB2 ⎟ ⎝ ⎠ −TBC +

0

0

c ⎡ ⎤T = CD ⎢ 2 2⎥ + y − e c (B )⎦ ⎣

yB − e ⎡⎢ ⎥⎤ TCD − P = ⎢ c2 + ( yB − e) 2⎥ ⎣ ⎦

0

0

−c ⎡ ⎤T + ⎛ d ⎞T = CD ⎜ DE ⎢ 2 ⎥ 2 2 2⎟ + y − e + e c d (B )⎦ ⎝ ⎠ ⎣

0

⎡⎢ −( yB − e) ⎥⎤ e ⎞T − F = TCD + ⎛ ⎜ ⎟ DE 2 2 2 2 2 ⎢ c + ( yB − e) ⎥ + d e ⎝ ⎠ ⎣ ⎦

0

⎛ yB ⎞ ⎜ ⎟ ⎜ P ⎟ ⎜ TAB ⎟ ⎜ ⎟ = Find ( yB , P , TAB , TBC , TCD , TDE) TBC ⎜ ⎟ ⎜ TCD ⎟ ⎜ ⎟ ⎝ TDE ⎠ 728

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Tmax = max ( TAB , TBC , TCD , TDE)

yB = 3.53 m P = 800.00 N Tmax = 8.17 kN

Problem 7-90 Cable ABCD supports the lamp of mass M1 and the lamp of mass M2. Determine the maximum tension in the cable and the sag of point B. Given: M1 = 10 kg M2 = 15 kg a = 1m b = 3m c = 0.5 m d = 2m Solution: Guesses

Given

yB = 1 m

TAB = 1 N

TBC = 1 N

b ⎛ −a ⎞ T + ⎡ ⎤T = AB BC ⎜ 2 ⎢ 2 2⎟ 2⎥ + y + y − d a b ( ) B B ⎝ ⎠ ⎣ ⎦

TCD = 1 N 0

yB yB − d ⎛⎜ ⎡ ⎞⎟ ⎥⎤ TBC − M1 g = TAB + ⎢ ⎜ a2 + yB2 ⎟ ⎢ b2 + ( yB − d) 2⎥ ⎝ ⎠ ⎣ ⎦ −b ⎡ ⎤T + ⎛ c ⎞T = BC ⎜ CD ⎢ 2 2⎥ 2 2⎟ + y − d + d b c ( ) ⎝ ⎠ B ⎣ ⎦

0

0

⎡⎢ −( yB − d) ⎥⎤ d ⎞T − M g = TBC + ⎛ CD 2 ⎜ ⎟ 2 2 ⎢ b2 + ( yB − d) 2⎥ + d c ⎝ ⎠ ⎣ ⎦

0

729

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

⎛ yB ⎞ ⎜ ⎟ ⎜ TAB ⎟ = Find ( y , T , T , T ) B AB BC CD ⎜ TBC ⎟ ⎜ ⎟ ⎝ TCD ⎠ Tmax = max ( TAB , TBC , TCD)

⎛ TAB ⎞ ⎛ 100.163 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 38.524 ⎟ N ⎜ T ⎟ ⎜⎝ 157.243 ⎟⎠ ⎝ CD ⎠ Tmax = 157.2 N

yB = 2.43 m

Problem 7-91 The cable supports the three loads shown. Determine the sags yB and yD of points B and D. Given: a = 4 ft

e = 12 ft

b = 12 ft

f = 14 ft

c = 20 ft

P 1 = 400 lb

d = 15 ft

P 2 = 250 lb

Solution: Guesses

Given

yB = 1 ft

yD = 1 ft

TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

TDE = 1 lb

c ⎛ −b ⎞ T + ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + y + f − y b c ( ) B ⎠ B ⎦ ⎝ ⎣

0

yB f − yB ⎛⎜ ⎡ ⎞⎟ ⎤⎥ TAB − ⎢ TBC − P 2 = ⎜ b2 + yB2 ⎟ ⎢ c2 + ( f − yB) 2⎥ ⎝ ⎠ ⎣ ⎦

0

−c d ⎡ ⎤T + ⎡ ⎤T = BC ⎢ CD ⎢ 2 ⎥ 2 2 2⎥ + f − y + f − y c d ( ) ( ) B D ⎣ ⎦ ⎣ ⎦

0

730

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

f − yB f − yD ⎡⎢ ⎥⎤ TBC + ⎡⎢ ⎥⎤ TCD − P1 = ⎢ c2 + ( f − yB) 2⎥ ⎢ d2 + ( f − yD) 2⎥ ⎣ ⎦ ⎣ ⎦ −d e ⎡ ⎤T + ⎡ ⎤T = CD ⎢ DE ⎢ 2 ⎥ ⎥ 2 2 2 + f − y + a + y d e ( D) ⎦ ( D) ⎦ ⎣ ⎣

0

a + yD ⎡⎢ −( f − yD) ⎥⎤ ⎡ ⎥⎤ TDE − P2 = TCD + ⎢ ⎢ d2 + ( f − yD) 2⎥ ⎢ e2 + ( a + yD) 2⎥ ⎣ ⎦ ⎣ ⎦

⎛⎜ TAB ⎞⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , T , y , y ) AB BC CD DE B D ⎜ TDE ⎟ ⎜ ⎟ ⎜ yB ⎟ ⎜ yD ⎟ ⎝ ⎠

0

0

⎛ TAB ⎞ ⎛ 675.89 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 566.90 ⎟ lb ⎜ TCD ⎟ ⎜ 603.86 ⎟ ⎜ ⎟ ⎜ 744.44 ⎟ ⎠ ⎝ TDE ⎠ ⎝ ⎛ yB ⎞ ⎛ 8.67 ⎞ ⎜ ⎟ =⎜ ⎟ ft ⎝ yD ⎠ ⎝ 7.04 ⎠

Problem 7-92 The cable supports the three loads shown. Determine the magnitude of P 1 and find the sag yD for the given data. Given: P 2 = 300 lb

c = 20 ft

yB = 8 ft

d = 15 ft

a = 4 ft

e = 12 ft

b = 12 ft

f = 14 ft

Solution: Guesses

P 1 = 1 lb

TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

731

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

TDE = 1 lb

Chapter 7

yD = 1 ft

Given c ⎛ −b ⎞ T + ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + y + f − y b c ( B) ⎦ B ⎠ ⎝ ⎣

0

yB f − yB ⎛⎜ ⎡ ⎞⎟ ⎤⎥ TAB − ⎢ TBC − P 2 = ⎜ b2 + yB2 ⎟ ⎢ c2 + ( f − yB) 2⎥ ⎝ ⎠ ⎣ ⎦

0

−c d ⎡ ⎤T + ⎡ ⎤T = BC CD ⎢ 2 ⎢ 2 2⎥ 2⎥ + f − y + f − y c d ( ) ( ) B D ⎣ ⎦ ⎣ ⎦

0

f − yB f − yD ⎡⎢ ⎥⎤ TBC + ⎡⎢ ⎥⎤ TCD − P1 = ⎢ c2 + ( f − yB) 2⎥ ⎢ d2 + ( f − yD) 2⎥ ⎣ ⎦ ⎣ ⎦ −d e ⎡ ⎤T + ⎡ ⎤T = CD DE ⎢ 2 ⎢ 2 2⎥ 2⎥ + f − y + a + y d e ( ) ( ) D D ⎣ ⎦ ⎣ ⎦

0

a + yD ⎡⎢ −( f − yD) ⎥⎤ ⎡ ⎥⎤ TDE − P2 = TCD + ⎢ ⎢ d2 + ( f − yD) 2⎥ ⎢ e2 + ( a + yD) 2⎥ ⎣ ⎦ ⎣ ⎦

⎛⎜ TAB ⎞⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , T , P , y ) AB BC CD DE 1 D ⎜ TDE ⎟ ⎜ ⎟ ⎜ P1 ⎟ ⎜ yD ⎟ ⎝ ⎠

0

0

⎛ TAB ⎞ ⎛ 983.33 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 854.21 ⎟ lb ⎜ TCD ⎟ ⎜ 916.11 ⎟ ⎜ ⎟ ⎜ 1084.68 ⎟ ⎠ ⎝ TDE ⎠ ⎝ P 1 = 658 lb yD = 6.44 ft

Problem 7-93 The cable supports the loading shown. Determine the distance xB the force at point B acts from A.

732

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Engineering Mechanics - Statics

Chapter 7

Given: P = 40 lb

c = 2 ft

F = 30 lb

d = 3 ft

a = 5 ft

e = 3

b = 8 ft

f = 4

Solution: The initial guesses: TAB = 10 lb

TCD = 30 lb

TBC = 20 lb

xB = 5 ft

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + P = ⎜ xB2 + a2 ⎟ ⎢ ( xB − d) 2 + b2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB BC ⎜ ⎢ 2 2⎟ 2 2⎥ + a − d + b x x ( ) B B ⎝ ⎠ ⎣ ⎦

0

0

xB − d ⎡⎢ d ⎤⎥ f ⎞T + ⎛ ⎞F = TBC − ⎛ CD ⎜ ⎜ ⎟ ⎟ 2 2 2 2 ⎢ ( xB − d) 2 + b2⎥ ⎝ c +d ⎠ ⎝ e +f ⎠ ⎣ ⎦

0

b ⎡ ⎤T − ⎛ c ⎞T − ⎛ e ⎞F = BC ⎜ CD ⎜ ⎢ 2 2⎥ 2 2⎟ 2 2⎟ − d + b + d + f x c e ( ) ⎝ ⎠ ⎝ ⎠ B ⎣ ⎦

0

⎛ TAB ⎞ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , x ) AB CD BC B ⎜ TBC ⎟ ⎜ ⎟ ⎝ xB ⎠

⎛ TAB ⎞ ⎛ 50.90 ⎞ ⎜ ⎟ ⎜ TCD ⎟ = ⎜ 36.70 ⎟ lb ⎜ T ⎟ ⎜⎝ 38.91 ⎟⎠ ⎝ BC ⎠

xB = 4.36 ft

Problem 7-94 The cable supports the loading shown. Determine the magnitude of the horizontal force P.

733

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Engineering Mechanics - Statics

Chapter 7

Given: F = 30 lb

c = 2 ft

xB = 6 ft

d = 3 ft

a = 5 ft

e = 3

b = 8 ft

f = 4

Solution: The initial guesses: TAB = 10 lb

TCD = 30 lb

TBC = 20 lb

P = 10 lb

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + P = ⎜ xB2 + a2 ⎟ ⎢ ( xB − d) 2 + b2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB BC ⎜ 2 ⎢ 2 2⎟ 2⎥ + x + x − d a b ( ) B B ⎝ ⎠ ⎣ ⎦

0

0

xB − d ⎤⎥ f ⎛ −d ⎞ T + ⎡⎢ ⎞F = TBC + ⎛ CD ⎜ 2 2⎟ ⎜ ⎟ 2 2 ⎢ b2 + ( xB − d) 2⎥ ⎝ c +d ⎠ ⎝ e +f ⎠ ⎣ ⎦

0

b ⎛ −c ⎞ T + ⎡ ⎤T − ⎛ e ⎞F = BC ⎜ ⎜ 2 2 ⎟ CD ⎢ 2 2⎥ 2 2⎟ + x − d + f b e ( ) ⎝ c +d ⎠ ⎝ ⎠ B ⎣ ⎦

0

⎛ TAB ⎞ ⎜ ⎟ T BC ⎜ ⎟ = Find ( T , T , T , P) AB BC CD ⎜ TCD ⎟ ⎜ ⎟ ⎝ P ⎠

⎛ TAB ⎞ ⎛ 70.81 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 48.42 ⎟ lb ⎜ T ⎟ ⎜⎝ 49.28 ⎟⎠ ⎝ CD ⎠

P = 71.40 lb

734

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Engineering Mechanics - Statics

Chapter 7

Problem 7-95 Determine the forces P1 and P2 needed to hold the cable in the position shown, i.e., so segment CD remains horizontal. Also, compute the maximum tension in the cable. Given:

3

kN = 10 N

F = 5 kN

d = 4m

a = 1.5 m

e = 5m

b = 1m

f = 4m

c = 2m Solution: Guesses F AB = 1 kN

F BC = 1 kN

F CD = 1 kN

F DE = 1 kN

P 1 = 1 kN

P 2 = 1 kN

Given

⎛ −c ⎞ F + ⎛ d ⎞ F = ⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ ⎝ b +d ⎠

0

735

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Engineering Mechanics - Statics

Chapter 7

⎛ a ⎞F − ⎛ b ⎞F − F = ⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ ⎝ b +d ⎠ ⎛ −d ⎞ F + F = ⎜ 2 2 ⎟ BC CD ⎝ b +d ⎠ ⎛ b ⎞F − P = ⎜ 2 2 ⎟ BC 1 ⎝ b +d ⎠ −F CD +

0

0

0

f ⎡ ⎤F = DE ⎢ 2 2⎥ ⎣ f + ( a + b) ⎦

a+b ⎡ ⎤F − P = DE 2 ⎢ 2 2⎥ + ( a + b ) f ⎣ ⎦

0

0

⎛⎜ FAB ⎟⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( F , F , F , F , P , P ) AB BC CD DE 1 2 ⎜ FDE ⎟ ⎜ ⎟ ⎜ P1 ⎟ ⎜ P2 ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎛ 12.50 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FBC ⎟ = ⎜ 10.31 ⎟ kN ⎜ FCD ⎟ ⎜ 10.00 ⎟ ⎜ ⎟ ⎜ 11.79 ⎟ ⎠ ⎝ FDE ⎠ ⎝

⎛ P1 ⎞ ⎛ 2.50 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ P2 ⎠ ⎝ 6.25 ⎠

Tmax = max ( FAB , F BC , F CD , FDE) F max = max ( F AB , FBC , FCD , F DE)

Tmax = 12.50 kN F max = 12.50 kN

736

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Engineering Mechanics - Statics

Chapter 7

Problem 7-96 The cable supports the loading shown. Determine the distance xB from the wall to point B. Given: W1 = 8 lb W2 = 15 lb a = 5 ft b = 8 ft c = 2 ft d = 3 ft Solution: Guesses TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

xB = 1 ft

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + W2 = ⎜ a2 + xB2 ⎟ ⎢ b2 + ( xB − d) 2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + x + x − d a b ( ) B ⎠ B ⎝ ⎣ ⎦ xB − d ⎡⎢ ⎤⎥ ⎛ d ⎞T = TBC − ⎜ CD 2 2⎟ ⎢ b2 + ( xB − d) 2⎥ + d c ⎝ ⎠ ⎣ ⎦

0

0

b ⎡ ⎤T − ⎛ c ⎞T − W = BC ⎜ CD 1 ⎢ 2 2⎥ 2 2⎟ + x − d + d b c ( ) ⎝ ⎠ B ⎣ ⎦

⎛ TAB ⎞ ⎜ ⎟ TBC ⎜ ⎟ = Find ( T , T , T , x ) AB BC CD B ⎜ TCD ⎟ ⎜ ⎟ ⎝ xB ⎠

0

0

⎛ TAB ⎞ ⎛ 15.49 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 10.82 ⎟ lb ⎜ T ⎟ ⎜⎝ 4.09 ⎟⎠ ⎝ CD ⎠

xB = 5.65 ft

Problem 7-97 Determine the maximum uniform loading w, measured in lb/ft, that the cable can support if it is 737

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

g capable of sustaining a maximum tension Tmax before it will break.

pp

Given: Tmax = 3000 lb a = 50 ft b = 6 ft Solution: 1 ⌠ ⌠

2

y=

wx ⎮ ⎮ w dx dx = FH ⌡ ⌡ 2 FH

y=

⎛ w ⎞ x2 ⎜ 2F ⎟ ⎝ H⎠

x=

2

a

y=b

2

wa 8b

FH =

⎛ dy ⎞ = tan ( θ ) = w ⎛ a ⎞ = 4b ⎜ ⎟ ⎜ ⎟ max a FH ⎝ 2 ⎠ ⎝ dx ⎠ Tmax =

FH

cos ( θ max)

θ max = atan ⎛⎜

⎟ ⎝a⎠

2

=

wa

(

8 b cos θ max

)

4b ⎞

w =

Tmax8 b cos ( θ max) a

2

θ max = 25.64 deg

w = 51.93

lb ft

Problem 7-98 The cable is subjected to a uniform loading w. Determine the maximum and minimum tension in the cable. Units Used: 3

kip = 10 lb Given: w = 250

lb ft

a = 50 ft b = 6 ft

738

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Solution: 2

y=

⎛ a⎞ b= ⎜ ⎟ 2FH ⎝ 2 ⎠

wx

w

2FH

tan ( θ max) =

Tmax =

d dx

y ⎛⎜ x =

⎝

2

FH =

w ⎛ a⎞ = ⎟ ⎜ ⎟ 2⎠ FH ⎝ 2 ⎠

a⎞

FH

wa

2

F H = 13021 lb

8b

⎞ θ max = atan ⎛⎜ ⎟ 2 F ⎝ H⎠ wa

θ max = 25.64 deg

Tmax = 14.44 kip

cos ( θ max)

The minimum tension occurs at

θ = 0 deg

Tmin = FH

Tmin = 13.0 kip

Problem 7-99 The cable is subjected to the triangular loading. If the slope of the cable at A is zero, determine the equation of the curve y = f(x) which defines the cable shape AB, and the maximum tension developed in the cable. Units Used: kip = 103 lb Given: w = 250

lb ft

a = 20 ft b = 30 ft Solution: ⌠ ⌠

1 ⎮ ⎮ y= FH ⎮ ⎮

⌡ ⌡

y=

wx b

dx dx

⎞ ⎛ w x3 ⎜ + c1 x + c2⎟ FH ⎝ 6 b ⎠ 1

739

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Apply boundary conditions y = x = 0 and

d

y = 0,x = 0

3

C1 = C2 = 0

3

wx y= 6FH b

set

y=a

x=b

2

wb FH = 6a Tmax =

Thus

dx

F H = 1.875 kip FH

cos ( θ max)

θ max

wb a= 6FH b

⎛ w b2 ⎞ = atan ⎜ ⎟ ⎝ 2 FH b ⎠

θ max = 63.43 deg

Tmax = 4.19 kip

Problem 7-100 The cable supports a girder which has weight density γ. Determine the tension in the cable at points A, B, and C. Units used: 3

kip = 10 lb Given:

γ = 850

lb ft

a = 40 ft b = 100 ft c = 20 ft

Solution:

y=

1 ⌠ ⌠

⎮ ⎮ γ dx dx FH ⌡ ⌡ 2

y=

γx

2FH

γx d y = FH dx

x1 = 1 ft

Guesses

F H = 1 lb

740

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Engineering Mechanics - Statics

c=

Given

tan ( θ A) =

γ FH

tan ( θ C) =

TA =

γ x1

Chapter 7

2

2 FH

a=

γ

θ A = atan ⎢

⎛ γ x⎞ 1⎟ ⎝ FH ⎠

θ C = atan ⎜

x1

FH

cos ( θ A)

2FH

⎛ x1 ⎞ ⎜ ⎟ = Find ( x1 , FH) ⎝ FH ⎠

(b − x1)2

⎡ γ x − b⎤ ( 1 )⎥ ⎣FH ⎦

(x1 − b)

FH

γ

TB = F H

TC =

F H = 36.46 kip

θ A = −53.79 deg

θ C = 44.00 deg

FH

cos ( θ C)

⎛ TA ⎞ ⎛ 61.71 ⎞ ⎜ ⎟ ⎜ TB ⎟ = ⎜ 36.46 ⎟ kip ⎜ T ⎟ ⎜⎝ 50.68 ⎟⎠ ⎝ C⎠

Problem 7-101 The cable is subjected to the triangular loading. If the slope of the cable at point O is zero, determine the equation of the curve y = f(x) which defines the cable shape OB, and the maximum tension developed in the cable. Units used: kip = 103 lb Given: w = 500

lb ft

b = 8 ft

a = 15 ft

741

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Engineering Mechanics - Statics

Chapter 7

Solution:

⎛⌠ ⎜⎮ y= FH ⎜⎮ ⎝⌡

⎞

⌠ ⎮ ⎮ ⌡

1

wx dx dx⎟ a ⎟

⎠

⎤ ⎡w ⎛ x3 ⎞ ⎢ ⎜ ⎟ + C1 x + C2⎥ y= FH ⎣ a ⎝ 6 ⎠ ⎦ 1

2⎞

1 ⎛ wx d ⎜ y = FH ⎝ 2a dx

At x = 0,

d dx

⎛ C1 ⎞ ⎟+⎜ ⎟ ⎠ ⎝ FH ⎠

y = 0, C1 = 0

At x = 0, y = 0, C2 = 0 3

y=

2

wx d y = 2a FH dx

wx 6a FH

3

At x = a, y = b

b=

wa 6a FH

FH =

1 6

⎛ a2 ⎞ ⎟ ⎝b⎠

w⎜

F H = 2343.75 lb

742

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

2

wa d y = tan ( θ max) = 2a FH dx Tmax =

(θ max)

FH

⎛ w a2 ⎞ = atan ⎜ ⎟ ⎝ 2a FH ⎠

(θ max) = 57.99 deg

Tmax = 4.42 kip

cos ( θ max)

Problem 7-102 The cable is subjected to the parabolic loading w = w0(1− (2x/a)2). Determine the equation y = f(x) which defines the cable shape AB and the maximum tension in the cable. Units Used: 3

kip = 10 lb Given:

⎡

w = w0 ⎢1 −

⎣

a = 100 ft

2 ⎛ 2x ⎞ ⎥⎤ ⎜ ⎟ ⎝a⎠⎦

w0 = 150

lb ft

b = 20 ft Solution: y=

1 ⌠ ⌠

⎮ ⎮ w ( x) dx dx FH ⌡ ⌡ ⌠

y=

⎛⎜ 4 x ⎞⎟ + C1 dx ⎮ w0 x − 2⎟ FH ⎜ ⎮ ⎝ 3a ⎠ ⌡ 1 ⎮

3

⎞ ⎛ w0 x2 x4w0 ⎜ ⎟ y= − + C1 x + C2 ⎜ ⎟ 2 2 FH 3a ⎝ ⎠ 1

⎞ ⎛ 4 w0 x dy 1 ⎜ ⎟ = w0 x − + C1 ⎟ 2 dx FH ⎜ 3a ⎝ ⎠ 3

743

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

dy

x=0

At

dx

At x = 0

=0

y=0

Chapter 7

C1 = 0 C2 = 0

Thus

⎛ w0 x2 x4w0 ⎞ ⎜ ⎟ y= − ⎜ ⎟ 2 2 FH 3a ⎠ ⎝

3 ⎛ 4 w0 x ⎞ ⎜ ⎟ = w0 x − ⎜ ⎟ 2 FH dx 3a ⎠ ⎝

dy

1

x=

At

a

we have

2

⎡⎢w0 ⎛ a ⎞ 2 w0 ⎛ a ⎞ 4⎥⎤ b= ⎜ ⎟ − 2 ⎜2⎟ = FH ⎢ 2 ⎝ 2 ⎠ ⎥ 3a ⎝ ⎠ ⎦ ⎣ 1

tan ( θ max)

⎛⎜ w0 a2 ⎟⎞ ⎟ 48 ⎜ ⎝ FH ⎠ 5

FH =

5 w0 a

2

F H = 7812.50 lb

48b

3 w0 a a⎞ 1 ⎡ a ⎞ 4 w0 ⎛ a ⎞ ⎥⎤ ⎛ ⎛ ⎢ = y⎜ ⎟ = w0 ⎜ ⎟ − = ⎜ ⎟ FH ⎢ ⎝ 2 ⎠ 3 a2 ⎝ 2 ⎠ ⎥ 3 FH dx ⎝ 2 ⎠ ⎣ ⎦

d

⎛ w0 a ⎞ ⎟ ⎝ 3 FH ⎠

θ max = atan ⎜

Tmax =

1

FH

cos ( θ max)

θ max = 32.62 deg

Tmax = 9.28 kip

Problem 7-103 The cable will break when the maximum tension reaches Tmax. Determine the minimum sag h if it supports the uniform distributed load w. Given:

kN = 103 N

Tmax = 10 kN w = 600

N m

a = 25 m Solution: The equation of the cable: y=

1 ⌠ ⌠

⎮ ⎮ w dx dx FH ⌡ ⌡

y=

⎛ w x2 ⎞ ⎜ + C1 x + C2⎟ FH ⎝ 2 ⎠ 1

dy dx

=

1

FH

(w x + C1)

744

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Boundary Conditions: y = 0 at x = 0, then from Eq.[1]

0 =

d

y = 0 at x = 0, then from Eq.[2]

0 =

w ⎞ 2 y = ⎛⎜ ⎟x ⎝ 2 FH ⎠

w

dx Thus,

tan ( θ max) =

Tmax =

dy dx

2FH

cos ( θ max)

Guess

h = 1m

Given

Tmax =

FH

FH 1

FH

(C2)

C2 = 0

(C1)

C1 = 0

⎛ a⎞ h= ⎜ ⎟ 2FH ⎝ 2 ⎠ w

x

2

4 F H + ( w a)

=

wa 2

2

FH +

( w a) 4

2

=

a

wa 2

2

⎛ a⎞ FH = ⎜ ⎟ 2h ⎝ 2 ⎠ w

2

2 FH

cos ( θ max) =

wa

FH

=

1

2

2 2

+1

16h

2

a

16h

2

+1

h = Find ( h)

h = 7.09 m

Problem 7-104 A fiber optic cable is suspended over the poles so that the angle at the supports is θ. Determine the minimum tension in the cable and the sag. The cable has a mass density ρ and the supports are at the same elevation. Given:

θ = 22 deg ρ = 0.9

kg m

a = 30 m g = 9.81

m 2

s Solution:

θ max = θ 745

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

w0 = ρ g a ⎞ ⎛ ⎜ 2 ⎟ d y = tan ( θ ) = sinh ⎜ w0 ⎟ dx ⎝ FH ⎠

w0 ⎛⎜ FH =

a⎞

⎟ ⎝2⎠

F H = 336 N

asinh ( tan ( θ ) )

Tmax =

FH

Tmax = 363 N

cos ( θ )

⎛ ⎛w a ⎞ ⎞ ⎜ ⎜ 02⎟ ⎟ FH h = ⎜ cosh ⎜ ⎟ − 1⎟ w0 ⎝ ⎝ FH ⎠ ⎠

h = 2.99 m

Problem 7-105 A cable has a weight density γ and is supported at points that are a distance d apart and at the same elevation. If it has a length L, determine the sag. Given:

γ = 3

lb

d = 500 ft

ft

L = 600 ft

Solution: Guess

Given

h =

F H = 100 lb L 2

−

⎡FH ⎡ γ ⎛ d ⎞⎤⎤ ⎢ sinh ⎢ ⎜ 2 ⎟⎥⎥ = ⎣γ ⎣FH ⎝ ⎠⎦⎦

F H = Find ( FH)

0

⎛1 γ ⎞ ⎞ ⎜ cosh ⎜ 2 F d⎟ − 1⎟ γ ⎝ ⎝ H ⎠ ⎠

FH ⎛

F H = 704.3 lb

h = 146 ft

Problem 7-106 Show that the deflection curve of the cable discussed in Example 7.15 reduces to Eq. (4) in Example 7.14 when the hyperbolic cosine function is expanded in terms of a series and only the 746

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

p yp f p y first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.) Solution: 2

cosh ( x) = 1 +

x

2!

+ ..

Substituting into 2 2 ⎞ w0 x2 ⎛ w0 ⎞ ⎞ FH ⎛⎜ w0 x ⎟ y= 1+ + .. − 1 = ⎜cosh ⎜ x⎟ − 1⎟ = 2 w0 ⎝ ⎟ 2 FH ⎝ FH ⎠ ⎠ w0 ⎜⎝ 2FH ⎠

FH ⎛

Using the boundary conditions y = h at

h=

⎛ L⎞ ⎜ ⎟ 2FH ⎝ 2 ⎠ w0

2

FH =

w0 L 8h

x=

L 2

2

We get

y=

4h 2

L

2

x

Problem 7-107 A uniform cord is suspended between two points having the same elevation. Determine the sag-to-span ratio so that the maximum tension in the cord equals the cord's total weight. Solution: s=

y=

FH w0

⎛ w0 ⎞ x⎟ ⎝ FH ⎠

sinh ⎜

FH ⎛

⎛ w0 ⎞ ⎞ ⎜cosh ⎜ x⎟ − 1⎟ w0 ⎝ ⎝ FH ⎠ ⎠

At x =

L 2

⎛ w0 L ⎞ d y = tan ( θ max) = sinh ⎜ ⎟ dx max ⎝ 2FH ⎠ cos ( θ max) =

1

⎛ w0 L ⎞ ⎟ ⎝ 2 FH ⎠

cosh ⎜

747

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

⎛ w0 L ⎞ ⎟ ⎝ 2 FH ⎠

FH

Tmax =

Chapter 7

w0 2 s = F H cosh ⎜

cos ( θ max)

⎛ w0 L ⎞ ⎛ w0 L ⎞ ⎟ = FH cosh ⎜ ⎟ ⎝ 2 FH ⎠ ⎝ 2FH ⎠

⎛ w0 L ⎞ ⎟= ⎝ 2 FH ⎠

2 F H sinh ⎜

k1 = atanh ( 0.5)

when x =

L

⎛ FH ⎞ ⎟ ⎝ w0 ⎠

h=

⎛ 2 Fh ⎞ ⎟ ⎝ w0 ⎠

h = k2 ⎜

w0 L

k1 = 0.55

y=h

2

tanh ⎜

L = k1 ⎜

W0

(cosh (k1) − 1)

ratio =

2

= k1

2 FH

FH

1

k2 h = 2 k1 L

k2 = cosh ( k1 ) − 1

ratio =

k2 = 0.15

k2 2 k1

ratio = 0.14

Problem 7-108 A cable has a weight denisty γ. If it can span a distance L and has a sag h determine the length of the cable. The ends of the cable are supported from the same elevation. Given:

γ = 2

lb ft

L = 100 ft

h = 12 ft

Solution: From Eq. (5) of Example 7-15 :

⎡⎢⎛ γ L ⎞ 2⎤⎥ ⎜ ⎟ FH ⎢⎝ 2 FH ⎠ ⎥ h= ⎢ 2 ⎥ γ ⎣ ⎦

FH =

1 8

⎛ L2 ⎞ ⎟ ⎝h⎠

γ⎜

F H = 208.33 lb

From Eq. (3) of Example 7-15: l 2

=

⎛ FH ⎞ ⎡ γ ⎛ L ⎞⎤ ⎜ ⎟ sinh ⎢ ⎜ ⎟⎥ ⎝ γ ⎠ ⎣FH ⎝ 2 ⎠⎦

⎛ FH ⎞ ⎛ 1 L ⎞ ⎟ sinh ⎜ γ ⎟ ⎝ γ ⎠ ⎝ 2 FH ⎠

l = 2⎜

l = 104 ft

748

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-109 The transmission cable having a weight density γ is strung across the river as shown. Determine the required force that must be applied to the cable at its points of attachment to the towers at B and C. Units Used: 3

kip = 10 lb Given:

γ = 20

lb ft

b = 75 ft

a = 50 ft

c = 10 ft

Solution: From Example 7-15,

y=

FH ⎡ ⎛ γ ⎞ ⎤ ⎢cosh⎜ F x⎟ − 1⎥ γ ⎣ ⎝ H ⎠ ⎦

Guess

Given

dy ⎛ γx ⎞ = sinh ⎜ ⎟ dx ⎝ FH ⎠

F H = 1000 lb c=

At B:

⎛ γa ⎞ ⎞ ⎜ cosh ⎜ − F ⎟ − 1⎟ γ ⎝ ⎝ H⎠ ⎠

FH ⎛

⎛ γa ⎞ ⎟ ⎝ FH ⎠

θ B = atan ⎜ sinh ⎜

⎛ γb ⎞ ⎟ ⎝ FH ⎠

θ C = atan ⎜ sinh ⎜

tan ( θ B) = sinh ⎜ −

tan ( θ C) = sinh ⎜

TB =

FH

cos ( θ B)

TC =

FH

cos ( θ C)

F H = Find ( FH)

F H = 2.53 kip

⎛ ⎝

⎛ −γ a ⎞ ⎞ ⎟⎟ ⎝ FH ⎠⎠

θ B = −22.06 deg

⎛ ⎝

⎛ γ b ⎞⎞ ⎟⎟ ⎝ FH ⎠⎠

θ C = 32.11 deg

⎛ TB ⎞ ⎛ 2.73 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ TC ⎠ ⎝ 2.99 ⎠

Problem 7-110 Determine the maximum tension developed in the cable if it is subjected to a uniform load w.

749

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Engineering Mechanics - Statics

Chapter 7

Units Used: MN = 106 N Given: N

w = 600

m

a = 100 m b = 20 m

θ = 10 deg Solution: The Equation of the Cable:

⎛ w x2 ⎞ ⎜ y= ⎮ ⎮ w( x) dx dx = + C1 x + C2⎟ FH ⌡ ⌡ FH ⎝ 2 ⎠ 1 ⌠ ⌠

dy dx

=

1

FH

1

(w x + C1)

Initial Guesses: Given

C1 = 1 N

FH = 1 N

Boundary Conditions: 0 =

x=0

1

FH

tan ( θ ) =

C2

b=

y = b at x = a

tan ( θ max) =

1

FH FH

(w a + C1)

cos ( θ max)

1

FH

(C1)

⎛ w ⎞ a2 + ⎛ C1 ⎞ a ⎜ ⎟ ⎜ 2F ⎟ ⎝ H⎠ ⎝ FH ⎠

⎛ C1 ⎞ ⎜ ⎟ ⎜ C2 ⎟ = Find ( C1 , C2 , FH) ⎜F ⎟ ⎝ H⎠

Tmax =

C2 = 1 N⋅ m

C1 = 0.22 MN

C2 = 0.00 N⋅ m

⎛ w a + C1 ⎞ ⎟ ⎝ FH ⎠

θ max = atan ⎜

F H = 1.27 MN

θ max = 12.61 deg

Tmax = 1.30 MN

750

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-111 A chain of length L has a total mass M and is suspended between two points a distance d apart. Determine the maximum tension and the sag in the chain. Given: L = 40 m

M = 100 kg

d = 10 m

g = 9.81

m 2

s Solution: w = M

s=

g L

⎛ FH ⎞ ⎛ w ⎞ ⎜ ⎟ sinh ⎜ x⎟ ⎝ w ⎠ ⎝ FH ⎠ F H = 10 N

Guesses

Given

L

=

2

y=

θ max = atan ⎛⎜ sinh ⎛⎜

F H = 37.57 N

d dx

y = sinh ⎛⎜

w

⎝ FH

x⎞⎟

⎠

h = 10 m

⎛ FH ⎞ ⎛ w ⎜ ⎟ sinh ⎜ ⎝ w ⎠ ⎝ FH ⎝

⎛ FH ⎞ ⎛ w ⎜ ⎟ ⎜cosh ⎛⎜ x⎟⎞ − 1⎞⎟ ⎝ w ⎠⎝ ⎝ FH ⎠ ⎠

d⎞ 2⎟ ⎠

h=

FH ⎛

w cosh ⎛⎜ ⎜ w ⎝ ⎝ FH

w d ⎞⎞

Tmax =

⎟⎟ ⎝ FH 2 ⎠⎠

h = 18.53 m

d⎞

− 1⎞⎟ 2⎟ ⎠ ⎠

⎛ FH ⎞ ⎜ ⎟ = Find ( FH , h) ⎝ h ⎠

FH

cos ( θ max)

Tmax = 492 N

Problem 7-112 The cable has a mass density ρ and has length L. Determine the vertical and horizontal components of force it exerts on the top of the tower. Given:

ρ = 0.5

kg m

L = 25 m

θ = 30 deg d = 15 m g = 9.81

m 2

s

751

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Engineering Mechanics - Statics

Chapter 7

Solution: ⌠ ⎮ x=⎮ ⎮ ⎮ ⎮ ⎮ ⌡

1 1+

⎞ ⎛⌠ ⎜ ⎮ ρ g ds⎟ 2⎜ ⎟ FH ⎝⌡ ⎠

ds 2

1

Performing the integration yields:

⎞ ⎛ ρ g s + C1 ⎞ ⎜ asinh ⎜ ⎟ + C2⎟ ρg ⎝ ⎝ FH ⎠ ⎠

FH ⎛

x=

dy dx

=

1 ⌠

⎮ ρ g ds = (ρ g s + C1) FH ⌡ FH dy

At s = 0; dy dx

=

1

dx

ρg s FH

= tan( θ )

C1 = F H tan ( θ )

Hence

+ tan ( θ )

Applying boundary conditions at x = 0; s = 0 to Eq.[1] and using the result C1 = F H tan ( θ ) yields C2 = −asinh ( tan ( θ ) ). Hence Guess

FH = 1 N

Given

d=

⎛ FH ⎞ ⎡ 1 ⎜ ⎟ ⎢asinh ⎡⎢⎛⎜ ⎞⎟ ( ρ g L + FH tan ( θ ) )⎥⎤ − ( asinh ( tan ( θ ) ) )⎤⎥ ⎝ ρg ⎠⎣ ⎣⎝ FH ⎠ ⎦ ⎦

F H = Find ( FH)

At A

FA =

F H = 73.94 N

tan ( θ A) =

ρg L

FH

F Ax = F A cos ( θ A)

cos ( θ A)

FH

+ tan ( θ )

⎛ ρ g L + tan ( θ )⎞ ⎟ ⎝ FH ⎠

θ A = atan ⎜

F Ay = F A sin ( θ A)

θ A = 65.90 deg

⎛ FAx ⎞ ⎛ 73.94 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FAy ⎠ ⎝ 165.31 ⎠

752 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-113 A cable of length L is suspended between two points a distance d apart and at the same elevation. If the minimum tension in the cable is Tmin, determine the total weight of the cable and the maximum tension developed in the cable. Units Used:

kip = 103 lb

Given:

L = 50 ft

d = 15 ft

Tmin = 200 lb

Solution:

Tmin = F H

F H = Tmin

F H = 200 lb

s=

From Example 7-15:

w0 = 1

Guess

L

Given

2

=

⎛ FH ⎞ ⎛ w0 x ⎞ ⎜ ⎟ sinh ⎜ ⎟ ⎝ w0 ⎠ ⎝ FH ⎠

lb ft

⎛ FH ⎞ ⎛ w0 d ⎞ ⎜ ⎟ sinh ⎜ ⎟ ⎝ w0 ⎠ ⎝ FH 2 ⎠

w0 = Find ( w0 )

Totalweight = w0 L

Totalweight = 4.00 kip

w0 L

⎡w ⎛ L ⎞⎤ ⎢ 0⎜⎝ 2 ⎟⎠⎥ = atan ⎢ ⎥ ⎣ FH ⎦

tan ( θ max) =

FH 2

θ max

w0 = 79.93

lb ft

θ max = 84.28 deg

Then, Tmax =

FH

cos ( θ max)

Tmax = 2.01 kip

753

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-114 The chain of length L is fixed at its ends and hoisted at its midpoint B using a crane. If the chain has a weight density w, determine the minimum height h of the hook in order to lift the chain completely off the ground. What is the horizontal force at pin A or C when the chain is in this position? Hint: When h is a minimum, the slope at A and C is zero.

Given: L = 80 ft d = 60 ft w = 0.5

lb ft

Solution: F H = 10 lb

Guesses

Given

h=

FH ⎛ w

h = 1 ft

⎛w ⎝ FH

⎜cosh ⎜ ⎝

⎛ h ⎞ ⎜ ⎟ = Find ( h , FH) ⎝ FH ⎠

d⎞ 2⎟ ⎠

⎞ ⎠

L

− 1⎟

FA = FH

2

FC = FH

=

⎛ FH ⎞ ⎛ w ⎜ ⎟ sinh ⎜ ⎝ w ⎠ ⎝ FH

d⎞ 2⎟ ⎠

⎛ FA ⎞ ⎛ 11.1 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FC ⎠ ⎝ 11.1 ⎠ h = 23.5 ft

Problem 7-115 A steel tape used for measurement in surveying has a length L and a total weight W. How much horizontal tension must be applied to the tape so that the distance marked on the ground is a? In practice the calculation should also include the effects of elastic stretching and temperature changes on the tape’s length. Given: L = 100 ft W = 2 lb a = 99.90 ft

754

© 2007 R. C.

Chapter 1

Problem 1-1 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m/ms (b) μkm (c) ks/mg (d) km⋅ μN Units Used: μN = 10

−6

μkm = 10

N

−6

km

9

Gs = 10 s 3

ks = 10 s mN = 10

−3

−3

ms = 10

N s

Solution: ( a)

m 3m = 1 × 10 ms s m km =1 ms s

( b)

μkm = 1 × 10

−3

m

μkm = 1 mm ( c)

ks 9 s = 1 × 10 mg kg ks Gs =1 mg kg

( d)

−3

km⋅ μN = 1 × 10

mN

km⋅ μN = 1 mm⋅ N

1

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-2 Wood has a density d. What is its density expressed in SI units? Units Used: Mg = 1000 kg Given: d = 4.70

slug ft

3

Solution: 1slug = 14.594 kg d = 2.42

Mg m

3

Problem 1-3 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm (b) mN/μs (c) μm⋅ Mg Solution: 3

( a)

6

Mg 10 kg 10 kg Gg = = = −3 mm m m 10 m Mg Gg = mm m

( b)

mN μs mN μs

( c)

=

10

−3

10 =

N

−6

3

=

s

10 N kN = s s

kN s

(

−6

μm⋅ Mg = 10

)(

3

)

m 10 kg = 10

−3

m⋅ kg

μm⋅ Mg = mm⋅ kg

2

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-4 Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/mm, (c) mN/( kg⋅ μs). Solution: 3

( a)

6

Mg 10 kg 10 kg Gg = = = −3 ms s s 10 s Mg Gg = ms s N = mm

( b)

1N 10

−3

= 10

3N

m

m

=

kN m

N kN = mm m mN

( c)

kg⋅ μs

10

=

mN kg⋅ μs

−3

−6

10 =

N

=

kg⋅ s

kN kg⋅ s

kN kg⋅ s

Problem 1-5 Represent each of the following with SI units having an appropriate prefix: (a) S1, (b) S2, (c) S3. Units Used: kg = 1000 g

−3

ms = 10

s

3

kN = 10 N

Given: S1 = 8653 ms S2 = 8368 N S3 = 0.893 kg Solution: ( a)

S1 = 8.653 s

3

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

( b)

S2 = 8.368 kN

( c)

S3 = 893 g

Chapter 1

Problem 1-6 Represent each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, and (c) z. Units Used: 6

MN = 10 N μg = 1 × 10

−6

gm

3

kN = 10 N Given: x = 45320 kN

(

y = 568 × 10

5

) mm

z = 0.00563 mg Solution: ( a)

x = 45.3 MN

( b)

y = 56.8 km

( c)

z = 5.63 μg

Problem 1-7 Evaluate ( a⋅ b)/c to three significant figures and express the answer in SI units using an appropriate prefix. Units Used: μm = 10

−6

m

Given: a = ( 204 mm) 4

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

b = ( 0.00457 kg) c = ( 34.6 N) Solution: l =

ab

l = 26.945

c

μm⋅ kg N

Problem 1-8 If a car is traveling at speed v, determine its speed in kilometers per hour and meters per second. Given: v = 55

mi hr

Solution: v = 88.514

km hr

m

v = 24.6

s

Problem 1-9 Convert: (a) S1 to N ⋅ m , (b) S2 to kN/m3, (c) S3 to mm/s. Express the result to three significant figures. Use an appropriate prefix. Units Used: 3

kN = 10 N Given: S1 = 200g lb⋅ ft S2 = 350g

lb ft

S3 = 8

3

ft hr

5

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Solution: ( a)

S1 = 271 N⋅ m

( b)

S2 = 55.0

kN m

S3 = 0.677

( c)

3

mm s

Problem 1-10 What is the weight in newtons of an object that has a mass of: (a) m1, (b) m2, (c) m3? Express the result to three significant figures. Use an appropriate prefix. Units Used: 3

Mg = 10 kg mN = 10

−3

N

3

kN = 10 N Given: m1 = 10 kg m2 = 0.5 gm m3 = 4.50 Mg Solution: ( a)

W = m1 g W = 98.1 N

( b)

W = m2 g W = 4.90 mN

( c)

W = m3 g W = 44.1 kN

6

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-11 If an object has mass m, determine its mass in kilograms. Given: m = 40 slug Solution: m = 584 kg

Problem 1-12 The specific weight (wt./vol.) of brass is ρ. Determine its density (mass/vol.) in SI units. Use an appropriate prefix. Units Used: 3

Mg = 10 kg Given: lb

ρ = 520

ft

3

Solution: Mg

ρ = 8.33

m

3

Problem 1-13 A concrete column has diameter d and length L. If the density (mass/volume) of concrete is ρ, determine the weight of the column in pounds. Units Used: 3

Mg = 10 kg 3

kip = 10 lb Given: d = 350 mm L = 2m

7

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Engineering Mechanics - Statics

ρ = 2.45

Chapter 1

Mg m

3

Solution: 2

d V = π ⎛⎜ ⎟⎞ L ⎝ 2⎠ W = ρ V

V = 192.423 L W = 1.04 kip

Problem 1-14 The density (mass/volume) of aluminum is ρ. Determine its density in SI units. Use an appropriate prefix. Units Used: Mg = 1000 kg Given:

ρ = 5.26

slug ft

3

Solution:

ρ = 2.17

Mg m

3

Problem 1-15 Determine your own mass in kilograms, your weight in newtons, and your height in meters. Solution: Example W = 150 lb m = W

m = 68.039 kg W g = 667.233 N

h = 72 in

h = 1.829 m

8

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-16 Two particles have masses m1 and m2, respectively. If they are a distance d apart, determine the force of gravity acting between them. Compare this result with the weight of each particle. Units Used: G = 66.73 × 10

− 12

m

3 2

kg⋅ s nN = 10

−9

N

Given: m1 = 8 kg m2 = 12 kg d = 800 mm Solution: F =

G m1 m2 d

2

F = 10.0 nN W1 = m1 g

W1 = 78.5 N

W2 = m2 g

W2 = 118 N

W1 F

W2 F

= 7.85 × 10

9

= 1.18 × 10

10

Problem 1-17 Using the base units of the SI system, show that F = G(m1m2)/r2 is a dimensionally homogeneous equation which gives F in newtons. Compute the gravitational force acting between two identical spheres that are touching each other. The mass of each sphere is m1, and the radius is r. Units Used: μN = 10

−6

N

G = 66.73 10

− 12

m

3 2

kg⋅ s

9

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Given: m1 = 150 kg r = 275 mm Solution: F =

G m1 ( 2r)

2

2

F = 4.96 μN Since the force F is measured in Newtons, then the equation is dimensionally homogeneous.

Problem 1-18 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, (c) z. Units Used: 6

MN = 10 N 3

kN = 10 N μm = 10

−6

m

Given: x = ( 200 kN)

2

y = ( 0.005 mm) z = ( 400 m)

2

3

Solution: 2

( a)

x = 0.040 MN

( b)

y = 25.0 μm

( c)

z = 0.0640 km

2

3

10

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Problem 1-19 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) a 1/b1, (b) a2b2/c2, (c) a3b3. Units Used: μm = 10

−6

6

m

Mm = 10 m

Mg = 10 gm

kg = 10 gm

6

−3

ms = 10

3

s

Given: a1 = 684 μm b1 = 43 ms a2 = 28 ms b2 = 0.0458 Mm c2 = 348 mg a3 = 2.68 mm b3 = 426 Mg Solution: a1

( a)

b1

= 15.9

a2 b2

( b)

c2

mm s

= 3.69 Mm

s kg

a3 b3 = 1.14 km⋅ kg

( c)

Problem 1-20 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) a1/b12 (b) a22b23. Units Used: 6

Mm = 10 m

11

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 1

Given: a1 = 0.631 Mm b1 = 8.60 kg a2 = 35 mm b2 = 48 kg Solution:

( a)

a1 b1

( b)

2

2

= 8.532

km kg

3

2

3

a2 b2 = 135.48 kg ⋅ m

2

12

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-1 Determine the magnitude of the resultant force FR = F1 + F 2 and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 600 N F 2 = 800 N F 3 = 450 N

α = 45 deg β = 60 deg γ = 75 deg

Solution:

ψ = 90 deg − β + α FR =

F1 + F 2 − 2 F1 F 2 cos ( ψ) 2

2

F R = 867 N FR

sin ( ψ)

=

F2

sin ( θ )

⎛ ⎝

θ = asin ⎜F 2 θ = 63.05 deg

sin ( ψ) ⎞ ⎟ FR ⎠

φ = θ+α φ = 108 deg

Problem 2-2 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 80 lb F 2 = 60 lb 13

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

θ = 120 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 180 deg − θ ) 2

2

F R = 72.1 lb

⎛ ⎝

β = asin ⎜ F1

sin ( 180 deg − θ ) ⎞ ⎟ FR ⎠

β = 73.9 deg

Problem 2-3 Determine the magnitude of the resultant force F R = F1 + F 2 and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 250 lb F 2 = 375 lb

θ = 30 deg φ = 45 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 90 deg + θ − φ ) 2

2

F R = 178 kg FR

sin ( 90 deg + θ − φ )

⎛ F1

β = asin ⎜

⎝ FR

=

F1

sin ( β )

⎞

sin ( 90 deg + θ − φ )⎟

⎠

β = 37.89 deg 14

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Angle measured ccw from x axis 360 deg − φ + β = 353 deg

Problem 2-4 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive u axis. Given: F 1 = 300 N F 2 = 500 N

α = 30 deg β = 45 deg γ = 70 deg Solution: FR =

F1 + F 2 − 2 F1 F 2 cos ( 180 deg − β − γ + α ) 2

2

F R = 605 N FR

sin ( 180 deg − β − γ + α )

⎛ ⎝

θ = asin ⎜F 2

=

F2

sin ( θ )

sin ( 180 deg − β − γ + α ) ⎞ ⎟ FR ⎠

θ = 55.40 deg φ = θ+α φ = 85.4 deg Problem 2-5 Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 300 N

α = 30 deg 15

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F 2 = 500 N

Chapter 2

β = 45 deg

γ = 70 deg Solution: F1u

=

sin ( γ − α )

F1

sin ( 180 deg − γ ) sin ( γ − α )

F 1u = F1

sin ( 180 deg − γ )

F 1u = 205 N F 1v

sin ( α )

=

F 1v = F 1

F1

sin ( 180 deg − γ ) sin ( α )

sin ( 180 deg − γ )

F 1v = 160 N

Problem 2-6 Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 300 N F 2 = 500 N

α = 30 deg β = 45 deg γ = 70 deg Solution:

⎛ sin ( β ) ⎞ ⎟ ⎝ sin ( γ ) ⎠

F 2u = F2 ⎜

16

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 2u = 376.2 N

⎡ sin ⎡⎣180 deg − ( β + γ )⎤⎦⎤ ⎥ sin ( γ ) ⎣ ⎦

F 2v = F 2 ⎢

F 2v = 482.2 N

Problem 2-7 Determine the magnitude of the resultant force F R = F 1 + F 2 and its direction measured counterclockwise from the positive u axis. Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg Solution:

α = 180 deg − ( θ 3 + θ 1 ) FR =

F2 + F 1 − 2 F1 F 2 cos ( α ) 2

2

F R = 61.4 lb sin ( θ' ) sin ( α ) = F2 FR

⎛

F2 ⎞

⎝

FR ⎠

θ' = asin ⎜ sin ( α )

⎟

θ' = 51.8 deg θ = θ' − θ 3 θ = 6.8 deg

Problem 2-8 Resolve the force F 1 into components acting along the u and v axes and determine the components. 17

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg Solution: −F u

=

sin ( θ 3 − θ 2 )

Fu =

F1

sin ( θ 2 )

−F1 sin ( θ 3 − θ 2 ) sin ( θ 2 )

F u = −12.9 lb Fv

sin ( 180 deg − θ 3 )

Fv =

=

F1

sin ( θ 2 )

F 1 sin ( 180 deg − θ 3 ) sin ( θ 2 )

F v = 35.4 lb

Problem 2-9 Resolve the force F2 into components acting along the u and v axes and determine the components. Given: F 1 = 25 lb F 2 = 50 lb

θ 1 = 30 deg 18

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Engineering Mechanics - Statics

Chapter 2

θ 2 = 30 deg θ 3 = 45 deg Solution: Fu

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

=

F2

sin ( θ 2 )

F2 sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fu =

sin ( θ 2 )

F u = 86.6 lb −Fv

=

sin ( θ 1 ) Fv =

F2

sin ( θ 2 )

−F 2 sin ( θ 1 ) sin ( θ 2 )

F v = −50 lb

Problem 2-10 Determine the components of the F force acting along the u and v axes. Given:

θ 1 = 70 deg θ 2 = 45 deg θ 3 = 60 deg F = 250 N Solution: Fu

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fu =

=

F sin ( θ 2 )

F sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ sin ( θ 2 )

F u = 320 N

19

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Fv

sin ( θ 1 )

Fv =

=

Chapter 2

F sin ( θ 2 )

F sin ( θ 1 )

F v = 332 N

sin ( θ 2 )

Problem 2-11 The force F acts on the gear tooth. Resolve this force into two components acting along the lines aa and bb. Given: F = 20 lb

θ 1 = 80 deg θ 2 = 60 deg Solution: F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fa =

Fa

sin ( θ 1 )

F sin ( θ 1 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

Fb =

=

F sin ( θ 2 )

=

F a = 30.6 lb

Fb

sin ( θ 2 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

F b = 26.9 lb

Problem 2-12 The component of force F acting along line aa is required to be Fa. Determine the magnitude of F and its component along line bb.

20

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F a = 30 lb

θ 1 = 80 deg θ 2 = 60 deg

Solution: Fa

=

sin ( θ 1 )

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

⎛ sin ( 180 deg − θ 1 − θ 2) ⎞ ⎜ ⎟ sin ( θ 1 ) ⎝ ⎠

F = Fa

Fa

=

sin ( θ 1 ) Fb =

F

F = 19.6 lb

Fb

sin ( θ 2 )

Fa sin ( θ 2 ) sin ( θ 1 )

F b = 26.4 lb

Problem 2-13 A resultant force F is necessary to hold the ballon in place. Resolve this force into components along the tether lines AB and AC, and compute the magnitude of each component.

Given: F = 350 lb

θ 1 = 30 deg θ 2 = 40 deg Solution: F AB

sin ( θ 1 )

=

F

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦

21

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

sin ( θ 1 ) ⎡ ⎤ ⎢ ⎥ ⎣ sin ⎡⎣180 deg − ( θ 1 + θ 2)⎤⎦⎦

F AB = F

F AB = 186 lb F AC

sin ( θ 2 )

F

=

sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ sin ( θ 2 ) ⎡ ⎤ ⎢ ⎥ ⎣sin ⎡⎣180 deg − ( θ 1 + θ 2 )⎤⎦ ⎦

F AC = F

F AC = 239 lb

Problem 2-14 The post is to be pulled out of the ground using two ropes A and B. Rope A is subjected to force F 1 and is directed at angle θ1 from the horizontal. If the resultant force acting on the post is to be F R, vertically upward, determine the force T in rope B and the corresponding angle θ. Given: F R = 1200 lb F 1 = 600 lb

θ 1 = 60 deg Solution: T =

F 1 + FR − 2 F1 F R cos ( 90 deg − θ 1 ) 2

2

T = 744 lb sin ( 90 − θ 1 ) sin ( θ ) = FR T

⎛ ⎝

θ = asin ⎜ sin ( 90 deg − θ 1 )

F1 ⎞ T

⎟ ⎠

θ = 23.8 deg

22

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-15 Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 250 N F 2 = 150 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 105 deg Solution: F 1v

sin ( θ 1 )

=

F1

sin ( θ 3 )

⎛ sin ( θ 1 ) ⎞ ⎜ ⎟ ⎝ sin ( θ 3 ) ⎠

F 1v = F 1

F 1v = 129 N F 1u

sin ( 180 deg − θ 1 − θ 3 ) F 1u = F1

=

F1

sin ( θ 3 )

⎛ sin ( 180 deg − θ 1 − θ 3 ) ⎞ ⎜ ⎟ sin ( θ 3 ) ⎝ ⎠

F 1u = 183 N

Problem 2-16 Resolve the force F 2 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 250 N 23

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 2 = 150 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 105 deg Solution: F 1v

sin ( θ 1 )

F2

=

F 1v = F 2

sin ( 180 deg − θ 3 ) sin ( θ 1 ) ⎛ ⎞ ⎜ ⎟ ⎝ sin ( 180 deg − θ 3 ) ⎠

F 1v = 77.6 N F2u

sin ( 180 deg − θ 3 ) F 2u = F2

=

F2

sin ( 180 deg − θ 3 )

⎛ sin ( 180 deg − θ 3) ⎞ ⎜ ⎟ ⎝ sin ( 180 deg − θ 3) ⎠

F 2u = 150 N

Problem 2-17 Determine the magnitude and direction of the resultant force F R. Express the result in terms of the magnitudes of the components F 1 and F 2 and the angle φ.

Solution: F R = F 1 + F2 − 2F1 F 2 cos ( 180 deg − φ ) 2

2

2

24

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Since cos ( 180 deg − φ ) = −cos ( φ ), F 1 + F2 + 2 F 1 F2 cos ( φ ) 2

FR =

2

From the figure, tan ( θ ) =

F1 sin ( φ )

F 2 + F 1 cos ( φ )

Problem 2-18 If the tension in the cable is F1, determine the magnitude and direction of the resultant force acting on the pulley. This angle defines the same angle θ of line AB on the tailboard block. Given: F 1 = 400 N

θ 1 = 30 deg

Solution: FR =

F1 + F 1 − 2F 1 F 1 cos ( 90 deg − θ 1 ) 2

2

F R = 400 N sin ( θ 1 ) sin ( 90 deg − θ ) = FR F1

⎛ FR

θ = 90 deg − asin ⎜

⎝ F1

⎞

sin ( θ 1 )⎟

⎠

θ = 60 deg

25

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-19 The riveted bracket supports two forces. Determine the angle θ so that the resultant force is directed along the negative x axis. What is the magnitude of this resultant force? Given: F 1 = 60 lb F 2 = 70 lb

θ 1 = 30 deg Solution: sin ( θ 1 ) sin ( θ ) = F1 F2

⎛

F1 ⎞

⎝

F2 ⎠

θ = asin ⎜ sin ( θ 1 )

⎟

θ = 25.4 deg φ = 180 deg − θ − θ 1 φ = 124.6 deg F1 + F 2 − 2F 1 F 2 cos ( φ ) 2

R =

2

R = 115 lb

Problem 2-20 The plate is subjected to the forces acting on members A and B as shown. Determine the magnitude of the resultant of these forces and its direction measured clockwise from the positive x axis.

Given: F A = 400 lb F B = 500 lb

26

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

θ 1 = 30 deg θ = 60 deg Solution: Cosine law: FR =

FB + FA − 2F B FA cos ( 90 deg − θ + θ 1 ) 2

2

F R = 458 lb Sine law: sin ( 90 deg − θ + θ 1 ) FR

=

sin ( θ − α ) FA

⎛

FA ⎞

⎝

FR ⎠

α = θ − asin ⎜ sin ( 90 deg − θ + θ 1 )

⎟

α = 10.9 deg

Problem 2-21 Determine the angle θ for connecting member B to the plate so that the resultant of FA and FB is directed along the positive x axis. What is the magnitude of the resultant force? Given: F A = 400 lb F B = 500 lb

θ 1 = 30 deg 27

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution: Sine law: sin ( 90 deg − θ 1 ) sin ( θ ) = FA FB

⎛

FA ⎞

⎝

FB ⎠

θ = asin ⎜ sin ( 90 deg − θ 1 )

⎟

θ = 43.9 deg FR

sin ( 90 deg + θ 1 − θ )

FR = FA

=

FA

sin ( θ )

sin ( 90 deg − θ + θ 1 ) sin ( θ )

F R = 561 lb

Problem 2-22 Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by first finding the resultant F' = F1 + F 2 and then forming F R = F' + F 3. Given: F 1 = 30 N F 2 = 20 N F 3 = 50 N

θ = 20 deg c = 3 d = 4

28

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution: c α = atan ⎛⎜ ⎟⎞

⎝ d⎠

F' =

F1 + F 2 − 2F 1 F 2 cos ( 90 deg + θ − α ) 2

2

F' = 30.9 N F'

sin ( ( 90 deg − θ + α ) )

=

⎛ ⎝

F1

sin ( 90 deg − θ − β )

β = 90 deg − θ − asin ⎜F 1

sin ( 90 deg − θ + α ) ⎞ ⎟ F' ⎠

β = 1.5 deg Now add in force F3. FR =

F' + F3 − 2F' F3 cos ( β ) 2

2

F R = 19.2 N FR

sin ( β )

F'

=

sin ( φ )

⎛ ⎝

φ = asin ⎜F'

sin ( β ) ⎞ ⎟ FR ⎠

φ = 2.4 deg

Problem 2-23 Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F' = F 2 + F 3 and then forming F R = F' + F 1. Given: F 1 = 30 N F 2 = 20 N 29

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 3 = 50 N

θ = 20 deg c = 3 d = 4 Solution: F' =

F2 + F 3 − 2F 2 F 3 cos ( ( 90 deg − θ ) ) 2

2

F' = 47.07 N F2

sin ( β )

F'

=

sin ( 90 deg − θ )

⎛ ⎝

β = asin ⎜ F2

sin ( 90 deg − θ ) ⎞ ⎟ F' ⎠

β = 23.53 deg c α = atan ⎛⎜ ⎟⎞

⎝ d⎠

γ = α−β FR =

F' + F1 − 2F' F 1 cos ( γ ) 2

2

F R = 19.2 N FR

sin ( γ )

=

F1

sin ( φ )

⎛ ⎝

φ = asin ⎜F 1

sin ( γ ) ⎞ ⎟ FR ⎠

φ = 21.16 deg ψ = β−φ ψ = 2.37 deg

30

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-24 Resolve the force F into components acting along (a) the x and y axes, and (b) the x and y' axes.

Given: F = 50 lb

α = 65 deg β = 45 deg γ = 30 deg Solution: (a)

F x = F cos ( β ) F x = 35.4 lb F y = F sin ( β ) F y = 35.4 lb

(b)

Fx

sin ( 90 deg − β − γ )

=

F

sin ( 90 deg + γ )

sin ( 90 deg − β − γ )

Fx = F

sin ( 90 deg + γ )

F x = 14.9 lb Fy'

sin ( β )

=

F y' = F

F

sin ( 90 deg + γ ) sin ( β )

sin ( 90 deg + γ )

31

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F y' = 40.8 lb

Problem 2-25 The boat is to be pulled onto the shore using two ropes. Determine the magnitudes of forces T and P acting in each rope in order to develop a resultant force F1, directed along the keel axis aa as shown. Given:

θ = 40 deg θ 1 = 30 deg F 1 = 80 lb

Solution: F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

T = F1

=

T

sin ( θ )

sin ( θ )

sin ( 180 deg − θ − θ 1 )

T = 54.7 lb F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ P = sin ( θ 1 )

=

P

sin ( θ 1 )

F1

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

P = 42.6 lb

32

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-26 The boat is to be pulled onto the shore using two ropes. If the resultant force is to be F1, directed along the keel aa as shown, determine the magnitudes of forces T and P acting in each rope and the angle θ of P so that the magnitude of P is a minimum. T acts at θ from the keel as shown.

Given:

θ 1 = 30 deg F 1 = 80 lb

Solution: From the figure, P is minimum when

θ + θ 1 = 90 deg θ = 90 deg − θ 1 θ = 60 deg P

sin ( θ 1 )

P =

F1

=

sin ( 90 deg)

F 1 sin ( θ 1 ) sin ( 90 deg)

P = 40 lb T

sin ( θ ) T = F1

=

F1 sin ( 90 deg)

sin ( θ ) sin ( 90 deg)

T = 69.3 lb

33

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-27 The beam is to be hoisted using two chains. Determine the magnitudes of forces FA and FB acting on each chain in order to develop a resultant force T directed along the positive y axis.

Given: T = 600 N

θ 1 = 30 deg θ = 45 deg

Solution: FA

sin ( θ ) FA =

T

=

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ T sin ( θ )

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

F A = 439 N

FB

sin ( θ 1 )

FB = T

=

T

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦ sin ( θ 1 )

sin ⎡⎣180 deg − ( θ + θ 1 )⎤⎦

F B = 311 N

34

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-28 The beam is to be hoisted using two chains. If the resultant force is to be F, directed along the positive y axis, determine the magnitudes of forces FA and F B acting on each chain and the orientation θ of F B so that the magnitude of FB is a minimum. Given: F = 600 N

θ 1 = 30 deg Solution: For minimum FB, require

θ = 90 deg − θ 1 θ = 60 deg F A = F cos ( θ 1 ) F A = 520 N F B = F sin ( θ 1 ) F B = 300 N

Problem 2-29 Three chains act on the bracket such that they create a resultant force having magnitude F R. If two of the chains are subjected to known forces, as shown, determine the orientation θ of the third chain,measured clockwise from the positive x axis, so that the magnitude of force F in this chain is a minimum. All forces lie in the x-y plane.What is the magnitude of F ? Hint: First find the resultant of the two known forces. Force F acts in this direction.

35

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F R = 500 lb F 1 = 200 lb F 2 = 300 lb

φ = 30 deg

Solution: Cosine Law: F R1 =

F 1 + F2 − 2 F 1 F2 cos ( 90 deg − φ ) 2

2

F R1 = 264.6 lb Sine Law:

Make F parallel to FR1

sin ( φ + θ ) sin ( 90 deg − φ ) = F1 F R1

⎛

θ = −φ + asin ⎜ sin ( 90 deg − φ )

⎝

⎞ ⎟ FR1 ⎠ F1

θ = 10.9 deg When F is directed along F R1, F will be minimum to create the resultant forces. F = F R − FR1 F = 235 lb

Problem 2-30 Three cables pull on the pipe such that they create a resultant force having magnitude F R. If two of the cables are subjected to known forces, as shown in the figure, determine the direction θ of the third cable so that the magnitude of force F in this cable is a minimum. All forces lie in the 36

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

g x–y plane.What is the magnitude of F ? Hint: First find the resultant of the two known forces.

Given: F R = 900 lb F 1 = 600 lb F 2 = 400 lb

α = 45 deg β = 30 deg

Solution:

F' =

F1 + F 2 − 2F 1 F 2 cos ( 90 deg + α − β ) 2

2

F' = 802.64 lb F = F R − F' F = 97.4 lb sin ( φ ) F1

=

sin ( 90 deg + α − β ) F'

⎛ ⎝

φ = asin ⎜ sin ( 90 deg + α − β )

F1 ⎞ F'

⎟ ⎠

φ = 46.22 deg θ = φ−β θ = 16.2 deg

Problem 2-31 Determine the x and y components of the force F.

37

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Given: F = 800 lb

α = 60 deg β = 40 deg

Solution: F x = F sin ( β ) F y = −F cos ( β ) F x = 514.2 lb F y = −612.8 lb

Problem 2-32 Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis. Given: F 1 = 70 N F 2 = 50 N F 3 = 65 N

θ = 30 deg φ = 45 deg Solution: + →

F Rx = ΣFx;

F RX = F 1 + F 2 cos ( θ ) − F 3 cos ( φ )

38

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑

FR =

F RY = −F2 sin ( θ ) − F 3 sin ( φ )

F Ry = ΣF y; 2

FRX + FRY

Chapter 2

2

⎛ FRY ⎞ ⎟ ⎝ FRX ⎠

θ = atan ⎜

F R = 97.8 N

θ = 46.5 deg

Problem 2-33 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 50 lb F 2 = 35 lb

α = 120 deg β = 25 deg

Solution: + F Rx = ΣF x; →

F Rx = F 1 sin ( α ) − F2 sin ( β ) F Rx = 28.5 lb

+

↑ FRy = ΣFy;

F Ry = −F 1 cos ( α ) − F 2 cos ( β )

39

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F Ry = −6.7 lb FR =

2

FRx + FRy

2

F R = 29.3 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ' = atan ⎜

θ' = 13.3 deg θ = 360 deg − θ' θ = 347 deg

Problem 2-34 Determine the magnitude of the resultant force and its direction , measured counterclockwise from the positive x axis.

Given: F 1 = 850 N F 2 = 625 N F 3 = 750 N

θ = 45 deg φ = 30 deg c = 3 d = 4

Solution: + →

F Rx = SF x;

F RX = F 1

d 2

c +d

2

− F 2 sin ( φ ) − F3 sin ( θ )

40

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Engineering Mechanics - Statics

+

↑

F Ry = SF y;

F RY = −F1

c 2

2

− F2 cos ( φ ) + F3 cos ( θ )

c +d

F RX = −162.8 N FR =

Chapter 2

2

FRX + FRY

F RY = −520.9 N 2

F R = 546 N

⎛ FRY ⎞ ⎟ ⎝ FRX ⎠

α = atan ⎜

α = 72.64 deg β = α + 180 deg β = 252.6 deg

Problem 2-35 Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the resultant force is directed along the positive x' axis and has a magnitude of FR. Units Used: 3

kN = 10 N Given: F R = 1 kN F 2 = 450 N F 3 = 200 N

α = 45 deg β = 30 deg Solution: + →

F Rx = SF x;

F R cos ( β ) = F 3 + F 2 cos ( α ) + F 1 cos ( θ + β )

+

F Ry = SF y;

−F R sin ( β ) = F2 sin ( α ) − F 1 sin ( θ + β )

↑

41

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 1 cos ( θ + β ) = FR cos ( β ) − F3 − F2 cos ( α ) F 1 sin ( θ + β ) = F 2 sin ( α ) + FR sin ( β )

⎛

F2 sin ( α ) + F R sin ( β )

⎞ ⎟−β ⎝ FR cos ( β ) − F3 − F2 cos ( α ) ⎠

θ = atan ⎜

θ = 37 deg

(FR cos (β ) − F3 − F2 cos (α ))2 + (F2 sin (α ) + FR sin (β ))2

F1 =

F 1 = 889 N

Problem 2-36 Determine the magnitude and direction, measured counterclockwise from the x' axis, of the resultant force of the three forces acting on the bracket.

Given: F 1 = 300 N F 2 = 450 N F 3 = 200 N

α = 45 deg β = 30 deg θ = 20 deg Solution: F Rx = F 1 cos ( θ + β ) + F3 + F2 cos ( α )

F Rx = 711.03 N

F Ry = −F 1 sin ( θ + β ) + F 2 sin ( α )

F Ry = 88.38 N

42

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Engineering Mechanics - Statics

FR =

2

FRx + FRy

Chapter 2

2

F R = 717 N

φ (angle from x axis)

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 7.1 deg

φ' (angle from x' axis) φ' = β + φ

φ' = 37.1 deg

Problem 2-37 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

Given: F 1 = 800 N F 2 = 600 N

θ = 40 deg c = 12 d = 5 Solution: + F Rx = ΣF x; →

F Rx = F 1 cos ( θ ) − F 2 ⎜

2

+

F Ry = F 1 sin ( θ ) + F2 ⎜

⎛

d

↑

F Ry = ΣF y;

⎛

c

⎞

2⎟

⎝ c +d ⎠ 2

⎞

2⎟

⎝ c +d ⎠

FR =

2

FRx + FRy

2

F Rx = 59 N

F Ry = 745 N

F R = 747 N

43

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 85.5 deg

Problem 2-38 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Units Used: 3

kN = 10 N Given: F 1 = 30 kN F 2 = 26 kN

θ = 30 deg c = 5 d = 12

Solution: + F Rx = ΣF x; →

F Rx = −F 1 sin ( θ ) −

c ⎛ ⎞ ⎜ 2 2 ⎟ F2 ⎝ c +d ⎠

F Rx = −25 kN

+

F Ry = −F 1 cos ( θ ) +

⎛ d ⎞F ⎜ 2 2⎟ 2 ⎝ c +d ⎠

F Ry = −2 kN

↑ FRy = ΣFy;

FR =

2

FRx + FRy

2

F R = 25.1 kN

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 4.5 deg

β = 180 deg + φ

β = 184.5 deg

44 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-39 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 60 lb F 2 = 70 lb F 3 = 50 lb

θ 1 = 60 deg θ 2 = 45 deg c = 1 d = 1 Solution: d θ 3 = atan ⎛⎜ ⎟⎞

⎝c⎠

F Rx = −F 1 cos ( θ 3 ) − F2 sin ( θ 1 )

F Rx = −103 lb

F Ry = F 1 sin ( θ 3 ) − F 2 cos ( θ 1 ) − F3

F Ry = −42.6 lb

FR =

2

FRx + FRy

2

F R = 111.5 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = 180 deg + atan ⎜

θ = 202 deg

Problem 2-40 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis by summing the rectangular or x, y components of the forces to obtain the resultant force. 45

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 600 N F 2 = 800 N F 3 = 450 N

θ 1 = 60 deg θ 2 = 45 deg θ 3 = 75 deg Solution: + F Rx = ΣF x; → +

↑ FRy = ΣFy;

F Rx = F 1 cos ( θ 2 ) − F2 sin ( θ 1 )

F Rx = −268.556 N

F Ry = F 1 sin ( θ 2 ) + F 2 cos ( θ 1 )

F Ry = 824.264 N

FR =

2

FRx + FRy

2

F R = 867 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = 180 deg − atan ⎜

θ = 108 deg

Problem 2-41 Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by summing the rectangular or x, y components of the forces to obtain the resultant force. Given: F 1 = 30 N F 2 = 20 N

46

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 3 = 50 N

θ = 20 deg c = 3 d = 4

Solution: F Rx = −F 1

F Ry = F 1

FR =

⎛ d ⎞ − F ( sin ( θ ) ) + F 2 3 ⎜ 2 2⎟ ⎝ c +d ⎠

F Rx = 19.2 N

c ⎛ ⎞ ⎜ 2 2 ⎟ − F2 cos ( θ ) ⎝ c +d ⎠ 2

FRx + FRy

F Ry = −0.8 N

2

F R = 19.2 N

⎛ −FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 2.4 deg

Problem 2-42 Determine the magnitude and orientation, measured counterclockwise from the positive y axis, of the resultant force acting on the bracket. Given: F A = 700 N

47

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F B = 600 N

θ = 20 deg φ = 30 deg

Solution: Scalar Notation: Suming the force components algebraically, we have F Rx = ΣF x;

F Rx = F A sin ( φ ) − F B cos ( θ ) F Rx = −213.8 N

F Ry = ΣF y;

F Ry = F A cos ( φ ) + FB sin ( θ ) F Ry = 811.4 N

The magnitude of the resultant force F R is 2

FR =

FRx + FRy

2

F R = 839 N The directional angle θ measured counterclockwise from the positive x axis is

⎛ FRx ⎞ ⎟ ⎝ FRy ⎠

θ = atan ⎜

θ = 14.8 deg

Problem 2-43 Determine the magnitude and direction, measured counterclockwise from the positive x' axis, of the resultant force of the three forces acting on the bracket. Given: F 1 = 300 N

48

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 2 = 200 N F 3 = 180 N

θ = 10 deg θ 1 = 60 deg c = 5 d = 12

Solution: + →

+

↑

F Rx = ΣF x;

F Ry = ΣF y;

F Rx = F 1 sin ( θ 1 + θ ) −

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

F Ry = F 1 cos ( θ 1 + θ ) + F2 +

FR =

2

FRx + FRy

2

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

F Rx = 115.8 N

F Ry = 371.8 N

F R = 389 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = atan ⎜

φ = 72.7 deg

ψ = φ − ( 90 deg − θ 1 )

ψ = 42.7 deg

Problem 2-44 Determine the x and y components of F 1 and F 2.

49

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 200 N F 2 = 150 N

θ = 45 deg φ = 30 deg Solution: F 1x = F 1 sin ( θ ) F 1x = 141.4 N F 1y = F 1 cos ( θ ) F 1y = 141.4 N F 2x = −F 2 cos ( φ ) F 2x = −129.9 N F 2y = F 2 sin ( φ ) F 2y = 75 N

Problem 2-45 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 200 N F 2 = 150 N

θ = 45 deg φ = 30 deg

50

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution: + F Rx = ΣF x; → +

↑ FRy = ΣFy;

F Rx = F 1 sin ( θ ) − F2 cos ( φ )

F Rx = 11.5 N

F Ry = F 1 cos ( θ ) + F 2 sin ( φ )

F Ry = 216.4 N

F =

2

FRx + FRy

2

F = 217 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

β = atan ⎜

β = 87 deg

Problem 2-46 Determine the x and y components of each force acting on the gusset plate of the bridge truss. Given: F 1 = 200 lb

c = 3

F 2 = 400 lb

d = 4

F 3 = 300 lb

e = 3

F 4 = 300 lb

f = 4

Solution: F 1x = −F 1 F 1x = −200 lb F 1y = 0 lb F 2x = F 2 ⎛ ⎜

d 2

⎞

2⎟

⎝ c +d ⎠

F 2x = 320 lb F 2y = −F 2 ⎛

c ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠

F 2y = −240 lb F 3x = F 3 ⎛ ⎜

e 2

⎞

2⎟

⎝ e +f ⎠ 51

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 3x = 180 lb F 3y = F 3 ⎛ ⎜

f 2

⎞

2⎟

⎝ e +f ⎠

F 3y = 240 lb F 4x = −F 4 F 4x = −300 lb F 4y = 0 lb

Problem 2-47 Determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis. Units Used: 3

kN = 10 N

Given: F 1 = 20 kN F 2 = 40 kN F 3 = 50 kN

θ = 60 deg c = 1 d = 1 e = 3 f = 4 Solution: + F Rx = ΣF x; →

F Rx = F 3 ⎛ ⎜

⎞ + F ⎛ d ⎞ − F cos ( θ ) 2⎜ 1 2 2⎟ 2 2⎟ ⎝ e +f ⎠ ⎝ c +d ⎠ f

F Rx = 58.28 kN

52

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Engineering Mechanics - Statics

+

Chapter 2

⎛

c ⎞−F ⎛ ⎞ 2 ⎜ 2 2 ⎟ − F1 sin ( θ ) 2 2⎟ ⎝ e +f ⎠ ⎝ c +d ⎠

↑ FRy = ΣFy;

F Ry = F 3 ⎜

e

F Ry = −15.6 kN F =

2

FRx + FRy

2

F = 60.3 kN

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 15 deg

Problem 2-48 Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the resultant force is directed along the positive x' axis and has magnitude FR. Given: F 2 = 200 N F 3 = 180 N

θ 1 = 60 deg F R = 800 N c = 5 d = 12

Solution: Initial Guesses:

F 1 = 100 N

θ = 10 deg

Given + →

F Rx = ΣF x;

F R sin ( θ 1 ) = F1 sin ( θ 1 + θ ) −

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

53

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑

F Ry = ΣF y;

Chapter 2

F R( cos ( θ 1 ) ) = F 1 cos ( θ 1 + θ ) + F2 +

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , θ ) ⎝θ ⎠ F 1 = 869 N

θ = 21.3 deg

Problem 2-49 Determine the magnitude and direction, measured counterclockwise from the positive x' axis, of the resultant force acting on the bracket. Given: F 1 = 300 N F 2 = 200 N F 3 = 180 N

θ 1 = 60 deg θ = 10 deg c = 5 d = 12

Solution: Guesses

F Rx = 100 N

F Ry = 100 N

Given + →

F Rx = ΣF x;

F Rx = F1 sin ( θ 1 + θ ) −

+

F Ry = ΣF y;

F Ry = F1 cos ( θ 1 + θ ) + F 2 +

↑

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠ c ⎛ ⎞ ⎜ 2 2 ⎟ ( F3 ) ⎝ c +d ⎠

54

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

⎛ FRx ⎞ ⎜ ⎟ = Find ( FRx , FRy) ⎝ FRy ⎠ FR =

2

FRx + FRy

2

⎛ FRx ⎞ ⎛ 115.8 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FRy ⎠ ⎝ 371.8 ⎠ F R = 389 N

φ = atan ⎜

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

φ = 72.7 deg

φ' = ⎡⎣φ − ( 90 deg − θ 1 )⎤⎦

φ' = 42.7 deg

Problem 2-50 Express each of the three forces acting on the column in Cartesian vector form and compute the magnitude of the resultant force. Given: F 1 = 150 lb

θ = 60 deg

F 2 = 275 lb

c = 4

F 3 = 75 lb

d = 3

Solution: Find the components of each force.

⎛

⎞ ⎟ 2 2 ⎝ c +d ⎠

F 1x = F 1 ⎜

F 1v =

d

⎛ F1x ⎞ ⎜ ⎟ ⎝ F1y ⎠

F 2x = 0 lb

F 2v =

⎛ F2x ⎞ ⎜ ⎟ ⎝ F2y ⎠

⎛

⎞ ⎟ 2 2 ⎝ c +d ⎠

F 1y = F 1 ⎜

F 1v =

−c

⎛ 90 ⎞ ⎜ ⎟ lb ⎝ −120 ⎠

F 2y = −F 2

F 2v =

⎛ 0 ⎞ ⎜ ⎟ lb ⎝ −275 ⎠ 55

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F 3x = −F 3 cos ( θ )

F 3v =

Chapter 2

F 3y = −F 3 sin ( θ )

⎛ F3x ⎞ ⎜ ⎟ ⎝ F3y ⎠

F 3v =

⎛ −37.5 ⎞ ⎜ ⎟ lb ⎝ −65 ⎠

Now find the magnitude of the resultant force. F R = F 1v + F 2v + F 3v

F R = 462.9 lb

Problem 2-51 Determine the magnitude of force F so that the resultant F R of the three forces is as small as possible. What is the minimum magnitude of FR? Units Used: kN = 1000 N Given: F 1 = 5 kN F 2 = 4 kN

θ = 30 deg Solution: Scalar Notation: Suming the force components algebrically, we have + → +

↑

F Rx = ΣF x;

F Rx = F1 − F sin ( θ )

F Ry = ΣF y;

F Ry = F cos ( θ ) − F 2

The magnitude of the resultant force F R is FR =

2

2

F Rx + F Ry =

(F1 − F sin(θ ))2 + (F cos (θ ) − F2)2 56

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F R = F 1 + F2 + F − 2FF1 sin ( θ ) − 2F2 Fcos( θ ) 2

2FR

2

dF R dF

2

= 2F − 2F1 sin ( θ ) − 2F2 cos ( θ )

If F is a minimum, then FR =

2

⎞ ⎛ dFR = 0⎟ ⎜ ⎝ dF ⎠

F = F 1 sin ( θ ) + F2 cos ( θ )

(F1 − F sin(θ ))2 + (F cos (θ ) − F2)2

F = 5.96 kN F R = 2.3 kN

Problem 2-52 Express each of the three forces acting on the bracket in Cartesian vector form with respect to the x and y axes. Determine the magnitude and direction θ of F1 so that the resultant force is directed along the positive x' axis and has magnitude F R. Units Used: kN = 1000 N Given: F R = 600 N F 2 = 350 N F 3 = 100 N

φ = 30 deg Solution: F 2v =

⎛ F2 ⎞ ⎜ ⎟ ⎝0 ⎠

F 3v =

⎛ 0 ⎞ ⎜ ⎟ ⎝ −F 3 ⎠

F 1v =

⎛ F1 cos ( θ ) ⎞ ⎜ ⎟ ⎝ F1 sin ( θ ) ⎠

F 2v =

⎛ 350 ⎞ ⎜ ⎟N ⎝ 0 ⎠

F 3v =

⎛ 0 ⎞ ⎜ ⎟N ⎝ −100 ⎠

57

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 1 = 20 N θ = 10 deg

The initial guesses: Given

⎛ F1 cos ( θ ) ⎞ ⎜ ⎟ + F2v + F3v = ( ) F sin θ 1 ⎝ ⎠ ⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , θ ) ⎝θ ⎠

⎛ FR cos ( φ ) ⎞ ⎜ ⎟ ⎝ FR sin ( φ ) ⎠

F 1 = 434.5 N

θ = 67 deg

Problem 2-53 The three concurrent forces acting on the post produce a resultant force F R = 0. If F 2 = (1/2)F 1, and F 1 is to be 90° from F 2 as shown, determine the required magnitude F 3 expressed in terms of F1 and the angle θ. Solution:

Use the primed coordiates.

ΣFRx = 0

F 3 cos ( θ − 90 deg) − F1 = 0

ΣFRy = 0

−F 3 sin ( θ − 90 deg) + F 2 = 0 tan ( θ − 90 deg) =

F2 F1

=

1 2

1 θ = 90 deg + atan ⎛⎜ ⎟⎞

⎝ 2⎠

θ = 117 deg k =

1

cos ( θ − 90 deg)

k = 1.1

F3 = k F1

58 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-54 Three forces act on the bracket. Determine the magnitude and orientation θ of F2 so that the resultant force is directed along the positive u axis and has magnitude F R. Given: F R = 50 lb F 1 = 80 lb F 3 = 52 lb

φ = 25 deg c = 12 d = 5

Solution: Guesses F 2 = 1 lb

θ = 120 deg

Given F R cos ( φ ) = F1 + F2 cos ( φ + θ ) +

−F R sin ( φ ) = −F 2 sin ( φ + θ ) +

⎛ F2 ⎞ ⎜ ⎟ = Find ( F2 , θ ) ⎝θ ⎠

⎛ d ⎞F ⎜ 2 2⎟ 3 ⎝ c +d ⎠

c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠

θ = 103.3 deg

F 2 = 88.1 lb

59

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Problem 2-55 Determine the magnitude and orientation, measured clockwise from the positive x axis, of the resultant force of the three forces acting on the bracket. Given: F 1 = 80 lb F 2 = 150 lb F 3 = 52lb

θ = 55 deg φ = 25 deg c = 12 m d = 5m

Solution:

⎛

⎞ + F cos ( θ + φ ) 2 ⎝ c +d ⎠

F Rx = F 1 + F 3 ⎜

⎛

d

2

F Rx = 126.05 lb

2⎟

⎞ − F sin ( θ + φ ) 2 ⎟ 2 2 ⎝ c +d ⎠ c

F Ry = F 3 ⎜

FR =

2

FRx + FRy

F Ry = −99.7 lb

2

F R = 161 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

β = atan ⎜

β = 38.3 deg

Problem 2-56 Three forces act on the ring. Determine the range of values for the magnitude of P so that the magnitude of the resultant force does not exceed F. Force P is always directed to the right.

60

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Units Used: 3

kN = 10 N Given: F = 2500 N F 1 = 1500 N F 2 = 600 N

θ 1 = 60 deg θ 2 = 45 deg Solution: Initial Guesses:

F Rx = 100 N

F Ry = 100 N

P = 100 N

Given + →

F Rx = ΣF x;

F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 )

+

F Ry = ΣF y;

F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 )

↑

F=

2

F Rx + F Ry

⎛ FRx ⎞ ⎜ ⎟ ⎜ FRy ⎟ = Find ( FRx , FRy , P) ⎜P ⎟ ⎝ max ⎠ Initial Guesses:

2

P max = 1.6 kN

F Rx = −100 N

F Ry = 100 N

P = −2000 N

Given + →

F Rx = ΣF x;

F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 )

+

F Ry = ΣF y;

F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 )

↑

F=

⎛ FRx ⎞ ⎜ ⎟ ⎜ FRy ⎟ = Find ( FRx , FRy , P) ⎜P ⎟ ⎝ min ⎠

2

F Rx + F Ry

2

P min = −1.7 kN

61

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Engineering Mechanics - Statics

Since P > 0

we conclude that

Chapter 2

0 <= P <= P max = 1.6 kN

Problem 2-57 Determine the magnitude and coordinate direction angles of F1 and F2. Sketch each force on an x, y, z reference. Given:

⎛ 60 ⎞ F 1 = ⎜ −50 ⎟ N ⎜ ⎟ ⎝ 40 ⎠ ⎛ −40 ⎞ F 2 = ⎜ −85 ⎟ N ⎜ ⎟ ⎝ 30 ⎠ Solution: F 1 = 87.7 N

⎛ α1 ⎞ ⎜ ⎟ ⎛ F1 ⎞ ⎜ β 1 ⎟ = acos ⎜ ⎟ ⎝ F1 ⎠ ⎜γ ⎟ ⎝ 1⎠ ⎛ α 1 ⎞ ⎛ 46.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 1 ⎟ = ⎜ 124.7 ⎟ deg ⎜ γ ⎟ ⎝ 62.9 ⎠ ⎝ 1⎠ F 2 = 98.6 N

62

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Engineering Mechanics - Statics

Chapter 2

⎛ α2 ⎞ ⎜ ⎟ ⎛ F2 ⎞ ⎜ β 2 ⎟ = acos ⎜ ⎟ ⎝ F2 ⎠ ⎜γ ⎟ ⎝ 2⎠ ⎛ α 2 ⎞ ⎛ 113.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 2 ⎟ = ⎜ 149.5 ⎟ deg ⎜ γ ⎟ ⎝ 72.3 ⎠ ⎝ 2⎠

Problem 2-58 Express each force in Cartesian vector form. Units Used: 3

kN = 10 N Given: F 1 = 5 kN F 2 = 2 kN

θ 1 = 60 deg θ 2 = 60 deg θ 3 = 45 deg Solution:

F 1v

⎛ cos ( θ 2) ⎞ ⎜ ⎟ = F1 ⎜ cos ( θ 3 ) ⎟ ⎜ cos θ ⎟ ⎝ ( 1) ⎠

⎛ 2.5 ⎞ F 1v = ⎜ 3.5 ⎟ kN ⎜ ⎟ ⎝ 2.5 ⎠

F 2v

⎛0⎞ = F2 ⎜ −1 ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ F 2v = ⎜ −2 ⎟ kN ⎜ ⎟ ⎝0⎠

63

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Engineering Mechanics - Statics

Chapter 2

Problem 2-59 Determine the magnitude and coordinate direction angles of the force F acting on the stake. Given: F h = 40 N

θ = 70 deg c = 3 d = 4

Solution:

⎛ c2 + d2 ⎞ ⎟ F = Fh ⎜ ⎝ d ⎠ F = 50 N c ⎛ ⎞ ⎜ 2 2 ⎟F ⎝ c +d ⎠

F x = F h cos ( θ )

F y = F h sin ( θ )

Fz =

F x = 13.7 N

F y = 37.6 N

F z = 30 N

⎛ Fx ⎞ ⎟ ⎝F⎠

⎛ Fy ⎞ ⎟ ⎝F⎠

⎛ Fz ⎞ ⎟ ⎝F⎠

α = acos ⎜

β = acos ⎜

γ = acos ⎜

α = 74.1 deg

β = 41.3 deg

γ = 53.1 deg

Problem 2-60 Express each force in Cartesian vector form.

64

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 400 lb F 2 = 600 lb

θ 1 = 45 deg θ 2 = 60 deg θ 3 = 60 deg θ 4 = 45 deg θ 5 = 30 deg

Solution:

F 1v

⎛ −cos ( θ 2) ⎞ ⎜ ⎟ = F1 ⎜ cos ( θ 3 ) ⎟ ⎜ cos θ ⎟ ( 1) ⎠ ⎝

⎛ −200 ⎞ F 1v = ⎜ 200 ⎟ lb ⎜ ⎟ ⎝ 282.8 ⎠

F 2v

⎛ cos ( θ 5) cos ( θ 4 ) ⎞ ⎜ ⎟ = F2 ⎜ cos ( θ 5 ) sin ( θ 4 ) ⎟ ⎜ −sin θ ( 5) ⎟⎠ ⎝

⎛ 367.4 ⎞ F 2v = ⎜ 367.4 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠

Problem 2-61 The stock S mounted on the lathe is subjected to a force F , which is caused by the die D. Determine the coordinate direction angle β and express the force as a Cartesian vector. Given: F = 60 N

α = 60 deg γ = 30 deg 65

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

Solution: cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

(

2

β = acos 1 − cos ( α ) − cos ( γ ) 2

⎛⎜ −cos ( α ) ⎞⎟ F v = F⎜ −cos ( β ) ⎟ ⎜ −cos ( γ ) ⎟ ⎝ ⎠

2

)

β = 90 deg

⎛ −30 ⎞ Fv = ⎜ 0 ⎟ N ⎜ ⎟ ⎝ −52 ⎠

Problem 2-62 Determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 = 80 lb F 2 = 130 lb

θ = 40 deg φ = 30 deg Solution:

F 1v

⎛⎜ cos ( φ ) cos ( θ ) ⎟⎞ = F1 ⎜ −cos ( φ ) sin ( θ ) ⎟ ⎜ ⎟ sin ( φ ) ⎝ ⎠

⎛ 53.1 ⎞ F 1v = ⎜ −44.5 ⎟ lb ⎜ ⎟ ⎝ 40 ⎠

F 2v

⎛0⎞ = F2 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

⎛ 0 ⎞ F 2v = ⎜ 0 ⎟ lb ⎜ ⎟ ⎝ −130 ⎠

F R = F1v + F2v

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

F R = 113.6 lb

⎛⎜ α ⎞⎟ ⎛ 62.1 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 113.1 ⎟ deg ⎜ γ ⎟ ⎝ 142.4 ⎠ ⎝ ⎠

66

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Engineering Mechanics - Statics

Chapter 2

Problem 2-63 Specify the coordinate direction angles of F 1 and F 2 and express each force as a cartesian vector. Given: F 1 = 80 lb F 2 = 130 lb

φ = 30 deg θ = 40 deg Solution:

F 1v

⎛⎜ cos ( φ ) cos ( θ ) ⎟⎞ = F1 ⎜ −cos ( φ ) sin ( θ ) ⎟ ⎜ ⎟ sin ( φ ) ⎝ ⎠

⎛ α1 ⎞ ⎜ ⎟ ⎛ F1v ⎞ ⎜ β 1 ⎟ = acos ⎜ ⎟ ⎝ F1 ⎠ ⎜γ ⎟ ⎝ 1⎠

F 2v

⎛0⎞ ⎜ ⎟ = F2 0 ⎜ ⎟ ⎝ −1 ⎠

⎛ α2 ⎞ ⎜ ⎟ ⎛ F2v ⎞ ⎜ β 2 ⎟ = acos ⎜ ⎟ ⎝ F2 ⎠ ⎜γ ⎟ ⎝ 2⎠

⎛ 53.1 ⎞ ⎜ ⎟ F 1v = −44.5 lb ⎜ ⎟ ⎝ 40 ⎠ ⎛ α 1 ⎞ ⎛ 48.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 1 ⎟ = ⎜ 123.8 ⎟ deg ⎜ γ ⎟ ⎝ 60 ⎠ ⎝ 1⎠ ⎛ 0 ⎞ ⎜ 0 ⎟ lb F 2v = ⎜ ⎟ ⎝ −130 ⎠ ⎛ α 2 ⎞ ⎛ 90 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 2 ⎟ = ⎜ 90 ⎟ deg ⎜ γ ⎟ ⎝ 180 ⎠ ⎝ 2⎠

Problem 2-64 The mast is subjected to the three forces shown. Determine the coordinate angles α1, β1, γ1 of F1 so that the resultant force acting on the mast is FRi. Given: F R = 350 N

67

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Engineering Mechanics - Statics

Chapter 2

F 1 = 500 N F 2 = 200 N F 3 = 300 N Solution: Guesses

α = 20 deg β = 20 deg γ = 20 deg

Given

⎛⎜ cos ( α ) ⎟⎞ ⎛0⎞ ⎛0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ F 1 ⎜ cos ( β ) ⎟ + F2 0 + F3 −1 = FR⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ cos ( γ ) ⎟ ⎝ −1 ⎠ ⎝0⎠ ⎝0⎠ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎜ β ⎟ = Find ( α , β , γ ) ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 45.6 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 53.1 ⎟ deg ⎜ γ ⎟ ⎝ 66.4 ⎠ ⎝ ⎠

Problem 2-65 The mast is subjected to the three forces shown. Determine the coordinate angles α1, β1, γ1 of F1 so that the resultant force acting on the mast is zero . Given: F 1 = 500 N F 2 = 200 N F 3 = 300 N Solution: Guesses

α = 20 deg β = 20 deg γ = 20 deg

68

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Engineering Mechanics - Statics

Chapter 2

⎛⎜ cos ( α ) ⎞⎟ ⎛0⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ F 1 ⎜ cos ( β ) ⎟ + F2 0 + F3 −1 = 0 ⎜ ⎟ ⎜ ⎟ ⎜ cos ( γ ) ⎟ ⎝ −1 ⎠ ⎝0⎠ ⎝ ⎠

Given

⎛⎜ α ⎞⎟ ⎜ β ⎟ = Find ( α , β , γ ) ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 90 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 53.1 ⎟ deg ⎜ γ ⎟ ⎝ 66.4 ⎠ ⎝ ⎠

Problem 2-66 The shaft S exerts three force components on the die D. Find the magnitude and direction of the resultant force. Force F2 acts within the octant shown. Given: F 1 = 400 N F 2 = 300 N F 3 = 200 N

α 2 = 60 deg γ 2 = 60 deg c = 3 d = 4 Solution: cos ( α 2 ) + cos ( β 2 ) + cos ( γ 2 ) = 1 2

2

2

Solving for the positive root, 2 2 β 2 = acos ⎛⎝ 1 − cos ( α 2 ) − cos ( γ 2 ) ⎞⎠

F 1v

⎛⎜ F1 ⎞⎟ = ⎜ 0 ⎟ ⎜0 ⎟ ⎝ ⎠

F 2v

β 2 = 45 deg

⎛ cos ( α 2) ⎞ ⎜ ⎟ = F2 ⎜ cos ( β 2 ) ⎟ ⎜ cos γ ⎟ ⎝ ( 2) ⎠

F 3v =

⎛0⎞ ⎜ −d ⎟ ⎟ 2 2⎜ c +d ⎝ c ⎠ F3

69

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Engineering Mechanics - Statics

Chapter 2

F R = F1v + F2v + F3v

F R = 615 N

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 26.6 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 85.1 ⎟ deg ⎜ γ ⎟ ⎝ 64.0 ⎠ ⎝ ⎠

Problem 2-67 The beam is subjected to the two forces shown. Express each force in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 = 630 lb

α = 60 deg

F 2 = 250 lb

β = 135 deg

c = 24

γ = 60 deg

d = 7

Solution:

F 1v =

F 2v

⎛0⎞ ⎜d⎟ 2 2⎜ ⎟ c + d ⎝ −c ⎠ F1

⎛⎜ cos ( α ) ⎟⎞ = F2 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

F R = F1v + F2v

⎛ 0 ⎞ ⎜ ⎟ F 1v = 176.4 lb ⎜ ⎟ ⎝ −604.8 ⎠ ⎛ 125 ⎞ ⎜ ⎟ F 2v = −176.8 lb ⎜ ⎟ ⎝ 125 ⎠ F R = 495.8 lb

70

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Engineering Mechanics - Statics

Chapter 2

⎛ αR ⎞ ⎜ ⎟ ⎛ FR ⎞ ⎜ β R ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ R⎠

⎛ α R ⎞ ⎛ 75.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β R ⎟ = ⎜ 90 ⎟ deg ⎜ γ ⎟ ⎝ 165.4 ⎠ ⎝ R⎠

Problem 2-68 Determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 = 350 N

α = 60 deg

F 2 = 250N

β = 60 deg

c = 3

γ = 45 deg

d = 4

θ = 30 deg

Solution:

F 1v

⎛⎜ cos ( α ) ⎟⎞ = F1 ⎜ cos ( β ) ⎟ ⎜ −cos ( γ ) ⎟ ⎝ ⎠

F 2h = F2 ⎛

d ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠

F 2v

⎛ F2h cos ( θ ) ⎞ ⎜ ⎟ = ⎜ −F 2h sin ( θ ) ⎟ ⎜ F ⎟ 2y ⎝ ⎠

F R = F1v + F2v

⎛ αR ⎞ ⎜ ⎟ ⎛ FR ⎞ ⎜ β R ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ R⎠

⎛ 175 ⎞ F 1v = ⎜ 175 ⎟ N ⎜ ⎟ ⎝ −247.5 ⎠ F 2y = F 2 ⎛

c ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠

⎛ 173.2 ⎞ F 2v = ⎜ −100 ⎟ N ⎜ ⎟ ⎝ 150 ⎠ F R = 369.3 N

⎛ α R ⎞ ⎛ 19.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β R ⎟ = ⎜ 78.3 ⎟ deg ⎜ γ ⎟ ⎝ 105.3 ⎠ ⎝ R⎠

Problem 2-69 Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three 71

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Engineering Mechanics - Statics

Chapter 2

forces acts along the positive y axis and has magnitude F. Given: F = 600 lb F 1 = 180 lb F 2 = 300 lb

α 1 = 30 deg α 2 = 40 deg Solution: Initial guesses:

α = 40 deg β = 50 deg

γ = 50 deg F 3 = 45 lb

Given F Rx = ΣF x;

0 = −F1 + F2 cos ( α 1 ) sin ( α 2 ) + F 3 cos ( α )

F Ry = ΣF y;

F = F2 cos ( α 1 ) cos ( α 2 ) + F3 cos ( β )

F Rz = ΣF z;

0 = −F2 sin ( α 1 ) + F3 cos ( γ ) cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

2

2

F 3 = 428 lb

⎛⎜ α ⎞⎟ ⎛ 88.3 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 20.6 ⎟ deg ⎜ γ ⎟ ⎝ 69.5 ⎠ ⎝ ⎠

Problem 2-70 Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces is zero.

72

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 180 lb

α 1 = 30 deg

F 2 = 300 lb

α 2 = 40 deg

Solution: Initial guesses:

α = 40 deg

γ = 50 deg

β = 50 deg

F 3 = 45 lb

Given F Rx = ΣF x;

0 = −F1 + F2 cos ( α 1 ) sin ( α 2 ) + F 3 cos ( α )

F Ry = ΣF y;

0 = F2 cos ( α 1 ) cos ( α 2 ) + F3 cos ( β )

F Rz = ΣF z;

0 = −F2 sin ( α 1 ) + F3 cos ( γ ) cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

2

2

F 3 = 250 lb

⎛⎜ α ⎞⎟ ⎛ 87.0 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 142.9 ⎟ deg ⎜ γ ⎟ ⎝ 53.1 ⎠ ⎝ ⎠

Problem 2-71 Specify the magnitude F 3 and directions α3, β3, and γ3 of F 3 so that the resultant force of the three forces is FR. Units Used: 3

kN = 10 N Given: F 1 = 12 kN

c = 5

F 2 = 10 kN

d = 12

θ = 30 deg

73

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Engineering Mechanics - Statics

Chapter 2

⎛0⎞ F R = ⎜ 9 ⎟ kN ⎜ ⎟ ⎝0⎠ Solution: F 3x = 1 kN

Initial Guesses:

F 3y = 1 kN

⎛ F3x ⎞ ⎛ 0 ⎞ ⎜ ⎟ F R = ⎜ F3y ⎟ + F 1 ⎜ cos ( θ ) ⎟ + ⎜ ⎟ ⎜F ⎟ ( ) − sin θ ⎝ ⎠ ⎝ 3z ⎠

Given

⎛ F3x ⎞ ⎜ ⎟ ⎜ F3y ⎟ = Find ( F3x , F3y , F3z) ⎜F ⎟ ⎝ 3z ⎠ ⎛ α3 ⎞ ⎜ ⎟ ⎛ F3 ⎞ ⎜ β 3 ⎟ = acos ⎜ ⎟ ⎝ F3 ⎠ ⎜γ ⎟ ⎝ 3⎠

F 3z = 1 kN

⎛ −d ⎞ ⎜0⎟ ⎟ 2 2⎜ c +d ⎝ c ⎠ F2

⎛ F3x ⎞ ⎜ ⎟ F 3 = ⎜ F 3y ⎟ ⎜F ⎟ ⎝ 3z ⎠

⎛ 9.2 ⎞ F 3 = ⎜ −1.4 ⎟ kN ⎜ ⎟ ⎝ 2.2 ⎠

F 3 = 9.6 kN

⎛ α 3 ⎞ ⎛ 15.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 3 ⎟ = ⎜ 98.4 ⎟ deg ⎜ γ ⎟ ⎝ 77.0 ⎠ ⎝ 3⎠

Problem 2-72 The pole is subjected to the force F , which has components acting along the x,y,z axes as shown. Given β and γ, determine the magnitude of the three components of F . Units Used: kN = 1000 N Given: F = 3 kN

β = 30 deg γ = 75 deg Solution: cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

(

2

)

74

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Engineering Mechanics - Statics

α = acos

(

Chapter 2

)

−cos ( β ) − cos ( γ ) + 1 2

2

α = 64.67 deg F x = F cos ( α )

F y = F cos ( β )

F z = F cos ( γ )

F x = 1.28 kN

F y = 2.60 kN

F z = 0.8 kN

Problem 2-73 The pole is subjected to the force F which has components F x and Fz. Determine the magnitudes of F and Fy. Units Used: kN = 1000 N Given: F x = 1.5 kN F z = 1.25 kN

β = 75 deg Solution: cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

2

2

2

⎛ Fx ⎞ 2 ⎛ Fz ⎞ ⎜ ⎟ + cos ( β ) + ⎜ ⎟ = 1 ⎝F⎠ ⎝F⎠ 2

F =

2

Fx + Fz

2 1 − cos ( β )

F y = F cos ( β )

F = 2.02 kN

F y = 0.5 kN

Problem 2-74 The eye bolt is subjected to the cable force F which has a component F x along the x axis, a component F z along the z axis, and a coordinate direction angle β. Determine the magnitude of F .

75

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Engineering Mechanics - Statics

Chapter 2

Given: F x = 60 N F z = −80 N

β = 80 deg Solution: F y = F cos ( β ) F x + F z + Fy cos ( β ) 2

Fy =

2

2

Fy =

2

Fx + Fz

1 − cos ( β ) 2

F =

2

2

2

cos ( β )

2

Fx + Fy + Fz

F y = 17.6 N

F = 102 N

Problem 2-75 Three forces act on the hook. If the resultant force FR has a magnitude and direction as shown, determine the magnitude and the coordinate direction angles of force F3. Given: F R = 120 N F 1 = 80 N F 2 = 110 N c = 3 d = 4

θ = 30 deg φ = 45 deg

76

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Engineering Mechanics - Statics

Chapter 2

Solution:

F 1v

⎛d⎞ ⎛ F1 ⎞ ⎜ ⎟ = ⎜ ⎟ 0 2 2 ⎜ ⎟ c + d ⎝ ⎠⎝ c ⎠

⎛ 64 ⎞ F 1v = ⎜ 0 ⎟ N ⎜ ⎟ ⎝ 48 ⎠

F 2v

⎛0⎞ = F2 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

⎛ 0 ⎞ F 2v = ⎜ 0 ⎟ N ⎜ ⎟ ⎝ −110 ⎠

F Rv

⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟ = F R⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ sin ( φ ) ⎟ ⎝ ⎠

⎛ 42.4 ⎞ F Rv = ⎜ 73.5 ⎟ N ⎜ ⎟ ⎝ 84.9 ⎠

F 3v = FRv − F 1v − F 2v

⎛ −21.6 ⎞ F 3v = ⎜ 73.5 ⎟ N ⎜ ⎟ ⎝ 146.9 ⎠

⎛⎜ α ⎞⎟ ⎛ F3v ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ F3v ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 97.5 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 63.7 ⎟ deg ⎜ γ ⎟ ⎝ 27.5 ⎠ ⎝ ⎠

F 3v = 165.6 N

Problem 2-76 Determine the coordinate direction angles of F 1 and FR. Given: F R = 120 N F 1 = 80 N F 2 = 110 N c = 3 d = 4

θ = 30 deg φ = 45 deg

77

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Engineering Mechanics - Statics

Chapter 2

Solution:

F 1v

⎛d⎞ ⎛ F1 ⎞ ⎜ ⎟ = ⎜ ⎟ 0 2 2 ⎜ ⎟ c + d ⎝ ⎠⎝ c ⎠

⎛ 64 ⎞ F 1v = ⎜ 0 ⎟ N ⎜ ⎟ ⎝ 48 ⎠

⎛ α1 ⎞ ⎜ ⎟ ⎛ F1v ⎞ ⎜ β 1 ⎟ = acos ⎜ ⎟ ⎝ F1v ⎠ ⎜γ ⎟ ⎝ 1⎠

F Rv

⎛ α 1 ⎞ ⎛ 36.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 1 ⎟ = ⎜ 90 ⎟ deg ⎜ γ ⎟ ⎝ 53.1 ⎠ ⎝ 1⎠

⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟ = F R⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ sin ( φ ) ⎟ ⎝ ⎠

⎛ 42.4 ⎞ F Rv = ⎜ 73.5 ⎟ N ⎜ ⎟ ⎝ 84.9 ⎠

⎛ αR ⎞ ⎜ ⎟ ⎛ FRv ⎞ ⎜ β R ⎟ = acos ⎜ ⎟ ⎝ FRv ⎠ ⎜γ ⎟ ⎝ R⎠

⎛ α R ⎞ ⎛ 69.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β R ⎟ = ⎜ 52.2 ⎟ deg ⎜ γ ⎟ ⎝ 45 ⎠ ⎝ R⎠

Problem 2-77 The pole is subjected to the force F , which has components acting along the x, y, z axes as shown. Given the magnitude of F and the angles α and γ , determine the magnitudes of the components of F . Given: F = 80 N α = 60 deg

γ = 45 deg

Solution:

(

β = acos − 1 − cos ( α ) − cos ( γ ) 2

2

)

β = 120 deg F x = F cos ( α )

F y = F cos ( β )

F z = F cos ( γ )

F x = 40 N

F y = 40 N

F z = 56.6 N

78

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Engineering Mechanics - Statics

Chapter 2

Problem 2-78 Two forces F1 and F 2 act on the bolt. If the resultant force F R has magnitude F R and coordinate direction angles α and β, as shown, determine the magnitude of F2 and its coordinate direction angles. Given: F 1 = 20 lb F R = 50 lb

α = 110 deg β = 80 deg Solution: cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

(

2

γ = acos − 1 − cos ( α ) − cos ( β ) 2

Initial Guesses

Given

F 2x = 1 lb

2

)

γ = 157.44 deg

F 2y = 1 lb

F 2z = 1 lb

⎛⎜ cos ( α ) ⎟⎞ ⎛ 0 ⎞ ⎛⎜ F2x ⎟⎞ ⎜ ⎟ F R⎜ cos ( β ) ⎟ = F 1 0 + ⎜ F2y ⎟ ⎜ ⎟ ⎜ cos ( γ ) ⎟ ⎝ −1 ⎠ ⎜⎝ F2z ⎟⎠ ⎝ ⎠

⎛ F2x ⎞ ⎜ ⎟ F 2 = ⎜ F 2y ⎟ ⎜F ⎟ ⎝ 2z ⎠

⎛ F2x ⎞ ⎜ ⎟ ⎜ F2y ⎟ = Find ( F2x , F2y , F2z) ⎜F ⎟ ⎝ 2z ⎠

⎛ −17.1 ⎞ ⎜ 8.7 ⎟ lb F2 = ⎜ ⎟ ⎝ −26.2 ⎠

F 2 = 32.4 lb

⎛ α2 ⎞ ⎜ ⎟ ⎛ F2 ⎞ ⎜ β 2 ⎟ = acos ⎜ ⎟ ⎝ F2 ⎠ ⎜γ ⎟ ⎝ 2⎠

⎛ α 2 ⎞ ⎛ 121.8 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 2 ⎟ = ⎜ 74.5 ⎟ deg ⎜ γ ⎟ ⎝ 143.8 ⎠ ⎝ 2⎠

Problem 2-79 Given r1, r2, and r3, determine the magnitude and direction of r = 2r1 − r2 + 3r3. 79

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Engineering Mechanics - Statics

Chapter 2

Given:

⎛3⎞ ⎜ ⎟ r1 = −4 m ⎜ ⎟ ⎝3⎠

⎛4⎞ ⎜ ⎟ r2 = 0 m ⎜ ⎟ ⎝ −5 ⎠

⎛3⎞ ⎜ ⎟ r3 = −2 m ⎜ ⎟ ⎝5⎠

Solution: r = 2r1 − r2 + 3r3

⎛ 11 ⎞ ⎜ ⎟ r = −14 m ⎜ ⎟ ⎝ 26 ⎠ ⎛⎜ α ⎞⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠

r = 31.5 m

⎛⎜ α ⎞⎟ ⎛ 69.6 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 116.4 ⎟ deg ⎜ γ ⎟ ⎝ 34.4 ⎠ ⎝ ⎠

Problem 2-80 Represent the position vector r acting from point A(a, b, c) to point B(d, e, f) in Cartesian vector form. Determine its coordinate direction angles and find the distance between points A and B. Given: a = 3m b = 5m c = 6m d = 5m e = −2 m f = 1m Solution:

⎛d − a⎞ ⎜ ⎟ r = e−b ⎜ ⎟ ⎝ f− c⎠

⎛2⎞ ⎜ ⎟ r = −7 m ⎜ ⎟ ⎝ −5 ⎠

r = 8.8 m

80

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Engineering Mechanics - Statics

Chapter 2

α = acos ⎛⎜

d − a⎞

α = 76.9 deg

β = acos ⎛⎜

e − b⎞

β = 142 deg

γ = acos ⎛⎜

f − c⎞

γ = 124 deg

⎟ ⎝ r ⎠

⎟ ⎝ r ⎠

⎟ ⎝ r ⎠

Problem 2-81 A position vector extends from the origin to point A(a, b, c). Determine the angles α, β, γ which the tail of the vector makes with the x, y, z axes, respectively. Given: a = 2m b = 3m

c = 6m

Solution:

⎛a⎞ r = ⎜b⎟ ⎜ ⎟ ⎝c ⎠

⎛2⎞ r = ⎜3⎟ m ⎜ ⎟ ⎝6⎠

⎛⎜ α ⎞⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 73.4 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 64.6 ⎟ deg ⎜ γ ⎟ ⎝ 31.0 ⎠ ⎝ ⎠

Problem 2-82 Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate direction angles. Given: a = 4m

81

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Engineering Mechanics - Statics

Chapter 2

b = 8m c = 3m d = 4m

Solution:

⎛ −c ⎞ r = ⎜ −d − b ⎟ ⎜ ⎟ ⎝ a ⎠ ⎛⎜ α ⎞⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠

⎛ −3 ⎞ r = ⎜ −12 ⎟ m ⎜ ⎟ ⎝ 4 ⎠

r = 13 m

⎛⎜ α ⎞⎟ ⎛ 103.3 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 157.4 ⎟ deg ⎜ γ ⎟ ⎝ 72.1 ⎠ ⎝ ⎠

Problem 2-83 Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate direction angles. Given: a = 8 ft b = 2 ft c = 5 ft

θ = 30 deg φ = 20 deg Solution:

⎛⎜ −c cos ( φ ) sin ( θ ) ⎞⎟ r = ⎜ a − c cos ( φ ) cos ( θ ) ⎟ ⎜ b + c sin ( φ ) ⎟ ⎝ ⎠

⎛ −2.35 ⎞ r = ⎜ 3.93 ⎟ ft ⎜ ⎟ ⎝ 3.71 ⎠

r = 5.89 ft

82

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Engineering Mechanics - Statics

Chapter 2

⎛⎜ α ⎞⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 113.5 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 48.2 ⎟ deg ⎜ γ ⎟ ⎝ 51 ⎠ ⎝ ⎠

Problem 2-84 Determine the length of the connecting rod AB by first formulating a Cartesian position vector from A to B and then determining its magnitude. Given: b = 16 in a = 5 in

α = 30 deg

Solution:

r =

⎛ a sin ( α ) + b ⎞ ⎜ ⎟ ⎝ −a cos ( α ) ⎠

r=

⎛ 18.5 ⎞ ⎜ ⎟ in ⎝ −4.3 ⎠

r = 19 in

Problem 2-85 Determine the length of member AB of the truss by first establishing a Cartesian position vector from A to B and then determining its magnitude. Given: a = 1.2 m b = 0.8 m c = 0.3 m d = 1.5 m

θ = 40 deg 83

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Engineering Mechanics - Statics

Chapter 2

Solution:

⎛ c + d cot ( θ ) ⎞ ⎜ ⎟ ⎝ d−a ⎠

r =

r=

⎛ 2.09 ⎞ ⎜ ⎟m ⎝ 0.3 ⎠

r = 2.11 m

Problem 2- 86 The positions of point A on the building and point B on the antenna have been measured relative to the electronic distance meter (EDM) at O. Determine the distance between A and B. Hint: Formulate a position vector directed from A to B; then determine its magnitude. Given: a = 460 m b = 653 m

α = 60 deg β = 55 deg θ = 30 deg φ = 40 deg Solution:

rOA

⎛⎜ −a cos ( φ ) sin ( θ ) ⎟⎞ = ⎜ a cos ( φ ) cos ( θ ) ⎟ ⎜ ⎟ a sin ( φ ) ⎝ ⎠

rOB

⎛⎜ −b cos ( β ) sin ( α ) ⎟⎞ = ⎜ −b cos ( β ) cos ( α ) ⎟ ⎜ ⎟ b sin ( β ) ⎝ ⎠

rAB = rOB − rOA

⎛ −148.2 ⎞ ⎜ ⎟ rAB = −492.4 m ⎜ ⎟ ⎝ 239.2 ⎠

rAB = 567.2 m

Problem 2-87 Determine the lengths of cords ACB and CO. The knot at C is located midway between A and B.

84

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Engineering Mechanics - Statics

Chapter 2

Given: a = 3 ft b = 6 ft c = 4 ft

Solution:

rAB

⎛c ⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝ −a ⎠

rAC =

rOA

⎛0⎞ ⎜ ⎟ = 0 ⎜ ⎟ ⎝a⎠

rAB 2

rAB = 7.8 ft rOC = rOA + rAC rOC = 3.91 ft

Problem 2-88 Determine the length of the crankshaft AB by first formulating a Cartesian position vector from A to B and then determining its magnitude.

85

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Engineering Mechanics - Statics

Chapter 2

Given: a = 400 b = 125

θ = 25 deg

Solution:

rAB

⎡a + b sin ( θ ) ⎤ ⎢ ⎥ = −( b cos ( θ ) ) mm ⎢ ⎥ 0 ⎣ ⎦

rAB = 467 mm

Problem 2-89 Determine the length of wires AD, BD, and CD. The ring at D is midway between A and B. Given: a = 0.5 m b = 1.5 m c = 2m d = 2m e = 0.5 m

86

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Engineering Mechanics - Statics

Chapter 2

Solution:

rAD

⎛ −c ⎞ ⎜ 2 ⎟ ⎜ ⎟ d ⎟ ⎜ = ⎜ 2 ⎟ ⎜e b⎟ ⎜ − ⎟ ⎝2 2⎠

⎛ ⎜ ⎜ = ⎜ ⎜ ⎜ ⎜a + ⎝

rCD

rAD

⎛ −1 ⎞ ⎜ 1 ⎟m = ⎜ ⎟ ⎝ −0.5 ⎠

rAD = 1.5 m

rBD = −rAD

rBD = 1.5 m

⎞ ⎟ 2 ⎟ d ⎟ 2 ⎟ b e⎟ − ⎟ 2 2⎠ c

rCD

⎛1⎞ ⎜ ⎟ = 1 m ⎜ ⎟ ⎝1⎠

rCD = 1.7 m

Problem 2-90 Express force F as a Cartesian vector; then determine its coordinate direction angles. Given: F = 600 lb

c = 3 ft

a = 1.5 ft

φ = 60 deg

b = 5 ft

Solution: r = bi + ( a + c sin ( φ ) ) j + ( 0 − c cos ( φ ) ) k r =

b + ( a + c sin ( φ ) ) + ( c cos ( φ ) ) 2

2

2

r=2m

87

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Engineering Mechanics - Statics

Chapter 2

⎡ −( c cos ( φ ) )⎤ ⎥ r ⎣ ⎦

b d = F r

⎛ a + c sin ( φ ) ⎞ e = F⎜ ⎟ r ⎝ ⎠

f = F⎢

d = 452 lb

e = 370 lb

f = −136 lb

F = ( di + ej + fk) lb d α = acos ⎛⎜ ⎞⎟

α = 41.1 deg

e β = acos ⎛⎜ ⎞⎟

β = 51.9 deg

f γ = acos ⎛⎜ ⎞⎟

γ = 103 deg

⎝F⎠

⎝ F⎠

⎝ F⎠

Problem 2-91 Express force F as a Cartesian vector; then determine its coordinate direction angles. Given: a = 1.5 ft b = 5 ft c = 3 ft

θ = 60 deg F = 600 lb

Solution: b ⎛ ⎞ ⎜ ⎟ r = a + c sin ( θ ) ⎜ ⎟ ⎝ −c cos ( θ ) ⎠

F = F

⎛ 452 ⎞ ⎜ ⎟ F = 370 lb ⎜ ⎟ ⎝ −136 ⎠

r r

⎛⎜ α ⎞⎟ ⎛ F ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ F ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 41.1 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 51.9 ⎟ deg ⎜ γ ⎟ ⎝ 103.1 ⎠ ⎝ ⎠

88

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Engineering Mechanics - Statics

Chapter 2

Problem 2-92 Determine the magnitude and coordinate direction angles of the resultant force acting at point A. Given: F 1 = 150 N F 2 = 200 N a = 1.5 m b = 4m c = 3m d = 2m e = 3m

θ = 60 deg Solution: Define the position vectors and then the forces

rAB

⎛ e cos ( θ ) ⎞ = ⎜ a + e sin ( θ ) ⎟ ⎜ ⎟ −b ⎝ ⎠

rAC

⎛ c ⎞ = ⎜a − d⎟ ⎜ ⎟ ⎝ −b ⎠

F 1v = F1

F 2v = F2

⎛ 38 ⎞ F 1v = ⎜ 103.8 ⎟ N ⎜ ⎟ ⎝ −101.4 ⎠

rAB rAB

⎛ 119.4 ⎞ F 2v = ⎜ −19.9 ⎟ N ⎜ ⎟ ⎝ −159.2 ⎠

rAC rAC

Add the forces and find the magnitude of the resultant F R = F1v + F2v

⎛ 157.4 ⎞ F R = ⎜ 83.9 ⎟ N ⎜ ⎟ ⎝ −260.6 ⎠

F R = 316 N

Find the direction cosine angles

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 60.1 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 74.6 ⎟ deg ⎜ γ ⎟ ⎝ 145.6 ⎠ ⎝ ⎠

89

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Engineering Mechanics - Statics

Chapter 2

Problem 2-93 The plate is suspended using the three cables which exert the forces shown. Express each force as a Cartesian vector. Given: F BA = 350 lb F CA = 500 lb F DA = 400 lb a = 3 ft b = 3 ft c = 6 ft d = 14 ft e = 3 ft f = 3 ft g = 2 ft Solution:

rBA

⎛ −e − g ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ d ⎠

rCA

⎛f⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝d⎠

rDA

⎛ −g ⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝d⎠

F BAv = FBA

F CAv = FCA

F DAv = FDA

rBA rBA

rCA rCA

rDA rDA

F BAv

⎛ −109.2 ⎞ ⎜ 131 ⎟ lb = ⎜ ⎟ ⎝ 305.7 ⎠

⎛ 102.5 ⎞ ⎜ ⎟ F CAv = 102.5 lb ⎜ ⎟ ⎝ 478.5 ⎠ ⎛ −52.1 ⎞ ⎜ ⎟ F DAv = −156.2 lb ⎜ ⎟ ⎝ 364.5 ⎠

90

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Engineering Mechanics - Statics

Chapter 2

Problem 2-94 The engine of the lightweight plane is supported by struts that are connected to the space truss that makes up the structure of the plane. The anticipated loading in two of the struts is shown. Express each of these forces as a Cartesian vector. Given: F 1 = 400 lb F 2 = 600 lb a = 0.5 ft b = 0.5 ft c = 3.0 ft d = 2.0 ft e = 0.5 ft f = 3.0 ft Solution:

rCD

⎛c ⎞ ⎜ ⎟ = −b ⎜ ⎟ ⎝a⎠

rAB

⎛ −c ⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝ −e ⎠

F 1v = F1

F 2v = F2

rCD rCD

rAB rAB

⎛ 389.3 ⎞ ⎜ ⎟ F 1v = −64.9 lb ⎜ ⎟ ⎝ 64.9 ⎠ ⎛ −584.0 ⎞ ⎜ 97.3 ⎟ lb F 2v = ⎜ ⎟ ⎝ −97.3 ⎠

Problem 2-95 The window is held open by cable AB. Determine the length of the cable and express the force F acting at A along the cable as a Cartesian vector. 91

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Engineering Mechanics - Statics

Chapter 2

Given: a = 300 mm b = 500 mm c = 150 mm d = 250 mm

θ = 30 deg F = 30 N

Solution:

rAB

⎛ −a cos ( θ ) ⎞ ⎜ c−b ⎟ = ⎜ ⎟ ⎝ d + a sin ( θ ) ⎠

Fv = F

rAB rAB

rAB = 591.6 mm

⎛ −13.2 ⎞ ⎜ ⎟ F v = −17.7 N ⎜ ⎟ ⎝ 20.3 ⎠

Problem 2-96 The force acting on the man, caused by his pulling on the anchor cord, is F. If the length of the cord is L, determine the coordinates A(x, y, z) of the anchor.

92

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Engineering Mechanics - Statics

Chapter 2

Given:

⎛ 40 ⎞ F = ⎜ 20 ⎟ N ⎜ ⎟ ⎝ −50 ⎠ L = 25 m Solution:

r = L

F F

⎛ 14.9 ⎞ r = ⎜ 7.5 ⎟ m ⎜ ⎟ ⎝ −18.6 ⎠

Problem 2-97 Express each of the forces in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 = 80 lb

c = 4 ft

93

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Engineering Mechanics - Statics

Chapter 2

F 2 = 50 lb

d = 2.5 ft

a = 6 ft

e = 12

b = 2 ft

f = 5

Solution:

rAC

⎛⎜ −d ⎞⎟ −c ⎟ = ⎜ ⎜ d⎟ ⎜e f ⎟ ⎝ ⎠

rAB

⎛b⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝ −a ⎠

F R = F1v + F2v

F 1v = F1

F 2v = F2

⎛ −26.2 ⎞ F 1v = ⎜ −41.9 ⎟ lb ⎜ ⎟ ⎝ 62.9 ⎠

rAC rAC

⎛ 6.1 ⎞ F 2v = ⎜ −12.1 ⎟ kg ⎜ ⎟ ⎝ −18.2 ⎠

rAB rAB

⎛ −12.8 ⎞ F R = ⎜ −68.7 ⎟ lb ⎜ ⎟ ⎝ 22.8 ⎠

F R = 73.5 lb

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 1.7 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 2.8 ⎟ ⎜ γ ⎟ ⎝ 1.3 ⎠ ⎝ ⎠

Problem 2-98 The cable attached to the tractor at B exerts force F on the framework. Express this force as a Cartesian vector

94

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Engineering Mechanics - Statics

Chapter 2

Given: F = 350 lb a = 35 ft b = 50 ft

θ = 20 deg

Solution: Find the position vector and then the force

rAB

⎛ b sin ( θ ) ⎞ ⎜ ⎟ = b cos ( θ ) ⎜ ⎟ ⎝ −a ⎠

Fv = F

⎛ 98.1 ⎞ ⎜ ⎟ F v = 269.4 lb ⎜ ⎟ ⎝ −200.7 ⎠

rAB rAB

Problem 2-99 The cable OA exerts force F on point O. If the length of the cable is L, what are the coordinates (x, y, z) of point A? Given:

⎛ 40 ⎞ ⎜ ⎟ F = 60 N ⎜ ⎟ ⎝ 70 ⎠ L = 3m Solution:

r = L

F F

95

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Engineering Mechanics - Statics

Chapter 2

⎛ 1.2 ⎞ r = ⎜ 1.8 ⎟ m ⎜ ⎟ ⎝ 2.1 ⎠

Problem 2-100 Determine the position (x, y, 0) for fixing cable BA so that the resultant of the forces exerted on the pole is directed along its axis, from B toward O, and has magnitude F R. Also, what is the magnitude of force F 3? Given: F 1 = 500 N F 2 = 400 N F R = 1000 N a = 1m b = 2m c = 2m d = 3m

Solution: Initial Guesses F3 = 1 N

x = 1m

y = 1m

Given

⎛0⎞ ⎛ −c ⎞ ⎛a⎞ ⎛x ⎞ F3 F2 ⎛ F1 ⎞ ⎜ ⎟ ⎛ ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟ ⎜ ⎟ FR 0 = ⎜ ⎟ 0 ⎟ + ⎜ 2 2 2 ⎟ ⎜ −b ⎟ + ⎜ 2 2 2 ⎟ ⎜ y ⎟ ⎜ ⎟ 2 2 ⎜ c + d ⎠ ⎝ −d ⎠ ⎝ a + b + d ⎠ ⎝ −d ⎠ ⎝ x + y + d ⎠ ⎝ −d ⎠ ⎝ −1 ⎠ ⎝ ⎛⎜ F3 ⎞⎟ ⎜ x ⎟ = Find ( F3 , x , y) ⎜ y ⎟ ⎝ ⎠

⎛ x ⎞ ⎛ 1.9 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ y ⎠ ⎝ 2.4 ⎠

F 3 = 380 N

96

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Engineering Mechanics - Statics

Chapter 2

Problem 2-101 The cord exerts a force F on the hook. If the cord is length L, determine the location x, y of the point of attachment B, and the height z of the hook. Given:

⎛ 12 ⎞ F = ⎜ 9 ⎟ lb ⎜ ⎟ ⎝ −8 ⎠ L = 8 ft a = 2 ft

Solution: Initial guesses

Given

x = 1 ft

⎛x − a⎞ ⎜ y ⎟=L F ⎜ ⎟ F ⎝ −z ⎠

y = 1 ft

z = 1 ft

⎛x⎞ ⎜ y ⎟ = Find ( x , y , z) ⎜ ⎟ ⎝z⎠

⎛ x ⎞ ⎛ 7.65 ⎞ ⎜ y ⎟ = ⎜ 4.24 ⎟ ft ⎜ ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 3.76 ⎠

Problem 2-102 The cord exerts a force of magnitude F on the hook. If the cord length L, the distance z, and the x component of the force, Fx, are given, determine the location x, y of the point of attachment B of the cord to the ground. Given: F = 30 lb L = 8 ft z = 4 ft F x = 25 lb a = 2 ft

Solution: Guesses x = 1 ft 97

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Engineering Mechanics - Statics

Chapter 2

y = 1 ft

Given Fx =

⎛ x − a ⎞F ⎜ ⎟ ⎝ L ⎠

2

2

2

2

L = ( x − a) + y + z

⎛x⎞ ⎜ ⎟ = Find ( x , y) ⎝ y⎠

⎛ x ⎞ ⎛ 8.67 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ y ⎠ ⎝ 1.89 ⎠

Problem 2-103 Each of the four forces acting at E has magnitude F. Express each force as a Cartesian vector and determine the resultant force. Units used: 3

kN = 10 N Given: F = 28 kN a = 4m b = 6m c = 12 m

Solution: Find the position vectors and then the forces

rEA

⎛b⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝ −c ⎠

F EA = F

rEA rEA

⎛ 12 ⎞ F EA = ⎜ −8 ⎟ kN ⎜ ⎟ ⎝ −24 ⎠ 98

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Engineering Mechanics - Statics

rEB

⎛b⎞ = ⎜a ⎟ ⎜ ⎟ ⎝ −c ⎠

rEC

⎛ −b ⎞ = ⎜ a ⎟ ⎜ ⎟ ⎝ −c ⎠

rED

⎛ −b ⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝ −c ⎠

F EB = F

F EC = F

F ED = F

Chapter 2

rEB rEB

rEC rEC

rED rED

Find the resultant sum F R = FEA + F EB + F EC + FED

⎛ 12 ⎞ F EB = ⎜ 8 ⎟ kN ⎜ ⎟ ⎝ −24 ⎠ ⎛ −12 ⎞ F EC = ⎜ 8 ⎟ kN ⎜ ⎟ ⎝ −24 ⎠ ⎛ −12 ⎞ F ED = ⎜ −8 ⎟ kN ⎜ ⎟ ⎝ −24 ⎠ ⎛ 0 ⎞ F R = ⎜ 0 ⎟ kN ⎜ ⎟ ⎝ −96 ⎠

Problem 2-104 The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles α, β, γ of the resultant force. Units Used: 3

kN = 10 N

Given: x = 20 m

a = 16 m

y = 15 m

b = 18 m

F 1 = 600 N

c = 6m

F 2 = 400 N

d = 4m

F 3 = 800 N

e = 24 m

99

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Engineering Mechanics - Statics

Chapter 2

Solution: Find the position vectors, then the force vectors

rDC

⎛a⎞ = ⎜ −b ⎟ ⎜ ⎟ ⎝ −e ⎠

rDA

⎛x⎞ = ⎜ y ⎟ ⎜ ⎟ ⎝ −e ⎠

rDB

⎛ −c ⎞ = ⎜d ⎟ ⎜ ⎟ ⎝ −e ⎠

F 1v = F1

F 2v = F2

F 3v = F3

⎛ 282.4 ⎞ F 1v = ⎜ −317.6 ⎟ N ⎜ ⎟ ⎝ −423.5 ⎠

rDC rDC

⎛ 230.8 ⎞ F 2v = ⎜ 173.1 ⎟ N ⎜ ⎟ ⎝ −277 ⎠

rDA rDA

⎛ −191.5 ⎞ F 3v = ⎜ 127.7 ⎟ N ⎜ ⎟ ⎝ −766.2 ⎠

rDB rDB

Find the resultant, magnitude, and direction angles

F R = F1v + F2v + F3v

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛ 0.322 ⎞ F R = ⎜ −0.017 ⎟ kN ⎜ ⎟ ⎝ −1.467 ⎠

F R = 1.502 kN

⎛⎜ α ⎞⎟ ⎛ 77.6 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 90.6 ⎟ deg ⎜ γ ⎟ ⎝ 167.6 ⎠ ⎝ ⎠

Problem 2-105 The chandelier is supported by three chains which are concurrent at point O. If the force in each chain has magnitude F, express each force as a Cartesian vector and determine the magnitude and coordinate direction angles of the resultant force.

100

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Engineering Mechanics - Statics

Chapter 2

Given: F = 60 lb a = 6 ft b = 4 ft

θ 1 = 120 deg θ 2 = 120 deg Solution:

θ 3 = 360 deg − θ 1 − θ 2

rOA

⎛ b sin ( θ 1 ) ⎞ ⎜ ⎟ = ⎜ b cos ( θ 1 ) ⎟ ⎜ ⎟ ⎝ −a ⎠

rOB

⎛ b sin ( θ 1 + θ 2 ) ⎞ ⎜ ⎟ = ⎜ b cos ( θ 1 + θ 2 ) ⎟ ⎜ ⎟ −a ⎝ ⎠

rOC

⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝ −a ⎠

FA = F

FB = F

FC = F

rOA rOA

rOB rOB

rOC rOC

FR = FA + FB + FC

⎛ 28.8 ⎞ ⎜ ⎟ F A = −16.6 lb ⎜ ⎟ ⎝ −49.9 ⎠ ⎛ −28.8 ⎞ ⎜ ⎟ F B = −16.6 lb ⎜ ⎟ ⎝ −49.9 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F C = 33.3 lb ⎜ ⎟ ⎝ −49.9 ⎠ F R = 149.8 lb

⎛⎜ α ⎞⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 90 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 90 ⎟ deg ⎜ γ ⎟ ⎝ 180 ⎠ ⎝ ⎠

101

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Engineering Mechanics - Statics

Chapter 2

Problem 2-106 The chandelier is supported by three chains which are concurrent at point O. If the resultant force at O has magnitude F R and is directed along the negative z axis, determine the force in each chain assuming F A = F B = F C = F. Given: a = 6 ft b = 4 ft F R = 130 lb Solution: 2

2

a +b FR 3a

F =

F = 52.1 lb

Problem 2-107 Given the three vectors A, B, and D, show that A⋅ ( B + D) = ( A⋅ B) + ( A⋅ D).

Solution: Since the component of (B + D) is equal to the sum of the components of B and D, then A⋅ ( B + D) = A⋅ B + A⋅ D

(QED)

Also, A⋅ ( B + D) = ( A xi + A yj + Azk) ⎡⎣( B x + Dx) i + ( By + Dy) j + ( Bz + Dz) k⎤⎦ =

Ax ( B x + Dx) + Ay ( B y + Dy) + Az( B z + Dz) 102

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Engineering Mechanics - Statics

Chapter 2

=

( Ax Bx + Ay By + Az Bz) + ( Ax Dx + Ay Dy + Az Dz)

=

( A⋅ B) + ( A⋅ D)

(QED)

Problem 2-108 Cable BC exerts force F on the top of the flagpole. Determine the projection of this force along the z axis of the pole. Given: F = 28 N a = 12 m b = 6m c = 4m Solution:

rBC

⎛b⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝ −a ⎠

Fv = F

⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠

rBC rBC

F z = −Fv k

F z = 24 N

Problem 2-109 Determine the angle θ between the tails of the two vectors.

103

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Engineering Mechanics - Statics

Chapter 2

Given: r1 = 9 m r2 = 6 m

α = 60 deg β = 45 deg γ = 120 deg φ = 30 deg ε = 40 deg Solution: Determine the two position vectors and use the dot product to find the angle

r1v

⎛⎜ sin ( ε ) cos ( φ ) ⎟⎞ = r1 ⎜ −sin ( ε ) sin ( φ ) ⎟ ⎜ cos ( ε ) ⎟ ⎝ ⎠ ⎛ r1v⋅ r2v ⎞ ⎟ ⎝ r1v r2v ⎠

θ = acos ⎜

r2v

⎛⎜ cos ( α ) ⎟⎞ = r2 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

θ = 109.4 deg

Problem 2-110 Determine the magnitude of the projected component of r1 along r2, and the projection of r2 along r1. Given: r1 = 9 m r2 = 6 m

α = 60 deg β = 45 deg γ = 120 deg φ = 30 deg ε = 40 deg

104

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Engineering Mechanics - Statics

Chapter 2

Solution: Write the vectors and unit vectors

r1v

⎛⎜ sin ( ε ) cos ( φ ) ⎟⎞ = r1 ⎜ −sin ( ε ) sin ( φ ) ⎟ ⎜ cos ( ε ) ⎟ ⎝ ⎠

⎛ 5.01 ⎞ r1v = ⎜ −2.89 ⎟ m ⎜ ⎟ ⎝ 6.89 ⎠

r2v

⎛⎜ cos ( α ) ⎟⎞ = r2 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

⎛ 3 ⎞ r2v = ⎜ 4.24 ⎟ m ⎜ ⎟ ⎝ −3 ⎠

u1 =

r1v r1v

u2 =

r2v r2v

⎛ 0.557 ⎞ u1 = ⎜ −0.321 ⎟ ⎜ ⎟ ⎝ 0.766 ⎠

⎛ 0.5 ⎞ u2 = ⎜ 0.707 ⎟ ⎜ ⎟ ⎝ −0.5 ⎠

The magnitude of the projection of r1 along r2.

r1v⋅ u2 = 2.99 m

The magnitude of the projection of r2 along r1.

r2v⋅ u1 = 1.99 m

Problem 2-111 Determine the angles θ and φ between the wire segments. Given: a = 0.6 b = 0.8 c = 0.5 d = 0.2 Solution: e = a−d

105

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Engineering Mechanics - Statics

rBA

⎛0⎞ = ⎜ −e ⎟ m ⎜ ⎟ ⎝ −c ⎠

rCB

⎛ −b ⎞ = ⎜ −d ⎟ m ⎜ ⎟ ⎝c ⎠

Chapter 2

rBC

⎛b⎞ = ⎜ d ⎟ ft ⎜ ⎟ ⎝ −c ⎠

θ = acos ⎜

rCD

⎛ −b ⎞ = ⎜ 0 ⎟ ft ⎜ ⎟ ⎝0⎠

φ = acos ⎜

⎛ rBA⋅ rBC ⎞ ⎟ ⎝ rBA rBC ⎠

θ = 74.0 deg

⎛ rCB⋅ rCD ⎞ ⎟ ⎝ rCB rCD ⎠

φ = 33.9 deg

⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠

θ = 64.6 deg

Problem 2-112 Determine the angle θ between the two cords. Given: a = 3m b = 2m c = 6m d = 3m e = 4m

Solution:

rAC

⎛b⎞ = ⎜ a ⎟ ft ⎜ ⎟ ⎝c⎠

rAB

⎛0⎞ = ⎜ −d ⎟ ft ⎜ ⎟ ⎝e ⎠

θ = acos ⎜

Problem 2-113 Determine the angle θ between the two cables. Given: a = 8 ft b = 10 ft 106

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Engineering Mechanics - Statics

Chapter 2

c = 8 ft d = 10 ft e = 4 ft f = 6 ft F AB = 12 lb

Solution:

rAC

⎛a − f⎞ = ⎜ −c ⎟ ft ⎜ ⎟ ⎝ b ⎠

rAB

⎛ −f ⎞ = ⎜ d − c ⎟ ft ⎜ ⎟ ⎝ e ⎠

⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠

θ = acos ⎜

θ = 82.9 deg

Problem 2-114 Determine the projected component of the force F acting in the direction of cable AC. Express the result as a Cartesian vector. Given: F = 12 lb

107

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Engineering Mechanics - Statics

Chapter 2

a = 8 ft b = 10 ft c = 8 ft d = 10 ft e = 4 ft f = 6 ft

Solution:

rAC

⎛a − f⎞ ⎜ −c ⎟ m = ⎜ ⎟ ⎝ b ⎠

rAB

⎛ −f ⎞ ⎜ ⎟ = d−c ⎜ ⎟ ⎝ e ⎠

F AC = ( F AB⋅ uAC) uAC

uAC =

rAC rAC

F AB = F

F AC

rAB rAB

uAC

⎛ 0.2 ⎞ ⎜ ⎟ = −0.6 ⎜ ⎟ ⎝ 0.8 ⎠

⎛ −9.6 ⎞ ⎜ ⎟ F AB = 3.2 lb ⎜ ⎟ ⎝ 6.4 ⎠

⎛ 0.229 ⎞ ⎜ ⎟ = −0.916 lb ⎜ ⎟ ⎝ 1.145 ⎠

Problem 2-115 Determine the components of F that act along rod AC and perpendicular to it. Point B is located at the midpoint of the rod.

108

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Engineering Mechanics - Statics

Chapter 2

Given: F = 600 N c = 4 m a = 4m

d = 3m

b = 6m

e = 4m

Solution: Find the force vector and the unit vector uAC.

rBD

⎛c + d ⎞ ⎜ 2⎟ ⎜ ⎟ e⎟ ⎜ = b− ⎜ 2⎟ ⎜ −a ⎟ ⎜ ⎟ ⎝ 2 ⎠

rAC

⎛ −d ⎞ ⎜ ⎟ = e ⎜ ⎟ ⎝ −a ⎠

⎛ 5.5 ⎞ ⎜ ⎟m rBD = 4 ⎜ ⎟ ⎝ −2 ⎠

rAC

Fv = F

⎛ −3 ⎞ ⎜ ⎟ = 4 m ⎜ ⎟ ⎝ −4 ⎠

uAC =

rBD rBD

rAC rAC

⎛ 465.5 ⎞ ⎜ ⎟ F v = 338.6 N ⎜ ⎟ ⎝ −169.3 ⎠

uAC

⎛ −0.5 ⎞ ⎜ ⎟ = 0.6 ⎜ ⎟ ⎝ −0.6 ⎠

Now find the component parallel to AC. F parallel = Fv ⋅ uAC

F parallel = 99.1 N

The perpendicular component is now found F perpendicular =

2

Fv ⋅ F v − F parallel

F perpendicular = 591.8 N

Problem 2-116 Determine the components of F that act along rod AC and perpendicular to it. Point B is located a distance f along the rod from end C.

109

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Engineering Mechanics - Statics

Chapter 2

Given: F = 600 N

c = 4m

a = 4m

d = 3m

b = 6m

e = 4m

f = 3m Solution: f

r = 2

2

d +e +a

2

Find the force vector and the unit vector uAC.

rBD

⎡c + d( 1 − r) ⎤ ⎢ ⎥ = b − e( 1 − r) ⎢ ⎥ ⎣ −a r ⎦

Fv = F

rAC

⎛ 5.5944 ⎞ ⎜ ⎟ rBD = 3.8741 m ⎜ ⎟ ⎝ −1.8741 ⎠ ⎛ 475.6 ⎞ ⎜ ⎟ F v = 329.3 N ⎜ ⎟ ⎝ −159.3 ⎠

rBD rBD

⎛ −d ⎞ ⎜ ⎟ = e ⎜ ⎟ ⎝ −a ⎠

rAC

⎛ −3 ⎞ ⎜ ⎟ = 4 m ⎜ ⎟ ⎝ −4 ⎠

uAC =

rAC rAC

uAC

⎛ −0.5 ⎞ ⎜ ⎟ = 0.6 ⎜ ⎟ ⎝ −0.6 ⎠

Now find the component parallel to AC. F parallel = Fv ⋅ uAC

F parallel = 82.4 N

The perpendicular component is now found F perpendicular =

2

Fv ⋅ F v − F parallel

F perpendicular = 594.3 N

Problem 2-117 Determine the magnitude of the projected component of the length of cord OA along the Oa axis.

110

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Engineering Mechanics - Statics

Chapter 2

Given: a = 10 ft b = 5 ft c = 15 ft d = 5 ft

θ 1 = 45 deg θ 2 = 60 deg

Solution:

rOA

⎛ cos ( θ 1) cos ( θ 2 ) ⎞ ⎜ ⎟ = d ⎜ cos ( θ 1 ) sin ( θ 2 ) ⎟ ⎜ ⎟ sin ( θ 1 ) ⎝ ⎠

rOa = rOA ⋅ uOa

rOa

⎛c ⎞ = ⎜ a ⎟ ⎜ ⎟ ⎝ −b ⎠

uOa =

rOa rOa

rOa = 2.1 ft

111

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Engineering Mechanics - Statics

Chapter 2

Problem 2-118 Force F acts at the end of the pipe. Determine the magnitudes of the components F1 and F 2 which are directed along the pipe's axis and perpendicular to it. Given:

a = 5 ft

⎛ 0 ⎞ ⎜ 0 ⎟ lb F = ⎜ ⎟ ⎝ −40 ⎠

b = 3 ft c = 3 ft

Solution:

⎛b⎞ ⎜ ⎟ r = a ⎜ ⎟ ⎝ −c ⎠

u =

F1 = F⋅ u F2 =

F⋅ F − F1

r r

F 1 = 18.3 lb 2

F 2 = 35.6 lb

Problem 2-119 Determine the projected component of the force F acting along the axis AB of the pipe.

112

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Engineering Mechanics - Statics

Chapter 2

Given: F = 80 N a = 4m b = 3m c = 12 m d = 2m e = 6m

Solution: Find the force and the unit vector

⎛ −e ⎞ ⎜ ⎟ rA = −a − b ⎜ ⎟ ⎝ d−c ⎠

rAB

⎛ −e ⎞ ⎜ ⎟ = −b ⎜ ⎟ ⎝d⎠

⎛ −6 ⎞ ⎜ ⎟ rA = −7 m ⎜ ⎟ ⎝ −10 ⎠ ⎛ −6 ⎞ ⎜ ⎟ rAB = −3 m ⎜ ⎟ ⎝2⎠

Fv = F

uAB =

rA rA

rAB rAB

⎛ −35.3 ⎞ ⎜ ⎟ F v = −41.2 N ⎜ ⎟ ⎝ −58.8 ⎠ ⎛ −0.9 ⎞ ⎜ ⎟ uAB = −0.4 ⎜ ⎟ ⎝ 0.3 ⎠

Now find the projection using the Dot product. F AB = F v ⋅ uAB

F AB = 31.1 N

Problem 2-120 Determine the angles θ and φ between the axis OA of the pole and each cable, AB and AC. Given: F 1 = 50 N

113

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Engineering Mechanics - Statics

Chapter 2

F 2 = 35 N a = 1m b = 3m c = 2m d = 5m e = 4m f = 6m g = 4m Solution:

rAO

⎛0⎞ ⎜ ⎟ = −g ⎜ ⎟ ⎝−f ⎠

rAB

⎛e⎞ ⎜ ⎟ = a ⎜ ⎟ ⎝−f⎠

rAC

⎛ rAO⋅ rAB ⎞ ⎟ ⎝ rAO rAB ⎠

θ = 52.4 deg

⎛ rAO⋅ rAC ⎞ ⎟ ⎝ rAO rAC ⎠

φ = 68.2 deg

θ = acos ⎜

φ = acos ⎜

⎛ −c ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ −f ⎠

Problem 2-121 The two cables exert the forces shown on the pole. Determine the magnitude of the projected component of each force acting along the axis OA of the pole. Given: F 1 = 50 N F 2 = 35 N a = 1m b = 3m c = 2m 114

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Engineering Mechanics - Statics

Chapter 2

d = 5m e = 4m f = 6m g = 4m Solution:

rAB

⎛e⎞ ⎜ ⎟ = a ⎜ ⎟ ⎝−f⎠

F 1v = F1

F 2v = F2

rAC

rAC rAC rAB rAB

⎛ −c ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ −f ⎠

rAO

⎛0⎞ ⎜ ⎟ = −g ⎜ ⎟ ⎝−f ⎠

F 1AO = F 1v⋅ uAO

F 1AO = 18.5 N

F 2AO = F 2v⋅ uAO

F 2AO = 21.3 N

uAO =

rAO rAO

Problem 2-122 Force F is applied to the handle of the wrench. Determine the angle θ between the tail of the force and the handle AB.

115

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Engineering Mechanics - Statics

Chapter 2

Given: a = 300 mm b = 500 mm F = 80 N

θ 1 = 30 deg θ 2 = 45 deg Solution:

⎛ −cos ( θ 1) sin ( θ 2) ⎞ ⎜ ⎟ F v = F ⎜ cos ( θ 1 ) cos ( θ 2 ) ⎟ ⎜ ⎟ sin ( θ 1 ) ⎝ ⎠ ⎛ Fv⋅ uab ⎞ ⎟ ⎝ F ⎠

θ = acos ⎜

uab

⎛0⎞ ⎜ ⎟ = −1 ⎜ ⎟ ⎝0⎠

θ = 127.8 deg

Problem 2-123 Two cables exert forces on the pipe. Determine the magnitude of the projected component of F 1 along the line of action of F2. Given: F 1 = 30 lb

β = 30 deg

F 2 = 25 lb

γ = 60 deg

α = 30 deg

ε = 60 deg

Solution: We first need to find the third angle ( > 90 deg) that locates force F 2. Initial Guess:

φ = 120 deg

116

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Engineering Mechanics - Statics

Chapter 2

Given cos ( ε ) + cos ( γ ) + cos ( φ ) = 1 2

2

φ = Find ( φ )

2

φ = 135 deg

Find the force F 1v and the unit vector u2.

F 1v

⎛⎜ cos ( α ) sin ( β ) ⎞⎟ = F1 ⎜ cos ( α ) cos ( β ) ⎟ ⎜ −sin ( α ) ⎟ ⎝ ⎠

⎛⎜ cos ( φ ) ⎞⎟ u2 = ⎜ cos ( ε ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠ Now find the projection

⎛ 13 ⎞ F 1v = ⎜ 22.5 ⎟ lb ⎜ ⎟ ⎝ −15 ⎠ ⎛ −0.7 ⎞ u2 = ⎜ 0.5 ⎟ ⎜ ⎟ ⎝ 0.5 ⎠

F 12 = F1v⋅ u2

F 12 = 5.4 lb

Problem 2-124 Determine the angle θ between the two cables attached to the pipe. Given: F 1 = 30 lb

β = 30 deg

F 2 = 25 lb

γ = 60 deg

α = 30 deg

ε = 60 deg

Solution: We first need to find the third angle ( > 90 deg) that locates force F 2. Initial Guesses:

φ = 120 deg

117

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Engineering Mechanics - Statics

Chapter 2

Given cos ( ε ) + cos ( γ ) + cos ( φ ) = 1 2

2

2

φ = Find ( φ )

φ = 135 deg

Find the unit vectors u1 and u2.

⎛⎜ cos ( α ) sin ( β ) ⎞⎟ u1 = ⎜ cos ( α ) cos ( β ) ⎟ ⎜ −sin ( α ) ⎟ ⎝ ⎠

⎛ 0.4 ⎞ u1 = ⎜ 0.8 ⎟ ⎜ ⎟ ⎝ −0.5 ⎠

⎛⎜ cos ( φ ) ⎞⎟ u2 = ⎜ cos ( ε ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

⎛ −0.7 ⎞ u2 = ⎜ 0.5 ⎟ ⎜ ⎟ ⎝ 0.5 ⎠

Find the angle using the dot product

θ = acos ( u1 ⋅ u2 )

θ = 100.4 deg

Problem 2-125 Determine the angle θ between the two cables. Given: a = 7.5 ft b = 2 ft c = 3 ft d = 2 ft e = 3 ft f = 3 ft F 1 = 60 lb F 2 = 30 lb

118

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Engineering Mechanics - Statics

Chapter 2

Solution:

rAC

⎛ −d − b ⎞ = ⎜ −c − f ⎟ ⎜ ⎟ ⎝ a−e ⎠

rAB

⎛ −d = ⎜ −c − ⎜ ⎝ −e

⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠

θ = acos ⎜

⎞ f⎟ ⎟ ⎠

θ = 59.2 deg

Problem 2-126 Determine the projection of the force F 1 along cable AB. Determine the projection of the force F 2 along cable AC. Given: a = 7.5 ft b = 2 ft c = 3 ft d = 2 ft e = 3 ft f = 3 ft F 1 = 60 lb F 2 = 30 lb Solution:

rAC

⎛ −d − b ⎞ = ⎜ −c − f ⎟ ⎜ ⎟ ⎝ a−e ⎠

rAB

⎛ −d = ⎜ −c − ⎜ ⎝ −e

⎞ ⎟ f ⎟ ⎠

uAC =

rAC rAC

F 1v = F1 uAC

F 1AB = F1v⋅ uAB

F 1AB = 30.8 lb

F 2v = F2 uAB

F 2AC = F2v⋅ uAC

F 2AC = 15.4 lb

uAB =

rAB rAB

119

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Engineering Mechanics - Statics

Chapter 2

Problem 2-127 Determine the angle θ between the edges of the sheet-metal bracket. Given: a = 50 mm b = 300 mm c = 250 mm d = 400 mm Solution: Find the unit vectors and use the dot product

⎛d⎞ ⎜ ⎟ r1 = 0 ⎜ ⎟ ⎝c⎠

u1 =

⎛a⎞ ⎜ ⎟ r2 = b ⎜ ⎟ ⎝0⎠

u2 =

r1 r1

r2 r2

⎛ 0.848 ⎞ ⎛ 0.164 ⎞ ⎜ ⎟ ⎜ ⎟ u1 = 0.000 u2 = 0.986 ⎜ ⎟ ⎜ ⎟ ⎝ 0.530 ⎠ ⎝ 0.000 ⎠

θ = acos ( u1 ⋅ u2 )

θ = 82 deg

Problem 2-128 Determine the magnitude of the projected component of the force F acting along the axis BC of the pipe.

120

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Engineering Mechanics - Statics

Chapter 2

Given: F = 100 lb a = 2 ft b = 8 ft c = 6 ft d = 4 ft e = 2 ft

Solution:

rCD

⎛ −c ⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝e⎠

uCD =

F BC = ( FuCD) ⋅ uCB

rCD

rCB

rCD

⎛ −c ⎞ ⎜ ⎟ = −d ⎜ ⎟ ⎝e ⎠

uCB =

rCB rCB

F BC = 10.5 lb

Problem 2-129 Determine the angle θ between pipe segments BA and BC. Given: F = 100 lb a = 3 ft b = 8 ft c = 6 ft d = 4 ft e = 2 ft Solution:

rBC

⎛c⎞ ⎜ ⎟ = d ⎜ ⎟ ⎝ −e ⎠

rBA

⎛ −a ⎞ ⎜ ⎟ = 0 ⎜ ⎟ ⎝0⎠

⎛ rBC⋅ rBA ⎞ ⎟ ⎝ rBC rBA ⎠

θ = acos ⎜

θ = 143.3 deg

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Engineering Mechanics - Statics

Chapter 2

Problem 2-130 Determine the angles θ and φ made between the axes OA of the flag pole and AB and AC, respectively, of each cable. Given: F B = 55 N

c = 2m

F c = 40 N

d = 4m

a = 6m

e = 4m

b = 1.5 m

f = 3m

Solution:

rAO

⎛0⎞ ⎜ ⎟ = −e ⎜ ⎟ ⎝−f⎠

rAB

⎛ b ⎞ ⎜ −e ⎟ = ⎜ ⎟ ⎝a − f⎠

rAC

⎛ −c ⎞ ⎜ −e ⎟ = ⎜ ⎟ ⎝d − f⎠

⎛ rAB⋅ rAO ⎞ ⎟ ⎝ rAB rAO ⎠

θ = 74.4 deg

⎛ rAC⋅ rAO ⎞ ⎟ ⎝ rAC rAO ⎠

φ = 55.4 deg

θ = acos ⎜ φ = acos ⎜

Problem 2-131 Determine the magnitude and coordinate direction angles of F3 so that resultant of the three forces acts along the positive y axis and has magnitude FR.

122

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Engineering Mechanics - Statics

Chapter 2

Given: F R = 600 lb F 1 = 180 lb F 2 = 300 lb

φ = 40 deg θ = 30 deg

Solution: The initial guesses: F 3 = 100 lb

β = 30 deg

α = 10 deg

γ = 60 deg

Given F Rx = ΣF x;

−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0

F Ry = ΣF y;

F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = FR

F Rz = ΣF z;

−F 2 sin ( θ ) + F3 cos ( γ ) = 0 cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

2

Solving:

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

⎛⎜ α ⎞⎟ ⎛ 88.3 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 20.6 ⎟ deg ⎜ γ ⎟ ⎝ 69.5 ⎠ ⎝ ⎠

F 3 = 428.3 lb

123

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Engineering Mechanics - Statics

Chapter 2

Problem 2-132 Determine the magnitude and coordinate direction angles of F3 so that resultant of the three forces is zero. Given: F 1 = 180 lb F 2 = 300 lb

φ = 40 deg θ = 30 deg

Solution: The initial guesses:

α = 10 deg β = 30 deg γ = 60 deg F 3 = 100 lb

Given F Rx = ΣF x;

−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0

F Ry = ΣF y;

F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = 0

F Rz = ΣF z;

−F 2 sin ( θ ) + F3 cos ( γ ) = 0 cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

2

Solving:

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

⎛⎜ α ⎞⎟ ⎛ 87 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 142.9 ⎟ deg ⎜ γ ⎟ ⎝ 53.1 ⎠ ⎝ ⎠

F 3 = 249.6 lb

124

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Engineering Mechanics - Statics

Chapter 2

Problem 2-133 Resolve the force F into two components, one acting parallel and the other acting perpendicular to the u axis. Given: F = 600 lb

θ 1 = 60 deg θ 2 = 20 deg Solution: F perpendicular = F cos ( θ 1 − θ 2 ) F perpendicular = 460 lb F parallel = F sin ( θ 1 − θ 2 ) F parallel = 386 lb

Problem 2-134 The force F has a magnitude F and acts at the midpoint C of the thin rod. Express the force as a Cartesian vector.

125

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Engineering Mechanics - Statics

Chapter 2

Given: F = 80 lb a = 2 ft b = 3 ft c = 6 ft Solution:

rCO

Fv = F

⎛ −b ⎞ ⎜ 2 ⎟ ⎜ ⎟ a ⎟ = ⎜ ⎜ 2 ⎟ ⎜ −c ⎟ ⎜ ⎟ ⎝ 2 ⎠

rCO rCO

⎛ −34.3 ⎞ ⎜ ⎟ F v = 22.9 lb ⎜ ⎟ ⎝ −68.6 ⎠

Problem 2-135 Determine the magnitude and direction of the resultant FR = F 1 + F2 + F 3 of the three forces by first finding the resultant F' = F1 + F 3 and then forming FR = F ' + F 2. Specify its direction measured counterclockwise from the positive x axis.

126

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Engineering Mechanics - Statics

Chapter 2

Given: F 1 = 80 N F 2 = 75 N F 3 = 50 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg

Solution:

⎛ −sin ( θ 1 ) ⎞ ⎟ ⎝ cos ( θ 1) ⎠

F 1v = F1 ⎜

⎛ cos ( θ 2 + θ 3 ) ⎞ ⎟ ⎝ sin ( θ 2 + θ 3 ) ⎠

⎛ cos ( θ 3) ⎞ ⎟ ⎝ sin ( θ 3) ⎠

F 2v = F2 ⎜

⎛ −4.6 ⎞ ⎜ ⎟N ⎝ 104.6 ⎠

F' = F1v + F3v

F' =

F R = F' + F2v

FR =

i =

⎛ 14.8 ⎞ ⎜ ⎟N ⎝ 177.1 ⎠

F 3v = F3 ⎜

⎛1⎞ ⎜ ⎟ ⎝0⎠

j =

⎛0⎞ ⎜ ⎟ ⎝1⎠

F R = 177.7 N

⎛ FR j ⎞ ⎟ ⎝ FR i ⎠

θ = atan ⎜

θ = 85.2 deg

Problem 2-136 The leg is held in position by the quadriceps AB, which is attached to the pelvis at A. If the force exerted on this muscle by the pelvis is F , in the direction shown, determine the stabilizing force component acting along the positive y axis and the supporting force component acting along the negative x axis. Given: F = 85 N

127

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Engineering Mechanics - Statics

Chapter 2

θ 1 = 55 deg θ 2 = 45 deg

Solution: F x = F cos ( θ 1 − θ 2 ) F x = 83.7 N F y = F sin ( θ 1 − θ 2 ) F y = 14.8 N

Problem 2-137 Determine the magnitudes of the projected components of the force F in the direction of the cables AB and AC . Given:

⎛ 60 ⎞ ⎜ ⎟ F = 12 N ⎜ ⎟ ⎝ −40 ⎠ a = 3m b = 1.5 m c = 1m d = 0.75 m e = 1m

128

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Engineering Mechanics - Statics

Chapter 2

Solution: Find the unit vectors, then use the dot product

rAB

rAC

⎛ −a ⎞ ⎜ ⎟ = −d ⎜ ⎟ ⎝e ⎠ ⎛ −a ⎞ ⎜ ⎟ = c ⎜ ⎟ ⎝b⎠

⎛ −3 ⎞ ⎜ ⎟ rAB = −0.8 m ⎜ ⎟ ⎝ 1 ⎠ ⎛ −3 ⎞ ⎜ ⎟ rAC = 1 m ⎜ ⎟ ⎝ 1.5 ⎠ ⎛ −78.5 ⎞ ⎜ ⎟ F AB = −19.6 N ⎜ ⎟ ⎝ 26.2 ⎠

F AB = FuAB

uAB =

uAC =

rAB rAB rAC rAC

F AC = FuAC

⎛ −0.9 ⎞ ⎜ ⎟ uAB = −0.2 ⎜ ⎟ ⎝ 0.3 ⎠ ⎛ −0.9 ⎞ ⎜ ⎟ uAC = 0.3 ⎜ ⎟ ⎝ 0.4 ⎠ F AC

⎛ −72.9 ⎞ ⎜ ⎟ = 24.3 N ⎜ ⎟ ⎝ 36.4 ⎠

Problem 2-138 Determine the magnitude and coordinate direction angles of F3 so that resultant of the three forces is zero. Given: F 1 = 180 lb

φ = 40 deg

F 2 = 300 lb

θ = 30 deg

Solution: The initial guesses:

α = 10 deg β = 30 deg γ = 60 deg F 3 = 100 lb

Given F Rx = ΣFx;

−F 1 + F 2 cos ( θ ) sin ( φ ) + F3 cos ( α ) = 0

F Ry = ΣF y;

F 2 cos ( θ ) cos ( φ ) + F 3 cos ( β ) = 0 129

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Engineering Mechanics - Statics

Chapter 2

−F 2 sin ( θ ) + F3 cos ( γ ) = 0

F Rz = ΣF z;

cos ( α ) + cos ( β ) + cos ( γ ) = 1 2

2

2

Solving:

⎛ F3 ⎞ ⎜ ⎟ ⎜ α ⎟ = Find ( F , α , β , γ ) 3 ⎜β ⎟ ⎜ ⎟ ⎝γ ⎠

⎛⎜ α ⎞⎟ ⎛ 87 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 142.9 ⎟ deg ⎜ γ ⎟ ⎝ 53.1 ⎠ ⎝ ⎠

F 3 = 249.6 lb

Problem 2-139 Determine the angles θ and φ so that the resultant force is directed along the positive x axis and has magnitude FR. . Given: F 1 = 30 lb F 2 = 30 lb F R = 20 lb Solution: Initial Guesses:

θ = 20 deg

φ = 20 deg

Given F1

sin ( φ )

=

F2

sin ( θ )

F R = F 1 + F2 − 2 F 1 F2 cos ( 180 deg − θ − φ ) 2

2

2

130

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Engineering Mechanics - Statics

Chapter 2

⎛θ⎞ ⎜ ⎟ = Find ( θ , φ ) ⎝φ⎠ θ = 70.5 deg φ = 70.5 deg

Problem 2-140 Determine the magnitude of the resultant force and its direction measured counterclockwise from the x axis. Given: F 1 = 300 lb F 2 = 200 lb

θ 1 = 40 deg θ 2 = 100 deg

Solution: F Rx = F 1 cos ( 180 deg − θ 2 ) + F2 cos ( θ 1 )

F Rx = 205.3 lb

F Ry = F 1 sin ( 180 deg − θ 2 ) − F 2 sin ( θ 1 )

F Ry = 166.9 lb

FR =

2

FRx + FRy

2

F R = 265 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 39.1 deg

131

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Engineering Mechanics - Statics

Chapter 3

Problem 3-1 Determine the magnitudes of F1 and F2 so that the particle is in equilibrium. Given: F = 500 N

θ 1 = 45 deg θ 2 = 30deg Solution: Initial Guesses F 1 = 1N

F 2 = 1N

Given + ΣF x = 0; →

F 1 cos ( θ 1 ) + F2 cos ( θ 2 ) − F = 0

+

F 1 sin ( θ 1 ) − F 2 sin ( θ 2 ) = 0

↑ ΣFy = 0;

⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , F2) ⎝ F2 ⎠

⎛ F1 ⎞ ⎛ 259 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ F2 ⎠ ⎝ 366 ⎠

Problem 3-2 Determine the magnitude and direction θ of F so that the particle is in equilibrium. Units Used: 3

kN = 10 N Given: F 1 = 7 kN F 2 = 3 kN c = 4 d = 3 132

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Engineering Mechanics - Statics

Chapter 3

Solution: F = 1kN

The initial guesses:

θ = 30deg

Given Equations of equilibrium: + Σ F x = 0; →

+

↑

Σ F y = 0;

⎛ −d ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ c +d ⎠ c ⎛ ⎞ ⎜ 2 2 ⎟ F1 − F2 − F sin ( θ ) = 0 ⎝ c +d ⎠

⎛F⎞ ⎜ ⎟ = Find ( F , θ ) ⎝θ ⎠

F = 4.94 kN

θ = 31.8 deg

Problem 3-3 Determine the magnitude of F and the orientation θ of the force F 3 so that the particle is in equilibrium. Given: F 1 = 700 N F 2 = 450 N F 3 = 750 N

θ 1 = 15 deg θ 2 = 30 deg

Solution: Initial Guesses:

F = 1N

θ = 10deg

Given + ΣF x = 0; →

F 1 cos ( θ 1 ) − F2 sin ( θ 2 ) − F3 cos ( θ ) = 0 133

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Engineering Mechanics - Statics

+

↑ ΣFy = 0;

Chapter 3

F + F2 cos ( θ 2 ) + F 1 sin ( θ 1 ) − F 3 sin ( θ ) = 0

⎛F⎞ ⎜ ⎟ = Find ( F , θ ) ⎝θ ⎠

F = 28.25 N

θ = 53.02 deg

Problem 3-4 Determine the magnitude and angle θ of F so that the particle is in equilibrium. Units Used: 3

kN = 10 N Given: F 1 = 4.5 kN F 2 = 7.5 kN F 3 = 2.25 kN

α = 60 deg φ = 30 deg

Solution: Guesses: F = 1 kN

θ = 1

Given Equations of Equilibrium: + → +

Σ F x = 0;

↑ Σ F y = 0;

⎛F⎞ ⎜ ⎟ = Find ( F , θ ) ⎝θ ⎠

F cos ( θ ) − F2 sin ( φ ) − F 1 + F 3 cos ( α ) = 0 F sin ( θ ) − F 2 cos ( φ ) − F 3 sin ( α ) = 0 F = 11.05 kN

θ = 49.84 deg

134

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Engineering Mechanics - Statics

Chapter 3

Problem 3-5 The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium. Units Used: 3

kN = 10 N Given: F 1 = 8 kN F 2 = 5 kN

θ 1 = 45 deg θ = 30 deg

Solution: + ΣF x = 0; →

−T cos ( θ ) + F 1 + F 2 sin ( θ 1 ) = 0

T =

F1 + F2 sin ( θ 1 ) cos ( θ )

T = 13.3 kN +

↑ ΣFy = 0;

F − T sin ( θ ) − F 2 cos ( θ 1 ) = 0 F = T sin ( θ ) + F2 cos ( θ 1 ) F = 10.2 kN

Problem 3-6 The gusset plate is subjected to the forces of four members. Determine the force in member B and its proper orientation θ for equilibrium. The forces are concurrent at point O. Units Used: 3

kN = 10 N 135

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Engineering Mechanics - Statics

Chapter 3

Given: F = 12 kN F 1 = 8 kN F 2 = 5 kN

θ 1 = 45 deg Solution: Initial Guesses

T = 1kN

θ = 10deg

Given + ΣF x = 0; →

F 1 − T cos ( θ ) + F 2 sin ( θ 1 ) = 0

+

−T sin ( θ ) − F2 cos ( θ 1 ) + F = 0

↑ ΣFy = 0;

⎛T⎞ ⎜ ⎟ = Find ( T , θ ) ⎝θ⎠

T = 14.31 kN

θ = 36.27 deg

Problem 3-7 Determine the maximum weight of the engine that can be supported without exceeding a tension of T1 in chain AB and T2 in chain AC. Given:

θ = 30 deg T1 = 450 lb T2 = 480 lb Solution: Initial Guesses F AB = T1 F AC = T2 W = 1lb 136

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Engineering Mechanics - Statics

Chapter 3

Given Assuming cable AB reaches the maximum tension F AB = T1. + ΣF x = 0; →

F AC cos ( θ ) − F AB = 0

+

F AC sin ( θ ) − W = 0

↑ ΣFy = 0;

⎛ FAC1 ⎞ ⎜ ⎟ = Find ( FAC , W) ⎝ W1 ⎠ Given

W1 = 259.81 lb

Assuming cable AC reaches the maximum tension FAC = T2.

+ ΣF x = 0; →

F AC cos ( θ ) − F AB = 0

+

F AC sin ( θ ) − W = 0

↑ ΣFy = 0;

⎛ FAB2 ⎞ ⎜ ⎟ = Find ( FAB , W) ⎝ W2 ⎠ W = min ( W1 , W2 )

W2 = 240.00 lb

W = 240.00 lb

Problem 3-8 The engine of mass M is suspended from a vertical chain at A. A second chain is wrapped around the engine and held in position by the spreader bar BC. Determine the compressive force acting along the axis of the bar and the tension forces in segments BA and CA of the chain. Hint: Analyze equilibrium first at A, then at B. Units Used: 3

kN = 10 N Given: M = 200 kg

θ 1 = 55 deg g = 9.81

m 2

s

137

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Engineering Mechanics - Statics

Chapter 3

Solution: Initial guesses: Given

F BA = 1 kN

F CA = 2 kN

Point A

+ ΣF x = 0; →

F CA cos ( θ 1 ) − FBA cos ( θ 1 ) = 0

+

M( g) − FCA sin ( θ 1 ) − FBA sin ( θ 1 ) = 0

↑ ΣFy = 0;

⎛ FBA ⎞ ⎜ ⎟ = Find ( FBA , FCA) ⎝ FCA ⎠

⎛ FBA ⎞ ⎛ 1.20 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ FCA ⎠ ⎝ 1.20 ⎠

At point B: + ΣF x = 0; →

F BA cos ( θ 1 ) − F BC = 0 F BC = FBA cos ( θ 1 )

F BC = 687 N

Problem 3-9 Cords AB and AC can each sustain a maximum tension T. If the drum has weight W, determine the smallest angle θ at which they can be attached to the drum. Given: T = 800 lb W = 900 lb Solution: +

↑ Σ Fy = 0;

W − 2T sin ( θ ) = 0 W⎞ ⎟ ⎝ 2T ⎠

θ = asin ⎛⎜

θ = 34.2 deg

138

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Engineering Mechanics - Statics

Chapter 3

Problem 3-10 The crate of weight W is hoisted using the ropes AB and AC. Each rope can withstand a maximum tension T before it breaks. If AB always remains horizontal, determine the smallest angle θ to which the crate can be hoisted. Given: W = 500 lb T = 2500 lb Solution: TAB = T

Case 1: Assume The initial guess

θ = 30 deg

TAC = 2000 lb

Given + Σ F x = 0; →

TAB − TAC cos ( θ ) = 0

+

TAC sin ( θ ) − W = 0

↑ Σ F y = 0;

⎛ TAC1 ⎞ ⎜ ⎟ = Find ( TAC , θ ) ⎝ θ1 ⎠

θ 1 = 11.31 deg

TAC1 = 2550 lb

TAC = T

Case 1: Assume The initial guess

θ = 30 deg

TAB = 2000 lb

Given + Σ F x = 0; →

TAB − TAC cos ( θ ) = 0

+

TAC sin ( θ ) − W = 0

↑ Σ F y = 0;

⎛ TAB2 ⎞ ⎜ ⎟ = Find ( TAB , θ ) ⎝ θ2 ⎠ θ = max ( θ 1 , θ 2 )

θ 2 = 11.54 deg

TAB2 = 2449 lb

θ = 11.54 deg

139

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Engineering Mechanics - Statics

Chapter 3

Problem 3-11 Two electrically charged pith balls, each having mass M, are suspended from light threads of equal length. Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured distance between them is r. Given: M = 0.2 gm r = 200 mm l = 150 mm d = 50 mm Solution: The initial guesses : T = 200 N

F = 200 N

Given +

↑

Σ F x = 0;

⎛ r − d⎞ = 0 ⎟ ⎝ 2l ⎠

F − T⎜

2 ⎡⎢ 2 r − d ⎞ ⎥⎤ ⎛ ⎢ l − ⎜⎝ 2 ⎟⎠ ⎥ + → Σ Fy = 0; T⎢⎣ ⎥ − Mg = 0 l ⎦

⎛T ⎞ ⎜ ⎟ = Find ( T , F) ⎝F⎠

F = 1.13 × 10

−3

T = 0.00 N

N

Problem 3-12 The towing pendant AB is subjected to the force F which is developed from a tugboat. Determine the force that is in each of the bridles, BC and BD, if the ship is moving forward with constant velocity. Units Used: 3

kN = 10 N

140

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Engineering Mechanics - Statics

Chapter 3

Given: F = 50 kN

θ 1 = 20 deg θ 2 = 30 deg Solution: Initial guesses:

TBC = 1 kN

TBD = 2 kN

Given + ΣF x = 0; →

TBC sin ( θ 2 ) − TBD sin ( θ 1 ) = 0

+

TBC cos ( θ 2 ) + TBD cos ( θ 1 ) − F = 0

↑ ΣFy = 0;

⎛ TBC ⎞ ⎜ ⎟ = Find ( TBC , TBD) ⎝ TBD ⎠

⎛ TBC ⎞ ⎛ 22.32 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ TBD ⎠ ⎝ 32.64 ⎠

Problem 3-13 Determine the stretch in each spring for equilibrium of the block of mass M. The springs are shown in the equilibrium position.

141

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Engineering Mechanics - Statics

Chapter 3

Given: M = 2 kg a = 3m b = 3m c = 4m kAB = 30

N

kAC = 20

N

kAD = 40

N

g = 9.81

m

m

m m 2

s Solution:

The initial guesses: F AB = 1 N F AC = 1 N Given +

→

Σ F x = 0;

+

Σ F y = 0;

↑

F AB⎛

c b ⎛ ⎞ ⎞ ⎜ 2 2 ⎟ − FAC⎜ 2 2 ⎟ = 0 ⎝ a +c ⎠ ⎝ a +b ⎠

F AC⎛

a ⎛ a ⎞ ⎞ ⎜ 2 2 ⎟ + FAB⎜ 2 2 ⎟ − M g = 0 ⎝ a +b ⎠ ⎝ a +c ⎠

⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , FAB) ⎝ FAB ⎠ xAC =

xAB =

F AC kAC FAB kAB

⎛ FAC ⎞ ⎛ 15.86 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FAB ⎠ ⎝ 14.01 ⎠

xAC = 0.79 m

xAB = 0.47 m 142

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Engineering Mechanics - Statics

Chapter 3

Problem 3-14 The unstretched length of spring AB is δ. If the block is held in the equilibrium position shown, determine the mass of the block at D. Given:

δ = 2m a = 3m b = 3m c = 4m kAB = 30

N

kAC = 20

N

kAD = 40

N

g = 9.81

m

m

m m 2

s Solution:

F AB = kAB

(

2

2

a +c −δ

)

The initial guesses: mD = 1 kg

F AC = 1 N

Given + Σ F x = 0; →

+

↑

Σ F y = 0;

F AB⎛

c b ⎛ ⎞ ⎞ ⎜ 2 2 ⎟ − FAC⎜ 2 2 ⎟ = 0 ⎝ a +c ⎠ ⎝ a +b ⎠

F AC⎛

⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , mD) ⎝ mD ⎠

a ⎛ a ⎞ ⎞ ⎜ 2 2 ⎟ + FAB⎜ 2 2 ⎟ − mD g = 0 ⎝ a +b ⎠ ⎝ a +c ⎠ F AC = 101.8 N

mD = 12.8 kg

143

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Engineering Mechanics - Statics

Chapter 3

Problem 3-15 The springs AB and BC have stiffness k and unstretched lengths l/2. Determine the horizontal force F applied to the cord which is attached to the small pulley B so that the displacement of the pulley from the wall is d. Given: l = 6m k = 500

N m

d = 1.5 m

Solution:

⎡

2

⎛ l ⎞ + d2 − ⎜ ⎟ ⎣ ⎝ 2⎠

T = k⎢

l⎤

⎥

T = 177.05 N

2⎦

+ Σ F x = 0; →

d 2

d +

F =

d

( 2T)

( 2T) − F = 0

⎛l⎞ ⎜ ⎟ ⎝ 2⎠

2

F = 158.36 N

2

⎛ l ⎞ + d2 ⎜ ⎟ ⎝ 2⎠

Problem 3-16 The springs AB and BC have stiffness k and an unstretched length of l. Determine the displacement d of the cord from the wall when a force F is applied to the cord.

144

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Engineering Mechanics - Statics

Chapter 3

Given: l = 6m k = 500

N m

F = 175 N Solution: The initial guesses: d = 1m

T = 1N

Given + Σ F x = 0; →

d

−F + ( 2T) 2

d +

⎡

Spring

⎛T⎞ ⎜ ⎟ = Find ( T , d) ⎝d ⎠

T = k⎢ d +

⎣

2

=0

⎛l⎞ ⎜ ⎟ ⎝ 2⎠

2

2 ⎤ ⎛ l ⎞ − l⎥ ⎜ ⎟ 2⎦ ⎝ 2⎠

T = 189.96 N

d = 1.56 m

Problem 3-17 Determine the force in each cable and the force F needed to hold the lamp of mass M in the position shown. Hint: First analyze the equilibrium at B; then, using the result for the force in BC, analyze the equilibrium at C. Given: M = 4 kg

θ 1 = 30 deg θ 2 = 60 deg θ 3 = 30 deg Solution: Initial guesses: 145

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Engineering Mechanics - Statics

TBC = 1 N

Chapter 3

TBA = 2 N

Given At B: + ΣF x = 0; →

TBC cos ( θ 1 ) − TBA cos ( θ 2 ) = 0

+

TBA sin ( θ 2 ) − TBC sin ( θ 1 ) − M g = 0

↑ ΣFy = 0;

⎛ TBC ⎞ ⎜ ⎟ = Find ( TBC , TBA) ⎝ TBA ⎠ At C:

TCD = 1 N

⎛ TBC ⎞ ⎛ 39.24 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ TBA ⎠ ⎝ 67.97 ⎠ F = 2N

Given + ΣF x = 0; →

−TBC cos ( θ 1 ) + TCD cos ( θ 3 ) = 0

+

TBC sin ( θ 1 ) + TCD sin ( θ 3 ) − F = 0

↑ ΣFy = 0;

⎛ TCD ⎞ ⎜ ⎟ = Find ( TCD , F) ⎝ F ⎠

⎛ TCD ⎞ ⎛ 39.24 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ F ⎠ ⎝ 39.24 ⎠

Problem 3-18 The motor at B winds up the cord attached to the crate of weight W with a constant speed. Determine the force in cord CD supporting the pulley and the angle θ for equilibrium. Neglect the

146

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Engineering Mechanics - Statics

Chapter 3

size of the pulley at C. Given: W = 65 lb

c = 12

d = 5

Solution: The initial guesses: θ = 100 deg

F CD = 200 lb

Given Equations of Equilibrium: + Σ F x = 0; →

+

↑Σ Fy = 0;

⎛

⎞=0 2 2⎟ ⎝ c +d ⎠

F CD cos ( θ ) − W⎜

⎛

d

⎞−W=0 ⎟ 2 2 ⎝ c +d ⎠

F CD sin ( θ ) − W⎜

c

⎛ θ ⎞ ⎜ ⎟ = Find ( θ , FCD) ⎝ FCD ⎠ θ = 78.69 deg F CD = 127.5 lb

Problem 3-19 The cords BCA and CD can each support a maximum load T. Determine the maximum weight of the crate that can be hoisted at constant velocity, and the angle θ for equilibrium.

147

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Engineering Mechanics - Statics

Chapter 3

Given: T = 100 lb c = 12 d = 5 The maximum will occur in CD rather than in BCA. Solution : The initial guesses: θ = 100 deg

W = 200 lb

Given Equations of Equilibrium: + Σ F x = 0; →

+

↑

⎛

⎞=0 ⎝ c +d ⎠

T cos ( θ ) − W⎜

⎛ Σ F y = 0; T sin ( θ ) − W⎜

d

2

2⎟

⎞−W=0 ⎝ c +d ⎠ c

2

2⎟

⎛θ⎞ ⎜ ⎟ = Find ( θ , W) ⎝W⎠ θ = 78.69 deg W = 51.0 lb

148

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Engineering Mechanics - Statics

Chapter 3

Problem 3-20 The sack has weight W and is supported by the six cords tied together as shown. Determine the tension in each cord and the angle θ for equilibrium. Cord BC is horizontal. Given: W = 15 lb

θ 1 = 30 deg θ 2 = 45 deg θ 3 = 60 deg Solution: Guesses

TBE = 1 lb

TAB = 1 lb

θ = 20deg

TBC = 1 lb

TAC = 1 lb

TCD = 1 lb

TAH = 1 lb

Given At H: +

↑ ΣFy = 0;

TAH − W = 0

At A: + ΣF x = 0; →

−TAB cos ( θ 2 ) + TAC cos ( θ 3 ) = 0

+

TAB sin ( θ 2 ) + TAC sin ( θ 3 ) − W = 0

↑ ΣFy = 0;

At B: + ΣF x = 0; →

TBC − TBE cos ( θ 1 ) + TAB cos ( θ 2 ) = 0

+

TBE sin ( θ 1 ) − TAB sin ( θ 2 ) = 0

↑ ΣFy = 0;

At C: + ΣF x = 0; →

TCD cos ( θ ) − TBC − TBE cos ( θ 3 ) = 0

+

TCD sin ( θ ) − TAC sin ( θ 3 ) = 0

↑ ΣFy = 0;

149

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Engineering Mechanics - Statics

Chapter 3

⎛ TBE ⎞ ⎜ ⎟ ⎜ TAB ⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TAC ⎟ = Find ( TBE , TAB , TBC , TAC , TCD , TAH , θ ) ⎜T ⎟ ⎜ CD ⎟ ⎜ TAH ⎟ ⎜ ⎟ ⎝ θ ⎠

⎛⎜ TBE ⎞⎟ ⎛⎜ 10.98 ⎞⎟ T ⎜ AB ⎟ 7.76 ⎟ ⎜ ⎟ ⎜ ⎜ TBC ⎟ = ⎜ 4.02 ⎟ lb ⎜ TAC ⎟ ⎜ 10.98 ⎟ ⎟ ⎜ ⎟ ⎜ 13.45 ⎜ ⎟ ⎜ TCD ⎟ ⎜ ⎜ TAH ⎟ ⎝ 15.00 ⎟⎠ ⎝ ⎠ θ = 45.00 deg

Problem 3-21 Each cord can sustain a maximum tension T. Determine the largest weight of the sack that can be supported. Also, determine θ of cord DC for equilibrium. Given: T = 200 lb

θ 1 = 30 deg θ 2 = 45 deg θ 3 = 60 deg

Solution: Solve for W = 1 and then scale the answer at the end.

Guesses

TBE = 1

TAB = 1

TBC = 1

TAC = 1

TCD = 1

TAH = 1

150

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Engineering Mechanics - Statics

Chapter 3

θ = 20 deg Given At H: +

↑ ΣFy = 0;

TAH − W = 0

At A: + ΣF x = 0; →

−TAB cos ( θ 2 ) + TAC cos ( θ 3 ) = 0

+

TAB sin ( θ 2 ) + TAC sin ( θ 3 ) − W = 0

↑ ΣFy = 0;

At B: + ΣF x = 0; →

TBC − TBE cos ( θ 1 ) + TAB cos ( θ 2 ) = 0

+

TBE sin ( θ 1 ) − TAB sin ( θ 2 ) = 0

↑ ΣFy = 0;

At C: + ΣF x = 0; →

TCD cos ( θ ) − TBC − TBE cos ( θ 3 ) = 0

+

TCD sin ( θ ) − TAC sin ( θ 3 ) = 0

↑ ΣFy = 0;

⎛ TBE ⎞ ⎜ ⎟ ⎜ TAB ⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TAC ⎟ = Find ( TBE , TAB , TBC , TAC , TCD , TAH , θ ) ⎜T ⎟ ⎜ CD ⎟ ⎜ TAH ⎟ ⎜ ⎟ ⎝ θ ⎠ W =

T

⎛⎜ TBE ⎞⎟ ⎛⎜ 0.73 ⎞⎟ T ⎜ AB ⎟ 0.52 ⎟ ⎜ ⎟ ⎜ TBC ⎜ ⎟ = ⎜ 0.27 ⎟ ⎜ TAC ⎟ ⎜ 0.73 ⎟ ⎟ ⎜ ⎟ ⎜ 0.90 ⎜ ⎟ ⎜ TCD ⎟ ⎜ ⎜ TAH ⎟ ⎝ 1.00 ⎟⎠ ⎝ ⎠ W = 200.00 lb

max ( TBE , TAB , TBC , TAC , TCD , TAH )

θ = 45.00 deg

151

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Engineering Mechanics - Statics

Chapter 3

Problem 3-22 The block has weight W and is being hoisted at uniform velocity. Determine the angle θ for equilibrium and the required force in each cord. Given: W = 20 lb

φ = 30 deg Solution: The initial guesses:

θ = 10 deg

TAB = 50 lb

Given TAB sin ( φ ) − W sin ( θ ) = 0 TAB cos ( φ ) − W − W cos ( θ ) = 0

⎛ θ ⎞ ⎜ ⎟ = Find ( θ , TAB) ⎝ TAB ⎠ θ = 60.00 deg TAB = 34.6 lb

Problem 3-23 Determine the maximum weight W of the block that can be suspended in the position shown if each cord can support a maximum tension T. Also, what is the angle θ for equilibrium?

152

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Engineering Mechanics - Statics

Chapter 3

Given: T = 80 lb

φ = 30 deg The maximum load will occur in cord AB. Solution: TAB = T The initial guesses:

θ = 100deg W = 200lb Given +

↑Σ Fy = 0;

+ Σ F x = 0; →

TAB cos ( φ ) − W − W cos ( θ ) = 0 TAB sin ( φ ) − W sin ( θ ) = 0

⎛θ⎞ ⎜ ⎟ = Find ( θ , W) ⎝W⎠ W = 46.19 lb

θ = 60.00 deg

Problem 3-24 Two spheres A and B have an equal mass M and are electrostatically charged such that the repulsive force acting between them has magnitude F and is directed along line AB. Determine the angle θ, the tension in cords AC and BC, and the mass M of each sphere.

153

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Engineering Mechanics - Statics

Chapter 3

Unit used: mN = 10

−3

N

Given: F = 20 mN g = 9.81

m 2

s

θ 1 = 30 deg θ 2 = 30 deg Solution: Guesses

TB = 1 mN

M = 1 gm

TA = 1 mN

θ = 30 deg

Given For B: + ΣF x = 0; →

F cos ( θ 2 ) − TB sin ( θ 1 ) = 0

+

F sin ( θ 2 ) + TB cos ( θ 1 ) − M g = 0

↑ ΣFy = 0;

For A: + ΣF x = 0; →

TA sin ( θ ) − F cos ( θ 2 ) = 0

+

TA cos ( θ ) − F sin ( θ 2 ) − M g = 0

↑ ΣFy = 0;

⎛ TA ⎞ ⎜ ⎟ ⎜ TB ⎟ = Find ( T , T , θ , M) A B ⎜θ ⎟ ⎜ ⎟ ⎝M⎠

⎛ TA ⎞ ⎛ 52.92 ⎞ ⎜ ⎟=⎜ ⎟ mN ⎝ TB ⎠ ⎝ 34.64 ⎠

θ = 19.11 deg

M = 4.08 gm

Problem 3-25 Blocks D and F weigh W1 each and block E weighs W2. Determine the sag s for equilibrium. Neglect the size of the pulleys. 154

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Engineering Mechanics - Statics

Chapter 3

Given: W1 = 5 lb W2 = 8 lb a = 4 ft Solution: Sum forces in the y direction Guess

s = 1 ft

Given

2⎜

⎛

⎞W − W = 0 1 2 ⎝ s +a ⎠ s

2

2⎟

s = Find ( s)

s = 5.33 ft

Problem 3-26 If blocks D and F each have weight W1, determine the weight of block E if the sag is s. Neglect the size of the pulleys. Given: W1 = 5 lb s = 3 ft a = 4 ft Solution: Sum forces in the y direction

⎛

⎞W − W = 0 1 ⎝ s +a ⎠

2⎜

s

2

2⎟

155

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Engineering Mechanics - Statics

⎛

Chapter 3

⎞W 1 ⎟ 2 2 + a s ⎝ ⎠ s

W = 2⎜

W = 6.00 lb

Problem 3-27 The block of mass M is supported by two springs having the stiffness shown. Determine the unstretched length of each spring. Units Used: 3

kN = 10 N Given: M = 30 kg l1 = 0.6 m l2 = 0.4 m l3 = 0.5 m kAC = 1.5

kN m

kAB = 1.2

kN m

g = 9.81

m 2

s Solution:

Initial guesses:

F AC = 20 N

F AB = 30 N

Given + ΣF x = 0; →

+

↑

ΣF y = 0;

l2 FAB 2

l2 + l3

− 2

l3 FAB 2

l2 + l3

2

l1 + l3

+ 2

l1 F AC

=0 2

l3 F AC 2

l1 + l3

−Mg=0 2

156

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Engineering Mechanics - Statics

Chapter 3

⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , FAB) ⎝ FAB ⎠ LAB = 0.1 m

Then guess

Given

⎛ FAC ⎞ ⎛ 183.88 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FAB ⎠ ⎝ 226.13 ⎠ LAC = 0.1 m

F AC = kAC ⎛⎝ l1 + l3 − LAC⎞⎠ 2

2

F AB = kAB⎛⎝ l2 + l3 − LAB⎞⎠ 2

2

⎛ LAB ⎞ ⎜ ⎟ = Find ( LAB , LAC) L AC ⎝ ⎠

⎛ LAB ⎞ ⎛ 0.452 ⎞ ⎜ ⎟=⎜ ⎟m 0.658 L ⎝ ⎠ AC ⎝ ⎠

Problem 3-28 Three blocks are supported using the cords and two pulleys. If they have weights of WA = WC = W, WB = kW, determine the angle θ for equilibrium. Given: k = 0.25

Solution: + ΣF x = 0; →

W cos ( φ ) − k W cos ( θ ) = 0

+

W sin ( φ ) + k W sin ( θ ) − W = 0

↑ ΣFy = 0;

cos ( φ ) = k cos ( θ ) sin ( φ ) = 1 − k sin ( θ )

157

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Engineering Mechanics - Statics

Chapter 3

1 = k cos ( θ ) + ( 1 − k sin ( θ ) ) = 1 + k − 2k sin ( θ ) 2

2

k θ = asin ⎛⎜ ⎟⎞

⎝ 2⎠

2

2

θ = 7.18 deg

Problem 3-29 A continuous cable of total length l is wrapped around the small pulleys at A, B, C, and D. If each spring is stretched a distance b, determine the mass M of each block. Neglect the weight of the pulleys and cords. The springs are unstretched when d = l/2. Given: l = 4m k = 500

N m

b = 300 mm g = 9.81

m 2

s

Solution: Fs = k b

F s = 150.00 N

Guesses T = 1N

θ = 10 deg

M = 1 kg

Given 2T sin ( θ ) − F s = 0 −2T cos ( θ ) + M g = 0 b+

l 4

sin ( θ ) =

l 4

⎛T⎞ ⎜ θ ⎟ = Find ( T , θ , M) ⎜ ⎟ ⎝M⎠ 158

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Engineering Mechanics - Statics

Chapter 3

θ = 44.43 deg

T = 107.14 N

M = 15.60 kg

Problem 3-30 Prove Lami's theorem, which states that if three concurrent forces are in equilibrium, each is proportional to the sine of the angle of the other two; that is, P/sin α = Q/sin β = R/sin γ.

Solution: Sine law: R

sin ( 180deg − γ )

=

However, in general R

sin ( γ )

=

Q

sin ( β )

=

Q

sin ( 180deg − β )

=

P

sin ( 180deg − α )

sin ( 180deg − φ ) = sin ( φ ) , hence P

sin ( α )

Q.E.D.

Problem 3-31 A vertical force P is applied to the ends of cord AB of length a and spring AC. If the spring has an unstretched length δ, determine the angle θ for equilibrium. Given: P = 10 lb

δ = 2 ft k = 15

lb ft

a = 2 ft

159

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Engineering Mechanics - Statics

Chapter 3

b = 2 ft Guesses

θ = 10 deg

φ = 10 deg

T = 1 lb

F = 1 lb

x = 1ft Given −T cos ( θ ) + F cos ( φ ) = 0 T sin ( θ ) + F sin ( φ ) − P = 0 F = k( x − δ ) a sin ( θ ) = x sin ( φ ) a cos ( θ ) + x cos ( φ ) = a + b

⎛θ ⎞ ⎜ ⎟ ⎜φ ⎟ ⎜ T ⎟ = Find ( θ , φ , T , F , x) ⎜F⎟ ⎜ ⎟ ⎝x⎠

⎛ T ⎞ ⎛ 10.30 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ F ⎠ ⎝ 9.38 ⎠

θ = 35 deg

Problem 3-32 Determine the unstretched length δ of spring AC if a force P causes the angle θ for equilibrium. Cord AB has length a. Given: P = 80 lb

θ = 60 deg k = 50

lb ft

a = 2 ft 160

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Engineering Mechanics - Statics

Chapter 3

b = 2 ft Guesses

δ = 1ft

φ = 10 deg

T = 1 lb

F = 1 lb

x = 1ft Given −T cos ( θ ) + F cos ( φ ) = 0 T sin ( θ ) + F sin ( φ ) − P = 0 F = k( x − δ ) a sin ( θ ) = x sin ( φ ) a cos ( θ ) + x cos ( φ ) = a + b

⎛δ ⎞ ⎜ ⎟ ⎜φ ⎟ ⎜ T ⎟ = Find ( δ , φ , T , F , x) ⎜F⎟ ⎜ ⎟ ⎝x⎠

⎛ T ⎞ ⎛ 69.28 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ F ⎠ ⎝ 40.00 ⎠

δ = 2.66 ft

Problem 3-33 The flowerpot of mass M is suspended from three wires and supported by the hooks at B and C. Determine the tension in AB and AC for equilibrium. Given: M = 20 kg l1 = 3.5 m l2 = 2 m l3 = 4 m 161

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Engineering Mechanics - Statics

Chapter 3

l4 = 0.5 m g = 9.81

m 2

s Solution: Initial guesses:

TAB = 1 N

TAC = 1 N

θ = 10 deg

φ = 10 deg

Given −TAC cos ( φ ) + TAB cos ( θ ) = 0 TAC sin ( φ ) + TAB sin ( θ ) − M g = 0 l1 cos ( φ ) + l2 cos ( θ ) = l3 l1 sin ( φ ) = l2 sin ( θ ) + l4

⎛ TAB ⎞ ⎜ ⎟ ⎜ TAC ⎟ = Find ( T , T , θ , φ ) AB AC ⎜ θ ⎟ ⎜ ⎟ ⎝ φ ⎠

⎛ θ ⎞ ⎛ 53.13 ⎞ ⎜ ⎟=⎜ ⎟ deg ⎝ φ ⎠ ⎝ 36.87 ⎠

⎛ TAB ⎞ ⎛ 156.96 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ TAC ⎠ ⎝ 117.72 ⎠

Problem 3-34 A car is to be towed using the rope arrangement shown. The towing force required is P. Determine the minimum length l of rope AB so that the tension in either rope AB or AC does not exceed T. Hint: Use the equilibrium condition at point A to determine the required angle θ for attachment, then determine l using trigonometry applied to triangle ABC. Given: P = 600 lb T = 750 lb

φ = 30 deg 162

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Engineering Mechanics - Statics

Chapter 3

d = 4 ft Solution: The initial guesses TAB = T TAC = T

θ = 30 deg l = 2 ft Case 1:

Assume TAC = T

Given + Σ F x = 0; →

TAC cos ( φ ) − TAB cos ( θ ) = 0

+

P − TAC sin ( φ ) − TAB sin ( θ ) = 0

↑

Σ F y = 0;

l

sin ( φ )

=

⎛ TAB ⎞ ⎜ ⎟ ⎜ θ ⎟ = Find ( TAB , θ , l) ⎜ l ⎟ ⎝ 1 ⎠ Case 2:

d

sin ( 180deg − θ − φ )

TAB = 687.39 lb

θ = 19.11 deg

l1 = 2.65 ft

θ = 13.85 deg

l2 = 2.89 ft

Assume TAB = T

Given + Σ F x = 0; →

TAC cos ( φ ) − TAB cos ( θ ) = 0

+

P − TAC sin ( φ ) − TAB sin ( θ ) = 0

↑

Σ F y = 0;

l

sin ( φ )

⎛ TAC ⎞ ⎜ ⎟ ⎜ θ ⎟ = Find ( TAC , θ , l) ⎜ l ⎟ ⎝ 2 ⎠ l = min ( l1 , l2 )

=

d

sin ( 180deg − θ − φ ) TAC = 840.83 lb

l = 2.65 ft

163

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Engineering Mechanics - Statics

Chapter 3

Problem 3-35 Determine the mass of each of the two cylinders if they cause a sag of distance d when suspended from the rings at A and B. Note that s = 0 when the cylinders are removed. Given: d = 0.5 m l1 = 1.5 m l2 = 2 m l3 = 1 m k = 100

N m m

g = 9.81

2

s Solution: TAC = k⎡⎣

( l 1 + d) 2 + l 2 2 −

2

l1 + l2

2⎤

⎦

TAC = 32.84 N

⎛ l1 + d ⎞ ⎟ ⎝ l2 ⎠

θ = atan ⎜

θ = 45 deg

M =

TAC sin ( θ ) g

M = 2.37 kg

164

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Engineering Mechanics - Statics

Chapter 3

Problem 3-36 The sling BAC is used to lift the load W with constant velocity. Determine the force in the sling and plot its value T (ordinate) as a function of its orientation θ , where 0 ≤ θ ≤ 90° .

Solution: W − 2T cos ( θ ) = 0 T=

⎛ W ⎞ ⎜ ⎟ 2 ⎝ cos ( θ ) ⎠ 1

Problem 3-37 The lamp fixture has weight W and is suspended from two springs, each having unstretched length L and stiffness k. Determine the angle θ for equilibrium. Units Used: 3

kN = 10 N Given: W = 10 lb L = 4 ft k = 5

lb ft

a = 4 ft

165

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Engineering Mechanics - Statics

Chapter 3

Solution: The initial guesses: T = 200 lb

θ = 10 deg

Given Spring +

↑Σ Fy = 0;

⎛T⎞ ⎜ ⎟ = Find ( T , θ ) ⎝θ⎠

⎛ a − L⎞ ⎟ ⎝ cos ( θ ) ⎠

T = k⎜

2T sin ( θ ) − W = 0

T = 7.34 lb

θ = 42.97 deg

Problem 3-38 The uniform tank of weight W is suspended by means of a cable, of length l, which is attached to the sides of the tank and passes over the small pulley located at O. If the cable can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cable.What is this tension? Given: W = 200 lb l = 6 ft a = 1 ft b = 2 ft c = b d = 2a Solution: Free Body Diagram: By observation, the force F has to support the entire weight of the tank. Thus, F = W. The tension in

166

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Engineering Mechanics - Statics

Chapter 3

cable is the same throughout the cable.

Equations of Equilibrium W − 2T sin ( θ ) = 0

ΣF y = 0; Attached to CD

θ 1 = acos ⎛⎜

2 a⎞

Attached to AB

θ 2 = acos ⎛⎜

2 b⎞

⎟ ⎝ l ⎠ ⎟ ⎝ l ⎠

θ 1 = 70.53 deg θ 2 = 48.19 deg

We choose the largest angle (which will produce the smallest force)

θ = max ( θ 1 , θ 2 ) T =

1⎛ W

⎜

⎞ ⎟

2 ⎝ sin ( θ ) ⎠

θ = 70.53 deg T = 106 lb

Problem 3-39 A sphere of mass ms rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass mB of block B needed to hold it in the equilibrium position shown. Given: ms = 4 kg a = 0.4 m b = 0.4 m

θ = 60 deg g = 9.81

m 2

s Solution: k =

a 2

b

Geometry: The angle θ1 which the surface make with the horizontal is to be determined first.

167

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Engineering Mechanics - Statics

tan ( θ 1 ) =

dy dx

=2kx

Chapter 3

evaluated at x = a

θ 1 = atan ( 2 k a)

θ 1 = 63.43 deg

Free Body Diagram : The tension in the cord is the same throughout the cord and is equal to the weight of block B, mBg. The initial guesses: mB = 200 kg

F N = 200 N

Given + Σ F x = 0; → +

↑Σ Fy = 0;

mB g cos ( θ ) − FN sin ( θ 1 ) = 0 mB g sin ( θ ) + F N cos ( θ 1 ) − ms g = 0

⎛ mB ⎞ ⎜ ⎟ = Find ( mB , FN) ⎝ FN ⎠

F N = 19.66 N mB = 3.58 kg

Problem 3-40 The pipe of mass M is supported at A by a system of five cords. Determine the force in each cord for equilibrium. Given: M = 30 kg g = 9.81

m

c = 3 d = 4

2

s

θ = 60 deg Solution: Initial guesses: TAB = 1 N

TAE = 1 N

TBC = 1 N

TBD = 1 N

Given TAB sin ( θ ) − M g = 0 TAE − TAB cos ( θ ) = 0 168

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Engineering Mechanics - Statics

Chapter 3

TBD⎛

c ⎞ ⎜ 2 2 ⎟ − TAB sin ( θ ) = 0 ⎝ c +d ⎠

TBD⎛

d ⎞ ⎜ 2 2 ⎟ + TAB cos ( θ ) − TBC = 0 ⎝ c +d ⎠

⎛ TAB ⎞ ⎜ ⎟ ⎜ TAE ⎟ = Find ( T , T , T , T ) AB AE BC BD ⎜ TBC ⎟ ⎜ ⎟ ⎝ TBD ⎠ ⎛ TAB ⎞ ⎛ 339.8 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TAE ⎟ = ⎜ 169.9 ⎟ N ⎜ TBC ⎟ ⎜ 562.3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ TBD ⎠ ⎝ 490.5 ⎠

Problem 3-41 The joint of a space frame is subjected to four forces. Strut OA lies in the x-y plane and strut OB lies in the y-z plane. Determine the forces acting in each of the three struts required for equilibrium. Units Used: 3

kN = 10 N Given: F = 2 kN

θ 1 = 45 deg θ 2 = 40 deg Solution: ΣF x = 0;

−R sin ( θ 1 ) = 0 R = 0

169

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Engineering Mechanics - Statics

Chapter 3

P sin ( θ 2 ) − F = 0

ΣF z = 0;

P =

F

sin ( θ 2 )

P = 3.11 kN Q − P cos ( θ 2 ) = 0

ΣF y = 0;

Q = P cos ( θ 2 ) Q = 2.38 kN

Problem 3-42 Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle. Units Used: 3

kN = 10 N Given: F 4 = 800 N

α = 60 deg β = 30 deg γ = 30 deg c = 3 d = 4

Solution: The initial guesses: F 1 = 100 N

F 2 = 100 N

F 3 = 100 N

Given

⎛ cos ( α ) ⎞ ⎜ 0 ⎟+ F1 ⎜ ⎟ ⎝ sin ( α ) ⎠

⎛ −cos ( γ ) ⎞ ⎛ 0 ⎞ ⎛c ⎞ ⎜ −d ⎟ + F ⎜ −sin ( γ ) ⎟ + F ⎜ sin ( β ) ⎟ = 0 3 4 ⎟ ⎜ ⎟ ⎜ ⎟ 2 2⎜ c +d ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ −cos ( β ) ⎠ F2

170

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Engineering Mechanics - Statics

⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠

Chapter 3

⎛ F1 ⎞ ⎛ 800 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 147 ⎟ N ⎜ F ⎟ ⎝ 564 ⎠ ⎝ 3⎠

Problem 3-43 Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle. Units Used: kN = 1000 N Given: F 4 = 8.5 kN F 5 = 2.8 kN

α = 15 deg β = 30 deg c = 7 d = 24

Solution: Initial Guesses:

F 1 = 1 kN

F 2 = 1 kN

F 3 = 1 kN

Given

⎛ −cos ( β ) ⎞ ⎜ 0 ⎟+ F1 ⎜ ⎟ ⎝ sin ( β ) ⎠

⎛ −sin ( α ) ⎞ ⎛ −c ⎞ ⎛1⎞ ⎛0⎞ ⎜ −d ⎟ + F ⎜ 0 ⎟ + F ⎜ cos ( α ) ⎟ + F ⎜ 0 ⎟ = 0 3 4 5 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 2⎜ c +d ⎝ 0 ⎠ ⎝0⎠ ⎝ −1 ⎠ ⎝ 0 ⎠ F2

171

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Engineering Mechanics - Statics

⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠

Chapter 3

⎛ F1 ⎞ ⎛ 5.60 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 8.55 ⎟ kN ⎜ F ⎟ ⎝ 9.44 ⎠ ⎝ 3⎠

Problem 3-44 Determine the magnitudes of F1, F 2 and F3 for equilibrium of the particle F = {- 9i - 8j - 5k}. Units Used: 3

kN = 10 N

Given:

⎛ −9 ⎞ F = ⎜ −8 ⎟ kN ⎜ ⎟ ⎝ −5 ⎠ a = 4m b = 2m c = 4m

θ 1 = 30 deg θ 2 = 60 deg θ 3 = 135 deg θ 4 = 60 deg θ 5 = 60 deg Solution: Initial guesses:

F 1 = 8 kN F 2 = 3 kN F 3 = 12 kN

172

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Engineering Mechanics - Statics

Chapter 3

Given

⎛ cos ( θ 2 ) cos ( θ 1) ⎞ ⎛ cos ( θ 3 ) ⎞ ⎜ ⎟ ⎜ ⎟ F 1 ⎜ −cos ( θ 2 ) sin ( θ 1 ) ⎟ + F 2 ⎜ cos ( θ 5 ) ⎟ + ⎜ ⎟ ⎜ cos θ ⎟ sin ( θ 2 ) ⎝ ⎠ ⎝ ( 4) ⎠ ⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠

⎛a⎞ ⎜ c ⎟+F=0 ⎟ 2 2 2⎜ a + b + c ⎝ −b ⎠ F3

⎛ F1 ⎞ ⎛ 8.26 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 3.84 ⎟ kN ⎜ F ⎟ ⎝ 12.21 ⎠ ⎝ 3⎠

Problem 3-45 The three cables are used to support the lamp of weight W. Determine the force developed in each cable for equilibrium. Units Used: 3

kN = 10 N Given: W = 800 N

b = 4m

a = 4m

c = 2m

Solution: Initial Guesses: F AB = 1 N

F AC = 1 N

F AD = 1 N

173

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Engineering Mechanics - Statics

Chapter 3

Given

⎛0⎞ ⎛1⎞ ⎜ ⎟ F AB 1 + FAC ⎜ 0 ⎟ + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠

⎛ −c ⎞ ⎛0⎞ ⎜ −b ⎟ + W⎜ 0 ⎟ = 0 ⎟ ⎜ ⎟ 2 2 2⎜ a +b +c ⎝ a ⎠ ⎝ −1 ⎠ F AD

⎛ FAB ⎞ ⎜ ⎟ F ⎜ AC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

⎛ FAB ⎞ ⎛ 800 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ 400 ⎟ N ⎜ F ⎟ ⎝ 1200 ⎠ ⎝ AD ⎠

Problem 3-46 Determine the force in each cable needed to support the load W. Given: a = 8 ft b = 6 ft c = 2 ft d = 2 ft e = 6 ft W = 500 lb

Solution: Initial guesses: F CD = 600 lb

F CA = 195 lb

F CB = 195 lb

174

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Engineering Mechanics - Statics

Chapter 3

Given

⎛c⎞ ⎜ −e ⎟ + 2 2⎜ ⎟ c +e ⎝0 ⎠ FCA

⎛ −d ⎞ ⎜ −e ⎟ + ⎟ 2 2⎜ d +e ⎝ 0 ⎠ F CB

⎛0⎞ ⎛0⎞ ⎜ b ⎟ + W⎜ 0 ⎟ = 0 ⎜ ⎟ 2 2⎜ ⎟ a + b ⎝a⎠ ⎝ −1 ⎠ F CD

⎛ FCD ⎞ ⎜ ⎟ F ⎜ CA ⎟ = Find ( FCD , FCA , FCB) ⎜F ⎟ ⎝ CB ⎠

⎛ FCD ⎞ ⎛ 625 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ CA ⎟ = ⎜ 198 ⎟ lb ⎜ F ⎟ ⎝ 198 ⎠ ⎝ CB ⎠

Problem 3-47 Determine the stretch in each of the two springs required to hold the crate of mass mc in the equilibrium position shown. Each spring has an unstretched length δ and a stiffness k. Given: mc = 20 kg

δ = 2m k = 300

N m

a = 4m b = 6m c = 12 m Solution: Initial Guesses F OA = 1 N F OB = 1 N F OC = 1 N Given

⎛0⎞ ⎛ −1 ⎞ ⎜ ⎟ F OA −1 + F OB⎜ 0 ⎟ + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠

⎛b⎞ ⎛0⎞ ⎜ a ⎟ + m g⎜ 0 ⎟ = 0 c ⎜ ⎟ 2 2 2⎜ ⎟ a + b + c ⎝c⎠ ⎝ −1 ⎠ F OC

175

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Engineering Mechanics - Statics

Chapter 3

⎛ FOA ⎞ ⎜ ⎟ ⎜ FOB ⎟ = Find ( FOA , FOB , FOC) ⎜F ⎟ ⎝ OC ⎠ δ OA =

δ OB =

F OA k F OB k

⎛ FOA ⎞ ⎛ 65.40 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FOB ⎟ = ⎜ 98.10 ⎟ N ⎜ F ⎟ ⎝ 228.90 ⎠ ⎝ OC ⎠

δ OA = 218 mm

δ OB = 327 mm

Problem 3-48 If the bucket and its contents have total weight W, determine the force in the supporting cables DA, DB, and DC. Given: W = 20 lb a = 3 ft b = 4.5 ft c = 2.5 ft d = 3 ft e = 1.5 ft f = 1.5 ft

176

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Engineering Mechanics - Statics

Chapter 3

Solution: The initial guesses: F DA = 40 lb

F DB = 20 lb

F DC = 30 lb

Given Σ F x = 0;

b−e e ⎡ ⎤F − ⎡ ⎤F = 0 DA ⎢ DC ⎢ ⎥ 2 2 2 2 2 2⎥ ⎣ ( b − e) + f + a ⎦ ⎣ e + d + ( c − f) ⎦

Σ F y = 0;

−f c− f ⎡ ⎤F + ⎡ ⎤F − F = 0 DA ⎢ DC DB ⎢ ⎥ 2 2 2 2 2 2⎥ + f + a + d + ( c − f ) ( b − e ) e ⎣ ⎦ ⎣ ⎦

Σ F z = 0;

a d ⎡ ⎤F + ⎡ ⎤F − W = 0 DA ⎢ DC ⎢ ⎥ 2 2 2 2 2 2⎥ ⎣ ( b − e) + f + a ⎦ ⎣ e + d + ( c − f) ⎦

⎛ FDA ⎞ ⎜ ⎟ ⎜ FDB ⎟ = Find ( FDA , FDB , FDC) ⎜F ⎟ ⎝ DC ⎠

⎛ FDA ⎞ ⎛ 10.00 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDB ⎟ = ⎜ 1.11 ⎟ lb ⎜ F ⎟ ⎝ 15.56 ⎠ ⎝ DC ⎠

Problem 3-49

The crate which of weight F is to be hoisted with constant velocity from the hold of a ship using the cable arrangement shown. Determine the tension in each of the three cables for equilibrium. 177

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Engineering Mechanics - Statics

Chapter 3

Units Used: 3

kN = 10 N Given: F = 2.5 kN a = 3m b = 1m c = 0.75 m d = 1m e = 1.5 m f = 3m Solution: The initial guesses F AD = 3 kN

F AC = 3 kN −c

Given 2

2

c +b +a

2

b 2

2

c +b +a

2

F AD +

2

c +b +a

d 2

2

d +e +a

F AD +

2

F AD +

⎛ FAD ⎞ ⎜ ⎟ F ⎜ AC ⎟ = Find ( FAD , FAC , FAB) ⎜F ⎟ ⎝ AB ⎠

2

e 2

2

d +e +a

−a 2

F AB = 3 kN

2

F AC +

2

d +e +a

2

2

d + f +a

F AC +

2

F AC +

2

−f 2

2

d + f +a

−a 2

d

2

F AB = 0

F AB = 0

−a 2

2

d + f +a

2

F AB + F = 0

⎛ FAD ⎞ ⎛ 1.55 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ 0.46 ⎟ kN ⎜ F ⎟ ⎝ 0.98 ⎠ ⎝ AB ⎠

178

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Engineering Mechanics - Statics

Chapter 3

Problem 3-50 The lamp has mass ml and is supported by pole AO and cables AB and AC. If the force in the pole acts along its axis, determine the forces in AO, AB, and AC for equilibrium. Given: ml = 15 kg

d = 1.5 m

a = 6m

e = 4m

b = 1.5 m

f = 1.5 m

c = 2m

g = 9.81

m 2

s Solution: The initial guesses: F AO = 100 N F AB = 200 N F AC = 300 N Given Equilibrium equations: c 2

2

c +b +a

2

F AO −

2

2

c +b +a a 2

2

c +b +a

2

2

2

( c + e) + ( b + f) + a

b

−

c+e

2

2

2

( c + e) + ( b + f) + a a 2

2

( c + e) + ( b + f) + a

⎛ FAO ⎞ ⎜ ⎟ ⎜ FAB ⎟ = Find ( FAO , FAB , FAC) ⎜F ⎟ ⎝ AC ⎠

2

c 2

2

c + ( b + d) + a

b+ f

F AO +

F AO −

2

F AB −

2

F AC = 0

b+d

F AB +

F AB −

2

2

2

c + ( b + d) + a a 2

2

c + ( b + d) + a

2

2

F AC = 0

F AC − ml g = 0

⎛ FAO ⎞ ⎛ 318.82 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ 110.36 ⎟ N ⎜ F ⎟ ⎝ 85.84 ⎠ ⎝ AC ⎠

179

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Engineering Mechanics - Statics

Chapter 3

Problem 3-51 Cables AB and AC can sustain a maximum tension Tmax, and the pole can support a maximum compression Pmax. Determine the maximum weight of the lamp that can be supported in the position shown. The force in the pole acts along the axis of the pole. Given: Tmax = 500 N

c = 2m

Pmax = 300 N

d = 1.5 m

a = 6m

e = 4m

b = 1.5 m

f = 1.5 m

Solution: Lengths 2

2

2

AO =

a +b +c

AB =

a + ( c + e) + ( b + d)

AC =

a + c + ( b + d)

2

2

2

2

2

2

The initial guesses: F AO = Pmax

F AB = Tmax

F AC = Tmax

W = 300N

Case 1 Assume the pole reaches maximum compression Given

⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞ ⎛0⎞ AB ⎜ AC ⎜ ⎟ ⎟ ⎟ −b + b+ f + b + d + W⎜ 0 ⎟ = 0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AO AB AC ⎝a⎠ ⎝ −a ⎠ ⎝ −a ⎠ ⎝ −1 ⎠

F AO ⎜

⎛ W1 ⎞ ⎜ ⎟ ⎜ FAB1 ⎟ = Find ( W , FAB , FAC) ⎜F ⎟ ⎝ AC1 ⎠

⎛ W1 ⎞ ⎛ 138.46 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB1 ⎟ = ⎜ 103.85 ⎟ N ⎜F ⎟ ⎝ AC1 ⎠ ⎝ 80.77 ⎠

180

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Engineering Mechanics - Statics

Chapter 3

Case 2 Assume that cable AB reaches maximum tension Given

⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞ ⎛0⎞ AB ⎜ AC ⎜ ⎟ ⎟ ⎟ −b + b+ f + b + d + W⎜ 0 ⎟ = 0 ⎟ AC ⎜ ⎟ ⎜ ⎟ AO ⎜ ⎟ AB ⎜ ⎝a⎠ ⎝ −a ⎠ ⎝ −a ⎠ ⎝ −1 ⎠

F AO ⎜

⎛ W2 ⎞ ⎜ ⎟ ⎜ FAO2 ⎟ = Find ( W , FAO , FAC) ⎜F ⎟ ⎝ AC2 ⎠

⎛ W2 ⎞ ⎛ 666.67 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAO2 ⎟ = ⎜ 1444.44 ⎟ N ⎜F ⎟ ⎝ AC2 ⎠ ⎝ 388.89 ⎠

Case 3 Assume that cable AC reaches maximum tension Given

⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞ ⎛0⎞ AB ⎜ AC ⎜ ⎟ ⎟ ⎟ −b + b+ f + b + d + W⎜ 0 ⎟ = 0 ⎟ AC ⎜ ⎟ ⎜ ⎟ AO ⎜ ⎟ AB ⎜ ⎝a⎠ ⎝ −a ⎠ ⎝ −a ⎠ ⎝ −1 ⎠

F AO ⎜

⎛ W3 ⎞ ⎜ ⎟ ⎜ FAO3 ⎟ = Find ( W , FAO , FAB) ⎜F ⎟ ⎝ AB3 ⎠ Final Answer

⎛ W3 ⎞ ⎛ 857.14 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAO3 ⎟ = ⎜ 1857.14 ⎟ N ⎜F ⎟ ⎝ AB3 ⎠ ⎝ 642.86 ⎠

W = min ( W1 , W2 , W3 )

W = 138.46 N

Problem 3-52 Determine the tension in cables AB, AC, and AD, required to hold the crate of weight W in equilibrium. Given: W = 60 lb a = 6 ft b = 12 ft c = 8 ft 181

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Engineering Mechanics - Statics

Chapter 3

d = 9 ft e = 4 ft f = 6 ft

Solution: The initial guesses: TB = 100 lb TC = 100 lb TD = 100 lb Given Σ F x = 0;

b

TB −

2

2

b +c +d Σ F y = 0;

d 2

2

b +c +d Σ F z = 0;

2

2

TC −

2

2

2

2

2

2

b +e + f e 2

b +e + f

c

−W +

b

TC −

b +c +d

2

TC +

2

TD = 0

TD = 0

f 2

2

b +e + f

2

TD = 0

Solving

⎛ TB ⎞ ⎜ ⎟ ⎜ TC ⎟ = Find ( TB , TC , TD) ⎜T ⎟ ⎝ D⎠

⎛ TB ⎞ ⎛ 108.84 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TC ⎟ = ⎜ 47.44 ⎟ lb ⎜ T ⎟ ⎝ 87.91 ⎠ ⎝ D⎠

Problem 3-53 The bucket has weight W. Determine the tension developed in each cord for equilibrium. Given: W = 20 lb

182

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Engineering Mechanics - Statics

Chapter 3

a = 2 ft b = 2 ft c = 8 ft d = 7 ft e = 3 ft f = a Solution: F DA = 20 lb

Initial Guesses:

F DB = 10 lb

F DC = 15 lb

Given c−e

ΣF x = 0;

2

2

( c − e) + b + a

2

−b

ΣF x = 0;

2

2

( c − e) + b + a

2

a

ΣF y = 0;

2

2

( c − e) + b + a

2

F DA +

−e 2

2

e + ( d − b) + f

F DA +

d−b 2

2

e + ( d − b) + f F DA +

2

2

f 2

2

e + ( d − b) + f

2

F DC +

−e 2

2

e +b + f

F DC +

−b 2

2

2

2

2

e +b + f F DC +

2

f 2

e +b + f

F DB = 0

F DB = 0

F DB − W = 0

⎛ FDA ⎞ ⎜ ⎟ ⎜ FDB ⎟ = Find ( FDA , FDB , FDC) ⎜F ⎟ ⎝ DC ⎠ ⎛ FDA ⎞ ⎛ 21.54 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDB ⎟ = ⎜ 13.99 ⎟ lb ⎜ F ⎟ ⎝ 17.61 ⎠ ⎝ DC ⎠

Problem 3-54 The mast OA is supported by three cables. If cable AB is subjected to tension T, determine the tension in cables AC and AD and the vertical force F which the mast exerts along its axis on the collar at A. Given: T = 500 N 183

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Engineering Mechanics - Statics

Chapter 3

a = 6m b = 3m c = 6m d = 3m e = 2m f = 1.5 m g = 2m Solution: Initial Guesses:

F AC = 90 N

F AD = 350 N

F = 750 N

Given Σ F x = 0;

e 2

2

e +d +a Σ F y = 0;

2

2

−a 2

2

2

e +d +a

2

f +g +a

2

2

f +g +a

⎛ FAC ⎞ ⎜ ⎟ F ⎜ AD ⎟ = Find ( FAC , FAD , F) ⎜ F ⎟ ⎝ ⎠

2

a

T− 2

2

g

T+

e +d +a Σ F z = 0;

2

d 2

f

T−

2

2

f +g +a

2

F AC −

b 2

2

2

2

2

2

2

b +c +a F AC −

c 2

b +c +a F AC −

a 2

b +c +a

F AD = 0

F AD = 0

F AD + F = 0

⎛ FAC ⎞ ⎛ 92.9 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AD ⎟ = ⎜ 364.3 ⎟ N ⎜ F ⎟ ⎝ 757.1 ⎠ ⎝ ⎠

Problem 3-55 The ends of the three cables are attached to a ring at A and to the edge of the uniform plate of mass M. Determine the tension in each of the cables for equilibrium. Given: M = 150 kg

e = 4m 184

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Engineering Mechanics - Statics

Chapter 3

a = 2m

f = 6m

b = 10 m

g = 6m

c = 12 m

h = 6m

d = 2m

i = 2m

gravity = 9.81

m 2

s Solution: The initial guesses: F B = 15 N F C = 16 N F D = 16 N Given Σ F x = 0;

F B( f − i ) 2

2

+ 2

2

( f − i) + h + c

Σ F y = 0;

F B ( − h) 2

2

F B( −c) 2

2

( f − i) + h + c

⎛ FB ⎞ ⎜ ⎟ ⎜ FC ⎟ = Find ( FB , FC , FD) ⎜F ⎟ ⎝ D⎠

2

2

+ 2

2

FC( −c) 2

2

( d + e) + ( h − a) + c

2

=0 2

FD( g) 2

2

=0 2

e +g +c

+ 2

FD( −e) e +g +c

( d + e) + ( h − a) + c

+ 2

2

F C [ − ( h − a) ] 2

( f − i) + h + c

Σ F z = 0;

+

( d + e) + ( h − a) + c

+ 2

F C( −d − e)

FD( −c) 2

2

+ M gravity = 0 2

e +g +c

⎛ FB ⎞ ⎛ 858 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FC ⎟ = ⎜ 0 ⎟ N ⎜ F ⎟ ⎝ 858 ⎠ ⎝ D⎠

Problem 3-56 The ends of the three cables are attached to a ring at A and to the edge of the uniform plate. Determine the largest mass the plate can have if each cable can support a maximum tension of T.

185

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Engineering Mechanics - Statics

Chapter 3

3

kN = 10 N Given: T = 15 kN

e = 4m

a = 2m

f = 6m

b = 10 m

g = 6m

c = 12 m

h = 6m

d = 2m

i = 2m

gravity = 9.81

m 2

s

Solution: The initial guesses:

FB = T

FC = T

FD = T

M = 1 kg

Case 1: Assume that cable B reaches maximum tension Given Σ F x = 0;

F B( f − i ) 2

2

+ 2

2

( f − i) + h + c

Σ F y = 0;

F B ( − h) 2

2

F B( −c) 2

2

( f − i) + h + c

⎛ M1 ⎞ ⎜ ⎟ ⎜ FC1 ⎟ = Find ( M , FC , FD) ⎜F ⎟ ⎝ D1 ⎠

2

2

FC( −c) 2

2

( d + e) + ( h − a) + c

⎛ FC1 ⎞ ⎛ −0.00 ⎞ ⎜ ⎟=⎜ ⎟ kN 15.00 F ⎝ ⎠ D1 ⎝ ⎠

=0 2

FD( g) 2

2

=0 2

e +g +c

+ 2

2

e +g +c

+ 2

FD( −e) 2

( d + e) + ( h − a) + c

+ 2

2

F C [ − ( h − a) ] 2

( f − i) + h + c

Σ F z = 0;

+

( d + e) + ( h − a) + c

+ 2

F C( −d − e)

FD( −c) 2

2

+ M gravity = 0 2

e +g +c

M1 = 2621.23 kg

186

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Engineering Mechanics - Statics

Chapter 3

Case 2: Assume that cable D reaches maximum tension Given F B( f − i )

Σ F x = 0;

2

2

2

2

( f − i) + h + c F B ( − h)

Σ F y = 0;

2

2

F B( −c) 2

2

2

( f − i) + h + c

⎛ M2 ⎞ ⎜ ⎟ ⎜ FB2 ⎟ = Find ( M , FB , FC) ⎜F ⎟ ⎝ C2 ⎠

2

2

FC( −c) 2

2

( d + e) + ( h − a) + c

⎛ FB2 ⎞ ⎛ 15.00 ⎞ ⎜ ⎟=⎜ ⎟ kN 0.00 F ⎝ ⎠ C2 ⎝ ⎠

=0 2

FD( g) 2

2

=0 2

e +g +c

+ 2

2

e +g +c

+ 2

FD( −e) 2

( d + e) + ( h − a) + c

+ 2

2

F C [ − ( h − a) ]

+ 2

+

( d + e) + ( h − a) + c

( f − i) + h + c

Σ F z = 0;

F C( −d − e)

+

FD( −c) 2

2

+ M gravity = 0 2

e +g +c

M2 = 2621.23 kg

For this set of number FC = 0 for any mass that is applied. For a different set of numbers it would be necessary to also check case 3: Assume that the cable C reaches a maximum. M = min ( M1 , M2 )

M = 2621.23 kg

Problem 3-57 The crate of weight W is suspended from the cable system shown. Determine the force in each segment of the cable, i.e., AB, AC, CD, CE, and CF. Hint: First analyze the equilibrium of point A, then using the result for AC, analyze the equilibrium of point C. Units Used: kip = 1000 lb Given: W = 500 lb a = 10 ft b = 24 ft c = 24 ft

187

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Engineering Mechanics - Statics

Chapter 3

d = 7 ft e = 7 ft

θ 1 = 20 deg θ 2 = 35 deg

Solution:

At A:

Initial guesses:

F AC = 570 lb

F AB = 500 lb

Given + ΣF x = 0; →

F AB cos ( θ 1 ) − F AC cos ( θ 2 ) = 0

+

F AB sin ( θ 1 ) + FAC sin ( θ 2 ) − W = 0

↑ ΣFy = 0;

⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , FAB) ⎝ FAB ⎠ At C:

⎛ FAC ⎞ ⎛ 574 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FAB ⎠ ⎝ 500 ⎠

Initial Guesses

F CD = 1 lb

F CE = 1 lb

F CF = 1 lb

Given

⎛ cos ( θ 2 ) ⎞ ⎜ ⎟ F AC⎜ 0 ⎟+ ⎜ −sin ( θ ) ⎟ 2 ⎠ ⎝

⎛ −a ⎞ ⎜ 0⎟+ ⎟ 2 2⎜ a +b ⎝ b ⎠ FCD

⎛0⎞ ⎜d⎟+ 2 2⎜ ⎟ c + d ⎝ −c ⎠ F CE

⎛0⎞ ⎜ −e ⎟ = 0 2 2⎜ ⎟ c + e ⎝ −c ⎠ F CF

188

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Engineering Mechanics - Statics

Chapter 3

⎛ FCD ⎞ ⎜ ⎟ F ⎜ CE ⎟ = Find ( FCD , FCE , FCF ) ⎜F ⎟ ⎝ CF ⎠

F CD = 1.22 kip

⎛ FCE ⎞ ⎛ 416 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FCF ⎠ ⎝ 416 ⎠

Problem 3-58 The chandelier of weight W is supported by three wires as shown. Determine the force in each wire for equilibrium. Given: W = 80 lb r = 1 ft h = 2.4 ft

Solution: The initial guesses: F AB = 40 lb F AC = 30 lb F AD = 30 lb

Given r

Σ F x = 0;

2

2

FAC −

r +h Σ F y = 0;

−r 2

2

r +h

FAD +

r cos ( 45 deg) 2

r +h

2

r cos ( 45 deg) 2

r +h

2

FAB = 0

FAB = 0

189

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Engineering Mechanics - Statics

Chapter 3

h

Σ F z = 0;

2

2

FAC +

r +h

h 2

r +h

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

2

F AD +

h 2

2

FAB − W = 0

r +h

⎛ FAB ⎞ ⎛ 35.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC ⎟ = ⎜ 25.4 ⎟ lb ⎜ F ⎟ ⎝ 25.4 ⎠ ⎝ AD ⎠

Problem 3-59 If each wire can sustain a maximum tension Tmax before it fails, determine the greatest weight of the chandelier the wires will support in the position shown. Given: Tmax = 120 lb r = 1 ft h = 2.4 ft

Solution: The initial guesses: F AB = Tmax F AC = Tmax F AD = Tmax W = Tmax Case 1 Assume that cable AB has maximum tension Given r

Σ F x = 0;

2

2

FAC −

r +h Σ F y = 0;

−r 2

2

r +h

FAD +

r cos ( 45deg) 2

r +h

2

r cos ( 45deg) 2

r +h

2

F AB = 0

F AB = 0

190

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Engineering Mechanics - Statics

Chapter 3

h

Σ Fz = 0;

2

2

FAC +

r +h

h 2

r +h

⎛ W1 ⎞ ⎜ ⎟ ⎜ FAC1 ⎟ = Find ( W , FAC , FAD) ⎜F ⎟ ⎝ AD1 ⎠

2

F AD +

h 2

2

FAB − W = 0

r +h

⎛ W1 ⎞ ⎛ 267.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC1 ⎟ = ⎜ 84.9 ⎟ lb ⎜F ⎟ ⎝ AD1 ⎠ ⎝ 84.9 ⎠

Case 2 Assume that cable AC has maximum tension Given r

Σ F x = 0;

2

2

FAC −

r cos ( 45 deg) 2

r +h Σ F y = 0;

−r 2

r +h

FAD +

2

r cos ( 45 deg) 2

r +h

r +h

h

Σ F z = 0;

2

2

2

FAC +

r +h

2

h 2

r +h

⎛ W2 ⎞ ⎜ ⎟ ⎜ FAB2 ⎟ = Find ( W , FAB , FAD) ⎜F ⎟ ⎝ AD2 ⎠

2

FAB = 0

FAB = 0

F AD +

h 2

2

FAB − W = 0

r +h

⎛ W2 ⎞ ⎛ 378.2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB2 ⎟ = ⎜ 169.7 ⎟ lb ⎜F ⎟ ⎝ AD2 ⎠ ⎝ 120 ⎠

Case 3 Assume that cable AD has maximum tension Given r

Σ F x = 0;

2

2

FAC −

r +h Σ F y = 0;

−r 2

2

2

r +h

h 2

r +h

2

r cos ( 45deg)

r +h Σ F z = 0;

2

r +h

FAD +

2

r cos ( 45deg)

FAC +

2

h 2

r +h

2

F AB = 0

F AB = 0

F AD +

h 2

2

FAB − W = 0

r +h

191

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Engineering Mechanics - Statics

Chapter 3

⎛ W3 ⎞ ⎜ ⎟ ⎜ FAB3 ⎟ = Find ( W , FAB , FAC) ⎜F ⎟ ⎝ AC3 ⎠ W = min ( W1 , W2 , W3 )

⎛ W3 ⎞ ⎛ 378.2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB3 ⎟ = ⎜ 169.7 ⎟ lb ⎜F ⎟ ⎝ AC3 ⎠ ⎝ 120 ⎠ W = 267.42 lb

Problem 3-60 Determine the force in each cable used to lift the surge arrester of mass M at constant velocity. Units Used: 3

kN = 10 N 3

Mg = 10 kg Given: M = 9.50 Mg a = 2m b = 0.5 m

θ = 45 deg

Solution: Initial guesses:

F B = 50 kN

F C = 30 kN

Given b

ΣF x = 0;

FB

ΣF y = 0;

FC

ΣF z = 0;

M g − FB

2

b +a

− FC

2

2

=0

2

b +a

−b sin ( θ ) 2

b cos ( θ )

F D = 10 kN

+ FD

2

b +a

b 2

=0

b +a a

2

2

b +a

− FC

2

a 2

b +a

2

− FD

a 2

=0 2

b +a

192

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Engineering Mechanics - Statics

Chapter 3

⎛ FB ⎞ ⎜ ⎟ ⎜ FC ⎟ = Find ( FB , FC , FD) ⎜F ⎟ ⎝ D⎠

⎛ FB ⎞ ⎛ 28.13 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FC ⎟ = ⎜ 39.78 ⎟ kN ⎜ F ⎟ ⎝ 28.13 ⎠ ⎝ D⎠

Problem 3-61 The cylinder of weight W is supported by three chains as shown. Determine the force in each chain for equilibrium.

Given: W = 800 lb r = 1 ft d = 1 ft Solution: The initial guesses: F AB = 1 lb F AC = 1 lb F AD = 1 lb Given Σ F x = 0;

r 2

2

FAC −

r cos ( 45 deg) 2

r +d Σ F y = 0;

−r 2

2

r +d FAD +

r +d d Σ F z = 0;

2

2

2

r sin ( 45 deg) 2

2

FAB = 0

F AB = 0

r +d FAD +

r +d

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

d 2

r +d

2

F AC +

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ ⎜F ⎟ = ⎝ AD ⎠

d 2

2

FAB − W = 0

r +d

⎛ 469 ⎞ ⎜ 331 ⎟ ⎜ ⎟ ⎝ 331 ⎠ lb

193

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Engineering Mechanics - Statics

Chapter 3

Problem 3-62 The triangular frame ABC can be adjusted vertically between the three equal-length cords. If it remains in a horizontal plane, determine the required distance s so that the tension in each of the cords, OA, OB, and OC, equals F . The lamp has a mass M. Given: F = 20 N M = 5 kg m

g = 9.81

2

s d = 0.5 m Solution:

3F cos ( γ ) = M g

ΣF z = 0;

γ = acos ⎛⎜

M g⎞

⎟ ⎝ 3F ⎠

2d cos ( 30 deg)

Geometry

3 s =

γ = 35.16 deg = s tan ( γ )

2d cos ( 30 deg) 3 tan ( γ )

s = 410 mm

Problem 3-63 Determine the force in each cable needed to support the platform of weight W. Units Used: 3

kip = 10 lb

194

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Engineering Mechanics - Statics

Chapter 3

Given: W = 3500 lb

d = 4 ft

a = 2 ft

e = 3 ft f = 3 ft

b = 4 ft c = 4 ft

g = 10 ft

Solution: The initial guesses: F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

Given −b 2

2

2

c−d

FAD +

2

g + ( e − a) + b e−a 2

2

2

2

2

2

g + ( e − a) + b

2

( c − d) + e + g e

FAD +

2

g + ( e − a) + b −g

2

2

2

g 2

2

2

e + ( c − d) + g

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

2

2

2

2

2

c + f +g f

FAC −

e + ( c − d) + g

FAD −

c

FAC +

FAC −

2

c + f +g

g 2

2

2

F AB = 0

F AB = 0

F AB + W = 0

g + f +c

⎛ FAB ⎞ ⎛ 1.467 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC ⎟ = ⎜ 0.914 ⎟ kip ⎜ F ⎟ ⎝ 1.42 ⎠ ⎝ AD ⎠

195

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Engineering Mechanics - Statics

Chapter 3

Problem 3-64 A flowerpot of mass M is supported at A by the three cords. Determine the force acting in each cord for equilibrium. Given: M = 25 kg g = 9.81

m 2

s

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 60 deg θ 4 = 45 deg Solution: Initial guesses: F AB = 1 N F AD = 1 N F AC = 1 N Given ΣF x = 0;

F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0

ΣF y = 0;

−F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0

ΣF z = 0;

F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − M g = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠

⎛ FAB ⎞ ⎛ 219.89 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC ⎟ = ⎜ 103.65 ⎟ N ⎜ F ⎟ ⎝ 103.65 ⎠ ⎝ AD ⎠

Problem 3-65 If each cord can sustain a maximum tension of T before it fails, determine the greatest weight of the flowerpot the cords can support.

196

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Engineering Mechanics - Statics

Chapter 3

Given: T = 50 N

θ 1 = 30 deg θ 2 = 30 deg θ 3 = 60 deg θ 4 = 45 deg Solution: Initial guesses: F AB = T F AD = T F AC = T W = T Case 1

Assume that AB reaches maximum tension

Given ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0 ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0 ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0

⎛ W1 ⎞ ⎜ ⎟ ⎜ FAC1 ⎟ = Find ( W , FAC , FAD) ⎜F ⎟ ⎝ AD1 ⎠ Case 2

⎛ W1 ⎞ ⎛ 55.77 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAC1 ⎟ = ⎜ 23.57 ⎟ N ⎜F ⎟ ⎝ AD1 ⎠ ⎝ 23.57 ⎠

Assume that AC reaches maximum tension

197

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Engineering Mechanics - Statics

Chapter 3

Given ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0 ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0 ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0

⎛ W2 ⎞ ⎜ ⎟ ⎜ FAB2 ⎟ = Find ( W , FAB , FAD) ⎜F ⎟ ⎝ AD2 ⎠ Case 3

⎛ W2 ⎞ ⎛ 118.30 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB2 ⎟ = ⎜ 106.07 ⎟ N ⎜F ⎟ ⎝ AD2 ⎠ ⎝ 50.00 ⎠

Assume that AD reaches maximum tension

Given ΣF x = 0; F AD sin ( θ 1 ) − F AC sin ( θ 2 ) = 0 ΣF y = 0; −F AD cos ( θ 1 ) sin ( θ 3 ) − FAC cos ( θ 2 ) sin ( θ 3 ) + F AB sin ( θ 4 ) = 0 ΣF z = 0; F AD cos ( θ 1 ) cos ( θ 3 ) + F AC cos ( θ 2 ) cos ( θ 3 ) + F AB cos ( θ 4 ) − W = 0

⎛ W3 ⎞ ⎜ ⎟ ⎜ FAB3 ⎟ = Find ( W , FAB , FAC) ⎜F ⎟ ⎝ AC3 ⎠ W = min ( W1 , W2 , W3 )

⎛ W3 ⎞ ⎛ 118.30 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB3 ⎟ = ⎜ 106.07 ⎟ N ⎜F ⎟ ⎝ AC3 ⎠ ⎝ 50.00 ⎠ W = 55.77 N

198

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Engineering Mechanics - Statics

Chapter 3

Problem 3-66 The pipe is held in place by the vice. If the bolt exerts force P on the pipe in the direction shown, determine the forces F A and F B that the smooth contacts at A and B exerton the pipe. Given: P = 50 lb

θ = 30 deg c = 3 d = 4

Solution: Initial Guesses

F A = 1 lb

F B = 1 lb

Given + Σ F x = 0; → +

↑Σ Fy = 0;

⎛ FA ⎞ ⎜ ⎟ = Find ( FA , FB) ⎝ FB ⎠

⎛

⎞=0 ⎟ 2 2 ⎝ c +d ⎠

F B − FA sin ( θ ) − P⎜

⎛

d

⎞=0 ⎟ 2 2 ⎝ c +d ⎠

−F A cos ( θ ) + P⎜

c

⎛ FA ⎞ ⎛ 34.6 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FB ⎠ ⎝ 57.3 ⎠

Problem 3-67 When y is zero, the springs sustain force F0. Determine the magnitude of the applied vertical forces F and -F required to pull point A away from point B a distance y1. The ends of cords CAD and CBD are attached to rings at C and D.

199

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Engineering Mechanics - Statics

Chapter 3

Given: F 0 = 60 lb k = 40

lb ft

d = 2 ft y1 = 2 ft Solution: Initial spring stretch: s1 =

F0

s1 = 1.50 ft

k

Initial guesses: F s = 1 lb

T = 1 lb

F = 1 lb

Given 2

⎛ y1 ⎞ d −⎜ ⎟ ⎝ 2⎠ T−F =0 s 2

F−

y1 2d

T=0

⎛⎜ Fs ⎞⎟ ⎜ T ⎟ = Find ( Fs , T , F) ⎜F⎟ ⎝ ⎠

d

⎛ Fs ⎞ ⎛ 70.72 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ T ⎠ ⎝ 81.66 ⎠

2 ⎡ ⎤ 2 ⎛ y1 ⎞ ⎢ F s = k d − d − ⎜ ⎟ + s1⎥ ⎣ ⎝2⎠ ⎦

F = 40.83 lb

Problem 3-68 When y is zero, the springs are each stretched a distance δ. Determine the distance y if a force F is applied to points A and B as shown. The ends of cords CAD and CBD are attached to

200

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Engineering Mechanics - Statics

Chapter 3

pp p rings at C and D. Given:

δ = 1.5 ft k = 40

lb ft

d = 2 ft F = 60 lb Solution: Initial guesses: F s = 1 lb

T = 1 lb

y = 1 ft

Given F−

y T=0 2d 2

y d − ⎛⎜ ⎟⎞ ⎝ 2⎠ T − F = 0 s 2

d

⎛⎜ Fs ⎞⎟ ⎜ T ⎟ = Find ( Fs , T , y) ⎜y⎟ ⎝ ⎠

⎡

F s = k⎢d −

⎣

2

d −

2 ⎤ ⎛ y ⎞ + δ⎥ ⎜ ⎟ ⎝ 2⎠ ⎦

⎛ Fs ⎞ ⎛ 76.92 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ T ⎠ ⎝ 97.55 ⎠

y = 2.46 ft

Problem 3-69 Cord AB of length a is attached to the end B of a spring having an unstretched length b. The other end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If a weight W is suspended from B, determine the angle θ of cord AB for equilibrium.

201

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Engineering Mechanics - Statics

Chapter 3

Given: a = 5 ft b = 5 ft k = 10

lb ft

W = 10 lb Solution: Initial Guesses F BA = 1 lb F sp = 1 lb

θ = 30 deg Given F sp − FBA cos ( θ ) = 0 F BA sin ( θ ) − W = 0 F sp = k( a − a cos ( θ ) )

⎛ Fsp ⎞ ⎜ ⎟ ⎜ FBA ⎟ = Find ( Fsp , FBA , θ ) ⎜ ⎟ ⎝ θ ⎠

⎛ Fsp ⎞ ⎛ 11.82 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FBA ⎠ ⎝ 15.49 ⎠

θ = 40.22 deg

Problem 3-70 The uniform crate of mass M is suspended by using a cord of length l that is attached to the sides of the crate and passes over the small pulley at O. If the cord can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cord and specify the cord tension in this case.

202

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Engineering Mechanics - Statics

Chapter 3

Given: M = 50 kg g = 9.81

m 2

s a = 0.6 m b = 1.5 m l = 2m c =

a

d =

b

2

2

Solution: Case 1 Attached at A and B

Given

⎡⎢ l 2 2 ⎥⎤ ⎛ ⎞ −d ⎢ ⎜⎝ 2 ⎟⎠ ⎥ Mg − ⎢ ⎥ 2T = 0 l ⎢ ⎥ 2 ⎣ ⎦

Case 2 Attached at C and D

Given

Guess

Guess

⎡⎢ l 2 2 ⎥⎤ ⎛ ⎞ −c ⎢ ⎜⎝ 2 ⎟⎠ ⎥ Mg − ⎢ ⎥ 2T = 0 l ⎢ ⎥ 2 ⎣ ⎦

Choose the arrangement that gives the smallest tension.

T = 1N

T1 = Find ( T)

T1 = 370.78 N

T = 1N

T2 = Find ( T)

T = min ( T1 , T2 )

T2 = 257.09 N

T = 257.09 N

Problem 3-71 The man attempts to pull the log at C by using the three ropes. Determines the direction θ in which he should pull on his rope with a force P, so that he exerts a maximum 203

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Engineering Mechanics - Statics

Chapter 3

p p force on the log. What is the force on the log for this case ? Also, determine the direction in which he should pull in order to maximize the force in the rope attached to B.What is this maximum force? Given: P = 80 lb

φ = 150 deg

Solution: + Σ F x = 0; → +

↑Σ Fy = 0;

F AC =

F AB + P cos ( θ ) − F AC sin ( φ − 90 deg) = 0 P sin ( θ ) − F AC cos ( φ − 90 deg) = 0

P sin ( θ )

In order to maximize FAC we choose

cos ( φ − 90deg)

Thus

θ = 90 deg

F AC =

P sin ( θ )

cos ( φ − 90deg)

sin ( θ ) = 1.

F AC = 160.00 lb

Now let's find the force in the rope AB. F AB = −P cos ( θ ) + F AC sin ( φ − 90 deg) F AB = −P cos ( θ ) +

F AB = P

P sin ( θ ) sin ( φ − 90 deg) cos ( φ − 90 deg)

sin ( θ ) sin ( φ − 90 deg) − cos ( θ ) cos ( φ − 90 deg) cos ( φ − 90 deg)

= −P

cos ( θ + φ − 90deg) cos ( φ − 90 deg)

204

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Engineering Mechanics - Statics

Chapter 3

cos ( θ + φ − 90 deg) = −1

In order to maximize the force we set

θ + φ − 90 deg = 180 deg θ = 270 deg − φ

F AB = −P

θ = 120.00 deg

cos ( θ + φ − 90 deg)

F AB = 160.00 lb

cos ( φ − 90 deg)

Problem 3-72 The "scale" consists of a known weight W which is suspended at A from a cord of total length L. Determine the weight w at B if A is at a distance y for equilibrium. Neglect the sizes and weights of the pulleys.

Solution: +

↑ ΣFy = 0;

Geometry

2W sin ( θ ) − w = 0

h=

⎛L − ⎜ ⎝ 2

y⎞

⎟ ⎠

2

2

−

⎛ d ⎞ = 1 ( L − y) 2 − d2 ⎜ ⎟ 2 ⎝ 2⎠

205

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Engineering Mechanics - Statics

Chapter 3

⎡ 1 ( L − y) 2 + d2 ⎤ ⎢2 ⎥ w = 2W⎢ ⎥ L− y ⎢ ⎥ 2 ⎣ ⎦

w=

2W L−y

2

2

( L − y) − d

Problem 3-73 Determine the maximum weight W that can be supported in the position shown if each cable AC and AB can support a maximum tension of F before it fails. Given:

θ = 30 deg F = 600 lb c = 12 d = 5

Solution: Initial Guesses

F AB = F

F AC = F

W = F

Case 1 Assume that cable AC reaches maximum tension

Given

F AC sin ( θ ) −

d 2

FAB = 0

2

c +d F AC cos ( θ ) +

c 2

2

FAB − W = 0

c +d

⎛ W1 ⎞ ⎜ ⎟ = Find ( W , FAB) ⎝ FAB1 ⎠

⎛ W1 ⎞ ⎛ 1239.62 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FAB1 ⎠ ⎝ 780.00 ⎠

Case 2 Assume that cable AB reaches maximum tension

206

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Engineering Mechanics - Statics

F AC sin ( θ ) −

Given

Chapter 3

d 2

FAB = 0

2

c +d F AC cos ( θ ) +

c 2

2

FAB − W = 0

c +d

⎛ W2 ⎞ ⎜ ⎟ = Find ( W , FAC) ⎝ FAC2 ⎠ W = min ( W1 , W2 )

⎛ W2 ⎞ ⎛ 953.55 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FAC2 ⎠ ⎝ 461.54 ⎠ W = 953.6 lb

Problem 3-74 If the spring on rope OB has been stretched a distance δ. and fixed in place as shown, determine the tension developed in each of the other three ropes in order to hold the weight W in equilibrium. Rope OD lies in the x-y plane. Given: a = 2 ft b = 4 ft c = 3 ft d = 4 ft e = 4 ft f = 4 ft xB = −2 ft yB = −3 ft zB = 3 ft

θ = 30 deg k = 20

lb in

δ = 2 in W = 225 lb

207

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Engineering Mechanics - Statics

Chapter 3

Solution: F OA = 10 lb

Initial Guesses

F OC = 10 lb

F OD = 10 lb

Given c 2

2

2

a +b +c −b 2

2

2

2

2

2

a +b +c

2

2

2

yB

F OA + kδ

2

2

2

⎛ FOA ⎞ ⎜ ⎟ ⎜ FOC ⎟ = Find ( FOA , FOC , FOD) ⎜F ⎟ ⎝ OD ⎠

2

2

2

2

2

f 2

d +e + f e

+

xB + yB + zB

2

d +e + f

2

zB 2

2

+

xB + yB + zB F OA + kδ

−d

+

xB + yB + zB

a +b +c a

xB

F OA + kδ

2

2

d +e + f

FOC + F OD sin ( θ ) = 0

FOC + F OD cos ( θ ) = 0

FOC − W = 0

⎛ FOA ⎞ ⎛ 201.6 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FOC ⎟ = ⎜ 215.7 ⎟ lb ⎜ F ⎟ ⎝ 58.6 ⎠ ⎝ OD ⎠

Problem 3-75 The joint of a space frame is subjected to four member forces. Member OA lies in the x - y plane and member OB lies in the y - z plane. Determine the forces acting in each of the members required for equilibrium of the joint. Given: F 4 = 200 lb

θ = 40 deg φ = 45 deg Solution: The initial guesses : F 1 = 200 lb

F 2 = 200 lb

F 3 = 200 lb

208

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Engineering Mechanics - Statics

Chapter 3

Given Σ F y = 0;

F 3 + F 1 cos ( φ ) − F 2 cos ( θ ) = 0

Σ F x = 0;

−F 1 sin ( φ ) = 0

Σ F z = 0;

F 2 sin ( θ ) − F4 = 0

⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠ ⎛ F1 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 311.1 ⎟ lb ⎜ F ⎟ ⎝ 238.4 ⎠ ⎝ 3⎠

209

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Engineering Mechanics - Statics

Chapter 4

Problem 4-1 If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e., A × ( B + D) = ( A × B) + ( A × D). Solution: Consider the three vectors; with A vertical. Note triangle obd is perpendicular to A. od = A × ( B + D) = A

(

B + D ) sin ( θ 3 )

ob = A × B = A B sin ( θ 1 ) bd = A × D = A B sin ( θ 2 ) Also, these three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product is proportional to the length of each side of the triangle. The three vector cross - products also form a closed triangle o'b'd' which is similar to triangle obd. Thus from the figure, A × ( B + D) = A × B + A × D

(QED)

Note also, A = Axi + Ayj + A zk B = Bxi + B yj + B zK D = Dxi + Dyj + Dzk j k ⎞ ⎛ i ⎜ ⎟ Ay Az ⎟ A × ( B + D) = ⎜ Ax ⎜B + D B + D B + D ⎟ x y y z z⎠ ⎝ x =

=

⎡⎣Ay( Bz + Dz) − Az( By + Dy)⎤⎦ i − ⎡⎣Ax( Bz + Dz) − Az( Bx + Dx)⎤⎦ j + ⎡⎣Ax( By + Dy) − Ay( Bx − Dx)⎤⎦ k ⎡⎣( Ay Bz − Az By) i − ( Ax Bz − Az Bx) j + ( Ax By − Ay Bx) k⎤⎦ ... + ⎡⎣( Ay Dz − A z Dy) i − ( Ax Dz − A z Dx) j + ( A x Dy − A y Dx) k⎤⎦ 210

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Engineering Mechanics - Statics

Chapter 4

⎛ i j k⎞ ⎛ i j k⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ Ax Ay Az ⎟ + ⎜ Ax Ay Az ⎟ ⎜B B B ⎟ ⎜D D D ⎟ ⎝ x y z⎠ ⎝ x y z⎠

( A × B) + ( A × D)

=

(QED)

Problem 4-2 Prove the triple scalar product identity A⋅ ( B × C) = ( A × B) ⋅ C. Solution: As shown in the figure Area = B ( C sin ( θ ) ) = B × C Thus, Volume of parallelopiped is B × C h But, h = A⋅ u

B× C

⎛ B× C ⎞ ⎟ ⎝ B× C ⎠

= A⋅ ⎜

Thus, Volume = A⋅ ( B × C) Since A × B⋅ C represents this same volume then A⋅ ( B × C) = ( A × B) ⋅ C

(QED)

Also, LHS = A⋅ ( B × C)

⎛ i j k⎞ ⎜ ⎟ = ( A xi + A yj + Azk) ⎜ Bx By Bz ⎟ ⎜C C C ⎟ ⎝ x y z⎠ = Ax( By Cz − B z Cy) − Ay( Bx Cz − B z Cx) + Az( B x Cy − B y Cx) = Ax B y Cz − A x Bz Cy − Ay B x Cz + A y Bz Cx + Az Bx Cy − A z B y Cx RHS = ( A × B) ⋅ C 211

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Engineering Mechanics - Statics

Chapter 4

⎛ i j k⎞ ⎜ ⎟ A A A x y z ⎜ ⎟ ( Cxi + Cyj + Czk) ⎜B B B ⎟ ⎝ x y z⎠

=

= Cx( Ay B z − Az By) − Cy( A x Bz − A z B x) + Cz( A x By − Ay B x) =

Ax B y Cz − A x Bz Cy − Ay B x Cz + A y Bz Cx + Az Bx Cy − A z B y Cx

Thus,

LHS = RHS

A⋅ B × C = A × B⋅ C

(QED)

Problem 4-3 Given the three nonzero vectors A, B, and C, show that if A⋅ ( B × C) = 0, the three vectors must lie in the same plane. Solution: Consider, A⋅ ( B × C) = A B × C cos ( θ ) =

(

=

A cos ( θ ) ) B × C h

B× C

= BC h sin ( φ ) = volume of parallelepiped. If A⋅ ( B × C) = 0, then the volume equals zero, so that A, B, and C are coplanar.

Problem 4-4 Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point O. Given: F 1 = 40 lb F 2 = 60 lb

212

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Engineering Mechanics - Statics

Chapter 4

θ 1 = 30 deg θ 2 = 45 deg a = 5 in b = 13 in c = 3 in d = 6 in e = 3 in f = 6 in Solution: MRO =ΣMO; MRO = F 1 cos ( θ 2 ) e − F1 sin ( θ 2 ) f − F2 cos ( θ 1 ) b − a − F2 sin ( θ 1 ) a 2

MRO = −858 lb⋅ in

2

MRO = 858 lb⋅ in

Problem 4-5 Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point P. Units Used: kip = 1000 lb Given: F 1 = 40 lb

b = 13 in

F 2 = 60 lb

c = 3 in

θ 1 = 30 deg

d = 6 in

θ 2 = 45 deg

e = 3 in

a = 5 in

f = 6 in

Solution: MRP = ΣMP; MRP = F1 cos ( θ 2 ) ( e + c) − F 1 sin ( θ 2 ) ( d + f) − F2 cos ( θ 1 ) + −F2 sin ( θ 1 ) ( a − c)

(

2

2

)

b − a + d ...

213

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Engineering Mechanics - Statics

Chapter 4

MRP = −1165 lb⋅ in

MRP = 1.17 kip⋅ in

Problem 4-6 Determine the magnitude of the force F that should be applied at the end of the lever such that this force creates a clockwise moment M about point O. Given: M = 15 N m

φ = 60 deg θ = 30 deg a = 50 mm b = 300 mm Solution: M = F cos ( θ ) ( a + b sin ( φ ) ) − F sin ( θ ) ( b cos ( φ ) )

F =

M

cos ( θ ) ( a + b sin ( φ ) ) − sin ( θ ) ( b cos ( φ ) )

F = 77.6 N

Problem 4-7 Determine the angle θ (0 <= θ <= 90 deg) so that the force F develops a clockwise moment M about point O. Given: F = 100 N

φ = 60 deg

214

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Engineering Mechanics - Statics

Chapter 4

M = 20 N⋅ m a = 50 mm

θ = 30 deg

b = 300 mm

Solution:

θ = 30 deg

Initial Guess

Given M = F cos ( θ ) ( a + b sin ( φ ) ) − F sin ( θ ) ( b cos ( φ ) )

θ = Find ( θ )

θ = 28.6 deg

Problem 4-8 Determine the magnitude and directional sense of the moment of the forces about point O. Units Used: 3

kN = 10 N Given: F B = 260 N

e = 2m

a = 4m

f = 12

b = 3m

g = 5

c = 5m

θ = 30 deg

d = 2m

F A = 400 N

Solution: Mo = F A sin ( θ ) d + FA cos ( θ ) c + F B

f 2

( a + e) 2

f +g Mo = 3.57 kN⋅ m

(positive means counterclockwise)

215

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Engineering Mechanics - Statics

Chapter 4

Problem 4-9 Determine the magnitude and directional sense of the moment of the forces about point P.

Units Used: 3

kN = 10 N Given: F B = 260 N

e = 2m

a = 4m

f = 12

b = 3m

g = 5

c = 5m

θ = 30 deg

d = 2m F A = 400 N Solution: Mp = F B

g 2

f +g

2

b + FB

Mp = 3.15 kN⋅ m

f 2

2

e − FA sin ( θ ) ( a − d) + FA cos ( θ ) ( b + c)

f +g

(positive means counterclockwise)

Problem 4-10 A force F is applied to the wrench. Determine the moment of this force about point O. Solve the problem using both a scalar analysis and a vector analysis.

216

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Engineering Mechanics - Statics

Chapter 4

Given: F = 40 N

θ = 20 deg a = 30 mm b = 200 mm Scalar Solution MO = −F cos ( θ ) b + F sin ( θ ) a MO = −7.11 N⋅ m

MO = 7.11 N⋅ m

Vector Solution

⎛ b ⎞ ⎛ −F sin ( θ ) ⎞ ⎜ ⎟ ⎜ ⎟ MO = a × −F cos ( θ ) ⎜ ⎟ ⎜ ⎟ 0 ⎝0⎠ ⎝ ⎠

⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −7.11 ⎠

MO = 7.107 N⋅ m

Problem 4-11 Determine the magnitude and directional sense of the resultant moment of the forces about point O. Units Used: 3

kip = 10 lb Given: F 1 = 300 lb

e = 10 ft

F 2 = 250 lb

f = 4

a = 6 ft

g = 3

b = 3 ft

θ = 30 deg

c = 4 ft

φ = 30 deg

d = 4 ft

217

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Engineering Mechanics - Statics

Chapter 4

Solution: Mo = F 2

f 2

f +g Mo = 2.42 kip⋅ ft

2

e sin ( φ ) + F2

g 2

2

e cos ( φ ) + F1 sin ( θ ) a − F1 cos ( θ ) b

f +g

positive means clockwise

Problem 4-12 To correct a birth defect, the tibia of the leg is straightened using three wires that are attached through holes made in the bone and then to an external brace that is worn by the patient. Determine the moment of each wire force about joint A. Given: F1 = 4 N

d = 0.15 m

F2 = 8 N

e = 20 mm

F3 = 6 N

f = 35 mm

a = 0.2 m

g = 15 mm

b = 0.35 m

θ 1 = 30 deg

c = 0.25 m

θ 2 = 15 deg

Solution:

Positive means counterclockwise

MA1 = F 1 cos ( θ 2 ) d + F1 sin ( θ 2 ) e

MA1 = 0.6 N⋅ m

MA2 = F 2 ( c + d)

MA2 = 3.2 N⋅ m

MA3 = F 3 cos ( θ 1 ) ( b + c + d) − F 3 sin ( θ 1 ) g MA3 = 3.852 N⋅ m

Problem 4-13 To correct a birth defect, the tibia of the leg is straightened using three wires that are attached through holes made in the bone and then to an external brace that is worn by the patient. Determine the moment of each wire force about joint B. Given: F1 = 4 N

d = 0.15 m

218

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Engineering Mechanics - Statics

F2 = 8 N

e = 20 mm

F3 = 6 N

f = 35 mm

a = 0.2 m

g = 15 mm

b = 0.35 m

θ 1 = 30 deg

c = 0.25 m

θ 2 = 15 deg

Chapter 4

Solution: Positive means clockwise MB1 = F 1 cos ( θ 2 ) ( a + b + c) − F 1 sin ( θ 2 ) e MB1 = 3.07 N⋅ m MB2 = F 2 ( a + b)

MB2 = 4.4 N⋅ m

MB3 = F 3 cos ( θ 1 ) a + F 3 sin ( θ 1 ) g

MB3 = 1.084 N⋅ m

Problem 4-14 Determine the moment of each force about the bolt located at A. Given: F B = 40 lb

a = 2.5 ft

α = 20 deg

F C = 50 lb

b = 0.75 ft

β = 25 deg

γ = 30 deg

219

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Engineering Mechanics - Statics

Chapter 4

Solution: MB = FB cos ( β ) a MB = 90.6 lb⋅ ft MC = F C cos ( γ ) ( a + b) MC = 141 lb⋅ ft

Problem 4-15 Determine the resultant moment about the bolt located at A. Given: F B = 30 lb F C = 45 lb a = 2.5 ft b = 0.75 ft

α = 20 deg β = 25 deg γ = 30 deg Solution: MA = FB cos ( β ) a + FC cos ( γ ) ( a + b)

MA = 195 lb⋅ ft

Problem 4-16 The elbow joint is flexed using the biceps brachii muscle, which remains essentially vertical as the arm moves in the vertical plane. If this muscle is located a distance a from the pivot point A on the humerus, determine the variation of the moment capacity about A if the constant force developed by the muscle is F. Plot these results of M vs. θ for −60 ≤ θ ≤ 80.

220

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N

Given: a = 16 mm F = 2.30 kN

θ = ( −60 .. 80) Solution: MA ( θ ) = F ( a) cos ( θ deg)

N.m

50

MA( θ )

0

50

0

50

100

θ

Problem 4-17 The Snorkel Co.produces the articulating boom platform that can support weight W. If the boom is in the position shown, determine the moment of this force about points A, B, and C. Units Used: 3

kip = 10 lb

221

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Engineering Mechanics - Statics

Chapter 4

Given: a = 3 ft b = 16 ft c = 15 ft

θ 1 = 30 deg θ 2 = 70 deg W = 550 lb

Solution: MA = W a

MA = 1.65 kip⋅ ft

MB = W( a + b cos ( θ 1 ) )

MB = 9.27 kip⋅ ft

MC = W( a + b cos ( θ 1 ) − c cos ( θ 2 ) )

MC = 6.45 kip⋅ ft

Problem 4-18 Determine the direction θ ( 0° ≤ θ ≤ 180°) of the force F so that it produces (a) the maximum moment about point A and (b) the minimum moment about point A. Compute the moment in each case. Given: F = 40 lb a = 8 ft b = 2 ft

222

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Engineering Mechanics - Statics

Chapter 4

Solution: The maximum occurs when the force is perpendicular to the line between A and the point of application of the force. The minimum occurs when the force is parallel to this line.

( a)

MAmax = F

2

a +b

2

b φ a = atan ⎛⎜ ⎟⎞

φ a = 14.04 deg

θ a = 90 deg − φ a

θ a = 76.0 deg

MAmin = 0 lb⋅ ft

MAmin = 0 lb⋅ ft

b φ b = atan ⎛⎜ ⎟⎞

φ b = 14.04 deg

θ b = 180 deg − φ b

θ b = 166 deg

⎝ a⎠

( b)

MAmax = 329.848 lb⋅ ft

⎝ a⎠

Problem 4-19 The rod on the power control mechanism for a business jet is subjected to force F. Determine the moment of this force about the bearing at A. Given: F = 80 N

θ 1 = 20 deg

a = 150 mm θ 2 = 60 deg

Solution: MA = F cos ( θ 1 ) ( a) sin ( θ 2 ) − F sin ( θ 1 ) ( a) cos ( θ 2 )

MA = 7.71 N⋅ m

Problem 4-20 The boom has length L, weight Wb, and mass center at G. If the maximum moment that can be developed by the motor at A is M, determine the maximum load W, having a mass center at G', that can be lifted. 223

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Engineering Mechanics - Statics

Chapter 4

Given: L = 30 ft Wb = 800 lb a = 14 ft b = 2 ft

θ = 30 deg 3

M = 20 × 10 lb⋅ ft Solution: M = Wb ( L − a) cos ( θ ) + W ( L cos ( θ ) + b) W =

M − Wb ( L − a) cos ( θ )

W = 319 lb

L cos ( θ ) + b

Problem 4-21 The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench. If a force P is applied to the wrench at B in the direction shown, determine the moment it creates about the nut at C. What is the magnitude of force F at A so that it creates the opposite moment about C ? Given: P = 50 N

θ = 60 deg a = 400 mm

b = 300 mm c = 5 d = 12

Solution: (a)

MA = P sin ( θ ) b MA = 13.0 N⋅ m

(b)

MA − F

d 2

c +d

a=0 2

224

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Engineering Mechanics - Statics

Chapter 4

⎛ c2 + d2 ⎞ ⎟ F = MA ⎜ ⎝ da ⎠ F = 35.2 N

Problem 4-22 Determine the clockwise direction θ ( 0 deg ≤ θ ≤ 180 deg) of the force F so that it produces (a) the maximum moment about point A and (b) no moment about point A. Compute the moment in each case. Given: F = 80 lb a = 4 ft b = 1 ft

Solution: (a)

2

MAmax = 330 lb⋅ ft

b φ = atan ⎛⎜ ⎟⎞

φ = 14.0 deg

θ a = 90 deg + φ

θ a = 104 deg

⎝ a⎠

( b)

2

MAmax = F a + b

MAmin = 0 b θ b = atan ⎛⎜ ⎟⎞

⎝ a⎠

θ b = 14.04 deg

Problem 4-23 The Y-type structure is used to support the high voltage transmission cables. If the supporting cables each exert a force F on the structure at B, determine the moment of each force about point A. Also, by the principle of transmissibility, locate the forces at points C and D and determine the moments.

225

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Engineering Mechanics - Statics

Chapter 4

Units Used: kip = 1000 lb Given: F = 275 lb a = 85 ft

θ = 30 deg Solution: MA1 = F sin ( θ ) a

MA1 = 11.7 kip⋅ ft

MA2 = F sin ( θ ) a

MA2 = 11.7 kip⋅ ft

Also

b = ( a)tan ( θ )

MA1 = F cos ( θ ) b

MA1 = 11.7 kip⋅ ft

MA2 = F cos ( θ ) b

MA2 = 11.7 kip⋅ ft

Problem 4-24 The force F acts on the end of the pipe at B. Determine (a) the moment of this force about point A, and (b) the magnitude and direction of a horizantal force, applied at C, which produces the same moment. Given: F = 70 N a = 0.9 m b = 0.3 m c = 0.7 m

θ = 60 deg Solution:

(a)

MA = F sin ( θ ) c + F cos ( θ ) a

MA = 73.9 N⋅ m 226

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Engineering Mechanics - Statics

Chapter 4

F C ( a) = MA

(b)

FC =

MA

F C = 82.2 N

a

Problem 4-25 The force F acts on the end of the pipe at B. Determine the angles θ ( 0° ≤ θ ≤ 180° ) of the force that will produce maximum and minimum moments about point A. What are the magnitudes of these moments? Given: F = 70 N a = 0.9 m b = 0.3 m c = 0.7 m Solution: MA = F sin ( θ ) c + F cos ( θ ) a

For maximum moment

d dθ

MA = c F cos ( θ ) − a F sin ( θ ) = 0

c θ max = atan ⎛⎜ ⎟⎞

θ max = 37.9 deg

MAmax = F sin ( θ max) c + F cos ( θ max) a

MAmax = 79.812 N⋅ m

⎝ a⎠

For minimum moment

MA = F sin ( θ ) c + F cos ( θ ) a = 0 −a ⎞ ⎟ ⎝c⎠

θ min = 180 deg + atan ⎛⎜

θ min = 128 deg

MAmin = F c sin ( θ min ) + F ( a) cos ( θ min )

MAmin = 0 N⋅ m

Problem 4-26 The towline exerts force P at the end of the crane boom of length L. Determine the placement 227

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Engineering Mechanics - Statics

Chapter 4

g p x of the hook at A so that this force creates a maximum moment about point O. What is this moment? Unit Used: 3

kN = 10 N Given: P = 4 kN L = 20 m

θ = 30 deg a = 1.5 m Solution: Maximum moment, OB ⊥ BA Guesses

x = 1m

d = 1 m (Length of the cable from B to A)

Given

L cos ( θ ) + d sin ( θ ) = x a + L sin ( θ ) = d cos ( θ )

⎛x⎞ ⎜ ⎟ = Find ( x , d) ⎝d⎠

x = 23.96 m

Mmax = P L

Mmax = 80 kN⋅ m

Problem 4-27 The towline exerts force P at the end of the crane boom of length L. Determine the position θ of the boom so that this force creates a maximum moment about point O. What is this moment? Units Used: 3

kN = 10 N

228

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Engineering Mechanics - Statics

Chapter 4

Given: P = 4 kN x = 25 m L = 20 m a = 1.5 m Solution: Maximum moment, OB ⊥ BA Guesses

θ = 30 deg

Given

L cos ( θ ) + d sin ( θ ) = x

d = 1m

(length of cable from B to A)

a + L sin ( θ ) = d cos ( θ )

⎛θ⎞ ⎜ ⎟ = Find ( θ , d) ⎝d ⎠

θ = 33.573 deg

Mmax = P L

Mmax = 80 kN⋅ m

Problem 4-28 Determine the resultant moment of the forces about point A. Solve the problem first by considering each force as a whole, and then by using the principle of moments.

229

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: F 1 = 250 N

a = 2m

F 2 = 300 N

b = 3m

F 3 = 500 N

c = 4m

θ 1 = 60 deg

d = 3

θ 2 = 30 deg

e = 4

Solution Using Whole Forces:

Geometry

d α = atan ⎛⎜ ⎟⎞

⎝e⎠

L = ⎛⎜ a + b −

d

⎝

e

c⎟⎞

e

⎠ e2 + d2

MA = −F1 ⎡⎣( a)cos ( θ 2 )⎤⎦ − F2 ( a + b) sin ( θ 1 ) − F3 L

MA = −2.532 kN⋅ m

Solution Using Principle of Moments: MA = −F1 cos ( θ 2 ) a − F2 sin ( θ 1 ) ( a + b) + F3

d 2

2

d +e

c − F3

e 2

( a + b) 2

d +e

3

MA = −2.532 × 10 N⋅ m

Problem 4-29 If the resultant moment about point A is M clockwise, determine the magnitude of F 3.

230

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: M = 4.8 kN⋅ m a = 2 m F 1 = 300 N

b = 3m

F 2 = 400 N

c = 4m

θ 1 = 60 deg

d = 3

θ 2 = 30 deg

e = 4

Solution: Initial Guess

F3 = 1 N

Given −M = −F 1 cos ( θ 2 ) a − F 2 sin ( θ 1 ) ( a + b) + F 3 F 3 = Find ( F3 )

e ⎛ d ⎞c − F ⎛ ⎞ 3 ⎜ 2 2⎟ ⎜ 2 2 ⎟ ( a + b) ⎝ d +e ⎠ ⎝ d +e ⎠

F 3 = 1.593 kN

Problem 4-30 The flat-belt tensioner is manufactured by the Daton Co. and is used with V-belt drives on poultry and livestock fans. If the tension in the belt is F, when the pulley is not turning, determine the moment of each of these forces about the pin at A.

231

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Engineering Mechanics - Statics

Chapter 4

Given: F = 52 lb a = 8 in b = 5 in c = 6 in

θ 1 = 30 deg θ 2 = 20 deg

Solution: MA1 = F cos ( θ 1 ) ( a + c cos ( θ 1 ) ) − F sin ( θ 1 ) ( b − c sin ( θ 1 ) ) MA1 = 542 lb⋅ in MA2 = F cos ( θ 2 ) ( a − c cos ( θ 2 ) ) − F ( sin ( θ 2 ) ) ( b + c sin ( θ 2 ) ) MA2 = −10.01 lb⋅ in

Problem 4-31 The worker is using the bar to pull two pipes together in order to complete the connection. If he applies a horizantal force F to the handle of the lever, determine the moment of this force about the end A. What would be the tension T in the cable needed to cause the opposite moment about point A. Given: F = 80 lb

θ 1 = 40 deg

θ 2 = 20 deg

a = 0.5 ft

b = 4.5 ft

232

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Engineering Mechanics - Statics

Chapter 4

Solution: MA = F( a + b) cos ( θ 1 ) MA = 306 lb⋅ ft Require

MA = T cos ( θ 2 ) ( a) cos ( θ 1 ) + T sin ( θ 2 ) ( a) sin ( θ 1 ) T =

MA

( a) ( cos ( θ 2 ) cos ( θ 1 ) + sin ( θ 2 ) sin ( θ 1 ) )

T = 652 lb

Problem 4-32 If it takes a force F to pull the nail out, determine the smallest vertical force P that must be applied to the handle of the crowbar. Hint: This requires the moment of F about point A to be equal to the moment of P about A. Why? Given: F = 125 lb a = 14 in b = 3 in c = 1.5 in

θ 1 = 20 deg θ 2 = 60 deg

233

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Engineering Mechanics - Statics

Chapter 4

Solution: MF = F sin ( θ 2 ) ( b)

MF = 325 lb⋅ in

P ⎡⎣( a)cos ( θ 1 ) + ( c)sin ( θ 1 )⎤⎦ = MF

P =

MF

( a) cos ( θ 1 ) + ( c) sin ( θ 1 )

P = 23.8 lb

Problem 4-33 The pipe wrench is activated by pulling on the cable segment with a horizantal force F . Determine the moment MA produced by the wrench on the pipe at θ. Neglect the size of the pulley. Given: F = 500 N a = 0.2 m b = 0.5 m c = 0.4 m

θ = 20 deg

Solution: Initial Guesses

φ = 20 deg MA = 1 N⋅ m Given b − c sin ( θ )

c cos ( θ ) − a

= tan ( φ − θ )

MA = F c sin ( φ )

⎛ φ ⎞ ⎜ ⎟ = Find ( φ , MA) ⎝ MA ⎠

φ = 84.161 deg

MA = 199 N⋅ m

234

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Engineering Mechanics - Statics

Chapter 4

Problem 4-34 Determine the moment of the force at A about point O. Express the result as a Cartesian vector. Given:

⎛ 60 ⎞ ⎜ ⎟ F = −30 N ⎜ ⎟ ⎝ −20 ⎠ a = 4m

d = 4m

b = 7m

e = 6m

c = 3m

f = 2m

Solution:

rOA

⎛ −c ⎞ ⎜ ⎟ = −b ⎜ ⎟ ⎝a⎠

MO = rOA × F

⎛ 260 ⎞ ⎜ ⎟ MO = 180 N⋅ m ⎜ ⎟ ⎝ 510 ⎠

Problem 4-35 Determine the moment of the force at A about point P. Express the result as a Cartesian vector. Given: a = 4m

b = 7m

c = 3m

d = 4m

e = 6m

f = 2m

⎛ 60 ⎞ ⎜ ⎟ F = −30 N ⎜ ⎟ ⎝ −20 ⎠ Solution:

rPA

⎛ −c − d ⎞ ⎜ ⎟ = −b − e ⎜ ⎟ ⎝ a+ f ⎠

MP = rPA × F

⎛ 440 ⎞ ⎜ ⎟ MP = 220 N⋅ m ⎜ ⎟ ⎝ 990 ⎠

235

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Engineering Mechanics - Statics

Chapter 4

Problem 4-36 Determine the moment of the force F at A about point O. Express the result as a cartesian vector. Units Used: 3

kN = 10 N Given: F = 13 kN a = 6m b = 2.5 m c = 3m d = 3m e = 8m f = 6m g = 4m h = 8m Solution:

rAB

⎛b − g⎞ ⎜ ⎟ = c+d ⎜ ⎟ ⎝h − a⎠

MO = rOA × F1

rOA

⎛ −b ⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝a⎠

F1 = F

rAB rAB

⎛ −84 ⎞ ⎜ ⎟ MO = −8 kN⋅ m ⎜ ⎟ ⎝ −39 ⎠

Problem 4-37 Determine the moment of the force F at A about point P. Express the result as a Cartesian vector. Units Used:

3

kN = 10 N

236

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Engineering Mechanics - Statics

Chapter 4

Given: F = 13 kN a = 6m b = 2.5 m c = 3m d = 3m e = 8m f = 6m g = 4m h = 8m Solution:

rAB

⎛b − g⎞ ⎜ ⎟ = c+d ⎜ ⎟ ⎝h − a⎠

MO = rPA × F 1

rPA

⎛ −b − f ⎞ ⎜ ⎟ = −c − e ⎜ ⎟ ⎝ a ⎠

F1 = F

rAB rAB

⎛ −116 ⎞ ⎜ 16 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −135 ⎠

Problem 4-38 The curved rod lies in the x-y plane and has radius r. If a force F acts at its end as shown, determine the moment of this force about point O. Given: r = 3m

a = 1m

θ = 45 deg

F = 80 N b = 2 m

237

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Engineering Mechanics - Statics

Chapter 4

Solution:

rAC

⎛a⎞ ⎜ ⎟ = −r ⎜ ⎟ ⎝ −b ⎠

Fv = F

rOA

rAC

⎛1⎞ ⎜ ⎟ = −3 m ⎜ ⎟ ⎝ −2 ⎠

⎛ 21.381 ⎞ ⎜ ⎟ F v = −64.143 N ⎜ ⎟ ⎝ −42.762 ⎠

rAC rAC

⎛r⎞ ⎜ ⎟ = r ⎜ ⎟ ⎝0⎠

rOA

MO = rOA × Fv

⎛3⎞ ⎜ ⎟ = 3 m ⎜ ⎟ ⎝0⎠

⎛ −128.285 ⎞ ⎜ ⎟ MO = 128.285 N⋅ m ⎜ ⎟ ⎝ −256.571 ⎠

Problem 4-39 The curved rod lies in the x-y plane and has a radius r. If a force F acts at its end as shown, determine the moment of this force about point B. Given: F = 80 N

c = 3m

a = 1m

r = 3m

238

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Engineering Mechanics - Statics

Chapter 4

θ = 45 deg

b = 2m

Solution:

rAC

⎛a⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝ −b ⎠

Fv = F

rAC rBA

rAC

MB = rBA × F v

⎛ r cos ( θ ) ⎞ ⎜ ⎟ = r − r sin ( θ ) ⎜ ⎟ 0 ⎠ ⎝

⎛ −37.6 ⎞ ⎜ 90.7 ⎟ N⋅ m MB = ⎜ ⎟ ⎝ −154.9 ⎠

Problem 4-40 The force F acts at the end of the beam. Determine the moment of the force about point A. Given:

⎛ 600 ⎞ ⎜ ⎟ F = 300 N ⎜ ⎟ ⎝ −600 ⎠ a = 1.2 m b = 0.2 m c = 0.4 m Solution:

rAB

⎛b⎞ ⎜ ⎟ = a ⎜ ⎟ ⎝0⎠

MA = rAB × F

⎛ −720 ⎞ ⎜ ⎟ MA = 120 N⋅ m ⎜ ⎟ ⎝ −660 ⎠

239

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Engineering Mechanics - Statics

Chapter 4

Problem 4-41 The pole supports a traffic light of weight W. Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A. Given: W = 22 lb

a = 12 ft

θ = 30 deg

Solution:

⎡ ( a)sin ( θ ) ⎤ ⎢ ⎥ r = ( a)cos ( θ ) ⎢ ⎥ ⎣ 0 ⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ F = ⎜ ⎟ ⎝ −W ⎠ MA = r × F

⎛ −229 ⎞ ⎜ ⎟ MA = 132 lb⋅ ft ⎜ ⎟ ⎝ 0 ⎠

Problem 4-42 The man pulls on the rope with a force F. Determine the moment that this force exerts about the base of the pole at O. Solve the problem two ways, i.e., by using a position vector from O to A, then O to B. Given: F = 20 N a = 3m

240

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Engineering Mechanics - Statics

Chapter 4

b = 4m c = 1.5 m d = 10.5 m

Solution:

rAB

rOB

⎛ b ⎞ ⎜ −a ⎟ = ⎜ ⎟ ⎝c − d⎠ ⎛b⎞ ⎜ ⎟ = −a ⎜ ⎟ ⎝c ⎠

MO1 = rOA × F v

MO2 = rOB × Fv

rOA

⎛0⎞ ⎜ ⎟ = 0 ⎜ ⎟ ⎝d⎠

Fv = F

rAB rAB

MO1

⎛ 61.2 ⎞ ⎜ ⎟ = 81.6 N⋅ m ⎜ ⎟ ⎝ 0 ⎠

MO2

⎛ 61.2 ⎞ ⎜ ⎟ = 81.6 N⋅ m ⎜ ⎟ ⎝ −0 ⎠

Problem 4-43 Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has radius r, to fail at the support C. This requires a moment to be developed at C of magnitude M.

241

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Engineering Mechanics - Statics

Chapter 4

Given: r = 5 ft M = 80 lb⋅ ft

θ = 60 deg a = 7 ft b = 6 ft

Solution:

rAB

b ⎞ ⎛ ⎜ = a − r sin ( θ ) ⎟ ⎜ ⎟ ⎝ −r cos ( θ ) ⎠

Guess Given

uAB =

rAB

rCB

rAB

⎛b⎞ = ⎜a⎟ ⎜ ⎟ ⎝ −r ⎠

F = 1 lb rCB × ( F uAB) = M

F = Find ( F)

F = 18.6 lb

Problem 4-44 The pipe assembly is subjected to the force F . Determine the moment of this force about point A.

242

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Engineering Mechanics - Statics

Chapter 4

Given: F = 80 N a = 400 mm b = 300 mm c = 200 mm d = 250 mm

θ = 40 deg φ = 30 deg Solution:

rAC

⎛b + d⎞ ⎜ a ⎟ = ⎜ ⎟ ⎝ −c ⎠

MA = rAC × Fv

⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟ F v = F ⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ −sin ( φ ) ⎟ ⎝ ⎠ ⎛ −5.385 ⎞ ⎜ ⎟ MA = 13.093 N⋅ m ⎜ ⎟ ⎝ 11.377 ⎠

Problem 4-45 The pipe assembly is subjected to the force F . Determine the moment of this force about point B.

243

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Engineering Mechanics - Statics

Chapter 4

Given: F = 80 N a = 400 mm b = 300 mm c = 200 mm d = 250 mm

θ = 40 deg φ = 30 deg Solution:

rBC

⎛b + d⎞ = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎛ 550 ⎞ rBC = ⎜ 0 ⎟ mm ⎜ ⎟ ⎝ −200 ⎠

⎛⎜ cos ( φ ) sin ( θ ) ⎞⎟ F v = F ⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ −sin ( φ ) ⎟ ⎝ ⎠ MB = rBC × F v

⎛ 44.534 ⎞ F v = ⎜ 53.073 ⎟ N ⎜ ⎟ ⎝ −40 ⎠ ⎛ 10.615 ⎞ MB = ⎜ 13.093 ⎟ N⋅ m ⎜ ⎟ ⎝ 29.19 ⎠

Problem 4-46 The x-ray machine is used for medical diagnosis. If the camera and housing at C have mass M and a mass center at G, determine the moment of its weight about point O when it is in the position shown. Units Used: 3

kN = 10 N

244

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Engineering Mechanics - Statics

Chapter 4

Given: M = 150 kg a = 1.2 m b = 1.5 m

θ = 60 deg g = 9.81

m 2

s Solution:

⎡ −( b) cos ( θ ) ⎤ ⎛ 0 ⎞ ⎢ ⎥×⎜ 0 ⎟ MO = a ⎢ ⎥ ⎜ ⎟ ⎣ ( b)sin ( θ ) ⎦ ⎝ −M g ⎠

⎛ −1.77 ⎞ ⎜ ⎟ MO = −1.1 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠

Problem 4-47 Using Cartesian vector analysis, determine the resultant moment of the three forces about the base of the column at A. Units Used : 3

kN = 10 N Given:

⎛ 400 ⎞ ⎜ ⎟ F 1 = 300 N ⎜ ⎟ ⎝ 120 ⎠ ⎛ 100 ⎞ ⎜ ⎟ F 2 = −100 N ⎜ ⎟ ⎝ −60 ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟N F3 = ⎜ ⎟ ⎝ −500 ⎠ a = 4m b = 8m 245

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Engineering Mechanics - Statics

Chapter 4

c = 1m Solution:

rAB

⎛ 0 ⎞ ⎜ 0 ⎟ = ⎜ ⎟ ⎝a + b⎠

rA3

⎛0⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝b⎠

The individual moments MA1 = rAB × F1

MA2 = rAB × F2

MA3 = rA3 × F3

⎛ −3.6 ⎞ ⎜ ⎟ MA1 = 4.8 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠

⎛ 1.2 ⎞ ⎜ ⎟ MA2 = 1.2 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠

⎛ 0.5 ⎞ ⎜ ⎟ MA3 = 0 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠

The total moment MA = MA1 + MA2 + MA3

⎛ −1.9 ⎞ ⎜ 6 ⎟ kN⋅ m MA = ⎜ ⎟ ⎝ 0 ⎠

Problem 4-48 A force F produces a moment MO about the origin of coordinates, point O. If the force acts at a point having the given x coordinate, determine the y and z coordinates. Units Used :

3

kN = 10 N

Given:

⎛6⎞ ⎜ ⎟ F = −2 kN ⎜ ⎟ ⎝1⎠ ⎛ 4 ⎞ ⎜ 5 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −14 ⎠ x = 1m Solution: The initial guesses: y = 1m

z = 1m 246

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Engineering Mechanics - Statics

Chapter 4

Given

⎛x⎞ ⎜ y⎟ × F = M O ⎜ ⎟ ⎝z⎠

⎛ y⎞ ⎜ ⎟ = Find ( y , z) ⎝z⎠

⎛ y⎞ ⎛2⎞ ⎜ ⎟=⎜ ⎟m ⎝ z⎠ ⎝1⎠

Problem 4-49 The force F creates a moment about point O of MO. If the force passes through a point having the given x coordinate, determine the y and z coordinates of the point. Also, realizing that MO = Fd, determine the perpendicular distance d from point O to the line of action of F . Given:

⎛6⎞ F = ⎜8 ⎟ N ⎜ ⎟ ⎝ 10 ⎠ ⎛ −14 ⎞ MO = ⎜ 8 ⎟ N⋅ m ⎜ ⎟ ⎝ 2 ⎠ x = 1m Solution: The initial guesses:

y = 1m

z = 1m

Given

⎛x⎞ ⎜ y⎟ × F = M O ⎜ ⎟ ⎝z⎠ d =

MO F

⎛ y⎞ ⎜ ⎟ = Find ( y , z) ⎝z⎠

⎛ y⎞ ⎛1⎞ ⎜ ⎟=⎜ ⎟m ⎝ z⎠ ⎝3⎠

d = 1.149 m

Problem 4-50 The force F produces a moment MO about the origin of coordinates, point O. If the force acts at a point having the given x-coordinate, determine the y and z coordinates.

247

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: x = 1m

⎛6⎞ ⎜ ⎟ F = −2 kN ⎜ ⎟ ⎝1⎠ ⎛ 4 ⎞ ⎜ 5 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −14 ⎠ Solution: y = 1m

Initial Guesses:

Given

⎛x ⎞ ⎜ ⎟ MO = y × F ⎜ ⎟ ⎝z⎠

z = 1m

⎛ y⎞ ⎜ ⎟ = Find ( y , z) ⎝z⎠

⎛ y⎞ ⎛2⎞ ⎜ ⎟=⎜ ⎟m ⎝ z⎠ ⎝1⎠

Problem 4-51 Determine the moment of the force F about the Oa axis. Express the result as a Cartesian vector. Given:

⎛ 50 ⎞ ⎜ ⎟ F = −20 N ⎜ ⎟ ⎝ 20 ⎠ a = 6m b = 2m c = 1m d = 3m e = 4m

248

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Engineering Mechanics - Statics

Chapter 4

Solution:

rOF

⎛c ⎞ = ⎜ −b ⎟ ⎜ ⎟ ⎝a⎠

MOa =

rOa

⎛0⎞ = ⎜e⎟ ⎜ ⎟ ⎝d⎠

⎡⎣( rOF × F) ⋅ uOa⎤⎦ uOa

uOa =

MOa

rOa rOa

⎛ 0 ⎞ = ⎜ 217.6 ⎟ N⋅ m ⎜ ⎟ ⎝ 163.2 ⎠

Problem 4-52 Determine the moment of the force F about the aa axis. Express the result as a Cartesian vector. Given: F = 600 lb a = 6 ft b = 3 ft c = 2 ft d = 4 ft e = 4 ft f = 2 ft

Solution:

Fv =

Maa =

⎛ −d ⎞ ⎜e ⎟ ⎟ 2 2 2⎜ c +d +e ⎝ c ⎠

⎛d⎞ r = ⎜0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎡⎣( r × Fv) ⋅ uaa⎤⎦ uaa

⎛ −441 ⎞ Maa = ⎜ −294 ⎟ lb⋅ ft ⎜ ⎟ ⎝ 882 ⎠

F

uaa =

⎛ −b ⎞ ⎜−f ⎟ ⎟ 2 2 2⎜ a +b + f ⎝ a ⎠ 1

249

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Engineering Mechanics - Statics

Chapter 4

Problem 4-53 Determine the resultant moment of the two forces about the Oa axis. Express the result as a Cartesian vector. Given: F 1 = 80 lb F 2 = 50 lb

α = 120 deg β = 60 deg γ = 45 deg a = 5 ft b = 4 ft c = 6 ft

θ = 30 deg φ = 30 deg Solution:

F 1v

⎛⎜ cos ( α ) ⎟⎞ = F1 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

⎡ ( b)sin ( θ ) ⎤ ⎢ ⎥ r1 = ( b)cos ( θ ) ⎢ ⎥ ⎣ c ⎦

uaa

⎛ cos ( φ ) ⎞ ⎜ ⎟ = −sin ( φ ) ⎜ ⎟ ⎝ 0 ⎠

F 2v

⎛⎜ 0 ⎞⎟ = ⎜ 0 ⎟ ⎜ F2 ⎟ ⎝ ⎠

0 ⎤ ⎡ ⎢ r2 = −( a) sin ( φ ) ⎥ ⎢ ⎥ 0 ⎣ ⎦

Maa =

⎡⎣( r1 × F1v + r2 × F2v) uaa⎤⎦ uaa

⎛ 26.132 ⎞ ⎜ ⎟ Maa = −15.087 lb⋅ ft ⎜ ⎟ ⎝ 0 ⎠

250

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Engineering Mechanics - Statics

Chapter 4

Problem 4-54 The force F is applied to the handle of the box wrench. Determine the component of the moment of this force about the z axis which is effective in loosening the bolt. Given: a = 3 in b = 8 in c = 2 in

⎛8⎞ F = ⎜ −1 ⎟ lb ⎜ ⎟ ⎝1⎠ Solution:

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠

⎛c ⎞ r = ⎜ −b ⎟ ⎜ ⎟ ⎝a⎠

Mz = ( r × F ) ⋅ k

Mz = 62 lb⋅ in

Problem 4-55 The force F acts on the gear in the direction shown. Determine the moment of this force about the y axis. Given: F = 50 lb a = 3 in

θ 1 = 60 deg θ 2 = 45 deg θ 3 = 120 deg

251

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛0⎞ j = ⎜1⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ r = ⎜0⎟ ⎜ ⎟ ⎝a⎠

⎛ −cos ( θ 3) ⎞ ⎜ ⎟ F v = F ⎜ −cos ( θ 2 ) ⎟ My = ( r × Fv ) ⋅ j ⎜ −cos θ ⎟ ( 1) ⎠ ⎝

My = 75 lb⋅ in

Problem 4-56 The RollerBall skate is an in-line tandem skate that uses two large spherical wheels on each skate, rather than traditional wafer-shape wheels. During skating the two forces acting on the wheel of one skate consist of a normal force F 2 and a friction force F1. Determine the moment of both of these forces about the axle AB of the wheel. Given:

θ = 30 deg F 1 = 13 lb F 2 = 78 lb a = 1.25 in Solution:

⎛ F1 ⎞ ⎛0⎞ ⎜ ⎟ F = ⎜ F2 ⎟ r = ⎜ −a ⎟ ⎜ ⎟ ⎜0 ⎟ ⎝0⎠ ⎝ ⎠ ⎛ cos ( θ ) ⎞ ab = ⎜ −sin ( θ ) ⎟ ⎜ ⎟ ⎝ 0 ⎠

Mab = ( r × F ) ⋅ ab

Mab = 0 lb⋅ in

Problem 4-57 The cutting tool on the lathe exerts a force F on the shaft in the direction shown. Determine the moment of this force about the y axis of the shaft.

252

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given:

⎛6⎞ ⎜ ⎟ F = −4 kN ⎜ ⎟ ⎝ −7 ⎠ a = 30 mm

θ = 40 deg Solution:

⎛ cos ( θ ) ⎞ ⎜ 0 ⎟ r = a ⎜ ⎟ ⎝ sin ( θ ) ⎠

⎛0⎞ ⎜ ⎟ j = 1 ⎜ ⎟ ⎝0⎠

My = ( r × F) ⋅ j

My = 0.277 kN⋅ m

Problem 4-58 The hood of the automobile is supported by the strut AB, which exerts a force F on the hood. Determine the moment of this force about the hinged axis y. Given: F = 24 lb

a = 2 ft

b = 4 ft

c = 2 ft

d = 4 ft

253

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛b⎞ ⎜ ⎟ rA = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ j = 1 ⎜ ⎟ ⎝0⎠

rAB

⎛ −b + c ⎞ ⎜ a ⎟ = ⎜ ⎟ ⎝ d ⎠

Fv = F

My = ( rA × F v ) ⋅ j

rAB rAB

⎛ −9.798 ⎞ ⎜ ⎟ F v = 9.798 lb ⎜ ⎟ ⎝ 19.596 ⎠

My = −78.384 lb⋅ ft

Problem 4-59 The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F at A. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. If the force F can be applied at A in any other direction, will it be possible to turn the nut?

254

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Engineering Mechanics - Statics

Chapter 4

Given: F = 30 N M = 14 N⋅ m a = 0.25 m b = 0.3 m c = 0.5 m d = 0.1 m Solution: 2

Mx = F c − b

2

Mx = 12 N⋅ m Mx < M

No

For Mxmax, apply force perpendicular to the handle and the x-axis. Mxmax = F c Mxmax = 15 N⋅ m Mxmax > M

Yes

Problem 4-60 The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F . Assume that the cheater pipe AB is slipped over the handle of the wrench and the F force can be applied at any point and in any direction on the assembly. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. Given: F 1 = 30 N

M = 14 N⋅ m

a = 0.25 m

b = 0.3 m

c = 0.5 m

d = 0.1 m

255

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Engineering Mechanics - Statics

Chapter 4

Solution: Mx = F1

a+c c

2

2

c −b

Mx = 18 N⋅ m Mx > M

Yes

Mxmax occurs when force is applied perpendicular to both the handle and the x-axis. Mxmax = F1 ( a + c) Mxmax = 22.5 N⋅ m Mxmax > M

Yes

Problem 4-61 The bevel gear is subjected to the force F which is caused from contact with another gear. Determine the moment of this force about the y axis of the gear shaft. Given: a = 30 mm b = 40 mm

⎛ 20 ⎞ ⎜ 8 ⎟N F = ⎜ ⎟ ⎝ −15 ⎠

256

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛ −b ⎞ ⎜ ⎟ r = 0 ⎜ ⎟ ⎝a⎠

⎛0⎞ ⎜ ⎟ j = 1 ⎜ ⎟ ⎝0⎠

My = ( r × F) ⋅ j

My = 0 N⋅ m

Problem 4-62 The wooden shaft is held in a lathe. The cutting tool exerts force F on the shaft in the direction shown. Determine the moment of this force about the x axis of the shaft. Express the result as a Cartesian vector. The distance OA is a. Given: a = 25 mm

θ = 30 deg

⎛ −5 ⎞ ⎜ ⎟ F = −3 N ⎜ ⎟ ⎝8⎠ Solution:

⎡ 0 ⎤ ⎢ ⎥ r = ( a)cos ( θ ) ⎢ ⎥ ⎣ ( a)sin ( θ ) ⎦

⎛1⎞ ⎜ ⎟ i = 0 ⎜ ⎟ ⎝0⎠

Mx =

⎡⎣( r × F) ⋅ i⎤⎦ i

⎛ 0.211 ⎞ ⎜ 0 ⎟ N⋅ m Mx = ⎜ ⎟ ⎝ 0 ⎠

Problem 4-63 Determine the magnitude of the moment of the force F about the base line CA of the tripod. Given:

⎛ 50 ⎞ ⎜ ⎟ F = −20 N ⎜ ⎟ ⎝ −80 ⎠

257

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Engineering Mechanics - Statics

Chapter 4

a = 4m b = 2.5 m c = 1m d = 0.5 m e = 2m f = 1.5 m g = 2m

Solution:

rCA

⎛ −g ⎞ = ⎜ e ⎟ ⎜ ⎟ ⎝0⎠

uCA =

rCA rCA

rCD

⎛b − g⎞ = ⎜ e ⎟ ⎜ ⎟ ⎝ a ⎠

MCA = ( rCD × F) ⋅ uCA

MCA = 226 N⋅ m

Problem 4-64 The flex-headed ratchet wrench is subjected to force P, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A. Given: P = 16 lb

a = 10 in

θ = 60 deg b = 0.75 in Solution: M = P ⎡⎣b + ( a)sin ( θ )⎤⎦

M = 150.564 lb⋅ in

258

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Engineering Mechanics - Statics

Chapter 4

Problem 4-65 If a torque or moment M is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench. Given: M = 80 lb⋅ in

θ = 60 deg a = 10 in b = 0.75 in Solution: M = P ⎡⎣b + ( a)sin ( θ )⎤⎦

P =

M

b + ( a)sin ( θ )

P = 8.50 lb

Problem 4-66 The A-frame is being hoisted into an upright position by the vertical force F . Determine the moment of this force about the y axis when the frame is in the position shown. Given: F = 80 lb a = 6 ft b = 6 ft

θ = 30 deg φ = 15 deg Solution: Using the primed coordinates we have

259

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Engineering Mechanics - Statics

⎛ −sin ( θ ) ⎞ ⎜ ⎟ j = cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

Chapter 4

⎛0⎞ ⎜ ⎟ Fv = F 0 ⎜ ⎟ ⎝1⎠

My = ( rAC × F v ) ⋅ j

rAC

⎛ −b cos ( φ ) ⎞ ⎜ ⎟ a ⎟ = ⎜ 2 ⎜ ⎟ ⎜ b sin ( φ ) ⎟ ⎝ ⎠

My = 281.528 lb⋅ ft

Problem 4-67 Determine the moment of each force acting on the handle of the wrench about the a axis. Given:

⎛ −2 ⎞ ⎜ ⎟ F 1 = 4 lb ⎜ ⎟ ⎝ −8 ⎠

⎛3⎞ ⎜ ⎟ F 2 = 2 lb ⎜ ⎟ ⎝ −6 ⎠

b = 6 in c = 4 in d = 3.5 in

θ = 45 deg

Solution:

⎛ cos ( θ ) ⎞ ⎜ 0 ⎟ ua = ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎜ 0 ⎟ r1 = b 0 + ( c + d) ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠ M1a = ( r1 × F1 ) ⋅ ua

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ + c⎜ 0 ⎟ r2 = b ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

M1a = 30 lb⋅ in

260

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Engineering Mechanics - Statics

M2a = ( r2 × F2 ) ⋅ ua

Chapter 4

M2a = 8 lb⋅ in

Problem 4-68 Determine the moment of each force acting on the handle of the wrench about the z axis. Given:

⎛ −2 ⎞ ⎜ ⎟ F 1 = 4 lb ⎜ ⎟ ⎝ −8 ⎠

⎛3⎞ ⎜ ⎟ F 2 = 2 lb ⎜ ⎟ ⎝ −6 ⎠

b = 6 in c = 4 in d = 3.5 in

θ = 45 deg

Solution:

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎜ 0 ⎟ r1 = b 0 + ( c + d) ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ + c⎜ 0 ⎟ r2 = b ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠

M1z = ( r1 × F 1 ) ⋅ k

M1z = 38.2 lb⋅ in

M2z = ( r2 × F 2 ) ⋅ k

M2z = 14.1 lb⋅ in

⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠

Problem 4-69 Determine the magnitude and sense of the couple moment. Units Used: 3

kN = 10 N 261

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Engineering Mechanics - Statics

Chapter 4

Given: F = 5 kN

θ = 30 deg a = 0.5 m b = 4m c = 2m d = 1m Solution: MC = F cos ( θ ) ( a + c) + F sin ( θ ) ( b − d) MC = 18.325 kN⋅ m

Problem 4-70 Determine the magnitude and sense of the couple moment. Each force has a magnitude F. Given: F = 65 lb a = 2 ft b = 1.5 ft c = 4 ft d = 6 ft e = 3 ft

Solution: Mc = ΣMB;

⎡ ⎛

⎞ ( d + a)⎤ + ⎡F⎛ e ⎞ ( c + a)⎤ ⎥ ⎢ ⎜ 2 2⎟ ⎥ 2 2⎟ ⎣ ⎝ c +e ⎠ ⎦ ⎣ ⎝ c +e ⎠ ⎦

MC = ⎢F⎜

c

MC = 650 lb⋅ ft

(Counterclockwise)

Problem 4-71 Determine the magnitude and sense of the couple moment. 262

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kip = 10 lb Given: F = 150 lb a = 8 ft b = 6 ft c = 8 ft d = 6 ft e = 6 ft f = 8 ft Solution: MC = ΣMA;

MC = F

d 2

d + f MC = 3120 lb⋅ ft

f

( a + f) + F 2

2

d + f

( c + d) 2

MC = 3.120 kip⋅ ft

Problem 4-72 If the couple moment has magnitude M, determine the magnitude F of the couple forces. Given: M = 300 lb⋅ ft a = 6 ft b = 12 ft c = 1 ft d = 2 ft e = 12 ft f = 7 ft

263

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Engineering Mechanics - Statics

Chapter 4

Solution:

M=F

( f − d) ( b + e) ⎤ ⎡ e( f + a) − ⎢ 2 2 2 2⎥ ( f − d) + e ⎦ ⎣ ( f − d) + e

M

F =

e( f + a) 2

−

F = 108 lb

( f − d) ( b + e)

2

2

( f − d) + e

2

( f − d) + e

Problem 4-73 A clockwise couple M is resisted by the shaft of the electric motor. Determine the magnitude of the reactive forces −R and R which act at supports A and B so that the resultant of the two couples is zero. Given: a = 150 mm

θ = 60 deg M = 5 N⋅ m

Solution: MC = −M +

2R a

tan ( θ )

=0

R =

M tan ( θ ) 2

a

R = 28.9 N

Problem 4-74 The resultant couple moment created by the two couples acting on the disk is MR. Determine the magnitude of force T.

264

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kip = 10 lb Given:

⎛0⎞ ⎜ ⎟ MR = 0 kip⋅ in ⎜ ⎟ ⎝ 10 ⎠ a = 4 in b = 2 in c = 3 in Solution: Initial Guess

Given

T = 1 kip

⎛ a ⎞ ⎛ 0 ⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛ −b − c ⎞ ⎛ 0 ⎞ ⎜ 0 ⎟ × ⎜ T ⎟ + ⎜ 0 ⎟ × ⎜ −T ⎟ + ⎜ 0 ⎟ × ⎜ −T ⎟ = M R ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠

T = Find ( T)

T = 0.909 kip

Problem 4-75 Three couple moments act on the pipe assembly. Determine the magnitude of M3 and the bend angle θ so that the resultant couple moment is zero.

Given:

θ 1 = 45 deg M1 = 900 N⋅ m M2 = 500 N⋅ m

265

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Engineering Mechanics - Statics

Chapter 4

Solution: Initial guesses:

θ = 10 deg

M3 = 10 N⋅ m

Given + ΣMx = 0; →

M1 − M3 cos ( θ ) − M2 cos ( θ 1 ) = 0

+

M3 sin ( θ ) − M2 sin ( θ 1 ) = 0

↑ ΣMy = 0;

⎛ θ ⎞ ⎜ ⎟ = Find ( θ , M3) ⎝ M3 ⎠

θ = 32.9 deg

M3 = 651 N⋅ m

Problem 4-76 The floor causes couple moments MA and MB on the brushes of the polishing machine. Determine the magnitude of the couple forces that must be developed by the operator on the handles so that the resultant couple moment on the polisher is zero. What is the magnitude of these forces if the brush at B suddenly stops so that MB = 0? Given: a = 0.3 m MA = 40 N⋅ m MB = 30 N⋅ m Solution:

MA − MB − F 1 a = 0

F1 =

MA − F 2 a = 0

F2 =

MA − MB a MA

F 1 = 33.3 N

F 2 = 133 N

a

266

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Engineering Mechanics - Statics

Chapter 4

Problem 4-77 The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is M clockwise. Given: F 1 = 600 N F 2 = 250 N a = 1m

θ = 40 deg M = 400 N⋅ m Solution: Initial Guess

Given

F1

F = 1N

⎛ a ⎞ − F a − F ⎛ a ⎞ = −M ⎜ ⎟ 2 ⎜ ⎟ ⎝ 2 cos ( θ ) ⎠ ⎝ 2 cos ( θ ) ⎠

F = Find ( F)

F = 830 N

Problem 4-78 Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is M counterclockwise. Where on the beam does the resultant couple moment act? Given: M = 450 lb⋅ ft P = 200 lb a = 1.5 ft b = 1.25 ft c = 2 ft

θ = 30 deg

267

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Engineering Mechanics - Statics

Chapter 4

Solution: MR = Σ M

M = F b cos ( θ ) + P a

F =

M − Pa

b cos ( θ )

F = 139 lb

The resultant couple moment is a free vector. It can act at any point on the beam.

Problem 4-79 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O. Given:

⎛0⎞ F = ⎜0 ⎟ N ⎜ ⎟ ⎝ 25 ⎠ a = 300 mm b = 150 mm c = 400 mm d = 200 mm e = 200 mm

Solution:

( a)

( b)

rAB

⎛ −e − b ⎞ = ⎜ −c + d ⎟ ⎜ ⎟ ⎝ 0 ⎠

M = rAB × F

rOB

⎛a⎞ = ⎜d⎟ ⎜ ⎟ ⎝0⎠

⎛a + b + e⎞ ⎟ c = ⎜ ⎜ ⎟ ⎝ 0 ⎠

rOA

⎛ −5 ⎞ M = ⎜ 8.75 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

⎛ −5 ⎞ M = ⎜ 8.75 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = rOB × F + rOA × ( −F)

268

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Engineering Mechanics - Statics

Chapter 4

Problem 4-80 If the couple moment acting on the pipe has magnitude M, determine the magnitude F of the vertical force applied to each wrench. Given: M = 400 N⋅ m a = 300 mm b = 150 mm c = 400 mm d = 200 mm e = 200 mm

Solution:

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠

rAB

⎛ −e − b ⎞ = ⎜ −c + d ⎟ ⎜ ⎟ ⎝ 0 ⎠

Guesss

F = 1N rAB × ( Fk) = M

Given

F = Find ( F)

F = 992.278 N

Problem 4-81 Determine the resultant couple moment acting on the beam. Solve the problem two ways: (a) sum moments about point O; and (b) sum moments about point A. Units Used: 3

kN = 10 N

269

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 2 kN

θ 1 = 30 deg

F 2 = 8 kN

θ 2 = 45 deg

a = 0.3 m b = 1.5 m c = 1.8 m Solution: MR = ΣMO;

( a)

MRa = ( F1 cos ( θ 1 ) + F 2 cos ( θ 2 ) ) c + ( F2 cos ( θ 2 ) − F 1 sin ( θ 1 ) ) a ... + −( F2 cos ( θ 2 ) + F 1 cos ( θ 1 ) ) ( b + c) MRa = −9.69 kN⋅ m MR = ΣMA;

( b)

MRb = ( F2 sin ( θ 2 ) − F1 sin ( θ 1 ) ) a − ( F 2 cos ( θ 2 ) + F1 cos ( θ 1 ) ) b MRb = −9.69 kN⋅ m

Problem 4-82 Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is M counterclockwise. Where on the beam does the resultant couple act? Given: M = 300 lb⋅ ft a = 4 ft b = 1.5 ft P = 200 lb c = 3 d = 4

270

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Engineering Mechanics - Statics

Chapter 4

Solution: c

M=

Fa +

2

2

2

2⎛ M + P b ⎞

c +d F =

d

c +d

F = 167 lb

2

c +d

Fb − Pb 2

⎜ ⎟ ⎝ ca + d b⎠

Resultant couple can act anywhere.

Problem 4-83 Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the couple forces F 1. Given: F 1 = 80 lb F 2 = 50 lb a = 1 ft b = 3 ft c = 2 ft e = 3 ft f = 3 g = 4

θ = 30 deg Solution: g ⎡−F cos ( θ ) e + ⎛ ⎞ ⎤ ⎢ 2 ⎜ 2 2 ⎟ F1 d⎥ = 0 ⎣ ⎝ g +f ⎠ ⎦

d =

F2 F1

⎛ g2 + f 2 ⎞ ⎟ g ⎝ ⎠

cos ( θ ) e ⎜

d = 2.03 ft

271

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Engineering Mechanics - Statics

Chapter 4

Problem 4-84 Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4-13) and (b) summing the moments of all the force components about point A. Given: F 1 = 80 lb

c = 2 ft

g = 4

F 2 = 50 lb

d = 4 ft

θ = 30 deg

a = 1 ft

e = 3 ft

b = 3 ft

f = 3

Solution: ( a)

M = Σ ( r × F)

⎛ e ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −g ⎞⎤ 1 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ M = 0 × F 2 −cos ( θ ) + d × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦

⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft M= ⎜ ⎟ ⎝ 126.096 ⎠

( b)

Summing the moments of all force components aboout point A.

M1 =

g ⎛ −g ⎞ F b + ⎛ ⎞ 1 ⎜ 2 2⎟ ⎜ 2 2 ⎟ F 1 ( b + d) ⎝ f +g ⎠ ⎝ f +g ⎠

M2 = F 2 cos ( θ ) c − F 2 sin ( θ ) ( a + b + d) − F 2 cos ( θ ) ( c + e) + F 2 sin ( θ ) ( a + b + d) M = M1 + M2

M = 126.096 lb⋅ ft

Problem 4-85 Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4 -13) and (b) summing the moments of all the force components about point B.

272

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 80 lb

d = 4ft

F 2 = 50 lb

e = 3 ft

a = 1 ft

f = 3 ft

b = 3 ft

g = 4 ft

c = 2 ft

θ = 30 deg

Solution: ( a)

M = Σ ( r × F)

⎛ e ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −g ⎞⎤ 1 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ M = 0 × F 2 −cos ( θ ) + d × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ( b)

⎛ 0 ⎞ ⎟ lb⋅ ft M=⎜ 0 ⎜ ⎟ ⎝ 126.096 ⎠

Summing the moments of all force components about point B. M1 =

g g ⎛ ⎛ ⎞ ⎞ ⎜ 2 2 ⎟ F1( a + d) − ⎜ 2 2 ⎟ F1 a ⎝ f +g ⎠ ⎝ f +g ⎠

M2 = F 2 cos ( θ ) c − F 2 cos ( θ ) ( c + e) M = M1 + M2

M = 126.096 lb⋅ ft

Problem 4-86 Determine the couple moment. Express the result as a Cartesian vector.

273

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Engineering Mechanics - Statics

Chapter 4

Given:

⎛8⎞ F = ⎜ −4 ⎟ N ⎜ ⎟ ⎝ 10 ⎠ a = 5m b = 3m c = 4m d = 2m e = 3m Solution:

⎛ −b − e ⎞ r = ⎜ c+d ⎟ ⎜ ⎟ ⎝ −a ⎠

M = r× F

⎛ 40 ⎞ M = ⎜ 20 ⎟ N⋅ m ⎜ ⎟ ⎝ −24 ⎠

Problem 4-87 Determine the couple moment. Express the result as a Cartesian vector. Given: F = 80 N a = 6m b = 10 m c = 10 m d = 5m e = 4m f = 4m

274

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Engineering Mechanics - Statics

Chapter 4

Solution: u =

⎛ b ⎞ ⎜c + d⎟ ⎟ 2 2 2⎜ a + b + ( c + d) ⎝ − a ⎠ 1

F v = Fu

⎛−f − b⎞ r = ⎜ −d − c ⎟ ⎜ ⎟ ⎝ e+a ⎠

⎛ −252.6 ⎞ M = ⎜ 67.4 ⎟ N⋅ m ⎜ ⎟ ⎝ −252.6 ⎠

M = r × Fv

Problem 4-88 If the resultant couple of the two couples acting on the fire hydrant is MR = { −15i + 30j} N ⋅ m, determine the force magnitude P. Given: a = 0.2 m b = 0.150 m

⎛ −15 ⎞ M = ⎜ 30 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠ F = 75 N Solution: Initial guess

P = 1N

Given

⎛ −F a ⎞ M = ⎜ Pb ⎟ ⎜ ⎟ ⎝ 0 ⎠

P = Find ( P)

P = 200 N

Problem 4-89 If the resultant couple of the three couples acting on the triangular block is to be zero, determine the magnitude of forces F and P . 275

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 150 N a = 300 mm b = 400 mm d = 600 mm

Solution:

Initial guesses:

Given

F = 1N

P = 1N

⎛ d ⎞ ⎛ 0 ⎞ ⎛ d ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ −F1 ⎟⎞ ⎛ 0 ⎞ ⎛⎜ F1 ⎞⎟ ⎜ 0 ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ −P ⎟ + ⎜ b ⎟ × ⎜ ⎟ + 0 × =0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜0 ⎟ ⎝ a ⎠ ⎝ F ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ a ⎠ ⎜⎝ 0 ⎟⎠

⎛F⎞ ⎜ ⎟ = Find ( F , P) ⎝P⎠

⎛ F ⎞ ⎛ 75 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ P ⎠ ⎝ 100 ⎠

Problem 4-90 Determine the couple moment that acts on the assembly. Express the result as a Cartesian vector. Member BA lies in the x-y plane.

276

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Engineering Mechanics - Statics

Chapter 4

Given:

⎛ 0 ⎞ ⎜ 0 ⎟N F = ⎜ ⎟ ⎝ 100 ⎠ a = 300 mm b = 150 mm c = 200 mm d = 200 mm

θ = 60 deg

Solution:

⎡−( c + d) sin ( θ ) − b cos ( θ ) ⎤ ⎢ ⎥ r = −b sin ( θ ) + ( c + d)cos ( θ ) ⎢ ⎥ 0 ⎦ ⎣

M = r× F

⎛ 7.01 ⎞ ⎜ ⎟ M = 42.14 N⋅ m ⎜ ⎟ ⎝ 0.00 ⎠

Problem 4-91 If the magnitude of the resultant couple moment is M, determine the magnitude F of the forces applied to the wrenches. Given: M = 15 N⋅ m

c = 200 mm

a = 300 mm

d = 200 mm

b = 150 mm

θ = 60 deg

277

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎡−( c + d) sin ( θ ) − b cos ( θ ) ⎤ ⎢ ⎥ r = −b sin ( θ ) + ( c + d)cos ( θ ) ⎢ ⎥ 0 ⎦ ⎣ Guess Given

⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠

F = 1N r × ( Fk) = M

F = Find ( F)

F = 35.112 N

Problem 4-92 The gears are subjected to the couple moments shown. Determine the magnitude and coordinate direction angles of the resultant couple moment. Given: M1 = 40 lb⋅ ft M2 = 30 lb⋅ ft

θ 1 = 20 deg θ 2 = 15 deg θ 3 = 30 deg

Solution:

⎛ M1 cos ( θ 1) sin ( θ 2) ⎞ ⎜ ⎟ M1 = ⎜ M1 cos ( θ 1 ) cos ( θ 2 ) ⎟ ⎜ −M sin θ ( 1) ⎟⎠ 1 ⎝ MR = M1 + M2

⎛⎜ α ⎞⎟ ⎛ MR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ MR ⎠ ⎜γ ⎟ ⎝ ⎠

⎛ −M2 sin ( θ 3) ⎞ ⎜ ⎟ M2 = ⎜ M2 cos ( θ 3 ) ⎟ ⎜ ⎟ 0 ⎝ ⎠ MR = 64.0 lb⋅ ft

⎛⎜ α ⎞⎟ ⎛ 94.7 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 13.2 ⎟ deg ⎜ γ ⎟ ⎝ 102.3 ⎠ ⎝ ⎠

278

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Engineering Mechanics - Statics

Chapter 4

Problem 4-93 Express the moment of the couple acting on the rod in Cartesian vector form. What is the magnitude of the couple moment? Given:

⎛ 14 ⎞ ⎜ ⎟ F = −8 N ⎜ ⎟ ⎝ −6 ⎠ a = 1.5 m b = 0.5 m c = 0.5 m d = 0.8 m Solution:

⎛d⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ M = a × F + 0 × ( −F) ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝b⎠

⎛ −17 ⎞ ⎜ ⎟ M = −9.2 N⋅ m ⎜ ⎟ ⎝ −27.4 ⎠

M = 33.532 N⋅ m

Problem 4-94 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O.

279

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Engineering Mechanics - Statics

Chapter 4

Given: a = 0.3 m b = 0.4 m c = 0.6 m

⎛ −6 ⎞ F = ⎜ 2 ⎟ N ⎜ ⎟ ⎝3⎠ Solution:

⎛0⎞ M = ⎜ c ⎟×F ⎜ ⎟ ⎝ −b ⎠

(a)

⎛ 2.6 ⎞ M = ⎜ 2.4 ⎟ N⋅ m ⎜ ⎟ ⎝ 3.6 ⎠

⎛0⎞ ⎛ 0 ⎞ ⎜ ⎟ M = 0 × ( −F ) + ⎜ c ⎟ × F ⎜ ⎟ ⎜ ⎟ ⎝ −a ⎠ ⎝ −a − b ⎠

(b)

⎛ 2.6 ⎞ M = ⎜ 2.4 ⎟ N⋅ m ⎜ ⎟ ⎝ 3.6 ⎠

Problem 4-95 A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment. Given: F 1 = 35 N θ = 60 deg F 2 = 25 N r1 = 175 mm r2 = 175 mm Solution:

⎛⎜ −F1 2 r1 ⎞⎟ ⎛⎜ −F2 2r2 cos ( θ ) ⎞⎟ M = ⎜ 0 ⎟ + ⎜ −F2 2 r2 sin ( θ ) ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ 0 ⎠

⎛ −16.63 ⎞ M = ⎜ −7.58 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = 18.3 N⋅ m

280

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Engineering Mechanics - Statics

Chapter 4

⎛⎜ α ⎞⎟ ⎛ M ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ M ⎠ ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 155.496 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 114.504 ⎟ deg ⎜ γ ⎟ ⎝ 90 ⎠ ⎝ ⎠

Problem 4-96 Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment? Given: F = 125 N a = 150 mm b = 150 mm c = 200 mm d = 600 mm

Solution:

⎛ c ⎞ ⎛0⎞ M = ⎜a + b⎟ × ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝F⎠

⎛ 37.5 ⎞ M = ⎜ −25 ⎟ N⋅ m ⎜ ⎟ ⎝ 0 ⎠

M = 45.1 N⋅ m

Problem 4-97 If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the forces applied to the wrenches.

281

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Engineering Mechanics - Statics

Chapter 4

Given: M = 300 N⋅ m a = 150 mm b = 150 mm c = 200 mm d = 600 mm Solution: F = 1N

Initial guess:

Given

⎛ c ⎞ ⎛0⎞ ⎜a + b⎟ × ⎜ 0 ⎟ = M ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝F⎠

F = Find ( F)

F = 832.1 N

Problem 4-98 Replace the force at A by an equivalent force and couple moment at point O. Given: F = 375 N a = 2m b = 4m c = 2m d = 1m

θ = 30 deg

282

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛ sin ( θ ) ⎞ ⎜ ⎟ F v = F −cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

⎛ 187.5 ⎞ ⎜ ⎟ F v = −324.76 N ⎜ ⎟ ⎝ 0 ⎠

⎛ −a ⎞ ⎜ ⎟ MO = b × Fv ⎜ ⎟ ⎝0⎠

⎛ 0 ⎞ ⎜ ⎟ N⋅ m 0 MO = ⎜ ⎟ ⎝ −100.481 ⎠

Problem 4-99 Replace the force at A by an equivalent force and couple moment at point P . Given: F = 375 N a = 2m b = 4m c = 2m d = 1m

θ = 30 deg Solution:

⎛ sin ( θ ) ⎞ ⎜ ⎟ F v = F −cos ( θ ) ⎜ ⎟ ⎝ 0 ⎠

⎛ 187.5 ⎞ ⎜ ⎟ F v = −324.76 N ⎜ ⎟ ⎝ 0 ⎠

⎛ −a − c ⎞ ⎜ ⎟ MP = b − d × Fv ⎜ ⎟ ⎝ 0 ⎠

⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MP = ⎜ ⎟ ⎝ 736.538 ⎠

Problem 4-100 Replace the force system by an equivalent resultant force and couple moment at point O.

283

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 60 lb

a = 2 ft

F 2 = 85 lb

b = 3 ft

F 3 = 25 lb

c = 6 ft

θ = 45 deg

d = 4 ft

e = 3 f = 4 Solution:

⎛⎜ 0 ⎟⎞ F = ⎜ −F1 ⎟ + ⎜ 0 ⎟ ⎝ ⎠

⎛ cos ( θ ) ⎞ ⎛ −e ⎞ ⎜ − f ⎟ + F ⎜ sin ( θ ) ⎟ 3 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝ 0 ⎠ F2

⎛ −33.322 ⎞ F = ⎜ −110.322 ⎟ lb ⎜ ⎟ 0 ⎝ ⎠

F = 115.245 lb

⎛ −c ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0 ⎞ ⎡ F ⎛ −e ⎞⎤ ⎛ d ⎞ ⎡ ⎛ cos ( θ ) ⎞⎤ 2 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ + ⎜ b ⎟ × ⎢F ⎜ sin ( θ ) ⎟⎥ MO = 0 × ⎜ −F1 ⎟ + a × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 3⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ MO = ⎜ 0 ⎟ lb⋅ ft ⎜ ⎟ ⎝ 480 ⎠

MO = 480 lb⋅ ft

Problem 4-101 Replace the force system by an equivalent resultant force and couple moment at point P.

284

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 60 lb

a = 2 ft

F 2 = 85 lb

b = 3 ft

F 3 = 25 lb

c = 6 ft

θ = 45 deg

d = 4 ft

e = 3

f = 4

Solution:

⎛⎜ 0 ⎟⎞ F = ⎜ −F1 ⎟ + ⎜ 0 ⎟ ⎝ ⎠

⎛ cos ( θ ) ⎞ ⎛ −e ⎞ ⎜ − f ⎟ + F ⎜ sin ( θ ) ⎟ 3 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝ 0 ⎠ F2

⎛ −33.322 ⎞ ⎜ ⎟ F = −110.322 lb ⎜ ⎟ 0 ⎝ ⎠

F = 115.245 lb

⎛ −c − d ⎞ ⎛⎜ 0 ⎟⎞ ⎛ −d ⎞ ⎡ F ⎛ −e ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ cos ( θ ) ⎞⎤ 2 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ − f ⎟⎥ + ⎜ b ⎟ × ⎢F ⎜ sin ( θ ) ⎟⎥ 0 MP = × ⎜ −F 1 ⎟ + a × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 3 ⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MP = ⎜ ⎟ ⎝ 921 ⎠

MP = 921 lb⋅ ft

Problem 4-102 Replace the force system by an equivalent force and couple moment at point O. Units Used: 3

kip = 10 lb Given: F 1 = 430 lb a = 2 ft

F 2 = 260 lb e = 5 ft 285

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Engineering Mechanics - Statics

Chapter 4

b = 8 ft

f = 12

c = 3 ft

g = 5

d = a

θ = 60 deg

Solution:

⎛ −sin ( θ ) ⎞ ⎜ ⎟ F R = F1 −cos ( θ ) + ⎜ ⎟ ⎝ 0 ⎠

⎛g⎞ ⎜f⎟ 2 2⎜ ⎟ g + f ⎝0⎠

⎛ −272 ⎞ ⎜ 25 ⎟ lb FR = ⎜ ⎟ ⎝ 0 ⎠

F2

⎛ −d ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ e ⎞ ⎡ F ⎛ g ⎞⎤ 2 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ f ⎟⎥ MO = b × F 1 −cos ( θ ) + 0 × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + f ⎝ 0 ⎠⎦

F R = 274 lb

⎛ 0 ⎞ ⎜ 0 ⎟ kip⋅ ft MO = ⎜ ⎟ ⎝ 4.609 ⎠

Problem 4-103 Replace the force system by an equivalent force and couple moment at point P. Units Used: 3

kip = 10 lb Given: F 1 = 430 lb

F 2 = 260 lb

a = 2 ft

e = 5 ft

b = 8 ft

f = 12

c = 3 ft

g = 5

d = a

θ = 60 deg

Solution:

⎛ −sin ( θ ) ⎞ ⎜ ⎟ F R = F1 −cos ( θ ) + ⎜ ⎟ ⎝ 0 ⎠

⎛g⎞ ⎜f⎟ 2 2⎜ ⎟ g + f ⎝0⎠

⎛ −272 ⎞ ⎜ 25 ⎟ lb FR = ⎜ ⎟ ⎝ 0 ⎠

F2

F R = 274 lb

286

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Engineering Mechanics - Statics

Chapter 4

⎛ 0 ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛ d + e ⎞ ⎡ F ⎛ g ⎞⎤ 2 ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ c ⎟ × ⎢ ⎜ f ⎟⎥ MP = b + c × F 1 −cos ( θ ) + ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + f ⎝ 0 ⎠⎦

⎛ 0 ⎞ ⎜ 0 ⎟ kip⋅ ft MP = ⎜ ⎟ ⎝ 5.476 ⎠

Problem 4-104 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point O. Given: F 1 = 30 lb

a = 1 ft

d = 3

F 2 = 40 lb

b = 3 ft

e = 4

F 3 = 60 lb

c = 2 ft

Solution:

⎛0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 −1 + F2 0 + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎛ 4 ⎞ ⎜ ⎟ F R = −78 lb ⎜ ⎟ ⎝ 0 ⎠

⎛ −d ⎞ ⎜ −e ⎟ ⎟ 2 2⎜ d +e ⎝ 0 ⎠ F3

F R = 78.1 lb

⎛ 0 ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ 1 ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −d ⎞⎤ 3 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ MO = a + b + c × F1 −1 + c × F 2 0 + b + c × ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ d + e ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MO = ⎜ ⎟ ⎝ 100 ⎠

Problem 4-105 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point P.

287

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 30 lb F 2 = 40 lb F 3 = 60 lb a = 1 ft b = 3 ft c = 2 ft d = 3 e = 4 Solution:

⎛0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 −1 + F2 0 + ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎛ 4 ⎞ ⎜ ⎟ F R = −78 lb ⎜ ⎟ ⎝ 0 ⎠

⎛ −d ⎞ ⎜ −e ⎟ ⎟ 2 2⎜ d +e ⎝ 0 ⎠ F3

F R = 78.1 lb

⎡⎛ 0 ⎞ ⎤ ⎡ ⎛ 0 ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ 1 ⎞⎤ ⎛ 0 ⎞ ⎡ F ⎛ −d ⎞⎤ 3 ⎢ ⎜ ⎟ ⎥ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ MP = 0 ft × F1 −1 + −a − b × F2 0 + −a × ⎢⎜ ⎟ ⎥ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎣⎝ 0 ⎠ ⎦ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ d + e ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ lb⋅ ft MP = ⎜ ⎟ ⎝ 124 ⎠

Problem 4-106 Replace the force and couple system by an equivalent force and couple moment at point O. Units Used: 3

kN = 10 N 288

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 4

Given: M = 8 kN m

θ = 60 deg

a = 3m

f = 12

b = 3m

g = 5

c = 4m

F 1 = 6 kN

d = 4m

F 2 = 4 kN

e = 5m Solution:

FR =

⎛ −cos ( θ ) ⎞ ⎛g⎞ ⎜ f ⎟ + F ⎜ −sin ( θ ) ⎟ 2 ⎜ ⎟ 2 2⎜ ⎟ f + g ⎝0⎠ ⎝ 0 ⎠ F1

⎛ 0.308 ⎞ ⎜ ⎟ F R = 2.074 kN ⎜ ⎟ ⎝ 0 ⎠

F R = 2.097 kN

⎛ 0 ⎞ ⎛ −c ⎞ ⎡ F ⎛ g ⎞⎤ ⎛ 0 ⎞ ⎡ ⎛ −cos ( θ ) ⎞⎤ 1 ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ f ⎟⎥ + ⎜ −d ⎟ × ⎢F ⎜ −sin ( θ ) ⎟⎥ MO = 0 + −e × ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 ⎜ ⎟⎥ ⎝ M ⎠ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −10.615 ⎠

Problem 4-107 Replace the force and couple system by an equivalent force and couple moment at point P. Units Used: 3

kN = 10 N

289

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Engineering Mechanics - Statics

Chapter 4

Given: M = 8 kN⋅ m

θ = 60 deg

a = 3m

f = 12

b = 3m

g = 5

c = 4m

F 1 = 6 kN

d = 4m

F 2 = 4 kN

e = 5m Solution:

FR =

⎛ −cos ( θ ) ⎞ ⎛g⎞ ⎜ f ⎟ + F ⎜ −sin ( θ ) ⎟ 2 ⎜ ⎟ 2 2⎜ ⎟ f + g ⎝0⎠ ⎝ 0 ⎠ F1

⎛ 0.308 ⎞ F R = ⎜ 2.074 ⎟ kN ⎜ ⎟ ⎝ 0 ⎠

F R = 2.097 kN

⎛ 0 ⎞ ⎛ −c − b ⎞ ⎡ F ⎛ g ⎞⎤ ⎛ −b ⎞ ⎡ ⎛ −cos ( θ ) ⎞⎤ 1 ⎜ f ⎟⎥ + ⎜ −d ⎟ × ⎢F ⎜ −sin ( θ ) ⎟⎥ MP = ⎜ 0 ⎟ + ⎜ −e ⎟ × ⎢ ⎜ ⎟ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2⎜ ⎟⎥ ⎝ M ⎠ ⎝ 0 ⎠ ⎣ f + g ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎛ 0 ⎞ ⎟ kN⋅ m 0 MP = ⎜ ⎜ ⎟ ⎝ −16.838 ⎠

Problem 4-108 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point O. Given: F 1 = 125 lb F 2 = 350 lb F 3 = 850 lb

290

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Engineering Mechanics - Statics

Chapter 4

a = 2 ft b = 6 ft c = 3 ft d = 4 ft

Solution: F Ry = F 3 − F 2 − F 1

F Ry = 375 lb

F Ry x = F3 ( b + c) − F 2 ( b) + F1 ( a) x =

F3 ( b + c) − F 2 ( b) + F1 a FRy

x = 15.5 ft

Problem 4-109 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point P. Given: F 1 = 125 lb

a = 2 ft

F 2 = 350 lb

b = 6 ft

F 3 = 850 lb

c = 3 ft d = 4 ft

Solution: F Ry = F 3 − F 2 − F 1

F Ry = 375 lb

F Ry x = F2 ( d + c) − F 3 ( d) + F1 ( a + b + c + d) x =

F2 d + F2 c − F3 d + F1 a + F1 b + F1 c + F1 d

x = 2.47 ft

F Ry (to the right of P)

291

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Engineering Mechanics - Statics

Chapter 4

Problem 4-110 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment acting at point O. Express the results in Cartesian vector form.

Given: a = 120 mm b = 800 mm

Solution:

⎛ −50 ⎞ F t = ⎜ 80 ⎟ N ⎜ ⎟ ⎝ −158 ⎠

⎛ −20 ⎞ F h = ⎜ 60 ⎟ N ⎜ ⎟ ⎝ −250 ⎠

⎛ −6 ⎞ Mt = ⎜ 4 ⎟ N⋅ m ⎜ ⎟ ⎝2⎠

⎛ −20 ⎞ Mh = ⎜ 8 ⎟ N⋅ m ⎜ ⎟ ⎝ 3 ⎠

FR = Ft + Fh

⎛ −70 ⎞ F R = ⎜ 140 ⎟ N ⎜ ⎟ ⎝ −408 ⎠ r0Ft

⎛a⎞ = ⎜0⎟ ⎜ ⎟ ⎝0⎠

MRP = ( r0Ft × F t) + Mt + Mh

⎛ −26 ⎞ MRP = ⎜ 31 ⎟ N⋅ m ⎜ ⎟ ⎝ 14.6 ⎠

Problem 4-111 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment 292

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Engineering Mechanics - Statics

Chapter 4

acting at point P. Express the results in Cartesian vector form. Given: a = 120 mm b = 800 mm

⎛ −50 ⎞ ⎜ 80 ⎟ N Ft = ⎜ ⎟ ⎝ −158 ⎠ ⎛ −6 ⎞ ⎜ ⎟ Mt = 4 N⋅ m ⎜ ⎟ ⎝2⎠ ⎛ −20 ⎞ ⎜ 60 ⎟ N Fh = ⎜ ⎟ ⎝ −250 ⎠ ⎛ −20 ⎞ ⎜ 8 ⎟ N⋅ m Mh = ⎜ ⎟ ⎝ 3 ⎠ Solution: FR = Ft + Fh

⎛ −70 ⎞ ⎜ ⎟ F R = 140 N ⎜ ⎟ ⎝ −408 ⎠

⎛b⎞ ⎛a + b⎞ ⎜ ⎟ ⎜ 0 ⎟×F MP = Mt + Mh + 0 × F h + ⎜ ⎟ ⎜ ⎟ t ⎝0⎠ ⎝ 0 ⎠

⎛ −26 ⎞ ⎜ ⎟ MP = 357.4 N⋅ m ⎜ ⎟ ⎝ 126.6 ⎠

Problem 4-112 Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B.

293

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 500 lb F 2 = 200 lb F 3 = 260 lb a = 5 ft

e = 3

b = 3 ft

f = 4

c = 2 ft

g = 12

d = 4 ft

h = 5

Solution: FR =

⎛−f⎞ ⎛0⎞ ⎜ −e ⎟ + F ⎜ −1 ⎟ + 2 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝0⎠ F1

Initial guess:

⎛h⎞ ⎜ −g ⎟ ⎟ 2 2⎜ g +h ⎝ 0 ⎠ F3

⎛ −300 ⎞ F R = ⎜ −740 ⎟ lb ⎜ ⎟ ⎝ 0 ⎠

F R = 798 lb

x = 1 ft

Given

⎛a⎞ ⎡ F ⎛ − f ⎞⎤ ⎛ a + b ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ a + b + c ⎞ ⎡ F ⎛ h ⎞⎤ ⎛ −x ⎞ 1 3 ⎜0⎟ × ⎢ ⎜ −e ⎟⎥ + ⎜ 0 ⎟ × ⎢F ⎜ −1 ⎟⎥ + ⎜ 0 ⎟ × ⎢ ⎜ −g ⎟⎥ = ⎜ 0 ⎟ × F ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ R ⎝ 0 ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + h ⎝ 0 ⎠⎦ ⎝ 0 ⎠ x = Find ( x)

x = −7.432 ft

Problem 4-113 Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B. Given: F 1 = 500 lb F 2 = 200 lb F 3 = 260 lb a = 5 ft

e = 3

294

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Engineering Mechanics - Statics

b = 3 ft

f = 4

c = 2 ft

g = 12

d = 4 ft

h = 5

Chapter 4

Solution: FR =

⎛−f⎞ ⎛0⎞ ⎜ −e ⎟ + F ⎜ −1 ⎟ + 2 ⎜ ⎟ 2 2⎜ ⎟ e + f ⎝0 ⎠ ⎝0⎠ F1

Initial guess:

⎛h⎞ ⎜ −g ⎟ ⎟ 2 2⎜ g +h ⎝ 0 ⎠ F3

⎛ −300 ⎞ F R = ⎜ −740 ⎟ lb ⎜ ⎟ ⎝ 0 ⎠

F R = 798 lb

x = 1ft

Given

⎛ −b − c − d ⎞ ⎡ F ⎛ − f ⎞⎤ ⎛ −c − d ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ −d ⎞ ⎡ F ⎛ h ⎞⎤ ⎛ −x ⎞ 1 3 ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ −g ⎟⎥ = ⎜ 0 ⎟ × F 0 −e + × 0 × F 2 −1 + 0 × ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 2 ⎜ ⎟⎥ ⎜ ⎟ 0 ⎝ ⎠ ⎣ e + f ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ g + h ⎝ 0 ⎠⎦ ⎝ 0 ⎠ x = Find ( x)

x = 6.568 ft

measured to the left of B

Problem 4-114 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. Given: F 1 = 300 lb F 2 = 200 lb F 3 = 400 lb F 4 = 200 lb

M = 600 lb⋅ ft a = 3 ft b = 4 ft c = 2 ft d = 7 ft

Solution: F Rx = −F 4

F Rx = −200 lb

F Ry = −F 1 − F 2 − F 3

F Ry = −900 lb

295

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Engineering Mechanics - Statics

2

F =

FRx + FRy

Chapter 4

2

F = 922 lb

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ = atan ⎜

θ = 77.5 deg

F Ry x = −F2 a − F 3 ( a + b) − F 4 c + M x = −

F2 ( a) + F 3 ( a + b) + F 4 c − M

x = 3.556 ft

F Ry

Problem 4-115 Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end A. Given: F 1 = 450 N

a = 2m

F 2 = 300 N

b = 4m

F 3 = 700 N

c = 3m

θ = 60 deg

M = 1500 N⋅ m

φ = 30 deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ )

F Rx = −125 N

F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2

F Ry = −1.296 × 10 N

F =

2

FRx + FRy

3

3

2

F = 1.302 × 10 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 84.5 deg

F Ry( x) = −F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M x =

−F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M F Ry

x = 7.36 m

Problem 4-116 Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end B. 296

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 450 N

a = 2m

F 2 = 300 N

b = 4m

F 3 = 700 N

c = 3m

θ = 60 deg

M = 1500 N⋅ m

φ = 30deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ )

F Rx = −125 N

F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2

F Ry = −1.296 × 10 N

F =

2

FRx + FRy

3

3

2

F = 1.302 × 10 N

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 84.5 deg

F Ry x = F1 sin ( θ ) b − F3 cos ( φ ) c − M x =

F1 sin ( θ ) b − F3 cos ( φ ) c − M

x = 1.36 m

F Ry

(to the right)

Problem 4-117 Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. Given: F 1 = 200 N F 2 = 450 N M = 200 N⋅ m a = 0.2 m b = 1.5 m c = 2m d = 1.5 m

297

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Engineering Mechanics - Statics

Chapter 4

θ = 30 deg Solution:

⎛ −sin ( θ ) ⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ F R = F1 1 + F2 −cos ( θ ) ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝ 0 ⎠ ⎛ −225 ⎞ ⎜ ⎟ F R = −190 N ⎜ ⎟ ⎝ 0 ⎠

F R = 294 N

⎛ b + c ⎞ ⎡ ⎛ 0 ⎞⎤ ⎛ b ⎞ ⎡ ⎛ −sin ( θ ) ⎞⎤ ⎛0⎞ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ a MO = × F 1 + a × F −cos ( θ ) + M 0 ⎜ ⎟ ⎢ 1⎜ ⎟⎥ ⎜ ⎟ ⎢ 2 ⎜ ⎟⎥ ⎜ ⎟ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ 0 ⎠ ⎣ ⎝ 0 ⎠⎦ ⎝ −1 ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −39.6 ⎠

Problem 4-118 Determine the magnitude and direction θ of force F and its placement d on the beam so that the loading system is equivalent to a resultant force FR acting vertically downward at point A and a clockwise couple moment M. Units Used: 3

kN = 10 N Given: F 1 = 5 kN

a = 3m

F 2 = 3 kN

b = 4m

F R = 12 kN

c = 6m

M = 50 kN⋅ m

e = 7

f = 24

Solution: Initial guesses:

F = 1 kN

θ = 30 deg

d = 2m

298

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Engineering Mechanics - Statics

Given

Chapter 4

⎛ −e ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ e +f ⎠ ⎛ − f ⎞ F − F sin ( θ ) − F = −F 2 R ⎜ 2 2⎟ 1 ⎝ e +f ⎠ f ⎛ ⎞ ⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M ⎝ e +f ⎠

⎛F⎞ ⎜ θ ⎟ = Find ( F , θ , d) ⎜ ⎟ ⎝d⎠

θ = 71.565 deg

F = 4.427 kN

d = 3.524 m

Problem 4-119 Determine the magnitude and direction θ of force F and its placement d on the beam so that the loading system is equivalent to a resultant force F R acting vertically downward at point A and a clockwise couple moment M. Units Used: 3

kN = 10 N Given: F 1 = 5 kN

a = 3m

F 2 = 3 kN

b = 4m

F R = 10 kN

c = 6m

M = 45 kN⋅ m e = 7 f = 24 Solution: Initial guesses:

Given

F = 1 kN

θ = 30 deg

d = 1m

⎛ −e ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ e +f ⎠ ⎛ − f ⎞ F − F sin ( θ ) − F = −F 2 R ⎜ 2 2⎟ 1 ⎝ e +f ⎠ 299

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Engineering Mechanics - Statics

Chapter 4

f ⎛ ⎞ ⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M ⎝ e +f ⎠

⎛F⎞ ⎜ θ ⎟ = Find ( F , θ , d) ⎜ ⎟ ⎝d⎠

θ = 57.529 deg

F = 2.608 kN

d = 2.636 m

Problem 4-120 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. Given: F 1 = 500 N

a = 3m b = 2m

F 2 = 300 N

c = 1m

F 3 = 250 N

d = 2m

M = 400 N⋅ m

e = 3m

θ = 60 deg

f = 3 g = 4

Solution:

⎛

⎞ − F ( cos ( θ ) ) 1 ⎟ 2 2 ⎝ g +f ⎠ g

F Rx = −F 3 ⎜

F Rx = −450 N

⎛

f

F Ry = −883.0127 N

FRx + FRy

2

F R = 991.066 N

⎞ − F sin ( θ ) 1 ⎟ 2 2 ⎝ f +g ⎠

F Ry = −F 2 − F 3 ⎜

FR =

2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 62.996 deg

300

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Engineering Mechanics - Statics

Chapter 4

−F Rx( y) = M + F 1 cos ( θ ) a + F3

g 2

g + f M + F 1 cos ( θ ) a + F3

g 2

g + f

y =

2

2

( b + a) − F2 ( d) − F 3 ⎛ ⎜

f 2

⎞ ( d + e)

2⎟

⎝ g +f ⎠

( b + a) − F2 ( d) − F 3 ⎛

f ⎞ ⎜ 2 2 ⎟ ( d + e) ⎝ g +f ⎠

−FRx y = 1.78 m

Problem 4-121 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C. Given: F 1 = 500 N

a = 3m b = 2m

F 2 = 300 N

c = 1m

F 3 = 250 N

d = 2m

M = 400 N⋅ m

e = 3m

θ = 60 deg

f = 3 g = 4

Solution: F Rx = −F 3 ⎛ ⎜

⎞ − F ( cos ( θ ) ) 1 ⎝ g +f ⎠ g

2

F Ry = −F 2 − F 3 ⎛ ⎜

F Rx = −450 N

2⎟

⎞ − F sin ( θ ) 1 ⎝ f +g ⎠

FR =

2

f

2

FRx + FRy

F Ry = −883.0127 N

2⎟

2

F R = 991.066 N

301

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Engineering Mechanics - Statics

Chapter 4

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

θ 1 = 62.996 deg

⎛

⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ ) 1 1 ⎝ g +f ⎠

F Ry( x) = M − F2 ( d + c) − F 3 ⎜

⎛

f

2

2⎟

⎞ ( c + d + e) − F ( b) cos ( θ ) − F c sin ( θ ) 1 1 ⎝ g +f ⎠

M − F2 ( d + c) − F 3 ⎜ x =

f

2

2⎟

F Ry x = 2.64 m

Problem 4-122 Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member AB, measured from point A. Given: F 1 = 35 lb a = 2 ft F 2 = 20 lb b = 4 ft F 3 = 25 lb c = 3 ft

θ = 30 deg d = 2 ft Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = −F 1 cos ( θ ) − F 2

F Ry = −50.311 lb

FR =

2

FRx + FRy

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

2

F R = 65.9 lb

θ 1 = −49.8 deg

F Ry x = −F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

302

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Engineering Mechanics - Statics

x =

Chapter 4

−F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

x = 2.099 ft

FRy

Problem 4-123 Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member BC, measured from point B. Given: F 1 = 35 lb F 2 = 20 lb F 3 = 25 lb

θ = 30 deg a = 2 ft b = 4 ft c = 3 ft d = 2 ft

Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = −F 1 cos ( θ ) − F 2

F Ry = −50.311 lb

FR =

2

FRx + FRy

2

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

F R = 65.9 lb

θ 1 = −49.8 deg

F Rx y = F 1 cos ( θ ) b + F3 ( c)

303

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Engineering Mechanics - Statics

y =

F 1 cos ( θ ) b + F3 ( c) FRx

Chapter 4

y = 4.617 ft

(Below point B)

Problem 4-124 Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A. Given: F 1 = 35 lb

a = 2 ft

F 2 = 20 lb

b = 4 ft

F 3 = 25 lb

c = 3 ft

θ = 30 deg

d = 2 ft

Solution: F Rx = F 1 sin ( θ ) + F3

F Rx = 42.5 lb

F Ry = F 1 cos ( θ ) + F 2

F Ry = 50.311 lb

FR =

2

FRx + FRy

⎛ FRy ⎞ ⎟ ⎝ FRx ⎠

θ 1 = atan ⎜

2

F R = 65.9 lb

θ 1 = 49.8 deg

MRA = −F1 cos ( θ ) a − F 2 ( a + b) + F 3 ( c)

MRA = −106 lb⋅ ft

Problem 4-125 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point O. Express the results in Cartesian vector form.

304

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠

a = 3m b = 3m

⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠

c = 4m

e = 5m f = 6m g = 5m

d = 6m

Solution:

⎛−f⎞ ⎜ ⎟ MR = M + e × F ⎜ ⎟ ⎝g⎠

FR = F

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −10 ⎞ ⎜ ⎟ MR = 18 kN⋅ m ⎜ ⎟ ⎝ −56 ⎠

Problem 4-126 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠ ⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠ a = 3m b = 3m

e = 5m

c = 4m

f = 6m 305

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Engineering Mechanics - Statics

d = 6m

Chapter 4

g = 5m

Solution:

⎛ −f ⎞ ⎜ e ⎟×F MR = M + ⎜ ⎟ ⎝d + g⎠

FR = F

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −46 ⎞ ⎜ ⎟ MR = 66 kN⋅ m ⎜ ⎟ ⎝ −56 ⎠

Problem 4-127 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point Q. Express the results in Cartesian vector form. Units Used: 3

kN = 10 N Given:

⎛8⎞ ⎜ ⎟ F = 6 kN ⎜ ⎟ ⎝8⎠ ⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠ a = 3m b = 3m

e = 5m

c = 4m

f = 6m

d = 6m

g = 5m

Solution:

FR = F

⎛0⎞ ⎜ ⎟ MR = M + e × F ⎜ ⎟ ⎝g⎠

⎛8⎞ ⎜ ⎟ F R = 6 kN ⎜ ⎟ ⎝8⎠

⎛ −10 ⎞ ⎜ ⎟ MR = −30 kN⋅ m ⎜ ⎟ ⎝ −20 ⎠

Problem 4-128 The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the −k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in 306

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Engineering Mechanics - Statics

Chapter 4

p y Cartesian vector form.

q

p

p

p

Given: F 1 = 40 N

r = 80 mm

F 2 = 40 N

a = 300 mm

θ = 0 deg Solution:

F 1v

F 2v

⎛0⎞ = F1 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

⎛ 0 ⎞ = F2 ⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ −sin ( θ ) ⎠

⎛ −a ⎞ r2 = ⎜ −r sin ( θ ) ⎟ ⎜ ⎟ ⎝ r cos ( θ ) ⎠

⎛ −a ⎞ r1 = ⎜ r ⎟ ⎜ ⎟ ⎝0⎠

F R = F1v + F2v

MA = r1 × F1v + r2 × F2v

⎛ 0 ⎞ F R = ⎜ −40 ⎟ N ⎜ ⎟ ⎝ −40 ⎠

⎛ 0 ⎞ MA = ⎜ −12 ⎟ N⋅ m ⎜ ⎟ ⎝ 12 ⎠

Problem 4-129 The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the −k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form.

307

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 40 N F 2 = 40 N

θ = 0 deg r = 80 mm a = 300 mm

θ = 45 deg

Solution:

F 1v

⎛0⎞ = F1 ⎜ 0 ⎟ ⎜ ⎟ ⎝ −1 ⎠

F R = F1v + F2v

⎛ 0 ⎞ F R = ⎜ −28.28 ⎟ N ⎜ ⎟ ⎝ −68.28 ⎠

F 2v

⎛ 0 ⎞ = F2 ⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ −sin ( θ ) ⎠

⎛ −a ⎞ r1 = ⎜ r ⎟ ⎜ ⎟ ⎝0⎠

⎛ −a ⎞ r2 = ⎜ −r sin ( θ ) ⎟ ⎜ ⎟ ⎝ r cos ( θ ) ⎠

MA = r1 × F1v + r2 × F2v

⎛ 0 ⎞ MA = ⎜ −20.49 ⎟ N⋅ m ⎜ ⎟ ⎝ 8.49 ⎠

Problem 4-130 Replace this system by an equivalent resultant force and couple moment acting at O. Express the results in Cartesian vector form. Given: F 1 = 50 N F 2 = 80 N

308

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Engineering Mechanics - Statics

Chapter 4

F 3 = 180 N a = 1.25 m b = 0.5 m c = 0.75 m

Solution:

⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎟⎞ ⎛⎜ 0 ⎟⎞ FR = ⎜ 0 ⎟ + ⎜ 0 ⎟ + ⎜ 0 ⎟ ⎜ F 1 ⎟ ⎜ −F2 ⎟ ⎜ −F 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟N FR = ⎜ ⎟ ⎝ −210 ⎠ ⎛ a + c ⎞ ⎛⎜ 0 ⎞⎟ ⎛ a ⎞ ⎛⎜ 0 ⎟⎞ ⎛ a ⎞ ⎛⎜ 0 ⎟⎞ ⎜ b ⎟ × 0 + ⎜b⎟ × 0 + ⎜0⎟ × 0 MO = ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 0 ⎠ ⎝ F1 ⎠ ⎝ 0 ⎠ ⎝ −F2 ⎠ ⎝ 0 ⎠ ⎝ −F3 ⎟⎠

⎛ −15 ⎞ ⎜ ⎟ MO = 225 N⋅ m ⎜ ⎟ ⎝ 0 ⎠

Problem 4-131 Handle forces F 1 and F 2 are applied to the electric drill. Replace this system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form. Given: a = 0.15 m b = 0.25 m c = 0.3 m

⎛ 6 ⎞ ⎜ ⎟ F 1 = −3 N ⎜ ⎟ ⎝ −10 ⎠ ⎛0⎞ ⎜ ⎟ F2 = 2 N ⎜ ⎟ ⎝ −4 ⎠ 309

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Engineering Mechanics - Statics

Chapter 4

Solution: FR = F1 + F2

⎛ 6 ⎞ ⎜ ⎟ F R = −1 N ⎜ ⎟ ⎝ −14 ⎠

⎛a⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ MO = 0 × F1 + −b × F 2 ⎜ ⎟ ⎜ ⎟ ⎝c⎠ ⎝c ⎠

⎛ 1.3 ⎞ ⎜ 3.3 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −0.45 ⎠

Problem 4-132 A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of F R for the rectus, FO for the oblique, FL for the lumbar latissimus dorsi, and F E for the erector spinae. These loadings are symmetric with respect to the y - z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form. Given: F R = 35 N

a = 75 mm

F O = 45 N

b = 45 mm

F L = 23 N

c = 15 mm

F E = 32 N

d = 50 mm

e = 40 mm

f = 30 mm

Solution: F Res = ΣFi;

F Res = 2( FR + F O + F L + F E)

F Res = 270 N

MROx = ΣMOx;

MRO = −2FR a + 2F E c + 2F L b

MRO = −2.22 N⋅ m

Problem 4-133 The building slab is subjected to four parallel column loadings.Determine the equivalent resultant force and specify its location (x, y) on the slab.

310

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: F 1 = 30 kN

a = 3m

F 2 = 40 kN

b = 8m

F 3 = 20 kN

c = 2m

F 4 = 50 kN

d = 6m e = 4m

Solution: +

↑ FR = ΣFx;

FR = F1 + F2 + F3 + F4 F R = 140 kN

MRx = ΣMx;

−F R( y) = −( F4 ) ( a) −

y =

⎡⎣( F1 ) ( a + b)⎤⎦ − ⎡⎣( F2 ) ( a + b + c)⎤⎦

F4 a + F1 a + F1 b + F2 a + F2 b + F2 c FR

y = 7.14 m MRy = ΣMy;

(FR)x = (F4)( e) + (F3)( d + e) + (F2)( b + c) x =

F4 e + F3 d + F3 e + F2 b + F2 c FR

x = 5.71 m

Problem 4-134 The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab.

311

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Engineering Mechanics - Statics

Chapter 4

Units Used: 3

kN = 10 N Given: F 1 = 20 kN

a = 3m

F 2 = 50 kN

b = 8m

F 3 = 20 kN

c = 2m

F 4 = 50 kN

d = 6m e = 4m

Solution: FR = F1 + F2 + F3 + F4

F R = 140 kN

F R x = F2 e + F1 ( d + e) + F 2 ( d + e) x =

2 F2 e + F1 d + F1 e + F2 d

x = 6.43 m

FR

−F R y = −F 2 a − F3 ( a + b) − F2 ( a + b + c)

y =

2 F2 a + F3 a + F3 b + F2 b + F2 c FR

y = 7.29 m

Problem 4-135 The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the magnitude F of the couple forces so that the system can be simplified to a wrench acting at point C. Given: a = 0.6 m

312

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Engineering Mechanics - Statics

Chapter 4

b = 0.8 m c = 0.25 m d = 0.7 m e = 0.3 m f = 0.3 m g = 0.5 m h = 0.25 m P = 60 N Q = 40 N Solution: Initial Guess

F = 1N

MC = 1 N⋅ m

Given

⎛⎜ −MC ⎞⎟ ⎡ −P( c + h) ⎤ ⎡ 0 ⎢ ⎥+⎢ 0 ⎜ 0 ⎟=⎢ 0 ⎥ ⎢ ⎜ 0 ⎟ ⎣ 0 ⎦ ⎣ −F ( e + ⎝ ⎠ ⎛ F ⎞ ⎜ ⎟ = Find ( F , MC) ⎝ MC ⎠

⎤ ⎛ a ⎞ ⎛ −Q ⎞ ⎥ + ⎜b⎟ × ⎜ 0 ⎟ ⎥ ⎜ ⎟ ⎜ ⎟ f) ⎦ ⎝ 0 ⎠ ⎝ 0 ⎠

MC = 30 N⋅ m

F = 53.3 N

Problem 4-136 The three forces acting on the block each have a magnitude F 1 = F2 = F3. Replace this system by a wrench and specify the point where the wrench intersects the z axis, measured from point O. Given: F 1 = 10 lb

a = 6 ft

F2 = F1

b = 6 ft

F3 = F1

c = 2 ft

Solution: The vectors

313

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Engineering Mechanics - Statics

F 1v =

⎛b⎞ ⎜ −a ⎟ ⎟ 2 2⎜ b +a ⎝ 0 ⎠ F1

Chapter 4

F 2v

⎛⎜ 0 ⎞⎟ = ⎜ −F 2 ⎟ ⎜ 0 ⎟ ⎝ ⎠

F 3v =

M = 1 lb⋅ ft

R x = 1 lb

⎛ −b ⎞ ⎜a⎟ ⎟ 2 2⎜ b +a ⎝ 0 ⎠ F3

Place the wrench in the x - z plane. x = 1ft z = 1ft

Guesses

R y = 1 lb

R z = 1 lb

⎛ Rx ⎞ ⎜ ⎟ Given ⎜ Ry ⎟ = F1v + F2v + F3v ⎜R ⎟ ⎝ z⎠ ⎛ Rx ⎞ ⎛ 0 ⎞ ⎛ x ⎞ ⎛⎜ Rx ⎟⎞ ⎛0⎞ ⎛b⎞ ⎜ ⎟ M ⎜0⎟ × R + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ry ⎟ = ⎜ a ⎟ × F2v + ⎜ a ⎟ × F1v + ⎜ 0 ⎟ × F3v ⎜ ⎟ ⎜ y⎟ 2 2 2 Rx + Ry + Rz ⎜ R ⎟ ⎝ c ⎠ ⎝ z ⎠ ⎜⎝ Rz ⎟⎠ ⎝0⎠ ⎝c ⎠ ⎝ z⎠ ⎛x ⎞ ⎜ ⎟ ⎜ z ⎟ ⎛ Rx ⎞ ⎜ ⎟ ⎜M ⎟ M Mv = ⎜ R ⎟ = Find ( x , z , M , Rx , Ry , Rz) ⎜ Ry ⎟ 2 2 2 ⎜ x⎟ Rx + Ry + Rz ⎜ R ⎟ ⎝ z⎠ R ⎜ y⎟ ⎜ ⎟ ⎝ Rz ⎠ ⎛ Rx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎛x⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎟ Mv = −14.142 lb⋅ ft ⎜ ⎟=⎜ ⎟ ft ⎜ Ry ⎟ = ⎜ −10 ⎟ lb ⎜ ⎟ ⎝ z ⎠ ⎝ 0.586 ⎠ ⎜R ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ z⎠ Problem 4-137 Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate. Units Used: 3

kN = 10 N

314

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Engineering Mechanics - Statics

Chapter 4

Given: F A = 500 N F B = 800 N F C = 300 N a = 4m b = 6m

Solution:

⎛ FA ⎞ ⎜ ⎟ FR = ⎜ FC ⎟ ⎜F ⎟ ⎝ B⎠ Guesses

F R = 0.9899 kN

x = 1m

Given

M

FR FR

y = 1m

M = 100 N⋅ m

⎛x⎞ + ⎜ y ⎟ × FR = ⎜ ⎟ ⎝0⎠

⎛M⎞ ⎜ x ⎟ = Find ( M , x , y) ⎜ ⎟ ⎝y⎠

⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0 ⎞ ⎛⎜ 0 ⎞⎟ ⎜a⎟ × F + ⎜a⎟ × 0 ⎜ ⎟ ⎜ C⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ FB ⎟⎠ ⎛ x ⎞ ⎛ 1.163 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ y ⎠ ⎝ 2.061 ⎠

M = 3.07 kN⋅ m

Problem 4-138 Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(y, z) where its line of action intersects the plate. Given: F A = 80 lb

a = 12 ft

F B = 60 lb

b = 12 ft

F C = 40 lb

315

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Engineering Mechanics - Statics

Chapter 4

Solution:

⎛ −F C ⎞ ⎜ ⎟ F R = ⎜ −F B ⎟ ⎜ −F ⎟ ⎝ A⎠

F R = 108 lb

y = 1 ft

Guesses

Given

M

FR FR

z = 1 ft

⎛0⎞ + ⎜ y ⎟ × FR = ⎜ ⎟ ⎝z⎠

⎛M⎞ ⎜ y ⎟ = Find ( M , y , z) ⎜ ⎟ ⎝z⎠

M = 1 lb⋅ ft

⎛ 0 ⎞ ⎛⎜ −FC ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜a⎟ × ⎜ ⎟ ⎜ 0 ⎟ + a × ⎜ −FB ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ b ⎠ ⎜⎝ 0 ⎟⎠ ⎛ y ⎞ ⎛ 0.414 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ z ⎠ ⎝ 8.69 ⎠

M = −624 lb⋅ ft

Problem 4-139 The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. Given: w1 = 2

lb ft

w2 = 3.5

lb ft

a = 2.75 ft b = 4 ft c = 1.5 ft R = 1 lb

Solution: Guesses Given

d = 1ft

w1 b + w2 c = R w1 b⎛⎜ a −

⎝

⎛R⎞ ⎜ ⎟ = Find ( R , d) ⎝d⎠

b⎞

⎞ ⎛c ⎟ − w2 c⎜ + b − a⎟ = −d R 2⎠ ⎝2 ⎠ R = 13.25 lb

d = 0.34 ft

316

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Engineering Mechanics - Statics

Chapter 4

Problem 4-140 Replace the loading by an equivalent resultant force and couple moment acting at point A. Units Used: 3

kN = 10 N Given: w1 = 600

N

w2 = 600

N

m

m

a = 2.5 m b = 2.5 m Solution: F R = w1 a − w2 b MRA = w1 a⎛⎜

a + b⎞

⎟ ⎝ 2 ⎠

FR = 0 N MRA = 3.75 kN⋅ m

Problem 4-141 Replace the loading by an equivalent force and couple moment acting at point O. Units Used: 3

kN = 10 N Given: w = 6

kN m

F = 15 kN M = 500 kN⋅ m a = 7.5 m b = 4.5 m

317

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Engineering Mechanics - Statics

Chapter 4

Solution: FR =

1 2

w( a + b) + F

MR = −M −

F R = 51.0 kN

⎛ 1 w a⎞ ⎛ 2 a⎞ − ⎛ 1 w b⎞ ⎛a + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝

b⎞

⎟ − F( a + b)

3⎠

MR = −914 kN⋅ m

Problem 4-142 Replace the loading by a single resultant force, and specify the location of the force on the beam measured from point O. Units Used: 3

kN = 10 N Given: w = 6

kN m

F = 15 kN M = 500 kN⋅ m a = 7.5 m b = 4.5 m Solution: Initial Guesses:

F R = 1 kN

d = 1m

Given FR =

1 2

w( a + b) + F

−F R d = −M −

⎛ 1 w a⎞ ⎛ 2 a⎞ − ⎛ 1 w b⎞ ⎛ a + b ⎞ − F( a + b) ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝ 3 ⎠

⎛ FR ⎞ ⎜ ⎟ = Find ( FR , d) ⎝ d ⎠

F R = 51 kN

d = 17.922 m

Problem 4-143 The column is used to support the floor which exerts a force P on the top of the column. The effect of soil pressure along its side is distributed as shown. Replace this loading by an 318

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Engineering Mechanics - Statics

Chapter 4

p g p g y equivalent resultant force and specify where it acts along the column, measured from its base A. 3

kip = 10 lb

Units Used: Given:

P = 3000 lb w1 = 80

lb ft

w2 = 200

lb ft

h = 9 ft Solution: F Rx = w1 h +

1 (w2 − w1)h 2

F Rx = 1260 lb

F Ry = P

2

FR =

FRx + P

2

⎞ θ = atan ⎛⎜ ⎟ F ⎝ Rx ⎠ P

θ = 67.2 deg

1 h h w2 − w1 ) h + w1 h ( 2 3 2

F Rx y =

y =

F R = 3.25 kip

1 2 w2 + 2 w1 h FRx 6

y = 3.86 ft

Problem 4-144 Replace the loading by an equivalent force and couple moment at point O. Units Used: 3

kN = 10 N Given: kN m

w1 = 15

w2 = 5

kN m 319

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 4

d = 9m Solution: FR =

1 (w1 + w2)d 2

MRO = w2 d

F R = 90 kN

d 1 d + ( w1 − w2 ) d 2 2 3

MRO = 338 kN⋅ m

Problem 4-145 Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C. Units Used: 3

kip = 10 lb Given: w = 800

lb ft

a = 15 ft b = 15 ft

θ = 30 deg Solution: FR = w a + FR x = w a

wa x =

a 2

wb 2

F R = 18 kip

a wb⎛ b⎞ + ⎜a + ⎟ 2 ⎝ 2 3⎠ +

b⎞ ⎜a + ⎟ 2 ⎝ 3⎠

wb⎛ FR

x = 11.7 ft

Problem 4-146 The beam supports the distributed load caused by the sandbags. Determine the resultant force on the beam and specify its location measured from point A.

320

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Engineering Mechanics - Statics

Units Used:

Chapter 4

3

kN = 10 N

Given: w1 = 1.5 w2 = 1

kN

a = 3m

m

kN

b = 3m

m

w3 = 2.5

kN

c = 1.5 m

m

Solution: F R = w1 a + w2 b + w3 c MA = w1 a

a 2

+ w2 b⎛⎜ a +

⎝

F R = 11.25 kN b⎞

c⎞ ⎛ ⎟ + w3 c⎜ a + b + ⎟ 2⎠ 2⎠ ⎝

MA = 45.563 kN⋅ m

d =

MA

d = 4.05 m

FR

Problem 4-147 Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is M clockwise. Units Used: 3

kN = 10 N Given: w1 = 4

kN

w2 = 2.5

m

M = 8 kN⋅ m

kN m

c = 9m

Solution: Initial Guesses: Given

−1 2

a = 1m w1 b +

1 2

b = 1m

w2 c = 0

321

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Engineering Mechanics - Statics

1 2

w1 b⎛⎜ a +

⎝

⎛a⎞ ⎜ ⎟ = Find ( a , b) ⎝b⎠

Chapter 4

2b ⎞

1

2c

⎟ − w2 c = −M 3⎠ 2 3 ⎛ a ⎞ ⎛ 1.539 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ b ⎠ ⎝ 5.625 ⎠

Problem 4-148 Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A. Units Used: 3

kN = 10 N Given: w1 = 800

N

w2 = 200

N

m

m

a = 2m b = 3m Solution: F R = w2 b + w1 a + x F R = w1 a

w1 a x =

a 2

+

1 2

(w1 − w2)b

F R = 3.10 kN

b b w1 − w2 ) b⎛⎜ a + ⎞⎟ + w2 b⎛⎜ a + ⎞⎟ ( 2 3 2 1

⎝

⎠

⎝

a 1 ⎛ b⎞ ⎛ b⎞ + ( w1 − w2 ) b⎜ a + ⎟ + w2 b⎜ a + ⎟ 2 2 ⎝ 3⎠ ⎝ 2⎠ FR

⎠

x = 2.06 m

Problem 4-149 The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O. Units Used: 322

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 4

3

kip = 10 lb Given: w1 = 50

lb ft

w2 = 300

lb

w3 = 100

lb

ft

ft

a = 12 ft b = 9 ft Solution :

(w2 − w1)a + 2 (w2 − w3)b + w3 b 2 1

F R = w1 a + F R d = w1 a

a 2

1

+

(w2 − w1)a 2 1

2

d =

2a 3

+

F R = 3.9 kip

b b w2 − w3 ) b⎛⎜ a + ⎞⎟ + w3 b⎛⎜ a + ⎞⎟ ( 2 3 2 1

2

⎝

2

⎠

3 w3 b a + 2 w3 b + w1 a + 2 a w2 + 3 b w2 a + w2 b 6FR

⎝

⎠

2

d = 11.3 ft

Problem 4-150 The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero. Given: w1 = 40

lb

w2 = 60

lb

ft

c = 10ft d = 6 ft

ft

Solution: Initial Guesses:

a = 1 ft

b = 1ft

323

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Engineering Mechanics - Statics

Chapter 4

Given 1 2

1

w2 d − w1 b = 0

⎛a⎞ ⎜ ⎟ = Find ( a , b) ⎝b⎠

2

w2 d⎛⎜ c +

⎝

d⎞

⎛ ⎟ − w1 b⎜ a + 3⎠ ⎝

b⎞

⎟=0

2⎠

⎛ a ⎞ ⎛ 9.75 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ b ⎠ ⎝ 4.5 ⎠

Problem 4-151 Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point B. Units Used: 3

kip = 10 lb Given: w1 = 800

lb

w2 = 500

lb

ft

ft

a = 12 ft b = 9 ft Solution: FR =

1 2

a w1 +

1 2

(w1 − w2)b + w2 b

F R = 10.65 kip

a 1 1 b b F R x = − a w1 + ( w1 − w2 ) b + w2 b 3 2 2 3 2

x =

a 1 b b 1 − a w1 + ( w1 − w2 ) b + w2 b 3 2 3 2 2 FR

x = 0.479 ft ( to the right of B )

Problem 4-152 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A.

324

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Engineering Mechanics - Statics

Chapter 4

Given: w1 = 200

N m

w2 = 100

N m

w3 = 200

N m

a = 5m b = 6m Solution: F Rx = −w3 a F Ry =

F Rx = −1000 N

−1 (w1 + w2)b 2

− y FRx = w3 a

w3 a y =

a 2

F Ry = −900 N

a b 1 b − w2 b − ( w1 − w2 ) b 2 2 2 3

− w2 b

b

−

(w1 − w2)b 3 2 1

b

2 −FRx

y = 0.1 m

Problem 4-153 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C. Units Used: 3

kN = 10 N

325

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Engineering Mechanics - Statics

Chapter 4

Given: w1 = 200

N

w2 = 100

N

w3 = 200

N

m

m

m

a = 5m b = 6m Solution : F Rx = −w3 a F Ry =

−1 2

(w1 + w2)b

−x F Ry = −w3 a

−w3 a x =

F Rx = −1000 N

a 2

+ w2 b

F Ry = −900 N b 2

+

(w1 − w2)b 2 1

2b 3

a b 1 2b + w2 b + ( w1 − w2 ) b 2 2 2 3

x = 0.556 m

−FRy

⎛ FRx ⎞ ⎜ ⎟ = 1.345 kN ⎝ FRy ⎠ Problem 4-154 Replace the loading by an equivalent resultant force and couple moment acting at point O. Units Used: 3

kN = 10 N Given: w1 = 7.5

kN m

326

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Engineering Mechanics - Statics

w2 = 20

Chapter 4

kN m

a = 3m b = 3m c = 4.5 m

Solution:

(w2 − w1)c + w1 c + w1 b + 2 1

FR =

1 2

w1 a

F R = 95.6 kN MRo = −

c c b w2 − w1 ) c − w1 c − w1 b⎛⎜ c + ⎞⎟ − ( 2 3 2 2 1

⎝

⎠

1 2

w1 a⎛⎜ b + c +

⎝

a⎞

⎟

3⎠

MRo = −349 kN⋅ m

Problem 4-155 Determine the equivalent resultant force and couple moment at point O. Units Used: 3

kN = 10 N Given: a = 3m wO = 3

kN m

x w ( x) = wO ⎛⎜ ⎟⎞ ⎝ a⎠

2

Solution: ⌠a F R = ⎮ w ( x) dx ⌡0

F R = 3 kN

⌠a MO = ⎮ w ( x) ( a − x) dx ⌡0

MO = 2.25 kN⋅ m

327

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Engineering Mechanics - Statics

Chapter 4

Problem 4-156 Wind has blown sand over a platform such that the intensity of the load can be approximated by 3

⎛ x ⎞ . Simplify this distributed loading to an equivalent resultant force and ⎟ ⎝d⎠

the function w = w0 ⎜

specify the magnitude and location of the force, measured from A.

Units Used: 3

kN = 10 N Given: w0 = 500

N m

d = 10 m

⎛x⎞ ⎟ ⎝ d⎠

3

w ( x) = w0 ⎜ Solution: d

⌠ F R = ⎮ w ( x) dx ⌡0

d =

⌠d ⎮ x w ( x) dx ⌡0 FR

F R = 1.25 kN

d=8m

Problem 4-157 Determine the equivalent resultant force and its location, measured from point O.

328

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Engineering Mechanics - Statics

Chapter 4

Solution: L

⌠ 2w0 L ⎛ πx⎞ F R = ⎮ w0 sin ⎜ ⎟ dx = ⎮ π ⎝L⎠ ⌡ 0

L

d=

⌠ ⎮ x w sin ⎛ π x ⎞ dx ⎜ ⎟ 0 ⎮ ⎝L⎠ ⌡ 0

FR

=

L 2

Problem 4-158 Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and specify where it acts, measured from point A. Given: a = 3 ft k = 86

lb ft

3 2

w ( x) = k x Solution: a

⌠ F R = ⎮ w ( x) dx ⌡0

x =

⌠a ⎮ x w ( x) dx ⌡0 FR

F R = 774 lb

x = 2.25 ft

Problem 4-159 Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To 329

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Engineering Mechanics - Statics

Chapter 4

y g y p j y alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A. Given: a = 0.5 ft w0 = 12

lb ft

lb

k = 24

ft

3

2

w ( x) = w0 + kx

Solution: ⌠a F R = ⎮ w ( x) dx ⌡0

F R = 7 lb

a

x =

⌠ ⎮ x w ( x) dx ⌡0 FR

x = 0.268 ft

Problem 4-160 Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integrals using Simpson's rule.

Units Used: 3

kN = 10 N Given: c1 = 5 c2 = 16 a = 3

330

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Engineering Mechanics - Statics

Chapter 4

b = 1 Solution: a+ b

⌠ FR = ⎮ ⌡0

2

c1 x +

c2 + x dx

x c1 x +

c2 + x dx

F R = 14.9

a+ b

d =

⌠ ⎮ ⌡0

2

d = 2.27

FR

Problem 4-161 Determine the coordinate direction angles of F, which is applied to the end A of the pipe assembly, so that the moment of F about O is zero. Given: F = 20 lb a = 8 in b = 6 in c = 6 in d = 10 in Solution: Require Mo = 0. This happens when force F is directed either towards or away from point O.

⎛ c ⎞ r = ⎜a + b⎟ ⎜ ⎟ ⎝ d ⎠

u =

r r

⎛ 0.329 ⎞ u = ⎜ 0.768 ⎟ ⎜ ⎟ ⎝ 0.549 ⎠

If the force points away from O, then

⎛⎜ α ⎞⎟ ⎜ β ⎟ = acos ( u) ⎜γ ⎟ ⎝ ⎠

⎛⎜ α ⎞⎟ ⎛ 70.774 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 39.794 ⎟ deg ⎜ γ ⎟ ⎝ 56.714 ⎠ ⎝ ⎠

If the force points towards O, then 331

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Engineering Mechanics - Statics

⎛⎜ α ⎞⎟ ⎜ β ⎟ = acos ( −u) ⎜γ ⎟ ⎝ ⎠

Chapter 4

⎛⎜ α ⎞⎟ ⎛ 109.226 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 140.206 ⎟ deg ⎜ γ ⎟ ⎝ 123.286 ⎠ ⎝ ⎠

Problem 4-162 Determine the moment of the force F about point O. The force has coordinate direction angles α, β, γ. Express the result as a Cartesian vector. Given: F = 20 lb

a = 8 in

α = 60 deg

b = 6 in

β = 120 deg

c = 6 in

γ = 45 deg

d = 10 in

Solution:

⎛ c ⎞ r = ⎜a + b⎟ ⎜ ⎟ ⎝ d ⎠

⎛⎜ cos ( α ) ⎟⎞ F v = F⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠

M = r × Fv

⎛ 297.99 ⎞ M = ⎜ 15.147 ⎟ lb⋅ in ⎜ ⎟ ⎝ −200 ⎠

Problem 4-163 Replace the force at A by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used :

332

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Engineering Mechanics - Statics

Chapter 4

3

kN = 10 N Given: a = 4m b = 6m c = 8m d = 4m

⎛ −300 ⎞ ⎜ ⎟ F = 200 N ⎜ ⎟ ⎝ −500 ⎠ Solution : FR = F

⎛ −300 ⎞ ⎜ ⎟ F R = 200 N ⎜ ⎟ ⎝ −500 ⎠

⎛ −a − c ⎞ ⎜ b ⎟×F MP = ⎜ ⎟ ⎝ d ⎠

⎛ −3.8 ⎞ ⎜ ⎟ MP = −7.2 kN⋅ m ⎜ ⎟ ⎝ −0.6 ⎠

Problem 4-164 Determine the moment of the force FC about the door hinge at A. Express the result as a Cartesian vector.

333

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Engineering Mechanics - Statics

Chapter 4

Given: F = 250 N b = 1m c = 2.5 m d = 1.5 m e = 0.5 m

θ = 30 deg

Solution:

rCB

⎛ c−e ⎞ ⎜ ⎟ = b + d cos ( θ ) ⎜ ⎟ ⎝ −d sin ( θ ) ⎠

MA = rAB × F v

rAB

⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝0⎠

Fv = F

rCB rCB

⎛ −59.7 ⎞ ⎜ 0.0 ⎟ N⋅ m MA = ⎜ ⎟ ⎝ −159.3 ⎠

Problem 4-165 Determine the magnitude of the moment of the force FC about the hinged axis aa of the door.

334

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Engineering Mechanics - Statics

Chapter 4

Given: F = 250 N b = 1m c = 2.5 m d = 1.5 m e = 0.5 m

θ = 30 deg

Solution:

rCB

⎛ c−e ⎞ ⎜ ⎟ = b + d cos ( θ ) ⎜ ⎟ ⎝ −d sin ( θ ) ⎠

Maa = ( rAB × Fv) ⋅ ua

rAB

⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝0⎠

Fv = F

rCB rCB

⎛1⎞ ⎜ ⎟ ua = 0 ⎜ ⎟ ⎝0⎠

Maa = −59.7 N⋅ m

Problem 4-166 A force F1 acts vertically downward on the Z-bracket. Determine the moment of this force about the bolt axis (z axis), which is directed at angle θ from the vertical.

335

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 80 N a = 100 mm b = 300 mm c = 200 mm

θ = 15 deg Solution:

⎛ −b ⎞ ⎜ ⎟ r = a+c ⎜ ⎟ ⎝ 0 ⎠ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ F = F1 ⎜ ⎟ ⎝ −cos ( θ ) ⎠ ⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠ Mz = ( r × F ) k Mz = −6.212 N⋅ m

Problem 4-167 Replace the force F having acting at point A by an equivalent force and couple moment at point C. Units Used:

3

kip = 10 lb

Given: F = 50 lb a = 10 ft b = 20 ft c = 15 ft

336

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Engineering Mechanics - Statics

Chapter 4

d = 10 ft e = 30 ft

Solution :

rAB

⎛d⎞ = ⎜ c ⎟ ⎜ ⎟ ⎝ −e ⎠

Fv = F

rCA

rAB rAB

⎛ 0 ⎞ = ⎜a + b⎟ ⎜ ⎟ ⎝ e ⎠

FR = Fv

⎛ 14.286 ⎞ F R = ⎜ 21.429 ⎟ lb ⎜ ⎟ ⎝ −42.857 ⎠

MR = rCA × Fv

⎛ −1.929 ⎞ MR = ⎜ 0.429 ⎟ kip⋅ ft ⎜ ⎟ ⎝ −0.429 ⎠

Problem 4-168 The horizontal force F acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis? Given: F = 30 N a = 50 mm b = 200 mm c = 10 mm

θ = 45 deg Solution:

⎛ sin ( θ ) ⎞ F v = F⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ 0 ⎠

rOA

⎛ −c ⎞ = ⎜b ⎟ ⎜ ⎟ ⎝a⎠

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠ 337

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Engineering Mechanics - Statics

Mz = ( rOA × F v ) k

Chapter 4

Mz = −4.03 N⋅ m

Problem 4-169 The horizontal force F acts on the handle of the wrench. Determine the moment of this force about point O. Specify the coordinate direction angles α, β, γ of the moment axis. Given: F = 30 N

c = 10 mm

a = 50 mm θ = 45 deg b = 200 mm Solution:

⎛ sin ( θ ) ⎞ F v = F⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ 0 ⎠ MO = rOA × Fv

⎛⎜ α ⎞⎟ ⎛ MO ⎞ ⎜ β ⎟ = acos ⎜ ⎟ MO ⎠ ⎝ ⎜γ ⎟ ⎝ ⎠

rOA

⎛ −c ⎞ = ⎜b ⎟ ⎜ ⎟ ⎝a⎠

⎛ 1.06 ⎞ MO = ⎜ 1.06 ⎟ N⋅ m ⎜ ⎟ ⎝ −4.03 ⎠ ⎛⎜ α ⎞⎟ ⎛ 75.7 ⎞ ⎜ ⎟ ⎜ β ⎟ = ⎜ 75.7 ⎟ deg ⎜ γ ⎟ ⎝ 159.6 ⎠ ⎝ ⎠

Problem 4-170 If the resultant couple moment of the three couples acting on the triangular block is to be zero, determine the magnitudes of forces F and P.

338

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Engineering Mechanics - Statics

Chapter 4

Given: F 1 = 10 lb a = 3 in b = 4 in c = 6 in d = 3 in

θ = 30 deg

Solution: Initial Guesses:

F = 1 lb

P = 1 lb

Given

⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ −F c ⎟ + ⎜ 0 ⎟ + ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ −P c ⎠

⎛0⎞ ⎜a⎟ = 0 2 2⎜ ⎟ a + b ⎝b⎠

⎛F⎞ ⎜ ⎟ = Find ( F , P) ⎝P⎠

F1 d

⎛F⎞ ⎛3⎞ ⎜ ⎟ = ⎜ ⎟ lb ⎝P⎠ ⎝4⎠

339

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Engineering Mechanics - Statics

Chapter 5

Problem 5-1 Draw the free-body diagram of the sphere of weight W resting between the smooth inclined planes. Explain the significance of each force on the diagram.

Given: W = 10 lb

θ 1 = 105 deg θ 2 = 45 deg

Solution:

NA, NB force of plane on sphere. W force of gravity on sphere.

Problem 5-2 Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B. Given: F = 8 lb a = 1.5 ft b = 0.2 ft c = 2 ft

340

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Engineering Mechanics - Statics

Chapter 5

Solution:

Problem 5-3 Draw the free-body diagram of the beam supported at A by a fixed support and at B by a roller. Explain the significance of each force on the diagram. Given: w = 40

lb ft

a = 3 ft b = 4 ft

θ = 30 deg

Solution:

A x, A y, MA effect of wall on beam. NB force of roller on beam. wa 2

resultant force of distributed load on beam.

Problem 5-4 Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by member (link) BC.

341

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: F = 8 kN a = 3m b = 4m c = 0.4 m d = 3 e = 4

Solution:

Problem 5-5 Draw the free-body diagram of the C-bracket supported at A, B, and C by rollers. Explain the significance of each forcce on the diagram. Given: a = 3 ft b = 4 ft

θ 1 = 30 deg θ 2 = 20 deg F = 200 lb 342

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Engineering Mechanics - Statics

Chapter 5

Solution: NA , NB , NC force of rollers on beam.

Problem 5-6 Draw the free-body diagram of the smooth rod of mass M which rests inside the glass. Explain the significance of each force on the diagram. Given: M = 20 gm a = 75 mm b = 200 mm

θ = 40 deg

Solution: A x , A y , NB force of glass on rod. M(g) N force of gravity on rod.

Problem 5-7 Draw the free-body diagram of the “spanner wrench” subjected to the force F. The support at A can be considered a pin, and the surface of contact at B is smooth. Explain the significance of 343

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Engineering Mechanics - Statics

Chapter 5

p each force on the diagram.

p

g

Given: F = 20 lb a = 1 in b = 6 in

Solution: A x, A y, NB force of cylinder on wrench.

Problem 5-8 Draw the free-body diagram of the automobile, which is being towed at constant velocity up the incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are free to roll. Explain the significance of each force on the diagram. Units Used: 3

Mg = 10 kg Given: M = 5 Mg

d = 1.50 m

a = 0.3 m

e = 0.6 m

b = 0.75 m

θ 1 = 20 deg

c = 1m

θ 2 = 30 deg

g = 9.81

m 2

s

344

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Engineering Mechanics - Statics

Chapter 5

Solution: NA, NB force of road on car. F force of cable on car. Mg force of gravity on car.

Problem 5-9 Draw the free-body diagram of the uniform bar, which has mass M and center of mass at G. The supports A, B, and C are smooth. Given: M = 100 kg a = 1.75 m b = 1.25 m c = 0.5 m d = 0.2 m g = 9.81

m 2

s Solution:

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 5

Problem 5-10 Draw the free-body diagram of the beam, which is pin-connected at A and rocker-supported at B. Given: F = 500 N M = 800 N⋅ m a = 8m b = 4m c = 5m

Solution:

Problem 5-11 The sphere of weight W rests between the smooth

inclined planes. Determine the reaactions at the supports. Given: W = 10 lb

θ 1 = 105 deg θ 2 = 45 deg Solution: Initial guesses 346

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Engineering Mechanics - Statics

NA = 1 lb

Chapter 5

NB = 1 lb

Given NB cos ( θ 1 − 90 deg) − NA cos ( θ 2 ) = 0 NA sin ( θ 2 ) − NB sin ( θ 1 − 90 deg) − W = 0

⎛ NA ⎞ ⎜ ⎟ = Find ( NA , NB) ⎝ NB ⎠

⎛ NA ⎞ ⎛ 19.3 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ NB ⎠ ⎝ 14.1 ⎠

Problem 5-12 Determine the magnitude of the resultant force acting at pin A of the handpunch. Given: F = 8 lb a = 1.5 ft b = 0.2 ft c = 2 ft

Solution:

Σ F x = 0; Σ M = 0;

Ax − F = 0

Ax = F

F a − Ay c = 0

Ay = F

Ax = 8 lb a c

Ay = 6 lb

347

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Engineering Mechanics - Statics

FA =

Chapter 5

2

Ax + Ay

2

F A = 10 lb

Problem 5-13 The C-bracket is supported at A, B, and C by rollers. Determine the reactions at the supports. Given: a = 3 ft b = 4 ft

θ 1 = 30 deg θ 2 = 20 deg F = 200 lb Solution: Initial Guesses: NA = 1 lb NB = 1 lb NC = 1 lb Given NA a − F b = 0 NB sin ( θ 1 ) − NC sin ( θ 2 ) = 0 NB cos ( θ 1 ) + NC cos ( θ 2 ) − NA − F = 0

⎛ NA ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NA , NB , NC) ⎜N ⎟ ⎝ C⎠

⎛ NA ⎞ ⎛ 266.7 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NB ⎟ = ⎜ 208.4 ⎟ lb ⎜ N ⎟ ⎝ 304.6 ⎠ ⎝ C⎠

348

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Engineering Mechanics - Statics

Chapter 5

Problem 5-14 The smooth rod of mass M rests inside the glass. Determine the reactions on the rod. Given: M = 20 gm a = 75 mm b = 200 mm

θ = 40 deg g = 9.81

m 2

s Solution: Initial Guesses: Ax = 1 N

Ay = 1 N

NB = 1 N

Given Ax − NB sin ( θ ) = 0 Ay − M g + NB cos ( θ ) = 0 −M g

a+b cos ( θ ) + NB b = 0 2

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , NB) ⎜N ⎟ ⎝ B⎠

⎛ Ax ⎞ ⎛ 0.066 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0.117 ⎟ N ⎜ N ⎟ ⎝ 0.103 ⎠ ⎝ B⎠

Problem 5-15 The “spanner wrench” is subjected to the force F. The support at A can be considered a pin, and the surface of contact at B is smooth. Determine the reactions on the spanner wrench.

349

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Engineering Mechanics - Statics

Chapter 5

Given: F = 20 lb a = 1 in b = 6 in Solution: Initial Guesses: Ax = 1 lb Ay = 1 lb NB = 1 lb Given − Ax + NB = 0

Ay − F = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , NB) ⎜N ⎟ ⎝ B⎠

−F ( a + b) + Ax a = 0

⎛ Ax ⎞ ⎛ 140 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 20 ⎟ lb ⎜ N ⎟ ⎝ 140 ⎠ ⎝ B⎠

Problem 5-16 The automobile is being towed at constant velocity up the incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are free to roll. Determine the reactions on both wheels at A and B and the tension in the cable at C. Units Used: 3

Mg = 10 kg

3

kN = 10 N

350

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Engineering Mechanics - Statics

Chapter 5

Given: M = 5 Mg

d = 1.50 m

a = 0.3 m

e = 0.6 m

b = 0.75 m

θ 1 = 20 deg

c = 1m

θ 2 = 30 deg

g = 9.81

m 2

s Solution: Guesses

F = 1 kN

NA = 1 kN

NB = 1 kN

Given NA + NB + F sin ( θ 2 ) − M g cos ( θ 1 ) = 0 −F cos ( θ 2 ) + M g sin ( θ 1 ) = 0 F cos ( θ 2 ) a − F sin ( θ 2 ) b − M g cos ( θ 1 ) c − M g sin ( θ 1 ) e + NB( c + d) = 0

⎛F ⎞ ⎜ ⎟ ⎜ NA ⎟ = Find ( F , NA , NB) ⎜ NB ⎟ ⎝ ⎠

⎛ F ⎞ ⎛ 19.37 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NA ⎟ = ⎜ 13.05 ⎟ kN ⎜ NB ⎟ ⎝ 23.36 ⎠ ⎝ ⎠

Problem 5-17 The uniform bar has mass M and center of mass at G. The supports A, B, and C are smooth. Determine the reactions at the points of contact at A, B, and C. Given: M = 100 kg a = 1.75 m b = 1.25 m

351

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Engineering Mechanics - Statics

Chapter 5

c = 0.5 m d = 0.2 m

θ = 30 deg g = 9.81

m 2

s Solution:

The initial guesses: NA = 20 N NB = 30 N NC = 40 N Given

ΣMA = 0; +

↑Σ Fy = 0; +

↑Σ Fy = 0;

−M g cos ( θ ) a − M g sin ( θ )

d + NB sin ( θ ) d + NC( a + b) = 0 2

NB − M g + NC cos ( θ ) = 0 NA − NC sin ( θ ) = 0

⎛ NC ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NC , NB , NA) ⎜N ⎟ ⎝ A⎠

⎛ NC ⎞ ⎛ 493 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NB ⎟ = ⎜ 554 ⎟ N ⎜ N ⎟ ⎝ 247 ⎠ ⎝ A⎠

Problem 5-18 The beam is pin-connected at A and rocker-supported at B. Determine the reactions at the pin A and at the roller at B. Given: F = 500 N M = 800 N⋅ m a = 8m

352

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Engineering Mechanics - Statics

Chapter 5

b = 4m c = 5m Solution:

⎞ ⎟ ⎝ a + b⎠

α = atan ⎛⎜

c

ΣMA = 0;

−F

a

cos ( α )

By =

− M + By a = 0

F a + M cos ( α ) cos ( α ) a

B y = 642 N + Σ F x = 0; → +

↑Σ Fy = 0;

− Ax + F sin ( α ) = 0

Ax = F sin ( α )

Ax = 192 N

− Ay − F cos ( α ) + B y = 0

Ay = −F cos ( α ) + By

Ay = 180 N

Problem 5-19 Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam. Given: F 1 = 600 N F 2 = 400 N

θ = 15 deg a = 4m b = 8m c = 3 d = 4 Solution:

ΣMA = 0;

B y( a + b) − F2 cos ( θ ) ( a + b) − F1 a = 0 353

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Engineering Mechanics - Statics

By = + Σ F x = 0; →

Chapter 5

F 2 cos ( θ ) ( a + b) + F 1 a a+b

Ax − F 2 sin ( θ ) = 0 Ax = F2 sin ( θ )

+

↑Σ Fy = 0;

B y = 586 N

Ax = 104 N

Ay − F 2 cos ( θ ) + B y − F1 = 0 Ay = F2 cos ( θ ) − B y + F1

FA =

2

Ax + Ay

2

Ay = 400 N

F A = 413 N

Problem 5-20 Determine the reactions at the supports. Given: w = 250

lb ft

a = 6 ft b = 6 ft c = 6 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

B y = 1 lb

Given Ax = 0

wa

Ay + B y − w( a + b) −

⎛ a ⎞ − w b⎛ b ⎞ − 1 w c⎛b + ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 2⎠ ⎝ 2⎠ 2 ⎝

1 2

wc = 0

c⎞

⎟ + By b = 0

3⎠

354

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Engineering Mechanics - Statics

Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , By) ⎜B ⎟ ⎝ y⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 2750 ⎟ lb ⎜ B ⎟ ⎝ 1000 ⎠ ⎝ y⎠

Problem 5-21 When holding the stone of weight W in equilibrium, the humerus H, assumed to be smooth, exerts normal forces FC and F A on the radius C and ulna A as shown. Determine these forces and the force F B that the biceps B exerts on the radius for equilibrium. The stone has a center of mass at G. Neglect the weight of the arm. Given: W = 5 lb

θ = 75 deg a = 2 in b = 0.8 in c = 14 in

Solution:

ΣMB = 0;

− W ( c − a) + F A a = 0 FA = W

⎛ c − a⎞ ⎜ ⎟ ⎝ a ⎠

F A = 30 lb +

↑Σ Fy = 0;

F B sin ( θ ) − W − FA = 0

FB =

W + FA sin ( θ )

F B = 36.2 lb + Σ F x = 0; →

F C − F B cos ( θ ) = 0 355

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Engineering Mechanics - Statics

Chapter 5

F C = FB cos ( θ ) F C = 9.378 lb

Problem 5-22 The uniform door has a weight W and a center of gravity at G. Determine the reactions at the hinges if the hinge at A supports only a horizontal reaction on the door, whereas the hinge at B exerts both horizontal and vertical reactions. Given: W = 100 lb a = 3 ft b = 3 ft c = 0.5 ft d = 2 ft

Solution: ΣMB = 0;

W d − A x ( a + b) = 0 Ax = W

⎛ d ⎞ ⎜ ⎟ ⎝ a + b⎠

Ax = 33.3 lb

ΣF x = 0;

Bx = Ax

B x = 33.3 lb

ΣF y = 0;

By = W

B y = 100 lb

Problem 5-23 The ramp of a ship has weight W and center of gravity at G. Determine the cable force in CD needed to just start lifting the ramp, (i.e., so the reaction at B becomes zero). Also, determine the horizontal and vertical components of force at the hinge (pin) at A.

356

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Engineering Mechanics - Statics

Chapter 5

Given: W = 200 lb

a = 4 ft

θ = 30 deg

b = 3 ft

φ = 20 deg

c = 6 ft

Solution:

ΣMA = 0; −F CD cos ( θ ) ( b + c) cos ( φ ) + FCD sin ( θ ) ( b + c) sin ( φ ) + W c cos ( φ ) = 0

F CD =

W c cos ( φ )

( b + c) ( cos ( θ ) cos ( φ ) − sin ( θ ) sin ( φ ) )

+ Σ F x = 0; →

F CD sin ( θ ) − A x = 0 Ax = FCD sin ( θ )

+

↑Σ Fy = 0;

F CD = 195 lb

Ax = 97.5 lb

Ay − W + F CD cos ( θ ) = 0 Ay = W − FCD cos ( θ )

Ay = 31.2 lb

Problem 5-24 The drainpipe of mass M is held in the tines of the fork lift. Determine the normal forces at A and B as functions of the blade angle θ and plot the results of force (ordinate) versus θ (abscissa) for 0 ≤ θ ≤ 90 deg.

357

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Engineering Mechanics - Statics

Chapter 5

Units used: 3

Mg = 10 kg Given: M = 1.4 Mg a = 0.4 m g = 9.81

m 2

s Solution:

θ = 0 .. 90 NA ( θ ) =

M g sin ( θ deg) 3

10 NB ( θ ) =

M g cos ( θ deg) 3

10

Force in kN

15 NA( θ )10 NB( θ )

5

0

0

20

40

60

80

100

θ

Angle in Degrees

Problem 5-25 While slowly walking, a man having a total mass M places all his weight on one foot. Assuming that the normal force NC of the ground acts on his foot at C, determine the resultant vertical compressive force F B which the tibia T exerts on the astragalus B, and the vertical tension FA in the achilles tendon A at the instant shown.

358

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: M = 80 kg a = 15 mm b = 5 mm c = 20 mm d = 100 mm

Solution: NC = M g NC = 785 N ΣMA = 0;

−F B c + NC ( c + d) = 0 F B = NC

⎛ c + d⎞ ⎜ ⎟ ⎝ c ⎠

F B = 4.71 kN ΣF y = 0;

F A − FB + NC = 0 F A = F B − NC F A = 3.92 kN

Problem 5-26 Determine the reactions at the roller A and pin B.

359

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Engineering Mechanics - Statics

Chapter 5

Given: M = 800 lb ft

c = 3 ft

F = 390 lb

d = 5

a = 8 ft

e = 12

b = 4 ft

θ = 30 deg

Solution: Guesses

R A = 1 lb B x = 1 lb

B y = 1 lb

Given R A sin ( θ ) + B x −

⎛ d ⎞F = 0 ⎜ 2 2⎟ ⎝ e +d ⎠

R A cos ( θ ) + By −

e ⎛ ⎞ ⎜ 2 2 ⎟F = 0 ⎝ e +d ⎠

M − R A cos ( θ ) ( a + b) + Bx c = 0

⎛ RA ⎞ ⎜ ⎟ ⎜ Bx ⎟ = Find ( RA , Bx , By) ⎜B ⎟ ⎝ y⎠

R A = 105.1 lb

⎛ Bx ⎞ ⎛ 97.4 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ By ⎠ ⎝ 269 ⎠

Problem 5-27 The platform assembly has weight W1 and center of gravity at G1. If it is intended to support a maximum load W2 placed at point G2,,determine the smallest counterweight W that should be placed at B in order to prevent the platform from tipping over. Given: W1 = 250 lb

a = 1 ft

c = 1 ft

e = 6 ft

W2 = 400 lb

b = 6 ft

d = 8 ft

f = 2 ft

360

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Engineering Mechanics - Statics

Chapter 5

Solution: When tipping occurs, R c = 0

ΣMD = 0;

−W2 f + W1 c + WB ( b + c) = 0

WB =

W2 f − W1 c b+c

WB = 78.6 lb

Problem 5-28 The articulated crane boom has a weight W and mass center at G. If it supports a load L, determine the force acting at the pin A and the compression in the hydraulic cylinder BC when the boom is in the position shown. Units Used: 3

kip = 10 lb Given: W = 125 lb

361

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Engineering Mechanics - Statics

Chapter 5

L = 600 lb a = 4 ft b = 1 ft c = 1 ft d = 8 ft

θ = 40 deg

Solution: Guesses

Given

Ax = 1 lb

Ay = 1 lb

F B = 1 lb

− Ax + F B cos ( θ ) = 0

− Ay + F B sin ( θ ) − W − L = 0

F B cos ( θ ) b + F B sin ( θ ) c − W a − L( d + c) = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , FB) ⎜F ⎟ ⎝ B⎠

F B = 4.19 kip

⎛ Ax ⎞ ⎛ 3.208 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ Ay ⎠ ⎝ 1.967 ⎠

Problem 5-29 The device is used to hold an elevator door open. If the spring has stiffness k and it is compressed a distnace δ, determine the horizontal and vertical components of reaction at the pin A and the resultant force at the wheel bearing B. Given: N m

b = 125 mm

δ = 0.2 m

c = 100 mm

a = 150 mm

θ = 30 deg

k = 40

362

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Engineering Mechanics - Statics

Solution:

ΣMA = 0;

Chapter 5

F s = kδ −F s a + F B cos ( θ ) ( a + b) − F B sin ( θ ) c = 0 FB = Fs

a

cos ( θ ) ( a + b) − sin ( θ ) c

F B = 6.378 N + Σ F x = 0; →

Ax − F B sin ( θ ) = 0 Ax = FB sin ( θ ) Ax = 3.189 N

+

↑Σ Fy = 0;

Ay − F s + F B cos ( θ ) = 0 Ay = Fs − FB cos ( θ ) Ay = 2.477 N

Problem 5-30 Determine the reactions on the bent rod which is supported by a smooth surface at B and by a collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod. Given: F = 100 lb M = 200 lb ft a = 3 ft b = 3 ft c = 2ft

363

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Engineering Mechanics - Statics

Chapter 5

d = 3 e = 4 f = 12 g = 5 Solution: Initial Guesses: NA = 20 lb

NB = 10 lb

MA = 30 lb ft

Given f g ⎞ ⎞ ⎛ ⎛ ⎜ 2 2 ⎟ ( a + b) − NB ⎜ 2 2 ⎟ c = 0 ⎝ f +g ⎠ ⎝ f +g ⎠

ΣMA = 0;

MA − F a − M + NB

ΣF x = 0;

NA

e g ⎞ ⎞ ⎛ ⎛ ⎜ 2 2 ⎟ − NB ⎜ 2 2 ⎟ = 0 ⎝ e +d ⎠ ⎝ f +g ⎠

ΣF y = 0;

NA

f ⎞−F=0 ⎛ d ⎞+N ⎛ B⎜ ⎜ 2 2⎟ ⎟ 2 2 ⎝ e +d ⎠ ⎝ f +g ⎠

⎛ NA ⎞ ⎜ ⎟ ⎜ NB ⎟ = Find ( NA , NB , MA) ⎜M ⎟ ⎝ A⎠

⎛ NA ⎞ ⎛ 39.7 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ NB ⎠ ⎝ 82.5 ⎠

MA = 106 lb⋅ ft

Problem 5-31 The cantilevered jib crane is used to support the load F. If the trolley T can be placed anywhere in the range x1 ≤ x ≤ x2, determine the maximum magnitude of reaction at the supports A and B. Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, whereas the one at A does not. Units Used: kip = 1000 lb Given: F = 780 lb

364

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Engineering Mechanics - Statics

Chapter 5

a = 4 ft b = 8 ft x1 = 1.5 ft x2 = 7.5 ft Solution: The maximum occurs when x = x2

ΣMA = 0;

−F x2 + B x a = 0

Bx = F

x2 a 3

B x = 1.462 × 10 lb + Σ F x = 0; → +

↑Σ Fy = 0; FB =

Ax − Bx = 0 By − F = 0 2

Bx + By

Ax = Bx By = F

2

3

Ax = 1.462 × 10 lb B y = 780 lb

F B = 1.657 kip

Problem 5-32 The uniform rod AB has weight W. Determine the force in the cable when the rod is in the position shown. Given: W = 15 lb L = 5 ft

365

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Engineering Mechanics - Statics

Chapter 5

θ 1 = 30 deg θ 2 = 10 deg

Solution:

ΣMA = 0;

NB L sin ( θ 1 + θ 2 ) − W

NB =

⎛ L ⎞ cos ( θ + θ ) = 0 ⎜ ⎟ 1 2 ⎝2⎠

W cos ( θ 1 + θ 2 ) 2 sin ( θ 1 + θ 2 )

NB = 8.938 lb ΣF x = 0;

T cos ( θ 2 ) − NB

T =

NB

cos ( θ 2 )

T = 9.08 lb

Problem 5-33 The power pole supports the three lines, each line exerting a vertical force on the pole due to its weight as shown. Determine the reactions at the fixed support D. If it is possible for wind or ice to snap the lines, determine which line(s) when removed create(s) a condition for the greatest moment reaction at D. Units Used: 3

kip = 10 lb

366

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Engineering Mechanics - Statics

Chapter 5

Given: W1 = 800 lb W2 = 450 lb W3 = 400 lb a = 2ft b = 4 ft c = 3 ft Solution: + Σ F x = 0; → +

↑Σ Fy = 0;

Dx = 0 Dy − ( W1 + W2 + W3 ) = 0 Dy = W1 + W2 + W3

ΣMD = 0;

Dy = 1.65 kip

−W2 b − W3 c + W1 a + MD = 0 MD = W2 b + W3 c − W1 a

MD = 1.4 kip⋅ ft

Examine all cases. For these numbers we require line 1 to snap. MDmax = W2 b + W3 c

MDmax = 3 kip⋅ ft

Problem 5-34 The picnic table has a weight WT and a center of gravity at GT . If a man weighing WM has a center of gravity at GM and sits down in the centered position shown, determine the vertical reaction at each of the two legs at B.Neglect the thickness of the legs. What can you conclude from the results? Given: WT = 50 lb

367

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Engineering Mechanics - Statics

Chapter 5

WM = 225 lb a = 6 in b = 20 in c = 20 in

Solution: ΣMA = 0;

2 NB( b + c) + WM a − WT b = 0

NB =

WT b − WM a 2 ( b + c)

NB = −4.37 lb Since NB has a negative sign, the table will tip over.

Problem 5-35 If the wheelbarrow and its contents have a mass of M and center of mass at G, determine the magnitude of the resultant force which the man must exert on each of the two handles in order to hold the wheelbarrow in equilibrium. Given: M = 60 kg a = 0.6 m b = 0.5 m c = 0.9 m d = 0.5 m g = 9.81

m 2

s

368

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Engineering Mechanics - Statics

Chapter 5

Solution:

ΣMB = 0;

− Ay ( b + c) + M g c = 0 Ay =

Mgc b+c

Ay = 378.386 N + Σ F x = 0; →

B x = 0N

+

Ay − M g + 2 By = 0

↑Σ Fy = 0;

By =

Bx = 0

M g − Ay 2

B y = 105.107 N

Problem 5-36 The man has weight W and stands at the center of the plank. If the planes at A and B are smooth, determine the tension in the cord in terms of W and θ.

Solution: ΣMB = 0;

W

L cos ( φ ) − NA L cos ( φ ) = 0 2

NA = ΣF x = 0;

W 2

T cos ( θ ) − NB sin ( θ ) = 0

(1)

369

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Engineering Mechanics - Statics

Chapter 5

T sin ( θ ) + NB cos ( θ ) + NA − W = 0

ΣF y = 0;

(2)

Solving Eqs. (1) and (2) yields:

T=

W sin ( θ ) 2

NB =

W cos ( θ ) 2

Problem 5-37 When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is F , determine the unstretched length of the spring if the stiffness of the spring is k. Given: F = 3N k = 80

N m

a = 100 mm b = 50 mm c = 40 mm d = 10 mm

θ = 30 deg Solution: ΣMD = 0; F s b − F cos ( θ ) c − F sin ( θ ) d = 0 Fs = F

cos ( θ ) c + sin ( θ ) d b

Fs = k x

x =

L0 = a − x

Fs k

F s = 2.378 N

x = 29.73 mm L0 = 70.3 mm

370 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 5

Problem 5-38 The telephone pole of negligible thickness is subjected to the force F directed as shown. It is supported by the cable BCD and can be assumed pinned at its base A. In order to provide clearance for a sidewalk right of way, where D is located, the strut CE is attached at C, as shown by the dashed lines (cable segment CD is removed). If the tension in CD' is to be twice the tension in BCD, determine the height h for placement of the strut CE. Given: F = 80 lb

θ = 30 deg a = 30 ft b = 10 ft

Solution:

+ ΣMA = 0;

−F cos ( θ ) a +

b ⎛ ⎞ ⎜ 2 2 ⎟ TBCD a = 0 ⎝ a +b ⎠

TBCD = F cos ( θ )

2

2

a +b b

TBCD = 219.089 lb

Require TCD' = 2 TBCD + ΣMA = 0;

TCD' = 438.178 lb

TCD' d − F cos ( θ ) a = 0

d = Fa

⎛ cos ( θ ) ⎞ ⎜T ⎟ ⎝ CD' ⎠

d = 4.7434 ft

Geometry: a−h a = d b

h = a−a

⎛ d⎞ ⎜ ⎟ ⎝ b⎠

h = 15.8 ft

371

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Engineering Mechanics - Statics

Chapter 5

Problem 5-39 The worker uses the hand truck to move material down the ramp. If the truck and its contents are held in the position shown and have weight W with center of gravity at G, determine the resultant normal force of both wheels on the ground A and the magnitude of the force required at the grip B. Given: W = 100 lb

e = 1.5 ft

a = 1 ft

f = 0.5 ft

b = 1.5 ft

θ = 60 deg

c = 2 ft

φ = 30 deg

d = 1.75 ft

Solution: ΣMB = 0; NA cos ( θ − φ ) ( b + c + d) + NA sin ( θ − φ ) ( a − f) − W cos ( θ ) ( b + c) − W sin ( θ ) ( e + a) = 0

NA =

W cos ( θ ) ( b + c) + W sin ( θ ) ( e + a)

cos ( θ − φ ) ( b + c + d) + sin ( θ − φ ) ( a − f)

NA = 81.621 lb

ΣF x = 0;

−B x + NA sin ( φ ) = 0

B x = NA sin ( φ )

B x = 40.811 lb

ΣF y = 0;

B y + NA ( cos ( φ ) − W = 0)

B y = W − NA cos ( φ )

B y = 29.314 lb

FB =

2

Bx + By

2

F B = 50.2 lb

372

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Engineering Mechanics - Statics

Chapter 5

T sin ( θ ) + NB cos ( θ ) + NA − W = 0

ΣF y = 0;

(2)

Solving Eqs. (1) and (2) yields:

T=

W sin ( θ ) 2

NB =

W cos ( θ ) 2

Problem 5-37 When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is F , determine the unstretched length of the spring if the stiffness of the spring is k. Given: F = 3N k = 80

N m

a = 100 mm b = 50 mm c = 40 mm d = 10 mm

θ = 30 deg Solution: ΣMD = 0; F s b − F cos ( θ ) c − F sin ( θ ) d = 0 Fs = F

cos ( θ ) c + sin ( θ ) d b

Fs = k x

x =

L0 = a − x

Fs k

F s = 2.378 N

x = 29.73 mm L0 = 70.3 mm 373

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Engineering Mechanics - Statics

Chapter 5

The shelf supports the electric motor which has mass m1 and mass center at Gm. The platform upon which it rests has mass m2 and mass center at Gp. Assuming that a single bolt B holds the shelf up and the bracket bears against the smooth wall at A, determine this normal force at A and the horizontal and vertical components of reaction of the bolt B on the bracket. Given: m1 = 15 kg

c = 50 mm

m2 = 4 kg

d = 200 mm

a = 60 mm

e = 150 mm

b = 40 mm g = 9.81

m 2

s Solution: ΣMA = 0;

B x a − m2 g d − m1 g( d + e) = 0

Bx = g + Σ F x = 0; →

m2 d + m1 ( d + e)

B x = 989 N

a

Ax − Bx = 0 Ax = Bx

+

↑Σ Fy = 0;

Ax = 989 N

B y − m2 g − m1 g = 0 B y = m2 g + m1 g

B y = 186 N

Problem 5-42 A cantilever beam, having an extended length L, is subjected to a vertical force F. Assuming that the wall resists this load with linearly varying distributed loads over the length a of the beam portion inside the wall, determine the intensities w1 and w2 for equilibrium.

374

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: F = 500 N a = 0.15 m L = 3m Solution: The initial guesses w1 = 1

kN m

w2 = 1

kN m

Given +

↑Σ Fy = 0; ΣMA = 0;

⎛ w1 ⎞ ⎜ ⎟ = Find ( w1 , w2) ⎝ w2 ⎠

1

1 w1 a − w2 a − F = 0 2 2 −F L −

1 2

1 ⎛ 2 a⎞ ⎟ + w2 a⎜ ⎟ = 0 ⎝ 3⎠ 2 ⎝ 3 ⎠

w1 a⎛⎜

a⎞

⎛ w1 ⎞ ⎛ 413 ⎞ kN ⎜ ⎟=⎜ ⎟ ⎝ w2 ⎠ ⎝ 407 ⎠ m

Problem 5-43 The upper portion of the crane boom consists of the jib AB, which is supported by the pin at A, the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If the load F is supported by the hoist line, which passes over the pulley at B, determine the magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has a radius of r.

375

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: F = 5 kN r = 0.1 m a = r b = 1.5 m c = 5m Solution: From pulley, tension in the hoist line is ΣMB = 0;

T( a) − F( r) = 0

T = F

r T = 5 kN

a

From the jib, ΣMA = 0;

b+a

−F ( c) + TBC

c=0

2

c + ( b + a) 2

c + ( b + a)

TBC = F

+

↑Σ Fy = 0;

2

2

TBC = 16.406 kN

b+a b+a ⎤−F=0 ⎢ 2 2⎥ ⎣ c + ( b + a) ⎦

− Ay + TBC ⎡

b+a ⎤−F ⎢ 2 2⎥ ⎣ c + ( b + a) ⎦

Ay = TBC⎡

+ Σ F x = 0; →

Ay = 0 kN

⎤−F=0 ⎣ c + ( b + a) ⎦

Ax − TBC ⎡ ⎢

c

2

2⎥

376

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Engineering Mechanics - Statics

Chapter 5

c

Ax = TBC

2

c + ( b + a) FA =

2

Ax + Ay

+F 2

2

Ax = 20.6 kN

F A = 20.6 kN

Problem 5-44 The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center of gravity at G2. Determine the smallest angle of tilt θ of the boom, without causing the crane to overturn if the suspended load has weight W. Neglect the thickness of the tracks at A and B. Given: W1 = 120000 lb W2 = 30000 lb W = 40000 lb a = 4 ft b = 6 ft c = 3 ft d = 12 ft e = 15 ft Solution: When tipping occurs, R A = 0 ΣMB = 0;

−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ + W1 ( b + c) = 0

⎡W2 c + W c + W1 ( b + c) ⎤ ⎥ ⎣ W2 d + W ( d + e) ⎦

θ = acos ⎢

θ = 26.4 deg

377

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Engineering Mechanics - Statics

Chapter 5

Problem 5-45 The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center of gravity at G2. If the suspended load has weight W determine the normal reactions at the tracks A and B. For the calculation, neglect the thickness of the tracks . Units Used: 3

kip = 10 lb Given: W1 = 120000 lb

a = 4 ft

W2 = 30000 lb

b = 6 ft

W = 16000 lb

c = 3 ft

θ = 30 deg

d = 12 ft e = 15 ft

Solution: ΣMB = 0; −W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ − RA( a + b + c) + W1 ( b + c) = 0

RA = +

↑

−W2 ( d cos ( θ ) − c) − W⎡⎣( d + e)cos ( θ ) − c⎤⎦ + W1 ( b + c)

Σ F y = 0;

a+b+c

R A = 40.9 kip

R A + RB − W1 − W2 − W = 0 R B = −R A + W1 + W2 + W

R B = 125 kip

Problem 5-46 The man attempts to support the load of boards having a weight W and a center of gravity at G. If he is standing on a smooth floor, determine the smallest angle θ at which he can hold them up in the position shown. Neglect his weight .

378

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Engineering Mechanics - Statics

Chapter 5

Given: a = 0.5 ft b = 3 ft c = 4 ft d = 4 ft

Solution: ΣMB = 0;

−NA( a + b) + W( b − c cos ( θ ) ) = 0

As θ becomes smaller, NA goes to 0 so that, cos ( θ ) =

b c

b θ = acos ⎛⎜ ⎟⎞

⎝c⎠

θ = 41.4 deg

Problem 5-47 The motor has a weight W. Determine the force that each of the chains exerts on the supporting hooks at A, B, and C. Neglect the size of the hooks and the thickness of the beam.

379

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Engineering Mechanics - Statics

Chapter 5

Given: W = 850 lb a = 0.5 ft b = 1 ft c = 1.5 ft

θ 1 = 10 deg θ 2 = 30 deg θ 3 = 10 deg

Solution: Guesses F A = 1 lb

F B = 1 lb

F C = 1 lb

Given ΣMB = 0;

F A cos ( θ 3 ) b + W a − FC cos ( θ 1 ) ( a + c) = 0

ΣF x = 0;

F C sin ( θ 1 ) − F B sin ( θ 2 ) − FA sin ( θ 3 ) = 0

ΣF y = 0;

W − F A cos ( θ 3 ) − F B cos ( θ 2 ) − F C cos ( θ 1 ) = 0

⎛ FA ⎞ ⎜ ⎟ ⎜ FB ⎟ = Find ( FA , FB , FC) ⎜F ⎟ ⎝ C⎠ ⎛ FA ⎞ ⎛ 432 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FB ⎟ = ⎜ −0 ⎟ lb ⎜ F ⎟ ⎝ 432 ⎠ ⎝ C⎠

380

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Engineering Mechanics - Statics

Chapter 5

Problem 5-48 The boom supports the two vertical loads. Neglect the size of the collars at D and B and the thickness of the boom, and compute the horizontal and vertical components of force at the pin A and the force in cable CB. Given: F 1 = 800 N F 2 = 350 N a = 1.5 m b = 1m c = 3 d = 4

θ = 30 deg Solution: ΣMA = 0; −F 1 a cos ( θ ) − F2 ( a + b) cos ( θ ) +

d 2

2

FCB ( a + b) sin ( θ ) +

c +d

F CB

d 2

c +d Ax =

2

FCB ( a + b) cos ( θ ) = 0

F CB = 0

d 2

2

F CB = 782 N

d sin ( θ ) ( a + b) + c cos ( θ ) ( a + b) Ax −

2

c +d

2 2 ⎡⎣F1 a + F2 ( a + b)⎤⎦ cos ( θ ) c + d =

+ Σ F x = 0; →

c

2

Ax = 625 N

F CB

c +d +

↑

Σ F y = 0;

Ay − F1 − F2 +

c 2

c +d Ay = F1 + F2 −

2

F CB = 0

c 2

2

FCB

Ay = 681 N

c +d

Problem 5-49 The boom is intended to support two vertical loads F1 and F 2. If the cable CB can sustain a 381

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Engineering Mechanics - Statics

Chapter 5

maximum load Tmax before it fails, determine the critical loads if F1 = 2F2. Also, what is the magnitude of the maximum reaction at pin A? Units Used: 3

kN = 10 N Given: Tmax = 1500 N a = 1.5 m b = 1m c = 3 d = 4

θ = 30deg Solution: ΣMA = 0;

F1 = 2 F2

−2 F 2 a cos ( θ ) − F2 ( a + b) cos ( θ ) +

d 2

2

Tmax( a + b) sin ( θ ) +

c +d

F2 =

( a + b)Tmax( d sin ( θ ) + c cos ( θ ) ) 2

F1 = 2 F2 + ΣF x = 0; →

Ax −

d 2

c +d Ax =

2

2

Tmax( a + b) cos ( θ ) = 0

F 1 = 1.448 kN

Tmax = 0

d 2

2

c +d

F 2 = 724 N

c + d cos ( θ ) ( 3 a + b) 2

c

2

Ax = 1.20 kN

Tmax

c +d

+

↑Σ Fy = 0;

Ay − F2 − F1 +

c 2

c +d

2

Tmax = 0

382

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Engineering Mechanics - Statics

Chapter 5

c

Ay = F2 + F1 −

2

2

Ay = 1.27 kN

Tmax

c +d

2

FA =

Ax + Ay

2

F A = 1.749 kN

Problem 5-50 The uniform rod of length L and weight W is supported on the smooth planes. Determine its position θ for equilibrium. Neglect the thickness of the rod.

Solution: L

cos ( θ ) + NA cos ( φ − θ ) L = 0

ΣMB = 0;

−W

ΣMA = 0;

W

ΣF x = 0;

NB sin ( ψ) − NA sin ( φ ) = 0

W cos ( θ )

2 cos ( ψ + θ )

L 2

2

NA =

cos ( θ ) − NB cos ( ψ + θ ) L = 0

sin ( ψ) −

W cos ( θ )

2 cos ( φ − θ )

NB =

W cos ( θ )

2 cos ( φ − θ ) W cos ( θ )

2 cos ( ψ + θ )

sin ( φ ) = 0

sin ( ψ) cos ( φ − θ ) − sin ( φ ) cos ( ψ + θ ) = 0 sin ( ψ) ( cos ( φ ) cos ( θ ) + sin ( φ ) sin ( θ ) ) − sin ( φ ) ( cos ( ψ) cos ( θ ) − sin ( ψ) sin ( θ ) ) = 0 2 sin ( ψ) sin ( φ ) sin ( θ ) = ( sin ( φ ) cos ( ψ) − sin ( ψ) cos ( φ ) ) cos ( θ ) tan ( θ ) =

sin ( φ ) cos ( ψ) − sin ( ψ) cos ( φ ) 2 sin ( ψ) sin ( φ )

=

cot ( ψ) − cot ( φ ) 2

⎛ cot ( ψ) − cot ( φ ) ⎞ ⎟ 2 ⎝ ⎠

θ = atan ⎜

383 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 5

Problem 5-51 The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in place by the spring which has unstretched length δ. Determine the magnitude of the resultant force at A and the normal force on the peg at B when the lever is in the position shown. Given:

δ = 200 mm k = 5

N m

a = 100 mm b = 300 mm c = 300 mm

θ = 30 deg

Solution: Using the law of cosines and the law of sines c + ( a + b) − 2 c( a + b) cos ( 180 deg − θ ) 2

l =

sin ( φ ) c

=

2

sin ( 180 deg − θ ) l

Fs = k s = k (l − δ )

ΣMA = 0;

−F s sin ( φ ) ( a + b) + NB a = 0

⎛ sin (180 deg − θ ) ⎞ ⎟ l ⎝ ⎠

φ = asin ⎜c

φ = 12.808 deg

F s = k( l − δ )

F s = 2.3832 N

NB = F s sin ( φ )

a+b a

NB = 2.11 N

ΣF x = 0;

Ax − F s cos ( φ ) = 0

Ax = Fs cos ( φ )

Ax = 2.3239 N

ΣF y = 0;

Ay + NB − Fs sin ( φ ) = 0

Ay = Fs sin ( φ ) − NB

Ay = −1.5850 N

FA =

2

Ax + Ay

2

F A = 2.813 N

384 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 5

Problem 5-52 The rigid beam of negligible weight is supported horizontally by two springs and a pin. If the springs are uncompressed when the load is removed, determine the force in each spring when the load P is applied. Also, compute the vertical deflection of end C. Assume the spring stiffness k is large enough so that only small deflections occur. Hint: The beam rotates about A so the deflections in the springs can be related. Solution: ΣMA = 0; 3 F B L + FC2 L − P L = 0 2 F B + 2 FC = 1.5 P ΔC = 2 ΔB FC k

=

2 FB k

F C = 2FB 5 F B = 1.5 P F B = 0.3 P F c = 0.6 P ΔC =

0.6 P k

Problem 5-53 The rod supports a weight W and is pinned at its end A. If it is also subjected to a couple moment of M, determine the angle θ for equilibrium.The spring has an unstretched length δ and a stiffness k.

385

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Engineering Mechanics - Statics

Chapter 5

Given: W = 200 lb M = 100 lb ft

δ = 2 ft k = 50

lb ft

a = 3 ft b = 3 ft c = 2 ft Solution: Initial Guess:

θ = 10 deg

Given k⎡⎣( a + b)sin ( θ ) + c − δ⎤⎦ ( a + b) cos ( θ ) − W a cos ( θ ) − M = 0

θ = Find ( θ )

θ = 23.2 deg

Problem 5-54 The smooth pipe rests against the wall at the points of contact A, B, and C. Determine the reactions at these points needed to support the vertical force F . Neglect the pipe's thickness in the calculation. Given: F = 45 lb

386

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Engineering Mechanics - Statics

Chapter 5

θ = 30 deg a = 16 in b = 20 in c = 8 in

Solution: R A = 1 lb

Initial Guesses:

R B = 1 lb

R C = 1 lb

Given F cos ( θ ) ( a + b) − F sin ( θ ) c − RC b + R B c tan ( θ ) = 0

ΣMA = 0;

R C cos ( θ ) − R B cos ( θ ) − F = 0

+

↑Σ Fy = 0;

+ Σ F x = 0; →

R A + RB sin ( θ ) − RC sin ( θ ) = 0

⎛ RA ⎞ ⎜ ⎟ ⎜ RB ⎟ = Find ( RA , RB , RC) ⎜R ⎟ ⎝ C⎠

⎛ RA ⎞ ⎛ 25.981 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ RB ⎟ = ⎜ 11.945 ⎟ lb ⎜ R ⎟ ⎝ 63.907 ⎠ ⎝ C⎠

Problem 5-55 The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the stiffness of the springs at A and B is k, and the strip is originally horizontal when the springs are unstretched, determine the smallest force needed to close the contact gap at C. Units Used: mN = 10

−3

N

387

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Engineering Mechanics - Statics

Chapter 5

Given: a = 50 mm b = 50 mm c = 10 mm k = 5

N m

Solution: Initial Guesses:

F = 0.5 N

yA = 1 mm

yB = 1 mm

Given c − yA a+b

=

yB − yA a

⎛ yA ⎞ ⎜ ⎟ ⎜ yB ⎟ = Find ( yA , yB , F) ⎜F⎟ ⎝ ⎠

k yA + k yB − F = 0

⎛ yA ⎞ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ ⎟ mm ⎝ yB ⎠ ⎝ 4 ⎠

k yB a − F( a + b) = 0

F = 10 mN

Problem 5-56 The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine the maximum stiffness k of the springs at A and B so that the contact at C closes when the vertical force developed there is F . Originally the strip is horizontal as shown. Units Used: mN = 10

−3

N

Given: a = 50 mm b = 50 mm c = 10 mm F = 0.5 N

388

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Engineering Mechanics - Statics

Chapter 5

Solution:

Initial Guesses: k = 1

N

yA = 1 mm

m

yB = 1 mm

Given c − yA a+b

=

yB − yA

k yA + k yB − F = 0

a

⎛ yA ⎞ ⎜ ⎟ ⎜ yB ⎟ = Find ( yA , yB , k) ⎜k ⎟ ⎝ ⎠

⎛ yA ⎞ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ ⎟ mm ⎝ yB ⎠ ⎝ 4 ⎠

k yB a − F( a + b) = 0

k = 250

N m

Problem 5-57 Determine the distance d for placement of the load P for equilibrium of the smooth bar in the position θ as shown. Neglect the weight of the bar.

Solution: +

↑Σ Fy = 0; Σ MA = 0;

R cos ( θ ) − P = 0 −P d cos ( θ ) + R R d cos ( θ ) = R 2

d=

a

cos ( θ )

=0

a

cos ( θ )

a cos ( θ )

3

Problem 5-58 The wheelbarrow and its contents have mass m and center of mass at G. Determine the greatest 389

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Engineering Mechanics - Statics

Chapter 5

angle of tilt θ without causing the wheelbarrow to tip over.

g

Solution: Require point G to be over the wheel axle for tipping. Thus b cos ( θ ) = a sin ( θ ) − 1⎛ b ⎞

θ = tan

⎜ ⎟ ⎝ a⎠

Problem 5-59 Determine the force P needed to pull the roller of mass M over the smooth step. Given: M = 50 kg a = 0.6 m b = 0.1 m

θ = 60 deg θ 1 = 20 deg g = 9.81

m 2

s

390

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Engineering Mechanics - Statics

Chapter 5

Solution:

φ = acos ⎛⎜

a − b⎞

⎟ ⎝ a ⎠

φ = 33.56 deg ΣMB = 0,

M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) ... = 0 + P cos ( θ ) ( a − b) − P sin ( θ ) a sin ( φ )

⎡sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤ ⎥ ⎣ cos ( θ ) ( a − b) + sin ( θ ) a sin ( φ ) ⎦

P = M g⎢

P = 441 N

Problem 5-60 Determine the magnitude and direction θ of the minimum force P needed to pull the roller of mass M over the smooth step. Given: a = 0.6 m b = 0.1 m

θ 1 = 20 deg M = 50 kg g = 9.81

m 2

s Solution:

For Pmin, NA tends to 0

φ = acos ⎛⎜

a − b⎞

⎟ ⎝ a ⎠

φ = 33.56 deg ΣMB = 0 M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) ... = 0 + ⎡⎣−P cos ( θ ) ( a − b)⎤⎦ − P sin ( θ ) a sin ( φ )

391

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Engineering Mechanics - Statics

P=

Chapter 5

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦ cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )

For Pmin : dP dθ

=

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦

⎡⎣cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )⎤⎦ θ = atan ⎛⎜ sin ( φ ) ⋅

which gives,

P =

⎝

a

2

⎡⎣a sin ( φ ) cos ( θ ) − ( a − b)sin ( θ )⎤⎦ = 0

⎞ ⎟

θ = 33.6 deg

a − b⎠

M g⎡⎣sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤⎦ cos ( θ ) ( a − b) + a sin ( φ ) sin ( θ )

P = 395 N

Problem 5-61 A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a radius r. Determine the angle of inclination θ for equilibrium.

Solution: By Observation φ = θ. Equilibirium : ΣMA = 0; NB2 r cos ( θ ) − W NB =

L 2

cos ( θ ) = 0

WL 4r

ΣF x = 0; NA cos ( θ ) − W sin ( θ ) = 0

NA = W tan ( θ ) 392

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Engineering Mechanics - Statics

Chapter 5

ΣF y = 0; W tan ( θ ) sin ( θ ) +

WL 4r

− W cos ( θ ) = 0

sin ( θ ) − cos ( θ ) = 1 − 2 cos ( θ ) = 2

2

2 cos ( θ ) − 2

cos ( θ ) =

L 4r

L+

2

−L 4r

cos ( θ )

cos ( θ ) − 1 = 0 2

⎛ L + L2 + 128 r2 ⎞ ⎟ 16 r ⎝ ⎠

2

L + 128 r

θ = acos ⎜

16 r

Problem 5-62 The disk has mass M and is supported on the smooth cylindrical surface by a spring having stiffness k and unstretched length l0. The spring remains in the horizontal position since its end A is attached to the small roller guide which has negligible weight. Determine the angle θ to the nearest degree for equilibrium of the roller. Given: M = 20 kg k = 400

N m

l0 = 1 m r = 2m g = 9.81

m 2

s a = 0.2 m Guesses

F = 10 N

Solution:

R = 10 N θ = 30 deg

Given

+ Σ F y = 0; →

R sin ( θ ) − M g = 0

+

R cos ( θ ) − F = 0

↑Σ Fx = 0;

Spring

F = k⎡⎣( r + a)cos ( θ ) − l0⎤⎦ 393

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Engineering Mechanics - Statics

⎛F⎞ ⎜ R ⎟ = Find ( F , R , θ ) ⎜ ⎟ ⎝θ ⎠

Chapter 5

⎛ F ⎞ ⎛ 163.633 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ R ⎠ ⎝ 255.481 ⎠

θ = 50.171 deg

There is also another answer that we can find by choosing different starting guesses. Guesses

F = 200 N R = 200 N θ = 20 deg

Solution: + Σ F y = 0; → +

↑Σ Fx = 0;

Spring

Given R sin ( θ ) − M g = 0 R cos ( θ ) − F = 0 F = k⎡⎣( r + a)cos ( θ ) − l0⎤⎦

⎛F⎞ ⎜ R ⎟ = Find ( F , R , θ ) ⎜ ⎟ ⎝θ ⎠

⎛ F ⎞ ⎛ 383.372 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ R ⎠ ⎝ 430.66 ⎠

θ = 27.102 deg

Problem 5-63 Determine the x, y, z components of reaction at the fixed wall A.The force F2 is parallel to the z axis and the force F1 is parallel to the y axis. Given: a = 2m

d = 2m

b = 1m

F 1 = 200 N

c = 2.5 m

F 2 = 150 N

394

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Engineering Mechanics - Statics

Chapter 5

Solution: ΣF x = 0;

Ax = 0

ΣF x = 0;

Ay = −F1 Ay = −200 N

ΣF x = 0;

Az = F2 Az = 150 N

ΣΜx = 0;

MAx = −F2 a + F 1 d MAx = 100 N⋅ m

ΣΜy = 0;

MAy = 0

ΣMz = 0;

MAz = F 1 c MAz = 500 N⋅ m

Problem 5-64 The wing of the jet aircraft is subjected to thrust T from its engine and the resultant lift force L. If the mass of the wing is M and the mass center is at G, determine the x, y, z components of reaction where the wing is fixed to the fuselage at A. Units Used: 3

Mg = 10 kg 3

kN = 10 N g = 9.81

m 2

s

395

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Engineering Mechanics - Statics

Chapter 5

Given: T = 8 kN L = 45 kN M = 2.1 Mg a = 2.5 m b = 5m 395

c = 3m d = 7m Solution: ΣF x = 0;

− Ax + T = 0 Ax = T

Ax = 8 kN

ΣF y = 0;

Ay = 0

Ay = 0

ΣF z = 0;

− Az − M g + L = 0 Az = L − M g

ΣMy = 0;

Az = 24.4 kN

M y − T ( a) = 0 My = T a

ΣMx = 0;

My = 20.0 kN⋅ m

L( b + c + d) − M g b − Mx = 0 M x = L ( b + c + d) − M g b

ΣMz = 0;

Mx = 572 kN⋅ m

Mz − T( b + c) = 0 Mz = T( b + c)

Mz = 64.0 kN⋅ m

Problem 5-65 The uniform concrete slab has weight W. Determine the tension in each of the three parallel supporting cables when the slab is held in the horizontal plane as shown.

396

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kip = 10 lb Given: W = 5500 lb a = 6 ft b = 3 ft c = 3 ft d = 6 ft

Solution: Equations of Equilibrium : The cable tension TB can be obtained directly by summing moments about the y axis.

ΣMy = 0;

ΣMx = 0;

W

d 2

− TB d = 0

TC a + TB( a + b) − W⎛⎜

⎝

TC =

ΣF z = 0;

TB =

W 2

TB = 2.75 kip

a + b + c⎞

⎟=0 ⎠

2

1⎡ a + b + c − TB( a + b)⎥⎤ ⎢W a⎣ 2 ⎦

TA + TB + TC − W = 0

TA = − TB − TC + W

TC = 1.375 kip

TA = 1.375 kip

397

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Engineering Mechanics - Statics

Chapter 5

Problem 5-66 The air-conditioning unit is hoisted to the roof of a building using the three cables. If the tensions in the cables are TA, TB and TC, determine the weight of the unit and the location (x, y) of its center of gravity G. Given: TA = 250 lb TB = 300 lb TC = 200 lb a = 5 ft b = 4 ft c = 3 ft d = 7 ft e = 6 ft Solution: ΣF z = 0; TA + TB + TC − W = 0 W = TA + TB + TC ΣMy = 0;

W x − T A ( c + d) − T C d = 0 x =

ΣMx = 0;

W = 750 lb

TA( c + d) + TC d W

x = 5.2 ft

TA a + TB( a + b − e) + TC( a + b) − W y = 0

y =

TA a + TB( a + b − e) + TC( a + b) W

y = 5.267 ft

Problem 5-67 The platform truck supports the three loadings shown. Determine the normal reactions on each of its three wheels. 398

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Engineering Mechanics - Statics

Chapter 5

Given: F 1 = 380 lb F 2 = 500 lb F 3 = 800 lb a = 8 in

b = 12 in c = 10 in d = 5 in e = 12 in f = 12 in

Solution: The initail guesses are

F A = 1 lb

F B = 1 lb

F C = 1 lb

Given ΣMx = 0;

F 1 ( c + d) + F2 ( b + c + d) + F3 d − F A( a + b + c + d) = 0

ΣMy = 0;

F1 e − FB e − F2 f + FC f = 0

ΣF y = 0;

FB + FC − F2 + FA − F1 − F3 = 0

⎛ FA ⎞ ⎜ ⎟ ⎜ FB ⎟ = Find ( FA , FB , FC) ⎜F ⎟ ⎝ C⎠

⎛ FA ⎞ ⎛ 663 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FB ⎟ = ⎜ 449 ⎟ lb ⎜ F ⎟ ⎝ 569 ⎠ ⎝ C⎠

Problem 5-68 Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights WA, WB and WC, determine the normal reactions of the wheels D, E, and F on the ground.

399

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kip = 10 lb Given: WA = 45000 lb WB = 8000 lb WC = 6000 lb a = 8 ft

e = 20 ft

b = 6 ft

f = 4 ft

c = 8 ft

g = 3 ft

d = 6 ft Solution: Initial guesses: R D = 1 kip

R E = 1 kip

R F = 1 kip

Given ΣMx = 0;

WB b − RD( a + b) − WC c + R E( c + d) = 0

ΣMy = 0;

W B f + W A( g + f ) + W C f − R F( e + g + f ) = 0

ΣF z = 0;

R D + R E + RF − WA − WB − WC = 0

⎛ RD ⎞ ⎜ ⎟ ⎜ RE ⎟ = Find ( RD , RE , RF) ⎜R ⎟ ⎝ F⎠

⎛ RD ⎞ ⎛ 22.6 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ RE ⎟ = ⎜ 22.6 ⎟ kip ⎜ R ⎟ ⎝ 13.7 ⎠ ⎝ F⎠

Problem 5-69 If the cable can be subjected to a maximum tension T, determine the maximum force F which may be applied to the plate. Compute the x, y, z components of reaction at the hinge A for this loading.

400

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Engineering Mechanics - Statics

Chapter 5

Given: a = 3 ft b = 2 ft c = 1 ft d = 3 ft e = 9 ft T = 300 lb Solution: Initial guesses: F = 10 lb

MAx = 10 lb ft

MAz = 10 lb ft

Ax = 10 lb

Ay = 10 lb

Az = 10 lb

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ = 0 ⎜ A ⎟ ⎝T − F⎠ ⎝ z⎠

⎛ MAx ⎞ ⎛ a ⎞ ⎛ 0 ⎞ ⎛ e ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ −c ⎟ × ⎜ 0 ⎟ + ⎜ −b − c ⎟ × ⎜ 0 ⎟ = 0 ⎜ MAz ⎟ ⎝ 0 ⎠ ⎝ −F ⎠ ⎝ 0 ⎠ ⎝ T ⎠ ⎝ ⎠

⎛ F ⎞ ⎜ A ⎟ ⎜ x ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( F , Ax , Ay , Az , MAx , MAz) Az ⎜ ⎟ ⎜ MAx ⎟ ⎟ ⎜ ⎝ MAz ⎠

⎛ MAx ⎞ ⎛ 0 ⎞ ⎜ ⎟ = ⎜ ⎟ lb ft ⎝ MAz ⎠ ⎝ 0 ⎠ ⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ lb ⎜ A ⎟ ⎝ 600 ⎠ ⎝ z⎠

F = 900 lb

Problem 5-70 The boom AB is held in equilibrium by a ball-and-socket joint A and a pulley and cord system as shown. Determine the x, y, z components of reaction at A and the tension in cable DEC.

401

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Engineering Mechanics - Statics

Chapter 5

Given:

F

⎛ 0 ⎞ = ⎜ 0 ⎟ lb ⎜ ⎟ ⎝ −1500 ⎠

a = 5 ft b = 4 ft c = b d = 5 ft e = 5 ft f = 2 ft Solution:

α = atan ⎛⎜

a

⎞ ⎟

⎝ d + e⎠

L =

2

a + ( d + e)

β = atan ⎛⎜

2

⎞ ⎟ fL ⎜L − ⎟ d+ e⎠ ⎝

Guesses

b

TBE = 1 lb TDEC = 1 lb

Ax = 1 lb Ay = 1 lb Given

Az = 1 lb

2 TDEC cos ( β ) = TBE

0 ⎞ ⎛0⎞ ⎛ 0 ⎞ ⎛⎜ ⎜ d ⎟ × F + ⎜ d + e ⎟ × −TBE cos ( α ) ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝0⎠ ⎝ 0 ⎠ ⎝ TBE sin ( α ) ⎟⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + F + TBE⎜ −cos ( α ) ⎟ = 0 ⎜A ⎟ ⎝ sin ( α ) ⎠ ⎝ z⎠

402

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Engineering Mechanics - Statics

Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ Az ⎟ = Find ( A , A , A , T , T x y z BE DEC) TDEC = 919 lb ⎜ ⎟ ⎜ TBE ⎟ ⎜T ⎟ ⎝ DEC ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 1.5 × 103 ⎟ lb ⎜A ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 750 ⎠

Problem 5-71 The cable CED can sustain a maximum tension Tmax before it fails. Determine the greatest vertical force F that can be applied to the boom. Also, what are the x, y, z components of reaction at the ball-and-socket joint A? Given: Tmax = 800 lb a = 5 ft b = 4 ft c = b d = 5 ft e = 5 ft f = 2 ft Solution: a ⎞ ⎟ ⎝ d + e⎠

α = atan ⎛⎜ L =

2

a + ( d + e)

β = atan ⎛⎜

2

⎞ fL ⎟ ⎜L − ⎟ d+ e⎠ ⎝

Guesses

b

TBE = 1 lb

Ax = 1 lb Ay = 1 lb Given

TDEC = Tmax

F = 1 lb Az = 1 lb

2 TDEC cos ( β ) = TBE

403

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Engineering Mechanics - Statics

Chapter 5

0 ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ ⎜ d ⎟ × ⎜ 0 ⎟ + ⎜ d + e ⎟ × −TBE cos ( α ) ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝ 0 ⎠ ⎝ −F ⎠ ⎝ 0 ⎠ ⎝ TBE sin ( α ) ⎟⎠

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ = Find ( A , A , A , T , F) x y z BE ⎜ z ⎟ ⎜ TBE ⎟ ⎜ ⎟ ⎝ F ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ + TBE⎜ −cos ( α ) ⎟ = 0 ⎜ A ⎟ ⎝ −F ⎠ ⎝ sin ( α ) ⎠ ⎝ z⎠

F = 1306 lb

0 ⎛ Ax ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 1.306 × 103 ⎟ lb ⎜A ⎟ ⎜ ⎟ ⎝ z ⎠ ⎝ 653.197 ⎠

Problem 5-72 The uniform table has a weight W and is supported by the framework shown. Determine the smallest vertical force P that can be applied to its surface that will cause it to tip over. Where should this force be applied? Given: W = 20 lb a = 3.5 ft b = 2.5 ft c = 3 ft e = 1.5 ft f = 1 ft

Solution: f θ = atan ⎛⎜ ⎟⎞

θ = 33.69 deg

d = e sin ( θ )

d = 0.832 ft

⎛ a −e⎞ ⎜ 2 ⎟ φ = atan ⎜ ⎟ ⎜ b ⎟ ⎝ 2 ⎠

φ = 11.31 deg

⎝ e⎠

404

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Engineering Mechanics - Statics

2

d' =

⎛ a − e⎞ + ⎛ b ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠

Chapter 5

2

d' = 1.275 ft

Tipping will occur about the g - g axis. Require P to be applied at the corner of the table for Pmin. W d = P d' sin ( 90 deg − φ + θ ) P = W

d

P = 14.1 lb

d' sin ( 90 deg − φ + θ )

Problem 5-73 The windlass is subjected to load W. Determine the horizontal force P needed to hold the handle in the position shown, and the components of reaction at the ball-and-socket joint A and the smooth journal bearing B. The bearing at B is in proper alignment and exerts only force reactions perpendicular to the shaft on the windlass. Given: W = 150 lb a = 2 ft b = 2 ft c = 1 ft d = 1 ft e = 1 ft f = 0.5 ft Solution: ΣMy = 0;

W f−P d=0 P =

W f d

ΣF y = 0;

Ay = 0 lb

ΣMx = 0;

−W a + B z( a + b) = 0

P = 75 lb Ay = 0

405

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Engineering Mechanics - Statics

Bz = ΣF z = 0;

Chapter 5

Wa

B z = 75 lb

a+b

Az + Bz − W = 0 Az = W − Bz

ΣMz = 0;

B x( a + b) − ( a + b + c + e)P = 0 Bx =

ΣF x = 0;

Az = 75 lb

P ( a + b + c + e) a+b

B x = 112 lb

Ax − Bx + P = 0 Ax = Bx − P

Ax = 37.5 lb

Problem 5-74 A ball of mass M rests between the grooves A and B of the incline and against a vertical wall at C. If all three surfaces of contact are smooth, determine the reactions of the surfaces on the ball. Hint: Use the x, y, z axes, with origin at the center of the ball, and the z axis inclined as shown. Given: M = 2 kg

θ 1 = 10 deg θ 2 = 45 deg

Solution: ΣF x = 0; F c cos ( θ 1 ) − M g sin ( θ 1 ) = 0 F c = M g⋅ tan ( θ 1 ) F c = 0.32 kg⋅ m

406

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Engineering Mechanics - Statics

Chapter 5

ΣF y = 0; NA cos ( θ 2 ) − NB cos ( θ 2 ) = 0 NA = NB ΣF z = 0; 2 NA sin ( θ 2 ) − M g cos ( θ 1 ) − F c sin ( θ 1 ) = 0

NA =

1 M g⋅ cos ( θ 1 ) + F c⋅ sin ( θ 1 ) ⋅ 2 sin ( θ 2 )

NA = 1.3 kg⋅ m N = NA = NB

Problem 5-75 Member AB is supported by cable BC and at A by a square rod which fits loosely through the square hole at the end joint of the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the cylinder of weight W in equilibrium. Units Used: 3

kip = 10 lb Given: W = 800 lb a = 2 ft b = 6 ft c = 3 ft Solution:

⎛

⎞=0 ⎝ c +b +a ⎠

ΣF x = 0

F BC ⎜

ΣF y = 0

Ay = 0

c

2

2

2⎟

Ay = 0lb

F BC = 0 lb

Ay = 0 lb

407

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Engineering Mechanics - Statics

Chapter 5

ΣF z = 0

Az − W = 0

ΣMx = 0

MAx − W b = 0

ΣMy = 0 ΣMz = 0

Az = W

Az = 800 lb

MAx = W b

MAx = 4.80 kip⋅ ft

MAy = 0 lb ft

MAy = 0 lb⋅ ft

MAz = 0 lb ft

MAz = 0 lb⋅ ft

Problem 5-76 The pipe assembly supports the vertical loads shown. Determine the components of reaction at the ball-and-socket joint A and the tension in the supporting cables BC and BD. Units Used: 3

kN = 10 N Given: F 1 = 3 kN d = 2 m F 2 = 4 kN e = 1.5 m a = 1m

g = 1m

b = 1.5 m

h = 3m

c = 3m

i = 2m

f = c−e

j = 2m

Solution: The initial guesses are: TBD = 1 kN

TBC = 1 kN

Ax = 1 kN

Ay = 1 kN

Az = 1 kN

The vectors

408

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Engineering Mechanics - Statics

⎛ 0 r1 = ⎜ a + b + ⎜ ⎝ d

rBC

⎞ f⎟ ⎟ ⎠

⎛ i ⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝h − g⎠

uBC =

Given

rBC rBC

Chapter 5

⎛ 0 ⎞ r2 = ⎜ a + b + c ⎟ ⎜ ⎟ ⎝ d ⎠

rBD

⎛ −j ⎞ = ⎜ −a ⎟ ⎜ ⎟ ⎝h − g⎠

uBD =

rBD rBD

rAB

⎛0⎞ = ⎜a⎟ ⎜ ⎟ ⎝g⎠

⎛1⎞ i = ⎜0⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ j = ⎜1⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ k = ⎜0⎟ ⎜ ⎟ ⎝1⎠

Ax i + Ay j + Az k − F1 k − F2 k + TBDuBD + TBC uBC = 0 rAB × ( TBDuBD + TBCuBC) + r1 × ( −F 1 k) + r2 × ( −F 2 k) = 0

⎛ TBD ⎞ ⎜ ⎟ ⎜ TBC ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BD BC x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBD ⎞ ⎛ 17 ⎞ ⎜ ⎟ = ⎜ ⎟ kN ⎝ TBC ⎠ ⎝ 17 ⎠ ⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 11.333 ⎟ kN ⎜ A ⎟ ⎝ −15.667 ⎠ ⎝ z⎠

Problem 5-77 The hatch door has a weight W and center of gravity at G. If the force F applied to the handle at C has coordinate direction angles of α, β and γ, determine the magnitude of F needed to hold the door slightly open as shown. The hinges are in proper alignment and exert only force reactions on the door. Determine the components of these reactions if A exerts only x and z components of force and B exerts x, y, z force components. Given: W = 80 lb

α = 60 deg

409

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Engineering Mechanics - Statics

Chapter 5

β = 45 deg γ = 60 deg a = 3 ft b = 2 ft c = 4 ft d = 3 ft Solution: Initial Guesses: Ax = 1 lb

Az = 1 lb

F = 1 lb

B x = 1 lb

B y = 1 lb

B z = 1 lb

Given

⎛ Ax ⎞ ⎛ Bx ⎞ ⎛⎜ cos ( α ) ⎟⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ By ⎟ + F⎜ cos ( β ) ⎟ + ⎜ 0 ⎟ = 0 ⎜ Az ⎟ ⎜ B ⎟ ⎜ cos ( γ ) ⎟ ⎝ −W ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ z⎠ ⎛ a + b ⎞ ⎢⎡ ⎛⎜ cos ( α ) ⎟⎞⎥⎤ ⎛ a ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎜ 0 ⎟ × F cos ( β ) + ⎜ c ⎟ × ⎜ 0 ⎟ + ⎜ c + d ⎟ × B = 0 ⎟⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎢ ⎜ ⎟ ⎜ y⎟ ⎢ ⎜ ⎟ ⎥ ⎝ 0 ⎠ ⎣ ⎝ cos ( γ ) ⎠⎦ ⎝ 0 ⎠ ⎝ −W ⎠ ⎝ 0 ⎠ ⎜⎝ Bz ⎟⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Az ⎟ ⎜ Bx ⎟ ⎜ ⎟ = Find ( Ax , Az , Bx , By , Bz , F) ⎜ By ⎟ ⎜ Bz ⎟ ⎜ ⎟ ⎝F ⎠

⎛ Ax ⎞ ⎛ −96.5 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Az ⎠ ⎝ −13.7 ⎠ ⎛ Bx ⎞ ⎛ 48.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ By ⎟ = ⎜ −67.9 ⎟ lb ⎜ B ⎟ ⎝ 45.7 ⎠ ⎝ z⎠ F = 96 lb

410

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Engineering Mechanics - Statics

Chapter 5

Problem 5-78 The hatch door has a weight W and center of gravity at G. If the force F applied to the handle at C has coordinate direction angles α, β, γ determine the magnitude of F needed to hold the door slightly open as shown. If the hinge at A becomes loose from its attachment and is ineffective, what are the x, y, z components of reaction at hinge B? Given: W = 80 lb

α = 60 deg β = 45 deg γ = 60 deg a = 3 ft b = 2 ft c = 4 ft d = 3 ft Solution: a

ΣMy = 0;

F = W

ΣF x = 0;

B x + F cos ( α ) = 0

cos ( γ ) ( a + b)

B x = −F cos ( α ) ΣF y = 0;

B y = −67.9 lb

B z − W + F cos ( γ ) = 0 B z = W − F cos ( γ )

ΣMx = 0;

B x = −48 lb

B y + F cos ( β ) = 0 B y = −F cos ( β )

ΣF z = 0;

F = 96 lb

B z = 32 lb

MBx + W d − F cos ( γ ) ( c + d) = 0 MBx = −W d + F cos ( γ ) ( c + d)

ΣMz = 0;

MBx = 96 lb⋅ ft

MBz + F cos ( α ) ( c + d) + F cos ( β ) ( a + b) = 0 MBz = −F cos ( α ) ( c + d) − F cos ( β ) ( a + b)

MBz = −675 lb⋅ ft

411

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Engineering Mechanics - Statics

Chapter 5

Problem 5-79 The bent rod is supported at A, B, and C by smooth journal bearings. Compute the x, y, z components of reaction at the bearings if the rod is subjected to forces F 1 and F 2. F1 lies in the y-z plane. The bearings are in proper alignment and exert only force reactions on the rod. Given: F 1 = 300 lb

d = 3 ft

F 2 = 250 lb

e = 5 ft

a = 1 ft

α = 30 deg

b = 4 ft

β = 45 deg

c = 2 ft

θ = 45 deg

Solution: The initial guesses: Ax = 100 lb

Ay = 200 lb

B x = 300 lb

B z = 400 lb

Cy = 500 lb

Cz = 600 lb

Given Ax + B x + F2 cos ( β ) sin ( α ) = 0 Ay + Cy − F1 cos ( θ ) + F2 cos ( β ) cos ( α ) = 0 B z + Cz − F 1 sin ( θ ) − F2 sin ( β ) = 0 F 1 cos ( θ ) ( a + b) + F 1 sin ( θ ) ( c + d) − B z d − A y b = 0 Ax b + Cz e = 0 Ax( c + d) + Bx d − Cy e = 0

412

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Engineering Mechanics - Statics

Chapter 5

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) x y x z y z ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cy ⎟ ⎜ Cz ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 632.883 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ −141.081 ⎠ ⎛ Bx ⎞ ⎛ −721.271 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Bz ⎠ ⎝ 895.215 ⎠ ⎛ Cy ⎞ ⎛ 200.12 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Cz ⎠ ⎝ −506.306 ⎠

Problem 5-80 The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F2 which will cause the reaction Cy at the bearing C to be equal to zero. The bearings are in proper alignment and exert only force reactions on the rod. Given: F 1 = 300 lb

d = 3 ft

Cy = 0 lb

e = 5 ft

a = 1 ft

α = 30 deg

b = 4 ft

β = 45 deg

c = 2 ft

θ = 45 deg

Solution: The initial guesses: Ax = 100 lb

Ay = 200 lb

B x = 300 lb

B z = 400 lb

F 2 = 500 lb

Cz = 600 lb

Given Ax + B x + F2 cos ( β ) sin ( α ) = 0

413

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Engineering Mechanics - Statics

Chapter 5

Ay + Cy − F1 cos ( θ ) + F2 cos ( β ) cos ( α ) = 0 B z + Cz − F 1 sin ( θ ) − F2 sin ( β ) = 0 F 1 cos ( θ ) ( a + b) + F 1 sin ( θ ) ( c + d) − B z d − A y b = 0 Ax b + Cz e = 0 Ax( c + d) + Bx d − Cy e = 0

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , F ) x y x z z 2 ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cz ⎟ ⎜ F2 ⎟ ⎝ ⎠

F 2 = 673.704 lb

Problem 5-81 Determine the tension in cables BD and CD and the x, y, z components of reaction at the ball-and-socket joint at A. Given: F = 300 N a = 3m b = 1m c = 0.5 m d = 1.5 m Solution:

rBD

⎛ −b ⎞ ⎜ ⎟ = d ⎜ ⎟ ⎝a⎠

414

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Engineering Mechanics - Statics

rCD

Chapter 5

⎛ −b ⎞ = ⎜ −d ⎟ ⎜ ⎟ ⎝a⎠

Initial Guesses:

TBD = 1 N

TCD = 1 N

Ax = 1 N

Ay = 1 N

Az = 1 N

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ rBD rCD + TCD +⎜ 0 ⎟ =0 ⎜ Ay ⎟ + TBD ⎜ ⎟ rBD rCD ⎜A ⎟ ⎝ −F ⎠ ⎝ z⎠ ⎛d⎞ ⎛d⎞ ⎛d − c⎞ ⎛ 0 ⎞ rBD ⎞ ⎜ ⎟ ⎛ rCD ⎞ ⎜ ⎜ −d ⎟ × ⎛ T + d × ⎜ TCD + 0 ⎟×⎜ 0 ⎟ = 0 ⎜ ⎟ ⎟ BD ⎜ ⎟ ⎝ ⎟ ⎜ ⎟ rBD ⎠ ⎜ ⎟ ⎝ rCD ⎠ ⎜ ⎝0⎠ ⎝0⎠ ⎝ 0 ⎠ ⎝ −F ⎠ ⎛ TBD ⎞ ⎜ ⎟ ⎜ TCD ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BD CD x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBD ⎞ ⎛ 116.7 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ TCD ⎠ ⎝ 116.7 ⎠ ⎛ Ax ⎞ ⎛ 66.7 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ N ⎜ A ⎟ ⎝ 100 ⎠ ⎝ z⎠

Problem 5-82 Determine the tensions in the cables and the components of reaction acting on the smooth collar at A necessary to hold the sign of weight W in equilibrium. The center of gravity for the sign is at G. Given: W = 50 lb

f = 2.5 ft

a = 4 ft

g = 1 ft

b = 3 ft

h = 1 ft

415

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Engineering Mechanics - Statics

Chapter 5

c = 2 ft

i = 2 ft

d = 2 ft

j = 2 ft

e = 2.5 ft

k = 3 ft

Solution: The initial guesses are: TBC = 10 lb

Ax = 10 lb

MAx = 10 lb⋅ ft

TDE = 10 lb

Ay = 10 lb

MAy = 10 lb⋅ ft

Given TDE

( a − k)

2

2

( a − k) + i + j −i

TDE 2

( a − k) + i + j

j

TBC 2

2

2

2

TBC

2

2

2

+ Ax = 0

+ Ay = 0

−W=0 2

2

( d − b) + i + c

i 2

2

( a − k) + i + j

MAy − TDE k

2

( d − b) + i + c

+c

( a − k) + i + j

MAx + TDE j

2

( d − b) + i + c

2

TDE 2

2

−i 2

TBC

+ ( −b + d)

2

j 2

2

( a − k) + i + j

2

i

+ c TBC

2

− Wi = 0 2

2

2

2

( d − b) + i + c

d

+ TBC c

2

+ W ( k − f) = 0

( d − b) + i + c

416

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Engineering Mechanics - Statics

Chapter 5

i

−TDE a

2

2

( a − k) + i + j

2

+ TBC b

i 2

=0 2

2

( d − b) + i + c

⎛⎜ MAx ⎞⎟ ⎜ MAy ⎟ ⎜ ⎟ ⎜ TBC ⎟ = Find ( M , M , T , T , A , A ) Ax Ay BC DE x y ⎜ TDE ⎟ ⎜ ⎟ ⎜ Ax ⎟ ⎜ Ay ⎟ ⎝ ⎠

⎛ TBC ⎞ ⎛ 42.857 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ TDE ⎠ ⎝ 32.143 ⎠ ⎛ Ax ⎞ ⎛ 3.571 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ 50 ⎠ ⎛ MAx ⎞ ⎛ 2.698 × 10− 13 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAy ⎠ ⎝ −17.857 ⎠

Problem 5-83 The member is supported by a pin at A and a cable BC. If the load at D is W, determine the x, y, z components of reaction at these supports. Units Used: 3

kip = 10 lb Given: W = 300 lb a = 1 ft b = 2 ft c = 6 ft d = 2 ft e = 2 ft f = 2 ft

417

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Engineering Mechanics - Statics

Chapter 5

Solution: Initial Guesses: TBC = 1 lb

Ax = 1 lb

Ay = 1 lb

Az = 1 lb

MAy = 1 lb ft

MAz = 1 lb ft

Given

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ + ⎜A ⎟ ⎝ z⎠

⎛e + f − a⎞ ⎛ 0 ⎞ ⎜ −c ⎟ + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎟ 2 2 2⎜ b + c + ( e + f − a) ⎝ b ⎠ ⎝ −W ⎠ TBC

⎛ 0 ⎞ ⎛ −e ⎞ ⎛ 0 ⎞ ⎛ −a ⎞ ⎡ ⎛ e + f − a ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ TBC ⎜ −c ⎟⎥ = 0 ⎜ MAy ⎟ + ⎜ c ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎢ ⎜ ⎟⎥ ⎜ MAz ⎟ ⎝ 0 ⎠ ⎝ −W ⎠ ⎝ b ⎠ ⎣ b2 + c2 + ( e + f − a) 2 ⎝ b ⎠⎦ ⎝ ⎠ ⎛⎜ TBC ⎞⎟ ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ = Find ( T , A , A , A , M , M ) BC x y z Ay Az ⎜ Az ⎟ ⎜ ⎟ ⎜ MAy ⎟ ⎜ MAz ⎟ ⎝ ⎠

TBC = 1.05 kip

⎛ Ax ⎞ ⎛ −450 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 900 ⎟ lb ⎜A ⎟ ⎝ 0 ⎠ ⎝ z⎠ ⎛ MAy ⎞ ⎛ −600 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAz ⎠ ⎝ −900 ⎠

Problem 5-84 Determine the x, y, z components of reaction at the pin A and the tension in the cable BC necessary for equilibrium of the rod.

418

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Engineering Mechanics - Statics

Chapter 5

Given: F = 350 lb

e = 12 ft

a = 4 ft

f = 4 ft

b = 5 ft

g = 10 ft

c = 4 ft

h = 4 ft

d = 2 ft

i = 10 ft

Solution: Initial Guesses: F BC = 1 lb Ay = 1 lb MAy = 1 lb⋅ ft Ax = 1 lb Az = 1 lb MAz = 1 lb⋅ ft Given

⎛ 0 ⎞ ⎛d⎞ ⎡ ⎛ g − d ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎢ F ⎜ e − c ⎟⎥ ... = 0 ⎜ MAy ⎟ + ⎜ c ⎟ × ⎢ ⎟⎥ 2 2 2⎜ ⎜ MAz ⎟ ⎝ 0 ⎠ ⎣ ( g − d) + ( e − c) + f ⎝ − f ⎠⎦ ⎝ ⎠ ⎛ −a ⎞ ⎡ ⎛ a − h ⎞⎤ F BC ⎜ ⎟ ⎢ ⎜ −e ⎟⎥ + 0 × ⎜ ⎟ ⎢ ⎟⎥ 2 2 2⎜ ⎝ b ⎠ ⎣ ( a − h) + e + b ⎝ b ⎠⎦ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ + ⎜A ⎟ ⎝ z⎠

⎛g − d⎞ ⎜e− c⎟ + ⎟ 2 2 2⎜ ( g − d) + ( e − c) + f ⎝ − f ⎠ F

⎛a − h⎞ ⎜ −e ⎟ = 0 ⎟ 2 2 2⎜ ( a − h) + e + b ⎝ b ⎠ F BC

419

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Engineering Mechanics - Statics

Chapter 5

⎛⎜ FBC ⎞⎟ ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ = Find ( F , A , A , A , M , M ) BC x y z Ay Az ⎜ Az ⎟ ⎜ ⎟ ⎜ MAy ⎟ ⎜ MAz ⎟ ⎝ ⎠

F BC = 101 lb

⎛ Ax ⎞ ⎛ −233.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ −140 ⎟ lb ⎜ A ⎟ ⎝ 77.8 ⎠ ⎝ z⎠ ⎛ MAy ⎞ ⎛ −388.9 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAz ⎠ ⎝ 93.3 ⎠

Problem 5-85 Rod AB is supported by a ball-and-socket joint at A and a cable at B. Determine the x, y, z components of reaction at these supports if the rod is subjected to a vertical force F as shown. Given: F = 50 lb a = 2 ft

c = 2 ft

b = 4 ft

d = 2 ft

Solution: TB = 10 lb

Ax = 10 lb

Ay = 10 lb

Az = 10 lb

B y = 10 lb Given ΣF x = 0;

− TB + A x = 0

ΣF y = 0;

Ay + By = 0

420

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Engineering Mechanics - Statics

ΣF z = 0;

Chapter 5

−F + A z = 0

ΣMAx = 0;

F ( c) − By( b) = 0

ΣMAy = 0;

F ( a) − TB( b) = 0

ΣMAz = 0;

B y( a) − TB( c) = 0

Solving,

⎛ TB ⎞ ⎜ ⎟ ⎜ Ax ⎟ ⎜ Ay ⎟ = Find ( T , A , A , A , B ) B x y z y ⎜ ⎟ ⎜ Az ⎟ ⎜B ⎟ ⎝ y⎠

⎛ TB ⎞ ⎛ 25 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ax ⎟ ⎜ 25 ⎟ ⎜ Ay ⎟ = ⎜ −25 ⎟ lb ⎜ ⎟ ⎜ ⎟ ⎜ Az ⎟ ⎜ 50 ⎟ ⎜ B ⎟ ⎝ 25 ⎠ ⎝ y⎠

Problem 5-86 The member is supported by a square rod which fits loosely through a smooth square hole of the attached collar at A and by a roller at B. Determine the x, y, z components of reaction at these supports when the member is subjected to the loading shown. Given: M = 50 lb⋅ ft

⎛ 20 ⎞ ⎜ ⎟ F = −40 lb ⎜ ⎟ ⎝ −30 ⎠ a = 2 ft b = 1 ft c = 2 ft Solution: Initial Guesses Ax = 1 lb

Ay = 1 lb

421

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Engineering Mechanics - Statics

Chapter 5

MAx = 1 lb ft

MAy = 1 lb ft

MAz = 1 lb ft

B z = 1 lb

Given

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ 0 ⎟ + F = 0 ⎜ 0 ⎟ ⎜ Bz ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ MAx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎡⎛ 0 ⎞ ⎛ 0 ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎜ MAy ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟ + ⎢⎜ a + b ⎟ × F + ⎜ 0 ⎟⎥ = 0 ⎜ M ⎟ ⎝ 0 ⎠ ⎜ Bz ⎟ ⎣⎝ −c ⎠ ⎝ −M ⎠⎦ ⎝ ⎠ ⎝ Az ⎠ ⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ MAx ⎟ = Find ( A , A , M , M , M , B ) x y Ax Ay Az z ⎜ MAy ⎟ ⎜ ⎟ ⎜ MAz ⎟ ⎜ Bz ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ −20 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ 40 ⎠ B z = 30 lb

⎛ MAx ⎞ ⎛ 110 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ MAy ⎟ = ⎜ 40 ⎟ lb⋅ ft ⎜ M ⎟ ⎝ 110 ⎠ ⎝ Az ⎠

Problem 5-87 The platform has mass M and center of mass located at G. If it is lifted using the three cables, determine the force in each of these cables. Units Used: 3

Mg = 10 kg

3

kN = 10 N

g = 9.81

m 2

s

422

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Engineering Mechanics - Statics

Chapter 5

Given: M = 3 Mg a = 4m b = 3m c = 3m d = 4m e = 2m Solution: The initial guesses are: F AC = 10 N

F BC = 10 N

F DE = 10 N

Given b ( F AC) 2

2

a +b

−

c ( F BC) 2

=0

2

a +c

M g e − ( F AC) a

d+e 2

2

− FBC

a +b a 2

2

a( d + e) 2

=0

2

a +c

F BC( b + c) − M g b + FDE b = 0

a +c

⎛ FAC ⎞ ⎜ ⎟ F ⎜ BC ⎟ = Find ( FAC , FBC , FDE) ⎜F ⎟ ⎝ DE ⎠

⎛ FAC ⎞ ⎜ ⎟ F ⎜ BC ⎟ = kN ⎜F ⎟ ⎝ DE ⎠

423

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Engineering Mechanics - Statics

Chapter 5

Problem 5-88 The platform has a mass of M and center of mass located at G. If it is lifted using the three cables, determine the force in each of the cables. Solve for each force by using a single moment equation of equilibrium. Units Used: Mg = 1000 kg 3

kN = 10 N g = 9.81

m 2

s Given: M = 2 Mg

c = 3m

a = 4m

d = 4m

b = 3m

e = 2m

Solution:

rBC

⎛0⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝a⎠

rAD

⎛ −e − d ⎞ = ⎜ b ⎟ ⎜ ⎟ ⎝ 0 ⎠

rAC

⎛0⎞ = ⎜b⎟ ⎜ ⎟ ⎝a⎠

rBD

⎛ −d − e ⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝ 0 ⎠

First find FDE. ΣMy' = 0;

F DE( d + e) − M g d = 0

Next find FBC.

Guess

⎡⎛ e ⎞ ⎛ 0 Given ⎢⎜ 0 ⎟ × ⎜ 0 ⎢⎜ ⎟ ⎜ ⎣⎝ 0 ⎠ ⎝ −M

F DE =

Mgd d+e

2

F DE = 4.1 s kN

F BC = 1 kN

⎞ ⎛e + d⎞ rBC ⎟ + ⎜ c ⎟ × ⎛F ⎜ BC ⎟ ⎜ ⎟ ⎝ rBC g⎠ ⎝ 0 ⎠

⎤ ⎞⎥ ⎟⎥ rAD = 0 ⎠ ⎦

F BC = Find ( FBC )

F BC = kN Now find FAC.

Guess

F AC = 1 kN

424

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Engineering Mechanics - Statics

⎡⎛ e ⎞ ⎛ 0 Given ⎢⎜ 0 ⎟ × ⎜ 0 ⎢⎜ ⎟ ⎜ ⎣⎝ 0 ⎠ ⎝ −M

Chapter 5

⎞ ⎛e + d⎞ rAC ⎟ + ⎜ −b ⎟ × ⎛ F ⎜ AC ⎟ ⎜ ⎟ ⎝ rAC g⎠ ⎝ 0 ⎠

⎤ ⎞⎥ ⎟⎥ rBD = 0 ⎠ ⎦

F AC = Find ( FAC )

F AC = kN

Problem 5-89 The cables exert the forces shown on the pole. Assuming the pole is supported by a ball-and-socket joint at its base, determine the components of reaction at A. The forces F 1 and F2 lie in a horizontal plane. Given: F 1 = 140 lb F 2 = 75 lb

θ = 30 deg a = 5 ft b = 10 ft c = 15 ft

Solution: The initial guesses are TBC = 100 lb

TBD = 100 lb

Ax = 100 lb

Ay = 100 lb

Az = 100 lb

Given c ⎞ − T a⎛ ⎞=0 BD ⎜ ⎟ ⎟ 2 2 2 2 2 ⎝ a +b +c ⎠ ⎝ a +c ⎠

(F1 cos (θ ) + F2)c − TBC a⎛⎜

c

⎞=0 ⎝ a +b +c ⎠

F 1 sin ( θ ) c − b TBC ⎛ ⎜

c

2

2

2⎟

425

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Engineering Mechanics - Statics

⎛

Chapter 5

⎞ ⎟=0 2 2 2 a + b + c ⎝ ⎠

Ax + F 1 sin ( θ ) − b⎜

TBC

TBC ⎞ ⎞ + a⎛ =0 ⎜ ⎟ ⎟ 2 2 2 2 2 + c a + b + c a ⎝ ⎠ ⎝ ⎠

⎛

Ay − F 1 cos ( θ ) − F 2 + TBD⎜

a

TBC ⎛ TBD ⎞ ⎛ ⎞ − c =0 ⎟ ⎜ ⎟ 2 2 2 2 2 a + c a + b + c ⎝ ⎠ ⎝ ⎠

Az − c⎜

⎛ TBC ⎞ ⎜ ⎟ ⎜ TBD ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BC BD x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠

⎛ TBC ⎞ ⎛ 131.0 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ TBD ⎠ ⎝ 509.9 ⎠ ⎛ Ax ⎞ ⎛ −0.0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0.0 ⎟ lb ⎜ A ⎟ ⎝ 588.7 ⎠ ⎝ z⎠

Problem 5-90 The silo has a weight W, a center of gravity at G and a radius r. Determine the vertical component of force that each of the three struts at A, B, and C exerts on the silo if it is subjected to a resultant wind loading of F which acts in the direction shown. Given: W = 3500 lb F = 250 lb

θ 1 = 30 deg θ 2 = 120 deg θ 3 = 30 deg r = 5 ft b = 12 ft c = 15 ft

426

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Engineering Mechanics - Statics

Chapter 5

Solution: Initial Guesses:

Az = 1 lb

B z = 2 lb

Cz = 31 lb

Given ΣMy = 0; B z r cos ( θ 1 ) − Cz r cos ( θ 1 ) − F sin ( θ 3 ) c = 0

[1]

ΣMx = 0; −B z r sin ( θ 1 ) − Cz r sin ( θ 1 ) + Az r − F cos ( θ 3 ) c = 0

[2]

ΣF z = 0; Az + Bz + Cz = W

[3]

Solving Eqs.[1], [2] and [3] yields:

⎛ Az ⎞ ⎜ ⎟ ⎜ Bz ⎟ = Find ( Az , Bz , Cz) ⎜C ⎟ ⎝ z⎠

⎛ Az ⎞ ⎛ 1600 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Bz ⎟ = ⎜ 1167 ⎟ lb ⎜ C ⎟ ⎝ 734 ⎠ ⎝ z⎠

Problem 5-91 The shaft assembly is supported by two smooth journal bearings A and B and a short link DC. If a couple moment is applied to the shaft as shown, determine the components of force reaction at the bearings and the force in the link. The link lies in a plane parallel to the y-z plane and the bearings are properly aligned on the shaft. Units Used: 3

kN = 10 N Given: M = 250 N⋅ m a = 400 mm b = 300 mm c = 250 mm d = 120 mm

θ = 30 deg 427

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Engineering Mechanics - Statics

Chapter 5

φ = 20 deg Solution: Initial Guesses: Ay = 1 kN

Az = 1 kN

B z = 1 kN

F CD = 1 kN

B y = 1 kN

Given 0 ⎞ ⎛0 ⎞ ⎛0⎞ ⎛ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ By ⎟ + ⎜ −FCD cos ( φ ) ⎟ = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ F sin ( φ ) ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ CD ⎠ 0 ⎞ ⎛ −a − b ⎞ ⎛ 0 ⎞ ⎛ −M ⎞ ⎛ d − a ⎞ ⎛⎜ ⎜ c sin ( θ ) ⎟ × −FCD cos ( φ ) ⎟ + ⎜ 0 ⎟ × ⎜ By ⎟ + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ c cos ( θ ) ⎠ ⎝ FCD sin ( φ ) ⎠ ⎝ 0 ⎠ ⎜⎝ Bz ⎟⎠ ⎝ 0 ⎠

⎛ Ay ⎞ ⎜ ⎟ A z ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , B , F ) y z y z CD ⎜ ⎟ ⎜ Bz ⎟ ⎜F ⎟ ⎝ CD ⎠

⎛ Ay ⎞ ⎛ 573 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ Az ⎠ ⎝ −208 ⎠ ⎛ By ⎞ ⎛ 382 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ Bz ⎠ ⎝ −139 ⎠ F CD = 1.015 kN

Problem 5-92 If neither the pin at A nor the roller at B can support a load no greater than F max, determine the maximum intensity of the distributed load w, so that failure of a support does not occur.

428

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Engineering Mechanics - Statics

Chapter 5

Units Used: 3

kN = 10 N Given: F max = 6 kN a = 3m b = 3m

Solution: The greatest reaction is at A. Require ΣMB = 0;

−F max( a + b) + w a

w =

F max( a + b) a⎛⎜

a

⎝2

+ b⎞⎟ +

⎠

2

⎛ a + b⎞ + 1 w b 2 b = 0 ⎜ ⎟ 3 ⎝2 ⎠ 2 w = 2.18

b

kN m

3

Problem 5-93 If the maximum intensity of the distributed load acting on the beam is w, determine the reactions at the pin A and roller B. Units Used: 3

kN = 10 N Given: F = 6 kN a = 3m b = 3m

429

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Engineering Mechanics - Statics

w = 4

Chapter 5

kN m

Solution: ΣF x = 0;

Ax = 0

ΣMA = 0;

−w a

By =

a 2 1 6

−

1 2

w b⎛⎜ a +

⎝

⎟ + By( a + b) = 0

3⎠

2

w

3a + 3 a b + b

Ay + By − w a −

ΣF y = 0;

b⎞

2

B y = 7 kN

a+b 1 2

Ay = −By + w a +

wb=0 1 2

Ay = 11 kN

wb

Problem 5-94 Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member. Units Used: 3

kN = 10 N Given: F 1 = 10 kN F 2 = 6 kN a = 0.6 m b = 0.6 m c = 0.8 m d = 0.4 m

θ = 60 deg

430

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Engineering Mechanics - Statics

Chapter 5

Solution: Initial Guesses: NA = 1 kN B x = 1 kN B y = 1 kN Given B x − F2 sin ( θ ) = 0 B y + NA − F 1 − F 2 cos ( θ ) = 0 F 2 d + F1 ⎡⎣b + ( c + d)cos ( θ )⎤⎦ − NA⎡⎣a + b + ( c + d)cos ( θ )⎤⎦ = 0

⎛ NA ⎞ ⎜ ⎟ ⎜ Bx ⎟ = Find ( NA , Bx , By) ⎜B ⎟ ⎝ y⎠

⎛ NA ⎞ ⎛ 8 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 5.196 ⎟ kN ⎜B ⎟ ⎝ 5 ⎠ ⎝ y⎠

Problem 5-95 The symmetrical shelf is subjected to uniform pressure P. Support is provided by a bolt (or pin) located at each end A and A' and by the symmetrical brace arms, which bear against the smooth wall on both sides at B and B'. Determine the force resisted by each bolt at the wall and the normal force at B for equilibrium. Units Used: 3

kPa = 10 Pa Given: P = 4 kPa a = 0.15 m b = 0.2 m c = 1.5 m Solution: ΣMA = 0;

431

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Engineering Mechanics - Statics

NB a − P ⎛⎜ b

Chapter 5

c⎞ b

⎟ =0 ⎝ 2⎠ 2 2

NB = P

b c

NB = 400 N

4a

ΣF x = 0; Ax = NB

Ax = 400 N

ΣF y = 0; Ay = P b

FA =

c

Ay = 600 N

2 2

Ax + Ay

2

F A = 721 N

Problem 5-96 A uniform beam having a weight W supports a vertical load F . If the ground pressure varies linearly as shown, determine the load intensities w1 and w2 measured in lb/ft, necessary for equilibrium. Given: W = 200 lb F = 800 lb a = 7 ft b = 6 ft Solution: Initial Guesses: w1 = 1

lb ft

w2 = 1

lb ft

432

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Engineering Mechanics - Statics

Chapter 5

Given 1

w1 ( a + b) +

w1 ( a + b)

2

(w2 − w1)( a + b) − F − W = 0

a+b 2

+

(w2 − w1)( a + b) 3 ( a + b) − W 2 1

⎛ w1 ⎞ ⎜ ⎟ = Find ( w1 , w2) ⎝ w2 ⎠

2

a+b 2

− Fa = 0

⎛ w1 ⎞ ⎛ 62.7 ⎞ lb ⎜ ⎟=⎜ ⎟ ⎝ w2 ⎠ ⎝ 91.1 ⎠ ft

Problem 5-97 The uniform ladder rests along the wall of a building at A and on the roof at B. If the ladder has a weight W and the surfaces at A and B are assumed smooth, determine the angle θ for equilibrium. Given: a = 18 ft W = 25 lb

θ 1 = 40 deg Solution: Initial guesses: R A = 10 lb R B = 10 lb

θ = 10 deg Given ΣMB = 0;

−R A a sin ( θ ) + W

ΣF x = 0;

R A − RB sin ( θ 1 ) = 0

ΣF y = 0;

R B cos ( θ 1 ) − W = 0

a 2

cos ( θ ) = 0

Solving,

433

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Engineering Mechanics - Statics

Chapter 5

⎛ RB ⎞ ⎜ ⎟ ⎜ RA ⎟ = Find ( RB , RA , θ ) ⎜ ⎟ ⎝θ ⎠

⎛ RA ⎞ ⎛ 21 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ RB ⎠ ⎝ 32.6 ⎠

θ = 30.8 deg

Problem 5-98 Determine the x, y, z components of reaction at the ball supports B and C and the ball-and-socket A (not shown) for the uniformly loaded plate. Given: P = 2

lb ft

2

a = 4 ft b = 1 ft c = 2 ft d = 2 ft

Solution: The initial guesses are

Ax = 1 lb

Ay = 1 lb

Az = 1 lb

B z = 1 lb

Cz = 1 lb

Given ΣF x = 0;

Ax = 0

ΣF y = 0;

Ay = 0

ΣF z = 0; Az + Bz + Cz − P a c = 0 c⎞ ⎟ + Cz b = 0 ⎝ 2⎠

ΣMx = 0; c B z − P a c⎛⎜

a⎞ ⎟ − Cz a = 0 ⎝ 2⎠

ΣMy = 0; −B z ( a − d) + P a c⎛⎜

434

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Engineering Mechanics - Statics

Chapter 5

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ Az ⎟ = Find ( A , A , A , B , C ) x y z z z ⎜ ⎟ ⎜ Bz ⎟ ⎜C ⎟ ⎝ z⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0 ⎟ lb ⎜ A ⎟ ⎝ 5.333 ⎠ ⎝ z⎠

⎛ Bz ⎞ ⎛ 5.333 ⎞ ⎜ ⎟=⎜ ⎟ lb 5.333 C ⎝ ⎠ z ⎝ ⎠

Problem 5-99 A vertical force F acts on the crankshaft. Determine the horizontal equilibrium force P that must be applied to the handle and the x, y, z components of force at the smooth journal bearing A and the thrust bearing B. The bearings are properly aligned and exert the force reactions on the shaft. Given: F = 80 lb a = 10 in b = 14 in c = 14 in d = 8 in e = 6 in f = 4 in

Solution: ΣMy = 0;

P d−F a= 0 P = F

ΣMx = 0;

P = 100 lb

B z( b + c) − F c = 0 Bz = F

ΣMz = 0;

⎛ a⎞ ⎜ ⎟ ⎝ d⎠ ⎛ c ⎞ ⎜ ⎟ ⎝ b + c⎠

B z = 40 lb

−B x( b + c) − P ( e + f) = 0

435

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Engineering Mechanics - Statics

B x = −P ΣF x = 0;

⎛ e + f ⎞ B = −35.7 lb ⎜ ⎟ x ⎝ b + c⎠

Ax + Bx − P = 0 Ax = −Bx + P

Ax = 135.7 lb By = 0

ΣF y = 0; ΣF z = 0;

Chapter 5

Az + Bz − F = 0 Az = −B z + F

Az = 40 lb

Problem 5-100 The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is kA , determine the required stiffness of the spring at B so that if the beam is loaded with the force F , it remains in the horizontal position both before and after loading. Units Used: 3

kN = 10 N Given: kA = 5

kN m

F = 800 N

a = 1m b = 2m

Solution: Equilibrium: ΣMA = 0;

F B ( a + b) − F a = 0 FB = F

⎛ a ⎞ ⎜ ⎟ ⎝ a + b⎠

F B = 266.667 N ΣMB = 0;

F b − FA( a + b) = 0 436

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Engineering Mechanics - Statics

Chapter 5

FA = F

⎛ b ⎞ ⎜ ⎟ ⎝ a + b⎠

F A = 533.333 N Spring force formula: xA = xB

FA kA

=

FB kB

kB =

FB FA

kA

kB = 2.5

kN m

437

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Engineering Mechanics - Statics

Chapter 6

Problem 6-1 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 7 kN P 2 = 7 kN Solution:

θ = 45 deg Initial Guesses: F AB = 1 kN F DC = 1 kN

F AD = 1 kN

F DB = 1 kN

F CB = 1 kN

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint D:

F DB cos ( θ ) − F AD cos ( θ ) + F DC cos ( θ ) = 0

(FAD + FDB − FDC)sin(θ ) − P2 = 0 Joint C:

F CB + FDC sin ( θ ) = 0

⎛⎜ FAB ⎞⎟ ⎜ FAD ⎟ ⎜ ⎟ ⎜ FDB ⎟ = Find ( FAB , FAD , FDB , FDC , FCB) ⎜F ⎟ ⎜ DC ⎟ ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ 7 ⎞ ⎜ FAD ⎟ ⎜ −9.9 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDB ⎟ = ⎜ 4.95 ⎟ kN ⎜ F ⎟ ⎜ −14.85 ⎟ ⎜ DC ⎟ ⎜ ⎟ 10.5 ⎝ ⎠ ⎜ FCB ⎟ ⎝ ⎠

Positive means Tension, Negative means Compression

438

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Engineering Mechanics - Statics

Chapter 6

Problem 6-2 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 8 kN P 2 = 10 kN Solution:

θ = 45 deg Initial Guesses: F AB = 1 kN F DC = 1 kN

F AD = 1 kN

F DB = 1 kN

F CB = 1 kN

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint D:

F DB cos ( θ ) − F AD cos ( θ ) + F DC cos ( θ ) = 0

(FAD + FDB − FDC)sin(θ ) − P2 = 0 Joint C:

F CB + FDC sin ( θ ) = 0

⎛⎜ FAB ⎞⎟ ⎜ FAD ⎟ ⎜ ⎟ F ⎜ DB ⎟ = Find ( FAB , FAD , FDB , FDC , FCB) ⎜F ⎟ ⎜ DC ⎟ ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ 8 ⎞ ⎜ FAD ⎟ ⎜ −11.31 ⎟ ⎟ ⎜ ⎟ ⎜ F 7.07 = ⎟ kN ⎜ DB ⎟ ⎜ ⎜ F ⎟ ⎜ −18.38 ⎟ ⎜ DC ⎟ ⎜ ⎟ ⎜ FCB ⎟ ⎝ 13 ⎠ ⎝ ⎠

Positive means Tension, Negative means Compression

439

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Engineering Mechanics - Statics

Chapter 6

Problem 6-3 The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Units Used: 3

kip = 10 lb Given: P 1 = 600 lb P 2 = 400 lb a = 4 ft

θ = 45 deg Solution: Initial Guesses F AB = 1 lb

F AD = 1 lb

F DC = 1 lb

F BC = 1 lb

F BD = 1 lb

F DE = 1 lb

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint B:

F BC − F AB = 0 −P 2 − F BD = 0

Joint D:

(FDC − FAD)cos (θ ) + FDE = 0 (FDC + FAD)sin(θ ) + FBD = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AD BC BD DC DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDC ⎟ ⎜ ⎟ ⎝ FDE ⎠

440

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Engineering Mechanics - Statics

⎛ FAB ⎞ ⎜ ⎟ ⎛ 600 ⎞ F AD ⎜ ⎟ ⎜ −849 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BC ⎟ = 600 ⎟ lb ⎜ FBD ⎟ ⎜ −400 ⎟ ⎟ ⎜ ⎟ ⎜ 1414 ⎜ ⎟ ⎜ FDC ⎟ ⎜ ⎜ ⎟ ⎝ −1600 ⎟⎠ F ⎝ DE ⎠

Chapter 6

Positive means Tension, Negative means Compression

Problem 6-4 The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Units Used: 3

kip = 10 lb Given: P 1 = 800 lb P 2 = 0 lb a = 4 ft

θ = 45 deg Solution: Initial Guesses F AB = 1 lb

F AD = 1 lb

F DC = 1 lb

F BC = 1 lb

F BD = 1 lb

F DE = 1 lb

Given Joint A:

F AB + F AD cos ( θ ) = 0 −P 1 − F AD sin ( θ ) = 0

Joint B:

F BC − F AB = 0 −P 2 − F BD = 0

441

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Engineering Mechanics - Statics

Joint D:

Chapter 6

(FDC − FAD)cos (θ ) + FDE = 0 (FDC + FAD)sin(θ ) + FBD = 0

⎛ FAB ⎞ ⎜ ⎟ F ⎜ AD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AD BC BD DC DE) ⎜ FBD ⎟ ⎜ ⎟ F ⎜ DC ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 800 ⎞ ⎜ FAD ⎟ ⎜ −1131 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BC ⎟ = 800 ⎟ lb ⎜ FBD ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDC ⎟ ⎜ 1131 ⎟ ⎜ ⎟ ⎜⎝ −1600 ⎟⎠ F ⎝ DE ⎠

Positive means Tension, Negative means Compression

Problem 6-5 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 20 kN P 2 = 10 kN a = 1.5 m e = 2m Solution:

e θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

442

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Engineering Mechanics - Statics

Chapter 6

Initial Guesses: F AB = 1 kN F AG = 1 kN F CF = 1 kN F BC = 1 kN F BG = 1 kN F DE = 1 kN F CG = 1 kN F FG = 1 kN F EF = 1 kN F CD = 1 kN F DF = 1 kN Given Joint B

F BC − F AB cos ( θ ) = 0 −F BG − F AB sin ( θ ) = 0

Joint G

F FG + F CG cos ( θ ) − F AG = 0 F CG sin ( θ ) + FBG − P1 = 0

Joint C

(

)

−F BC + F CD + FCF − F CG cos ( θ ) = 0

(

)

− FCG + FCF sin ( θ ) = 0 Joint D

−F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0

Joint F

F EF − F FG − F CF cos ( θ ) = 0 F DF + F CF sin ( θ ) − P 2 = 0

Joint E

−F DE cos ( θ ) − F EF = 0

443

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCG ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FAG ⎟ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBC , FCG , FCD , FAG , FBG , FFG , FDF , FCF , FDE , FEF) ⎜F ⎟ ⎜ FG ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛⎜ −21.88 ⎟⎞ ⎜ FBC ⎟ ⎜ −13.13 ⎟ kN ⎜ ⎟=⎜ ⎟ F 3.13 CG ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ −9.37 ⎠ ⎝ CD ⎠

⎛ FAG ⎞ ⎜ ⎟ ⎛⎜ 13.13 ⎞⎟ ⎜ FBG ⎟ ⎜ 17.5 ⎟ kN ⎜ ⎟=⎜ ⎟ F 11.25 FG ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ 12.5 ⎠ ⎝ DF ⎠

⎛ FCF ⎞ ⎛ −3.13 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ DE ⎟ = ⎜ −15.62 ⎟ kN ⎜ F ⎟ ⎝ 9.37 ⎠ ⎝ EF ⎠

Positive means Tension, Negative means Compression

Problem 6-6 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 40 kN P 2 = 20 kN a = 1.5 m e = 2m

444

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

e θ = atan ⎛⎜ ⎞⎟

Solution:

⎝ a⎠

Initial Guesses: F AB = 1 kN F AG = 1 kN F CF = 1 kN F BC = 1 kN F BG = 1 kN F DE = 1 kN F CG = 1 kN F FG = 1 kN F EF = 1 kN F CD = 1 kN F DF = 1 kN Given Joint B

F BC − F AB cos ( θ ) = 0 −F BG − F AB sin ( θ ) = 0

Joint G

F FG + F CG cos ( θ ) − F AG = 0 F CG sin ( θ ) + FBG − P1 = 0

Joint C

(

)

−F BC + F CD + FCF − F CG cos ( θ ) = 0

(

)

− FCG + FCF sin ( θ ) = 0 Joint D

−F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0

Joint F

F EF − F FG − F CF cos ( θ ) = 0 F DF + F CF sin ( θ ) − P 2 = 0

Joint E

−F DE cos ( θ ) − F EF = 0

445

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCG ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FAG ⎟ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBC , FCG , FCD , FAG , FBG , FFG , FDF , FCF , FDE , FEF) ⎜F ⎟ ⎜ FG ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛⎜ −43.75 ⎟⎞ F ⎜ BC ⎟ ⎜ −26.25 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCG ⎟ ⎜ 6.25 ⎟ ⎜ F ⎟ ⎝ −18.75 ⎠ ⎝ CD ⎠

⎛ FAG ⎞ ⎜ ⎟ ⎛⎜ 26.25 ⎞⎟ F ⎜ BG ⎟ ⎜ 35 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FFG ⎟ ⎜ 22.5 ⎟ ⎜ F ⎟ ⎝ 25 ⎠ ⎝ DF ⎠

⎛ FCF ⎞ ⎛ −6.25 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −31.25 ⎟ kN ⎜ F ⎟ ⎝ 18.75 ⎠ ⎝ EF ⎠

Positive means Tension, Negative means Compression

Problem 6-7 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 3 kN F 2 = 8 kN F 3 = 4 kN F 4 = 10 kN

446

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

a = 2m b = 1.5 m b θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ a⎠

Initial Guesses F BA = 1 kN

F BC = 1 kN

F AF = 1 kN

F CD = 1 kN

F DF = 1 kN

F ED = 1 kN

F AC = 1 kN F CF = 1 kN F EF = 1 kN

Given Joint B

F 1 + F BC = 0 −F 2 − F BA = 0

Joint C

F CD − F BC − F AC cos ( θ ) = 0 −F 3 − F AC sin ( θ ) − FCF = 0

Joint E

−F EF = 0

Joint D

−F CD − F DF cos ( θ ) = 0 −F 4 − F DF sin ( θ ) − FED = 0

Joint F

−F AF + F EF + F DF cos ( θ ) = 0 F CF + FDF sin ( θ ) = 0

⎛ FBA ⎞ ⎜ ⎟ ⎜ FAF ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( FBA , FAF , FDF , FBC , FCD , FED , FAC , FCF , FEF) ⎜F ⎟ ⎜ ED ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎟ ⎜ ⎝ FEF ⎠

447

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ −8 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 4.167 ⎟ ⎜ DF ⎟ ⎜ 5.208 ⎟ ⎜ FBC ⎟ ⎜ −3 ⎟ ⎟ ⎜ ⎟ ⎜ F − 4.167 = ⎟ kN ⎜ CD ⎟ ⎜ ⎜ F ⎟ ⎜ −13.125 ⎟ ⎟ ⎜ ED ⎟ ⎜ ⎜ FAC ⎟ ⎜ −1.458 ⎟ ⎜ ⎟ ⎜ −3.125 ⎟ F ⎜ CF ⎟ ⎜ ⎟ ⎟ ⎝ 0 ⎠ ⎜ ⎝ FEF ⎠

Positive means tension, Negative means compression.

Problem 6-8 Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

Solution: ΣMA = 0;

−P a + Cy2a − P a = 0 Cy = P

Joint C: ΣF x = 0;

1

ΣF y = 0;

P+

2

F BC − 1 17

4 17

F CD −

FCD = 0 1 2

F BC = 0

448

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

F BC =

4 2P

F CD =

17 P

1

FCD +

3

3

= 1.886 P (C) = 1.374 P (T)

Joint B: ΣF x = 0;

P−

ΣFy = 0;

1

2

2

F CD +

2

2P

F AB = F BD =

1

3 5P 3

1 2

FAB = 0

F AB − F BD = 0

= 0.471P

= 1.667P

( C)

( T)

Joint D: ΣF x = 0;

F DA = F CD = 1.374P

(T)

Problem 6-9 The maximum allowable tensile force in the members of the truss is Tmax, and the maximum allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads that can be applied to the truss. Given: Tmax = 1500 lb Cmax = 800 lb Solution: Set

P = 1 lb

Initial Guesses F AB = 1 lb

F AD = 1 lb

F BC = 1 lb

F CD = 1 lb

F BD = 1 lb

449

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given Joint B

(FBC − FAB)

1

+P=0

2

(

−F BD − FAB + FBC Joint D

(FCD − FAD)

4

)

1

=0

2

=0

17

(

)

F BD − P − F AD + F CD Joint C

−F BC

1 2

− F CD

4

1

=0

17

=0

17

⎛⎜ FAB ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FAD ⎟ = Find ( FAB , FBC , FAD , FCD , FBD) ⎜F ⎟ ⎜ CD ⎟ ⎜ FBD ⎟ ⎝ ⎠

⎛⎜ FAB ⎞⎟ ⎛ −0.471 ⎞ ⎜ FBC ⎟ ⎜ −1.886 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAD ⎟ = ⎜ 1.374 ⎟ lb ⎜ F ⎟ ⎜ 1.374 ⎟ ⎜ CD ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎝ 1.667 ⎠ ⎝ ⎠

Now find the critical load P1 = P

Tmax

(

max F AB , F BC , F AD , F CD , F BD

P2 = P

)

P 1 = 900 lb

)

P 2 = 424.264 lb

Cmax

(

min F AB , F BC , F AD , F CD , F BD

(

)

P = min P 1 , P 2

P = 424.3 lb

Problem 6-10 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 0 lb

450

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

P 2 = 1000 lb a = 10 ft b = 10 ft Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AG = 1 lb

F BG = 1 lb

F BC = 1 lb

F DC = 1 lb

F DE = 1 lb

F EG = 1 lb

F EC = 1 lb

F CG = 1 lb

Given Joint B

F BC − F AB = 0 F BG − P 1 = 0

Joint G

(FCG − FAG)cos (θ ) + FEG = 0 (

)

− FCG + FAG sin ( θ ) − FBG = 0 Joint C

F DC − FBC − FCG cos ( θ ) = 0 F EC + F CG sin ( θ ) − P2 = 0

Joint E

F DE cos ( θ ) − F EG = 0 −F EC − F DE sin ( θ ) = 0

Joint D

−F DE cos ( θ ) − F DC = 0

451

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ FAG ⎟ ⎜ ⎟ F ⎜ DC ⎟ = Find ( FAB , FBC , FEG , FAG , FDC , FEC , FBG , FDE , FCG) ⎜F ⎟ ⎜ EC ⎟ ⎜ FBG ⎟ ⎜ ⎟ F ⎜ DE ⎟ ⎟ ⎜ ⎝ FCG ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 333 ⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 333 ⎟ ⎜ EG ⎟ ⎜ −667 ⎟ ⎜ FAG ⎟ ⎜ −471 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDC ⎟ = ⎜ 667 ⎟ lb ⎜ F ⎟ ⎜ 667 ⎟ ⎟ ⎜ EC ⎟ ⎜ ⎜ FBG ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ −943 ⎟ ⎜ FDE ⎟ ⎜ ⎟ ⎟ ⎝ 471 ⎠ ⎜ ⎝ FCG ⎠

Positive means tension, Negative means compression.

Problem 6-11 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 500 lb P 2 = 1500 lb

452

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

a = 10 ft b = 10 ft Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AG = 1 lb

F BG = 1 lb

F BC = 1 lb

F DC = 1 lb

F DE = 1 lb

F EG = 1 lb

F EC = 1 lb

F CG = 1 lb

Given Joint B

F BC − F AB = 0 F BG − P 1 = 0

Joint G

(FCG − FAG)cos (θ ) + FEG = 0 (

)

− FCG + FAG sin ( θ ) − FBG = 0 Joint C

F DC − FBC − FCG cos ( θ ) = 0 F EC + F CG sin ( θ ) − P2 = 0

Joint E

F DE cos ( θ ) − F EG = 0 −F EC − F DE sin ( θ ) = 0

Joint D

−F DE cos ( θ ) − F DC = 0

453

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ FAG ⎟ ⎜ ⎟ F ⎜ DC ⎟ = Find ( FAB , FBC , FEG , FAG , FDC , FEC , FBG , FDE , FCG) ⎜F ⎟ ⎜ EC ⎟ ⎜ FBG ⎟ ⎜ ⎟ F ⎜ DE ⎟ ⎟ ⎜ ⎝ FCG ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 833 ⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 833 ⎟ ⎜ EG ⎟ ⎜ −1167 ⎟ ⎜ FAG ⎟ ⎜ −1179 ⎟ ⎟ ⎜ ⎟ ⎜ F 1167 = ⎟ lb ⎜ DC ⎟ ⎜ ⎜ F ⎟ ⎜ 1167 ⎟ ⎟ ⎜ EC ⎟ ⎜ ⎜ FBG ⎟ ⎜ 500 ⎟ ⎜ ⎟ ⎜ −1650 ⎟ F ⎜ DE ⎟ ⎜ ⎟ ⎟ ⎝ 471 ⎠ ⎜ ⎝ FCG ⎠

Positive means tension, Negative means compression.

Problem 6-12 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 10 kN P 2 = 15 kN

454

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

a = 2m b = 4m c = 4m Solution:

c α = atan ⎛⎜ ⎟⎞

⎝ a⎠

c β = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses: F AB = 1 kN

F AF = 1 kN

F GB = 1 kN

F BF = 1 kN

F FC = 1 kN

F FE = 1 kN

F BC = 1 kN

F EC = 1 kN

F CD = 1 kN

F ED = 1 kN Given Joint B

−F GB + F BC − F AB cos ( α ) = 0

Joint F

−F AB sin ( α ) − FBF = 0 −F AF + F FE + F FC cos ( β ) = 0 F BF + F FC sin ( β ) − P 1 = 0

Joint C

−F BC − F FC cos ( β ) + FCD cos ( α ) = 0 −F FC sin ( β ) − F CD sin ( α ) − FEC = 0

Joint E

−F FE + F ED = 0 F EC − P 2 = 0

Joint D

−F CD cos ( α ) − F ED = 0 F CD sin ( α ) = 0

455

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎜ ⎟ = Find ( FAB , FBF , FBC , FED , FAF , FFC , FEC , FGB , FFE , FCD) F FC ⎜ ⎟ ⎜F ⎟ ⎜ EC ⎟ ⎜ FGB ⎟ ⎜ ⎟ F ⎜ FE ⎟ ⎜F ⎟ ⎝ CD ⎠ ⎛⎜ FAB ⎞⎟ ⎛ −27.951 ⎞ ⎜ FBF ⎟ ⎜ 25 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FBC ⎟ = ⎜ 15 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ ED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎝ −15 ⎠ ⎝ ⎠

⎛⎜ FFC ⎞⎟ ⎛ −21.213 ⎞ ⎜ FEC ⎟ ⎜ 15 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGB ⎟ = ⎜ 27.5 ⎟ kN Positive means Tension, Negative means Compression ⎜F ⎟ ⎜ 0 ⎟ FE ⎜ ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎝ 0 ⎠ ⎝ ⎠

Problem 6-13 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 0 kN P 2 = 20 kN a = 2m b = 4m c = 4m

456

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Solution:

c α = atan ⎛⎜ ⎞⎟

⎝ a⎠

Chapter 6

c β = atan ⎛⎜ ⎞⎟

⎝ b⎠

Initial Guesses: F AB = 1 kN

F AF = 1 kN

F GB = 1 kN

F BF = 1 kN

F FC = 1 kN

F FE = 1 kN

F BC = 1 kN

F EC = 1 kN

F CD = 1 kN

F ED = 1 kN Given Joint B

−F GB + F BC − F AB cos ( α ) = 0

Joint F

−F AB sin ( α ) − FBF = 0 −F AF + F FE + F FC cos ( β ) = 0 F BF + F FC sin ( β ) − P 1 = 0

Joint C

−F BC − F FC cos ( β ) + FCD cos ( α ) = 0 −F FC sin ( β ) − F CD sin ( α ) − FEC = 0

Joint E

−F FE + F ED = 0 F EC − P 2 = 0

Joint D

−F CD cos ( α ) − F ED = 0 F CD sin ( α ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FAF ⎟ ⎜ ⎟ = Find ( FAB , FBF , FBC , FED , FAF , FFC , FEC , FGB , FFE , FCD) FFC ⎜ ⎟ ⎜F ⎟ ⎜ EC ⎟ ⎜ FGB ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎜F ⎟ ⎝ CD ⎠

457

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛⎜ FAB ⎞⎟ ⎛ −22.361 ⎞ ⎜ FBF ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FBC ⎟ = ⎜ 20 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ ED ⎟ ⎜ ⎟ − 20 ⎝ ⎠ ⎜ FAF ⎟ ⎝ ⎠

⎛⎜ FFC ⎞⎟ ⎛ −28.284 ⎞ ⎜ FEC ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGB ⎟ = ⎜ 30 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ FE ⎟ ⎜ ⎟ 0 ⎝ ⎠ ⎜ FCD ⎟ ⎝ ⎠

Positive means Tension, Negative means Compression

Problem 6-14 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 100 lb P 2 = 200 lb P 3 = 300 lb a = 10 ft b = 10 ft

θ = 30 deg Solution:

b φ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AF = 1 lb

F BC = 1 lb

F BF = 1 lb

F FC = 1 lb

F FE = 1 lb

F ED = 1 lb

F EC = 1 lb

F CD = 1 lb

458

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given Joint B

F BC − F AB cos ( φ ) = 0 −F BF − F AB sin ( φ ) = 0

Joint F

−F AF + F FE + F FC cos ( φ ) = 0 −P 2 + F BF + F FC sin ( φ ) = 0

Joint C

−F BC + F CD cos ( φ ) − F FC cos ( φ ) = 0 −F EC − F CD sin ( φ ) − FFC sin ( φ ) = 0

Joint E

−F FE + F ED = 0 F EC − P 3 = 0

Joint D

−F ED cos ( θ ) − F CD cos ( φ + θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ F AF ⎜ ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBF ⎟ ⎜ ⎟ F ⎜ FC ⎟ = Find ( FAB , FAF , FBC , FBF , FFC , FFE , FED , FEC , FCD) ⎜F ⎟ ⎜ FE ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FEC ⎟ ⎜ ⎟ ⎝ FCD ⎠

459

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ −330.0 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 79.4 ⎟ ⎜ BC ⎟ ⎜ −233.3 ⎟ ⎜ FBF ⎟ ⎜ 233.3 ⎟ ⎟ ⎜ ⎟ ⎜ F − 47.1 = ⎟ lb ⎜ FC ⎟ ⎜ ⎜ F ⎟ ⎜ 112.7 ⎟ ⎟ ⎜ FE ⎟ ⎜ ⎜ FED ⎟ ⎜ 112.7 ⎟ ⎜ ⎟ ⎜ 300.0 ⎟ F ⎜ EC ⎟ ⎜ ⎟ ⎟ ⎝ −377.1 ⎠ ⎜ ⎝ FCD ⎠

Positive means Tension, Negative means Compression

Problem 6-15 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 = 400 lb P 2 = 400 lb P 3 = 0 lb a = 10 ft b = 10 ft

θ = 30 deg Solution:

b φ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Initial Guesses: F AB = 1 lb

F AF = 1 lb

F BC = 1 lb

F BF = 1 lb

F FC = 1 lb

F FE = 1lb

F ED = 1 lb

F EC = 1 lb

F CD = 1 lb

460

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given Joint B

F BC − F AB cos ( φ ) = 0 −F BF − F AB sin ( φ ) = 0

Joint F

−F AF + F FE + F FC cos ( φ ) = 0 −P 2 + F BF + F FC sin ( φ ) = 0

Joint C

−F BC + F CD cos ( φ ) − F FC cos ( φ ) = 0 −F EC − F CD sin ( φ ) − FFC sin ( φ ) = 0

Joint E

−F FE + F ED = 0 F EC − P 3 = 0

Joint D

−F ED cos ( θ ) − F CD cos ( φ + θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAF ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBF ⎟ ⎜ ⎟ ⎜ FFC ⎟ = Find ( FAB , FAF , FBC , FBF , FFC , FFE , FED , FEC , FCD) ⎜F ⎟ ⎜ FE ⎟ ⎜ FED ⎟ ⎜ ⎟ ⎜ FEC ⎟ ⎜ ⎟ ⎝ FCD ⎠

461

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Engineering Mechanics - Statics

⎛ FAB ⎞ ⎜ ⎟ ⎛ −377.1 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 189.7 ⎟ ⎜ BC ⎟ ⎜ −266.7 ⎟ ⎜ FBF ⎟ ⎜ 266.7 ⎟ ⎟ ⎜ ⎟ ⎜ F 188.6 = ⎟ lb ⎜ FC ⎟ ⎜ ⎜ F ⎟ ⎜ 56.4 ⎟ ⎟ ⎜ FE ⎟ ⎜ ⎜ FED ⎟ ⎜ 56.4 ⎟ ⎜ ⎟ ⎜ 0.0 ⎟ F ⎜ EC ⎟ ⎜ ⎟ ⎟ ⎝ −188.6 ⎠ ⎜ ⎝ FCD ⎠

Chapter 6

Positive means Tension, Negative means Compression

Problem 6-16 Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

Solution: Support reactions: ΣME = 0;

3 Ax d − P d = 0 2

Ax =

2P 3

462

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Engineering Mechanics - Statics

Chapter 6

2P

ΣF x = 0;

Ax − Ex = 0

Ex =

ΣF y = 0;

Ey − P = 0

Ey = P

3

Joint E: 2

ΣF x = 0;

F EC

ΣF y = 0;

P − FED − FEC

13

− Ex = 0

13

F EC =

3

=0

13

3

P = 1.20P (T)

F ED = 0

Joint A: 1

− F AD

1

2

=0

ΣF y = 0;

F AB

ΣF x = 0;

Ax − 2FAB

5

=0

5

F AB = F AD

F AB = F AD =

5

Joint D: ΣF x = 0;

F AD

ΣF y = 0;

2FAD

2 5

− F DC

1 5

2

=0

5

− FDB = 0

F DC = F DB =

5 6 P

5 6

P = 0.373P (C)

P = 0.373P (C)

(T)

3

Joint B: ΣF x = 0; F AB

1 5

− F BC

1 5

=0

F BC =

5 6

P = 0.373P (C)

463

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Engineering Mechanics - Statics

Chapter 6

Problem 6-17 The maximum allowable tensile force in the members of the truss is Tmax and the maximum allowable compressive force is Cmax. Determine the maximum magnitude of the load P that can be applied to the truss. Units Used: 3

kN = 10 N Given: Tmax = 5 kN Cmax = 3 kN d = 2m Solution: P = 1 kN

Set

Initial Guesses: F AD = 1 kN

F AB = 1 kN

F BC = 1 kN

F BD = 1 kN

F CD = 1 kN

F CE = 1 kN

F DE = 1 kN

Given Joint A

Joint B

Joint D

F AD F BC

1 5 2 5

− F AB

1

− F AB

2

=0

5 =0

5

(FBC + FAB)

1

(FCD − FAD)

2

5

+ F BD = 0 =0

5

(

)

1

F DE − F BD − FAD + FCD Joint C

− FCD + FBC

(

)

(FCD − FBC )

1

2 5

5

2

− F CE

+ F CE

=0

5 =0

13 3

−P=0

13

464

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Engineering Mechanics - Statics

Chapter 6

⎛ FAD ⎞ ⎜ ⎟ ⎜ FAB ⎟ ⎜ ⎟ FBC ⎜ ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F ( AD AB BC BD CD CE DE) ⎜ BD ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ FCE ⎟ ⎜F ⎟ ⎝ DE ⎠ ⎛ FAD ⎞ ⎜ ⎟ ⎛ −0.373 ⎞ ⎟ ⎜ FAB ⎟ ⎜ ⎜ ⎟ ⎜ −0.373 ⎟ ⎜ FBC ⎟ ⎜ −0.373 ⎟ ⎜ F ⎟ = ⎜ 0.333 ⎟ kN ⎟ ⎜ BD ⎟ ⎜ ⎜ FCD ⎟ ⎜ −0.373 ⎟ ⎜ ⎟ ⎜ 1.202 ⎟ ⎟ ⎜ FCE ⎟ ⎜ 0 ⎝ ⎠ ⎜F ⎟ ⎝ DE ⎠ Now Scale the answer P1 = P

P2 = P

Tmax

(

max F AD , F AB , F BC , F BD , F CD , F CE , FDE

)

Cmax min F AD , F AB , F BC , F BD , F CD , F CE , FDE

(

(

)

P = min P 1 , P 2

)

P = 4.16 kN

Problem 6-18 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The horizontal force component at A must be zero. Why?

465

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Engineering Mechanics - Statics

Chapter 6

Units Used: 3

kip = 10 lb Given: F 1 = 600 lb F 2 = 800 lb a = 4 ft b = 3 ft

θ = 60 deg Solution: Initial Guesses F BA = 1 lb

F BD = 1 lb

F CB = 1 lb

F CD = 1 lb

Given Joint C

Joint B

−F CB − F2 cos ( θ ) = 0 F CB + FBD

b 2

a +b

−F CD − F 2 sin ( θ ) = 0 =0

2

⎛ FBA ⎞ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ = Find ( FBA , FBD , FCB , FCD) ⎜ FCB ⎟ ⎜F ⎟ ⎝ CD ⎠

−F BA − F BD

a 2

2

a +b

− F1 = 0

⎛ FBA ⎞ ⎛ 3 ⎜ ⎟ −1.133 × 10 ⎞ ⎜ ⎟ Positive means Tension ⎜ FBD ⎟ 666.667 ⎟ lb Negative means ⎜ ⎟=⎜ ⎜ ⎟ Compression F −400 ⎜ CB ⎟ ⎜ ⎜ F ⎟ ⎝ −692.82 ⎟⎠ ⎝ CD ⎠

Problem 6-19 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The resultant force at the pin E acts along member ED. Why? Units Used: 3

kN = 10 N

466

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 3 kN F 2 = 2 kN a = 3m b = 4m Solution: Initial Guesses: F CB = 1 kN

F CD = 1 kN

F BA = 1 kN

F BD = 1 kN

F DA = 1 kN

F DE = 1 kN

Given Joint C

−F 2 − F CD

Joint B

2a

−F CB − FCD

=0

2

( 2 a) + b b 2

2

=0 2

( 2 a) + b

−F BA + F CB = 0 −F 1 − F BD = 0

Joint D

(FCD − FDA − FDE) (

2a 2

=0 2

( 2 a) + b

F BD + FCD + FDA − FDE

)

b 2

=0 2

( 2 a) + b

467

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FCB ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ BA ⎟ = Find F , F , F , F , F , F ( CB CD BA BD DA DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FCB ⎞ ⎜ ⎟ ⎛ 3 ⎞ F ⎜ CD ⎟ ⎜ −3.606 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ BA 3 ⎜ ⎟= ⎜ ⎟ kN ⎜ FBD ⎟ −3 ⎟ ⎜ ⎜ ⎟ ⎜ FDA ⎟ ⎜ 2.704 ⎟ ⎜ ⎟ ⎜⎝ −6.31 ⎟⎠ F ⎝ DE ⎠

Positive means Tension, Negative means Compression

Problem 6-20 Each member of the truss is uniform and has a mass density ρ. Determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member. Given:

ρ = 8

kg m

g = 9.81

m 2

s F1 = 0 N F2 = 0 N a = 3m b = 4m

468

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Solution: Initial Guesses: F CB = 1 N

F CD = 1 N

F BA = 1 N

F BD = 1 N

F DA = 1 N

F DE = 1 N

Given Joint C

−F CB − FCD

−F 2 − F CD

Joint B

2a 2

( 2 a) + b

=0 2

−F BA + F CB = 0

⎛ b⎞ −F 1 − F BD − ρ g⎜ a + ⎟ = 0 ⎝ 4⎠ Joint D

2 2⎤ ⎛ a⎞ + ⎛ b⎞ ⎥ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 4⎠ ⎦

⎡a − ρ g⎢ + 2 2 ⎣2 ( 2 a) + b b

(

F BD + FCD + FDA − FDE

(FCD − FDA − FDE)

⎡b − ρ g⎢ + 3 2 2 ⎣4 ( 2 a) + b

)

b

2a 2

2 2⎤ ⎛ a⎞ + ⎛ b⎞ ⎥ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 4⎠ ⎦

=0 2

( 2 a) + b

⎛ FCB ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ BA ⎟ = Find F , F , F , F , F , F ( CB CD BA BD DA DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎟ ⎝ FDE ⎠

469

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Engineering Mechanics - Statics

⎛ FCB ⎞ ⎜ ⎟ ⎛ 389 ⎞ F CD ⎜ ⎟ ⎜ −467 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BA ⎟ = 389 ⎟ N ⎜ FBD ⎟ ⎜ −314 ⎟ ⎟ ⎜ ⎟ ⎜ 736 ⎜ ⎟ ⎜ FDA ⎟ ⎜ ⎜ ⎟ ⎝ −1204 ⎟⎠ F ⎝ DE ⎠

Chapter 6

Positive means Tension, Negative means Compression

Problem 6-21 Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

Solution: Joint B: +

↑Σ Fy = 0;

F BA sin ( 2 θ ) − P = 0 F BA = P csc ( 2 θ )

+ Σ F x = 0; →

( C)

F BAcos( 2 θ ) − FBC = 0 F BC = Pcot( 2 θ )

( C)

Joint C: + Σ F x = 0; →

P cot ( 2 θ ) + P + F CD cos ( 2 θ ) − F CA cos ( θ ) = 0

+

F CD sin ( 2 θ ) − FCA sin ( θ ) = 0

↑Σ Fy = 0;

470

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F CA =

Chapter 6

cot ( 2 θ ) + 1

cos ( θ ) − sin ( θ ) cot( 2 θ )

P

F CA = ( cot ( θ ) csc ( θ ) − sin ( θ ) + 2 cos ( θ ) ) P

( T)

F CD = ( cot ( 2 θ ) + 1) P

( C)

Joint D: + Σ F x = 0; →

F DA − ⎡⎣cot( 2 θ ) + 1⎤⎦ ⎡⎣cos( 2 θ )⎤⎦ P = 0 F DA = ⎡⎣cot( 2 θ ) + 1⎤⎦ ⎡⎣cos( 2 θ )⎤⎦ P

( C)

Problem 6-22 The maximum allowable tensile force in the members of the truss is Tmax, and the maximum allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads that can be applied to the truss. Units Used: 3

kN = 10 N Given: Tmax = 2 kN Cmax = 1.2 kN L = 2m

θ = 30 deg Solution: Initial guesses (assume all bars are in tension). Use a unit load for P and then scale the answer later. F BA = 1 kN

F BC = 1 kN

F CA = 1 kN

F CD = 1 kN

F DA = 1 kN

P = 1 kN

471

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Engineering Mechanics - Statics

Chapter 6

Given Joint B +

↑ Σ Fy = 0; + Σ F x = 0; →

−F BA sin ( 2 θ ) − P = 0 −F BA cos ( 2 θ ) + F BC = 0

Joint C +

↑Σ Fy = 0;

−F CA sin ( θ ) − F CD sin ( 2 θ ) = 0

+ Σ F x = 0; →

Joint D

−F BC + P − FCD cos ( 2 θ ) − FCA cos ( θ ) = 0

+ Σ F x = 0; →

−F DA + F CD cos ( 2 θ ) = 0

⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCA ⎟ = Find ( FBA , FBC , FCA , FCD , FDA) ⎜F ⎟ ⎜ CD ⎟ ⎜ FDA ⎟ ⎝ ⎠

⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ans = ⎜ FCA ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FDA ⎟ ⎝ ⎠

⎛ −1.155 ⎞ ⎜ −0.577 ⎟ ⎜ ⎟ ans = ⎜ 2.732 ⎟ kN ⎜ −1.577 ⎟ ⎜ ⎟ ⎝ −0.789 ⎠

Now find the biggest tension and the biggest compression. T = max ( ans)

T = 2.732 kN

C = min ( ans)

C = −1.577 kN

Decide which is more important and scale the answer

⎡⎛ Tmax ⎞ ⎤ ⎢⎜ ⎟ ⎥ T ⎢ ⎜ ⎟ P⎥ P = min ⎢⎜ −Cmax ⎟ ⎥ ⎢⎜ ⎟ ⎥ ⎣⎝ C ⎠ ⎦

P = 732.051 N

Problem 6-23

472

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Engineering Mechanics - Statics

Chapter 6

The Fink truss supports the loads shown. Determine the force in each member and state if the members are in tension or compression. Approximate each joint as a pin. Units Used: 3

kip = 10 lb Given: F 1 = 500 lb

a = 2.5 ft

F 2 = 1 kip

θ = 30 deg

F 3 = 1 kip Solution: Entire truss:

(

)

ΣF x = 0;

E x = F1 + F2 + F3 + F2 + F1 sin ( θ )

ΣME = 0;

− Ay 4a cos ( θ ) + F 14a + F 23a + F 32a + F 2 a = 0 Ay =

ΣF y = 0;

2 F1 + 2 F2 + F3

E x = 2000 lb

Ay = 2309.4 lb

2 cos ( θ )

E y = − Ay + 2 cos ( θ ) F1 + 2 cos ( θ ) F 2 + cos ( θ ) F3

E y = 1154.7 lb

Joint A: ΣF y = 0;

F AB =

−cos ( θ ) F 1 + Ay sin ( θ )

F AB = 3.75 kip ΣF x = 0;

(C)

F AH = −sin ( θ ) F 1 + F AB cos ( θ ) F AH = 3 kip

(T)

473

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Engineering Mechanics - Statics

Chapter 6

Joint B: ΣF x = 0;

F BC = FAB

F BC = 3.75 kip

ΣF y = 0;

F BH = F 2

F BH = 1 kip

(C)

(C)

Joint H: Σ F y = 0; ΣF x = 0;

F HC = F 2

F HC = 1 kip

(T)

F GH = −F2 cos ( −90 deg + θ ) − FHC cos ( −90 deg + θ ) + F AH F GH = 2 kip

(T)

Joint E: ΣF y = 0;

F EF =

(

− F 1 − E x sin ( θ ) − Ey cos ( θ )

F EF = 3 kip ΣF x = 0;

sin ( θ )

)

(T)

F ED = −Ey sin ( θ ) + Ex cos ( θ ) + F EF cos ( θ ) F ED = 3.75 kip

(C)

Joint D: ΣF x = 0;

F DC = F ED F DC = 3.75 kip

ΣF y = 0;

(C)

F DF = F2 F DF = 1 kip

(C)

Joint C: ΣF x = 0;

F CF = F HC F CF = 1 kip

ΣF y = 0;

(T)

F CG = F3 + FHC cos ( 90 deg − θ ) ( 2)

F CG = 2 kip

(C)

F FG = FEF − FCF cos ( 90 deg − θ ) ( 2)

F FG = 2 kip

(T)

Joint F: ΣF x = 0;

474

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Engineering Mechanics - Statics

Chapter 6

Problem 6-24 Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression.

Solution: ΣΜ A = 0; +

↑ ΣF y = 0;

P

L 2L +P − Dy L = 0 3 3

Ay + Dy − 2 P = 0

Joint F: +

1

↑ ΣF y = 0;

F FB

+ ΣF x = 0; →

F FD − F FE − F FB

−P=0

2 1

=0

2

475

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Joint E: +

↑ ΣF y = 0;

+ ΣF x = 0; →

1

F EC

−P=0

2 1

F EF − F EA + F EC

=0

2

Joint B: +

↑ ΣF y = 0;

F BA

+ ΣF x = 0; →

F BA

1 2 1 2

+ F BD

1

+ F FB

1

1

− F FB

5

=0

2

⎛ 2 ⎞=0 ⎟ ⎝ 5⎠

− F BD ⎜

2

Joint C: +

↑ ΣF y = 0;

F CA

+ ΣF x = 0; →

F CA

1 5 2 5

+ FCD

1

− FEC

1

2

2

− FEC

1

− FCD

1

=0

2 =0

2

476

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Joint A: + ΣF x = 0; →

F AE − F BA

1 2

− F CA

2

=0

5

Solving we find F EF = 0.667 P ( T ) F FD = 1.67 P ( T) F AB = 0.471 P ( C) F AE = 1.67 P ( T) F AC = 1.49 P ( C) F BF = 1.41 P ( T) F BD = 1.49 P ( C) F EC = 1.41 P ( T) F CD = 0.471 P ( C)

Problem 6-25 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The vertical component of force at C must equal zero. Why? Units Used: 3

kN = 10 N Given: F 1 = 6 kN

477

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

F 2 = 8 kN a = 1.5 m b = 2m c = 2m

Solution: Initial Guesses: F AB = 1 kN

F AE = 1 kN

F EB = 1 kN

F BC = 1 kN

F BD = 1 kN

F ED = 1 kN

Given Joint A

F AB

F AB

Joint E

a 2

2

a +c c 2

2

a +c

+ F AE = 0 − F1 = 0

F ED − F AE = 0 F EB − F 2 = 0

Joint B

F BC + F BD −F EB − F BD

b 2

2

b +c c 2

2

b +c

− F AB − F AB

a 2

=0 2

a +c c 2

=0 2

a +c

⎛ FAB ⎞ ⎜ ⎟ F AE ⎜ ⎟ ⎜F ⎟ ⎜ EB ⎟ = Find F , F , F , F , F , F ( AB AE EB BC BD ED) ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎝ FED ⎠

478

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ 7.5 ⎞ F AE ⎜ ⎟ ⎜ −4.5 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ 8 ⎜ EB ⎟ = kN ⎜ FBC ⎟ ⎜ 18.5 ⎟ ⎟ ⎜ ⎟ ⎜ − 19.799 ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎜ ⎟ ⎝ −4.5 ⎟⎠ F ⎝ ED ⎠

Positive means Tension, Negative means Compresson.

Problem 6-26 Each member of the truss is uniform and has a mass density ρ. Remove the external loads F1 and F 2 and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member. Given: F1 = 0 F2 = 0

ρ = 8

kg m

a = 1.5 m b = 2m c = 2m g = 9.81

m 2

s Solution:

Find the weights of each bar. 2

2

WAB = ρ g a + c

WBC = ρ g b

WBE = ρ g c

WAE = ρ g a

WBD = ρ g b + c

2

2

WDE = ρ g b

479

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Engineering Mechanics - Statics

Guesses

Chapter 6

F AB = 1 N

F AE = 1 N

F BC = 1 N

F BE = 1 N

F BD = 1 N

F DE = 1 N

Given Joint A

a

F AE +

2

a +c

c 2

2

a +c Joint E

F AB −

F AB = 0 WAB + WAE 2

=0

F DE − F AE = 0 F BE −

Joint B

2

F BC +

WAE + WBE + WDE 2 b 2

2

b +c

−c

F BD −

F AB − F BE − 2 2 a +c

=0

a 2

2

a +c c 2

2

b +c

F AB = 0

F BD −

WAB + WBE + WBD + WBC 2

=0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAE ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AB AE BC BD BE DE) ⎜ FBD ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ ⎝ FDE ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 196 ⎞ F ⎜ AE ⎟ ⎜ −118 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ BC 857 ⎜ ⎟= ⎜ ⎟N ⎜ FBD ⎟ −1045 ⎜ ⎟ ⎜ ⎟ 216 ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ ⎜⎝ −118 ⎟⎠ FDE ⎝ ⎠

Positive means tension, Negative means Compression.

480

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Engineering Mechanics - Statics

Chapter 6

Problem 6-27 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 4 kN P 2 = 0 kN a = 2m

θ = 15 deg Solution: Take advantage of the symetry. Initial Guesses: F BD = 1 kN

F CD = 1 kN

F CA = 1 kN

F BC = 1 kN

Given Joint D Joint B

−P 1 2

F AB = 1 kN

− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0

−P 2 cos ( 2 θ ) − F BC = 0 F BD − F AB − P 2 sin ( 2 θ ) = 0

Joint C

F CD cos ( θ ) − F CA cos ( θ ) = 0

(FCD + FCA)sin(θ ) + FBC = 0 ⎛⎜ FBD ⎞⎟ ⎜ FCD ⎟ ⎜ ⎟ F ⎜ AB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC ) ⎜F ⎟ ⎜ CA ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛⎜ FBD ⎞⎟ ⎜ FED ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ ⎟ F F = ⎜ GF ⎟ ⎜ AB ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ CA ⎟ ⎜ FFE ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎝ ⎠

481

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Engineering Mechanics - Statics

⎛⎜ FBD ⎞⎟ ⎛ −4 ⎞ ⎜ FCD ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ −4 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ CA ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ 0 ⎠ ⎝ ⎠

Chapter 6

⎛⎜ FFD ⎞⎟ ⎛ −4 ⎞ ⎜ FED ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ −4 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ EG ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎝ 0 ⎠ ⎝ ⎠

Positvive means Tension, Negative means Compression

Problem 6-28 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: P 1 = 2 kN P 2 = 4 kN a = 2m

θ = 15 deg Solution: Take advantage of the symmetry. Initial Guesses: F BD = 1 kN

F CD = 1 kN

F CA = 1 kN

F BC = 1 kN

Given Joint D Joint B

−P 1 2

F AB = 1 kN

− F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0

−P 2 cos ( 2 θ ) − F BC = 0 F BD − F AB − P 2 sin ( 2θ ) = 0

Joint C

F CD cos ( θ ) − F CA cos ( θ ) = 0

(FCD + FCA)sin(θ ) + FBC = 0 482

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ FBD ⎞⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ FAB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC ) ⎜F ⎟ ⎜ CA ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎛⎜ FBD ⎞⎟ ⎛ −11.46 ⎞ ⎜ FCD ⎟ ⎜ 6.69 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ −13.46 ⎟ kN ⎜ F ⎟ ⎜ 6.69 ⎟ ⎜ CA ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ −3.46 ⎠ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛⎜ FBD ⎞⎟ ⎜ FED ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ FAB ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ CA ⎟ ⎜ FFE ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎝ ⎠

⎛⎜ FFD ⎞⎟ ⎛ −11.46 ⎞ ⎜ FED ⎟ ⎜ 6.69 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ −13.46 ⎟ kN ⎜ F ⎟ ⎜ 6.69 ⎟ ⎜ EG ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎝ −3.46 ⎠ ⎝ ⎠

Positvive means Tension, Negative means Compression

Problem 6-29 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 2 kip F 2 = 1.5 kip F 3 = 3 kip F 4 = 3 kip a = 4 ft b = 10 ft Solution:

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses

483

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Engineering Mechanics - Statics

Chapter 6

F AB = 1 lb

F BC = 1 lb

F CD = 1 lb

F DE = 1 lb

F AI = 1 lb

F BI = 1 lb

F CI = 1 lb

F CG = 1 lb

F CF = 1 lb

F DF = 1 lb

F EF = 1 lb

F HI = 1 lb

F GI = 1 lb

F GH = 1 lb

F FG = 1 lb

Given Joint A

F AI cos ( θ ) + FAB = 0

Joint B

F BC − F AB = 0 F BI = 0

Joint C

(

)

F CD − F BC + FCF − F CI cos ( θ ) = 0

(

)

F CG + FCF + F CI sin ( θ ) = 0 Joint D

F DE − F CD = 0 F DF = 0

Joint I

(

)

F 2 + FGI + F CI − F AI cos ( θ ) = 0

(

)

F HI − FBI + FGI − F AI − FCI sin ( θ ) = 0 Joint H

F GH cos ( θ ) + F 1 = 0 −F GH sin ( θ ) − FHI = 0

Joint G

Joint F

(FFG − FGH − FGI )cos (θ ) = 0 −F 3 − F CG + ( FGH − FFG − FGI ) sin ( θ ) = 0 (FEF − FFG − FCF)cos (θ ) = 0 (FFG − FCF − FEF)sin(θ ) − F4 − FDF = 0

484

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜F ⎟ ⎜ DE ⎟ ⎜ FAI ⎟ ⎜ ⎟ ⎜ FBI ⎟ ⎜ ⎟ ⎜ FCI ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F , F , F , F , F , F , F , F , F ( AB BC CD DE AI BI CI CG CF DF EF HI GI GH ⎜ CG ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎜F ⎟ ⎜ EF ⎟ ⎜ FHI ⎟ ⎜ ⎟ ⎜ FGI ⎟ ⎜F ⎟ ⎜ GH ⎟ ⎜ FFG ⎟ ⎝ ⎠ ⎛⎜ FAB ⎞⎟ ⎛ 3.75 ⎞ ⎜ FBC ⎟ ⎜ 3.75 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCD ⎟ = ⎜ 7.75 ⎟ kip ⎜ F ⎟ ⎜ 7.75 ⎟ ⎜ DE ⎟ ⎜ ⎟ ⎜ FAI ⎟ ⎝ −4.04 ⎠ ⎝ ⎠

⎛⎜ FBI ⎞⎟ ⎛ 0 ⎞ ⎜ FCI ⎟ ⎜ 0.27 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCG ⎟ = ⎜ 1.4 ⎟ kip ⎜ F ⎟ ⎜ −4.04 ⎟ ⎜ CF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎝ 0 ⎠ ⎝ ⎠

⎛⎜ FEF ⎟⎞ ⎛ −12.12 ⎞ ⎜ FHI ⎟ ⎜ 0.8 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGI ⎟ = ⎜ −5.92 ⎟ kip ⎜ F ⎟ ⎜ −2.15 ⎟ ⎜ GH ⎟ ⎜ ⎟ ⎜ FFG ⎟ ⎝ −8.08 ⎠ ⎝ ⎠

Positive means Tension, Negative means Compression

Problem 6-30 The Howe bridge truss is subjected to the loading shown. Determine the force in members DE, EH, and HG, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

485

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 30 kN F 2 = 20 kN F 3 = 20 kN F 4 = 40 kN a = 4m b = 4m Solution: −F 2 a − F 3( 2a) − F 4( 3a) + Gy( 4a) = 0 Gy =

F2 + 2F 3 + 3F4 4

Gy = 45 kN Guesses

F DE = 1 kN

F EH = 1 kN

F HG = 1 kN

Given −F DE − F HG = 0

Gy − F 4 − F EH = 0

F DE b + Gy a = 0

⎛ FDE ⎞ ⎜ ⎟ F ⎜ EH ⎟ = Find ( FDE , FEH , FHG) ⎜F ⎟ ⎝ HG ⎠

⎛ FDE ⎞ ⎛ −45 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ EH ⎟ = ⎜ 5 ⎟ kN ⎜ F ⎟ ⎝ 45 ⎠ ⎝ HG ⎠

Positive (T) Negative (C)

Problem 6-31 The Pratt bridge truss is subjected to the loading shown. Determine the force in members LD, LK, CD, and KD, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

486

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 50 kN F 2 = 50 kN F 3 = 50 kN a = 4m b = 3m Solution: Ax = 0 Ay =

3F 3 + 4F2 + 5F 1 6

Guesses F LD = 1 kN

F LK = 1 kN

F CD = 1 kN

F KD = 1 kN

Given F 2 b + F 1( 2b) − Ay( 3b) − F LK a = 0 F CD a + F1 b − Ay( 2b) = 0 Ay − F1 − F2 −

a ⎞ ⎛ ⎜ 2 2 ⎟ FLD = 0 ⎝ a +b ⎠

−F 3 − F KD = 0

⎛ FLD ⎞ ⎜ ⎟ ⎜ FLK ⎟ ⎜ ⎟ = Find ( FLD , FLK , FCD , FKD) ⎜ FCD ⎟ ⎜F ⎟ ⎝ KD ⎠

⎛ FLD ⎞ ⎜ ⎟ ⎛⎜ 0 ⎟⎞ F ⎜ LK ⎟ ⎜ −112.5 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCD ⎟ ⎜ 112.5 ⎟ ⎜ F ⎟ ⎝ −50 ⎠ ⎝ KD ⎠

Positive (T) Negative (C)

487

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Engineering Mechanics - Statics

Chapter 6

Problem 6-32 The Pratt bridge truss is subjected to the loading shown. Determine the force in members JI, JE, and DE, and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 50 kN F 2 = 50 kN F 3 = 50 kN a = 4m b = 3m Solution: Initial Guesses Gy = 1 kN F JI = 1 kN F JE = 1 kN F DE = 1 kN Given Entire Truss −F 1 b − F2( 2b) − F3( 3b) + Gy( 6b) = 0 Section −F DE − F JI = 0

F JE + Gy = 0

Gy( 2b) − F DE a = 0 488

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Engineering Mechanics - Statics

Chapter 6

⎛ Gy ⎞ ⎜ ⎟ ⎜ FJI ⎟ ⎜ ⎟ = Find ( Gy , FJI , FJE , FDE) Gy = 50 kN F JE ⎜ ⎟ ⎜F ⎟ ⎝ DE ⎠

⎛ FJI ⎞ ⎛ −75 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JE ⎟ = ⎜ −50 ⎟ kN ⎜ F ⎟ ⎝ 75 ⎠ ⎝ DE ⎠ Positive means Tension, Negative means Compression

Problem 6-33 The roof truss supports the vertical loading shown. Determine the force in members BC, CK, and KJ and state if these members are in tension or compression.

Units Used: 3

kN = 10 N Given: F 1 = 4 kN F 2 = 8 kN a = 2m b = 3m

489

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Engineering Mechanics - Statics

Chapter 6

Solution: Initial Guesses Ax = 1 kN

Ay = 1 kN

F BC = 1 kN

F CK = 1 kN

F KJ = 1 kN Given Ax = 0 F 2( 3a) + F 1( 4a) − Ay( 6a) = 0

⎛ 2b ⎞ ⎟ + Ax⎜ ⎟ − Ay( 2a) = 0 ⎝3⎠ ⎝3⎠

F KJ ⎛⎜

2b ⎞

F KJ + A x +

3a ⎞F = 0 ⎛ BC ⎜ 2 ⎟ 2 ⎝ b + 9a ⎠

F CK + A y +

b ⎞F = 0 ⎛ BC ⎜ 2 2⎟ ⎝ b + 9a ⎠

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ F ⎜ KJ ⎟ = Find ( Ax , Ay , FKJ , FCK , FBC ) ⎜F ⎟ ⎜ CK ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 6.667 ⎟ ⎜ ⎟ ⎜ ⎟ F 13.333 = ⎜ KJ ⎟ ⎜ ⎟ kN Positive (T) Negative (C) ⎜F ⎟ ⎜ 0 ⎟ CK ⎜ ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ −14.907 ⎠ ⎝ ⎠

Problem 6-34 Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the deck of a bridge. State if these members are in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 4000 lb F 2 = 8000 lb

490

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Engineering Mechanics - Statics

Chapter 6

F 3 = 5000 lb a = 9 ft b = 12 ft

Solution Initial Guesses:

F DJ = 1 kip

Ay = 1 kip

F CD = 1 kip

F CJ = 1 kip

F KJ = 1 kip

Given F 3 a + F 2( 4a) + F 1( 5a) − Ay( 6a) = 0 − Ay( 2a) + F 1 a + FKJ b = 0 F CD + F KJ +

a ⎛ ⎞ ⎜ 2 2 ⎟ FCJ = 0 ⎝ a +b ⎠

Ay − F1 − F2 −

b ⎛ ⎞ ⎜ 2 2 ⎟ FCJ = 0 ⎝ a +b ⎠

−F DJ = 0

⎛⎜ Ay ⎞⎟ ⎜ FKJ ⎟ ⎜ ⎟ ⎜ FCJ ⎟ = Find ( Ay , FKJ , FCJ , FDJ , FCD) ⎜F ⎟ ⎜ DJ ⎟ ⎜ FCD ⎟ ⎝ ⎠

⎛⎜ Ay ⎞⎟ ⎛ 9.5 ⎞ ⎜ FKJ ⎟ ⎜ 11.25 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCJ ⎟ = ⎜ −3.125 ⎟ kip ⎜F ⎟ ⎜ 0 ⎟ ⎜ DJ ⎟ ⎜ ⎟ − 9.375 ⎝ ⎠ ⎜ FCD ⎟ ⎝ ⎠

Positive (T) Negative (C)

491

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Engineering Mechanics - Statics

Chapter 6

Problem 6-35 Determine the force in members EI and JI of the truss which serves to support the deck of a bridge. State if these members are in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 4000 lb F 2 = 8000 lb F 3 = 5000 lb a = 9 ft b = 12 ft Solution: Initial Guesses: Gy = 1 kip

F EI = 1 kip

F JI = 1 kip

Given −F 1 a − F22a − F35a + Gy6a = 0 Gy2a − F 3 a − FJI b = 0 F EI − F3 + Gy = 0

⎛ Gy ⎞ ⎜ ⎟ F ⎜ JI ⎟ = Find ( Gy , FJI , FEI ) ⎜F ⎟ ⎝ EI ⎠

⎛ Gy ⎞ ⎛ 7.5 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JI ⎟ = ⎜ 7.5 ⎟ kip ⎜ F ⎟ ⎝ −2.5 ⎠ ⎝ EI ⎠

Positive (T) Negative (C)

Problem 6-36 Determine the force in members BE, EF, and CB, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

492

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 5 kN

F 4 = 10 kN

F 2 = 10 kN

a = 4m

F 3 = 5 kN

b = 4m

a θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ b⎠

Inital Guesses F CB = 1 kN F BE = 1 kN

F EF = 1 kN

Given F 1 + F 2 − F BE cos ( θ ) = 0 −F CB − FEF − FBE sin ( θ ) − F3 = 0 −F 1 a + F CB b = 0

⎛ FCB ⎞ ⎜ ⎟ F ⎜ BE ⎟ = Find ( FCB , FBE , FEF) ⎜F ⎟ ⎝ EF ⎠ ⎛ FCB ⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FBE ⎟ = ⎜ 21.2 ⎟ kN ⎜ F ⎟ ⎝ −25 ⎠ ⎝ EF ⎠

Positive (T) Negative (C)

Problem 6-37 Determine the force in members BF, BG, and AB, and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 5 kN

F 4 = 10 kN

493

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Engineering Mechanics - Statics

F 2 = 10 kN

a = 4m

F 3 = 5 kN

b = 4m

Solution:

Chapter 6

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Inital Guesses F AB = 1 kN F BG = 1 kN

F BF = 1 kN

Given F 1 + F 2 + F 4 + F BG cos ( θ ) = 0 −F 1 3a − F 22a − F 4 a + FAB b = 0 −F BF = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FBG ⎟ = Find ( FAB , FBG , FBF) ⎜F ⎟ ⎝ BF ⎠ ⎛ FAB ⎞ ⎛ 45 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ BG ⎟ = ⎜ −35.4 ⎟ kN Positive (T) Negative (C) ⎜F ⎟ ⎝ 0 ⎠ BF ⎝ ⎠

Problem 6-38 Determine the force developed in members GB and GF of the bridge truss and state if these members are in tension or compression. Given: F 1 = 600 lb

494

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Engineering Mechanics - Statics

Chapter 6

F 2 = 800 lb a = 10 ft b = 10 ft c = 4 ft Solution: Initial Guesses Ax = 1 lb

Ay = 1 lb

F GB = 1 lb

F GF = 1 lb

Given F 2 b + F1( b + 2c) − A y2( b + c) = 0 Ax = 0 Ay − F GB = 0 − Ay b − FGF a = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , FGB , FGF) ⎜ FGB ⎟ ⎜F ⎟ ⎝ GF ⎠ ⎛ Ax ⎞ ⎞ ⎜ ⎟ ⎛⎜ 0 ⎟ ⎜ Ay ⎟ ⎜ 671.429 ⎟ lb Positive (T) ⎜ ⎟=⎜ FGB 671.429 ⎟ Negative (C) ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ −671.429 ⎠ ⎝ GF ⎠

Problem 6-39 Determine the force members BC, FC, and FE, and state if the members are in tension or compression. Units Used: 3

kN = 10 N

495

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 6 kN F 2 = 6 kN a = 3m b = 3m Solution:

a θ = atan ⎛⎜ ⎟⎞

⎝ b⎠

Initial Guesses Dy = 1 kN

F BC = 1 kN

F FC = 1 kN

F FE = 1 kN

Given −F 1 b − F2( 2b) + Dy( 3b) = 0 Dy b − FFE cos ( θ ) a = 0

(

)

−F FC − FBC + FFE cos ( θ ) = 0

(

)

−F 2 + Dy + FFE + FBC sin ( θ ) = 0

⎛ Dy ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( Dy , FBC , FFC , FFE) ⎜ FFC ⎟ ⎜F ⎟ ⎝ FE ⎠ ⎛ Dy ⎞ ⎜ ⎟ ⎛⎜ 6 ⎟⎞ F ⎜ BC ⎟ ⎜ −8.49 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FFC ⎟ ⎜ 0 ⎟ ⎜ F ⎟ ⎝ 8.49 ⎠ ⎝ FE ⎠

Positive (T) Negative (C)

496

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Engineering Mechanics - Statics

Chapter 6

Problem 6-40 Determine the force in members IC and CG of the truss and state if these members are in tension or compression. Also, indicate all zero-force members. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN a = 1.5 m b = 2m Solution: By inspection of joints B, D, H and I. AB, BC, CD, DE, HI, and GI are all zero-force members. Guesses

Ay = 1 kN

Given

− Ay( 4a) + F 1( 2a) + F 2 a = 0 − Ay( 2a) − −a 2

2

a +b −b 2

2

a +b

F IC = 1 kN

b 2

2

a +b

FIC +

FIC −

FIC a −

a 2

a +b

2

b 2

a +b

2

⎛ Ay ⎞ ⎜ ⎟ ⎜ FIC ⎟ ⎜ ⎟ = Find ( Ay , FIC , FCG , FCJ ) ⎜ FCG ⎟ ⎜F ⎟ ⎝ CJ ⎠

F CG = 1 kN

a 2

2

a +b

F CJ = 1 kN

FIC b = 0

F CJ = 0

F CJ − FCG = 0

⎛ Ay ⎞ ⎜ ⎟ ⎛⎜ 4.5 ⎟⎞ F ⎜ IC ⎟ ⎜ −5.625 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCG ⎟ ⎜ 9 ⎟ ⎜ F ⎟ ⎝ −5.625 ⎠ ⎝ CJ ⎠

Positive (T) Negative (C)

Problem 6-41 Determine the force in members JE and GF of the truss and state if these members are in tension or compression. Also, indicate all zero-force members. 497

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Engineering Mechanics - Statics

Chapter 6

Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN a = 1.5 m b = 2m Solution: By inspection of joints B, D, H and I. AB, BC, CD, DE, HI, and GI are all zero-force members. E y = 1 kN

Guesses Given

F JE = 1 kN

F GF = 1 kN

−F 1( 2a) − F 2( 3a) + E y( 4a) = 0 Ey +

b 2

a +b

−a 2

2

a +b

2

F JE = 0

FJE − FGF = 0

⎛ Ey ⎞ ⎜ ⎟ F ⎜ JE ⎟ = Find ( Ey , FJE , FGF) ⎜F ⎟ ⎝ GF ⎠

⎛ Ey ⎞ ⎛ 7.5 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ JE ⎟ = ⎜ −9.375 ⎟ kN Positive (T) Negative (C) ⎜ F ⎟ ⎝ 5.625 ⎠ ⎝ GF ⎠

Problem 6-42 Determine the force in members BC, HC, and HG. After the truss is sectioned use a single equation of equilibrium for the calculation of each force. State if these members are in tension or compression. Units Used: 3

kN = 10 N

498

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 2 kN F 4 = 5 kN a = 5 m F 2 = 4 kN F 5 = 3 kN b = 2 m F 3 = 4 kN

c = 3m

Solution: Guesses Ax = 1 kN

Ay = 1 kN

F BC = 1 kN

F HC = 1 kN

F HG = 1 kN

d = 1m

Given c b = a+d a

− Ax = 0

(F1 − Ay)( 4a) + F2( 3a) + F3( 2a) + F4( a) = 0 (F1 − Ay)( a) + Ax( c) − FBC ( c) = 0 (F1 − Ay)( 2a) + F2( a) + ( A y − F 1) ( d) − F 2( a + d ) +

a 2

a +b

2

F HG( c) +

c 2

2

a +c

b 2

a +b

FHC ( a + d) +

⎛ Ay ⎞ ⎜ ⎟ ⎜ Ax ⎟ ⎜ ⎟ FBC ⎜ ⎟ = Find A , A , F , F , F , d ( y x BC HC HG ) ⎜F ⎟ HC ⎜ ⎟ ⎜ FHG ⎟ ⎜ ⎟ ⎝ d ⎠

2

F HG( a) = 0 a 2

2

a +c

F HC ( c) = 0

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ = kN ⎜ Ay ⎟ ⎜⎝ 8.25 ⎟⎠ ⎝ ⎠ d = 2.5 m

⎛ FBC ⎞ ⎛ −10.417 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FHC ⎟ = ⎜ 2.235 ⎟ kN ⎜ F ⎟ ⎝ 9.155 ⎠ ⎝ HG ⎠ Positive (T) Negative (C)

Problem 6-43 Determine the force in members CD, CF, and CG and state if these members are in tension or 499

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Engineering Mechanics - Statics

Chapter 6

compression. Units Used: 3

kN = 10 N Given: F 1 = 2 kN F 4 = 5 kN a = 5 m F 2 = 4 kN F 5 = 3 kN b = 2 m F 3 = 4 kN

c = 3m

Solution: Guesses E y = 1 kN

F CD = 1 kN

F CF = 1 kN

F CG = 1 kN

F FG = 1 kN

F GH = 1 kN

Given

(

)

−F 2( a) − F3( 2a) − F4( 3a) + E y − F5 ( 4a) = 0

(

)

(

)

F CD( c) + Ey − F 5 ( a) = 0

a

−F 4( a) − F 5 − E y ( 2a) −

a 2

2

a +b b 2

2

a +b

(F5 − Ey)

FFG −

a 2

a +b

2

2

2

a +b

FFG( b + c) = 0

F GH = 0

(FFG + FGH ) + FCG = 0 a( c − b) b

+ F4⎡⎢a +

⎣

a( c − b) ⎤ b

⎥− ⎦

c 2

2

a +c

F CF⎡⎢2 a +

⎣

a( c − b) ⎤ b

⎥=0 ⎦

500

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Engineering Mechanics - Statics

Chapter 6

⎛ Ey ⎞ ⎜ ⎟ F CD ⎜ ⎟ ⎜F ⎟ ⎜ CF ⎟ = Find E , F , F , F , F , F ( y CD CF CG FG GH ) ⎜ FCG ⎟ ⎜ ⎟ ⎜ FFG ⎟ ⎜ ⎟ ⎝ FGH ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛ 9.75 ⎞ F CD ⎜ ⎟ ⎜ −11.25 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ CF ⎟ = 3.207 ⎟ kN Positive (T) ⎜ FCG ⎟ ⎜ −6.8 ⎟ Negative (C) ⎟ ⎜ ⎟ ⎜ ⎜ FFG ⎟ ⎜ 9.155 ⎟ ⎜ ⎟ ⎜⎝ 9.155 ⎟⎠ F ⎝ GH ⎠

Problem 6-44 Determine the force in members OE, LE, and LK of the Baltimore truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 2 kN

a = 2m

F 2 = 2 kN

b = 2m

F 3 = 5 kN F 4 = 3 kN Solution: Ax = 0 kN Initial Guesses Ay = 1 kN

F OE = 1 kN

F DE = 1 kN F LK = 1 kN F LE = 1 kN Given F LE = 0 F 4( 3b) + F 3( 4b) + F 2( 5b) + F 1( 6b) − Ay( 8b) = 0 F LK + FDE + FOE

b 2

a +b

=0 2

501

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Engineering Mechanics - Statics

Ay − F 1 − F 2 − F OE

Chapter 6

a 2

=0 2

a +b

−F LK ( 2a) + F2( b) + F 1( 2b) − Ay( 4b) = 0

⎛⎜ Ay ⎞⎟ ⎜ FOE ⎟ ⎜ ⎟ ⎜ FDE ⎟ = Find ( Ay , FOE , FDE , FLK , FLE) ⎜F ⎟ ⎜ LK ⎟ ⎜ FLE ⎟ ⎝ ⎠ ⎛⎜ Ay ⎞⎟ ⎛ 6.375 ⎞ = kN ⎜ FDE ⎟ ⎜⎝ 7.375 ⎟⎠ ⎝ ⎠

⎛ FOE ⎞ ⎛ 3.36 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FLE ⎟ = ⎜ 0 ⎟ kN ⎜ F ⎟ ⎝ −9.75 ⎠ ⎝ LK ⎠

Positive (T) Negative (C)

Problem 6-45 Determine the force in member GJ of the truss and state if this member is in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 1000 lb F 2 = 1000 lb F 3 = 1000 lb F 4 = 1000 lb a = 10 ft

θ = 30 deg

502

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= Chapter 6

Engineering Mechanics - Statics

Solution: Guess

E y = 1 lb

F GJ = 1 lb

Given −F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = 0 −F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = 0

⎛⎜ Ey ⎟⎞ = Find ( Ey , F GJ ) ⎜ FGJ ⎟ ⎝ ⎠

⎛⎜ Ey ⎟⎞ ⎛ 1.5 ⎞ = kip ⎜ FGJ ⎟ ⎜⎝ −2 ⎟⎠ ⎝ ⎠

Positive (T) Negative (C)

Problem 6-46 Determine the force in member GC of the truss and state if this member is in tension or compression. Units Used: 3

kip = 10 lb Given: F 1 = 1000 lb F 2 = 1000 lb F 3 = 1000 lb F 4 = 1000 lb a = 10 ft

θ = 30 deg Solution: Guess

E y = 1 lb

F GJ = 1 lb

F HG = 1 lb

F GC = 1 lb

Given −F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = 0 −F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = 0 503

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Engineering Mechanics - Statics

Chapter 6

−F HG cos ( θ ) + FGJ cos ( θ ) = 0

(

)

−F 3 − F GC − F HG + FGJ sin ( θ ) = 0

⎛ Ey ⎞ ⎜ ⎟ ⎜ FGJ ⎟ ⎜ ⎟ = Find ( Ey , FGJ , FGC , FHG) F GC ⎜ ⎟ ⎜F ⎟ ⎝ HG ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛⎜ 1.5 ⎟⎞ ⎜ FGJ ⎟ ⎜ −2 ⎟ ⎜ ⎟ = ⎜ ⎟ kip F GC ⎜ ⎟ ⎜ 1 ⎟ ⎜ F ⎟ ⎝ −2 ⎠ ⎝ HG ⎠

Positive (T) Negative (C)

Problem 6-47 Determine the force in members KJ, JN, and CD, and state if the members are in tension or compression. Also indicate all zero-force members. Units Used: 3

kip = 10 lb Given: F = 3 kip a = 20 ft b = 30 ft c = 20 ft Solution:

Ax = 0 2c ⎞ ⎟ ⎝ 2a + b ⎠

θ = atan ⎛⎜

2c ⎞ ⎟ ⎝b⎠

φ = atan ⎛⎜

Initial Guesses: Ay = 1 lb

F CD = 1 lb

F KJ = 1 lb

F JN = 1 lb

504

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Engineering Mechanics - Statics

Chapter 6

Given

⎛ ⎝

F⎜a +

b⎞ ⎟ − Ay( 2a + b) = 0 2⎠

⎛ b⎞ F CD c − Ay⎜ a + ⎟ = 0 ⎝ 2⎠ F CD + F JN cos ( φ ) + F KJ cos ( θ ) = 0 Ay + F JN sin ( φ ) + FKJ sin ( θ ) = 0

⎛ Ay ⎞ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ = Find ( Ay , FCD , FJN , FKJ ) ⎜ FJN ⎟ ⎜F ⎟ ⎝ KJ ⎠

Ay = 1.5 kip

⎛ FCD ⎞ ⎛ 2.625 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FJN ⎟ = ⎜ 0 ⎟ kip ⎜ F ⎟ ⎝ −3.023 ⎠ ⎝ KJ ⎠ Positive (T), Negative (C)

Problem 6-48 Determine the force in members BG, HG, and BC of the truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 7 kN F 3 = 4 kN a = 3m b = 3m c = 4.5 m

505

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Engineering Mechanics - Statics

Chapter 6

Solution:

Initial Guesses

F BG = 1 kN

Ax = 1 kN

F HG = 1 kN

Ay = 1 kN

F BC = 1 kN

Given − Ax = 0 − Ay( a) −

a ⎡ ⎤ F ( b) = 0 HG ⎢ 2 2⎥ ⎣ ( c − b) + a ⎦

F 3( a) + F2( 2a) + F1( 3a) − A y( 4a) = 0 F BC +

a ⎡ ⎤F + ⎛ a ⎞F − A = 0 HG ⎜ BG x ⎢ 2 2⎥ 2 2⎟ ⎣ ( c − b) + a ⎦ ⎝ a +c ⎠

Ay − F1 +

c−b c ⎡ ⎤F + ⎛ ⎞F = 0 HG ⎜ BG ⎢ ⎥ ⎟ 2 2 2 2 ⎣ ( c − b) + a ⎦ ⎝ a +c ⎠ 506

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ FHG ⎟ = Find ( Ax , Ay , FHG , FBG , FBC ) ⎜F ⎟ ⎜ BG ⎟ ⎜ FBC ⎟ ⎝ ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 9 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FHG ⎟ = ⎜ −10.062 ⎟ kN ⎜ F ⎟ ⎜ 1.803 ⎟ ⎜ BG ⎟ ⎜ ⎟ 8 ⎝ ⎠ ⎜ FBC ⎟ ⎝ ⎠

Positive (T) Negative (C)

Problem 6-49 The skewed truss carries the load shown. Determine the force in members CB, BE, and EF and state if these members are in tension or compression. Assume that all joints are pinned.

Solution: ΣMB = 0;

−P d − F EF d = 0

ΣME = 0;

−P d +

+ Σ F x = 0; →

2

F CB d = 0

F CB =

FCB − F BE = 0

P BE =

5 P−

1 5

F EF = −P 5 P 2 P 2

F EF = P

( C)

F CB = 1.12P

( T)

F BE = 0.5P

( T)

Problem 6-50 The skewed truss carries the load shown. Determine the force in members AB, BF, and EF and state if these members are in tension or compression. Assume that all joints are pinned. 507

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Engineering Mechanics - Statics

Chapter 6

Solution: ΣMF = 0;

−P 2d + P d + FAB d = 0

F AB = P

F AB = P

( T)

ΣMB = 0;

−P d − F EF d = 0

F EF = −P

F EF = P

( C)

F BE = − 2 P

F BF = 1.41P

( C)

+ Σ F x = 0; →

P + FBF

1

=0

2

Problem 6-51 Determine the force developed in members BC and CH of the roof truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 1.5 kN F 2 = 2 kN

508

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Engineering Mechanics - Statics

Chapter 6

a = 1.5 m b = 1m c = 2m d = 0.8 m Solution: a θ = atan ⎛⎜ ⎟⎞

⎝c⎠

a ⎞ ⎟ ⎝ c − b⎠

φ = atan ⎛⎜

Initial Guesses: E y = 1 kN

F BC = 1 kN

F CH = 1 kN

Given −F 2( d) − F1( c) + Ey( 2c) = 0 F BC sin ( θ ) ( c) + FCH sin ( φ ) ( c − b) + E y( c) = 0 −F BC sin ( θ ) − FCH sin ( φ ) − F1 + Ey = 0

⎛ Ey ⎞ ⎜ ⎟ ⎜ FBC ⎟ = Find ( Ey , FBC , FCH ) Ey = 1.15 kN ⎜F ⎟ ⎝ CH ⎠

⎛⎜ FBC ⎞⎟ ⎛ −3.25 ⎞ = kN ⎜ FCH ⎟ ⎜⎝ 1.923 ⎟⎠ ⎝ ⎠

Positive (T) Negative (C)

Problem 6-52 Determine the force in members CD and GF of the truss and state if the members are in tension or compression. Also indicate all zero-force members. Units Used: 3

kN = 10 N Given: F 1 = 1.5 kN F 2 = 2 kN 509

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Engineering Mechanics - Statics

Chapter 6

a = 1.5 m b = 1m c = 2m d = 0.8 m Solution: a θ = atan ⎛⎜ ⎟⎞

⎝c⎠

⎞ ⎟ ⎝ c − b⎠

φ = atan ⎛⎜

a

Initial Guesses: E y = 1 kN

F CD = 1 kN

F GF = 1 kN

Given −F 2( d) − F1( c) + Ey( 2c) = 0 E y( b) + F CD sin ( θ ) ( b) = 0 E y( c) − FGF( a) = 0

⎛ Ey ⎞ ⎜ ⎟ F ⎜ CD ⎟ = Find ( Ey , FCD , FGF) Ey = 1.15 kN ⎜F ⎟ ⎝ GF ⎠

⎛⎜ FCD ⎞⎟ ⎛ −1.917 ⎞ Positive (T) = kN Negative (C) ⎜ FGF ⎟ ⎜⎝ 1.533 ⎟⎠ ⎝ ⎠ DF and CF are zero force members.

Problem 6-53 Determine the force in members DE, DL, and ML of the roof truss and state if the members are in tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 12 kN F 3 = 12 kN

510

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Engineering Mechanics - Statics

Chapter 6

F 4 = 12 kN a = 4m b = 3m c = 6m Solution:

θ = atan ⎛⎜

c − b⎞

⎟ ⎝ 3a ⎠

⎡b + ⎢ φ = atan ⎢ ⎣

2 ⎤ ( c − b) ⎥ 3 a

⎥ ⎦

Initial Guesses: Ay = 1 kN

F ML = 1 kN

F DL = 1 kN

F DE = 1 kN

Given F 2( a) + F3( 2a) + F4( 3a) + F3( 4a) + F2( 5a) + F1( 6a) − A y( 6a) = 0 2 F 1( 2a) + F 2( a) − A y( 2a) + FML⎡⎢b + ( c − b)⎤⎥ = 0 ⎣ 3 ⎦ Ay − F 1 − F 2 − F 3 + F DE sin ( θ ) − FDL sin ( φ ) = 0 F ML + F DL cos ( φ ) + FDE cos ( θ ) = 0

⎛ Ay ⎞ ⎜ ⎟ ⎜ FML ⎟ ⎜ ⎟ = Find ( Ay , FML , FDE , FDL) ⎜ FDE ⎟ ⎜F ⎟ ⎝ DL ⎠

Ay = 36 kN

⎛ FML ⎞ ⎛ 38.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −37.1 ⎟ kN ⎜ F ⎟ ⎝ −3.8 ⎠ ⎝ DL ⎠ Positive (T), Negative (C)

Problem 6-54 Determine the force in members EF and EL of the roof truss and state if the members are in 511

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

tension or compression. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 12 kN F 3 = 12 kN F 4 = 12 kN a = 4m b = 3m c = 6m Solution:

θ = atan ⎛⎜

c − b⎞

⎟ ⎝ 3a ⎠

Initial Guesses: Iy = 1 kN

F EF = 1 kN

F EL = 1 kN Given −F 2( a) − F3( 2a) − F4( 3a) − F3( 4a) − F2( 5a) − F1( 6a) + Iy( 6a) = 0 −F 3( a) − F2( 2a) − F1( 3a) + Iy( 3a) + F EF cos ( θ ) ( c) = 0 −F 4 − F EL − 2F EF sin ( θ ) = 0

⎛ Iy ⎞ ⎜ ⎟ ⎜ FEF ⎟ = Find ( Iy , FEF , FEL) ⎜F ⎟ ⎝ EL ⎠

Iy = 36 kN

⎛⎜ FEF ⎞⎟ ⎛ −37.108 ⎞ = kN ⎜ FEL ⎟ ⎜⎝ 6 ⎟⎠ ⎝ ⎠

Positive (T) Negative (C)

Problem 6-55 Two space trusses are used to equally support the uniform sign of mass M. Determine the force developed in members AB, AC, and BC of truss ABCD and state if the members are in tension or compression. Horizontal short links support the truss at joints B and D and there is a ball-and512

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Engineering Mechanics - Statics

Chapter 6

p socket joint at C.

pp

j

Given: M = 50 kg

g = 9.81

m 2

s

a = 0.25 m b = 0.5 m c = 2m Solution:

h =

2

2

b −a

⎛ −a ⎞ AB = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠

⎛a⎞ AD = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠

⎛ 2a ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛a⎞ BC = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −h ⎠

⎛0⎞ AC = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 N

F AD = 1 N

F AC = 1 N

F BC = 1 N

F BD = 1 N

By = 1 N

Given

⎛⎜ 0 ⎟⎞ 0 ⎟ AB AD AC F AB + FAD + FAC +⎜ =0 ⎜ −M g ⎟ AB AD AC ⎜ 2 ⎟ ⎝ ⎠ ⎛⎜ 0 ⎟⎞ F AB + FBD + F BC + ⎜ −B y ⎟ = 0 AB BD BC ⎜ 0 ⎟ ⎝ ⎠ −AB

BD

BC

513

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ F AD ⎜ ⎟ ⎜F ⎟ ⎜ AC ⎟ = Find F , F , F , F , F , B ( AB AD AC BC BD y) ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎝ By ⎠

⎛ By ⎞ ⎛ 566 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AD ⎟ = ⎜ 584 ⎟ N ⎜F ⎟ ⎝ 0 ⎠ ⎝ BD ⎠ ⎛ FAB ⎞ ⎛ 584 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ −1133 ⎟ N ⎜ F ⎟ ⎝ −142 ⎠ ⎝ BC ⎠ Positive (T), Negative (C)

Problem 6-56 Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at B, C, and D. Given: F = 600 N a = 3m b = 1m c = 1.5 m Solution:

⎛b⎞ ⎜ ⎟ AB = −c ⎜ ⎟ ⎝ −a ⎠ ⎛0⎞ ⎜ ⎟ AC = c ⎜ ⎟ ⎝ −a ⎠ ⎛ −b ⎞ ⎜ ⎟ AD = −c ⎜ ⎟ ⎝ −a ⎠

⎛ −b ⎞ ⎜ ⎟ CD = −2c ⎜ ⎟ ⎝ 0 ⎠

514

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Engineering Mechanics - Statics

⎛ b ⎞ CB = ⎜ −2c ⎟ ⎜ ⎟ ⎝ 0 ⎠

Chapter 6

⎛ −2b ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝ 0 ⎠

Guesses F BA = 1 N

F BC = 1 N

F CA = 1 N

F DA = 1 N

F BD = 1 N

F DC = 1 N

By = 1 N

Bz = 1 N

Cz = 1 N

Given

⎛ 0 ⎞ F BA + FCA + F DA +⎜ 0 ⎟ =0 ⎜ ⎟ AB AC AD ⎝ −F ⎠ AB

AC

AD

⎛⎜ 0 ⎟⎞ F CA + FDC + F BC +⎜0 ⎟ =0 AC CD CB ⎜ Cz ⎟ ⎝ ⎠ −AC

CD

CB

⎛0⎞ ⎜ ⎟ F BC + FBD + F BA + ⎜ By ⎟ = 0 CB BD AB ⎜B ⎟ ⎝ z⎠ −CB

BD

−AB

⎛ FBA ⎞ ⎜ ⎟ F BC ⎜ ⎟ ⎜F ⎟ ⎜ CA ⎟ ⎜ FDA ⎟ ⎜ ⎟ F ⎜ BD ⎟ = Find ( FBA , FBC , FCA , FDA , FBD , FDC , By , Bz , Cz) ⎜F ⎟ ⎜ DC ⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Bz ⎟ ⎟ ⎜ ⎝ Cz ⎠

515

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Engineering Mechanics - Statics

⎛ By ⎞ ⎛ 1.421 × 10− 14 ⎞ ⎜ ⎟ ⎜ ⎟ B = ⎜ z⎟ ⎜ ⎟N 150 ⎜C ⎟ ⎜ ⎟ 300 ⎠ ⎝ z⎠ ⎝

Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ −175 ⎞ F BC ⎜ ⎟ ⎜ 79.1 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ CA ⎟ = −335.4 ⎟ N ⎜ FDA ⎟ ⎜ −175 ⎟ ⎟ ⎜ ⎟ ⎜ 25 ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎜ ⎟ ⎝ 79.1 ⎟⎠ F ⎝ DC ⎠

Positive (T), Negative (C)

Problem 6-57 Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at A, B, and C. Given: a = 4 ft b = 2 ft c = 3 ft d = 2 ft e = 8 ft

⎛ 0 ⎞ ⎜ ⎟ F = 500 lb ⎜ ⎟ ⎝ 0 ⎠ Solution:

⎛ −c ⎞ ⎜ ⎟ AD = a ⎜ ⎟ ⎝e⎠

⎛ −c ⎞ ⎜ ⎟ BD = −b ⎜ ⎟ ⎝e ⎠

⎛d⎞ ⎜ ⎟ CD = −b ⎜ ⎟ ⎝e ⎠

⎛ 0 ⎞ ⎜ ⎟ AB = a + b ⎜ ⎟ ⎝ 0 ⎠

⎛ −c − d ⎞ ⎛ −c − d ⎞ ⎜ ⎟ ⎜ 0 ⎟ AC = a + b BC = ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ Guesses

516

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Engineering Mechanics - Statics

Chapter 6

F BA = 1 lb

F BC = 1 lb

F BD = 1 lb

F AD = 1 lb

F AC = 1 lb

F CD = 1 lb

Ay = 1 lb

Az = 1 lb

B x = 1 lb

B z = 1 lb

Cy = 1 lb

Cz = 1 lb

Given F + FAD

−AD AD

+ FBD

−BD BD

+ F CD

−CD CD

=0

⎛0 ⎞ ⎜ ⎟ F AD + F BA + FAC + ⎜ Ay ⎟ = 0 AD AB AC ⎜A ⎟ ⎝ z⎠ AD

AB

AC

⎛ Bx ⎞ ⎜ ⎟ −AB BC BD F BA + FBC + F BD +⎜ 0 ⎟ =0 AB BC BD ⎜B ⎟ ⎝ z⎠ ⎛0⎞ CD −AC −BC ⎜ C ⎟ F CD + F AC + F BC + ⎜ y⎟ = 0 CD AC BC ⎜C ⎟ ⎝ z⎠ ⎛⎜ FBA ⎞⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜F ⎟ ⎜ AD ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ = Find ( FBA , FBC , FBD , FAD , FAC , FCD , Ay , Az , Bx , Bz , Cy , Cz) ⎜ Ay ⎟ ⎜ A ⎟ ⎜ z ⎟ ⎜ Bx ⎟ ⎜ ⎟ ⎜ Bz ⎟ ⎜ C ⎟ ⎜ y ⎟ ⎜ Cz ⎟ ⎝ ⎠

517

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Engineering Mechanics - Statics

⎛ Ay ⎞ ⎜ ⎟ ⎛ −200 ⎞ ⎜ Az ⎟ ⎜ −667 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ x ⎟ = 0 ⎟ lb ⎜ Bz ⎟ ⎜ 667 ⎟ ⎟ ⎜ ⎟ ⎜ − 300 ⎜ ⎟ ⎜ Cy ⎟ ⎜ ⎜ ⎟ ⎝ 0 ⎟⎠ ⎝ Cz ⎠

Chapter 6

⎛ FBA ⎞ ⎜ ⎟ ⎛ 167 ⎞ F BC ⎜ ⎟ ⎜ 250 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ BD ⎟ = −731 ⎟ lb ⎜ FAD ⎟ ⎜ 786 ⎟ ⎟ ⎜ ⎟ ⎜ − 391 ⎜ ⎟ ⎜ FAC ⎟ ⎜ ⎜ ⎟ ⎝ 0 ⎟⎠ F ⎝ CD ⎠

Positive (T) Negative (C)

Problem 6-58 The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Given:

⎛ 0 ⎞ ⎜ 0 ⎟ lb F1 = ⎜ ⎟ ⎝ −500 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F 2 = 400 lb ⎜ ⎟ ⎝ 0 ⎠ a = 4 ft b = 3 ft c = 3 ft Solution: Find the external reactions Guesses E y = 1 lb

Cy = 1 lb

Cz = 1 lb

Dx = 1 lb

Dy = 1 lb

Dz = 1 lb

518

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given

⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0 ⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ D ⎛0⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛⎜ x ⎞⎟ ⎛ −b ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + ⎜ 0 ⎟ × Cy = 0 ⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎝ c ⎠ ⎜⎝ Dz ⎟⎠ ⎝ c ⎠ ⎜⎝ Cz ⎟⎠

⎛ Ey ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜C ⎟ ⎜ z ⎟ = Find E , C , C , D , D , D ( y y z x y z) ⎜ Dx ⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎝ Dz ⎠

⎛ Ey ⎞ ⎜ ⎟ ⎛ 266.667 ⎞ ⎜ Cy ⎟ ⎜ −400 ⎟ ⎟ ⎜C ⎟ ⎜ ⎜ ⎟ 0 ⎜ z⎟ = ⎟ lb ⎜ Dx ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ − 266.667 ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎜ ⎟ ⎝ 500 ⎟⎠ ⎝ Dz ⎠

Now find the force in each member.

⎛ −b ⎞ ⎛ −b ⎞ ⎜ ⎟ AB = 0 AC = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝c ⎠

⎛0⎞ AD = ⎜ −a ⎟ ⎜ ⎟ ⎝c ⎠

⎛0⎞ AE = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛b⎞ ⎜ ⎟ BC = −a BE = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ c ⎝ ⎠ ⎝0⎠

⎛0⎞ BF = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛b⎞ CD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛0⎞ ⎜ ⎟ CF = 0 DE = ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝ −c ⎠

⎛ −b ⎞ DF = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎛ −b ⎞ EF = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F BC = 1 lb

F BE = 1 lb

F BF = 1 lb

F CD = 1 lb

519

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F CF = 1 lb

Chapter 6

F DE = 1lb

F DF = 1 lb

F EF = 1 lb

Given F 1 + F AB

AB

F 2 + F BC

BC

AB

BC

+ FAC

AC

+ FBF

BF

AC

BF

+ FAD

AD

+ FBE

BE

AD

BE

+ FAE + F AB

AE AE −AB AB

=0

=0

⎛⎜ 0 ⎟⎞ −AE −BE EF −DE + FBE + F EF + F DE =0 ⎜ Ey ⎟ + FAE AE BE EF DE ⎜0⎟ ⎝ ⎠ F BF

−BF BF

+ F CF

−CF CF

+ F DF

−DF DF

+ FEF

−EF EF

=0

⎛0⎞ ⎜ ⎟ −BC −AC CD CF + FAC + FCD + FCF =0 ⎜ Cy ⎟ + FBC BC AC CD CF ⎜C ⎟ ⎝ z⎠ ⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF) ⎜ FBF ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

520

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Engineering Mechanics - Statics

⎛ FAB ⎞ ⎜ ⎟ ⎛ −300 ⎞ F AC ⎜ ⎟ ⎜ 583.095 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎜ AD ⎟ = 333.333 ⎟ lb ⎜ FAE ⎟ ⎜ −666.667 ⎟ ⎟ ⎜ ⎟ ⎜ 0 ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎜ ⎟ ⎝ 500 ⎟⎠ F ⎝ BE ⎠

Chapter 6

⎛ FBF ⎞ ⎜ ⎟ ⎛ 0 ⎞ F CD ⎜ ⎟ ⎜ −300 ⎟ ⎟ ⎜F ⎟ ⎜ Positive (T) ⎜ ⎜ CF ⎟ = −300 ⎟ lb Negative (C) ⎜ FDE ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ 424.264 ⎜ ⎟ ⎜ FDF ⎟ ⎜ ⎜ ⎟ ⎝ −300 ⎟⎠ F ⎝ EF ⎠

Problem 6-59 The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Given:

⎛ 200 ⎞ ⎜ ⎟ F 1 = 300 lb ⎜ ⎟ ⎝ −500 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F 2 = 400 lb ⎜ ⎟ ⎝ 0 ⎠ a = 4 ft b = 3 ft c = 3 ft Solution: Find the external reactions Guesses E y = 1 lb

Cy = 1 lb

Cz = 1 lb

Dx = 1 lb

Dy = 1 lb

Dz = 1 lb

521

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given

⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0 ⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ D ⎛0⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛⎜ x ⎞⎟ ⎛ −b ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + ⎜ 0 ⎟ × Cy = 0 ⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎝ c ⎠ ⎜⎝ Dz ⎟⎠ ⎝ c ⎠ ⎜⎝ Cz ⎟⎠

⎛ Ey ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜C ⎟ ⎜ z ⎟ = Find E , C , C , D , D , D ( y y z x y z) ⎜ Dx ⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎝ Dz ⎠

⎛ Ey ⎞ ⎞ ⎜ ⎟ ⎛ −33.333 ⎜ ⎟ C ⎜ y⎟ −666.667 ⎟ ⎜C ⎟ ⎜ 200 ⎜ ⎟ z ⎜ ⎟= ⎜ ⎟ lb −200 ⎜ Dx ⎟ ⎜ ⎟ ⎜ ⎟ − 13 ⎟ ⎜ −1.253 × 10 ⎜ Dy ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 300 ⎠ Dz ⎝ ⎠

Now find the force in each member.

⎛ −b ⎞ ⎛ −b ⎞ ⎜ ⎟ AB = 0 AC = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝c ⎠

⎛0⎞ AD = ⎜ −a ⎟ ⎜ ⎟ ⎝c ⎠

⎛0⎞ AE = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛b⎞ ⎜ ⎟ BC = −a BE = ⎜ −a ⎟ ⎜ ⎟ ⎜ ⎟ ⎝c ⎠ ⎝0⎠

⎛0⎞ BF = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠

⎛b⎞ CD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎛0⎞ ⎜ ⎟ CF = 0 DE = ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝ −c ⎠

⎛ −b ⎞ DF = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠

⎛ −b ⎞ EF = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F BC = 1 lb

F BE = 1 lb

F BF = 1 lb

F CD = 1 lb

F CF = 1 lb

F DE = 1lb

F DF = 1 lb

F EF = 1 lb

Given 522

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F 1 + F AB

AB

F 2 + F BC

BC

AB

BC

Chapter 6

+ FAC

AC

+ FBF

BF

AC

BF

+ FAD

AD

+ FBE

BE

AD

BE

+ FAE + F AB

AE AE −AB AB

=0

=0

⎛⎜ 0 ⎟⎞ −AE −BE EF −DE + FBE + F EF + F DE =0 ⎜ Ey ⎟ + FAE AE BE EF DE ⎜0⎟ ⎝ ⎠ F BF

−BF BF

+ F CF

−CF CF

+ F DF

−DF DF

+ FEF

−EF EF

=0

⎛0⎞ ⎜ ⎟ −BC −AC CD CF + FAC + FCD + FCF =0 ⎜ Cy ⎟ + FBC BC AC CD CF ⎜C ⎟ ⎝ z⎠ ⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF) F BF ⎜ ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

523

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎛ FBF ⎞ −300 ⎜ ⎟ ⎛ ⎜⎞ ⎟ ⎛ 0 ⎞ ⎜ ⎟ F F AC CD ⎜ ⎟ ⎜ ⎟ ⎜ −500 ⎟ 971.825 ⎜ ⎟ ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ Positive (T) − 11 ⎜ ⎟ ⎜ ⎜ AD ⎟ = 1.121 × 10 ⎜ lbCF ⎟ = −300 ⎟ lb Negative (C) ⎜ FAE ⎟ ⎜ −366.667 ⎜⎟ FDE ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜⎟ ⎟ ⎜ 424.264 ⎜ ⎟ ⎜ ⎟ 0 ⎜ FBC ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎝ ⎜⎠ ⎟ ⎝ −300 ⎟⎠ 500 F F ⎝ BE ⎠ ⎝ EF ⎠

Problem 6-60 Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by a ball-and-socket joints at A, B, and E. Hint: The support reaction at E acts along member EC. Why? Given:

⎛ −200 ⎞ ⎜ ⎟ F = 400 N ⎜ ⎟ ⎝ 0 ⎠

a = 2m b = 1.5 m c = 5m d = 1m e = 2m

Solution:

⎛ 0 ⎞ ⎜ ⎟ AC = a + b ⎜ ⎟ ⎝ 0 ⎠

⎛d⎞ ⎜ ⎟ AD = a ⎜ ⎟ ⎝e ⎠

⎛ −c − d ⎞ ⎜ 0 ⎟ BC = ⎜ ⎟ ⎝ 0 ⎠

⎛ −c ⎞ ⎜ ⎟ BD = −b ⎜ ⎟ ⎝e ⎠

Guesses

Given

⎛d⎞ ⎜ ⎟ CD = −b ⎜ ⎟ ⎝e ⎠

F AC = 1 N

F AD = 1 N

F BC = 1 N

F CD = 1 N

F EC = 1 N

F BD = 1 N

F + FAD

−AD AD

+ FBD

−BD BD

+ F CD

−CD CD

=0

524

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛⎜ 0 ⎞⎟ F CD + F BC + FAC +⎜ 0 ⎟ =0 CD BC AC ⎜ −FEC ⎟ ⎝ ⎠ CD

−BC

−AC

⎛ FAC ⎞ ⎜ ⎟ F ⎜ AD ⎟ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AC AD BC BD CD EC ) ⎜ FBD ⎟ ⎜ ⎟ F ⎜ CD ⎟ ⎜ ⎟ ⎝ FEC ⎠

⎛ FAC ⎞ ⎜ ⎟ ⎛ 221 ⎞ F ⎜ AD ⎟ ⎜ 343 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ BC ⎟ = ⎜ 148 ⎟ N ⎜ FBD ⎟ ⎜ 186 ⎟ ⎟ ⎜ ⎟ ⎜ − 397 ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎜ ⎟ ⎝ −295 ⎟⎠ F ⎝ EC ⎠

Positive (T) Negative (C)

Problem 6-61 Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at C, D, E, and G. Units Used: 3

kN = 10 N Given: F = 3 kN a = 2m b = 1.5 m c = 2m d = 1m e = 1m Solution: Fv =

⎛0⎞ ⎜b⎟ 2 2⎜ ⎟ b + c ⎝ −c ⎠ F

525

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

uAG =

⎛ −e ⎞ ⎜ −a ⎟ ⎟ 2 2⎜ a +e ⎝ 0 ⎠

uAE =

⎛d⎞ ⎜ −a ⎟ ⎟ 2 2⎜ a +d ⎝ 0 ⎠

1

1

⎛0⎞ ⎜ ⎟ uAB = 0 ⎜ ⎟ ⎝ −1 ⎠

uBC = uAE uBD = uAG

⎛d⎞ ⎜ −a ⎟ ⎟ 2 2 2⎜ a +c +d ⎝ c ⎠ 1

uBE =

⎛ −e ⎞ ⎜ −a ⎟ ⎟ 2 2 2⎜ a +e +c ⎝ c ⎠ 1

uBG =

Guesses F AB = 1 kN

F AE = 1 kN

F BC = 1 kN

F BD = 1 kN

F BE = 1 kN

F BG = 1 kN

F AG = 1 kN

Given −c 2

c +b

2

e 2

a

F ( a) −

2

a +e

2

2

a +d F BD( a) −

FBC ( c) −

d 2

a +d

2

a 2

2

a +e

FBD( c) = 0

F BC ( a) = 0

F v + F AEuAE + FAGuAG + F ABuAB = 0 −F AB uAB + FBGuBG + FBEuBE + F BC uBC + FBDuBD = 0

526

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAE ⎟ ⎜ ⎟ FAG ⎜ ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F ( AB AE AG BC BD BE BG) ⎜ BC ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜F ⎟ ⎝ BG ⎠

⎛ FAB ⎞ ⎜ ⎟ ⎛ −2.4 ⎞ ⎟ ⎜ FAE ⎟ ⎜ ⎜ ⎟ ⎜ 1.006 ⎟ ⎜ FAG ⎟ ⎜ 1.006 ⎟ ⎜ F ⎟ = ⎜ −1.342 ⎟ kN ⎟ ⎜ BC ⎟ ⎜ ⎜ FBD ⎟ ⎜ −1.342 ⎟ ⎜ ⎟ ⎜ 1.8 ⎟ ⎟ ⎜ FBE ⎟ ⎜ 1.8 ⎝ ⎠ ⎜F ⎟ ⎝ BG ⎠ Positive (T) Negative (C)

Problem 6-62 Determine the force in members BD, AD, and AF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F. Given:

⎛ 0 ⎞ ⎜ ⎟ F = 250 lb ⎜ ⎟ ⎝ −250 ⎠ a = 6 ft b = 6 ft

θ = 60 deg Solution: Find the external reactions h = b sin ( θ ) Guesses Ax = 1 lb

B y = 1 lb

B z = 1 lb

Dy = 1 lb

F y = 1 lb

F z = 1 lb

Given

⎛

⎞

⎛

⎞ 527

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ Ax ⎞⎟ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎞⎟ F + ⎜ 0 ⎟ + ⎜ By ⎟ + ⎜ Dy ⎟ + ⎜ F y ⎟ = 0 ⎜ 0 ⎟ ⎜B ⎟ ⎜ 0 ⎟ ⎜F ⎟ ⎝ ⎠ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ ⎛ 0.5b ⎞ ⎛ b ⎞ ⎛⎜ Ax ⎞⎟ ⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0.5b ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜a⎟ × ⎜ ⎟ B ⎜ 0 ⎟ × D + ⎜ a ⎟ × Fy = 0 ⎜ 0 ⎟ + 0 × ⎜ y⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ h ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ h ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ Fz ⎟⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜B ⎟ ⎜ z ⎟ = Find A , B , B , D , F , F ( x y z y y z) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ ⎝ Fz ⎠

⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ By ⎟ ⎜ 72 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ z ⎟ = 125 ⎟ lb ⎜ Dy ⎟ ⎜ −394 ⎟ ⎟ ⎜ ⎟ ⎜ 72 ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎜ ⎟ ⎝ 125 ⎟⎠ ⎝ Fz ⎠

Now find the forces in the members

⎛0⎞ ⎜ ⎟ AB = −a ⎜ ⎟ ⎝0⎠

⎛ −b ⎞ ⎜ ⎟ AC = −a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ −a ⎟ AD = ⎜ ⎟ ⎝ h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ AE = ⎜ ⎟ ⎝ h ⎠ ⎛ −0.5b ⎞ ⎜ 0 ⎟ BD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ AF = 0 ⎜ ⎟ ⎝0⎠ ⎛ 0.5b ⎞ ⎜ 0 ⎟ CD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ BC = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ CF = a ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎜ ⎟ DE = a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ a ⎟ DF = ⎜ ⎟ ⎝ −h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ EF = ⎜ ⎟ ⎝ −h ⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F AF = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

F CF = 1 lb

528

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Engineering Mechanics - Statics

F DE = 1 lb

F DF = 1 lb

Chapter 6

F EF = 1 lb

Given −DE −AE EF F + FDE + F AE + FEF =0 DE AE EF F CF

CD −BC −AC + F CD + F BC + FAC =0 CF CD BC AC CF

⎛⎜ 0 ⎞⎟ F DE + FDF + F AD + F BD + FCD + ⎜ Dy ⎟ = 0 DE DF AD BD CD ⎜0 ⎟ ⎝ ⎠ ⎛0⎞ ⎜ ⎟ −AB BC BD F AB + FBC + F BD + ⎜ By ⎟ = 0 AB BC BD ⎜B ⎟ ⎝ z⎠ DE

DF

−AD

−BD

−CD

⎛⎜ Ax ⎞⎟ F AB + FAC + FAF + F AD + F AE +⎜ 0 ⎟ =0 AB AC AF AD AE ⎜0 ⎟ ⎝ ⎠ AB

AC

AF

AD

AE

⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FAF ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FAF , FBC , FBD , FCD , FCF , FDE , FDF , FEF) ⎜ FBD ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

529

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Engineering Mechanics - Statics

⎛ FBD ⎞ ⎛ −144.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAD ⎟ = ⎜ 204.1 ⎟ lb ⎜ F ⎟ ⎝ 72.2 ⎠ ⎝ AF ⎠

Chapter 6

Positive (T) Negative (C)

Problem 6-63 Determine the force in members CF, EF, and DF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F. Given:

⎛ 0 ⎞ ⎜ ⎟ F = 250 lb ⎜ ⎟ ⎝ −250 ⎠ a = 6 ft b = 6 ft

θ = 60 deg Solution: Find the external reactions h = b sin ( θ ) Guesses Ax = 1 lb

B y = 1 lb

B z = 1 lb

Dy = 1 lb

F y = 1 lb

F z = 1 lb

Given

⎛⎜ Ax ⎞⎟ ⎛⎜ 0 ⎟⎞ ⎛⎜ 0 ⎞⎟ ⎛⎜ 0 ⎟⎞ F + ⎜ 0 ⎟ + ⎜ By ⎟ + ⎜ Dy ⎟ + ⎜ F y ⎟ = 0 ⎜ 0 ⎟ ⎜B ⎟ ⎜ 0 ⎟ ⎜F ⎟ ⎝ ⎠ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ ⎛ 0.5b ⎞ ⎛ b ⎞ ⎛⎜ Ax ⎞⎟ ⎛ b ⎞ ⎛⎜ 0 ⎟⎞ ⎛ 0.5b ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ 0 ⎟⎞ ⎜ a ⎟ × F + ⎜a⎟ × ⎜ ⎟ ⎜ 0 ⎟ × D + ⎜ a ⎟ × Fy = 0 + 0 × By + ⎜ ⎟ ⎜ ⎟ ⎜0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ h ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ h ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ Fz ⎟⎠

530

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Engineering Mechanics - Statics

Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜B ⎟ ⎜ z ⎟ = Find A , B , B , D , F , F ( x y z y y z) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ ⎝ Fz ⎠

⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ By ⎟ ⎜ 72 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ ⎜ z ⎟ = 125 ⎟ lb ⎜ Dy ⎟ ⎜ −394 ⎟ ⎟ ⎜ ⎟ ⎜ 72 ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎜ ⎟ ⎝ 125 ⎟⎠ ⎝ Fz ⎠

Now find the forces in the members

⎛0⎞ ⎜ ⎟ AB = −a ⎜ ⎟ ⎝0⎠

⎛ −b ⎞ ⎜ ⎟ AC = −a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ −a ⎟ AD = ⎜ ⎟ ⎝ h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ AE = ⎜ ⎟ ⎝ h ⎠ ⎛ −0.5b ⎞ ⎜ 0 ⎟ BD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ AF = 0 ⎜ ⎟ ⎝0⎠ ⎛ 0.5b ⎞ ⎜ 0 ⎟ CD = ⎜ ⎟ ⎝ h ⎠

⎛ −b ⎞ ⎜ ⎟ BC = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ CF = a ⎜ ⎟ ⎝0⎠

⎛0⎞ ⎜ ⎟ DE = a ⎜ ⎟ ⎝0⎠

⎛ −0.5b ⎞ ⎜ a ⎟ DF = ⎜ ⎟ ⎝ −h ⎠

⎛ −0.5b ⎞ ⎜ 0 ⎟ EF = ⎜ ⎟ ⎝ −h ⎠

Guesses F AB = 1 lb

F AC = 1 lb

F AD = 1 lb

F AE = 1 lb

F AF = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

F CF = 1 lb

F DE = 1 lb

F DF = 1 lb

F EF = 1 lb

Given F + FDE

F CF

CF CF

−DE DE

+ F AE

+ F CD

−AE

CD CD

AE

+ FEF

+ F BC

EF EF

−BC BC

=0

+ FAC

−AC AC

=0

531

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ 0 ⎞⎟ F DE + FDF + F AD + F BD + FCD + ⎜ Dy ⎟ = 0 DE DF AD BD CD ⎜0 ⎟ ⎝ ⎠ DE

DF

−AD

−BD

−CD

⎛0⎞ ⎜ ⎟ F AB + FBC + F BD + ⎜ By ⎟ = 0 AB BC BD ⎜B ⎟ ⎝ z⎠ −AB

BC

BD

⎛⎜ Ax ⎞⎟ F AB + FAC + FAF + F AD + F AE +⎜ 0 ⎟ =0 AB AC AF AD AE ⎜0 ⎟ ⎝ ⎠ AB

AC

AF

AD

AE

⎛⎜ FAB ⎞⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FAF ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FAF , FBC , FBD , FCD , FCF , FDE , FDF , FEF) ⎜ FBD ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠

⎛ FCF ⎞ ⎛ 72.2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FEF ⎟ = ⎜ −144.3 ⎟ lb ⎜F ⎟ ⎝ 0 ⎠ ⎝ DF ⎠

Positive (T) Negative (C)

532

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Engineering Mechanics - Statics

Chapter 6

Problem 6-64 Determine the force developed in each member of the space truss and state if the members are in tension or compression. The crate has weight W. Given: W = 150 lb a = 6 ft b = 6 ft c = 6 ft

Solution: h =

2

Unit Vectors

c −

⎛ a⎞ ⎜ ⎟ ⎝ 2⎠

2

uAD =

⎛⎜ −a ⎟⎞ 1 ⎜ 2 ⎟ 2⎜ 0 ⎟ 2 ⎛ a⎞ ⎜ ⎟ h +⎜ ⎟ ⎝ 2⎠ ⎝ h ⎠

uBD =

⎛⎜ a ⎟⎞ 1 ⎜2⎟ 2⎜ 0 ⎟ a 2 h + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

uAC =

⎛⎜ − a ⎟⎞ 1 ⎜ 2⎟ 2⎜ b ⎟ a 2 2 h + b + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

uBC =

⎛⎜ a ⎟⎞ 1 ⎜2⎟ 2⎜ b ⎟ a 2 2 h + b + ⎛⎜ ⎟⎞ ⎜ h ⎟ ⎝ 2⎠ ⎝ ⎠

Guesses F AB = 1 lb

B y = 1 lb

Ax = 1 lb

F AC = 1 lb

Ay = 1 lb

F AD = 1 lb

F BC = 1 lb

F BD = 1 lb

F CD = 1 lb

533

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Engineering Mechanics - Statics

Chapter 6

Given

⎛⎜ 0 ⎟⎞ ⎜ −FCD ⎟ − FAC uAC − FBC uBC = 0 ⎜ −W ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎜ ⎟ B ⎜ y ⎟ + FBC uBC + FBDuBD = 0 ⎜ ⎟ ⎝ 0 ⎠ ⎛ Ax − FAB ⎞ ⎜ ⎟ A ⎜ ⎟ + FAC uAC + FADuAD = 0 y ⎜ ⎟ 0 ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ B ⎟ ⎜ y ⎟ ⎜ FAB ⎟ ⎜ ⎟ ⎜ FAC ⎟ = Find ( Ax , Ay , By , FAB , FAC , FAD , FBC , FBD , FCD) ⎜F ⎟ ⎜ AD ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎟ ⎛ FAB ⎞ ⎜ ⎜ ⎟ ⎛ 0.0 ⎞ ⎝ FCD ⎠ F ⎜ AC ⎟ ⎜ −122.5 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ AD 86.6 Positive (T) ⎜ ⎟= lb ⎜ ⎟ Negative (C) ⎜ FBC ⎟ −122.5 ⎜ ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ 86.6 ⎟ ⎜ ⎟ ⎜⎝ 173.2 ⎟⎠ FCD ⎝ ⎠

Problem 6-65 The space truss is used to support vertical forces at joints B, C, and D. Determine the force in each member and state if the members are in tension or compression.

534

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Engineering Mechanics - Statics

Chapter 6

Units Used: 3

kN = 10 N Given: F 1 = 6 kN

a = 0.75 m

F 2 = 8 kN

b = 1.00 m

F 3 = 9 kN

c = 1.5 m

Solution: Assume that the connections at A, E, and F are rollers Guesses F BC = 1 kN

F CF = 1 kN

F CD = 1 kN

F AD = 1 kN

F DF = 1 kN

F DE = 1 kN

F BD = 1 kN

F BA = 1 kN

F EF = 1 kN

F AE = 1 kN

F AF = 1 kN Given Joint C F BC = 0

F CD = 0

−F 2 − F CF = 0 Joint D a 2

2

a +b

FBD +

a 2

2

2

a +b +c

FAD = 0

535

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Engineering Mechanics - Statics

b

−F CD −

2

−b

+ 2

a +b

2

2

2

2

2

b +c

−c 2

F AD −

c

−F 3 − F DE − +

FBD ...

2

a +b +c

2

a +b +c

Chapter 6

=0 b 2

2

b +c

F DF

F DF ... = 0

F AD

Joint B a

−F BC −

2

2

a +b

FBD = 0

b 2

2

a +b

FBD = 0

−F 1 − F BA = 0

Joint E a 2

2

a +b

FAE = 0

−F EF −

b 2

2

a +b

FAE = 0

⎛ FBC ⎞ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜F ⎟ ⎜ AD ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FDE ⎟ = Find ( FBC , FCF , FCD , FAD , FDF , FDE , FBD , FBA , FEF , FAE , FAF) ⎜F ⎟ ⎜ BD ⎟ ⎜ FBA ⎟ ⎜ ⎟ ⎜ FEF ⎟ ⎜ ⎟ ⎜ FAE ⎟ ⎜F ⎟ ⎝ AF ⎠

536

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Engineering Mechanics - Statics

⎛ FBC ⎞ ⎜ ⎟ ⎜ FCF ⎟ ⎛⎜ 0.00 ⎟⎞ ⎜ ⎟ ⎜ −8.00 ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 0.00 ⎟ AD ⎜ ⎟ ⎜ 0.00 ⎟ ⎜ FDF ⎟ ⎜ 0.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −9.00 ⎟ kN ⎜ F ⎟ ⎜ 0.00 ⎟ ⎜ BD ⎟ ⎜ ⎟ ⎜ FBA ⎟ ⎜ −6.00 ⎟ ⎜ ⎟ ⎜ 0.00 ⎟ ⎜ FEF ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0.00 ⎟ F AE ⎜ ⎟ ⎝ 0.00 ⎠ ⎜F ⎟ ⎝ AF ⎠

Chapter 6

Positive (T) Negative (C)

Problem 6-66 A force P is applied to the handles of the pliers. Determine the force developed on the smooth bolt B and the reaction that pin A exerts on its attached members. Given: P = 8 lb a = 1.25 in b = 5 in c = 1.5 in Solution: ΣMA = 0;

−R B c + P b = 0 RB = P

b c

R B = 26.7 lb ΣF x = 0;

Ax = 0

ΣF y = 0;

Ay − P − RB = 0 537

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Ay = P + RB Ay = 34.7 lb

Problem 6-67 The eye hook has a positive locking latch when it supports the load because its two parts are pin-connected at A and they bear against one another along the smooth surface at B. Determine the resultant force at the pin and the normal force at B when the eye hook supports load F . Given: F = 800 lb a = 0.25 in b = 3 in c = 2 in

θ = 30 deg

Solution: Σ MA = 0;

−F B cos ( 90 deg − θ ) ( b) − F B sin ( 90 deg − θ ) ( c) + F a = 0 FB = F

+

↑Σ Fy = 0;

+ Σ F x = 0; →

a

cos ( 90 deg − θ ) b + sin ( 90 deg − θ ) c

F B = 61.9 lb

−F − FB sin ( 90 deg − θ ) + A y = 0 Ay = F + F B sin ( 90 deg − θ )

Ay = 854 lb

Ax − F B cos ( 90 deg − θ ) = 0 Ax = FB cos ( 90 deg − θ ) FA =

2

2

Ax + Ay

Ax = 30.9 lb F A = 854 lb

Problem 6-68 Determine the force P needed to hold the block of mass F in equilibrium. 538

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Engineering Mechanics - Statics

Chapter 6

Given: F = 20 lb Solution: Pulley B:

ΣF y = 0;

2P − T = 0

Pulley A:

ΣF y = 0;

2T − F = 0

T =

1 F 2

2P = T

T = 10 lb P =

1 T 2

P = 5 lb

Problem 6-69 The link is used to hold the rod in place. Determine the required axial force on the screw at E if the largest force to be exerted on the rod at B, C or D is to be F max. Also, find the magnitude of the force reaction at pin A. Assume all surfaces of contact are smooth. Given: F max = 100 lb a = 100 mm b = 80 mm c = 50 mm

θ = 45 deg

539

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Engineering Mechanics - Statics

Chapter 6

Solution: Assign an initial value for R E. This will be scaled at the end of the problem. Guesses

Given

Ax = 1 lb

Ay = 1 lb

R B = 1 lb

R C = 1 lb

R D = 1 lb

R E = 1 lb

− Ax + R E − R B cos ( θ ) = 0

Ay − R B sin ( θ ) = 0

R E a − RB cos ( θ ) ( a + b + c cos ( θ ) ) − RB sin ( θ ) c sin ( θ ) = 0 R B cos ( θ ) − R D = 0

R B sin ( θ ) − RC = 0

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ RB ⎟ = Find ( Ax , Ay , RB , RC , RD) ⎜R ⎟ ⎜ C⎟ ⎜ RD ⎟ ⎝ ⎠

⎛⎜ Ax ⎟⎞ ⎛ 0.601 ⎞ ⎜ Ay ⎟ ⎜ 0.399 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ RB ⎟ = ⎜ 0.564 ⎟ lb ⎜ R ⎟ ⎜ 0.399 ⎟ ⎜ C⎟ ⎜ ⎟ 0.399 ⎝ ⎠ ⎜ RD ⎟ ⎝ ⎠

Now find the critical load and scale the problem

⎛ RB ⎞ ⎜ ⎟ ans = ⎜ R C ⎟ ⎜R ⎟ ⎝ D⎠

F scale =

Fmax

R E = Fscale RE

max ( ans )

F A = Fscale

2

2

Ax + Ay

R E = 177.3 lb

F A = 127.9 lb

Problem 6-70 The man of weight W1 attempts to lift himself and the seat of weight W2 using the rope and pulley system shown. Determine the force at A needed to do so, and also find his reaction on the seat.

540

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Engineering Mechanics - Statics

Chapter 6

Given: W1 = 150 lb W2 = 10 lb Solution: Pulley C: ΣF y = 0;

3T − R = 0

Pulley B: ΣF y = 0;

3R − P = 0

Thus,

P = 9T

Man and seat: ΣF y = 0;

T + P − W1 − W2 = 0 10T = W1 + W2 T =

W1 + W2 10

P = 9T

T = 16 lb P = 144 lb

Seat: ΣF y = 0;

P − N − W2 = 0 N = P − W2

N = 134 lb

Problem 6-71 Determine the horizontal and vertical components of force that pins A and C exert on the frame. Given: F = 500 N a = 0.8 m

d = 0.4 m

541

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Engineering Mechanics - Statics

Chapter 6

b = 0.9 m

e = 1.2 m

c = 0.5 m

θ = 45deg

Solution: BC is a two-force member Member AB : ΣMA = 0;

e

−F c + FBC

2

2

a +e

b + F BC

2

a 2

( c + d) = 0 2

a +e

2

a +e F BC = F c eb+a c+a d

F BC = 200.3 N

Thus, Cx = F BC

Cy = F BC

ΣF x = 0;

ΣF y = 0;

Ax − F BC

e 2

Cx = 167 N

2

a +e a 2

Cy = 111 N

2

a +e e 2

=0

Ax = FBC

2

a +e

Ay − F + FBC

a 2

=0 2

a +e

e 2

Ax = 167 N

2

a +e

Ay = F − F BC

a 2

2

a +e

Ay = 389 N

Problem 6-72 Determine the horizontal and vertical components of force that pins A and C exert on the frame. Units Used: 3

kN = 10 N Given: F 1 = 1 kN

542

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Engineering Mechanics - Statics

Chapter 6

F 2 = 500 N

θ = 45 deg a = 0.2 m b = 0.2 m c = 0.4 m d = 0.4 m Solution: Guesses Ax = 1 N

Ay = 1 N

Cx = 1 N

Cy = 1 N

Given Ax − Cx = 0

Ay + Cy − F1 − F2 = 0

F1 a − Ay 2 a + Ax d = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠

−F 2 b + Cy( b + c) − Cx d = 0

⎛ Ax ⎞ ⎜ ⎟ ⎛⎜ 500 ⎞⎟ ⎜ Ay ⎟ ⎜ 1000 ⎟ N ⎜ ⎟=⎜ ⎟ C 500 x ⎜ ⎟ ⎜ ⎟ ⎜ C ⎟ ⎝ 500 ⎠ ⎝ y⎠

Problem 6-73 The truck exerts the three forces shown on the girders of the bridge. Determine the reactions at the supports when the truck is in the position shown. The girders are connected together by a short vertical link DC. Units Used: 3

kip = 10 lb

543

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Engineering Mechanics - Statics

Chapter 6

Given: a = 55 ft

f = 12 ft

b = 10 ft

F 1 = 5 kip

c = 48 ft

F 2 = 4 kip

d = 5 ft

F 3 = 2 kip

e = 2 ft Solution:

Member CE: ΣMC = 0;

−F 3 e + E y( e + c) = 0

e Ey = F3 e+c

E y = 80 lb

ΣF y = 0;

Cy − F3 + Ey = 0

Cy = F 3 − E y

Cy = 1920 lb

Member ABD: ΣMA = 0;

−F 1 a − F2( d + a) − Cy( a + d + b) + By( a + d) = 0 By =

ΣF y = 0;

F 1 a + F2( d + a) + Cy( a + d + b) d+a

B y = 10.8 kip

Ay − F 1 + B y − F2 − Cy = 0 Ay = Cy + F 1 − B y + F2

Ay = 96.7 lb

Problem 6-74 Determine the greatest force P that can be applied to the frame if the largest force resultant acting at A can have a magnitude F max. Units Used: 3

kN = 10 N

544

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Engineering Mechanics - Statics

Chapter 6

Given: F max = 2 kN a = 0.75 m b = 0.75 m c = 0.5 m d = 0.1 m Solution: Σ MA = 0;

T( c + d) − P( a + b) = 0

+ Σ F x = 0; →

Ax − T = 0

+

↑Σ Fy = 0;

Ay − P = 0

Thus,

T=

a+b c+d

P

Ay = P

Ax =

a+b c+d

P

Require, F max = P =

2

2

Ax + Ay F max 2

⎛ a + b⎞ + 1 ⎜ ⎟ ⎝c + d⎠ P = 743 N

Problem 6-75 The compound beam is pin supported at B and supported by rockers at A and C. There is a hinge (pin) at D. Determine the reactions at the supports. Units Used: 3

kN = 10 N

545

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 7 kN

a = 4m

F 2 = 6 kN

b = 2m

F 3 = 16 kN

c = 3m

θ = 60 deg

d = 4m

Solution: Member DC : ΣMD = 0;

−F 1 sin ( θ ) ( a − c) + Cy a = 0 Cy = F 1 sin ( θ )

ΣF y = 0;

a−c

Cy = 1.52 kN

a

Dy − F 1 sin ( θ ) + Cy = 0 Dy = F1 sin ( θ ) − Cy

ΣF x = 0;

Dy = 4.55 kN

Dx − F 1 cos ( θ ) = 0 Dx = F1 cos ( θ )

Dx = 3.5 kN

Member ABD : ΣMA = 0;

−F 3 a − F2( 2 a + b) − Dy( 3 a + b) + By2 a = 0 By =

ΣF y = 0;

F 3 a + F2( 2a + b) + Dy ( 3a + b) 2a

Ay − F 3 + B y − F2 − Dy = 0 Ay = Dy + F3 − By + F 2

ΣF x = 0;

B y = 23.5 kN

Ay = 3.09 kN

B x − F1 cos ( θ ) = 0 B x = F 1 cos ( θ )

B x = 3.5 kN

Problem 6-76 The compound beam is fixed supported at A and supported by rockers at B and C. If there are hinges (pins) at D and E, determine the reactions at the supports A, B, and C.

546

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Engineering Mechanics - Statics

Chapter 6

Units Used: 3

kN = 10 N Given: a = 2 m M = 48 kN⋅ m b = 4m

kN w1 = 8 m

c = 2m

kN w2 = 6 d = 6m m e = 3m

Solution: Guesses Ax = 1 N

Ay = 1 N

MA = 1 N m

Dx = 1 N

Dy = 1 N

By = 1 N

Ey = 1 N

Ex = 1 N

Cy = 1 N

Given Ay − w2 a − Dy = 0 a MA − w2 a − Dy a = 0 2 −w1

( b + c) 2

E y − w1 −w1⎛⎜

2

d+e 2

+ B y b − E y( b + c) = 0 + Cy = 0

− Ax − Dx = 0 Dy − w1( b + c) + By − E y = 0 Dx + E x = 0 −E x = 0

d + e⎞⎛ d + e⎞

⎟⎜ ⎟ + Cy d − M = 0 ⎝ 2 ⎠⎝ 3 ⎠

547

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Engineering Mechanics - Statics

Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜M ⎟ ⎜ A⎟ ⎜ Dx ⎟ ⎜ ⎟ D ⎜ y ⎟ = Find ( Ax , Ay , MA , Dx , Dy , By , Ey , Ex , Cy) ⎜B ⎟ ⎜ y⎟ ⎜ Ey ⎟ ⎜ ⎟ E ⎜ x⎟ ⎟ ⎜ ⎝ Cy ⎠

⎛⎜ Ax ⎞⎟ ⎛ 0 ⎞ = kN ⎜ Ay ⎟ ⎜⎝ 19 ⎟⎠ ⎝ ⎠ MA = 26 kN m B y = 51 kN Cy = 26 kN

Problem 6-77 Determine the reactions at supports A and B. Units Used: 3

kip = 10 lb Given: lb w1 = 500 ft lb w2 = 700 ft a = 6 ft b = 8 ft c = 9 ft d = 6 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

B x = 1 lb

B y = 1 lb

F CD = 1 lb MA = 1 lb⋅ ft

548

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Given Ay − w1 a −

MA − w1 a a −

d 2

b +d

2

F CD = 0

d 2

d +b

2

Ax −

b 2

b +d

b

F CD2 a = 0

2

2

b +d

d

2

FCD = 0

FCD − Bx = 0

c c B y c − w2 =0 2 3

c FCD − w2 + B y = 0 2 2 2 b +d

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ B ⎟ ⎜ x ⎟ = Find A , A , B , B , F , M ( x y x y CD A) ⎜ By ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎝ MA ⎠

⎛⎜ Ax ⎞⎟ ⎛ 2.8 ⎞ = kip ⎜ Ay ⎟ ⎜⎝ 5.1 ⎟⎠ ⎝ ⎠ MA = 43.2 kip⋅ ft

⎛⎜ Bx ⎟⎞ ⎛ 2.8 ⎞ = kip ⎜ By ⎟ ⎜⎝ 1.05 ⎟⎠ ⎝ ⎠

Problem 6-78 Determine the horizontal and vertical components of force at C which member ABC exerts on member CEF. Given: F = 300 lb a = 4 ft b = 6 ft c = 3 ft r = 1 ft Solution: Guesses Ax = 1 lb

Ay = 1 lb

F y = 1 lb

549

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Engineering Mechanics - Statics

Cx = 1 lb

Chapter 6

Cy = 1 lb

Given b b Ax a − A y − Cx a − Cy = 0 2 2 Cx a − F r = 0 Ax = 0 Ay + Fy − F = 0 F y b − F ( b + c + r) = 0

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Fy ⎟ = Find ( Ax , Ay , Fy , Cx , Cy) ⎜C ⎟ ⎜ x⎟ ⎜ Cy ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ −200 ⎟ lb ⎜ F ⎟ ⎝ 500 ⎠ ⎝ y⎠

⎛⎜ Cx ⎟⎞ ⎛ 75 ⎞ = lb ⎜ Cy ⎟ ⎜⎝ 100 ⎟⎠ ⎝ ⎠

Problem 6-79 Determine the horizontal and vertical components of force that the pins at A, B, and C exert on their connecting members. Units Used: 3

kN = 10 N Given: F = 800 N a = 1m r = 50 mm b = 0.2 m Solution: −F ( a + r) + A x b = 0 550

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Engineering Mechanics - Statics

Ax = F

a+r b

Chapter 6

Ax = 4.2 kN

− Ax + Bx = 0 Bx = Ax

B x = 4.2 kN

−F r − Ay b + A x b = 0 Ay =

−F r + A x b b

Ay = 4 kN

Ay − By − F = 0 By = Ay − F

B y = 3.2 kN

− Ax + F + Cx = 0 Cx = Ax − F

Cx = 3.4 kN

Ay − Cy = 0 Cy = Ay

Cy = 4 kN

Problem 6-80 Operation of exhaust and intake valves in an automobile engine consists of the cam C, push rod DE, rocker arm EFG which is pinned at F, and a spring and valve, V. If the spring is compressed a distance δ when the valve is open as shown, determine the normal force acting on the cam lobe at C. Assume the cam and bearings at H, I, and J are smooth.The spring has a stiffness k. Given: a = 25 mm b = 40 mm

δ = 20 mm k = 300

N m

551

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Engineering Mechanics - Statics

Chapter 6

Solution: F s = kδ

Fs = 6 N

ΣF y = 0; −F G + Fs = 0 FG = Fs

FG = 6 N

ΣMF = 0; FG b + T a = 0 b T = FG a

T = 9.60 N

552

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Engineering Mechanics - Statics

Chapter 6

Problem 6-81 Determine the force P on the cord, and the angle θ that the pulley-supporting link AB makes with the vertical. Neglect the mass of the pulleys and the link. The block has weight W and the cord is attached to the pin at B. The pulleys have radii of r1 and r2. Given: W = 200 lb r1 = 2 in r2 = 1 in

φ = 45 deg Solution: The initial guesses are

θ = 30 deg +

↑Σ Fy = 0;

F AB = 30 lb 2P − W = 0 P =

1 2

W

P = 100 lb

Given + Σ F x = 0; →

P cos ( φ ) − FAB sin ( θ ) = 0

+

F AB cos ( θ ) − P − P − P sin ( φ ) = 0

↑Σ Fy = 0;

⎛ θ ⎞ ⎜ F ⎟ = Find ( θ , FAB) ⎝ AB ⎠

θ = 14.6 deg

F AB = 280 lb

Problem 6-82 The nail cutter consists of the handle and the two cutting blades. Assuming the blades are pin connected at B and the surface at D is smooth, determine the normal force on the fingernail when a force F is applied to the handles as shown.The pin AC slides through a smooth hole at A and is attached to the bottom member at C.

553

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Engineering Mechanics - Statics

Chapter 6

Given: F = 1 lb a = 0.25 in b = 1.5 in Solution: Handle : ΣMD = 0;

FA a − F b = 0 FA = F

ΣF y = 0;

⎛ b⎞ ⎜ ⎟ ⎝ a⎠

F A = 6 lb

ND − FA − F = 0 ND = F A + F

ND = 7 lb

Top blade : ΣMB = 0;

ND b − F N( 2 a + b) = 0 b ⎞ F N = ND ⎛⎜ ⎟ ⎝ 2 a + b⎠

F N = 5.25 lb

Problem 6-83 The wall crane supports load F. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W? Units Used:

3

kip = 10 lb

Given: F = 700 lb

554

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Engineering Mechanics - Statics

Chapter 6

a = 4 ft b = 4 ft c = 4 ft

θ = 60 deg Solution: Pulley E: +

↑Σ Fy = 0; T =

1 2

2T − F = 0 T = 350 lb

F

This is the force in the cable at the winch W a φ = atan ⎛⎜ ⎟⎞

Member ABC:

⎝ b⎠

Σ MA = 0;

(

)

−F ( b + c) + TBD sin ( φ ) − T sin ( θ ) b = 0

⎛ b + c ⎞ + T sin ( θ ) ⎟ ⎝ b ⎠

F⎜ TBD =

sin ( φ ) 3

TBD = 2.409 × 10 lb +

↑Σ Fy = 0; − Ay + TBD sin ( φ ) − T sin ( θ ) − F = 0 Ay = TBD sin ( φ ) − T sin ( θ ) − F

Ay = 700 lb

+ Σ F x = 0; →

Ax − TBD cos ( φ ) − T cos ( θ ) = 0 Ax = TBD cos ( φ ) + T cos ( θ )

Ax = 1.878 kip

At D: 555

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Engineering Mechanics - Statics

Chapter 6

Dx = TBD cos ( φ )

Dx = 1.703 kip

Dy = TBD sin ( φ )

Dy = 1.703 kip

Problem 6-84 Determine the force that the smooth roller C exerts on beam AB. Also, what are the horizontal and vertical components of reaction at pin A? Neglect the weight of the frame and roller. Given: M = 60 lb⋅ ft a = 3 ft b = 4 ft c = 0.5 ft Solution: Σ MA = 0;

−M + Dx c = 0 M Dx = c Dx = 120 lb

+ Σ F x = 0; →

+

↑Σ Fy = 0; Σ MB = 0;

Ax − Dx = 0 Ax = Dx

Ax = 120 lb

Ay = 0

Ay = 0 lb

−Nc b + Dx c = 0 c Nc = Dx b

Nc = 15.0 lb

Problem 6-85 Determine the horizontal and vertical components of force which the pins exert on member ABC.

556

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Engineering Mechanics - Statics

Chapter 6

Given: W = 80 lb a = 6 ft b = 9 ft c = 3 ft r = 0.5 ft Solution: + Σ F x = 0; →

− Ax + W = 0 Ax = W

+

↑Σ Fy = 0;

Ax = 80 lb

Ay − W = 0 Ay = W

Σ MC = 0;

Ay = 80 lb

Ay( a + b) − B y b = 0 By = Ay

a+b b

B y = 133 lb Σ MD = 0;

−W( c − r) + By b − Bx c = 0 Bx =

+ Σ F x = 0; →

B y b − W ( c − r)

B x = 333 lb

c

Ax + B x − Cx = 0 Cx = Ax + B x

+

↑Σ Fy = 0;

Cx = 413 lb

− Ay + B y − Cy = 0 Cy = B y − A y

Cy = 53.3 lb

Problem 6-86 The floor beams AB and BC are stiffened using the two tie rods CD and AD. Determine the force along each rod when the floor beams are subjected to a uniform load w. Assume the three contacting members at B are smooth and the joints at A, C, and D are pins. Hint: 557

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Engineering Mechanics - Statics

Chapter 6

g j Members AD, CD, and BD are two-force members

p

3

kip = 10 lb

Units Used: Given: w = 80

lb ft

b = 5 ft a = 12 ft Solution: Due to summetry: Cy =

w( 2a) 2

Cy = 960 lb

Member BC : ΣMB = 0;

a Cy( a) − w a⎛⎜ ⎟⎞ − T⎛ ⎜ 2

⎝ ⎠

a T = ⎛⎜ Cy − w ⎟⎞ 2⎠ ⎝

⎞a = 0 2 2⎟ ⎝ a +b ⎠

2

b

2

a +b

T = 1.248 kip

b

Problem 6-87 Determine the horizontal and vertical components of force at pins B and C. Given: F = 50 lb

c = 6 ft

a = 4 ft

d = 1.5 ft

558

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

b = 4 ft Solution:

Chapter 6

r = 0.5 ft Guesses

Cx = 1 lb

Cy = 1 lb

B x = 1 lb

B y = 1 lb

Given F ( a − r) + Cx c − Cy( a + b) = 0 −F ( a − r) − F( d + r) + Cy( a + b) = 0 −B x + F + Cx = 0 B y − F + Cy = 0

⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜ ⎟ = Find ( Bx , By , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠ ⎛ Bx ⎞ ⎜ ⎟ ⎛⎜ 66.667 ⎞⎟ ⎜ By ⎟ ⎜ 15.625 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Cx ⎟ ⎜ 16.667 ⎟ ⎜ C ⎟ ⎝ 34.375 ⎠ ⎝ y⎠

559

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Engineering Mechanics - Statics

Chapter 6

Problem 6-88 The skid steer loader has a mass M1, and in the position shown the center of mass is at G1. If there is a stone of mass M2 in the bucket, with center of mass at G2 determine the reactions of each pair of wheels A and B on the ground and the force in the hydraulic cylinder CD and at the pin E. There is a similar linkage on each side of the loader. Units Used: 3

Mg = 10 kg 3

kN = 10 N Given: M1 = 1.18 Mg M2 = 300 kg a = 1.25 m

d = 0.15 m

b = 1.5 m

e = 0.5 m

c = 0.75 m

θ = 30 deg

Solution:

Entire System:

ΣMA = 0;

M2 g b − M1 g( c − d) + NB c = 0 NB =

ΣF y = 0;

M1 g( c − d) − M2 g b c

NB = 3.37 kN

(Both wheels)

NA = 11.1 kN

(Both wheels)

NB − M2 g − M1 g + NA = 0 NA = −NB + M2 g + M1 g

Upper member: ΣME = 0;

M2 g( a + b) − 2 F CD sin ( θ ) a = 0

560

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

F CD = ΣF x = 0; ΣF y = 0;

Chapter 6

M 2 g( a + b)

F CD = 6.5 kN

2 sin ( θ ) a

E x = F CD( cos ( θ ) ) M2 g Ey − + FCD sin ( θ ) = 0 2 Ey = FR =

M2 g 2

E x = 5607 N

− F CD sin ( θ )

2

E y = −1766 N

2

Ex + Ey

F R = 5.879 kN

Problem 6-89 Determine the horizontal and vertical components of force at each pin. The suspended cylinder has a weight W. Given: W = 80 lb

d = 6 ft

a = 3 ft

e = 2 ft

b = 4 ft

r = 1 ft

c = 4 ft Solution: Guesses Ax = 1 lb

B x = 1 lb

B y = 1 lb

F CD = 1 lb

E x = 1 lb

E y = 1 lb

Given Ex − Bx + W = 0 −E y + By = 0

561

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

−W r + Ey a − Ex d = 0 −B y +

−B y a +

c 2

FCD − W = 0

2

c +d c 2

c +d

− Ax + Bx +

2

F CD d − W( d + e − r) = 0

d 2

2

c +d

FCD − W = 0

⎛ Ax ⎞ ⎜ ⎟ B x ⎜ ⎟ ⎜ B ⎟ ⎜ y ⎟ = Find A , B , B , F , E , E ( x x y CD x y) ⎜ FCD ⎟ ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎟ ⎝ Ey ⎠ Cx = F CD

d 2

c +d

2

Cy = F CD

c 2

c +d

2

Dx = −Cx

Dy = −Cy

⎛ Ax ⎞ ⎜ ⎟ ⎛ 160 ⎞ ⎜ Bx ⎟ ⎜ ⎟ 80 ⎜B ⎟ ⎜ ⎟ 26.667 ⎜ y⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ 160 ⎜ ⎟ ⎜ ⎟ 106.667 C = ⎜ y⎟ ⎜ ⎟ lb ⎜D ⎟ ⎜ ⎟ −160 ⎜ x⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ −106.667 ⎟ ⎜ ⎟ ⎜ −8.694 × 10− 13 ⎟ ⎜ Ex ⎟ ⎜ ⎟ 26.667 ⎠ ⎜ ⎟ ⎝ Ey ⎝ ⎠

562

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Engineering Mechanics - Statics

Chapter 6

Problem 6-90 The two-member frame is pin connected at C, D, and E. The cable is attached to A, passes over the smooth peg at B, and is attached to a load W. Determine the horizontal and vertical reactions at each pin. Given: a = 2 ft b = 1 ft c = 0.75 ft W = 100 lb Solution: d =

c

( a + 2 b) b Initial guesses: Cx = 1 lb

Cy = 1 lb

Dx = 1 lb

Dy = 1 lb

E x = 1 lb

E y = 1 lb

Given −Dx + Cx − W = 0 −Dy + Cy − W = 0 −Cx c + Cy b + W d − W( a + 2 b) = 0 W − Cx + Ex = 0 E y − Cy = 0 −W d + Cx c + Cy b = 0

562

563

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Engineering Mechanics - Statics

Chapter 6

⎛ Cx ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜D ⎟ ⎜ x ⎟ = Find C , C , D , D , E , E ( x y x y x y) ⎜ Dy ⎟ ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎟ ⎝ Ey ⎠

⎛ Cx ⎞ ⎜ ⎟ ⎛ 133 ⎞ ⎜ Cy ⎟ ⎜ 200 ⎟ ⎟ ⎜D ⎟ ⎜ ⎜ ⎜ x ⎟ = 33 ⎟ lb ⎜ Dy ⎟ ⎜ 100 ⎟ ⎟ ⎜ ⎟ ⎜ 33 ⎜ ⎟ ⎜ Ex ⎟ ⎜ ⎜ ⎟ ⎝ 200 ⎟⎠ ⎝ Ey ⎠

Problem 6-91 Determine the horizontal and vertical components of force which the pins at A, B, and C exert on member ABC of the frame. Given: F 1 = 400 N F 2 = 300 N F 3 = 300 N a = 1.5 m b = 2m c = 1.5 m d = 2.5 m f = 1.5 m g = 2m e = a+b+c−d Solution: Guesses Ay = 1 N F BD = 1 N

Cx = 1 N

Cy = 1 N

F BE = 1 N

564

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Engineering Mechanics - Statics

Chapter 6

Given F 1 g + F2( a + b) + F3 a − Ay( f + g) = 0 F 1 g − Cy( f + g) = 0 Cx e = 0 −Cx −

f+g 2

e + ( f + g)

Ay − Cy −

2

FBD +

e 2

e + ( f + g)

2

f+g 2

d + ( f + g)

FBD −

2

F BE = 0

d 2

d + ( f + g)

2

F BE = 0

⎛⎜ Ay ⎞⎟ ⎜ Cx ⎟ ⎜ ⎟ C ⎜ y ⎟ = Find ( Ay , Cx , Cy , FBD , FBE) ⎜F ⎟ ⎜ BD ⎟ ⎜ FBE ⎟ ⎝ ⎠ Bx = −

By =

f+g 2

e + ( f + g) e 2

e + ( f + g)

2

2

F BD +

F BD +

f+g 2

d + ( f + g) d 2

d + ( f + g)

2

2

FBE

FBE

Ay = 657 N

⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ = N ⎜ By ⎟ ⎜⎝ 429 ⎟⎠ ⎝ ⎠ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎞ = N ⎜ Cy ⎟ ⎜⎝ 229 ⎟⎠ ⎝ ⎠

565

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Engineering Mechanics - Statics

Chapter 6

Problem 6-92 The derrick is pin-connected to the pivot at A. Determine the largest mass that can be supported by the derrick if the maximum force that can be sustained by the pin at A is Fmax. Units Used: 3

kN = 10 N m

g = 9.81

2

s 3

Mg = 10 kg Given: F max = 18 kN L = 5m

θ = 60 deg Solution: AB is a two-force member. Require F AB = Fmax +

↑

Σ F y = 0;

F AB sin ( θ ) − M = 2

Mg

⎛ FAB ⎞ ⎜ ⎟ ⎝ g ⎠

2

sin ( θ ) − W = 0

⎛ sin ( θ ) ⎞ ⎜ ⎟ ⎝ sin ( θ ) + 2 ⎠

M = 5.439

1 2

Mg

s

Problem 6-93 Determine the required mass of the suspended cylinder if the tension in the chain wrapped around the freely turning gear is T. Also, what is the magnitude of the resultant force on pin A? Units Used: 3

kN = 10 N g = 9.8

m 2

s

566

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Engineering Mechanics - Statics

Chapter 6

Given: T = 2 kN L = 2 ft

θ = 30 deg φ = 45 deg

Solution: Σ MA = 0;

−2 T L cos ( θ ) + M g cos ( φ ) L cos ( θ ) + M g sin ( φ ) L sin ( θ ) = 0 M =

2 T cos ( θ )

(cos ( φ ) cos ( θ ) + sin (φ ) sin ( θ )) g

M = 1793 +

→ Σ Fx = 0;

1 2

kg

s

2 T − M g cos ( φ ) − A x = 0 Ax = 2 T − M g cos ( φ )

+

↑Σ Fy = 0;

M g sin ( φ ) − Ay = 0 Ay = M g sin ( φ ) FA =

2

2

Ax + Ay

F A = 2.928 kN

567

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Problem 6-94 The tongs consist of two jaws pinned to links at A, B, C, and D. Determine the horizontal and vertical components of force exerted on the stone of weight W at F and G in order to lift it. Given: a = 1 ft b = 2 ft c = 1.5 ft d = 1 ft W = 500 lb Solution: Guesses F x = 1 lb F y = 1 lb F AD = 1 lb F BE = 1 lb Given

2 Fy − W = 0 F AD − F x −

F AD b − Fx( b + c) = 0 a ⎞ ⎛ ⎜ 2 2 ⎟ FBE = 0 ⎝ a +d ⎠

⎛ Fx ⎞ ⎜ ⎟ ⎜ Fy ⎟ ⎜ ⎟ = Find ( Fx , Fy , FAD , FBE) ⎜ FAD ⎟ ⎜F ⎟ ⎝ BE ⎠

−F y +

d ⎞ ⎛ ⎜ 2 2 ⎟ FBE = 0 ⎝ a +d ⎠

⎛⎜ Gx ⎞⎟ ⎛⎜ Fx ⎟⎞ = ⎜ Gy ⎟ ⎜ Fy ⎟ ⎝ ⎠ ⎝ ⎠

⎛ Fx ⎞ ⎜ ⎟ ⎛⎜ 333 ⎟⎞ ⎜ Fy ⎟ ⎜ 250 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Gx ⎟ ⎜ 333 ⎟ ⎜ G ⎟ ⎝ 250 ⎠ ⎝ y⎠

Problem 6-95 Determine the force P on the cable if the spring is compressed a distance δ when the mechanism is in the position shown. The spring has a stiffness k.

568

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Engineering Mechanics - Statics

Chapter 6

Given:

δ = 0.5 in

c = 6 in

lb

d = 6 in

k = 800

ft

a = 24 in

e = 4 in

b = 6 in

θ = 30 deg

Solution: F E = kδ

F E = 33.333 lb

The initial guesses are P = 20 lb

B x = 11 lb

B y = 34 lb

F CD = 34 lb

Given ΣMA = 0;

B x b + B y c − F E ( a + b) = 0

ΣMD = 0;

By d − P e = 0

+ Σ F x = 0; →

ΣMB = 0;

−B x + FCD cos ( θ ) = 0 F CD sin ( θ ) d − P ( d + e) = 0

⎛⎜ FCD ⎞⎟ ⎜ Bx ⎟ ⎜ ⎟ = Find ( FCD , Bx , By , P) B ⎜ y ⎟ ⎜ P ⎟ ⎝ ⎠ B x = 135.398 lb

B y = 31.269 lb

F CD = 156.344 lb

P = 46.903 lb

569

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Engineering Mechanics - Statics

Chapter 6

Problem 6-96 The scale consists of five pin-connected members. Determine the load W on the pan EG if a weight F is suspended from the hook at A. Given: F = 3 lb

b = 3 in

a = 5 in

c = 4 in

d = 6 in

f = 2 in

e = 8 in Solution: Guesses

TC = 10 lb

TD = 10 lb

TG = 10 lb

W = 10 lb

Given Member ABCD:

ΣMB = 0;

F a − TC b − TD( b + e − d) = 0 Member EG: ΣMG = 0;

− TC e + W c = 0

ΣF y = 0;

TG − W + TC = 0

Member FH: ΣMH = 0;

−TD( d + f) + TG f = 0

⎛⎜ TC ⎟⎞ ⎜ TD ⎟ ⎜ ⎟ = Find ( TC , TD , TG , W) ⎜ TG ⎟ ⎜W⎟ ⎝ ⎠ W = 7.06 lb

Problem 6-97 The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF, 570

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Engineering Mechanics - Statics

Chapter 6

g p gg which are operated by the hydraulic cylinder H. The toggles push the movable bar G forward, pressing the plate p into the cavity. If the force which the plate exerts on the head is P, determine the force F in the hydraulic cylinder for the given angle θ. Units Used: 3

kN = 10 N Given: P = 12 kN a = 200 mm

θ = 30 deg Solution: Member EF: ΣME = 0;

−F y a cos ( θ ) + Fy =

ΣF x = 0;

Ex − Ex =

ΣF y = 0;

2

2

2

a sin ( θ ) = 0

tan ( θ )

P

P

P

F y = 3.464 kN

=0

P

E x = 6 kN

2

Ey − Fy = 0 Ey = Fy

E y = 3.464 kN

Joint E: ΣF x = 0;

−F DE cos ( θ ) − E x = 0 F DE =

−E x

cos ( θ )

F DE = −6.928 kN

F − Ey + F DE sin ( θ ) = 0

ΣF y = 0;

F = E y − FDE sin ( θ )

F = 6.93 kN

Problem 6-98 Determine the horizontal and vertical components of force at pins A and C of the two-member frame.

571

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Engineering Mechanics - Statics

Chapter 6

Given: N w1 = 500 m N w2 = 400 m N w3 = 600 m a = 3m b = 3m

Solution: Guesses Ax = 1 N

Ay = 1 N Cx = 1 N

Cy = 1 N

Given 1 Ay + Cy − w1 a − w2 a = 0 2

Ax a −

1 − Ax + Cx + w3 b = 0 2

− Ay a +

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠

1 2a 1 b a w1 a − w3 b − w2 a = 0 2 3 2 3 2 1 a w1 a = 0 2 3

⎛ Ax ⎞ ⎜ ⎟ ⎛⎜ 1400 ⎞⎟ ⎜ Ay ⎟ ⎜ 250 ⎟ N ⎜ ⎟=⎜ ⎟ C 500 x ⎜ ⎟ ⎜ ⎟ ⎜ C ⎟ ⎝ 1700 ⎠ ⎝ y⎠

572

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Engineering Mechanics - Statics

Chapter 6

Problem 6-99 The truck rests on the scale, which consists of a series of compound levers. If a mass M1 is placed on the pan P and it is required that the weight is located at a distance x to balance the “beam” ABC, determine the mass of the truck. There are pins at all lettered points. Is it necessary for the truck to be symmetrically placed on the scale? Explain. Units Used: 3

Mg = 10 kg

g = 9.81

m 2

s Given: M1 = 15 kg

FD = 3 m

x = 0.480 m

EF = 0.2 m

a = 0.2 m HI = 0.1 m

GH = 2.5 m

KJ = HI

KG = GH

Solution: Member ABC : ΣMB = 0;

−M1 g x + F AD a = 0

x F AD = M1 g a Member EFD : ΣME = 0;

2

F AD = 72 s N

−F y EF + F AD( FD + EF) = 0

⎛ FD + EF ⎞ ⎟ ⎝ EF ⎠

F y = F AD⎜

2

F y = 1152 s N

573

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Engineering Mechanics - Statics

Chapter 6

Member GHI : ΣMI = 0;

Hy HI − GY( GH + HI ) = 0

Member JKG : ΣMJ = 0;

(Fy − Gy)( KJ + GH ) − Ky( KJ ) = 0 Ky + Hy = Fy

KJ + KG HI

Scale Platform : ΣF y = 0;

Ky + Hy = W W = Fy⎛⎜

⎝

M =

W g

KJ + KG ⎞ HI

⎟ ⎠

M = 14.98 Mg

Because KJ = HI and KG = GH it doesn't matter where the truck is on the scale.

Problem 6-100 By squeezing on the hand brake of the bicycle, the rider subjects the brake cable to a tension T If the caliper mechanism is pin-connected to the bicycle frame at B, determine the normal force each brake pad exerts on the rim of the wheel. Is this the force that stops the wheel from turning? Explain. Given: T = 50 lb a = 2.5 in b = 3 in

574

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Engineering Mechanics - Statics

Chapter 6

Solution: Σ MB = 0;

−N b + T a = 0

N = T

a b

N = 41.7 lb

This normal force does not stop the wheel from turning. A frictional force (see Chapter 8), which acts along the wheel's rim stops the wheel.

Problem 6-101 If a force of magnitude P is applied perpendicular to the handle of the mechanism, determine the magnitude of force F for equilibrium. The members are pin-connected at A, B, C, and D. Given: P = 6 lb a = 25 in b = 4 in c = 5 in d = 4 in e = 5 in f = 5 in g = 30 in Solution: Σ MA = 0;

F BC b − P a = 0 F BC =

Pa b

575

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Engineering Mechanics - Statics

Chapter 6

F BC = 37.5 lb + Σ F x = 0; →

− Ax + P = 0 Ax = P

+

↑Σ Fy = 0;

Ax = 6 lb

− Ay + F BC = 0 Ay = FBC

Σ MD = 0;

Ay = 37.5 lb

−e Ax − Ay( b + c) + ( g + b + c)F = 0 F =

e Ax + Ay( b + c)

F = 9.423 lb

g+b+c

Problem 6-102 The pillar crane is subjected to the load having a mass M. Determine the force developed in the tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied in the position shown. Units Used: 3

kN = 10 N Given: M = 500 kg a = 1.8 m b = 2.4 m

θ 1 = 10 deg θ 2 = 20 deg g = 9.81

m 2

s Solution:

initial guesses: F CB = 10 kN

F AB = 10 kN

576

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Engineering Mechanics - Statics

Chapter 6

Given −M

( )

( )

( )

( )

g cos θ 1 − F AB cos θ 2 + FCB 2

−M

g sin θ 1 − FAB sin θ 2 + FCB 2

⎛ FAB ⎞

⎜ ⎟ = Find F , F ( AB CB) ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ Cx ⎟⎞ ⎜ Cy ⎟ ⎝ ⎠

b

=0

2

2

a +b a 2

− Mg = 0 2

a +b

=

⎛b⎞ ⎜ ⎟ 2 2 a a +b ⎝ ⎠ FCB

⎛ FAB ⎞ ⎛ 9.7 ⎞ ⎜ ⎟ ⎜ ⎟ C ⎜ x ⎟ = ⎜ 11.53 ⎟ kN ⎜ C ⎟ ⎝ 8.65 ⎠ ⎝ y ⎠

Problem 6-103 The tower truss has a weight W and a center of gravity at G. The rope system is used to hoist it into the vertical position. If rope CB is attached to the top of the shear leg AC and a second rope CD is attached to the truss, determine the required tension in BC to hold the truss in the position shown. The base of the truss and the shear leg bears against the stake at A, which can be considered as a pin. Also, compute the compressive force acting along the shear leg. Given: W = 575 lb

θ = 40 deg a = 5 ft b = 3 ft c = 10 ft d = 4 ft e = 8 ft Solution: Entire system:

ΣMA = 0;

TBC cos ( θ ) ( d + e) − TBC sin ( θ ) a − W( a + b) = 0 TBC =

W( a + b)

cos ( θ ) ( d + e) − sin ( θ ) a

TBC = 769 lb 577

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

CA is a two-force member.

At C:

e ⎞ ⎟ ⎝ b + c⎠

φ = atan ⎛⎜ initial guesses:

F CA = 500 lb

TCD = 300 lb

Given ΣF x = 0;

ΣF y = 0;

F CA

F CA

a 2

a + ( d + e)

2

d+e 2

a + ( d + e)

⎛⎜ FCA ⎟⎞ = Find ( FCA , TCD) ⎜ TCD ⎟ ⎝ ⎠

2

+ TCD cos ( φ ) − TBC cos ( θ ) = 0

− TCD sin ( φ ) − TBC sin ( θ ) = 0

TCD = 358 lb

F CA = 739 lb

Problem 6-104 The constant moment M is applied to the crank shaft. Determine the compressive force P that is exerted on the piston for equilibrium as a function of θ. Plot the results of P (ordinate) versus θ (abscissa) for 0 deg ≤ θ ≤ 90 deg. Given: a = 0.2 m b = 0.45 m M = 50 N⋅ m Solution: a cos ( θ ) = b sin ( φ )

φ = asin ⎛⎜ cos ( θ )⎟⎞ a ⎝b

⎠

−M + F BC cos ( θ − φ ) a = 0

578

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Engineering Mechanics - Statics

F BC =

Chapter 6

M

a cos ( θ − φ )

P = FBC cos ( φ ) =

M cos ( φ )

a cos ( θ − φ )

This function goes to infinity at θ = 90 deg, so we will only plot it to θ = 80 deg.

θ = 0 , 0.1 .. 80 φ ( θ ) = asin ⎛⎜ cos ( θ deg)⎟⎞

P (θ) =

a

⎝b

⎠

M cos ( φ ( θ ) )

a cos ( θ deg − φ ( θ ) )

Newtons

1500

P( θ )

1000 500 0

0

20

40

60

80

θ

Degrees

Problem 6-105 Five coins are stacked in the smooth plastic container shown. If each coin has weight W, determine the normal reactions of the bottom coin on the container at points A and B. Given: W = 0.0235 lb a = 3 b = 4

579

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Engineering Mechanics - Statics

Chapter 6

Solution: All coins : ΣF y = 0;

NB = 5W NB = 0.1175 lb

Bottom coin : ΣF y = 0;

⎛

⎞=0 ⎝ a +b ⎠ b

NB − W − N⎜

(

N = NB − W

2⎟

2

⎛ a2 + b2 ⎞ )⎜⎝ b ⎟⎠

N = 0.1175 lb ΣF x = 0;

⎛

NA = N⎜

a 2

⎞

2⎟

⎝ a +b ⎠

NA = 0.0705 lb

Problem 6-106 Determine the horizontal and vertical components of force at pin B and the normal force the pin at C exerts on the smooth slot. Also, determine the moment and horizontal and vertical reactions of force at A. There is a pulley at E. Given: F = 50 lb a = 4 ft b = 3 ft Solution: Guesses B x = 1 lb B y = 1 lb NC = 1 lb 580

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Ax = 1 lb

Ay = 1 lb

Chapter 6

MA = 1 lb⋅ ft

Given Bx +

a ⎛ ⎞ ⎜ 2 2 ⎟ NC − F = 0 ⎝ a +b ⎠

By −

b ⎛ ⎞ ⎜ 2 2 ⎟ NC − F = 0 ⎝ a +b ⎠

(F + Bx)a − (F + By)b = 0 F−

a ⎛ ⎞ ⎜ 2 2 ⎟ NC − Ax = 0 ⎝ a +b ⎠

b ⎛ ⎞ ⎜ 2 2 ⎟ NC − Ay = 0 ⎝ a +b ⎠ −F 2 a +

a ⎛ ⎞ ⎜ 2 2 ⎟ NC a + MA = 0 ⎝ a +b ⎠

⎛ Bx ⎞ ⎜ ⎟ B y ⎜ ⎟ ⎜N ⎟ ⎜ C ⎟ = Find B , B , N , A , A , M ( x y C x y A) ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎝ MA ⎠

NC = 20 lb

⎛⎜ Bx ⎟⎞ ⎛ 34 ⎞ = lb ⎜ By ⎟ ⎜⎝ 62 ⎟⎠ ⎝ ⎠ ⎛⎜ Ax ⎞⎟ ⎛ 34 ⎞ = lb ⎜ Ay ⎟ ⎜⎝ 12 ⎟⎠ ⎝ ⎠ MA = 336 lb⋅ ft

Problem 6-107 A force F is applied to the handles of the vise grip. Determine the compressive force developed on the smooth bolt shank A at the jaws. 581

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Engineering Mechanics - Statics

Chapter 6

Given: F = 5 lb

b = 1 in

a = 1.5 in

c = 3 in

d = 0.75 in

e = 1 in

θ = 20 deg Solution: From FBD (a) ΣME = 0;

⎡

⎤b = 0 ⎣ c + ( d + e) ⎦

F ( b + c) − F CD⎢

d+e

2

2⎥

⎡ c2 + ( d + e) 2⎤ ⎥ ⎣ b( d + e) ⎦

F CD = F( b + c) ⎢ ΣF x = 0;

⎡

F CD = 39.693 lb

⎤ ⎥ 2 2 ⎣ c + ( d + e) ⎦

E x = F CD⎢

c

E x = 34.286 lb From FBD (b) ΣMB = 0;

NA sin ( θ ) d + NA cos ( θ ) a − Ex( d + e) = 0

NA = Ex

d+e ⎛ ⎜ ( ) ⎝ sin θ d + cos ( θ )

⎞ ⎟

a⎠

NA = 36.0 lb

Problem 6-108 If a force of magnitude P is applied to the grip of the clamp, determine the compressive force F that the wood block exerts on the clamp.

582

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Engineering Mechanics - Statics

Chapter 6

Given: P = 10 lb a = 2 in b = 2 in c = 0.5 in d = 0.75 in e = 1.5 in

Solution: Define

b φ = atan ⎛⎜ ⎟⎞

φ = 69.444 deg

⎝ d⎠

From FBD (a), ΣMB = 0;

F CD cos ( φ ) c − P ( a + b + c) = 0 F CD =

+

↑Σ Fy = 0;

P( a + b + c)

F CD = 256.32 lb

cos ( φ ) ( c)

F CD sin ( φ ) − By = 0 B y = F CD sin ( φ )

B y = 240 lb

From FBD (b), ΣMA = 0;

By d − F e = 0 F =

By d

F = 120 lb

e

583

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Engineering Mechanics - Statics

Chapter 6

Problem 6-109 The hoist supports the engine of mass M. Determine the force in member DB and in the hydraulic cylinder H of member FB. Units Used: 3

kN = 10 N Given: M = 125 kg

d = 1m

a = 1m

e = 1m

b = 2m

f = 2m

c = 2m

g = 9.81

m 2

s Solution: Member GFE: ΣME = 0; −F FB⎡ ⎢

⎤ b + M g ( a + b) = 0 2 2⎥ ⎣ ( c + d) + ( b − e) ⎦

F FB = M g

c+d

⎡ a + b ⎤ ( c + d) 2 + ( b − e) 2 ⎢ ⎥ ⎣ b( c + d) ⎦

F FB = 1.94 kN ΣF x = 0;

⎡

⎤=0 ⎣ ( c + d) + ( b − e) ⎦ b−e

E x − FFB⎢

2⎥

2

⎡

⎤

b−e

E x = F FB⎢

2⎥

2

⎣ ( c + d) + ( b − e) ⎦

Member EDC: ΣΜc = 0;

⎛

⎞d = 0 ⎟ 2 2 ⎝ e +d ⎠

E x( c + d) − F DB⎜

F DB = Ex

⎛ c + d⎞ ⎜ ⎟ ⎝ ed ⎠

e

2

2

e +d

F DB = 2.601 kN

584

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Engineering Mechanics - Statics

Chapter 6

Problem 6-110 The flat-bed trailer has weight W1 and center of gravity at GT . It is pin-connected to the cab at D. The cab has a weight W2 and center of gravity at GC. Determine the range of values x for the position of the load L of weight W3 so that no axle is subjected to a force greater than FMax. The load has a center of gravity at GL . Given: W1 = 7000 lb

a = 4 ft

W2 = 6000 lb

b = 6 ft

W3 = 2000 lb

c = 3 ft

F max = 5500 lb d = 10 ft e = 12 ft Solution: Case 1:

Assume

Guesses

Ay = Fmax x = 1 ft

Given

Ay = Fmax B y = F max

Cy = F max

Dy = Fmax

Ay + B y − W2 − Dy = 0 −W2 a − Dy( a + b) + B y( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3 x + W1 e − Dy( c + d + e) = 0

⎛⎜ By ⎞⎟ ⎜ Cy ⎟ ⎜ ⎟ = Find ( By , Cy , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠ ⎛ Ay ⎞ ⎛⎜ 5.5 × 103 ⎟⎞ ⎜ ⎟ x1 = 30.917 ft ⎜ By ⎟ = ⎜ 6.333 × 103 ⎟ lb x1 = x ⎟ ⎜C ⎟ ⎜ Since By > Fmax then this solution is no good. ⎝ y ⎠ ⎜⎝ 3.167 × 103 ⎟⎠ Case 2:

Assume

Guesses

Ay = Fmax

B y = F max B y = F max

Cy = F max 585

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Engineering Mechanics - Statics

x = 1 ft Given

Chapter 6

Dy = Fmax

Ay + B y − W2 − Dy = 0 −W2 a − Dy( a + b) + B y( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3 x + W1 e − Dy( c + d + e) = 0

⎛⎜ Ay ⎞⎟ ⎜ Cy ⎟ ⎜ ⎟ = Find ( Ay , Cy , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠

⎛ Ay ⎞ ⎛⎜ 5.25 × 103 ⎟⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 5.5 × 103 ⎟ lb x2 = x ⎟ ⎜C ⎟ ⎜ 3 ⎜ ⎝ y ⎠ ⎝ 4.25 × 10 ⎟⎠

x2 = 17.375 ft

Since Ay < Fmax and Cy < F max then this solution is good. Cy = F max

Case 3:

Assume

Guesses

Ay = Fmax

B y = F max

x = 1 ft

Dy = Fmax

Given

Cy = F max

Ay + B y − W2 − Dy = 0 −W2 a − Dy( a + b) + B y( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3 x + W1 e − Dy( c + d + e) = 0

⎛⎜ Ay ⎞⎟ ⎜ By ⎟ ⎜ ⎟ = Find ( Ay , By , Dy , x) ⎜ Dy ⎟ ⎜ x ⎟ ⎝ ⎠

⎛ Ay ⎞ ⎛⎜ 4.962 × 103 ⎟⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 4.538 × 103 ⎟ lb x3 = x ⎟ ⎜C ⎟ ⎜ 3 ⎜ ⎝ y ⎠ ⎝ 5.5 × 10 ⎟⎠

x3 = 1.75 ft

Since Ay < Fmax and B y < F max then this solution is good. We conclude that x3 = 1.75 ft < x < x2 = 17.375 ft

586

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Engineering Mechanics - Statics

Chapter 6

Problem 6-111 Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovel in equilibrium. The shovel load has a mass W and a center of gravity at G. All joints are pin connected. Units Used: 3

Mg = 10 kg 3

kN = 10 N Given: a = 0.25 m θ 1 = 30 deg b = 0.25 m θ 2 = 10 deg c = 1.5 m

θ 3 = 60 deg

d = 2m

W = 1.25 Mg

e = 0.5 m Solution: Assembly FHG :

(

( )) = 0

ΣMH = 0; −[ W g( e) ] + FEF c sin θ 1 F EF = W g

⎛ e ⎞ F = 8.175 kN (T) ⎜ c sin θ ⎟ EF ( ) 1 ⎝ ⎠

Assembly CEFHG:

(

)

( )

ΣMC = 0; F AD cos θ 1 + θ 2 b − W g⎡( a + b + c)cos θ 2 + e⎤ = 0 ⎣ ⎦ F AD = W g

⎛ cos ( θ 2) a + cos ( θ 2) b + cos ( θ 2) c + e ⎞ ⎜ ⎟ cos ( θ 1 + θ 2) b ⎝ ⎠

F AD = 158 kN (C)

587

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Engineering Mechanics - Statics

Chapter 6

Problem 6-112 The aircraft-hangar door opens and closes slowly by means of a motor which draws in the cable AB. If the door is made in two sections (bifold) and each section has a uniform weight W and length L, determine the force in the cable as a function of the door's position θ. The sections are pin-connected at C and D and the bottom is attached to a roller that travels along the vertical track.

587

Solution: L

⎛ θ ⎞ − 2L sin ⎛ θ ⎞ N = 0 ⎟ ⎜ ⎟ A ⎝2⎠ ⎝2⎠

cos ⎜

ΣMD = 0;

2W

ΣΜC = 0;

T L cos ⎜

2

⎛ θ ⎞ − N L sin ⎛ θ ⎞ − W L cos ⎛ θ ⎞ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ A 2 ⎝2⎠ ⎝2⎠ ⎝2⎠

W ⎛θ⎞ NA = cot ⎜ ⎟ 2 ⎝2⎠ T=W

Problem 6-113 A man having weight W attempts to lift himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight of the platform. Given: W = 175 lb

588

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Engineering Mechanics - Statics

Chapter 6

Solution:

(a) Bar: +

↑Σ Fy = 0;

2

⎛ F⎞ − 2 ⎛ W⎞ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝2⎠

F = W

F = 175 lb

Man: +

↑Σ Fy = 0;

NC − W − 2

⎛ F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W + F

NC = 350 lb

( b) Bar: +

↑

Σ F y = 0;

2

⎛ W⎞ − 2 F = 0 ⎜ ⎟ 2 ⎝4⎠

F =

W 2

Man: +

↑Σ Fy = 0;

NC − W + 2

F = 87.5 lb

⎛ F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W − F

NC = 87.5 lb

589

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Engineering Mechanics - Statics

Chapter 6

Problem 6-114 A man having weight W1 attempts to lift himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has weight W2.

Given: W1 = 175 lb W2 = 30 lb Solution: (a) Bar: +

↑Σ Fy = 0;

2

F 2

(

)

− W1 + W2 = 0

F = W1 + W2 F = 205 lb Man: +

↑

Σ F y = 0;

F NC − W1 − 2 =0 2 NC = F + W1 NC = 380 lb

( b) Bar: +

↑

Σ F y = 0;

−2

⎛ F ⎞ + 2 ⎛⎜ W1 + W2 ⎞⎟ = 0 ⎜ ⎟ ⎝ 2⎠ ⎝ 4 ⎠

590

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Engineering Mechanics - Statics

F =

Chapter 6

W1 + W2 2

F = 102 lb Man: +

↑

Σ F y = 0;

NC − W1 + 2

⎛F⎞ = 0 ⎜ ⎟ ⎝ 2⎠

NC = W1 − F NC = 72.5 lb

Problem 6-115 The piston C moves vertically between the two smooth walls. If the spring has stiffness k and is unstretched when θ = 0, determine the couple M that must be applied to AB to hold the mechanism in equilibrium. Given: k = 15

lb in

θ = 30 deg a = 8 in b = 12 in

Solution: Geometry: b sin ( ψ) = a sin ( θ )

ψ = asin ⎛⎜ sin ( θ )

⎝

a⎞ ⎟ b⎠

φ = 180 deg − ψ − θ

ψ = 19.471 deg φ = 130.529 deg

591

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Engineering Mechanics - Statics

rAC

sin ( φ )

=

b

Chapter 6

rAC = b

sin ( θ )

⎛ sin ( φ ) ⎞ ⎜ ⎟ ⎝ sin ( θ ) ⎠

rAC = 18.242 in

Free Body Diagram: The solution for this problem will be simplified if one realizes that member CB is a two force member. Since the spring stretches x = ( a + b) − rAC

x = 1.758 in F sp = k x

the spring force is

F sp = 26.371 lb

Equations of Equilibrium: Using the method of joints +

↑

F CB cos ( ψ) − Fsp = 0

Σ F y = 0;

F CB =

Fsp

cos ( ψ)

F CB = 27.971 lb

From FBD of bar AB F CB sin ( φ ) a − M = 0

+ ΣMA = 0;

M = FCB sin ( φ ) a

M = 14.2 lb⋅ ft

Problem 6-116 The compound shears are used to cut metal parts. Determine the vertical cutting force exerted on the rod R if a force F is applied at the grip G. The lobe CDE is in smooth contact with the head of the shear blade at E. Given: F = 20 lb

e = 0.5 ft

a = 1.4 ft

f = 0.5 ft

b = 0.2 ft

g = 0.5 ft

c = 2 ft

h = 2.5 ft

d = 0.75 ft

θ = 60 deg

Solution: Member AG: ΣMA = 0;

F ( a + b) − FBC b sin ( θ ) = 0

592

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Engineering Mechanics - Statics

F BC = F

⎛ a+b ⎞ ⎜ ⎟ ⎝ b sin ( θ ) ⎠

Chapter 6

F BC = 184.75 lb

Lobe: ΣMD = 0;

F BC f − NE e = 0 f NE = FBC e

NE = 184.75 lb

Head: ΣMF = 0;

−NE( h + g) + h NR = 0 NR = NE

⎛ h + g⎞ ⎜ ⎟ ⎝ h ⎠

NR = 222 lb

Problem 6-117 The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear C. Note that AB is pinned at its ends to gear C and the underside of the table EF, which is allowed to move vertically due to the smooth guides at E and F. If the gears exert tangential forces between them, determine the compressive force developed on the cylinder S when a vertical force F is applied to the handle of the press. Given: F = 40 N a = 0.5 m b = 0.2 m c = 1.2 m d = 0.35 m e = 0.65 m Solution: Member GD: ΣMG = 0;

−F a + F CG b = 0

593

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Engineering Mechanics - Statics

F CG = F

Chapter 6

a

F CG = 100 N

b

Sector gear :

⎞d = 0 2 2⎟ ⎝ c +d ⎠

ΣMH = 0; F CG( d + e) − FAB⎛ ⎜

c

⎛ c2 + d2 ⎞ ⎟ F = 297.62 N F AB = FCG ( d + e) ⎜ ⎝ c d ⎠ AB Table: ΣF y = 0;

c ⎞ ⎜ 2 2 ⎟ − Fs = 0 ⎝ c +d ⎠

F AB⎛

F s = FAB

c ⎞ ⎛ ⎜ 2 2⎟ ⎝ c +d ⎠

F s = 286 N

Problem 6-118 The mechanism is used to hide kitchen appliances under a cabinet by allowing the shelf to rotate downward. If the mixer has weight W, is centered on the shelf, and has a mass center at G, determine the stretch in the spring necessary to hold the shelf in the equilibrium position shown. There is a similar mechanism on each side of the shelf, so that each mechanism supports half of the load W. The springs each have stiffness k. Given: W = 10 lb k = 4

a = 2 in

lb

b = 4 in

in

φ = 30 deg

c = 15 in

θ = 30 deg

d = 6 in

Solution: ΣMF = 0;

W 2

b − a F ED cos ( φ ) = 0

594

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Engineering Mechanics - Statics

F ED =

Wb

2 a cos ( φ )

+ Σ F x = 0; →

↑Σ Fy = 0;

Fy =

W 2

F ED = 11.547 lb

−F x + FED cos ( φ ) = 0

F x = F ED cos ( φ ) +

Chapter 6

−W 2

F x = 10 lb + F y − FED sin ( φ ) = 0

+ F ED⋅ sin ( φ )

F y = 10.774 lb

Member FBA: ΣMA = 0;

F s = k s;

F y( c + d) cos ( φ ) − F x( c + d) sin ( φ ) − Fs sin ( θ + φ ) d = 0 Fy( c + d) cos ( φ ) − Fx( c + d) sin ( φ ) Fs = F s = 17.5 lb d sin ( θ + φ ) Fs = k x

x =

Fs

x = 4.375 in

k

Problem 6-119 If each of the three links of the mechanism has a weight W, determine the angle θ for equilibrium.The spring, which always remains horizontal, is unstretched when θ = 0°.

595

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Engineering Mechanics - Statics

Chapter 6

Given: W = 25 lb k = 60

lb ft

a = 4 ft b = 4 ft Solution: Guesses

θ = 30 deg

B x = 10 lb

B y = 10 lb

Cx = 10 lb

Cy = 10 lb

Given

⎛ a ⎞ sin ( θ ) − C a sin ( θ ) + C a cos ( θ ) = 0 ⎟ y x ⎝ 2⎠

−W⎜

⎛ b⎞ + C b = 0 ⎟ y ⎝ 2⎠

−W⎜

B y + Cy − W = 0 −B x + Cx = 0

⎛ a⎞ ⎛ a⎞ ⎛ a⎞ −B x a cos ( θ ) − B y a sin ( θ ) − W⎜ ⎟ sin ( θ ) + k⎜ ⎟ sin ( θ ) ⎜ ⎟ cos ( θ ) = 0 ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜ C ⎟ = Find B , B , C , C , θ (x y x y ) ⎜ x⎟ ⎜ Cy ⎟ ⎜ ⎟ ⎝θ ⎠

⎛ Bx ⎞ ⎜ ⎟ ⎛⎜ 16.583 ⎞⎟ ⎜ By ⎟ ⎜ 12.5 ⎟ lb ⎜ ⎟=⎜ ⎟ ⎜ Cx ⎟ ⎜ 16.583 ⎟ ⎜ C ⎟ ⎝ 12.5 ⎠ ⎝ y⎠

θ = 33.6 deg

Problem 6-120 Determine the required force P that must be applied at the blade of the pruning shears so that the blade exerts a normal force F on the twig at E.

596

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Engineering Mechanics - Statics

Chapter 6

Given: F = 20 lb a = 0.5 in b = 4 in c = 0.75 in d = 0.75 in e = 1 in Solution: initial guesses: Ax = 1 lb

Ay = 1 lb

Dy = 10 lb P = 20 lb

Dx = 10 lb F CB = 20 lb

Given − P ( b + c + d) − A x a + F e = 0 Dy − P − A y − F = 0 Dx − Ax = 0 − Ay( d) − A x( a) + ( b + c)P = 0

⎞=0 2 2⎟ ⎝ c +a ⎠

Ax − F CB⎛ ⎜

c

⎞=0 ⎟ 2 2 ⎝ a +c ⎠

Ay + P − FCB⎛ ⎜

a

597

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ Dx ⎜ ⎟ = Find A , A , D , D , P , F ( x y x y CB) ⎜ D ⎟ y ⎜ ⎟ ⎜ P ⎟ ⎜F ⎟ ⎝ CB ⎠

⎛⎜ Ax ⎟⎞ ⎛ 13.333 ⎞ ⎜ Ay ⎟ ⎜ 6.465 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dx ⎟ = ⎜ 13.333 ⎟ lb ⎜ D ⎟ ⎜ 28.889 ⎟ ⎜ y ⎟ ⎜ ⎟ 16.025 ⎝ ⎠ ⎜ FCB ⎟ ⎝ ⎠

P = 2.424 lb

Problem 6-121 The three power lines exert the forces shown on the truss joints, which in turn are pin-connected to the poles AH and EG. Determine the force in the guy cable AI and the pin reaction at the support H. Units Used: 3

kip = 10 lb Given: F 1 = 800 lb

d = 125 ft

F 2 = 800 lb

e = 50 ft

a = 40 ft

f = 30 ft

b = 20 ft

g = 30 ft

c = 20 ft Solution: AH is a two-force member. c θ = atan ⎛⎜ ⎟⎞

φ = atan ⎛⎜

d β = atan ⎛⎜ ⎟⎞

e γ = atan ⎛⎜ ⎟⎞

⎝ b⎠

⎝ f⎠

⎞ ⎟ ⎝ a + b⎠ c

⎝ d⎠

α = 90 deg − β + γ

Guesses F AB = 1 lb

F BC = 1 lb

F CA = 1 lb

F AI = 1 lb

F H = 1 lb

598

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Engineering Mechanics - Statics

Chapter 6

Given F BC − F AB cos ( θ ) = 0 F AB sin ( θ ) − F1 = 0 2 F CA sin ( φ ) − F 2 = 0 −F AI sin ( α ) + F AB sin ( β − θ ) + F CA sin ( β − φ ) = 0 −F AI cos ( α ) − FAB cos ( β − θ ) − F CA cos ( β − φ ) + F H = 0

⎛⎜ FAB ⎞⎟ ⎜ FCA ⎟ ⎜ ⎟ F ⎜ BC ⎟ = Find ( FAB , FCA , FBC , FAI , FH) ⎜F ⎟ ⎜ AI ⎟ ⎜ FH ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎛ 1.131 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FCA ⎟ = ⎜ 1.265 ⎟ kip ⎜ F ⎟ ⎝ 0.8 ⎠ ⎝ BC ⎠

⎛⎜ FAI ⎞⎟ ⎛ 2.881 ⎞ = kip ⎜ FH ⎟ ⎜⎝ 3.985 ⎟⎠ ⎝ ⎠

Problem 6-122 The hydraulic crane is used to lift the load of weight W. Determine the force in the hydraulic cylinder AB and the force in links AC and AD when the load is held in the position shown. Units Used: 3

kip = 10 lb Given:

W = 1400 lb a = 8 ft

c = 1 ft

b = 7 ft

γ = 70 deg

599

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Solution: ΣMD = 0; F CA sin ( 60 deg)c − W a = 0 F CA =

Wa sin ( 60 deg) c

+

↑Σ Fy = 0;

F CA sin ( 60 deg) − FAB sin ( γ ) = 0

F AB = FCA + Σ F x = 0; →

F CA = 12.9 kip

sin ( 60 deg) sin ( γ )

F AB = 11.9 kip

−F AB cos ( γ ) + FCA cos ( 60 deg) − FAD = 0

F AD = −FAB cos ( γ ) + F CA cos ( 60 deg)

F AD = 2.39 kip

Problem 6-123 The kinetic sculpture requires that each of the three pinned beams be in perfect balance at all times during its slow motion. If each member has a uniform weight density γ and length L, determine the necessary counterweights W1, W2 and W3 which must be added to the ends of each member to keep the system in balance for any position. Neglect the size of the counterweights. Given:

γ = 2

lb ft

L = 3 ft a = 1 ft

Solution: ΣMA = 0;

W1 a cos ( θ ) − γ L cos ( θ ) ⎛⎜

L

⎝2

− a⎟⎞ = 0

⎠

600

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

γ L⎛⎜

⎝2

W1 = +

↑Σ Fy = 0;

L

− a⎞⎟

⎠

a

W1 = 3 lb

R A − W1 − γ L = 0 R A = W1 + γ L

R A = 9 lb

ΣMB = 0;

⎛ L − a⎞ − R ( L − a) cos ( φ ) = 0 ⎟ A ⎝2 ⎠

W2 a cos ( φ ) − γ L cos ( φ ) ⎜

γ L⎛⎜ W2 =

L

⎝2

+

↑Σ Fy = 0;

− a⎟⎞ + R A( L − a)

⎠

a

W2 = 21 lb

R B − W2 − RA − γ L = 0 R B = W2 + R A + γ L

R B = 36 lb

L ΣMC = 0; R B( L − a) cos ( φ ) + γ L⎛⎜ − a⎟⎞ cos ( φ ) − W3 a cos ( φ ) = 0 2

⎝ ⎠ L R B( L − a) + γL⎛⎜ − a⎟⎞ ⎝2 ⎠ W3 = a

W3 = 75 lb

Problem 6-124 The three-member frame is connected at its ends using ball-and-socket joints. Determine the x, y, z components of reaction at B and the tension in member ED. The force acting at D is F. Given:

⎛ 135 ⎞ ⎜ ⎟ F = 200 lb ⎜ ⎟ ⎝ −180 ⎠ a = 6 ft

e = 3 ft

b = 4 ft

f = 1 ft

601

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Engineering Mechanics - Statics

d = 6 ft

Chapter 6

g = 2 ft

c = g+ f Solution: AC and DE are two-force members. Define some vectors

⎛ −e ⎞ ⎜ ⎟ rDE = −b − g ⎜ ⎟ ⎝ a ⎠

uDE =

⎛ −d − e ⎞ ⎜ −b ⎟ rAC = ⎜ ⎟ ⎝ 0 ⎠

uAC =

⎛e⎞ ⎜ ⎟ rBD = − f ⎜ ⎟ ⎝0⎠

⎛e + d⎞ ⎜ −c ⎟ rBA = ⎜ ⎟ ⎝ 0 ⎠

rDE rDE

rAC rAC

Guesses B x = 1 lb

B y = 1 lb

B z = 1 lb

F DE = 1 lb

F AC = 1 lb

Given

⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ + FDE uDE + FAC uAC + F = ⎜B ⎟ ⎝ z⎠

0

⎛⎜ Bx ⎞⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Bz ⎟ = Find ( Bx , By , Bz , FDE , FAC ) ⎜F ⎟ ⎜ DE ⎟ ⎜ FAC ⎟ ⎝ ⎠

(

)

(

)

rBD × FDEuDE + F + rBA × FAC uAC = 0

⎛ Bx ⎞ ⎛ −30 ⎞ F ⎜ ⎟ ⎜ ⎟ ⎛⎜ DE ⎞⎟ ⎛ 270 ⎞ = lb ⎜ By ⎟ = ⎜ −13.333 ⎟ lb⎜ ⎟ ⎜ 16.415 ⎟⎠ F ⎜B ⎟ ⎜ − 12 ⎟ ⎝ AC ⎠ ⎝ ⎝ z ⎠ ⎝ 3.039 × 10 ⎠

602

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Engineering Mechanics - Statics

Chapter 6

Problem 6-125 The four-member "A" frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is F max, determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pon at G only exert force components on the frame. Given: F max = 800 N a = 300 mm b = 600 mm c = 600 mm Solution: ΣMx = 0; b

−P 2 c +

2

2

b +c

Fmax c = 0

F max b

P = 2

2

P = 282.843 N 2

b +c

c

B z + Dz − Fmax

2

b +c

F max c

Bz = 2

2

=0 2

Dz = B z

2

b +c

Dz = Bz

B z = 283 N Dz = 283 N

b

B y + Dy − Fmax

2

b +c

F max b

By = 2

2

=0 2

2

b +c

Dy = By

Dy = B y

B y = 283 N

Dy = 283 N

B x = Dx = 0

Problem 6-126 The structure is subjected to the loading shown. Member AD is supported by a cable AB and a roller at C and fits through a smooth circular hole at D. Member ED is supported by a roller at 603

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Engineering Mechanics - Statics

Chapter 6

g pp y D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of reaction at E and the tension in cable AB. Units Used: 3

kN = 10 N Given:

⎛ 0 ⎞ F = ⎜ 0 ⎟ kN ⎜ ⎟ ⎝ −2.5 ⎠ a = 0.5 m

d = 0.3 m

b = 0.4 m

e = 0.8 m

c = 0.3 m Solution: Guesses

⎛ −c − d ⎞ AB = ⎜ 0 ⎟ ⎜ ⎟ ⎝ e ⎠

F AB = 1 kN

Dx = 1 kN

Dz = 1 kN

Dz2 = 1 kN

E x = 1 kN

E y = 1 kN

MDx = 1 kN⋅ m

MDz = 1 kN⋅ m

Cx = 1 kN

MEx = 1 kN⋅ m

MEy = 1⋅ kN m

Given

⎛⎜ Cx ⎟⎞ ⎛⎜ Dx ⎞⎟ F + FAB +⎜ 0 ⎟+⎜ 0 ⎟ =0 AB ⎜ 0 ⎟ ⎜D ⎟ ⎝ ⎠ ⎝ z⎠ AB

⎛ MDx ⎞ ⎛ 0 ⎞ ⎛ Cx ⎞ ⎛ d ⎞ ⎛c + d⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB ⎞ ⎜ ⎟ ⎛ ⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × F + ⎜ b ⎟ × ⎜ FAB ⎟=0 AB ⎠ ⎝ ⎜M ⎟ ⎝0⎠ ⎜ 0 ⎟ ⎝0⎠ ⎝ 0 ⎠ ⎝ ⎠ ⎝ Dz ⎠

⎛ −Dx ⎞ ⎛ Ex ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ Ey ⎟ = 0 ⎜D − D ⎟ ⎜ ⎟ z⎠ ⎝ 0 ⎠ ⎝ z2

⎛ −MDx ⎞ ⎛ MEx ⎞ ⎛ 0 ⎞ ⎛ Dx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + ⎜ MEy ⎟ + ⎜ a ⎟ × ⎜ 0 ⎟ = 0 ⎜ −M ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ Dz ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ Dz − Dz2 ⎠ 604

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Engineering Mechanics - Statics

Chapter 6

⎛ Cx ⎞ ⎜ ⎟ ⎜ Dx ⎟ ⎜ ⎟ D z ⎜ ⎟ ⎜D ⎟ ⎜ z2 ⎟ ⎜ Ex ⎟ ⎜ ⎟ E ⎜ y ⎟ = Find ( Cx , Dx , Dz , Dz2 , Ex , Ey , FAB , MDx , MDz , MEx , MEy) ⎜F ⎟ ⎜ AB ⎟ ⎜ MDx ⎟ ⎜ ⎟ ⎜ MDz ⎟ ⎜ ⎟ ⎜ MEx ⎟ ⎜M ⎟ ⎝ Ey ⎠ ⎛ Cx ⎞ ⎜ ⎟ ⎛⎜ 0.937 ⎞⎟ D ⎜ x⎟ ⎜ 0 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ Dz ⎟ ⎜ 1.25 ⎟ ⎜ D ⎟ ⎝ 1.25 ⎠ ⎝ z2 ⎠ ⎛⎜ Ex ⎟⎞ ⎛ 0 ⎞ = kN ⎜ Ey ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

⎛⎜ MDx ⎞⎟ ⎛ 0.5 ⎞ = kN⋅ m ⎜ MDz ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

⎛⎜ MEx ⎟⎞ ⎛ 0.5 ⎞ = kN⋅ m ⎜ MEy ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

F AB = 1.562 kN

Problem 6-127 The structure is subjected to the loadings shown.Member AB is supported by a ball-and-socket at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C.

605

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Engineering Mechanics - Statics

Chapter 6

Given: a = 2m

M = 800 N⋅ m

b = 1.5 m

F = 250 N

c = 3m

θ 1 = 60 deg

d = 4m

θ 2 = 45 deg

Solution:

θ 3 = 60 deg

Guesses Bx = 1 N

By = 1 N

Ax = 1 N

Ay = 1 N

Az = 1 N

Cx = 1 N

Cy = 1 N

Cz = 1 N

MBx = 1 N⋅ m

MBy = 1 N⋅ m

MCy = 1 N⋅ m

MCz = 1 N⋅ m

Given

⎛ Ax ⎞ ⎛ Bx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ + ⎜ By ⎟ = 0 ⎜A ⎟ ⎜ ⎟ ⎝ z⎠ ⎝ 0 ⎠ ⎛ c ⎞ ⎛⎜ Bx ⎟⎞ ⎛ M ⎞ ⎛⎜ −MBx ⎞⎟ ⎜ a ⎟ × B + ⎜ 0 ⎟ + −M =0 ⎜ ⎟ ⎜ y ⎟ ⎜ ⎟ ⎜ By ⎟ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎜⎝ 0 ⎟⎠ ⎛ cos ( θ 1) ⎞ ⎛ −Bx ⎞ ⎛ Cx ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ F ⎜ cos ( θ 2) ⎟ + ⎜ −B y ⎟ + ⎜ Cy ⎟ = 0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ cos ( θ 3) ⎠ ⎝ 0 ⎠ ⎝ Cz ⎠ ⎡ ⎛ cos θ ⎞⎤ ⎛ 0 ⎞ ⎢ ⎜ ( 1) ⎟⎥ ⎛ 0 ⎞ ⎛⎜ −Bx ⎞⎟ ⎛⎜ MBx ⎟⎞ ⎛⎜ 0 ⎞⎟ ⎜ 0 ⎟ × F cos ( θ ) + ⎜ 0 ⎟ × −B + M + MCy ⎟ = 0 2 ⎟⎥ ⎜ ⎟ ⎢ ⎜ ⎜ ⎟ ⎜ y ⎟ ⎜ By ⎟ ⎜ ⎝ b + d ⎠ ⎢ ⎜ cos ( θ 3) ⎟⎥ ⎝ b ⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ MCz ⎟⎠ ⎣ ⎝ ⎠⎦

606

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Engineering Mechanics - Statics

Chapter 6

⎛⎜ Ax ⎞⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Az ⎟ ⎜ C ⎟ ⎜ x ⎟ ⎜ Cy ⎟ ⎜ ⎟ ⎜ Cz ⎟ ⎜ ⎟ = Find ( Ax , Ay , Az , Cx , Cy , Cz , Bx , By , MBx , MBy , MCy , MCz ) ⎜ Bx ⎟ ⎜ B ⎟ ⎜ y ⎟ ⎜ MBx ⎟ ⎜ ⎟ ⎜ MBy ⎟ ⎜M ⎟ ⎜ Cy ⎟ ⎜ MCz ⎟ ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎛ −172.3 ⎞ ⎜ Ay ⎟ ⎜ −114.8 ⎟ ⎟ ⎜A ⎟ ⎜ ⎜ ⎟ z 0 ⎜ ⎟= ⎜ ⎟N ⎜ Cx ⎟ 47.3 ⎜ ⎟ ⎜ ⎟ − 61.9 ⎜ ⎟ ⎜ Cy ⎟ ⎜ ⎜ ⎟ ⎝ −125 ⎟⎠ ⎝ Cz ⎠

⎛⎜ MCy ⎞⎟ ⎛ −429 ⎞ = N⋅ m ⎜ MCz ⎟ ⎜⎝ 0 ⎟⎠ ⎝ ⎠

Problem 6-128 Determine the resultant forces at pins B and C on member ABC of the four-member frame. Given: w = 150

lb ft

a = 5 ft b = 2 ft

607

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Engineering Mechanics - Statics

Chapter 6

c = 2 ft e = 4 ft d = a+b−c

Solution: The initial guesses are F CD = 20 lb

F BE = 40 lb

Given F CD( c + d) − FBE

⎛ a + b⎞ + ⎟ ⎝ 2 ⎠

−w( a + b) ⎜

ec

=0 2

2

( d − b) + e F BE e 2

2

( d − b) + e

⎛⎜ FCD ⎞⎟ = Find ( F CD , F BE) ⎜ FBE ⎟ ⎝ ⎠

a − FCD( a + b) = 0

⎛⎜ FCD ⎞⎟ ⎛ 350 ⎞ = lb ⎜ FBE ⎟ ⎜⎝ 1531 ⎟⎠ ⎝ ⎠

Problem 6-129 The mechanism consists of identical meshed gears A and B and arms which are fixed to the gears. The spring attached to the ends of the arms has an unstretched length δ and a stiffness k. If a torque M is applied to gear A, determine the angle θ through which each arm rotates. The gears are each pinned to fixed supports at their centers.

608

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Engineering Mechanics - Statics

Chapter 6

Given:

δ = 100 mm k = 250

N m

M = 6 N⋅ m r =

δ 2

a = 150 mm Solution: ΣMA = 0;

−F r − P a cos ( θ ) + M = 0

ΣMB = 0;

P a cos ( θ ) − F r = 0 2P a cos ( θ ) = M 2k( 2a) sin ( θ ) a cos ( θ ) = M 2k a sin ( 2θ ) = M 2

θ =

1 2

⎞ ⎟ 2 ⎝ 2ka ⎠

asin ⎛⎜

M

θ = 16.1 deg

Problem 6-130 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: kN = 1000 N

609

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Engineering Mechanics - Statics

Chapter 6

Given: F 1 = 20 kN F 2 = 10 kN a = 1.5 m b = 2m Solution: b θ = atan ⎛⎜ ⎟⎞

⎝ a⎠

Guesses

F AG = 1 kN

F BG = 1 kN

F GC = 1 kN

F GF = 1 kN

F AB = 1 kN

F BC = 1 kN

F CD = 1 kN

F CF = 1 kN

F DF = 1 kN

F DE = 1 kN

F EF = 1 kN

Given −F AB cos ( θ ) + F BC = 0 −F AB sin ( θ ) − FBG = 0 F GC cos ( θ ) + FGF − FAG = 0 F GC sin ( θ ) + F BG − F 1 = 0 −F BC + F CD − F GC cos ( θ ) + FCF cos ( θ ) = 0 −F GC sin ( θ ) − F CF sin ( θ ) = 0 −F CD + F DE cos ( θ ) = 0 −F DF − F DE sin ( θ ) = 0 −F GF − F CF cos ( θ ) + FEF = 0

610

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Engineering Mechanics - Statics

Chapter 6

F DF + F CF sin ( θ ) − F 2 = 0 −F DE cos ( θ ) − F EF = 0

611

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Engineering Mechanics - Statics

Chapter 6

⎛ FAG ⎞ ⎜ ⎟ ⎜ FBG ⎟ ⎜ ⎟ ⎜ FGC ⎟ ⎜F ⎟ ⎜ GF ⎟ ⎜ FAB ⎟ ⎜ ⎟ ⎜ FBC ⎟ = Find ( FAG , FBG , FGC , FGF , FAB , FBC , FCD , FCF , FDF , FDE , FEF) ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎝ EF ⎠ ⎛ FAG ⎞ ⎜ ⎟ ⎜ FBG ⎟ ⎛⎜ 13.13 ⎟⎞ ⎜ ⎟ ⎜ 17.50 ⎟ F GC ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 3.13 ⎟ ⎜ GF ⎟ ⎜ 11.25 ⎟ ⎜ FAB ⎟ ⎜ −21.88 ⎟ ⎜ ⎟ ⎜ ⎟ F ⎜ BC ⎟ = ⎜ −13.13 ⎟ kN ⎜ F ⎟ ⎜ −9.37 ⎟ ⎜ CD ⎟ ⎜ ⎟ ⎜ FCF ⎟ ⎜ −3.13 ⎟ ⎜ ⎟ ⎜ 12.50 ⎟ F ⎜ DF ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −15.62 ⎟ ⎜ FDE ⎟ ⎝ 9.37 ⎠ ⎜F ⎟ ⎝ EF ⎠

Positive (T) Negative (C)

Problem 6-131 The spring has an unstretched length δ. Determine the angle θ for equilibrium if the uniform links each have a mass mlink.

612

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Engineering Mechanics - Statics

Chapter 6

Given: mlink = 5 kg

δ = 0.3 m N

k = 400

m

a = 0.1 m b = 0.6 m g = 9.81

m 2

s Solution: Guesses

θ = 10 deg F BD = 1 N Ex = 1 N Given mlink g

a+b

cos ( θ ) − FBD b cos ( θ ) + E x b sin ( θ ) = 0

2

−2 mlink g

a+b 2

cos ( θ ) + Ex 2 b sin ( θ ) = 0

F BD = k( 2 b sin ( θ ) − δ )

⎛ FBD ⎞ ⎜ ⎟ ⎜ Ex ⎟ = Find ( FBD , Ex , θ ) ⎜ ⎟ ⎝ θ ⎠

θ = 21.7 deg

Problem 6-132 The spring has an unstretched length δ. Determine the mass mlink of each uniform link if the angle for equilibrium is θ. Given:

δ = 0.3 m

613

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Engineering Mechanics - Statics

Chapter 6

θ = 20 deg k = 400

N m

a = 0.1 m b = 0.6 m g = 9.81

m 2

s Solution: Guesses

Ey = 1 N

mlink = 1 kg

Fs = 1 N Given F s = ( 2 b sin ( θ ) − δ ) k mlink g

a+b cos ( θ ) − Fs b cos ( θ ) + E y b sin ( θ ) = 0 2

−2mlink g

a+b cos ( θ ) + E y2b sin ( θ ) = 0 2

⎛ mlink ⎞ ⎜ ⎟ ⎜ Fs ⎟ = Find ( mlink , Fs , Ey) ⎜ E ⎟ ⎝ y ⎠ mlink = 3.859 kg y = sin ( θ ) 2( b) y = 2 b sin ( θ ) F s = ( y − δ ) ( k) F s = 44.17 N

614

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Engineering Mechanics - Statics

ΣMA = 0;

Chapter 6

a+b

( cos (θ ) ) = 0 ⎡⎣Ey ( 2) ( a + b) sin ( θ ) − 2( M) ( g)⎤⎦ 2 ⎡E ( 2) sin ( θ ) − 2 ( m) ⎢ y ⎣

g⎤

⎥ ( cos ( θ ) = 0

2⎦

⇒ E y ( 2 sin ( θ ) ) = m ( g) ( cos ( θ ) ) Ey = ΣMC = 0;

m( g) ( cos ( θ ) ) 2 sin ( θ )

m( g) ( cos ( θ ) ) 2 sin ( θ )

( a + b) sin ( θ ) + m( g) ⎛⎜

a + b⎞

⎟ cos ( θ ) − Fs( b cos ( θ ) ) = 0 ⎝ 2 ⎠

b m = Fs g ( a + b) m = 3.859 kg

Problem 6-133 Determine the horizontal and vertical components of force that the pins A and B exert on the two-member frame. Given: w = 400

N m

a = 1.5 m b = 1m c = 1m F = 0N

θ = 60 deg Solution: Guesses Ax = 1 N

Ay = 1 N

Bx = 1 N

615

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Engineering Mechanics - Statics

By = 1 N

Cx = 1 N

Chapter 6

Cy = 1 N

Given −w a

a + Cx a sin ( θ ) − Cy a cos ( θ ) = 0 2

F c − Cx c − Cy b = 0

Ay − Cy − w a cos ( θ ) = 0

− Ax − Cx + w a sin ( θ ) = 0

Cx − Bx − F = 0 Cy + By = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜B ⎟ ⎜ x ⎟ = Find A , A , B , B , C , C ( x y x y x y) ⎜ By ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ ⎝ Cy ⎠

⎛⎜ Ax ⎞⎟ ⎛ 300.0 ⎞ = N ⎜ Ay ⎟ ⎜⎝ 80.4 ⎟⎠ ⎝ ⎠

⎛⎜ Bx ⎟⎞ ⎛ 220 ⎞ = N ⎜ By ⎟ ⎜⎝ 220 ⎟⎠ ⎝ ⎠

Problem 6-134 Determine the horizontal and vertical components of force that the pins A and B exert on the two-member frame. Given: w = 400

N m

a = 1.5 m b = 1m c = 1m F = 500 N

θ = 60 deg

616

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Engineering Mechanics - Statics

Chapter 6

Solution: Guesses Ax = 1 N

Ay = 1 N

Bx = 1 N

By = 1 N

Cx = 1 N

Cy = 1 N

Given −w a

a 2

+ Cx a sin ( θ ) − Cy a cos ( θ ) = 0

Ay − Cy − w a cos ( θ ) = 0 Cx + Bx − F = 0

F c − Cx c − Cy b = 0 − Ax − Cx + w a sin ( θ ) = 0

⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜B ⎟ ⎜ x ⎟ = Find A , A , B , B , C , C ( x y x y x y) ⎜ By ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ ⎟ ⎝ Cy ⎠

Cy − By = 0

⎛⎜ Ax ⎞⎟ ⎛ 117.0 ⎞ = N ⎜ Ay ⎟ ⎜⎝ 397.4 ⎟⎠ ⎝ ⎠

⎛⎜ Bx ⎟⎞ ⎛ 97.4 ⎞ = N ⎜ By ⎟ ⎜⎝ 97.4 ⎟⎠ ⎝ ⎠

Problem 6-135 Determine the force in each member of the truss and indicate whether the members are in tension or compression. Units Used: kip = 1000 lb Given: F 1 = 1000 lb

b = 8 ft

F 2 = 500 lb

c = 4 ft

a = 4 ft

Solution: Joint B: 617

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 6

Initial Guesses: Given F BC = 100 lb

F BA = 150 lb

ΣF x = 0;

b c F 1 − F BC cos ⎛⎜ atan ⎛⎜ ⎟⎞ ⎞⎟ − FBA cos ⎜⎛ atan ⎛⎜ ⎟⎞ ⎞⎟ = 0 a a

ΣF y = 0;

−F BC sin ⎜⎛ atan ⎛⎜

⎝

⎝ ⎠⎠

⎝

⎝ ⎠⎠

⎛ ⎛ c ⎞⎞ ⎟ ⎟ + FBA sin ⎜atan ⎜ ⎟ ⎟ − F2 = 0 ⎝ a ⎠⎠ ⎝ ⎝ a ⎠⎠

⎝

b ⎞⎞

⎛⎜ FBC ⎞⎟ = Find ( F BC , F BA) ⎜ FBA ⎟ ⎝ ⎠ F BC = 373 lb(C) F BA = 1178.51 lb F BA = 1.179 kip (C) Joint A: ΣF y = 0;

c F AC − FBA sin ⎜⎛ atan ⎛⎜ ⎟⎞ ⎞⎟ = 0 a

(

)

⎝

c

F AC = FBA

⎝ ⎠⎠

2

a

2

a +c 2

a F AC = 833 lb (T)

Problem 6-136 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: 3

kip = 10 lb Given: F = 1000 lb a = 10 ft b = 10 ft

618

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Engineering Mechanics - Statics

Solution:

Chapter 6

b θ = atan ⎛⎜ ⎞⎟

⎝ a⎠

Guesses F AB = 1 lb

F AG = 1 lb

F BC = 1 lb

F BG = 1 lb

F CD = 1 lb

F CE = 1 lb

F CG = 1 lb

F DE = 1 lb

F EG = 1 lb Given F AB + F AG cos ( θ ) = 0 −F AB + F BC = 0 F BG = 0 F CD − F BC − F CG cos ( θ ) = 0 F CE − F + F CG sin ( θ ) = 0 −F CD − F DE cos ( θ ) = 0 F DE cos ( θ ) − F EG = 0 −F CE − FDE sin ( θ ) = 0 F EG − F AG cos ( θ ) + F CG cos ( θ ) = 0 −F AG sin ( θ ) − FBG − FCG sin ( θ ) = 0

⎛ FAB ⎞ ⎜ ⎟ ⎜ FAG ⎟ ⎜F ⎟ ⎜ BC ⎟ ⎜ FBG ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( FAB , FAG , FBC , FBG , FCD , FCE , FCG , FDE , FEG) ⎜F ⎟ ⎜ CE ⎟ ⎜ FCG ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎟ ⎜ ⎝ FEG ⎠

619

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Engineering Mechanics - Statics

Chapter 6

⎛ FAB ⎞ ⎜ ⎟ ⎛ 333 ⎞ ⎜ FAG ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ −471 ⎟ ⎜ BC ⎟ ⎜ 333 ⎟ ⎜ FBG ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ F 667 = ⎟ lb ⎜ CD ⎟ ⎜ ⎜ F ⎟ ⎜ 667 ⎟ ⎟ ⎜ CE ⎟ ⎜ ⎜ FCG ⎟ ⎜ 471 ⎟ ⎜ ⎟ ⎜ −943 ⎟ F ⎜ DE ⎟ ⎜ ⎟ ⎟ ⎝ −667 ⎠ ⎜ ⎝ FEG ⎠

Positive (T), Negative (C)

Problem 6-137 Determine the force in members AB, AD, and AC of the space truss and state if the members are in tension or compression. The force F is vertical. Units Used: 3

kip = 10 lb Given: F = 600 lb a = 1.5 ft b = 2 ft c = 8 ft Solution:

⎛a⎞ AB = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

⎛ −a ⎞ AC = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ AD = ⎜ −c ⎟ ⎜ ⎟ ⎝b⎠ Guesses

F AB = 1 lb

F AC = 1 lb

F AD = 1 lb 620

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Given

Chapter 6

⎛ 0 ⎞ F AB + FAC + FAD +⎜ 0 ⎟ =0 ⎜ ⎟ AB AC AD ⎝ −F ⎠ AB

AC

AD

⎛ FAB ⎞ ⎜ ⎟ F ⎜ AC ⎟ = Find ( FAB , FAC , FAD) ⎜F ⎟ ⎝ AD ⎠ ⎛ FAB ⎞ ⎛ −1.221 ⎞ ⎜ ⎟ ⎜ ⎟ F ⎜ AC ⎟ = ⎜ −1.221 ⎟ kip ⎜ F ⎟ ⎝ 2.474 ⎠ ⎝ AD ⎠

Positive (T) Negative (C)

621

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Engineering Mechanics - Statics

Chapter 7

Problem 7-1 The column is fixed to the floor and is subjected to the loads shown. Determine the internal normal force, shear force, and moment at points A and B. Units Used: 3

kN = 10 N Given: F 1 = 6 kN F 2 = 6 kN F 3 = 8 kN a = 150 mm b = 150 mm c = 150 mm Solution: Free body Diagram: The support reaction need not be computed in this case. Internal Forces: Applying equations of equillibrium to the top segment sectioned through point A, we have + Σ F x = 0; →

VA = 0

+

NA − F1 − F2 = 0

NA = F 1 + F 2

NA = 12.0 kN

F 1 a − F2 b − MA = 0

MA = F1 a − F 2 b

MA = 0 kN⋅ m

↑Σ Fy = 0; ΣMA = 0;

VA = 0

Applying equations of equillibrium to the top segment sectioned through point B, we have + Σ F x = 0; →

VB = 0

+

NB − F1 − F2 − F3 = 0

NB = F 1 + F 2 + F 3

NB = 20.0 kN

F 1 a − F 2 b − F3 c + MB = 0

MB = −F1 a + F 2 b + F3 c

MB = 1.20 kN⋅ m

↑Σ Fy = 0;

+ΣMB = 0;

VB = 0

Problem 7-2 The axial forces act on the shaft as shown. Determine the internal normal forces at points A and B. 622

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Engineering Mechanics - Statics

Chapter 7

Given: F 1 = 20 lb F 2 = 50 lb F 3 = 10 lb

Solution: Section A: ΣF z = 0;

F 2 − 2 F1 − NA = 0 NA = F 2 − 2 F 1 NA = 10.00 lb

Section B: ΣF z = 0;

F 2 − 2 F1 − NA + NB = 0 NB = −F 2 + 2 F1 + NA NB = 0.00 lb

Problem 7-3 The shaft is supported by smooth bearings at A and B and subjected to the torques shown. Determine the internal torque at points C, D, and E. Given: M1 = 400 N⋅ m M2 = 150 N⋅ m M3 = 550 N⋅ m

623

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Engineering Mechanics - Statics

Chapter 7

Solution: Section C: ΣMx = 0;

TC = 0

Section D: ΣMx = 0;

TD − M 1 = 0 TD = M 1 TD = 400.00 N⋅ m

Section E: ΣMx = 0;

M 1 + M 2 − TE = 0 TE = M 1 + M 2 TE = 550.00 N⋅ m

Problem 7-4 Three torques act on the shaft. Determine the internal torque at points A, B, C, and D. Given: M1 = 300 N⋅ m M2 = 400 N⋅ m M3 = 200 N⋅ m Solution: Section A: ΣΜx = 0;

− TA + M 1 − M 2 + M 3 = 0 TA = M 1 − M 2 + M 3 TA = 100.00 N⋅ m

Section B: ΣMx = 0;

TB + M 3 − M 2 = 0

624

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Engineering Mechanics - Statics

Chapter 7

TB = − M 3 + M 2 TB = 200.00 N⋅ m Section C: ΣΜx = 0;

− TC + M 3 = 0 TC = M 3 TC = 200.00 N⋅ m

Section D: ΣΜx = 0;

TD = 0

Problem 7-5 The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the normal force, shear force, and moment at a section passing through (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the force F 2. Units Used: 3

kip = 10 lb Given: F 1 = 2.5 kip F 2 = 3 kip w = 75

lb ft

a = 6 ft

b = 12 ft c = 2 ft

Solution: Σ MB = 0;

625

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

b − Ay( b + c) + F1 ( a + b + c) + w b⎛⎜ c + ⎞⎟ + F2 c = 0 ⎝ 2⎠ F1 ( a + b + c) + w b⎛⎜ c +

⎝

Ay =

b⎞

⎟ + F2 c

2⎠

b+c

Ay = 4514 lb + Σ F x = 0; → +

↑Σ Fy = 0;

B x = 0 lb Ay − F1 − w b − F2 + By = 0 By = − Ay + F1 + w b + F2

Σ MC = 0; + Σ F x = 0; → +

↑

Σ F y = 0; Σ MD = 0;

+ → +

↑

Σ F x = 0; Σ F y = 0;

F 1 a + Mc = 0

B y = 1886 lb

MC = −F 1 a

NC = 0 lb

MC = −15.0 kip⋅ ft NC = 0.00 lb

−F 1 + Ay − V C = 0 −MD + By c = 0

VC = Ay − F1

V C = 2.01 kip

MD = B y c

MD = 3.77 kip⋅ ft

ND = 0 lb

ND = 0.00 lb

VD − F2 + By = 0

VD = F2 − By

V D = 1.11 kip

Problem 7-6 Determine the internal normal force, shear force, and moment at point C. Given: M = 400 lb⋅ ft a = 4 ft b = 12 ft

626

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Engineering Mechanics - Statics

Chapter 7

Solution: Beam: ΣMB = 0;

M − Ay( a + b) = 0 Ay =

M a+b

Ay = 25.00 lb

Segment AC: ΣF x = 0;

NC = 0

ΣF y = 0;

Ay − VC = 0 VC = Ay V C = 25.00 lb

ΣMC = 0;

− Ay a + MC = 0 MC = Ay a MC = 100.00 lb⋅ ft

Problem 7-7 Determine the internal normal force, shear force, and moment at point C. Units Used: kN = 103 N Given: F 1 = 30 kN F 2 = 50 kN F 3 = 25 kN a = 1.5 m b = 3m

θ = 30 deg Solution: ΣF x = 0;

−NC + F 3 cos ( θ ) = 0 627

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

NC = F3 cos ( θ ) NC = 21.7 kN V C − F 2 − F 3 sin ( θ ) = 0

ΣF y = 0;

V C = F2 + F3 sin ( θ ) V C = 62.50 kN ΣMC = 0; −MC − F2 b − F 3 sin ( θ ) 2 b = 0 MC = −F 2 b − F3 sin ( θ ) 2b MC = −225.00 kN⋅ m

Problem 7-8 Determine the normal force, shear force, and moment at a section passing through point C. Assume the support at A can be approximated by a pin and B as a roller. Units used: 3

kip = 10 lb Given: F 1 = 10 kip

a = 6 ft

F 2 = 8 kip

b = 12 ft

kip ft

c = 12 ft

w = 0.8

d = 6 ft

Solution: ΣMA = 0;

⎛ b + c ⎞ − F ( b + c + d) + B ( b + c) + F a = ⎟ 2 y 1 ⎝ 2 ⎠

−w( b + c) ⎜

w

( b + c)

By = + Σ F x = 0; →

2

0

2

+ F 2 ( b + c + d) − F 1 a B y = 17.1 kip

b+c NC = 0 628

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Engineering Mechanics - Statics

+

↑Σ Fy = 0;

Chapter 7

VC − w c + By − F2 = 0 VC = w c − By + F2

V C = 0.5 kip

⎛ c ⎞ + B c − F ( c + d) = ⎟ y 2 ⎝ 2⎠

−MC − w c⎜

ΣMC = 0;

⎛ c2 ⎞ ⎟ + B y c − F 2 ( c + d) ⎝2⎠

MC = −w⎜

0

MC = 3.6 kip⋅ ft

Problem 7-9 The beam AB will fail if the maximum internal moment at D reaches Mmax or the normal force in member BC becomes Pmax. Determine the largest load w it can support. Given: Mmax = 800 N⋅ m P max = 1500 N a = 4m b = 4m c = 4m d = 3m Solution:

⎛ a + b ⎞ − A ( a + b) = ⎟ y ⎝ 2 ⎠

w ( a + b) ⎜ Ay =

w( a + b) 2

⎛ a⎞ − A a + M = ⎟ y D ⎝2⎠

w a⎜

MD = w

0

⎛ a b⎞ ⎜2 ⎟ ⎝ ⎠

Ay − w( a + b) +

T=

0

w( a + b)

⎛ d ⎞T = ⎜ 2 2⎟ ⎝ c +d ⎠ 2

c +d

0

2

2d

Assume the maximum moment has been reached 629

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Engineering Mechanics - Statics

Chapter 7

MD = Mmax

2 MD

w1 =

ab

w1 = 100

N m

w2 = 225

N m

Assume that the maximum normal force in BC has been reached T = Pmax

T2 d

w2 =

2

( a + b) c + d

2

w = min ( w1 , w2 )

Now choose the critical load

w = 100

N m

Problem 7-10 Determine the shear force and moment acting at a section passing through point C in the beam. Units Used: 3

kip = 10 lb Given: w = 3

kip ft

a = 6 ft b = 18 ft Solution: ΣMB = 0;

− Ay b + Ay =

ΣMC = 0;

1 6

− Ay a +

1 2

wb

⎛ b⎞ = ⎜3⎟ ⎝ ⎠

0

Ay = 9 kip

wb

a⎞ ⎛ a⎞ w ⎟ a ⎜ ⎟ + MC = ⎜ 2 ⎝ b⎠ ⎝ 3 ⎠ 1⎛

0

3

wa MC = Ay a − 6b MC = 48 kip⋅ ft +

↑

Σ F y = 0;

Ay −

1 2

⎛w a ⎞ a − V = ⎜ ⎟ C ⎝ b⎠

0

2

wa VC = Ay − 2b

630

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Engineering Mechanics - Statics

Chapter 7

V C = 6 kip

Problem 7-11 Determine the internal normal force, shear force, and moment at points E and D of the compound beam. Given: M = 200 N⋅ m

c = 4m

F = 800 N

d = 2m

a = 2m

e = 2m

b = 2m Solution: Segment BC : M d+e

−M + Cy( d + e) = 0

Cy =

−B y + Cy = 0

B y = Cy

Segment EC : −NE = 0

NE = 0 N

V E + Cy = 0

NE = 0.00

V E = −Cy

−ME − M + Cy e = 0

V E = −50.00 N

ME = Cy e − M

ME = −100.00 N⋅ m

Segment DB : −ND = 0

ND = 0 N

ND = 0.00

VD − F + By = 0

VD = F − By

V D = 750.00 N

−MD − F b + By( b + c) = 0 MD = −F b + B y( b + c)

MD = −1300 N⋅ m

631

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Engineering Mechanics - Statics

Chapter 7

Problem 7-12 The boom DF of the jib crane and the column DE have a uniform weight density γ. If the hoist and load have weight W, determine the normal force, shear force, and moment in the crane at sections passing through points A, B, and C. Treat the boom tip, beyond the hoist, as weightless. Given: W = 300 lb

γ = 50

lb ft

a = 7 ft b = 5 ft c = 2 ft d = 8 ft e = 3 ft Solution: + ΣF x = 0; →

−NA = 0

+

VA − W − γ e = 0

↑ ΣFy = 0;

NA = 0 lb

VA = W + γ e ΣΜΑ = 0;

+ → +

ΣF x = 0;

↑ ΣFy = 0; ΣΜΒ = 0;

NA = 0.00 lb

V A = 450 lb

⎛e⎞ − We = ⎟ ⎝ 2⎠ ⎛ e2 ⎞ MA = γ ⎜ ⎟ + W e ⎝2⎠ MA − γ e⎜

0

−NB = 0

MA = 1125.00 lb⋅ ft NB = 0 lb

V B − γ ( d + e) − W = 0

V B = γ ( d + e) + W

⎛ d + e ⎞ − W( d + e) = ⎟ ⎝ 2 ⎠

MB − γ ( d + e) ⎜ MB =

1 2

NB = 0.00 lb

2

γ ( d + e) + W( d + e)

V B = 850 lb

0

MB = 6325.00 lb⋅ ft

632

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Engineering Mechanics - Statics

Chapter 7

+ →

ΣF x = 0;

VC = 0

+

ΣF y = 0;

NC − ( c + d + e) γ − W − γ ( b) = 0

↑

V C = 0 lb

V C = 0.00 lb

NC = γ ( c + d + e + b) + W ΣΜC = 0;

NC = 1200.00 lb

MC − ( c + d + e) γ

⎛ c + d + e ⎞ − W( c + d + e) = ⎜ 2 ⎟ ⎝ ⎠

MC = ( c + d + e) γ

⎛ c + d + e ⎞ + W( c + d + e) ⎜ 2 ⎟ ⎝ ⎠

0

MC = 8125.00 lb⋅ ft

Problem 7-13 Determine the internal normal force, shear force, and moment at point C. Units Used: 3

kip = 10 lb Given: a = 0.5 ft

d = 8 ft

b = 2 ft

e = 4 ft

c = 3 ft

w = 150

lb ft

Solution: Entire beam: ΣMA = 0;

⎛ d + e ⎞ + T ( a + b) = ⎟ ⎝ 2 ⎠

−w( d + e) ⎜

w ( d + e) T = 2 ( a + b) ΣF x = 0;

2

T = 4.32 kip

Ax − T = 0 Ax = T

ΣF y = 0;

0

Ax = 4.32 kip

Ay − w( d + e) = 0 Ay = w( d + e)

Ay = 1.80 kip

Segment AC: 633

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

ΣF x = 0;

Ax + NC = 0

NC = − A x

NC = −4.32 kip

ΣF y = 0;

Ay − w c − VC = 0

VC = Ay − w c

V C = 1.35 kip

ΣMC = 0;

− Ay c + w c⎜

⎛c⎞ + M = ⎟ C ⎝2⎠

MC = Ay c − w

0

⎛ c2 ⎞ ⎜ ⎟ ⎝2⎠

MC = 4.72 kip⋅ ft

Problem 7-14 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: 3

kN = 10 N Given: w = 400

N m

a = 2.5 m b = 3m c = 6m

Solution: ΣMA = 0;

−1 2

⎛ 2 c⎞ + F ⎛ a ⎞ c = ⎜3 ⎟ BC ⎜ ⎝ ⎠ 2 2⎟ ⎝ a +c ⎠

wc

F BC = +

→ Σ Fx = 0;

↑Σ Fy = 0;

3

wc

⎛ a2 + c2 ⎞ ⎜ ⎟ ⎝ a ⎠

⎛ c ⎞F − A = x ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ Ax =

+

1

Ay −

⎛ c ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ 1 2

wc +

0

F BC = 2080 N

0

Ax = 1920 N

⎛ a ⎞F = ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

0

634

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Ay = + Σ F x = 0; → +

↑Σ Fy = 0;

Chapter 7

1 2

wc −

⎛ a ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

ND − Ax = 0 Ay −

Ay = 400 N

ND = A x

b⎞ ⎜w ⎟ b − VD = c⎠

1⎛ 2⎝

0

2⎞

1 ⎛b V D = A y − w⎜ 2 ⎝ c

ΣF y = 0;

− Ay b +

⎟ ⎠

V D = 100 N

b b w ⎞⎟ b⎛⎜ ⎟⎞ + MD = ⎜ 2⎝ c⎠ ⎝3⎠ 1⎛

MD = Ay b −

1 6

ND = 1.920 kN

0

⎛ b3 ⎞ w⎜ ⎟ ⎝c⎠

MD = 900 N m

Problem 7-15 The beam has weight density γ. Determine the internal normal force, shear force, and moment at point C. Units Used: kip = 103 lb Given:

γ = 280

lb ft

a = 3 ft b = 7 ft c = 8 ft d = 6 ft

635

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Solution: c θ = atan ⎛⎜ ⎟⎞

W = γ ( a + b)

⎝ d⎠

Guesses Ax = 1 lb

Ay = 1 lb

B x = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb ft

Given Entire beam: Ax − Bx = 0

Ay − W = 0

Bx c − W

⎛ d⎞ = ⎜ 2⎟ ⎝ ⎠

0

Bottom Section Ax − NC cos ( θ ) + V C sin ( θ ) = 0 Ay − W

⎛ a ⎞ − N sin ( θ ) − V cos ( θ ) = ⎜ ⎟ C C ⎝ a + b⎠

MC − VC a − W

⎛ a ⎞ ⎜ ⎟ ⎝ a + b⎠

⎛ a ⎞ cos ( θ ) = ⎜ 2⎟ ⎝ ⎠

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , N , V , M ) x y x C C C ⎜ NC ⎟ ⎜ ⎟ ⎜ VC ⎟ ⎜ MC ⎟ ⎝ ⎠

0

0

⎛ Ax ⎞ ⎛ 1.05 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 2.80 ⎟ kip ⎜ B ⎟ ⎜⎝ 1.05 ⎟⎠ ⎝ x⎠

⎛ NC ⎞ ⎛ 2.20 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ VC ⎠ ⎝ 0.34 ⎠ MC = 1.76 kip⋅ ft

Problem 7-16 Determine the internal normal force, shear force, and moment at points C and D of the beam.

636

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Units Used: 3

kip = 10 lb Given: w1 = 60

lb ft

a = 12 ft b = 15 ft

lb w2 = 40 ft

c = 10 ft d = 5 ft

F = 690 lb e = 12

f = 5 e θ = atan ⎛⎜ ⎟⎞

Solution:

⎝ f⎠

Guesses B y = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb⋅ ft

ND = 1 lb

V D = 1 lb

MD = 1 lb⋅ ft Given

⎛ b ⎞ ... = ⎟ ⎝ 2⎠

B y b − F sin ( θ ) ( b + c) − w2 b⎜ +

−1 2

(w1 − w2)b⎛⎜ 3 ⎟⎞

0

b

⎝ ⎠

−NC − F cos ( θ ) = 0 b − a⎞ ⎟ ( b − a) ... = 0 ⎝ b ⎠ + B y − w2 ( b − a) − F sin ( θ ) VC −

(w1 − w2) ⎛⎜ 2 1

−ND − F cos ( θ ) = 0 V D − F sin ( θ ) = 0 −MD − F sin ( θ ) d = 0

⎛ b − a⎞ − ⎟ ⎝ 2 ⎠

−MC − w2 ( b − a) ⎜

b − a⎞ ⎛ b − a⎞ ⎟ ( b − a) ⎜ 3 ⎟ ... = 0 ⎝ ⎠ ⎝ b ⎠

⎛ w1 − w2 ) ⎜ ( 2 1

+ B y ( b − a) − F sin ( θ ) ( c + b − a)

637

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

⎛ By ⎞ ⎜ ⎟ ⎜ NC ⎟ ⎜ VC ⎟ ⎜ ⎟ M ⎜ C ⎟ = Find ( By , NC , VC , MC , ND , VD , MD) ⎜N ⎟ ⎜ D⎟ ⎜ VD ⎟ ⎜ ⎟ ⎝ MD ⎠

⎛ NC ⎞ ⎛ −265 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ VC ⎟ = ⎜ −649 ⎟ lb ⎜ ND ⎟ ⎜ −265 ⎟ ⎜ ⎟ ⎜ 637 ⎟ ⎠ ⎝ VD ⎠ ⎝ ⎛ MC ⎞ ⎛ −4.23 ⎞ ⎜ ⎟=⎜ ⎟ kip⋅ ft ⎝ MD ⎠ ⎝ −3.18 ⎠

Problem 7-17 Determine the normal force, shear force, and moment acting at a section passing through point C. 3

kip = 10 lb

Units Used: Given:

F 1 = 800 lb F 2 = 700 lb F 3 = 600 lb

θ = 30 deg a = 1.5 ft b = 1.5 ft c = 3 ft d = 2 ft e = 1 ft f = a+b+c−d−e Solution: Guesses

B y = 1 lb

Ax = 1 lb

Ay = 1 lb

NC = 1 lb

V C = 1 lb

MC = 1 lb⋅ ft

638

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given − Ax + V C sin ( θ ) + NC cos ( θ ) = 0 Ay − V C cos ( θ ) + NC sin ( θ ) = 0 MC − A x ( a) sin ( θ ) − Ay ( a) cos ( θ ) = 0 − Ax + F 1 sin ( θ ) − F3 sin ( θ ) = 0 Ay + B y − F2 − F1 cos ( θ ) − F3 cos ( θ ) = 0 −F 1 ( a + b) − F 2 ( a + b + c) cos ( θ ) − F 3 cos ( θ ) ( a + b + c + d + e) cos ( θ ) ... = 0 + F 3 sin ( θ ) f sin ( θ ) + B y2 ( a + b + c) cos ( θ )

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , N , V , M ) x y y C C C ⎜ NC ⎟ ⎜ ⎟ ⎜ VC ⎟ ⎜ MC ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 100 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 985 ⎟ lb ⎜ B ⎟ ⎜⎝ 927 ⎟⎠ ⎝ y⎠ ⎛ NC ⎞ ⎛ −406 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ VC ⎠ ⎝ 903 ⎠ MC = 1.355 kip⋅ ft

Problem 7-18 Determine the normal force, shear force, and moment acting at a section passing through point D. Units Used:

kip = 103 lb

639

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Engineering Mechanics - Statics

Chapter 7

Given: F 1 = 800 lb F 2 = 700 lb F 3 = 600 lb

θ = 30 deg a = 1.5 ft b = 1.5 ft c = 3 ft d = 2 ft e = 1 ft f = a+b+c−d−e Solution: Guesses

B y = 1 lb

Ax = 1 lb

Ay = 1 lb

ND = 1 lb

V D = 1 lb

MD = 1 lb⋅ ft

Given V D sin ( θ ) − ND cos ( θ ) − F3 sin ( θ ) = 0 B y + VD cos ( θ ) + ND sin ( θ ) − F 3 cos ( θ ) = 0 −MD − F3 e + By( e + f) cos ( θ ) = 0 Ay + B y − F2 − F1 cos ( θ ) − F3 cos ( θ ) = 0 − Ax + F 1 sin ( θ ) − F3 sin ( θ ) = 0 −F 1 ( a + b) − F 2 ( a + b + c) cos ( θ ) − F 3 cos ( θ ) ( a + b + c + d + e) cos ( θ ) ... = 0 + F 3 sin ( θ ) f sin ( θ ) + B y2 ( a + b + c) cos ( θ )

⎛⎜ Ax ⎟⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ By ⎟ = Find ( A , A , B , N , V , M ) x y y D D D ⎜ ND ⎟ ⎜ ⎟ ⎜ VD ⎟ ⎜ MD ⎟ ⎝ ⎠

⎛ Ax ⎞ ⎛ 100 ⎞ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 985 ⎟ lb ⎜ B ⎟ ⎜⎝ 927 ⎟⎠ ⎝ y⎠ ⎛ ND ⎞ ⎛ −464 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ VD ⎠ ⎝ −203 ⎠ MD = 2.61 kip⋅ ft

640

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Engineering Mechanics - Statics

Chapter 7

Problem 7-19 Determine the normal force, shear force, and moment at a section passing through point C. Units Used: 3

kN = 10 N Given: P = 8 kN

c = 0.75 m

a = 0.75m

d = 0.5 m

b = 0.75m

r = 0.1 m

Solution: ΣMA = 0;

−T ( d + r) + P( a + b + c) = 0 T = P

⎛ a + b + c⎞ T = ⎜ ⎟ ⎝ d+r ⎠

30 kN

+ Σ F x = 0; →

Ax = T

Ax = 30 kN

+

Ay = P

Ay = 8 kN

↑Σ Fy = 0;

+ Σ F x = 0; →

−NC − T = 0 NC = −T

+

↑Σ Fy = 0;

VC + P = 0 V C = −P

ΣMC = 0;

NC = −30 kN

V C = −8 kN

−MC + P c = 0 MC = P c

MC = 6 kN⋅ m

Problem 7-20 The cable will fail when subjected to a tension Tmax. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading. Units Used: 3

kN = 10 N 641

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given: Tmax = 2 kN a = 0.75 m b = 0.75 m c = 0.75 m d = 0.5 m r = 0.1 m Solution: ΣMA = 0;

−Tmax( r + d) + P ( a + b + c) = 0

⎛ d+r ⎞ ⎟ ⎝ a + b + c⎠

P = Tmax⎜

P = 0.533 kN

+ Σ F x = 0; →

Tmax − Ax = 0

Ax = Tmax

Ax = 2 kN

+

Ay − P = 0

Ay = P

Ay = 0.533 kN

+ Σ F x = 0; →

−NC − Ax = 0

NC = − A x

NC = −2 kN

VC = Ay

V C = 0.533 kN

MC = Ay c

MC = 0.400 kN⋅ m

↑Σ Fy = 0;

+

↑Σ Fy = 0; ΣMC = 0;

−V C + Ay = 0 −MC + A y c = 0

Problem 7-21 Determine the internal shear force and moment acting at point C of the beam. Units Used: 3

kip = 10 lb Given: w = 2

kip ft

a = 9 ft

642

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Engineering Mechanics - Statics

Chapter 7

Solution: ΣF x = 0;

NC = 0

ΣF y = 0;

wa

ΣMC = 0;

MC −

2

NC = 0

wa

−

− VC = 0

2

⎛ w a⎞a + ⎛ w a ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠

VC = 0

⎛ a⎞ = ⎜3⎟ ⎝ ⎠

0

2

MC =

wa

MC = 54.00 kip⋅ ft

3

Problem 7-22 Determine the internal shear force and moment acting at point D of the beam. Units Used: kip = 103 lb Given: w = 2

kip ft

a = 6 ft b = 9 ft Solution: ΣF x = 0; ΣF y = 0;

ND = 0 wb 2

− w⎛⎜

VD =

a⎞

⎛ a⎞ ⎟ ⎜ 2 ⎟ − VD = ⎝ b⎠ ⎝ ⎠

wb 2

− w⎛⎜

0

a⎞

⎛ a⎞ ⎟ ⎜ 2⎟ ⎝ b⎠ ⎝ ⎠

V D = 5.00 kip ΣMD = 0;

⎛ w b ⎞a + ⎛ w a ⎞⎛ a ⎞ ⎜ 2 ⎟ ⎜ ⎟⎜ 2 ⎟ ⎝ ⎠ ⎝ b ⎠⎝ ⎠ ⎛ w a3 ⎞ wb⎞ ⎟ MD = ⎛⎜ a−⎜ ⎟ ⎝ 2 ⎠ ⎝ 6b ⎠ MD −

⎛ a⎞ = ⎜ 3⎟ ⎝ ⎠

0

MD = 46.00 kip⋅ ft

643

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Engineering Mechanics - Statics

Chapter 7

Problem 7-23 The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the internal normal force, shear force, and moment at (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the F2 force. Units Used: kip = 103 lb Given: F 1 = 2500 lb

a = 6 ft

F 2 = 3000 lb

b = 12 ft

w = 75

lb ft

c = 2 ft

Solution: ΣMB = 0;

− Ay( b + c) + F1 ( a + b + c) + w b

Ay =

(2

2

b+c

B x = 0 lb

ΣF y = 0;

Ay − F1 − w b − F2 + By = 0 By = − Ay + F1 + w b + F2

ΣMC = 0;

ΣF x = 0; ΣF y = 0;

)

1 2 F 1 ( a + b + c) + w b + 2 c b + 2 F 2 c

ΣF x = 0;

Segment AC :

⎛ b + c⎞ + F c = ⎜2 ⎟ 2 ⎝ ⎠

0

Ay = 4514 lb

B y = 1886 lb

F 1 a + MC = 0 MC = −F 1 a

MC = −15 kip⋅ ft

NC = 0

NC = 0

−F 1 + Ay − V C = 0 VC = Ay − F1

V C = 2.01 kip

Segment BD: ΣMD = 0;

−MD + By c = 0 MD = B y c

MD = 3.77 kip⋅ ft

ΣF x = 0;

ND = 0

ND = 0

ΣF y = 0;

VD − F2 + By = 0 VD = F2 − By

V D = 1.11 kip

644

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-24 The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam. Neglect the weight of the beams. Units Used: kip = 103 lb Given: P = 5000 lb a = 2 ft b = 10 ft Solution: Segment: ΣMC = 0; MC +

⎛ P ⎞b = ⎜2⎟ ⎝ ⎠

MC = −

P 2

0

b

MC = −25.00 kip⋅ ft

Problem 7-25 The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam. Assume that each beam has a uniform weight density γ. Units Used: kip = 103 lb Given: P = 5000 lb 645

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

γ = 150

Chapter 7

lb ft

a = 2 ft b = 10 ft Solution: Beam: +

↑Σ Fy = 0;

P − 2 γ ( a + b) − 2 R = 0 R =

P 2

− γ ( a + b)

R = 700 lb Segment: ΣMC = 0;

M C + R b + γ ( a + b) MC = −R b − γ

⎛ a + b⎞ = ⎜ 2 ⎟ ⎝ ⎠

( a + b)

0

2

2

MC = −17.8 kip⋅ ft

Problem 7-26 Determine the normal force, shear force, and moment in the beam at sections passing through points D and E. Point E is just to the right of the F load. Units Used: kip = 103 lb Given: w = 1.5

a = 6 ft kip ft

F = 3 kip

b = 6 ft c = 4 ft d = 4 ft

646

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Engineering Mechanics - Statics

Chapter 7

Solution: Σ MB = 0;

1 2

⎛ a + b ⎞ − A ( a + b) = ⎟ y ⎝ 3 ⎠

w( a + b) ⎜ 1 2

Ay =

w( a + b)

0

⎛ a + b⎞ ⎜ 3 ⎟ ⎝ ⎠

a+b

Ay = 3 kip + Σ F x = 0; →

Bx = 0

+

By + Ay −

↑

Σ F y = 0;

1

w( a + b) = 0

2

1

By = − Ay +

2

w( a + b)

B y = 6 kip + Σ F x = 0; →

ND = 0

+

Ay −

↑

Σ F y = 0;

⎞a − V = ⎜ ⎟ D 2 ⎝ a + b⎠ 1⎛ aw

VD = Ay −

Σ MD = 0;

MD +

1⎛ aw

⎜

⎞a ⎟

V D = 0.75 kip

2 ⎝ a + b⎠

⎞a ⎛ a ⎞ − A a = ⎜ ⎟ ⎜ ⎟ y 2 ⎝ a + b⎠ ⎝ 3⎠ 1⎛ aw

MD =

−1 ⎛ a w ⎞ ⎛ a ⎞ ⎜ ⎟ a ⎜ ⎟ + Ay a 2 ⎝ a + b⎠ ⎝ 3 ⎠

+ Σ F x = 0; →

NE = 0

+

−V E − F − B y = 0

↑Σ Fy = 0;

0

V E = −F − By ΣME = 0;

0

MD = 13.5 kip⋅ ft

V E = −9 kip

ME + B y c = 0 ME = −By c

ME = −24.0 kip⋅ ft

647

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-27 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: kN = 103 N Given: w1 = 200

N

w2 = 400

N

m

m

a = 2.5 m b = 3m c = 6m Solution: Σ MA = 0;

F BC + Σ F x = 0; → +

↑Σ Fy = 0;

−w1 c⎛⎜

c⎞

→ Σ Fx = 0; +

↑

Σ F y = 0;

2

(w2 − w1)c ⎛⎜ 3 ⎟⎞ + 2c

Ax =

2

1 2

2

2

(FBC c) = 0

a +c

2

a +c

F BC = 2600 N

ac

⎛ c ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠

Ay − w1 c −

a

⎝ ⎠

2 ⎡ c2 c⎤ ⎢ = w1 + ( w2 − w1 ) ⎥ 3⎦ ⎣ 2

Ay = w1 c + +

1

⎟− ⎝ 2⎠

Ax = 2400 N

(w2 − w1)c + ⎛⎜

⎞F = BC ⎟ 2 2 + c a ⎝ ⎠

(w2 − w1)c − ⎛⎜ 2 1

a

⎞F

a

2⎟

2

⎝ a +c ⎠

− Ax + ND = 0

0

BC

ND = A x

ND = 2.40 kN

1 b Ay − w1 b − ( w2 − w1 ) ⎛⎜ ⎟⎞ b − V D = 0 2 ⎝c⎠ 2⎞

⎛b 1 V D = A y − w1 b − ( w2 − w1 ) ⎜ 2 ⎝c Σ MD = 0;

Ay = 800 N

− Ay b + w1 b⎛⎜

b⎞

⎟+ ⎝2⎠

⎟ ⎠

V D = 50 N

(w2 − w1)⎛⎜ c ⎟⎞ b ⎛⎜ 3 ⎟⎞ + MD = 0 2 1

⎛ b2 ⎞ MD = Ay( b) − w1 ⎜ ⎟ − ⎝2⎠

b

b

⎝ ⎠

⎝ ⎠

⎛ b3 ⎞ (w2 − w1)⎜ 3c ⎟ 2 ⎝ ⎠ 1

648

MD = 1.35 kN⋅ m

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-28 Determine the normal force, shear force, and moment at sections passing through points E and F. Member BC is pinned at B and there is a smooth slot in it at C. The pin at C is fixed to member CD. Units Used: 3

kip = 10 lb Given: M = 350 lb⋅ ft w = 80

lb ft

c = 2 ft

F = 500 lb

d = 3 ft

θ = 60 deg

e = 2 ft

a = 2 ft

f = 4 ft

b = 1 ft

g = 2 ft

Solution: Σ MB = 0; −1 2

wd

⎛ 2d ⎞ − F sin ( θ ) d + C ( d + e) = ⎜3⎟ y ⎝ ⎠

⎛ w d2 ⎞ ⎜ ⎟ + F sin ( θ ) d 3 ⎠ ⎝ Cy =

Cy = 307.8 lb

d+e

+ Σ F x = 0; →

B x − F cos ( θ ) = 0

+

By −

↑

Σ F y = 0;

By = + Σ F x = 0; →

0

1 2

B x = F cos ( θ )

B x = 250 lb

w d − F sin ( θ ) + Cy = 0

1 2

w d + F sin ( θ ) − Cy

−NE − Bx = 0

NE = −B x

B y = 245.2 lb NE = −250 lb

649

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑Σ Fy = 0;

Chapter 7

VE − By = 0

VE = By

V E = 245 lb

−ME − B y c = 0

ME = −By c

ME = −490 lb⋅ ft

+ Σ F x = 0; →

NF = 0

NF = 0 lb

NF = 0.00 lb

+

−Cy − VF = 0

V F = −Cy

V F = −308 lb

Cy( f) + MF = 0

MF = − f Cy

MF = −1.23 kip⋅ ft

Σ ME = 0

↑Σ Fy = 0; Σ MF = 0;

Problem 7-29 The bolt shank is subjected to a tension F. Determine the internal normal force, shear force, and moment at point C. Given: F = 80 lb a = 6 in

Solution: ΣF x = 0;

NC + F = 0 NC = −F NC = −80.00 lb

ΣF y = 0;

VC = 0

ΣMC = 0;

MC + F a = 0 MC = −F a MC = −480.00 lb⋅ in

Problem 7-30 Determine the normal force, shear force, and moment acting at sections passing through points B and C on the curved rod.

650

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Units Used: kip = 103 lb Given: F 1 = 300 lb F 2 = 400 lb

θ = 30 deg

r = 2 ft

φ = 45 deg

Solution: Σ F x = 0;

F 2 sin ( θ ) − F1 cos ( θ ) + NB = 0 NB = −F 2 sin ( θ ) + F1 cos ( θ ) NB = 59.8 lb

Σ F y = 0;

V B + F2 cos ( θ ) + F1 sin ( θ ) = 0 V B = −F 2 cos ( θ ) − F 1 sin ( θ ) V B = −496 lb

Σ MB = 0;

MB + F 2 r sin ( θ ) + F 1 ( r − r cos ( θ ) ) = 0 MB = −F2 r sin ( θ ) − F1 r( 1 − cos ( θ ) ) MB = −480 lb⋅ ft

+ Σ F x = 0; →

F2 − Ax = 0

Ax = F2

Ax = 400 lb

+

Ay − F1 = 0

Ay = F1

Ay = 300 lb

↑Σ Fy = 0; Σ MA = 0;

−MA + F 1 2 r = 0

MA = 2 F 1 r

Σ F x = 0;

NC + Ax sin ( φ ) + A y cos ( φ ) = 0

MA = 1200 lb⋅ ft

NC = − A x sin ( φ ) − Ay cos ( φ ) Σ F y = 0;

V C − Ax cos ( φ ) + Ay sin ( φ ) = 0 V C = A x cos ( φ ) − A y sin ( φ )

Σ MC = 0;

NC = −495 lb

V C = 70.7 lb

−MC − MA − A x r sin ( φ ) + A y( r − r cos ( φ ) ) = 0

651

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

MC = −MA − Ax r sin ( φ ) + Ay r( 1 − cos ( φ ) )

MC = −1.59 kip⋅ ft

Problem 7-31 The cantilevered rack is used to support each end of a smooth pipe that has total weight W. Determine the normal force, shear force, and moment that act in the arm at its fixed support A along a vertical section. Units Used: 3

kip = 10 lb Given: W = 300 lb r = 6 in

θ = 30 deg

Solution: Pipe: +

↑

Σ F y = 0;

NB cos ( θ ) −

NB =

1 2

W 2

=0

⎛ W ⎞ ⎜ ( )⎟ ⎝ cos θ ⎠

NB = 173.205 lb

Rack: + Σ F x = 0; →

−NA + NB sin ( θ ) = 0 NA = NB sin ( θ )

+

↑Σ Fy = 0;

NA = 86.6 lb

V A − NB cos ( θ ) = 0 V A = NB cos ( θ )

Σ MA = 0;

MA − NB

⎛ r + r sin ( θ ) ⎞ = ⎜ ⎟ ⎝ cos ( θ ) ⎠

MA = NB

⎛ r + r sin ( θ ) ⎞ ⎜ ⎟ ⎝ cos ( θ ) ⎠

V A = 150 lb 0

MA = 1.800 kip⋅ in

652

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-32 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: 3

kN = 10 N

Given: kN m

w = 0.75 F = 4 kN a = 1.5 m

d = 1.5 m

b = 1.5 m

e = 3

c = 2.5 m

f = 4

Solution: Σ MC = 0; −B x( c + d) +

f ⎛ ⎞ ⎜ 2 2 ⎟F d = ⎝ e +f ⎠ fdF

Bx =

2

2

0

B x = 1.2 kN

e + f ( c + d) Σ MA = 0;

⎛ c + d ⎞ + B ( a + b) + B ( c + d) = ⎟ y x ⎝ 2 ⎠

−w( c + d) ⎜

⎡ ( c + d) 2⎤ ⎥ − B x ( c + d) w⎢ 2 ⎣ ⎦ By = a+b

0

B y = 0.40 kN

+ Σ F x = 0; →

−ND − B x = 0

ND = −Bx

ND = −1.2 kN

+

VD + By = 0

V D = −By

V D = −0.4 kN

↑Σ Fy = 0;

653

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Σ MD = 0;

−MD + By b = 0

Chapter 7

MD = B y b

MD = 0.6 kN⋅ m

Problem 7-33 Determine the internal normal force, shear force, and moment acting at point A of the smooth hook. Given:

θ = 45 deg a = 2 in F = 20 lb Solution: ΣF x = 0;

NA − F cos ( θ ) = 0 NA = F cos ( θ )

ΣF y = 0;

V A − F sin ( θ ) = 0 V A = F sin ( θ )

ΣMB = 0;

NA = 14.1 lb

V A = 14.1 lb

MA − NA a = 0 MA = NA a

MA = 28.3 lb⋅ in

Problem 7-34 Determine the internal normal force, shear force, and moment acting at points B and C on the curved rod.

654

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Engineering Mechanics - Statics

Chapter 7

Units Used: kip = 103 lb Given:

θ 2 = 30 deg

F = 500 lb r = 2 ft

a = 3

θ 1 = 45 deg

b = 4

Solution: ΣF N = 0;

⎛ F b ⎞ sin ( θ ) − ⎛ F a ⎞ cos ( θ ) + N = 2 2 B ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ a +b ⎠ ⎝ a +b ⎠ ⎡( a)cos ( θ 2 ) − b sin ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

NB = F ⎢

ΣF V = 0;

VB +

ΣMB = 0;

NB = 59.8 lb

⎛ F b ⎞ cos ( θ ) + ⎛ F a ⎞ sin ( θ ) = 2 2 ⎜ 2 2⎟ ⎜ 2 2⎟ + b + b a a ⎝ ⎠ ⎝ ⎠

V B = −F

MB +

0

⎡⎢ b cos ( θ 2) + ( a)sin ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

0

V B = −496 lb

⎛ F b ⎞ r sin ( θ ) + F⎛ a ⎞ ( r − r cos ( θ ) ) = 2 2 ⎜ 2 2⎟ ⎜ 2 2⎟ + b + b a a ⎝ ⎠ ⎝ ⎠

MB = F r

⎡⎢−b sin ( θ 2) − a + ( a)cos ( θ 2 )⎥⎤ 2 2 ⎢ ⎥ a +b ⎣ ⎦

0

MB = −480 lb⋅ ft

Also, ΣF x = 0;

Fb

− Ax +

=0

2

a +b Ax = F

ΣF y = 0;

Ay −

2

⎛ b ⎞ ⎜ 2 2⎟ ⎝ a +b ⎠ Fa 2

a +b

Ax = 400.00 lb

=0 2

655

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Engineering Mechanics - Statics

Ay = F

ΣMA = 0;

−MA +

MA =

Chapter 7

⎛ a ⎞ ⎜ 2 2⎟ ⎝ a +b ⎠

⎛ F a ⎞ 2r = ⎜ 2 2⎟ ⎝ a +b ⎠ 2F r a 2

2

Ay = 300.00 lb

0

MA = 1200 lb⋅ ft

a +b ΣF x = 0;

NC + Ax sin ( θ 1 ) + Ay cos ( θ 1 ) = 0 NC = − A x sin ( θ 1 ) − A y cos ( θ 1 )

ΣF y = 0;

NC = −495 lb

V C − Ax cos ( θ 1 ) + A y sin ( θ 1 ) = 0 V C = A x cos ( θ 1 ) − Ay sin ( θ 1 )

ΣMC = 0;

V C = 70.7 lb

−MC − MA + A y( r − r cos ( θ 1 ) ) − Ax r sin ( θ 1 ) = 0 MC = −MA + Ay( r − r cos ( θ 1 ) ) − A x r sin ( θ 1 )

MC = −1.59 kip⋅ ft

Problem 7-35 Determine the ratio a/b for which the shear force will be zero at the midpoint C of the beam.

Solution: Find Ay:

ΣMB = 0;

656

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Engineering Mechanics - Statics

Chapter 7

( 2 a + b) w ⎡⎢ ( b − a)⎤⎥ − Ay b = 0

1

1

⎣3

2

Ay =

⎦

w ( 2 a + b) ( b − a) 6b

This problem requires

V C = 0.

Summing forces vertically for the section, we have +

↑Σ Fy = 0; w 1 ( 2a + b) ( b − a) − 2 6b

⎛a + ⎜ ⎝

b⎞ w

⎟

2⎠ 2

=0

w w ( 2a + b) ( b − a) = ( 2a + b) 8 6b 4 ( b − a) = 3 b

b = 4a a 1 = 4 b

Problem 7-36 The semicircular arch is subjected to a uniform distributed load along its axis of w0 per unit length. Determine the internal normal force, shear force, and moment in the arch at angle θ. Given:

θ = 45 deg

Solution: Resultants of distributed load:

657

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Engineering Mechanics - Statics

Chapter 7

θ ⌠ F Rx = ⎮ w0 r dθ sin ( θ ) = r w0 ( 1 − cos ( θ ) ) ⌡0

F Rx = r w0 ( 1 − cos ( θ ) ) θ ⌠ F Ry = ⎮ w0 r dθ cos ( θ ) = r w0 sin ( θ ) ⌡0

F Rx = r w0 ( sin ( θ ) ) θ ⌠ 2 MRo = ⎮ w0 r dθ r = r w0 θ ⌡0

Σ F x = 0;

−V + FRx cos ( θ ) − F Ry sin ( θ ) = 0

V = ⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ cos ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ sin ( θ ) V = w0 r( cos ( θ ) − 1 ) a = cos ( θ ) − 1 a = −0.293 Σ F y = 0;

V = a r w0

N + FRy cos ( θ ) + F Rx sin ( θ ) = 0 N = −⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ sin ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ cos ( θ ) N = −w0 r sin ( θ ) b = −sin ( θ ) b = −0.707

ΣMo = 0;

N = w0 r b

−M + r w0 ( θ ) + b r w0 r = 0 2

M = w0 r ( θ + b) 2

c = θ+b c = 0.0783

2

M = c r w0

658

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Engineering Mechanics - Statics

Chapter 7

Problem 7-37 The semicircular arch is subjected to a uniform distributed load along its axis of w0 per unit length. Determine the internal normal force, shear force, and moment in the arch at angle θ. Given:

θ = 120 deg Solution: Resultants of distributed load: θ ⌠ F Rx = ⎮ w0 r dθ sin ( θ ) = r w0 ( 1 − cos ( θ ) ) ⌡0

F Rx = r w0 ( 1 − cos ( θ ) ) θ ⌠ F Ry = ⎮ w0 r dθ cos ( θ ) = r w0 sin ( θ ) ⌡0

F Rx = r w0 ( sin ( θ ) ) θ ⌠ 2 MRo = ⎮ w0 r dθ r = r w0 θ ⌡0

Σ F x = 0;

−V + FRx cos ( θ ) − F Ry sin ( θ ) = 0 V = ⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ cos ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ sin ( θ ) V = w0 r( cos ( θ ) − 1 ) a = cos ( θ ) − 1 a = −1.500

Σ F y = 0;

V = a r w0

N + FRy cos ( θ ) + F Rx sin ( θ ) = 0 N = −⎡⎣r w0 ( 1 − cos ( θ ) )⎤⎦ sin ( θ ) − ⎡⎣r w0 ( sin ( θ ) )⎤⎦ cos ( θ ) N = −w0 r sin ( θ ) b = −sin ( θ ) 659

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Engineering Mechanics - Statics

Chapter 7

b = −0.866 Σ Mo = 0;

N = w0 ⋅ r⋅ b

−M + r w0 ( θ ) + b r w0 r = 0 2

M = w0 r ( θ + b) 2

c = θ+b c = 1.2284

2

M = c r w0

Problem 7-38 Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe. Units Used: kip = 103 lb Given:

⎛ 0 ⎞ F 1 = ⎜ 350 ⎟ lb ⎜ ⎟ ⎝ −400 ⎠ ⎛ 150 ⎞ F 2 = ⎜ 0 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠ a = 1.5 ft

b = 2 ft

c = 3 ft

Solution:

⎛c⎞ r1 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ r2 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

F C = −F1 − F2

⎛ −150.00 ⎞ F C = ⎜ −350.00 ⎟ lb ⎜ ⎟ ⎝ 700.00 ⎠

660

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Engineering Mechanics - Statics

MC = −r1 × F1 − r2 × F2

Chapter 7

⎛ 1400.00 ⎞ MC = ⎜ −1200.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −750.00 ⎠

Problem 7-39 Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe. Units Used: kip = 103 lb Given:

⎛ −80 ⎞ F 1 = ⎜ 200 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠ ⎛ 250 ⎞ F 2 = ⎜ −150 ⎟ lb ⎜ ⎟ ⎝ −200 ⎠ a = 1.5 ft

b = 2 ft

c = 3 ft

Solution:

⎛c⎞ r1 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

⎛0⎞ r2 = ⎜ b ⎟ ⎜ ⎟ ⎝0⎠

F C = −F1 − F2

⎛ −170.00 ⎞ F C = ⎜ −50.00 ⎟ lb ⎜ ⎟ ⎝ 500.00 ⎠

MC = −r1 × F1 − r2 × F2

⎛ 1000.00 ⎞ MC = ⎜ −900.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −260.00 ⎠

Problem 7-40 Determine the x, y, z components of internal loading in the rod at point D.

661

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Engineering Mechanics - Statics

Chapter 7

Units Used: kN = 103 N Given: M = 3 kN⋅ m

⎛ 7 ⎞ F = ⎜ −12 ⎟ kN ⎜ ⎟ ⎝ −5 ⎠ a = 0.75 m b = 0.2 m c = 0.2 m d = 0.6 m e = 1m Solution: Guesses Cx = 1 N

Cy = 1 N

Bx = 1 N

Bz = 1 N

Ay = 1 N

Az = 1 N

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ Bx ⎟⎞ ⎛⎜ Cx ⎟⎞ ⎜ Ay ⎟ + ⎜ 0 ⎟ + ⎜ Cy ⎟ + F = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ −e ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎛ 0 ⎞ ⎞ ⎜ b + c + d ⎟ × ⎜ Ay ⎟ + ⎜ b + c ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎜ C ⎟ + ⎜ b + c + d ⎟ × F + ⎜ 0 ⎟ = 0 y ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ 0 a ⎠ ⎝ Bz ⎠ ⎝ ⎠ ⎝ 0 ⎟⎠ ⎝ 0 ⎠ ⎝ −M ⎠ ⎝ 0 ⎠ ⎝ Az ⎠ ⎝

662

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Engineering Mechanics - Statics

⎛⎜ Ay ⎞⎟ ⎜ Az ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) y z x z x y ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ Cy ⎟ ⎝ ⎠

Chapter 7

⎛⎜ Ay ⎞⎟ ⎛ −53.60 ⎞ ⎜ Az ⎟ ⎜ 87.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 109.00 ⎟ kN ⎜ Bz ⎟ ⎜ −82.00 ⎟ ⎜ ⎟ ⎜ −116.00 ⎟ ⎜ Cx ⎟ ⎜ ⎟ 65.60 ⎝ ⎠ ⎜ Cy ⎟ ⎝ ⎠

Guesses V Dx = 1 N

NDy = 1 N

V Dz = 1 N

MDx = 1 N⋅ m

MDy = 1 N⋅ m

MDz = 1 N⋅ m

Given

⎛⎜ Cx ⎟⎞ ⎛⎜ VDx ⎟⎞ ⎜ Cy ⎟ + ⎜ NDy ⎟ = 0 ⎜ 0 ⎟ ⎜V ⎟ ⎝ ⎠ ⎝ Dz ⎠ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛⎜ MDx ⎟⎞ ⎛ 0 ⎞ ⎜ −b ⎟ × ⎜ C ⎟ + ⎜ MDy ⎟ + ⎜ 0 ⎟ = 0 ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ ⎝ a ⎠ ⎝ 0 ⎠ ⎝ MDz ⎟⎠ ⎝ −M ⎠ ⎛⎜ VDx ⎟⎞ ⎜ NDy ⎟ ⎜ ⎟ ⎜ VDz ⎟ = Find ( V , N , V , M , M , M ) Dx Dy Dz Dx Dy Dz ⎜ MDx ⎟ ⎜ ⎟ ⎜ MDy ⎟ ⎜ MDz ⎟ ⎝ ⎠

⎛ VDx ⎞ ⎛ 116.00 ⎞ ⎜ ⎟ ⎜ NDy ⎟ = ⎜ −65.60 ⎟ kN ⎜ V ⎟ ⎜⎝ 0.00 ⎟⎠ ⎝ Dz ⎠ ⎛ MDx ⎞ ⎛ 49.20 ⎞ ⎜ ⎟ ⎜ MDy ⎟ = ⎜ 87.00 ⎟ kN⋅ m ⎜ M ⎟ ⎜⎝ 26.20 ⎟⎠ ⎝ Dz ⎠

663

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Engineering Mechanics - Statics

Chapter 7

Problem 7-41 Determine the x, y, z components of internal loading in the rod at point E. Units Used: 3

kN = 10 N Given: M = 3 kN⋅ m

⎛ 7 ⎞ F = ⎜ −12 ⎟ kN ⎜ ⎟ ⎝ −5 ⎠ a = 0.75 m b = 0.4 m c = 0.6 m d = 0.5 m e = 0.5 m Solution: Guesses Cx = 1 N

Cy = 1 N

Bx = 1 N

Bz = 1 N

Ay = 1 N

Az = 1 N

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ Bx ⎟⎞ ⎛⎜ Cx ⎟⎞ ⎜ Ay ⎟ + ⎜ 0 ⎟ + ⎜ Cy ⎟ + F = 0 ⎜ Az ⎟ ⎜ Bz ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ −d − e ⎞ ⎛⎜ 0 ⎞⎟ ⎛ 0 ⎞ ⎛⎜ Bx ⎟⎞ ⎛ 0 ⎞ ⎛⎜ Cx ⎟⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ b + c ⎟ × ⎜ Ay ⎟ + ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜ 0 ⎟ × ⎜ C ⎟ + ⎜ b + c ⎟ × F + ⎜ 0 ⎟ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y⎟ ⎜ 0 ⎝ −M ⎠ ⎝ ⎠ ⎝ Az ⎠ ⎝ 0 ⎠ ⎝ Bz ⎠ ⎝ a ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠

664

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

⎛⎜ Ay ⎞⎟ ⎜ Az ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) y z x z x y ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cx ⎟ ⎜ Cy ⎟ ⎝ ⎠

Chapter 7

⎛⎜ Ay ⎞⎟ ⎛ −53.60 ⎞ ⎜ Az ⎟ ⎜ 87.00 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 109.00 ⎟ kN ⎜ Bz ⎟ ⎜ −82.00 ⎟ ⎜ ⎟ ⎜ −116.00 ⎟ ⎜ Cx ⎟ ⎜ ⎟ 65.60 ⎝ ⎠ ⎜ Cy ⎟ ⎝ ⎠

Guesses NEx = 1 N

V Ey = 1 N

V Ez = 1 N

MEx = 1 N⋅ m

MEy = 1 N⋅ m

MEz = 1 N⋅ m

Given

⎛⎜ 0 ⎞⎟ ⎛⎜ NEx ⎟⎞ ⎜ Ay ⎟ + ⎜ VEy ⎟ = 0 ⎜ Az ⎟ ⎜ V ⎟ ⎝ ⎠ ⎝ Ez ⎠ ⎛ −e ⎞ ⎛⎜ 0 ⎞⎟ ⎛⎜ MEx ⎞⎟ ⎜ 0 ⎟ × Ay + M =0 ⎜ ⎟ ⎜ ⎟ ⎜ Ey ⎟ ⎝ 0 ⎠ ⎜⎝ Az ⎟⎠ ⎜⎝ MEz ⎟⎠ ⎛⎜ NEx ⎞⎟ ⎜ VEy ⎟ ⎜ ⎟ ⎜ VEz ⎟ = Find ( N , V , V , M , M , M ) Ex Ey Ez Ex Ey Ez ⎜ MEx ⎟ ⎜ ⎟ ⎜ MEy ⎟ ⎜ MEz ⎟ ⎝ ⎠

⎛ NEx ⎞ ⎛ 0.00 ⎞ ⎜ ⎟ ⎜ VEy ⎟ = ⎜ 53.60 ⎟ kN ⎜ V ⎟ ⎜⎝ −87.00 ⎟⎠ ⎝ Ez ⎠ ⎛ MEx ⎞ ⎛ 0.00 ⎞ ⎜ ⎟ ⎜ MEy ⎟ = ⎜ −43.50 ⎟ kN⋅ m ⎜ M ⎟ ⎜⎝ −26.80 ⎟⎠ ⎝ Ez ⎠

Problem 7-42 Draw the shear and moment diagrams for the shaft in terms of the parameters shown; There is 665

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

g a thrust bearing at A and a journal bearing at B.

p

Units Used: 3

kN = 10 N Given: P = 9 kN a = 2m L = 6m Solution: P ( L − a) − Ay L = 0

Ay = P

L−a L

x1 = 0 , 0.01a .. a Ay − V 1 ( x) = 0

V 1 ( x) =

M1 ( x) − Ay x = 0

M1 ( x) =

Ay kN

Ay x kN⋅ m

x2 = a , 1.01a .. L Ay − P − V2 ( x) = 0

V 2 ( x) =

Ay − P kN

M2 ( x) − Ay x + P ( x − a) = 0

M2 ( x) =

A y x − P ( x − a) kN⋅ m

666

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (kN)

10

V1( x1) 5 V2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Moment (kN-m)

15 10

M1( x1) M2( x2)

5 0 5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Problem 7-43 Draw the shear and moment diagrams for the beam in terms of the parameters shown. Given: P = 800 lb

a = 5 ft

L = 12 ft

Solution:

667

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

x1 = 0 , 0.01a .. a

Chapter 7

x2 = a , 1.01a .. L − a

x3 = L − a , 1.01( L − a) .. L

V 2 ( x) = 0

V 3 ( x) =

P

V 1 ( x) =

lb Px

M1 ( x) =

M2 ( x) =

lb⋅ ft

Pa

M3 ( x) =

lb⋅ ft

−P lb P( L − x) lb⋅ ft

Force (lb)

1000

V1( x1)

500

V2( x2)

0

V3( x3)

500

1000

0

2

4

6

8

10

8

10

12

x1 x2 x3 , , ft ft ft

Distance (ft)

Moment (lb-ft)

4000

M1( x1) M2( x2)

M3( x3)2000

0

0

2

4

6

12

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-44 Draw the shear and moment diagrams for the beam (a) in terms of the parameters 668

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

g shown; (b) set M0 and L as given.

( )

p

Given: M0 = 500 N⋅ m L = 8m Solution: For

0 ≤ x≤

L 3

+

↑Σ Fy = 0;

V1 = 0

ΣMx = 0;

For

L 3

≤x≤

M1 = 0

2L 3

+

↑Σ Fy = 0; ΣMx = 0;

For

2L 3

M2 = M0

≤x≤L

+

↑Σ Fy = 0; ΣMx = 0;

( b)

V2 = 0

V3 = 0 M3 = 0

x1 = 0 , 0.01L ..

L 3

x2 =

L L 2L , 1.01 .. 3

3

3

x3 =

2L 2L 3

,

3

V 1 ( x1 ) = 0

V 2 ( x2 ) = 0

V 3 ( x3 ) = 0

M1 ( x1 ) = 0

M2 ( x2 ) = M0

M3 ( x3 ) = 0

1.01 .. L

669

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

1

Shear force in N

0.5

V1( x1) V2( x2)

0

V3( x3)

0.5

1

0

1

2

3

4

5

6

7

8

9

x1 , x2 , x3

Distance in m

600

Moment in N - m

400

M1( x1) M2( x2)

M3( x3)200

0

0

1

2

3

4

5

6

7

8

9

x1 , x2 , x3

Distance in m

Problem 7-45 The beam will fail when the maximum shear force is V max or the maximum bending moment is Mmax. Determine the magnitude M0 of the largest couple moments it will support.

670

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Engineering Mechanics - Statics

Chapter 7

Units Used: 3

kN = 10 N Given: L = 9m V max = 5 kN Mmax = 2 kN⋅ m

Solution: The shear force is zero everywhere in the beam. The moment is zero in the first third and the last third of the bam. In the middle section of the beam the moment is

M = M0

M0 = Mmax

Thus the beam will fail when

M0 = 2.00 kN⋅ m

Problem 7-46 The shaft is supported by a thrust bearing at A and a journal bearing at B. Draw the shear and moment diagrams for the shaft in terms of the parameters shown. Given: w = 500

lb ft

L = 10 ft Solution:

For

ΣF y = 0;

wL

ΣM = 0;

2

0 ≤ x< L

− wx − V = 0

−w L 2

⎛x⎞ + M = ⎟ ⎝ 2⎠

x + w x⎜

V ( x) =

0

w 2

( L − 2 x)

M ( x) =

1

lb

1 2 ( Lx − x ) 2 lb⋅ ft

w

671

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (lb)

5000

V( x )

0

5000

0

2

4

6

8

10

x ft

Distance (ft) 1 .10

Moment (lb-ft)

4

M( x)

0

1 .10

4

0

2

4

6

8

10

x ft

Distance (ft)

Problem 7-47 The shaft is supported by a thrust bearing at A and a journal bearing at B. The shaft will fail when the maximum moment is Mmax. Determine the largest uniformly distributed load w the shaft will support. Units Used: 3

kip = 10 lb Given: L = 10 ft Mmax = 5 kip⋅ ft

672

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Engineering Mechanics - Statics

Chapter 7

Solution: wL 2

−

− wx − V = 0

wL 2

V = −w x +

⎛x⎞ x + wx ⎜ ⎟ + M = ⎝2⎠

wL 2

2 wL⎞ wx ⎛ M=⎜ ⎟x − 2 ⎝ 2 ⎠

0

From the Moment Diagram, wL

Mmax =

2

w =

8

8 Mmax

L

2

w = 400.00

lb ft

Problem 7-48 Draw the shear and moment diagrams for the beam. 3

kN = 10 N

Units Used: Given: w = 2

kN m

L = 5m MB = 5 kN⋅ m Solution: ΣF y = 0; −V ( x) + wL − wx = 0

ΣM = 0;

V ( x) = ( w L − w x)

⎛ L ⎞ + M − w L x + w x⎛ x ⎞ = ⎟ ⎜ 2⎟ B ⎝2⎠ ⎝ ⎠

M ( x) + w L⎜

⎡ ⎣

1

kN

0

⎛ x ⎞ − w L⎛ L ⎞ − M ⎤ 1 ⎟ ⎜2⎟ B⎥ ⎝2⎠ ⎝ ⎠ ⎦ kN⋅ m

M ( x) = ⎢w L x − w x⎜

673

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (kN)

10

V( x ) 5

0

0

1

2

3

4

5

4

5

x

Distance (m)

Moment (kN-m)

0

M( x) 20

40

0

1

2

3

x

Distance (m)

Problem 7-49 Draw the shear and moment diagrams for the beam. 3

kN = 10 N

Units Used: Given: w = 3

kN m

F = 10 kN

L = 6m Solution: V ( x) − w( L − x) − F = 0 V ( x) = [ w( L − x) + F ]

1

kN

⎛ L − x ⎞ − F( L − x) = ⎟ ⎝ 2 ⎠

−M ( x) − w( L − x) ⎜

0

674

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Engineering Mechanics - Statics

⎡

( L − x)

⎣

2

M ( x) = ⎢−w

2

Chapter 7

⎤

− F ( L − x)⎥

1

⎦ kN⋅ m

Force (kN)

40

V( x) 20

0

0

1

2

3

4

5

6

x

Distance (m) Moment (kN-m)

0 50

M( x) 100 150

0

1

2

3

4

5

6

x

Distance (m)

Problem 7-50 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: a = 2m

b = 4m

w = 1.5

kN m

Solution:

⎛ b − a⎞ − A b = ⎟ y ⎝ 2 ⎠

w( b − a) ⎜

0

Ay =

w ( b − a) 2b

2

Ay = 0.75 kN

x1 = 0 , 0.01a .. a Ay − V 1 ( x) = 0

V 1 ( x) = Ay

1

kN 675

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

− Ay x + M1 ( x) = 0

M1 ( x) = A y x

1

kN⋅ m

x2 = b − a , 1.01( b − a) .. b 1

Ay − w( x − a) − V 2 ( x) = 0

V 2 ( x) = ⎡⎣A y − w( x − a)⎤⎦

⎛ x − a ⎞ + M ( x) = − Ay x + w( x − a) ⎜ ⎟ 2 ⎝ 2 ⎠

2 ⎡ ( x − a) ⎤ ⎢ ⎥ 1 M2 ( x) = Ay x − w 2 ⎣ ⎦ kN⋅ m

0

kN

Force (kN)

2

V1( x1) 0 V2( x2)

2

4

0

0.5

1

1.5

2

2.5

3

2.5

3

3.5

4

x1 , x2

Distance (m)

Moment (kN-m)

2

M1( x1) 1 M2( x2) 0

1

0

0.5

1

1.5

2

3.5

4

x1 , x2

Distance (m)

676

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-51 Draw the shear and moment diagrams for the beam. Given: L = 20 ft

w = 250

lb ft

M1 = 150 lb⋅ ft

Solution:

⎛ L⎞ − M − A l = ⎟ 2 y ⎝2⎠

M 1 + w L⎜

V ( x) = ( Ay − w x)

M2 = 150 lb⋅ ft

M1 − M2 + w Ay =

0

⎛ L2 ⎞ ⎜ ⎟ ⎝2⎠

L

Ay = 2500 lb

⎡ ⎛ x2 ⎞ ⎤ 1 ⎢ M ( x) = A y x − w⎜ ⎟ − M1⎥ ⎣ ⎝2⎠ ⎦ lb⋅ ft

1

lb

Force in lb

2000

V( x )

0

2000 0

5

10

15

20

x ft

Distance in ft

677

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

1.5 .10

Moment in lb ft

4

1 .10

4

M( x) 5000

0 0

5

10

15

20

x ft

Distance in ft

Problem 7-52 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: w = 40

kN m

F = 20 kN

M = 150 kN⋅ m

Solution: − Ay a + w a

⎛ a⎞ − F b − M = ⎜ 2⎟ ⎝ ⎠

0

x1 = 0 , 0.01a .. a 1

kN

⎛x

M1 ( x) = ⎢Ay x − w⎜

⎣

b = 3m

⎡ ⎛ a2 ⎞ ⎤ ⎟ − F b − M⎥ ⎛⎜ 1 ⎟⎞ ⎣ ⎝2⎠ ⎦⎝ a ⎠

Ay = ⎢w⎜

Ay = 133.75 kN

x2 = a , 1.01a .. a + b

V 1 ( x) = ( A y − w x)

⎡

a = 8m

2 ⎞⎤

⎟⎥ 1 ⎝ 2 ⎠⎦ kN⋅ m

V 2 ( x) = F

1

kN

M2 ( x) = [ −F( a + b − x) − M]

1

kN⋅ m

678

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force in kN

200

V1( x1) V2( x2)

0

200

0

2

4

6

8

10

x1 , x2

Distance in m

Moment (kN-m)

400

M1( x1)

200

M2( x2)

0 200 400

0

2

4

6

8

10

x1 , x2

Distance (m)

Problem 7-53 Draw the shear and moment diagrams for the beam.

679

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Engineering Mechanics - Statics

Chapter 7

Solution: 0 ≤ x< a ΣF y = 0; −V − w x = 0

V = −w x ΣM = 0;

M + w x⎛⎜

x⎞

M = −w

x

⎟= ⎝2⎠

0

2

2

a < x ≤ 2a ΣF y = 0; −V + 2 w a − w x = 0

V = w( 2 a − x) ΣM = 0;

M + w x⎛⎜

x⎞

⎟ − 2 w a( x − a) = ⎝2⎠

0

2

M = 2 w a( x − a) −

wx 2

Problem 7-54 Draw the shear and bending-moment diagrams for beam ABC. Note that there is a pin at B.

680

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Engineering Mechanics - Statics

Chapter 7

Solution: Support Reactions: From FBD (b), wL ⎛ L⎞

⎛ L⎞ ⎜ 4 ⎟ − B y⎜ 2 ⎟ = ⎝ ⎠ ⎝ ⎠

2

wL

By =

4

Ay −

From FBD (a),

wL 2

−

B y

=0

3w L

Ay =

−B y

0

4

L 2

− w⎛⎜

MA = w

⎛ L⎞ + M = ⎟ ⎜ ⎟ A ⎝2⎠ ⎝4⎠ L⎞

0

⎛ L2 ⎞ ⎜ ⎟ ⎝4⎠

Shear and Moment Functions: From FBD (c) For 0 ≤ x≤ L

V=

w 4

Ay − w x − V = 0

( 3 L − 4 x)

x MA − Ay x + w x ⎛⎜ ⎟⎞ + M = 0 ⎝ 2⎠

M=

w 4

(3L x − 2x2 − L2)

681

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-55 The beam has depth a and is subjected to a uniform distributed loading w which acts at an angle θ from the vertical as shown. Determine the internal normal force, shear force, and moment in the beam as a function of x. Hint:The moment loading is to be determined from a point along the centerline of the beam (x axis). Given: a = 2 ft L = 10 ft

θ = 30 deg w = 50

lb ft

Solution: 0 ≤ x≤ L

N + w sin ( θ ) x = 0

ΣF x = 0;

N ( x) = −w sin ( θ ) x −V − w cos ( θ ) x = 0

ΣF y = 0;

V = −w cos ( θ ) x

⎛ x ⎞ − w sin ( θ ) x⎛ a ⎞ + M = ⎟ ⎜ 2⎟ ⎝2⎠ ⎝ ⎠

w cos ( θ ) x⎜

ΣM = 0;

M ( x) = −w cos ( θ )

0

⎛ x2 ⎞ ⎜ ⎟ + w sin ( θ ) ⎛⎜ x a ⎞⎟ ⎝2⎠ ⎝2⎠

Problem 7-56 Draw the shear and moment diagrams for the beam. Given: w = 250

lb ft

L = 12 ft

682

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Engineering Mechanics - Statics

Chapter 7

Solution: 1

2

wx =0 L

ΣF y = 0;

−V −

ΣM = 0;

x ⎛ w x⎞ 1 M + ⎜ ⎟ ⎛⎜ x⎟⎞ = 0 2 ⎝ L ⎠⎝ 3 ⎠

2

x

V ( x) = −

wx 1 2 L lb 3

−w x M ( x) = 6L

1

lb⋅ ft

Force (lb)

0

V( x) 1000

2000

0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (lb-ft)

0

M( x)

5000

1 .10

4

0

2

4

6

8

10

12

x ft

Distance (ft)

Problem 7-57 The beam will fail when the maximum shear force is V max or the maximum moment is Mmax. Determine the largest intensity w of the distributed loading it will support.

683

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given: L = 18 ft V max = 800 lb Mmax = 1200 lb⋅ ft Solution: For 0 ≤ x ≤ L 2

V=

M=

2L

w1 = 2

6L

2

⎛ Vmax ⎞ ⎜ ⎟ ⎝ L ⎠

Mmax =

−w x

wL

V max =

w2 = 6

3

−w x

w2 L

w1 = 88.9

lb ft

2

6

⎛ Mmax ⎞ ⎜ 2 ⎟ ⎝ L ⎠

w2 = 22.2

Now choose the critical case

lb ft

w = min ( w1 , w2 )

w = 22.22

lb ft

Problem 7-58 The beam will fail when the maximum internal moment is Mmax. Determine the position x of the concentrated force P and its smallest magnitude that will cause failure.

684

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Engineering Mechanics - Statics

Chapter 7

Solution: For ξ < x,

M1 =

For ξ > x,

M2 =

Pξ ( L − x) L P x( L − ξ ) L

Note that M1 = M2 when x = ξ Mmax = M1 = M2 =

d dx

P x( L − x) L

(L x − x2) = L − 2x

Thus,

Mmax =

x=

=

P L

(L x − x2)

L 2

P ⎛ L ⎞⎛

L⎞ P ⎛ L⎞ ⎜ 2 ⎟ ⎜L − 2 ⎟ = 2 ⎜ 2 ⎟ L ⎝ ⎠⎝ ⎠ ⎝ ⎠

P=

4 Mmax

L

Problem 7-59 Draw the shear and moment diagrams for the beam. Given: w = 30

lb

MC = 180 lb⋅ ft

ft

a = 9 ft

b = 4.5 ft

Solution: − Ay a +

1 2

wa

⎛ a⎞ − M = ⎜ 3⎟ C ⎝ ⎠

0

685

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Ay =

wa 6

Ay + By −

By =

wa 2

MC

− 1 2

Chapter 7

a wa = 0

− Ay

x1 = 0 , 0.01a .. a Ay −

1 2

⎛x⎞ w ⎜ ⎟ x − V1 ( x) = ⎝ a⎠

2 ⎛ wx ⎞ 1 ⎜ ⎟ V 1 ( x) = A y − 2 a ⎠ lb ⎝

0

⎛x⎞ ⎛x⎞ − Ay x + w⎜ ⎟ x⎜ ⎟ + M1 ( x) = 2 ⎝ a⎠ ⎝ 3 ⎠ 1

3 ⎛ wx ⎞ 1 ⎜ ⎟ M1 ( x) = Ay x − 6 a ⎠ lb⋅ ft ⎝

0

x2 = a , 1.01a .. a + b Ay −

1 2

w a + By − V 2 ( x) = 0

− Ay x +

1 2

⎛ ⎝

w a ⎜x −

V 2 ( x) =

2a ⎞ 3

⎟ − By( x − a) + M2 ( x) = ⎠

⎛A + B − wa⎞ 1 ⎜ y y 2 ⎟ ⎝ ⎠ lb

0

⎡ ⎣

M2 ( x) = ⎢Ay x + B y( x − a) −

wa 2

⎛ ⎝

⋅ ⎜x −

2 a ⎞⎤ 3

1

⎟⎥ ⎠⎦ lb⋅ ft

Force (lb)

100

V1( x1) V2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Distance (ft)

686

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (lb-ft)

100

M1( x1) M2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Diastance (ft)

Problem 7-60 The cantilevered beam is made of material having a specific weight γ. Determine the shear and moment in the beam as a function of x. Solution: By similar triangles y h = x d

y=

h x d

W = γV = γ

⎛⎜ 1 y x t⎟⎞ = γ ⎡⎢ 1 ⎛⎜ h x⎟⎞ x t⎤⎥ = ⎛⎜ γ h t ⎟⎞ x2 ⎝2 ⎠ ⎣2 ⎝ d ⎠ ⎦ ⎝ 2d ⎠

ΣF y = 0;

V−

⎛ γ h t ⎞ x2 = ⎜ ⎟ ⎝ 2d ⎠

V=

ΣM = 0;

−M −

0

⎛ γ h t ⎞ x2 ⎜ ⎟ ⎝ 2d ⎠ ⎛ γ h t ⎞ x2 ⎛ x ⎞ = ⎜ ⎟ ⎜3⎟ ⎝ 2d ⎠ ⎝ ⎠

0

687

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

M=−

Chapter 7

γh t 3 6d

x

Problem 7-61 Draw the shear and moment diagrams for the beam. 3

kip = 10 lb

Given:

w1 = 30

lb ft

w2 = 120

lb ft

L = 12 ft Solution:

w1 L

⎛ L⎞ + ⎜2⎟ ⎝ ⎠

(w2 − w1)L ⎛⎜ 3 ⎞⎟ − Ay L = 0 2 L

1

⎝ ⎠

⎛ L ⎞ + ⎛⎜ w2 − w1 ⎞⎟ ⎛ L ⎞ ⎟ ⎜ ⎟ ⎝2⎠ ⎝ 2 ⎠ ⎝3⎠

Ay = w1 ⎜

Ay − w1 x −

(w2 − w1)⎛⎜ L ⎞⎟ x − V ( x) 2 x

1

⎝ ⎠

⎡ V ( x) = ⎢Ay − w1 x − ⎣ ⎛x⎞ + ⎟ ⎝ 2⎠

− Ay x + w1 x⎜

=0

⎛ x2 ⎞⎤ 1 (w2 − w1)⎜ L ⎟⎥ lb 2 ⎝ ⎠⎦ 1

(w2 − w1)⎛⎜ L ⎞⎟ x ⎛⎜ 3 ⎟⎞ + M ( x) 2 1

x

x

⎝ ⎠

⎝ ⎠

=0

⎡ ⎛ x2 ⎞ ⎛ x3 ⎞⎤ 1 ⎜ ⎟ M ( x) = ⎢A y x − w1 − ( w2 − w1 ) ⎜ ⎟⎥ ⎝2⎠ ⎣ ⎝ 6 L ⎠⎦ kip⋅ ft

688

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (lb)

500

0

V( x )

500 0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (kip-ft)

2

M( x) 1

0

0

2

4

6

8

10

12

x ft

Distance (ft)

Problem 7-62 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb Given: F 1 = 20 kip

F 2 = 20 kip

w = 4

kip ft

a = 15 ft

b = 30 ft

c = 15 ft

Solution:

⎛ b⎞ − F c − A b = ⎟ 2 y ⎝ 2⎠

F 1 ( a + b) + w b⎜

⎛ a + b⎞ + wb − F ⎛ c ⎞ ⎟ 2⎜ ⎟ ⎝ b ⎠ 2 ⎝ b⎠

Ay = F1 ⎜

0

Ay + By − F1 − F2 − w b = 0

By = F1 + F2 + w b − Ay 689

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

x1 = 0 , 0.01a .. a 1

−F 1 − V 1 ( x) = 0

V 1 ( x) = −F 1

F 1 x + M1 ( x) = 0

M1 ( x) = −F1 x

kip 1

kip⋅ ft

x2 = a , 1.01a .. a + b 1

V 2 ( x) = ⎡⎣−F1 − w( x − a) + A y⎤⎦

−F 1 − w( x − a) + Ay − V 2 ( x) = 0

⎛ x − a ⎞ + M ( x) = ⎟ 2 ⎝ 2 ⎠

F 1 x − Ay( x − a) + w( x − a) ⎜

kip

0

⎡

( x − a)

⎣

2

M2 ( x) = ⎢−F 1 x + Ay( x − a) − w

2⎤

⎥ 1 ⎦ kip⋅ ft

x3 = a + b , 1.01( a + b) .. a + b + c 1

V 3 ( x) − F2 = 0

V 3 ( x) = F 2

−M3 ( x) − F 2 ( a + b + c − x) = 0

M3 ( x) = −F2 ( a + b + c − x)

kip 1

kip⋅ ft

Force (kip)

100

V1( x1) V2( x2) V3( x3)

50

0

50

100

0

10

20

30

40

50

60

x1 x2 x3 , , ft ft ft

Distance (ft)

690

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

200

Moment (kip-ft)

M1( x1)

0

M2( x2) M3( x3)

200

400

0

10

20

30

40

50

60

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-63 Express the x, y, z components of internal loading in the rod at the specific value for y, where 0 < y< a Units Used: 3

kip = 10 lb Given:

y = 2.5 ft

w = 800

lb F = 1500 lb ft

a = 4 ft

b = 2 ft

Solution:

In general we have

⎡ F ⎤ ⎥ V = ⎢ 0 ⎢ ⎥ ⎣w( a − y) ⎦

691

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

.

Engineering Mechanics - Statics

⎡⎢ w( a − y) ⎛ a − ⎜ 2 ⎝ ⎢ M = ⎢ −F b ⎢ −F( a − y) ⎣

Chapter 7

y⎞⎤

⎟⎥ ⎠⎥ ⎥ ⎥ ⎦

For these values we have

⎛ 1500.00 ⎞ V = ⎜ 0.00 ⎟ lb ⎜ ⎟ ⎝ 1200.00 ⎠

⎛ 900.00 ⎞ M = ⎜ −3000.00 ⎟ lb⋅ ft ⎜ ⎟ ⎝ −2250.00 ⎠

Problem 7-64 Determine the normal force, shear force, and moment in the curved rod as a function of θ. Given: c = 3 d = 4

Solution: For

0 ≤ θ ≤ π

ΣF x = 0;

ΣF y = 0;

N−

⎛ d ⎞ P cos ( θ ) − ⎛ c ⎞ P sin ( θ ) = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠

N=

⎛ P ⎞ ( d cos ( θ ) + c sin ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

V−

⎛ d ⎞ P sin ( θ ) + ⎛ c ⎞ P cos ( θ ) = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠

V=

⎛ P ⎞ ( d sin ( θ ) − c cos ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

0

0

692

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

ΣM = 0;

Chapter 7

⎛ −d ⎞ P( r − r cos ( θ ) ) + ⎛ c ⎞ P r sin ( θ ) + M = ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎝ c +d ⎠ M=

0

⎛ P r ⎞ ( d − d cos ( θ ) − c sin ( θ ) ) ⎜ 2 2⎟ ⎝ c +d ⎠

Problem 7-65 The quarter circular rod lies in the horizontal plane and supports a vertical force P at its end. Determine the magnitudes of the components of the internal shear force, moment, and torque acting in the rod as a function of the angle θ.

Solution: ΣF z = 0;

V= P

ΣMx = 0;

M + P r cos ( θ ) = 0 M = −P r cos ( θ ) M = P r cos ( θ )

ΣMy = 0;

T + P r( l − sin ( θ ) ) = 0 T = −P r ( l − sin ( θ ) T = P r ( 1 − sin ( θ )

693

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-66 Draw the shear and moment diagrams for the beam. Given: MB = 800 lb⋅ ft a = 5 ft F = 100 lb

b = 5 ft

Solution: V 1 ( x) = F

x2 = a , 1.01a .. a + b

V 2 ( x) = F

Force (lb)

x1 = 0 , 0.01a .. a

1

M1 ( x) = ⎡⎣−F ( a + b − x) − MB⎤⎦

lb 1

M2 ( x) = −F( a + b − x)

lb

1

lb⋅ ft

1

lb⋅ ft

V1( x1) 100 V2( x2)

99.9

99.8

0

2

4

6

8

10

x1 x2 , ft ft

Distance (ft)

Moment (lb-ft)

0

M1( x1) M2( x2)

1000

2000

0

2

4

6

8

10

x1 x2 , ft ft

Distane (ft)

694

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-67 Draw the shear and moment diagrams for the beam. kN = 103 N

Units Used: Given: w = 3

kN

F = 10 kN

m

L = 6m Solution: V ( x) = [ w( L − x) + F ]

⎡

( L − x)

⎣

2

M ( x) = ⎢−w

2

1

kN

⎤

− F ( L − x)⎥

1

⎦ kN⋅ m

Force (kN)

40

V( x) 20

0

0

1

2

3

4

5

6

x

Distance (m)

Moment (kN-m)

0 50

M( x) 100 150

0

1

2

3

4

5

6

x

Distance (m)

695

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-68 Draw the shear and moment diagrams for the beam. 3

kN = 10 N

Units Used: Given:

F = 7 kN

Solution: Given

M = 12 kN⋅ m A = 1N

Guesses A+B−F=0

a = 2m

b = 2m

c = 4m

B = 1N

⎛A⎞ ⎜ ⎟ = Find ( A , B) ⎝B ⎠

−F a − M + B ( a + b + c) = 0

x1 = 0 , 0.01a .. a

x2 = a , 1.01a .. a + b

x3 = a + b , 1.01( a + b) .. a + b + c

V 1 ( x1 ) = A

V 2 ( x2 ) = ( A − F)

V 3 ( x3 ) = −B

M1 ( x1 ) =

1

kN A x1

kN⋅ m

M2 ( x2 ) =

1

kN

A x2 − F( x2 − a) kN⋅ m

1

kN

M3 ( x3 ) = B ( a + b + c − x3 )

1

kN⋅ m

Force (kN)

5

V1( x1) V2( x2) V3( x3)

0

5

0

1

2

3

4

5

6

7

x1 , x2 , x3

Distance (m)

696

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8

Engineering Mechanics - Statics

Chapter 7

Moment (kN-m)

15

M1( x1)

10

M2( x2)

5

M3( x3)

0

5

0

1

2

3

4

5

6

7

8

x1 , x2 , x3

Distance (m)

Problem 7-69 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: a = 2m

b = 4m

w = 1.5

kN m

Solution:

⎛ b − a⎞ − A b = ⎟ y ⎝ 2 ⎠

w( b − a) ⎜

x1 = 0 , 0.01a .. a

V 1 ( x) = Ay

x2 = b − a , 1.01( b − a) .. b

Ay =

0 1

kN

w ( b − a) 2b

2

Ay = 0.75 kN

M1 ( x) = A y x

V 2 ( x) = ⎡⎣A y − w( x − a)⎤⎦

1

kN⋅ m

2 ⎡ ( x − a) ⎤ 1 ⎢ ⎥ kN M2 ( x) = Ay x − w 2 ⎣ ⎦ kN⋅ m

1

697

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (kN)

2

V1( x1) 0 V2( x2)

2

4

0

0.5

1

1.5

2

2.5

3

2.5

3

3.5

4

x1 , x2

Distance (m)

Moment (kN-m)

2

M1( x1) 1 M2( x2) 0

1

0

0.5

1

1.5

2

3.5

4

x1 , x2

Distance (m)

Problem 7-70 Draw the shear and moment diagrams for the beam. Given: lb ft

w = 30

MC = 180 lb⋅ ft

a = 9 ft

b = 4.5 ft

Solution: − Ay a +

Ay =

1 2

wa 6

⎛ a⎞ − M = ⎟ C ⎝ 3⎠

w a⎜

−

0

MC a 698

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Ay + By − By =

wa 2

1

Chapter 7

wa = 0

2

− Ay

x1 = 0 , 0.01a .. a 2 ⎛ wx ⎞ 1 ⎜ Ay − ⎟ 2 a ⎠ lb ⎝

V 1 ( x) =

M1 ( x) =

3 ⎛ wx ⎞ 1 ⎜ Ay x − ⎟ 6 a ⎠ lb⋅ ft ⎝

x2 = a , 1.01a .. a + b

⎛A + B − wa⎞ 1 ⎜ y y 2 ⎟ ⎝ ⎠ lb

V 2 ( x) =

⎡ ⎣

M2 ( x) = ⎢Ay x + B y( x − a) −

wa⎛ 2

⎜x − ⎝

2 a ⎞⎤ 1 3

⎟⎥ ⎠⎦ lb⋅ ft

Force (lb)

100

V1( x1) V2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Distance (ft)

699

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (lb-ft)

100

M1( x1) M2( x2)

0

100

200

0

2

4

6

8

10

12

x1 x2 , ft ft

Diastance (ft)

Problem 7-71 Draw the shear and moment diagrams for the beam. 3

kip = 10 lb

Given:

w1 = 30

lb ft

w2 = 120

lb ft

L = 12 ft Solution:

⎛ L⎞ + ⎟ ⎝2⎠

w1 L⎜

(w2 − w1)L⎛⎜ 3 ⎞⎟ − Ay L = 0 2 L

1

⎝ ⎠

⎛ L ⎞ + ⎛⎜ w2 − w1 ⎞⎟ ⎛ L ⎞ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 3 ⎠

Ay = w1 ⎜

⎡ ⎣

⎦

⎡

x

⎣

2

M ( x) = ⎢A y x − w1

2⎤

x 1 w2 − w1 ) ⎥ ( 2 L lb 1

V ( x) = ⎢Ay − w1 x −

2

− ( w2 − w1 )

x ⎤ 1 ⎥ 6 L⎦ kip⋅ ft 3

700

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Force (lb)

500

0

V( x )

500 0

2

4

6

8

10

12

x ft

Distance (ft)

Moment (kip-ft)

2

M( x) 1

0

0

2

4

6

8

10

12

x ft

Distance (ft)

Problem 7-72 Draw the shear and moment diagrams for the shaft. The support at A is a journal bearing and at B it is a thrust bearing. Given: F 1 = 400 lb w = 100

Solution:

lb in

F 2 = 800 lb a = 4 in

b = 12 in

c = 4 in

⎛ b ⎞ − F b + B( b + c) = ⎟ 2 ⎝2⎠

F 1 a − w b⎜

⎛ b2 ⎞ ⎟ + F2 b − F1 a ⎝2⎠ B=

w⎜ 0

B =

b+c

950.00 lb

701

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

x1 = 0 , 0.01a .. a

Chapter 7

V 1 ( x) = −F 1

x2 = a , 1.01a .. a + b M2 ( x2 )

1

M1 ( x) = −F1 x

lb

V 2 ( x2 ) = ⎡⎣−B + F2 + w( a + b − x2 )⎤⎦

1

lb⋅ in

1

lb

2 ⎡ a + b − x2 ) ⎤ 1 ( ⎥ = ⎢B( a + b + c − x2 ) − F2 ( a + b − x2 ) − w⋅ 2 ⎣ ⎦ lb⋅ in

V 3 ( x3 ) =

x3 = a + b , 1.01( a + b) .. a + b + c

−B lb

M3 ( x3 ) = B ( a + b + c − x3 )

1

lb⋅ in

Force (lb)

2000

V1( x1) V2( x2)

1000

V3( x3)

0

1000

0

5

10

15

20

x1 x2 x3 , , in in in

Distance (in)

Moment (lb-in)

4000

M1( x1) M2( x2) M3( x3)

2000

0

2000

0

5

10

15

x1 x2 x3 , , in in in

Distance (ini)

702

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

20

Engineering Mechanics - Statics

Chapter 7

Problem 7-73 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: F 1 = 10 kN

F 2 = 10 kN

M0 = 12 kN⋅ m

a = 2m

b = 2m

c = 2m

d = 2m

Solution: F 1 ( b + c + d) − M0 + F2 d − Ay( a + b + c + d) = 0

Ay =

F 1 ( b + c + d) − M 0 + F 2 d

Ay = 8.50 kN

a+b+c+d

Ay + By − F1 − F2 = 0

By = F1 + F2 − Ay

B y = 11.50 kN

x1 = 0 , 0.01a .. a V 1 ( x) = Ay

1

M1 ( x) = A y x

kN

1

kN⋅ m

x2 = a , 1.01a .. a + b V 2 ( x) = ( A y − F1 )

1

kN

M2 ( x) = ⎡⎣Ay x − F 1 ( x − a)⎤⎦

1

kN⋅ m

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = ( A y − F1 )

1

kN

M3 ( x) = ⎡⎣Ay x − F 1 ( x − a) + M0⎤⎦

1

kN⋅ m

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x) = −B y

1

kN

M4 ( x) = By( a + b + c + d − x)

1

kN⋅ m

703

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

10

Force (kN)

V1( x1)

5

V2( x2) 0 V3( x3) 5

V4( x4)

10 15

0

1

2

3

4

5

6

7

5

6

7

8

x1 , x2 , x3 , x4

Distance (m)

Moment (kN-m)

30

M1( x1) M2( x2)

20

M3( x3) M4( x4)10

0

0

1

2

3

4

8

x1 , x2 , x3 , x4

Distance (m)

Problem 7-74 Draw the shear and moment diagrams for the shaft. The support at A is a journal bearing and at B it is a thrust bearing.

704

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Given: F = 200 lb

w = 100

lb

M = 300 lb⋅ ft

ft

Solution: F ( a + b) − A b + w b⎛⎜

F( a + b) +

b⎞

⎟−M= ⎝ 2⎠

x1 = 0 , 0.01a .. a

a = 1 ft

A =

0

V 1 ( x1 ) = −F

⎛ w b2 ⎞ ⎜ ⎟−M ⎝ 2 ⎠

M1 ( x1 ) = −F x1

1

lb

M2 ( x2 )

c = 1 ft

A = 375.00 lb

b

V 2 ( x2 ) = ⎡⎣−F + A − w( x2 − a)⎤⎦

x2 = a , 1.01a .. a + b

b = 4 ft

1

lb⋅ ft

1

lb

⎡ (x2 − a)2⎥⎤ 1 ⎢ = −F x2 + A ( x2 − a) − w 2 ⎣ ⎦ lb⋅ ft V 3 ( x3 ) = 0

x3 = a + b , 1.01( a + b) .. a + b + c

1

lb

M3 ( x3 ) = −M

1

lb⋅ ft

Force (lb)

200

V1( x1) V2( x2) V3( x3)

0

200

400

0

1

2

3

4

5

x1 x2 x3 , , ft ft ft

Distance (ft)

705

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6

Engineering Mechanics - Statics

Chapter 7

0

Moment (lb-ft)

M1( x1)

100

M2( x2) M3( x3)

200

300

0

1

2

3

4

5

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-75 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given:

F = 8 kN c = 2m

M = 20 kN⋅ m

w = 15

kN m

a = 2m

b = 1m

d = 3m

Solution:

⎛ d⎞ = ⎟ ⎝2⎠

− A( a + b + c) − M + F c − w d⎜

A =

0

x1 = 0 , 0.01a .. a

V 1 ( x1 ) = A

x2 = a , 1.01a .. a + b

V 2 ( x2 ) = A

x3 = a + b , 1.01( a + b) .. a + b + c

⎛ d2 ⎞ ⎟ ⎝2⎠

F c − M − w⎜

1

kN 1

kN

A = −14.30 kN

a+b+c M1 ( x1 ) = A x1

1

kN⋅ m

M2 ( x2 ) = ( A x2 + M)

V 3 ( x3 ) = ( A − F)

1

kN⋅ m

1

kN

M3 ( x3 ) = ⎡⎣A x3 + M − F ( x3 − a − b)⎤⎦

1

kN⋅ m

706

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6

Engineering Mechanics - Statics

Chapter 7

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x4 ) = w( a + b + c + d − x4 )

M4 ( x4 ) = −w

1

kN

(a + b + c + d − x4)2

1

2

kN⋅ m

60

Force (kN)

V1( x1)

40

V2( x2) 20 V3( x3) 0

V4( x4)

20 40

0

1

2

3

4

5

6

7

8

5

6

7

8

x1 , x2 , x3 , x4

Distance (m)

Moment (kN-m)

0

M1( x1) 20 M2( x2) M3( x3) 40 M4( x4) 60

80

0

1

2

3

4

x1 , x2 , x3 , x4

Distance (m)

707

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-76 Draw the shear and moment diagrams for the shaft. The support at A is a thrust bearing and at B it is a journal bearing. Units Used: 3

kN = 10 N Given:

w = 2

kN m

F = 4 kN

⎛ a⎞ = ⎟ ⎝2⎠

B ( a + b ) − F a − w a⎜

Solution:

a = 0.8 m

x2 = a , 1.01a .. a + b

B =

0

a+b

A = wa + F − B

V 1 ( x1 ) = ( A − w x1 ) V 2 ( x2 ) = −B

⎛ a2 ⎞ ⎟ ⎝2⎠

F a + w⎜

A + B − wa − F = 0 x1 = 0 , 0.01a .. a

b = 0.2 m

M1 ( x1 )

1

kN

B = 3.84 kN A = 1.76 kN

⎡ ⎛ x1 2 ⎞⎤ 1 ⎢ ⎟⎥ = A x1 − w⎜ ⎣ ⎝ 2 ⎠⎦ kN⋅ m

M2 ( x2 ) = B ( a + b − x2 )

1

kN

1

kN⋅ m

Force (kN)

5

V1( x1) V2( x2)

0

5

0

0.2

0.4

0.6

0.8

x1 , x2

Distance (m)

Moment (kN-m)

1

M1( x1) 0.5 M2( x2)

0

0.5

0

0.2

0.4

0.6

0.8

x1 , x2

Distance (m)

708 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-77 Draw the shear and moment diagrams for the beam. Given:

w = 20

lb ft

MB = 160 lb⋅ ft

a = 20 ft

b = 20 ft

Solution:

⎛ a⎞ −w a⎜ ⎟ − MB + Cy( a + b) = ⎝ 2⎠

0

Ay − w a + Cy = 0

⎛ a2 ⎞ 1 ⎞ Cy = ⎜ w + MB⎟ ⎛⎜ ⎟ ⎝ 2 ⎠⎝ a + b ⎠

Cy = 104.00 lb

Ay = w a − Cy

Ay = 296.00 lb

x1 = 0 , 0.01a .. a V 1 ( x) = ( A y − w x)

2 ⎛ x ⎞ 1 ⎜ M1 ( x) = Ay x − w ⎟ 2 ⎠ lb⋅ ft ⎝

1

lb

x2 = a , 1.01a .. a + b V 2 ( x) = −Cy

1

M2 ( x) = Cy( a + b − x)

lb

1

lb⋅ ft

Force (lb)

400

V1( x1) 200 V2( x2)

0

200

0

5

10

15

20

25

30

35

40

x1 x2 , ft ft

Distance (ft)

709

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (lb-ft)

3000

M1( x1)2000 M2( x2) 1000

0

0

10

20

30

40

x1 x2 , ft ft

Distance (ft)

Problem 7-78 The beam will fail when the maximum moment is Mmax or the maximum shear is Vmax. Determine the largest distributed load w the beam will support. 3

kip = 10 lb

Units used: Given:

Mmax = 30 kip⋅ ft

Solution:

−A b + w

kip ft

w = 1

Set

V max = 8 kip

b ⎞ + b⎟ + w b = ⎜ 2⎝3 2 ⎠ a 2

b = 6 ft

and then scale the answer at the end

a⎛a

A + B − wb − w

a = 6 ft

w A =

0

b ⎞ + b⎟ + w ⎜ 2⎝3 2 ⎠ b

⎛ ⎝

B = w⎜ b +

=0

2

a⎛a

a⎞

⎟−A

2⎠

A = 7.00 kip

B = 2.00 kip

Shear limit - check critical points to the left and right of A and at B

⎛ ⎝

V big = max ⎜ B , w

wshear =

⎛ Vmax ⎞ ⎜ ⎟w ⎝ Vbig ⎠

a 2

, w

a 2

−A

⎞ ⎟ ⎠

wshear = 2.00

V big = 4.00 kip

kip ft

710

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment limit - check critical points at A and betwen A and B B ⎛ a⎞⎛ a ⎞ MA = −w⎜ ⎟ ⎜ ⎟ x = w ⎝ 2 ⎠⎝ 3 ⎠ Mbig = max ( MA , MAB wmoment =

MAB

)

⎛ x2 ⎞ = B x − w⎜ ⎟ ⎝2⎠

⎛ MA ⎞ ⎛ −6.00 ⎞ ⎜ ⎟=⎜ ⎟ kip⋅ ft ⎝ MAB ⎠ ⎝ 2.00 ⎠

Mbig = 6.00 kip⋅ ft

⎛ Mmax ⎞ ⎜ ⎟w ⎝ Mbig ⎠

wmoment = 5.00

kip ft

wans = min ( wshear , wmoment)

Choose the critical case

wans = 2.00

kip ft

Problem 7-79 The beam consists of two segments pin connected at B. Draw the shear and moment diagrams for the beam.

Given: F = 700 lb

w = 150

lb ft

Solution:

⎛ c ⎞ − M + Cc = ⎟ ⎝2⎠

−w c⎜

B + C − wc = 0 x1 = 0 , 0.01a .. a x2 = a , 1.01a .. a + b

M = 800 lb ft

⎛ c2 ⎞ ⎟+M ⎝2⎠

a = 8 ft

b = 4 ft

c = 6 ft

w⎜ 0

C =

C = 583.33 lb

c

B = wc − C V 1 ( x1 ) = ( B + F ) V 2 ( x2 ) = B

x3 = a + b , 1.01( a + b) .. a + b + c M3 ( x3 )

B = 316.67 lb M1 ( x1 ) = ⎡⎣−F( a − x1 ) − B ( a + b − x1 )⎤⎦

1

lb

M2 ( x2 ) = −B ( a + b − x2 )

1

lb

V 3 ( x3 ) = ⎡⎣−C + w( a + b + c − x3 )⎤⎦

1

lb⋅ ft 1

lb

2 ⎡ ⎤ 1 a + b + c − x3 ) ( = ⎢C( a + b + c − x3 ) − w − M⎥ 2 ⎣ ⎦ lb⋅ ft

711

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

lb⋅ ft

Engineering Mechanics - Statics

Chapter 7

Force (lb)

1500

V1( x1)

1000

V2( x2)

500

V3( x3)

0 500 1000

0

2

4

6

8

10

12

14

16

18

20

x1 x2 x3 , , ft ft ft

Distance (ft)

Moment (lb-ft)

5000

M1( x1)

0

M2( x2) M3( x3)

5000

1 .10

4

0

2

4

6

8

10

12

14

16

18

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-80 The beam consists of three segments pin connected at B and E. Draw the shear and moment diagrams for the beam. 3

Units Used:

kN = 10 N

Given:

MA = 8 kN⋅ m

F = 15 kN

w = 3

c = 2m

d = 2m

e = 2m

kN m

a = 3m

b = 2m

f = 4m

712

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

20

Engineering Mechanics - Statics

Guesses

Chapter 7

Ay = 1 N

By = 1 N

Cy = 1 N

Dy = 1 N

Ey = 1 N

Fy = 1 N

Given Ay + Cy + Dy + Fy − F − w f = 0

F b − MA − Ay( a + b) = 0

−w f⎛⎜

M A + F a + B y ( a + b) = 0

f⎞

⎟ + Fy f = ⎝2⎠

0

B y + Cy + Dy + E y = 0

−B y c + Dy d + E y( d + e) = 0

⎛⎜ Ay ⎞⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Cy ⎟ = Find ( A , B , C , D , E , F ) y y y y y y ⎜ Dy ⎟ ⎜ ⎟ ⎜ Ey ⎟ ⎜ Fy ⎟ ⎝ ⎠

⎛⎜ Ay ⎞⎟ ⎛ 4.40 ⎞ ⎜ By ⎟ ⎜ −10.60 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ Cy ⎟ = ⎜ 15.20 ⎟ kN ⎜ Dy ⎟ ⎜ 1.40 ⎟ ⎜ ⎟ ⎜ −6.00 ⎟ ⎟ ⎜ Ey ⎟ ⎜ ⎜ Fy ⎟ ⎝ 6.00 ⎠ ⎝ ⎠

x1 = 0 , 0.01a .. a M1 ( x) = ( Ay x + MA)

1

V 1 ( x) = Ay

kN x2 = a , 1.01a .. a + b V 2 ( x) = ( A y − F)

1

kN

1

kN⋅ m

M2 ( x) = ⎡⎣Ay x + MA − F ( x − a)⎤⎦

1

kN⋅ m

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = B y

1

M3 ( x) = By( x − a − b)

kN

1

kN⋅ m

x4 = a + b + c , 1.01( a + b + c) .. a + b + c + d V 4 ( x) = ( By + Cy)

1

kN

M4 ( x) = ⎡⎣B y( x − a − b) + Cy( x − a − b − c)⎤⎦

1

kN⋅ m

x5 = a + b + c + d , 1.01( a + b + c + d) .. a + b + c + d + e V 5 ( x) = −E y

1

kN

M5 ( x) = Ey( a + b + c + d + e − x)

1

kN⋅ m

x6 = a + b + c + d + e , 1.01( a + b + c + d + e) .. a + b + c + d + e + f 713

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

V 6 ( x) = ⎡⎣−Fy + w( a + b + c + d + e + f − x)⎤⎦

1

kN

2 ⎡ ( a + b + c + d + e + f − x) ⎤ 1 ⎢ ⎥ M6 ( x) = F y( a + b + c + d + e + f − x) − w 2 ⎣ ⎦ kN⋅ m

10

V1( x1) 5

Force (kN)

V2( x2) V3( x3) 0 V4( x4) V5( x5) 5 V6( x6) 10

15

0

2

4

6

8

10

12

14

x1 , x2 , x3 , x4 , x5 , x6

Distance (m)

714

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

30

M1( x1) 20

Moment (kN-m)

M2( x2) M3( x3)

10

M4( x4) 0 M5( x5) M6( x6)

10

20

30

0

2

4

6

8

10

12

14

x1 , x2 , x3 , x4 , x5 , x6

Distance (m)

Problem 7-81 Draw the shear and moment diagrams for the beam.

Solutions: Support Reactions: ΣMx = 0;

⎛ L ⎞ − w0 L ⎛ 4 L ⎞ = ⎟ ⎜ ⎟ 2 ⎝ 3 ⎠ ⎝2⎠

B y L − w0 L⎜

0

By =

7 w0 L 6

715

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

+

↑

Σ F y = 0;

Ay +

Chapter 7

⎛ 7w0 L ⎞ ⎛ w0 L ⎞ ⎜ ⎟ − w0 L − ⎜ ⎟= ⎝ 6 ⎠ ⎝ 2 ⎠

0

Ay =

w0 L 3

716

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-82 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb Given: F = 2000 lb

a = 9 ft

lb ft

b = 9 ft

w = 500 Solution:

⎛ 2b ⎞ ⎛ a⎞ w b⎜ ⎟ + F b + w a⎜ b + ⎟ − A y ( a + b ) = 2 ⎝3⎠ ⎝ 2⎠ 1

Ay + By − F − w a − Ay = 5.13 kip

1 2

0

wb = 0

⎛ w b2 ⎞ ⎜ ⎟ + F b + w a⎛⎜b + 3 ⎠ ⎝ ⎝ Ay =

a⎞

a+b

By = w a +

1 2

⎟

2⎠

w b − Ay + F

B y = 3.63 kip

x1 = 0 , 0.01a .. a

717

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

V 1 ( x) = ( A y − w x)

Chapter 7

x M1 ( x) = ⎡⎢Ay x − w x⎛⎜ ⎞⎟⎥⎤ 2

1

⎣

kip

1

⎝ ⎠⎦ kip⋅ ft

x2 = a , 1.01a .. a + b V 2 ( x) = ⎡⎢−By +

⎣

1 2

w⎛⎜

⎝

a + b − x⎞ b

M2 ( x) = ⎡⎢B y( a + b − x) −

⎣

⎤ 1 ⎟ ( a + b − x)⎥ ⎠ ⎦ kip

1 2

w⎛⎜

⎝

a + b − x⎞ b

⎛ a + b − x ⎞⎤ 1 ⎟ ( a + b − x) ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kip⋅ ft ⎠

Force (kip)

10

V1( x1) 5 V2( x2) 0

5

0

5

10

15

x1 x2 , ft ft

Distance (ft)

Moment (kip-ft)

30

M1( x1)

20

M2( x2)

10 0 10

0

5

10

15

x1 x2 , ft ft

Distance (ft)

718

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-83 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given: w = 3

kN m

a = 3m

b = 3m

Solution:

wb

− Ay( a + b) + w

Ay + By −

2

⎛ w b ⎞ ⎛ 2b ⎞ ⎜ ⎝

⎟⎜ ⎠⎝

2

3

⎛ w a ⎞⎛

⎟+⎜ ⎠ ⎝

2

a⎞

⎟ ⎜b + 3 ⎟ = ⎠⎝ ⎠

0

( a + b) = 0

+

3

Ay = By =

2

⎛ w a ⎞ ⎛b + a ⎞ ⎜ 2 ⎟⎜ 3 ⎟ ⎝ ⎠⎝ ⎠ a+b

w 2

( a + b) − A y

x1 = 0 , 0.01a .. a

⎡ ⎣

V 1 ( x) = ⎢A y −

1 2

⎛ x ⎞ x⎤ 1 ⎟⎥ ⎝ a ⎠ ⎦ kN

⎡ ⎣

w⎜

M1 ( x) = ⎢Ay x −

1 2

⎛ x ⎞ x⎛ x ⎞⎤ 1 ⎟ ⎜ ⎟⎥ ⎝ a ⎠ ⎝ 3 ⎠⎦ kN⋅ m

w⎜

x2 = a , 1.01a .. a + b

⎡ ⎣

V 2 ( x) = ⎢−By +

1 2

⎛ a + b − x ⎞ ( a + b − x)⎤ 1 ⎟ ⎥ ⎝ b ⎠ ⎦ kN

w⎜

⎡ ⎣

M2 ( x) = ⎢B y( a + b − x) −

1 2

⎛ a + b − x ⎞ ( a + b − x) ⎛ a + b − x ⎞⎤ 1 ⎟ ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kN⋅ m ⎝ b ⎠

w⎜

Force (kN)

5

V1( x1) V2( x2)

0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

719

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (kN-m)

10

M1( x1) 5 M2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Problem 7-84 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb Given: w = 100 Guesses: Given

lb ft

M0 = 9 kip⋅ ft

Ay = 1 lb

1 2

⎛ Ay ⎞ ⎜ ⎟ = Find ( Ay , By) ⎝ By ⎠

b = 6 ft

c = 4 ft

B y = 1 lb

−M0 + By( a + b) − Ay + By −

a = 6 ft

1 2

⎛ 2 a ⎞ − 1 w b⎛ a + ⎟ ⎜ ⎝3⎠ 2 ⎝

w a⎜

b⎞

⎟=

0

M1 ( x) = ⎢Ay x −

1

3⎠

w( a + b) = 0

⎛ Ay ⎞ ⎛ −0.45 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ By ⎠ ⎝ 1.05 ⎠

x1 = 0 , 0.01a .. a

⎡ ⎣

V 1 ( x) = ⎢A y −

1 2

⎛ x ⎞ x⎤ 1 ⎟⎥ ⎝ a ⎠ ⎦ lb

w⎜

⎡ ⎣

2

⎛ x ⎞ x⎛ x ⎞⎤ 1 ⎟ ⎜ ⎟⎥ ⎝ a ⎠ ⎝ 3 ⎠⎦ kip⋅ ft

w⎜

x2 = a , 1.01a .. a + b

720

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

V 2 ( x) = ⎡⎢−By + w⎛⎜ 2 ⎝ ⎣ 1

Chapter 7

a + b − x⎞ b

⎤1 ⎟ ( a + b − x)⎥ ⎠ ⎦ lb

M2 ( x) = ⎡⎢B y( a + b − x) − M0 −

⎣

1 2

w⎛⎜

⎝

a + b − x⎞

⎛ a + b − x ⎞⎤ 1 ⎟ ( a + b − x) ⎜ 3 ⎟⎥ ⎝ ⎠⎦ kip⋅ ft ⎠

b

x3 = a + b , 1.01( a + b) .. a + b + c V 3 ( x) = 0

M3 ( x) = −M0

1

kip⋅ ft

V1( x1) V2( x2)

500

V3( x3) 1000

1500

0

2

4

6

8

10

12

14

16

x1 x2 x3 , , ft ft ft

Distance (ft) 0

Moment (kip-ft)

Force (lb)

0

M1( x1) M2( x2) M3( x3)

5

10

0

5

10

15

x1 x2 x3 , , ft ft ft

Distance (ft)

721

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-85 Draw the shear and moment diagrams for the beam.

Solution:

722

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-86 Draw the shear and moment diagrams for the beam. Units Used: 3

kN = 10 N Given:

kN m

w = 2

a = 3m

b = 3m

Solution: x1 = 0 , 0.01a .. a V 1 ( x) =

⎡ 1 w b + 1 w⎛ a − x ⎞ ( a − x)⎤ 1 ⎢2 ⎜ ⎟ ⎥ 2 ⎝ a ⎠ ⎣ ⎦ kN

M1 ( x) =

⎡−1 w b⎛ 2 b + a − x⎞ − 1 w⎛ a − x ⎞ ( a − x) ⎛ a − x ⎞⎤ 1 ⎢2 ⎜3 ⎟ 2 ⎜ ⎟ ⎜ 3 ⎟⎥ ⎝ ⎠ ⎝ ⎠⎦ kN⋅ m ⎣ ⎝ a ⎠

x2 = a , 1.01a .. a + b V 2 ( x) =

⎡ 1 w b − 1 w⎛ x − a ⎞ ( x − a)⎤ 1 ⎢2 ⎜ ⎟ ⎥ 2 ⎝ a ⎠ ⎣ ⎦ kN

M2 ( x) =

⎡−1 w b⎛a + 2b − x⎞ − 1 w⎛ x − a ⎞ ( x − a) ⎛ x − a ⎞⎤ 1 ⎢2 ⎜ ⎟ 2 ⎜ ⎟ ⎜ 3 ⎟⎥ 3 ⎝ ⎠ ⎝ ⎠⎦ kN⋅ m ⎣ ⎝ b ⎠

Force (kN)

10

V1( x1) 5 V2( x2) 0

5

0

1

2

3

4

5

6

x1 , x2

Distance (m)

723

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Moment (kN-m)

0

M1( x1) M2( x2)

10

20

0

1

2

3

4

5

6

x1 , x2

Distance (m)

Problem 7-87 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb

Given: w = 5

kip ft

M1 = 15 kip⋅ ft

M2 = 15 kip⋅ ft

a = 6 ft

b = 10 ft

c = 6 ft

Solution:

⎛ a ⎞ ⎛b + ⎟⎜ ⎝ 2 ⎠⎝

M1 − A b − M2 + w⎜

a⎞

⎛ b⎞ ⎛ c ⎞⎛ c ⎞ + w b⎜ ⎟ − w⎜ ⎟ ⎜ ⎟ = ⎟ 3⎠ ⎝2⎠ ⎝ 2 ⎠⎝ 3 ⎠

0

⎛ a + c⎞ = ⎟ ⎝ 2 ⎠

A + B − w b − w⎜

2 2 ⎛ a ⎞ ⎛b + a ⎞ + w⎛⎜ b ⎟⎞ − w⎛⎜ c ⎞⎟ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 3 ⎠ ⎝2⎠ ⎝6⎠

M1 − M2 + w⎜ A =

b

⎛ ⎝

B = w⎜ b +

a + c⎞ 2

x1 = 0 , 0.01a .. a

⎟−A ⎠

⎛ A ⎞ ⎛ 40.00 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ B ⎠ ⎝ 40.00 ⎠ ⎛ ⎝

V 1 ( x) = ⎜ −w

M1p ( x) =

x⎞ x 1 ⎟ a ⎠ 2 kip

⎡⎛ −w x ⎞ x x − M ⎤ 1 ⎢⎜ ⎟ 1⎥ ⎣⎝ a ⎠ 2 3 ⎦ kip⋅ ft 724

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

0

Engineering Mechanics - Statics

Chapter 7

x2 = a , 1.01a .. a + b

⎡ ⎣

V 2 ( x) = ⎢A − w

⎡ ⎣

a 2

⎤ 1 ⎦ kip

− w( x − a)⎥

M2p ( x) = ⎢−M1 − w

a⎛

⎜x − 2⎝

2a ⎞ 3

⎛ x − a ⎞⎤ 1 ⎟ + A( x − a) − w( x − a) ⎜ 2 ⎟⎥ ⎠ ⎝ ⎠⎦ kip⋅ ft

x3 = a + b , 1.01( a + b) .. a + b + c

⎛ a + b + c − x⎞⎛ a + b + c − x⎞ 1 ⎟⎜ ⎟ 2 c ⎠ kip ⎝ ⎠⎝

V 3 ( x) = w⎜

⎡ ⎛ a + b + c − x⎞⎛ a + b + c − x⎞⎛ a + b + c − x⎞ − M ⎤ 1 ⎟⎜ ⎟⎜ ⎟ 2⎥ 2 3 c ⎠⎝ ⎠ ⎣ ⎝ ⎠⎝ ⎦ kip⋅ ft

M3p ( x) = ⎢−w⎜

Force (lb)

40

V1( x1) V2( x2) V3( x3)

20

0

20

40

0

5

10

15

20

x1 x2 x3 , , ft ft ft

Distance (ft)

725

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

20

Moment (kip-ft)

M1p( x1) M2p( x2) M3p( x3)

0

20

40

60

0

5

10

15

20

x1 x2 x3 , , ft ft ft

Distance (ft)

Problem 7-88 Draw the shear and moment diagrams for the beam. Units Used: 3

kip = 10 lb

Given: w1 = 2

kip ft

w2 = 1

kip ft

a = 15 ft

Solution: x = 0 , 0.01a .. a

⎡ ⎣

⎛ ⎝

V ( x) = ⎢w2 x − ⎜ w1

x ⎞ x⎤ 1 ⎟ ⎥ a ⎠ 2⎦ kip

⎡ ⎣

M ( x) = ⎢w2 x

x 2

⎛ ⎝

− ⎜ w1

x ⎞ x x⎤ 1 ⎟ ⎥ a ⎠ 2 3⎦ kip⋅ ft

726

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Engineering Mechanics - Statics

Chapter 7

Force (kip)

4

V( x ) 2

0

0

2

4

6

8

10

12

14

x ft

Distance (ft) Moment (kip-ft)

40

M( x) 20

0

0

2

4

6

8

10

12

14

x ft

Distance (ft)

Problem 7-89 Determine the force P needed to hold the cable in the position shown, i.e., so segment BC remains horizontal. Also, compute the sag yB and the maximum tension in the cable. Units Used: 3

kN = 10 N

727

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Engineering Mechanics - Statics

Chapter 7

Given: a = 4m

F 1 = 4 kN

b = 6m

F 2 = 6 kN

c = 3m d = 2m e = 3m Solution: Initial guesses: yB = 1 m P = 1 kN

TAB = 1 kN TBC = 1 kN TCD = 1 kN TDE = 1 kN

Given

⎛ −a ⎞ T + T = AB BC ⎜ 2 2⎟ + y a B ⎠ ⎝ yB ⎛⎜ ⎞⎟ TAB − F1 = ⎜ a2 + yB2 ⎟ ⎝ ⎠ −TBC +

0

0

c ⎡ ⎤T = CD ⎢ 2 2⎥ + y − e c (B )⎦ ⎣

yB − e ⎡⎢ ⎥⎤ TCD − P = ⎢ c2 + ( yB − e) 2⎥ ⎣ ⎦

0

0

−c ⎡ ⎤T + ⎛ d ⎞T = CD ⎜ DE ⎢ 2 ⎥ 2 2 2⎟ + y − e + e c d (B )⎦ ⎝ ⎠ ⎣

0

⎡⎢ −( yB − e) ⎥⎤ e ⎞T − F = TCD + ⎛ ⎜ ⎟ DE 2 2 2 2 2 ⎢ c + ( yB − e) ⎥ + d e ⎝ ⎠ ⎣ ⎦

0

⎛ yB ⎞ ⎜ ⎟ ⎜ P ⎟ ⎜ TAB ⎟ ⎜ ⎟ = Find ( yB , P , TAB , TBC , TCD , TDE) TBC ⎜ ⎟ ⎜ TCD ⎟ ⎜ ⎟ ⎝ TDE ⎠ 728

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Engineering Mechanics - Statics

Chapter 7

Tmax = max ( TAB , TBC , TCD , TDE)

yB = 3.53 m P = 800.00 N Tmax = 8.17 kN

Problem 7-90 Cable ABCD supports the lamp of mass M1 and the lamp of mass M2. Determine the maximum tension in the cable and the sag of point B. Given: M1 = 10 kg M2 = 15 kg a = 1m b = 3m c = 0.5 m d = 2m Solution: Guesses

Given

yB = 1 m

TAB = 1 N

TBC = 1 N

b ⎛ −a ⎞ T + ⎡ ⎤T = AB BC ⎜ 2 ⎢ 2 2⎟ 2⎥ + y + y − d a b ( ) B B ⎝ ⎠ ⎣ ⎦

TCD = 1 N 0

yB yB − d ⎛⎜ ⎡ ⎞⎟ ⎥⎤ TBC − M1 g = TAB + ⎢ ⎜ a2 + yB2 ⎟ ⎢ b2 + ( yB − d) 2⎥ ⎝ ⎠ ⎣ ⎦ −b ⎡ ⎤T + ⎛ c ⎞T = BC ⎜ CD ⎢ 2 2⎥ 2 2⎟ + y − d + d b c ( ) ⎝ ⎠ B ⎣ ⎦

0

0

⎡⎢ −( yB − d) ⎥⎤ d ⎞T − M g = TBC + ⎛ CD 2 ⎜ ⎟ 2 2 ⎢ b2 + ( yB − d) 2⎥ + d c ⎝ ⎠ ⎣ ⎦

0

729

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Engineering Mechanics - Statics

Chapter 7

⎛ yB ⎞ ⎜ ⎟ ⎜ TAB ⎟ = Find ( y , T , T , T ) B AB BC CD ⎜ TBC ⎟ ⎜ ⎟ ⎝ TCD ⎠ Tmax = max ( TAB , TBC , TCD)

⎛ TAB ⎞ ⎛ 100.163 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 38.524 ⎟ N ⎜ T ⎟ ⎜⎝ 157.243 ⎟⎠ ⎝ CD ⎠ Tmax = 157.2 N

yB = 2.43 m

Problem 7-91 The cable supports the three loads shown. Determine the sags yB and yD of points B and D. Given: a = 4 ft

e = 12 ft

b = 12 ft

f = 14 ft

c = 20 ft

P 1 = 400 lb

d = 15 ft

P 2 = 250 lb

Solution: Guesses

Given

yB = 1 ft

yD = 1 ft

TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

TDE = 1 lb

c ⎛ −b ⎞ T + ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + y + f − y b c ( ) B ⎠ B ⎦ ⎝ ⎣

0

yB f − yB ⎛⎜ ⎡ ⎞⎟ ⎤⎥ TAB − ⎢ TBC − P 2 = ⎜ b2 + yB2 ⎟ ⎢ c2 + ( f − yB) 2⎥ ⎝ ⎠ ⎣ ⎦

0

−c d ⎡ ⎤T + ⎡ ⎤T = BC ⎢ CD ⎢ 2 ⎥ 2 2 2⎥ + f − y + f − y c d ( ) ( ) B D ⎣ ⎦ ⎣ ⎦

0

730

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Engineering Mechanics - Statics

Chapter 7

f − yB f − yD ⎡⎢ ⎥⎤ TBC + ⎡⎢ ⎥⎤ TCD − P1 = ⎢ c2 + ( f − yB) 2⎥ ⎢ d2 + ( f − yD) 2⎥ ⎣ ⎦ ⎣ ⎦ −d e ⎡ ⎤T + ⎡ ⎤T = CD ⎢ DE ⎢ 2 ⎥ ⎥ 2 2 2 + f − y + a + y d e ( D) ⎦ ( D) ⎦ ⎣ ⎣

0

a + yD ⎡⎢ −( f − yD) ⎥⎤ ⎡ ⎥⎤ TDE − P2 = TCD + ⎢ ⎢ d2 + ( f − yD) 2⎥ ⎢ e2 + ( a + yD) 2⎥ ⎣ ⎦ ⎣ ⎦

⎛⎜ TAB ⎞⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , T , y , y ) AB BC CD DE B D ⎜ TDE ⎟ ⎜ ⎟ ⎜ yB ⎟ ⎜ yD ⎟ ⎝ ⎠

0

0

⎛ TAB ⎞ ⎛ 675.89 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 566.90 ⎟ lb ⎜ TCD ⎟ ⎜ 603.86 ⎟ ⎜ ⎟ ⎜ 744.44 ⎟ ⎠ ⎝ TDE ⎠ ⎝ ⎛ yB ⎞ ⎛ 8.67 ⎞ ⎜ ⎟ =⎜ ⎟ ft ⎝ yD ⎠ ⎝ 7.04 ⎠

Problem 7-92 The cable supports the three loads shown. Determine the magnitude of P 1 and find the sag yD for the given data. Given: P 2 = 300 lb

c = 20 ft

yB = 8 ft

d = 15 ft

a = 4 ft

e = 12 ft

b = 12 ft

f = 14 ft

Solution: Guesses

P 1 = 1 lb

TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

731

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Engineering Mechanics - Statics

TDE = 1 lb

Chapter 7

yD = 1 ft

Given c ⎛ −b ⎞ T + ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + y + f − y b c ( B) ⎦ B ⎠ ⎝ ⎣

0

yB f − yB ⎛⎜ ⎡ ⎞⎟ ⎤⎥ TAB − ⎢ TBC − P 2 = ⎜ b2 + yB2 ⎟ ⎢ c2 + ( f − yB) 2⎥ ⎝ ⎠ ⎣ ⎦

0

−c d ⎡ ⎤T + ⎡ ⎤T = BC CD ⎢ 2 ⎢ 2 2⎥ 2⎥ + f − y + f − y c d ( ) ( ) B D ⎣ ⎦ ⎣ ⎦

0

f − yB f − yD ⎡⎢ ⎥⎤ TBC + ⎡⎢ ⎥⎤ TCD − P1 = ⎢ c2 + ( f − yB) 2⎥ ⎢ d2 + ( f − yD) 2⎥ ⎣ ⎦ ⎣ ⎦ −d e ⎡ ⎤T + ⎡ ⎤T = CD DE ⎢ 2 ⎢ 2 2⎥ 2⎥ + f − y + a + y d e ( ) ( ) D D ⎣ ⎦ ⎣ ⎦

0

a + yD ⎡⎢ −( f − yD) ⎥⎤ ⎡ ⎥⎤ TDE − P2 = TCD + ⎢ ⎢ d2 + ( f − yD) 2⎥ ⎢ e2 + ( a + yD) 2⎥ ⎣ ⎦ ⎣ ⎦

⎛⎜ TAB ⎞⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , T , P , y ) AB BC CD DE 1 D ⎜ TDE ⎟ ⎜ ⎟ ⎜ P1 ⎟ ⎜ yD ⎟ ⎝ ⎠

0

0

⎛ TAB ⎞ ⎛ 983.33 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 854.21 ⎟ lb ⎜ TCD ⎟ ⎜ 916.11 ⎟ ⎜ ⎟ ⎜ 1084.68 ⎟ ⎠ ⎝ TDE ⎠ ⎝ P 1 = 658 lb yD = 6.44 ft

Problem 7-93 The cable supports the loading shown. Determine the distance xB the force at point B acts from A.

732

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Engineering Mechanics - Statics

Chapter 7

Given: P = 40 lb

c = 2 ft

F = 30 lb

d = 3 ft

a = 5 ft

e = 3

b = 8 ft

f = 4

Solution: The initial guesses: TAB = 10 lb

TCD = 30 lb

TBC = 20 lb

xB = 5 ft

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + P = ⎜ xB2 + a2 ⎟ ⎢ ( xB − d) 2 + b2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB BC ⎜ ⎢ 2 2⎟ 2 2⎥ + a − d + b x x ( ) B B ⎝ ⎠ ⎣ ⎦

0

0

xB − d ⎡⎢ d ⎤⎥ f ⎞T + ⎛ ⎞F = TBC − ⎛ CD ⎜ ⎜ ⎟ ⎟ 2 2 2 2 ⎢ ( xB − d) 2 + b2⎥ ⎝ c +d ⎠ ⎝ e +f ⎠ ⎣ ⎦

0

b ⎡ ⎤T − ⎛ c ⎞T − ⎛ e ⎞F = BC ⎜ CD ⎜ ⎢ 2 2⎥ 2 2⎟ 2 2⎟ − d + b + d + f x c e ( ) ⎝ ⎠ ⎝ ⎠ B ⎣ ⎦

0

⎛ TAB ⎞ ⎜ ⎟ ⎜ TCD ⎟ = Find ( T , T , T , x ) AB CD BC B ⎜ TBC ⎟ ⎜ ⎟ ⎝ xB ⎠

⎛ TAB ⎞ ⎛ 50.90 ⎞ ⎜ ⎟ ⎜ TCD ⎟ = ⎜ 36.70 ⎟ lb ⎜ T ⎟ ⎜⎝ 38.91 ⎟⎠ ⎝ BC ⎠

xB = 4.36 ft

Problem 7-94 The cable supports the loading shown. Determine the magnitude of the horizontal force P.

733

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Engineering Mechanics - Statics

Chapter 7

Given: F = 30 lb

c = 2 ft

xB = 6 ft

d = 3 ft

a = 5 ft

e = 3

b = 8 ft

f = 4

Solution: The initial guesses: TAB = 10 lb

TCD = 30 lb

TBC = 20 lb

P = 10 lb

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + P = ⎜ xB2 + a2 ⎟ ⎢ ( xB − d) 2 + b2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB BC ⎜ 2 ⎢ 2 2⎟ 2⎥ + x + x − d a b ( ) B B ⎝ ⎠ ⎣ ⎦

0

0

xB − d ⎤⎥ f ⎛ −d ⎞ T + ⎡⎢ ⎞F = TBC + ⎛ CD ⎜ 2 2⎟ ⎜ ⎟ 2 2 ⎢ b2 + ( xB − d) 2⎥ ⎝ c +d ⎠ ⎝ e +f ⎠ ⎣ ⎦

0

b ⎛ −c ⎞ T + ⎡ ⎤T − ⎛ e ⎞F = BC ⎜ ⎜ 2 2 ⎟ CD ⎢ 2 2⎥ 2 2⎟ + x − d + f b e ( ) ⎝ c +d ⎠ ⎝ ⎠ B ⎣ ⎦

0

⎛ TAB ⎞ ⎜ ⎟ T BC ⎜ ⎟ = Find ( T , T , T , P) AB BC CD ⎜ TCD ⎟ ⎜ ⎟ ⎝ P ⎠

⎛ TAB ⎞ ⎛ 70.81 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 48.42 ⎟ lb ⎜ T ⎟ ⎜⎝ 49.28 ⎟⎠ ⎝ CD ⎠

P = 71.40 lb

734

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Engineering Mechanics - Statics

Chapter 7

Problem 7-95 Determine the forces P1 and P2 needed to hold the cable in the position shown, i.e., so segment CD remains horizontal. Also, compute the maximum tension in the cable. Given:

3

kN = 10 N

F = 5 kN

d = 4m

a = 1.5 m

e = 5m

b = 1m

f = 4m

c = 2m Solution: Guesses F AB = 1 kN

F BC = 1 kN

F CD = 1 kN

F DE = 1 kN

P 1 = 1 kN

P 2 = 1 kN

Given

⎛ −c ⎞ F + ⎛ d ⎞ F = ⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ ⎝ b +d ⎠

0

735

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Engineering Mechanics - Statics

Chapter 7

⎛ a ⎞F − ⎛ b ⎞F − F = ⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ ⎝ b +d ⎠ ⎛ −d ⎞ F + F = ⎜ 2 2 ⎟ BC CD ⎝ b +d ⎠ ⎛ b ⎞F − P = ⎜ 2 2 ⎟ BC 1 ⎝ b +d ⎠ −F CD +

0

0

0

f ⎡ ⎤F = DE ⎢ 2 2⎥ ⎣ f + ( a + b) ⎦

a+b ⎡ ⎤F − P = DE 2 ⎢ 2 2⎥ + ( a + b ) f ⎣ ⎦

0

0

⎛⎜ FAB ⎟⎞ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( F , F , F , F , P , P ) AB BC CD DE 1 2 ⎜ FDE ⎟ ⎜ ⎟ ⎜ P1 ⎟ ⎜ P2 ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎛ 12.50 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FBC ⎟ = ⎜ 10.31 ⎟ kN ⎜ FCD ⎟ ⎜ 10.00 ⎟ ⎜ ⎟ ⎜ 11.79 ⎟ ⎠ ⎝ FDE ⎠ ⎝

⎛ P1 ⎞ ⎛ 2.50 ⎞ ⎜ ⎟=⎜ ⎟ kN ⎝ P2 ⎠ ⎝ 6.25 ⎠

Tmax = max ( FAB , F BC , F CD , FDE) F max = max ( F AB , FBC , FCD , F DE)

Tmax = 12.50 kN F max = 12.50 kN

736

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Engineering Mechanics - Statics

Chapter 7

Problem 7-96 The cable supports the loading shown. Determine the distance xB from the wall to point B. Given: W1 = 8 lb W2 = 15 lb a = 5 ft b = 8 ft c = 2 ft d = 3 ft Solution: Guesses TAB = 1 lb

TBC = 1 lb

TCD = 1 lb

xB = 1 ft

Given xB − d ⎛⎜ −xB ⎟⎞ ⎡ ⎤⎥ TAB − ⎢ TBC + W2 = ⎜ a2 + xB2 ⎟ ⎢ b2 + ( xB − d) 2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB ⎢ BC ⎜ 2 ⎟ 2 2 2⎥ + x + x − d a b ( ) B ⎠ B ⎝ ⎣ ⎦ xB − d ⎡⎢ ⎤⎥ ⎛ d ⎞T = TBC − ⎜ CD 2 2⎟ ⎢ b2 + ( xB − d) 2⎥ + d c ⎝ ⎠ ⎣ ⎦

0

0

b ⎡ ⎤T − ⎛ c ⎞T − W = BC ⎜ CD 1 ⎢ 2 2⎥ 2 2⎟ + x − d + d b c ( ) ⎝ ⎠ B ⎣ ⎦

⎛ TAB ⎞ ⎜ ⎟ TBC ⎜ ⎟ = Find ( T , T , T , x ) AB BC CD B ⎜ TCD ⎟ ⎜ ⎟ ⎝ xB ⎠

0

0

⎛ TAB ⎞ ⎛ 15.49 ⎞ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 10.82 ⎟ lb ⎜ T ⎟ ⎜⎝ 4.09 ⎟⎠ ⎝ CD ⎠

xB = 5.65 ft

Problem 7-97 Determine the maximum uniform loading w, measured in lb/ft, that the cable can support if it is 737

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

g capable of sustaining a maximum tension Tmax before it will break.

pp

Given: Tmax = 3000 lb a = 50 ft b = 6 ft Solution: 1 ⌠ ⌠

2

y=

wx ⎮ ⎮ w dx dx = FH ⌡ ⌡ 2 FH

y=

⎛ w ⎞ x2 ⎜ 2F ⎟ ⎝ H⎠

x=

2

a

y=b

2

wa 8b

FH =

⎛ dy ⎞ = tan ( θ ) = w ⎛ a ⎞ = 4b ⎜ ⎟ ⎜ ⎟ max a FH ⎝ 2 ⎠ ⎝ dx ⎠ Tmax =

FH

cos ( θ max)

θ max = atan ⎛⎜

⎟ ⎝a⎠

2

=

wa

(

8 b cos θ max

)

4b ⎞

w =

Tmax8 b cos ( θ max) a

2

θ max = 25.64 deg

w = 51.93

lb ft

Problem 7-98 The cable is subjected to a uniform loading w. Determine the maximum and minimum tension in the cable. Units Used: 3

kip = 10 lb Given: w = 250

lb ft

a = 50 ft b = 6 ft

738

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Solution: 2

y=

⎛ a⎞ b= ⎜ ⎟ 2FH ⎝ 2 ⎠

wx

w

2FH

tan ( θ max) =

Tmax =

d dx

y ⎛⎜ x =

⎝

2

FH =

w ⎛ a⎞ = ⎟ ⎜ ⎟ 2⎠ FH ⎝ 2 ⎠

a⎞

FH

wa

2

F H = 13021 lb

8b

⎞ θ max = atan ⎛⎜ ⎟ 2 F ⎝ H⎠ wa

θ max = 25.64 deg

Tmax = 14.44 kip

cos ( θ max)

The minimum tension occurs at

θ = 0 deg

Tmin = FH

Tmin = 13.0 kip

Problem 7-99 The cable is subjected to the triangular loading. If the slope of the cable at A is zero, determine the equation of the curve y = f(x) which defines the cable shape AB, and the maximum tension developed in the cable. Units Used: kip = 103 lb Given: w = 250

lb ft

a = 20 ft b = 30 ft Solution: ⌠ ⌠

1 ⎮ ⎮ y= FH ⎮ ⎮

⌡ ⌡

y=

wx b

dx dx

⎞ ⎛ w x3 ⎜ + c1 x + c2⎟ FH ⎝ 6 b ⎠ 1

739

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Apply boundary conditions y = x = 0 and

d

y = 0,x = 0

3

C1 = C2 = 0

3

wx y= 6FH b

set

y=a

x=b

2

wb FH = 6a Tmax =

Thus

dx

F H = 1.875 kip FH

cos ( θ max)

θ max

wb a= 6FH b

⎛ w b2 ⎞ = atan ⎜ ⎟ ⎝ 2 FH b ⎠

θ max = 63.43 deg

Tmax = 4.19 kip

Problem 7-100 The cable supports a girder which has weight density γ. Determine the tension in the cable at points A, B, and C. Units used: 3

kip = 10 lb Given:

γ = 850

lb ft

a = 40 ft b = 100 ft c = 20 ft

Solution:

y=

1 ⌠ ⌠

⎮ ⎮ γ dx dx FH ⌡ ⌡ 2

y=

γx

2FH

γx d y = FH dx

x1 = 1 ft

Guesses

F H = 1 lb

740

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Engineering Mechanics - Statics

c=

Given

tan ( θ A) =

γ FH

tan ( θ C) =

TA =

γ x1

Chapter 7

2

2 FH

a=

γ

θ A = atan ⎢

⎛ γ x⎞ 1⎟ ⎝ FH ⎠

θ C = atan ⎜

x1

FH

cos ( θ A)

2FH

⎛ x1 ⎞ ⎜ ⎟ = Find ( x1 , FH) ⎝ FH ⎠

(b − x1)2

⎡ γ x − b⎤ ( 1 )⎥ ⎣FH ⎦

(x1 − b)

FH

γ

TB = F H

TC =

F H = 36.46 kip

θ A = −53.79 deg

θ C = 44.00 deg

FH

cos ( θ C)

⎛ TA ⎞ ⎛ 61.71 ⎞ ⎜ ⎟ ⎜ TB ⎟ = ⎜ 36.46 ⎟ kip ⎜ T ⎟ ⎜⎝ 50.68 ⎟⎠ ⎝ C⎠

Problem 7-101 The cable is subjected to the triangular loading. If the slope of the cable at point O is zero, determine the equation of the curve y = f(x) which defines the cable shape OB, and the maximum tension developed in the cable. Units used: kip = 103 lb Given: w = 500

lb ft

b = 8 ft

a = 15 ft

741

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Engineering Mechanics - Statics

Chapter 7

Solution:

⎛⌠ ⎜⎮ y= FH ⎜⎮ ⎝⌡

⎞

⌠ ⎮ ⎮ ⌡

1

wx dx dx⎟ a ⎟

⎠

⎤ ⎡w ⎛ x3 ⎞ ⎢ ⎜ ⎟ + C1 x + C2⎥ y= FH ⎣ a ⎝ 6 ⎠ ⎦ 1

2⎞

1 ⎛ wx d ⎜ y = FH ⎝ 2a dx

At x = 0,

d dx

⎛ C1 ⎞ ⎟+⎜ ⎟ ⎠ ⎝ FH ⎠

y = 0, C1 = 0

At x = 0, y = 0, C2 = 0 3

y=

2

wx d y = 2a FH dx

wx 6a FH

3

At x = a, y = b

b=

wa 6a FH

FH =

1 6

⎛ a2 ⎞ ⎟ ⎝b⎠

w⎜

F H = 2343.75 lb

742

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

2

wa d y = tan ( θ max) = 2a FH dx Tmax =

(θ max)

FH

⎛ w a2 ⎞ = atan ⎜ ⎟ ⎝ 2a FH ⎠

(θ max) = 57.99 deg

Tmax = 4.42 kip

cos ( θ max)

Problem 7-102 The cable is subjected to the parabolic loading w = w0(1− (2x/a)2). Determine the equation y = f(x) which defines the cable shape AB and the maximum tension in the cable. Units Used: 3

kip = 10 lb Given:

⎡

w = w0 ⎢1 −

⎣

a = 100 ft

2 ⎛ 2x ⎞ ⎥⎤ ⎜ ⎟ ⎝a⎠⎦

w0 = 150

lb ft

b = 20 ft Solution: y=

1 ⌠ ⌠

⎮ ⎮ w ( x) dx dx FH ⌡ ⌡ ⌠

y=

⎛⎜ 4 x ⎞⎟ + C1 dx ⎮ w0 x − 2⎟ FH ⎜ ⎮ ⎝ 3a ⎠ ⌡ 1 ⎮

3

⎞ ⎛ w0 x2 x4w0 ⎜ ⎟ y= − + C1 x + C2 ⎜ ⎟ 2 2 FH 3a ⎝ ⎠ 1

⎞ ⎛ 4 w0 x dy 1 ⎜ ⎟ = w0 x − + C1 ⎟ 2 dx FH ⎜ 3a ⎝ ⎠ 3

743

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Engineering Mechanics - Statics

dy

x=0

At

dx

At x = 0

=0

y=0

Chapter 7

C1 = 0 C2 = 0

Thus

⎛ w0 x2 x4w0 ⎞ ⎜ ⎟ y= − ⎜ ⎟ 2 2 FH 3a ⎠ ⎝

3 ⎛ 4 w0 x ⎞ ⎜ ⎟ = w0 x − ⎜ ⎟ 2 FH dx 3a ⎠ ⎝

dy

1

x=

At

a

we have

2

⎡⎢w0 ⎛ a ⎞ 2 w0 ⎛ a ⎞ 4⎥⎤ b= ⎜ ⎟ − 2 ⎜2⎟ = FH ⎢ 2 ⎝ 2 ⎠ ⎥ 3a ⎝ ⎠ ⎦ ⎣ 1

tan ( θ max)

⎛⎜ w0 a2 ⎟⎞ ⎟ 48 ⎜ ⎝ FH ⎠ 5

FH =

5 w0 a

2

F H = 7812.50 lb

48b

3 w0 a a⎞ 1 ⎡ a ⎞ 4 w0 ⎛ a ⎞ ⎥⎤ ⎛ ⎛ ⎢ = y⎜ ⎟ = w0 ⎜ ⎟ − = ⎜ ⎟ FH ⎢ ⎝ 2 ⎠ 3 a2 ⎝ 2 ⎠ ⎥ 3 FH dx ⎝ 2 ⎠ ⎣ ⎦

d

⎛ w0 a ⎞ ⎟ ⎝ 3 FH ⎠

θ max = atan ⎜

Tmax =

1

FH

cos ( θ max)

θ max = 32.62 deg

Tmax = 9.28 kip

Problem 7-103 The cable will break when the maximum tension reaches Tmax. Determine the minimum sag h if it supports the uniform distributed load w. Given:

kN = 103 N

Tmax = 10 kN w = 600

N m

a = 25 m Solution: The equation of the cable: y=

1 ⌠ ⌠

⎮ ⎮ w dx dx FH ⌡ ⌡

y=

⎛ w x2 ⎞ ⎜ + C1 x + C2⎟ FH ⎝ 2 ⎠ 1

dy dx

=

1

FH

(w x + C1)

744

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Boundary Conditions: y = 0 at x = 0, then from Eq.[1]

0 =

d

y = 0 at x = 0, then from Eq.[2]

0 =

w ⎞ 2 y = ⎛⎜ ⎟x ⎝ 2 FH ⎠

w

dx Thus,

tan ( θ max) =

Tmax =

dy dx

2FH

cos ( θ max)

Guess

h = 1m

Given

Tmax =

FH

FH 1

FH

(C2)

C2 = 0

(C1)

C1 = 0

⎛ a⎞ h= ⎜ ⎟ 2FH ⎝ 2 ⎠ w

x

2

4 F H + ( w a)

=

wa 2

2

FH +

( w a) 4

2

=

a

wa 2

2

⎛ a⎞ FH = ⎜ ⎟ 2h ⎝ 2 ⎠ w

2

2 FH

cos ( θ max) =

wa

FH

=

1

2

2 2

+1

16h

2

a

16h

2

+1

h = Find ( h)

h = 7.09 m

Problem 7-104 A fiber optic cable is suspended over the poles so that the angle at the supports is θ. Determine the minimum tension in the cable and the sag. The cable has a mass density ρ and the supports are at the same elevation. Given:

θ = 22 deg ρ = 0.9

kg m

a = 30 m g = 9.81

m 2

s Solution:

θ max = θ 745

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

w0 = ρ g a ⎞ ⎛ ⎜ 2 ⎟ d y = tan ( θ ) = sinh ⎜ w0 ⎟ dx ⎝ FH ⎠

w0 ⎛⎜ FH =

a⎞

⎟ ⎝2⎠

F H = 336 N

asinh ( tan ( θ ) )

Tmax =

FH

Tmax = 363 N

cos ( θ )

⎛ ⎛w a ⎞ ⎞ ⎜ ⎜ 02⎟ ⎟ FH h = ⎜ cosh ⎜ ⎟ − 1⎟ w0 ⎝ ⎝ FH ⎠ ⎠

h = 2.99 m

Problem 7-105 A cable has a weight density γ and is supported at points that are a distance d apart and at the same elevation. If it has a length L, determine the sag. Given:

γ = 3

lb

d = 500 ft

ft

L = 600 ft

Solution: Guess

Given

h =

F H = 100 lb L 2

−

⎡FH ⎡ γ ⎛ d ⎞⎤⎤ ⎢ sinh ⎢ ⎜ 2 ⎟⎥⎥ = ⎣γ ⎣FH ⎝ ⎠⎦⎦

F H = Find ( FH)

0

⎛1 γ ⎞ ⎞ ⎜ cosh ⎜ 2 F d⎟ − 1⎟ γ ⎝ ⎝ H ⎠ ⎠

FH ⎛

F H = 704.3 lb

h = 146 ft

Problem 7-106 Show that the deflection curve of the cable discussed in Example 7.15 reduces to Eq. (4) in Example 7.14 when the hyperbolic cosine function is expanded in terms of a series and only the 746

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

p yp f p y first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.) Solution: 2

cosh ( x) = 1 +

x

2!

+ ..

Substituting into 2 2 ⎞ w0 x2 ⎛ w0 ⎞ ⎞ FH ⎛⎜ w0 x ⎟ y= 1+ + .. − 1 = ⎜cosh ⎜ x⎟ − 1⎟ = 2 w0 ⎝ ⎟ 2 FH ⎝ FH ⎠ ⎠ w0 ⎜⎝ 2FH ⎠

FH ⎛

Using the boundary conditions y = h at

h=

⎛ L⎞ ⎜ ⎟ 2FH ⎝ 2 ⎠ w0

2

FH =

w0 L 8h

x=

L 2

2

We get

y=

4h 2

L

2

x

Problem 7-107 A uniform cord is suspended between two points having the same elevation. Determine the sag-to-span ratio so that the maximum tension in the cord equals the cord's total weight. Solution: s=

y=

FH w0

⎛ w0 ⎞ x⎟ ⎝ FH ⎠

sinh ⎜

FH ⎛

⎛ w0 ⎞ ⎞ ⎜cosh ⎜ x⎟ − 1⎟ w0 ⎝ ⎝ FH ⎠ ⎠

At x =

L 2

⎛ w0 L ⎞ d y = tan ( θ max) = sinh ⎜ ⎟ dx max ⎝ 2FH ⎠ cos ( θ max) =

1

⎛ w0 L ⎞ ⎟ ⎝ 2 FH ⎠

cosh ⎜

747

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

⎛ w0 L ⎞ ⎟ ⎝ 2 FH ⎠

FH

Tmax =

Chapter 7

w0 2 s = F H cosh ⎜

cos ( θ max)

⎛ w0 L ⎞ ⎛ w0 L ⎞ ⎟ = FH cosh ⎜ ⎟ ⎝ 2 FH ⎠ ⎝ 2FH ⎠

⎛ w0 L ⎞ ⎟= ⎝ 2 FH ⎠

2 F H sinh ⎜

k1 = atanh ( 0.5)

when x =

L

⎛ FH ⎞ ⎟ ⎝ w0 ⎠

h=

⎛ 2 Fh ⎞ ⎟ ⎝ w0 ⎠

h = k2 ⎜

w0 L

k1 = 0.55

y=h

2

tanh ⎜

L = k1 ⎜

W0

(cosh (k1) − 1)

ratio =

2

= k1

2 FH

FH

1

k2 h = 2 k1 L

k2 = cosh ( k1 ) − 1

ratio =

k2 = 0.15

k2 2 k1

ratio = 0.14

Problem 7-108 A cable has a weight denisty γ. If it can span a distance L and has a sag h determine the length of the cable. The ends of the cable are supported from the same elevation. Given:

γ = 2

lb ft

L = 100 ft

h = 12 ft

Solution: From Eq. (5) of Example 7-15 :

⎡⎢⎛ γ L ⎞ 2⎤⎥ ⎜ ⎟ FH ⎢⎝ 2 FH ⎠ ⎥ h= ⎢ 2 ⎥ γ ⎣ ⎦

FH =

1 8

⎛ L2 ⎞ ⎟ ⎝h⎠

γ⎜

F H = 208.33 lb

From Eq. (3) of Example 7-15: l 2

=

⎛ FH ⎞ ⎡ γ ⎛ L ⎞⎤ ⎜ ⎟ sinh ⎢ ⎜ ⎟⎥ ⎝ γ ⎠ ⎣FH ⎝ 2 ⎠⎦

⎛ FH ⎞ ⎛ 1 L ⎞ ⎟ sinh ⎜ γ ⎟ ⎝ γ ⎠ ⎝ 2 FH ⎠

l = 2⎜

l = 104 ft

748

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-109 The transmission cable having a weight density γ is strung across the river as shown. Determine the required force that must be applied to the cable at its points of attachment to the towers at B and C. Units Used: 3

kip = 10 lb Given:

γ = 20

lb ft

b = 75 ft

a = 50 ft

c = 10 ft

Solution: From Example 7-15,

y=

FH ⎡ ⎛ γ ⎞ ⎤ ⎢cosh⎜ F x⎟ − 1⎥ γ ⎣ ⎝ H ⎠ ⎦

Guess

Given

dy ⎛ γx ⎞ = sinh ⎜ ⎟ dx ⎝ FH ⎠

F H = 1000 lb c=

At B:

⎛ γa ⎞ ⎞ ⎜ cosh ⎜ − F ⎟ − 1⎟ γ ⎝ ⎝ H⎠ ⎠

FH ⎛

⎛ γa ⎞ ⎟ ⎝ FH ⎠

θ B = atan ⎜ sinh ⎜

⎛ γb ⎞ ⎟ ⎝ FH ⎠

θ C = atan ⎜ sinh ⎜

tan ( θ B) = sinh ⎜ −

tan ( θ C) = sinh ⎜

TB =

FH

cos ( θ B)

TC =

FH

cos ( θ C)

F H = Find ( FH)

F H = 2.53 kip

⎛ ⎝

⎛ −γ a ⎞ ⎞ ⎟⎟ ⎝ FH ⎠⎠

θ B = −22.06 deg

⎛ ⎝

⎛ γ b ⎞⎞ ⎟⎟ ⎝ FH ⎠⎠

θ C = 32.11 deg

⎛ TB ⎞ ⎛ 2.73 ⎞ ⎜ ⎟=⎜ ⎟ kip ⎝ TC ⎠ ⎝ 2.99 ⎠

Problem 7-110 Determine the maximum tension developed in the cable if it is subjected to a uniform load w.

749

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Engineering Mechanics - Statics

Chapter 7

Units Used: MN = 106 N Given: N

w = 600

m

a = 100 m b = 20 m

θ = 10 deg Solution: The Equation of the Cable:

⎛ w x2 ⎞ ⎜ y= ⎮ ⎮ w( x) dx dx = + C1 x + C2⎟ FH ⌡ ⌡ FH ⎝ 2 ⎠ 1 ⌠ ⌠

dy dx

=

1

FH

1

(w x + C1)

Initial Guesses: Given

C1 = 1 N

FH = 1 N

Boundary Conditions: 0 =

x=0

1

FH

tan ( θ ) =

C2

b=

y = b at x = a

tan ( θ max) =

1

FH FH

(w a + C1)

cos ( θ max)

1

FH

(C1)

⎛ w ⎞ a2 + ⎛ C1 ⎞ a ⎜ ⎟ ⎜ 2F ⎟ ⎝ H⎠ ⎝ FH ⎠

⎛ C1 ⎞ ⎜ ⎟ ⎜ C2 ⎟ = Find ( C1 , C2 , FH) ⎜F ⎟ ⎝ H⎠

Tmax =

C2 = 1 N⋅ m

C1 = 0.22 MN

C2 = 0.00 N⋅ m

⎛ w a + C1 ⎞ ⎟ ⎝ FH ⎠

θ max = atan ⎜

F H = 1.27 MN

θ max = 12.61 deg

Tmax = 1.30 MN

750

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-111 A chain of length L has a total mass M and is suspended between two points a distance d apart. Determine the maximum tension and the sag in the chain. Given: L = 40 m

M = 100 kg

d = 10 m

g = 9.81

m 2

s Solution: w = M

s=

g L

⎛ FH ⎞ ⎛ w ⎞ ⎜ ⎟ sinh ⎜ x⎟ ⎝ w ⎠ ⎝ FH ⎠ F H = 10 N

Guesses

Given

L

=

2

y=

θ max = atan ⎛⎜ sinh ⎛⎜

F H = 37.57 N

d dx

y = sinh ⎛⎜

w

⎝ FH

x⎞⎟

⎠

h = 10 m

⎛ FH ⎞ ⎛ w ⎜ ⎟ sinh ⎜ ⎝ w ⎠ ⎝ FH ⎝

⎛ FH ⎞ ⎛ w ⎜ ⎟ ⎜cosh ⎛⎜ x⎟⎞ − 1⎞⎟ ⎝ w ⎠⎝ ⎝ FH ⎠ ⎠

d⎞ 2⎟ ⎠

h=

FH ⎛

w cosh ⎛⎜ ⎜ w ⎝ ⎝ FH

w d ⎞⎞

Tmax =

⎟⎟ ⎝ FH 2 ⎠⎠

h = 18.53 m

d⎞

− 1⎞⎟ 2⎟ ⎠ ⎠

⎛ FH ⎞ ⎜ ⎟ = Find ( FH , h) ⎝ h ⎠

FH

cos ( θ max)

Tmax = 492 N

Problem 7-112 The cable has a mass density ρ and has length L. Determine the vertical and horizontal components of force it exerts on the top of the tower. Given:

ρ = 0.5

kg m

L = 25 m

θ = 30 deg d = 15 m g = 9.81

m 2

s

751

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Solution: ⌠ ⎮ x=⎮ ⎮ ⎮ ⎮ ⎮ ⌡

1 1+

⎞ ⎛⌠ ⎜ ⎮ ρ g ds⎟ 2⎜ ⎟ FH ⎝⌡ ⎠

ds 2

1

Performing the integration yields:

⎞ ⎛ ρ g s + C1 ⎞ ⎜ asinh ⎜ ⎟ + C2⎟ ρg ⎝ ⎝ FH ⎠ ⎠

FH ⎛

x=

dy dx

=

1 ⌠

⎮ ρ g ds = (ρ g s + C1) FH ⌡ FH dy

At s = 0; dy dx

=

1

dx

ρg s FH

= tan( θ )

C1 = F H tan ( θ )

Hence

+ tan ( θ )

Applying boundary conditions at x = 0; s = 0 to Eq.[1] and using the result C1 = F H tan ( θ ) yields C2 = −asinh ( tan ( θ ) ). Hence Guess

FH = 1 N

Given

d=

⎛ FH ⎞ ⎡ 1 ⎜ ⎟ ⎢asinh ⎡⎢⎛⎜ ⎞⎟ ( ρ g L + FH tan ( θ ) )⎥⎤ − ( asinh ( tan ( θ ) ) )⎤⎥ ⎝ ρg ⎠⎣ ⎣⎝ FH ⎠ ⎦ ⎦

F H = Find ( FH)

At A

FA =

F H = 73.94 N

tan ( θ A) =

ρg L

FH

F Ax = F A cos ( θ A)

cos ( θ A)

FH

+ tan ( θ )

⎛ ρ g L + tan ( θ )⎞ ⎟ ⎝ FH ⎠

θ A = atan ⎜

F Ay = F A sin ( θ A)

θ A = 65.90 deg

⎛ FAx ⎞ ⎛ 73.94 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FAy ⎠ ⎝ 165.31 ⎠

752 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-113 A cable of length L is suspended between two points a distance d apart and at the same elevation. If the minimum tension in the cable is Tmin, determine the total weight of the cable and the maximum tension developed in the cable. Units Used:

kip = 103 lb

Given:

L = 50 ft

d = 15 ft

Tmin = 200 lb

Solution:

Tmin = F H

F H = Tmin

F H = 200 lb

s=

From Example 7-15:

w0 = 1

Guess

L

Given

2

=

⎛ FH ⎞ ⎛ w0 x ⎞ ⎜ ⎟ sinh ⎜ ⎟ ⎝ w0 ⎠ ⎝ FH ⎠

lb ft

⎛ FH ⎞ ⎛ w0 d ⎞ ⎜ ⎟ sinh ⎜ ⎟ ⎝ w0 ⎠ ⎝ FH 2 ⎠

w0 = Find ( w0 )

Totalweight = w0 L

Totalweight = 4.00 kip

w0 L

⎡w ⎛ L ⎞⎤ ⎢ 0⎜⎝ 2 ⎟⎠⎥ = atan ⎢ ⎥ ⎣ FH ⎦

tan ( θ max) =

FH 2

θ max

w0 = 79.93

lb ft

θ max = 84.28 deg

Then, Tmax =

FH

cos ( θ max)

Tmax = 2.01 kip

753

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 7

Problem 7-114 The chain of length L is fixed at its ends and hoisted at its midpoint B using a crane. If the chain has a weight density w, determine the minimum height h of the hook in order to lift the chain completely off the ground. What is the horizontal force at pin A or C when the chain is in this position? Hint: When h is a minimum, the slope at A and C is zero.

Given: L = 80 ft d = 60 ft w = 0.5

lb ft

Solution: F H = 10 lb

Guesses

Given

h=

FH ⎛ w

h = 1 ft

⎛w ⎝ FH

⎜cosh ⎜ ⎝

⎛ h ⎞ ⎜ ⎟ = Find ( h , FH) ⎝ FH ⎠

d⎞ 2⎟ ⎠

⎞ ⎠

L

− 1⎟

FA = FH

2

FC = FH

=

⎛ FH ⎞ ⎛ w ⎜ ⎟ sinh ⎜ ⎝ w ⎠ ⎝ FH

d⎞ 2⎟ ⎠

⎛ FA ⎞ ⎛ 11.1 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FC ⎠ ⎝ 11.1 ⎠ h = 23.5 ft

Problem 7-115 A steel tape used for measurement in surveying has a length L and a total weight W. How much horizontal tension must be applied to the tape so that the distance marked on the ground is a? In practice the calculation should also include the effects of elastic stretching and temperature changes on the tape’s length. Given: L = 100 ft W = 2 lb a = 99.90 ft

754

© 2007 R. C.