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Kogan, ,6. Yu .. I rhe Appli~ation ~f Me~anics to
AUTH~R
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Mathe.atic~. Univ.~ Ill. "Dept.
Lecture$ in INSTlrurION
Chica9~ N~tiona 1
SPONS AGENCY
74
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DATE GRANr
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of Mathe.at s. Scietlce Foun,da tion, Wa,t$ington,', D. C. '
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6Sp.: for related docuaentsl see SE 030 q~'465. Not a vailab le in ha rd cop y t1 \Ie t:): cop yright ,restr lct ion s. rranslated and adapted from the Rus~ian edition. The University of Chicago Press, Chica~o, IL',60637, , (Order No. 450163: $~.50'. " i, ,. l
, '8P01 Plus postage.' peNot 'Available fios ED~S. DES:RrPrORS 4cColleqe !!athematics: Force: Geometric conc~pts: " *Geolletry: Bigher Education: Lecture '!!et:hod; "*Mat~e.aticaI Applications: *Mathema~ic~: *Mech!ni:s
BDRS PIU CE
(Physi: s)
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Pr,esented in this trl\nsHltion are three chapters. , Chapter I discusses the comp"os!~l~n of forces !ind several theorels !)f g~OIetry ~re pr!)vpd asing the'fundamental concepts and c~rtain l~ws of statics. Cha~pter II discusses the ~rpetu9.1 mot.ion postqlate; sever!l ;eometri:::. th20rems are proved, u$ing th~ postulate th~t p~rp~ual motion is impossib~e. !nCnapter IIl, the center o~ ... Grav~, Potential E!lergy. an4work. are, discussed. (HK) , . '

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... Popular Lectures in ~thematiCS' SolIoII;t~ Survey of Recent East European Literature •
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A projeCt conducted oy lzaak Wirszup, Department of Mathetltatics. the University of Chicago, under a .grant from the National Science Foundation .
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.Mechatiics" to .Geometry
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Translated and •
adapted from the Russian by
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David J. Sookne and Robert A. Hummel
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• De University of Chicago Press Chicago and London
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• The University of Chicago Press, Chicago 6(}637 Press, Ltd .• London
1lle University of Chicago
lC 1974 by The University of Chicago All rights reserved. Published 1974 Printed in the Un.it~d States of America Intornational Standard Book Number: (}"'226450!63 • Libiafy of Congress Cataloi Card Number: 7389789 .
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1.1. 1.2. 1.3. '1 A 1.5. 1.6. . 1.7. 1.8. 1.9. 1.10.
Principal ~ssumptions A Theorem On the Angle Bisectors of a TriangleAnother Theorem on the Angle Bisectors of a Triangle A Theorem on the Altitudes of a Triangle A Theorem on the Medians of a Triangle A Generalization. of the Theorem on the Bi,sectors of the I nterior Angles at" a Triangle Ceva's Theorem The Resultant ~nd Its Point of Application A Third Theorem on the Angle Bisectors of a Triangle A Fourth Theorem on .the Angle Bisectors of a Triangle
2. The Perpetual Motion Postulate 2.1. The Moment of Force . 2.2. A Theorem on the Perpendicular Bisectors of t~ Sides of a Triangle 2.3: The PythagQrean Theorem 2.4. A Theorem on Tangents and Secants 2.5. A Theorem on Two Interse<;ting Chords of a Circle
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'1
1. The Composition of Forces
1 4· 4 S ·6 r
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9
11 12 13
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14 14 16
16 18 19
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3. The ~_~ter of Gravity, Potential Energy, and Work 3.1. The Center of Gravity 3.2. Potential Energy 3.3. The Centers of Gravity of Certain Figures and Curves 3.4. The Volume of a Cylindrict:}.l Region 3.5. The Volume of a Pyramid 3.6. The Volume of a Body of Revolution · (Guldin's First Theorem)
21 21 24 24 31
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'37
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3,7. The Volume of a Sphere 3.8. The Volumes of Certain Other Bodies of Rotation 3.9. The Surface a Body of Rotation .
of
49
{Guldin's Second Theorem) 3.10. The Surface of a Sphere . 3.11. The Surface of Certain Other Bodies of Rotation 3.12. Conclusion ....
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54 54
56
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°t..e Composition' df Forces
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.. 1.1. PriDdpal Asswn~ODS
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" In this chapter w..e shall prove several theorems of geometry using the I fundamental concepts and ~in laws of statics. We will defiQe the terms immediately. 1. Force is a vector and is characterized by magnitude, a direction, and a point of application. The line along which a force actS is called its T
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'tne of action.
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2. A bOdy which cannot be deformedthat is, which always keeps its ~ and shapeis said to be absolutely rigid.' <;P Ac!'uaUy, every body is capable of being deformed to some extent, but, these deformations are frequently so small that they can be neglected. The concept of an ~olutelY rigid body is an idealization. One fre'quently omits the wor absolutely and speaks Simply 'of a rigid body. 3. A collection of acting on a body is called a system offorces. A system of .forces is said to be in equilibrium or an equilibrium' system if no motion is 'caused when the system is applied tOo an ,absolutely ri8id body at r e s t . ' •. . 4. Two systems of forces are'saip to be equii'alent if they cause the same motion when applied to aljl absolutely rigid body, . From this definition it follows that, for all practical purposes, a system of forces acting on a rigid body can be replaced by anyequivalent syste'm without altering the discussion. 5. If a system of forces is equivalent to a single force R, we say that the force R is the resultant of this syste'm. 'Note that not eve~ system of forces has a resultant. The simplest example of such a system of forces is called a coupl"e of forces, as illustrated in figure'l.1. "In addition to the above concepts, we use the following rules (axioms) , of statics: "' ,
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" The €om~on of.Forces
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• R~ 1.1. Two forces Fl ~F'J acting at the same point have a re.sultant R which acts at the sante and is represerued by the Diagonal of the parallelogram lummg the jorces Fl and F'J as k1jacent sides {fig. 1.2}. . r ' •
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1.1
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This construction is often called the parallelogram law jar forces. The rule allows one to exchange the forces Fl and Fli for the force R .' and, conversely, to e~chan~e a.given force R for forces Fl and Fli • In the first case one speaks of the composition'offorces, and if! the second, of the resolution of the force R into the components F l and F 'J. (This resolution can be carried out in an infinite number of wa~; since it is\ possible to construct infinitely many parallelograms with a given diagonal R.)
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RULE 1.2. /fwe add any equilibrium system to a system offorces,'or if we remove all equilibrium sysum from a system of jQrces, lhe reSulling system will be equivalenlto the original one. .
In particular, 'this implies that a collection of equilibrium systems is an equilibrium system.
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1.3. Two forces are in ffjuilibrillm if and only if they have the ,same. magnitude, opposite directions, anJ a common line 0/ action (figs . 1.3 and 1.4). RULE
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104. A force at:lihg along its line of action. RULE
on a rigid body can be arbitrarily shifted
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The ;;mposilion
fo~
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In other w<>rds, if F and F' have the magnitude and direction and a common line of action. they are equivalent (fig. '1.5). The converse is also lru.: If the forces F and F' are equivalent, they have the same magnitude and direction and 11 common nne of action. 1 ' Rule 1.4 implies that for forces acting on a .. rigid.£. b6d'Y.'",the point of application is unimportant; rather, the line of action determ'ines ' • the res,ultant force .. The Vector of.a force acting on a rigid body is therefore called' a
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providing their lines of action intersect. Su~ pose that we have to aad forces Fl and F~ (fig. 1.6). Since the vectors of these· fo~ are sliding, we can translate them to the point 0 '. and then' use rule 1.1 to obtain the resultant R of the forces Fl and F~, by completing the panlllelogram. . . From rules 1.3 and 1.4 we deduce the following important proposition:
•
Fig. 1.5
PROPOSITION 1.1. If three non,arallel and coplanar forces acting on a rigid body are in equilibrium, then their lines of action intersect at a single point. . " . For suppose that the forces PI' p~, and Pa are in equilibrium with one another (fig. 1.7). Translating the forces PI and p~ to the point 0, we
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Fig. 1.7
Fig. 1.6
1. Rule 1.4 can be deduced from rule 1.3. We have not done this; however since both rules are equally intuitive.
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sliding vector. • Rule 1~4 thus enables one. to add forces ._ whose points 'of appiication 'are diiferent,
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.' obtain their rcsultant.Ru.' The fo~ P s and Rill. are now in equilibrium.  But this is possible only if they have a common line of action. Thus, the line of action of the force P s passes through the point Othat is, the' lin~ of action of all three forces ~t one another,at this point. " \ USing this propOsition;'we shall' now prove some theorems of geometrj. ~ ,

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'1.2. A,1bi:orem GO the ADale Bisectors of • ~
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, Let'us consider six equal forces Flo F~ •.. '~'FG acting along the sides er a triangle, as shown iII figure 1.8. Since these forces cancel onc another in pairs, they are clearly in ~uilibrium, an~ therefore, the resultants ltl~' i.23 • and ~5 are 'also'1n equilibrium.'Bu( the forces RUh R~s! and ~~ arc, directed ~ong thebisect.ors of the interior angles 'T • A.,' B, and C. (The parallelograms are rhqm~i, and the' ~iagona1 is an , angle bisector.) This leads, consequently. to the following theorem: . , \ THEoREllt 1.1. The bisectors of the interior. angles of a triangle inter
sect at a point. 8
B
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Fig, 1.8
•
1.9
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1.3. ADother Theorem on the Angle B'isectors of a Triangle
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Let us consider the six·e'qulli forces Flo Fill' , ., Fs shown in figure 1.9. These forces are in equilibriuIl! since each of the three' l!airs of forces, taken conse
11
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'~ The ~ompositi01t of ForC/!s
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bisector of interior angle C. The resultant ofF2 and Fa is directed along the bisector of int~tior angle B.1nerefore, the following theorem.holc1s: , THEOREM 1.2. Tlie bisectors of two exttrwr angles 'and an interior . " angle of a triangle intersect at a point.'
, 1.4. A: Theorem
OD
the Altitudes of a 1'riaD&Ie
In figure 1.10 we have drawn a triangle ABC, with the forces Flt,F." ... , Fe acting along the sides, We have chosen these forces so that the hold: following . eqUalities , \
. wnere
F1 Fa F5
= F':J = F cOs A , = F~.:::. F cos B.
= Fe =
F cos C t.
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(1.1) 
Fi~
sOQ'le convenient unit 4imension for force. (Note'that we arc using the convention where an F denotes a vector, and Fits corre: sponding magnitude.) Since the forces Fl. F~ •... , Fe are in eqUilibrium, ,the lines of action of the resultants R A • Ra. and Rc shown in ,the figure must intersect. We shall find the directions of these resultants. , 8
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B
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Fig. 1.11
Fig. 1.10 (
For example,. let us, add the forces ~l and Fe, which act at.,the vertex B (fig.' 1.11). To do this, we resolve each of these forces into two com• ponents, one paralle.l to the side' A C, and {he other perpendicular to it. ~ The first of these components we shall can'lYorizpntal, and the second, vertical. From' figure.. i.l1 it is clear that the htlizontal components of the forces Fl and Fa are equal to Fl cos C and Fe cos A. But from (1.1) it follows that ' " Fl cos A F8 ~ cos C·
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The Com~sition
6
or Forces .
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• Thus, ~e hprizontal components of the forces Fi and Fe are the same. From this faLt we conclude thtiUhey cancel one aflotherand, therefore, the resultan.t~fthe forces Fl aqd Fe is ~rpendicular to side AC. Therefore, the force Rs is directed along the altitude perpendicular to AC.· Analogously, we may deduce that forces R ... and Rc lie wong the two· other altitudes of the triangle ABC. thus arrive at the followiBg;
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THEOREM
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1.3.
T~ altitudes qf ~ triangle .miersect fit Q singl~ point.·
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1.5. A Theorem 011 the Medians of a Triangle .
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Let us consider forces Fh Fli •.••• F s , a¢ng as shown in figureJ.l2. Suppose that each of these fo~ has a magnitude, equal to onehalf the " length o~ the correspon.ding sid,e of the triangle. Then the resultant of the " . forceS Fl and Fa will be represented by the median dra~n to side BC; the resultant .of fo~ F ~ an~ F~ will be represented by the meqian drawn to side IfC; and tJii resultant of forces F 4 ~nd F 5 will be represented by • • the median drawn to side AB, since similar triangles are fonned by the parallelograms of forces shown in' fi,$ure 1..12. The forces Fl' F lI , ••. , Fe are in equilibrium, and this leads to the foilQwing theorem:
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"THEOREM
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1.4. The medians of tr.iangleintersec! at a single point. 8
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Fig. 1.12 .j
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1.6. A Generalization of the Theorem on the Bisectors' of Angl~ of a Triangle ,
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Interior
Suppose that we are'given the triangle ~ Be. Let us draw a straight line a dividi,ng angle A into parts (Xl and CX2, a straight line b dividing angle B into parts fil and fi2' and a straight line c dividing angle C into parts Yi
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and')'11 (fig. 1.13) We apply'to point Aan arbitrary force Rl ~ ~ng the line a and resolve t4is force into components P 1 and Ql dircc~ ted along the.sides A.C and A.B. Similarly; we..apply forces Kg and 1ls directed along the lines band c to the points B and C~ and resolve these forces into components P g, Qg and P a, Qa. 'We require, howeVer, that the component P g cancel the component Q1 and that the component 1»8 cancel the component Ql1' In this way we obtain a system 9f forces , (Rt, R~, Rs) equivalent to the 'system (P1, Qa). • ~ons1der n~w~ tl)~wing ratios:' ' sin 111 . sin
sin al
pi}.
sin as '
sin )'1 sin )':.1
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From the parallelograms at the vertices A, B, and C, we deduce..that . sin a1 Ql = 'Sin ~ PI'
B
sin /31 Q2 = &in#.C,P';1, ,
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sin"l Q9 sin = Pa '
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and, therefore, tJtat
sin a{ sin 111 sin )'1 sm ~ sin /3l1. sin ')'2
\~ Ql Q2 Qa
A
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P,
~=
PI P;. P a •
Fig. 1.13
But since P'l. = Ql and P a = Q'l.'
(1.2)
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Two cases are possible. Case I.
sin (Xl sin PI sin ')'1 = 1 sin (X2 sin f32 sin ')'11 •
(1.3)
Thoo PI = Qs. That is: the forces P l and Qs are in equilibrium; consequently the forces 'R l , R~'h Ra which are equivalent tQ ~ and Qs. are in , equilibrium. Fro.m this"ract we then ~nclude that the lines a, b, and c intersect at a single point.
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The Composition of Forces
Case 2.
~n "1 ~n}l ~n Yl ~ 1 :
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Then, according to equation (1.2), Pl =F ~. We shall prove that in this .~ the li.a, b, and c cannot intersect at a single point. Suppose that they intersect at a single point 0 (fig~ 1.14). Then, translating the forces R 1 , R~, and Rs to the point Q, we find their resultant R: which will also . act at (). FurthermOI:e, since the system (Rl t R:z. Rs) is equivalent to the system (Pl. Qs), the ~ultant It must be equivalent to the nonzero resultant of the fortes'~ and Qa. This is Impossible since th~ resultant of the forces P l , Qa lies on the line AC, and the line of action of 'the force R cannot coincide with the line AC (since the point does not lie on this .line). From this contradiction, we conc1udQ that .the lines a, h,.and c do not intersect at a single point. Thus, the lines shown in figure 1.13 intersect at a single point only when equality (1.3) is valid. In other words, the Jines a, b, and c intersect at a single point if and only if condition. (\.3) is satisfied. . ' The theorem just .proved may be regarded as a generalization of the theorem on the bisectors of the interior .angles of a triangle: [In that theorem, not only does condition (1.3) hold, but alsoeaeh of .ilie individual fact{,)rs (sin ~/sin ~, sin Pl/sin fJ~, sin "Iljsin yg) is equal to one.] . . ,
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This theorem also implies the theorem on the altitudes of a triangle (fig. 1.15). If lines Q, b, an6 c are drawn as altitudes, then sin fJl sin fJ~
= cOs A c<)s C '
and
sin Yl = cos B . sin Y:z cos A
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Fig. 1.14
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The CDmposition of Forces . '/:aking the~r~uct,. we ~d that , s m .CXl sm PI sm • sin ~ sin fJ~ sin
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COS C cos 4 cos B_1 cos B eos C cos A •
Consequently, the altitu~ of a triang1einte~t at a single point. .
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1.7. Ceva'& Theorem , , Consider t¥ triangle ABC (fig. 1.16). Suppose that forces Fl and F2 act wong the sides AC and AB, and that theiM"esultant acts along line AA. l • We draw· the line DE parallel to side BC and resolve Fl into. compQnents F/ and Fl ", and F2 into components F:/and'g". From the; . similar , triangles fanned, it is evident that ...
Therefore, d F' F.lil· BA.l F 1' = F1· Ale CA an :1 = AB . But since the resultant of the forces Fl and Flil is directed alOIlE AA 1 • F/ = Fg'; consequently,
or F1 Flil
'=
CA BAl AB AlC·
(1.4) .
This relation will be needed later. (It is ~asy to remember because the right side of this equation can be obtained by circling the triangle CAB clockwise.) Let us now determine At. 81> and elan triangle ABC'(fig.J.17). At the .points A, B. and C we apply forces R1 •.Rlil • and Rs directed a~ong lines AAlo BBh and Cel' and we resolve these forces into components directed along the sides of the triangle. The force Rl is chosen arbitrarily, but the forces Rlil and Ra are chosen so that the equalities ...
P'J = Ql'
P a = Q'J
are satisfied. Applying relation 0.4) to each vertex. we have that
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(1.5)
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Fig. 1.16
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We next multiply these equalities ."
to obtain
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Case 1.
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Then PI = Qa, that is, thes6 rces arc in equilibrium. Consequently, the forces Rit R~h and Rs are n equilibrium; and, thus, the lines AAlt BBh and eCl intersect at a , ngle point. , Case 2. . ,
Then according to (1.6) the forces PI and Q3 are different. Repeating the argument of the preceding theorem, we deduce that the lines A A 1, BBl ; and eCl.do not intersect at a poin~. ) Thus the following ttft!orem: ~ THEoREM
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1.5. For
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lines AA t , BEt. and ce l to intersect at a single
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point, it is neceSSQl'Y and sujJiclenJ tluzJ equation (1.7) be valid. This result is' known as Ceva's theorem. ~ . , .
The theorem on the medians pfta triangle is a specIU case of ~va's
~' since, 'ur the case for " "
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AC1 C1B
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tc;B
= AC BC'
BAl = AB' AlC AC'
CB1 BIA
and, consequently,
.=
BC AB'
,
AC1 BA'1 CB; _ AC AB BC _ 1 I CI'B AlC BIA.  BC AC A.B •
.
• Equality (1. 7) is satisfied, and we may apply Ceva's theorem. 1.8. The ResultaDt aDd Its PoiDt , of Application
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,
We should mak~ one more rerI}ark concerning the concept of the resultant. Suppose that the force R is the resultant of forces applied at different points of a rigid body. Since the vector R is a sliding vector, we can change its point of application by tra~slating it along its line of action. But since the force R haMlo actual point of application (as it is not directly apf}lied), any point on its line .. of action may be. taken as i~ point of application. Thus, the' resultant of forces applied at various points
Fig. 1.18
2. By extending the lines forming the sides of the: triangle, Cevats tneorem may
be generalized to the case where the lines AAt. RBI> and eel intersccC outside the
.
triangle ABC. The same is true, by the way, of the thct>rem of section 1.6.
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The Composition of Forces
de~nite
of a riiid body has a definite of action, but not a point of application. To illustrate this statement, let us consider forces F l , F~, and Fa as shown in figure'l.I8. To fi~d their resultant, we first add the forces Fl' and FlI~nd then add their resultant RIll to the force Fs. In this way we , finally end up wJth a resultapt R applied at the point'C. We now proceed in another waf: First we add the forces Fl and Fa. and then we add their resultant R,ls to the force F!:I' We t~en.obtain a ~s'.lltant R' acting at the point D. Thus, different methods of adding the forces Fl , F:;z,' ~nd 'Fa ~ give re~~tafts \\lith different. points of ~pptication. (One can.,....assert, ,.. . however, that the forces and IJ.' have the same lin~ of actiQn and that' R = R'.) , . /" . , ~rom this arguqlent we deduce the following'rule: '
a
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"t,5. Suppose that by adding f<JJces 'in various. order'S we obtain different points of applicatianf~ their resultant. Then tJrese po~s will be coflinear, and the line formed will coincide, with the line of action of the resultant. RULE
We shall
QOW
use this rule to prove two theorems .
.'
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1.9. A Third Theorem on the Angle Bisectors of a Triangle
Suppose that the forces Flo J;'2. and Fs have equal magnitudes and act along the side~ of triangle ABC (fig. 1.19). We shalf find their resultant. "
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....
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F2 A
K
F, Fig.' 1.19
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v i ...
Composing the forces F1 and F:.a, we obtain the resultant R 12 • which is directed .along the bisector )f D. Composing the force Rl~ with the force F 3; we then find the resultant of the forces J;\, F 2, and Fa, and this, resultant will act at the point D. , , Ifwe add the forces F1 and F3 first, we get tire resultant R13 • which will lie on the extension of the bisector CEo Next we add the forces R ls and F:., and again obtain the resultant of the forces Flj 1'2, and 1'3' This time, however. the resultant acts at the point E.
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.~ Composition 0/ FOlces.
. Suppose we combine the forces F:3 .and Fa ,firSt.· We thcn have ~c resultant R2t}, which will lie along thc bisector 'BK of the exterior angle B. Composing the forta R~ andFi, we obtain a resultant which acts at thc point K• . ' '.' . Th~. adding the forces Fl' F'h and Rs in. three different ways., we ,ob~ reSultants acting..at thc points D, E, and K.·Consequently, the 'pO~ p, E, and K are collinear. By dcpning the ,base of an aqg1e bi ~ sector to ~ its point of intersection with' the opposlte side, we have the . theorem:
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. 'IHEoRBM 1.6:' T¥ bases oJ the ;,;sectors oj .two interior angles and,# one exterior angle of a. triangle70mJ a straight line. a 'it • < •
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1.10. A Fourth Theorem
OD
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, aTriangle
the Angle Bisectol'!iof .
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By carrying out a similar argumeI}tfor three forces of equal magni ". tude, F l , F 2 • and Fa, acting as shown in figure 1.20, we may then prove the theorem: 1.7. The bases of the bisectors oj the three exterior angles o/.,a triangle are collinear (fig. 1.2~). ,,' THEOREM
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Fig. 1.20
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Fig. 1.21
3. We assume that the bi!;eCtor of the cxterior angle intersects the oppDsite side that is, is not parallel to this side. This remark is also applicable to the next ' theorem.
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geo~etric
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It is to certain theorems " that perpetual motion is impossihle. This chapter will demo~te sev~ra1 theo~ o~ tllis sort:: \ .. ' ' .'. .
, l.L TIae Momeat of FaRe
,
In 'addition
to the postulate on the impossibility of perpetual motion,
s~ first $lie this law. Suppose tJlat a body is under the: inz ftueilce of a force F and can revolve 'about the zaxiS (fig. 2.1) .. The rotational motion caused by, a force F is determined by its moment with reSpect to the zaxis. To compute this moment, we resolve the force F into components y' and F', with the first component lying in a plane perpendicular to the zaxis, and the second parallel to the axis. The rotational motion caused by the component F"' is clearly equal to zero, ami the rotational action of the component F' is measured by the product of vector F' by Fig. 2.1 'the scalar d, where d is the distancetbetween the zaxis and the' line of action of the force F'. This product, deqoted by F'd, is sometimes referred to as the: torque, and this text (for clarity), will be called the moment of the force F With respect to the ZoXif . . Since the force F' is the projection· of the force F onto the plane P, we can give the following definition of moment:
we will need ,the law of moments. We .
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The PerpetUQ/ Motion Post..u}QJe
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'. DEfINITION 2.1. The moment of the force F with'respect' to tIu! z~ . is the product F~ where'F' is the projection o/thejorce F onto the p/tJile perpendicular to the zDXis, and d is the distance between the za:ds.and the line 0/ action, of.. the prOje~tion F.·
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Thus,
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M.em the
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where is mom~nt of the force F ~pect tt) the. zaxis. ., .It folloWs from the defipition that the moment of force is equaJ to , zero in only two caSeso: w'henthe line of action of th6lforce F intersects 'the zaxis, or whei\ it is parallel to the axis. . ~.) .' If....as k.mY,ently Occurs, the force 'Fhas' a line of action that lies in a ... plane 'perpendicular 1'0 the taxis, then F~ = F and. therefore,
with
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M.(F). =Fd. ..
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In this case the distance, d is called the arm of the for~) "
.,.1 We assign a definite sign to the moment of force.',WO'r this purpose ~  we designate one of the directions of rotation positi~, 'and the other as negative. Then if the force tends to rota the body in the positive direction, we consider its momentum positiv and in the opposite case, negative. Therefore. we can write
•
Mi/l(F) = ±F'd, where the sign is determined by the directi The following two rules will be needed:
of rotation,
••
~ULE 2. ~. If R is the resultant of the system (F 1. F~. " .• F,,). the moment offorce R is equal to the vector sum of the individual moments oJ., forces Fl. F li •••• , F".l This rule may be written in the fonn
(2.1)
.,
.......
where MJR) denotes the moment of the force R with respect to the • zaxis. ....
I. This proposition is known as Vur;gnon's theorem. Varignon's name is also given to the theorem about seaments joining midpoints of the sides of a quadrilateral.
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16
til
P~l J"tion PostuJate . (The law ofmom~ SlgJpose that a rigid body can ~otate The
RULE
2.2.
abot}t a fixed axis. In order for folce~acting on it not,to cause rolation, it is necessary and sufficient that ("I.e vector sum of their moments equal zero ...
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(In other words. the moment of the forces tending to rotate the body in the poSitive direction must have the same magnitude as the moment of the forces tending 'to rotate it in the negative direction.) .'~.
.
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2~ A Theorem on the Perpendicular Bisedors of the Sides of •
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Consider a container having the form of a right triangular prism AlB'l CIA~B~C~ . . (ijg. 2.2): Imagine that \ it is filled with gas and. that no external foI'Cef', noleven gravity, act on It. (We cali assume, for example, that it is located far from the earth and the hC1lvenly bo4tes.)· If we consider the forces ~xerted on the container, we conclude that the net resultanr must be constant,. since no .external force is applied. Since we have postulated that perpetual motion is impossible, this constant resultant must be zero. Thus, the container will #main in' its initial state at rest. Consequently, the fqrces that the gas~xerts on the walls must be inthe equilibrium. But ~nce the pressures on the two paranel faces clearly baiance one .another, the forces exerted by 'the gas against the side walls of the container must be in equilibrium. We may represent these forces by FAB, F Be, and F ..(c, which lie in the plane defined by triangle ABC, and have points of application at the midpointsof~eir respective sides. Since these forces are in equilibrium, by our firsJ;proposition, their lines of action . must intersect at a single point. Noting that the vectors F AB. F Be, and FAc are perpendicular to the sides of the triangle ABC. we have the following:
alI
2.1. The perpendicular bisect(JTs oj the sides of a triangle intersect at a single PQill~ . THEOREM
...
2.3. The
Pythagor~
I l'beorem
Consider now a right triangular prism whose base is the right triangle A Be (fig. 2.3). We f1I1the container with gas and allow it to rotate about the vertical axis' AD' (the A Be plane is considered horizontal). Since perpetual motion is impossible. the container will remain in its initial state at rest, and the forces caused by the gas on the side walls of the container must be in equilibrium. Each of these forces tends to rotate the container about the.OO' axis: the forces Fl and F~ counterclockwise, I
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Fig. 2.3 
Fa
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and the force clockwise. Therefore, the sum of the rotational moments of the forces W 1 and F~ must equal the rotational mon;:tent of the force .F3 • Since the arms of these forces are equal to ABj2, BC/2, and AC/2, respectively, we may use our formulas for rotational momcn~, . and equate magnitudes, to obtain
(2.2) But
Fl = p(AB·h) ,
.
F2 = p(BC·h), Fa = p(AC·h) •
where p is the pressure of the gas and h is the height of the container. Substituting, we find that equation (2.2; now takes the form. AB p{AB·h) T
BC
+ p~BC.h) T =
AC p(AC.h)"2'
I
Multiplying by the constant 2/ph, we have . AB2 ,'"~
+ BC2
= AC2.
TIius, we have proved the following theorem: THEQREM
2.2. The sum of the squares of the legs of a right triangle is
., ~qual to the square of the hypotenuse . .
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.~
"The Perpetual Motion Postulate .
18
It is possible to 8lneralize to the law of cosines by substituting an arbitrary triangle for the right trianpe used in this proof. Ho",~ver,'WC shall now turn to a couple of geometric theorems dealing with circles. ..
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Suppose that a gasfilled vessel has a base whose shape is thefigwy ABC (figure 2.4 ,.shows the view from abQve; the plane ABC is horizontal). The vessel ,has height h as pltaSured along any of the vertical ~
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sides joined to the lines A B, A C, 0;'; the arc;. Be. Thus the. vessel is a boxlike container. wi~h ~osssectional sha.BCWuppose, fu~her more, that the vesselts"'tlghtly fas~encd to the rod ofT, and that thiS rod ,is fastened ,to the vertical axis O. In this way, we allow the vessel to rotate about this axis. As in the preceding section, the vessel will remain at rest, and, therefore, the sum of the moments of all of the forces acting upon the container must be equal.to zero. But only two of these forces create rotational moments; the forces Fl and F2 of the pressure of the gas on the walls ABan.d AC. (The foif ofthc'gas on the curv,.ri wall Be do not contribute to the rof'at~~oment, since each force has a line of action that passe...... through the axis 0,) Using the fact that the moments of forces 1<\ and I<~:l have opposite signs and that the arms of these forces are equal to BK and LM, we know that .
F1
·BK=4 .. r
But~BK = A E/2 and
LM
=
LCVL1.. = ~D
2
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+ 2
LA = AD
2 .
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'The Perpetual Motion PQstuJate
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19
Consequently, .,
_AB
P1 · 
2
=
AD
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that is, (2.3) .•
• Furthermore, we h?ve that '.~
..
. Fl = p(AB·h) and F..: == p(AC·h) ,
Y.
I
where p urthe pressure of the gas and h the height of the vessel. ~ubsii . tbting these expressions in (2.3), we ,obtain' • . ". .
\
p<:AB.h)AB,='p(AC.h)A'D. ~
hence, .
(2.4) Returning to the circle,shown in figure 2.4, we see that the segment AB is tangent to the circle from point A; segment A D is a secant; and the segment AC is the external part of secant AD. Thus, equation (2.4) ex. wellkno~
Pfesses the
geometric theorem:
THEOREM 2.3. The square of the tangent to Q circle from Q point is equal 16 the product of a secant from that point times the external part of the secant. 
~.
A Theorem 011 Two Intersecting Chords of a Circle
Let. us replace the vessel just considered by the vessel ABD shown . in figure ~y~ng the argu":nts used in the beginning of the
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preceding theorem, we obtain the relation
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The Perpetual MQ.tion ,Postulate
20
Furthcnnorc, since
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== AL _ A.K _ AC _ AB """ AC  AB _ Be ,
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2
MN == DM ' DN ...
2
2
. ,,2 . .:::,
D8 "~"]fE 2  2 '
EE. _ DB :.. DE 2
2
.,
that is, I
As before, however, Fl  p(AB·h) , F~;'" p(DB·h) ,. .
and, consequently, . p(AB·h)·1C == p(DB·h)·BE,'
• hence,
AB·BC"", DB· BE .•
,(2.5)
Equation (2.5) expresses a wellknown geometric theorem on two intersecting chords of a circle. o
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27
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'The Center ,. of Gravity, Potential Energy, and Work
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1n this chapter we shall compute the volumes and surfaces of certain bodies. OUf discussion will make use of ccncepts involving potential energy, work, and properties of the CciPter of gravity. J
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3.1. The Center of Gral'ity
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The resultant of two parallel forces Fl and F3 (fig. 3.1) is represented by a force R' acting in the same direction, with a line of action passing through point C. This point, which may be considered as the point of application of the resultant R, is defined by the relation (3.1) I"
Now let u~ consider a system of sevoral parallelfofces. For example, we shall finl,the resultant ofJ"our forces (fig. 3.2) by adding them one by one. Upon combining the forces .Fl and Fill we obtain a force R'
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Fa
R'
Fl.
'W,' R
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Fi i.3.1
R
Fig. 3.2 ')
21
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22
The Center of Gravity. Potential Energy, and Work. ,
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acting at some point C'. which may be determined by relation (3.1); adding the forces R' and Fa, we obtain a resultant R· acting ~t a new point C"; and, finally. addingR" and F., we have the resultant R. ~cting at some point C whose position may be determined by repeatoo usc of equation (3..1). The torce R will have magnitude equal to F1 +: F~ + Fa + Fl,. In this manner, it is possible to add together any Dumber of parallel ·forces to find their .resu~tant an4 a point C at which the resultant acts. H can be proved that the position of C is independent of the order in which one adds the forces. Tpis point is called the center of the given system of parallel forces. Let us consider a rigid boqy located in the vicinity of the earth (fig. 3.3). If the dimensions of this body are sman in comparison with the radius of the earth • the forces of gravity acting upon its particles are essentially parallel. We may therefore tlnd a limit point C of a sequence of centen; of arbitrarily large selections of parallel forces. This point is· Fig. 3.3 called the c,enter of gravity of th~ given body. We may consider it to be the point of application of the force P, which represents the) weight of the . body. . We should make a couple of observatio s that follow directly from this definition of the center of gravity. Fi t, if we translate· or rotate a body, the ceoter point C is similarly tran ated; hence, relative to the body. the center of gravity remains fixed or'" firmly attached ,; to the body under rigid transformations. Second, it follows from equation .1) that a proportional increase or decrease of all the forces in a stem of parallel forces results in no change in the position of thc center C. I n othe~ words, the center of gravity of a homogeneous body depends only on its size and shape. For this rcason, the center of gravity is somctimes called the center of mass. I., The center (?l gravity of a line and a square. In mechanics, one speaks ot'a matcrial point wruch is analogous to the geometric concept of a point. A material poillf is a point possessing a definite mass. The theoretical idea ofa material point is a body having some measurable mass, but no dimensions. (Sometimes these concepts. actually occur. For example, in studying the motion of'thc earth about the sun, we can treat the earth as a material point.) j
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Sinillarlyt we can introduce the concept of a nuJJeriDlline. By this we shall mean a curve of finite Imgth .possessing a certain mass.·1 We shall represent this mass by its distribution along the length of the curve. 'In the same manner, we can speak of a moter(aIfigure. By this we shall
mean aplane figure possessing a definitemass distributed throughout ilaarea.
•
A material1ine may be visuaJiud 'as a thin wire and a material figure as a thin plate. The thinner 'the wire or the plate, the closer it .approximates the material line or material fi~. : '..: .~ , , '. operties to our discussion of , We may extend. certain physical material lines and figures., By dete .nins the weight of' a .material ,.object per unit I~ngth, or per unit ~" we may speak of the specific weight· of the body. If the mass of a material CUl'VC is distributed , uniformly along its length, the specific weight of this curve will be the same at allpaints. We shall call such a material curve homogeneous. In tho same sense we can speak of a homogeneous material figure. A thin 'homogeneous wire of constant diameter serves as a prototype of a material curve. Analogously, a thin homogene&s plate of constant Ihickness is a protot~ of a material figure. Since a material ~urve and a material figure possess mass and con~ quently weight. one can speak of the center of gravity" of a material line or a material figure. If the material curve or the material figure is hom'ogeneous, the position of its center of gravity depends solely on its shape, and is independent of its specific weight. Therefore, the center of gravity of a homogeneous material curve can be called the center of gravity of the curve, and the Center of gravity of a homogeneous material figure can, be called the center gravity of the area (or the center of gravity of lh~ figure). ' • 2. The center of pressure. It is not necessary' to limit ourselves to forces of gravity when discussing the center of gravity of a thin plate. Suppose that the pressure p acts on one side of a figure of area S (fig. 3.4). Since the forces associated with this pressure are parallel, their resultant is equal to
tr
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F
Fia.3.4
= pS,
and acts at some point C, which is the center of these forces. This point is, therefore, called the center of Pfessure. To determine its position, 'we,:.n;ust know the magnitude of the
30 rt
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\ The Center of Gravity, Potential Energy, and Work
'24
force on any arbitrarily small area. But this force is equal to p~S, where ~S is the area of the cell on whlCh the pressure acts (sometimes called ~ gaussian area). This is numerically equal to the weight of the cell, given that the material figure has a specific weight equal to p. Consequently, any partition. of the figure yields forces of pressure having the ~me magnitudes as the forces of weight for a homogeneous material figure S. From this we may conclude that the 'point C coincides with the center of gravity of the figure S. , Thus, the center of a uniform pressure on a material figure is the same point as the center of gravity of\ the figure. This result will be needed in . the proof of Guldin's first theorem.
'~EDergy
~
We shall assume the following propositions about tktential' energy in a gravitational force field:
•
PROPOSITION 3.1. The potential energy of a material point is equal to Ph, where P is its weight and h its height.
3.2. The potential energy of a material system is equal to the sum of the potential energies of its points. PROPOSITION
3.3. The potential energy of a rigid bod~quallo Ph e • where P is the weight of the body and he is the height of its center of PROPOSITION
g~avity.
The first two of these propositions are the definitions of the potential :·energy of a material point and a material system. It is important to note that the third proposition may be applied to material curves or materiat figures since these bodies have weight.
,
3,3. The Centers of Gravity of Certain Figures and Curves
To find the center of gravity using the method suggested by the definition, one must carry out the addition of a large number of parallel forces. In certain cascs, however, the center of gravity can be found by an indirect method. We shall do this for several simple ligures and curves. 1. A rectangle. It is known that jf a homogeneous body has a plane of symmetry, its center of gravity lies on this plane. Similarly. if a figure or a curve has an axis of symmetry, its center of gravity lies on this axis. Consequently, the center of gravity of a rectangle is located at its g£'o
ml.'tric cenLer. 1 l. Consequently. the ('ellter of pressure of a rectangle is its geometric center. This fact was used several times in chapter 2 (for example, in the proof of Pythagoras' theorem). .
31
.,J, .. II
.
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'I'M CMi~r 0/ GraVity, Poten#al EMr6Y, and Work
.2S
2. A circle. By a similar argument, the center of gravity of a circl( lies at its center. 3. A triangulm region. Before generalizing to·· the case' of a triQ.Ilg1e, we will 'decompose a ..... • trapezoid into a large number of I I I , narrow strips of uniform width I I (fig. 3.5). Each strip has a center of I D o A gravity lying on the segment PQ . , which jpin$ .the mi4Points of lite fii.3.S bases AD "and Be: Making' the width of each strip infinitesimally small, we deduce that the center of gravity of the trapezoidal regioillies on the line PQ . . Now suppose that the length of the upper base of the trapezoid~on. verges to zero. Then the trapezoid approaches a triangle, and the line PQ becomes a median (fig. 3.6). Consequently, the center of gravity of a triangle lies on its median. But since this is true for each median of the triangle. its center of gravity C coincides with the point of intersection of its medians. . 4. A circular sector. We now consider a circula1' sector ABO, viewed as a material figure lying in the vertical plane (fig. 3.7). Suppose that~ C \..
B
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v '
Fig. 3.6
Fig. 3.7
we have rotated the sector A BO about the center 0 by the angle S, and
that its new position is A' B'O. Let us calculate the change in its potential , energy. Suppose that the point C is the center of gravity of the region ABO, and the point C' the center of gravity of the region A' B'O. The difference in potential energy between the positions A' 8'0 ood ABO is expressed by the formula ,.
,. (3.2)
32
/
The Center. of Gtavity. Potential Energy, and Work
26
where PABO is the weight of the sector and Ho' is the height of the point C' above the horizontal line. OD. (We have assumed that OC was originally horizontal.) But He'
===
OC'·sin 8 = DC· sin 8 and P ABO = .
11'
J(J
a.
(3.3)
JlSB1o,
(3.4)
W ABO = .RlI«y·OC·sin
On the ol:hcr hand, \
From these equations we get WA'B'O 
WABO =
WA'AO 
But since the regions A'AO and B'B~ are congruent, the difference W"'AO WS'BO represents the change in potential energy of the sector B' BO after being translat¢· to the position A' A O. From the symmetry of the dia~, we have . .\' (3.5)
•
where P A' AD is the weight of the sector A' AD, S is its center of gravitYt and Hs is the height of the point S above the line OD. Furthermor~ ! since (S is in radians) and
Hs
=
os,sin( + ~) , IX
"'equation " (3.5) takes the form
,
3.1
\
22«·1"= RliI.· (t.y , '11
where I' is the specific weight of ·the sector, and 2a measures the arc length of the sector, in radians. Therefore, . . , WA'B'O 
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'I'M C~1IlU 11/ G,lWiIy, POl~ E1Jergy, tmd Work
.
21
Combinlng this raUlt ~, equation (3.4) and (3.3), we have
...
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R1lary·OC·sin 8'" ~ay·OS·sin ("
+~) ,
o
or,
., . OC"", OS. sin (t% +:8/2)..!,
ex
..
sma
.
(3.6)
This equation allows us to compute ~C. Since equati~n (3.6) is valid for arbitrary a.and, inparticu1ar. for 8 as small as ~/p1oasc, we may , write " DC
= lim*[os.sin (ex + 812).~]. sm 8
ex
.... 0
(3.7) t , .
But
t
lim sin (0: .... 0
8/2) .... ~ ,
0:
and from calculus we know that
lim
.... 0
0:
"
J=1 Slll 8
.'
.!Z
•
Therefore, equation (3.7) becomes .
OC =
(lim os) sinex ex •
(3.8)
' .... 0
I
Furthermore, as 8 '" 0 we can substitute the chord A.A.' for the arc A' A, and the triangle A'AD for the sector ,A'AD. The~fore, as 8+0, the point S approaches the point of intersection of the medians of this triangle. Since the medians of a triangle intersect at a point onethird of the distance from the base to the opposite vertex along the median, we may conclude that lim as
..... 0
=~3 R .
2. This equation is contained in many tex.tbooks on tri~onometry (it is usually written in the form Iim.+o ([sin 8]/8) = 1). It expresses therotuitive fact that as an arc of a circle converges to zero, the ratio of its length to the length of the chord IPannin& it converaes to unity .
..
,
,
....
~
"
The Center of Gravity, Potential Energy, and Work.
28 "
Equation (3.8) now takes the form
DC = ~ R. sin a • 3 tc
(3.9)
Formula (3.9) defines t~e position of the center of gravity of a circular
sector.
"
5. A halfdisc. Setting"ct = 1T/2 in foqnula. (3.9), we obtain 4R 31T
OC=.  . This
(3.10)
~tion defines the position of the center of graVity of a halfdisc
(fig. 3.8).
Fig. 3.8
Fig. 3.9
6: A circular segment. Suppose that a homogeneous material figure has the form of. a circular sector and is located in a vertical plane (fig. 3'.9). The sector OA DB is divided into the triangle OAB and the segment under investigation~ A DB. , We allow the sector to rotate about the horizontal ails 0 D, and compute the moment of the forces of gravity acting upon it. Denoting the moments of the corresponding figures by MOADBt MADBt and M OABt we may write M oADs =
MADB
+
MOAB'
(3.11 )
But the moment of any system of forces and, in particular, of the forces of gravity, is equal to the moment of the resultant of the system (refer to rule 2.1 ; specifically, equation (2.1), on p. 15). We employ the formula M.(F) = Fd, to obtain MOADB MADR MdAB"
= ySa.4DB"OC",
=
OC t = ySOAB' OC', ySADB'
c:. ,3 .)
(3.12)
\
The Celller of Gravity. Potential £nergy, tmd Work
29
where S is the area, y is the specific weight, and C, C ' , and c." are the centers of gravity of the segment, the triangle, and the sector, respectively. Substituting (3.12) into (3.11) and canceling y, we hav\
SOADS'OC" = SADB'OC
+ SOAn' ~C' .
(3.13)
. II
Furthermore, we know that .'
, 
DC'
= iR cos ex,
and DC" = tR sin ex . ex
. In the last equation we have used formula (3.9). The others are derived from simple geometry. Relation (3.13) now takes the form
. R~"( iR si: ex)
=z SADB'
DC + R2 sin ex cos cx(tR cos a) ,
or, simplifying, ' • SADS·DC = iRS sin a(l  coslZ ex) = iRS sinS ex,
DC=
2Rs sin 3
0:
3SADB
(3.14)
.
The equation obtained defines the position of the center of gravity of the kgment. Since Rsin IX = ABri, we may write our formula in the form
OC = (AB)3 , 12S"wB ~I
or, more briefly,
J3
DC = 12S
(3.15)
where S is the area of the segment and J is its chord. 7. An arc of a circle. The center of gravity of a circular arc lies somewh~re between the arc and the center of the circle defined by the arc. We may locate its position in the same way we found the center of gravity for a circular sector. Let us rotate a homogeneous maD terial arc A B lying in the vertical plane about the centerpoint 0 by the angle S (fig. 3.10). The new position of the arc is . A' B', and its center of grwity shifts from C to Ct. The potential energy of the arc is increased by an ~mount Fig. 3,10 (3.16)
..
r
,
,
30
The Center of Gravity, Potential Energy, and Work
where PAB is the weight of the arc A.B and He' is the height of the point , C ' above the line OD. But ,,'
H c' = DC"sin 8 = DC· sin S and PAll = 2Ray, where y is the specific weight of the material arc. Therefore, equation (3.16) may be written in the form ,
As before,
W.tB =
WA'B' 
2Ray· DC· sin a .
(3.17)
we also have that
and
hence, (3.18) But since the arcs ,A' A. and B' B are congruent, the change in potential energy may be expressed as ,
.
where PA'A is.1he weight ofthe arc A' A., Sis its center 9f'~vity, artd Hs is the'height of the poi~t S above the line OD. However, PA'A
= R8y. and
Hs = OS·sin
(ex + ~) ~
Therefore, from equation (3.18), our fonnula for potential energy ~~ " WA'B'  WAB = 2R3y·OS·sin
(ex +~).
Equating (3.17) and (3.19), we have
2Ray·OC·sin
~=
2RSy·OS·sjn (ex
+
(3.19)
n'
or
DC =
as. sin (ex '+
S/2).~ . sm
a
Taking the limit as
~~
~
0, we obtain
DC = (lim 6+0
as) sin ex • a
(3.20)
'/ 'l1#
But as
a+ 0
.
Cem~r
o/Gravily,Pote1Jlial Energy, and Work
31'
'
e point S converges to A.
an~
therefore,
lim OS = R.
•
...........
OC=R.~.
(3.21)
ex
8. A semicircle. In the case when
R CJi
(
a
= w/2.,.formula (3.21)
c
becomes
2R OC=· .....
I
(3.22)
1T
Equation (3.22) defines the position of .the center of gravity of a semicircle (fig. Fi&, 3.11 3.11). We have now derived a number of formulas for the center of gravity of some simple figures. We will usc these to compute surface area and volume of certain bodies.
3.4. Tbe Volume of a CyliDdrlcal Reeton Suppose we generate a cylinder perpendicular to ~ closed curve lying in a plane. If another plane intersects the cylinder so as to bound a single region, we have a rather special cylindrical solid with at least one base perpendicular to the generator. We wish to find a formula for the volume,ophis type o(solid: . In figure 3.12, we have drawn a cylindrical solid 0' of this type, with a vertical generator and a horizontal base. Suppose that we have lifted it by a certain height h, so that it now occupies the position A' B' D' F'. Let us calculate the change in potential energy. Denoting the potential energy in the original position by Wand in the final position by W', we have' ; A' ,,., ..... , F'
___ F
W'  W = Ph,
where P is the weight of the solid and h the increase in the height of its center of gravity. Clearly, h = AA' = BB'.
Fig. 3.12
•
,
.
.
32
The Center of Gravity, Potential Encray, and Work
Furthermore, supposing the solid to be homogeneous, we can write
p = Vi'. <~
where V is tile volume of the solid and i' its specific weight. Consequently,
W'  W=.vyh. On the other
(3.23)
~d,
'W' = W""BDF' +. WBB'D'D, W == W""BD" + W.tA,,,,', and, therefote.
.
W'  W
W,"'F"
= Wall'D'D 
(3.24)
t
of
, , that is, W'  W is equal to the differetce the potential energy of the bodies BB' D' D and AA. 'F' F. The volumes of these two regions are equal, and we may denote this volume by v. We will then have
WliB'D'» = vyHea , ....
,"...
where C1 and ell are the centers of gravity of the volumes AA.' F' F and BB' D' D, and HOl and HOg are the heights of the points C 1 and ell above, the plane AF. Substituting the expressions (3.25) into equation (3.24), we get ~ W'  W = vy(H02
.

He) .
Furthennore, since the body AA'F'Fis a right cylinder, v = Sh, where S is the area of the base AF. Therefore, (3.26)
We
or
v = S(Hca  lIe).
3t)
(3.27)
The Cl'nter 0/ Gravity. Potential Energy, and Work
33
We are not finished, however, since equatioQ (3.27) is valid for arbitrarily small h, and V does not pepend on h. We proceed by taking a
limit: .
'.. Y
= lim [S(Holi .~O
.
Hel )] = S(lim Heg  lim ~~o
h~O
HCI) .
(3.28)
Wc shall compute the last two Ilmits. First of all, it isclear that (3.29)
.
Furthermore, if h converges to zero, the points 1( and D:. converge to the points B and D, and the body BB' D' D co~ges to a plate of constant thickn~ constructed on the BD. Therefore, as h + 0, the point C2 converges to the center of gravity of a homogen~ous material figure BD, or, in other words, to the center of gravity of the figure BD. Denoting this center of gravity by C, we get
base
lim He g
..... 0
= 'He,
..
(3.30)
'II
where He is the height of the point (above the plane AF. Substitut~ng (3.29) and (3.30) into (3.28), we now find .
v=
(3.31)
SHe.
In other words, the volume of the solid is equal to the area 'If its base . multiplied by the neight of the center of gravity of the figure bounl;iing the solid on top. ' This relation 'will be useful for the computation of certain volumes. We should make one further remark concerning the derivation of . equation (3.31). In this derivation we have assumed that h can be chosen so small that the points of the figure A' F' lie below the points of tfiefigure BD. There are, howeveJ,: cylindrical regions for which it is impossible to do this for any h'>; O. This will be the case when the: figure ED has a point in common with the base AF. An example ofsucJl a solid, labeled A BDF, is shown in figure 3.13. The proof given cannot be applied to this tyI?C of solid, and this ~s~ must be cons(dered sepa• .~. .,. rately. , However, we may extend A BDFto the solid A' B'BDF' (fig. 3.1"4). We will then h a v e · .r ' _. I
VABDI.'
.. ,
'.,
=
VA'B'BnF'
~
,
VA'B'BF}"
s· ee'
"'"
.
..
S· C'C~ = .
,. .,'.
.'
,
'40
s· cc· . \
(3.32)
..
,
.
..
The Center of Oravity, Poten~ &crsY, and Work ,.
34
,
,
.. 8 A'
Pi&. 3.13
FiB 3.14
.
•
where S == .SAB, = S~'B'F" and C is the center of gravity of the figure ABD. But CC' is th~ height of the point C above the plane ABF. Con
sequently, eqilatiQn (3.32) b e c o m e s  ·
•
....
(
"
. where He rl:enotes the height of the center of gravity C abo\t¢ the base . of the solid ABDF. Thus, relation (3.31) is valid for cylindr:icaJ regions of the form shown in figure 3.13, as well as those previously considered. Computing the volume of..a solid by means of equation (3.31),poses a particular problemw~ must locate the center of gravity of the upper ., ,~ace. The following th~rem may help make this task Casier. .
!')rnoREM 3.1. Suppose that a cylinder is Itounded by a surface per
~dicu/ar to the cylinder, with a center of gravily at C, and by a second surface witll a cemer of gravity at C'. Then the line e . to the gener(ltor of tlie cylinder..
ee'
will be parallel
ProD/ ~uppose that the line ce' is n?t parane1 to the generator. Then the cy1indricaJ ,region may be positioned so that the generator is horizontal and the line ee' inclined as shown for the solid ABDF in
...
,
,
r
... _ _
t
... _ ....,4 \
\
B
.,4: _
t
B'
\,
\
\
\
+I
,
I
F'
Fig. 3.15
/
j
• \ .TM C~lIt~' of Gravity. PotellliDJ &ln6Y, and Work '
3S

...
,"I.'
denote the heights of thopoints C and C 1 above some horizontal p~ tflen, by this' construction.
figure 3.15. If He and
Hi;I
(3.33) l.ct us now displace the cylinder by a distance I in the direction of its . generator. The new position will be A' B' D' F', and since the shift is horizon£al, the potential energy of the solid remains unchanged. Thus,
(3.34) But
and and, therefore, equation (3.34) .1akcs the form WBB'D'D
•
=. WAA','F •
f
(3.35) .
•
Using the notation as before, we have that •
WSS'D'D
=
PUB';;;
and
W&A'r, = PAA'F'P!ls' . . ,
where Sand S' are the centers of gravity of the bodies BB' lY D and . AA'F'F, respectively. Since PBB'D'D = P&A'F'P, substituting these . expressions into (3.35) yields
..
(3.3Q)
lls = H s"
Now let I be very small. As J converges to zero, the body BB' D' D will approach· a homogeneous· material figure BD, and the center of gravity C will converge to the center of gravity S of the figure BD. Con'. sequently.
and from equations (3.36), (3.37), and (3.38) we conclude that
.
12
•
v
t." ,
36
The Center of Gravity, Potential Energy, and Work
.
But since this result contradicts relation (3.33); our original assumption that the line ec' isnot parallel to the generator of the cylindroid must be'false. . ~ This proves that the center of gt;'avity of the upper base of a cylindricaf • solid lies dire'ctty above the ce~ ~f gravity of the bottom base. s We shall now compute the volume of two special cylindrical solids. ,. 8' 1. We shall compute the volume of a prism; which may be viewed as a cylindrica} region with a D' . tnangular base. Let e and C' denote the center's of gravity of its bases. Then, by equation (3.31), we will have
v = S·CC' ;
(3.39)
where SO/is the area of triangle ABD. Furthermore; since the center'of gravity of a triangle is located at the l?oint of intersection of its medians,
8 A F~.
C'F' = tBIP' ,
3.16
I
'
and, consequently, Le' = tKB'. Therefore, '"
,
, ec'
/
= FE'
+ t(BB'
 FF')
= 2FF't
BB' .
'"
But since FF' = 1{AA'
+
DD') ,
we have that
ee , =
AA'
+ BB' +
•
3
DD' ,.
Substituting this expression into (3.39), our formula becomes
V=S
,
AA'
+ BB'\ +
DD'
(3.40)
3
3, This proof iii not applicable to the solid shown in figure 3.13. However, the same trick. used in figure 3.14 allows us to claim the result for the general case.
"
'J'.'M Cenler of Gravity. Potential Energy. and Work
37
In general. for a triangular prism with at least onc of the two end faces perpendicular to the length of the body, if S is the area of the , cross section, and H 1 : Hg , and H3 are the measufei.of the heights along th~ len~ of the body, then t~c desired f~rrnu~r volume may 'be
wnltcn as
•
/.. V = S Hi
+
i
.1"
3.
. (3.41)
2. We next consider a• regi,,\ obtained frolJ1 • a circular cylinder by a plane section, passing through the diameter of the lower.base (fig. 3.17). Its volume is equalt,?
, '11 Rg '11 R~ V == ·CS = ·OC·lana 2 2
..
where C is the center of gravity of the halfdisc which forms the.base ofthe solid. But, as we found ea~lier.
Fi&.3.17
OC = 4R
:m
(see formula (3.10) and fig. 3.8). Consequelltly,
'11R'i1.4R V == T 3'11 tan a =
i R3 tan a .
(3.42)
It is interesting to note that this formula for volume does not use the . value of pi. 3.5. The Volume of a Pyramid
,
Let us consider a triangular pyramid, one of whose sides is perpendicular to the plane of the base (fig. 3.18). Since this is merely a special case of the prism shown in figme 3.1'6, we may find its volume by using formula (3.41). Setting 111 = 1/.,) ~ 0 and II~ = 1/ in this formula, we
have
. ,
v=
iSH.
,
pyrarrri~f,~i;""a
th~t
'Now suppose that v.:e have a more general base, is, any arbitrary polygon lying in a' plane (fig. 3.19). We may t~en divide
)
..
:"
~.
... "~'
.
. . ;,
38 .
The CeDtu of Gravity. Potential , . , ..
•
EncraY, and Work v .
8'
.~
0'
D· "','
 A
..
.
A ~
'F1&. .3.18
Pia. 3.19
~n
it into selral triangular py.ramids of the form shown figure 3.18. The pyramid shown in figure ~ for exa.mpI~, can be divided into four \. such pyramids having the edge 00' in common. Computing the volume of each of these pYr~ds, we have .. ,
...
.. Therefore,
or, .
v = iSH,
(3.43)
where S is the area of the base ABeD. Formula (3.43) is the familiar expression for the volume of a pyramid.· \
.3.6. The Volume
Let us consider a body obtained by the rotation of a plane figure Q about an axis lying in its plane (half of such a body is shown in figure 3.20). We assume the axis 00' to be vertical. Let us' construct a new structure by adjoining to the body shown in figure 3.20 a cylindrical '\ pipe, and assume the entire p,ipe structure to be hollow and connected
. ... '. ",'
<&'
~
""
of. Body of RevolU#ou (Guldin's First Theorem)
•
4. We divided the pyramid shown in figure 3.19 into four pyramids of the type shown in figure 3,18. However, if a pyramid is very" oblique," the point 0 can lie outsid~hc base ABC D and such a decomposition will not be possible. We will then need to consider th~ II algebraic" sum of pyramids rather than the" arithmetic ,sum" (tbat is, take the volwnes of several of the pyramids with a nesative sign) . •<
"
.'
'
'~.
,
t,"
.;'
.~
,
'I"
•
.ihe Cent!r of Gravity, Potential Energy. and WOI'k
39
by a narrow neck (figure 3.21 shows the top view; the axis 00' is repre.. sented by the point 0). ~e insert a piston ABDE in the cylindrical part ofth'e pipe and the piston KLMN in. the circu1ar part. We fill the cavity between the pistons with an incompressible liquid. Suppose that the force F acts on the piston ABDE, forcing it to assume the position A' B' D' £'. We compute the work performed by this force. t.

o·
o Fig. 3.21
Fig. 3.20 "
Since the p~th traveled by the point of application of the force equal to BB', the work is equal to
A = P·BB' = F·AA'.
F is
(3.44)
:7
The forc~ F is equal to the pressure of the liquid on the wall AD. There.
fure,
\
• (3.45)
P = pSAD' ,
,J
where p is the pressure of the liquid and SAD is the area of the piston face A D. Substituting this expression in (3.44), We have (3.46) and since SAv·AA' represents the volume of the section A'AlYD, (3.47)
A = P VA'AD'V .
This is the expression for the work performed by the force F .. Let us now compute this work in another way. We shall consider it to be the work performed by the force R pressing against the piston KLMN. We then g~t
.
A=R·CC',
... •
;,.
..
" .40
The Center of Gravity, Potential Energy, and Work
where C is the point of application of the force R and ec ' is the length . of the arc described by this point during the motion of the piston. But
..,.
R = pSl,N,
where SLNis the area of the pistonface LN. Consequently, ~A
Comparing expressions
(3,:4~
= pSLir ce' .
(3.48)
and (3.48), we will have (3.49)
But since the volume described by the piston ABDE is equal to the volume descr~ by the piston KLMN. equation (3.49) can he written in the form VZ.'£NN'
=
Sur CC ' .
(3.50)
Note that SLN is simply the area of the figure Q in figu{e 3.20, and is the volume of the region obtained by the rotation of this figure. Since fo~ula (3.50) is valid for any arc length ec', in general, VL'LNN'
.
. V= S·CC '
. '(3.51)
'\
where V is the volume of the region generated, and S is the area of the figure Q. " 'The value of ec' is d~termined by the length of the arc described by the point C as the figure is revolved. But the point C has a simple geometric interpretation. Since it is the center of pressure of the wall LN and the pressure is uniform at all po;nts of this wall, the point e coincides with the center of gravity of the'figure Q. Consequently, the"arc CC ' is the arc described by the center of gravity of this figure. Let u~ply equation (3,51) to the body drawn in figure 3.20 (half of a body orrotation). In this case the arc CC ' is equal to 7TR c , where Rc is the distance of the' point C from the axis of rotation. Consequently, the volume of the body under consideration is equal to S7T' Re. But since this volume is half that of the complete body of rotation, we can . write
"
,
.
" Tm Center o/Gravlty,
POlent~
v=
2'1J'RcS.
Energy, and Work'
,," 41
hence, (3.52)
We now have the following theorem: THEOREM' 3.2. The volume of a body of revolution is, equal to the area of the figure from which it i~' obtained multiplied by the length of the circumferen~e of the cirCle described by the center of gravity of this figure.
This theorem is known as Guldin's first theorem. 5 Another proofofGuldin'sfirst theorem. Suppose that the plate Q is free to rotate , about the horizontal zaxis (figure 3.22 shows the view from above). We shall assume that Q is a material figure of weight P. The plane' of the plate is horizontal and passes through the zaxis. Rods 1 and 2 are assumed weightless. . The weight of the plate creates a rotational moment with respect to the zaxis. This moment is equal to
z
Fig. 3.22
MJ.P) = PRe',
'where Rc is the distance f~om the center of gravity of the figure Q to the zaxis. But
p = yS,
(3.54)
where S is the area of the figure Q and y is its specific weight. (We assume the figure is homogeneous.) Consequently, MAP)
=
(3.55)
ySR c .'
The product S Rc is called the static moment of the area S with respect to the zaxis. We shall denote it hy A'I AS), t ha tis, (3.56)
•
5, Paul Guldin (15771643), Swiss mathematician known for introducing the centro baric method,
,10 ..... 0
t,
. The Center of ~~1ty, Potential Energy, and Work
42 ,
,
)
Before proceeding with ~ p'roof, we should investigate this new term carefully. Comparing equations (3.53) and (3.56). we see that formula (3.56) is obtained from fonnula (3.53) by the substitution of S for P . . Loosely speaking, therefore, the static moment of area is the moment "created by the area" of the figure under consideration. Moreover, from equations (3.55) and (3.56), it is clear that the static moment of , area can be viewed as the moment created by the weight of a figure for • which y = 1. From equations (3.S5) and (3.56) it fo~ows that
M.(P)
=
rM.,(S) .
(3.57)
This relation serves as a conversion formula for the mo~ent of weight and the moment of area. Let us divide the plate into several parts. We will then be able to write
MJ.P)
= MJ,P
1}
..j. MJ.P~
+ ... +
MJPJ,
(3.58)
where PI' P"J" . .• , Pn are the weights of these parts. (Note that we have used the rule that the moment of a resultant is eQual to the sum of the moments of the forces of the system.). Using the conversion fonnula (3.57) for each moment, we obtain '
,
•
yM.,(S) = yM11.(SJ
+ yM1.S';}.) + . " + yMz(S.) .
Finally, dividing this equation by y (or. setting y '" l)"wc arrive at the following rule: RULE
3.1. If the art'a S consists of the areas St.
S~,
... , S1\,
then
We shaH now ~se this cquation for the proof of Guldin's theorem. '.
Suppose in figure 3.22 that the plate Q is a rectangle with one of its sides parallel to the zaxis (fig. 3.23). The volume of the body of rotation which is obtaincd is expressed by
or (3.60)
.
"
('
," '
.. ':"" .... ,..
.
~
'1'1se Center 0/ GTtlVity, Polen/lal Energy, tmd Work
But
43
".
:.tl¢'
and
where. S is the area of the rectangle and C its center of gravity. Consequentl~
V = 2'1TSRc.
that is, V = 2'11MJ.,S) ,
(3.61l
I
where M J.,S) is the moment of area of the rectangle with respect to the axis of rotation. ~ Let us now substitute an arbitrary figure Q for this rectangle (fig. 3.24). We divide this figure into a large number of Darrow strips,and" .". 2
4
z
,. h
C
He
P
H, R2
Fig. 3.24
Fig.3.23
approximate each of these strips by the rectangle inscribed in each strip. If n denotes the number of strips, and we allow this number without. bound, the approximations become successively better. We then have " v = lim (,Vl + Vll + ... + Vn) ,
t'
" .... gO
whe~e
V is the'volume of body obtained by the rotation of the figure Q and V1 , Vll , , •• , Vn are the volumes of the bodies,obtained by the rotation of each of the rectangles, But according to (3.61), •
51 J
•
,The Center of. G.~vity. Potential Energy. and Work
44
Consequently,
..
..
V.:. 217" lim [M.(Sl) .... 110
+ MiI(S'll + ... +
(3.62)
M ..(Sll)]
where Sh S'l' ... , S1l are the areas of the rectangles. But accOrding ro (3.59) the sum inside the square brackets i~ equal to the static moment of the area of a cross section (the are'h bounded by the boldface lines in figu~ 3.24). But since this figure converges to the figure Q n + ~,
i
lim [M,,(Sl)
+ M.(S,J + ... + M.,(S,,)]
= M.(S),.
where M;:{S) is the static moment of the area boundtjd by the figure Q. Therefore, equation (3,,62) takes the form '
v = 271M.(S) •
(3.63)
Formula (3.63) shows that the) volume 'of the body obtained by. the rotation of a plane figure is equal to the static moment of its area·multiplied by 2'11. Using expression (3.56), this impljes
V = 217'RcS ,
.
which proves Guldin's theorem.
~.
.,.'
3.7. The Volume of a Sphere ,.
The region determined. by the rotation of a semicircle about its diameter forms a sphere. Substituting into the formula given above, • the volume of the sp~ere may be expressed as
'" 71R'l V = 211'OC, \ 2
.'
where Cis the center of gravity of the semicircular plate (see figure But according to formula (3.10)
OC
= 4R, 311
and, consequently,
,.
4R 7TR'l V = 271'''
3'11
2
3.if
'I'M Cenltr,.o/ Gravity, Potential Energy, and Work
3.8. The
4S .
v01umes of. CerWu Odler Bodies of Rotatioa
The str~ngth of. Guldin's theOrem lies in its applicability to a large number of differe~ We will give several further c~pJes. ' '·It.'
,ff. ",T:
1. A circular cylinder. Referring to figure 3.25, it is clear that the volume of a ~ar cylinder may be given by
\
R
:v
V = 21tRcS = 21T 2' RH =, 1TRfJH,
where R is the radius'of the
CYlind.~ and H i$ i;:~
• Rc H
H
C
~3_!_~5
Fig. 3.26
.
•
. 2. A cone. A cfyfular ~e can be viewed as a body obtained by the rotation of a rigA( triangle about one of its legs (fig. 3.26). The center of gravity of a triangle is located at the point of intersection of its medians. Consequently, Rc is equal to Rj3, and we obtain the familiar expression for the Volume of a cone: V
R RlJ
= 27TRcS = 21T32 =
~
7TRfJ}J
3'
3. A torus. A torus is a body obtained by tl,le rotation of a circle about an axis lying in the same plane as the circle (fig. 3.27): In accordance with Guldin's theorem. the volume of a torus is equal to
v ""'" 21TReS = 21TR1Tr~ =
2~Rr~
.
,
.",~.
. ,I.,
The Center of 'Gravity, Potential EnetaYt and wait
46
'/
• ," ,
•
o

FJa. 3.27
Fia 3.28
4. Suppose that a circular segment rotates about the dlameter parand to its chord (fig: 3.28). The volume of the ringshaped body'Obtained is
~ual
to
,.
V = 21T·OC·S ,
where OC is the distance from th~ center of the circle to the center of gravity of the segment. But according to (3.15).
OC
::0:::
13 118'
where 1 is the length of the chord of the segment Therefore.
v· '"6 TTJS
'"
".
,.
~
"
It is intereslag to note that the volume w1uch we have found depends only on I. We have, by no means, exhausted the applications or' Guldin's theorem. Fqr example, figure 3.29 shows a solid obtained by the rotation of a circular gegment about its chord. Since we know the location of the center of gravity of a segment, the reader migllt want to determine the volume of this solid. 5. Guldin's theorem is often useful to substantially reduce the num· her of necessary calculations. Suppose, for example, that the square ABDE can rotate about the axis 00' (fig. 3.30). The volume of the co~ponding solid of rotation can be found by computing the difference between the volumes of the two inclined edges which form slices from a cone. This, however, is comparatively complicated, sinCe by means of Guldin's.tbcorem, we can immediately obtain
V
=
2rrRcS
=
21T(a +
wbefO' a is the diagonal of the square.
~) 2
cf = 2
~2 'TTa s ,
"
•
I" •..
,'
,'F "
"
"
I
',.'"
~
... , .. . '
:
.0
'
,f
..~ CI~I G'~Y, PoulUiDJ EM",. GIld Wo,k
..
. O·
.. ?
,.
,',
.
•
•
}'f
~.
.
47
.....
'.
8
A~c>~,D
.; "
E '
...
o
.,
•
Fi& 3.30
A similar ~ample is 1Uustrated by t~fonowing problem. A triangle . f . rotates once'about the axis Zl and once about the axis %2' Given that the axis' Z2 is parallel to 4l,. (fig. 3.31), 'how are the volumes of the corresponding bodies of rotatiog related 1 . Guidln;s theqrem allows one to solve this problem witho~rrying out any computations:,Siiice tllt~ medians of the triang1emtersect at a • point which is twice as far from the axis Z2 as it is from Zlo VI: V2 = 1: 2, ~{i. SUPP9se that h6mogtileous material figure Q lies in a horizontal plane (fig. 3.32). If we allow it to ro~te about the horizontal axis ~O' passing through its...center of Sravity, it will remain in cquilib~um, Consequently.' .' ' • .
I
a
.

/
where y is the specific weig~t of the figure, Sl and Sri are the areas of the 0'
0
'(
Fig. 3.31
Fig. 3.32 (
\ 54'
I I
\ I
'.".,.~
I
\
6,1.
.., I
',...
48
I'
'The Center of GravitY,ifotential Energy, and Work ~
~
two parts· divided by the line 00', and Rl and R2 are the distances of the centers of these parts from the line 00', Multiplyip.g the above, equation .by 217/Y, we will have
t
(3.64) . Equation (3.64) shows that the volumes of the two distinct solids . obtained by rotating the left and right halves of this figure about the axis 00' are the same'. ., This result is valid for any plane flgureand any straight line passing through its center of gravity. Suppose, for example, that the triangle ABC rotates about the median BD (fig. 3.33). Then the volumes of the bodies described by the ttiangles ABD and BDC are equal. ' 7. In aU,..tlie examples up to this point, Guldin's theo"'rem has been used strictly for the calculation of volumes. It is possiQ,l.e, however, to use it in another way. Knowing the volume of a body of revolution, we ,may find the center of gravity of the figure from which the body is obtained. Let us consider two examples.
•
,
,
. B
..
...
8
. A
fig. 3.34
Fig. 3.33
a. Suppose that a triangle with sides a, b, and c rotates about an axis lying along the side a (fig. 3.34). The volume of the resulting solid can be easily computed by adding together the volumes of the two cones. In this way we obtain the formula V = trrhu. 2a , ~erc
h(1 is the altitulle dropped to the side a. Applying Guldin's theorem to this triangle,.we now have ,
frrh" ~a
= 271 Rdah .. ,
or,
, '~
. I
..
...
'
"
."
l
',1,.\".
,
",'
"
,'" .", " " ,"
,
\
~
The Center 0/ Gravity, Potentifl/ ~rgy. and Work
49
Thus, the center of gravity of this triangle lies at a disfance ihg from side a. Arguing in the same way, we may deduce that it lies at distances thb and !he from sides band c, reSpectively, But only one point of the triangle possesses this propertyf,he point of intersection of its medians. h. As a second example, we shall find the center of gravity of a halfdisc. Applying Guldin's lheorem to the sphere, we obtain ' , 4 ' '1TR2 '1TlfJ = 2'1T·OC·, 3 2 &
where 0 is the center of the disc and C the center of gravity of the halfdisc. Consequently, v ,
I
4R
OC=·
•
..
'
(3.65)
3?T
,Formula (3.65) defines the position of the center of gravity, of a halfdisc. z Of course,,.if we derive formuJa (3.65)'in this way, we may be accused of circular reasoning if we use the formula to compute the volume of a sphere. We can, however, use it to compute the volumes of certain ,; other'bodies, for example, 'the body show~ figure t~.o:>I 0 .. 3.17 (see equation (3.42) on p. 37). Thus, Guldin's theorem allows us to compute the volume of this body starting from the formula for the volume ot a sphere. We can find other examples of this kind. Suppose; for example, that a halfdisc rotates about the zaxis (fig, 3.35). The volume of the resulting body ofrotat{on can' Fig. 3,35 be found by computing the difference Rc = R  ~C, ' applying formula (3.65), Consequently, b~ assuming the formula for the vt>Iumc. we can compute the volume of this lrody of rotation. (Note that this body is not the H sum" or .. difference" of solids whose volumes ai'b known. For this reason, a direct computation \ of its volume turns out to be difficult.)
in
",
I
3.9. The Surface of a Body of Rotation (Guldin's Second Theorem) ,
Let us first introduce two new con~pts. 1. The tangent. Let us take points M and M' on the curve ;.1 Band draw the secant M M' (fig. 3.36). We now'fix the point M and bring the '.
.. "
56 , ,
:.'
,
~'
.
..
. : ...
".,"
'"
so
The Center of Gravity, Potential Enugy, and Work
. , A:'" .
. '
~~
M
...
A
8
..
p
N
Fia. 3.36
B
Fig. 3.37
.{.
\
'0.
point M' arbitrarily close~o M. The secant MM' will then approach, asa limi~ the position MP. The line MP is called the tangent to the CIp'Vt! AB at the point M. 2. The normal. Let us ta~c a point M on a plane curve (fig. 3.37). We draw· the tangent MP and the line MN perpendicular to MP through .• the point M. The line MN is called the normoJ to the given curve at ~e point M. Loosely speaking, we may refer to it as the perpendiculatto
the curve A B at the poi~t M. Let us now consider the arc AB.of a plane curve (fig. 3.38). We take an arbitary pOint C on the curve, draw the normal through this point, and mark off a small segment ec' of given length d along the normal. Drawing such segments at each point of the arc AB, we may define a curve A' B' as the locus of the endpoints of these segments. It is possible to prove that each of the se8tnents CC' is normal not only to the. curve A.B, but also to the curve· A'B'. Therefore, the distance CC' "can be viewe4 a~· the width of' th~ strip AA' B' B. Since this width is the 'Same· at all points along the curve, we \ViIl say that the strip AA' B' B h~ . constant width. . . , SuppOse thatanarrow;,strip AA' B'.B has constant width d and..l'otates . about the .axi~ 00' (fig: 3.39). Denoting the volume of the resulting body by V; we may apply Guldin's first theo~m to write
~
,
I
V = 21TRc ,S,(A'B'B'
..
where C' is the center of gravity of the figure AA' B' B. Let us now fix the arc AB and begin to dt.'Crease d. We will then have
•
V
lim d d ... O
21T(lim
.=
.
<1·'0
Rc')
(3.66)
.'
But ,as d becomes very small, the folIo
.
•
"
TIut C,nur {J/Gravi,y, Potentitil Energy.,and Work
..
51
o· A
.A
J
.
,
Rc'
He
.'
o Fia.3.39
where SAS is the area of the surface obtained by the rotation of the arc AB and JAS is the length of this arc. From these relations we conclude
that .
V
.
: 11m
11m d = SAB, d ... O
d .... O
SAA,'B'B
d
=
I
AB,
and equation (3.66) takes the form .• I
(3.61) .Furthermore, since thtl.'Strip AA' B' B has constant 'width, for very small ,d its center of gravi.ty will be close to the center of gravity of the arc AB:'
Therefore,
as a limit we have ....
, lim Ret = Rc , d ... O
I'
(3.68)
where C ~ the center of gravity of 'the arc A B. Substituting (3.68) into (3.67), we have ..
or, dropping the index AB,
.
Equation (3.69) cxprCSSt..'S the
(3.69) fonow~g
theorem:
,
,l
".'..
,
.~
""",
'
•
.
The Center of Gravity, Potential Energy, and Work
S2
"
3.3. The surface of a body of rotation, is etjila/ to the length of the CUrl'e from which this surface. is obtained multiplied by the"cifcwn " THEOREM
terence of the circle described by the center 'bf gravity of the curve. The theoremjust proved is called Guldi"s second theorem. Another proof ofGbldin'Y"$econd Iheor£.m. Just as in section 3.6 we introduced the concept of the static moment of area, we can introduce the concept of the static moment of length. ... .' Let us consider a material plane cutvc L (fig. 3.40). We place it iita . horizontal plane andpnit"'to the horizontal zaxis Jying in this plane ~, by mean~ of weightless rods 1 and 2. We~ allow it to rotate about this axis. The force of the weight of this curve creates a certain moment with respect to the zaxis. This moment is equal to
where P is the weight of the curve L anQ .Re is the distance from its center of gravity to the zaxis. But
P = "II where y is the specific weight of the curve and I is its length. Consequently, .•
M.{P) = yl~c. The product IRe is caBed the" static moment of the length I with respect to the zaxis. Denoting it by M.,(/), we c,m write
Mz (')
.. ., """,
'
'
.
.,..,
IRe.
(3.70)
The static moment of length is equivalent to the moment created by the weight of a curve when y '" 1. If we divide the curve into parts I~, /'}., ... , In, we will then have IfIY
(3.71)
ltr~ti~1n "'
.
(3.71) allows one to prove Guldin's second theorem without
difficulty. We will use a formula for the surface area of the side o'f a trum:ated cone to compute the area of the surface generated by the segment A B (fig. 3.41). This yields the equa'ion
5.<) "
',:''';''
,'.'"
'~
. \ ,
~
The Cemer of GiO'Dity, PoleJllial Energy, ti8d Work ...
53
.. "
z
z,
R,
fie
t'
Re 2
"
~
J ,~
">"1'.
~
•
Fig. 3.40
.
R2 Fig. 3.41
~
This may be wdtten ih the .form '
SAB = 21TIAB , and since
\
Rl
+ 2
we have that
Rl + 2
.
.
.<
It'
}
~,
" =
Re.
.,
'.
SAB = 21TIABRc , i~,
that
,
~,
" R!J
R~
..
, SAB = 21TMiIAB) .. '
..
"'
"
p.72}
'.
where MZ(/AB) is the static moment of the segment AB with res~t~to , the axis of rotation. ". Let us now substitute an arbitrai;r ~ane curVe for th~, segment A B (fig. 3.42). We approximate • it. by a broke~ line consistin~of n segments and. z suppose that n goes to inn'nity and that the '" . lengths of the segments converge'lo zero. We then have '"
where S, S)o,S2' .. " S)1 arc the areas of the sur" faecs obtained ~y rotating the given' curve ,and ~ e~ch seg~nt approximating thf'eurve. The ~rcas SH S2, ... , Sn, however, cart. be r~presented in the formJ3.72), which parniits us to write (3.73)
•
"
as
Fig. 3.42.
S =
277
lim [Mz(/1 ) +Ma(l!l)
,
" He ,
. .
I
~
.• .
+ '" + b~(I!\)], (.. /'
~
•
r
,.
Th~
54
Center of Gravity, Potential Energy, and Work ;
that is,
. ,
.
(3.74) where Mz(l) = Iim,pao [M"(/l) + .. : + M z(/,,)] is the static moment.of the given curve with respect to !he zpis. Substituting (3.70) and (3.74), we can now write
S = 27TRcl .. "
rhis equation
expr~ssos
Guldin's second theorem.
3.10. The Surface of a Spbue" , Suppose that a semicircle is rotated about its diameter. Applying GlJlrlin's second theorem, we may write ..~ •
where 0 is the center of the circle and C is the center of gravity of the (tlg. 3.11)., Furthermore, we derived in equation (3.22) that ._H semici~cle
,
<
"
','
2R OC=, 7T
and, therefdre,
,

S
~
.
~
'2R 211  .  11 7T
..,
". ..."Ilk""" V
R.
.~,
,
/j
,
Consequently, the surface area of a sphere is equal ttl
~
.. ) ", 
r
oS .:::: 4w R'). •
•
"~:~
..
..
, 3.11. :fhe Surfaces of Certain Other Bodies of Rotation

"
'. Using Guldin's theorem, we can c()rnputc t,hc area of a {lUmber .of surfaces of rotation. Let us consider severa.4 examples. ;)
"
.
1. A torus. Since the center of"'gravitY;Qf a drde lies at its geometric center, the surface area of a torus (£ig." 3.21) is equal to ....,. . to
"I
S
=
211R· 211'_ = 4Jr2 Rr •
'WIi;a,.
.
,~
,
'.
,
~
The Cemer of GUlVity, PotMtW EnozY. and Work ,,';
,'
p
......... .
55
c' B
o
s=
2'f1"C'C,/,
where Cis the center of gravity oftbe arc
C'C = OC·sin fJ and' 1= 2Ra.
.,
Therefore,
.. ,
.
S = ,21T' DC· sin fJ· 2Ra . Equation (~.21), however, tells US that DC
s=
= R(sin a/a). Consequently,
21TR sin ex sin,8. 2Ra = 21TR· 2R sin a sin,8 . a
.
Since , "
2R sin a = I,
we now have
Again referring to the sketch, we note that the 'second factor of this product is. equal tp the altitude of the spherical strip (that is, the projectron of the chdtd"A B onto the diameter PQ). Denoting this altitude by H, we finally' obtain' the formula ' , " ...
,
.
3. Suppose that the square shown in figure 3.30 rotates about the axis 00'. The surface area ,of the resulting solid'is equal to ,
"
'
( + a)2 ·4 ay'(2) ~ '= (my'(2~ .
,.~ S = 27TRcI = 27T ,a
4. Guldin's theorem allows one to determine the center of gravity of certain curves. For example, knowing the surface area of a sphere, we can . find tnc center of gravity of a semicircle. In the same way, by assuming .\
.
.
.AB and I is its length. Furthermore~ as is clear from the sketch,
..
a
,',
2. A spherical strip. This surface is o&' tained by the r~Wion of the material arc AB about the diameter PQ (fig. 3.43). From Guldiii'~ theorem we have
A
• 'fiB. 3.43
'.
'
.
56
The Center of Gravity. Potentiai Energy. add Work
.
....
the formula for the surface area of a spherical strip. we can easily derive the center of gravity of a circular arc. This permits us to co1llpute the area of the surface formed by the rotation of this arc about an arbitrary axis. I n this W?y, in particular, it is possible to find the surface area of the body shown in' figure 3.29.,
3.12.' Conclusion
, .'
•
Tie proofs presented in this book 'might suggest certain questions, \ nir exaf!1plc, one might ask whether or not we have used circular '\ reasoning in any of the arguments. In the proof of Pythagoras' the.orem we assumed the law of moments. This law is deriyed in physics with the htlP. of certain physical and also geom~tr~c Cfncepts. We can therefore ask whether or not .pythagotas' theor/lm was used in its derivation. . Forru!~ely. an ~nalysi~ of the.usual derivation of tIle law of moments shows that it is based only. on the axioms of 'statics and on 'certain theorems of similar triangles. In ~his instan&, at)east, 'we have avoided . circular reasoning in the proof. We need to say the same thing about the . other physical 'laws which we have assumed in this book~ We have not \ . assumed a law tbat is based on any the theprems which were proved . by using th<1tlaw. ' .A 'he mjght asi/ a. second .,question: To wh,it extent are the idealizations whieh we have \!sed orr v'arious occasions pe.,rmissi ble? I n the third ' chapter, for example, w~,h.an with thcconcept'of a line having weight but no thickness, wftich ~s admittedly inlpossiblc. To' answer this· question, we should point4 but that the~e idealizations are essentially the same as those used in·ge"Oni~try, \\there one speaks of ~l point" without length or width," .and of a liac" without thivkness." A 'ine having labstractio'n of the same kind, arising weight but not thickness i~ " I . from the representation of a thin t:"rv~ (o.d with a definite weight, but a thickness so'sligl1t that it, cal,1'bc neglected. In this respect, it is pos~iblc to make further abstractio~~, and rather th:in usirfg weight as the ccnt;al property, assign to ,lline some otlier physical property, such as , flexibility or elasticity. I n ,this sense it would, for ~lJl!Inl'le, be possible to speak of .i' line having'no thickness but having chlstr,:.lpropcrtics. A prototype of such a line' would be a thin rubber lilamcnL li .
.
,
of
an
,.
1\
//".I'il1/,.
(It.:Cllr~
in V. 1\. Uspenskii's book; Some App/iwork: Blaisdell Publishing ('~)mpaIlY, 1961), The t.:oncept of ;HI l'Ia.l/i, ' line O\:Cllr ill L. A. LYllslernik's hook, ,','hoftt'st Paths: Vari(Jlior~1,/1 Pf!{lhll'ms (London: p. gamon Press, 19M) [translated .rnd adtlptcd by the Surv~y of I~cent Last r lropclln ,Ma~hcl11ati~'al literal/Hc]. In .th~sc books these cont.:epts rfe used for t e pmof of certain rCol1letril.: llTCorcms. 6. The idea of a
line
liol/.\' of Mcchanip /0 MIl/iIol/a/in (New
.
'
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",
The Center of Gravily, Potenlial.Energy, and Work
57
We might, finally, ask a, third question: tlo we have it right to uSe such nongeome1(ic axioms as the rule of the parallelogtiim of forces or , the postulafe on lhe impossibility of perpetual motion? Since we'introduce nongeometric objects (such as forces), however, we must introduCe axioms describing the properties of these objects. Therefore. the use of nongeometric axioms in this ca,sC is natural. We can say that the proofs presented in this book are based on a system of concepts and postulates from the reafm of mechanics,: rather than the usual system from the rea1m of geometry. The fact that we are able to prove purely geometric . theorems from, these 'Unusual postulates testifi,es to the coasistency of oor ideas oj the physical ~orld .
.
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\ The University of Chicago Press
Paper
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65
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ISBN: 0226450163 .......