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a^c^ Lj:.*t-*v^ 
 
 ELEMENTS OF MECHANIC! 
 
 INCLUDmO 
 
 KINEMATICS, KINETICS 
 AND STATICS 
 
 WITH APPLICATIONS 
 
 BY 
 
 THOMAS WALLACE WRIGHT 
 
 M.A., Ph.D. 
 Professor in Union College 
 
 n/^yri^.^.^--^ /yi- 
 
 jCjz. 
 
 SECOND EDITION , ^ -^^J^ 
 
 NEW YORK 
 
 D. VAN NOSTRAND COMPANY 
 
 London: E. and F. N. SPON 
 
 1898 
 
Copyright, 1896, 
 
 BY 
 
 THOMAS WALLACE WRIGHT. 
 
 fiOBERT DBUMHOin}, ELECTBOTTPER AND PRINTER, NEW TOBK. 
 

 PREFACE TO THE SECOND EDITION. 
 
 The main features of this book are the following : 
 
 The subject is treated in an elementary manner. No 
 mathematics have been introduced beyond the elementary 
 principles of geometry, trigonometry, and the calculus. The 
 calculus has been introduced because in many cases its 
 methods are of marked advantage for clearness and compact- 
 ness of demonstration. The book is so arranged, however, 
 that the sections involving the calculus may be omitted with- 
 out disturbing the continuity, and thus a more elementary 
 course be taken. By introducing the calculus and giving 
 alternative proofs the student does not form the idea, as is 
 often the case, that there is a kind of mechanics called ele- 
 mentary, another analytical, a third theoretical, and so on. 
 He sees better the oneness of the subject. 
 
 Great care has been taken to indicate clearly the nature of 
 the units in which the various mechanical quantities are 
 expressed. Both the common (British) and the metric gravi- 
 tation systems of units are explained and fully illustrated. 
 The C. G. S. system employed in Astronomy and Physics, with 
 the related practical units employed in Electrical Engineering, 
 are also very fully treated. Synoptical tables for facilitating 
 the passage from one system to another are appended (pp. 364, 
 365). 
 
 Though the metric system has been legalized in the United 
 States for over thirty years, there is little disposition among 
 people in general to use it. It is probable that commercial 
 
 ill 
 
 ivi304883 
 
IV PREFACE. 
 
 influences may in time become sufficiently powerful to cause 
 the general adoption of the metric system, the only civilized 
 countries not now using it being Great Britain, Russia, and 
 the United States. While, however, the British system is in 
 almost universal use in English-speaking countries, it is 
 necessary, in a text-book at all events, to conform to it — in 
 fact there is no choice. 
 
 As regards nomenclature, I have followed prevailing cus- 
 tom. With the gravitation system of units the terms in 
 general use are given, and with the C. G. S. system the ter- 
 minology employed by physicists and electricians. 
 
 The term mass has been defined in its strict sense (p. 46), 
 and its use consequently confined to the absolute system of 
 units only. The term weight has been defined in accordance 
 with the sense in which it is employed in the official descrip- 
 tion of the standard weights and measures (p. 62), which is 
 the sense understood by people in general. 
 
 In this way the relation w = mg is dispensed with, and the 
 passage from one system of units to another becomes clear 
 and straightforward. Unless this course is followed an 
 uncouth phraseology must be introduced, which is used 
 nowhere but in the lecture-room. Thus we have "a force of 
 10 lbs. weight " for " a force of 10 pounds," " a box of masses '' 
 for " a box of weights," and the like. 
 
 To each chapter is appended a series of examination ques- 
 tions on the text preceding or capable of being easily deduced 
 from it. The number of illustrative problems is very large — 
 over eight hundred in all. These have been made of as inter- 
 esting and concrete a character as possible. Short historical 
 notes have been freely interspersed, as tending to add a more 
 lively interest. 
 
 For convenience of reference Tables of the units and of the 
 symbols and abbreviations used in the formulas of the gravi- 
 tation and 0. G. S. systems have been added (pp. 366, 367). 
 The dimensions of the C. G. S. units have been placed side by 
 
PREFACE. Y 
 
 side with the symbols. The tables of symbols and abbrevia- 
 tions follow in the main the recommendations of the com- 
 mittee of the International Electrical Congress held at 
 Chicago, August 24, 1893, so far as its recommendations 
 related to mechanical quantities in the C. G. S. system of 
 units. It is desirable to have an international notation, sym- 
 bols, and abbreviations in mechanics, and the recommendations 
 of this committee deserve attention from writers and teachers 
 everywhere. 
 
 My thanks are due to many friends who have favored me 
 with corrections and suggestions, 
 
 T. W. Wright. 
 
 Schenectady, N. Y., Feb. 1. 1888. 
 
CONTENTS. 
 
 PAOS 
 
 Introduction 1 
 
 CHAPTER I. 
 Kinematics— Motion 6 
 
 CHAPTER II. 
 Matter in Motion — Newton's Laws of Motion 44 
 
 CHAPTER III. 
 Dynamics of a Particle 72 
 
 CHAPTER IV. 
 Statics of a Body 141 
 
 CHAPTER V. 
 Friction 206 
 
 CHAPTER VI. 
 Work and Energy 224 
 
 CHAPTER VII. 
 Dynamics of Rotation 271 
 
 CHAPTER VIII. 
 Elastic Solids— Impact 320 
 
 CHAPTER IX. 
 Metbic Units. Dimensions of Units 338 
 
 Tables 364 
 
 Index .,,,.,. 369 
 
 vii 
 
ELEMENTS OF MECHANICS. 
 
 INTRODUCTION. 
 
 1. Mechanical experiences are without doubt of great 
 antiquity. The earliest investigations concerning mechanic- 
 al principles are ascribed to Archytas of Tarentum (b.c. 
 400), He is said to have worked out the theory of the 
 pulley. Later, in the writings of Archimedes of Syracuse 
 (B.C. 287-212), are found applications of geometry to various 
 mechanical questions, including a treatise on levers and 
 other machines. From Archimedes to Galileo and Stevinus, 
 a period of nearly two thousand years, no marked advance 
 was made. It is to Galileo and Stevinus that we owe the 
 transition from Mechanics in its original signification as the 
 Science of Machines to Mechanics as the term is now under- 
 stood — in fact they are to be regarded as the founders of the 
 science of mechanics. 
 
 2. The qualities of natural phenomena become known 
 to us through our senses. Certain of these qualities are 
 assumed to be fundamental, in the sense that no one can be 
 expressed in terms of the others. They are incapable of 
 definition and include space, matter, and time. Closely 
 related to these fundamental ideas are the ideas of motion 
 and force. 
 
 But although it is not possible to define these qualities, 
 we may consider their mutual relations. In order to investi- 
 
2 INTRODUCTION. [§ 3 
 
 gate these relations it is necessary to compare by measure- 
 ment the quantities that enter. The science which treats of 
 the relations of matter, motion, and force and of their meas- 
 urement is called Mechanics. 
 
 3. The term body is applied to a limited portion of matter. 
 Bodies are said to occupy different positions relative to 
 neighboring bodies. We define the position of any point in 
 a body by reference to points in some other body chosen as 
 points of reference. 
 
 Thus the position of a point P relative to a chosen point 
 in the same plane is defined by either (1) the distance OP 
 
 and the angle POX made by OP with any known line OX in 
 the plane, — an east and west line, for example,— or (2) the 
 distances PM, PN Iyoth two perpendicular lines OX, OJ^in 
 the plane. 
 
 In the first case the distance OP and the angle POX are 
 the polar co-ordinates of P; in the second, PM and PXor 
 Oif are the rectangular co-ordinates of P. 
 
 4. When a body is changing its position it is said to be in 
 motion. The line drawn through the successive positions 
 occupied forms the path of the moving body. 
 
 Now as we contemplate the body moving in its path, 
 questions arise as to the influence of the body itself on the 
 motion. We may, however, consider the motion only, apart 
 from the body moving, and study the nature of the path 
 traced out as the body moves from one position to another. 
 Of course no such separation exists. It is a mere abstraction 
 introduced to reduce questions of motion to a purely mathe- 
 matical form and to serve as an introduction to the more 
 complex problem itself. 
 
§ 6] INTRODUCTION. 3 
 
 This science which investigates motion without consider- 
 ing the nature of the body moved or how the motion is in'o- 
 duced is called Phoronomics [= law of going] or, more com- 
 monly, but less properly, Kinematics. 
 
 Since in changing position a certain time is taken, the ele- 
 ments of a motion may be said to be distance, direction, and 
 time. Kinematics, therefore, deals with distance, direction, 
 and time, and may be regarded as an extension of geometry 
 by the introduction of tlie idea of time. Like geometry, 
 it is a purely abstract science resting upon certain ideal 
 assumptions. 
 
 The term Kinematics [cinematique] was first proposed by 
 Ampere [1775-1836]. The term Phoronomics ((pofjeco) ex- 
 presses the idea of mere motion. Kinematics (Ktveoo) in- 
 volves the idea of the cause of motion. 
 
 The tendency at present is rather to restrict the term 
 Kinematics to the geometry of machine parts. 
 
 5. When we consider not only the motion but the body 
 moving as well, we pass from kinematics to dynamics 
 (dvvajut?), the science of force. If a body at rest or in 
 motion has its condition of rest or motion changed, it is 
 usual to say that the change is produced by the action of 
 force. If the form of the body is changed — as in bending a 
 spring — we say that the change of form is due to the opera- 
 tion of force. In the popular sense a force is a push or a 
 pull. The idea of force seems to be derived from a sense of 
 resistance offered to the use of our muscles, and to muscular 
 effort, or to anything producing like effects we give the name 
 force. The science which treats of the different effects of 
 force on bodies is called Dynamics. 
 
 6. Each of the two effects of force — change of motion and 
 change of form — furnishes a means of measuring force. Of 
 the two the former mentioned is the more elementary and 
 will be considered first. Forces causing change of size or of 
 shape, or strain as it is called, will be discussed in Chapter 
 VIII. 
 
4 INTRODUCTION. [§ 7 
 
 Mechanics is thus divided into Kinematics and Dynamics. 
 For purposes of study each of these divisions has various 
 subdivisions, depending upon the circumstances of the mo- 
 tion, the nature of the body moved or of the forces acting, 
 and which will appear as we proceed. 
 
 7. Kinematical Units. — There can be no exact knowledge 
 in physical science without measurement. In order to meas- 
 ure any quantity we must choose some definite quantity of 
 the same kind as standard unit of measurement. Natural 
 standards first suggest themselves, as the palm, foot, span, 
 quadrant of the earth, etc., some variable and others whose 
 values are approached as our methods and instruments are 
 improved. Bfit a standard should as far as possible have a 
 constant value. Hence at present, artificial standards are in 
 general use; their invariability being as certain as that of any 
 natural standard, they can be chosen of dimensions conven- 
 ient for the purpose in hand and any number of copies can 
 be made with the greatest precision. (See Chapter IX.) 
 
 In kinematics the fundamental or independent units are 
 those of length or distance and time. All other units can be 
 expressed in terms of these two. 
 
 (a) The standards of length differ in different countries. 
 Two systems of units are in use in Great Britain and the 
 United States, the British and the metric. The British, 
 being the system of every-day life, will be explained first; the 
 metric is explained in Chapter IX. All formulas will be so 
 expressed as to be applicable to any system of units. 
 
 The British standard unit of length is the imperial stand- 
 ard yard, which is " the straight line or distance between the 
 centres of the transverse lines in the two gold plugs or pins 
 in the bronze bar declared to be the imperial standard" 
 when the bar is at the temperature 62° F.* This bronze bar 
 is deposited in the standards department of the Board of 
 
 * The Weights and Measures Act 1878, 41 & 42 Vict. c. 49. 
 
§ 7] KIXEMATICAL UNITS. 5 
 
 Trade, London. One third part of the imperial yard is a 
 foot, and one twelfth of a foot is an inch. 
 
 The Tronghton 82 - inch brass scale obtained by Mr. 
 Hassler for the Coast Survey in 1814 was formerly accepted 
 as a standard of length of tlie United States, the dis- 
 tance between the twenty-seventh and sixty-third divisions 
 being at 62° F., the standard yard. Direct comparison 
 showed that this distance was equal to the imperial yard at 
 59.62° F., instead of at 62° F. By reason of its faulty con- 
 struction and the inferiority of its graduation the Tronghton 
 scale " is entirely unsuitable for a standard, and for a long 
 time it has been of historic interest only." 
 
 In 1866 the metric system of measures was made lawful 
 throughout the United States, and the yard as known in the 
 Office of Weights and Measures at Washington is defined by 
 the relation 
 
 ^ y^'^ = 3937 ^^*^'' 
 
 this being the ratio legalized by Congress. 
 
 In the absence of any material normal standard of the yard 
 the value of the yard is derived from the standard of the 
 meter, in accordance with the above relation. (Arts. 65, 
 280.) 
 
 (b) The standard unit of time throughout the world is 
 the mean solar day, which is the average of the intervals 
 between successive transits of the sun's centre across the 
 same meridian. Familiarly it is the time given by two revo- 
 lutions of the hour-hand of a common clock. The one- 
 twenty-fourth part of a day is an hour, the one-sixtieth part 
 of an hour a minute, and the one-sixtieth part of a minute 
 a second. (See Art. 48.) 
 
CHAPTER I. 
 KINEMATICS— MOTION. 
 
 8. One body is said to be in motion relative to another 
 body when it changes its position with respect to that other. 
 Change of position implies change of distance or of direction 
 or of both distance and direction. Also in this displacement 
 a certain time is taken, so that the elements of motion are 
 distance, direction, and time. Although we cannot assert 
 that any body in the universe is at rest absolutely, yet it is in 
 most cases sufficient to consider motion referred to some 
 body assumed as fixed. Thus the motion of a railroad train 
 may be referred to the roadbed and depots as fixed, though 
 they all have the motion of the earth. 
 
 The more general case will be discussed later on (Art. 42). 
 
 9. When all points of a body describe paths equal in mag- 
 nitude and parallel in direction, the motion is said to be a 
 motion of translation. The path of the body is therefore 
 determined when the path of any point of it is determined. 
 
 This conception of dealing with 
 a point greatly simplifies the 
 treatment of problems of trans- 
 lation. 
 
 If a point proceeds continually 
 in the same direction, the path 
 is a straight line; if the direction 
 is continually changing, the path 
 is a curve. The direction of 
 motion at any point A ot a> curvilinear path ABB being 
 the line joining that point to the consecutive point in the 
 
 6 
 
§ 11] UEASUHE OP CONSTANT VELOCITY. 7 
 
 path, will be in the direction of the tangent AA^ to the curve 
 at A. Similarly, at B the direction of motion is along the 
 tangent BB^ , so that in moving from A to B the direction of 
 motion has changed from AA^ to BB^ , or through the angle 
 A,CB,. 
 
 10. Velocity. — When a point changes its position, displace- 
 ment takes place along some continuous path and occupies a 
 certain time. 
 
 The rate at which a moving point changes its position is 
 called its velocity. Velocity is thus rate of growth of dis- 
 tance. 
 
 If the point moves so as to pass over equal distances in 
 equal intervals of time it is said to have a constant speed. If 
 the direction also is constant, the point is said to move with 
 constant velocity. 
 
 The extremity of the minute-hand of a clock moves in a 
 circular path over equal distances in equal intervals, but its 
 direction is continually changing. Its speed is therefore 
 constant, but its velocity is not. 
 
 The term speed thus denotes the magnitude of a velocity. 
 However, the term velocity itself is ordinarily used in the 
 sense of speed as well as in the strict sense of speed and di- 
 rection. In fact in the great majority of cases the direction 
 is assumed to be known, and the magnitude of the velocity is 
 the important thing. 
 
 11. Measure of Constant Velocity. — Velocity, if constant, is 
 measured by the number of units of distance described in 
 the unit of time. Thus if a distance s ft were described in t 
 sec, the velocity v would be expressed 
 
 V = s/t, 
 
 giving the number of feet described in one second. 
 
 To express the measure of a velocity it is necessary to 
 assume a unit of velocity. Since when 5 = 1 and ^ = 1 we 
 have V = 1, we must take for unit of velocity the velocity of 
 
8 KINEMATICS — MOTION. [§ 13 
 
 a point which describes unit distance in unit time. The 
 above velocity, which describes ?; ft in 1 sec, contains v 
 units and may be written 
 
 V ft per sec or v ft/sec. 
 
 Its numerical measure is v or s/t. 
 
 The unit of velocity of 1 ft per sec being expressed in 
 terms of the fundamental kinematical units of length and 
 time, is a derived unit. It has no single name generally 
 adopted. The names vel, veto, fas, etc., have been proposed. 
 
 12. Distance passed over.— If s ft denote the distance 
 passed over in t sec by a point moving with constant velocity, 
 then s/t is the distance passed over in one second and the 
 velocity v is found from 
 
 V = s/t. 
 
 Conversely, if v were known we should determine s from 
 the relation 
 
 s = vt. 
 
 This is otherwise evident. For if the point has a constant 
 velocity v ft/sec, it will describe ?; ft in 1 sec, 2v ft in 2 sec, 
 and so on. Hence if s ft is the distance described in t sec, 
 
 s = vt ft. 
 
 13. Graphical Representation. — A velocity having magni- 
 tude and direction as its components may be appropriately 
 represented by a straight line whose length on an assigned scale 
 
 will show the magnitude and 
 whose direction as indicated 
 by an arrow will show the 
 direction of the velocity. 
 
 We may hence give a geo- 
 metrical representation of uni- 
 form motion. Along a straight 
 o A B c D X line OX lay oft' equal dis- 
 
 tances OAj AB, ... to any convenient scale (as 10 sec = 1 
 
 vel.curve 
 
 tine 
 
§13] GRAPHICAL REPRESENTATIOK. 9 
 
 in) for as many seconds as the motion has continued. Let 
 the velocities at the points 0, A, B, . . . be represented by 
 V, Aa, ... to any scale (as 10 ft = 1 in). Now since the 
 velocity is constant, the lines OYy Aa, ... are equal to one 
 another, and the curve of velocity Yd is a straight line par- 
 allel to OX, the time line. Also 
 
 Distance passed over = OY x OD 
 = area YD, 
 
 That is, the number of feet in the distance described would 
 be represented by the number of square feet in the area of the 
 rectangle YD. 
 
 Ex. 1. A velocity of GO miles per hour is 88 feet per second. 
 For 
 
 60 m 60 X 5280 ft ^ _ ft _ _ ,. , 
 
 V = —^r- = r.r. V. ar^ = ^^ = 88 f t/sCC* 
 
 1 h 60 X 60 sec sec ^ 
 
 t-2. In a thunder-storm the clap was heard 6 seconds after the 
 flash was seen. Find the distance of the discharge, the ve- 
 locity of sound being 1100 ft/sec. Ans. 1.25 miles. 
 
 3. A passenger sitting in a railroad -car counts 45 telegraph 
 poles (distant 100 ft) passed in one minute: show that the 
 train is running at 50 miles an hour. 
 
 4. The minute-lnind of a clock is 7 in long: find the linear 
 velocity of its extremity. A7is. 1 1/900 in/sec. 
 
 ^ 5. Compare the velocities of two points one of which passes 
 over a ft in b sec, and the other b ft in a sec. An^t. a^ : b"^. 
 J^ 6. A knot being a sea-mile per liour, find how far apart the 
 knots on the log line of a vessel must be placed so that the 
 number of knots whicli pass over the talfrail in half a minute 
 may give the speed of the vessel in knots. Aiis. 50 ft 8 in. 
 
 [The value of the nautical mile adopted by the U. S. Coast 
 and Geodetic Survey is 6080.27 ft; by the English Hydro- 
 graphic Office, 6080 ft.] 
 
 7. A man a ft in height walks along a level street at the 
 rate of c miles an hour in a straight line from an electric light 
 b ft in height: find the velocity of the end of his shadow. 
 
 A71S. bc/{b — a) miles/hour. 
 / 8. The diameter of the earth being 8000 miles, show that 
 the velocity of a body at the equator due to the earth's rota- 
 tion is about 17.5 miles/minute. 
 
10 KINEMATICS— MOtlOK. [§ 14 
 
 9. Plot a velocity of 60 miles an hour on a scale of 
 
 11 ft/sec = 1 in. 
 
 [For plotting it is convenient to use cross-section paper.] 
 
 10. AB, AC are any two straight lines. Two bodies P, Q 
 move along these lines with uniform velocities ti, v. Find 
 the path of a body E which lies at the middle point of the 
 line PQ. 
 
 14. Variable Velocity. — If a point in motion does not pass 
 over equal distances in equal intervals of time the velocity is 
 said to be variable. 
 
 The expression s/t will then represent the average velocity 
 with which s ft are described in t sec. Thus a train in pass- 
 ing over 150 miles in 3 hours would have an average velocity 
 of 50 miles an hour, though its actual velocity at any instant 
 might often differ from this. 
 
 The actual velocity at any instant would be determined 
 by finding the limiting value of the average velocity for an 
 indefinitely small distance ^s described during an indefinitely 
 small time At^ including the instant. Thus 
 
 ,. ., As ds 
 
 V = limit -r; = -V7. 
 At dt 
 
 This being the general expression for the velocity at a point 
 includes the case of constant velocity. For 
 
 ds = V dt, 
 
 and the total distance passed over in the time t is found by 
 summing the distances ds, that is by the integration ( = 
 summation) of the expression v dt between the proper limits. 
 AVe have, if 2' is constant, 
 
 s = v I dt 
 
 = Vt + Cy 
 
 where c is the constant of integration. 
 
 If the time is measured from the starting-point of the 
 motion, or ^ = when 5 = 0, then c = 0, and 
 
 s = vt 
 as found above. 
 
16] 
 
 VARIABLE VELOCITY. 
 
 11 
 
 time 
 
 16. Variable motiou may be illustrated geometrically. For 
 the velocities Oy, Aa, Bb, ... at the times 0, OA, OBy . . . 
 if plotted and their extremities y, a,h . . . joined would form 
 a line yah . . . not parallel to the time line OX, as in the case 
 
 of uniform motion. This line 
 would be the velocity curve. 
 The points A, B,. . . may bo con- 
 ceived to be taken so close to- 
 gether that the.velocities between 
 may be considered uniform and 
 the figures Oa, Ah, . . . formed 
 to be rectangles. Each rectangle represents a distance passed 
 over, and therefore the total distance would be represented 
 by the sum of these rectangles, that is, by the area 01. 
 
 Conceive now a point to start from 0, and moving with 
 uniform velocity Oz, to describe the same distance s in the 
 time t as the point moving with the variable velocity. The 
 distance would be represented by the rectangle OM. But 
 with the variable velocities it is represented by the figure 01. 
 Hence the rectangle OM is equal to the figure 01, which can 
 happen only when Oz is the average (mean) of the values 
 Oy, Aa, Bb, . . . Hence, if we can find the average velocity 
 V, or the velocity of a point which, moving uniformly, passes 
 over the same distance in the same time t as the point moving 
 with variable velocity, we can find the distance s described, 
 from the equation 
 
 s = vt, 
 and conversely. 
 
 16. A velocity curve may also be constructed by laying 
 off OA, OB, . . . along 
 the line OX to represent 
 the distances passed over, 
 and Aa, Bb, ... at right 
 angles to OX to represent 
 the corresponding velocities. 
 The line through the points 
 a,h,.., would represent the curve of velocities. 
 
 distance 
 
12 KINEMATICS— MOTIOK. [§16 
 
 A familiar illustration is afforded by the motion of the 
 piston of a steam-engine. At the beginning and end of its 
 stroke the velocity is zero, at the middle of the stroke the 
 velocity is greatest, and it varies from this value to the end 
 values. The curve of velocity, if plotted, is found to be of a 
 form such as Oah ... X in the figure. 
 
 / Ex. 1. In September 1895 the Empire State Express made 
 the trip from New York to East Buffalo, 436^ miles, in 407 
 minutes: find the average speed. 
 
 Ans. 64.35 miles/hour, or 94.4 ft/sec. 
 / r 2. A train runs 29 miles for 2 hours, 30 miles for 3 hours, 
 Nand 32 miles for 1 hour: find its average velocity. 
 
 Ans. 44 ft/sec. 
 ^ 3. An engine makes 100 revolutions a minute. The stroke 
 is 2 ft: find the average piston speed. A7is. 400 ft/min. 
 [Piston speed is usually stated in feet per minute.] 
 
 4. A speed of 16.5 knots is equivalent to a speed of 19 miles 
 an hour. 
 
 5. In October 1894 the S.S. Lucania made the trip from 
 Queenstown to Sandy Hook Lightship, 2779 nautical miles, in 
 5 d 7 h 28 min: find the average speed in knots. 
 
 Ans. 21.80 knots. 
 
 V^e. Two trains running on parallel tracks pass at a certain 
 place in 6J seconds. If each train lias the same velocity, and 
 consists of 8 coaches of 52 ft 9 in in length, find the rate per 
 hour. A71S. 46 miles/hour. 
 
 ^7. In one of the shafts of the Comstock lode, Nevada, a 
 36-inch water-wheel is run under a head- of 2100 ft at 1150 
 revolutions per minute : find the peripheral velocity in ft/min. 
 (;r = 3|) Ans. 10842.9 ft/min. 
 
 8. The tunnel of the Cataract Construction Co. at Niagara 
 Falls is 6700 ft long. A chip thrown into the water at the 
 wheel-pit will pass out of the portal in 3.5 minutes. Show 
 that the water has a velocity of about 21 miles an hour. 
 
 9. The law of velocity being v = at -\- b, to find the law of 
 distance, distance being reckoned from the initial position. 
 
 An^. s --= af/% -f ft. 
 
 10. The law of velocity being ?; = c 4^5, to find the law of 
 distance. Ans. s = &f l^. 
 
 11. A point describes a diameter (2r) of a circle with a 
 velocity proportional to the corresponding ordinate, to find 
 the law of distance. Ans. 5 = r sin ct. 
 
§ 17] COMPOSITION OF VELOCITIES. 13 
 
 17. Composition of Velocities. — A body in motion has one 
 definite path and a definite velocity at each point of its path. 
 But tliis velocity may be due to several velocities separately 
 communicated to it. Thus a rowboat crosses a stream with 
 a velocity made up of the velocity of rowing and the velocity 
 of the current. 
 
 To the single velocity to which the path is due the name 
 of resultant velocity is given, and to the separate velocities 
 the name of component velocities. 
 
 (a) Velocities i7i the Same Line. — Suppose a point P in 
 
 motion along a straight line ABy as a ring sliding along a 
 
 rod, and that at the same time the rod 
 
 1 is moved in the same straight line AB. 
 
 The point thus receives two simultaneous velocities and has 
 a single velocity equal to the sum of the separate velocities 
 of ring and rod if in the same direction, and equal to their 
 difference if in the opposite direction. 
 
 Ex. A balloon rises with a velocity of 15 ft/sec for one 
 minute, it then descends at the rate of 5 ft/sec for 10 sec and 
 10 ft/sec for one minute: how far is it now above ground ? 
 
 Arts. 250 ft. 
 
 {h) Velocities Not in the Same Line. — Suppose the point P 
 has two simultaneous velocities w, v in fixed directions AC, 
 AB not in the same line. For ex- 
 ample, if a ring slide along a rod AC 
 with velocity w, and at the same time 
 AC moves parallel to itself with ve- 
 locity V along AB. 
 
 At the beginning of the motion let 
 the ring be at A. After t sec the rod has moved to BD and 
 the ring has moved a distance BD along the rod and is 
 found at D, Then 
 
 BD = ut, AB = vt, 
 and therefore 
 
 BD/AB = u/v. 
 
14 KINEMATICS— MOTION. [§ 18 
 
 But u and v are constant. Hence the ratio of BD io AB \& 
 constant and the ring describes a straight line AD passing 
 through A. 
 
 At the end of the first second let BD be in the position hd, 
 then Al) = u^ M=^, and the ring will be at d on the line AD. 
 Also 
 
 AD /Ad = BD/hd = iit/u = t/1, 
 
 or the distances passed over by the ring are proportional to 
 the times. Hence the distance passed over in one second, or 
 the resultant velocity of the ring, is constant. 
 
 Again, since Ad represents the resultant velocity, if we 
 draw da parallel to bA, then Aa is equal to id and Abda is a 
 parallelogram with sides Aa = u, Ab = v. Hence 
 . If a point possess two simultaneous velocities represented 
 by two atraiglit lines Aa, Ab, their resulta^it is represented 
 in magnitude and direction by the diagonal Ad of the paral- 
 lelogram constructed on Aa, Ab as adjacent sides. 
 
 This proposition is known as the parallelogram of veloci- 
 ties. 
 
 18. Motion in a curvilinear path is a result of the paral- 
 lelogram law on the assumption that each of the elements of 
 the path is rectilinear. (See Art. 9.) The direction at any 
 point of the path is along the tangent at that point. 
 
 Ex. 1. Show that the resultant of two velocities OA, OB is 
 represented by 20 D when D is the middle point of AB. 
 
 2. A BCD is a square and is the middle point of BC. 
 Find the resultant of the velocities AB, AO, AC. 
 
 Ans. ^AO. 
 
 3. Velocities of 8 ft/sec and 10 ft/sec are impressed upon 
 a particle. Find the greatest and least values of their result- 
 ant. A71S. 18 ft/sec; 2 ft/sec. 
 
 4. Show that the resultant velocity diminishes as the angle 
 between the directions of the two component velocities in- 
 creases. 
 
 5. The velocity along AB is 9 ft/sec and along AC 12 
 ft/sec. If the angle BAC = 90", find the resultant velocity. 
 
 [Draw AB, AC Sit right angles. Plot on a scale of 12 ft = 
 
§19] 
 
 COMPOSITION OF VELOCITIES. 
 
 15 
 
 1 in. Then AB = | in, AC =1 in. Complete the parallelo- 
 gram A BCD. Scale off AD. It measnres 1^ in. Hence 
 resultant velocity (= li X 12) = 15 ft/sec] 
 
 6. A particle has a velocity of 5 ft/sec and an equal 
 velocity at an angle of 120° is coniniunicated to it. Show 
 that the final velocity is 5 ft/sec and that we have an ex- 
 ample of change of direction without change of speed. 
 
 7. A ball moving with a velocity of 15 ft/sec is struck so 
 as to move off at right angles to its path with a velocity of 20 
 ft/sec: find the velocity given to it. Ans. 25 ft/sec. 
 
 8. Show that the direction of motion of any point B on 
 the rim of a wheel running 
 with velocity z; on a straight 
 track is perpendicular to ^^ 
 when A is the point of the 
 wheel in contact with the track 
 at the instant considered. 
 
 [For B has two velocities 
 each = V, one along the tangent 
 BE and the other along BD 
 
 parallel to the path of C. The resultant BG bisects 
 angle BBD. .'. ABG = 90°. 
 
 The solution also follows at once from Art. 222. 
 
 The path of ^ is a cycloid. The line BG is a tangent to 
 the path of B at the point B. Hence we have a simple 
 method of drawing a tangent to a cycloid at a given point — a 
 method first suggested by Boscovich and Roberval. 
 
 9. Find the magnitude of the resultant velocity in (8) when 
 6/ = 60°. Afis. V. 
 
 10. Draw a tangent to an ellipse at a point P by the 
 method of Roberval. (See Scientific American Supplement, 
 No. 879.) 
 
 [The tangent bisects the angle between the focal dis- 
 tances.] 
 
 19. It is often convenient to find the resultant by comput- 
 ing the diagonal of the parallelogram instead of finding it by 
 a geometrical construction. Thus in Ex. 5 preceding, since 
 BD = AC, we find AD by computing the hypotenuse of the 
 right triangle ABD. We have 
 
 AD' = AB'-{-BD\ 
 
16 KINEMATICS — MOTION. [§ 20 
 
 EMi* 
 
 lT 
 
 rcs- 
 
 -MOTION. 
 
 R' 
 
 = 
 
 12' 
 
 + 9» 
 
 R 
 
 = 
 
 15, 
 
 as before 
 
 or 
 
 and 
 
 Ex. 1. Solve Ex. 6, 7 Art. 18 by computation. 
 
 2. Two velocities of 3 ft/sec, 4 ft/sec are inclined at an 
 angle of 60°. Find their resultant. Ans. V^i ft/sec. 
 
 3. Two velocities u, v are inclined at 120°. Find their 
 resultant. Ans. Vu^ — uv -\- v". 
 
 Show that the resultant makes with v an angle 
 tan-i uV3/{2v - n). 
 
 4. Two velocities 15 ft/sec and 30 ft/sec have a resultant 
 of 39 ft/sec. Find the angle between them. Ans. 90°. 
 
 5. If velocities u, v are inclined at 120° and the resultant 
 is inclined to u at 90°, show that v = 2u. 
 
 6. A point has velocities 3, 3, 5 inclined at 120° to each 
 other. Find their resultant. Ans. 2. 
 
 What is its direction ? 
 
 7. Two velocities u, v are inclined at an a ngle 6°. Show that 
 resultant velocity = Vii" + 2uv cos + v". 
 
 20. Resolution of Velocities. — Conversely, a velocity v rep- 
 resented by AD may be broken up or resolved into two com- 
 
 c D ponents AB, AC, being the adjacent sides 
 
 ^^\ of the parallelogram constructed on AD 
 I ^r \ as diagonal. This may be done in an in- 
 
 "^ ^^ I definitely great number of ways, as an in- 
 
 A com'p. B definitely great number of parallelograms 
 
 may be constructed on the same diagonal. Other conditions 
 must be added to render the problem determinate. 
 
 Suppose (1) that the components of AD are to be at right 
 angles and the angle BAD (= 0) is given. Then CAD is 
 known, and the components AB, ^Ccan be plotted. They 
 are thus determined graphically. 
 
 Since BD = AC,\i is evident that the magnitudes of the 
 components of AD could be represented by the two sides AB, 
 BD of the triangle ABD. Their values may be found by 
 solving the triangle ABD trigonometrically. Thus 
 
 AB = AD cos 6^ = z; cos 6, 
 AC = AD &m 6 = v sin d. 
 
20] 
 
 RESOLUTION OF VELOCITIES. 
 
 17 
 
 These are the rectangular components of the velocity v. 
 In general, if a particle is at the point P, whose co-ordi- 
 nates are x, y at the time /, and if ds/dt is the velocity of the 
 
 particle in its path and dx/dt, dy/dt the 
 
 velocities of the particle parallel to the 
 
 axes of X and Y respectively, and d the 
 
 angle which the direction of motion, that 
 
 is, the tangent at x, y, makes with the axis 
 
 of X, then, since 
 
 n _dx _dx Ids 
 
 ~Ts~~dil Te 
 
 cos 
 
 we have 
 
 dx 
 Tt 
 
 w^ ds ^ 
 
 -77 = -71 cos 6. 
 
 dt 
 
 Similarly, 
 
 dy ds . ^ 
 dt dt 
 
 These are the rectangular components of the velocity 
 ds/dt. 
 
 Also, 
 
 (dxV (dyV _ (d^y 
 \dt) '^Kdtj ~\dtj' 
 
 \ (2) Let the components AB, AC ot AD be not at right 
 ' angles. 
 
 Denote angle BAD by a and CAD by /?. 
 ^^ . Then 
 
 ^^ ' AB/AD = 8mADB/8mABD 
 
 \P^ i = sin /?/sin (a + /3), 
 
 Also, 
 
 or AB = V sin /J/sin (a + /?). 
 
 AG = V sin a/sin {a -f /3), 
 
 Hence the components of a velocity v in two directions, 
 making given angles with it, are found. 
 
 Ex. 1. A ship is sailing N. 80° E. at 8 miles an hour. Find 
 its easterly velocity and its noj'therly velocity. 
 
 Ans, 4 m, 41^3 m an hour. 
 
18 KINEMATICS— MOTION". [§ 21 
 
 2. Find the vertical velocity of a train when moving up a 
 Ifo grade at 30 m an hour. Ans. 0.3 m an hour. 
 
 [A 1^ grade is a rise of 1 ft in a horizontal distance of 100 
 ft. This is the engineering rule.] 
 
 4. Show that the components of a velocity v in two direc- 
 tions, making angles of 30°, 60° with it on opposite sides, are 
 
 vV3/2, v/2. 
 
 21. Acceleration. — In variable motion the velocity changes 
 from point to point of the path. The rate of change of 
 velocity is called acceleration. 
 
 If a point moves in a straight line with uniform velocity, 
 there is no change of velocity, and therefore no acceleration. 
 If the velocity changes uniformly with the time [grows], the 
 acceleration is constant; if it does not change uniformly 
 with the time, it is variable. 
 
 22. Measure of Acceleration. — Acceleration, if constant, 
 is measured by the number of units of velocity added per 
 second. 
 
 The unit of acceleration is the acceleration of a point 
 moving with constant velocity which has its velocity changed 
 by the unit of velocity per second. Now unit velocity being 
 unit distance per second, unit acceleration must be unit dis- 
 tance-per-second per second. Thus if an engine 3 minutes 
 after starting has a velocity of 30 ft per sec, 
 
 the total change of velocity is 30 ft per sec; 
 the rate of change of velocity is 30 ft per sec in 3 min 
 
 or ^ ft per sec in 1 sec; 
 and the acceleration is said to be | ft per sec per sec. 
 
 In general, if a point moves with a velocity that changes 
 uniformly so as to change v ft per sec in i sec, then v/t is the 
 change of velocity in one second, and the acceleration a 
 would be expressed 
 
 a = v/t ft per sec per sec. 
 
 The unit acceleration of 1 ft per sec per sec has no single 
 name in general use; the names accel, eel, celo,faspy etc., have 
 
§ 24] MOTION UNDER CONSTANT ACCELERATION. 19 
 
 been proposed. We shall use the abbreviation ft/sec' for ft 
 per sec per sec. 
 
 Like the unit velocity, the unit acceleration is a derived 
 unit. 
 
 23. The nature of an acceleration may be indicated by the 
 signs + and — , as the change of velocity is in the same sense 
 or in the opposite sense to that of the original velocity. 
 Hence an acceleration is -{- if the velocity increases algebraic- 
 ally, and — if it decreases algebraically. To a negative accel- 
 eration the name retardation is sometimes given. 
 
 24. Motion under Constant Acceleration. — A point moves in 
 a straight line with a constant acceleration a. If u denotes 
 its initial velocity, it is required to find its velocity v at the 
 end of a time ^, and the distance .<? passed over in this time. 
 
 The change of velocity in the time t being v — u, the rate 
 of change is {v — u)/t. Hence 
 
 a = (v — u)/t, 
 and .*. Vj=. U:^^at, (1) 
 
 giving the velocity at the end of the time t. 
 
 Also the distance s passed over is (Art. 15) equal to the 
 product of the average velocity by the time t. But the 
 growth of velocity being constant from beginning to end of 
 the motion, the average velocity is equal to that of the point 
 half-way, that is 
 
 average vel = ^ (initial vel. -f- final vel.) 
 = i{u^ v). 
 
 Hence ^ s = \ (u -\- v) X t; (2) 
 
 or, putting v = u -\- at, 
 
 we have s = ut -\- jat*, (3) 
 
 giving the distance passed over in the time t. 
 
 The two equations (1) and (3) contain relations between the 
 quantities involved which are independent of one another. 
 Other relations may be deduced from them which are con- 
 
20 KINEMATICS — MOTION. [§ 25 
 
 venient, but which contain no new principle. Thus, elimi- 
 nating t, we find 
 
 V' = u' -^ 2uat + aH' from (1) 
 
 X^ =20' ^ 2a (ut + iat') 
 
 = u' + 2as from (3) 
 or V = Vti^ -\- 2as (4) 
 
 which may be compared with (1) 
 
 It is sometimes useful to write (4) in the form 
 
 vy2 - uy2 = as (5) 
 
 25. If the point had started from rest, then u = 0, and the 
 equations become 
 
 V = at, \v^ = as, 
 
 s = ^af, s = ivt. 
 
 These results may be deduced independently. For if the 
 point has a constant acceleration a ft/sec^ there will be added 
 a ft/sec to the velocity in each second of the motion. In t 
 sec there will be added at ft/sec. Hence, if v denotes the 
 velocity acquired from rest in t sec, v will be equal to this 
 gain, or 
 
 v = at. . . , (1) 
 
 Also s = average velocity X time 
 = iv X t 
 = iat X t 
 = iat\ (2) 
 
 giving the distance passed over. 
 Eliminating t between (1) and (2), 
 
 iv' = as, (3) 
 
 as before. 
 
 26. Variable Acceleration. — If the change of velocity is 
 variable, then 
 
 a = (v — ti)/t 
 
 would give the average acceleration with which the velocity 
 changed from w to v in time t. 
 
§ 27] VARIABLE ACCELERATION. 21 
 
 The actual acceleration at any instant is the limiting value 
 of the average acceleration for an indefinitely small change of 
 velocity ^v during an indefinitely small time ^t, including 
 the instant. We have 
 
 ^v dv , . 
 
 a = lumt^^ = j^ 0) 
 
 This may be put in two other forms,* both of which are of 
 frequent application. Thus 
 
 (2) 
 
 dv ds dv dv 
 ^~di~TtTs~^d~s' ' ' 
 
 . . . 
 
 dv Adt) d's 
 ^~ dt~ dt ~ df ' • 
 
 ... 
 
 27. The formulas found in Art. 24 follow at once from these 
 equations. Thus (a) let a particle start from 
 
 a point with a velocity u : it is required to —^ — ? j • 
 
 find its velocity v and distance OP from at 
 
 the end of a time t, the acceleration a being constant. Let 
 
 OP = s; then 
 
 d's 
 
 dt 
 
 Integrating, 
 
 ds 
 
 ji = at + c, 
 
 c being the constant of integration. But when ^ = 0, ds/dt 
 ov V = u'y hence c — u, and 
 
 ds 
 
 -r. = ot + n, 
 
 which gives the velocity at the end of the time t. 
 Integrating a second time, 
 
 ^ s = iat^ -\- ut, 
 * First published by Varignon in 1700. 
 
22 
 
 KIKEMATICS— MOTION. 
 
 [§28 
 
 since when t = 0, s = 0, and therefore the constant of inte- 
 gration is 0. 
 
 (b) Let the particle start from with velocity u, and under 
 a constant acceleration a: required to find its velocity v after 
 passing over a distance s. Here 
 
 dv 
 
 a. 
 
 Integrating, 
 
 v'^/2 = as -\- c. 
 
 But when s = 0, v = u; 
 
 vy2 = as-{- uy2, 
 or v^/2 — u^/2 = as, 
 
 as found before. 
 
 28. Graphical Representation. — We have seen how to con- 
 struct the curve of velocity of a point in motion, either by 
 taking the times as abscissas or the distances as abscissas. 
 We proceed now to show how to construct the curve of accel- 
 eration when the curve of velocity has been plotted. 
 
 Take the case of a motion in which the times are plotted 
 as abscissas. Let the velocity change uniformly from ic to v 
 
 in the time t. The rate of 
 change of velocity being con- 
 stant, the velocity curve ah 
 . . . becomes a straight line. 
 The acceleration a=(v—ti)/t 
 is represented in the figure 
 by tan 6, that is, by the tan- 
 gent of inclination of the 
 line at . . . to OX, Hence, if the distances OA, AB, . . . 
 represent one second, the accelerations measured on the 
 velocity scale would be represented from second to second by 
 aa^ , bh^f . . . all of which are equal to one another. If, 
 therefore, a line be drawn parallel to OX, and at a distance 
 a from it measured on the velocity scale, it will represent the 
 curve of acceleration. 
 
dv 
 
 time 
 
 § 28] GRAPHICAL REPRESENTATION. 23 
 
 If the rate of velocity is not constant, so that the curve of 
 velocity is curvilinear, then since a = dv/diy the acceleration 
 at a point P (whose co-ordinates 
 are v, t) is represented by tan 6 
 when B is the inclination of the 
 tangent at P to OX. Hence, to | 
 plot the acceleration curve, draw 1 
 tangents to the velocity curve 
 from second to second, and lay off 
 as the ordinates the rise or fall of 
 the tangent measured on the velocity scale. The curve may 
 thus be plotted from point to point. 
 
 ^ Ex. 1. In 5 seconds the velocity of a point changes from 
 200 ft/sec to 100 ft/sec. Find the acceleration. 
 
 Ans. « = — 20 ft/sec'. 
 ■i 2. The velocity of a point changes from 20 ft/sec to 10 
 ft/sec in passing over 75 ft. Find the acceleration and time 
 of motion. Ann. a = — 2 ft/sec% t = b sec. 
 
 Draw a figure illustrating the motion. 
 f 3. A point starts from rest. Show that numerically the 
 acceleration, if constant, is equal to twice the distance de- 
 scribed in the first second. 
 
 4. A point moving with constant acceleration describes 
 160 ft in the first two seconds of its motion, and 50 ft in the 
 next second. When will it come to rest? When has it a 
 velocity of 20 ft/sec ? When of - 20 ft/sec ? 
 
 Ans. 5 sec; 4 sec; 6 sec. 
 y 5. A point starts with a velocity u and under a constant 
 acceleration — a. Show that it will come to rest in a time 
 u/a, after describing a distance 2c^/2a. 
 
 * 6. The distances described by a point uniformly accel- 
 erated in the first, second, . . . seconds of its motion form an 
 arithmetic progression whose common difference is a. 
 
 7. The velocity of a railway train increases uniformly for 
 the first 3 minutes after starting, and during this time the 
 train travels 1 mile. Find the velocity acquired. 
 
 A71S. 40 miles/hour. 
 > 8. It is observed that a body describes 40, 76, and 112 ft 
 in successive half seconds. Is the motion consistent with 
 uniform acceleration ? 
 
24 KINEMATICS — MOTION". [§ 29 
 
 9. A train uniformly accelerated takes 5 minutes to run 
 the first mile. How long will it take to run the next mile ? 
 
 Ans. 2.1 min, nearly. 
 
 10. In passing between two stations 6 miles apart a train 
 is first constantly accelerated and then equally retarded. 
 The time taken is 12 min. Find the greatest velocity at- 
 tained. ■ Ans. GO miles/hour. 
 
 ^ 11. In the Westinghouse air-brake trials (1887) on the 
 P. R. R. a train of 50 freight cars running at 36 miles an 
 hour was stopped in 593.5 ft. Find the acceleration of the 
 brake. Ans. a = — 2.3 ft/sec\ 
 
 29. Acceleration in Curvilinear Motion. — When a point 
 moves in a curve, it must have an acceleration not in the 
 direction of motion. For if the acceleration were in the 
 
 direction of motion, the path 
 would be a straight line. 
 
 Let v^ , v^ denote the velocities 
 at two points ^, ^ of a curve, 
 and t the time occupied in pass- 
 ing from A to B. The direc- 
 tion of the velocity at A and B 
 is along the tangent at A and B, 
 From any point draw Oa 
 parallel to the tangent at A to represent v^ in magnitude and 
 direction, and 01) parallel to the tangent at B to represent v,. 
 Complete the parallelogram OaM. 
 
 Then the velocities Oa and Ocl or ah are equivalent to 
 the velocity Ob (Art. 17). Hence ab represents the velocity 
 which must be combined with Oa to produce Ob, and there- 
 fore ab is the change of velocity in the time t in both magni- 
 tude and direction. The magnitude of the rate of change 
 would be ab/t and its sense is indicated by the direction 
 of ab» 
 
 In its most general signification, therefore, acceleration 
 involves change in both magnitude and direction. 
 
 If Oa and Ob were in the same straight line, acceleration 
 would be, strictly, rate of change of speed. The term Quick- 
 
§ 30] COMPOSITION AND RESOLUTION OF ACCELERATIONS. 25 
 
 ening has been suggested for this. But commonly the term 
 acceleration is used in a similar double sense as velocity 
 (Art. 10), and has been so used in the sections preceding. 
 
 30. Composition and Resolution of Accelerations. — An ac- 
 celeration, like a velocity, being a quantity which has magni- 
 tude and direction, may be represented by a straight line. 
 
 The same method of reasoning as 
 
 in Art. 17 may be applied to the com- ^^- * 7 
 
 bination of two accelerations. The / ^^^^ / 
 
 statement — called the par^elogram / ^^^ ''' 
 
 of accelerations — is this: / --^^' 
 
 c 
 
 If a point possess two simvltaneous 
 accelerations represeiited by two straight lines AB, AC, their 
 resultant is represented in magnitude and direction by the 
 diagonal AD of the parallelogram ABDC. 
 
 Conversely, an acceleration may be resolved into its com- 
 ponents after the manner of Art. 20. 
 
 In general an acceleration a at any point x, y of a curvi- 
 linear path may be resolved into two components in given 
 directions. The directions usually taken are along tlie tan- 
 gent and normal at the point, and in directions parallel to 
 rectangular axes OX, OY. The latter is in general the more 
 convenient. 
 
 With the notation of Art. 20 the components parallel to the 
 axes being the rates of increase of dx/dt and of dy/dt would 
 be denoted by d'x/df and d'^y/df, respectively. The resultant 
 acceleration a is the sq. root of the sum of the squares of these 
 components. 
 
 To find the component acceleration «, along the tangent at 
 P, we have 
 
 d'^x n ^ d'^y ' a 
 
 _ d^x dx d^y dy 
 ~drds '^Wds 
 
 = ^ [by differentiation of ^ = ^ + ^ J, 
 
 or, tlie tangential component of the acceleration is the same 
 as for rectilinear motion. 
 
26 KINEMATICS— MOTIOK. [§ 31 
 
 The method of resolving parallel to fixed axes the accelera- 
 tion of a particle and replacing the curvilinear motion by the 
 formulas for rectilinear motion was given by Maclaurin in 
 1742. This marks a great advance in algebraical mechanics, 
 or, as it is commonly called, analytical mechanics. 
 
 31. Notice that although the same law of combination 
 applies to velocities and accelerations, yet a velocity and an 
 acceleration cannot be combined directly. An acceleration 
 continuing for a certain time produces a change of velocity, 
 and this velocity combined with the initial velocity by the 
 parallelogram of velocities gives the final velocity. Or an 
 acceleration continuing for a certain time produces a certain 
 displacement, and this may be combined with the displace- 
 ment produced by the velocity continued for a certain time. 
 See for illustration Art. 96. 
 
 Ex. A particle moves so that the components of its velocity 
 parallel to the co-ordinate axes vary as the corresponding co- 
 ordinates X, ?/, respectively. Show that the axial accelerations 
 also vary as the co-ordinates. 
 
 [j^ = ax; then ^ = „=^, etc.] 
 
 APPLICATIONS AKD ILLUSTRATIONS. 
 
 The general principles developed in the preceding pages 
 will now be applied to various important special cases. 
 
 32. (1) Circular Motion. — When a point describes a circle 
 of radius r with constant velocity v, it experiences a con- 
 stant acceleration v^/r directed 
 towards the center of the circle. 
 For suppose a point to move 
 uniformly along a side AB of a 
 regular polygon with velocity v 
 in time t, and that when it ar- 
 rives at the angle B it receives a 
 velocity u which causes it to 
 
 ^ move with the same velocity v 
 
 along BC. Similarly, at C a velocity u which causes it to 
 move along CD, and so on. 
 
§ 32] CIRCULAR MOTION. 27 
 
 From any point P draw PQ parallel to AB and = v; PR 
 parallel to ^Cand = v\ complete the parallelogram PQRS. 
 
 Then (Art. 17) PS or QR represents the change of velocity 
 u at B. Also, if is the center of the polygon, then from 
 the geometry of the figure BO is parallel to QR, or the change 
 of velocity u at B is directed along the radius BO. 
 
 Similarly, the change of velocity u at C is directed along 
 the radius CO, and so on. 
 
 To find the magnitude of the change of velocity per second: 
 
 The triangles PQR, OBC are similar. 
 
 ... QR/PQ = BC/OB, 
 
 or u/v = vt/r, 
 
 or ti/t = v^/r, 
 
 giving the rate of change of velocity towards 0. 
 
 Now when the number of sides is indefinitely increased, the 
 polygon becomes a circle, and the constant velocity v in the 
 polygon becomes constant velocity in the circle. 
 
 Tlie changes of velocity which in the polygon took place 
 at the angles, and therefore occurred at intervals, take place 
 from point to point and become continuous in the circle. 
 The rate of change ti/i, that is, the acceleration, is equal to 
 the same quantity v^/r, and is always directed towards tlie 
 center 0, which proves the proposition. 
 
 The velocity being constant in magnitude, the acceleration 
 is expended in changing the direction of motion. Being 
 always directed towards the center of the circular path, it is 
 said to be centripetal. 
 
 If T is the time in which the circular path is described, 
 then 
 
 
 vT=27rr. 
 
 Also, 
 
 a = V* /r. 
 
 Hence 
 
 a = i7r'r/T\ 
 
28 
 
 KIK EMATICS— MOTION. 
 
 [§33 
 
 33. On account of the importance of this proposition 
 another demonstration will be given. 
 
 A point describes a circle of radius r with 
 uniform velocity v to find the resultant ac- 
 celeration at any position in its path. 
 
 Refer the circle to rectangular axes OX, 
 OY and let x, y be the co-ordinates of a 
 point P in the circle. Then 
 
 x' + f = r' (1) 
 
 Hence 
 
 dx , dy 
 
 dt 
 
 dt 
 
 0. 
 
 (2) 
 
 But by hypothesis, 
 
 ©■+(!)■='• (') 
 
 Solving (2) and (3), we find 
 
 dx 
 dt 
 
 vy_ 
 r ' 
 
 dy 
 
 vx 
 r 
 
 Again, 
 
 d'^x 
 df 
 
 V dy 
 
 ^, 
 
 d^y _ vdx _ 
 lF~~rdt ~ ' 
 
 giving the component accelerations, 
 the resultant accel. = 
 
 V 
 
 yy 
 
 r 
 Hence 
 
 V' 
 
 S/x'-\-f 
 
 = - v'/r, 
 as before. (For a third proof see art. 109.) 
 
 Ex. 1. Find the centripetal acceleration of a point which 
 moves in a circle of 5 ft diameter with a velocity of 5 ft/sec. 
 
 Ans. 10 ft/sec^ 
 
 2. Explain how it is that although the particle constantly 
 gains velocity along the radius it never possesses any such 
 velocity. 
 
§34] 
 
 MOTIOI^ OF OSCILLATION. 
 
 29 
 
 [" As fast as velocity along the radius is generated, so fast 
 does the direction of the radius change, in the same way that 
 a promise for to-morrow need never be fulfilled because to- 
 morrow never comes. "J 
 
 3. A man falls overboard from a vessel running at 20 miles 
 an hour. If the masts are 88 ft above the water, find when 
 they will cease to be visible. A7is. 204/3 min. 
 
 4. The earth if suddenly stopped in its orbit would fall to 
 the sun about one ninth of an inch in the first second. 
 
 f^b. A point describes a circle of radius r so that its pro- 
 jection on a diameter describes the diameter with uniform 
 velocity v: to find the acceleration perpendicular to diam- 
 eter. A71S. — r^v^ iii^. 
 
 1^6. A point describes a hyperbola xij = & with a uniform 
 velocity v\ to find the co*nponent accelerations parallel to 
 the asymptotes. 
 
 Ans. 2c*v'^/r'^Xy 2c*v'^/r'^y when r' = a;' + ^'. 
 
 34. (2) Motion of Oscillation. — Suppose that while a par- 
 ticle P moves in a circumference of radius r with uniform 
 velocity v, another particle Jf moves along any diameter AA^ 
 in such a way that PJf is always perpendicular to AA^\ then 
 as P describes the equal arcs AB, BC, . . . M describes the 
 
 B^ 
 
 
 D? 
 
 ' \ 
 
 f 
 
 i 
 
 ° I 
 
 \ 
 
 J 
 
 '•^^=: 
 
 v^ 
 
 distances Ab^ he, . . ., until A^ is reached. As P continues, 
 M returns along the diameter A^A to A. Thus while P 
 makes a complete revolution, M oscillates back and forth 
 along the diameter AOA^. The motion of J[f as it oscillates 
 along the line AOA^ about as a centre is called a simple 
 harmonic motion (S.H.M.). Unlike a motion of translation, 
 which is constant in direction, a motion of oscillation is 
 alternate in direction. 
 
30 KINEMATICS— MOTION. [§ 35 
 
 Tlie name harmonic is given to such oscillatory motion 
 because this motion is characteristic of the vibrations of 
 sounding bodies, as tuning-forks, piano-wires, etc. 
 
 S.H.M. is the simplest form of harmonic vibration, and is 
 of great importance in mathematical physics. 
 
 The time of oscillating from ^ to ^^ and back to A, being 
 the same as that required for passing once round the circum- 
 ference in the corresponding circular motion, is 27tr/v, and 
 is called the period of the S.H.M. Denote it by T, 
 
 The distance OM of the moving particle M at any time 
 from its mean position is called the displacement at that 
 time. Denote it by y. 
 
 The distance OA or OB of the extreme position of M from 
 the mean position is called the amplitude. It is the radius 
 of the circle of reference and may be denoted by r. 
 
 The fraction of the period of vibration which has elapsed 
 in the passage from ^ to if is called the phase of if at the 
 time considered. It is evidently the ratio of the angle FOA 
 to 360° or 6/27r if is the circular measure of the angle 
 POA. 
 
 The phase of vibration at is diiferent by 1/4 from that at 
 A, and the phase at^, differs by 1/2 from that at A, or A and 
 J J are said to be in opposite phases. 
 
 35. Let the time be counted from the instant when P is at 
 A. Then if 6 denotes the angle POA described iu t sec and 
 T the period, ^^^^ /NT -^ ^ 
 
 t:T=6:27r, 
 or 6 = 27rt/T 
 
 — Got, 
 
 where go = 27r/T is the angle described in one second ex- 
 pressed in circular measure. Hence 
 
 y = OM = OP cos = r cos cot, 
 
 giving the displacement in t seconds. 
 
§ 37] MOTION OF OSCILLATION. 31 
 
 If from P a perpendicular PN is let fall on OX, then 
 
 X = ON = r sin cot = r cos {odI — 2;r/4), 
 
 or the motion of ^ is a S.H.M. of the same amplitude and 
 period as My but differing 1/4 in phase. 
 
 X 36. The acceleration « of if is the component of the ac- 
 celeration of P in the direction AA^, But the acceleration 
 of P is along PO, and is equal to v'/r (Art. 32). Hence 
 
 n=z- cos POA 
 
 V r 
 
 v^ y 
 
 •= - X- 
 
 r r 
 
 = v'y/r\ 
 
 Also, since the circular path is described in 7^ seconds with 
 uniform velocity v, 
 
 But 
 
 or, the circular measure gd of the angle described in one sec- 
 ond being constant, the acceleration of M varies as the dis- 
 placement ;/ from 0. (See Art. 101.) 
 37. Again, since 
 
 T= In/ 00, 
 
 if G? is a constant quantity, it follows that the period of the 
 S.H.M. is independent of the amplitude 1\ In other words, 
 at whatever distance from the particle M is started it will 
 return to its starting-point in the same time. This property 
 of oscillating in the same time, whatever the amplitude, is 
 called isochronism, and the vibrations of M are said to be 
 isochronous. 
 
 
 vT= 27rr, 
 
 
 GoT=27r; 
 
 
 \ V = oar, 
 
 and hence 
 
 «»=_^!^ 
 
32 
 
 KINEMATICS — MOTION. 
 
 [§38 
 
 We may write 
 
 In/ 00 
 
 _ i /displacement at any point 
 ~ ^ acceleration at that point ' 
 
 a convenient formula for finding the period. 
 
 38. Compositio7i of 8.H.M/s. — Simple harmonic motions 
 may be combined into one resultant motion by the parallelo- 
 gram law as with motions of translation. For example, to 
 find the resultant of two equal S.H.M.'s in directions at right 
 angles to each other and in the same phase. 
 
 1. Graphically. Let both motions start from 0, one along 
 OX and the other along 01^ at right angles 
 to OX. 
 
 With the equal amplitudes OX, OF as 
 radii describe the semicircles ABX, BAY, 
 The periods being the same, let the semi- 
 circles be divided into the same number of 
 equal parts, say eight. The harmonic in- 
 tervals Oa, ab, . . . along OX, 01^ are equal. 
 
 Since the motions both start from 0, the actual position of 
 the particle at the end of the first interval will be at a^ , at the 
 end of the second interval at &, , and so on. Hence the result- 
 ant motion will be along the straight line OaJ)^c^ . . . and 
 will be a S.H.M. Since 
 
 Oa, : Ob, 
 
 = Oa : Ob : 
 
 the amplitude is evidently r V2. 
 
 2. Analytically. The equations to the two component 
 S.H.M.'s are 
 
 x = r COB not', 
 y — r GOB Got, 
 
§39] 
 
 HARMONIC CURVES. 
 
 33 
 
 Eliminating ty the equation to the resultant path of the par- 
 ticle is 
 
 a straight line inclined at 45° to the axes of X and Y, as 
 
 before. 
 
 39. Harmonic Curve. — When a S.H.M. along AA^ is com- 
 bined with a uniform motion at right angles to AA^, the 
 resultant path is called the harmonic curve. 
 
 To find this curve (1) graphically. Let 01, 1 2, . . . rep- 
 resent the distances passed over in equal time-intervals, and 
 Ah, lOf . . . the distances passed over in the same intervals 
 
 by the vibrating particle. At the end of the first interval 
 the particle will be at 6', at the end of the second interval at 
 c', and so on. Having reached e' it has made half a vibra- 
 tion, and at k' a complete vibration. The curve Ah'c'd'e' 
 . . . ^•' is the harmonic curve. 
 
 (2) Analytically. The equation of a S.H.M. along AA^ is 
 
 y = r cos cot, 
 and of uniform motion in a straight line 
 s = vt, 
 
 where v is the velocity of motion and s the displacement in 
 the time t. 
 
 To find the locus eliminate t, and 
 y =^ r cos oDs/v 
 = r cos 27rs/vT 
 = r cos 27rs/\, 
 
34 KINEMATICS — MOTION. [§ 40 
 
 if we place vT =X, the distance traveled in the period 1 
 in the straight path with velocity v. 
 Now when 
 
 5 = 0, y = r, _ 
 
 s = A/8, y = r/V2, 
 
 s = A/4, y = 0, 
 
 s = A/2, y=-r, 
 
 s = 3A/4, y = 0, 
 
 s = X, y = r, 
 
 and the vibration is completed. 
 
 The curve now repeats itself, and the periodicity of the 
 motion of the particle is shown. The time of a period is 
 27r/GD, as above. 
 
 40. Now instead of the single particle which has the two 
 motions which combined form the harmonic curve conceive 
 particles placed along the line OX extending from to 8, 
 and made to oscillate along lines perpendicular to OX, with 
 S.H.M.'s differing uniformly in phase. In the figure the 
 S.H.M/s differ 1/8 in phase. As before, call the period T. 
 
 At the end of a certain time the particle 8 is in the highest 
 position Jc' ; the other particles, being times T/8, 2 T/8, . . . 
 behind, will be in the positions A', ^', . . .shown in the figure. 
 The positions will plainly lie in the harmonic curve. 
 
 At the end of the next interval T/8, 7i' will be the highest 
 point, d' the lowest, and at the end of the period, that is, at 
 the end of a complete oscillation, the highest point will have 
 traveled to A. 
 
 The form of the curve assumed by the vibrating particles 
 is called a wave, and the distance passed over by the wave 
 during a complete oscillation is the wave-length, that is, Ak' 
 or its equal 08. The wave-length is thus the distance be- 
 tween two points in corresponding positions on the harmonic 
 curve. 
 
 If the oscillations had taken place along OX, we should 
 have had longitudinal instead of transverse waves. 
 
 41. Harmonic vibrations in the same line may be combined 
 
§41] 
 
 HARMONIC CURVES. 
 
 35 
 
 by first combining each with a perpendicular translation and 
 
 plotting the harmonic curves. 
 
 Then, since the motions are 
 
 in the same straight line, the 
 
 compound harmonic curve 
 
 will result from plotting the 
 
 curve whose ordinates are 
 
 equal to the algebraic sum of 
 
 the ordinates of the simple 
 
 curves. 
 
 Thus to compound two 
 equal S.H.M.'s in the same 
 line and having the same amplitude the resultant curve will 
 be a curve of twice the amplitude. 
 
 If the curves are of opposite phase, the resultant curve is a 
 straight line, or the motions destroy one another. This illus- 
 trates the interference of two waves of the same length, but 
 of opposite phase. 
 
 Ex. 1. A particle revolving uniformly in a circle will to an 
 eye in the plane of the circle appear to oscillate along a di- 
 ameter. 
 
 [The motion of Jupiter's moons as seen from the earth 
 appears to be approximately a S.H.M.] 
 
 2. Show that the upward and downward motion of the 
 connecting-rod of a locomotive is a S.H.M. 
 
 3. In t^he cross-head of an engine a slot is cut perpen- 
 dicular to the direction of the stroke. If the crank-arm 
 revolving uniformly works with one extremity C in the slot, 
 the motion of C is a S.H.M. 
 
 4. If the particle M (Art. 34) be projected on any line at Q, 
 the motion of Q will be a S.H.M. of the same period and 
 phase as M, but of amplitude the projection of the ampli- 
 tude of M. 
 
 [This may be illustrated by viewing the S.H.M. of M in 
 figure on p. 29 obliquely.] 
 
 5. Find the average velocity of the point J/ as it oscillates 
 from A to A^ and back again to Ay if the greatest velocity at- 
 tained is 1 ft/sec. A71S, 27t-^ ft./sec. 
 
36 KINEMATICS — MOTION. [§ 42 
 
 6. The velocity of M along the diameter AA^ is v sin cot, 
 the velocity of P in the circle of reference being v. 
 
 7. By taking the first and second differentials of the equa- 
 tion y =^ r cos oot show that 
 
 vel of Jf = — roD sin cot ■= — oox; 
 accel of if = — roo'^ cos cot = — ao^y; 
 
 X, y being the co-ordinates of P in the circle of reference. 
 
 8. At what point is the velocity of if a maximum, and what 
 is its value ? Ans. v at the center 0. 
 
 When is the acceleration greatest, and when least ? 
 
 9. Combine graphically and analytically two S.H.M.'s in 
 same line and of equal amplitude, period, and phase. 
 
 Ans. A S.H.M. of equal period and of twice the amplitude. 
 
 10. Compound two S.H.M. 's in tlie same line of the same 
 amplitude, but differing 1/4 in phase, into a S.H.M. in the 
 same line and of the same period. 
 
 [^ = r cos Got -\- r co^{oDt -\- 27t/4:) = r V'Z cos {aot -{- 7r/4)]. 
 
 11. Find the path of a point which has two equal S.H.M.^s 
 in directions at right angles to each other and differing 1/2 
 in phase. Ajis. A straight line. 
 
 12. If in (11) the two S.H.M.'s differ 1/4 in phase and also 
 in amplitude, find the path. Ans. An ellipse. 
 
 13. Uniform circular motion is equivalent to two S.H.M.'s 
 at right angles to each other of equal amplitude and period, 
 but differing 1/4 in phase. 
 
 14. A particle has two S.H.M.'s of the same amplitude and 
 phase and in directions at right angles, the periods being as 
 1 to 2. Find the path. Ans. A parabola. 
 
 Eliminate t between x = r cos cot and y = r j^os 2ojt. 
 Plot the curve after the manner of Art. 38. 
 
 This forms one of a series of curves known as Lissajous' 
 curves. Others may be formed by changing the ratio of the 
 periods. 
 
 [For a mechanical method of producing these curves see 
 Art. 118.] 
 
 15. If in Ex. 14 the phase differs by 1/4, find the equation 
 
 of the path. Ans. r'y' = 4:x\r' - x'). >P 
 
 '^42. Relative Motion. — The motion of a point P has been de- ' 
 fined by its change of position with reference to another point 
 regarded as fixed. This gave the alsolute motion of P, 
 But if the point is also in motion, or has an absolute motion 
 
§ 43] RELATIVE MOTION. 37 
 
 with respect to a third point, the motion of P is no longer 
 said to be absolute, but relative. This is really the case of 
 bodies in nature, as no point in space is known to be fixed 
 absolutely. Still, for the purpose considered, an assumed 
 point may be regarded as fixed, and in this sense motion is 
 said to be absolute. 
 
 The problem of relative motion is really an example of the 
 composition of motions. Suppose, for example, a point A to 
 move with an absolute velocity u, and B with an absolute 
 velocity v, both being referred to the same fixed point 0, and 
 in the same straight line; it is required to find the velocity 
 of A relative to B. Conceive both A and B to move in a 
 medium which itself moves with a velocity v, but in the oppo- 
 site direction to the motion of B. Then B is at rest with 
 reference to the fixed point 0. But as the motion of the 
 medium affects A and B alike, their relative motion is un- 
 changed. Hence, as a velocity v has been imparted to J, the 
 velocity of A relative to B will be w — t; if both were origi- 
 nally moving in the same direction and ic -\- v \i in opposite 
 directions. 
 
 As an illustration, take two men A and B walking on a 
 boat's deck from bow to stern, and that the velocity of the 
 boat is equal to the velocity of B. Then B is at rest relative 
 to the shore, and the motion of A relative to B is the same as 
 if the boat were at rest. 
 
 43. Consider next when the velocities w, v of the two 
 points A, B are not in the same straight line. Suppose the 
 lines OXy OF to represent these ve- 
 locities in magnitude and direction. V" ~~~P^\ 
 Let a velocity v in the direction YO A ^^ \ 
 
 and represented by OZ be imparted V/__^w \ 
 
 to both A and B. The relative mo- ° \^ ^-' 
 
 tion of A and B is unchanged, and vS^ ^.-^^ 
 
 the point B is now at rest. The V-'^^ 
 
 velocity of A is the resultant of the 
 
 velocities OZ, OX, that is, is equal to the diagonal* OPT, 
 which therefore represents the velocity of A relative to B. 
 
38 KINEMATICS — MOTION. [§ 43 
 
 The three velocities are represented by the sides of the tri- 
 angle OXW. Hence, if in a triangle one side OX represent 
 the velocity of A, XW a velocity equal to and opposite that 
 of B, and OX, XW are in the same sense around the tri- 
 angle, the third side W taken in the opposite sense around 
 the triangle will represent the velocity of A relative to B. 
 
 So, too, if a point X had a velocity XO relative to Z, and 
 at the same time Z had a velocity ZO relative to Y, then evi- 
 dently XY is the velocity of X relative to Y; that is. 
 
 If two sides of a triangle XO, OY taken the same way 
 round represent the velocity of X relative to Z and of Z rela- 
 tive to Y, the third side XY will represent the velocity of X 
 relative to Y. 
 
 Corresponding propositions hold for accelerations and for 
 linear displacements. 
 
 Any mechanism may be employed to illustrate relative 
 motion by putting a sheet of paper on one of its moving 
 pieces, and a pencil on another moving piece, when the curve 
 traced by the pencil on the paper will represent the relative 
 motion of the two pieces. 
 
 Ex. 1. Two trains A and B are running on parallel tracks 
 at 20 and 50 m/h respectively. Find the velocity of A rela- 
 tive to B, Ans. — 30 ra/h or — 70 m/h. 
 3. A steamer is sailing north at 16 miles an hour in an east 
 wind blowing 12 miles an hour. Find the apparent 
 direction of the wind to a passenger on the steamer. 
 [Let a velocity of 16 miles an hour south be im- 
 parted to steamer and wind. The relative motion 
 is unchanged. The steamer is at rest and the ve- 
 locity of the wind is composed of 12 miles an hour 
 "ifi west and 16 miles an hour south. .*. apparent di- 
 rection of wind = tan~^ 12/16 or tan"^ 3/4 east of 
 ' " north.] 
 
 3. In Ex. 2 find the direction of the vessel's smoke. 
 f 4. Two vessels start at the same time from the same har- 
 bor, one sailing east at 12 miles an hour, the other south at 
 9 miles an hour. Find the velocity of one relative to the 
 other. A?is. 15 miles an hour. 
 
 5. Two railroad tracks intersect at 60° and two trains 
 
§ 43] HELATIVE MOTIOK. 39 
 
 start at the same instant from the junction at 30 miles an 
 hour each. Find their relative velocity in magnitude and 
 direction. 
 
 vl 6. Two bodies A, B move with velocities w, v inclined at 
 an angle 0, Show that the velocity of B with respect 
 to A is Vii* -^ v^ — 2uv cos 6^, and inclined at an angle 
 tan~^ V sin 6/{v cos 6^ — w) to the direction of A. 
 -r 7. Two railroad tracks intersect at 90°. To a passenger in 
 one train traveling at the rate of 32 miles an hour the other 
 seems to have a velocity of 40 miles an hour. Find its abso- 
 lute velocity. Ans. 24 miles an hour. 
 ^■-8. The displacement of A relative to ^ is s ft south and 
 relative to C it is s ft west. Find the position of C relative 
 to^. Ans. 5i^3 S.E. 
 ^ 9. A boat is propelled at 12 miles an hour across a stream 
 flowing at 5 miles an hour in a direction perpendicular to the 
 current. Find the velocity of the boat with reference to the 
 bottom of the channel. Ans. 13 m/h. 
 
 10. The hour and minute hands of a clock are 6 in and 7 
 in respectively. Find the relative velocities of their ex- 
 tremities at (1) 6 A.M., (2) 9 A.M., (3) noon, and show that at 
 3 P.M. the direction of their relative velocity makes an angle 
 tan"* 1/14 with the horizontal. 
 
 11. A man traveling eastward at v miles an hour in a wind 
 apparently from the north doubles his speed when the wind 
 appears to blow from the northeast. Show that the wind is 
 really from the northwest and blowing with a velocity of 
 vi^2 miles an hour. 
 
 /x 12. A vessel sails due north at 10 knots in a wind dueteast. 
 If the vane on the mast is 30° east of south, find the velocity 
 of the wind. A7is. 6.65 m/h. 
 
 13. A point A moves in a straight line J C with uniform 
 velocity of 1 ft/sec. A second point B, starting from the 
 line A C, makes 1 revolution per second in a circle of radius 
 1 ft about A as center. Find the path of B. 
 
 Ans. A cycloid. 
 
 14. Find the linear displacement of the highest point A of 
 a 6-ft locomotive-wheel with reference to the lowest point B 
 while B makes a quarter revolution along a straight track. 
 
 Ans. 6^2 ft inclined at 45° to the track. 
 "^ 15. Is it correct to say that the earth actually goes round 
 the sun, or that the sun goes round the earth ? 
 
40 KINEMATICS— MOTION. [§ 43 
 
 [Either or neither. Relatively to the earth the^ sun de- 
 scribes an ellipse round the earth in a focus; relatively to the 
 sun the earth describes an ellipse about the sun in a focus. 
 Relatively to Jupiter neither statement is correct. Why, 
 then, do we say that it is more scientific to suppose the earth 
 to go round the sun ? Simply for this reason : the sun as 
 center of the planetary system enables us to describe in con- 
 ception the routine of our perceptions far more clearly and 
 briefly than the earth as center. Neither of these systems is 
 the description of an absolute motion actually occurring in 
 the world of phenomena. The symmetry of the planetary 
 system is thus seen to depend largely on the standpoint from 
 which we perceive it. — Pearson, Grammar of Science. 
 Hence, too, Milton's statement that it makes no difference 
 " Whether the sun, predominant in heaven. 
 Rise on the earth, or earth rise on the sun: 
 He from the east his flaming road begin, 
 Or she from west her silent course advance 
 With inoffensive pace," etc.] 
 
 EXAMINATION 
 
 1. How is the position of a point defined ? 
 
 2. What ideas are involved in the term displacement ? 
 
 3. In what sense is a body said to be at rest ? in motion ? 
 
 4. Mention some units of length. What is the British 
 standard of length ? 
 
 5. Mention some units of time. Mention some natural 
 st^idards of time besides the day. 
 
 "^ 6. Define the term velocity, and state how velocity is meas- 
 ured when uniform, and when variable. 
 
 i 7. Define the average velocity of a moving point in any 
 given time. 
 
 ^ 8. Explain the statement 
 
 distance = average velocity X time. 
 
 9. A knot is, roughly, a velocity of 100 ft/min. , ^ 
 
 10. Give examples of a body in motion the different parts 
 of which have different velocities. 
 
 — 11. The minute-hand of a clock is twice as long as the 
 hour-hand. Compare the speeds of their extremities. 
 
 Ans. 1 : 24. 
 
§ 43] EXAMlNATIOK. 41 
 
 '^ 12. Explain the statement 
 
 unit velocity = one foot/one second. •/ 
 How is velocity represented graphically ? K^}^ 
 >7l3. Explain the difference between uniform motion and 
 uniformly accelerated motion. 
 
 14. Give examples of bodies having accelerations (1) con- 
 stant in magnitude and direction, (2) constant in magnitude, 
 but not in direction, (3) variable in magnitude and direction. 
 
 15. Show that for uniform acceleration from rest 
 
 s = vt/2 = af/2. 
 
 16. Define acceleration at a given instant of time. 
 
 X 17. Illustrate the meaning of « = dv/dt geometrically. 
 y^l8. If a body is projected with the velocity lo in the direc- 
 tion of a uniform acceleration a, and v be the velocity and 8 
 the distance described at the end of the time t, prove 
 {v — u)/a = 2s /(v -f- u) = t. 
 
 19. A particle starts with a velocity of 12 ft/sec and the 
 motion is retarded 2 ft/sec^ Draw a diagram to find the 
 distance described in 6 seconds. 
 
 20. Show how the distance described may be represented 
 graphically when a particle moves with constant accelera- 
 tion, starting with a velocity u. 
 
 21. Show by a diagram that when a particle moves from 
 rest with constant acceleration the distance described is pro- 
 portional to the square of the time of motion. 
 
 — 22. The distance s described in the nth second of its mo- 
 tion by a particle having an initial velocity u and a uniform 
 acceleration a is given by 
 
 s = u-[- (2n — \)a/2. 
 
 Compare an acceleration of 1200 yds/min' with an accelera- 
 tion of 1 ft/sec', the unit of acceleration. 
 
 23. Show how two coexistent velocities or accelerations 
 may be combined. 
 
 24. Show how to resolve a velocity in two directions at 
 right angles to each other. 
 
42 KIKEMATICS— MOTION. [§ 43 
 
 25. state the principle of Robervars method of drawing / 
 tangents to a curve. 
 
 26. Find the direction and magnitude of the acceleration 
 of a particle which describes a circle of radius r with uni- 
 form velocity v. 
 
 27. From a point draw lines to represent in magnitude and 
 direction the velocity at the different points of the path of a 
 particle moving uniformly in a circle. 
 
 28. Define a simple harmonic motion. Is it a rectilinear 
 motion ? 
 
 29. A point P moves uniformly in a circle. Show that the 
 velocity and acceleration of the projection if of P on any di- 
 ameter are proportional to PM and OM respectively, being 
 the center of the circle. 
 
 30. Oscillatory motion is the projection of circular motion 
 upon a diameter. 
 
 31. When is motion said to be periodic ? 
 
 — 32. The motion of a point along a straight line OX is de- 
 fined by the equation 
 
 a; = a cos Qoty 
 
 show by finding dx/dt, d'^x/df that the motion is vibratory 
 and of a period ^n/oo, 
 
 -■''3^ Given the velocities of two points A and B to deter- 
 mine the velocity of A relative to B. 
 
 34. A person walking rapidly in a vertical rain holds his 
 umbrella towards the front. Explain. 
 
 35. Two trains are on parallel tracks. Why is it difficult 
 for a passenger to tell whether his own train or the other is 
 in motion ? 
 
 -36. Explain why the smoke of a steamer is in a parallel 
 plane to the vane on the mast ? 
 
 -^ 37. From a car window a man fires at a buffalo running in 
 a parallel direction with the train. Show that he must aim 
 in front if his velocity is < that of the buffalo. 
 
 38. Find the velocity with which a man must jump back- 
 ward from a car in motion to fall vertically. 
 
§ 43] EXAMINATION. 43 
 
 39. Two points move with velocities u and v in opposite 
 directions round a circle. Find their greatest and least rela- 
 tive velocities. 
 
 40. Define phorononiics, kinematics. 
 
 41. What are the fundamental independent units used in 
 linear kinematics ? [Length and time.J 
 
 42. Mention some derived units. 
 
 43. Discuss the argument of Zeno of Elea (b.c. 500) 
 against motion : " Since an arrow cannot move where it is 
 not and since it cannot move where it is, it therefore follows 
 that it cannot move at all." (See Carlyle, Friedrich the Sec- 
 ond, Yol. iv. p. 263.) 
 
44 MATTER IN MOTION. [§ 44 
 
 CHAPTER 11. 
 MATTER IN MOTION. NEWTON'S LAWS OF MOTION. 
 
 44. Hitherto we have considered motion in the abstract — 
 how represented and how jneasiired. No reference has been 
 made to the nature of the body moving, and the problem has 
 been as purely ideal as a problem of geometry. 
 
 We shall now consider motion with reference to the body 
 moving and the force acting, that is, pass from kinematics to 
 dynamics. This brings us to the region of sense and makes 
 our results capable of verification, or such that we can test 
 computation by observation and measurement. In order to 
 verify results certain fundamental postulates are taken as 
 bases of operation, which will now be stated. 
 
 45. The relations of matter, motion, and force, which con- 
 stitute the science of dynamics, may* be based upon three 
 postulates known as Newton's laws of motion. These laws 
 were known to Galileo and other forerunners of Newton, but 
 were first stated by Newton in concise terms. They are not 
 axiomatic in the sense of a geometrical axiom, because when 
 stated they are not at once assented to. They do not com- 
 mend themselves to the mind either as true or as false. 
 
 In stating Newton's laws certain rude experiments will be 
 indicated which are sufficient to suggest the truth of the 
 laws, but not to establish it. No direct proof is possible. 
 The proof is indirect and is made in this way. Assume the 
 laws true, and certain consequences follow which can be 
 
 * See Art. 217. 
 
§ 47] NEWTON'S LAWS OF MOTION. 45 
 
 tested experimentally. This has been done in so many ways 
 and by so many independent observers, particularly in astro- 
 nomical work, that we are justified in accepting them as true. 
 For example, the Ephemeris or Nautical Almanac is pub- 
 lished several years beforehand, and the predictions made in 
 it and based on these laws are always found to agree with 
 the occurrences when observed. Such, for instance, are the 
 predictions of the times of eclipses of the sun and moon, the 
 positions of the planets, etc. 
 
 46. Law I. Every body [particle] continues in its state 
 of rest or of uniform ^notion in a straight line except in so 
 far as it may he compelled iy external force to change that 
 state. 
 
 The law lies beyond our experience, as we have no experi- 
 ence of one body not acted upon by another. Our direct 
 experience goes, however, a certain distance in confirmation 
 of the law. Thus, as suggested by Galileo, consider a body 
 placed on a level surface. If at rest it will remain at rest; 
 if in motion it will come to rest after going a distance de- 
 pending upon the smoothness of the surface. The smoother 
 the surface the farther it goes and the more nearly in a 
 straight line. Conceive a surface perfectly smooth and the 
 air to have no influence on the motion, and we cannot think 
 of any reason why the body should not continue to move uni- 
 formly in a straight line. 
 
 47. From the law we learn that rest and motion are equally 
 states of a body, the body being wholly without influence on 
 its rest or motion. This property of matter is called inertia 
 [the vis insita of Newton], and the law itself is often named 
 the law of inertia. 
 
 From the law we also learn that by the term force is meant 
 a cause of change in motion, not in the sense of moving 
 agent, but in the sense of antecedent. Force is thus not to 
 be regarded as the cause of a state of motion, but of a change 
 of state, from rest to motion, motion to rest, or to an altera- 
 tion of motion — in a word, of acceleration. Whenever force 
 
46 MATTER IK MOTION. [§ 48 
 
 acts, an acceleration of the motion of the body acted upon is 
 produced. 
 
 48. The law guides us in finding a timekeeper. A body in 
 motion and not acted upon by external forces would afford a 
 means of measuring times. For the distances passed over by 
 such a body in equal times are equal. 
 
 We know of no permanent motion that is at the same time 
 uniform and rectilinear. The standard motion for the meas- 
 urement of time is the rotation of the earth on its axis. We 
 assume that the earth revolves uniformly or through equal 
 angles in equal times, and find that predictions of astronom- 
 ical phenomena made on this hypothesis agree closely with 
 subsequent observation. 
 
 There is no essential reason why the rotation of some other 
 planet s^.ioiild not be adopted as a standard of uniformity, 
 and the results would not necessarily be the same in the two 
 cases; for there are certain causes (notably the tides) which 
 tend to make our planet rotate more slowly, so that after the 
 lapse of many centuries the period of its rotation may have 
 appreciably increased, and the extent of such retarding in- 
 fluences would be different for different planets.* 
 
 49. Law II. Having learned that a characteristic mani- 
 festation of force is acceleration, our next inquiry is as to the 
 relation between force acting, body acted upon, and accelera- 
 tion produced — in a word, as to how force is measured. 
 
 Now it is found that when the same body is exposed to 
 action of the same force it has the same change of motion. 
 Thus the same pull of a spring- balance — equal pulls being 
 measured by equal stretch of spring — gives the same body 
 the same acceleration at all times and places. The same 
 general result is found no matter how the manner of making 
 the experiment is varied. 
 
 If two bodies exposed to the action of the same force receive 
 the same acceleration we say that they are of the same mass, 
 and if the accelerations are not the same ^ve say that the 
 bodies are of different mass. The term mass is thus applied 
 
 * Bmtou, Dynamics t p. 102. 
 
§ 50] NEWTON'S LAWS OF MOTION. 47 
 
 to that physical quality of a body that determines its accel- 
 eration. Experiment shows that mass is a definite entity 
 altogether independent of the physical state of the body. 
 
 Newton defines " the quantity of any matter as the meas- 
 ure of it by its density and volume conjointly," and states 
 that tliis quantity is what he shall understand by the term 
 mass or body,* 
 
 50. The quantitative relation between force, mass, and ac- 
 celeration is given by Newton's second law of motion. 
 
 Change of motion is 2^foportional to the im^jressed force 
 and takes place in the direction of the straight line in which 
 the force is impressed; or, in modern phraseology [see Art. 
 53 for other statements], 
 
 ForceX^s, that which produces acceleration Midi] i&- propor- 
 tional to the mass m of the lodg and the acceleration a pro- 
 duced jointly. 
 
 Expressed in symbols, the law gives the relation 
 
 F = cma, 
 where c is a constant. 
 
 In this expression tlie only unit whose value has been 
 already defined is the unit of acceleration. If one of the two, 
 unit force or unit mass, is assumed, the other is fixed by the 
 equation. It is usual to assume nnit mass as the mass of a 
 certain piece of metal carefully preserved as standard of 
 reference. (A fuller account of this unit will be given later on.) 
 
 Having assumed the unit mass, the unit force may be 
 chosen such that c = 1, so that we may write 
 
 p =z ma 
 
 as the expression of the law. 
 
 In this equation when a = \ and m = 1 we have F =\, 
 or nnit force is that force which acting on unit mass pro- 
 duces unit acceleration. 
 
 * Of this definitiou Mach remarks [Die Mechanik, p. 181] : " It is to 
 be observed that the formulation given by Newton which defines mass 
 as the quantity of matter of a ])ody measured by the product of its vol- 
 ume and density, is unfortunate. Since we can only define density as 
 the mass of unit volume, the circle is manifest." 
 
 See also Pearson, Orammar of Science, p. 358. 
 
48 MATTER IN" MOTION". [§ 51 
 
 61. It follows that with this system of units the equation 
 
 F=ma 
 
 means that the force which produces an acceleration « in a 
 body of mass m is expressed by the product ma, called the 
 mass-accele ration. 
 Also, since a = d ""s/df or a = v dv/ds, 
 
 we have F=m-T-. 
 
 df 
 
 or F — mv-T-, 
 
 ds 
 
 as the expression of the law. 
 
 Any one of these forms is the general equation of motion 
 of a particle free to move, the two latter forms being some- 
 times called the differential equations of motion. 
 
 The general equation of motion, which is the algebraic 
 statement of the second law of motion, is the connecting link 
 between motion and force. It enables us to pass from the 
 kinematical properties of motion already laid down to ques- 
 tions involving force and mass. 
 •52. Inferences from the law: 
 
 (1) If i^ units of force produce in a mass 7n an acceleration 
 a, and F^ units produce in a mass m^ an acceleration a^ , then 
 
 F/F^ = ma/m^a^. 
 
 {a) Now when m = m^, the forces are proportional to the 
 accelerations. Hence 
 
 Equal forces are such as produce equal accelerations in 
 equal masses. 
 
 (b) When F= F^, the masses are inversely proportional to 
 the accelerations. Hence 
 
 Equal masses are such as are equally accelerated hy equal 
 forces. 
 
 The law thus enables us to express all forces in terms of 
 
§ 53] NEWTON'S LAWS OF MOTION. 49 
 
 the unit force, and all masses in terms of the unit mass. (See 
 question 24, p. 68.) 
 
 (2) The law implies that when two or more forces act on a 
 particle at the same time, each produces an acceleration in its 
 own direction without reference to the others. In other 
 words, the acceleration produced by a force on a particle is 
 independent of any motion the particle may have and inde- 
 pendent of motions produced by other forces acting simul- 
 taneously. This is known as the principle of the independ- 
 ence of forces. It was first pointed out by Galileo. 
 
 That this principle is not axiomatic is evident from the 
 opinion of Descartes. "It is certain that a stone is not 
 equally disposed to receive a new motion or increase of veloc- 
 ity when it is already moving very quickly and when it is 
 moving slowly." 
 
 63. 'The law may be stated in other forms. 
 (a) Let a force F act on a mass m for a time ^, and let v 
 be the velocity acquired. Then if a is the acceleration of 
 motion, 
 
 F = ma, ' 
 
 Also (Art. 25), v = at. 
 
 Eliminating a, we find ^^ "" (r^^ 
 
 Ft = mv (t) 
 
 as the expression of the law. 
 
 The product mv of the mass m and its velocity v at any 
 instant is called the momentum. The unit of momentum is 
 the momentum of unit mass moving with unit velocity. 
 
 The product Ft is called the impulse of the force F during 
 the time t. It is expressed in the same unit as mi\ 
 Hence the statement : 
 
 27ie momenttcm acquired by a body in any time is numeri- 
 cally equal to the impulse of the force which produces I't; or, 
 in a wordf 
 
 change of momentum = impulse. 
 
50 MATTER IN MOTION. [§ 54 
 
 The term momentum (Lat. mo?nentum = movimentum, a 
 movement) is the modern equivalent of Newton's phrase 
 "quantity of motion" (quantitas mohts). 
 
 The term impulse was proposed by Belanger (1790-1874) 
 in his Cours de Mecanique (1847). 
 
 {h) Suppose the distance passed over by the mass m under 
 the action of the force i^to be s, and that v is the velocity ac- 
 quired. Then (Art. 25) 
 
 vy2 = as, ^ , ^'' 
 But F = ma. 
 
 Eliminating a, we have 
 
 Fs = mvy2 
 
 as the expression of the law. 
 
 The product Fs is called the work done by the force i^ act- 
 ing through the distance s. The unit work is the work done 
 by unit force acting through unit distance. 
 
 The expression mv^/2 is the energy of the moving body in 
 "units of the same name as the unit work. 
 
 Hence the statement; 
 
 The energy acquired by a 7noving lody is Jiumerically 
 equal to the work which produces it, or 
 
 energy change = tvork done. 
 
 The term work was proposed by Coriolis (1792-1843) in his 
 Traite de Mecanique (1829) ; the term energy in this sense 
 by Thomas Young (1773-1829), physicist and Egyptologist. 
 
 54. The relations in the preceding article indicate differ- 
 ent aspects of the measure of force. The first two given by 
 the equations 
 
 F — ma. 
 Ft = mv, 
 
 in which force is measured by the mass-acceleration or by the 
 rate of change of momentum with time, are due to Newton. 
 All dynamical questions are by him disposed of with the aid 
 of the ideas of force, mass, and momentum. 
 
§ 56] NEWTON'S LAWS OF MOTION. 61 
 
 The third equation, 
 
 Fs = mvy2, 
 
 introduces a distinct measure of force, and is the method of 
 Huygens. 
 
 The first method is the one most commonly used in ele- 
 mentary mechanics and will be employed at first; the other 
 will be discussed in Chapters VI and VIII. 
 
 65. If the acting force F be variable, it may be considered 
 to consist of a succession of constant forces acting for in- 
 definitely small intervals ^t and through indefinitely small 
 distances ^s. The total impulse would be found by the sum- 
 mation of FJt throughout the whole time of motion t, and 
 would be represented by 
 
 / 
 
 
 
 The total work would be found by the summation of i^z/.9 
 throughout the whole distance of motion s, and would be 
 represented by 
 
 X 
 
 Fds, 
 
 
 
 An impulse may therefore be defined as the time-integral 
 of the force, and work may be defined as the line-integral of 
 the force. 
 
 66. Law III. In order to exert force the agent acting 
 must meet a resistance. Thus the hand in motion does not 
 exert force uiitil it meets some object. The object reacts on 
 the hajid. Press the table and the table will press the hand. 
 Force is always a mutual action: in other words, forces are 
 never single, but act in pairs — one the action and the other 
 the reaction. This pair of actions between two bodies or two 
 parts of the same body is known as a stress.. If it is of the 
 nature of a push, preventing approach of the two bodies, it 
 is called compressio/i or pressure; if of the nature of a pull, 
 preventing separation, it is called tensi07i; if of the nature 
 
52 MATTER IN MOTION. [§ 57 
 
 of a shear, preventing sliding, it is called a shearing stress or 
 shear. 
 
 When we speak of a force acting on a body we consider 
 only one of the two bodies between which stress exists. The 
 force is the component of the stress on the body — the action. 
 This was the case in discussing the preceding two laws. 
 
 But since a force cannot exist by itself, — forces being dval, 
 — the view given in laws I and II is only partial and re- 
 quires to be supplemented. This is done by the law of 
 stress, or Newton's third law of motion, which is: 
 
 When one hody acts on another, the reacting force (reac- 
 tion) is equal in magnitude and opposite in direction to the 
 acting force {action), or, as it may be expressed: 
 
 The mutual actions of tivo bodies are always equal and act 
 in oppodte directions. 
 
 57. In some cases the relation between the action of the 
 agent and the reaction of the resistance is sufficiently evi- 
 dent. Thus if one body rests upon another it will be granted 
 that the pressure exerted by the upper is equal to the counter- 
 pressure exerted by the lower : if a horse hauls a canal-boat 
 to which he is attached by a rope, the pull of the rope on the 
 horse is equal to its pull on the boat, and so on. But when a 
 stone falls from a height it is not evident whether the action 
 of the earth on the stone is equal to the action of the stone 
 on the earth. Nor is the relation evident between the actions 
 of a magnet and a piece of iron,* nor between bodies widely 
 separated, as the earth and the moon. But the law asserts 
 that in all cases the acting force and reacting force are equal. 
 
 Newton points out the consequence of denying the truth 
 of the law: " For instance, if the attraction of any part of the 
 earth, say a mountain, upon the remainder of the earth were 
 
 * Newtou, by placing a magnet in one vessel and the iron in another, 
 and floating both vessels in water so as to touch each other, showed 
 tliat, as neither vessel was able to propel the other along with itself 
 through the water, the attraction of the iron on the magnet mu.st be 
 equal and opposite to that of the magnet on the iron, both beiug equal 
 to the pressure between the two vessels. (Maxwell.) 
 
§ 60] Newton's laws op motion. 53 
 
 greater or less than that of tlie remainder of the earth upon 
 the mountain, there would l)e a residual force acting upon 
 the system of tlie eartli and the mountain as a whole which 
 would cause it to move off with an ever-increasing velocity 
 through infinite space. This is contrary to the first law of 
 motion, wliich asserts that a body does not change its state 
 ot motion unless acted upon by external force." 
 
 58. Newton interprets the terms action and reaction in 
 more than one sense. These interpretations are contained 
 in the statements — 
 
 When two bodies act upon one another (1) the impressed 
 force on the one is equal and opposite to the impressed force 
 on the other; (2) the gain of momentum by the one is equal 
 to the loss of momentum by the other; (3) the gain of en- 
 ergy by the one is equal to the loss of energy by the other. 
 
 Notice that action and reaction take place between differ- 
 ent bodies. Thus in a tug-of-war the pull along the rope of 
 A on B is equal to the pull of B on A. No matter where 
 the rope is cut, the force on one side of the section is equal 
 to that on the other side. 
 ^ 59. Summary of the Laios. — The laws of motion may be 
 summarized in these statements: 
 
 (1) Every force is one component of a stress, and the two 
 forces of a stress are equal and opposite. 
 
 (2) There is no acceleration without force, and the mass- 
 acceleration produced is directly proportional to the measure 
 of the force acting. In symbols : If a force F acts on a mass 
 m and produces an acceleration a, then 
 
 F = ma, 
 and if v be the velocity acquired after a time t. 
 
 Ft = mv, 
 or if V be the velocity acquired in describing a distance s, 
 
 Fs = \mv^^~^ 
 
 60. Which of the laws of motion do the following state- 
 ments illustrate? ^ ^\ 
 
54 MATTER IK MOTION. f§ 61 
 
 (a) A circns-rider to jump through a hooji and alight on 
 his horse springs vertically upward. 
 
 (b) " If a stone be dropped from the top of the mast of a 
 ship in motion, the stone will fall at the foot of the mast not- 
 withstanding the motion of the ship." 
 
 (c) When an omnibus turns a sharp corner there is a tend- 
 ency to throw the driver from his seat. 
 
 (d) When a horse tows a canal-boat the pull backward on 
 the horse is equal to the pull forward on the boat. 
 
 {e) When a train stops suddenly the passengers receive a 
 jerk. 
 
 ( /') A cannon-ball has a different effect on a granite wall 
 and on an earth wall. 
 
 ■ (g) A severe jar is received from a step downward when 
 one expects to step on the level. 
 
 (h) In suburban -passenger traffic the trains must stop and 
 start quickly. The boiler and machinery are placed over the 
 driving-wheels, 
 
 (i) Speaking of the third law of motion, Dr. Lodge says: 
 
 " Action and reaction are equal and opposite. Sometimes 
 an absurd difficulty is felt with regard to this, even by engi- 
 neers. They say: "If the cart pulls against the horse with 
 precisely the same force as the horse pulls the cart, why does 
 the cart move?" Why on earth not? The cart moves 
 because the horse pulls it, and because nothing else is pulling 
 it back. " Yes," they say, " the cart is pulling back, but what 
 is it pulling back, not itself surely ? " " No, the horse." Yes, 
 certainly the cart is pulling the horse; if the cart offered no 
 resistance, what would be the good of the horse ? That is 
 what he is for, to overcome the pull-back of the cart; but 
 nothing is pulling the cart back. There is no puzzle at all 
 when once you realize that there are two bodies and two 
 forces acting, and that one force acts on each body." 
 
 It has been objected to this that if the cart pulls the horse 
 as much as the horse pulls the cart, there is just as much 
 reason that the cart drag the horse as the horse drag the cart 
 after it. 
 
 Does the objection appear to you a valid one ? 
 
 61. Dynamical Units. — In Art. 7 were given the names 
 of the kinematical units of length and time. Before giving 
 the names of the units 'of dynamics some explanations are 
 necessary. 
 
§ 62] NEWTON'S LAWS OF MOTION. 55 
 
 The methods of comparing masses and of comparing forces 
 given in Art. 52 are theoretically sufficient. A system of 
 units, known as an absolute system, based upon these con- 
 siderations and used in theoretical investigations in scientific 
 laboratories, is given in Chapter IX. 
 
 But these methods, though capable of being described in 
 succinct terms, are difficult of performance in practice. To 
 compare masses by comparing their accelerations produced 
 by forces assumed equal, or to compare forces by comparing 
 their effects upon masses assumed equal, though useful as a 
 laboratory experiment, is an experiment not capable of much 
 precision with any apparatus at present in existence. Ac- 
 cordingly this dynamical method is in practice replaced by 
 another more easily put in operation and much more precise.* 
 
 62. It is a fact of common observation that a body free to 
 move falls towards the earth. It acts as if the earth attracted 
 it. It is assumed as a convenient explanation of the observed 
 phenomenon that there exists a stress between the earth and 
 the body, and to one component of this stress the Jia>me force 
 of gravity is given. 
 
 A body free to move if exposed to the action of the force 
 of gravity is uniformly accelerated. Experiment shows that 
 at the same place this force acts on all bodies in the same 
 way; that is, the acceleration g [initial of gravity] produced 
 by it has no relation to the magnitude or form of the bodies 
 or to the material of which they are composed. 
 
 Experiment shows, too, that the value of g is constant so 
 long as we keep to the same place on the earth's surface. 
 It is found to vary with the latitude and the height above 
 sea-level. It is greatest at the poles and least at the equator. 
 It decreases as the height above sea-level increases. (See Arts. 
 101, 111.) 
 
 * For the comparison of masses by the inertia method see text-books 
 of laboratory physics. For the method by the ballistic balance see 
 Hicks' Dynamics, pp. 23-26. 
 
56 MATTER IN MOTION". [§ 63 
 
 At New York g = 32.16, at San Francisco g = 32.16, at 
 London g = 32.19, at Paris g = 32.18, at the equator g ~ 
 32.09 ft/sec'. The approximate value 32 ft/sec' is very often 
 used for convenience of computation and as being close 
 enough in ordinary cases. 
 
 The experiments referred to are carried out with Atwood's 
 Machine (Art. 70); or after Galileo's method, with falling 
 bodies and the inclined plane (Art. 105); or best of all with 
 the pendulum, as first pointed out by Newton (Art. 116). 
 
 We add a familiar illustration. If we let drop say a coin 
 and a piece of paper, the coin will reach the floor first; but if 
 we place both in a vessel without a lid and drop the vessel, the 
 bodies inside will reach the floor together. This suggests that 
 when the resistance offered by the air is removed, the time of 
 falling the same distance and consequently the acceleration 
 of gravity on both bodies is the same. 
 
 A better method is to let drop the bodies simultaneously 
 from a shelf in a tall glass cylinder from which the air has 
 been pumped out. Both will be seen to strike the bottom at 
 the same instant. 
 
 63. Let two bodies m, m' be exposed to the action of the 
 force of gravity. If connected together, as by a thread pass- 
 ing over a smooth pulley, and in equilibrium, the 
 action and reaction through the thread being the 
 same, we have 
 
 mg = m'g'. 
 But from the preceding article 
 
 9=9'- 
 Hence 
 
 m = m', 
 
 and in this way masses may be compared. 
 
 No great accuracy of measurement is to be looked for with 
 this apparatus, simple as it appears (see Ex. 1, Art. 70). It, 
 however, is theoretically the simplest form of balance, and 
 suggests that instrument. 
 
§ 65] GRAVITATION UNITS. 5'J' 
 
 In making measurements of precision the connection be- 
 tween the bodies is made by placing them on the pans of 
 a beam-balance. If the balance remains in equilibrium the 
 bodies are of the same mass. AVhen the comparison is made 
 in this way the term weight is by prevailing custom used 
 instead of mass, and it is usual to say that bodies which 
 equipoise in the beam-balance are of the same weight. The 
 process is called weighing; and by assuming a standard unit 
 the weights of all bodies may be expressed in terms of the 
 standard. Instead, therefore, of comparing masses by the 
 dynamical metliod as required by the fundamental concep- 
 tion of mass, we may make the comparison by the beam- 
 balance — an operation capable of extreme precision. With 
 our present appliances and methods weighings can be carried 
 out within one part in ten millions. " It is claimed that two 
 avoirdupois pounds can be compared with an error not ex- 
 ceeding 0.0002 of a grain." 
 
 64. Xotice that this method of comparing bodies by the 
 balance is entirely independent of the locality where the ex- 
 periment is made. The beam-balance does not show what 
 the forces on the two bodies are, but only that these forces 
 are equal at any place. With a spring-balance, however, 
 assuming that it takes the same force to produce the same 
 deflection of the spring, we may compare forces at the same 
 place or at different places. 
 
 Thus, at any place, let two bodies be compared by beam- 
 balance and also by spring-balance. If the whole apparatus 
 is carried to another locality of different elevation, the bodies 
 will still equipoise on the beam-balance, but the deflection of 
 the spring-balance will be different from what it was before, 
 though both bodies will produce the same deflection. The 
 weight [or the mass] of a body is therefore constant, but the 
 force of gravity on the body varies from place to place. 
 
 65. Gravitation Units — (a) Unit Weiglit. — The standard 
 unit used in comparing bodies by weighing is the pound (lb). 
 The British standard of weight is the imperial standard 
 
5g MATTER t^ MOTtOlS'. [§ 65 
 
 pound, which is defined by the Weights and Measures Act 
 1878, 41 and 42 Vict. c. 49, as "of phitinum, the form being 
 that of a cylinder . . . marked P. S. [Parliamentary Stand- 
 ard] 1844, 1 lb/' Like the imperial yard it is deposited in 
 the standards department of the Board of Trade, London. 
 
 In the United States the weights and measures in common 
 use were introduced from England before the Revolution. 
 In 1828 Congress legalized a troy pound for purposes of 
 coinage. In 1830 the Treasury Department, for custom- 
 house purposes, adopted the Troughton 82-inch brass scale as 
 standard unit of length (Art. 7), the pound avoirdupois de- 
 rived from the troy pound as unit of weight, the wine gallon 
 of 231 cubic inches as unit of liquid measure, and the Win- 
 chester bushel of 2150.42 cubic inches as unit of dry meas- 
 ure. 
 
 In 1836 Congress authorized the Secretary of the Treasury 
 to send copies of all weights and measures adopted as stand- 
 ards by the Treasury Department to every State for the use 
 of the State. Some of the States had already legalized 
 standards, but all of them formally adopted these standards, 
 and in this way a practically uniform system of weights and 
 measures was secured throughout the Union. Strictly speak- 
 ing, however, each State has its own standards, and these 
 standards are entirely independent of the standards of the 
 Office of Weights and Measures at Washington, though copies 
 of them. 
 
 In 1866 the metric system of weights and measures was 
 made legal throughout the United States, and the pound was 
 defined in terms of the kilogram by the relation 
 
 1 pound avoirdupois = 1/2.2046 kilogram. 
 
 There being no material normal standard of the pound 
 avoirdupois, its value is derived from the standard of the 
 kilogram in accordance with this relation. (Art. 280.) 
 
 The act of 1873 reads: "For the purpose of securing a 
 due conformity in weight of the coins of the United States 
 
§ 66] RELATIOK OF THE tTKlTS OF FOftCE. 69 
 
 the brass troy pound weight procured by the Minister of the 
 United States at London in 1827 for the use of the mint and 
 now in the custody of the mint at Philadelphia shall be the 
 standard troy pound of the mint of the United States, con- 
 formably to which the coinage thereof shall be regulated/' 
 
 /{b) The unit force is naturally taken to be equal to the 
 force with which the earth attracts the unit weight, the 
 standard pound. The unit force is named a pound. 
 
 The pound is called the gravitation unit of force, because 
 its value depends on the force of gravity. This force is not 
 constant over the earth's surface, but varies from place to 
 place. The unit of force defined in Art. 50, being entirely 
 independent of the force of gravity, is known as the absolute 
 unit of force. ><^ 
 
 With the gravitation system of units the abbreviation 
 lb will be used to indicate mass as found by weighing, and 
 the word pound to indicate force; thus we speak of a body 
 weighing m or M lb, a force of F pounds. It is customary 
 to use the letter w or W instead of m or M with gravitation 
 measures. 
 
 66. Relation of the Units of Force. — The fundamental 
 units of distance, time, and mass or weight are the same in 
 both the absolute and gravitation systems. The divergence 
 occurs at the derived unit force and related quantities. 
 
 A relation between the absolute and gravitation units of 
 force is readily found. For 
 
 1 grav. unit force produces in unit weight ^ units accel. 
 1 abs. unit force produces in unit mass 1 unit accel. 
 
 .'. 1 gravitation unit of force = g absolute units of force. 
 
 We may therefore convert absolute measures of force to 
 gravitation measures by dividing by the value of g at the 
 place in question, and vice versa. 
 
 Thus a force acting upon a body weighing W and causing 
 an acceleration a is equivalent to 
 
 Wa absolute units of force, 
 Wa/g gravitation units of force. 
 
60 MATTER IK MOTIOK. [§ 67 
 
 Similarly, the momentum of a body weighing W and mov- 
 ing with velocity v is 
 
 Wv absolute units of momentum, 
 Wv/g gravitation units of momentum. 
 
 The energy of a body weighing W and moving with veloc- 
 ity V is 
 
 Wv^/2 absolute units of energy, 
 Wv^/2g gravitation units of energy. 
 
 In all cases g is 7iol to be regarded as a divisor of either one 
 of the factors W or a or v, but as a divisor of the product 
 JVa or Wv or Wv"^, and with a corresponding change of unit 
 when the division is performed. Keeping this in mind, there 
 can be no confusion in the use of g in passing from one sys- 
 tem of units to another. 
 
 Hence rules such as the following are misleading: "The 
 number of units of mass in a body is found by dividing the 
 weight in pounds by the value of g at the place where the 
 weight is ^etermined.^' 
 
 67. Restatement of Newton" s Second Laio. — Comparing Arts. 
 50, QQ, it is evident that, using gravitation measures, the state- 
 ment of Newton's second law would be 
 
 i^F= Wa/g ....... (1) 
 
 when a force impounds acting on a body weighing W lb pro- 
 duces an acceleration a ft/sec^; or, if v ft/sec is the velocity 
 acquired, 
 
 4 Ft = Wv/g, (2) 
 
 yi\iQXQ FtJisJihSL-impul&e in second-pounda, and Wv/g is the 
 momentum in second-pounds; or, if s ft is the distance passed 
 over, 
 
 \ Fsr^ Wv'/^g (3) 
 
 where Fs is the loorlc done in foot-pounds, and Wv^/2g is the 
 energy acquired in foot-pounds. 
 
§ 68] WEIGHT. 61 
 
 Equation (1) may be written in the differential form by 
 putting 
 
 a = d "^s/iWy or a = v dv/ds, 
 
 68. Weiglit. — Wit?i the gravitation system of units, which 
 is the system of every-day life and of the more usual com- 
 mercial and engineering questions, it is necessary to conform 
 to the terms sanctioned by prevailing custom. Here there 
 is some confusion. Thus the term weight is used in the 
 double sense of quantity and of force* — or rather in a 
 triple sense. (1) In commerce it is used to denote quantity 
 as measured by the beam-balance. For example, when we 
 say the weight of a barrel of flour is 196 lb we indicate 
 quantity. This, too, is the legal sense of the term. The 
 Weights and Measures Act, 18T8, in defining the standard 
 pound, states that "the weight in vacuo of the platinum 
 weight declared to be the imperial standard shall be the legal 
 standard of weight, and of measures having reference to 
 weight, and shall be called the imperial standard pound, and 
 shall be the only unit or standard measure of weight from 
 which all other weights and all measures having reference to 
 weight shall be ascertained." 
 
 (2) Again, we speak of the pressure of a weight, lifting a 
 weight, etc. A pressure may be balanced by the pressure of 
 a weight. The law of action and reaction is the formulation 
 of the equality of pressure and counterpressure in the dy- 
 namical sense. Thus the term weight is extended to mean 
 the pull or force of the earth on the body.f A body which 
 weighs W lb is attracted with a force of W pounds. 
 
 * lu French, poids and pesanteur. 
 
 f Some writers who define mass as " quantity of matter" introduce 
 into the gravitation system of units the term mass in the sense of quan- 
 tity, and restrict the term weight to mean the force with which the 
 earth attracts the body. Thus a barrel of flour has a mnss of 196 lb, 
 and its weight is expressed as "the weight of 196 lb," or as "196 lb 
 weight." The weight of a mass of 1 cwt is 112 lb weiglit, etc. 
 
 The term mass, however, belongs strictl}' to an absolute system of 
 units only, and the innovation adds confusion to a system already con- 
 fused enough. See, for a discussiou of this question. Nature, vols. 
 
62 MATTER IN MOTION. [§ 69 
 
 The distinction is stated by Maxwell [TJieory of Heat, p. 
 80] as follows: "In fact, the only occasions in common life 
 in which it is required to estimate weight considered as a 
 force is when we have to determine the strength required to 
 lift or carry things, or when we have to make a structure 
 strong enough to support their weight. In all other cases 
 the word weight must be understood to mean tlie quantity of 
 the tiling as deterinined by the pi^occss of weighing against 
 * standard tueights.' " 
 
 (3) The term is also used to denote the determinate body 
 employed in the beam-balance as "a weight," "a set of 
 weights," etc. 
 
 The word pound has a similar variety of meanings. We 
 speak of a pound weight, a pound force, and of a certain 
 body itself as " a pound." 
 
 At the close of the last century, in different parts of the 
 world, the word pound was applied to 391 units of weight, 
 and the word/bo^ to 292 different units of length. Not only 
 were no two of these identical, but in only a few cases were 
 their relative values known with anything like precision. 
 Most of these have since been swept away. (Mendenhall.) 
 
 69. It is to be clearly understood that in mechanics we are 
 compelled to recognize two classes of problems, each with its 
 nomenclature and units. 
 
 To the problems that ordinarily present themselves in daily 
 life — local questions — the units of the gravitation system are 
 exclusively used. But many questions in electricity and 
 magnetism, in physical science and in astronomy, compel us 
 to drop local considerations. Eor the purpose of comparing 
 measurements the need of standards universal, and not local, 
 becomes imperative. This was indicated by Newton and 
 clearly pointed out by Gauss, who introduced the absx>lute 
 system of units (Arts. 50, 65). 
 
 There are two series of gravitation units in use in Great 
 Britain and the United States — the British and the metric. 
 
 xxxvii-xLVi — particularly a paper by Prof. Grcenhill, vol. xlvi, pp. 
 247-253, 
 
§ 69] WEIGHT. 63 
 
 At present the British gravitation system is in most common 
 use. The metric system, however, is legalized by both coun- 
 tries, and is the only system legalized by the United States, 
 the British system having been "inherited from England 
 with the common law." 
 
 The absolute system of units is founded on metric meas- 
 ures only — that is, the only absolute system that has obtained 
 universal recognition, the C.G.S. system. (See Chapter IX.) 
 
 The student must be familiar with all three sets of meas- 
 ures and with the methods of passing from one to another. 
 For clearness the British system and the metric system are 
 in this book presented separately. So far the British system 
 only has been made use of. But at this point Chapter IX 
 may be read for an account of the metric system. The 
 metric system may then, if thought advisable, be introduced 
 and carried side by side with the British system, or may be 
 used altogether. The tables on pp. 364-367 have been ar- 
 ranged to give a synoptical view of the various nomenclatures 
 and to facilitate conversions from one system to another. 
 
 Ex. 1. A force of 5 pounds acts on a weight of 40 lb. Find 
 the acceleration produced. A?is. a = 4 ft/sec'. 
 
 2. A 60-lb shot is fired with a velocity of 1200 ft/sec. Find 
 the impulse communicated by the powder. 
 
 Ans, 2250 second-pounds. 
 : 3. A 16-lb weight falling from a height strikes the ground 
 with a velocity of 16 ft/sec. Find the momentum destroyed. 
 
 A71S. 8 second-pounds. 
 
 4. In what senses is the term weight used in the following 
 statements ? 
 
 {a) " Every person who sells coal in quantities of less than 
 half a ton in weight shall keep scales and weights of the legal 
 standard, and shall weigh such coal before delivery." — [The 
 Weights and Measures Act.] 
 
 (b) "It will therefore be understood that the expression 
 'weight of the earth' has no physical meaning. This weight 
 may be anything. It is an indeterminate quantity." — [Nipher's 
 Electricity, p. 16.] 
 
 5. Explain how to pass from gravitation to absolute meas- 
 ures of force, momentum, and energy; and vice versa (Art. 66). 
 
64 MATTER IN- MOTION'. [§ 69 
 
 6. A baseball weighing 5^ oz has a velocity of 50 ft/sec. 
 What constant force must the catcher apply to bring it to rest 
 in 1 ft ? A71S. 13.4 pounds. 
 
 7. A baseball has a horizontal velocity of 66 ft/sec. An 
 impulse causes it to travel back on the same line with a ve- 
 locity of 62 ft/sec. Find the change of momentum. 
 
 A)is, 11/8 second-pounds. 
 
 8. A railroad train which weighs 150 tons and has a ve- 
 locity of 60 miles/hour is brought to rest in half a minute 
 by the action of the brakes. Find the average force exerted. 
 
 A71S. 27,500 pounds. 
 
 9. A 4-lb ball is moving with a velocity of 40 ft/sec and 
 in 10 sec afterwards it is moving with the same speed in the 
 opposite direction. What force has acted during this time ? 
 
 Ans. 1 pound. 
 
 10. A 50-ton train acquires a speed from rest of 10 
 miles/hour in 5 minutes. How long would it take a 75-ton 
 train drawn by the same engine to acquire a speed of 12 
 miles/hour ? Ans. 9 minutes. 
 
 11. " In computing the energy of a railroad train I am di- 
 rected to use the formula Wv''/2g. Now why use 2g, since 
 that quantity is exclusively an element of falling bodies ? Is 
 it not possible to compute the energy by a process entirely 
 independent of gravity considerations?" — [Query in Set. 
 Amer., 1893.] 
 
 12. The 66-ton Canet guns made (1890) for the Japanese 
 navy use projectiles weighing 1034 lb with an initial velocity 
 of 2262 ft/sec. Find the velocity of recoil of the gun. 
 
 Ans, 17.7 ft/sec. 
 
 13. A 1000-lb shot strikes a target directly with a velocity of 
 700 ft/sec and rebounds with a velocity of 100 ft/sec. Find 
 the measure of the impulse. A?is. 25,000 second-pounds. 
 
 14. A ball weighing 1 lb falls on a level floor with a velocity 
 of 100 ft/sec and rebounds with a velocity of 76 ft/sec. Find 
 the average force exerted between ball and floor, supposing 
 the time of impact to be 1/100 sec. Ans. 550 pounds. 
 
 15. A man who weighs 150 lb moves from the bow to the 
 stern of a boat, a length of 10 ft. If the boat weighs 225 lb 
 find how far it will have moved forward, supposing the resist- 
 ance of the water not taken into account. Ans. 4 ft. 
 
 16. A spring-balance is graduated at New York. Find the 
 true weight of a body which weighs 67 lb in this balance at 
 London. Ans. 67 lb 1 oz. 
 
 [At New York g = 32.16, at London y == 32.19 ft/sec\J 
 
§70] 
 
 TEST OF THE LAWS OF MOTION. 
 
 65 
 
 - 17. A body weighing W\h acted on by a force i^ pounds, de- 
 scribes the distance iS ft in t sec. Show that 
 
 velocity acquired = Fgt/ W ft/sec, 
 distance passed over = Fgl"^ ft W ft, 
 momentum acquired = Ft second-pounds. 
 
 70. Test of the Laws of Motion. — A laboratory contrivance 
 in common use for ilhistrating the laws of motion is known 
 as Aticood's machine,* the essential features of 
 which are as follows : 
 
 Two bodies weighing TF, , PT, lb are fastened 
 to a light inextensible thread, which passes over 
 a pulley mounted so as to oppose the motion as 
 little as possible. The effect of the pulley is to 
 change the direction, but not to alter the mag- 
 nitude of the pull. A 
 
 Let P denote the pull of the thread expressed 
 in pounds. The one body PF, falls and the other 
 W^ rises with the same acceleraHon a along the 
 motion of the thread, and therefore of the same 
 sign. 
 
 The forfce causing an acceleration a ft/sec' on W, lb is 
 W\ — P pounds. 
 
 
 Vr 
 
 -p 
 ^4. 
 
 W.-P 
 
 Wfi/g. 
 
 f.. 
 
 ''t^* 
 
 The force causing an acceleration a ft/sec' on TT, lb is 
 P — }\\ pounds. 
 
 P - TF, = W^a/g. 
 
 Hence, eliminating P, 
 
 a = g{W,- W,)/(W, + W,) ft/sec'. 
 
 By taking W^ and W^ of considerable magnitude, but nearly 
 equal, the acceleration a may be reduced to so small a quan- 
 
 * Invented by George Atwood, F.R.8. (1746-1807), Cambridge, Eng- 
 land. 
 
66 MATTER INT MOTION. [§ 70 
 
 tity that the distance s fallen through in a given time t is 
 easily observed on a vertical scale attached. The time of 
 motion may be taken from a clock or from a pendulum at- 
 tached to the apparatus and beating seconds. 
 
 The manner of making experiments with the machine will 
 be found described in any book of laboratory physics. 
 
 Ex. 1. In an Atwood machine W^ =21 oz, W^ = 20 oz, 
 and it was observed that in 5 sec the weight W^ descended 
 9.5 ft. Find the value of g the acceleration of gravity. 
 
 Ajis. 31 -f ft/sec'. 
 
 [Hint. — Equate the values of a found from s = at'^/% and 
 a=g(W,- lf,)/(Tr,+ W:). 
 
 The result would seem to show that there is something 
 wrong either with the apparatus or with the observed values 
 of distance or time. The fact is that on account of friction 
 and the resistance of the^ir the distance and time observed 
 will not be the ideal quantities demanded by the problem, so 
 that no very exact measure of g can be made with this ap- 
 paratus. The Atwood machine is now regarded as of historic 
 interest mainly.] 
 
 2. In an Atwood machine find the pull of the cord if the 
 weights are 3 lb and 2 lb. Ans. 2.4 pounds. 
 
 ' 3. In an Atwood machine the weights are P -\- Q and 
 P — Q. Show that if a is the acceleration of motion, then 
 Pa = Qg. 
 
 4. In an Atwood machine the weights are each equal to 31 
 oz, and a weight of 2 oz is placed on one of them and removed 
 after it has fallen 2 ft. Find the time of fall and the dis- 
 tance through which each weight will move in the next second. 
 
 Ans. 2 sec; 2 ft. 
 
 5. In an Atwood machine, the sum of the weights being 
 given, show that the pull of the thread will be greatest when 
 the weights are equal. 
 
 \ 6. In an Atwood machine the pull of the thread is an har- 
 monic mean between the moving weights. 
 ^7. If the thread breaks at any instant, show that when the 
 ascending motion of one of the weights ceases the other will 
 have descended through three times the distance through, 
 which the former has ascended in the interval. 
 
70] EXAMINATION. 67 
 
 EXAMINATION. 
 
 1. A body is set iu motion and left wholly to itself. What 
 is the nature of the subsequent motion ? 
 
 2. What is meant by inertia ? 
 
 [The tendency of a body to continue in its state of rest or 
 of uniform motion in a straight line.] 
 
 3. Give ilkistrations of inertia. 
 
 [Ex. The taking up of water by a moving train from a long 
 trough situated between the tracks.] 
 
 4. Deduce from the laws of motion a definition of force. 
 Is force necessary to maintain motion ? 
 
 5. State the law of motion which gives a quantitative state- 
 ment as to the effect produced on a particle by a force. 
 
 6. A constant force produces a uniform acceleration. 
 
 7. How may force be measured ? 
 
 [By deformation caused, as of a spring; by acceleration 
 contributed (Art. 50) ; by opposing it to the pull of the earth 
 (Art. 65).] 
 
 8. Define an impulse. What is the unit? 
 
 9. State the parallelogram of momenta. 
 
 ^ 10. Prove that the average force which, acting for t sec 
 through ? ft on a body weighing PTlb, is required to produce 
 a velocity of v ft/sec is Wv/gt, or Wv'^/2gl pounds. 
 
 If the body is moving with the velocity v ft/sec, what force 
 is required to stop it ? 
 
 11. Show that force may be defined as: 
 
 (1) Rate of change of momentum with time. 
 
 (2) Rate of change of energy with distance. 
 
 [These questions suggest the answer to a question keenly 
 debated soon after the time of Galileo as to whether force 
 was to be regarded as proportional to the velocity produced 
 or to the square of the velocity. The first view was held 
 by Descartes and his school; the second by Leibnitz. The 
 dispute lasted for nearly sixty years, until D'Alembert dissi- 
 pated the misunderstanding by showing that with respect to 
 
68 MATTER IN MOTIOlf. [§ 70 
 
 time force is proportional to the velocity, but with respect to 
 distance, to the square of the velocity. Thus F = mv/t, or 
 F = mvy2s.] 
 
 12. Compare the momentum of a 20-lb shot moving with a 
 velocity of 500 ft/sec with that of a 2-oz bullet moving at the 
 rate of 500 yds/sec. Ans. 160 : 3. 
 
 13. What velocity will the force of gravity give to a pound 
 weight in one second ? A^is. g ft/sec. 
 
 14. *' One's weight is given, but not — or not in the same 
 way and degree — one's velocity. Weight given, it is only by 
 doubling or trebling his velocity that a man can make his 
 momentum Rouble or treble as needed." (Carlyle, Friedrich 
 the 8eco7icl. ) Explain. 
 
 15. Show that it necessarily follows from the second law of 
 motion that forces can be represented by straight lines. 
 
 16. What are the tests of the equality (1) of two forces, (2) 
 of two weights ? 
 
 17. A man with a hod on his shoulder falls oif a ladder. 
 Find the pressure on his shoulder during the fall. 
 
 18. Describe an experiment which involves the principle 
 of the second law of motion, and show how the probability of 
 the law may be inferred from it. 
 
 19. State the units of weight, force, momentum, and work 
 in the gravitation system. 
 
 20. "A force of 100 pounds is the force which would bal- 
 ance a weight of 100 lb if acted upon by gravity only." 
 
 — 21. AVould it be advantageous for a merchant to buy 
 groceries in New York to sell in Cuba if he used the same 
 spring-balance in both places ? How would a pair of scales 
 answer ? 
 
 ~^22. A force of 10 pounds acts upon a weight of 10 lb for 10 
 sec. Find the momentum acquired in second-pounds. " 
 
 23. " When you weigh a thing in an ordinary balance, do 
 you find its weight ? " 
 
 24. It is sometimes said that we reason in a circle in stating 
 the foundations of dynamics, defining force by mass and mass 
 
§ 70] EXAMINATION. 69 
 
 by force. We say in effect : " Equal forces are those which 
 produce equal accelerations in equal masses; equal masses 
 those in which equal accelerations are produced by equal 
 forces." 
 
 [" If we assume constancy of mass of a body and of the 
 physical properties say of a spiral spring, there is no diffi- 
 culty in getting out of this circle of definition. These are 
 assumptions we are entitled to make as the result of experi- 
 ence."] 
 
 25. Explain the theory of Atwood^s machine for illustrat- 
 ing the laws of motion. 
 
 26. Masses may be compared by the impulses required to 
 produce unit velocity. 
 
 -*^ 27. Weights Tf, , PT, lb hang at the ends of a string which 
 passes over a smooth pulley. Find the pull of the string. 
 
 Ans. 2 W, W\/( W, + W,) pounds, 
 -f 28. A bucket weighing 25 lb is let down into a well with 
 uniform speed. Find the pull of the rope. 
 
 Ans, 25 pounds. 
 
 29. " Why when we place an ax upon wood, and on the 
 ax place a heavy weight, is the wood but slightly indented; 
 whereas if we raise the ax without the weight and strike 
 upon the wood the wood is split, although the falling weight 
 is much less than the resting and pressing weight ? " (Aris- 
 totle.) 
 
 [In Aristotle's day the conception that force was measured 
 by mv was unsuspected. — Lewes.] 
 
 30. Describe any method of testing whether the accelera- 
 tion produced by gravity is uniform ? 
 
 31. Whicli of Newton's -laws implies the principle of the 
 " independence of forces " ? 
 
 32. What authority have we for saying that the same force 
 will produce the same change of motion in a particle whether 
 the particle is at rest or in motion ? 
 
 33. Write down the kinematical equation of rectilinear 
 motion. 
 
 [dv/df = a, or d*s/dt'' = a.] 
 
70 MATTEH liT MOTIOlf. [§ 70 
 
 34. Write down the dynamical equation of rectilinear 
 motion. 
 
 [mdv/dt = F, or md's/dt' = F.] 
 
 35. Write down the equation of motion of a particle acted 
 upon by (1) a vertical force only, (2) a horizontal force only. 
 
 36. When is a train said to be going at full speed ? 
 
 37. If A pulls against B and action and reaction are equal, 
 how is it that A may overcome B ? 
 
 —38. Explain the kick of a gun. 
 
 39. Explain how an ice-boat can sail with greater velocity 
 than that of the wind propelling it, assuming that the ice 
 offers no resistance to the boat's motion. 
 
 (1) Consider the case of the wind abaft the beam. 
 As the velocity of the boat increases the resistance of the 
 air on the sail increases, until a velocity is reached when the 
 force of the wind behind the sail is 
 balanced by the reaction of the air in 
 front of the sail. The action and re- 
 action balancing, the boat moves with 
 --"g uniform velocity and at full speed. 
 Call this velocity v. 
 Let AB represent the keel of the boat, CD the sail, FF the 
 direction of the wind, n its velocity, and /?, y the inclinations 
 of sail and wind to the keel AFB. 
 
 Let a velocity — v he imparted to both boat and wind. 
 The boat will be at rest. The velocity — v, combined with 
 the velocity u of the wind, will give the velocity direction of 
 the appare^it wind along EB (Art. 43), which must be paral- 
 lel to the sail, since at full speed the sail is in a calm. Hence 
 
 v/ii = sin (;/ — y5)/sin /3. 
 
 For a given direction of wind and a given position of sail u 
 and P are fixed. The value of v is, then, evidently greatest 
 when sin {y — /3) is greatest, that is, when sin (y — ^) = 1, 
 and then y - /3 = 90°, or y = 90" + /i. When this is the 
 case, 
 
 V sin /3 = II, 
 
 Now sin /? is always less than unity, and therefore 
 
 V > n, 
 
 or the velocity of the boat exceeds that of the wind. 
 
§ 70] EXAMINATION-. 71 
 
 Let V, denote the component of the boat's velocity in the 
 direction of the wind. Then 
 
 v^ = V cos (180 — y) 
 
 = u sin (;/ — /3) cos (180 — ;/)/sin /3. 
 
 This is a maximum when y = 135° -f fi/2, and then 
 
 v^ = w(l + sin /3)/2 sin /3, 
 
 Now v^> u 
 
 if 1 + sin )S > 2 sin /?, 
 
 or if sin /? < 1, which it is. 
 
 Hence the component in the direction of the wind is 
 greater than the velocity of the wind, or the boat can run to 
 leeward faster than by sailing directly before the wind. 
 
 If the wind is before the beam, it may be shown in a similar 
 manner that if y^ < 19.5° and y = 45° + y5/2, the component 
 velocity of the boat in the direction of the wind is greater 
 than the velocity of the wind, or it is possible to run to wind- 
 ward faster than the wind. 
 
 40. If the wind is nearly abeam, or y=Sb° say, and the sail 
 is set so that ft = 20°, show that v = 2.Qu. 
 
 41. An ice-boat runs directly before the wind. What is 
 the greatest velocity it can attain ? 
 
 42. Which of Newton's laws corrects the opinion held by 
 the ancients that circular motion is perfect and natural ? 
 
7*3 DYNAMICS OF A PAKTICLE. [§71 
 
 CHAPTEE III. 
 DYNAMICS OF A PARTICLE. 
 
 71. Having considered the geometrical properties of motion, 
 and also the methods of measuring force, we are ready to 
 study the motion produced in a body by forces of given mag- 
 nitude. 
 
 When a body is acted on by forces, experience shows that 
 various forms of motion may arise. If all of the component 
 particles of the body move through equal distances in the 
 same direction, the motion is a motion of translation. The 
 motion of any particle would in this case give the motion of 
 the body. In a motion of rotation the particles do not move 
 through equal distances in the same direction, those nearest 
 the axis of rotation moving the shortest distance. If a body 
 consisted of a single particle, it would, in its rotation about 
 an axis passing through it, remain in the same position. We 
 shall therefore exclude rotation, and be able to study the 
 translation of a body if we consider the motion of a single 
 particle only. 
 
 72. Composition of Forces.— The number of forces acting 
 on a particle may be one or more than one. If we can com- 
 bine the separate forces into an equivalent single force, we 
 can reduce all cases to that of the action of a single force. 
 The method of combining forces will be our first step, and 
 next we shall consider the motion of a particle under the 
 action of forces. 
 
 73. Representation of Force. — The elements of a force that 
 completely determine it are : 
 
 (1) Its magnitude or the number of units of force in it; 
 
 (2) Its position or line of action ; 
 
 (3) Its direction along the line of action or its sense. 
 
§ 74] EEPRESENTATIOK OF FORCE. 73 
 
 A force acting on a particle, or at a point in a body, may 
 therefore be represented by a straight line AB in the line of 
 action of the force, the length of AB representing the magni- 
 tude of the force, the direction from A to B the direction of 
 tlie force, and the point A the point of application. ^Each 
 unit of length of AB will represent unit force. But the 
 length of the unit is arbitrary. Hence we may plot forces to 
 •my scale we please, as 1 pound = 1 inch, 10 pounds = 1 
 inch, etc. 
 
 In a diagram the direction of a force is conveniently indi- 
 cjited by an arrow-head. When different forces act in the 
 same line, their directions may be indi- 
 cated by the signs + and — . Thus a — J + ^ 
 force of 3 pounds acting at A along AB 
 
 may be written + 2, and an equal force at A along ^C* would 
 be written — 2. The choice of signs is of course arbitrary. 
 
 74. A force acting on a body may be considered to act at 
 any point in the body in the line of action of the force. 
 
 Thus let the force F act at A in the direction A 0, and let 
 
 ^ be a fixed point in the line of action. The action of F at 
 
 A has an equal and opposite reaction F^ 
 
 A B c ^^ ^j according to Newton's third law. 
 
 "* F F,' Fj But F^ at B would be balanced by F^ at 
 
 B, equal and opposite to it, and therefore 
 
 in the same direction as F. Hence F at A may be replaced 
 
 by an equal force F^ at B. 
 
 This principle is known as the transmissibility of force, 
 
 Ex. 1. On a scale of 100 pounds per inch what force would 
 be represented by a line 20 inches long ? 
 
 Ans. 2000 pounds. 
 
 2. If a force of P pounds be represented by a straight line 
 a inches long, by what straight line will a force of Q pounds 
 be represented ? Ans. Qa/P inches. 
 
 3. Find the force equivalent to the forces 3, 5, — 7 pounds 
 acting downward in the same line. 
 
 4. The forces a -\-l), a — b, 2a, in the same line, can be 
 made to balance. How ? 
 
74 DYNAMICS OF A PARTIOLS. [§ 75 
 
 u 75. Composition of Tioo Forces (Graphical MetJiod).—Su-p- 
 
 pose that F^ , F^ are two forces which act on a particle w. 
 Let Ab, Ac represent the accelerations a, , a^ due to F^ , F^. 
 The resultant a of these accelerations is 
 represented by the diagonal Ad of the 
 parallelogram AMc (Art. 30). 
 
 But F^ being equal to loajg and i^^ 
 to wajg, may be represented by two lines 
 AB, ^ C in the directions of the accelera- 
 tions «, , a^ produced by them. Complete the parallelogram 
 ABD and join AD, Then 
 
 
 AB : BD = ivajg : wajg 
 
 
 = a, : a. 
 
 
 = Ab : Id, 
 
 and therefore Ad, 
 
 AD coincide in direction. 
 
 Also, from similar triangles Adh, ADB, 
 
 
 AD:DB = Ad: db, 
 
 or 
 
 AD : waJg = a : a„ 
 
 that is, 
 
 AD = wa/g. 
 
 Hence AD represents .the force which produces the acceler- 
 ation a, and is equivalent to the two forces F^, F^ represented 
 by AB, A C. It is therefore called the 
 
 •^ CD 
 
 resultant force. Hence / ~Z^ 
 
 If Urn forces F^ , F^ acting on a par- / ^^-""^^^ / 
 
 tide A [or at a point A) be represented / ^^-^^^ / 
 
 by two lines AB, AC, their resultant 1^^^ j 
 
 R is represented by the diagonal AD of 
 
 the parallelogram ABDG constructed 07i AB, AC as adja- 
 cent sides. 
 
 This principle is called the parallelogram of forces. 
 The problem of finding the resultant of two forces is by 
 this proposition reduced to the solution of a geometrical 
 problem. 
 
%] 
 
 COMPOSITION OF tWO FOtlCfiS. 
 
 n 
 
 The parallelogram of forces was first clearly formulated by 
 Newton (IG42-1727). It is a direct consequence of the second 
 law of motion, and may be employed to test the truth of the 
 law. The principle had been previously employed by Ste- 
 vinus of Bruges, who derived it from the inclined plane. 
 
 76. Consider in the parallelogram figure the manner in 
 
 which the resultant is formed. The forces F, 
 
 F^, drawn to 
 
 scale, are represented by the lines AB, AC. From B the line 
 BD is drawn parallel to AC, and from C the line CD is drawn 
 parallel to AB. The diagonal AD of the parallelogram rep- 
 resents the resultant in magnitude, direc- 
 tion, and position. 
 
 This construction is equivalent to the 
 following : Plot the forces F^ , F^ as 
 before. From B draw BD parallel and 
 equal to AC. Join AD, which repre- 
 sents the resultant. 
 
 Still better, by breaking the figure into two parts. Let 
 F^ , F^ be the forces acting at 0. From any point A draw 
 
 AB to scale equal and par- 
 allel to F^. From B draw 
 BD to scale equal and 
 parallel to F^. Join AD, 
 which will represent the 
 resultant in magnitude and 
 direction. To find its position: We know that it must pass 
 through 0, and hence, if through we draw a line equal and 
 parallel to AD, we have R in magnitude, direction, and po- 
 sition. We have therefore sl force diagram and a construction 
 diagram purely geometrical. In simple cases there is no con- 
 fusion when the two overlap and are included in one figure; 
 but in complicated cases it is better to keep them separate, as 
 we shall see later on. 
 
 •^Ex. 1. If forces of 10 pounds and 12 pounds act on a par- 
 ticle, find the greatest and least possible resultants. 
 
 Ans, 22 pounds; 2 pounds. 
 
 <\^^i.-^ 
 
76 
 
 DYNAMICS OF A PARTICLE. 
 
 [§77 
 
 2. Show by a drawing that the value of the resultant de- 
 creases as the angle between the forces increases. 
 y^S. ABCDEF is a regular hexagon. Show that the result- 
 '^nt of the system of forces represented by AB, AD, AE is 
 
 4. Two equal forces F act at an angle of 120°. Find R. 
 
 Ans, B = F. 
 
 5. Two equal forces act at a point. Through what angle 
 must one of them be turned that their resultant may be 
 turned through a right angle ? Ans. 180°. 
 
 6. Show by a drawing that the resultant of P and F -{- Q 
 acting at 120° is equal to the resultant of Q and Q -\- F act- 
 ing at 120°. 
 
 + 7. If i? is the resultant of two forces F and Q, and S the 
 resultant of F and B, show that the resultant of S and Q is 
 2B, 
 
 77. We may express the resultant B of two forces F^, F^ in 
 terms of the forces and their included angle 0. This method 
 of finding B is often more conven- 
 ient than the graphical method of 
 measuring the diagonal. 
 
 LetAB=F^,AC=F^, ZBAC=^6, 
 Then from trigonometry we have in 
 the triangle ABD [note AC= BD; 
 A BAC-^ A J^Z> = 180°] 
 
 AD' = AB' - 2AB . BD cos ABD + BD' 
 = AB' + 2AB . AC cos BAG + AC\ 
 or B' = F,' + 2F^F^ cos 8 + F,% 
 
 which gives the magnitude of the resultant. 
 
 The line of action of B may be found by solving the tri- 
 angle ABD to find the angle BAD. 
 Thus from D let fall DF perpendicu- 
 lar to AB. Then if Z DAB = a, we 
 have 
 
 tan a = DF/(AB + BE) 
 
 = F^ sin d/{F^ + F^ cos 6). 
 
 Hence B is completely determined. 
 
§ 78] COMPOSITION OF MORE THAN TWO FORCES. 77 
 
 The special case of the forces acting in directions at right 
 angles is important. The parallelogram becomes a rectangle, 
 and from the figure 
 
 R' = F,' + F,\ • 
 i^na=FJF,, 
 
 whence the resultant is determined. 
 
 Ex. 1. If two equal forces F, i^are at right angles to one 
 another, then R = FV2 and a = 45°. 
 
 i^. Find the resultant of two equal forces, each of 10 
 pounds, acting at an angle of 30°. Ans. 19.3 pounds. 
 
 3. If two equal forces F, Fare inclined at an angle 26, then 
 R = 2F cos d anda = 6, 
 
 4. Show from the general formula that the value of R in- 
 creases as the angle between the forces diminishes, and vice 
 versa. 
 
 5. When the angle BAC \^ 0° or 180° the forces are in the 
 same straight line, and the formula for finding i?, if correct, 
 should reduce to the sum or difference of the forces. Exam- 
 ine, and see if it does. 
 
 a^6. Find the resultant of P and P -{- Q acting at 120°, and 
 also the resultant of Q and Q -\- P acting at 120°. 
 
 Ans. R = ^P' -i- PQ-\- Q' in both cases. 
 
 7. The resultant of two forces P and Q at right angles to 
 each other is R. If each force is increased by R, show that 
 the new resultant makes with R an angle 
 
 t^n-' (P-Q)/(P+Q^-R). 
 
 8. The resultant R oi P and Q is equal to P. If P is 
 doubled, show that the resultant R^ of 2P and Q is at right 
 angles to Q, and find its magnitude in terms of P and Q. 
 
 78. Composition of More than Two Forces {Graphical 
 Method). — Let i^„ F^, F,, F^ represent forces acting on a 
 particle at 0; it is required to find their resultant. 
 
 Following the method of Art. 76, from a point A we draw 
 AB equal and parallel to F^ , BC equal and parallel to F^,CD 
 equal and parallel to i^,, BE equal and parallel to F^. Join 
 AE, which will represent the resultant in magnitude and di- 
 rection. For join AC, AD. Then ^C represents the result- 
 ant of F^, F^ in magnitude and direction; AD the resultant 
 
78 
 
 DYKAMICS OF A PARTICLE. 
 
 [§78 
 
 ot AC and F^, that is, of F^, F^, F^-, and AE the resultant 
 of AD and F, , that is, of F^,F^, F, , F^, 
 
 AE represents, therefore, the resultant of all the forces at 
 
 From in the force diagram 
 
 in magnitude and direction. 
 
 draw a line R equal and parallel to AE. Then will R repre- 
 sent the resultant of F^, F^, F^, F^ in magnitude, direction, 
 and position, and be completely determined (see Art. 73). 
 
 We mi'j:ht have combined the two diagrams, or we might 
 have derived the resultant directly from the parallelogram of 
 forces, as indicated in the figures below. 
 
 \ 
 
 ^^ 
 
 / 
 
 \ ~~ 
 
 
 
 --> 
 
 \ 
 
 /j 
 
 \ 
 
 ^-'' 
 
 
 / 
 
 
 V-'' 
 
 ,/ 
 
 / 
 
 V 
 
 
 / 
 
 
 / 
 
 •^v.^ \ 
 
 / 
 
 
 Notice that in any method we may take the forces in any 
 order and we shall always find the same value of the result- 
 ant AE. Test this by making drawings to a large scale. 
 
 Ex. 1. Three forces of 6, 8, 10 pounds act at angles of 120° 
 with each other. Find their resultant. Draw to scale by 
 different methods, and compare results. Vary order, and 
 compare. 
 
 2. Forces of 1, 2, 3, 4, 5, 6 act at angles of 60°. Find R. 
 Test as in Ex. 1. 
 
 3. Forces of 20, 20, 21 pounds act at a point. The angle 
 between the first and second is 120°, and between the second 
 and third 30°. Find R. Ans. 29 pounds. 
 
 4. Is it necessary that force and construction diagrams be 
 drawn to the same scale ? 
 
§ 79] RESOLUTIOX OF FORCES. 79 
 
 5. In the coustruction diagram the lines are drawn parallel 
 to the forces. Would it be allowable to draw them perpen- 
 dicular to the forces, or in(;lined at any (the same) angle, d 
 for example ? Test by a drawing. 
 
 6. If the forces are in the same straight line, what does the 
 force polygon become ? what the construction diagram ? 
 
 7. 6 is any point in the plane of the triangle J 5 C, and 
 />, E, Faro the middle points of the sides. Show that the 
 system of forces OA, OB, OC is equivalent to the system OD, 
 OE, OF. 
 
 ^ 79. Resolution of Forces. — By means of the parallelogram 
 of forces, a force R can be found equivalent to two forces 
 F^y i^,, acting on a particle A. Conversely, 
 the force R acting at A may be resolved into / "yj 
 
 two component forces F^ , F.^ acting at A, by f^/ ^^r / 
 constructing on AD ( = i?) as diagonal a par- 1^^ j 
 
 allelogram, and taking the sides AB, AC to k Fi b 
 
 represent the components. The problem is 
 similar to that already discussed in Art 20. 
 
 When the two components are at right angles their values 
 
 may readily be found analytically. For example, to resolve 
 
 a force R represented by the line AD into 
 
 . ^y^Q components X, Y at right angles along 
 
 o^ I AB, AC. 
 ^y*^ ! Complete the rectangle AD. 
 
 — — j( 1 The components are represented by the 
 
 sides AB, AC ol this rectangle. Then 
 X= AB= AD cos f^ = 7? cos 0; 
 
 r = AC = AD cos (90 - 6) = R cos (90 - 0) or R sin 6. 
 
 Hence the recta^igtdar component {or resolved part) of a 
 
 force R in a given direction is equal to the product of the 
 
 force and the cosine of the angle between the force and the 
 
 given direction. 
 
 As a check, 
 
 X' + F' = R' cos'' 6 4- R' sin^ 6 
 = R\ 
 which is also evident from the figure. 
 
80 DYN^AMICS OF A PARTICLE. [§ 79 
 
 Examples to be Solved Graphically/. 
 
 1. If a force is resolved into two components, prove that 
 the greater component always makes the smaller angle with 
 the force. 
 
 2. Eesolve a force of 20 pounds into two components eacii 
 of which makes an angle of 60° with it. 
 
 J71S. Each = 20 pounds, 
 
 3. If two forces acting at a point be represented in magni- 
 tude and direction by the diagonals of a quadrilateral, their 
 resultant will coincide with that of the forces represented by 
 two opposite sides. 
 
 4. Explain the boatmen^s saying that there is greater 
 "powder" in a horse hauling a canal-boat with a long rope 
 than with a short one. Is the same true of a steam-tug ? 
 
 5. Discuss the action of the wind in propelling a sailing- 
 vessel. 
 
 [Let AB be the keel, CD the sail. Let the 
 force of the wind be represented in magni- 
 tude and direction by EF, The component 
 GF of FF, perpendicular to the sail, is the 
 effective component in propelling the ship; 
 the other component, FG, parallel to the 
 sail, is useless. But GF drives the ship 
 forward and sidewise. The component 
 GH of GF, perpendicular to AB, produces 
 side motion or leeway, and the other component, HF, along 
 the keel produces forward motion or headway.] 
 
 6. Discuss the action of the rudder of a vessel in counter- 
 acting leeway. 
 
 7. Show that one eifect of the action of the rudder is to 
 diminish the vessel's motion. 
 
 Examples to he Solved Analytically, 
 
 8. Find the rectangular components of a force 10 'when 
 e = 60°, 90°,_120°, 180°, 240°, 270°, ^00°. Ans. 5, 5 V3',_0, 
 10; - 5, 5 Vd; - 10, 0; -5,-5 \^3; 0, - 10; 5,-5 V3. 
 
 [Draw a figure for each case, and explain the sign of the 
 result.] 
 
§ 80] COMPOSITION OF FORCES. 81 
 
 9. Resolve a force of 10 pounds into two equal components, 
 one of them making an angle of 45° with the force. 
 
 Ans. 7 pounds, nearly. 
 
 10. The force on a smooth surface from a N. E. wind of 
 strength i^is equal to the force from a north wind of strength 
 F/V2. 
 
 11. Find that rectangular component of a force of 10 
 pounds which makes an angle of 90° with the force. 
 
 12. The pull on the rope of a canal-boat is 100 pounds and 
 the direction of the rope makes an angle of 60° with the par- 
 allel banks. Find the force urging the boat forward. 
 
 A71S. 50 pounds. 
 
 13. Show that the components of a force F in two direc- 
 tions making angles of 30°, 45° with it on opposite sides are 
 2 F/{1 + i^) and V2F/{1 + VS). 
 
 It the angles are /?, y, show that the components are 
 Fsin j3/sin (fi -{■ y) and i^sin ;//sin (ft + y). 
 
 14. In a direct-acting steam-engine the piston-pressure P 
 is equivalent to P tan d perpendicular to its line of action 
 and P sec d along the connecting-rod, d being the angle of 
 inclination of the connecting-rod to the line of action of the 
 piston. 
 
 15. If F be the force of the wind, and /?, y the inclinations 
 of wind and sail to the keel AB of a boat, then (see ex. 5) 
 
 headway force = i^sin {y — /?) sin y, 
 leeway force = i^sin [y — /3) cos y, 
 
 16. Show that the headway force is greatest when the sail 
 bisects the angle between the boat and the wind. 
 
 80. Co7nposition of Forces (A7ialytical Method). — The ana- 
 lytical method of resolving a force into its / 
 components leads us to a method of combin- 
 ing forces which is often more convenient 
 than the graphical method given in Art. 78. 
 The two methods may be used to check one 
 another. 
 
 Take three forces i^, , F^, F^, acting on 
 a particle 0. Through draw any two ^ 'j ^""'^ 
 lines OX, OY at right angles to each other, and let 6^, , 0^, 6^ 
 
82 DYNAMICS OF A PARTICLE. [§ 81 
 
 denote the angles which the directions of F^, F^, F^ make 
 with OX. The components of 
 
 F^ are F^ cos 6^ along OX, F^ sin 6^ along OF; 
 F^ are F^ cos 6- along OX, F^ sin ^^ along OF; 
 i^3 are i^'3 cos 6^ along OX, i^3 sin 6^ along OF. 
 
 The components along OX being in the same straight line, 
 may be combined by addition into a single force X; that is, 
 
 F, cos 6^ + F^ cos e^ + F^ cos 6^ = X. . . (1) 
 
 Similarly, the components along OY, being in the same 
 straight line, may be combined into a single force F, or 
 
 F^ sin 6^ + i^, sin 0^ + i^3 sin 6^3 = F. . . (2) 
 
 Hence the original forces are equivalent to two forces X, F 
 acting in directions OX, OF at right angles to each other. 
 The resultant of X, F must therefore be the resultant of the 
 original forces. Call it R, and let d be the angle it makes 
 with the axis of X ; then 
 
 i2 cos 6> = X, i? sin ^ = F. ... (3) 
 
 Square and add (remembering that cos" 6 + sin' 6=1), and 
 
 R = VX' + F', (4) 
 
 which gives the 7nagnitude of the resultant. 
 
 Divide the second of equations (3) by the first, and 
 
 tan 6 = F/X, (5) 
 
 which gives the direction of the* resultant. 
 
 Hence, since the resultant acts at 0, it is known in position, 
 magnitude, and direction, and is completely determined. 
 
 81. If we equate the values of X, F in equations (1), (2), 
 (3), we find 
 
 Rco8e= F, cos 6*, + F^ cos B^ + F, cos B^-, 
 Rsind = F^ sin 6^ + F^ sin 6^ + F^ sin 6^, 
 
81] 
 
 RESOLVED PART OF A FORCE. 
 
 83 
 
 Now OX, OK are any two rectangular axes. Hence 
 
 The component in any direction of the resultant of a num- 
 ber of forces is equal to the sum of their components in the 
 same direction. 
 
 This principle may readily be proved geometrically. 
 
 For let F^y I\, t\ be three forces acting at 0. Find the 
 resultant R[= AD) by constructing the polygon A BCD 
 (Art. 78). 
 
 Let OX be the line along which the forces are to be re- 
 solved. From A, B, C, D let fall perpendiculars on OX. 
 Then evidently 
 
 ad = db -\- he -\- cd. 
 
 But ad is the component along OX of i?, ah of F^ , he of 
 F^i and cd of i^,. Hence the proposition is proved. 
 
 Ex. 1. Three forces of 6, 8, 10 pounds act on a particle at 
 angles of 120'" to each other. Find the resultant in magni- 
 tude and direction. (Solved graphically, p. 78.) 
 
 Since the direction of OJTis arbitrary, we may take it along 
 one of the forces. 
 
 (a) Take OX to fall along the force 6. 
 Then 
 
 X= 6 cos 0° + 8 cos 120° + 10 cos 240° 
 = 6 - 8 cos 60° - 10 cos 60° 
 = -3; 
 
 r = 6 sin 0° + 8 sin 120° + 10 sin 240° 
 = 6 sin 0° -f- 8 sin 60° - 10 sin 60° 
 
 K^ 
 
 'O « 
 
 .-. R- i^G-f 3 = 21^3; 
 
 tan 6/ = - -/S/ - 3 = l/f^S 
 
 or the resultant is perpendicular to the force 8. Plot it. 
 
 and e = 30° or 210' 
 
84 DYNAMICS OF A PARTICLE. [§ 83 
 
 (b) Take OX along the force 8. Then 
 
 X= 8 cos 0° - 6 cos 60° ~ 10 cos 60° 
 = 0; 
 
 Y=8 sin 0° - 6 sin 60° + 10 sin 60° 
 = 21/3. 
 
 .-. E - 4^0^12 = 2^3, as before; 
 
 tan 6 = 21^3/0 = ^ and 6 = 90°, 
 
 or the resultant is perpendicular to the force 8, and in the 
 same position relative to the forces as before. 
 (6-) Take OX along the force 10 and solve. 
 2. The forces of 2, 1, 4 pounds are inclined to the axis of ^Y 
 at angles of 0°, 60°, and 180°. Find the inclination of their 
 resultant to the same axis. Ajis. 150°. 
 
 ^ 3. Two forces of 1 and 2 pounds act at an angle of 120°. 
 Show that the direction of the resultant is perpendicular to 
 that of the smaller force. 
 
 > 4. A BCDEF is a regular hexagon. Show that the result- 
 ant of the forces represented by AB, AC, AD, AE, AF \% 
 6AB. 
 
 ' 5. A particle placed at the centre of an octagon is acted on 
 by forces in directions tending to each of the angles of the 
 figure and of magnitudes taken in order of 4, 6, 8, 10, 12, 14, 
 16, 18 pounds. Find the magnitude of the resultant and the 
 angle it makes with the force 8. 
 
 Ans. 8 ^^4 + 21^2 pounds; tan-^ (1 + V2), 
 6. Three smooth pegs are driven into a vertical wall and 
 form an equilateral triangle whose base is horizontal. Two 
 equal weights of 10 lb are connected by a thread which 
 is hung over the pegs; find the pressure on each peg. 
 
 Ans. 10 V3, 10\^2 - I^S" pounds. 
 ^^ 7. Three forces P, Q, R act at angles a, §, y. Find their 
 resultant. 
 Ans. \/{P'^Q''^R''-^2QB. cos af+2i2Pcos I3-^2PQ cos y). 
 
 82. We have now the means of combining the separate 
 forces that act on a particle into a single force producing the 
 same motion as the separate forces. If the forces act so as 
 to neutralize each other, this single force vanishes and the 
 condition of the particle as to rest or motion will remain 
 
§ 84] EQUILtBRItJM tJKDER THREE FORCES. 85 
 
 unchanged. If at rest before the forces begin to act it will 
 remain at rest; if in motion, it will continue to move with 
 uniform velocity in a straight line (Art. 4G). 
 
 Thus suppose a particle acted on by a number of forces 
 whose resultant is U. If a force equal and opposite to R 
 be applied, the resultant of the 
 forces now acting on the par- 
 ticle is nil and the acceleration 
 produced is nil. The forces are 
 said to eqiiilibratd one another, and the particle is said to be 
 in equilibrium. 
 
 A particle is in equilibrium so long as its condition of rest 
 or motion remains unchanged. Equilibrium therefore does 
 not imply rest, but rest implies equilibrium. That branch of 
 dynamics which considers the circumstances for which equi- 
 librium is possible is called Statics. 
 
 When the forces do not equilibrate an acceleration arises 
 from the resultant force, and the particle has a motion com- 
 pounded of the motion in its original path and that due to 
 the resultant. That branch of dynamics which considers the 
 circumstances under which change of motion takes place is 
 called Kinetics. 
 
 STATICS OF A PARTICLE. 
 
 83. The condition of equilibrium of forces acting on a par- 
 ticle is, in general terms, that the resultant of these forces 
 shall be zero, or, in other words, that any one of the forces 
 shall be equal and opposite to the resultant of the others. In 
 most cases, however, more specific rules are necessary. The 
 more important of these we shall now consider. 
 
 84. A. Equilibrium tinder Three Forces — LamVs Tlieorem, 
 — Let three forces i^,, F^, F^ not in the same straight line act 
 on the particle A, Let F^ , F^ be represented hy AB, A C, re- 
 spectively. Find the resultant {R= DA) of F^, F^ by com- 
 pleting the parallelogram ABDC. For equilibrium to exist 
 R and F^ must be equal and opposite. 
 
86 DYNAMICS OF A PARTICLE. [§ 84 
 
 Now in the parallelogram the side DC i^ equal and paral- 
 lel to BA. Hence the sides DC, CA, AD of the triangle Z> C'^ 
 are proportional to the forces. 
 
 Notice that the directions of the forces are the same way 
 round the triangle DC A. Thus if the first force is in the 
 direction DC, the second must be in the direction CA, and 
 the third in the direction AD, Notice also that the ratios of 
 the sides of the triangle are the same as those of any similar 
 triangle. Hence 
 
 If three forces in the same plane acting on a particle keep 
 it in equilihrium they may he represented in ^nagnitude and 
 direction {but not in position) by the three sides of a triangle 
 taken the same tuay round. 
 
 The converse of this proposition is known as the tricmgh^ 
 of forces. 
 
 By means of the triangle DC A we may find a relation be- 
 tween the forces and their included angles. If a, jS, y be 
 the angles between the directions of the forces, then in the 
 triangle the angles are evidently 180° — a, 180° — /?, 180° — ;/; 
 and since the sides of a triangle are as the sines of the oppo- 
 site angles, 
 
 DC/sin (180° - a) = CA/sin (180° - /3) 
 = AD/s'm (180° - y), 
 
 or 
 
 FJsin a = FJs'm /3 = FJsin y; 
 
84] 
 
 EQUILIBRIUM UNDER THREE FORCES. 
 
 87 
 
 that is, when three forces acting 07i a particle keep it in equi- 
 librium, each is proportional to the sine of the angle between 
 the directions of the other two forces. 
 
 This is known as Lami's Theorem. It was first j^ublished 
 by Father Lami in his Mecanique in 1687, the same year that 
 Newton's Principia appeared. 
 
 Illustration. — Take a piece of board and drive in three 
 smooth pegs. A, B^ C, or place 
 three pulleys at A, B, C. Run 
 strings over the pegs, and knot 
 together as at 0. Suspend 
 weights from the strings. Draw 
 lines along the strings on the 
 board, and plot the triangle abc 
 with sides parallel to these lines. 
 The sides of this triangle will 
 be found to be proportional to 
 the weights. Thus make the 
 weights 3, 4, 5 oz., and it will be found that one angle of the 
 triangle abc will be 90°. 
 
 Ex. 1. If two forces acting at a point are represented in 
 magnitude and direction by two sides of a triangle, show that 
 the third side represents the resultant. 
 
 2. Can forces of 5, 6, 13 pounds acting on a particle keep it 
 at rest ? 
 
 What is the least force that will produce rest ? 
 
 Ans. 2 pounds. 
 
 3. If three forces acting at a point are represented by the 
 sides of a triangle taken the same way round, show that they 
 equilibrate. [Triangle of forces.] 
 
 4. If three forces acting at a point equilibrate, show that 
 they may be represented by the three sides of a triangle per- 
 pendicular to the directions of the forces and taken the same 
 way round. 
 
 5. Three forces acting at a point are determined by the 
 three median lines of a triangle; show that they equilibrate. 
 
 6. Show that any force may be resolved into three forces 
 of given magnitude, the direction of one of these forces being 
 assumed. 
 
 7. Three forces P, Q, R acting on a particle equilibrate. 
 The angle between P and Q is 90°, between Q and R 120°; 
 show that 
 
 P:Q: R=V^:1:2. 
 
88 DYNAMICS OF A PARTICLE. [§ 85 
 
 85. B. Equilihriiim under any Forces {Graphical Con- 
 dition). — If in the force diagram (Art. 78) the direction of 
 R is reversed, the particle will be in equilibrium under the 
 action of F^^ F^, F^, F^, S, where S is a force equal and op- 
 posite to E. Indicate the directions of these forces on the 
 construction diagram and notice that they are the same way 
 round. Notice, too, that ABCDEA is a closed polygon. 
 Hence 
 
 If any number of forces m the same plane acting on a par- 
 ticle keep it in equilibrium, they may be represented in mag- 
 nitude and direction by the sides of a closed polygo7i taken 
 the same way round. 
 
 This is the graphical co7idition of equilibrium of forces 
 acting on a particle. 
 
 Ex. 1. A BCD is a quadrilateral, and the point of inter- 
 section of the lines bisecting the opposite sides. Show that 
 forces represented by OA, OB, OC^ OD acting at equili- 
 brate. 
 
 2. ABCDEF is a regular hexagon. Show that equal forces 
 acting along AB, CD, EF, AF, ED, CB equilibrate. 
 
 3. If forces acting at a point are represented by the sides 
 of a polygon taken the same way round, show that they equil- 
 ibrate. [Polygon of forces: converse of Art. 85.] 
 
 4. If four forces act at a point and are represented in magni- 
 tude and direction by the sides of a quadrilateral, either they 
 keep the point at rest or have a resultant which is double a 
 force represented by one of the sides or one of the diagonals, 
 or four times the line joining the middle points of the diagonals. 
 
 86. C. Equilibinum under any Forces {Analytical Con- 
 dition). — The analytical equivalent of Art. 85 may be deduced 
 from Art. 80. For if the forces acting at equilibrate, the 
 resultant R must be equal to zero. Hence 
 
 which, since X' and F' are both positive, can only be satis- 
 fied by X= 0, r = 0, that is, by 
 
 t F, cos ^, -f i^, cos 6', + . . . = 0, 
 ■^ F, sin 6^ + F, sin i9, + . . . = 0. 
 
§ 8G] EQUILIBRIUM UNDER ANY FORCES. 89 
 
 Ileuce if any number of forces in the same plane acting on 
 a particle keep it in equilibrium, the sums of the co77iponents 
 of the forces along any tivo straight lines at right angles to 
 each other through the particle are equal to zero. 
 
 These are the analytical conditions of 
 equilibrium when any number of forces 
 act oii a particle. 
 
 Ex. 1. A rod .45 whose weight may be 
 neglected is hinged at A, and supports i> 
 a weight W at B. It is lield up by a I 
 thread BC fastened to a fixed point C 
 vertically above A. It AB is horizontal 
 and angle ABC= 30", find the pull T^ 
 of the thread, and the thrust T^ along the rod AB. 
 
 [The point B is in equilibrium under tlie forces T„ 7\, W. 
 We have therefore the two relations, resolving along a hori- 
 zontal line through B as BA, w 7^, ^ 
 
 r, cos 0° - r, cos 30° - TTcos 90° = 0, ^''° '^'^'' ^ 
 and along a vertical line through B 2>& BW, ^ ^ - /V 
 
 r, sin 0° - r, sin 30° + If sin 90° = 0. 
 
 Solving these equations, 
 
 T, = 2 W, T, = V^W. ] 
 
 2. Solve Ex. 1 by Lami's theorem. 
 S3. A thread whose length is 21 is fastened at two points A i 
 
 and B in the same horizontal and distant I from each other. 
 The thread carries a smooth ring of weight W. Find the pull 
 of the thread. A^is. W/ Vx 
 
 y-4. In a canal with parallel banks a boat is moored by two 
 ropes attaclied to posts on the banks. If the ropes are in- 
 clined at angles of 30°, GO°to the banks, compare the pulls in 
 them, both ropes being in the same horizontal plane. 
 
 Ans. 1 : Vs. 
 -V 5. A man weighing 160 lb rests in a hammock suspended 
 by ropes which are inclined at 30° and 45° to two vertical 
 posts. Find the pull in each rope. 
 
 Ajis. 117.1 and 82.8 pounds. 
 
90 DYNAMICS OF A PARTICLE. [§ 87 
 
 6. A man weighing 160 lb is seated in a loop at the end of 
 a rope 10 ft 3 in long, the other end being fastened to a point 
 above. What horizontal force will pnll him 2 ft 3 in from 
 the vertical, and what will be the pull on the rope ? 
 
 Ans. 36 pounds; 164 pounds. 
 
 KINETICS OF A PARTICLE. 
 
 87. If a number of forces act on a particle and the result- 
 ant be found, a certain motion is due to this resultant. If 
 the particle has this motion, it is said to be free ; if it has 
 some other motion, the deviation must be owing to the en- 
 trance of some cause not accounted for, and the motion is 
 said to be constrained. In free motion the particle is iso- 
 lated from all causes tending to affect its motion except the 
 acting forces, while in constrained motion this is not the 
 case. 
 
 We have seen that the position of a particle is defined by 
 its coordinates with reference to certain axes assumed to be 
 fixed. A change in position is represented by changes in 
 these coordinates. Hence the coordinates being either a dis- 
 tance and two angles or three distances, a point is said to 
 have three degrees of freedom to move. 
 
 If the point is compelled to remain in an assigned plane 
 (as the plane of the paper), its position is defined by two co- 
 ordinates, and it is said to have two degrees of freedom and 
 one degree of constraint. Similarly, if compelled to remain 
 at the same distance from a fixed point it would move on the 
 surface of a sphere, and have two degrees of freedom and one 
 of constraint. 
 
 Again, if the point were compelled to remain in two planes, 
 that is, in their line of intersection, it would have one degree 
 of freedom and two of constraint: so also if compelled to re- 
 main in one plane and keep at the same distance from affixed 
 point, that is, to move in a circular path. 
 
 If compelled to remain in three planes, it can have only one 
 position, their point of intersection, and is therefore wholly 
 constrained. 
 
§ 89] FREE MOTIOK. 91 
 
 88. Free Motion. — By means of the second law of motion 
 (Art. 50) connecting force acting, particle acted upon, and 
 acceleration produced, we are able to extend the geometrical 
 properties of motion to particles acted upon by given forces. 
 We shall first of all consider the particle to be unconstrained 
 and have all degrees of freedom. 
 
 Various cases may arise depending on whether the force 
 acting is in the direction of motion of the particle, and 
 whether this force is or is not constant. 
 
 89. (A) Force Constant. — Let a particle weighing w lb be 
 acted on by a force F pounds in the direction of its motion. 
 The acceleration a produced is found 
 
 from F = wa/g, or ? ^ j* 
 
 a = Fg/to. 
 
 Let be the initial position of the particle, P its position 
 at the end of a time t, and let OP = s. It u is the initial 
 velocity of the particle, the velocity v at the end of the time t 
 is composed of that due to u and that due to the acceleration 
 a. Hence (Art. 24) 
 
 V = u -\- at = u -\- Fgt/w; (1) 
 
 and the distance 5 passed over is given by 
 
 s = ut-}- iaf = ut-{- iFgtyw (2) 
 
 The same results follow from the differential equation of 
 motion. For let a particle of weight tv be acted on by a force 
 F. If s is the distance of the particle from the starting-point 
 at time t, the equation of motion may be written (Art. 67) 
 
 w d^8 _ ^ 
 "gW^' 
 
 or, since w, F, g are constant, 
 
 d's 
 
 where the constant quantity Fg/iv is denoted by a. 
 
9,^ DYiTAMlCS 0^ A PARTICLE. [§ 90 
 
 This equation may be developed as in Art. 27, and the 
 values of v, s given above found. 
 
 / Ex. 1. An ice-boat weighing 1000 lb is driven for 30 sec 
 from rest by a wind force of 100 pounds. Find the velocity 
 acquired and the distance passed over. 
 
 Ans. 96 ft/sec; 1440 ft. 
 i- 2. A train of 100 tons is running at the rate of 45 miles an 
 hour. Find what constant force is required to bring it to 
 rest in one minute. Ans. 6875 pounds. 
 
 ;f 3. The pull of the engine on a train whose weight is 100 
 tons is 1000 pounds. In what time will the train acquire a 
 velocity of 45 miles an hour ? A^is, 6 min 52^ sec. 
 
 / 4. What pressure will a man weighing 150 pounds exert on 
 the floor of an elevator descending with an acceleration of 4 
 ft/sec'' ? 
 
 [The force causing the man to move is the resultant of the 
 force of gravity on the man (150 pounds) and the upward force 
 iVof the platform on the man, that is, to 150 — iV^ pounds. 
 
 But this force causes an acceleration 4 ft/sec" in a weight 
 of 150 lb. Hence, from Newton^s second law, 
 
 150 - AT = 150x4/32 
 
 and N= 131.25 pounds.] 
 
 5. Find the pressure when the elevator is ascending with 
 the same acceleration. 
 
 > 6. The scale-pans of a balance each weigh WVo and weights 
 W^, W^ are placed in them. Find the pressures on the pans 
 during the motion. 
 
 Ans. 2W,{W-{- Tr,)/(2 Tr+ W^ + IFJ pounds and 
 2 Tr,( r -h T^;)/(2 W+ W, -f W,) pounds. 
 
 7. A man who is just strong enough to lift 150 lb can lift 
 a barrel of flour of 200 lb from the floor of an elevator while 
 going down with an acceleration of 8 ft/sec\ 
 
 8. A chain 16 ft long is hung over a smooth pin with one 
 end 2 ft higher than the other end and then let* go. Show 
 that the chain will run off the pin in about 7/5 second. 
 
 90. Falling Bodies. — A case of special interest is that of 
 bodies falling freely towards the earth. The acting force is 
 the force of gravity which produces an acceleration g ft/sec" 
 vertically downward if the motion takes place in a vacuum, 
 and this no matter what the body may be. For places near 
 
§ 921 FALLING BODIES. 93 
 
 the earth's surface the value of g is practically constant. 
 (See Art. 62.) 
 
 The doctrine of Aristotle (B.C. 384-322) that the rate at 
 which a body falls depends upon its weight was held until 
 the time of Galileo (a.d. 1564-1642), who in 1590, by letting 
 fall weights from the leaning tower of Pisa, demonstrated its 
 falsity. By giving the true theory of falling bodies Galileo 
 laid the foundation of the modern science of motion. Its 
 development was carried forward by Huygens (1629-1693), 
 who first discussed the dynamics of a system of bodies, Gali- 
 leo having confined himself to that of a single body. It was, 
 Iiovvever, reserved for Newton to raise it to the rank of an 
 exact science. 
 
 91. In the case of a body falling in the open air it is not 
 true that the motion is uniformly accelerated. The air offers 
 a certain resistance to the motion. The exact law of this re- 
 sistance is not known. But for simplicity of treatment, and 
 as close enough for most purposes and for moderate heights, 
 it is assumed that a body falls in the air with a constant ac- 
 celeration g ft/sec'. 
 
 Hence if a body is projected vertically downward with a 
 velocity u, its velocity v at the end of a time t would be 
 found by putting ^ = « in equation (1), Art. 24, or 
 
 V = u-\- gt, 
 and the vertical distance y fallen through by putting g = a 
 in equation (3), or 
 
 y = ut-{-gty2. 
 
 Also by eliminating t, 
 
 v'^ft = wV2 -f gy, 
 
 which gives the velocity at any distance y fallen through. 
 
 Ex. Give the corresponding formulas when the body is pro- 
 jected vertically upward. [Write — g for ^.] 
 
 92. If the body falls from rest these equations reduce to 
 
 v = gty 
 
 y = gty^, 
 
 ^^ 
 
94 DYNAMICS OF A PARTICLE. [§ 93 
 
 The last equation gives the value acquired by a body fall- 
 ing through a height h to be 
 
 V =. V'Zgh, 
 
 and is sometimes said to be the velocity due to the head h. 
 Conversely, 
 
 h = vy2g, 
 
 which would give the head due to the velocity v. 
 
 93. The same results follow from the differential equa- 
 tion of motion. For let a body weighing w lb be projected 
 vertically downward with velocity u. The acting force being 
 w pounds, the equation of motion is 
 
 to cVy 
 
 — -—■==. in ' 
 
 9 dt' ' 
 d'y 
 dP = ^' 
 
 the axis of y being vertically downward. 
 
 This equation may be developed as in Art 27. The results 
 are 
 
 V or dy/dt — u -\- gt, 
 
 y = id + igt\ 
 as already found. 
 
 If the body were projected vertically upward the equation 
 of motion would be 
 
 dt' ~ ^' 
 and thence v =^ u — gt, 
 
 y — ut — igt', 
 
 94. Special Problems. — A body is projected vertically up- 
 ward with a velocity u ft/sec. 
 
 t{a) To find the time of reaching the highest point of its 
 ath. At the highest point the vertical velocity v = 0, or 
 
 u — gt = 0, 
 and t = u/g sec. 
 
§ 94] FALLING BODIES. 95 
 
 f^ (b) To find the time of flight. 
 
 This would be the time of reaching the starting-point. 
 At this time y = 0, or 
 
 ut — igt^ = 0, 
 and ^ = or 2u/g sec, 
 
 which shows that the body is twice at the starting-point, 
 once at the beginning of the motion when ^ = and again at 
 the end when t = 2ii/g. The latter being the time from the 
 beginning to the end of the motion is the time of flight. 
 The time of rising has been found = ti/g sec. 
 .-. time of falling = ^u/g — v/g = u/g sec, 
 or time of rising = time of falling.* 
 -^ (c) To find the greatest height reached. 
 This will be the value of y at the time u/g. 
 Substitute this value for t in 
 
 g = ut-igt% ^ ' ^t/ V^T>^ 
 and we find y = u^/2g ft, n "^ , , , ^ 
 
 the greatest height. V ^ J 
 
 Or it may be found by putting y in the form 
 
 w' r uVg 
 
 noting that y is greatest when t — u/g = 0, and then 
 
 y = uy2g ft 
 as before. 
 
 The result also follows by putting v = in 
 
 t;V2 = uy2 - gy. 
 
 SL^ (d) To find the velocity at any point of the path. Here 
 
 v = u — gt, (I 
 
 * Descartes wrote to Mersenne : " I am astonished at what you tell 
 me of having found by experiment that bodies thrown up in the air 
 take neither more nor less time to rise than to fall again." 
 
96 DYNAMICS OF A PARTICLE. [§ 95 
 
 which gives the velocity at any time f; 
 
 or vy2 = uy2 - gy, ^' 
 
 which gives the velocity at any height y. 
 
 96. Galileo^s experimental demonstration of the formula 
 y z=i gf/2 was as follows. 
 
 He caused balls to slide down grooves in inclined planes, 
 measured off the distances 1, 4, 9, . . . on the grooves, and 
 observed that the times of descent were represented by the 
 numbers 1, 2, 3, . . . He used inclined planes because by re- 
 tarding the descent he could observe the motion more accu- 
 rately than if the balls fell freely. He had to assume that 
 the law of descent was unaltered. 
 
 To measure small times he constructed a clock. This con- 
 sisted of a vessel of water of large cross-section and having a 
 small hole in the bottom. When a ball began to roll the 
 water was allowed to run out and fall on a balance. The 
 orifice was closed when the ball reached the end of its path. 
 The times were as the weights of water discharged. 
 
 Much more precise measurements can now be made with 
 Atwood^s machine and a modern clock. The leading idea in 
 the Atwood machine — that of diluting the acceleration of 
 gravity — is the same as Galileo's, which he effected by the 
 inclined plane. 
 
 Ex. 1. Find the distance passed over by a body falling 
 freely during the sixth second of its fall. 
 
 Dist. fallen in 6 sec = 32 X 672 = 16 X 36 ft; 
 Dist. fallen in 5 sec = 32 X 572 = 16 X 25 ft. 
 
 .-. Dist. fallen in the sixth second (= 16 X 11) = 176 ft. 
 
 1 2. Two bodies are dropped from a height at an interval of 
 2 sec. Find the distance between them at the end of the 
 next 2 sec. Arts. 192 ft. 
 
 4 3. Given that at New York a body falls through 48.24 ft 
 in the second second of its fall, find the value of the accelera- 
 tion of gravity there. A7is. 32.16 ft/sec'. 
 \ 4. A body is projected vertically upwards with a velocity of 
 160 ft/sec. Find (1) when it will come to rest, (2) the height 
 to which it will rise. A^is. 5 sec; 400 ft. 
 -V 5. In Ex. 4 find the velocity when the body is at a height 
 of 256 ft. Ans. 96 ft/sec. 
 f- 6. A body is projected vertically upwards with a velocity of 
 
§ 95] PALLING BODIES. 97 
 
 160 ft/sec. Find at what times it is 256 ft above the start- 
 ing-point and find the total time of flight. 
 
 Ans. 8 sec or 2 sec; 10 sec. 
 ^7. It is required to project a body vertically to a height of 
 36 ft. Find the velocity of projection. A7is. 48 ft/sec. 
 
 > 8. A stone thrown vertically upwards is observed to be at a 
 height of 96 ft in 2 sec. How much higher will it rise ? 
 
 Ans. 4 ft. 
 
 > 9. A stone after falling for one second strikes a pane of 
 glass, in breaking through which it loses one half of its ve- 
 locity. How far will it fall the next second ? Ans. 32 ft. 
 7^10. A body has fallen through 16 ft. With what velocity 
 must another be shot downwards so as to overtake the other 
 in 4 seconds ? Ans. 36 ft/sec. 
 
 11. If s, , 5,, ^3 are the distances described by a falling body 
 in the ;?th, ^'th, rth seconds of its fall, prove 
 
 ^Aq -r)+ 5,(r - p) + s,{ p-q)= 0. 
 
 12. A stone is dropped into a well and after 2 seconds is 
 heard to strike the water. Kequired the distance x to the 
 surface of the water, the velocity of sound being 1100 ft/sec. 
 
 [2 sec = time of fall of stone -|- time of rise of sound 
 = V2x/g-\- ic/llOO. .'.x = 60.5 ft.] 
 
 13. lih is the height fallen through by a body in the nih. 
 second of its fall, show that 2/^/// is an odd integer. 
 
 14. A body falls through the same distance at two differ- 
 ent places on the earth^s surface, and it is observed that the 
 time of falling is t sec less and the velocity acquired n ft/sec 
 greater at one place than at the other. If g^ , g^ be the accel- 
 erations of gravity at the two places, show that 
 
 >f 15. A two-ton hammer falls through 16 ft. Find its ve- 
 locity. Ans. 32 ft/sec. 
 A 16. Three bodies are thrown vertically downwards with 
 velocities w,, u^^ u^ from heights A,, 7?,, li^ and reach the 
 ground at the same instant. Show that 
 
 (A - K)l{u- ,,,) = {h- h,)/{u- «.) = {h.- h,)/{u~ u,). 
 
 17. A man in an elevator which is rising with a uniform 
 acceleration a tosses "a ball vertically upwards with velocity v 
 and' after t seconds overtakes it. Show that 
 
 a-{-g = 2v/t. 
 
98 DYNAMICS OF A PAKTICLE. [§ 96 
 
 96. Projectiles. — If a body be projected with a given ve- 
 locity in a direction not vertical and be acted upon by the 
 force of gravity only, it is called a projectile. Neglecting the 
 influence of the air, the body after projection is subject to a 
 vertical acceleration g duetto the force of gravity, as in the 
 case of a falling body. The path will result from a combina- 
 tion of the motions due to the velocity of projection and to 
 the vertical acceleration g. The name trajectory is often 
 given to it. 
 
 Let be the point of projection, OA the direction of pro- 
 jection, and u the velocity of 
 projection. 
 
 The body if subject to the 
 velocity u only would arrive in 
 time ^ at a point A, where 
 
 A<^. rY OA = ut; 
 
 and if subject to the accelera- 
 tion g only, would arrive at B, where 
 
 dt^'iyi 0B = gf/2, \HJt'(r^Od:^^ 
 But since the body receives both displacements simul- 
 taneously, its actual position is at P, the opposite vertex of 
 the parallelogram OP to 0. The locus of P for different 
 values of t will be the path. 
 
 A good illustration is afforded by a jet of water issuing 
 from a hose-pipe, or from an orifice in the side of a vessel. 
 
 97. In order to determine the properties of the path, it is 
 convenient to find its equation referred to rectangular axes 
 OX horizontal and 01^ vertical. 
 
 Let the velocity of projection u make an angle 6 with OX, 
 and let x, y denote the co-ordinates of P at the time t. Then 
 d_oo ii x = OM 
 
 ,^ = ut COB 6; (1) 
 
 = AM-AP(otOB) 
 
 = ut sin e - gty2 (2) 
 
§ 98] PROJECTILES. 99 
 
 Eliminating t between these two equations,, we find a rela- 
 tion between the co-ordinates x, y which holds for all values 
 of t and is therefore the equation of the path. This gives 
 
 y = x tan d — gx^l^u" cos" 0,^ x^ 
 
 which represents a parabola * — the parabola of Apollonius. 
 
 Ex. Refer to your Analytical Geometry and show that the 
 co-ordinates of the vertex of this parabola are ii^ sin 2d/2g, 
 n^ sin' 0/2g\ and the latus rectum is 2?^' cos' 6/g. What is 
 the distance of the directrix from the vertex ? 
 
 Ans. u" cos'' 0/2g. 
 
 98. Special Problems.— {a) To find the resultant velocity v 
 and the direction of motion at any point P of the path. 
 
 Resolve the initial velocity u into horizontal and vertical 
 components; that is, into u cos 6 along OX, and u sin 6 
 along OY. The horizontal acceleration is zero and the ver- 
 tical acceleration is -- g. Hence at the end of time t 
 
 horizontal velocity v^ — u cos B\ 
 vertical velocity i;, = «^ sin 6^ — gt, K 
 
 Then, x\ and v, being at right angles, 
 
 t;» = v^ + v^ 
 
 = {u COS (^y 4- (u sin 6 — gty 
 = u' - 2ut g sinS-^ gH\ 
 
 and the velocity at any time t is found. 
 This may be written 
 
 V* = u' — 2g(ut sin — gty2), 
 or vy2 = uy2-gy, 
 
 and the velocity at any height y is found. 
 
 * This was first demonstrated by Galileo. No attempt had been 
 made up to his time to explain curvilinear motion of any kind. (See 
 Art. 99.) 
 
100 DYNAMICS OF A PARTICLE. [§ 98 
 
 The direction of motion at P makes with the horizontal 
 OX an angle a such that 
 
 tan a = vjv^ 
 
 = {u sin — gf)/u cos 6 
 = tan — gt/u cos 6, 
 
 giving the direction at any time t; 
 
 or tan a =± yv.^/v^ 
 
 = \^(ti' sin" - 2gtj)/u' cos' d 
 = Vtan' d~^ '^gy/n'' cos" d, 
 
 giving the direction at any height y, 
 
 (b) To find the time of reaching the highest point of the 
 path. 
 
 At this point the vertical velocity v^ = 0. 
 
 u sin d — gt = 0, 
 and t = u sin 0/g, 
 
 the time of reaching the highest point. 
 
 (c) To find the greatest height reached. 
 
 This will be the value of y at the time u sin d/g. Substi- 
 tute this value of t in 
 
 y = ut sin 6 — igt^, 
 and y = u^ sin' 0/2g, 
 
 the greatest height. 
 
 (d) To find the time of flight, that is, the time in which 
 the particle will reach the line OX. At this time y = 0, and 
 
 = ut sin d — -J^r, 
 whence ^ = 0, t = 2u sin 0/g; 
 
 which shows that the particle is twice on the line OX, once 
 at 0, the beginning of the motion, when t = 0, and again at 
 A, the end of the motion, when t = 2u sin 6/g. The latter, 
 being the time from the beginning to the end of the motion, 
 is the time of flight. 
 
 (e) To find the range, that is, the distance OG. 
 
^ 
 
 § 99] PROJECTILES. 101 
 
 The distance OC is described with the constant velocity 
 u cos 6 in the time 2u sin 6/g. Hence 
 
 "^^ ^ - 00 = u coa 6 X 2u sin 6/g = w' sin 26/g, 
 
 the range. 
 
 The greatest value sin 26 can have is unity, and this occurs 
 when 26 =i 90° or ^ = 45°. Hence, for a given velocity, the 
 range is greatest when the angle of projection is 45° and its 
 value is u^/g. This result is not true in practice, as we have 
 not taken into account the resistance of the air. When the 
 resistance of the air is considered, experiment gives an angle of 
 about 34° instead of 45°. 
 
 99. It will be instructive to deduce the preceding results 
 from the differential equations of motion. 
 
 Take the axes of co-ordinates OX, OY horizontal and ver- 
 tically upward, respectively, and let Xy y be the co-ordinates 
 of the particle in its path at the end of the time t when re- 
 ferred to these axes. 
 
 The force acting is wholly vertical. Hence the horizontal 
 component is zero and the equations of motion along OX, 
 01^ are, respectively 
 
 w d*y 
 d'y 
 
 dr'="' ie = -3' 
 
 the minus sign entering since the axis of Y is drawn verti- 
 cally upward. 
 
 Integrating with respect to ^ and noting that u cos 6 is the 
 initial velocity along the axis of X and u sin 6 that along the 
 axis of Y, we have 
 
 ^7 = w cos 6y -Ti = u sm 6 — at, 
 
 dt dt ^ 
 
 Integrating a second time, 
 
 X = tu cos 6, y =1 tu sin 6 — igf\ 
 
 w 
 
 7 
 
 d'x 
 
 dt' 
 
 0, 
 
 
 d'x 
 
 dt' *" 
 
 0, 
 
102 DYNAMICS OP A PARTICLB. [§ 99 
 
 the constants of integration vanishing because when ^ = 0, 
 we have a; = 0, 3/ = 0. 
 
 These four equations completely determine the motion of 
 the particle. 
 
 Eliminating t between the last two equations, we have 
 
 y = 'x tan 6 — gx' /lu"^ cos'* B, 
 
 the equation to the parabolic path as already found (p. 99). 
 Special Problems. — {a) To find the velocity v at any time t. 
 
 -• =©•+(!)' 
 
 = {u COS By -)- {u sin 6 — gtf 
 — u^ — %ui ^ sin 6 -\- g'^f, 
 
 as already found (p. 99). 
 
 Also, if a is the angle which the velocity makes with the 
 horizontal, 
 
 tan a = dy/dx 
 
 — (u sin 6 — gt)/u cos 6 
 = tan B — gt/u cos B, 
 as before (p. 100). 
 
 (b) To find the greatest height and time of reaching it. 
 Make y a maximum in 
 
 y = tu sin B — Igf, 
 
 The first differential coefficient is zero, or 
 
 = dy/dt = u sin B — gt. 
 
 .'. t ~ u sin B/g, 
 
 which gives the time of reaching the greatest height. 
 This value of t makes 
 
 y = u' sin' B/2g, 
 
 the value of the greatest height, as before (p. 100). 
 (b) To find the range and its greatest value. 
 
§ 100] PROJECTILES. 103 
 
 Put y = in the equation to ihe path and 
 
 a; = tan 6^ X 2w' cos' Q/g = iC' sin 26?/^, 
 
 as before (p. 101). 
 
 This is a maximum when 
 
 dx/dB = or cos 2^ = 0, 
 
 which gives i) = 45°, as before. Hence 
 
 greatest range = u^g, as before (p. 101). 
 
 100. According to ancient theories of gunnery the trajec- 
 tory was composed of (i) the mollis violenkis when the ve- 
 locity was so great that the shot flew in a straight line, the 
 extent of the motus violentns being the point-blank range, 
 an error which prevails to this day; (ii) i\\Q motus viixtus, 
 the curvilinear path of the trajectory; (iii) the motus natu- 
 ralis, in which the body fell vertically downwards, as it tends 
 to in reality in the neighborhood of the vertical asymptote 
 when the resistance of the air is taken into account, so that 
 after all the ancient theory was a fair imitation of the true 
 trajectory. 
 
 Galileo combined the motus violentus OA and the motus 
 naturnlis AP into the motus mixtus OP [figure, p. 08] 
 throughout the trajectory, and thus obtained a parabola; 
 but when this theory was accepted by artillerists it was nec- 
 essary to suppose that the velocity of the shot was very much 
 less than its real value, a discrepancy that could not be de- 
 tected till Robins invented his Ballistic Pendulum, 1740. 
 (See Art. 255.) 
 
 When the resistance of the air is taken into account all the 
 simplicity of the preceding theory disappears. (Greenhill.) 
 K Ex. 1. Find the angle of elevation in order to attain a 
 range of 10,000 ft, the velocity of projection being 800 ft/sec. 
 
 Ans. 15°. 
 V 2. A man can throw a ball 100 ft vertically upwards. Find 
 the greatest distance he can throw on a level field. 
 
 Ans. 200 ft. 
 ■^-3. Prove that the range for an elevation of 30° is the same 
 as for an elevation of 60°. 
 
 4. Compare the greatest heights in the two cases in Ex. 3. 
 
 Ans, 1 : 3. 
 
DYNAMICS OF A PARTICLE. [§ 100 
 
 ^ 5. The range is four times the greatest height. Find the 
 angle of projection. A71S. 45°. 
 
 6. The champion eollege record (1893) for throwing a base- 
 ball is 349 ft. How high did the ball rise ? 
 
 A71S. 87 ft 3 in. 
 Show that the time of flight was about 4.7 sec. 
 
 7. Show that numerically 
 
 (twice time of flight in sec)'* = greatest height in ft. 
 
 '' 8. A shot is fired horizontally from a roof 16 ft in height. 
 Find the velocity with which it strikes the ground. 
 
 Aus. 32 ft/sec. 
 ■^ 9. A ball is fired at an angle of 45° so as just to pass over a 
 wall 10 ft high at a distance of 100 yards. How far from the 
 wall will it strike the ground ? A7is. 10.34 ft. 
 
 10. A body is projected horizontally from a given height h 
 with a velocity u. Prove that the equation to the path is 
 
 2u^y = gx^. 
 
 Show that the range is uV2h/g ft, and find the time of 
 flight. 
 
 11. A railroad train is crossing a bridge 64 ft above a river 
 at the rate of 30 miles an hour. At what distance from 
 the abutment must a passenger drop a coin so as just to land 
 at its foot ? Ans. 88 ft. 
 
 \- 1-12. An arrow shot horizontally from the top of a tree 64 ft 
 high strikes the ground 100 ft from the foot of the tree. 
 Find the time of flight and the initial velocity. 
 
 Ans. 2 sec; 50 ft/sec. 
 4- 13. A stone is thrown with a velocity of 80 ft/sec so as just 
 to pass over two trees 100 ft apart and each 50 ft in height. 
 Show that the time of passing between the trees is 2.5 sec- 
 onds and that the angle of throw is 60°. 
 • 14. When the elevation is <a' a projectile falls s ft short of 
 the mark aimed at, and when the elevation is /? it goes s ft 
 too far. Show that if is the elevation for hitting the mark, 
 
 then 2 sin 26^ = sin 2^ + sin 2j3, 
 
 15. Develop a formula for finding the distance to which a 
 stone could be hurled from an l-it sling swung in a horizon- 
 tal circle at n revolutions per second and h ft from the 
 ground. Ans. l.bllnVh. 
 
looT 
 
 PROJECTILES. 
 
 105 
 
 16. If t^ be the time from to any point P of the path of 
 a projectile, and t^ the time from P to A when OA is the 
 range on a horizontal plane, then the height of P above OA 
 
 >17. A ship sailing uniformly in a straight course with a 
 velocity u is fired at from a battery at the instant when she 
 passes nearest to it, her distance then being h. If v be the 
 velocity of the ball, it is necessary in order to hit that 
 
 v" — w' > gh. 
 
 ^18. The inclination of a slope OD to the horizontal is a. 
 From the foot of the slope a gun elevated at an angle d 
 above the horizontal is fired up 
 the slope with a velocity of u 
 ft/sec. Find the range OD and 
 time of flight t. 
 
 [Solved most readily by the 
 method of Art. 96. 
 
 Draw DU vertical to meet OE 
 at K Then 
 
 OE = ut, DE = igt\ 
 
 But the sides of a triangle being proportional to the sines 
 of the opposite angles, 
 
 DE/OE = sin {6 - «)/sin (90 + «)• 
 •'• igt^/'^i = sin {0 — «')/cos a 
 
 and t = 2u sin {6 — a)/g cos a. 
 
 Also OD/OE = sin (90 - 6^)/sin (90 + a), 
 
 and .*. OD = tit cos ^/cos a 
 
 = 2u^ sin ((^ — a) cos 6/g cos' a.] 
 
 19. Find the velocity of projection in order that a shot 
 fired at an elevation of 30° should hit an object half a mile 
 distant on a hillside sloping 1 in 40. A7is. 319 -f- ft/sec. 
 
 20. If in (18) the shot strikes the plane perpendicularly, 
 show that 
 
 2 tan (6 — a) tan a = 1. 
 
 21. In (18) show that the maximum range is u^/gO- + sin «'). 
 
106 I)YNAMICS OF A PARTICLE. [§ lOl 
 
 Show also that the angle of elevation for inaxironm range 
 bisects the angle between the slope OD and the vertical. 
 
 22. Particles projected from a point with velocity u and 
 in the same plane will at the end of any time t lie on a circle 
 whose radius is ut. Plot this circle. 
 
 -^ 23. A particle is projected with velocity u so as to reach a 
 given point a, h. If d^ , 6^ are the possible angles of eleva- 
 tion, show that 
 
 cot e^ + cot 6^ = 2auy{2I)u' + ga^), 
 
 24. A shell discharged from a mortar just touched in its 
 flight the top of a steeple, and in 4 seconds after fell at 
 the distance of 3262f feet from the bottom of the steeple, 
 from whence the report of its fall was heard at the mortar 
 just 12 seconds after the explosion. Find the time of flight 
 and the height of the steeple, taking the velocity of sound 
 1142 ft/sec. Ans. 7 sec ; 192 ft. 
 
 ' 25. ^' Swift of foot was Hiawatha: 
 
 He could shoot an arrow from him 
 
 And run forward with such fleetness 
 
 That the arrow fell behind him! 
 
 Strong of arm was Hiawatha: 
 
 He could shoot ten arrows upward. 
 
 Shoot them with such strength and swiftness 
 
 That the tenth had left the bowstring 
 
 Ere the first to earth had fallen." 
 If one second elapsed between the discharge of each of the ten 
 arrows and' Hiawatha shot at his greatest range every time, 
 he must have been able to run at least at the rate of 98 miles 
 an hour. 
 
 ~^^ 101. (B) Force Variable and in the Direction of Motion. — 
 
 (a) Let the force be directed to a fixed point as 
 Y center of force, and let its magnitude be propor- 
 
 Pf tional to the distance from 0. Let YPO be the 
 
 line of motion, and let F be the position of the 
 particle w at any time t. If the distance "OF is 
 
 denoted by y, the equation of motion along OY is 
 
 w d^y 
 
 -gw=-'y' 
 
§ 101] tHHE liOTlOlf. 107 
 
 where c is a constant, the minus sign being taken because the 
 acceleration tends to diminish y. 
 
 Put eg/ 10 = G?', and the equation of motion becomes 
 
 Multiply both sides of this equation by 2dy and integrate. 
 Then 
 
 {dy/dty = c- odY, 
 
 Let the motion start at a point distant /• from 0. Then 
 when y — Ty dy/dt or v = 0. Hence c = ooi'r^ and 
 
 (dy/dty = oo\r' - f), (2) 
 
 giving the velocity at any point of the path of a body attracted 
 from rest towards the center of force from a distance r. 
 
 To find the time of describing any distance. 
 
 Taking the square root of both members of eq. (2), we find 
 
 dt= - dy/oD Vr' - y\ 
 
 the minus sign being taken because the motion is towards 
 0, and as t increases y decreases. 
 Integrating, 
 
 oat = cos"^ y/Vy (3) 
 
 since ^ = when y = r. 
 This may be written 
 
 y = r cos Got, 
 
 the same equation as found in Art. 35. The motion is there- 
 fore a S.H.M. of period In/oo, of amplitude r, and whose 
 oscillations are isochronous. 
 
 This conclusion is also evident from eq. (2). For v — 
 when y = r ov y = — r, and v has its greatest value when 
 y = 0. Hence the particle starting from rest at P increases 
 in velocity until it reaches 0, decreases in velocity until Q is 
 reached, where OQ = — r, when it comes to rest. From Q 
 
108 DYNAMICS OF A PARTICLE. [§ 101 
 
 it returns under the action of the central force through to 
 P, when it again comes to rest. Hence it oscillates through 
 the distance 2r. 
 
 The time t of moving from P to is found by putting 
 y = in eq. (3), that is, 
 
 and cot = cos"^ = 7r/2, 
 
 t = 7r/2GD. 
 
 The time of an oscillation being four times this, is 2;r/G7. 
 
 Ex. 1. Suppose the earth a sphere 8000 miles in diameter, 
 and that a cannon-ball were dropped at F into a vertical 
 shaft POQ, passing through the center 0. The attraction of 
 the sphere on the i3all varies directly as the distance from the 
 center (Ex. 14, Art. 295). Hence, neglecting the resistance 
 of the air, the ball would oscillate between F and Q. Ee- 
 quired the time of an oscillation. 
 
 The acceleration of gravity at F = g ft/sec\ 
 Hence from equation (1), putting y = r, we have 
 
 and time = ^it /oa = 85 minutes, nearly. 
 
 2. Find the velocity of the ball at the earth's center. 
 
 Ans. 5 miles/sec nearly. 
 
 3. Show that the velocity at any time may be found from 
 V = — Gor sin cot. 
 
 4. Compare the time of falling down the first 2000 miles 
 with that of falling down the second 2000 miles. Ans. 2 : 1. 
 
 Tlie following paragraph shows ignorance of the real con- 
 ditions: 
 
 "Let us now conceive of a huge shaft of 8000 miles in 
 length extending entirely through the earth to our antipodes. 
 If a metal ball be dropped from either end of this shaft it is 
 evident that it will drop ^ down,' but we should not expect 
 to see the antipodean metallic sphere come falling *up' to 
 us, nor would the Celestials in China expect to see the ball 
 dropped from our side come falling 'up' to them." 
 
 (b) Let the force be inversely proportional to the square of 
 the distance from 0. 
 
§ 101] FREE MOTION. 109 
 
 With the same notation as before, the equation of motion 
 may be written 
 
 W-~y 
 
 r - - ..„ (1) 
 
 where c is a constant. The minus sign is taken, since y is 
 measured in a direction opposite to that of the motion. 
 
 Multiply both sides of the equation by 2dy and integrate. 
 Then 
 
 m- 
 
 -f- const. 
 
 Let the motion start at a point distant r from 0. Then 
 when y = r, V or dy/dt ~ 0, and 
 
 ©■-4-1). ■ ■ • • « 
 
 V or 
 
 which gives the velocity at any point of the path of a body 
 falling from rest towards the center of force from a dis- 
 tance r. 
 
 For example, assuming the earth a homogeneous sphere of 
 radius i?, and that its attraction on a body outside of it is the 
 same as if its weight were concentrated at its center (Art. 
 295), we have for the acceleration of a falling body at its 
 surface c/i2'. But this surface acceleration is known to be 
 equal to g, so that 
 
 c/R"" = g, or c = gR*. 
 
 Hence the velocity with which a body falling from a height 
 h would reach the earth's surface is found by putting y = R, 
 r = R -{- h in equation (2), and 
 
 v' = 2c\l/R-l/(R-{-h)\ 
 = 2gR^{l/R-l/(R + h)\ 
 =:2gh/(l + h/R) (3) 
 
110 DYNAMICS OF A PARTICLE. [§ 101 
 
 When the height h is small in comparison with the earth's 
 radius R, the term li/R may be neglected, and we have the 
 ordinary formula for the velocity of falling bodies already 
 given in Art. 92. 
 
 To find the time of reaching the center of force. 
 
 Take the square root of both members of eq. (2). Then 
 
 dt ^ l~^ y 
 
 V 
 
 dy V 2c Vry — y^ 
 
 the — sign being taken because y decreases as t increases. 
 Integrating between the limits y = r and i/ = 0, we find 
 
 t = 7tVP/2V2c (4) 
 
 Ex. 1. If a body fall from an indefinitely great distance it 
 will reach the earth with a velocity of about seven miles a 
 second. 
 
 2. Find the velocity of projection so that a body would not 
 return to the earth. 
 
 3. Show that the time of reaching a point distant y from 
 the center of force is 
 
 Vr/2c\Vry — y^ -\- r cos"-^ '^.^Af* 
 
 4. " Men called him Mulciber : and how he fell 
 
 From heaven they fabled, thrown by angry Jove 
 Sheer o'er the crystal battlement: from morn 
 To noon he fell, from noon to dewy eve, 
 A summer's day; and with the setting sun 
 Dropt from the zenith like a falling star. 
 On Leninos th' ^gean isle." 
 Taking the summer's day 15 hours, show that the distance 
 of Lemnos isle from heaven is about one-fourth of the dis- 
 tance to the moon. 
 
 5. If the earth were suddenly stopped in its orbit it would 
 fall into the sun in a little over two months, the eccentricity 
 of the orbit being neglected. 
 
§102] 
 
 FREE MOTION. 
 
 Ill 
 
 6. 
 
 102. (C) Force Variable and Not in the Direction of Mo- 
 tion. — Let the particle w be projected in any direction and 
 acted on by the attractive force F. The 
 path will lie in the plane passing through 
 the center of force aiid the direction of 
 projection. 
 
 In this plane let be the center of 
 force, OA", OF the axes of coordinates, and 
 X, y the coordinates of the particle P at a 
 time t. Let OP = r, and the angle POX 
 
 Let the force P vary inversely as the square of the distance 
 r from ov F — n/r'^ when n is a constant. 
 
 The equations of motion along OX and OY are (Art. 50) 
 
 to cPx ^ ^ 
 
 g df 
 
 w d'y -n, ' a 
 
 9 df 
 or 
 
 d^ _ 
 df ~ " 
 d^y _ 
 df ~ ■ r 
 when c = ng/w. 
 
 The relation between x and y will give the equation to the 
 path of the particle. To find it : 
 
 Multiply the first equation by y^ the second by x, and 
 subtract. 
 
 n X nx 
 -,T.x---7.. 
 
 
 » V y _ ».v 
 
 
 ex 
 
 w 
 
 • (1) 
 
 
 • (2) 
 
 d^y d^x 
 
 and by integration 
 
 dy dx J ^ , 
 
 x^ — yjj = k,a, constant. 
 
 Also, since x = r cos 0, y = r sin 6, 
 
112 DYNAMICS OF 
 
 A PARTICLE. 
 
 
 
 [§103 
 
 dy dx 
 
 
 • 
 
 • • 
 
 . (3) 
 
 Hence, eliminating r^, 
 
 
 
 
 
 d'x c je 
 
 -T75 = — 1" COS tf^rr, 
 
 df k dt ' 
 
 d'yj 
 df ~ k 
 
 sir 
 
 
 
 and by integration 
 
 
 
 
 
 dx c . a 
 
 ^y -c- 
 dt "'- 
 
 c 
 k 
 
 COS 
 
 
 = - cy/kr, 
 
 (4) 
 
 ■■ ex 
 
 ■ykr, 
 
 (5) 
 
 when Cj, ^2 are constants. 
 
 Multiply the fourth equation by y, the fifth by x, and sub- 
 tract, and we have 
 
 k -{- c^y — c^x = cr/k = cVx'^ -\- y^/k, 
 
 the equation to a conic section with the origin at the focus. 
 
 Hence the imth described hy a particle under the action of 
 a central force varying inversely as the square of the dis- 
 tance is a conic section lohose focus is at the center of force. 
 
 This is the case of planetary motion, the sun being at the 
 center of force. 
 
 The further discussion of this problem will be found in 
 works on mathematical astronomy. -^ 
 
 103. Constrained Motion. — To a particle acted on by a 
 force F in an assigned direction a certain path results. If 
 the path differs from this, it must be owing to some' cause 
 which changes the motion, that is, to the action of another 
 force. Hence if the path is prescribed, we may, by adding 
 forces which with the original force will give a resultant 
 which can produce this path, consider the motion free. The 
 discussion will therefore come under the principles already 
 laid down. 
 
§105] 
 
 MOTION ON A HORIZONTAL PLANE. 
 
 113 
 
 104. Motion on a Horizontal Plane.— Resolve the given 
 force i^into its two components F cos d along the plane AB 
 and i^sin 6 at right angles to it, 
 
 6 being the inclinaton of F to 
 the plane. To each of these 
 forces an acceleration is due. 
 But the particle is constrained 
 so as not to move in the direc- 
 tion of the force F sin 6. This can be brought about by as- 
 suming that the plane exerts an equal force F sin 6 in the 
 opposite direction, which force is known as the reaction of 
 the plane. 
 
 As regards the horizontal stress between the particle and 
 the plane, we can say nothing a priori. Experiment shows 
 that it depends on the nature of the surfaces in contact. We 
 shall for the present assume that the stress between the par- 
 ticle and the plane is normal only, or, as it is often expressed, 
 that the plane is smooth. 
 
 If therefore a particle slides on a smooth plane under the 
 action of a force F inclined at an angle 6, the reaction of the 
 plane is i^sin B and the force acting along the plane is F cos ^, 
 which latter, being that to which the motion is due, is the 
 effective force. 
 
 105. Motion on an Inclined Plane. — Suppose a particle P to 
 slide down a smooth inclined plane 
 AB under the action of gravity. 
 If the particle weighs lu lb, the 
 force acting is w pounds vertically 
 downward. 
 
 Resolve the vertical force w into 
 two components to sin 6 along the 
 plane and to cos at right angles 
 to the plane. The motion down the plane is evidently due to 
 the component 2v sin 6 only. 
 
 Hence if a denotes the acceleration down the plane, the 
 equation of motion is 
 
114 DYNAMICS OF A PARTICLE. [§ 105 
 
 w sin 6 = wa/g, 
 and .*. « = ^ sin Q. 
 
 The problem is therefore the same as that of a particle mov- 
 ing freely under an acceleration g sin 6*. 
 
 Let the particle start from rest at 
 to slide down the plane. If the point 
 P is reached in the time t, the velocity 
 
 V attained will be given by (Art. 25) 
 
 V = tg sin d. . . . (1) 
 
 The velocity may also be expressed in terms of the distance 
 OF passed over. Thus (Art. 25) 
 
 v' = 2g sin 6 X OP 
 = 2gx 00 
 
 if P (7 is let fall ± OB. 
 
 This velocity is the same as that attained at C by the par- 
 ticle falling freely from rest through the distance OC (Art. 
 92). 
 
 Hence the velocity acquired on reaching A, the foot of the 
 plane, is found from 
 
 V = V2gJi ....... (2) 
 
 when h = OB, the height of the plane. 
 
 If the particle at had an initial velocity u in the direc- 
 tion of the acceleration down the plane, then, corresponding 
 to eqs. (1) and (2), we should have 
 
 V = u -]- tg sin d; (3) 
 
 V = s/ii" + 2^/7. (4) 
 
 If t is the time of sliding down the plane from rest and I 
 represents the length OA, then 
 
 l = \gsine X t\ 
 
 Hence if I be measured and t be observed, we have a 
 jnethod of computing the value of g. On account of the im- 
 
§ 106] MOTION" ON AN INCLINED PLANE. 115 
 
 possibility of finding a smooth plane an accurate value of g is 
 not to be looked for by this method. Galileo, who first em- 
 ployed it, found g to be about 31 ft/sec\ See Art. 116. 
 
 106. The differential equation of motion of a particle P 
 sliding down an inclined plane is evidently 
 
 w d^s . ^ 
 
 d's . „ 
 
 or ^ = <7 sin e, 
 
 where s is the distance of F from at the time t. 
 
 This equation may be developed as in Art. 27 and the 
 above results obtained. 
 
 ■^^Ex. 1. If a body slide down an incline whose length is 
 twice its height, the acceleration of motion is one half that of 
 the same body falling freely. 
 
 2. The starting-point of a switchback railroad is 121 ft above 
 the terminus. Neglecting friction, find the terminal velocity 
 of a car which starts from rest. Ans. 60 m. an hour. 
 
 rx 3. A body starts from rest and slides down a smooth plane 
 of height h and length I ft. Prove that the distance passed 
 over in I sec is -^glh ft. 
 
 X 4. A car shunted up a 1^ grade with a velocity of 60 miles 
 an hour just reaches the top of the incline. Find the time oc- 
 cupied. Ami. 4 m 35 sec. 
 > 5. A car starts down a 1^ grade with a speed of 15 miles an 
 hour. Find the distance passed over in 2 minutes. 
 
 Ans. 4944 ft. 
 
 ^6. Find the time occupied in sliding down an inclined 
 plane of height h and length I ft. Ans. I V2/gh sec. 
 
 < 7. A weight of 4 lb is drawn up a smooth plane inclined at 
 30° by a weight of 4 lb joined to the first by a thread which 
 passes over a smooth peg at the top of the plane and which 
 descends vertically. Show that the acceleration is 8 ft/sec. 
 If the weight on the plane were 8 lb, what would the mo- 
 tion be? 
 
 . 8. A weight of W lb placed on a smooth plane inclined at 
 an angle ^ to the horizontal is acted on by a horizontal force 
 of W pounds. Show that the acceleration down the plane is 
 (/ V2 sin (& - tt/^). 
 
116 
 
 DYNAMICS OF A PARTICLE. 
 
 [§107 
 
 ^ 9. A body is projected np a smooth plane with velocity u. 
 Show that it will go a distance u^ /2g sin ^ in a time u/g sin 6 
 and then come to rest. 
 
 f 10. Show that the time of descent from a given point to the 
 center of a circle vertically below it is the 
 same as that to the circumference down a 
 tangent. 
 
 11. Prove that the time of descent of a 
 particle starting from the extremity J of a 
 vertical diameter JZ is the same along all 
 chords AB, AC, . . . of the circle. 
 
 Ans. time = 2 Vr/g, 
 AVould the same be true for chords of the 
 circle ending at the lowest point L ? 
 In this case show that the velocity attained on reaching the 
 lowest point varies as the length of the chord. 
 \ 12. Ex. 10 was first given by Galileo in this form : 
 
 Imagine gutters radiating in a vertical plane from a com- 
 mon point A at different inclinations. 
 Place at ^ a like number of heavy bodies 
 and let them descend the gutters simul- 
 taneously. The bodies will at any instant 
 lie on a circle. The radii of these circles 
 increase as the squares of the times. 
 [Show that in general r = gt'^/4:.'\ 
 13. Find the line of quickest descent 
 from a given point ^ to a given straight 
 line BC, 
 
 [Through A draw AB horizontal to meet BC at B. Make 
 BC= BA. Then AC is the line required. Proof?] 
 X 14. The axis of a parabola is vertical. Show that the focal 
 chord down which a particle would slide in the shortest time 
 is inclined to the axis at an angle tan"^ V2. 
 
 107. Equilibrium on an Inclined Plane.* — When a particle or 
 body weighing W lb, placed on a smooth inclined plane, is acted 
 on by gravity only, the acceleration down the plane is due to 
 
 * This is of special interest, as being one of the problems of oblique 
 forces first solved. The solution is due to Simon Stevinus of Bruges, 
 Belgium (1548-1620). It may be found in Whewell's Meclianics, p. 44. 
 
 The modem methods of statics date from Stevinus, 
 
§ 107] EQUILlBRItTM OK AN IKCLIKED PLANE. 
 
 117 
 
 the component along the plane W sin of the vertical force 
 Impounds. For equilibrium to exist an 
 equal acceleration must be applied in 
 the opposite direction; in other words, 
 the resultant force along the plane must 
 be nil. Hence if a force F be applied 
 parallel to the plane it will hold the par- 
 ticle in equilibrium if 
 
 F= Wsin 6. 
 
 Acceleration perpendicular to the plane is prevented by the 
 constraint of the plane, the normal reaction N being equal to 
 
 the normal component W cos 0. 
 Hence for equilibrium we have 
 the two conditions, 
 
 N= W cos 6, 
 
 (2) If a force F be applied par- 
 allel to the base of the plane, we shall evidently have equi- 
 librium provided 
 
 i^cos d= Fsin 6, 
 or F= PTtan (?, 
 
 and N= Wco^d-\- F^mO 
 
 = TTcos 6^4- If tan 6^ sin d 
 
 = Wsec 6. 
 
 These results may also be obtained 
 
 (a) Directly by means of Lami's Theorem. 
 
 (1) When the force is parallel to the plane : 
 
 F/sin (180° -6)^ iV/sin (90° -{- 0) = W/sin 90°. 
 
 (2) When the force is parallel to the base: 
 
 i^/sin(180° - 0) = JVysin 90° = W/sin (90 + 6), 
 
 which easily reduce to the same relations as before. 
 {b) By the method of Art. 86. 
 
118 
 
 DYKAMICS OP A ^ARTICLE. 
 
 r§io7 
 
 (1) When the force is parallel to the plane. 
 Resolve along AC and 
 
 F- TT cos (90 - e)=0. 
 Resolve along a line perpendicular to A 0, and 
 
 jsr- w cos = 0, 
 
 Whence F = W sin 6, JV = W cos 0, as before. 
 
 (2) Similarly, when the force is parallel to the base we 
 find 
 
 F= Wtan e, N= W sec 0, 
 
 (c) By the method of Art. 84. 
 For the particle is held in equilib- 
 rium by the three forces F^ W, N 
 pounds. Construct the triangle of 
 forces by drawing lines parallel to 
 the forces. Solving the right tri- 
 angles, we have 
 
 F= TFsin d, N = W cos 6, 
 F= Wtsiii 6, JSr= Wsecd, 
 
 or 
 
 as before. 
 
 -^ Ex. 1. What force acting horizontally will support a weight 
 of 10 lb on a plane inclined at 45° to the horizon ? 
 
 Ans. 10 pounds. 
 ^ 2. Find the angle of inclination when a force of 10 pounds 
 along the plane supports a weight of 20 lb. Ans. 30°. 
 
 3. A man pushes a garden roller weighing 80 lb up a 
 plank 10 ft 3 in long and with ^ne end 2 ft 3 in abov^e the 
 ground. If the handle is horizontal, find the force applied 
 and the pressure of the roller on the plank. 
 
 Ans. 18 pounds; 82 pounds. 
 \4. Show that a body of weight W resting on an inclined 
 plane will be held in equilibrium by a force F parallel to the 
 plane if 
 
 F: W= height : length, 
 
§ 108] MOTION IK A CIRCLE. 119 
 
 and parallel to the base if 
 
 F : W = height : base. 
 
 \ 5. A cask weighing 400 lb is lowered into a cellar down a 
 smooth slide inclined at 45° to the vertical. It is lowered by 
 two ropes passing under it, one end of each rope being fixed, 
 while two men pay out the other ends. Find the pull ex- 
 erted by each man. Ans. 100/ V2 pounds. 
 x6. Find the force required to haul a train weighing 100 
 tons up a 1^ grade if the force required on the level is 10 
 pounds per ton. Ans. 3000 pounds. 
 
 7. On an inclined plane, if N is the reaction when the force 
 acting is along the plane and iV, the reaction when the same 
 force is horizontal, show tliat the weight = VNN^. 
 
 8. On an inclined plane a force P parallel to the plane sup- 
 ports a weight W, and a horizontal force Q will also support 
 W, Show that 
 
 p-2 _ Q-^ = w-\ 
 
 9. On an inclined plane a force P acting parallel to the 
 plane can support a weight TFj , and acting horizontally a 
 weight W^; prove 
 
 w,' - w: = P\ 
 
 ^10. A body weighing Wlh is kept at rest on a plane whose 
 inclination is 6^ by a force P acting at an angle a with the 
 plane. Show that 
 
 P = Tf sin ^/cos a, 
 
 N = PF cos ( a-{- 6)/co8 a, 
 
 11. If two weights TF, , IT, support each other on a double 
 inclined plane by means of a thread passing over the common 
 vertex of the planes, and ^, , 0^ are the inclinations of the 
 planes, then 
 
 WJW^ = 8m 0,/Bm 0^, 
 
 108. Motion in a Circle. — Suppose a particle weighing w to 
 move with constant velocity v in the circumference of a circle 
 
120 DYNAMICS OF A PARTICLE. [§ 108 
 
 of radius r. It was shown in Art. 32 that the acceleration of 
 motion is always directed towards the center of the circular 
 path; in other words, the particle is constrained by a constant 
 force (7 directed towards 0, and therefore having its direction 
 always perpendicular to the direction of motion of the par- 
 ticle. To this force the name centripetal force is given. 
 
 The motion may be illustrated by supposing the particle 
 attached by a thread to an axis through and to revolve 
 about this axis with uniform velocity, the motion being due 
 to the initial velocity v and to the pull of the thread, and to 
 these alone. 
 
 Suppose that the thread is cut when the particle reaches 
 any point P. Since there is no force now acting, the particle 
 will move in the tangent to the circle at P and with velocity 
 V. This follows from the first law of motion. The pull of 
 the thread acts only in changing the motion from uniform 
 rectilinear to uniform circular motion; it acts towards the 
 center and constantly changes direction as the particle 
 changes place; it acts constantly, and produces an accelera- 
 tion a which changes at every moment the direction of the 
 velocity v without changing its magnitude. 
 
 Consider further. The pull in the thread is a stress, the 
 action on the particle towards the center and the reaction 
 of the particle from 0. The two are equal by Newton^s 
 third law. The first is the centripetal force, and to the other 
 the name centrifugal force is commonly given. 
 
 Notice that the centrifugal force does not act on the mov- 
 ing particle, but is the force with luliicli the moving particle 
 acts upon the constraint luhen it is constrained to move in a 
 circular path. In other words, the centrifugal force is the 
 tension of the thread directed outwards from 0, the center 
 of the circular path. 
 
 The pull of the thread acts as 
 
 a centrifugal force on 0, 
 a centripetal force on P, 
 
§ 109] MOTION- IN A CIRCLE. 121 
 
 y\ 109. To find the magnitude of the centripetal or centrifu- 
 ' gal force. 
 
 If the particle were free to move it would 
 
 proceed in tlie tangent PT with velocity v, 
 
 and after an interval t would be at 2^ when 
 
 PT=vt 
 
 But as it is found on the circumference at 
 Q, it must be deflected by the central force Ca distance TQ 
 in this interval. Let a be the acceleration due to this force. 
 Then 
 
 TQ = iat\ 
 
 Prolong TQ through the center to E. From geometry, 
 
 TQ X TR = TP\ 
 
 Proceeding to the limit, we have, since TR = 2r, 
 
 or a = v*/r , 
 
 the same result as already found in Art. 32. 
 Now, from Newton's second law, ^ 
 
 C = wa/g 
 = ivv'/gr, 
 
 the value of the centripetal or centrifugal force. 
 
 If the whole circumference is described in t seconds, then 
 
 w .¥ . 
 
 t v = 27r r y'^ '^]X p w ^^rr\^ 
 
 and C = 47r^tvr/gt^ pounds, 
 
 if zv is expressed in lb and r in ft. 
 
 If n is the number of revolutions per minute, 
 
 V = 27trn/Q0 ft/sec 
 
 and C = 0.00034wr?j' pounds, 
 
 a rule used by machinists. 
 
122 DYNAMICS OP A PARTICLE. [§ 110 
 
 The term centrifugal force was introduced by Huygens 
 (1G29-I695),and is very often misunderstood. See question 
 60, page 139. It is sometimes defined as the tendency of a 
 body in motion to continue to move in a straight line. This 
 definition may be compared with that given above. 
 
 110. A very remarkable application of the idea of cen- 
 tripetal force was made by Newton to test the truth of the 
 law that the acceleration of gravity 
 varies inversely as the square of tlie 
 distance from the earth's center.* 
 Observation shows that the moon 
 revolves round the earth in an or- 
 bit nearly circular and with uniform 
 velocity. It v = velocity of moon, 
 E = radius of orbit, then the accel- 
 eration of the moon directed to the 
 center of the earth is v^/E. It this acceleration is due to 
 gravity, we have 
 
 ff' = v'/B 
 
 when ^' is the value of g at the distance E from the earth's 
 center. Also, if the acceleration of gravity varies inversely 
 as the square of the distance from the earth's center, 
 
 g'/ff = rV^ 
 
 when r is the earth's radius. 
 
 Hence, eliminating g' , we have, as the condition to be sat- 
 isfied if the hypothesis is true, 
 
 v'E = r'g. 
 
 Now from measurement, E = 240,000 miles, r = 4000-miles, 
 g = 32 ft/sec% time of revolution of moon = 27 days 8 hours 
 
 * "In our day the principle is so familiar that we imagine it must 
 have been an easy step to generalize from terrestrial to celestial me- 
 chanics. Yet neither Kepler the bold nor Galileo the far-seeing had 
 the courage to make such a generalization. Even Newton was very 
 timid in extending terrestrial to celestial laws." — Lewes. 
 
§111] EFFECT OP THE EARTH*S ROTATlOlT. 123 
 
 = 2,300,000 seconds nearly, whence v = 3375 ft/sec. Sub- 
 stitute these values, and the expression will be found to 
 check. 
 
 111. Efect of the EarWs Rotation. — Consider a particle at 
 a point P on the earth's surface. Assuming the earth a 
 sphere, the force on the particle due to 
 gravity alone acts along PO towards 
 the center 0. But in consequence of 
 the earth's rotation the particle receives 
 a centripetal acceleration along PA 
 perpendicular to NS, the earth's axis. 
 Hence the force lu^ exerted by the par- 
 ticle on its support is the resultant of 
 IV along PO and the centripetal force 
 C reversed (Art. 108). 
 
 If the particle were supported by a thread, the direction of 
 the thread would be along w, , so that the direction of w^ is 
 that of a plumb-line at P. We may therefore say that the 
 particle P is in equilibrium under 
 
 w directed to the center of the earth; 
 
 — w, directed upward along the plumb-line; 
 
 — C the reversed centripetal force. 
 
 Let X = the latitude of P = Z POB; 
 r = the radius of the earth; 
 6 = inclination of plumb-line to radius OP. 
 
 Resolving the forces along OP and X to OP, we have 
 (Art. 86) 
 
 — w, cos f^ — C COS \ -{- to = 0, , . . . (1) 
 
 — w^ sin -{- CsinX = 0, .... (2) 
 
 from which to find w^ and ^. 
 
 Now C= i7t\u X PA/gt^ (Art. 109) 
 
 = iTt^wr cos X/gf. 
 
 ■\ 
 
124 DYKAMICS OP A PARTICLE. [§ 112 
 
 But t = 24 hours, r — 4000 miles; 
 
 .-. C=w cos V289, 
 and w^ = w(l — cos' A/289), 
 
 which shows how the attractive force is diminished in conse- 
 quence of the earth's rotation. 
 
 To put it in slightly different form: If g is the accelera- 
 tion due to gravity alone, and g^ is the resultant acceleration 
 in the direction of the plumb-line, then 
 
 and .-. g^ = g{l — cos' A/289), 
 
 which shows the variation of g^ with the latitude. 
 At the equator J where A = 0, experiment gives 
 
 g^ = 32.09 ft/sec'. 
 
 Hence at the equator g = (32.09 -f 0.11) ft/sec', 
 and at latitude A g^ = 32.09 + 0.11 - 0.11 cos' A 
 = 32.09 + 0.11 sin' A ft/sec'. 
 
 Also, from eqs. (1) and (2) 
 
 tan 6* = sin A cos A/289, 
 
 giving the inclination of the plumb-line at any point to the 
 radius of the earth through that point. 
 
 112. Attention was first called by Huygens to the influence 
 of centrifugal force on the acceleration of gravity. A pen- 
 dulum clock taken from Paris to Cayenne by Richer in 
 1671-1673 lost time unaccountably, and when adjusted and 
 brought back to Paris it gained an equal amount. AVhen 
 Halley in 1677 went to the island of St. Helena to observe 
 the stars of the southern hemisphere, he found his clock lost 
 so much that the screw at the bottom of the pendulum did 
 not enable him to shorten it sufficiently. Huygens, by demon- 
 strating the greater centrifugal acceleration of the earth near 
 the equator and the consequent diminution of g, cleared up the 
 mystery. (See Art. 115.) 
 
 Ex. 1. A stone weighing 4 ounces is whirled 90 times a 
 
§ 113] CONICAL PENDULUM. 125 
 
 minute at the end of a thread 3 ft 6 in long. Find the pull 
 of the thread. A71S. 2.4 pounds, nearly. 
 
 2. A locomotive weighing 60 tons is running at 15 miles an 
 hour on a level track round a curve of 3300 ft radius (about 
 1° 44'). Show that the lateral pressure on the rails is 550 
 pounds. 
 
 3. Show that the lateral pressure on the rails by a loco- 
 motive of weight W lb, and running at the rate of v miles 
 an hour on a curve of r ft radius is 0.067 IfV/r pounds. 
 
 4. Find the velocity of projection in order that a bullet 
 shot horizontally may travel round the earth continually. 
 
 Ans. 5 miles per sec. 
 
 5. The center of a steel crank-pin which weighs 16 lb is 
 12 in from the center of the engine-shaft. The shaft makes 
 180 revolutions per minute. Find the centrifugal force aris- 
 ing from the pin. Ans. 178 pounds nearly. 
 
 6. Show that the rubber tire of a bicycle becomes slack 
 when running at more than V7rgdT/W ft/sec, ^yhere W ia 
 the weight of the tire in lb, 7' the tension in pounds, and d 
 the diameter of the wheel in feet. 
 
 7. The attractive force of the earth is diminished in conse- 
 quence of the earth's rotation by 1/289 part. 
 
 8. "We know by the mere consideration of centrifugal 
 force that the whole sun attracts each ton of the earth with a 
 force of a little more than a pound, and that the whole earth 
 attracts each ton of the moon with a force of ten ounces." 
 
 9. Show that the acceleration of a body falling freely at 
 the equator is 1/9 ft/sec' less than it would be if the earth 
 did not revolve on its axis. 
 
 10. A spring-balance graduated at New Orleans, in lat. 30°, 
 will be in error 0.09^ as shown by the beam-balance on tlie 
 boundary between the U. S. and Canada in lat. 45°, and an 
 equal amount at the equator, near the mouth of the Amazon. 
 
 11. If the earth were to revolve on its axis 17 times as fast 
 as it does, no stress would exist at the equator between the 
 earth and a body resting on its surface. 
 
 Per Contra. — The universe is not twice given, with an 
 earth at rest and an earth in motion. It is accordingly not 
 permitted us to say how things would be if the earth did not 
 rotate. (Mach.) 
 
 113. Co)ncal Penduhim. — Suppose a particle of weight w 
 suspended by a thread from a point and caused to swing 
 
126 DYT^AMICS OF A PARTICLE. [§ 113 
 
 about the vertical axis OA with a uniform velocity v in a 
 circular path. Such an arrangement 
 V "^^ A is called a conical pendulum. 
 
 ^ vvs^"^ Al -^^^ ^ ^® *^® length of the thread 
 
 u =-vv^«. '. "y "v/ I^ ^-^^ and let T be the time of a com- 
 
 plete revolution, or the period. 
 
 Let i^ be the position of the par- 
 ^ — A J tide at any time. Denote the angle 
 
 between OB and the vertical OA by 
 6, and the radius AB by r. 
 tc y ,. ' The particle B is acted on by two 
 
 forces — the weight 7V vertically down- 
 wards and the pull P along the thread BO. Since the 
 resultant motion has a uniform velocity v in a circle with 
 center A and radius r, the resultant of the acting forces 
 must be a centripetal force directed to A. The magnitude 
 of this force is tvv'^/gr (Art. 109). 
 
 Hence the particle B would be in equilibrium under P, w, 
 and a force equal and opposite to the centripetal force wv^/gr. 
 Kesolving vertically and horizontally (Art. 86), 
 
 — ^- ^. »~ X^^u^^ - 10 + P cos 6=^0, . . . . (1) 
 ' — 2vv'/gr -{- P sm 6 = (2) 
 
 Also, since T is the period of revolution, 
 
 -, ^'-^* ^2'=^'^'- • • (3) 
 
 From these equations it readily follows that 
 
 tp'. ^V^e^ cos^ = ^yV4^7, (4) 
 
 a wV" ^^P = to/cos 0, (5) 
 
 ■j: - vy I •/-■ P ^ _^ , -^y v' = c/l sin tan (6) 
 
 It IS sometimes convenient to write these relations in a dif- 
 ferent form. Denote the vertical distance OA by li. Then 
 
 T=27tVh/g, (") 
 
 Ph = wl, (8) 
 
 W^gr\ ........ (9) 
 
§114] 
 
 CONICAL PENDULUM. 
 
 127 
 
 The relation (7) shows that the period is the same as that 
 of a simple pendulum of length h (Art. 115). 
 
 114. The conical pendulum may be used as a regulator of 
 mechanical motion. An apparatus 
 depending upon this principle, 
 known as the governor, was ap- 
 plied by James Watt to the steam- 
 engine.* 
 
 In the figure, which represents 
 one form of governor, as the speed 
 of the engine increases the spindle 
 AO revolves more quickly, and 
 the balls separate; as it diminishes, 
 the balls come together. The 
 slide rises and falls accordingly, 
 and by means of a set of levers, C, 
 the steam -valves of the engine are 
 acted ou, and the supply of steam 
 admitted to the cylinder regulated. 
 
 Ex. 1. Obtain the results (1) and (2) by means of Lami's 
 theorem. 
 
 2. By taking moments about A and in succession, show 
 that 
 
 Ph = wly liv^ = gr^. 
 
 3. Show that, roughly, 
 
 h = (twice period )y5. 
 
 4. Show how to find the value of the acceleration g of 
 gravity by means of a conical pendulum. 
 
 [Observe 7', measure h, and compute g from g = in^h/T^.] 
 
 5. A train is running around a horizontal curve of 5445 ft 
 radius at 30 miles an hour. Show that the surface of a basin 
 
 * " If a pair of common fire-tongs suspended by a cord from the top 
 be made to turn by the twislinuj or untwisting of the cord the legs will 
 separate from each other with force proportioned to the speed of rota- 
 tion. Mr. Watt adapted this fact most ingeniously to the regulation of 
 the speed of his steam-engine." 
 
128 DYNAMICS OF A PABTICLE. [§ 115 
 
 of water on the train will be inclined to the horizontal at an 
 angle tan'^ 1/90. 
 
 6. Find the length of a Watt governor that will run 60 
 revolutions per minute. Ans. 9.78 in. 
 
 7. If n is the number of revolutions per second of a conical 
 pendulum having a thread I ft long and a bob weighing w lb, 
 show that the pull of the thread is ATt^bi^w/g pounds. 
 
 8. A skater 5 ft 10 in in height in going round a ring 100 
 ft in diameter leans inward 5 in from the vertical. Find his 
 speed. ■ Ans. 10.7 ft/sec. 
 
 116. Simple Pendulum. — Consider a particle iv suspended 
 
 from a fixed point (7 by a thread of length I and moving in a 
 
 vertical arc under the action of the force 
 
 of gravity. The arrangement is called a 
 
 simple pendulum. (See Art. 119.) 
 
 Let P be any position of the particle. 
 Denote the angle between CP and the 
 vertical CO by 6. 
 
 The force iu on the particle acting ver- 
 tically downward may be resolved into two 
 rectangular components, w sin 6 along the tangent at P and 
 2V cos 6 along PC. The component along PC cannot affect 
 the motion in the arc. The motion of the particle therefore 
 depends on the tangential component only. 
 
 If a denotes the acceleration along the tangent at the point 
 P, the equation of motion is 
 
 w sin 6 = wa/g, 
 and therefore ^5 = ^ sin 0. 
 
 Now if d is expressed in circular measure and is a small 
 angle, we may replace sin 6 by 6. Hence 
 
 a = gdj approximately, if 6 be small; 
 = g (arc OP/radius CO), 
 
 = ^X OP, 
 
 or, since g and I are constant, the acceleration is proportional 
 to the displacement OP, and therefore the motion in the 
 
§ 116] SIMPLE PENDULUM. 129 
 
 small arc OP is a S. H. M. The time t of an oscillation 
 being one half the period, we have (Art. 37) 
 
 t = n Vdisplacement/acceleration. 
 
 a result independent of the length of the arc. Hence in all 
 small arcs the times of oscillation are the same, and the vibra- 
 tions of a pendulum are therefore said to be isochronous, 
 
 Galileo, it is said, first tested this by watching the swinging 
 of a bronze lamp (a masterpiece of Benvenuto Cellini) in the 
 cathedral at Pisa, Italy, and measuring the time by counting 
 the beats of his pulse. 
 
 116. A pendulum which makes one oscillation in one sec- 
 ond is called a seconds-pendulum. If I be its length, we have, 
 since ^ =- 1, 
 
 1 = n^TTg 
 or I — g/T^^. 
 
 An approximate value of g is 32.2 ft/sec'. Hence 
 
 I = 39.12 inches, 
 
 or a pendulum which beats seconds is about 3 ft 3| in in 
 length. [The length of the meter is 3 ft 3| in.] See Ex. lu, 
 p. 131. 
 
 Conversely, if the length I is known, we may find the value 
 of g, the acceleration due to gravity at any place. For then 
 
 g = 7iH = 10/, roughly. 
 
 To test whether the value of g was the same for all bodies, 
 Newton placed different materials in equal small boxes and 
 suspended them by equal long threads. He noted that the 
 time of oscillation was the same for each used as a pendu- 
 lum. The resistance of the air being the same in each case, 
 he concluded that the acceleration due to gravity was con- 
 stant at the same place for all substances, whatever their 
 chemical composition. 
 
130 DYNAMICS OF A PARTICLE. [§ 117 
 
 117. If a loendiilum of length I makes n oscillations in a 
 given time r at a place where the acceleration of gravity is g, 
 then 
 
 r/n — t— TtVl/g. 
 
 Suppose (1) the length of the pendulum changed to / + A, 
 when A is small. By differentiation, 
 
 and — dn = ndl/2l = n\/2ly 
 
 giving the loss in the number of oscillations. 
 
 Thus, if a seconds-pendulum loses 20 sec a day, its change 
 in length would be found from 
 
 20 = 86,400A/(2 X 39.12) 
 or A = 0.018 inch. 
 
 Suppose (2) the pendulum carried to a place where g has 
 the value g -{- y^ the change y being small and the length I 
 remaining the same. Then 
 
 rdn/n' = ^Vl/fdg 
 
 and dn = ndg/2g = 7iy/2g, 
 
 giving the gain in the number of oscillations. 
 
 Suppose (3) the pendulum carried to a height ^ above the 
 earth's surface. Then, since g varies as l/r'^, r being the dis- 
 tance from the earth's center, we have 
 
 r/?i = cr\^l, where c is a constant. 
 
 Hence — rdn/n^ = cVldr, 
 
 — dn = ndr/r = nh/r, 
 
 giving the loss in the number of oscillations. 
 
 (4) If carried to a depth h below the surface, g varies as 
 r, and the loss in the number of oscillations would be nh/2r. 
 
 Ex. 1. If a pendulum of length I vibrates n times in s sec- 
 onds, prove 
 
 l7z''7i'' = qs"^. 
 
§ 118] BLACKBURN PENDULUM. 131 
 
 2. Find the number of oscillations made by a pendulum a 
 yard long in one minute. A7is. G2.57. 
 
 3. A plummet attached to a fine wire vibrates 60 times in 
 3 minutes. P^ind the length of the wire. A71S. 29.36 ft. 
 
 4. A seconds-pendulum makes 11 oscillations more in 24 
 hours at the foot of a mountain than at the top. Find the 
 height h of the mountain. Ans. 1/2 mile nearly. 
 
 5. A pendulum which beats seconds at the surface when 
 carried to the bottom of a mine loses 5 beats in 24 hours. 
 Find the depth of the mine. 
 
 6. A seconds-pendulum is lengthened 1/100 inch. Show 
 that it will lose 11 sec per day. 
 
 7. A clock gains 30 min/week. How much should the 
 pendulum be lengthened for correct time ? 
 
 Ans. 0.006 of its length. 
 
 8. A clock keeps correct time at a place where g is 32.24 
 ft. Show that it will lose 3 min 21 sec/day at a place where 
 g is 32.09 ft. 
 
 9. A pendulum clock is in an elevator which is going down 
 with an acceleration of 1 ft/sec\ Show that the clock loses 
 15/16 sec/min. 
 
 10. To find the length a; of a seconds-pendulum. Count 
 the number of vibrations n made in a day by a pendulum of 
 known length I. Then compute x from 
 
 V^ = ;i/86,400. 
 
 118. The paths found analytically • in the examples of 
 combining S.H.M.'s (p. 36) and many others may be traced 
 mechanically by an apparatus known as 
 T7ie Blackhiirn pendtUum. — 
 
 Two threads ACP, BCP fastened at two 
 points AyB m B. horizontal line are brought 
 together by a small ring at C sliding over 
 the threads. At P is attached a funnel to 
 carry sand or ink to trace the curve made. 
 
 The bob P oscillates in the plane of the paper about C, and 
 also perpendicular to this plane about AB q& axis. Hence 
 for small disturbances P has two S.H.M.'s at right angles, 
 of which the period of that in the plane of the paper is 
 
 '^nVCP/g, and of the other 2n\^PD'/'g. If, for example, 
 CP = PD/4i, the curve traced out will be a parabola. 
 
132 
 
 BTNAMICS OF A TARTIOLE. 
 
 [§119 
 
 X 
 
 ^ -! 
 
 119. Motion on a Vertical Circle, — We shall now discuss 
 the motion of a particle w constrained to move on a smooth 
 vertical circle under the action of gravity. The problem of 
 the simple pendulum is a special case. 
 
 Since a circle may be regarded as the limit of a polygon of 
 an indefinitely great number of sides, the path on the circular 
 
 arc may be treated as a series of 
 inclined planes. The problem is 
 therefore similar to that of Art. 
 105. 
 
 Take the origin at 0, the lowest 
 point of the circle; OX, the axis of 
 X horizontal, and OF the axis of 
 Y vertical, and passing through 
 the center C. 
 
 Let A be the initial position of the particle and P its posi- 
 tion at the end of a time t-, a,l), the coordinates of ^; x, y, 
 the coordinates of P; AP = s, and angle OCP = 6, 
 Draw AB and PD parallel to OX. 
 
 The direction of motion at P being ultimately along the 
 tangent at P, the equation of motion is (Arts. 67, 20) 
 
 ' lu cVs 
 
 or 
 
 dy 
 g at as 
 
 d's _ dy 
 df~~ ^di' 
 
 (1) 
 
 the minus sign being taken because y decreases as s increases. 
 Multiply by 2ds and integrate. 
 
 .-. v' = (ds/dty ^ -%gy\ G, 
 
 But when y = a, v = 0, and .'. C = 2ga. 
 
 Hence v' = 2g{a — y) 
 
 = 2gxB'D, (2) 
 
 or the velocity at P is the same as would be acquired in fall- 
 
§ 121] MOTION ON A VERTICAL CIRCLE. 133 
 
 ing through the distance BD, the height of A above P (Art. 
 105). 
 If we put the angle OCA = /3, equation (2) may be written 
 
 V* = 2g(CB - CB) 
 
 = 2gr (cos - cos /3), (3) \ 
 
 where r is the radius of the circle. 
 
 120. To find the time of motion in the small arc AF. 
 We have, writing ds/dt for v, 
 
 l-j-f] = ^ff^ (cos ^— cos /3), 
 
 Also ds = rdO. 
 
 .'. V2g/rdt = - dO/i^cos 6 - cos /?, . . (4) 
 
 the — sign being taken because 6 decreases as t increases. 
 This equation cannot be integrated by the ordinary methods. 
 If, however, ft is so small that powers above the second may 
 be neglected, we have 
 
 cos ^ - cos /? = (1 - 6*72) - (1 - /?'/2) 
 
 Hence Vgjrdt = - dd/Vfi' - d\ (5) 
 
 Integrating, and noting that when t = 0, d = /3,yfe have 
 
 V^rt = cos-' (6/ /3), (6) 
 
 which gives the value of t, the time sought. 
 
 The time of making an oscillation, that is, the time of 
 moving through the arc AOA^, is found by putting 6 — — /3. 
 This gives 
 
 t = n^VTg, (' ) 
 
 the result already deduced in Art. 115. 
 
 121. We may also determine the pressure A^of the circle 
 on the particle w. 
 
 The centripetal force along PO = tvv'^/gr. 
 
134 DYNAMICS OF A PARTICLE. [§ 121 
 
 The resultant along PO of the force TV and the vertical 
 force tv is ]V — tv cos 0. Hence 
 
 N — w cos 6 = tvv*/gr 
 and N= lu cos -\- wv'^/gr 
 
 = io\gos 6 -\- 2(cos 6 — cos /?)[ 
 = w(3 COS ^ — 2 cos /3), 
 
 Ex. 1. Compare the times of a particle sliding down a 
 small arc ^0 of a vertical circle to the time of sliding down 
 the chord AO. Ans. 11 : 14. 
 
 2. A particle starts from the highest point of a smooth 
 vertical circle and slides down the convex side under the 
 action of gravity. Find where it leaves the circle. 
 
 Ans. At a depth =: radius/3. 
 
 3. A weight attached to a thread 2 ft long revolves in a 
 vertical circle. Find the velocity at the highest point of the 
 path that the thread may just remain taut. Aois. 8 ft/sec. 
 
 [At the highest point the centripetal accel. = the accel. of 
 gravity.] 
 
 4. A weight w hanging at the end of a thread of length I 
 is projected with a velocity u so as to describe a vertical 
 circle. Show that the pull P of the thread and the velocity 
 V at any point in the path whose vertical height above the 
 lowest point is h are found from 
 
 u'' = v'-{-2g7i; 
 Pl/w = ii'/g + I - U. 
 
 5. In (4) show that if ii"^ > bgl the particle will perform 
 complete revolutions. 
 
 In this case the pull at the lowest point is not less than 6w 
 pounds. 
 
 6. A pendulum is let go from a horizontal position. If If 
 is its weight, show that the pull on the thread when the bob 
 is in the lowest position is 3 IF. 
 
 7. If Z, is the length of the seconds-pendulum at latitude 
 A and I the length at the equator, then 
 
 IJl = 1 - cos'' A/289. 
 
§ 121] EXAMTKATIOIT. 135 
 
 EXAMINATION. 
 
 1. State the elements which specify a force. 
 
 2. Forces may be represented by straight lines drawn in 
 their direction and of lengths proportional to their magni- 
 tudes. 
 
 3. Which of Newton^s laws implies the principle of the 
 transmissibility of force ? 
 
 4. Show clearly from Newton's second law that the result- 
 ant of two concurrent forces may be found by the same proc- 
 ess as the resultant of two concurrent velocities. 
 
 5. Show that the parallelogram of forces may be illustrated 
 experimentally. 
 
 > 6. What is meant by a force resolved in a given direction ? 
 
 7. A force Tracts at in the line OA. Find its component 
 along the line OB. 
 
 8. A force may be resolved into two components in any 
 assigned directions. 
 
 ^9. What is meant by the resultant of a number of forces 
 acting at a point ? 
 
 10. Given two forces and their resultant, show how to find 
 the angle between their directions. 
 — 11. Since a force can have no component at right angles to 
 itself, how is it that a ship can be sailed at right angles to the 
 wind ? 
 
 y 12. To resolve a force P into two others such that the angle 
 between them is 60° and their sum is the greatest possible. 
 
 13. Explain the action of the forces by which an arrow is 
 discharged from a bow. 
 
 14. A body moves with uniform velocity in a straight line. 
 Find the relation between the acting forces. 
 
 15. When are forces said to equilibrate? 
 
 16. " Equilibrium is not a balancing of forces, but of the 
 effects of forces." Explain. 
 
 17. " Two or more forces can hardly be said to balance each 
 other unless they all act on the same body." Why ? 
 
'136 DYNAMICS OP A PARTICLE. [§ 121 
 
 ^ 18. State and prove the triangle of forces. 
 
 19. State and prove the polygon of forces. 
 
 20. State the grapJiical condition of equilibrium when 
 forces act at a point. 
 
 21. Three forces equilibrate. Show that no one of them 
 can be greater than the sum of the other two. 
 
 V 22. State and prove Lami's theorem. 
 
 23. State the analytical conditions of equilibrium when 
 two, three, , , . n forces act at a point. 
 
 24. What is meant by saying that the acceleration of a 
 falling body is g ft/sec^ ? 
 
 [" Velocity is poured into the body at that rate."] 
 
 25. What is the average velocity of a body during the first 
 second of its fall under the action of gravity ? 
 
 26. A particle is shot upwards with a velocity u ft/sec. 
 Find the height reached and the time of ascent. 
 
 ~^ 27. The times of falling from rest through two successive 
 equal distances are as V^ -\- 1 : 1. 
 
 ' 28. Prove that the distances passed over in the first, sec- 
 ond, . . . seconds by a body falling freely under gravity are as 
 the numbers 1, 3, 5, . . . respectively. 
 
 i 29. The velocity with which a body must be projected to 
 reach a height h ft is sVli ft/sec. 
 
 30. How did Galileo show experimentally that distances 
 fallen through are as the squares of the times of falling ? 
 
 31. Aristotle asserted that the time of falling of a body is 
 inversely as its weight. Show that this requires the accelera- 
 tion of gravity to be proportional to the square of the weight. 
 
 32. Find the momentum produced when a weight of 20 lb 
 falls through a distance of 25 ft. Ans. 25 second-pounds. 
 
 33. Discuss the motion of a body acted on by a constant 
 force (1) in the direction of motion; (2) not in the direction 
 of motion. 
 
 -\ 34. A shot is fired from an elevation in a horizontal direc- 
 tion with a velocity of 1000 ft/sec. Draw a figure represent- 
 ing its position at the end of 1, 1.5, 2, 2.5, 3 seconds. 
 
§ 121] EXAMlKAtlOlf. IS"? 
 
 35. Explain how it is that a body projected at a given angle 
 to the horizon describes a curve. 
 
 36. Find the path of a projectile in a vacuum. Show that 
 the problem is one of kinematics. 
 
 37. Find the position and velocity of a projectile at the 
 end of a given time t. 
 
 38. Find the range of a projectile on (1) a horizontal plane; 
 (2) an inclined plane. 
 
 4> — 39. A man running with uniform speed along a level road 
 throws a ball upward. In what direction must he throw it 
 that it may return to his hand ? 
 
 40. Two bodies are projected in any manner under the di- 
 rection of gravity. Show that their relative velocity is con- 
 stant throughout the motion. 
 
 -^ 41. From a balloon sailing horizontally at 60 miles an hour 
 a ball is let drop. Find its direction after 2| sec. 
 
 Ans, 45° to the horizontal. 
 
 42. Prove that the elevation required to attain a range R 
 with initial velocity u is given by 
 
 sin 26^ = gR/u\ 
 
 ^ 43. If R is the range and T the time of flight of a projec- 
 tile, the angle of elevation 6 is given by 
 
 i^ne = gT'/2R. 
 
 44. If ?/,, w, are the velocities at the ends of a focal chord 
 of the path of a projectile and v the horizontal velocity, show 
 that 
 
 ^ 45. Two projectiles are shot from two points in the same 
 horizontal plane with velocities u, v and at inclinations a, fi. 
 Show that if they meet, 
 
 w sin a = ?; sin /5. 
 
 46. Two bodies are projected from the same point at the 
 same time and in the same direction, but with different ve- 
 
1S8 t)YKAMICS OP A PARTICLl. [§ 12l 
 
 locities. Show that the direction of the line joining them at 
 any time is parallel to the line of projection. 
 
 [This forms a good illustration of the principle of the in- 
 dependence of forces. Art. 52.] 
 
 ^^47. The world's record for putting the shot is 47 ft. Find 
 the time the shot was in the air. 
 
 -48. Find the velocity acquired by a particle sliding from 
 rest down a length I of a smooth plane inclined at an angle 
 to the horizontal. 
 
 "^49. A body slides from rest down a series of smooth in- 
 clined planes whose total heights are h ft. Show that the 
 velocity at the bottom is V2gli ft/sec. 
 
 ^-^50. A series of inclined planes begin at the same point and 
 terminate in the same horizontal. Compare the velocities 
 acquired by bodies sliding down them. 
 
 51. The problem of finding the line of quickest descent 
 from a given point to a given line is equivalent to the geo- 
 metrical problem of describing a circle tangent to the given 
 line and whose highest point shall be the given point. 
 
 ~ 52. The angle of an incline is 1 5". If the pressure on the 
 plane is equal to the acting force, show that the inclination 
 of the force to the plane is 60". 
 
 ^ 53. Find the pressure exerted by a barrel of flour (196 lb) 
 on an elevator floor (1) rising with uniform speed, (2) falling 
 at a speed which increases 1 ft in each second. 
 
 Ans. 196; 189J pounds. 
 
 54. Give examples of a central stress as a tension and as a 
 pressure. 
 
 [Weight at the end of a thread; train passing round a 
 curve.] 
 
 55. Define centrifugal force. Illustrate by reference to the 
 preceding question. 
 
 56. " When a weight tied to a thread is wheeled about a 
 center the tension upon the thread is measured by the for- 
 mula Wvygr = pull.'' 
 
 57. A particle moves with uniform velocity in a circle. 
 
§ 121] EXAMIKATIOK. 139 
 
 Show that the centrifugal force varies as the radius directly 
 and as the square of the time of circuit inversely. 
 
 58. A fly-wheel in consequence of too rapid rotation goes 
 to pieces. In what direction do the pieces fly off ? 
 
 59. The centrifugal force of 1 lb making 1 revolution per 
 minute in a circle of 1 ft radius is 0.000341 pound. 
 
 60. " You have heard [said the lecturer] of the wonderful 
 centripetal force by which Divine Wisdom has retained the 
 planets in their orbits round the sun. But it must be clear 
 to you that if there were no other force in action the cen- 
 tripetal force would draw our earth and the other planets into 
 the sun, and universal ruin would ensue. To prevent such a 
 catastrophe the same Divine Wisdom has implanted a centrif- 
 ugal force of the same amount and directly opposite, etc." 
 Quoted by De Morgan, 
 
 What was the lecturer's probable meaning ? 
 
 61. How far will a body fall towards the earth in one min- 
 ute at the moon's distance ? Ans. 16 ft. 
 
 62. What is the "discount in the attraction of the earth 
 due to its rotation" at the equator ? 
 
 63. How is the law of gravitation verified by means of the 
 moon's motion ? 
 
 64. How is it that as an elevator comes to rest in its de- 
 scent a passenger feels as if he were being lifted up ? 
 
 65. Prove that a conical pendulum of which the bob de- 
 scribes a horizontal circle at a depth of // inches below the 
 point of support will make \SS/Vh revolutions per minute; 
 and that if the thread is I inches long and the bob weighs W 
 lb the tension of the thread is Wl/h pounds. 
 
 66. Find the time of oscillation of a pendulum of length I 
 inches. 
 
 67. Describe a method of finding the length of a seconds- 
 pendulum. 
 
 68. Show that the period of a conical pendulum is the 
 same as the time of oscillation of a certain simple pendu- 
 lum. 
 
140 . DYNAMICS OF A PAKTICLE. [§ 121 
 
 69. A pendulum-clock is carried in a balloon. Does it 
 gain or lose time as the balloon rises ? 
 
 70. Show that the lengths of pendulums vibrating seconds 
 at any place are inversely as the squares of the numbers of 
 oscillations. 
 
 71. A pendulum is let fall from a height h. Find the 
 height when the pull of the thread is equal to the weight of 
 the plummet. Ans. 2/i/3. 
 
 72. The vibrations of a simple pendulum are isochronous. 
 
 73. Show how the height of a mountain or the depth of a 
 mine may be found by counting the number of oscillations 
 lost by a pendulum which beats seconds on the earth's 
 surface. 
 
 74. A pendulum which beats seconds at Paris, where g = 
 32.18, is carried to New York, where g — 32.16. Find the 
 number of seconds lost per day. A7is. 27 sec. 
 
 75. A seconds pendulum if carried to the top of a moun- 
 tain 1 mile above sea-level would lose between 21 and 22 sec 
 per day. 
 
 76. The New York Central K.R. tracks between Albany 
 and Buffalo lie approximately along the 43d parallel of lati- 
 tude. The weight of the Empire State Express is about 280 
 tons. A speed of 75 miles an hour is often attained. At that 
 speed find the difference between the vertical pressures on 
 the rails of train east and train west. 
 
 Ans. About 400 pounds. 
 
§ 123] STATICS OP A BODY. 141 
 
 CHAPTER IV. 
 
 STATICS OF A BODY. 
 
 122. In the preceding chapter the behavior of a particle 
 under the action of forces has been considered. This includes 
 the case of a body under forces acting at the same point, and 
 such that the resulting motion is a motion of translation 
 only. 
 
 The directions of forces applied to a particle must necessa- 
 rily all pass through the particle. Applied to a body the 
 directions need not all pass through one point. Besides, 
 forces applied to a body may cause it to change its form, as 
 well as to change its position. To exclude the former we 
 shall assume that the body while under the action of the 
 forces retains an invariable form. It. is 7iot necessary to 
 assume that the body cannot be made to change its form, or, 
 as it is commonly stated, be rigid, but only that while the 
 forces act the form should remain unchanged. 
 -*' 123. Composition of Concurrent Forces. — Suppose two 
 forces F^ , F^ to act along the lines ABy CD at the points By 
 D of a. body. It is required to find their 
 resultant. 
 
 Prolong the lines of action to meet in 
 a point 0. The forces F^ , F, may be con- 
 sidered to act at this point (Art. 74). 
 Their resultant R is found by completing 
 the parallelogram OE as in Art. 75, and the line of action of 
 
142 
 
 STATICS OF A BODY. 
 
 [§125 
 
 R is along the line EO. The value of R may be computed 
 as in Art. 77. 
 
 If the number of forces is more than two, 
 combine two of them F^ , F^ into a resultant 
 R^ ; next combine R^ and F^ into a resultant 
 R^', and so on. The final resultant i? will 
 be the resultant of all the forces in magni- 
 tude and position. 
 ' 124. Composition of Parallel Forces.— If 
 
 in Art. 123 the lines of action of F^y F^ are parallel, the 
 
 construction for finding R fails. A 
 
 special artifice is necessary. 
 
 Let F^ be resolved into any two 
 
 components FA, BA, and F^ into the 
 
 two components FB, AB (Art. 79). 
 
 Since BA and AB are equal and oppo- 
 site forces, the two forces F^^ , F^ may 
 
 be replaced by FA, FB, whose resultant 
 
 is found as in Art. 123. Hence the resultant of F^ , F^ is 
 
 found. 
 
 125. With a number of parallel forces the resolutions 
 
 and compositions required to find R would produce a very 
 
 V,, 
 
 F/ R 
 
 /- 
 
 complicated force diagram. To avoid overlapping a construc- 
 tion diagram is introduced. 
 
 Let F^ , F^ be the parallel forces drawn to any convenient 
 scale. From any point a draw ab equal and parallel to F^ , 
 and from b draw be equal and parallel to F^ . 
 
§ 126] PARALLEL FORCES. 143 
 
 From any point draw the lines Oa, Ob, Oc to the points 
 a, b, c. 
 
 Now ab, or F^ , may be resolved into the two components 
 aO, Ob J and be, or F^ , into bO, Oc. 
 
 In order to determine the lines of action of these compo- 
 nents we transfer the components to the force diagram. Draw 
 any line AC parallel to aO to meet F^ in A, AB parallel to 
 bO to meet /', in B, and BC parallel to Oc. Then F^ is equiv- 
 alent to aO along AC, and Ob along BA ; /', is equivalent 
 to bO along AB, and Oc along BC But the forces along BA 
 and AB are equal and opposite. Hence F^ and F^ are equiv- 
 alent to flfO along A C, and Oc along BC. 
 
 Now the resultant of aO and Oc is ac, or F^ -\- F,, and the 
 lines of action AC, 5C intersect in C, which is therefore a 
 point on the resultant. 
 
 Hence, if tlirougli C a line equal and parallel to ac is 
 drawn, it will represent the resultant F^ -\- F^ of the two 
 parallel forces F^ , F^ in magnitude, direction, and line of ac- 
 tion. The resultant is therefore completely determined 
 (Art. 73). 
 
 The rule for plotting R that follows from this is given in 
 Art. 129. 
 
 126. We may readily compute the position of the point 
 D in which the resultant cuts AB. From similar triangles 
 Oab,ACD; Ocb, BCD, 
 
 CD/ AD = ab/Ob, CD/BD = cb/Ob, 
 
 Eliminate CD and Ob by dividing one equation by the other, 
 
 and 
 
 BD/AD = ab/cb = FJF„ 
 
 which, since the whole distance AB is known, gives the posi- 
 tion of the point D. 
 
 Notice that the position of the point D does not involve 
 the directions of the forces F^, F^. Being independent of 
 their directions, it is a fixed point for forces F^ , F^ acting at 
 
144 STATICS OF A BODY. [§ 127 
 
 assigned points A, B in any lines of action. (Test this by a 
 drawing.) 
 
 The point D is called the center of parallel forces — more 
 strictly it is the center of the points of application of the forces. 
 
 To sum up: When two parallel forces act, 
 
 (a) The magnitude of the resultant = the sum of the forces, 
 
 ih) The direction of the resultant = the direction of the 
 forces. 
 
 (c) The line of action of the resultant divides the line join- 
 ing the points of applicatio7i of the forces inversely as the 
 forces, 
 
 Ex. Show that AD = ABxFJ(F^-\-F^) 
 and BD = ABxFJ(F^-i-F^). 
 
 127. Galileo's demonstration of principle (c) is very in- 
 genious. 
 
 Imagine a uniform beam suspended at its middle point 
 and in a horizontal position. Let it be 
 divided into two parts by a vertical plane 
 so that the length of one part is 2a and 
 of the other 2b, the whole length of the 
 beam being 2a -{- 2b. 
 
 Imagine weights proportional to 2a, 2b 
 suspended at the middle points of the 
 two parts; then the distances of these weights from the 
 point of suspension are evidently b and a respectively. 
 
 Hence equilibrium exists if a weight a is suspended at a 
 distance b on one side of the point 0, and a weight b sus- 
 pended at a distance a on the other side of the same point. 
 
 128. General Method of Combining Forces (Graphical). — 
 We proceed now to show that the method of Art. 125 will in 
 general apply to combining forces in the same plane, whether 
 parallel or not. 
 
 Let F^, F^, F^ be the forces plotted to scale. From any 
 point a draw the line ab equal and parallel to F^, from b draw 
 be equal and parallel to F^, and from c draw cd equal and 
 parallel to F^, The line ad will represent the resultant E of 
 F^, F^, F^ in magnitude and direction (Art. 78). 
 
 To find the line of action of R\ From any convenient 
 
 
 h 
 
 a 
 
 
 
 
 
 
 
 2a 
 
 
 
 2b 
 
§129] 
 
 GENERAL GRAPHICAL METHOD. 
 
 145 
 
 point draw lines Oct, Ob, Oc, Od to the angular points of the 
 polygon abed. Draw any line Ip parallel to «0 to meet F^ in 
 
 r«0 
 
 PiPQ parallel to JO to meet F^ in q, qr parallel to cO to meet 
 F^ in r, and rs parallel to dO. 
 
 Then, remembering that the direction of a force is indi- 
 cated by the order of the letters on the line representing it, 
 
 jP,, or ah, is equivalent to aO along Ip and Ob along qp\ 
 F^f or be, is equivalent to bO along ^^^ and Oc along rq', 
 F^, or cd, is equivalent to cO along qr and Od along sr. 
 
 Adding, and noting that the forces along pq and qp are equal 
 and opposite, and that the forces along qj- and 7'q are equal 
 and opposite, we have 
 
 i^j, i^„ F^ equivalent to aO along Ip and Od along sr. 
 
 But the resultant of aO and Od is ad (or li), and the di- 
 rections Ip, sr of aO, Od intersect in t, which is therefore a 
 point on the resultant R. 
 
 Hence, if through t a line is drawn equal and parallel to ad, 
 it will represent the resultant R oi F^, F^, jP, in magnitude, 
 direction, and line of action. 
 
 129. We lience derive the following rule for finding graph- 
 ically the resultant of any number of forces in the same 
 plane : 
 
146 STATICS OF A BODY. [§ 130 
 
 (1) Construct a polygon alcd to scale whose sides are equal 
 and parallel to the forces; the closing side ad will represent 
 the resultant in magnitude and direction. 
 
 When the forces are parallel, this polygon becomes a 
 straight line (Art. 125), sometimes called the load-line. 
 
 (2) From any point 0, called the pole, draw lines Oa, Ob, 
 Oc, Od to the angular points of the polygon. 
 
 (3) Draw any line Ip parallel to aO,pq parallel to hO, qr 
 parallel to cO, and rs parallel to dO. 
 
 (4) A line through the intersection of Ip and sr equal and 
 parallel to ad will represent the resultant in magnitude, di- 
 rection, and position. 
 
 130. Condition of Equilibrium {Graphical). — It is evident 
 that a force equal and opposite to the resultant R would 
 equilibrate the forces F^, F^, F^. Hence /orces in a plane 
 which equilibrate may he represented hy the sides of a closed 
 polygon ahcd whose sides are parallel and in the sa7ne sense 
 as the forces. 
 
 The converse of this, that if forces acting in a plane can be 
 represented by the sides of a closed polygon which are paral- 
 lel to and in the same sense as the forces they equilibrate, is 
 not true. For the polygon would be the same, no matter 
 what the positions of the forces may be. This condition, in 
 fact, provides against translation only. An additional con- 
 dition to provide against rotation is necessary. 
 
 Examples. (1) Forces 7iot Parallel 
 
 1. Three forces are represented by AD, BO, DB in a par- 
 allelogram ABCD. Find their resultant. Ans. AO. 
 
 2. Four forces are represented by the sides AB, BC, CD, 
 AD of a rectangle ABCD. Find their resultant. 
 
 Alls. 2BC. 
 
 3. Show that if four forces be represented by the four sides 
 of a quadrilateral taken the same way round they cannot 
 equilibrate. 
 
 4. Forces 1, 2, 3, 4 pounds act along the sides AB, BO, CD 
 
§ 130] 
 
 EXAMPLES. 
 
 147 
 
 DA of a square, each side being 12 in long. Find their re- 
 sultant. 
 
 Ans. 24^2 pounds, parallel to CA and cutting CD in 
 E so that DE =\S in. 
 
 5. Forces 1, 2, 3 pounds act along the sides AB, BC, CA 
 of an equilateral triangle, each side being 12 in long. Find 
 their resultant. 
 
 Ans. V'6 pounds, perpendicular 5(7 and cutting 5(7 
 in D so that CD = 6 in. 
 
 6. Three forces P, Q, R are represented in direction by the 
 sides of an equilateral triangle taken the same way round. 
 Show that their resultant is 
 
 V{P" -\-Q'+Ii'-PQ-QR- EP). 
 
 7. If the lines of action of forces F^, F^, F,, F^ meet in a 
 point 0, the resultant E is completely determined by drawing 
 
 Find 
 
 through a line equal and opposite to ea, the closing side of 
 the polygon whose sides ab, be, , , . are parallel and equal to 
 the forces. 
 
 8. In a jib-crane a weight of 20 tons hangs from B, 
 the pull P of the tie-rod AB if AC = 12 ft, 
 AB = Q ft, and the weights of the parts are 
 neglected. 
 
 [The jib CB, neglecting its weight, is in 
 equilibrium under P, 20, and the reaction of 
 the hinge at C, which latter must be along 
 CB and equal to the thrust along the jib. 
 
 We may consider these three forces acting 
 at By and that this point is in equilibrium 
 under P, 20, and the thrust along the jib. 
 Hence the three forcei are proportional to 
 the sides of the triangle ABC (Art. 84). 
 Hence P = 10 tons.] 
 
Q 
 
 148 STATICS OF A BODT. [§ 130 
 
 9. The post AC of a jib-crane is 10 ft, the jib CB is in- 
 clined at 30°, and the tie ^^ at 60°, to the vertical. If the 
 weight lifted is 10 tons, find the stresses in AB, BO. 
 
 Ans. 10 tons; 101/ 3 tons. 
 
 (2) Forces Parallel. 
 
 + 10. Let B Che Si light rod suspended at A. Let weights of 
 3 oz, 4 oz be attached to it at the points B, C, 7 in apart. 
 Find the upward pull at A and the dis- 
 wt"/////////^////////////, tances BA, CA. 
 
 Ans. 7 oz; 4 in; 3 in. 
 [Make the apparatus and note if with 
 these weights and measurements it will 
 be in equilibrium.] 
 Xll. Draw the figure corresponding to 
 
 that in Art. 125 when F^ , F^ act in 
 
 opposite directions. Show that 
 R = F^- F^. 
 Illustrate by means of the above ap- 
 paratus. 
 ~ r 12. Two like parallel forces of 3 pounds and 5 pounds act 
 at points 2 ft apart. Find their resultant in magnitude and 
 position. Ans. 8 pounds; 9 in from 5. 
 
 + 13. Two unlike parallel forces of 3 pounds and 5 pounds 
 act at points 2 ft apart. Find their resultant in magnitude 
 and position. Ans. 2 pounds; 3 ft beyond 5. 
 
 f 14. Kesolve a force of 100 pounds into two like parallel 
 forces 10 ft apart, one of them being 2 ft from the given 
 force. Ans. 80 pounds; 20 pounds. 
 
 15. is any point within a triangle ABC If parallel 
 forces proportional to the areas of the triangles OBC, OCA, 
 OAB act at A, B, respectively, show that the resultant 
 must pass through 0. 
 
 ^^16. At the three vertices of a triangle are applied , three 
 parallel forces proportional to the opposite sides. Show that 
 the center of parallel forces is at the center of the inscribed 
 circle. 
 
 \ 17. The center of three parallel forces at the vertices of a 
 triangle ABC'\% at the intersection of the perpendiculars to 
 the opposite sides. Show that the forces must be proportional 
 to tan A, tan B, tan C. 
 
§131] 
 
 COtJPLE. 
 
 149 
 
 18. Show tliat the resultant R of two parallel forces i^, , F^ 
 may be found as follows: 
 
 At A, B introduce two equal and opposite forces S, — S. 
 Find the resultant R^ of F^, *V, and R^ 
 oiF,, -S. 
 
 Let i?, , /?, be transferred to 0. Re- 
 solve 7?, at into /S' and F^ and R^ into 
 — S and F^. Combine S and — S, F^ 
 and /;, at 0. 
 
 19. Draw the corresponding figure 
 when F^ , i^, act in opposite directions. 
 
 20. " AVe have a set of hay-scales, and 
 sometimes we have to weigh wagons that are too long to go 
 on them. Can we get the correct weight by weighing one end 
 at a time and then adding the two weights?" 
 
 131. Couple— Moment. — If in Art. 125 the two parallel 
 forces F^ , F^ are equal and act in opposite directions, the 
 points a and c in the construction diagram coincide, and the 
 line ao which represents the resultant becomes zero. Also 
 the lines AC, BO do not intersect, being parallel. Hence the 
 resultant of the forces is zero and its point of application is at 
 an infinite distance. In other words, the two forces cannot 
 be reduced to a single force in a definite position and direc- 
 tion. Another method of measuring the effect of force must 
 therefore be introduced. 
 
 r\ 
 
 F 
 b 
 
 i 
 
 cJ 
 
 To two equal parallel forces acting in opposite directions, 
 but not in the same straiglit line, the name couple is given. 
 A familiar example is seen in the use of a copying-press. The 
 
150 STATICS OF A BODY. [§ 132 
 
 handle is pushed at A and pulled at B, the push and pull if 
 equal forming a couple. In consequence the handle turns 
 about the axis of the screw. So, too, in winding a watch a 
 couple is employed. 
 
 132. The tendency of a couple on a body is to cause rota- 
 tion. Through any point in the plane of the couple draw 
 aOb perpendicular to the direction of the forces. Consider 
 one of the forces i^and let the point be fixed. The effect 
 of the force in producing rotation about will evidently 
 depend upon the magnitude of the force and 
 """^ upon the distance aO ot its line of action from 
 "^ 0. We may say, therefore, that the product 
 
 , F X aO is Si measure of the importance of the 
 
 ■^ force in producing turning. This product is 
 
 called the moment of the force F about the 
 point 0, the word moment being used in its 
 old-fashioned sense of importance or influence. Hence the 
 definition: 
 
 The mo7ne7it of a force about a point is the 2^roduct of the 
 7neasure of the force F and the perpendicular p let fall from 
 the poi7it upon the line of action of the force. 
 
 The term moment is from Lat. mome^itum, a particle suf- 
 ficient to turn the scales, a moving cause. The expression 
 " moment of a force " was used by Galileo to denote its effect 
 in setting a machine in motion. Many writers use the term 
 torque as the equivalent of " turning moment." 
 
 ^ 133. The Unit Moment is the moment of unit force act- 
 ing at an arm equal to unit length. It is named a foot- 
 pound, an inch-ton, etc., according to the units of length and 
 force employed. Thus a moment of 10 foot-pounds is ten 
 times the moment of a force of one pound acting at a dis- 
 tance of one foot from the point 0. 
 
 Less frequently the term pound-foot is used for unit of 
 moment. 
 
 134. Sign of the Moment. — It is evident that the direction 
 of turning about is as indicated by the arrow in the figure 
 
>■>. 
 
 § 136] varigkon's theorem. 151 
 
 the moment about being considered — or -f according as 
 the tendency of the force i^'to produce turning is in the di- 
 rection of motion of the hands of a clock [clockwise] or in 
 the opposite direction * [counter clockwise]. In the figure 
 the moment of F about = — Fp. 
 
 Or, better, since a moment has magnitude and direction, it 
 may be represented by a straight line. Lay off along a line 
 through normal to the plane of and F, that is, along the 
 axis of rotation, a length numerically equal to the moment. 
 That end of the line from which the rotation appears counter- 
 clockwise indicates the -\- direction, or 
 
 the + direction of the axis of a torque is that in which a 
 right-handed screw would move if driven hy the torque, 
 
 135. The moment of a force F with reference to a fixed 
 point may also be represented by an area 
 numerically equal to it. For if ah plotted to 
 scale represent the force F, and Oc the distance 
 pot from Fy then the moment Fp or ab X Oc, 
 is represented numerically by twice the area of 
 the triangle Gab, which has ab for base and Oc 
 for altitude. 
 ^136. M^uation. ofi Moments (Vfmgmii\s Theorem), — The 
 sum of the moments of two forces F^ , F^ about any point 
 in their plane is equal to the moment 
 
 ^-- — T "^ of their resultant E about the same 
 
 ^^7 ^^^^"^ point. 
 
 ^A~^ — ~^' W ^®^ ^^^® directions of the forces 
 
 Fj , F^ be along the lines AB and AC^ 
 and the direction of the resultant R be along AD. From 
 draw OD parallel to the direction AB of i^, , meeting the 
 directions of F, and R in G and B respectively; from D 
 draw DB parallel to GA : then AB, AG, AD represent the 
 forces F^, F^, R on the same scale. 
 
 Join OA and OB. Then 
 
 moment of F^ about = 2^0AB; 
 moment of F^ about = 2^0AG; 
 moment of R about = 2A0AD. 
 
 * This is the direction assumed by the whirl of cyclones. 
 
 X 
 

 
 a 
 
 I 
 
 b 
 
 F/ 
 
 ' 
 
 R ' 
 
 'h 
 
 152 STATICS OF A BODY. [§ 136 
 
 But evidently 
 
 .'. mom. of F^ abt. + mom. of F^ abt. = mom. of E abt. 0. 
 
 [b) If the forces F^ , F^ are parallel, from let fall Oab 
 
 perpendicular to their lines of action and 
 meeting the direction of the resultant R 
 in L Then (Art. 126) 
 
 F^Xal = F,X hi, 
 or F,{01 - Oa) = F/Ob - 01), 
 
 or F,X Oa + F,X Ob = (F^ + F,)Ol 
 = R X 01, 
 or mom. of F^ abt. + mom. of F^ abt. = mom. of R abt. 0. 
 
 (c) Generally if any number of forces in a plane act on a 
 body, the sum of the moments of the forces about any point 
 is equal to the moment of the resultant about the same point. 
 
 For, from the preceding, the sum of the moments of any 
 two forces is equal to the moment of their resultant about the 
 same point; of this resultant and a third force, that is, of the 
 three forces, to the moment of their resultant; and so on 
 until the last resultant is reached, which is the resultant of 
 all the forces. 
 
 This is known as Varignon's theorem of moments. 
 
 Ex. 1. Prove Art. 136 {fi) when the point lies between 
 F, and F^. 
 
 42. A force of 5 pounds acts along one side of an equilateral 
 triangle whose side is 2 ft long. Find the moment about the 
 vertex of the opposite angle. Ans. 5 1^3 foot-pounds. 
 
 — 3. A force of P pounds acts along the diagonal of a square 
 whose side is 25 ft. Find the moments of P about eat;h of 
 the four angular points. _ _ 
 
 A71S. 0, Ps V2, 0, - Ps V2 foot-pounds. 
 
 4. Find the moment of a force about any point in its line 
 of action. 
 
 5. Show that the moments of two forces about any point 
 on their resultant are equal and of opposite sign. 
 
§ 137] MOMENT OF A COUPLE. 153 
 
 5a. Hence find the algebraic sum of the moments of a sys- 
 tem of forces about any point on their resultant. 
 
 6. Apply the principle of Ex. 4 to find the position of the 
 resultant of forces 3, 4, 5, 6 acting along the sides AB, BC, 
 CD, DA of a square. 
 
 [Let the resultant cut AD, CD in E and F. Take mo- 
 ments about E and F and place each sum = 0. These two 
 relations determine the positions of E and F, It will be 
 found that the line ^i^is parallel to the diagonal AC^ 
 
 Get the same result by means of the polygon of forces. 
 tp 7. Forces 1, 2, 3 act along the sides of an equilateral tri- 
 angle. Show that the resultant is equal 1^3 and cuts the 
 direction of 2 at right angles. 
 
 Plot this resultant. 
 S 8. Forces P, Q act along the sides A B, AC of an equi- 
 lateral triangle ABC. Show that their moments about a 
 point D in BC are equal if BD = sQ/(P + Q), where 5 is a 
 side of the triangle. 
 
 -^9. If P is the thrust along the connecting rod of an engine, 
 r the crank radius, and the connecting-rod is inclined to the 
 crank axis at 150°, show that the moment of the thrust about 
 the crank-pin is one half the greatest moment possible, 
 f 10. At what height from the foot of a tree must one end of 
 a rope of length / ft be fastened so that a given force acting 
 at the other end may have the greatest tendency to overturn 
 the tree? Ans. 1/V2 ft. 
 
 -4-11. The post AC ot a jib-crane (page 147) is 10 ft, the jib 
 CB is inclined at 30°, and the tie AB at 60° to the vertical. 
 If the weight lifted is 10 tons, find the moment about C tend- 
 ing to upset the crane. Ans. 50 4/3 foot-tons. 
 
 X 137. Moment of a Couple. — In the couple represented by the 
 figure on page 149 the moments of the forces F, F about 
 are F X aO and F X bO units of moment. The total mo- 
 ment about is F(nO -\- hO) or F X ah units of moment. 
 
 The distance ah between the forces F, F is called the arm 
 of the couple. Hence the definition: 
 
 The moment of a couple [or the torque^ is the product of one 
 of the forces forming the couple and the arm of the couple. 
 
 The term torque is specially significant when applied to 
 "turning moments" in machine shafting. 
 
164 STATICS OF A BODY. [§ 138 
 
 The unit torque is the same as the unit moment. 
 
 The sign of the torque is negative if the couple tends to 
 cause rotation in the direction of driving a right-handed 
 screw (Art. 134). 
 
 138. Properties of a Couple. — The moment of a couple 
 depending only on the magnitude of the forces and the dis- 
 tance between them, the effect of a couple is not altered by 
 turning the arm through any angle about one end, nor by 
 moving the arm parallel to its former position in the plane of 
 the couple, nor by changing the couple into another couple 
 having the same moment. 
 
 It hence follows that the resultant of a number of couples 
 in a plane is a coiqjle ivliose moment is equal to the sum of 
 their momeyits. 
 
 It also follows that a single force F and a couple P, P act- 
 ing in the same plane on a body can- 
 not he in equilibrium. 
 
 For let the moment of the couple 
 be Pa, a being its arm. Replace the 
 couple P, P by a couple F, F of arm 
 b, so that Fb = Pa, and place it in 
 the plane so that one of its forces F 
 is opposite to the single force F. The two forces F, F at O 
 are in equilibrium, leaving the single force P at Z> unbal- 
 anced. Hence there cannot be equilibrium. 
 
 The theory of couples was first given by Poinsot in hk 
 Statique (1829). 
 
 Ex. 1. A force and a couple acting on a body are equiva- 
 lent to the single force acting in a direction parallel to its 
 original direction. 
 
 2. Two like couples of the same moment are together equal 
 to a single couple of twice the moment. 
 
 3. The side of a square A BCD is s inches long. Along 
 the sides AB, CD forces P act, and along AD, CB forces 
 2 P. Find the moment of the equivalent couple. 
 
 Ans, Ps inch-pounds. 
 
 PY 
 
§ 139J GENERAL ANALYTICAL METHOD. 155 
 
 4. Three forces are completely represented by the sides of 
 a triangle taken the same way round. Show that they form 
 a couple whose moment is represented by twice the area of 
 the triangle. 
 
 5. Show that the sum of the moments of the two forces 
 forming a couple about any point in their plane is equal to 
 the moment of the couple. 
 
 - . 139. General Method of Combining Forces {Analytical). — 
 It is evident from the preceding articles that — 
 
 {a) A7iy system of forces in one plane acting on a body 
 may he reduced to a single force or to a couple. 
 
 For (Art. 123) the forces may be combined two and two 
 until we arrive at a single force or at two forces forming a 
 couple. 
 
 This may be put in the slightly different form: 
 
 {b) Any system of forces in one plane acting on a body, 
 may be reduced to a single force acting at an assigned point, 
 and a couple. 
 
 For let the forces i^', , i^, , . . . act at the points A, B, , , . 
 of the body. At any point introduce two 
 forces F^ , F/, each equal to F^ and of oppo- 
 site directions. This will not disturb the 
 equilibrium. Hence 
 
 F,a,tA = F, at yl + i^, at + F,' at 
 = i^, at + i^, at ^ + F,' at 0. 
 
 But F^ at A and i^/ at form a couple *c \ 
 
 whose moment is F^p^ , where p, is the distance of from the 
 line of action of F^. Thus 
 
 i^, at ^ = F^ at and the couple F^p^. 
 Similarly, 
 
 i^, at J5 = F^ at and the couple F^p^, 
 and so on. 
 
 Adding, we have F, at ^, ^, at ^, . . . , equivalent to equal 
 and parallel forces F^ at 0^ F^ at 0, ... , together with the 
 couples whose moments are F^p^, Fj>^, , . . 
 
156 STATICS OP A BODY. [§ 140 
 
 The forces at may be combined into a single resultant R 
 by the method of Art. 78 or of Art. 80. The couples may be 
 combined into a single couple G by adding together their mo- 
 ments, so that (Art. 138) 
 
 G = F,p, + Fj,, + .. 
 
 where ^2 is the symbol of summation. 
 
 Hence the proposition is proved. 
 
 Cor. — Notice carefully the important principle employed 
 in the preceding demonstration : 
 
 A force F acting at a point A is equivalent to (1) an equal 
 and parallel force F acting at any point 0, and (2) a couple 
 ivhose mefnbers are + ^ acting at A and — F at ivith an 
 arm p, the perpe7idicular distance between the members, and a 
 moment Fp, 
 
 Ex. 1. Plot the single force equivalent to a given force of 
 10 pounds and a given couple consisting of two forces of 4 
 pounds each at a distance apart of 5 inches. 
 
 Ans. 10 pounds distant 2 inches from the given force. 
 
 2. A couple and a single force are equivalent to a single 
 force equal and parallel to the given force and at a distance 
 from it found by dividing the moment of the couple by the 
 single force. 
 
 3. Compare the above corollary with the second principle 
 of Art. 138. 
 
 4. Forces 1, 2, 3, 4 pounds act along the sides AB, BC, 
 CD, DA of a square 12 in long. Find the force passing 
 through the center and couple which are together equal to 
 these forces. 
 
 Ans. Force = 2 1^2 pounds along CA ; couple == 60 
 inch-pounds. 
 
 140. General Conditions of Equilibrium (Analytical). — 
 
 The general conditions of equilibrium may be stated in either 
 of two forms according as we start from one or other of the 
 methods of grouping forces just given. 
 
 {a) In general a system of forces acting in one plane on a 
 
§ 140] CONDITIONS OF EQUILIBRIUM. 157 
 
 body may be reduced to a single resultant force or to a couple 
 (Art. 139 (a) ). 
 
 Now a force cannot have zero moment unless the point of 
 moment is in its line of action. The line of action is fixed 
 by two points in it. Hence if the moment of the system 
 about each of three points not in one line is zero, the resultant 
 force must be zero. Also, since forces forming a couple never 
 have zero moment, if the moment of the system is zero no 
 couple can exist. Hence 
 X If forces act in one plane on a body so that the sum of their^ 
 ' moments about each of three points not in the same straight 
 line is zero, the body is in equilibrium. 
 
 (b) In general a system of forces acting in one plane on a 
 body may be reduced to a single force R referred to an arbi- 
 trary point and a single couple G (Art. 139 {b) ). 
 
 Now a single force and a single couple cannot equilibrate 
 (Art. 138). Hence, that the system may be in equilibrium, 
 we must have the two conditions: 
 
 1. The resultant force R =0; 
 
 2. The resultant couple G = 0. 
 
 That is, for equilibrium to exist there must be neither trans- 
 lation nor rotation. 
 
 These conditions may be put in a form more convenient for 
 computation. For the force R being the resultant of a sys- 
 tem of forces parallel to the given forces and acting at a point 
 0, the condition R=zO may be stated in the form given in Art. 
 86. Also if G — 0, then 2Fp = 0, or the sum of the mo- 
 ments of the forces about the point must = 0. 
 
 Hence the conditions of equilibrium may be stated: 
 
 \ (1) The sum of the components of the forces in any direc- 
 tion OX=0; 
 
 (2) Hie sum of the compo7ients of the forces in a JL direc- 
 tion OY=0', 
 
 (3) Tlie sum of the moments of the forces about any point 
 in their plane = 0. 
 
158 STATICS OF A BODY. [§141 
 
 This method is usually more convenient of application than 
 method (a). 
 
 Ex. State the conditions of equilibrium if the forces are 
 parallel. Ans. 2F= 0; 2Fp = 0. 
 
 Why only two conditions ? 
 
 141. Equililrium under Three Forces. — The case of equi- 
 librium under the action of three forces admits of special 
 simplification. For it is evident from Art. 124 that three 
 parallel forces acting on a body may equilibrate. 
 
 If three forces not parallel equilibrate they must meet in a 
 point. For two of the forces at least intersect. Their result- 
 ant acting at the point of intersection must equilibrate the 
 third force; that is, all three must meet in one point. Hence 
 
 If three forces in the same plane keep a hody in equilibrium 
 they must be parallel or meet in a point. 
 
 The great value of this principle in the solution of prob- 
 lems is that the meeting of- the lines of action enables us to 
 obtain a geometrical or trigonometrical statement. Also, we 
 can at once apply Lami's Theorem (Art. 84). 
 
 The results in any problem might also be found directly, 
 
 (1) by resolution along two rectangular axes (Art. 86); 
 
 (2) by taking moments (Art. 140); 
 
 (3) graphically (Art. 85). 
 
 142. In the application of the general conditions of equi- 
 librium of forces acting on a body (Arts. 140, 141) we are 
 met by a difficulty which did not appear when the equilibrium 
 of a particle only was considered (Art. 86). The force of 
 gravity on a particle acts vertically downwards, and its point 
 of application is the particle itself. As every particle of a 
 body is acted on by the force of gravity, we must be able to 
 find the position of the resultant force of gravity on the body 
 before we can apply the general conditions of equilibrium to 
 statical questions. This problem we now proceed to solve. 
 
 The subject proper will be resumed in Art. 152. 
 
§144] 
 
 CENTER OF GRAVITY. 
 
 159 
 
 
 . be a syst 
 
 em 
 
 B 
 
 G 
 
 M7j 
 
 _G2__ 
 
 c 
 
 ' F 
 
 + 
 
 143. Center of Gravity.— Let A, B, (7, 
 of particles in one plane rigidly con- 
 nected, and let these particles weigh 
 w, , w, , ?^3 , . . . lb respectively. The 
 forces due to gravity acting on the 
 particles are w, , tc^ , ii\, . . . pounds. 
 These forces are all directed towards w{ 
 the earth's center, which is so distant 
 that they may be considered parallel. 
 
 Then, as in Art. 12G, through whatever angle the system is 
 turned the vertical forces iv^ , w, at A, B will remain vertical 
 and may without altering their effect be supposed to act as 
 one force lu^ -\- w, at G^ , where 
 
 AGJBG, = wjw,. 
 
 Similarly w, + w, at (r, and w, at C, that is, w, at A, w, at 
 B, w^ at C, may be supposed to act as one force w^ -\- n\ + ^^'3 
 at G^ through whatever angle the system is turned, where 
 
 G,GJCG, = w,/(w,-i-ta,). 
 
 Adding thus particle to particle we see that the resultant 
 of the system of forcos will pass through a point G in the 
 body such that its position remains the same no matter how 
 the body is turned about. This point G is called the center 
 of gravity, or centroid, or mean center of the body, so that 
 
 The center of gravity of a body regarded as composed of a 
 system of particles rigidly connected is the point through 
 which the resultant force of gravity on the body passes no 
 matter how the body is turned about, ^ 
 
 The idea of the center of gravity is due to Archimedes 
 (B.C. 250). He determined the center of gravity of a triangle, 
 parallelogram, trapezium, parabola, etc. 
 
 144. It is evident that if a vertical force equal and opposite 
 
160 
 
 STATICS OF A BODY. 
 
 [§145 
 
 to the resultant force of gravity on the body be applied at the 
 center of gravity the body will be in equilib- 
 rium in any position. This suggests an ex- 
 perimental method of finding the center of 
 gravity. Thus conceive the body suspended 
 by a thread from a point P. The forces act- 
 ing are the resultant force of gravity at the 
 center of gravity and the pull of the thread. 
 The lines of action of these forces must lie in 
 the vertical through P. Hence, to find G, 
 
 suspend the body from P, and strike the vertical PH', next 
 
 suspend from any other point Q, and strike the vertical QK-, 
 
 the point of intersection of Pj^and ^^will be the center of 
 
 gravity G required. 
 
 145. Coordinates of the Center of Gravity. — Let a system 
 
 of particles A, B, . . .he referred to horizontal 
 
 and vertical axes OX, 01^ in their plane and 
 
 drawn through any fixed point 0. Also let 
 
 .T, y, be the coordinates of A; x^,y^, ot B; 
 
 . . . ; x,y,otG, the center of gravity. 
 
 The resultant w, + ^^^^ + . . . of the system 
 
 of parallel forces lu^ ,io^,... acts at G. Also, 
 
 since the moment of the resultant about a fixed point is 
 
 equal to the sum of the moments of the separate forces (Art. 
 
 136), we have 
 
 {lO^ ■\-lL\-\- . . , )X = W^X^ + 2U^X^ + . . . , 
 
 which may be written in the abbreviated form 
 
 X = 2tvx/^to. 
 Similarly, 
 
 y = ^wy/2tv, 
 
 i^Ex. 1. Weights 1, 2, 3, 4 oz are placed at the angles 0, X, 
 B, Y ot a, square OX^F whose sides are each 1 ft in length. 
 Find the coordinates of the 0. G. referred to OX, OF as 
 axes. Ans. 0.5 ft; 0.7 ft. 
 
 X ! o 
 
 
 — 1 
 
§ 147] CENTER OF GEAVITY. 161 
 
 ^ 2. Equal particles are placed at the angles ABC of an equi- 
 lateral triangle, and at the middle points D,B, Fot the sides. 
 Find the C. G. of the whole. 
 
 Ans. At G on AD wlien GD = AD/3. 
 ^ 3. Weights of 1, 2, 3 oz are placed at the angles of an equi- 
 lateral triangle whose sides are 6 inches long. Find the dis- 
 tances of their C. G. from the angles. 
 
 Ans. |/l9, i^l3, V? inches. 
 "' ia. If the velocities v, , v, , . . . of a system of particles 
 w^, w,, . . . parallel to a fixed line OY in their plane are 
 given at any instant, the velocity v of their C.G. is found from 
 
 V = 2wv/2w. 
 
 [For differentiate x = 2wx/2w with respect to t.] 
 ^ \h. Hence show that the momentum of the system collected 
 at the C.G. is equal to the sum of the momenta of the separate 
 particles. 
 
 4c. If the particles possess accelerations state the propo- 
 sition corresponding to (4«). 
 
 i^ 5. Two weights, 1 oz and 2 oz, are joined by a light thread 
 passing over a smooth peg and let go. Prove that during 
 the motion the acceleration of their center of gravity is ^/9. 
 
 6. Two bodies move with uniform speed along two straight 
 lines inclined at an angle B. Find the locus of their center 
 of gravity. Ans, A straight line. 
 
 146. It follows from the forms of the expressions 2wXj 
 2to, in Art. 145, that the finding of the center of gravity is a 
 problem of summation of molecular quantities and therefore 
 one of integration. It is a problem of the integral calculus. 
 In many cases, however, from the shape of the body or by 
 the introduction of some artifice, integration may be avoided. 
 
 We shall confine ourselves to bodies that are homogeneous 
 or of uniform density; that is, to bodies of such a nature that 
 the particles are equal and the same number of particles make 
 up the same volume in any and every part. 
 
 147. If a body of uniform density is symmetrical about a 
 point the C.G. must coincide with this point. Thus a sphere 
 being symmetrical about its center, the C.G. will be at the 
 
162 STATICS OF A BODY. [§ 148 
 
 center of the sphere. So the C.G. of a cube will be at the in- 
 tersection of the diagonals. 
 
 Again, with a straight rod of uniform cross-section, the 
 number of particles on one side of the central section being 
 equal to the number on the other side, the O.G. will be in 
 this section, and if we consider the area of the cross-section 
 to be indefinitely small, so that the rod becomes a straight 
 line, the C.G. will be at the middle point of the rod. 
 
 For a lamina or plate of uniform thickness the C.G. will 
 lie in a plane midway between the bounding planes. Con- 
 sider the lamina indefinitely thin, and the C.G. will lie in the 
 lamina itself and at its center of figure, if it has onco Thus 
 the C.G. of a circular lamina is at the centre of the circle, of 
 a rectangular lamina at the intersection of the diagonals, and 
 80 on. 
 
 148. In a triangular lamina not equilateral and therefore 
 without a center of figure we have to introduce a special 
 artifice in order to find the C.G. 
 
 Conceive the triangle divided into strips 
 parallel to one side ^5 and of indefinitely 
 small width. Each strip may be regarded 
 as a uniform rod whose C.G. is at its 
 middle point. These middle points all lie 
 in a line CD, joining C to the middle 
 point of AB. Hence the C.G. of the tri- 
 angle lies in the line CD. 
 
 Similarly the C.G. may be shown to lie in AE, the line 
 joining A to E, the middle point of BC. 
 
 Hence it is at G, the intersection of CD and AE. 
 
 To find the position of G on CD, 
 
 Join DE, Then DE is evidently parallel to AG, 
 
 Also, the triangles DGE and CGA are similar. 
 
 Hence DG/GC = DE/A G = 1/2, 
 
§ 148] CENTER OF GRAVITY. 163 
 
 and DG=GC/2 
 
 or DG^DC/Z, 
 
 Similarly, EG = EA/Z. 
 
 Hence the C. G. of a triangle is on a median line of the 
 triangle at ttvo thirds its length from the vertex. 
 
 A Ex. 1. Prove that the C.G. of three equal weights placed 
 at the angular points of a triangle coincides with that of the 
 triangle itself. 
 
 [For W at B and Tf at C are equivalent to 2 W at D, the 
 middle point of BC. Also, "IW sX D and W at A are equiva- 
 lent to 2.W 2ii G when DG = A G/1. But G is the C.G. of 
 the triangle.] 
 
 ^ 2. Three men support a heavy triangular board at its cor- 
 ners in a horizontal position. Compare the weights sup- 
 ported by each man. 
 
 -P 3. The sides of a triangle are 3, 4, 5. Find the distances 
 of the C.G. from the angles. Ans, Vn/S, 2VT3/3, 5/3. 
 "4. Equal weights are placed at the angular points of a tri- 
 angular board and also at the middle points of its sides. 
 Find the C.G. of the system. 
 -^ 5. If 6^ is the C.G. of a triangle ABC, prove 
 
 3(6^^^+ GB'-^ GC') = AB'-\-BC'-^CA\ 
 
 Jr -^6. A series of triangles of equal area are described on the 
 same base and on the same side of it. Show that their 
 centers of gravity lie in a straight line. 
 
 -f 7. A right triangle ABC is suspended from a point P in 
 the hypotenuse AC, and one side AB hangs vertical. Show 
 that 2AP = CP. 
 — 8. 6^ is the C.G. of a triangle ABC, any point in its 
 
 Elane. The forces represented by OA, OB, OC acting at 
 ave for resultant a force represented by 30G. 
 — ^9. A uniform wire is bent into the form of a triangle with 
 sides a, b, c. Show that the distances of the C.G. of the 
 whole from the sides are as 
 
 (b-\-c)/a:{c + a)/b:(a-\-h)/c. 
 
164 
 
 STATICS OF A BODY. 
 
 [§149 
 
 UO. A triangle ABC\ right-angled at C, is suspended suc- 
 cessively from A and B. If /?, y be the angles made by AC, 
 BG with the vertical in each position, show that 
 
 cot P cot y = 4:. 
 
 149. C. 0. of a System of Bodies. — We pass at once to the 
 center of gravity of a system of bodies rigidly connected, by 
 considering that each body may be 
 conceived concentrated into a par- 
 ticle of equal weight acting at the 
 center of gravity of the body. Thus 
 if G^ is the center of gravity of a 
 body weighing }\\ lb, and G^ the 
 center of gravity of a body weighing 
 TF, lb, then the vertical forces at G^ , 
 Wt G^ being If,, W^ pounds, the posi- 
 tion of the center of gravity G of 
 the system will be found from 
 
 Wi 
 
 i \ I 
 
 W,xG,G = W,xG,G. 
 
 Similarly, if the positions of G, G^ are given, that of G^ 
 may be found. 
 
 Ex. la. Weights of 1, 2, 3, 4, 5 lb are strung on a uniform 
 rod AB, whose weight is 3 lb at distances of 4 in from each 
 other. Find the point at which the rod 
 will balance. f^ 9 b 
 
 [The weight of the rod may be sup- f T { \ J 
 posed collected at the middle point (Art. 12345 
 148). We have then to find the O.G. of * 
 
 the weights 1, 2, 6, 4, 5 lb placed 4 in apart. Take moments 
 about A. 
 
 (1 + 2 + 6 + 4 + 5) X ^6^ = 2X4 + 6X8 + 4X12 + 5X16). 
 
 .-. AG= lOf in. 
 
 Check the result by taking moments about G; also about 
 other points.] 
 
§ 149] CENTER OF GRAVITY. 165 
 
 lb. What is the pressure on the point of support ? 
 
 2. A bicycle weighing w.^ lb has its wheels a in apart. The 
 C.G. of the rider, who weighs iv^ lb is ft, in behind the' front 
 wheel, and the C.G. of the bicycle is «, in behind the same 
 wheel. Find the pressure of the rear wheel on . 
 the ground. Ans. (w^a^ -f w^a,)/a pounds. jc b 
 
 3. A cylindrical jar 6 in deep weighs 3 lb and [^ 
 holds 2 lb of water. Its C.G. is 3^ in from the — 
 top when empty. Find its position when the jar 
 is full of water. Ans. 3.3 in from the top. 
 
 4. Find the C.G. of a T-iron whose depth of 
 flange = h, depth of web = A,, breadth of flange 
 = b, breadth of web = b^. 
 
 Ans. C.G. = (bW -f bji,' + 2bjih^)/2(b7i + bji^). 
 6. A common form of cross-section of a reservoir wall or 
 embankment wall is a trapezoid whose 
 top and bottom sides are parallel. If 
 top side = ft, bottom = b, and height 
 = h, show that 
 
 "■"'mi)- 
 
 [Divide the figure into a rectangle and a triangle or into 
 two triangles.] 
 
 ■iQ. Show that if one fourth part ADE of a triangle ABC is 
 cut off by a line DE parallel to the base BC, the distance of 
 the C.G. of the remainder from the vertex A is 7/9 of the 
 median AF. 
 
 7. CA and CB are the arms of a bent lever 2 ft and 4 ft in 
 length, respectively, and inclined at an angle of 60°. Find 
 the distance of their C.G. from C. j^jis, 1^21/3 ft. 
 
 "is. A triangular bracket projects 30 ft from a building aiid 
 weighs 2 tons. A load of 3 tons hangs from the vertex. Find 
 position of the C.G. of the whole. 
 
 A71S. 22 ft from the building. 
 — V 9. In a painter^s palette, formed by cutting a small circular 
 disk from a larger one, if the diameters of the disks are 1 to n 
 and the distance between their centers a, show that the dis- 
 
166 STATICS OF A BODY. [§ 150 
 
 tance of the C. G. of the palette from the center of the larger 
 disk is a/(7i^ — 1). 
 
 10. *Show that the cen- 
 ter of gravity 6r of a trape- E^__ c k 
 
 zoid lies in the straight ^'■^- V i^ 
 
 line ^^ joining the mid- / ~~/~~— 
 
 die points of AB and CD, L i 
 
 the parallel sides. a h « 
 
 11. Show in (10) that 
 
 HG : KG = AB + "ICD : 2AB + CD, 
 
 12. Prove the following construction for finding the C.G. 
 of a trapezoid A BCD. 
 
 Prolong AB, making BB = CD; prolong DC, making 
 CF= AB. The point of intersection of FF and HF, the 
 line joining the middle points of AB, CD, is the C.G. 
 
 [This construction is due to Poinsot.] 
 1 13. Prove the following rule for finding the C.G. of a qua- 
 drilateral v4^i)C: 
 
 Draw the diagonals AD, BC. Make 
 AF^DF. The C.G. of the triangle CFB 
 is also that of the quadrilateral. 
 
 14. Prove the following rule for finding 
 the C.G. of a quadrilateral. 
 
 Join each vertex with the middle point 
 of the opposite side. Join the intersec- 
 tions of the lines from opposite vertices. 
 The intersection of these lines is the C.G. 
 15. If x^ , y, ; x^, y^\ ^3 ? 2^3 5 ^4 ? Va, ^^® *^® coordinates of 
 the angular points of a quadrilateral, and x^ , y^ of the inter- 
 section of the diagonals, and x, y of the C.G, then 
 
 x = {x^-\-x^-\-x^-\-x^- 3:J/3; 
 
 160. General Method of Finding the C.G. — We now return 
 to the general formula of Art. 145, which includes all cases, 
 and shall show its application to some of the problems 
 already solved and to others which do not admit of any 
 special artifice in their solution. 
 
150] 
 
 C:fiNTER OP GRAVITY. 
 
 167 
 
 Ex. 1. To find the C.G. of a thin lamina 
 in the form of a quadrant of a circle of 
 radius ;*. 
 
 [Take the origin at the center 0, and let 
 the lamina h be divided into strips by lines 
 parallel to the axis of Y. 
 
 Let P^be any strip; x, y, the coordinates 
 of P; x-\- Axy y -\- Jy, the coordinates of ^; ° ^ ^ 
 
 and let the lamina weigh 6 per unit area. Then 
 
 area of strip = yAx\ 
 
 force of gravity on strip = SyAx, 
 
 The C.G. of the strip being ultimately at its middle point, 
 its coordinates are x, y/2. Hence if Xy y denote the coordi- 
 nates of the C.G. of the quadrant, 
 
 x X '26yAx = 2SxyAx; 
 "^ X 2dyAx = i^Sy'/Jx; 
 or in the notation of the calculus, 6 being constant, 
 
 xj ydx=J xydx\ yj ydx = \J y'dx. 
 
 Performing the integrations indicated, remembering that 
 x^ -^ y"^ — r', we have 
 
 X = 4r/37r = y, 
 
 and 0G= Vx -\-tf = 4r f^/3;r. 
 
 \ 2. To find the C.G. of a thin lamina in the form of a 
 quadrant of an ellipse whose semi-axes are a, b. 
 
 Ans. X — \a/Zn\ y = 46/3 ;r. 
 \ 3. Find the position of the centroid of a semicircle. 
 
 Ans.OG = \r/Z7t. 
 4. To find the C.G. of a triangle ACB whose sides are 
 «, 6, c, 
 
 [Take G as origin; CB, CA, as axes of 
 X,F. 
 
 Cut into strips parallel to CB. Let PQ 
 be any strip, and x, y the coordinates of P. 
 Then 
 
 Area of strip PQ = xAy sin C, 
 
168 
 
 STATICS OF A BODY. 
 
 t§150 
 
 The coordinates of the C.G. of the strip PQ are x/^, y, ulti- 
 mately. Hence 
 
 - r , r.' - r , r ' 
 
 X I xdy = I -^dy; y I xdy = I xydy. 
 
 But, from the equation of the straight line AB, 
 
 Performing the integrations, we find 
 
 X = a/d ; y = 5/3 ; 
 
 giving the same position of the C.G. as found in Art. 148.] 
 
 5. Solve Example 4 using rectangular coordinates. 
 -f- 6. To find the G. of a hemisphere of radius r and 
 center 0. 
 
 [From symmetry the C.G. must lie in 
 OX normal to the base. 
 
 Divide the hemisphere into slices by 
 planes normal to OX. 
 Volume of slice = ny'^Ax. 
 Coordinates of C.G. of slice are x^ 0. 
 
 .*. OG I Tty'^dx — I Ttxy^dXy 
 and 00 = 3r/8.] 
 
 Sometimes it is advantageous to introduce polar coordi- 
 nates. 
 
 -f Ex. 7. To find the centroid of a quadrant of a circle of 
 radius r (see Example 1). 
 
 ' [Take OX as initial line, and r, 6 the coordinates of Q. 
 Join OP, OQ, forming with PQ the element area OP^. 
 Area OPQ = iOP X PQ = ir'/iO ultimately. 
 Coordinates of C.G. of area OPQ are fr cos 6, fr sin 6, 
 ultimately. (Art. 148.) 
 
 TT n 
 
 Hence xJ^^r'dQ =J^ir cos e x ir'^dd, etc.] 
 
§ 151] CENTER OF GKAVITY. 169 
 
 ^ 8. Find after the manner of Ex. 7 the position of the cen- 
 troid of a semicircle. (Compare Ex. 2.) 
 ^9. To find the C.G. of a circular sector AOB if angle 
 A OB = 2/^ and r is the radius. 
 
 A71S. OG = ^r sin /3//3 = 2 radius X chord /3 arc. 
 "^10. Find the C.G. of a sector of a 
 circular ring, radius OA = r, , radius 
 0(7= r,, angle ^0^ = 2/?. 
 
 Ans. 
 0G = 2{r; - r,') sin /?/3(r,» - r;')/3. 
 
 [A practical illustration is the front 
 surface of a circular arch.] 
 
 11. Show that the distance of the C.G. of a segment of a 
 circle from the center is cyi2^ when c is the chord and A 
 the area of the segment. 
 
 " 12. To find the distance of the C.G. of a semi-circumfer- 
 ence from the center 0. Ans. OG = 2r/n. 
 
 13. To find the distance of the C.G. of a quadrantal arc 
 from the center 0. Ans. OG = 2V2r/7r. 
 
 ^4. To find the distance of the C.G. of a circular arc AB 
 subtending an angle 2/3 at the center. 
 
 Ans. OG = r sin /3//3, or OG = radius X chord /arc. 
 
 15. ABC is a triangle inscribed in a circle center 0, and 
 F, Gy Hare the centers of gravity of the sectors AOB, BOC, 
 CO A. Show that 
 
 AB/OF-^ BC/OG + CA/OH= Zn. 
 
 151. Having now found the position of the point of appli- 
 cation of the resultant force of gi'avity on a body, it is pos- 
 sible to apply the general conditions of equilibrium to bodies 
 acted on by forces (Art. 140). This application will form 
 the remainder of this chapter. 
 
 We shall confine ourselves to forces that lie in the same 
 plane. This will include the great majority of problems that 
 occur in practice; for in structures forces are in general so 
 applied as to be symmetrical about a plane, and therefore 
 their resultants lie in this plane, so that the forces may be 
 treated as if all acted in this plane. 
 
ItO S*ATtCS OF A iODT. [§ 15S 
 
 152. It is well to call to mind at this place how to find 
 
 (1) The resultant of a number of forces acting at a point 
 
 (a) graphically, Art. 78; 
 
 (b) analytically, Art. 80. 
 
 (2) The resultant of a number of forces acting at different 
 points 
 
 (a) graphically, Art. 128; 
 
 (b) analytically, Art. 139. 
 
 (3) The conditions of equilibrium when forces act at a 
 point 
 
 (a) graphically. Art. 85; 
 
 (b) analytically. Art. 86. 
 
 (4) The conditions of equilibrium when forces act on a 
 body 
 
 (a) graphically (three forces). Art. 141; 
 
 (b) analytically, Art. 140. 
 
 In many applications of Mechanics in Architecture and 
 Bridge-building both the graphical and analytical methods of 
 computation are used. At times one is more convenient 
 than the other, and always one may be used to check the 
 other. This will be illustrated as we proceed. 
 
 153. We shall consider first the equilibrium of a single 
 body variously supported, and then that of a system of bodies. 
 The same general principle runs through all cases. When 
 one body is in contact with another at one or more points, 
 the action of one body is equal to the reaction of the. other. 
 If, therefore, the support be removed and a force applied to 
 the body in magnitude and direction equal to the reaction of 
 the support on the body, the conditions of equilibrium may be 
 applied to the body as if acted on by the original forces and 
 this reaction. In this way, too, the equilibrium of a body as 
 part of a system may be studied. 
 
I 154] tlQtiLlBlltlTM. Itl 
 
 -A^64. Equilibrium of a Body Supported at One Point. — Con- 
 sider a uniform beam of length 21, depth 2^and weighing 
 If lb, suspended at its middle point 0. The force on the sup- 
 port is equal to W pounds and acts vertically along OG 
 through the center of gravity G of the beam. A body weigh- 
 
 
 
 
 
 r1 
 
 
 A B C^ , 
 
 4^ 
 
 G 
 
 A 
 
 fK 
 
 H 
 
 G 
 
 
 
 
 p 
 
 
 --^ 
 
 
 vy 
 
 w 
 
 ing P lb, if placed on the beam or suspended from it any- 
 where except on the line OG, will cause the beam to take an 
 inclined position. 
 
 Suppose P to be suspended at the extremity A^ and let R 
 denote the reaction of the support at 0, and 6 the angle of 
 deflection of the beam from the horizontal position. 
 
 We may consider the support removed and the beam in 
 equilibrium under the vertical parallel forces 
 P, W, R. Then (Art. 140) 
 
 (1) The sum of the forces is zero, or 
 
 P+ W- R = 0, 
 
 Hence the pressure R on the support is found. 
 
 (2) The sum of the moments about any point 
 is zero. Take moments about and 
 
 PX 00- Wx GH-\-RxO = 0, 
 or P X ^ cos ^ — W X h sin 6 = 0, 
 or tan 0= PI/ W7i, 
 
 and the inclination 6 is found. 
 
 By attaching a pointer to the beam free to move over a 
 graduated arc we have a means of comparing weights. An 
 example is afforded by the common letter-scale. 
 
172 STATICS OF A BODY. [§ 154 
 
 Ex. 1. Let AB represent a rigid rod (as a crowbar) turning 
 
 on a fixed support C. Let a force F 
 
 JR be applied at A, and let W be the re- 
 
 I sistance to be balanced at B. Given 
 
 } ^ 1 the lengths of AC, CB, it is required 
 
 j j to find the relation between F, W when 
 
 i I in equilibrium. 
 
 I i [Neglect for the present the weight 
 
 F ^l of the rod. Let F and W be vertical, 
 
 and let R denote the reaction at C. 
 We may consider the support removed and the rod in equi- 
 librium under F, W, R. Then the equations of equilibrium 
 are 
 
 F^W-R =0. 
 
 FxAG- WX BC=0. 
 
 Hence F/BC = W/A = R/AB, 
 
 the relation sought. 
 
 If W^ is the weight of the rod acting at its middle point G, 
 the equations of equilibrium become 
 
 F-i- W+ W,-R = 0, 
 Fx AC-i- W, X GC - W X BC=0, 
 
 from which the ratios of F, W, R may be found. 
 
 The rod AB is known as a Lever, the support C the fulcrum, 
 and the distances A C, CB the arms of the lever. The relation 
 
 F/BC =^ W/AC= R/AB 
 
 is sometimes called the pri?iciple of the lever. 
 
 The condition of equilibrium in the case of forces acting at 
 right angles to the arms of a straight lever was given by Ar- 
 chimedes (B.C. 287-212). But it was not until the end of the 
 fifteenth century that the case of forces acting obliquely was 
 solved. This was done by Leonardo da Vinci (1452-1519) of 
 Florence, painter and philosopher. 
 
 > 2. A lever is 2 ft long. Where must the fulcrum be placed 
 that 10 pounds at one end may balance 30 pounds at the 
 other end ? 
 
§155] 
 
 EQUItlBRIUM. 
 
 173 
 
 i^^-x^c 
 
 >3. From a pole resting on the shoulders of two men a 
 weight W is suspended. It is n times as far from one man 
 as from the other; what does each support ? 
 
 Ans. W/{n + 1), n W/(n + 1). 
 "^ 4. Find the relation between F and Wins, bell-crank lever. 
 A and B are the bell wires, C is the 
 pivot about which the lever turns. 
 
 [The directions of F, W, and the 
 reaction E of the pivot meet in a point 
 0. Hence take moments about C] 
 •^ 5. A pair of nut-crackers is a inches 
 in length, and a pressure of jo pounds 
 will crack a nut placed b iuches from 
 the hinge. What weight placed on 
 the nut would crack it ? 
 
 Ans. pa/b pounds. 
 -)• 6. The handle of a claw-hammer is 
 1 ft long and the length of the claw 
 is 2 in. A pressure of 25 pounds is 
 applied at the end of the handle. 
 Find what resistance offered by a nail would be overcome. 
 
 Ans. 150 pounds. 
 A: 7. A uniform wire bent into the form of three sides of a 
 square is hung up from one of the angles. Show that the in- 
 clination of the first side to the horizontal is tan"' 2/3. 
 
 8. ABC is a triangular board having AB = 5 in, BC = 4 
 in, AC = 3 in. Find the point on BC from which the board 
 if suspended will hang with AB horizo^ital. 
 
 Ans. 7/12 in from C. 
 ^ 9. A plank 16 ft long and weighing 120 lb projects over a 
 vertical wall in a horizontal plane to a distance of 6 ft • A 
 boy weighing 80 lb walks slowly along the plank. When will 
 it begin to topple over ? 
 
 A?is. When 3 ft from end. 
 "^10. Two weights P, Q balance at the ends of a lever whose 
 weight may be neglected. If when the weights are inter- 
 changed equilibrium is maintained by adding weights F^, Q^ 
 to /*, Q, respectively, show that 
 
 F'-Q^ = QQ^ - FP,. 
 
 166. Balance. — Consider a uniform beam hanging in equi- 
 librium in a horizontal position, being suspended from a 
 
174 
 
 STATICS OF A BODY. 
 
 [§155 
 
 point directly above the center of gravity G, and with the 
 
 extremities A, B equidistant from 
 and in the same straight line 
 with it. If from Ay B are sus- 
 pended two equal weights P, P, 
 the beam will still remain in a 
 horizontal position, since the mo- 
 ments about are equal. 
 Hence the apparatus, which is evidently a lever with equal 
 arms, may be used for comparing equal weights. Attaching 
 pans to Ay B, the weights for comparison may be placed in 
 these pans and the operation facilitated. Such an arrange- 
 ment is the common Balance. 
 
 If the attached weights are not equal, the beam will not rest 
 in a horizontal position. We proceed to find the position of 
 equilibrium in this case and thence to infer the requisites of 
 a good balance. 
 
 Let IF be the weight of the beam acting at G, and suppose 
 unequal weights P, Q suspended from A and B, P being the 
 greater. When the beam is again in equilibrium denote the 
 angle of inclination of ^^ to the horizontal CD by d. 
 Let A0= OB = 1 and OG = h. 
 
 The beam is in equilibrium under the parallel forces P, Q, 
 ^^^^W, and the reaction R at 0. Hence 
 
 P^Q-\-W-R = 0, . . 
 Take moments about 0, and 
 
 Fl cos - Wh sin - Ql cos 6 = 0, 
 or ts^nO =z{P - Q)l/Wi. 
 
 (1) 
 
 (2) - 
 
 Now the balance will indicate small differences P — Q the 
 more clearly the greater the angle 6 through which it swings 
 for these differences. But tan ^ or ^ is greatest when // W7i 
 is greatest, that is, when I is large or the beam has long arms, 
 when W is small or the beam light, when h is small or the 
 
§ 156] EQUILIBRIUM.. 175 
 
 center of gravity is just below the point of suspension. Such 
 a balance has great sensihility, and is suitable for delicate 
 investigations in Chemistry, Physics, Assaying, etc. 
 
 In scales for weighing large weights stability rather than 
 sensibility is wanted; that is, for small differences of P and 
 Q the angle of deviation of the beam from the horizontal as 
 shown by tan 6 should be small. This requires Wh to be 
 large or the beam to be heavy, with a long distance between 
 the center of gravity G and the point of suspension 0. By 
 making the arms long, a balance may be constructed which 
 shall possess in a measure both sensibility and stability. As 
 the two conditions are at variance, the amount of compro- 
 mise must be decided by the use to which the balance is to 
 be put. 
 
 For very accurate work the method called double weighing 
 is in use Place the body to be weighed in one pan, and bal- 
 ance with sand placed in the other pan. Remove the body, 
 and balance the sand with standard weights. The weight of 
 the body is then shown by the standard weights, for they pro- 
 duce the same effect as the body itself. 
 
 This method, which is due to Borda, is really a method of 
 substitution, and gives a correct result if the balance is not 
 properly adjusted. 
 
 166. To Test a Balance. — Suppose the beam to rest in a 
 horizontal position when the scale-pans are empty. Weigh a 
 body P first in one pan and then in the other. If the two 
 values obtained are equal to one another, the balance is true. 
 For let a, b denote the lengths of the arms and Wa, Wi, the 
 weiglits of the scale-pans. Then if W denotes the observed 
 weight of the body P, we have 
 
 Wa Xrt= TfftXS; (1) 
 
 (P+ Wa)Xa = (W^- W,)b; .... (2) 
 (W-\-W:)a = (P^W,)b. .... (3) 
 
 Hence a = b, Fa = Wi,, and P = W,ov the arms are equal, 
 
176 STATICS OF A BODY. [§ 156 
 
 the pans are of the same weight, and the observed weight is 
 the weight of the body. 
 
 Ex. la. The beam of an unloaded balance rests in a hori- 
 zontal position, and the arms are known to be of unequal 
 length. Show that the true weight W ot a body which appears 
 to weigh W^ when placed in one scale and W\ when placed 
 in the other scale is given by 
 
 W=VW\ W. 
 
 ^Ib. Show that if W^ , W\ are nearly equal, this may be 
 written closely enough 
 
 W={W, + W,)/2. 
 [For 2 VW\W, = (W,+ W^)-(VW,- VWy.] 
 
 V 2a. If the arms of a balance are of equal length but (from 
 wear) the pans are not of the same weight, show that the true 
 weight PT of a body which appears to weigh TF, when placed 
 in one pan and W^ when placed in the other pan is found from 
 
 W=(W,-{- W,)/2, 
 
 \ 2b. Show that the difference of the weights of the scale- 
 pans is equal to ( IF^ — W^)/2, 
 
 -' 3. If the C.Gr. of the beam is not directly under the point 
 of suspension, show that the correct weight may be found as 
 in Ex. 2a. 
 
 ^ 4. Is it fair to infer that in all cases 
 
 correct wt. = half-sum of apparent wts. in the two scales ? 
 
 \ 5. A balance has unequal arms. The apparent weights of a 
 body weighed first in one scale-pan and then in the otjier are 
 9 and 11 grains. What error is made by taking 10 grains as 
 the true weight ? A ?is. 0.05 grain. 
 
 \ 6. Why must not the C.G. of a beam-balance coincide with 
 the point of suspension ? 
 
 -\ 7. In a balance show that if the scale-pans hang freely no 
 error can arise from the weights not being placed in the 
 center of the pans. 
 
§ 157] EQUILIBRIUM. 177 
 
 -\ 8. A body whose weight is 10 lb when placed in one pan 
 of a false balance appears to weigh 9.5 lb. Find its apparent 
 weight when placed in the other pan. A7is. 10.526 lb. 
 
 -t 9. The arms of a balance are a in and b in. A grocer uses 
 the scale-pans in alternate order in serving customers. Find 
 lus gain or loss per lb. Ans. {a — by/2ab lb. 
 
 ^ 10. Discuss the balance if the point of suspension is at a 
 distance k above the beam. 
 
 157. Steelyard, — Consider a beam suspended from a point 
 directly above its center of gravity and hanging in equi- 
 librium in a horizontal posi- 
 tion. If from the beam we *2=T 
 suspend two bodies of un- 
 equal weight P, Q, it will still 
 remain in equilibrium in a 
 horizontal position if 
 
 P X AO = Q X BO. 
 
 I ^ k 
 
 Let the weight P suspended from ^ be a hook or a scale- 
 pan. If to P we add an unknown weight W, we shall still 
 have equilibrium, provided Q is shifted to a point C such that 
 
 (P-{-W)AO= Qx CO. 
 
 Subtract these equations, and 
 
 W X AO = Q X BC, 
 
 which gives the unknown Pf as soon sls BCis measured. 
 
 To save measurements of BC at every weighing of a body, 
 it is convenient to graduate the beam in the first place. 
 Thus suppose P = 1 lb, § = 2 lb, and AO = i in. Then 
 OB = 2 in, and a notch can be made at B, which, as the 
 weight Q then balances the pan P only, would be marked 0. 
 Let now W=l lb; then BC= A0xW/Q = 2 in, and C 
 would be the position at the 1-lb mark. Make W=2\h and 
 BJD = 4 in, giving D the 2-lb mark, auv^ so on. Hence in 
 weighing a body it is only necessary to place it in the pan 
 and move the weight Q until the notch is found where the 
 
178 STATICS OF A BODY. [§ 157 
 
 beam will remain horizontal. The number at the notch in- 
 dicates the weight. This instrument is called a Steelyard. 
 
 The advantages of a steelyard over the balance are: (1) 
 the exact adjustment of the instrument is made by moving a 
 single weight Q along the rod; (2) when the body to be 
 weighed is heavier than the fixed weight the pressure on the 
 point of support is less than in the balance. The steelyard 
 is therefore better adapted to measure large weights. There 
 is, on the other hand, this advantage in the balance, that by 
 using numerous small weights the reading can be effected 
 with greater precision than by subdividing the arm of the 
 steelyard. (Eouth.) 
 
 ^ Ex. 1. Graduate a steelyard to weigh half-pounds. 
 ■\ 2. If the point of suspension be not over the center of 
 gravity and the movable weight Q placed at a point H holds 
 the steelyard in a horizontal position, show that when the 
 weight P is attached HB = AOxP/Q, and hence show how 
 to graduate the steelyard. 
 
 [For AO and Q are constant. .'. HB varies as P. Hence 
 i^'is the point from which the graduations must be made. 
 If Q is at B when P = 1 lb, then by taking BD = BH and 
 placing Q at i), P will be 2 lb, that is, B is the 2-lb notch, 
 etc.] 
 
 > 3. A steelyard beam weighs 3 lb, the wt. § is 4 lb, arfd the 
 distance of the center of gravity from is 3 in, and of the 
 point of suspension of the scale A from 5 in. Show that 
 the 1-lb graduation-marks are at intervals of 5/4 in. 
 
 -t4. A steelyard weighs W\h and is correctly graduated for 
 a movable weight Q. Prove that a weight 2Q may be used 
 provided a fixed weight W is suspended at the center of grav- 
 ity of the steelyard, but the readings must be doubled. 
 
§ 158] 
 
 EQUILIBRIUM. 
 
 179 
 
 ^ 5. A piece is broken off the longer arm of a steelyard. 
 Show that the customer is defrauded. 
 
 -f 6. A grocer files the movable weight of his steelyard. 
 Show tliat he cheats his customers. 
 
 158. Equilibrium of a Body Supported at Two Points. — 
 
 Consider next a body in which two points are supported, as, 
 for example, a beam supported by 
 two smooth horizontal pins A and B, 
 or a beam resting on two props A 
 and B. 
 
 The forces acting are the weight 
 W vertically downwards through G, 
 the center of gravity of the beam, 
 and the reactions A^, , iV, of the sup- 
 ports A and B. Since the direction of PFis known, if from 
 the conditions of the question that of one of the two iVj or 
 iV, is given, the direction of the other is found; otherwise 
 the problem is indeterminate. 
 
 If the forces are not parallel they must meet in a point, 
 and the solution is given by Art. 141; if they are parallel it 
 is given by Art, 140. As this latter case is of special impor- 
 tance, it is here worked out in full. 
 
 Let the two props A, B be in the same horizontal plane. 
 Since the direction of Wis vertical, the directions of the reac- 
 tions JV^, A^, at A, B must also be 
 vertical. To find these reactions, 
 resolve vertically, and take mo- 
 ments about G'f then 
 
 A^. + Ar-Pr=0, 
 N,xAC-N^X BO=0, 
 
 from which A", , A^, are found. 
 Or take moments about B and A in succession, and 
 
 -JV,XAB + WXBC=0, 
 N,xAB-WxAC=0, 
 
 from which the same values of A', , A', result. 
 
 ^**i 
 
 iN, 
 
180 
 
 STATICS OF A BODY. 
 
 [§158 
 
 The pressures JV, , iV, on the supports may also be deter- 
 mined graphically. 
 Thus suppose the beam AB to carry besides its own weight 
 
 W a load W^ at (7. Draw ah, be 
 to scale to represent the vertical 
 forces Tf, , T^'in magnitude and 
 direction. The force repre- 
 sented by the line cl)a which 
 closes the polygon of forces 
 will hold W^y ]^^ in equilibrium 
 and will be the sum of the re- 
 actions JV, , JV,. 
 
 Take any pole and join 
 Oa, Oh, Oc. From any point 
 p in the direction of iV, draw 
 pq parallel to aO, from q draw 
 qr parallel to hO, and from r 
 draw rs parallel to cO. Join 
 spf and draw 01 parallel to sp. 
 Then cl, la are the reactions 
 N\,, N^ sought. 
 
 Proof. — The intersection t of 
 pq and sr gives the position of 
 the resultant ahc of W^ and W. 
 (Art. 128.) And cha, the re- 
 sultant of iVj and X^, being equal and opposite to ahc, must 
 act upwards through t. 
 
 Now cha acting at t is equiv. to cO along ts and Oa 
 along tp. 
 
 But cO along ts is equiv. to cl along sB and 10 along^jy^, 
 and Oa along ?fjt? is equiv. to 01 along 6?;? and la along ;? J^. 
 The forces 10 and 0^, being equal and opposite, balance. 
 Hence cha acting at t is equiv. to cl along sB and la along ;? J ; 
 that is, cl, la represent the reactions N^, N^dit A, B on the 
 scale of forces. 
 Hence the rule ; 
 
 •iM 
 
 j^^te 
 
§ 159] EQUILIBRIUM. 181 
 
 (a) Form the force polygon or load line aica by laying off 
 the forces to scale. 
 
 (6) Select a convenient pole and form the equilibrium 
 polygon pqrs. 
 
 {c) Draw 01 parallel to the closing line sp of the equilib- 
 rium polygon, dividing ca into parts la, cl, which will repre- 
 sent the reactions N^ , iV, at ^ and B respectively. (Compare 
 Art. 129.) 
 
 169. The graphical method may be employed in finding the 
 stresses in a mechanism. Take, for example, the steam- 
 engine. Let P be the 
 pressure exerted by the 
 piston on the pin A of 
 the cross-head. It is 
 transmitted by the con- 
 necting-rod to the crank- 
 pin By and thence to the 
 crank axis C. If now 
 the machinery is driven by a wheel (JU- on the axis G work- 
 ing in another wheel at Gy the resistance R would be tangent 
 to the pitch-circles of these wheels. 
 
 Consider the pin A. It is in equilibrium under the press- 
 ure P along the axis of the piston-rod, the thrust Q along 
 the connecting-rod, and the reaction N of the guide-bar of 
 the cross-head. Hence plot P to scale, and complete the tri- 
 angle of forces, from which scale off Q and N. 
 
 Again, the wheel CG is in equilibrium under the action of 
 Qy R and the reaction S of the crank axis C. All three must 
 meet in the point D, where Q and R meet. Hence plot the 
 triangle of forces, and scale off *S^ and R. The relation be- 
 tween Py the piston pressure, and Ry the force transmitted to 
 the mechanism, is therefore determined. 
 
 "We have neglected the weights of the pieces. The only 
 one important to consider is the weight resting on the crank 
 axis C. Call it W, It acts vertically. Combine Q and W 
 into one resultant J?,. The reaction S will now pass through 
 
162 
 
 STATICS OF A fiODY. 
 
 [§150 
 
 the intersection of R^ and R. Hence complete the triangle of 
 forces for /?, 8, 7?,, and scale off S and R. Thus the relation 
 between P and R is found. 
 
 Similarly, the weights of all the pieces may be taken into 
 account if desired. 
 
 Ex. 1. A spherical shot weighing 100 lb lies between two 
 smooth planes inclined at respectively 30° 
 and 60° to the horizontal. Find the press- 
 ure on each plane. 
 
 [The pressures of the shot on the planes 
 are perpendicular to the planes. These 
 pressures are balanced by the reactions 
 N^ , N^ of the planes, which reactions pass 
 through the center of the sphere. The 
 weight of the sphere acts vertically down- 
 ward at the center. 
 
 Hence the center may be considered in equilibrium 
 under the forces N^, iV,, 100 pounds, and the planes may be 
 removed. Finish in all the ways indicated in Art. 141. 
 
 Ans. N^ = 50 pounds; iV, = 50 VS pounds.] 
 ^"2. A rod AB whose weight may be neglected and which is 
 35 in long carries a weight IT at G 20 in from the end A, 
 
§ 159] EQUILIBRIUM. 183 
 
 The rod is carried by a thread 49 in long, tied to the ends 
 
 A, B and shing over a smooth peg C. Find 
 
 the pull on the thread and the inclination c 
 
 of the rod to the horizontal when it comes /n^\j 
 
 to rest. "y I N. 
 
 [The peg being smooth, the pull 7" is the / gI^-— ^^ 
 same on both sides of the peg. The rod ^^^r^^^_^ 
 is held in equilibrium by the three forces a 
 T, T, W. The directions of T, T pass 
 through C. Hence the direction of W, which is vertical, 
 must pass through C. (Art. 141.) 
 
 Let the angles at C be a, /?. 
 
 By Lami's theorem 
 
 T/sm a = r/sin /? = TF/sin (a + /?). 
 ,\ a = p and T = TF/2 cos a. 
 
 The value of the inclination 6 follows from the geometry of 
 the figure. 
 
 For AG: CB = AG: GB = 20: 15. 
 
 .-. AG =28 and 05 = 21. 
 
 Hence Z.AGB = 90° and « = 45^ 
 
 Also cos e = AH/AG 
 
 = 7/5 V2, and 8 is found. 
 
 Finally, T = W/2 cos 45 = TV/V2, the pull in the thread. 
 -<' 3. A uniform beam weighing W lb rests on two smooth 
 planes inclined at 30" and 60° to the horizontal. Find the 
 angle which the beam makes with the horizontal in the posi- 
 tion of equilibrium, and also the pi-essures on the planes. 
 
 A71S. 30"; W^3/2, PF/2 pounds. 
 -i~4. If in (3) the angles of inclination of the planes are /3, y, 
 and is the inclination of the beam to the horizontal, then 
 
 2 tan 6 = cot /3 — cot y. 
 
 J -5. If in (3-) the beam is not uniform, but has its C.G. at a 
 istance a from one end and b from the other end, then 
 
 (a -j- J) tan = a eot /3 — b cot y. 
 
184 STATICS OF A BODY. [§ 159 
 
 " 6. A cellar-door AB, hinged at the upper edge A, rests at 
 an angle of 45° with the horizontal. Its weight, If lb may be 
 taken to act at its middle point G. The door is raised by a 
 horizontal pull F applied at the lower edge B, Find F, and 
 also the reaction E at the hinge A. _ 
 
 Ans. F= PF/2; R = W V^/2. 
 -^ 7. A straight rod 4 in long is placed in a smooth hemi- 
 spherical cup, and when in equilibrium one inch projects 
 over the edge. Find the radius of the cup. Ans. VS in. 
 S 8. A rod 3 ft long is in equilibrium resting upon a smo.oth 
 pin and with one end against a smooth vertical wall. If the 
 pin is 1 ft from the wall, show that the inclination 8 to the 
 horizontal is given by 3 cos^ 6 = 2. 
 
 ' 9. One end of a heavy uniform rod rests against a smooth 
 vertical wall. A smooth ring whose weight may be neg- 
 lected, attached to a point in the wall by an inextensible 
 thread, slides on the rod. If 6 is the angle which the rod 
 makes with the wall when in equilibrium, and if the length 
 of the rod be 2n times that of the thread, prove 
 
 cot^ 6^ + cot ^ = ^. 
 
 -i 10. To a point A is fastened one end of a thread of length 
 Zjwith a smooth ring at the other end B. Through this ring 
 passes another thread with one end fastened at C, distant 21 
 from A and on the same horizontal line with it. A weight 
 W is attached to the other end of this thread. Show that in 
 the position of equilibrium, neglecting weight of thread and 
 ring, the inclination 6^ of ^^ to ^ C is given by 
 
 cos 6 = 2 cos 26, 
 
 and the inclination of CB to GA by 
 
 sec 0/2 + cosec 0/2 rn 4f^. 
 
 Examples. — To he Solved Grapliically (Art. 158). 
 
 1. A highway bridge 25 ft long weighs 6 tons. Find the 
 pressures on the abutments when a 2^-ton wagon is one fifth 
 of the distance across. 
 
 [Referring to figure of Art. 158, If = 6 tons, W^ = 2.5 tons. 
 
 Draw on a scale of forces of 1 ton = 1 in. Then ab = 2.5 
 in, be = Q in. 
 
 Complete the equilibrium polygon and draw 01. Then cl 
 
160] 
 
 EQUILIBRIUM. 
 
 185 
 
 will be found to measure 3.5 in and la 5 in, showing that the 
 pressures are 3.5 tons and 5 tons. 
 
 Solve also analytically by taking moments about the sup- 
 ports.] 
 
 2. A ladder 20 ft long and weighing 75 lb is carried by two 
 men, one at each end. If one man carries 30 lb, how far is 
 the C.G. from the end of the ladder ? Ans. 12 ft. 
 
 3. A beam of 40 ft span weighs 1 ton per running foot. 
 One half of it carries a uniform Toad (as a train of coal cars) 
 of 2 tons and the other of 3 tons per running foot. Find the 
 pressures on the end supports. Ans. 65 tons; 75 tons. 
 
 4. A truss of 60 ft span and weighing 100 tons carries an 
 
 OOP .-12- 
 
 A 10' 4'6' 4 6' 'h'2 
 
 8'r 
 
 Erie consolidation engine as in the figure. Find the press- 
 ures on the supports. Ans. 66.6 tons; 84.9 tons. 
 
 ^ 160. Equilibrium of a System of Bodies. — If bodies rest 
 against one another or are connected by threads, hinges, etc., 
 so that the system while in equilibrium under external forces 
 can be regarded as remaining unchanged in form, the condi- 
 tions of Art. 140 are at once applicable. The pulls (tensions) 
 of threads, reactions of hinges, etc., being in pairs and form- 
 ing stresses, are in equilibrium when the whole system is 
 considered. Hence in writing down the equations of equi- 
 librium these internal stresses may be neglected, the external 
 forces alone being considered. 
 
 Also, each body in the system being in equilibrium, we may 
 consider it as severed from the system provided we apply 
 forces equal to the resultant ac- 
 tions of the system on it. The 
 conditions of equilibrium may be 
 applied to each body in succes- 
 sion. 
 
 Thus take two equal uniform 
 beams AB, BC of weight W lb 
 each, hinged at B, and with the other ends A, C moving on 
 
186 STATICS OP A fiODY. [§ 160 
 
 hinges in the same horizontal plane, the beams being in a 
 vertical plane. 
 
 Put the length AB = I, the height BO = li, and the span 
 ^C = 2«. The beams being equal, the reactions N, N 2X B 
 will be horizontal. Also, if the beam BC were removed, AB 
 would be in equilibrium under the forces JV, N, and the reac- 
 tion N^ of the hinge A, which reaction must pass through 
 H, the intersection of W and N (Art. 141). Hence the di- 
 rection of N^ is known. Put its inclination to AC = ^. 
 
 The beam jlB is in equilibrium under the forces W, N, iV,. 
 Kesolve vertically and horizontally, and 
 
 W- N, sin/J = 0; 
 N- N,GO^ ft = 0, 
 
 or iVj = TTcosec /?; 
 
 W= Wcot /3 
 
 = Wa/2h; 
 
 and the forces JV, A^, are found. 
 
 Or, graphically, since the forces on the beam AB are par- 
 allel to the sides of the triangle AHE, if we take HK to 
 represent W, the values of A^, N^ will be represented by 
 AK, HK, and may be scaled off or computed. 
 
 Ex. 1. Find the values of A^, A", by taking moments about 
 A, Km succession. 
 
 2. Find the value of N^ by considering the system as a 
 whole in equilibrium under W, W, N^ , A", (the internal forces 
 A, A^ forming a stress) and resolving vertically. 
 
 3. Two beams AB, BC, of weight W lb 
 each are hinged at B, and with their mid- 
 dle points D, E joined by a beam I^E of 
 weight W^ lb hinged at D, E. The ends A, 
 C rest upon a smooth horizontal support; 
 to find the reactions at A, D, B, E, C. 
 
 [The reactions Y at A and C are vertical. 
 Let the reaction at B due to the weight W^ 
 be resolved into two components, X^ horizontal and Y^ verticaL 
 
§ 160] EQUILIBRIUM. W 
 
 The beam DE is in equilibrium, 
 The beam AB is in equilibrium, 
 
 r -{-¥,= w, 
 
 and taking moments about D, 
 
 JV^sin e= Fcos 0. 
 
 Hence Y, ]F, , X, , and N are found.] 
 
 4. What are the values of Y, F, , X, , and N in the above 
 example ? 
 
 What is the total reaction at D ? 
 
 5. In a row boat propelled by two oars the pulls exerted by 
 the rower are equal. Find the resistance of the water to the 
 motion of the boat and the pressure on the rowlock. 
 
 [Let P = pull on each oar; 
 
 Q = pressure of water on each blade; 
 
 B = reaction between each rowlock and oar; 
 
 S = resistance of water to motion of boat; 
 
 a = distance of hand of rower from rowlock; 
 
 b = distance of rowlock from blade of oar. 
 Now the system as a whole is in equilibrium under the ex- 
 ternal parallel forces Q, Q, S. Hence 
 
 Q+Q = S. 
 
 Also, since each oar is in equilibrium under the parallel 
 forces P, Q, R, by taking moments about the rowlock 
 
 Pa= Qb. 
 
 \ Pa = Sb/2, 
 
 and the resistance S is found. 
 
 The pressure R on the rowlock would be given by taking 
 moments about the blade, or 
 
 Rb = P(a + b). 
 As a check we have 
 
188 STATICS OF A BODY. [§ 160 
 
 the forces P, §, R being parallel. ] 
 
 6. In an 8-oar boat the oars are 10 ft long, and the dis- 
 tance of the hand of each rower from the rowlock is 2.5 ft. 
 If the pull of each rower is 75 pounds, find the force exerted 
 on the boat. Arts. 400 pounds. 
 
 \ la. Two weights P, Q rest on the outside of a smooth ver- 
 tical hoop and are connected by a thread which subtends a 
 right angle at the center of the hoop. Find the inclination 
 6 of the thread to the horizontal when in the position of equi- 
 librium. Ans. 6 = tan-i(P - Q)/(P + Q). 
 jL lb. Show that if T denotes the pull in the thread, then 
 
 rpyp. ^ rpyQ 
 
 2 — 9 
 
 A 
 
 j^ 8. Two smooth spheres, each 1 ft in diameter and weigh- 
 ing 10 lb, are placed inside a hollow cylinder of 20 in diameter, 
 open at both ends and resting on a horizontal table. Find 
 the weight of the cylinder when just on the point of over- 
 turning. Ans. 8 lb. 
 
 9. Two equal cylindrical sawlogs, of weight W^ lb each, are 
 in contact with each other and with the bottom and sides of 
 a truck in which they are placed. A third equal log is placed 
 on the other two- Find the pressure on the sides of the truck. 
 
 A ns. JV/2 V'S pounds. 
 
 10a. Find the proper elevation BE of the outer rail on a 
 
 railroad track for a given velocity v of engine weighing W 
 
 lb, and on a curve of radius r ft, in 
 
 -(^■^\ order that there may be no flange or 
 
 ur^\ \ v^ lateral pressure on the rails. 
 
 ^U \ \ Y^ ^ [The forces acting are W pounds 
 
 \ \ ^X ^---^— >>- yertically downward through G, the 
 
 \ '^^ [^-''^k^^' center of gravity of the engine, and 
 
 \^V|^^^^^~ tlie reactions N^, N^ of the rails. 
 
 ^^^^^-'- • Since there is no flange pressur-e, the 
 
 *^^ ^ \ reactions must be perpendicular to 
 
 i \ the track. 
 
 The resultant motion is due to a 
 force C directed to the center of the curve (Art. 108). Hence 
 the engine may be said to be in equilibrium under fF, N^,N^, 
 and — C. 
 
 Let Q be the inclination of ^P to the horizontal. 
 
§ 160] EQUILIBRIUM. 189 
 
 Resolve the forces along AB and 
 
 - Cco8 0-{-W&me = O, 
 
 But (Art. 109) 
 
 0= Wv*/gr -pounds, r:^ 
 
 /. i&Ti6=C/W=vygr. 
 
 If is small, tan = BE/AB = elevation/gauge of track. 
 
 .'. elevation of outer rail = gauge X v*/gr, ^?«^i» V 
 
 For standard gauge of 4 ft 8i in this gives 
 
 elevation of outer rail = 7v^/4:r inches, nearly, 
 
 V being expressed in feet per second, and 7' in feet.] 
 
 lOb. If the velocity Fis expressed in miles per hour, show 
 that 
 
 elevation of outer rail = 15 F''/4r inches, nearly, 
 
 the radius ?* being, as before, expressed in feet. 
 
 10c. If the velocity V is expressed in miles per hour, and 
 D is the degree of curve, then 
 
 elevation of outer rail = F'Z>/1530 inches, nearly. 
 
 11. A train is running round a curve of radius r with ve- 
 locity V. Show that the weight of a carriage is divided be- 
 tween the outer and inner rails in the ratio of 
 
 gra + ^Vi to gra — v''h, 
 
 where h is the height of the C.G. of the carriage above the 
 rails, and 2a is the distance between the rails. 
 
 12. A train is running round a level curve of 3025 ft 
 radius at 15 miles an hour. Show that the deflection of a 
 plumb-line suspended in a car from the vertical is tan"^ 1/200. 
 
 13. Find the greatest velocity v a locomotive can have to 
 be just on the point of overturning on a curved level track 
 
100 STATICS OF A BODY. [§ 161 
 
 of radius r ft, the center of gravity of the locomotive being 
 6 ft above the rails, and the gauge of the track 4 ft 8^ inches. 
 
 Ans. 3.55 Vr ft/sec, nearly. 
 
 'fl^lSl. Stability. — A body in equilibrium under the action 
 of forces may be at rest or move with uniform velocity in a 
 straight line. Consider it at rest in some one position. If 
 displaced from this position it may remain at rest under 
 the forces acting, or it may begin to move. Thus let a rod 
 suspended from a fixed point be in equilib- 
 rium. The force acting, the weight W verti- 
 cally downward, is balanced by the reaction of 
 the support. If the rod is displaced, the forces 
 at and G are no longer in the same line but 
 form a couple tending to turn the rod. If 
 
 'W jw is above G, the tendency of the rod is to re- 
 
 turn to its original position, and the rod in its 
 original position is said to be in stablE equilibrium. The sta- 
 bility is measured by the torque, that is, by the moment of W 
 about the point of turning 0. 
 
 If is below G, the tendency of the rod when disturbed is 
 to move farther from its original position, and the equilibrium 
 is unst-able. If is at G, the rod will remain in any position 
 and the equilibrium is neutruL 
 
 162. An important case is that in which a body rests on a 
 horizontal plane. If the body is symmetrical and is under 
 the action of gravity only, the resultant force of gravity acts 
 vertically through the center of gravity and must be balanced 
 by the resultant vertical reaction of the plane. The problem 
 may thus be treated as if all the forces acted in one plane. 
 
 If the body is not symmetrical and the resultant vertical 
 force falls within the base of the body resting on the plane, 
 equilibrium will exist; if it does not so fall, the weight down- 
 ward and equal reaction upward will form a couple and the 
 body will rotate and topple over. 
 
 Suppose, for example, A BCD to be the cross-section of a 
 
162] 
 
 STABILITY. 
 
 191 
 
 
 
 C 
 
 
 D 
 
 E -! 
 
 i 
 
 
 G 
 
 ^ 
 
 /k 
 
 
 H 
 
 
 B 
 
 
 
 "W 
 
 
 wall built to withstand the pressure of earth on one side, as 
 the wall of a railroad em- 
 bankment. Such a wall is 
 called a retaining wall. We 
 assume that the wall can- 
 not give way except in one 
 piece and by being over- 
 turned about the edge A. 
 Let P be the resultant of 
 the forces acting on the 
 side BD of the wall, and let 
 be the point of applica- 
 tion of P. The weight W of the wall acts vertically down- 
 ward through the center of gravity O, 
 
 The tendency to overturn about the edge A is measured by 
 the moment of P about A, that is, by P X AE. The ten- 
 dency to withstand overturning is measured by W X AH. 
 The stability against rotation depends upon the difference of 
 these two moments. 
 
 Hence in the position of limiting equilibrium when the 
 wall is just on the point of turning about A we have the 
 relation 
 
 PXAF= Wx AH. 
 
 For the condition of security against the shearing force 
 causing sliding see Ex. 21, p. 222. 
 
 The problem of stability is further discussed in Art. 219. 
 
 Ex. 1. A cubical block is placed on an inclined plane. If 
 kept from slipping, find the inclination of the plane when 
 just on the point of rolling over. A7is. 45°. 
 
 r~ 2. The silver dollar is 1.5 in diameter and 0.1 in thick. 
 Show that a cylindrical pile of 100 dollars, but no more, will 
 stand if placed on a desk sloping 3 in 20. 
 ■ — 3. A triangular board whose sides are 2, 5, VTs inches will 
 just stand if placed with the side 2 on a horizontal table. 
 ^ 4. A triangular lamina whose sides are a, b, c can just stand 
 on the side c when placed on a smooth table. Show that 
 
192 STATICS OF A BODY. [§ 163 
 
 "T 5a. A circular table weighing Wlh has three equal legs at 
 equidistant points on its circumference. The table is placed 
 on a level floor. Neglecting the weight of the legs, find the 
 smallest weight which, hung upon the edge, will be upon the 
 point of upsetting the table. A7is. Wlh. 
 
 -^ bl. If the table has four legs at equidistant points, find the 
 least weight that will upset it. Ayis. W{ V'Z + 1) lb. 
 
 -^ be. If the table is square and the legs are at the middle 
 points of the sides, show that any weight greater than W 
 placed at one of the corners will upset the table. 
 
 163. Geaphical Statics. — If in Art. 160 the beams, in- 
 stead of being connected at ^ by a hinge rigidly attached to 
 the beams, are connected by a pin distinct from both beams, 
 and also rest upon pins at A, C distinct from the beams, the 
 problem is much simplified. 
 
 Consider the beam AB. Suppose the forces acting on the 
 beam, including its weight, to be combined into a single force 
 F, and let F be resolved into two com- 
 ponents F^ at ^ and F^ at B. 
 
 Let N^ , X^ be the reactions of the 
 pins at A, B on the beam. The beam 
 is in equilibrium under the forces F^, 
 N^ at A, andi^2, N^ at B. Combine 
 F^ , N^ into a single force i?, , and F^ , 
 N^ into a single force R^. Then the 
 beam being in equilibrium, the forces 
 jRj, R^ must be equal and act in oppo- 
 site directions. Hence they form a stress. 
 
 Since i?, is the resultant of F^ and N^, a force R^, equal 
 and opposite 7?,, would keep F^ and A", in equilibrium. 
 
 Now since each pin in a framework connects two or^more 
 beams, the resultant action of these beams on the pin A, say, 
 is equal to A'j. Hence F^, the action of the component of 
 the external force on the pin, R^, the action of the beam AB 
 on the pin, and N^, the action of the other beams on the 
 pin Af are in equilibrium. 
 
 In general, each pin is in e(juilibrium under the action of 
 
§ 165] 
 
 GRAPHICAL STATICS. 
 
 193 
 
 the external force at the pin and the forces in the beams that 
 meet at the pin, the directions of these forces being along the 
 beams. 
 
 The external force F being known in magnitude and di- 
 rection, and the directions of the forces along the beams 
 being known from the form of the structure, the computation 
 of the magnitude of these forces (commonly called stresses) 
 forms a simple problem of forces meeting at a point, and 
 may be solved graphically or analytically. 
 
 A careful study of the examples will show that in some 
 cases the graphical solution is to be preferred, and in others 
 the analytical. The graphical solution is in general the more 
 convenient if the structure is at all complex. 
 
 164. The determination of stresses in beams pinned to- 
 gether and subjected to external forces generally in the form 
 of loads of some kind is very important in architecture and 
 engineering. The subject belongs to a special branch of 
 mechanics, known as graphical statics. We add a short 
 sketch of its application to simple framed structures. 
 
 165. Jointed Frames. — As the simplest possible example of 
 a jointed frame, let us consider three 
 
 beams hinged by pins at A, B, C, 
 and resting on supports at A, C in 
 the same horizontal plane. This is 
 known as a triangular truss. Sup- 
 pose the beams all alike and weigh- 
 ing W lb each. The reactions of the 
 supports A, C balance the weights 
 of the beams and act vertically up- 
 wards. Hence the external forces 
 acting and keeping the truss in equilibrium are as in the 
 figure. 
 
 Now transfer the weights to the pins. Thus, 
 
 Wat G, = iWsitA-\-iWat B, 
 TFat G^ = ^JV^t B-\-\W2^i G, 
 Fat (?, = ^rat^ + irat G\ 
 
194 
 
 STATICS OF A BODY. 
 
 [§165 
 
 -l-w- 
 
 rat G,+ Wiit G,-JrWat G,= W^t A-\-W at B+W at C, 
 
 or the weights at G^, G^, G^ are 
 replaced by weights PT at ^, TT at 
 i?, and Tfat C. 
 
 The total reactions at A, 
 being equal, each is one half the 
 total weight, or is equal to 3 W/2. 
 Combining the upward and 
 downward forces, we have, finally 
 the force Wat B vertically down- 
 ward, and the resultant forces (or reactions) W/2 at A and 
 TF/2 at C vertically upward, keeping the truss in equilibrium. 
 
 We have thus transferred the 
 weights of the beams to the joints, 
 and can now consider the beams 
 as without weight, and indicat- 
 ing direction only. The resulting 
 stresses in the pieces we next find. 
 
 Since the weights on each pin 
 are in equilibrium with the stresses 
 produced in the pieces meeting at 
 the pin, the condition of Art. 85 is at once applicable 
 sider the forces at each pin in order. 
 
 (a) Pin A, The forces acting are Tf/2 vertically upwards, 
 and the unknown stresses in BA, CA, and these three forces 
 keep the pin in equilibrium. Draw Oa to scale (fig. A) to 
 represent JV/2. From a draw ab parallel to BA, and bO par- 
 allel to AC, closing the triangle. Then ab, bO represent on 
 the same scale the stresses in AB,AC. The direction of pr/2 
 or Oa is known to be vertically upwards. And since for 
 equilibrum Oa, ab, bO must be taken the same way round, 
 their directions are as in the figure. 
 
 Transfer these directions to the truss diagram. The stress 
 ab in AB is towards A, showing that the piece AB is in com- 
 pi^ession, or is a Strut; the stress bO in AC from A, showing 
 that the piece ACis in tensmi, or is a Tie. 
 
 Con- 
 
§1G6] 
 
 GRAPHICAL STATICS. 
 
 195 
 
 (b) Pin B. The forces equilibrating are the stress in AB 
 towards B (being equal and opposite that towards A) and 
 the force W vertically downwards, which are known, and the 
 
 stress in CB, which is unknown. Draw (fig. B) ba to rep- 
 resent the stress in AB, ac to represent W; then cb will rep- 
 resent the stress in CB. Transferring the directions to the 
 stress diagram, we see that CB is a strut. 
 
 (c) Pin C. The forces equilibrating are the stress in BC, 
 W/2y and the stress in AC, all of which are known. For 
 check the diagram may be drawn as in Fig. C. 
 
 166. It is evident that we should have saved labor by add- 
 ing the second figure to the first and the third to the sum as 
 in the fourth figure, which is the complete stress diagram. 
 
 In practice it is convenient to consider the stress diagram 
 as in two parts. Thus the line ac is the polygon of external 
 forces, ac being the downward force at B balanced by the 
 upward forces cO at Cand Oa at A, and is complete in itself. 
 Tlie closing of this line shows that the external forces have 
 been properly estimated. On this force polygon as base the 
 stress diagram is added step by step by passing from pin to 
 pin as indicated. 
 
 In case the truss is symmetrical, as in our example, it is 
 only necessary to consider the first half of the pins. But it 
 is safer to consider all of them, as the symmetry of the draw- 
 ing will furnish a test of its accuracy. 
 
196 STATICS OF A BODY. [§ 168 
 
 Notice carefully that a study of the stress diagram shows 
 not only the amount of stress but the hind of stress in any 
 piece, and therefore whether a strut or a tie should be em- 
 ployed. 
 
 167. For tracing the connection between the pieces them- 
 selves and the stresses in them as shown by the stress dia- 
 gram, an exceedingly convenient system of notation, due to 
 Prof. Henrici, London, but usually known as Bow's notation, 
 is in common use. 
 
 A beam or a force is named by letters placed on either side 
 of it. Thus, in the second figure, p. 194, 01 is the tie AC, Oa 
 the reaction PF/2 at the left support, ah the strut AB, ac the 
 force W at B, and so on. These letters carried into the 
 stress diagram aOcb give us ah the stress in the piece ah, ch 
 the stress in ch, Oh the stress in the rod Oh. The letters 
 A, B, C at the pins do not enter the stress diagram and are 
 not necessary. 
 
 168. The following examples should be solved both graphi- 
 cally and analytically. The first will be worked out in detail 
 by both methods. 
 
 The general outline of the graphical solution is this: 
 
 (1) Draw the truss to scale from the dimensions given. 
 
 (2) Compute the pin loads. 
 
 (3) Compute the resultant reactions of the supports. 
 
 (4) Draw the force polygon to scale. 
 
 (5) Draw the stress diagram on the force polygon as base. 
 
 (6) Scale off the stresses and tabulate, indicating forces in 
 compression by the sign — , and pieces in tension by the 
 sign -f . 
 
 The analytical solution is carried out by equating th'e mo- 
 ments of the lateral forces and of the stresses developed in 
 the pieces. See page 198. 
 
 Ex. 1. In a roof of 32 ft span and height 12 ft the trusses 
 are 10 ft apart, and the pieces EF, GH come to the middle 
 points of the rafters. If the weight of the roof-covering is 
 25 lb/ft% draw the stress diagram and scale off the stresses. 
 
§168] 
 
 GRAPHICAL STATICS. 
 
 197 
 
 (a) Graphical Solutio7i. — (1) Draw the truss to scale from 
 the given dimensions. Take the scale of dimensions 5 f t = 
 lin. 
 
 
 
 
 ' 
 
 
 
 
 
 
 b/ 
 
 \c 
 
 
 
 
 
 A/ 
 ^^ E 
 
 y^ F 
 
 G^ 
 
 H 
 
 sD 
 
 
 
 O 
 Scale of Dimensions 
 
 
 
 1 
 
 Scale of Stresses 
 
 (2) The length of each rafter = VW -f 16' = 20 ft. 
 The load on each rafter = 20 X 25 x 10 = 5000 pounds. 
 The pin loads AB, BC, CD are each 2500 pounds and act 
 
 vertically downward as indicated in the figure. 
 
 (3) The resultant reaction at each pier = 7500/2 = 3750 
 pounds vertically upward. 
 
 (4) Take the scale of stresses 1000 pounds = 1 in. 
 
 Form the force polygon by laying off ab = 2500 pounds, 
 be = 2500 pounds, cd = 2500 pounds, do = 3750 pounds, 
 oa = 3750 pounds, thus closing the polygon or load line ad. 
 
 (5) Next the stress diagram. 
 
 Begin at the left-hand pin. The forces acting are OA = 
 3750 pounds and the unknown stresses in AE, BO. From 
 a draw a line ae parallel to the piece AE, and from o a line oe 
 parallel to the piece OE intersecting ae in e, thus forming the 
 stress triangle oae. Scale off ne,oey and we find the stresses in 
 the pieces AE, OE to be 6250 and 5000 pounds respectively. 
 
 To find the character of the stresses note that OA acts up- 
 ward. Carrying the direction the same way round the tri- 
 angle oae and transferring these directions to the pieces 
 themselves, we find the piece AE in compression and the 
 piece OE in tension. 
 
 Proceed to the next pin to the right. The stresses are rep- 
 resented by the sides of the quadrilateral ab/e, of which ab, ae 
 are known. The unknown stresses are bf, fe, which may be 
 scaled off and their directions determined as before. 
 
 Next take the pin at the vertex. The stresses in the pieces 
 
198 
 
 STATICS OP A BODY. 
 
 [§168 
 
 meeting here are represented by the sides of the crossed figure 
 hcfgh, of which he, bf are known. 
 
 (6) Finally, the results may be tabulated as follows: 
 
 Name of 
 Piece. 
 
 AE or DH 
 
 EO or HO 
 
 BFov CO 
 
 EFovOH 
 
 FQ 
 
 Stress 
 
 -6250 
 
 + 5000 
 
 -4170 
 
 -2080 
 
 + 2500 
 
 Note. — Instead of scaling off AE, EO, etc., we may com- 
 pute their values. For the triangle oae is similar to the tri- 
 angle formed by the left half of the truss diagram, whose 
 dimensions are known. Then 
 
 and hence 
 
 3750 (or oa) : a6 = 12: 20; 
 
 3750 \oYoa) : oe = 12 : 16; 
 
 ae = 6250; oe = 5000; as before. 
 
 The computed and measured stresses may differ a few 
 pounds, depending on the scale used. The allowable differ- 
 ence depends upon the character of the work. 
 
 (6) Analytical Solution, — Suppose the truss divided into 
 
 two parts by the plane /^/i. 
 x,/tv Consider the equilibrium of 
 
 fjK ^v the part to the left of this 
 
 plane. 
 
 The part to the right 
 may be removed if we con- 
 ceive forces X, y, z applied 
 equal and opposite to the 
 stresses in the pieces. Equilibrium will exist between the ex- 
 ternal forces 3750 pounds at a and 2500 pounds at h and these 
 forces. The directions in which the forces 3750, 2500 tend to 
 turn the truss is indicated by the arrows. To counteract this 
 tendency the forces X, Y, Z must act as shown in the figure. 
 In forming the equation of moments (Art. 136) labor is 
 saved if we take the point of moment for any unknown^force 
 at the intersection of the directions of the other two unknown 
 forces. Thus, to find x, take moments about and, noting 
 that the length of the perpendicular froui on the direction 
 of X is 9.6, we have 
 
 9.6X+ 2500 X 8 - 3750 X 16 = 0, 
 or X= 4167 pounds. 
 
168] 
 
 GRAPHICAL STATICS. 
 
 199 
 
 To find Y take moments about a, 
 
 then 
 
 9.6r- 2500 X 8 = 0, 
 or Y = 2083 pounds. 
 
 To find Z take moments about b, 
 
 then 6Z - 3750 X 8 = 0, 
 
 or Z = 5000 pounds, 
 
 the values already found. 
 
 The directions of the arrows show that Xand Y a.re forces 
 of compression and Z of tension. 
 
 Similarly, by taking sections in other places we may find 
 the stresses in all of the pieces. 
 
 (a) Why are a, b, the best points for taking moments 
 in finding X, Y, Z? Would other points do? Try and 
 see. 
 
 (/3) Explain clearly how it is that the taking of moments 
 about a, b, c gives the same results as the application of the 
 three conditions of equilibrium of Art. 140^. 
 
 2. In a triangular roof-truss the rafters are 2^ ft apart and 
 the roofing material weighs 20 lb/ft^ The span is 24 ft and 
 height 5 ft. Find the stresses in the rafters. 
 
 Ans. 845 pounds. 
 
 3. Show that the stress diagram for the truss represented 
 
 SCALE OF DIMENSIONS 
 
 SCALE OF STRESSES 
 
 in the figure, loaded at the center over the vertical piece ah, 
 known as the king-post, is as in the margin. 
 4. A foot-bridge 18 ft span and 6 ft breadth has a crowd 
 
200 
 
 STATICS OF A BODY. 
 
 [§168 
 
 of people on it equal to 100 Ib/ft^ floor surface. The king- 
 posts of the two trusses are 3 ft in flepth. Find the stresses. 
 Ans. Stress on post ab = 2700 pounds. 
 
 5. In (3) the span is 21, depth d. Show that the compres- 
 sion in Aa is Wl/2d, and find the tension in Oa and the stress 
 in the vertical ab. Ans. WVd'-\-r/2d; W. 
 
 6. Find the stresses in the roof-truss represented in the 
 
 figure. The span is 24 ft, rise of 
 ridge 16 ft, rise of hip 8 ft; the 
 trusses are 5 ft apart, and the 
 weight of the roof -covering is 20 
 lb/ft^ 
 
 Ans. 
 
 AE 
 or DH 
 
 BF 
 or CG 
 
 EF 
 or GH 
 
 EO 
 or HO 
 
 FG 
 
 -3750 
 
 -2500 
 
 -750 
 
 + 2700 
 
 + 3000 
 
 7. Draw a stress diagram for a queen-post truss. The 
 queen-posts ab, be divide the span into three equal parts, and 
 
 the truss is loaded at uie joints with weights W. 
 
 8. A foot-bridge (queen-post) of span 24 ft, breadth 7 ft, 
 length of queen-posts 3 ft, carries a load of 100 lb/ft'' of floor. 
 Find the stresses developed, the queen-posts dividing the 
 span into three equal parts. 
 
 Ans. Aa, Bb, ov Cc =i - 7467; Oa orOc=-{- 7970; 
 ab or be = - 2800; Ob = -{■ 7467 pounds. 
 
 EXAMINATION". 
 
 1. Show how to find graphically the resultant of any num- 
 ber of coplanar forces acting on a body. 
 
§ 168] EXAMINATION. 201 
 
 2. Is a system of coplanar forces equivalent to a single 
 force ? If not, state the exceptions. 
 
 3. Show how to find the resultant of two like parallel forces 
 and its line of action. When the forces are unlike and par- 
 allel, how then ? 
 
 4. Define the moment of a force with reference to an as- 
 signed point. 
 
 5. Why is it difficult to hold a heavy weight at arm's 
 length ? 
 
 [President Lincoln "could take a heavy ax and, grasping 
 it with his thumb and forefinger at the extreme end of the 
 handle, hold it out in a horizontal line from his body."] 
 "+6. Distinguish between the terms moment and momen- 
 tum, 
 -j- 7. Three men are to carry a stick 18 ft long and weighing 
 200 lb, each to sustain one third of the weight. One man is 
 to lift from the end and the other two by means of a cross- 
 bar. Where must the cross-bar be placed ? 
 
 8. An inch is taken as the unit of length. What is the 
 geometrical representation of the unit of moment ? 
 ^ 9. State and prove Varignon's theorem of moments (1) | 
 
 for forces not parallel, (2) for parallel forces. 
 
 4— 10. Is there any reason why a man should put his shoulder 
 to the spoke rather than to the body of a wagon in helping it 
 uphill ? 
 
 ^ 11. A couple can never bo balanced by a single force. 
 
 -^12. The moment of a couple can never be zero. 
 
 -U 13. State the changes which a couple may undergo without • ,/,< 
 altering its statical effect. f ^* 
 
 14. Show how to find a couple equivalent to a number of 
 couples having the same plane. 
 
 15. Forces acting along the sides of a polygon in order and 
 proportional to the sides in magnitude may be reduced to a 
 single couple. 
 
 16 Show that couples may be combined according to the 
 parallelogram law. 
 
S02 STATICS Of A BODY. [§ 1^^ 
 
 17. The sum of the moments of the two forces of a couple 
 about any point in their plane is constant. 
 
 18. The moment of a couple may be represented by the 
 area of the parallelogram formed by the two forces of the 
 couple as opposite sides. 
 
 19. Show how to combine analytically any number of co- 
 planar forces acting on a body. 
 
 20. If any number of forces acting at a point equilibrate, 
 the algebraic sum of the components of the forces in any two 
 directions must each be equal to zero. 
 
 21. Three forces in a plane equilibrate. Show that one of 
 two conditions must be satisfied. 
 
 22. State the conditions of equilibrium of any. number of 
 coplanar forces acting on a body. 
 
 23. The conditions to be satisfied for parallel forces in 
 equilibrium are 
 
 24. A body free to move in a plane has one point fixed. 
 State the condition of equilibrium. 
 
 25. Two torques [P, P), (§, Q) with arms «, h are in equi- 
 librium if Pa— Qh — 0. 
 
 26. " In problems of equilibrium it is necessary to con- 
 sider bodies to be perfectly rigid." Is this true ? Consider 
 a bridge, for example. 
 
 27. Find the pressures on the rails due to a given pressure 
 P on the locomotive crank-pin. 
 
 28. Three forces represented by the three median lines 
 AD, BE, CFoi a triangle J^ (7 equilibrate (Art. 140a). 
 
 29. Define the C.Gr. of a system of heavy particles,, and 
 show that in every case there exists only one such point. 
 
 [Suppose there are two such points G^ , G^. 
 
 The direction of the resultant force of gravity is towards 
 the earth's center no matter how the body is turned. 
 
 Turn the body so that the line 6^,, G^ is perpendicular to 
 the direction of this force. 
 
§ 168] EXAMINATION. 203 
 
 Tlie force cannot therefore pass through both G^ and 6^,.] 
 "^SO. A flat disk is not perfectly circular. How would you 
 find its C.G. ? 
 
 4-31. A body supported at its C.G. will remain at rest in any 
 position. 
 
 -f 32. A person going uphill appears to lean forwards and 
 going down to lean backwards. Explain. 
 "^ 33. State a rule for finding the centroid of a uniform tri- 
 angular (1) lamina, (2) wire. I ' ' " 
 A- 34. Find the centroid of a trapezoid, the parallel sides 
 being a and h respectively and their distance apart d. 
 "f 35. Given the weights and C.G. of a body and of one part 
 of it to find the C.G. of the remainder. 
 
 -^ 36. In an inclined plane AB n. weight P suspended by a 
 thread passing over the highest point B balances a weight Q 
 on the plane. Show that the C.G. of P and Q remains at 
 the same height above tlie base whatever the positions of the 
 weights. (Torricelli's principle.) 
 "f37. Find the C.G. of a quadrilateral. 
 ^38. P is a point without a triangle ABC. Show that the 
 resultant of the forces represented by PA, PB, PC passes 
 through the C.G. of the triangle. 
 
 -f 39. The three sections of a fishing-rod are 4 ft, 2 ft, and 
 1 ft long and weigh 8 oz, 6 oz, and 4 oz respectively. Find 
 the C.G. of the rod when drawn out to its full length. 
 
 40«. How does the property that every body has but one 
 C.G. help us to solve geometrical theorems ? 
 
 [Place weights at certain points. Combine in different 
 order to find the C.G. of the system. The various portions 
 of the C.G. found must coincide.] 
 
 40J. The median lines of a triangle pass through one point. 
 
 40c. The lines joining the middle points of the opposite 
 sides of a quadrilateral bisect each other. 
 
 40c?. The line joining the middle point of the two diago- 
 nals of a quadrilateral and the lines joining the middle points 
 of the opposite sides all intersect in one point. 
 
204 STATICS OF A BODY. [§ 168 
 
 ^41. Explain the positions of force, resistance, and fulcrum 
 
 in the following levers : Wheelbarrow, spade, pair of scissors, 
 
 the forearm. 
 
 J 42. A man carries a bundle at the end of a stick which 
 
 passes over his shoulder. Show that the pressure on his 
 
 shoulder varies inversely as the distance of his hand from his 
 
 shoulder. 
 
 43. Find the position in which a beam-balance will rest 
 when loaded with unequal weights. 
 
 -1^44. In a beam-balance if the C.G. coincides with the point 
 of suspension the balance remains in equilibrium in all posi- 
 tions. 
 
 -^45. Show that Borda's method of double weighing gives 
 the correct weight of a body no matter how false the balance 
 used. 
 
 ^ 46. In a beam-balance great sensitiveness and quick weigh- 
 ing are to a certain extent incompatible. 
 
 47, In a common steelyard the length of the graduations is 
 inversely proportional to the movable weight. 
 
 48. Two smooth spheres, each weighing W lb, rest in con- 
 tact between two smooth planes inclined at 30"" and 60° to 
 the horizontal. Find the position of equilibrium. 
 
 Ans. Inclination of line joining centers to horizon- 
 tal = tan-* 3/4. 
 T 49. A rod a yard long rests over the edge of a vertical 
 hollow cylinder 2^ inches diameter with one end against the 
 inner surface. Show that in the position of equilibrium the 
 rod is inclined at 60° to the horizontal. 
 
 -^50. Forces P, Q, R act at a point and keep it in equilib- 
 rium. If any line cut their directions in A, B, C, prove 
 
 P/OA + Q/OB + R/OC = 0. 
 
 51. If forces P, Q, R acting at the center of a circular 
 lamina along the radii OA, OB, OC equilibrate forces P\ Q\ 
 R' acting along the sides BC, CA, AB of the inscribed tri- 
 angle ABC, show that 
 
§ 168] EXAMiNATioiq'. 205 
 
 PF'/BC -f QQ'/CA + RR'/A B = 0. 
 
 52. If D is the degree of curve, show that the elevation of 
 outer rail required for an engine speed of 30 miles an hour in 
 passing round a circular curve is 
 
 3/)/5 inches, nearly. 
 
306 
 
 FRICTIOI^. 
 
 [§169 
 
 CHAPTEE V. 
 
 FRICTION. 
 
 169. As already stated, greater simplicity and clearness are 
 secured by considering the properties of bodies one at a time, 
 and thus leading up to the actual state of the case, which is 
 quite complicated, since bodies in nature possess many prop- 
 erties. Thus far the surface of a body has- been assumed to 
 be perfect iv smooth, that is (Art. 104), to offer no resistance 
 to the motion of a body in contact with it. But in reality we 
 know that if one body be moved along another (as a book 
 along a table) a certain resistance will be offered to the motion. 
 The resistance arises from irregularities in the surfaces in 
 contact, from elevations and depressions which fit more or 
 less closely into one another. To it the name Friction is 
 given. 
 
 Suppose a body weighing W lb to rest on a horizontal plane 
 and to be acted on by a vertical force of Q pounds. The 
 total vertical force (P = Q -{- W) 
 pounds is equilibrated by the ver- 
 tical reaction JV of the plane. If 
 now d small force is applied paral- 
 lel to the plane, the motion of the 
 body is prevented by the equal 
 force of friction called into play. 
 If the force be increased, the fric- 
 tion is equally increased until a • 
 certain limit is reached when a further increase of force 
 causes motion. The friction between two bodies thus accom- 
 modates itself to the acting force up to a certain limit when 
 
§ 171] LAWS OF FRICTION^. 207 
 
 motion takes place, and the amount called into play when the 
 body is just on the point of moving is called limiting friction. 
 Or, since the body is just on the point of moving and the act- 
 ing forces may be regarded as equilibrating one another, the 
 friction is named static friction. 
 
 If the surfaces in contact were smooth, the reaction JV would 
 be normal to the table. But now that the plane is not smooth 
 the reaction R is necessarily no longer normal. The com- 
 ponent of R normal to the plane would be the reaction N as 
 between smooth surfaces, and the component along the plane 
 would be the friction /between the two surfaces. 
 
 170. Laws of Static Friction. — From experiment it is found 
 that between surfaces with little or no lubrication, under mod- 
 erate loads, and just beginning to slide on one another, or 
 when the velocity of motion is small, the amount of static 
 friction is — 
 
 (1) Proportional to the normal stress hetweeii the surfaces 
 in contact. 
 
 (2) Indepejident of the areas in contact, the normal stress 
 remaining the same. 
 
 (3) Dependent on the material of lohich the bodies are com- 
 posed. 
 
 Hence from law (1) the friction / corresponding to the 
 normal stress iV is given by 
 
 /=//iv, or f/]sr=^, 
 
 where /^ is a constant depending on the material and character 
 of the surfaces in contact. It is called the coefficient of fric- 
 tion, and its value must be determined by experiment (Arts. 
 175, 179). 
 
 171. The above "laws" of friction were deduced from ex- 
 periments made with surfaces having little or no lubrication, 
 and moving with low velocities; and for such conditions only 
 are they to be depended on. In machines, however, surfaces 
 without lubrication and moving with low velocities are the 
 exception, and we there have an entirely dijfferent set of con- 
 
208 ^ FRICTION". [§ 172 
 
 ditions. The friction is now kinetic* Recent experiments 
 show that even with surfaces of the same material, the charac- 
 ter of the lubrication, the thickness of the lubricating film, 
 the load, the velocity, the form of the surfaces whether flat or 
 curved, the areas in contact, and the temperature of the sur- 
 faces have each great influence on the friction produced. 
 The films that coat the surfaces slide over one another when 
 the pressure is not excessive. No general relation between 
 the friction (/) and the pressure (i\^) producing it has yet 
 been deduced depending on these conditions. Special experi- 
 ments are necessary in all cases where the conditions differ 
 in any of)the points mentioned from those entering into ex- 
 periments already made. 
 
 We may, however, in general write 
 
 where the value of // is determined by the conditions of the 
 problem. These conditions will show whether we may assume 
 the so-called laws of statical friction or have recourse to special 
 experiment. Eoughly, the coefficient of friction /* for well- 
 lubricated surfaces is from J to yV that for dry surfaces. 
 ll72. When one surface rolls on another, the resistance en- 
 countered is termed 7'olling f7Hction. Consider a heavy wheel 
 rolling over a horizontal surface. The wheel from its weight 
 compresses the surface and is itself compressed at the place of 
 support. Thus the area in contact is greater than if both 
 were rigid, and the wheel is forced to climb an elevation in 
 order to move forward. The elevation occurs at every new 
 position of the wheel, and the resistance is continuous. The 
 amount of this frictional resistance depends on the nature of 
 the materials in contact. 
 
 Hence the importance of smooth hard roads for carriages 
 
 * The distinction between static and kinetic friction was first pointed 
 out by Coulomb (1736-1806). The laws of static friction were enun- 
 ciated by Coulomb, and confirmed by the later experiments of Moriu 
 at Metz in 1837-1838. 
 
§m] 
 
 ANGLE OF FRICTION. 
 
 209 
 
 and bicycles. Hence, too, the importance of tracks well bal- 
 lasted, and of heavy rails whether for loco- 
 motive or street-car traffic. In railway 
 work when the road is of heavy steel rails 
 the rolling friction is found to be small 
 in comparison with the bearing friction. 
 
 173. Angle of Friction— In the experi- 
 ment of Art. 169 the three forces F, P, R 
 hold the body in equilibrium and may be 
 represented by the sides of a right tri- 
 angle. The reaction R is equivalent to 
 the two forces / along the plane and N 
 normal to it. 
 The angle between the directions of 
 
 R and N is called the angle of friction. 
 Now, since the body is in equilibrium under F, JV,/, P, 
 
 F-f=0; 
 JV"-P = 0. 
 
 Also, since the body is in equilibrium under F, R, P, 
 
 F— Bsm (p = 0; . , 
 
 P — Rcos(f> = 0. . . 
 
 But by definition / = jjK. . . . 
 
 Hence X /< = //-^ = F/P = tan </>, 
 
 (1) 
 (2) 
 
 (3) 
 (4) 
 (5) 
 
 or the coefficient of friction is equal to the tangent of the angle 
 of friction, 
 
 Ex. 1. Find the coefficient of friction when the angle of 
 repose is 30°; 45°. Ans. 1/4/3 = .577; 1. 
 
 2. The coefficient of friction is 0.4. Find the angle of 
 friction. Ans. 21° 48'. 
 
 174. The fact that the reaction i?, which holds the forces 
 i^and P in equilibrium, makes with the normal to the sur- 
 faces in contact an angle equal to the angle of friction 0, 
 
210 
 
 FRICTION. 
 
 [§175 
 
 gives a key to the application of the graphical method of the 
 polygon of forces to problems involving friction. The method 
 is especially valuable in cases where -the forces are interlaced 
 as in mechanisms. 
 
 175. Measurement of the Coefficient of Friction.— Suppose a 
 
 body of weight W to rest 
 
 f on a table, and that the 
 
 table is tilted about the 
 edge A. We have now 
 the body resting on an 
 inclined plane ^(7. Con- 
 tinue tilting until an 
 inclination 6 is reached 
 when the body is just on 
 the point of moving down the plane. At this point the forces 
 holding it in equilibrium are PF vertically downward, iV^ nor- 
 mal to the plane, and /along the plane. Draw the triangle 
 of forces, and 
 
 //iV=tan^. 
 
 But since/ is the limiting friction, 
 
 f/]^ = tan 0. 
 
 Hence 6 = (p, or the angle of inclination of the plane, is 
 equal to 0, the angle of friction. For this reason the angle 
 of friction is called the Angle of Repose. 
 
 This gives an experimental method of measuring the angle 
 of friction, and thence the coefficient of friction between two 
 bodies F, Q. For form an inclined plane of one of the bodies, 
 F, and on this place the other, Q. Tilt the plane until Q just 
 begins to slide: the tangent of the angle of inclination is the 
 coefficient of friction for the two bodies. 
 
 176. The following coefficients of friction may be regarded 
 as average values, to be used when no special experiments 
 covering the cases under consideration are possible. The cir- 
 cumstances may be such that the tabular values are vjery far 
 
§178] 
 
 EQUILIBRIUM. 
 
 211 
 
 from the truth. Indeed, at present we may be said to be ac- 
 quainted with no quantitative laws of friction of much value. 
 
 Wood on wood or metal, surface dry 0.4 to 0.6 
 
 " « " " lubricated . 0.1 to 0.:^ 
 
 Metal on metal, " dry 0.2 
 
 " lubricated.. 0.075 
 
 Steel on ice 0.02 
 
 177. Equilibrium on a Rough Incline. — To find the limits 
 between which a force F must lie to hold in equilibrium a 
 body of weight W on the rough incline AC, the inclination B 
 of the plane to the horizon, the inclination ft of the force to 
 the plane, and the angle 
 of friction being given. 
 
 (a) Let the body be on 
 the point of moving up 
 the plane. 
 
 The friction/ acts down 
 the plane. 
 
 The forces acting are 
 Ff Wf JV, and /. Or, since the resultant R oi N and / makes 
 an angle with the normal to the plane, the forces acting may 
 be taken to be F, W, R, Then by Lami's theorem 
 
 i^/sin (^ + 0) = TT/sin (90 -f /? - 0), 
 
 or F= W sin {6 + 0)/cos (ft - 0), 
 
 {b) Let the body be on the point of moving down the plane* 
 Then similarly 
 
 F= W%in(d - 0)/cos (ft + 0). 
 
 Hence for any force lying between these values of F the 
 body remains in equilibrium and is not on the point of mov- 
 ing either up or down the plane. These values of F are the 
 limiting values required. 
 
 178. These results may also be obtained by the method of 
 Art, 86. For when the body is on the point of moving up the 
 
212 FRICTION. [§ 178 
 
 plane, we have, resolving along and perpendicular to the plane, 
 
 Fcos/S - f - Wsm e = 0, 
 
 Fsin /3-^]Sr- PT cos 6^ = 0. 
 
 Also, / = /iiV, and /i = tan 0. 
 
 The resulting value of F will be found to agree with that 
 just given in (a). 
 
 The value of the resultant reaction of the plane will be 
 found to be W cos {/3 + 6')/cos (/3 — 0). 
 
 V Ex. 1. A weight of 60 lb rests on a rough level floor. Find 
 the least horizontal force that will move it, the coefficient 
 of friction being 0.5. Ans. 30 pounds. 
 
 Find the resultant reaction of the floor. 
 
 A71S, 30 Vb pounds. 
 
 2. In Ex. 1 find the least force inclined at 45° to the floor 
 that will move the weight. A7is. 20 \^2 pounds. 
 
 Find the resultant reaction of the floor. 
 
 A71S. 20 V5 pounds. 
 What is its direction ? 
 
 3. The weight on the driving-wheels of a locomotive is 
 twenty tons, and the coefficient of friction is 0.2. Find the 
 greatest pull the engine is capable of. A7is. 4 tons. 
 
 V 4. Find the least force that will drag a body weighing 100 
 lb along a rough horizontal plane, the coefficient of friction 
 being 1/Vd. Ans, 50 pounds. 
 
 Find the resultant reaction of the plane. 
 
 Ans. 50 1^3 pounds. 
 • ba. Find the angle ^ which a given force F must make 
 with a horizontal plane that a weight W may just be on the 
 point of sliding on the plane, the angle of friction being 0. 
 
 Ans. /? is found from i^cos (/3 — (p) = W sin 0. 
 - bb. For what value of f3 is the force F the least possible, 
 and what is the value of the least force ? 
 
 Ans. /3 = cp; F= W sin 0. 
 be. What is the resultant reaction of the plane ? 
 
 A71S. W cos 0. 
 ;. 6. Find the least angle of inclination of a wooden incline 
 that stone blocks may slide down it under gravity only 
 (^ = 1/V3). Ans. 30°. 
 
§ 178] EQUILIBRIUM. 213 
 
 ' 7. The foot of a ladder of weight W and length I rests on 
 the ground at A, and the top at />' against a rough vertical 
 wall. Find its inclination when on the point of sliding, 
 the coefficient of friction in each case 
 being 0.5. 
 
 [The forces acting are W at G the 
 middle point of AB, the reaction ^Y of 
 the ground at A normal to AC, and the 
 friction / along AC; the reaction iV, 
 at J) normal to OB, and the friction /, 
 along CB. 
 
 The forces equilibrating, the con- 
 ditions of Art. Sij must be satisfied. 
 Hence, resolving vertically, horizontally, 
 and taking moments about G, 
 
 N^f,- r=0, (1) 
 
 f-N, =0, (2) 
 
 - Nl cos e -{-fl sin 6 + NJ, sin -]-fJ, cos 6 = 0,, (3) 
 
 Also, since the ladder is just on the point of sliding, we 
 have from the first of Morin's laws (Art. 170) 
 
 /=JVA /, = JV./2. 
 
 Hence, by substitution in (3), 
 
 tan e - 3/4. 
 
 From equations (1) and (2) we have' the reactions 
 
 N=4:W/b, A^, = 2^75.] 
 
 la. Is there any advantage in taking moments about G 
 m the preceding solution? 
 
 Ih, Take other points and find from which the most 
 simple equation results. Try A and B for 
 example. 
 
 7r. Deduce the relation tan 6 = 3/4 
 from geometrical considerations. 
 
 [The resultant 7? of A^ and / makes an 
 angle with the normal at A. 
 
 The resultant 11^ of A', and/, makes an 
 angle with the normal at B. 
 There being three forces R, R^ , W, they must meet in a 
 
214 FRICTIOK. [§ 178 
 
 point (Art. 141), which, since the direction and position of 
 W^are fixed, must lie on the vertical passing through G. 
 
 Evidently lAOB = 90°. But AG = GB. .'. GO =GB 
 and 90 - = (9 -h 0. Hence 
 
 tan 6^ = cot 20 = 3/4, since cot = 2. 
 
 The reactions iV^ and JV, may novr be found by Lamias 
 theorem. Find them.] 
 
 ^d. If the coefficients of friction at A and B are jj and /./, , 
 show that 
 
 tan 6= (I — MM,)/^M- 
 
 Obtain this in two ways as above. 
 
 7e. If the wall is smooth and the coefficient of friction 
 between ground and ladder is //, the inclination of the ladder 
 to the horizontal is cot~^ 2/i. 
 
 ' 8. Find the horizontal force necessary to push a body 
 weighing Wlh up a rough incline, the angle of inclination of 
 the plane to the horizon being 6 and the angle of friction 0. 
 
 Ans. irtan [6 -\- 0). 
 
 What is the force necessary to keep the body from sliding 
 down ? Ans. H'tan {6 — 0). 
 
 -f 9. A weight IF is just supported by friction on a plane in- 
 clined at an angle 6^ to the horizon. Show that it cannot be 
 moved up the plane by any horizontal force less than 
 H^tan 26. 
 
 Examine the special case of 6^ = 45°. 
 
 1 10. A ladder 20 ft long rests at 45° against a rough vertical 
 wall. A man whose weight is twice that of the ladder mounts 
 it. When will the ladder begin to slip, the coefficient of fric- 
 tion being 0.5 at either end? A7is. After going up 13 ft. 
 
 lla. A body weighing IF lb is placed on a rough plane in- 
 clined at an angle 6 to the horizon. Find the limits between 
 which a force acting parallel to the plane must lie in order to 
 keep the body from moving. 
 
 A71S. W (sin 6 ± /A cos 0) or W sin (6^ ± 0)/cos 
 if yu = tan 0. 
 
 11^. If the force necessary to pull the body up the plane 
 be n times that required to keep it from sliding down, then 
 
 {n — 1) tan 6 = {n -{- 1) tan 0. 
 
 12. A weight can just be sustained on a rough plane in- 
 
§ 178] EQUILIBRIUM. 216 
 
 clined at 45° by a force acting horizontally or by an equal 
 force acting along the plane. Show that 
 
 ;/= V2- 1. 
 
 13. On a hill sloping 1 in 50 a loaded sled weighing 1 ton 
 is kept from sliding down. Show that the pull of the horses 
 may vary from 3G0 to 440 pounds, the coefficient of friction 
 between sled and snow being 0.2. 
 
 14. A rod resting within a rough sphere subtends a right 
 angle at the center of the sphere. If the rod is on the point 
 of moving, show that its inclination to the horizontal is twice 
 the angle of friction. 
 
 15. If in (14) the rod subtends an angle 2>5 at the center, 
 then 
 
 2 tan 6^ = tan (/? + 0) - tan (/? - 0). 
 
 ~^16. The angle of a wooden incline is G8°. Show that it is 
 impossible to drag a wooden block up the plane by a horizon- 
 tal force, the coefficient of friction being 0.4. 
 
 17. The force necessary to haul a train at uniform speed 
 on a 1^ grade is 3.5 times that on the level. Show that the 
 coefficient of friction is 1/250. 
 
 18. A uniform beam rests on two rough inclined planes. 
 Show that in the position of equilibrium the inclination 6 of 
 the beam to the horizontal is given by 
 
 2 tan 6/ = cot (a + 0) - cot (ft - 0), 
 
 when a, /3 are the inclinations of the planes and is the 
 angle of friction. 
 
 19fl. Two weights IT, , IF,, joined by a thread, are placed on 
 a rough narrow inclined plane. If the coefficients of friction 
 between weights and plane are //, , //, , show that the weights 
 will just begin to slide down when the 
 plane is tilted to an angle 
 
 [This is a good illustration of Art. IGO. 
 
 Consider the system as a whole. Then 
 the tensions T, T are internal forces. For equilibrium, re- 
 solving along the plane and perpendicular to it. 
 
216 FRICTION-. [§ 179 
 
 W, sin 0-i- W,&m8 = //.iVr, + /i^JVr, 
 W, cos 6 + W, cos 6' = iV^, + iV,. 
 
 But since the first weight is in equilibrium under FT, , iV, T, 
 
 ]Sr^ = W, cos 6; 
 
 and since the second weight is in equilibrium under TF,, JV, T, 
 
 JV^ = W, cos 0, 
 
 Hence after a simple reduction 
 
 tan e = {IX, W, + //,rj/(r, + r,).] 
 
 195. Solve the problem by considering the weights in equi- 
 librium under W,, Ny T and W^ , N, T respectively, and 
 equating the two values of T. 
 
 19c. Show that the tension of the thread is 
 
 Pf sin 6 — p.W cos 6, 
 
 20. Three equal weights are joined as in Ex. 19. If the 
 coefficients of friction are 0.4, 0.5, 0.6, show that the slope of 
 the plane is as 1 to 2. 
 
 --179. Motion on a Rough Plane. — Suppose a body weighing 
 W lb to slide on a rough horizontal plane and let /i be the 
 coefficient of friction, to determine the motion. 
 
 The normal force being W pounds, the friction is yuTf 
 pounds. Hence the motion may be said to be due to a force 
 oi — fxW pounds. 
 
 Let a be the acceleration produced. Then (Art. 67) 
 
 - fiW= Wa/g, 
 and .*. a = — ^g. 
 
 If n is the initial velocity, we have (Art. 24) 
 
 v = u- ixgt, 
 s = ut — ipigf, 
 
 which will give the velocity acquired and distance passed over 
 at the end of a time t. 
 
§180] 
 
 MOTION ON A ROUGH PLANE. 
 
 217 
 
 Application, — Coulomb and Morin used an apparatus similar 
 to that shown in Art. 1G9 to determine the coefficient of fric- 
 tion. The coefficient of static friction was given by the force 
 F necessary to bring the body P from rest to motion. The 
 coefficient of kinetic friction was computed by noting the 
 time t taken by P to slide over a distance s. Here 
 
 moving force = F— /^P, 
 If a is the acceleration produced, 
 
 (F-\-P)a/g = F-MP' 
 
 s = -Jaf ; 
 
 But 
 
 ^ = F/P - 2(F-{- P)s/Pgt\ 
 
 and jj. is found. 
 
 180. Motion on a Rough Incline. — Let the plane be in- 
 clined to the horizontal at an angle 
 and let /< be the coefficient of friction. 
 
 If the body is moving doivn the plane, 
 the forces acting are Pf^ertically down- 
 wards, the reaction J^ normal to the 
 plane, and the friction juNup the plane. 
 
 The force causing motion down the 
 plane is the sum of the components 
 along the plane AC, or l^sin 6 — jjN. 
 Hence, if a is the acceleration of motion (Art. 67), 
 
 W sin 6 -/j]^=Wa/g (1) 
 
 Resolving the forces perpendicular to AC, we have 
 
 Ar= If cos 6^. (2) 
 
 Eliminating iV between (1) and (2), 
 
 a = ^(sin — jj. cos 6), 
 
 or, putting /a = tan 0, 
 
218 FRICTION^. t§ 1^^ 
 
 « = ^ sin (6 — 0)/cos 0, .... (3) 
 
 which gives the acceleration down the plane. 
 
 Similarly, if the body is projected up the plane, the acclera- 
 tion (found by changing the sign of //) is 
 
 ^ sin (^ + 0)/cos 0. 
 
 The acceleration being found, the velocity at any time and 
 the distance passed over are known from Art. 24. 
 
 -^ Ex. 1. A curling-stone is projected along ice with a veloc- 
 ity of 16 ft /sec. If the coefficient of friction is 0.1, find in 
 what time it will come to rest and how far it will travel. 
 
 Ans. 5 sec; 40 ft. 
 
 - 2. A weight of 8 lb rests on a rough floor. Find the least 
 horizontal force that will give it an acceleration of 4 ft/sec% 
 the coefficient of friction being 0.5. Ans. 5 pounds. 
 
 - 3. A body with initial velocity u slides along a rough hori- 
 zontal plane on which the coefficient of friction is ju. Show 
 that it will come to rest in a time u//yig after passing over a 
 distance uy2/Ag. 
 
 ^ 4. A train moving at 40 miles an hour on a level track is 
 brought to rest by friction in half a minute. Prove that the 
 coefficient of friction is 11/180. 
 
 -j- 5. A weight W\ lb is moved along a rough table by a weight 
 W^ lb attached to a thread passing over a smooth pin at the 
 edge of the table. Find the pull of the thread. 
 
 Ans. W, W,{1 + m)/{ W^ + W,) pounds. 
 -^ 6. The inclination of a plane is 45° and the coefficient of 
 friction 0.T5. Show that the time taken by a body to slide 
 down this plane is twice what ic would be if the plane were 
 smooth. 
 
 ~ 7. A body is projected up a rough plane inclined at an 
 angle 6 to the horizon. If t^ is the time of ascending and t^ 
 the time of descending, show that 
 
 {t^' + t,') cot 8 = (t,' - t^') cot 0, 
 
 where is the angle of friction. 
 
 ;^ 8. A train of 100 tons (excluding engine) runs up a 1^ grade 
 with an acceleration of 1 ft/sec\ If the friction is 10 pounds 
 per ton, find the pull on the drawbar between engine and 
 train. A7is. 4f tons. 
 
§181] 
 
 BELT FRICTION^. 
 
 219 
 
 181. Belt Friction. — This is an interesting and important 
 a})plication of the preceding principles. 
 Consider a belt or rope passing over (or 
 around) a cyFindrical block securely fas- 
 tened, and let P, , Q^ denote the forces at 
 the ends of the belt. Suppose -P, to be just 
 on the point of overcoming §,, or that the 
 belt is moving with uniform velocity. The 
 forces acting on the belt, its weight being 
 neglected, are P, , (J, , the reaction of the cylinder, and the 
 friction of the cylinder. In order to find the relation between 
 them conceive the arc of contact of the belt cut into ele- 
 ments of indefinitely small length, find the relation for one 
 element, and then sum up for the whole 
 arc. 
 
 The pull of the belt increases from 
 §, to Pj. Consider an element ab of 
 the belt and let P denote the pull at a 
 and P + dP the pull at b. The third 
 force acting is the reaction dR of the 
 pulley, which makes an angle equal 
 to the angle of friction with the radius (Art. 174). 
 
 The forces which hold ab in equilibrium Py P -\- dP, and 
 dR, being three in number, must intersect in one point c (Art. 
 141). Call the angle aOb = dd. 
 Then by Lamias theorem 
 
 P/sin (90 + -\-dd) = (P + <?P)/sin (90 - 0), 
 or P/(cos - dO sin 0) = (P + dP)/cos 0, 
 
 since d6 is a small angle, and therefore cos dd = 1, 
 sin dO = dd. 
 
 Hence, clearing of fractions, we have ultimately 
 
 dP = Ptfir\(pde 
 
 = fiPdd, 
 
 putting tan (p = i^ the coefficient of friction. 
 
220 FRICTION. [§ 182 
 
 Summing up all the elements ah of the rope in contact, 
 noting that the limits of P are P, and §, , and calling the 
 angle subtended by the arc of contact 6, we have, if 6 is ex- 
 pressed in circular measure, 
 
 / dP/P= / fxde, 
 
 . or loge PJQ, = ^e, 
 or P, = Q,e^^ 
 
 when e{ = 2.718) is the base of the natural system of 
 logarithms. 
 
 Transferring to the common system of logarithms, as more 
 convenient for computation, 
 
 log,oP,/Q, = OASiM^. 
 
 182. If the weight Q^ is on the point of slipping down in- 
 stead of being drawn up, the action is the same as if P^ and 
 Q^ changed places, and therefore Q^ = P^ef^^. 
 
 The conditions are the same and the relation between P, 
 and Q^ the same if the cylinder instead of being fixed is 
 capable of turning about 0, and no slipping occurs, axle fric- 
 tion being neglected. 
 
 Ex. 1. Obtain the relation 
 
 dP = fxPdd 
 
 by resolving the forces along and perpendicular to the tangent 
 at a. 
 
 2. A rope passing over a wooden cylinder supports a barrel 
 of flour weighing 196 lb. Find the force which willj'ust 
 raise the barrel, the coefficient of friction being 0.4. 
 
 Ans. 689 pounds. 
 
 3. In Ex. 2 find the force that will just keep the barrel 
 from slipping down. Ans. 56 pounds. 
 
 4. Find the number n of turns of rope round a snubbing- 
 post that a man pulling P pounds may just be able to hold a 
 canal-boat pulling Q pounds. Ans. "In^n — loge Q/P. 
 
§ 183] BXAMINATION. 221 
 
 5. A chain is wrapped twice round an iron drum. Find 
 the coefficient of friction if a pull of 100 pounds just supports 
 50 tons. Ans. 0.55. 
 
 6. 11 /A = 0.25, and a rope passes twice round a post, prove 
 that any force will balance another more than twenty times as 
 great. 
 
 7. If the coefficient of friction between belt and pulley is 
 0.3, and 0.4 of the pulley is embraced by the belt, show that 
 P = 2$, nearly. 
 
 EXAMINATION. 
 
 1. When is a body said to be smooth ? When rough ? 
 
 [As it exerts force on another body normal or not to the 
 surface in contact.] 
 
 2. " Friction is a self-adjusting force." Explain. 
 
 3. The slopes of railroad embankments vary. Explain. 
 
 4. What is the usual rule for slopes in a railroad cut 
 through sand and gravel ? [1^ to 1.] 
 
 5. State the laws of limiting friction and define the coeffi- 
 cient of friction. 
 
 6. Show that coefficient of friction = tan (angle of fric- 
 tion). 
 
 "^ 7. Give an experimental method of finding the coefficient 
 of friction between two surfaces. 
 
 8. What horizontal force will cause a state bordering on 
 motion in a weight of 10 lb lying on a level fioor if the coef- 
 ficient of friction is 0.5 ? 
 
 -- 9. The total resistance to motion of a body on a rough in- 
 cline and just on the point of motion is JVVl -f /i% where N 
 is the normal force. 
 
 10. A body is held in equilibrium on a rough inclined plane 
 by a force along the slope. Find the limits of this force. 
 
 11. The force required to haul a sled up hill is least when 
 the inclination of the tongue is equal to the angle of friction. 
 
 12. Simon Stevinus of Bruges (1548-1620) remarked that 
 the question, " What force will support a wagon on an inclined 
 plane ? is a statical question, but that the question. What force 
 
232 FRICTION. [§ 182 
 
 will move the wagon ? requires additional considerations to 
 be introduced." Discuss this statement. 
 
 13. A die rests on a rough board and the board is tilted. 
 Will the die slide or topple over ? 
 
 ^ 14. A cylindrical jar 3 in in diameter and 1 ft in height 
 rests on a rough table (yu = 1/3) and the table is tilted. Show 
 that the jar will topple over before it slides. 
 
 — 15. A sphere cannot rest upon an inclined plane however 
 rough. 
 
 — 16. A cubical box is half filled with water and placed upon 
 a rough incline. Show that it will slide or topple over as 
 /^< > 1. 
 
 17. What is meant by the "friction of adhesion" of a loco- 
 motive ? 
 
 [Force at circum. of driver/weight on driver.] 
 -^ 18. The weight on the driving-wheels of the St. Clair tun- 
 nel decapod locomotive is 195,000 lb. Taking the coefficient 
 of traction to be 600 pounds per ton, find the hauling force 
 on the drawbar. Ans. 58,500 pounds. 
 
 19. The tractive force required to propel a bicycle over a 
 smooth level surface is 0.01 of the load. Calling the load 150 
 lb, a force of one and a half pounds would be required to 
 move the wheel forward. 
 
 ---20. Explain the contrivance known as friction-2oheels. 
 What is the advantage of ball-bearings for bicycles ? Sketch 
 in section such a bearing. 
 
 21. In the figure of Art. 162 the condition of security of 
 the wall against sliding on the base ^^ is that the horizontal 
 component of the force P is less than pi W. 
 
 22. Account for girder-rails taking the place of thin rails 
 for economy in railroad traffic. 
 
 [The blows from the wheels would distort the rails so that 
 rolling friction would be enormous.] 
 
 -" 23. A tug-boat running at 15 miles/hour begins to turn on 
 a curve of 150 ft radius, Show that objects on the deck will 
 
§ 182] EXAMINATION. 223 
 
 slide unless the coefficient of friction exceeds 0.1, the deck 
 being supposed to remain horizontal. 
 
 24. Prove that starting from scratch the shortest time over 
 100 yards is about 15 seconds, taking the adhesion of the feet 
 as 1/12 the weight. 
 
 25. A drawer of length I can be pulled out by one handle 
 unless the distance of tlie handle from the center exceeds 
 Z/2/i, where fx is the coefficient of friction. 
 
 26. A rod I ft long and weighing 1 lb per foot lies on a 
 
 rough table, the coefficient of friction being jj. The rod is 
 pulled at right angles to its length by a force at one end. 
 Show that the point P about which it begins to turn is at a 
 distance l/V'Z ft from that end. 
 
224 WORK AND ENERGY. [§ 183 
 
 CHAPTER VI. 
 
 WORK AND ENERGY. 
 
 183. In Art. 53 it was shown that if a force i^acts upon a 
 particle to through a distance s, and v is the velocity acquired, 
 we have the relation 
 
 Fs = wv'^/'Z 
 F being expressed in absolute units, or 
 
 Fs = wvy2g 
 F being expressed in gravitation units. 
 
 This equation, which was first stated by Huygens, being 
 an algebraic statement of the second law of motion, may 
 consequently be used as the starting point of the science of 
 dj/namics. It was so used first by Lagrange, and often since 
 his time by other writers. (Art. 217.) 
 
 We proceed to discuss the separate terms of this equation 
 and then the equation itself. 
 
 184. Work. — When a particle acted on by a force is dis- 
 placed in the direction of the force, the force is said to do 
 worh in moving the particle, and the work done, depending 
 both on the force and the displacement, is expressed by their 
 product. Thus, if a weight of WVo falls ths-ough a height of 
 h ft, the acting force is Tf^ pounds, and the work done hi/ the 
 force of gravity is expressed by the product Wli. If the 
 weight is raised through h ft, the resisting force of gravity W 
 pounds is overcome through h ft, and the work done against 
 the force of gravity is expressed by — Wh. 
 
 In general, when work is being done by a force F, constant 
 in magnitude and line of action, the point of application of 
 
§185] 
 
 DEFINTTIOl^ OF WORK. 
 
 225 
 
 the force is being displaced through a distance s in the direc- 
 tion of the force, and Ihe ivorh done w by the force is meas- 
 ured by the product of the force F and the displacenieiit s, or 
 
 V4 = Fs. 
 
 Vvvt,V. 
 
 If the motion be uniform and E denote the resisting force 
 overcome, we may write 
 
 W = - i?5 
 
 as the expression for the work done 07i the resistance. 
 
 Work may therefore be defined as the overcoming of re- 
 sistance. 
 
 185. The definition of work done may be stated in an 
 equivalent form, which is often conven- 
 ient. Let the particle on which the force 
 i^acts at A be displaced to B so that AB 
 is the total displacement. Let fall BC 
 perpendicular to CA, and denote the 
 angle 5^ (7 by 6, 
 
 Then by definition 
 
 Vi = Fx AC. 
 But AC = AB COS e. 
 
 /. V^= FxAB cos 0, 
 or V4 = Fcios6xAB, 
 
 or the work done by a force acting obliquely to the path of a 
 particle is measured by the product of the force and the pro- 
 jection on its direction of the total displacement, or by the 
 product of the component of the force along the total dis- 
 placement by that disj^lacement. 
 
 When 6 = 90°, then cos 6 = and W = 0. Hence, when 
 the displacement is at right angles to the direction of the 
 force, the work done by the force is nil. 
 
 Ex. In a pendulum find the work done by the pull of the 
 rod on the bob as it swings to and fro. 
 
226 WORK AND ENERGY. [§ 186 
 
 186. In the definition nothing is said about the nature of 
 the path of the point of application of the force. "We shall 
 
 now show that the work done does not de- 
 ^ pend on the form of the path. 
 
 .*].2\^:^c Let the particle on which the force F 
 
 acts at A be moved to B either in the 
 straight path AB or in the broken path 
 ACB. Let fall Bha, Cc perpendicular to 
 OAy the direction of F. Then by definition 
 
 work done in straight path AB = F X Aa; 
 
 work done in broken path ACB = F X Ac -\- Fx ca 
 
 = F(Ac + ca) 
 = FxAa, 
 
 or the work done is the same in the two paths. 
 
 The same result evidently follows if the broken path con- 
 sists of any number of parts instead of two. And this no 
 matter what the magnitude of these parts. But a curve is 
 the limit of an indefinitely great number of indefinitely short 
 paths. Hence the result is true for a curvilinear path. So 
 that 
 
 The work done h.y a force on a particle in passing from 
 one position A to another position B is e7itirely independent 
 of the path of the particle between A and B. 
 
 187. Work under Variable Force. — If the acting force jPis 
 not constant in magnitude, we may conceive the path s de- 
 scribed by its point of application between two points A and B 
 divided into an indefinitely great number of parts, each part so 
 small that the force may be considered constant throughout 
 it. Let PQ(— ^s) denote one of these small portions of the 
 path. Then, the line of action of i^at P being along 'P§, 
 the work done through PQ is Px ^s, and the total work 
 done in moving through the distance s is found by summing 
 the elementary works F x ^s, and would be expressed by 
 
 X 
 
 Fds. 
 
§ 188] UNIT OF WORK. 227 
 
 More generally, if the line of action of i^' at P is not in the 
 direction of PQ, let ^p denote the projection of the dis- 
 placement of the point of application of jP along the direc- 
 tion of F\ then the elementary work done is Fx ^p, and the 
 total work done between two positions A and B would be ex- 
 pressed by 
 
 J^Fdp=J'J^ cos eds. 
 
 188. Unit of Work. — In the general formula for work 
 
 W = Fs; 
 
 taking i<^= 1, 5 = 1, we have w = 1; and therefore the U7iit 
 of work is taken to be the work done by unit force acting 
 through unit distance. 
 
 The engineering unit of work or unit of everyday life is 
 based on the gravitation measure of force, and is the work 
 done by a force of one pound acting through a distance of 
 one foot. This is usually called a foot-pound. Thus the 
 work done in raising a body weighing 100 lb vertically 
 through 6 feet is the work done in overcoming a force of 100 
 pounds through 6 feet, and is 600 foot-pounds. 
 
 "V Ex. 1. The tractive force of a locomotive is 10 tons. Find 
 the work done in hauling a train one mile. 
 
 Ans. 105,600,000 ft-pounds. 
 I 2. The steam-pressure on the piston of an engine is 50 
 pounds/in', the area of the piston 300 in', and the stroke is 
 4 ft. Find the work done in one stroke. 
 
 Ans. 60,000 ft-pounds. 
 'T 3. A hole is punched through a w rough t-iron plate 1/2 in 
 thick, the punch exerting a uniform pressure of 42 tons. 
 Find the work done. A ns. 3500 ft-pounds. 
 
 T 4. A horse hauli]ig a wagon exerts a constant pull of 100 
 pounds and travels at the rate of 2 miles an hour. How much 
 work will be done in 5 min ? Ans. 88,000 ft-pounds. 
 
 -f 5. Find the work done in lifting a chain which hangs ver- 
 tically, its length being I ft and its weight W lb. 
 
 Ans. Wl/2 ft-pounds. 
 
228 WOEK AND ENERGY. [§ 189 
 
 189. Graphical Rppresentatiou of Work. — Work done may 
 be represented graphically. 
 
 (1) Let the acting force F be constant in magnitude and 
 direction. Let AL plotted to scale rej)- 
 
 resent the displacement of the point 
 of application of the force, and let Aa 
 plotted to scale and perpendicular to AL 
 represent the force at A. Then, since the 
 force is uniform throughout, the work done AL X Aa is rep- 
 resented by the area of the rectangle Al. 
 
 Ex. Plot on a scale of 5 pounds = 1 in and 10 ft = 1 in 
 the work done in overcoming a resistance of 20 pounds 
 through a distance of 20 ft. 
 
 (2) If the force F is variable, the work done is expressed 
 as already explained in Art. 187. 
 
 Let AL represent the displacement, and let Aa, LI perpen- 
 dicular to AL represent the 
 
 r^f — initial and final values of the 
 
 \~\ ^=-^,;;^^^^ -1 fQYQQ jp in ^}^Q direction of mo- 
 
 I j ^^^ tion. Let AL be divided into a 
 
 g-^ 'l large number 7i of equal parts, 
 
 and let Bb, Cc, . . . represent 
 the corresponding forces. A line through the extremities 
 a, h, c, . . . I forms the curve of resistance. 
 
 The element work FJp is represented by the area BCcb 
 
 ultimately, and the total work / Fdp by the sum of these ele- 
 ment areas, that is, by the area ALla. 
 
 The value of the work done may be computed apj^roxi- 
 mately from the diagram. 
 
 For the work done from A to B may be taken to be that 
 due to the force \{Aa + Bh) acting through the distance AB 
 or AL/n, or 
 
 the work done through AB = i{Aa + Bl)AL/n. 
 
§ 190] WORK OF RAISING A SYSTEM. 229 
 
 Similarly, the work done through BC = i(Bh + Cc)AL/n, 
 
 Hence by addition 
 total work done = \{Aa + 2B1) + 2C*c + . . . + Ll)AL/n. 
 
 By means of certain contrivances the curve ahc . . . I may 
 be plotted mechanically by the force itself, as, for example, in 
 the steam-engine by means of the indicator. Having the 
 curve, the mean force ^0 may be found by stretching a thread 
 . so as to have equal areas above and below it. Or the area may 
 be read off at once by an Amsler polar planimeter, and the 
 work done found directly. Or the indicator drawing or 
 "card" may be divided up by drawing equidistant ordinates, 
 the lengths of these ordinates scaled off, and the formula 
 above applied. All of these methods are at times useful. 
 
 Ex. 1. Draw a diagram for the work done in raising the 
 chain in Ex. 5, p. 227. 
 
 Ans. A right triangle, the force decreasing from 
 WtoO. 
 
 2. Draw a diagram for the work done in lifting a chain 
 vertically by one end from the ground. 
 
 Ans, A rt. triangle. (Plot it.) 
 
 >f^ 190. Work of Raising a System of Particles from One 
 Position to Another. — Let the particles weigh w, , w, , . . . and 
 let a;, , a:, , . . . denote their respective heights above a fixed 
 horizontal plane. Let the system as a whole weigh W, and let 
 X denote the height of its C.G. above the plane. Then 
 
 Wx = w^x^ + tv^x^ + (1) 
 
 Suppose the system displaced so that w^ is raised through 
 a height /i, , w, through a height h.^ , . . . and the C.G. of the 
 whole through a height h. Then 
 
 W(h + x)=^w,(h,^x,)^w,(K-^x,)-\-... . (2) 
 Subtracting (1) from (2), 
 
 Wh = wfi, + wji^ + (3) 
 
230 WORK AND ENERGY. [§ 191 
 
 But by definition loji^ -\- wji^ + • • . is the work done in 
 raising the particles from one position to the other. And this 
 is equal to Wli, the work done in raising the total weight W 
 through the height li through which the O.G. of the system 
 has been raised. 
 
 The same method of treatment may readily be applied to 
 the case of raising a body in parts. 
 
 It is evident that in general it requires less work to com- 
 pute the value of Wli than that of 'Zn.)li. Hence the utility 
 of the proposition. 
 
 Ex. 1. A cylindrical shaft 14 ft in diameter has to be sunk 
 to a depth of 10 fathoms through chalk whose weight is 
 144 lb/ft\ Find the work done. 
 
 Ans. 39,916,800 foot-pounds. 
 2. How many foot-pounds must be expended in raising 
 from the ground the materials for building a column 66 ft 8 in 
 high and 21 ft square, the material weighing 112 lb/ft\ 
 
 Ans. 109,760,000 ft-pounds. 
 - 3. Find the work done in raising a Venetian blind having 
 11 slats, the weight of each slat being lo lb and their distance 
 apart a ft. Ans. aioniii + l)/2 ft-pounds. 
 
 -/4. In pumping 1000 gallons from a water-tank with vertical 
 sides the surface of the water is lowered 5 ft. Find the work 
 done, the discharge being 10 ft above the original surface. 
 (A gallon of water weighs 8^ lb.) 
 
 Ans. 104,167 foot-pounds. 
 Draw the work diagram. 
 
 \ 
 
 191. Principle of Work.— Let forces i^, , i^, , . . . act at a 
 
 point causing a displacement OA. 
 Let R be the resultant of the forces. 
 From A let fall perpendiculars on 
 the directions of the forces, and let 
 ^, , 6/^ , . . . be the inclinations of 
 these directions to OJ, and B the in- 
 clination of the resultant R. Then, 
 (Art. 81) the sum of the components of the forces along OA 
 
§ 192] PRINCIPLE OP WORK. 231 
 
 being equal to the component of the resultant along OA, we 
 have 
 
 F, cos e,-j-F^cosd,-\-... = R cos f^. ^ 
 or F, X Oa/OA -\-F,x Ob/OA + ... = Ex 01/ OA, 
 or F, X Oa-\-F,X Ob -{-,.. = Ex 01; 
 
 that is, the work done by a system of forces acting at a point 
 is equal to that done by their resultant. 
 
 The equation may be written 
 
 F^xOa^F^XOb + ,..- RxOl = 0', 
 
 which shows that if a system of forces acting at a point equili- 
 brate and the system receives a displacement the algebraic 
 sum of the works done by the forces is equal to zero. 
 
 Conversely, if any member of forces act at a point, the co7i- 
 dition of equilibrium is that the sum of the luorks done by 
 the forces for every disjjlacement shall be equal to zero. 
 
 For let F denote the resultant and p the resolved part of 
 the displacement parallel to the resultant. Then the work 
 done by the forces is equal to Fp. But this by hypotliesis is 
 zero. As p is not zero, F must be zero, or the forces must 
 equilibrate. 
 
 This is called the principle of work as applied to forces act- 
 ing at a point or to forces acting on a particle. 
 
 It is to be noticed that the displacement need not actually 
 be made. It may be conceived to be made, the geometrical 
 relations of the parts remaining the same. The displacement 
 is then said to be virtual [hypothetical], and the work done 
 by the forces virtual work. _J^ 
 
 192. Fquilibrium of a System. — Let a system of particles 
 forming a body be acted on by forces whose points of appli- 
 cation remain at invariable distances from each other. The 
 forces acting on the particles consist of the external forces 
 and of the forces arising from the connection of the particles 
 
232 WORK AND EKERGY. [§ 193 
 
 among themselves. The condition of equilibrium of any 
 particle is (Art. 191) that the sum of the works done by the 
 forces external and internal is for every displacement equal 
 to zero. Hence, summing from particle to particle, the con- 
 dition of equilibrium for the body is that the sum of the 
 works done by the forces external and internal is for every 
 displacement equal to zero. But the work done by the inter- 
 nal forces, being that due to the mutual actions and reactions 
 of the particles, necessarily vanishes for the whole system. 
 Hence it is necessary to consider the external forces only, and 
 the condition of equilibrium is that the sum of the ivorks 
 done hy the external forces is equal to zero for every displace- 
 ment of the lody. 
 
 The same principle will apply to a syste?n of bodies rigidly 
 connected, or so connected that the geometrical relations ex- 
 isting among the parts are not disturbed by a given displace- 
 ment, and hence to what is called a machine. 
 
 Ex. 1. If two particles of a system are joined by an inelas- 
 tic thread, the work of the pull of the thread is 7iil. 
 
 [For let the pull T a.t A be transferred to B. The pulls at 
 ^ being equal and opposite, the sum of the works done is nil.] 
 
 2. If a body turn about a fixed point, the work of the reac- 
 tion is nil. 
 
 [For there is no displacement.] 
 
 3. If a body slide on a smooth plane, the work of the reac- 
 tion is nil. 
 
 [Displacement is at right angles to force.] 
 
 193. Machines. — In a machine the separate pieces are so 
 constrained that when one piece moves every other receives 
 a determinate motion relative to it and to the others. The 
 motion of the pieces being thus constrained, the direction of 
 motion is altogether independent of that of the acting force. 
 Hence the paths of the parts and their relative velocities may 
 be treated as a kinematical problem when we consider the 
 mechanism only. 
 
 A machine is a mechanism to which a driving force is ap- 
 plied. The object of a machine is the overcoming of force 
 
§ 194] MACHINES. 233 
 
 at one place by means of a force applied at another place. 
 The acting force doing work on the machine is called the 
 driving force* and the force overcome is called the resist- 
 ance. The pieces in constraint are termed elements, and can 
 occur only in pairs. No single body can form a machine. 
 Thus a tow-line, though it transmits force, is not a machine. 
 Neither is a bridge, because the parts do not move relatively 
 to one another. 
 
 The condition of equilibrium for forces acting on a 
 machine is that the algebraic sum of the tvorks done hy the 
 external forces is equal to zero for any displacement which 
 does not disturb the geometrical relations of the parts. 
 
 The displacement must be so chosen as to be in conformity 
 with the motion of the parts when the machine is in opera- 
 tion. No work is then done by or against the internal 
 forces. 
 
 If, therefore, F is the driving iorce, R the resistance, x the 
 displacement of i^, and y i\\Q displacement of i?, the condition 
 of equilibrium may be written 
 
 Fx- Ry- 0. 
 
 This principle was first stated by Simon Stevinus of Bruges 
 in investigating the equilibrium of systems of pulleys. His 
 statement was: Ut spafium agentis ad spatium patientiSy sic 
 potentia patientis ad potentiani ageiitis. 
 
 It was also noted by Galileo in the case of the inclined 
 plane, and was given by him as the condition to be satisfied 
 for the equilibrium of a machine. 
 
 194. Machines (Friction Neglected). — In general in a 
 machine the force of friction enters as a resistance; but for 
 simplicity we shall, first of all, neglect friction and consider 
 the parts to be perfectly smooth. This is an ideal case, and 
 may be looked upon as a first approximation to the actual. 
 
 We proceed to find the relation between driving force and 
 
 * Often called "the power." It is best to reserve " power" to de- 
 note the rate of doing work. (Art. 209.) 
 
S34 WOUR AND ENERGt. [§ 195 
 
 resistance in some simple forms of machines, the motion 
 being uniform, that is, the acting forces are in equilibrium. 
 
 It is usually stated that "all machines are either certain 
 simple machines known as the mechanical powers, — the lever, 
 inclined plane, pulley, wedge, screw, — or combinations of two 
 or more of these mechanical powers." But as Prof. Kennedy 
 very Justly remarks in his Mechanics of Machinei^y, " It is 
 not worth while to discuss a theory so hopelessly inconsistent 
 with facts as this."" 
 
 See, for a full discussion, Reuleaux, Theoretische Kine- 
 matik, 
 
 - 195. Inclined Pla7ie. — A body weighing TT rests on a plane 
 inclined to the horizon, and is constrained so as to be capable 
 of sliding parallel to the greatest slope of the plane and of 
 no other motion — in a groove, for example. The mechanism 
 forms an example of what is called a slidi7ig pair. 
 
 {a) Let the body be held from sliding down the plane by a 
 force F acting parallel to the greatest 
 slope. 
 
 The force F may be regarded as the 
 acting force and the vertical force W as 
 the resistance. 
 
 In bringing the body from Aio O 
 
 work done hj F = F X AC; ^ 
 
 work done by W = —Wx OB, J 
 
 Hence, since there is equilibrium, 
 
 FxAC-WxCB = ^, 
 
 or W/F=AC/BC, 
 
 giving the force ratio, or mechanical advantage, or theoretical 
 advantage. 
 
 The same result could be found by considering the con- 
 straint of the plane replaced by the reaction JV and the body 
 
§195] 
 
 INCLINED PLAlTfi. 
 
 235 
 
 in equilibrium under Fy W, JV. 
 from A to C, 
 
 Then, if the displacement is 
 
 work done hy F = F x AC; 
 
 work done by ^ = 0; 
 
 work done by W = — Wx BC; 
 
 and for equilibrium (uniform motion) 
 
 FxAC+0-WxBC=0, 
 
 FxAC=WxBC, as before. 
 
 or 
 
 This equation also shows that tJie work done in moving a 
 locly up a smooth plane is equal to that done in raising it 
 through the vertical height of the plane. 
 
 "^ 
 
 To find the reaction N. 
 
 Let the body be displaced through a 
 distance DE perpendicular to the plane. 
 Then in the displacement 
 
 work done by i\^ = iV' X DE; 
 
 work done by i^ = 0; 
 
 work done by W = — W X EG. 
 
 Hence for equilibrium 
 
 NxDE -WxEG = 0, 
 
 and N/ W = EG/DE = AB/A C, 
 
 ih) If the force i^ which holds PTin equilibrium is parallel 
 to the base, then in moving from A to C 
 W acts through BCy and F acts through 
 AB, each in its line of action. Hence 
 
 FxAB-WxBC=Oy 
 
 or W/F=AB/BCy 
 giving the mechanical advantage. 
 
 Ex. 1. Show in {b) that N/W= AC/AB. 
 
 2. A weight of 100 lb is hauled up an incline of 1 in 100 
 and 1000 ft long. Find the work done. 
 
 Ans. 1000 ft-pounds. 
 
 3. Draw a diagram of the work done in moving a weight of 
 10 lb up a smooth incline 5 ft high. 
 
^6 
 
 WORK AND EKERGY. 
 
 [§196 
 
 c,--"'^ 
 
 J 
 
 Wt 
 
 V 
 
 196. 77^^ Straight Lever. — If a bar is constrained to move 
 about an axis or fulcrum, plane motion only being possible, 
 
 it is a lever, A lever — that is, 
 .B bar and fulcrum — forms an ex- 
 ample of a turning-pair. 
 
 Let the driving force F and 
 the resistance W be both ver- 
 tical. 
 
 While the point of applica- 
 tion of F descends a distance x 
 the resistance W will ascend a certain distance y. Then for 
 equilibrium 
 
 Fx-Wy = 0. 
 But from similar triangles 
 
 x:y = AC'.BG, 
 Hence FxAC=WxBC, 
 
 which is the " principle of the lever/' (Art. 154.) 
 
 Ex. 1. What is the mechanical advantage ? 
 2. If the force F and resistance B are not vertical, show 
 that 
 
 Fx AC=ExBC, 
 
 197. The Wheel and Axle. — Two cylinders fastened to- 
 gether move freely on a common axis C which is horizontal 
 and works in fixed bearings. A force F 
 
 acts by a cord coiled round the larger cyl- 
 inder (or wheel), and. balances a weight W 
 hanging from a cord coiled round the 
 smaller cylinder (or axle). 
 
 The mechanism is equivalent to a lever 
 with unequal arms, the axis' correspond- 
 ing to the fulcrum of the lever, and the 
 radii to the arms. It is called the luheel 
 and axle, and forms a turning-pair. 
 
 While the driving force F descends a distance equal to the 
 
198] 
 
 PULLEY AND CORD. 
 
 237 
 
 circumference of the wheel the weiglit Wis raised a distance 
 equal to the circumference of the axle. Hence for equi- 
 librium 
 
 F X 27rr, -Wx 2;rr, = 0, 
 
 or W/F=rjr^, 
 
 giving the mechanical advantage. 
 
 Ex. 1. Obtain the relation between Fand. Why the "prin- 
 ciple of moments." 
 
 2. Find the vertical force on the axis. 
 
 198. The Pulley and Cord.— A pulley is a wheel with a 
 groove round its outer edge [sheave], and capable of revolving 
 freely about an axis through its 
 center 0. This axis is fixed in a 
 frame or block to which a hook 
 is attached. 
 
 This mechanism consists of three links, the cord, sheave, 
 and frame, the term link being applied to bodies arranged so 
 as to give the required motion. 
 (a) The single fixed pulley. 
 The driving force F and the weight W to be 
 raised are both vertical, the constraint prevent- 
 ing cross-motion. 
 
 If F descend any distance x, W will ascend an 
 equal distance x. Hence for equilibrium 
 
 Fx-Wx = 0, 
 
 and W/F = 1. 
 
 If P is the force on the support and w the weight of the 
 pulley, then, the four parallel forces F, w, W, P being in 
 equilibrium, we have 
 
 P = F-{-W+w 
 = 2W-\-w. 
 
238 
 
 WORK AND ENERGY. 
 
 [§198 
 
 (b) The single movable pulley. 
 
 In the figure we have a single 
 fixed and a single movable pulley. 
 
 If F descends any vertical dis- 
 tance X, W will ascend a distance 
 x/%. Hence for equilibrium 
 
 Fx - Wx/2 = 0, 
 
 or W/F=2. 
 
 If the weight tu of the pulley is 
 taken into account, then evidently 
 
 W-\-w = 2F. 
 
 Ex. Find the pull on the hook C, 
 
 ^ (c) Pulley tackle. 
 
 In the figure is represented a tackle, the upper and lower 
 blocks each containing two sheaves, and the same rope pass- 
 ing round all. 
 
 The motion is not strictly in one plane, 
 but we may assume it to be so as a first 
 approximation. 
 
 While F descends a distance x^ W as- 
 cends a distance x/i. 
 
 /. Fx — Wx/4. = 0, 
 
 and W/F = 4. 
 
 v(^) The Weston differential pulley consists in the upper 
 block of two sheaves of slightly different radii a, b fastened 
 together, and in the lower block of a single sheave of diam- 
 eter a -\- b. Each block is a turning-pair. An endless chain 
 passes round the sheaves as in the figure. Notches are cut 
 or teeth set in the upper-block sheaves which fit the links of 
 the chain. 
 
§198] 
 
 THE DIFFERENTIAL PULLEY. 
 
 239 
 
 While the upper block makes one revolution the lower is 
 raised an amount equal to one half the dilference of the cir- 
 
 cumferences of the two pulleys in the upper block. Hence 
 for equilibrium 
 
 Fx27Tb-Wx(7rb- na) = 0, 
 or W/F ^ 2b/(b - a). 
 
 Ex. 1. In a Weston pulley the diameters of the pulleys in 
 the upper block are 7 and 8 inches. Find the theoretical 
 advantage. A7is. 16. 
 
 ." 2. Obtain the relation between i^and W by the method of 
 moments. 
 
 3. The diameters of the pulleys are 6 and 7 in and the 
 weight is to be raised at the rate of an inch a second. Find 
 the rate at which the chain must be pulled. 
 
 Ans. 70 ft/min. 
 
 Show that to raise the weight 1 ft between seven and eight 
 revolutions must be made. 
 
240 
 
 WORK AND EITERGY. 
 
 [§199 
 
 199. A common form of hoisting apparatus is the worm- 
 wheel gear, shown in the figure. A light end- 
 less chain passes over a pulley A, which has 
 teeth in the groove to fit the links. This is 
 the driving-chain. On the axle of the pulley 
 is a worm which works in the toothed wheel 
 fixed to the pulley B. Over the pulley B 
 passes the movable end of the heavy cliain 
 that runs under the block C. This is the lift- 
 ing-chain. 
 
 To C the weight to be raised is attached. 
 
 Ex. The driving and driven pulleys A and 
 B are each 1 ft in diameter and the number 
 of teeth in the toothed wheel is 24. What 
 weight could be raised by a pull of 10 pounds 
 applied to the endless chain ? A^is. 480 lb. 
 
 \200. TJie Screio. — In turning-pairs the mo- 
 tion is simply a motion of rotation. We may, however, con- 
 ceive the rotating body not only to revolve about the axis of 
 rotation,but to advance along the axis, the motions being in per- 
 pendicular planes. The combination of the two motions gives 
 rise to screw motion, and the two bodies form a twisting-pair. 
 
 Thus suppose iT to be a point on the surface of a cylinder 
 which is revolving in a bearing or nut and at the same time 
 advancing uniformly along its axis. 
 
§ 202] THE SCREW. 241 
 
 By the motion of rotation alone, while the cylinder makes 
 a revolution, H would describe a path equal to the circum- 
 ference. Develop this in the line HE, But the motion of 
 translation carries it a distance equal to EF, Hence it is 
 found at Kj and the path HK would be traced by wrapping 
 the triangle HEF about the cylinder. The path of the point 
 H is called the thread of the screw, and the distance HK or 
 -£'i^ between consecutive threads the pitch. 
 
 The inclination of the thread to the axis is given by the 
 angle EHF { = /3). Now 
 
 tan /3 = EF/HE 
 
 = pitch/circum. of cylinder. 
 
 201. Take the bell-bottom jack-screw (p. 249) with force F 
 applied at the end of a lever-arm I to raise a weight W with 
 uniform speed. 
 
 While the lever-arm makes one revolution, that is, while 
 the force moves through a distance 2;rZ, the weight is raised 
 a distance equal to the pitch p of the screw. Hence for 
 
 equilibrium 
 
 Fx27tl- Wxp = 0, 
 or F/W = p/27rl 
 
 202. In machine tools sliding motion is commonly pro- 
 duced by means of screws. The relation between the pitch, 
 number of revolutions, and rate of sliding is very simple. 
 Thus if p is the pitch in inches, n the number of revolutions 
 per minute, and v the velocity in ft/min, then evidently 
 
 np = 12v. 
 For example, if a screw moves a slide at 16 ft/min, and has 
 a pitch of 1.5 in, the number of revolutions required would be 
 128 per minute. 
 
 ^ Ex. 1. In a pulley tackle the driving force descends 1 ft 
 while the weight to be raised ascends 1 in. What force will 
 raise 1 ton ? Ans. 166f pounds. 
 
 >r 2. Find the relation between F and W in the copying-press 
 (p. 149), 21 being the length of the handle. 
 
 Ans. F= W X pitch of screw/4;r/. 
 
242 
 
 WORK AI^D ENERGY. 
 
 [§202 
 
 '' 3. A weight of 400 lb is being raised by a pair of double 
 pulley-blocks. The rope is fastened to the upper block, and 
 the parts of the rope (whose weight may be disregarded) are 
 considered vertical. Each block weighs 10 lb. Find the 
 pressure on the axle of the upper block. 
 
 Ans. 522.5 pounds. 
 ^ 4. The axle of a capstan is 2 ft in diameter. If four sail- 
 
 ors push with a force of 40 pounds each at the ends of spikes 
 4 ft long, find the weight of the anchor that is lifted. 
 ^ 6. A man has to raise a weight and has only one pulley at 
 his disposal. How must he apply it to gain the greatest ad- 
 vantage ? 
 
 4 6. In a differential pulley the diameters of the pulleys of 
 
 the compound sheave are a, 5 
 in. Find how many revolutions 
 are required to raise the weight 
 c in. A?is. 2c/7r{a — b). 
 
 j^^g gj^iii d'JiMhU ^' I^ ^ telescopic jack-screw a 
 
 I^^Mi^'' ' ififlB i smaller screw C works in a com- 
 
 panion nut cut in the larger 
 screw Df which latter works in 
 a nut in the fixed block B. The 
 block A being fixed, the upper 
 screw does not rotate. If I is 
 the length of the lever-arm, find 
 the relation between F and W. 
 A71S. Fx^ttI = W X diff. of 
 pitches of screws. 
 -^ 8. A screw moves the table of 
 a planing-machine 12.5 ft a min- 
 ute and makes 100 revolutions 
 per minute. Find the pitch. Ans. 1.5 inch. 
 
§ 202] THE SCREW. 243 
 
 •\ 9. Ill a derrick- winch the crank AB is I in leverage, the 
 
 /.*^ ^ y 
 
 gears w to 1, and the drum ^ is c? in diameter. Find the 
 two-maii-power capacity, each man exerting a force of p 
 pounds. Ans. ^jdn/d pounds. 
 
 In general show that for one man exerting a force F a 
 weight W will be raised, given by 
 
 F _ rad. drum no. of teeth in pinion 
 W ~ length arm no. of teeth in wheel ' 
 
 10. In a winch, given that the cranks have 18 in leverage, 
 the gears are 4 to 1, the drum is 6 in in diameter, and the 
 
 ^i 
 
 capacity with two-man power is 3 tons. Find the force exerted 
 by each man. Ans, 125 pounds. 
 
244 
 
 WORK AND ENERGY. 
 
 [§203 
 
 -Vl 1. In a hoisting-machine the gears are 36 .to 36 teeth, the 
 drum 21 in in diameter, and the load for one horse l^ tons. 
 Find the pull exerted by the horse at the end of a 7-ft hori- 
 zontal lever. Aiis. 375 pounds. 
 
 12. In a combination of single- 
 threaded worm and wheel used in 
 hoists the worm-wheel has n teeth, 
 the radius of the driving-wheel is I in 
 in length, and the radius of the drum 
 around which the lifting rope winds 
 is r in. Find the relation between 
 i^and W. Ans, Fhi = Wr. 
 
 203. Machines [Friction Consid- 
 ered). — To overcome frictional re- 
 sistance in a machine work must be 
 done. Part of the driving force is 
 taken up in doing this work, and is, 
 as it were, absorbed. Although it 
 serves no useful purpose, it is not lost. 
 ~^=^ (Art. 216.) 
 
 The work absorbed in overcoming friction is measured as 
 the work done on any other resistance. We have 
 
 work against friction = friction X distance described 
 — )jlN X s, 
 
 where N is the normal pressure, //the coefficient of friction, 
 and s the distance described in the direction of the resistance. 
 
 Thus the work done in overcoming the frictional resistance 
 in one revolution of a journal of diameter d ft and carrying a 
 load of IF lb is i^Wx nd ft-pounds. In case the journal 
 revolves n times per minute the work absorbed is 
 n y^ jx W X nd ft-pounds per minute. 
 
 In a machine, owing to friction between the pieces, part of 
 the work done by the driving force is wasted, so that the re- 
 sulting useful work is less than the total work done by the 
 driving force in the first place. We have, in fact, 
 total work = useful work -|- useless work. 
 
205] 
 
 THE INCLlKEt) PLANE. 
 
 245 
 
 The ratio of the useful work to the total work is known as the 
 efficieiicy of the machine, so that 
 
 efficiency = useful work/total work; 
 
 or, as sometimes expressed, 
 
 efficiency = work got out/work put in. 
 
 204. The Inclined Plane, — To find the force F necessary 
 to haul a weight W up a rough incline ACy the coefficient of 
 friction being // and the force F parallel 
 to the plane. ^ 
 
 Let N denote the normal reaction. 
 The friction jaN acts doivn the plane, /"N^ 
 the forces being as in the figure. 
 
 Then, if the displacement is from ^ to ^ 
 Cy we have for equilibrium 
 
 ♦ w 
 
 FxAC-WxBC-^NxAC=0, 
 Resolving the forces perpendicular to the plane, 
 N-Wgos = 0, . . , . 
 Eliminating N between (1) and (2), 
 
 FX AC =WxBC-{- mW cos Ox AC 
 
 (1) 
 
 (2) 
 
 (3) 
 
 = WxBC-{-mWxAB, . . . 
 
 or tJie work done in hnulijig the body up the plane A C is 
 equal to that done in raising it through the height BC and 
 171 hauling it along the base AB, the coefficient of friction 
 being the same on A C and A B. 
 The work wasted is (Art. 196) 
 
 ^iWxAB, 
 
 205. In case the force F acts doiun the plane, it may be 
 shown similarly that 
 
 FxAC = iAWxAB-WxBCy 
 
 and the work wasted is the same as before. 
 
246 
 
 WORK AKD EKERGY. 
 
 t§206 
 
 ^/yy/yy^^^^^y^y^^. 
 
 It jiiWx AB ==WX BC, then F = 0, and the body would 
 slide down the plane with uniform velocity. 
 
 ^Ex. 1. Find the work done in hauling a sled weighing 500 
 lb half a mile, the coefficient of friction being 0.2. 
 
 A71S. 264,000 ft-pounds. 
 
 2. Find the work done in hauling a train of 100 tons one 
 mile up a 1^ grade, the resistance being 8 pounds per ton. 
 
 Ans. 7392 foot-tons. 
 
 3. "On a grade of 1 in 10 a bicycle-rider, in addition to 
 the tractive force, actually lifts one tenth of his weight and 
 that of the machine." 
 
 206. The Pulley.— {a) The single fixed pulley. 
 Let F be the driving force and W be the weight to be 
 raised. The normal reaction is i^-|- If^, and 
 the friction fJL{F -[-W), if yw is the coefficient 
 of friction. 
 
 Let r — radius of axle, a = radius of sheave. 
 Let a complete revolution of the pulley be 
 made; then, summing the work done, we 
 have for equilibrium 
 
 F X 27ta - WX 27ra - m(^-{- W) X 27rr = 0, 
 
 or F{a — ^r) = W{a + jir), 
 
 and W/F = (a - jar)/ {a + ^r) 
 = 1 — 2//r/rt, nearly, 
 = 1/A, suppose, 
 
 the symbol X being introduced for convenience of writing. 
 
 In pulley tackle so much depends on the stiffness of the 
 ropes that it is of little use to determine the effect of axle- 
 friction only. The formulas in all practical cases are eppir- 
 ical and the investigation outside our province. 
 
 Ex. 1. If the axle were smooth, then F = W. 
 
 2. The effect of the friction is the same as the raising of 
 an additional weight {X — 1) W if the axle were smooth. 
 
 3. Find the tension of the cord supporting the pulley. 
 
 4. Show that the efficiency = 1/A. 
 
206] 
 
 DIFFERENTIAL PULLEY. 
 
 247 
 
 (b) The single movable pulley. 
 
 In a derrick where the rope passes under the movable 
 pulley and over the fixed pulley let F be the ^^ 
 force applied and It' the weight to be raised. 
 
 From the preceding it follows that the 
 pulls in the three parts of the cord are F, 
 F/\, and i^/A', respectively. 
 
 Hence, considering the cords which hold 
 the lower pulley to be parallel, 
 
 W=F/X-{-F/\\ 
 
 and W/F=(X-j-l)/X% 
 
 Ex. 1. If friction is neglected, show that 
 W=2F. 
 2. With two double-sheave blocks 
 
 W=F(\*-l)/(X'-r). 
 
 3. With two double - sheave blocks, neglecting friction, 
 W= 4:F. Hence find the loss due to friction. 
 
 4. In a double-sheave tackle the under block is fixed and 
 the upper movable. Find the pull on the support A and the 
 loss due to friction. 
 
 (c) The differential pulley. 
 
 Let F be the force applied and W the weight which is on 
 the point of being raised (figure, page 239). 
 
 Consider the lower pulley. Let F^ denote the pull of the 
 driving chain and F^ that of the driven. Then from (a) 
 
 XF„ 
 
 But 
 
 F,-\- F^= W. 
 
 (1) 
 (2) 
 
 In the upper block i^and F^ drive and F^ is driven; /'and 
 F^ act on the larger sheaf, and F^ reduced to this sheaf is 
 F^a/b. Then 
 
 F-\- F^a/h = XF,. 
 
 (3) 
 
248 
 
 WORK AND ENERGY. 
 
 [§207 
 
 Eliminating F^ , F^ from these equations, 
 W/F = h{l -^ X)/{X'h - a)y 
 which gives the mechanical advantage. 
 
 When the weight is on the point of slipping down, that is, 
 when W acts as driving force and F as resistance, then, writ- 
 ing 1/A for A, we have 
 
 F/W = {h - X'a)/b\{\ ^X) 
 
 as the condition to be fulfilled. 
 
 If the driving force F is removed, then whatever be the 
 weight we must have, in order that no back action may take 
 place, 
 
 b-Va = 0, 
 
 which gives the relation between the radii, that this may be 
 possible. If X = 1.1 then h/a = 1.2, and the diameters of 
 the pulleys are roughly as 5 to 6. For these dimensions 
 the chain will not slip, whatever weight is being raised. The 
 practical value of the pulley lies largely in this circumstance. 
 207. Hie Screiv. — Consider a screw-jack with the driving 
 force F acting at the end of a lever 
 I and a weight W on the point of 
 being raised, the coefficient of fric- 
 tion being //. 
 
 The screw unwrapped is an in- 
 clined plane with the force parallel 
 to the base (Art. 200). 
 Let a complete revolution of the screw be made; then, sum- 
 ming the works by the external forces, we have for equilibrium 
 
 \ 
 
 ( 
 
 ' 
 
 J^ 
 
 Fl/r 
 
 
 A 
 
 
 3 
 
 Fl 
 
 X AB - W X BC - M^ X AC = 0. 
 
 (1) 
 
 Resolving the forces perpendicular to A C, 
 
 FI 
 JV- }Vcos /3 -— sm/3 = 0; 
 
 (2) 
 
§ 208] THE SCREW. 249 
 
 /. by substituting for iV its value from (2) in (1), 
 
 / Til \ 
 
 = PT tan /? + yw sec J3\W cos /3 -{ sin /3j, 
 
 or Fl = Wr tsin (/3 -\- <p), (3) 
 
 the result required. 
 
 The useful work is W X AB tan /3, and the total work is 
 WX ABt&.n{/3-^(p). 
 
 ,\ efficiency = tan /3/tan (/3 -\- <p). 
 
 208. The relation between F and W may also be found 
 graphically. 
 
 The resultant 7? of ^ and ;aJV makes an angle with the 
 normal. Hence lay off W to scale, complete the triangle of 
 forces, and scale off Fl/r and R, 
 
 From the triangle we have at once 
 
 Fl/r= Wta.n {ft f (p). 
 
250 WORK AKD ENERGY. [§ 209 
 
 If the weight W is on the point of moving dotvn and the 
 acting force is denoted by F^ , then it may be shown in a 
 similar manner that 
 
 FJ/r = W tan (/3 - (P). 
 
 Hence if the force applied has any value between F and F^ 
 the screw will be in equilibrium when supporting a weight W. 
 
 Ex. 1. It /3 = (pwe have F^ = 0. What does this mean ? 
 
 2. The circumference of a screw is 4 in, there are 2 threads 
 to the inch, and the coefficient of friction is 0.2. Find the 
 force applied at the end of a lever 14 in long that will raise a 
 weight of 100 lb. Ans. 1.5 pounds. 
 
 What force would just support the weight ? 
 
 Ans. 1/3 pound. 
 
 3. It /3 = 0, the efficiency is null. Explain. 
 
 4. In what other case is the efficiency null ? 
 
 5. Show that the efficiency of a screw is greatest when the 
 pitch angle is 45°, nearly. 
 
 6. In a screw the pitch angle is 45° and the coefficient of 
 friction 0.16. Find the efficiency. Ans. 0.72. 
 
 7. If tan /3 = 0.1, tan = 0.01, prove efficiency = 0.9; 
 
 tan /3 = 0.1, tan = 0.1, " " = 0.5; 
 
 tan /3 = 0.1, tan = 0.2, " " = 0.3. 
 
 Hence show the importance of lubrication. 
 
 8. The efficiency of a rectangular screw is a maximum 
 when 
 
 ^ = (^ - 20)/4. 
 
 il209. Power. Activity. — It is plain from the definition of 
 work that any small force can do work of any magnitude pro- 
 vided sufficient time is given. Hence, in order to compare 
 agents which do work, it is necessary to take the time employed 
 into consideration. To indicate the rate of doing work or the 
 amount of work performed in a given interval the term'power 
 or activity is used. Thus to raise one ton of coal through 
 300 ft in 10 min would require an expenditure of 2000 x 300 
 ft-pounds of work in 10 min, or of 60,000 ft-pounds per min, 
 or of 1,000 ft-pounds per sec, either one of which would be 
 an expression for the power of the agent. 
 
§ 209] POWER. 251 
 
 In general, if a resistance of R pounds is overcome at a 
 velocity of v ft/sec, then 
 
 Rv ft-pounds/sec 
 
 will measure the power of the agent. 
 
 By the ^^n^t power or unit rate of working is meant the 
 power of an agent which can do unit work in unit time. Tlie 
 gravitation unit is the foot-pound per second or the foot- 
 pound per minute. 
 
 In engineering work the foot-pound per minute being 
 too small a unit for most purposes, the multiple unit of the 
 horse-power has been introduced. A horse-power is defined 
 as the power of an agent which can do a work of 33,000 foot- 
 pounds per minute or of 550 foot-pounds per second. It does 
 not give, the average power of a horse; but as a unit of this 
 size is convenient, the name introduced by Savery and defined 
 by James Watt has been retained. Carefully notice that the 
 term horse-power does not express an amount of work but the 
 rate of doing work. 
 
 Conversely, 33,000 X 60 (= 1,980,000) ft-pounds is called a 
 horse-power-hour, a term frequently employed in engineering. 
 The term horse-power-hour is thus a unit of work and not a 
 unit of power. 
 
 Watt derived his horse-power from actual experiments 
 carried out at Barclay and Perkins' brewery, London, with 
 certain of the very heavy horses belonging to tlie firm. These 
 horses were caused to raise a weight from the bottom of a 
 deep well by pulling horizontally on a rope passing over a 
 pulley. He found that a horse could walk 2.5 miles an hour 
 and at the same time raise a weight of 100 lb. This is equiva- 
 lent to 2.5 X 5280 X 100/60 = 22,000 foot-pounds per min- 
 ute. By adding 50 per cent to this he obtained 33,000. The 
 margin, 11,000 foot-pounds, he allowed for loss in engine- 
 friction, etc. 
 
 Gen. Morin's estimate of the power of a horse is 26,150 
 ft-pounds per minute. 
 
252 WORK AND ENERGY. [§ 210 
 
 210. For determining tlie power developed by a steam- 
 engine or other machine, some form of friction-brake is used. 
 The idea is to balance the work done by the machine by a 
 frictional resistance, compute this resistance, and thence find 
 the power of the machine. The brake absorbs the work to 
 be measured. 
 
 Let be a shaft of radius r, to which the brake ^^ is 
 fastened. By means of the screws a, b the friction of the 
 
 4] cr 
 
 brake on the shaft may be regulated. Suppose it adjusted so 
 that the engine develops a friction/, just sufficient to balance 
 a body of weight W placed at the end A of the beam. Then 
 the moments of/ and ^ about must be equal, or 
 
 fr = Wl 
 
 Suppose the shaft to revolve uniformly n times per min- 
 ute. Then, assuming that the friction for uniform motion 
 of the shaft is the same as at the point of just beginning to 
 move, we have 
 
 Work done in one min = friction developed in n revs. 
 
 = 27rnWl. 
 If Wis expressed in pounds and I in feet, then the 
 
 H.P. = 27r?nn/33,000 
 = 0.000l9}iWL 
 
§ 210] EFFICIENCY. 253 
 
 Since the total work done by a machine is given by the 
 indicated horse-power (Art. 189) and the useful work by 
 the braked horse-power, we may define the efficiency of the 
 machine to be the ratio of the B.H.P. to the I.H.P., or 
 
 eflaciency = B.H.P./I.H.P. 
 
 7 Ex. 1. Show that one H.P. = 396,000 inch-pounds/minute 
 
 = 198 inch-tons/minute. 
 ^ 2. 22 tons of coal are to be hoisted through 50 yards in 10 
 min. Find the H.P. of engine necessary. Ans, 20 HP. 
 J 3. A traction-engine weighing 5 tons hauls a load of 10 
 tons at 8 miles an hour, the resistance being 20 pounds per 
 ton. Find the H.P. exerted. Ans, 6.4 H.P. 
 
 4 4. A belt passing round two pulleys moves with a velocity 
 of 10 ft/sec. Find the H.P. transmitted if the difference of 
 tension of the belt above and below the pulleys is 1100 
 pounds. A71S. 20 H.P. 
 
 J 5. Find the H.P. required to propel a train weighing W 
 tons at V miles an hour on a level track, the resistance being 
 ;; pounds per ton. Avs. WpV/Zlb, 
 
 6. Find the H.P. required by a vessel to overcome a resist- 
 ance of R pounds at a speed of V knots. 
 
 Ans. i?F/325.66 horse-power. 
 Hence show that one horse-power, or 550 ft-pounds per 
 second, is about 326 knot-pounds. 
 
 7. A shaft 14 ft in diameter is sunk in gravel to a depth of 
 10 fathoms in 10 days of 10 hours each. Taking the weight 
 of the gravel at 100 lb/ft', find the H.P. expended in lifting 
 out the gravel. Ans. 0.14 H.P. 
 
 8. " At 80 miles an hour a pull of 4 pounds 11 ounces rep- 
 resents one horse-power." 
 
 9. Find the horse-power necessary to pump out the St. 
 MJiry's Falls Canal Lock, Sault Ste. Marie, in 24 hours, the 
 length of the lock '->eing 500 ft, width 80 ft, and depth of 
 water 18 ft, the water being delivered at a height of 42 ft 
 above the bottom of the lock. Ans. 31.2 H.P. 
 
 X 10. A belt can stand a pull of 100 pounds only. Find the 
 least speed at which it can be driven to transmit 20 H.P. 
 
 Ans. 110 ft/sec. 
 11, How many gallons of water would be raised per 
 
254 WORK AND ENERGY. [§ 210 
 
 minute from a mine 600 ft deep by an engine of 175 
 H.P., supposing a gallon of water to weigh 8^ lb.? 
 
 Ans. 1155 gals. 
 
 12. The average flow over Niagara Falls is 270,000 cubic 
 feet per second. The height of fall is 161 feet. Show that 
 the H.P. developed is in round numbers 5 million. 
 
 13. The U. S. war-ship Columbia has a speed of 23 knots 
 with an indicated horse-power of 22,000. Find the resistance 
 offered by the water to her passage. Atis. 156 tons. 
 
 14. In testing a Corliss engine running at 100 revolutions 
 per minute, the lever- arm of the brake employed was 10^ ft 
 and the weight attached 2000 lb. Find the H.P. developed. 
 
 Jns. 400 H.P. 
 
 15. "A C. and C. electric motor shows on a Prony brake a 
 pull of 5 ounces on a 1-ft lever, that is, 2 ft-pounds per 
 revolution, or about ^ H.P. at 1500 revolutions per minute." 
 Check the conclusions in this statement. 
 
 16. In a Corliss engine running at 100 pounds pressure per 
 sq in and 100 revolutions per minute, the diameter of the 
 cylinder is 18 in and length of stroke 42 in. If the brake 
 was used on a pulley 6 ft in diameter and keyed to the engine- 
 shaft, find the friction on the face of the pulley. 
 
 A71S. f = 9450 pounds. 
 
 17. A 6-ton fly-wheel on a 14-in axle makes 90 revolutions 
 per minute. Find the H. P. absorbed in friction, the coefifi- 
 cient of friction being 0.1. A?is. 12 H.P. 
 
 18. A steam-hoist of 3 H.P. is found to raise a weight of 
 10 tons to a height of 50 ft in 20 min. How many ft-pounds 
 of work are w^asted by friction in a day of 10 hours ? 
 
 Ans. 29,400,000 foot-pounds. 
 
 19. A pumping-engine of piston area 100 in% steam-press- 
 ure 60 pounds/in^, length of stroke 3 ft, and number of revo- 
 lutions per min 25, raises 500 gallons of water per minute a 
 height of 50 ft. Find the efficiency. A7is. 0.23, nearly. 
 
 20. A train weighing 100 tons runs at 42 miles an hour on 
 a level track, the resistance being 8 pounds per ton. Find 
 its speed up a Ifo grade (1 ft rise in 100 ft) if the engine- 
 power is unchanged. Ans. 12 miles/hour. 
 
 21. A traction-engine weighing 5 tons can haul 15 tons on 
 a level, the coefficient of friction being 0.02. Find the net 
 load it can haul up a 1^ grade. A7is. 8^ tons. 
 
 22. A train weighs W tons and the resistance is p pounds 
 
§210] 
 
 POWER. 
 
 255 
 
 per ton. Find the horse-power necessary to propel it at a 
 speed of V miles an hour up a grade of w^. 
 
 A71S. H'V{p -f- 20w)/375 horse-power. 
 
 23. Find the speed of a driving-pulley 3.5 ft iu diameter to 
 transmit 6 H.P., the driving 
 force of the belt being 150 
 pounds. Ans. 120 revolu- 
 tions/minute. 
 
 24a. The horse-power 
 transmitted by an endless 
 belt passing over two pul- 
 leys is 
 
 7tdn{P - 0/33000 
 
 when d = diameter of driving-pulley in ft, 
 
 n = no. of revolutions per minute of driving-pulley, 
 F, Q = the pulls in the belt on the taut and slack sides 
 expressed in pounds. 
 > 246. If T is the torque of the driver, then 
 
 H.P. = 7Vt/5250. 
 
 > 25. A train of 100 tons is hauled by an engine of 150 H.P. 
 The resistance is 14 pounds per ton. Find the greatest 
 velocity that the engine can attain. A?is. 40 miles/hour. 
 
 26. Check this statement: "55 pounds mean effective 
 pressure at 600 ft piston speed gives one H.P. for each sq in 
 of piston area." 
 
 > 27. Prove H.P. of an engine = SNAP/^dOOO where 
 
 S = length of stroke in ft; 
 iV= no. of strokes per minute; 
 A = piston area in square inches; 
 
 /' = mean steam-pressure in pounds per square inch of 
 piston area. 
 
 \28. Find the work done per hour at the crank-pin of an 
 engine revolving 40 times a minute and acting against a re- 
 sistance of 7000 pounds, the radius of the crank being 18 
 inches. Ans. 79,200 ft-tons. 
 
 29. Show that 
 
 tractive pull of a locomotive (in pounds) = d^s/D 
 
256 WORK AND ENERGY. [§ 211 
 
 where d = diameter of cylinders in inches; 
 
 p = mean effective pressure in pounds per sq. inch ; 
 s = stroke in inches; 
 
 D = diameter of driving-wheels in inches, 
 30. Show that the cylinder diameter of an engine that will 
 produce n horse-power at a piston velocity of s ft per minute 
 under a mean effective pressure of p pounds per sq in is 
 205 V?i/ps inches nearly, 
 
 211. Energy. — We have seen that when the forces acting 
 on a body are in equilibrium or the body moves with a uni- 
 form velocity, the sum of the works done is zero; that is, the 
 work done by the resultant driving force is equal to that done 
 on the resistance. Now the action of a force is to cause ac- 
 celeration. If the motion is uniform, the acceleration caused 
 by the driving force is balanced by the equal and opposite 
 acceleration caused by the resistance. But if the acceleration 
 caused by the driving force exceeds that caused by the re- 
 sistance, velocity is gained, and the motion is not uniform. 
 
 Thus let a force F pounds act on a body weighing W lb 
 through a distance s ft and let v ft/sec be the velocity ac- 
 quired. Then (Art. 183) 
 
 Fs = Wvy2g. 
 
 Now Fs is the work done by F in passing over a distance 
 s in its line of action, and therefore a body W in acquiring a 
 velocity v from rest must have Wv^/2g units of work done 
 upon it. 
 
 Conversely, the force i^ which will generate a velocity v in 
 acting through a distance s, will destroy the same velocity if 
 acting through the same distance in the opposite direction, or 
 the body by virtue of its velocity v can do Fs or Wv''/2g units 
 of work in giving up that velocity and coming to rest. This 
 capacity which a body possesses of doing work in consequence 
 of its velocity is known as energy. Hence the measure of 
 the energy of a body which weighs W and has a velocity v is 
 Wv^/2ff units of work. We may therefore state the equation 
 Fs = Wvy2g, 
 
§ 213] ENERGY. 257 
 
 If a body is in motion under the action of force, the work 
 do?ie in passing from one position to another is equal to the 
 energy produced. 
 
 212. More generjilly, if a force F acting on a body which 
 weighs W is opposed by a uniform resistance R in the same 
 line of action, then the net driving force is F — R. Let n 
 be the acceleration produced; then, from Newton's second 
 law, 
 
 F- R= Wa/g. "' 
 
 If s is the distance passed over and the velocity is changed 
 from u to V in passing over this distance, then 
 
 as = v'/'Z - wV2. 
 
 Eliminating a, 
 
 Fs = Rs+ Wvy2g - Wuy2g; 
 
 which may be stated : 
 
 T7ie work done by a force on a body {or system of bodies 
 with configuration remaining the same) is equal to the work 
 done against the resistance together with the change of energy 
 generated in the body (or system, of bodies). 
 
 213. The possession of energy as defined implies motion. 
 But the motion may not appeal directly to our senses. A 
 body may possess energy by virtue of its position. For ex- 
 ample, let a body be thrown upward. The body possesses 
 energy as shown by its capacity to overcome obstacles. But 
 its visible energy is continually diminishing, and becomes zero 
 when the highest point is reached. In its fall it acquires 
 visible energy until at the starting-point this energy is equal 
 to the initial energy. 
 
 In rising the energy was expended in overcoming the re- 
 sistance offered by gravity. The force of gravity is prepared 
 to restore in the return to the initial position the energy ab- 
 stracted. The body and the earth thus form a system of give 
 and take — a connected system in which the sum total of the 
 
258 WORK AKD EKERGY. [§ 214 
 
 energy remains the same. But, instead of saying that as one 
 member of a system loses energy some other member of that 
 system gains energy, it is convenient to confine our attention 
 to one member of the system only, and say that as it loses 
 energy of motion (kinetic energy) it stores or gains energy by 
 virtue of its position ('potential energy); or, as it is sometimes 
 expressed, by virtue of the configuration of the system. 
 
 The kinetic energy of a body or the energy which it pos- 
 sesses in virtue of its motion is measured by the work it can 
 do against resistance in parting with this motion. Thus the 
 kinetic energy of a body weighing W lb and moving with a 
 velocity v ft/sec is Wv^/2g foot-pounds of work. 
 
 The potential energy of a body, or the energy it possesses 
 by virtue of its configuration, is measured by the work it can 
 do in passing from its present configuration to some standard 
 configuration. Thus a weight W lb, if raised to a height h ft, 
 would possess a potential energy of Wli foot-pounds of work, 
 because it could do that amount of work in falling to its 
 original position — in driving a clock, for example. 
 
 The term potential energy was introduced by Prof. Mac- 
 quorn Eankine, and the term kinetic energy by Lord Kelvin. 
 
 214. Unit of Energy. — The energy expended being equal 
 to the work done, the unit of energy must be the same as the 
 unit of work. In the British system the unit is therefore 
 the foot-pound. 
 
 Note that unit kinetic energy is not the energy of unit 
 weight [1 lb] moving with unit velocity [1 ft/sec], but only 
 one half that quantity. 
 
 The expression Wv'^/g, introduced by Leibnitz (1646-1716) 
 under the name vis viva, or living force, is still made use of 
 by some writers. Its value is equal to twice the kinetic en- 
 ergy. In view of the prominence of the doctrine of energy 
 in modern physics, the term kinetic energy is to be preferred, 
 and vis viva, being quite superfluous, should be dropped. 
 The Leibnitz term vis mortua, as the name for pressure, is 
 long dead. 
 
§ 215] ENERGY. 259 
 
 215. For illustration, consider the relation between the 
 kinetic and potential energies in the following simple ex- 
 amples : 
 
 -> (1) A body weighing W lb falls from A to B through a 
 height h ft. 
 
 The p.e. at ^ = work in falling from ^ to ^ 
 = Wh foot-pounds. 
 
 The k.e. at ^ = 0. 
 
 .'. total energy at A = Wh foot-pounds. 
 
 Let C be any point in AB and let ^C = 5 ft. When the 
 weight W has fallen a distance .v it has acquired a velocity v, 
 so that v' = 2gs (Art. 91). Then 
 
 k.e. at6' = >fV/2^ = ff*; 
 p.e. at C = W{h - s). 
 
 ,', total energy at C = Ws -f- W(7i — s) 
 = Wh foot-pounds 
 = total energy at A. 
 
 But C is any point, and therefore the sum of the kinetic 
 and potential energies of the body is constant throughout 
 the fall. 
 
 (2) Consider a body sliding down a smooth inclined plane 
 of height h and length I under the action of gravity. 
 
 This is very similar to the preceding, and its development 
 is left to the student. 
 
 -i (3) A body sliding down a rough inclined plane AB under 
 the action of gravity. 
 
 With the same notation as before, /^ being the coefficient of 
 friction, 
 
 k.e. at ^ = 0; 
 
 p.e. at A = Wh 
 
 = Wl sin e. 
 
 .'. total energy at A = Wl sin 6 ft-pounds. 
 
 The acceleration down the plane = ^(sin 6 — fj. cos d) ; 
 
 velocity at B ^ 1^2^(8111 — j^ cos 0)1; 
 
260 WORK AND ENERGY. [§ 216 
 
 k.e. at 5 =i Wx 2^ (sin 6— pi cos 6)1/ g 
 = Wl{sin 6 — pi cos 6) 
 
 p.e. at ^ = 0; 
 
 .*. total energy at B = Wl(sm 8 — pi cos 6) foot-pounds. 
 
 216. In the last case, instead of the energies at A and B 
 being equal as in (1) and (2), there is a loss piWl cos ^ in 
 passing from A to B. It would thus appear that when en- 
 ergy passes from one form to another it is not always capable 
 of being changed into its original form — at least with the 
 same degree of readiness. The loss pi Wl cos 6 is explained 
 by saying that this energy has been converted into other 
 forms of energy, principally the molecular energy called heat. 
 Experiment shows that heat may be expressed as a definite 
 number of foot-pounds of work. Thus Joule found that 
 the amount of heat necessary to raise one pound of water 
 from 0° to 1° F. is capable of doing 778 foot-pounds of 
 work. 
 
 217. These examples are simple illustrations of a general 
 principle, known as the conservation of energy, which is 
 stated by Maxwell as follows : 
 
 The energy of a system is a quantity which can neither be 
 increased nor diminished hy any actions between the parts 
 of the system, though it may he transformed into any of the 
 forms of which energy is susceptible ; in a word. 
 
 The energy of a closed system is a constant quantity, or 
 
 Energy is indestructible. 
 
 218. The principle of the conservation of energy is the 
 greatest of all physical laws. It is founded upon experi- 
 mental evidence most extensive, so that no doubt of its appli- 
 cability to all forces in nature is now entertained. It includes 
 Newton's laws of motion and the principle of work, and hence 
 upon it the science of dynamics may be based. It cannot, 
 however, be deduced from Newton's laws of motion, as it is 
 more comprehensive than they. 
 
 It is perhaps more in' accordance with the spirit of modem 
 
§ 218] ENERGY. 261 
 
 science to consider space, time, matter, and energy as the fun- 
 damental concepts of Mechanics rather than space, time, mat- 
 ter, and force (Art. 2). The further principle necessary on 
 which to base the science would in the former case be that of 
 the conservation of energy, while with the latter Newton's 
 laws of motion are commonly employed. The hitter method, 
 which is that followed in this book, is given as being more 
 readily appreciated by the beginner. But as it involves diffi- 
 culties from which the first is free, it seems likely tliat the 
 notion of energy will in time be adopted universally as the 
 fundamental one rather than that of force. 
 
 J Ex. \a. Find the greatest velocity v attained by a car 
 weighing 20 tons if hauled along a level, straight track by a 
 pull of 1000 pounds through a distance of half a mile and 
 against a constant resistance of 15 pounds per ton. 
 
 Ans. 54.4 ft/sec. 
 -f- 1&. Show that the car will come to rest after traveling 1| 
 miles and in 3.77 min if the pull ceases at jthe end of half a 
 mile. 
 
 - 2. A weight of 25 lb has fallen through 25 ft. Find the 
 rate at which work is being done. Ans. 0.9 horse-power, 
 -f 3. A cannon when fired recoils with a velocity of 10 ft/sec 
 and runs up a platform having an incline of 1 in 4. Find the 
 distance it goes before coming to rest. Ans. 6 ft 3 in. 
 
 ^ 4. A body weighing G4 lb and moving east with a velocity 
 of 3 ft/sec receives a blow such that the velocity due to it is 4 
 ft/sec north. Find the resultant kinetic energy. 
 
 Ans. 25 foot-pounds. 
 [This resultant = the sum of the separate kinetic energies.] 
 -f-5. Find the work done in stopping a 100-lb shot moving 
 with a velocity of 1000 ft/sec. 
 
 Ans. 1,562,500 foot-pounds. 
 -^6. Find the force exerted in stopping a train of 250 tons in 
 1000 ft from a velocity of 30 miles an hour. 
 
 Ans. 15,125 pounds. 
 ^ 7. A shot pierces a target of a certain thickness. Show 
 that to pierce one of 4 times the thickness twice the velocity 
 is necessary. 
 
 ^8. What is the H.P. of an engine that will deliver 10,000 
 gallons of water per minute with a velocity of 10 ft/sec, if 10^ 
 is wasted by leakage? Ans, 4.4 H.P. 
 
262 WORK AND ENERGY. [§ 218 
 
 9. A blacksmith's helper using a 16-lb sledge strikes 20 
 times a minute and with a velocity of 30 ft/sec. Find his 
 rate of work. Ans. 3/22 H.P. 
 
 4 10. A stone is thrown with a horizontal velocity of 50 ft/sec. 
 Find the velocity with which it strikes the ground, which is 
 horizontal and G ft below the point of projection. 
 
 A71S. 53.7 ft/sec. 
 
 11. Show that to give a train a velocity of 20 miles an hour 
 requires tlie same energy as to lift it vertically through a 
 height of 13.4 feet. 
 
 12. A hoisting-engine lifts an elevator weighing 1 ton 
 through 50 ft when it attains a velocity of 4 ft/sec. If the 
 steam is shut off, how much higher will it rise ? Ans. 3 in. 
 
 13. In (12) find the time of rising 50 ft, supposing the 
 motion uniformly accelerated, and also find the H.P. of the 
 engine. Ans. 25 sec; 8.2 H.P. 
 
 ( 14. Show that the energy stored in a train of weight W lb 
 and moving with a velocity of V miles per hour is 1FFV30 
 foot-pounds. 
 
 I 15. A train of 100 tons is running at 30 miles an hour up 
 a 2% grade. Find the H.P. required, the resistance on a level 
 being 10 pounds per ton, due to axle-friction chiefly. 
 
 IQa. In a locomotive running on the level at 30 miles an 
 hour the tractive force is 8 tons. Taking the resistance of 
 friction as 10 pounds per ton, find the number of 20-ton cars 
 that can be hauled if engine and tender weigh 100 tons. 
 
 Ajis. 75 cars. 
 166. Find the number that would be hauled up a 2fo grade. 
 
 Ans. 11 cars. 
 16c. Find the H.P. exerted in the former case. 
 
 A71S. 1280 H.P. 
 ( 17«. An engine exerts on a car weighing 20,000 lb a net 
 pull of 2 pounds per ton. Find the energy stored in the car 
 after going 2^ miles. A?is. 264,000 foot-pounds. 
 
 A 17b. If shunted to a level side track when the frictional 
 resistance is 10 pounds per ton, find how far it will run before 
 coming to rest? Ans. 1/2 mile. 
 
 v 17c. If shunted on a side track with a 1^ grade, how far 
 will it run before coming to rept ? A?is. 1/6 mile. 
 
 yl7d. If there are brakes on half the wheels, and these are 
 applied with a pressure of half a wheel-load, how far will the 
 car run up a Ifo grade, the coefficient of friction between 
 wheel and brake-shoe being 0.2 ? A^is. 203 ft, nearly. 
 
§ 219] ' STABILITY. 263 
 
 18. In the Westinghoiise brake tests (Jan. 1887) at Wee- 
 hawken a passenger-train moving 22 miles an hour on a down 
 grade of 1^ was stopped in 91 ft. There was 94^ of the train 
 braked. Taking the frictional resistance as 8 pounds per ton, 
 find the net brake resistance per ton and the grade to which 
 this is equivalent. Afis. 367 pounds; 19.5 per cent. 
 
 19. The tractive force of an engine is F tons. If the 
 weight of engine and train is W tons and the frictional resist- 
 ance 71 pounds per ton, show that in going up an a^ gi'ade the 
 velocity acquired in t seconds from rest will be Qgt ft/sec, 
 and the energy 0.5 WQ'gt'' ft-tons, where 
 
 Q = P/W- a/100 - w/2000. 
 
 219. Stability of Equilibrium.— Conceive a body in equi- 
 librium to be slightly displaced: the forces acting will tend 
 either to restore the body to its original position or to move 
 it farther from that position, or will themselves equilibrate. 
 In the first case the original position is said to be one of 
 stable, in the second of unstable, and in the third of neutral, 
 equilibrium so far as this displacement is concerned. In the 
 special case when the forces continue to act in parallel direc- 
 tions, act at the same points, and remain of the same magni- 
 tude, the condition of neutral equilibrium is called astntia. 
 Thus a body supported at its center of gravity is in astatic 
 equilibrium. 
 
 We shall consider the force of gravity to be the only exter- 
 nal force acting. When displacement occurs and the body 
 tends to return, the force of gravity does work in this return. 
 Hence the potential energy is greater than in the position of 
 equilibrium, or the position of equilibrium is one of mini- 
 mum potential energy. But gravity alono acting, the poten- 
 tial energy depends on the height x of the center of gravity 
 above a fixed horizontal plane. Hence in a position of stable 
 equilibrium the center of gravity is in its lowest position. 
 
 Similarly, in a position of unstable equilibrium, the center 
 of gravity is in its highest position, and in neutral equilibrium 
 it remains in the same position after displacement. 
 
 The problem thus reduces itself to a determination of 
 
264 WORK AKD ENERGY. [§ 219 
 
 whether the Tieiglit of the C.G. above a fixed horizontal plane 
 is a minimum or a maximum, or is stationary. 
 
 The " Moving Stone '^ at Lexington is one of the most re- 
 markable freaks of nature in the State of Kentucky. In the 
 rear of the grounds attached to the home of the late Gov. 
 Gilmer is a huge boulder, standing alone on the edge of 
 a stream. Resting directly upon this boulder is another 
 weighing at least twenty tons. This upper boulder rests 
 upon a stone pinnacle not more than two fecu square, and so 
 evenly balanced that (although the slightest touch will cause 
 it to rock to and fro) a hundred horses could not pull it from 
 the socket. {St. Louis Republic.) 
 
 Ex. 1. A spoon rests in a hemispherical cup: to determine 
 the nature of the equilibrium. 
 
 Let r — radius of cup, 2Z = length 
 of spoon. 
 
 Let the C.G. of the spoon be at 
 its middle point G. The three 
 forces acting, the weight and the 
 reactions at A and 2) cut in a 
 point E. 
 
 Let X denote the de2Jth of G below OD, Then 
 
 X = r ^m 26 — I sin 0, 
 .*. dx/dd = 2r cos 26 — I cos 6. 
 
 Put dx/d6 = 0, and we have for the position of equi- 
 librium 
 
 cos 6 = (l-{-i^r-{- 32r^)/8r, .... (1) 
 
 the + sign of the radical being taken because 6 is less than a 
 right angle necessarily. 
 Also, in this position 
 
 d'x/d6' = - 4r sin 2^ + ? sin 6, 
 
 which is + or — according as 
 
 4r sin 26 < > I sin 6, 
 
 or cos ^ < > l/8r. 
 
§ 219] EXAMINATION. 265 
 
 But cos 6 > Z/8r from eq. (1). Hence (Vx/dd^ is negative, 
 or a: is a maximum and the equilibrium is stable. 
 
 2. A yard-stick rests over the edge of a cylindrical jar 2\ 
 in diameter with one end against the inner surface. Show 
 that in the position of equilibrium the stick is inclined at 60° 
 to the horizontal and that the equilibrium is unstable. 
 
 Show also that in order that the jar may not tumble over, 
 its weight must be at least six times that of the stick. 
 
 3. A hemisphere rests on a horizontal table. Show that 
 the equilibrium is stable. 
 
 4. A hemisphere, radius r, rests with its flat surface on the 
 top of a sphere, radius R. Show that the equilibrium is 
 stable or unstable as R > < 3r/8. 
 
 5. A cylindrical block of radius r with a hemispherical end 
 rests on a horizontal table. Find the height of the cylinder 
 for neutral equilibrium. Ans, r/V2. 
 
 6. A hemisphere, radius r, rests in neutral equilibrium with 
 its curved surface on the top of a fixed sphere, radius R. 
 Show that 
 
 5r = 3i?. 
 
 EXAMINATIOIS^. 
 
 1. "An unbalanced force always does work.^* 
 
 2. How is the work done by the force of gravity on a fall- 
 ing body measured ? 
 
 3. How is the work done against friction measured ? 
 
 4. Give examples of work done against resistances. 
 
 5. Define the terms foot-pound, foot-ton, and inch-pound. 
 
 6. Find the work done against gravity in going upstairs. 
 
 7. Equal forces act for the same time upon unequal weights 
 w^, w^. Find the relation between 
 
 (1) the momenta generated by the forces; 
 
 (2) the works done by the forces. 
 
 8. Define a lever, and find the conditions of equilibrium on 
 a straight lever. 
 
 9. " The reason why a force acting at a greater distance 
 from the fulcrum moves a weight more easily is because it 
 describes a larger circle." (Aristotle.) 
 
S66 Work akd energy. t§ ^^^ 
 
 10. Put a 54b weight into one of the pans of an ordinary 
 balance and drop a 4-lb weight into the other pan. The 
 beam will be tilted, and if the 4-lb weight is quickly removed 
 it may lead an onlooker to think it really weighs more than 
 the other. 
 
 11. In catching a flying ball the player lets his hand be 
 carried in the direction of motion of the ball. Why ? 
 
 12. Find the relation between the work done in dragging a 
 body up an inclined plane or in lifting it through the height 
 of the plane when {a) smooth, (b) rough. 
 
 Also find the condition that a body may slide down a rough 
 incline with uniform speed. 
 
 13. Find the condition of equilibrium in a system of 
 pulleys in which the same cord goes round all the sheaves. 
 
 14. The work of raising a body in sections is 
 
 Wh foot-pounds, 
 
 where W is the total weight in lb and h is the height in ft 
 through which the O.G. has been raised. 
 
 15. In a screw-jack the pitch of the screw is 1 inch, the 
 lever is 2 ft long, and the force applied at the end of the lever 
 25 pounds. Find the weight that can be lifted, friction being 
 neglected. 
 
 16. What is meant by backlash in screw gears ? 
 [Slack between screw and nut.] 
 
 17. Find the forces that will (1) just support, (2) just move 
 a rough screw when it carries a given weight. 
 
 18. The mechanical advantage of a smooth screw = (cir- 
 cumference described by acting force) /(pitch of screw). 
 
 19. Define the efficiency of a machine. In all cases the 
 efficiency must be a proper fraction. 
 
 20. In a certain machine the work that should be obtained 
 theoretically is 1600 ft-tons, but the amount actually obtained 
 is 1200 ft-tons. Find the efficiency. A^is. 75^. 
 
§ 219] Mamination. 267 
 
 21. Explain why in a letter-balance [Roberval balance] it 
 does not matter at what place on the scale-pans 
 the weights are placed. 
 
 [The Roberval balance consists of a jointed 
 parallelogram in which two opposite sides turn 
 freely about their middle points A and B. On 
 the other sides, which are always vertical, are 
 fixed the scale-pans. 
 
 A druggist's scales are an example.] 
 
 22. An ocean steamer is running at n knots when the 
 engines indicate JV horse-power. Show that the resistance 
 offered by the water is, roughly, K/Qn tons. 
 
 23. The weight on the drivers of an engine is W lb, the 
 adhesion is 1/n part, the diameter of the cylinders d in, the 
 diameter of the driver D in, the length of stroke s in, the 
 mean steam-pressure per sq in jt? pounds; then 
 
 WD = d'nps, 
 
 24. Give the units of activity. 
 
 25. What is meant by the term horse-power ? By horse- 
 power-hour ? 
 
 26. What is the ratio of man -power to horse -power? 
 [From 1/5 to 1/10.] 
 
 2Qa, May a man work at a higher rate than this ? 
 
 [Yes, for a short time. He may work at the rate of a 
 horse-power for a minute or two. 
 
 It does not follow that a one-horse-power engine could re- 
 place a horse in all cases, because a horse could make a spurt 
 and for a short time work at the rate of 3- or 4-horse power.] 
 
 27. Compute the horse-power necessary to propel a train 
 weighing IT tons at a speed of F miles an hour (1) on the level, 
 (2) up a grade of n^, if the resistance is p pounds per ton. 
 (See Examples, p. 254.) 
 
 Ans. H.P. on level =. 0027 WVp; 
 
 on grade = .0027 WV{p -f 20?i). 
 
 28. The tractive power of the engine [870, Empire State 
 
268 WORK AKD ENERGY. [§ 219 
 
 Express] is a fraction over 11 lb per pound of average effect- 
 ive cylinder-pressure. This represents at 52.1 miles/hour 
 very nearly 16 H.P. {Engineer, London.) 
 Check the conclusion. 
 
 29. If P pounds represents the driving force and v ft/sec 
 the velocity, show that the power is represented by Pv ft- 
 pounds/sec. 
 
 30. A turbine can utilize 75^ of the energy of falling 
 water. Show that the effective horse-power of a waterfall is 
 
 0^/700, nearly, 
 
 where Q = the number of cubic feet of water per minute, 
 h = the height of the fall in feet. 
 
 31. Give examples of bodies possessing kinetic energy. 
 
 32. A particle free to move is acted on by a force. Show 
 that the work done by the force is equal to the gain of kinetic 
 energy by the particle. 
 
 33. Show that force may be defined as rate of change of 
 kinetic energy with the distance. 
 
 34. " If we multiply one half the momentum of every par- 
 ticle of a body by its velocity and add the results, we shall get 
 the kinetic energy of the body." 
 
 35. How much kinetic energy does a body weighing Wlb 
 lose when its velocity changes from u to v ft/sec ? 
 
 36. An athlete can make a longer running jump than 
 standing jump. 
 
 37. The weight of a train is 95.5 tons, and the drawbar pull 
 is 6 pounds per ton. Find the H.P. required to keep the 
 train running at 25 miles an hour. Ans. 38.2 H.P. 
 
 37rt. Find the energy required to bring the train to full 
 speed from rest. A7is. 2000 foot-tons, nearly. 
 
 38. A train weighing Wlh resting on a straight level track 
 is acted on by a horizontal pull of F pounds for t seconds. 
 Show that the energy acquired is 
 
 i^'^^V2 If^ foot-pounds. 
 
§ 219] EXAMINATION. 269 
 
 39. The work done by an impulse of i second-pounds in 
 changing the velocity of a weight W\h from u to v ft/sec is 
 
 \(u -\- v)/2 foot-pounds. 
 
 40. The kinetic energy of a moving body is independent of 
 the direction of the motion. 
 
 41. Show that there is no parallelogram of kinetic energies. 
 
 42. Find the 11. P. of an engine that will discharge a gallons 
 of water per minute from a depth of b ft and with a velocity 
 of V ft/sec. (A gallon of water weighs 8^ lb.) 
 
 Ans. a(b + vy2g)/3960 H.P. 
 
 43. State what is meant by the conservation of energy. 
 
 44. Show from the principle of the conservation of energy 
 that if a weight slide from rest down a smooth inclined plane 
 of height h, 
 
 v' = 2gh, 
 
 45. A particle is shot up a smooth tube with velocity v. 
 Show that it will come to rest after reaching a point whose 
 height above the point of projection is v^/2g. 
 
 46. How is the potential energy of a body related to its 
 stability ? 
 
 47. A hemisphere rests on a rough inclined plane. Is the 
 equilibrium stable? 
 
 48. Point out the errors in the following : 
 
 (1) "A car of weight W and velocity V miles per hour 
 in passing from the tangent to a circular curve of radius E 
 has its direction suddenly changed by the impulse 
 
 wvyi6R, 
 
 causing a corresponding shock." 
 
 (2) "An equally accurate formula is the weight multiplied 
 by the fall in feet = the momentum of foot-pounds of work." 
 
 (3) " A collision between two bicycles, each with a 150-lb 
 rider, spinning at the moderate speed of seven miles an hour, 
 would result in a smash-up with a force of 3000 pounds." 
 
270 WORK AND ENERGY, [§ 219 
 
 (4) " How much energy is radiated by the sun is very well 
 known now, and is reckoned to be about 10,000 horse-power 
 per square foot of its surface." 
 
 49. In 1895 a train on the Lake Shore R.R. made a run of 
 86 miles at the rate of 73 miles an hour. The weight of the 
 train was 250 tons. Taking the train resistance on the 
 straight level track to be 15 pounds per ton at this speed, 
 show that the engine must have developed about 730 horse- 
 power. 
 
 50. The engine in (49) was a 10-wheeler, having drivers 
 5 ft 8 in in diameter, and cylinders 17 X 24 in. Show that 
 to develop 730 horse-power the average effective cylinder- 
 pressure must have been about 37 pounds per square inch. 
 
 [A pressure of over 60 lb/in' has been kept up at this speed. 
 See R,R, Gazette, Dec. 6, 1895.] 
 
§ 220] DYNAMICS OF ROTATION". . 271 
 
 CHAPTER VII. 
 DYNAMICS OF ROTATION. 
 
 220. The cases of the motion of a single particle and of a 
 motion of translation of a body, together with some of the 
 more simple problems connected with rotation about a fixed 
 axis, have been discussed in the preceding chapters. We now 
 proceed to treat with some degree of fullness the motion of a 
 body into wliich translation and rotation both enter. 
 
 The most important case and the only one we shall con- 
 sider is that in which the particles of the body move in 
 parallel planes. In this case the position of the body is de- 
 termined when the positions of two points in it in a plane 
 parallel to the plane of motion are known. For one point 
 being fixed, rotation only is possible about this point. But a 
 second point fixed fixes the body itself. Changes of position 
 of the body will be determined by considering the changes of 
 position of these two points fixed in the body. 
 
 221. Consider a body displaced from one position to an- 
 other. Let Ay B be the initial position of two fixed points in 
 tlie body and A^, B^ the new positions after , 
 displacement, the disi:)lacement taking place / 
 
 in the plane of the paper. ^\f b, 
 
 (I) If J,^, is parallel to AB, the body / 
 may be moved from its first position to its a b 
 
 second position by a motion parallel to AA^ or BB^. This a 
 motion of translation only. 
 
 (2) If A^ coincides with A, the body may be 
 
 moved from its first position to its second by a 
 
 ^ rotation about A. This a motion of rotation 
 
 ^ only. 
 
 (3) Let A^B^ assume any position relative to AB, This 
 
272 
 
 DYNAMICS OF ROTATION. 
 
 [§222 
 
 Z 
 
 / 
 
 displacement may be considered to occur in either of two 
 ways: 
 
 -i^a. From any point in the plane of the points let fall Oa 
 g perpendicular to the line AB joining 
 
 a' / a' b ' ^ and B. Let the body be rotated 
 about until AaB assumes the posi- 
 tion A'a'B' parallel to A^B^. A mo- 
 tion parallel to A^A^ will bring the 
 points A', B' into coincidence with 
 A^, B/, that is, the body may be 
 moved from the first position to the second by a rotation 
 about any assumed point and a translation. 
 
 b. Join AA^ and BB^. Bisect these lines at a, b and let the 
 perpendiculars aO, bO intersect in 0. / , 
 
 Then evidently OA = OA^, OB =0B,, 
 and I AOA^= I BOB^. Hence, if 
 the body be rotated about through 
 the angle AOA^, the point A will fall 
 on A^ and B on B^; that is, the body 
 will be moved from the first position 
 to the second by a rotation about 0. 
 
 It is evident from the construction 
 that the position of the point 
 changes from instant to instant with the change of position 
 of AB, In any position it is only the center of rotation 
 for an instant, so that is called the instantaneous center of__ 
 rotation. 
 
 222. This may be still further illustrated. 
 Suppose two points A, B of 'a. body to have any motion in 
 the plane of the paper. The points 
 A, B will each trace out a ' path. 
 Consider A. The line joining two 
 consecutive positions of A will give 
 the direction of motion in the path. 
 This line is the direction of the 
 tangent to the path at A, Since an indefinitely great 
 
 ^Bi 
 
§ 233] INSTANTANEOUS CENTER. 273 
 
 number of curves may have a common tangent at a point, it 
 follows that this tangent is quite independent of the form of 
 the path. Hence for the instant we may consider the path 
 to be a circular arc. The perpendicular ^0 to the tangent 
 will pass through the center of the circle, and, conversely, the 
 direction of motion at.l for the instant will be perpendicular 
 to the radius of the circle. Hence the instantaneous motion 
 of A is the same as if it took jilaoe in a circle with center 
 somewhere on A 0. Similarly, the motion of B is the same 
 as if in a circle with center somewhere on BO. But is 
 common to JO and BO. Hence the instantaneous motion of 
 A and B, and therefore of the line AB,is a motion of rota- 
 tion about a point as instantaneous center. 
 
 The points A, B are any two points in the body. Hence, 
 whatever the plane ^notion of the body, it is always possible to 
 find a point such that for the instant the motion about it 
 shall represent the actual motion; in other words, at any in- 
 stant one point is at rest, and the other points are moving 
 in directions perpendicular to the lines joining them to this 
 point. 
 
 If the radii AO, BO do not intersect, the tangents to the 
 paths at A, B are parallel, and the motion is a motion of 
 translation. The radii being parallel may be said to intersect 
 at infinity, and hence a motion of translation may be regarded 
 as a rotation about a center at an infinite distance. 
 
 223. In general, the instantaneous center will vary in 
 position from instant to instant. The locus or path de- 
 scribed by it is called a centrode. But in case the radii AO, 
 BO continue to intersect in the same point 0, as the motion 
 progresses the instantaneous center becomes a permanent or 
 fixed center. For example, a wagon-wheel revolves about the 
 axle as a permanent center, but with reference to the ground 
 it revolves about the point of contact as an instantaneous 
 center. The path traced by the wheel on the ground is the 
 centrode. 
 
274 
 
 DYNAMICS OF ROTATION. 
 
 [§224 
 
 Ex. 1. A ladder BC slides between a vertical wall and the 
 ground, which is horizontal. Find the in- 
 stantaneous center and the centrode. 
 
 [The paths are along AB,AC. Hence the 
 instantaneous center is at the intersection 
 of the perpendiculars BO, CO. It is evident 
 that AO = BC, the length of the ladder, 
 and tlierefore is at a constant distance 
 from A. Hence the centrode is a circle, with 
 A as center.] 
 
 2. What is the direction of motion of any point G in BC at 
 any instant ? Ans. _L to GO. 
 
 3. The Empire State Express engine 999 has driving-wheels 
 7 ft 2 in diameter. Find the height above the track of a 
 point on the circumference that'has half the velocity of the 
 highest point of the wheel. Ans. 1 ft 9.5 in. 
 
 224. In the case of moving bodies rigidly connected to- 
 gether the determination of the velocity of one with respect 
 
 to another may be based on the preceding. For illustration 
 take the ordinary steam-engine. The mechanism itself has 
 been already shown in Art. 159. 
 
 Suppose we wish to find the velocity v^ of the piston P 
 relative to the velocity v, of the crank-pin B. The velocity 
 of the piston is the same as that of the extremity A of the 
 connecting-rod AB. The velocity of the crank-pin B is the 
 same as that of the extremity B of the connecting-rod. 
 
§ 224] PISTON VELOCITY. 275 
 
 Hence the relation sought is the same as that between the 
 velocities of the extremities A and B of the connecting-rod. 
 
 The bed-plate is fixed. The extremity A of the connect- 
 ing-rod moves in a straight line PC, and, the direction of 
 motion being along PC, the instantaneous center is in a line 
 ^0 at right angles to PC, The extremity B moves in a 
 circle of center C, and therefore the instantaneous center is 
 in the line CB, Hence the connecting-rod is for the iristant 
 in the condition of a body turning about an axis through 0, 
 the intersection of ^ and CB, Consequently 
 
 v,\v^= OA: OB, 
 
 or the velocities of piston and crank-pin at any instant are as 
 the distances of the cross-head A and crank-piu B from the 
 instantaneous center 0. 
 
 If, therefore, the velocity of one of the two, piston or crank- 
 pin, is given, that of the other follows at once. Thus, suppose 
 the crank-pin to have a velocity of 10 ft/sec. Lay off to scale 
 a distance BH = 10 ft, and draw BG parallel to BA, Then 
 since 
 
 HB:GA = OB: OA, 
 
 we scale off GA as the velocity of the piston. 
 
 By drawing the crank in different positions, and finding 
 the corresponding positions of G, a curve will result, the or- 
 dinates of which will give the velocity of the piston through 
 its stroke. 
 
 Ex. 1. If gl> is the angular velocity of the crank B and v the 
 speed of the piston P, then 
 
 V = 00 X CR 
 
 where E is the intersection oi AB and the perpendicular CR 
 to CA at C. 
 2. If V denote the piston velocity, u the velocity of the 
 
276 DYKAMICS OF ROTATION. [§ 225 
 
 crank-pin B, I the length of AB, r the length of CB, and 6 
 the angle AUB, show that 
 
 v/u = sin ^(1 + r cos 6/1) nearly. 
 
 225. The relation between the piston -pressure P and the 
 crank-pin resistance Qy when the connecting-rod is inclined 
 
 at any angle, has already been found in Art. 159, but may be 
 solved more simply by aid of the instantaneous center of ro- 
 tation. 
 
 For if v^ is the velocity of the piston and v, that of the 
 crank-pin at any instant, then, being the instantaneous 
 center, we have (Art. 224) 
 
 v,',v^= OA: OB. 
 
 But from the principle of work 
 
 P Xv^ — QXv^=zO. 
 
 Hence P X OA = Q X OB, 
 
 the relation sought. 
 
 The value of Q for a given piston-pressure will thus vary 
 according to the position of the connecting-rod. It may be 
 represented graphically, as in the case of the indicator-diagram 
 (Art. 189). 
 
225] 
 
 CRANK-PIN RESISTANCK. 
 
 277 
 
 The average of the values of Q for a complete revolution of 
 the crank corresponding to a given piston-pressure P will be 
 found by equating the work done by each of the two forces. 
 We have, if r is the radius of the crank-arm and S the length 
 of the stroke, 
 
 Qx27cr = P X 2S. 
 
 But S = 2r. 
 
 Q = 2P/n, 
 
 the relation required. 
 
 Ex. 1. In an engine the diameter of the cylinder is 14 in, 
 and the steam-pressure 75 pounds/in'. Find the average 
 value of the force acting on the crank-pin. 
 
 Ans. 7350 pounds. 
 
 2. In (1) find the force acting when the crank stands at 
 60°, and the ratio of the connecting-rod to the crank is 5^. 
 
 3. In a steam-riveting machine the piston-pressure P is ap- 
 plied at the joint a, and the rivet squeezed between the jaws 
 c, d. Find the relation between P and the force Q exerted on 
 the rivet. 
 
 [The instantaneous center is 
 at 0, where la and the perpen- 
 dicular through c to the sliding 
 surface S intersect. Then 
 
 p xfO= QxcO 
 
 As a approaches g, cO dimin- 
 ishes ; and when a reaches 9, Q be- 
 comes indefinitely great. Hence 
 the advantage of the apparatus 
 in that an enormous pressure 
 may be produced by a moderate 
 force acting tlirough a small dis- 
 tance. 
 
 This is an example of the tog- 
 gle-joint, a mechanism of very 
 considerable importance. It is 
 applied, for example, in the railroad air-brake, in cider, oil. 
 
 s 
 
 
 l/\ ■ 
 
 ^ \ 
 
 ' Q 
 
278 
 
 DYKAMlCS OF HoTATIOK. 
 
 [§226 
 
 and other presses, etc. In the figure below is shown part of 
 a power-screw oil-press.] 
 
 226. Angular Velocity.— If the motion of a body is a mo- 
 tion of rotation about a fixed axis, each particle describes a 
 circumference whose center is in the axis. Since each cir- 
 cumference is described in the same time, the linear velocities 
 of the particles in the circular paths must be proportional to 
 the distances of the particles from the axis — the greater the 
 distance the greater the velocity. The linear velocity of the 
 body, however, cannot be said to be 
 equal to that of any particle, as in a mo- 
 tion of translation. 
 
 Suppose O to be an axis perpendicu- 
 lar to the plane of the paper, and A any 
 particle of the body. When the par- 
 ticle has reached the position Z>, it has 
 described the angle AOh. It is evident 
 that every particle of the body would describe an angle equal 
 to this. The velocity of rotation may therefore be defined in 
 
§ 228] ANGULAR VELOCITY. 279 
 
 terms of the angle of rotation. The rate at which the angle 
 is described is the angular velocity of the particle, and being 
 the same for every particle, is also the angular velocity of the 
 body. 
 
 Thus, if the motion is uniform and the angle AOa is 
 described in t sec, the angular velocity go of the body is 
 measured by the angle described in 1 sec, or 
 
 GD= zAOa/t 
 
 227. The unit of angular velocity is naturally taken to 
 be unit angle described in one second. The unit angle em- 
 ployed is the unit of circular measure, being the angle AOb, 
 which subtends an arc Ab equal in length to the radius AO. 
 This angle is called a radian,* so that the unit of angular 
 velocity is 07ie radian per second. 
 
 Thus, if a body makes n revolutions per second, the number 
 of radians described per second, or the angular velocity, is 
 27tn'y that is, 
 
 G? = 27tn radians/sec. 
 
 Notice that 
 
 1 revolution/sec = 2;r radians/sec. 
 
 228. Relation ofAngidar and Linear Velocity. — The angu- 
 lar velocity of every point of the body has the same value. 
 The relation between this and the linear velocity v of any 
 particle A situated at a distance r from the axis of rotation 
 follows at once. For the time of motion being t sec, we have 
 
 cD = /_ AOa/t 
 = arc Aa/rt 
 = v/r, 
 the relation sought. 
 
 * It is shown in treatises on trigonometiy that 
 one radian = 57.3°. 
 
280 
 
 DYNAMICS OF ROTATION. 
 
 [§229 
 
 229. The direction of motion of a particle A in a circular 
 
 path, or of a particle J of a body revolving 
 
 Y^ about an axis 0, is for an indefinitely small 
 
 ^ ^ arc perpendicular to AO, that is, along the 
 
 I tangent to the path at A. We may therefore 
 place 
 
 tangential velocity v = rao 
 
 if GO is the angular velocity of the particle about 0. 
 
 Let the particle be referred to fixed rectangular axes OX, 
 OY^ and let x, y be its coordinates when in y 
 any position P. Denote the angle POX by 
 6 and OP by r. 
 
 The components v^, Vy of the velocity v of 
 the particle parallel to OX, OF are 
 
 — V mi 6 = ~ Gor sin 6 
 
 coy. 
 
 Vy= V cos 6 = GDV COS 8 = GDX, 
 
 where go is the angular velocity about 0. 
 
 230. If the angular velocity is not constant, the actual 
 angular velocity at any instant is determined by finding the 
 limiting value of the average angular velocity for an indefi- 
 nitely small angle Ad described during an indefinitely small 
 time At, including the instant. We have 
 
 00 
 
 = limit J 6/ At 
 
 = dd/dt radians/sec. 
 
 231. Graphical Represe7itation. — An angular velocity about 
 an axis having magnitude and direction may be represented 
 by a straight line. This line is taken along the axis of'rota- 
 tion, and therefore perpendicular to the plane of rotation, 
 and of length proportional to the magnitude of the angular 
 velocity. The positive direction of the axis is taken to be 
 that in which a right-handed screw would move if placed 
 along the axis and turned with the body. 
 
§ 232] AKGULAR VELOCITY. 281 
 
 T232. Composition of Angular Velocities. — In the case of a 
 rigid body having a number of angular velocities one point 
 must be fixed, or may be considered fixed for the instant, and 
 the angular velocities may be combined in a similar way to 
 that in which linear velocities are combined by means of the 
 parallelogram law. 
 
 Thus let two concurrent angular velocities o)^, g9, about 
 two axes AB, A C be represented by the lines AB,AO, and let 
 
 F be any point lying in the di- c 
 
 agonal AD of the parallelogram r/\ q^^-^^^^ 
 
 ABDC. A^kcT^ / 
 
 Let fall the perpendiculars PQ, j^^^^\^^i — ^^/ ^ 
 PR on AB, AC. a q b 
 
 The linear velocity of P due to the angular velocity about 
 the axis AB \^ represented hy AB X PQ or 2JAPB. Also 
 the linear velocity of P due to the angular velocity about the 
 axis AC is represented hy AC X PR or 2/} A PC. Hence 
 
 resultant velocity of P = 2AAPB - 2JAPC 
 = 0, 
 
 or P is at rest. But P is any point in AD. Hence every 
 point in AD has a resultant velocity due to the two rotations 
 equal to zero; that is, AD is the direction of the resultant 
 axis of rotation. 
 
 To find the magnitude of the resultant; angular velocity co 
 about AD. 
 
 Consider a point Q of the body situated on AB. It has an 
 angular velocity co about AD, oo^ about A C, and about AB. 
 
 The linear vel. of Q about AD = go x A. from Q on AD 
 
 = GO X AQ sin QAD; 
 
 The linear vel. of Q about AC = &?, X X from Q on AC 
 
 = AC XAQ sin QAC; 
 
 The linear vel. of Q about AB = 0. 
 
282 DYKAMICS OP ROTATION. [§ 233 
 
 Hence, since AD is the resultant axis, 
 
 GO XAQ sm QAD = ACX AQ sin QAC, 
 
 and cD = ACxAI)/AO 
 = AD, 
 
 or GD is represented in magnitude by the diagonal AD of the 
 parallelogram. 
 
 If, then, 6 is the angle between the two axes AB and A C, 
 we have, as in Art. 19, 
 
 GJ^ = G9j' -f- 2Ce?,CB?3 cos 6 -\- GO^, 
 
 233. Resolution of Angular Velocities. — Any angular ve- 
 locity may be resolved into components after the manner of 
 linear velocities. 
 
 An interesting application is afforded by Foucault's pen- 
 dulum.* The apparatus consists of a long pendulum freely 
 suspended and set oscillating in a vertical plane. A horizon- 
 tal table placed below the pendulum appears to revolve in a 
 direction opposite that of the hands of a 
 watch. The problem is to find the an- 
 \^ gular velocity gd^ of the plane of oscillation 
 ^^ of the pendulum relative to the table. 
 1 Let P be the place of observation, X 
 J its latitude. If gd denotes the angular 
 \^ velocity of the earth about its axis ON, 
 then 
 
 00 — 'in radians/day. 
 
 Now GD about ON has two components — go sin \ about OPy 
 and GO cos \ about OB at right angles to OP, 
 
 The component go sin A gives the relative angular velocity 
 of the earth (or table) and the pendulum, the other compo- 
 
 * The experiment was first made by Foucault, the French physicist, 
 in the Pantheon in 1851. 
 
§ 235] AKGtJLAR ACCELERATION^. 2S3 
 
 nent, oo cos A, not producing any relative angular velocity. 
 The table appears to rotate from east to west. Hence 
 
 00^ = 00 sin X 
 
 = 2;r sin /I radians/day, from east to west. 
 
 The time in which the plane of oscillation of the pendulum 
 appears to describe a complete revolution is 
 
 27r/cL>, = 27t/27t sin A = cosec A days. 
 
 Ex. A Foucault pendulum is set vibrating at New Orleans 
 in lat. 30°. After what interval will it oscillate in the initial 
 plane of oscillation? jUis. 1 day. 
 
 What is the time of a complete revolution of the pendu- 
 lum? 
 
 234. Angular Acceleration. — The rate of change of angular 
 velocity about an axis is called the angular acceleration. 
 
 The unit of angular velocity being one radian per second, 
 the ^init of angular acceleration is taken one radian per 
 second per second, or, as it may be written, 1 radian/sec'. 
 
 235. If a body start from rest with a uniform angular ac- 
 celeration a, then after a time t the angular velocity oa would 
 be given by 
 
 CO = at (1) 
 
 Also the angle passed through is given by (Art. 15) 
 
 e = aty2 (2) 
 
 Eliminating t between (1) and (2), 
 
 (»72 = aB (3) 
 
 These results may be compared with those of Art. 25. 
 
284 DYiTAMtCS OP ftOTATIOir. [§ 236 
 
 Ex. If the body instead of starting from rest had an initial 
 angular velocity go^ , show that (compare Art. 24) 
 
 G? = G?„ + at, 
 
 e=GjJ-\- «f /2, 
 (wV2 - gd;/2 = ad. 
 
 236. If the angular acceleration is not constant, the actual 
 angular acceleration at any instant would be determined by 
 finding the limiting value of the average angular acceleration 
 for an indefinitely small angular velocity ^go acquired in an 
 indefinitely small time /It, including the instant. We have 
 
 a = limit Aoo/At 
 = doo/clt 
 = d'^d/cW^ radian s/sec% 
 
 or, as it may be written, 
 
 a = oodoo/dB radians/sec*. 
 
 These results may be compared with those of Art. 26. 
 
 237. Tangential and Normal Acceleration. — The accelera- 
 tion of a particle lu of a body revolving about an axis may be 
 resolved into two components at right angles to each other. 
 Let V be the linear velocity at any instant along the tangent to 
 the path, oo the angular velocity about the axis, and r the 
 radius of the path. Then 
 
 tangential acceleration = dv/dt 
 
 = d(oor)/df 
 = rdao/dt 
 = ra; 
 normal acceleration = v^/r 
 
 = GoVyr 
 = Go'r, 
 Hence we may write 
 
 tangential force = ivra/g, 
 centripetal force = wco'^r/ff, 
 
§ 239] CENTRIFUGAL FORCE. 285 
 
 where go and a may be expressed in any of the differential 
 forms given above. 
 
 238. As an illustration consider the centrifugal force de- 
 veloped by a plane lamina of weight W revolving about a 
 fixed axis perpendicular to its plane. 
 
 Since each particle in the lamina describes a circle about 
 0, the centrifugal forces all pass through 0. The centrifugal 
 force of a particle weighing w, and distant r, from is, at 
 the instant when the angular velocity is go, equal to w^oo^r^/g. 
 Hence for the whole lamina the total centrifugal force C is 
 given by 
 
 G— '2wGo^r/g 
 = oo^^iur/g 
 = Go'Wr/g 
 
 where r is the distance of the C.G. of the lamina from the 
 axis. 
 
 Thus the centrifugal force of rotation is the same as if the 
 whole weight were concentrated at the C.G. 
 
 For example, if a fly-wheel weighing 9.6 tons and making 
 60 revolutions per minute, has its C.G. I/IO inch out of the 
 center of rotation, 
 
 the centrifugal force = (2nY x (9.6 X 2000)/120 X 33 
 = 200 pounds. 
 
 239. It will be instructive to note the analogy between the 
 kinematical formulas for translation and the corresponding 
 formulas for rotation. Where there is no initial velocity and 
 the acceleration is constant, we have 
 
 Translation. Rotation. 
 
 V — at 00 = at 
 
 s = af'/2 6 = aty2 
 
 as = vys a0= G?y2 
 
 V = ds/dt GO = dB/dt 
 a = d's/dt' a = d'B/dt" 
 
 a = vdv/ds a = CO doo/dO 
 
286 DYNAMICS OF KOTATION. [§ 239 
 
 Ex. 1. If OD is expressed in degrees, show that 
 V = 27rG9r/360°. 
 
 2. A body makes n revolutions per second. Show that the 
 angular velocity is ^nn radians/sec. 
 
 3. A belt passes over a pulley d ft in diameter and making 
 n revolutions per min. Find its linear velocity. 
 
 Ans. 71 dn ft/min. 
 
 4. A wheel 4 ft in diameter revolves 420 times per minute. 
 Find the angular vt4ocity and the linear velocity of a point 
 1.5 ft from the center. Ans, 14;r radians/sec; 21 tt ft/sec. 
 
 5. The crank of an engine makes n revolutions per min. 
 Its radius is v ft. Find the linear velocity of the crank-pin 
 in ft/sec. Ans, nrn/ZO. 
 
 6. A locomotive is running at 45 miles an hour. The 
 driving-wheels are 6 ft in diameter and the stroke is 2 ft. 
 Find the average piston velocity in ft/sec. 
 
 Ans. \.^l7t ft/sec. 
 
 7. A wheel making 20 revolutions a second is brought 
 gradually to rest in 10 seconds. How many revolutions has 
 it made after the brake was applied ? 
 
 Ans, 100 revolutions. 
 
 8. One of the 12 spokes of a carriage-wheel is vertical. 
 Find the velocity of the extremity of the first spoke in ad- 
 vance if the velocity of the carriage is 7.5 miles an hour. 
 
 Ans. llV'2 + 1^3 ft/sec. 
 
 9. A particle describes a circle of radius r with uniform 
 velocity v. Find its angular velocity about any point on the 
 circumference. Ans. v/2r. 
 
 10. A coin, radius r, is rolled along a table. If v denotes 
 the linear velocity of its center and w its angular velocity 
 about the point of contact with the table, then 
 
 v = Gor, 
 
 l\a, A system has two component rotations of 2 and 3 
 radians/sec about axes inclined at 60°. Show that the re- 
 sultant rotation is 4^19 radians/sec. 
 
 lib. What is the position of the resultant axis of rotation ? 
 
 12«. If a body has two angular velocities g?, , oo^ about par- 
 allel axes through A, B, show that the resultant oo is found 
 from 
 
§ 241] EQUATIONS OF MOTION. 287 
 
 and is about an axis through a point C such that 
 cD^xAC=oD^xBa • 
 
 12ft. Give the corresponding proposition for parallel trans- 
 lations. 
 
 13. If w, V denote the component velocities of translation 
 of a point P parallel to rectangular axes OX, OY, and go the 
 angular velocity of P about 0, then 
 
 total velocity of P parallel to 0X= u — coy; 
 total velocity of P parallel to 0Y= v -{- oox. 
 
 240. Dynamical Equations of Motion. — We have seen (Art. 
 221) that the plane motion of a rigid body may be indicated 
 in two ways : 
 
 1. As a translation of the body and a rotation about an 
 axis through any point and perpendicular to the plane of 
 motion; or 
 
 2. As a rotation about an axis which changes from instant 
 to instant — the instantaneous axis of rotation. 
 
 The cases in which the motion is wholly a translation or 
 wholly a rotation about a fixed axis are evidently included. 
 
 There are two cases to be considered : (A) the motion due 
 to continuous forces, (B) the initial motion due to impulses. 
 
 241. Motion under Continuous Forces. — When a single par- 
 ticle w^ at rest and free to move is acted on by a force F^, the 
 acceleration a' produced is in the direction of F^, and its 
 value is found from (Art. 67) 
 
 F, = tvy/g. 
 
 But if the particle, instead of being free to move, forms 
 one of a rigid system, then, besides the external force i^,, 
 other forces act on the particle arising from the mutual 
 actions of the particles. Let /?, be the resultant of these 
 internal reactions on zv^. 
 
 The acceleration of w^ is now due not to F^, but to the 
 resultant of F^ and A\. Denote this acceleration by «,. 
 
288 DYNAMICS OF ROTATION. [§ 241 
 
 Then w^a /g, called the effective force on the particle, is the 
 resultant of F, and R^ , or, in other words, F^,R^, and w^ajg 
 reversed,* are in equilibrium. 
 
 Similarly, for a particle w^ the forces F^, i?,, and lo^ajg 
 reversed, are in equilibrium. 
 
 Summing up for the whole system, and assuming as a con- 
 sequence of the third law of motion that the internal actions 
 and reactions R^, R^, ... of the system are in equilibrium 
 among themselves, we conclude that 
 
 The external forces F^, F^, . . . and the reversed effective 
 forces for all the particles form a system of forces in 
 equilibrium. 
 
 Hence the solution of the problem is reduced to the statical 
 problem of Art. 140. 
 
 [The above principle is known as D'Alemberfs principle, 
 having been enunciated by D'Alembert in 1742. It is not 
 an independent principle, but an immediate consequence of 
 Newton^s laws of motion. Its great use is that it enables us 
 to state dynamical propositions in a statical form.] 
 
 For the first two conditions of equilibrium (Art. 140), that 
 the sum of the forces parallel to two rectangular axes should 
 each equal zero, let Xi, Yi; X^, Y^\ . . . de- 
 note the components of the external forces 
 F^, F^, . . . parallel to the rectangular axes 
 OX, OY drawn through a fixed point 0. 
 Also let w^aj/g, n\aj /g\ w^aj' /g, iu^ay"/g; 
 . . . denote the components of the effective 
 X forces parallel to these axes, aj, a/; aj' , 
 «y"; . . . being the component accelerations of the particles 
 w^, n\, ... 
 
 Then, resolving parallel to the axes, we have 
 
 :2X=2LuaJg, (1) 
 
 ::EY= 2way/g • • (2) 
 
 * The reversed effective force — loa/g is sometimes called the force of 
 inertia or the inertia remtance of the panicle. 
 
§ 241] BQUAtlONS OF MOTION". 289 
 
 
 '^1 
 
 For the third condition of equilibrium, that the sum of the 
 moments of the forces acting— that is, of the external forces 
 and the effective forces reversed — should 
 be equal to zero, let /?,, ;?,, . . . denote the 
 distances of the directions of the external 
 forces F^y F^y . . . from 0, and r,, r,, . . . 
 the distances of the particles F^, P^, ... 
 on which these forces act from 0. 
 
 Now, since each particle moves in a " ^ 
 
 circle about the axis 0, it is convenient in taking moments to 
 resolve the effective forces into components along and per- 
 pendicular to the radii OP,, OP,, . . . 
 
 The component forces at P, are (Art. 237) 
 
 w,r^GD'/g along OP^, 
 w,r^a/g±OF,, 
 
 where go is the angular velocity and a the angular accelera- 
 tion of the particle. 
 Hence, taking moments about for all the forces, we have 
 
 ^, P, + ^.Jf'. + • • • = ^i^>/^ + ^^.^>/ff + . . . , 
 
 the angular acceleration a being the same for all of the 
 particles. 
 This may be written 
 
 ^Fp = -2wr\ 
 9 
 
 since a/g is a constant factor in each term. 
 
 The left-hand member of this equation is the ordinary 
 expression for the statical moment or torque. The factor 
 2wr^ of the right-hand member, which is independent of 
 tjie angular acceleration a, is called the second moment or 
 moment of inertia of the body relative to the axis. It is usu- 
 ally denoted by the letter /, so that 
 
 2Fp = la/g (3) 
 
290 DYNAMICS OF ROTATIOIS'. [§ 242 
 
 Hence the angular acceleration o' of a body about the axis 
 
 of rotation is found; that is, 
 
 angnlar acceleration = torque /mo7nent of ine?'tia. 
 
 These three equations are the equations of motion of a 
 rigid body when acted on by continuous forces. 
 
 [The term moment of inertia was introduced by Euler in 
 order to express in analogous terms the formulas for linear 
 and angular acceleration. Thus for translation, using abso- 
 lute units (Art. 50), 
 
 F = ma, 
 or force = inertia X linear acceleration, ... (1) 
 
 if we use the term inertia instead of the term mass as Euler 
 did. 
 
 For rotation 
 
 2Fp = 2mr^ X a. 
 
 Here 2Fp is moment of force and a is angular accelera- 
 tion. Hence, if for ^m?-^ is written "moment of inertia," we 
 have, analogous to (1), 
 
 moment of force = moment of inertia X angular acceleration.] 
 
 242. The point selected as fixed point is usually the 
 O.G. of the body, and the motion is reduced to a translation 
 of this point and a rotation about it. 
 
 Thus if a^, ay are the accelerations of the C.Gr. of the body 
 parallel to the axes, then (Art. 146) 
 
 Wa^ = 2wa^, WUy = 2way. 
 
 Hence :SX= Wa^/g, 2r= Wajg, 
 
 or tlie C. G. of a body moves as if the ivJiole weight were con- 
 centrated at that point and the external forces were to act on 
 it parallel to their original directions. 
 
 The motion of rotation about the C.G. is given by 
 
 2Fp = Icx/g. 
 
§ 244] MOMENT OF INERTIA. 291 
 
 Being given the initial velocities of translation and rota- 
 tion, the displacement and resulting velocities after any term 
 may be found from Art. 24 for translation and from Art. 235 
 for rotation. Hence the motion is completely determined. 
 
 Ex. 1. A fly-wheel weighing W lb is rotating at n revolu- 
 tions per minute about its axle, whose radius is r ft. If 
 tlie driving-belt is slipped, in what time t will the wheel 
 come to rest, // being tlio coefficient of friction ? 
 
 Let GO = the angular velocity to be destroyed 
 = 27r)i/Q0 radians/sec; 
 a = the angular acceleration required to destroy g? 
 in t sec 
 = (A>/t radians/sec'. 
 
 Now, since the wheel rotates about a fixed axis, 
 
 torque = Jo(/g, 
 or j^iWx r = iTtn/^Ogt, 
 
 and t = l7tn/30/j. Wrg seconds. 
 
 2. Show that the number of revolutions made by the wheel 
 before it comes to rest is /7r?iy3600// Wrg. 
 
 243. It is evident that before we can find numerical results 
 in any case we must be able to compute the value of /, the 
 moment of inertia of the rotating body. The method of 
 doing this we proceed to explain in the following sections. 
 
 The sections treating of moments of inertia, like the sec- 
 tions on center of gravity (Arts. 143-151), form a sort of 
 interlude, and might be placed in a treatise on the integral 
 calculus — perhaps better placed there. The subject of me- 
 chanics proper is resumed in Art. 251. 
 
 244. Moment of Inertia. — The form of the expression for 
 the moment of inertia about an axis, 2'?^r', shows that we 
 may define it as the sum of the products of the particles 
 w,, w,, ... of a rigid body into the squares of their distances 
 r, J r,, . . . from the axis of rotation. The finding of moments 
 
292 DYNAMICS OF ROTATION. [§ 244 
 
 of inertia is therefore a problem of summation of indefinitely 
 small quantities, and not a mechanical question at all. This 
 summation may in some cases be made without the calculus,, 
 as in the first example following, which for illustration is 
 solved in both ways. 
 
 In the case of bodies of irregular shape it is in general most 
 convenient to determine the moment of. inertia experimen- 
 tally. A method of doing this is indicated in example 4, 
 page 314. 
 
 Ex. 1. To find the moment of inertia / of a thin uniform 
 rod, weight W, length I, about an axis 
 OY through its center 0, and at right 
 angles to the rod. 
 
 [Conceive the rod cut into elements 
 of indefinitely small length ^x, and let 
 
 ^ [^ xhe the distance of any one of these- 
 
 elements from 0. 
 Let each unit of length weigh d, then the length Zx will 
 weigh 6/}x, and the moment of inertia of this element about 
 the axis is S^x X x^. Hence for the whole rod 
 
 r: 
 
 dx'dx = dr/12 = wr/12. 
 
 Or thus : Suppose the rod divided into a large number 2n of 
 equal parts. The length of each part is l/27i and its weight 
 is dl/2n. The distances of these parts from may be taken 
 to be the distances of their centers of gravity from 0, that 
 is, Z/4w, dl/i7i, . . . Hence, taking half the rod, 
 
 ■i/= — — + ^r- h- +... to 7i terms 
 
 = 3^(1^ + 3' + . . . to ^ terms) 
 
 = Sr /24: when n is indefinitely great, 
 
 and / = WP/12, as before.] 
 
 la. Show that the moment of inertia about a perpendicular 
 axis through one extremity is If /y3. 
 
§ 246] MOMENT OF INERTIA. 293 
 
 245. Unit Moment of Inertia, — The unit moment of in- 
 ertia is the moment of inertia of a particle of unit weight 
 situated at unit distance from the axis of rotation. No 
 special name is in use, but it is necessary in giving moments 
 of inertia to state the units employed in computing them. 
 The name newton has been suggested as an appropriate name 
 for the unit moment in the British gravitation system. 
 
 If the pound weight were made into a thin cylinder of 1 ft 
 radius and caused to revolve about the central axis it would 
 have unit moment of inertia. 
 
 246. The computation of moments of inertia may be facili- 
 tated by the aid of the following two propositions : 
 
 (1) The moment of inertia of a tody about any axis is equal 
 to the momeyit of inertia about a parallel axis through the 
 center of gravity, together with the product of the weight of 
 the body into the square of the distance bettveen the two axes. 
 
 For suppose the two parallel axes through a point and 
 the center of gravity G to lie in a plane 
 perpendicular to the plane of the paper. 
 Take G as origin, the plane of the paper 
 the plane of Xy,and 06^Xthe axis of 
 X, Let X, y denote the coordinates of 
 any particle P- weighing w. Call the 
 distance GO = h. Then if /denote the -j— 
 moment of inertia about an axis through 
 
 0, we have 
 
 I=:Ew{fJr{x-\-hy\ 
 = 2iv(y' + x') + 2h2wx + ?i'2w, 
 since the distance h is constant. 
 
 Of the three expressions in the right-hand member, the 
 first is equal to /the moment of inertia about G; the second 
 is equal to zero, since G is the center of gravity (Art. 145); 
 and the third is equal to Wh*, where W is the total weight of 
 the body. Hence 
 
 I = 7j^Wh\ 
 which proves the proposition. 
 
 / 
 
 / 
 
294 
 
 DYNAMICS OF ROTATION. 
 
 [§246 
 
 This proposition is due to Lagrange. 
 
 (2) 2'he moment of inertia of a plane lamina about an axis 
 through any ijoint and perpendicular to its plane [polar 
 moment] is equal to the sum of the moments of inertia about 
 two rectang^Uar axes tlirouyh and in the plane [rectangular 
 moment]. 
 
 Let OX, OY hQ rectangular axes througli any point in 
 the plane of the lamina, x, y the coordi- 
 nates of P any particle weighing lu, and 
 r the distance OP. Then 
 
 r^z=x'-\- y\ 
 
 o ^ .'. lor^ = lox^ -{- luy"^ ', 
 
 and summing up for the particles in the whole lamina, 
 
 ^wr"^ = 2wx'^ + 2wy^, 
 /about = /about 0F+ /about OX, 
 
 or 
 
 or 
 
 1=1,-^1,, 
 
 which proves the proposition. 
 
 Ex. 2. Find the /of a thin rectan- 
 gular lamina or plate of breadth b 
 and depth //, about an axis through its 
 center of gravity and parallel to h, 
 
 [Conceive the lamina cut into 
 strips parallel to the axis, and of 
 breadth /]x. Let x denote the dis- 
 tance of one of these strips from the 
 axis, and let it weigh d per unit area. 
 The strip will weigh Sh^x. Hence 
 
 =/: 
 
 6hxdxxx'=^ i^dh¥ = j'sWb% 
 
 where the whole lamina weighs W.] 
 
§246J 
 
 MOMEKT OP INERTIA. 
 
 295 
 
 
 C D 
 
 A 
 
 G 
 
 B 
 
 h 
 
 b 
 
 
 3. If in (2) the axis is parallel to the side h show that 
 
 I = Whyi2. 
 
 4. Show that the 7 of a rectangular 
 lamina breadth b, depth h about the side 
 b is Wh'/^. (For example, a door about 
 its hinges.) 
 
 [Deduce from Ex. (3) by aid of Art. 
 24G, and also solve independently.] 
 
 5. Find the 7 of a rectangular lamina, 
 breadth ft, depth h, about an axis through 
 its center of gravity and perpendicular 
 to its plane. Aiis. I = W(b' + h')/12. 
 
 6. The / of a square plate of side a about a diagonal is 
 Wn'/12. 
 
 7. Show that the /of a square lamina about aiiy axis in 
 its plane and through its center is tlie same and equal to 
 ^^.^712. 
 
 8. Find the 7 of a rectangular plate about a diagonal, the 
 sides of the rectangle being a, b. 
 
 Ans. 7=rTIV&yG(a' + &'). 
 
 9. Find the 7 of a thin circular ring of radius r and weight 
 Tf about an axis through its center and perpendicular to its 
 plane. Ans. I — Wr^, 
 
 10. Find the 7 of a circular ring, radius r and weight W 
 about a diameter. 
 
 [Conceive the ring cut into elements 
 PQ subtending an angle dQ at the cen- / /\p 
 
 ter 0. Then weight of 7*^ = (J X rdd. 
 Hence 
 
 -r- 
 
 drdO X r' sin' 6 
 
 cos 26 
 
 de 
 
 = Ttdr^ 
 = TrrV2.] 
 
 11. Find the 7 of a circular ring or wire about a tangent. 
 
 Ans. 1= 3rrV2. 
 [Deduce from Ex. 10 by aid of Art. 246, and also solve in- 
 dependently.] 
 
296 
 
 DYNAMICS OF ROTATIOK. 
 
 [§246 
 
 12. Find the /of a quadrantiil wire about an axis through 
 one extremity and perpendicular to its plane. 
 
 A71S. I=2Wr'(l -2/7r). 
 13. Find the / of a thin circular 
 plate, radius r and weight W, about 
 a diameter. 
 
 [Conceive the plate cut into 
 strips parallel to the axis, and of 
 breadth /Jx. Then the equation 
 to the circle being x^ -\- y^ = r', 
 the length of a strip at a distance 
 X from is 2Vr'^ — x", and area 
 of strip = ^Vr"^ — x^Ax, Hence 
 
 I=r26x''Vr^-x''dx = Snr'/^ = Wr^/^.. 
 
 The simplest way of finding the value of this integral is 
 to put X = r sin i). The limits of integration then become 
 + 7r/2, - 7r/2.] 
 
 Solve the problem, using polar coordinates. 
 
 14. Find the / of a thin circular plate about a tangent. 
 
 Ans. 5 T^ry4. 
 
 15. Find the /of a circular plate of radius r and weight W 
 about a perpendicular axis through its center. 
 
 Ans. 1= Wry 2. 
 
 [May be deduced from Ex. 13 and Art. 246 (2), or inde- 
 pendently, as follows: 
 
 Conceive the plate composed of concentric rings. Let x 
 denote the distance of any ring from the center and ^x its 
 width. Then the ring weighs d X 27rx Ax. Hence 
 
 /z= r 27t6xdx X x^ 
 
 =-. nSr'1% 
 = Wry 2.] 
 
 Conversely, Ex. 13 may now be deduced from Ex. 15. 
 
 16. Find the moment of inertia of a grindstone 2 ft in di- 
 ameter and 6 inches thick about its axis, if the stone weighs 
 125 lb/ft'. Ans. 98.2. 
 
§246] 
 
 MOMEKT OF IKERTlA. 
 
 297 
 
 17. Find the / of a ring of radii r, , r, about a J_ axis 
 througli its center. Atis. 1= W(i\^ + r^)/^. 
 
 18. Find the / of a triangular lamina of 
 base h and height // about the base. v 
 
 [Divide the lamina into strips parallel to 
 the base. 
 
 Let y = distance of any strip from the 
 
 base; v. — i : 
 
 z/y = width of strip. 
 
 or 
 
 Then length of strip : b = h — y : h, 
 length of strip = b{h — y)/h, 
 area of strip = b{h—y)Ay/h, 
 
 J^6by\h-y)dy/h 
 
 = 6bh'/l2, 
 
 19. Find the / of a triangle of base b, height h, about an 
 axis through its center of gravity and parallel to the base. 
 
 Ans. /= Wih'/Q - /iVQ) = Wh'/lS. 
 
 20. Find the / of a triangle of base b, height 7i, about an 
 axis through its vertex and parallel to the base b. 
 
 Ans. /= Tf7iV2. 
 
 21. Find the /of a hexagon of side a about a diagonal. 
 
 Ans. I=bWay^, 
 
 22. Find the / of a T-iron, breadth of 
 flange = b, breadth of web = J, , depth of 
 flange = h, depth of web = /i, , about an axis J^^ 
 through the center of gravity. (See Ex. 4, 
 p. 165.) A?is. 
 
 ohyi2-^bfi;/i2-{-(h+h^y/4(b-'h-'-\-b['/i{'). 
 
 23. Find the / of a sphere of radius r 
 about a diameter as axis. 
 
 [Conceive the sphere divided into slices of 
 width dx by planes perpendicular to the axis 
 OX. Let the distance of any slice from the 
 center be x. Then radius of slice = Vr^ — x^, and volume 
 of slice = 7r(r' — z') Ja;. 
 
 hi 
 
 X 
 
S98 DVlf AMtCS of ROtATtOl^. [§ U1 
 
 The moment of inertia of the slice about OX is (Ex. 15) 
 
 Summing up for the whole sphere, that is, integrating be • 
 tween the limits x = r and x = — r, we have 
 
 = 2 Wry 5, 
 since weight of sphere = 4d;rry3.] 
 
 24. Find the /of a sphere, radius r, about a tangent line. 
 
 Ans. 1= r(2rV5 + r') = 7 Wr'/b. 
 
 25. Show, by differentiating the result of Ex. 15, that the 
 moment of inertia of an indefinitely thin ring of radius r 
 about an axis through its center and perpendicular to its 
 plane is Wr^. 
 
 26. Show, by differentiating the result of Ex. 23, that the 
 moment of inertia of an indefinitely thin spherical shell about 
 a diameter is 2 W^r^/'S. 
 
 27. Find the / of an elliptical plate about (1) its major 
 axis, (2) its minor axis, and (3) about an axis through its 
 center and perpendicular to its plane. 
 
 Ans. WV/i; Wa^/^-, W{a' ^¥)J^. 
 
 28. Find the / of a right cone, height li and radius of 
 base r, about an axis through the vertex parallel to the base 
 (solution similar to that of Ex. ll:!). 
 
 Ans. / = 3Tr(r' + 47i'')/20. 
 
 29. Find the / of a right cone, height li and radius of base 
 r, about an axis through its center of gravity and parallel to 
 the base. ^ Ans. 7=3 W^r" + /iV4)/20. 
 
 30. Find the /of a right cone, height li and radius of ba?e 
 r, about its own axis. Ans. I— ZWr'^/iO. 
 
 31. Prove that / is the same for all parallel axes situated 
 at equal distances from the center of gravity. 
 
 32. Of all parallel axes the / about that which passes 
 through the center of gravity is the least. 
 
 247. Radius of Gyration. — The general expression for the 
 moment of inertia of a series of jDarticles r.gidly connected 
 and weighing ^^, , w,, . . . about an axis situated r, , r,, . . . 
 from the particles is ^^^r^ The whole series of particles 
 
§ 24d] HADIITS OP GYRATIOlT. ^9d 
 
 forms a body weigliiiig "^tu or W. We may conceive the 
 body concentrated into a single particle weighing PF, and a 
 distance h from the axis may always be found, so that the 
 moment of inertia of the particle W about the axis is equal 
 to the sum of the moments of inertia of the separate particles 
 of the body, or 
 
 Wk'' = :Siur\ 
 
 To this distance k the name radius of gyration is commonly 
 given. 
 
 248. If the axis passes through the O.G. of the body, we 
 may write 
 
 Wlc' = 2tor' 
 
 when k is called the principal racliiis of gyration. 
 
 The relation between k and k follows at once. For (Art. 
 246) 
 
 2wr^ = 2wr -j- TF7i% 
 
 or Wk' = WV 4- Wi\ 
 
 or k^ = V -\- h\ 
 
 the relation sought. 
 
 249. Reduced Weight. — We may write 
 
 /= wjc:, 
 
 where k^ is any assumed distance from the axis of rotation 
 and W^ — Ilk^\ that is, we can replace the rotating body by 
 an equivalent particle of weight W^ at any distance ^•, from 
 the axis of rotation by dividing the moment of inertia of the 
 body about the axis of rotation by the square of the assumed 
 distance. The weight of this particle is called the reduced 
 weight. 
 
 The reduction is of frequent application in mechanisms 
 where many pieces have to be considered. For a simple illus- 
 tration take the case of a fly-wheel not properly centered. 
 
300 DYNAMICS OP ROTATION. [§ 250 
 
 Art. 238; also the compound pendulum, Art. 251. A more 
 complicated example will be found in Weisbach^s Hoisting 
 Machinery, p. 143. 
 
 250. The results in the following table for the square of 
 the radius of gyration about an axis through the center of 
 gravity should be carefully checked. 
 
 1. Straight rod, length l, 
 
 axis _L rod F = Zyi2 
 
 2. Rectangular lamina, breadth h, depth h, 
 
 axis II side h F = 5712 
 
 axis _L plane P = {¥-\-h')/l2 
 
 3. Square, side a, _ 
 
 axis a diagonal h"^ = ayi2 
 
 axis _L plane k^ = a'/S 
 
 4. Circular disk, diameter d, ^ 
 
 axis a diameter k^ = d^/16 
 
 axis _L plane F= dys 
 
 5. Circular ring, diameter d, _ 
 
 axis a diameter F = d^/8 
 
 axis J_ plane F = c?y4 
 
 6. Triangular lamina, base h, height h, _ 
 
 axis parallel base ^' = h'/18 
 
 7. A sphere, diameter ^7, _ 
 
 axis a diameter ^•' = ^'/IQ 
 
 8. A spherical shell, diameter d, _ 
 
 axis a diameter k^ = d^/6 
 
 9. Rectangular prism, sides a, b, c, _ 
 
 axis _L face Ic k' = (5'+c')/12 
 
§ 251] COMPOUND PENDULUM. 301 
 
 [Since the prism may be conceived to consist of an infinite 
 number of plates, each of which has the same radius of gyra- 
 tion with respect to an axis through the centre and perpen- 
 dicular to their planes, the radius of gyration of the prism is 
 the same as that of any plate.] 
 
 10. Circular cylinder, length Z, diameter d, 
 
 axis the axis of the cylinder ^' = 6?y8 
 
 axis _L axis of cylinder F = t?V16+r/12 
 
 11. Right cone, altitude li, diameter of base dy 
 
 axis X to axis of cone Ic^ = Z{d^-\-h^)/80 
 
 axis the axis of the cone k^ = 3^^/40. 
 
 12. Hollow cylinder, inner diameter d^ , outer diameter d^ , 
 
 axis the axis of the cylinder F = (d^^-^d^^)/8 
 
 251. The subject proper is now resumed. It was inter- 
 rupted at Arts. 245, 246 by the discussion of moments of 
 inertia. 
 
 Compound Pendulum. — A most important case on account 
 of its applications is that of a body oscillat- 
 ing about a horizontal axis under the action 
 of gravity. 
 
 Let G be the C.G. of the rotating body, 
 Wthe weight of the body, C the axis of ro- 
 tation, and 00 the angular velocity. 
 
 Let the distance CG = h, and the angle 
 of swing = 6 Q.t the instant considered. 
 
 The external forces acting are the weight 
 W at Gy and the reaction of the axis which 
 may be resolved into horizontal and vertical components X 
 and Y. 
 
 The effective forces on a particle w, distant r, from the 
 axis are w,r,fyy^ along r, ,and v)^r^a/g perpendicular to n 
 (Art. 237). The resultant of these effective forces acting at 
 G 18 evidently 
 
 Whoo^/g along CG and Wha/g J. CG, 
 
302 DYNAMICS OF ROTATION. [§ 251 
 
 Hence resolving horizontally, vertically, and taking moments 
 about (7, 
 
 Wha cos H/g - WJigj' sin 0/g = X, . , , . (1) 
 
 Wha sin 6/g + WJigo' cos 6/g= Y-W, . . (2) 
 
 lex/g - WJi sin = 0, . . . . (3) 
 
 the equations of motion. 
 
 The third equation gives the angular acceleration 
 
 a = Wgh sin O/I 
 = g sin 8/1, 
 
 if I/W7i is denoted by I 
 
 But the linear acceleration « at a distance I from C is la. 
 Hence 
 
 a = la 
 
 = g sin 6. 
 
 Now (Art. 115) this is the equation of motion of a simple 
 pendulum oscillating under the action of gravity. 
 
 Hence the angular motion of the body about the axis is the 
 same as that of a simple pendulum under the same initial cir- 
 cumstances and whose length is /. This length I or I/Wh is 
 called the length of the simple equivalent pendulum, and the 
 oscillating body is called a compound pendulum. 
 
 (a) Thne of Swing, — The time t of an oscillation (single 
 swing) will be given by (Art. 115) 
 
 t= nS/lfg 
 
 = 7rVl/Whg, 
 or, as it may be written, 
 
 t = 7tV(k'-i-h')/hg, 
 
 since /= Tf(F + h""), the first term being the moment of in- 
 ertia about C, and h the distance CG, 
 
§251] 
 
 COMPOUND PENDULUM. 
 
 303 
 
 {b) Principle of Reversion, — A point i) at a distance I from 
 Cy the point of suspension, is called the centre of oscillatio?i, 
 for the reason that the time of oscillation of the whole pen- 
 dulum is the same as tliat of a simple pendulum of length I 
 and swinging about 0. Denote the 
 distance BG by /*, , so that h -{-h^ = l. 
 
 Suppose now the pendulum in- 
 verted, and suspended from D instead 
 of from C. Now 
 
 CI) = OP 
 
 = (F + h')/h 
 = h + k'/h 
 and CG = 7i. 
 /. GD = P/h. 
 Hence CG . GD = k\ 
 
 This may be written 
 I)G.GG=k'; 
 
 which shows that if D is made the center of suspension, C 
 will become the centre of oscillation, as was first pointed out 
 by Huygens. 
 
 Hence the points of suspension and of oscillation can be 
 interchanged without changing the time of oscillation, and 
 appropriately therefore a pendulum with points of suspension 
 situated as C, D is known as a reversion pendnlum. 
 
 (c) Determinalion of g. — The pendulum furnishes one of 
 the most accurate methods of determining g^ the acceleration 
 due to gravity at the earth's surface. We have for the time 
 of an oscillation 
 
 t = n ^/TTg^ 
 whence g = ?r^l/t^. 
 
 If now t be observed, and / the length of the simple equiva- 
 lent pendulum, can be found, we may at once compute g. 
 
304 DYN-AMICS OF ROTATIOl^. [§ 251 
 
 Borda in 1792 constructed a pendulum as like a simple 
 pendulum as possible, of a platinum sphere suspended by a 
 very fine wire. Here, if r is the radius of the sphere 
 
 P = 2rV5, 
 and hence g = 7t'{li + 2ry5^)/r. 
 
 The time of swing t was observed, and the distance h from 
 the point of suspension to the center of the sphere measured. 
 Hence g is found. 
 
 The first pendulum constructed on the Huygens' principle 
 of reversion was Kater^s in 1818. Other forms by Repsold, 
 Mendenhall, etc., with methods of finding I and t, are described 
 in books of laboratory physics. 
 
 {d) Pressure on the Axis of Support. — Eeturning to the 
 equation of motion, we note that there are three equations 
 but four unknowns, X, Y, go, a. A fourth relation is afforded 
 by the pendulum motion (Art. 255), 
 
 iGoy^g = Wh(cos 6 - cos y5). 
 
 Hence X, Y may be computed. 
 
 For example, the pressure of a large bell when swinging. 
 
 Ex. 1. A rod of length I is suspended at one end and 
 caused to oscillate. Find the length of the equivalent simple 
 pendulum. Ans. 2//3. 
 
 2. A thin circular ring, diameter f/, and a small ball, sus- 
 pended by a fine thread of length d, are caused to oscillate 
 about a horizontal axis perpendicular to the plane of the 
 ring. Show that the two will oscillate together. 
 
 3. A circular disk IG in diameter makes small oscillations 
 about a horizontal tangent. Find the length of the equiva- 
 lent simple pendulum. Ans. lO in. 
 
 4. A sphere 20 in diameter makes small oscillations about 
 a horizontal tangent. Find the depth of the center of oscil- 
 lation below the axis. Ans. 14 in. 
 
 5. A rod 1 ft long is suspended from a point 3 in from one 
 end. Find the time of a small oscillation. 
 
 Ans, n^ljVlg sec. 
 
§ 251] COMPOUND PENDULUM. 305 
 
 6. In a compound pendulum the time of oscillation is a 
 minimum when h = k. 
 
 7. A sphere of radius r oscillates about a horizontal tan- 
 gent. Find the length of the equivalent simple pendulum. 
 
 Ans. 7r/5. 
 
 8. A cube of edge a oscillates about an edge. Find the 
 length of the equivalent simple pendulum. Ans, 2V2a/^. 
 
 9. A spoon of length r slips into a hemispherical bowl of 
 radius r. Show that the spoon will oscillate in the same 
 time as a pendulum of length 5?'/34/3. 
 
 10. A cylindrical disc of radius r and weighing IF lb rolls 
 without sliding down an inclined plane from rest under 
 the action of gravity. Determine the 
 
 motion. 
 
 The external forces are W pounds at 
 C vertically downwards, the reaction iV 
 normal to the plane, and the friction/ 
 up the plane. Unless there were fric- 
 tion the disc would slide. 
 
 The point of contact is the instan- 
 taneous center of rotation. Then, a 
 being the linear acceleration of tlie center of gravity C, the 
 resultant effective force along the plane is Wa/g. 
 
 Resolving along the plane, normal to the plane and taking 
 moments about C, the three equations of motion are 
 
 Wa/g = Wsm e -f; (1) 
 
 iV^=rcos^; (2) 
 
 fr = Ia/g (3) 
 
 Also I=Wry2; (4) 
 
 and since the disc rolls without sliding, 
 
 a = ra (5) 
 
 Hence we find 
 
 a — 2g sin ^/3; a = 2g sin f^J%r. 
 
306 DYNAMICS OF ROTATION-. [§ 252 
 
 The velocity v at any distance s from the starting-point is 
 given by 
 
 v'' = 2as 
 
 = 2s X 2g sin 6'/3 
 = 4//V3, 
 
 if h is the height of the starting-point above the point in 
 question. 
 
 The force of friction / = la/gr 
 
 = W sin 6^/3. 
 
 The reaction iV^ =W cos 6. 
 
 The angular acceleration a may be found at once by con- 
 sidering the disc as rotating for the instant about the instan- 
 taneous center of rotation 0. 
 
 Taking moments about 0, 
 
 Wr sin 6 = la/g — 3 Wr'a/lg, 
 and a — lg sin 6*73 r, as before. 
 
 If both rolling and sliding occur we cannot write a = ra, 
 but have the relation/ = /iJv^ instead, yu being the coefficient 
 of friction. 
 
 If the body slide without friction, ct = g sin 6. 
 
 11. A spherical shot rolls down a plane 70 ft long and in- 
 clined at 30° to the horizon. Find its velocity at the bottom. 
 
 Ans. 40 ft/sec. 
 
 12. A sphere will roll and not slide down an inclined plane 
 if the coefficient of friction is greater than 2 tan a/7 where 
 a is the inclination of the plane. 
 
 13. A ball is rolled up a 1^ incline with an initial velocity 
 of 4 ft/sec. How far will it run ? Atis. 37.5 ft. 
 
 14. Show that an empty keg will roll down an incline 
 slower than the keg filled solid — as with nails or sand. 
 
 252. Initial Motion Due to Impulse. — The impulse given to 
 a particle at rest is measured by the momentum of the par- 
 ticle. Eeasoning as in Art. 241, if an impulse be given to a 
 system the internal actions and reactions of the particles will 
 on the whole balance, and the external impulse will therefore 
 be measured by the momentum of the system. Considering, 
 
§ 252] IMPULSE. 307 
 
 then, the system in equilibrium under the external impulse 
 and the momentum reversed, the conditions of Art. 140 may 
 be applied. 
 
 In general, instantaneous motion of a body may be repre- 
 sented by a rotation about the instan- 
 taneous axis or by a translation and a 
 rotation about a fixed axis of the 
 body. Let the impulse I communi- 
 cated to the rod OGChe resolved into 
 components i^;, \y parallel to rectangu- \ 
 lar axes OX, OY. /ox, x 
 
 Let P, be a particle of the rod weighing w,; a;, , y, the co- 
 ordinates of P, ; and r, the distance of P, from 0. Let go 
 denote the angular velocity of the system of particles com- 
 posing the body. 
 
 The iiiitiiil velocity of P, is (wr, ±to OP, (Art. 229). Its 
 components along OX, OF are = — ooy, and = oox, respect- 
 ively. The momenta of the particle in these directions are 
 — w^ooyjg and iv^ooxjg. 
 
 The first and second conditions of equilibrium give 
 1^ = - w.QoyJg - w.ooyjg - . . . 
 
 = - Wooy/g (I) 
 
 1^ = WGoi/g (2) 
 
 where x, y are the coordinates of the C.G. of the body. 
 
 The resultant impulse is therefore Woor/g where r = OG 
 and is J_ to OG. 
 
 Hence the motion of the C.G. is the same as if the impulse 
 acted at the C.G. 
 
 For the third condition of equilibrium take moments about 
 the axis 0; then, if ji is the perpendicular from in the di- 
 rection of I, 
 
 \p = w^Goy^'/g H- w^Gox^yg -f . . . 
 = w,oor^/g + tu^oor^yg + . . . 
 = oo'2(wr^)/g 
 = M9 (3) 
 
308 DYNAMICS OF KOTATION". [§ 252 
 
 when / is the moment of inertia about 0, 
 Hence 
 
 angular velocity go = moment of impulse /moment of inertia. 
 
 As in the case of continuous forces, it is, in general, most 
 convenient to take the C.G. as origin. 
 
 Ex. 1. If a rod CGO at rest and free to move receives an 
 impulse I at and perpendicular to OG, the line joining to 
 the C.G. of the rod, it is required to determine 
 the motion. 
 
 Let W=ihe weight of the rod, u =the linear 
 velocity oiG,oD=i the angular velocity about G, 
 and h = the distance GO. 
 The equations of motion are: 
 
 r \ = Wu/ff; (1) 
 
 \h = loDJg 
 
 = WFGo/g (2) 
 
 Hence u and go are found. 
 
 The velocity of a point C on GO = u -\- velocity about G 
 
 z=U — GOX OG, 
 
 If the rod begins to move about 0, 
 
 = u — GOX OG, 
 and CG = u/go 
 
 = 'kyGO; 
 or CG.GO = k\ 
 
 The point is named the center of percussion, and the cor- 
 responding axis the axis of spontaneous rotation. 
 
 If I is the length of the rod, then k"" = r/12 and the center 
 of percussion is at a distance 2//3 from 0. 
 
 "A knowledge of the center of percussion, gained instinc- 
 tively or otherwise, enables the workman to wield his tools 
 with increased power, and gives greater force to the cut of 
 the swordsman^ so that with some physical strength he may 
 
§ 253] ENERGY OF MOTION. 309 
 
 perform the feat of cutting a slieop in half or severing, a la 
 Richard the Lion-liearted, an iron bar." 
 
 2. A rod of length / and free to rotate about 
 a fixed axis (J receives an impulse I perpendicu- 
 lar to the rod at 0, distant c from 6'. Find 
 the impulse on the axis. 
 
 A71S. |{ 1 - ch/(]i' + k')] where h = CG. 
 Hence deduce example 1. 
 
 3. A rod is suspended by two vertical < 
 threads at its ends. One thread is suddenly 
 cut. Show that the initial pull in the other is 
 halved. 
 
 4. A rod AB is swinging about the end J, and when hori- 
 zontal the end B is fixed. Show that the jerks at the ends 
 are as 1 to 2. 
 
 5. At what point must a rod AB & yard long be struck per- 
 pendicularly that one end A may be initially at rest ? 
 
 Ans. 2 ft from A, 
 
 263. Energy of Motion. — The kinetic energy of a body in 
 motion may be found for any insta7it by considering the 
 motion as given by an angular velocity about the axis of 
 motion, fixed or instantaneous. 
 
 Let 00 radians/sec be the angular velocity about the axis 
 and r ft the distance of any particle lu lb from the axis. 
 Then, if v ft/sec denote the linear velocity of the particle at 
 the given instant, 
 
 energy of the particle = wv*/2g 
 
 = tvoo^r^/2g foot-pounds. 
 
 Hence, by summing up the energies of all the particles and 
 noting that oo is the same for each particle, we have 
 
 energy of the body = 2wGo'r'*/2g 
 
 = (k)''2tvry2g 
 
 = lGo^/2g foot-pounds 
 
 where /is the moment of inertia about the axis of motion. 
 
 Take, for example, a railroad-car while in motion. The 
 wheel may be considered a disk of radius r ft. If its weight is 
 
310 DYlfAMTCS OF ROTATION. [§ 254 
 
 W lb, the moment of inertia about the point of contact with 
 the rail [the instantaneous center] is 3 H>y2. 
 
 Also, if the speed of the car, and therefore of the center of 
 the wheel, is u ft/sec, and oo radians/sec the angular velocity 
 of the wheel, then u = oor. 
 
 Hence energy of wheel = loa^/^g 
 
 = dWr'oDy4:g 
 
 = 3 Wic^'/i^g foot-pounds. 
 
 254. In finding the energy of motion it is in many cases 
 more convenient to refer to the center of gravity than to the 
 center of rotation. 
 
 Let p denote the distance between the center of gravity 
 
 and the axis of rotation, and / the moment of inertia about 
 the parallel axis through the center of gravity. Then (Art. 
 246) 
 
 I=T-{-Wp\ 
 
 and therefore 
 
 lGDy2g = lod'm + Tf/G?V2^. 
 
 But poo = the linear velocity of the center of gravity. Call 
 it u. Then 
 
 energy of motion = loo'' fig + Wu"^ /^g- 
 
 Hence the energy of motion is equal to the energy of rota- 
 tion about the center of gravity and the energy of translation 
 of the body moving with the linear velocity of the center of 
 gravity. 
 
 Thus in the preceding example I = Wr^/2, and 
 
 energy of motion = iWr''GDy2g + Wuy2g 
 = 3Wuy^g, 
 
 since u = aor, as found before. 
 
§ 254] t:Nt:RGY of motIoK^. 311 
 
 Ex. 1. A spherical disk (as a coin) rolls on its edge in a 
 vertical plane. Compare the rotation energy with the total 
 energy of motion. A7is. 1/3. 
 
 2. A hoop rolls in a vertical plane. Show that the energy 
 of rotation is 1/2 tlie total kinetic energy. 
 
 3. A loaded car weighs 40,000 lb, the eight wheels 4000 lb, 
 and the speed is 30 miles an hour. Find the kinetic energy 
 stored. Ans. 1,391,500 foot-pounds. 
 
 4. If / is the moment of inertia of a fly-wheel and n the 
 number of revolutions per second, then 
 
 energy of rotation = 5wV/8. 
 
 5. If the weight of the wheel in (4) is W lb and the greater 
 part is contained in a ring whose mean diameter is d ft, then 
 
 energy of rotation = 5d^n^ W/32 foot-pounds, 
 
 a working rule. 
 
 6. A fly-wheel weighs 15 tons, and its diameter is 20 ft; the 
 wheel makes 60 revolutions per minute. Find the energy 
 stored. Ans. 1,875,000 ft-pounds. 
 
 7. The axle of the wheel is 14 in in diameter. If the 
 wheel is disconnected, how many revolutions will it make 
 before coming to rest, the coefficient of friction being 0.8 ? 
 
 A71S. 21.31. 
 
 8. In order to control an engine against its own variations 
 and for external work it is necessary to call upon the fly-wheel 
 for 60,000 foot-pounds, and at tlie same time a change of ve- 
 locity from 160 to 140 revolutions per minute is allowable. 
 The wheel is to be 10 ft in diameter. Compute its weight. 
 
 Ans. 1.2 tons. 
 
 9. Find the energy of rotation of the earth, considering it a 
 uniform sphere of density 5.6 and of diameter 8000 miles. 
 
 Ans. 10''y5 foot-pounds. 
 
 10. The energy of a sphere rolling without sliding along a 
 plane is to that of an equal sphere sliding without rolling 
 and with the same velocity of the center of gravity as 7 to 5. 
 
 [Energy of rolling = Ico*/2g 
 
 = 7 Wr'oDyiOg, 
 the instantaneous axis being tangent to sphere and plane. 
 Energy of sliding = Wv^/'Zg, 
 
312 DYKAMICS OF KOTATION. [§ 255 
 
 Also, V = Gor, 
 
 Hence the result.] 
 
 11. The kinetic energy acquired by a sphere in moving from 
 rest down a smooth plane is to that acquired by an equal 
 sphere rolling without sliding down a rough plane of the 
 same inclination and length as 7 to 5. 
 
 12. The weight of a fly-wheel is W lb, and its diameter d^ 
 inches. If it is making n revolutions per minute, find in how 
 many revolutions it will be stopped by the friction of the axle 
 if its diameter is d^ inches and the coefficient of friction //. 
 
 A71S. 7rd'ny8Q400Md.^g. 
 
 13. Examine the following statement: "Every engineer 
 knows that a thing so balanced as to stand in any position is 
 not necessarily balanced for running; that a 4-lb weight at 3 
 in from the axis of rotation thougn balanced statically by a 
 1-lb weight at 12 in from the axis is not balanced by it dy- 
 namically. On the contrary, a 4-lb weight at 5 in is balanced 
 by a 1-lb weight at 10 in from the axis.^' 
 
 255. Equation of Energy. — When forces act on a system 
 of particles, the change of kinetic energy is equal to the work 
 done by the forces. But in the case where the particles are 
 rigidly connected the internal forces, being equal and op- 
 posite, do no work on the system as a whole. Hence the 
 change in kinetic energy is due to the external forces only, 
 and is equal to the work done by them. 
 
 If, then, F is the resultant external force and s the dis- 
 placement of its point of application, the equation of en- 
 ergy is 
 
 lGDy2g=Fs: (1) 
 
 If the force is not constant, then at any instant we have 
 
 d{lGoy2g) = Fds o (2) 
 
 as the expression of the equation of energy. 
 Expanding (2), 
 
 loodoo/g = Fds = Fjjdd 
 
 when dO is the angle of rotation and p is the distance of F 
 from the axis. That is, 
 
 i<^/g = Fp, (3) 
 
§ 255] EQUATION OF EKERGY. 813 
 
 the same equation as found in Art. 241 by D'Alembert's 
 principle. 
 
 Compare with the various forms of statement of Newton's 
 second law in Art. 67. 
 
 Ex. 1. To find the time of oscillation of a compound pen- 
 dulum (Art. 251). 
 
 Let be the point of suspension, OA the 
 vertical through 0, G the position of the center 
 of gravity when the pendulum is at the end of 
 its swing, 6r, the position of G when Z GfiA 
 = 0, and CO the angular velocity there. 
 
 Denote Z GO A by ^ and let OG^ = h. 
 
 The work done by gravity as G moves to (?, 
 
 = Wxab 
 = Wh{cos e-coa /3), 
 
 The kinetic energy acquired = lod'/lg. 
 
 Hence lod'/'Zg = Tf/i (cos 6 — cos /3), 
 
 and 00^ = 2 Wgh(co8 6 - cos /?)//, . . (1) 
 
 which gives the velocity in any position. 
 
 For a simple pendulum of length I and weight of bob Tfwe 
 have, since I=Wr, 
 
 Wroay^g = Wl(cos d - cos /?), 
 
 or G9' = 2^(cos d - cos /?)// (2) 
 
 Comparing the two equations (1) and (2), we see that if a 
 simple pendulum of length I/Wh be set oscillating simulta- 
 neously with the compound pendulum, the two will have the 
 same angular velocity for the same angle 6, and therefoi-w 
 the same time of oscillation. Hence the time of oscillation, 
 being that of the simple pendulum, is given by (Art. 115) 
 
 t = 7t Vlfg ^ 7t VT/W^i, 
 as found before (Art. 251). 
 
314 
 
 DYNAMICS OF ROTATION. 
 
 [§ 265 
 
 'Bi 
 
 2. Find the initial angular velocity of a simple pendulum 
 of length I that it may swing through a given angle fi. 
 
 A71S. GO = 2Vg/l sin i/3. 
 
 3. Find the initial angular velocity of a pendulum 39 in 
 long so that it shall swing through 30°. 
 
 Ans. 00 = 1.62 radians/sec. 
 
 4. A horizontal rod of length 2a hangs by two parallel 
 strings, each of length hy attached to its 
 ends. If it be twisted horizontally through 
 a small angle and allowed to oscillate, it is 
 required to find the time of an oscillation. 
 
 This is the problem of hifilar suspension, 
 and is of great importance in physical and 
 electrical investigations. 
 
 Let the angle of displacement of AB = 0, 
 and let the angle of displacement of the 
 thread from the vertical due to moving AB 
 = /?. Then ^' 
 
 aO = arc AA^ = /i/3 (1) 
 
 The height to which AB is raised in twisting through angle 
 6 is h — h cos /3. 
 
 The work done in raising AB = ]VJi{l — cos fi). 
 
 Hence lo^V^g = Wh{l - cos /3) 
 
 = Wh/3y2 (if /? is small) 
 = W7ia'6y2h\ from (1), 
 
 or go' = Wga'd'/Ih. 
 
 Tor a simple pendulum of length I and weight of bob W \ve 
 have, if the angle of displacement is & (Ex. 1, p. 313), 
 
 d.-' 
 
 go' = 2^(1 - cos 0)/l 
 
 = goyi 
 
 Hence 
 
 length of simple equivalent pendulum I = Ih/Wd', 
 
 and time of oscillation . = ^ V^/g 
 
 = 7t Vlh/Wa'g. 
 
§ 255] ENERGY OP MOTIOK. 315 
 
 Conversely, if the time of an oscillation be noted, the mo- 
 ment of inertia of the bar may be computed from this rela- 
 tion. This indeed is the main problem of bifilar suspension. 
 
 5. The bob of a pendulum consists of a box filled with 
 sand. The pendulum is deflected 
 from rest through an observed 
 angle 8 by a shot tired horizontally 
 and striking the box at a point P 
 when OP = x. If the pendulum 
 weighs ^Flb, the shot w lb, and the 
 distance OG = h where G is the 
 center of gravity, it is required to 
 find the velocity v with which the 
 shot strikes in terms of these observed quantities. 
 
 Ans. V = {W -\- 2v)JiV2gh{l — cos &)/wx ft/sec. 
 
 This instrument, called the ballistic pendulum, intro- 
 duced by Benjamin Robins in 1742, and at one time largely 
 used for finding the velocities of cannon-balls and rifle- 
 bullets, is now almost entirely discarded for more accurate 
 methods depending upon electric contacts. 
 
 5^. If there is no impulse on the axis, the shot must strike 
 at a point P such that OP = 1/ Wh. 
 
 [P is the center of percussion. (See Ex. 1, p. 308.)] 
 
 6. A disk of radius r and weight W rolls down an inclined 
 plane of height h under the action of gravity. To find the 
 velocity at the bottom. 
 
 Let 21 = velocity of center of gravity down the plane; 
 
 GO = the angular velocity. 
 Then 
 
 kinetic energy acquired = work done by gravity, 
 
 or TGjy2g + Wuy2g= Wh, 
 
 But 7= Wry 2, u = TOO. 
 
 .*. w" = 4^/^/3, as found before (p. 306). 
 
 This may be solved at once by referring to the instanta- 
 neous center of rotation instead of to the C.G. of the disk. 
 In this case 
 
 l(^y2g = Wh, 
 
 and 7=3 H>y2. 
 
 /. 21^ = -igh/S, as above. 
 
316 DYKAMICS OF noTATIOK. [§ 265 
 
 7. A solid sphere and a solid cylinder of equal radii roll 
 from rest down the same inclined plane. Compare their 
 times of descent. Ans. V\^/V\b. 
 
 8. A spherical disk of radius r rolls down an equal fixed 
 disk under the action of gravity. Show that the velocity 
 acquired when the line joining their centers makes an angle 
 6 with the vertical is found from 
 
 ^v' = Sgr{l - cos 6). 
 
 9. Where should a stop be placed behind a thin vertical 
 door, 6 feet high and 2.5 feet wide ? 
 
 Ans. 10 in from outer edge and 3 ft from top. 
 If the stop is placed on the floor (as with a railroad-car 
 door), account for the " twist." 
 
 10. If the diameter of Sisyphus' spherical stone be 2 ft, 
 which he continually rolls up the surface of a semi-globular 
 mountain half a mile high, what distance will the stoue have 
 rolled down under the force of gravity when it leaves the 
 mountain. Ans. 1088 ft. 
 
 How far from the foot of the mountain does the stone fall? 
 
 EXAMINATION. 
 
 1. When has a moving body a motion of translation ? 
 [When all points describe equal and similar paths.] 
 
 2. When has a moving body a motion of rotation ? 
 [When it has two points fixed.] 
 
 3. The motion of a particle in a plane may be represented 
 by a radial velocity dr/dt and a transverse velocity rdO/dt. 
 
 4. It is always possible to represent the plane motion of a 
 body at any instant by a motion of rotation about a certain 
 point. 
 
 5. Define the instantaneous center of rotation. 
 
 6. Define angular velocity and explain how it is meas- 
 ured. 
 
 7. Find the angular velocity of the extremity of the min- 
 ute-hand of a clock. Ans. tt/ISOO rad/sec. 
 
 8. A body makes 30 turns per minute. Show that its 
 average angular velocity is tt. 
 
§ 255] EXAMINATION. 317 
 
 9.. The angular velocity of a body in latitude A = ;r/12 
 radians/liour. 
 
 10. Show that 1 rev/min = 0.10472 radians/sec. 
 
 11. A particle starts from rest and moves in a circle with a 
 uniform acceleration of 8 radians/sec'. Find the ti me o f com- 
 pleting the first revolution. A7is. ^/n/l sec. 
 
 12. If a particle weighs w lb, the center-seeking force re- 
 quired to keep it moving with uniform angular velocity oo in 
 a circle of radius r is luroo'^/g pounds. 
 
 13. Show how to combine two angular velocities about par- 
 allel axes. 
 
 14. Give the unit of angular acceleration. 
 
 15. A Foucault pendulum is set up at the north pole. 
 Find the time of u complete revolution. Ans, 1 day. 
 
 If set up at the equator, how then ? 
 
 16. If a disk rolls on a plane, the velocity of translation of 
 its center is equal to the product of its angular velocity about 
 its center and its radius. 
 
 17. Show that the activity or power of a rotating arma- 
 ture is equal to torque X angular velocity. 
 
 18. Define the terms moment of inertia and radius of 
 gyration. 
 
 19. State two propositions which abbreviate computations 
 of moments of inertia. 
 
 20. The moment of inertia of a lamina about a central 
 axis perpendicular to its plane is 
 
 weight X sum of sqs of _L semi-axes/3, or 4, 
 
 according as the lamina is a rectangle or circle. 
 State the corresponding proposition for the sphere. 
 
 21. Show that the moment of inertia of a thin hollow 
 cylinder, 2 ft diameter and weight 1 lb, about the axis of the 
 cylinder is the unit moment of inertia. 
 
 22. The kinetical behavior of a body cannot be determined 
 until we know the value of the radius of gyration of the body. 
 
318 DY^TAMICS OF ROTATION-. [§ 255 
 
 23. State D'Alembert's principle and give an example of 
 its application. 
 
 24. Show how to find the angular acceleration of a body 
 about a fixed axis when the body is acted upon by an external 
 force. 
 
 25. Prove moment of impulse = moment of momentum. 
 26a. A circular cylinder with its axis horizontal rolls down 
 
 an inclined plane. Show that one third of the acceleration 
 of gravity is used in turning the cylinder. 
 
 26b. If the inclination of the plane is 30°, the distance 
 passed over by the cylinder in 6 seconds is 6g. 
 
 27. A sphere rolls down a plane of inclination 6. Show 
 that the motion of the center parallel to the plane is that of a 
 particle moving with a uniform acceleration 5g sin 6/7. 
 
 28. Show that change of momentum about a fixed axis is 
 equal to the moment of the impressed forces about the axis. 
 
 29. When is a machine said to be balanced P 
 
 [When the relative movements of its parts do not tend to 
 make it vibrate as a whole.] 
 
 30. Apply the principle of the conservation of energy to 
 find the time of swing of a compound pendulum. 
 
 31. What is meant by the center of oscillation ? center of 
 percussion ? 
 
 32. Find experimentally the center of percussion in a 
 baseball-bat. 
 
 33. Where is the centre of percussion in a hammer ? 
 
 34. A foot rule is held lightly at one end between finger 
 and thumb so as to hang vertically. If struck 4 inches from 
 the lower end, there is no pull on the fingers. If struck at 3 
 or 6 inches, what happens ? 
 
 35. Why is the lighter end of a baseball-bat held in the 
 hand ? 
 
 36. A semicircular wire, length Z, is bent at its middle 
 point into two quadrants having a common tangent. It os- 
 cillates about an axis through this middle point. Find the 
 length of the simple equivalent pendulum. 
 
 Ans, Z(;r — 2)/;r. 
 
§ 255] EXAMINATION. 319 
 
 37. In ringing a bell the swing is through an angle 6. 
 Compute the pressure on the supports. 
 
 38. Explain how to determine the moment of inertia of a 
 bar by using bifilar suspension. 
 
 39. An armor-plate suspended by chains is struck normally 
 by a shot at a point P vertically below the C.G. of the plate. 
 Find the axis about which the plate revolves. 
 
 40. A fly-wheel of a tons weight and b ft diameter makes 
 c revolutions per minute. Find the energy accumulated. 
 
 Ans. 0.087«^V ft-pounds, nearly. 
 
 41. A fly-wheel is not perfectly centered. How would you 
 compute the centrifugal force ? 
 
 42. "The outside diameter of an engine fly-wheel is 80 in, 
 width of face 26 in, average thickness of rim 5 in, revolutions 
 per minute 175 Show that the centrifugal force of the rim 
 is 260,136.35 pounds." (Exam, paper.) 
 
 [If the wheel is properly balanced, is not the centrifugal 
 force nilf'\ 
 
 43. The work stored in a fly-wheel is quadrupled if the 
 angular velocity is doubled. 
 
 44. "There is more energy stored in a ton of oaf- wheel 
 than in a ton of car-body." {R. R. Gazette^ 1892.) 
 
 How much more when the speed of the train is 30 
 miles/hour ? 
 
320 • ELASTIC SOLIDS. [§ 256 
 
 CHAPTEE VIII. 
 ELASTIC SOLIDS. 
 
 256. Observation and experiment show that forces acting 
 on a body may change its motion, its form, or its size. As 
 most simple, we have considered first of all changes of motion 
 only. The action of the external forces was conceived to be 
 resisted by the internal reactions of the particles on one an- 
 other in such a way that the particles retained their original 
 distances from one another, so that changes of form or of size 
 did not take place. The conditions of equilibrium and of 
 change of motion on this hypothesis have been developed in 
 the preceding chapters. 
 
 Experience shows that no perfectly rigid body exists in 
 nature; the body yields to the external forces, and the inter- 
 nal reactions do not prevent changes of form or of size. If 
 the body returns towards its original configuration on the re- 
 moval of the forces it is said to be elastic. 
 
 257. Conceive an elastic body under the action of forces to 
 be cut by a plane. The particles of the body act and react 
 on one another across the plane. The actions distributed 
 over one side of the plane may be regarded as combined into 
 one force whose place of application is the plane itself. The 
 action and reaction form a stress, though the term stress is 
 frequently used for either component. 
 
 The unit stress is unit force per unit area; as, for example, 
 1 pound per square inch, written 1 pound/in^ The average 
 stress over a surface (intensity of stress) is found by dividing 
 the total stress on the surface by the area of the surface. 
 
§ 259] hooke's law. 321 
 
 Ex. A wire of 0.1 in diameter is stretched by a 5 lb weight. 
 Find the stress across any section, neglecting the weight of 
 the wire. A7is. 2000/ tt pounds/in'. 
 
 258. When an elastic solid is subjected to the action of a 
 stress it suffers distortion. Any change of form or of size is 
 known as a strain.* Remove the stress and the solid returns 
 to its original dimensions. This is observed to be true for 
 stresses up to a certain amount. When that amount is ex- 
 ceeded the solid will not return to its original form on remov- 
 ing the stress, but will assume another form between the two, 
 or a permanent set, as it is called. Increase the stress, and 
 the solid will finally be ruptured. The limit of unit stress 
 up to which a body of unit cross-section may be subjected 
 without producing permanent sets is called the elastic limit 
 of the material in question. 
 
 Until this limit is reached experiment shows that strain is 
 proportional to the stress producing it. This is known as 
 Hooke's law, having been stated by Hooke in 1G78 under the 
 form Ut tensio sic vis. It is analogous to Newton's second 
 law of motion in that it connects stress and strain as the 
 second law connects force and motion. 
 
 259. The ratio of a stress to the strain which is produced 
 in any body is called a modulus of elasticity for that body. 
 
 (1) If the body is acted on equally in all directions by 
 stresses normal to its surface, the size is changed but not the 
 form, and the strain is a compression or an extension^ the 
 stress being compressive or tensile. Thus if V denotes the 
 volume when in the natural state, v the change of volume 
 [dilatation], and p the unit stress, then v/V\^ the unit strain, 
 and the ratio unit stress/unit strain is the modulus of elas- 
 ticity of vohime. 
 
 * lljinkine in 1850 introduced the term strain to denote the detinite 
 clianire in the size or shnj^e of a body produced by u stress. The term 
 iucbuies distortion, deflection, elongation, etc. 
 
 But the innovation has not been universally adopted. Many writers 
 continue to use strain in the sense of force and as synonymous with 
 stress. For example, architects and engineers use the term "strain- 
 sheet" rather than "stress sheet" for a diagram of stresses in structures. 
 
322 ELASTIC SOLIDS. [§ 259 
 
 It is usually denoted by the letter h, so that 
 
 h ^:^ p — v/V = 'p V/v. 
 
 It is evident that h is of the same name as unit stress; for 
 unit strain being the ratio of change of volume to original 
 volume is an abstract number. In the British system of units 
 k is usually expressed in pounds/in^ For steel, 
 
 ^ = 28 X 10" pounds/in'. 
 
 (2) If the external stress produces change of form but not 
 of size, it is called a shearing stress or shear 
 f* D p» and the distortion a shearing strain. 
 
 "^ / {a) Suppose the stress tangential. 
 
 / Let A BCD be a rectangular block on 
 
 the base BC, Conceive it built of infinitely 
 
 thin sheets placed on BC. It BC is fixed 
 
 and the sheets are moved in their own planes parallel to BC, 
 
 the block is displaced into the position A' BCD' say, and has 
 
 undergone a shear. 
 
 The unit strain is the difference of displacement of any 
 two planes divided by the distance between them. Thus 
 
 unit strain = AA^AB. 
 
 The shearing stress is parallel to BC. It p represent the 
 stress on unit area, then the ratio 
 
 unit stress/unit strain = p -^ AA'/AB, 
 = pXAB/AA' 
 
 is the modulus of elasticity of form., or the rigidity. 
 
 It is denoted by the letter n, so that n, like k, is expressed 
 
 in pounds/in^ For steel, ?i = 12 X 10' pounds/in''. 
 
 {h) Suppose the stress to twist the body about an axis. A 
 
 common example occurs in machine shafting, in which the 
 
 bodies twisted are circular cylinders, A similar case occurs 
 
§ 260] TORSION. 323 
 
 in many delicate physical instruments in which wires undergo 
 twist. 
 
 Consider a cylinder fastened at one end and acted on by 
 external forces. If the resultant of these forces is a couple in 
 a plane perpendicular to the axis, it will cause a twist about 
 the axis. This is balanced by the molecular reaction of the 
 particles of the body, giving rise to an equal and opposite 
 couple in a parallel plane. We therefore say that a cylinder 
 subject to two equal and opposite couples in planes perpen- 
 dicular to the axis is in a state of torsion. 
 
 If we suppose the cylinder built up of circular plates, we 
 may conceive them to slide upon one another in the twisting, 
 just as the leaves of a thick book when one cover is held 
 firmly and the other is pulled sideways. 
 
 Let the cylinder when deflected from its position of rest by 
 a torque T applied perpendicular to its length come to rest 
 when the angle of torsion is 6, Then it was shown by Cou- 
 .lomb experimentally that 
 
 T=he 
 
 where J is a constant, called the constant of torsion. 
 
 260. Consider a uniform wire, radius r, clamped at one 
 end and carryiug a weight attached to the free end. Let the 
 wire be deflected through an angle 6 from its position of rest 
 by the torque T. Then (Arts. 241, 259) 
 
 Ia= T=be 
 
 where a is the angular acceleration and / the moment of in- 
 ertia of the wire about the axis. 
 
 Hence a varies as 6 and ar varies as ^r, or the acceleration 
 of any point varies as the displacement of the point from the 
 position of no torsion. The motion of each point is therefore 
 a S.H.M., and the wire will oscillate about the position of no 
 torsion. The apparatus forms a torsion j)e7icltilum. 
 
324 ELASTIC SOLIDS. [§ 201 
 
 Hence the time of an oscillation is given by (Art. 115) 
 
 ^ = ;r Vdr/ar 
 = 7t V7/1. 
 
 261. The torsion pendulum may be used to determine the 
 constant of torsion. For let a weight of known moment of 
 inertia be suspended and the time of oscillation observed. 
 Then b may be computed. 
 
 Conversely, when h is known, the moment of inertia of any 
 suspended body may be computed. (See Ex. 7, p. 336.) 
 
 In many physical investigations the instruments employed 
 involve the use of apparatus suspended by a fiber and caused 
 to oscillate. Thus the Coulomb torsion balance is used for 
 the measurement of small forces and the Cavendish apparatus 
 for measuring the density of the earth. 
 
 Ill his classical researches (1894) on the constant of gravi- 
 tation, Prof. C. V. Boys found fibers of quartz to possess the 
 properties demanded in a suspension fiber to a greater degree 
 than any material hitherto used — metallic wires or silk fibers, 
 for example. 
 
 262. Let an elastic body be subjected to longitudinal com- 
 pression or extension. Let I denote the original length, A 
 the change of length, P the stress producing the change, and 
 A the cross-section. Then 
 
 longitudinal stress = P/A, 
 longitudinal strain = X/ly 
 and the ratio 
 
 long, stress/long, strain = P/A -r- \/l 
 
 is called Young's modulus of elasticity. 
 
 It is denoted by the letter E, so that 
 
 E = Pl/A\y 
 
§ 263] IMPACT. * 325 
 
 and is expressed in pounds/in'. For steel ^ = 30 x 10" 
 pounds/in*. 
 
 The work done by the unit stress in causing the strain A 
 is equal to the unit stress multiplied by the average strain 
 
 A/2, or 
 
 W = PA/2. 
 
 The work which a body can do in returning to its original 
 dimensions after it has been strained up to the elastic limit is 
 the ivork of resilience. 
 
 Ex. 1. A rod 0.1 in'' cross-section and 10 ft long is sus- 
 pended from one end. A weight of 1 ton is hung from the 
 lower end and the wire increases in length 0.1 in. Find the 
 unit stress. Ans. 20,000 pounds/in'. 
 
 Also show that the modulus of elasticity = 24,000,000 
 pounds/in^ 
 
 2. Can a steel rod 1/2 in X 1/2 in safely carry a load of 4 
 tons ? the elastic limit of steel being 50,000 pounds/in"" ? 
 
 Ans. Yes; the stress is within the elastic limit. 
 
 3. Show that Young's modulus may be defined as a stress 
 which would double the length of a bar of unit cross-section, 
 the bar remaining within the elastic limit. 
 
 4. A steel rod 50 ft. long and 2 in'' cross-section is stretched 
 1/25 in by a weight of 2 tons. Compute Young's modulus 
 for steel. Ans. 30 X 10" pounds/in^ 
 
 5. A steel bar 3 ft long and 2 in X 1 in section was sub- 
 jected to a tensile stress of 60 tons. The elongation was 0.05 
 m. Required the work of resilience. 
 
 Ans. 3000 inch-pounds. 
 
 6. A wrought-iron rod 50 ft long and 2 in' cross-section is 
 subjected to a pull of 25 tons. Taking the modulus of elas- 
 ticity = 30 X 10" pounds/in", find the elongation. 
 
 Ans. 0.5 inch. 
 The further development of this subject will be found in 
 treatises on the strength and elasticity of materials. 
 
 IMPACT. 
 
 263. In Arts. 53, 67 it was shown that the second law of 
 motion might be algebraically expressed 
 
 Ft-Wv, 
 
326 ELASTIC SOLIDS. [§ 264 
 
 i^ being in absolute units; or 
 
 Ft --= Wv/g, 
 
 F being in gravitation units. 
 In words, 
 
 the impulse imparted = the momentum acquired. 
 
 264. Impact of Two Bodies.— If two bodies come into con- 
 tact, a collision or impact is said to 
 take place. This impact is direct 
 J' — \ ^^^ ^^ ^^® bodies are moving in the di- 
 
 + ) y rection of the common normal at 
 
 V T^v.— ^ the point of contact; indirect if 
 
 they are not moving in this direc- 
 tion. 
 
 If the bodies are spheres, the 
 common normal is the line joining 
 their centers. 
 
 265. When two bodies impinge, 
 the impulse [action] received by 
 the one must be equal to the impulse [reaction] received by 
 the other. This follows from the law of stress. But the im- 
 pulses are measured by the momenta acquired. Hence, if by 
 the impact the velocity u ft/sec of the body weighing WVo 
 is changed to v ft/sec say, and the velocity u^ ft/sec of the 
 body weighing W^ lb is changed to v^ ft/sec, then 
 
 momentum lost by first = Wv/g — Wu/g second- 
 
 pounds; 
 
 momentum acquired by second = W^ujg — W^vjg second- 
 pounds. 
 Hence Wv/g — Wu/g = W^ujg — W^vjg, 
 
 or Wv/g-\-W,vJg = Wu/g -\-W,uJg', . . (1) 
 
 that is, the sum of the momenta after impact is equal to the 
 sum of the momenta before impact. 
 
§ 266] iMPACl*. 52'!' 
 
 In order to determine the unknowns v and v, we must have 
 another relation between them. Now it is found by experi- 
 ment that when two bodies impinge directly the difference of 
 velocities after impact bears a constant ratio to the difference 
 of their velocities before impact so long as the materials of 
 the bodies are the same, but these differences are in opposite 
 directions. If this constant ratio, which is called the coeffi- 
 cient of restitution of the two bodies, is denoted by the letter 
 e, we have 
 
 V — v^= — e{u — w,) (2) 
 
 as the second relation between v and v,. Solving (1) and (2), 
 we find 
 
 Wu + Wm^ — eWAu - uA 
 
 At — I 1 I Li Li. 
 
 _ Wu + W,u, -}-eW,(u- ti,) 
 ^»- W-^W, 
 
 giving the velocities after impact. 
 
 Also, the impulse i on the body TTis measured by the change 
 of momentum produced. Hence 
 
 I = W(u-v)/g 
 = WW,(l + e)(u - u,)/{W-\- W,)g. 
 
 The impulse on the other body is of course equal to this 
 and opposite in sign. 
 
 266. The value of the coefficient of restitution e depends 
 on the material composing the bodies. From its definition it 
 follows tliat the extreme values of e are and 1. If e = 0, or 
 the bodies are inelastic, then 
 
 V = v^, 
 
 or the bodies move together with a common velocity after im- 
 pact. If e = 1, then 
 
 V — i\ = - {u- w J, 
 
328 ELASTIC SOLIDS. [§ 267 
 
 or the difference of their velocities after impact is the same 
 as it was before impact, but in the opposite direction. In 
 this case the bodies are perfectly elastic. 
 
 No examples of either perfectly inelastic or of perfectly 
 elastic bodies occur in nature. But some bodies with very 
 little elasticity, as clay, for example, may be regarded as be- 
 longing to the first class, and others, as glass, to the second 
 class. For two glass balls e = 0.94; two ivory, e — 0.81 ; two 
 cast iron, e = 0.66; two lead, e = 0.2. 
 
 267. The special case of direct impact of a sphere on a 
 synooth fixed plane may be noticed. 
 
 The plane being fixed, we have 
 
 u^ = 0, V, = 0. 
 
 But from the experimental law 
 
 V — i^i = — e(w — Wj). 
 
 Hence there results 
 
 V =i — eu, 
 
 or the velocity of recoil is reversed in direction. 
 
 The impulse on the sphere is measured by the change of 
 momentum, and is = W{u — v)/g = JVu{l + e)/g. 
 The impulse on the plane is equal and opposite to this. 
 
 Ex. 1. A perfectly inelastic ball impinges on a plane per- 
 pendicularly. Show that there is no recoil. 
 
 2. Two inelastic balls are brought to rest by the impact. 
 Prove that they must have been moving in opposite direc- 
 tions with velocities inversely proportional to their weights. 
 
 3. Two balls of equal weight are perfectly elastic. Prove 
 that after impact they will exchange velocities. 
 
 4. A series of equal elastic balls are placed in contact in a 
 straight line. An equal ball impinges directly on them. 
 Show that all will remain at rest but the last, which will 
 fly off. 
 
 5. Find the elasticity of two balls of weights tu and TFin 
 order that if W impinges on iv at rest it will itself be brought 
 to rest. Ans. W/tv. 
 
 6. A ball falls from a height of 16 ft above a level floor. 
 
§ 268] IMPACT. 329 
 
 Find the velocity of rebound and the height to which the 
 ball will rebound if the coefficient of restitution is 0.75. 
 
 Ans. 24 ft/sec; 9 ft. 
 7. A ball falls from a height h above a level floor and re- 
 bounds to a height A,. Show that 
 
 h, = he' 
 
 where e is the coefficient of restitution. 
 
 For example, if the height is 64 ft and the ball hops 
 four times, show that the height of the fourth hop is 3 in, 
 the value of e being 1/2. 
 
 [Conversely, ball and floor being of the same material, we 
 can by this method find the value of e experimentally.] 
 \8. A ball falls from a height h above a level floor. Show 
 that the whole distance described before the body ceases to 
 rebound is /t(L -|- e')/(l — e') and the time taken is 
 
 V2h(l 4- er/g{i - ey. 
 
 [For distance = h -{- 2he' + 27ie* -h ... ad inf.] 
 "^ 9. A ball weighing 4 lb falls from a height of 9 ft on a level 
 floor and rebounds to a height of 4 ft. Find the impulse. 
 
 Ans. 5 second-pounds. 
 
 10. A sphere impinges directly on an equal sphere at rest. 
 Show that their velocities after impact are as 1 to 3, the co- 
 efficient of restitution being 0.5. 
 
 268. Oblique Impact of a Sphere on a Fixed Smooth Plane 
 AB. — \jQt u be the velocity before 
 impact, and v the velocity after im- 
 pact; the inclination of ti to the 
 normal, and y^ the inclination of v. 
 
 Resolve the velocities along and 
 normal to the plane. The plane 
 being smooth, it exerts a normal pressure only. Hence the 
 impact may be considered direct, with velocity w cos before 
 and V cos p after impact, and 
 
 .•. V cos /3 = — e X — u C08 d = eu cos 6. 
 Also, since the pressure is normal, the action along the plane 
 is unchanged by the impact, or 
 
 V sin fi = u sin 0, 
 Hence v and /? are found. 
 
330 ^tAStlC SOLIDS. L§ 26§ 
 
 ^ Ex. 1. What are the values of v and /3 above ? 
 
 A71S. V = u Vsin^ 6 -\- e^ cos"' 6\ tan /3 = e tan 0. 
 
 2. If the elasticity be perfect, show that the angle of inci- 
 dence 6 is equal to the angle of reflection /3 and the velocity- 
 is unchanged. 
 
 3. If the impact is direct, show that v = eu, as already 
 found in Art. 267. 
 
 4. Show that the impulse of the sphere on the plane is 
 
 Tr(l + e)2i cos 6/g sec-pounds 
 
 if the sphere weighs IF lb. 
 
 5. To hit a ball § by a ball P after reflection from the edge 
 CA of a billiard-table. **' Aim at a point B as far behind the 
 edge OA as Q is in front of it." Prove this. 
 
 6. A ball impinges on an equal ball at rest at an angle of 
 45° to the line of impact. Prove that if both are perfectly 
 elastic their velocities will be equal after impact. 
 
 7. A perfectly elastic ball falls from rest for one second and 
 strikes a smooth plane inclined at 45° to the vertical. After 
 what interval will it again strike the plane ? Ans. 2 sec. 
 
 • 8. Find the angle at which a ball must strike a plane so 
 that its direction after impact may be at right angles to its 
 former direction, the coefficient of restitution being 1/3. 
 
 A71S. 30° to the normal. 
 Show that a perfectly inelastic ball will after impact run 
 along the plane. 
 
 9. Two balls weighing 4 lb and 8 lb are moving with equal 
 velocities of 8 ft/sec in opposite parallel directions, and im- 
 pinge at an angle of 30° with the line joining their centers. 
 If e = 0.5, show that the impulse on the first ball = V3 
 second-pounds. 
 
 What is the impulse on the other ball ? 
 
 10. A shell is fired from a mortar on the ground, and after 
 once ricocheting rises so that at its greatest altitude it just 
 passes over a wall. If be the angle subtended by the wall 
 at the mortar, show that the angle of elevation a is given by 
 
 tan a = 20 tan 6, 
 
 the coefficient of restitution between ground and shell 
 being 0.5. 
 
 269. Change of Energy by Impact. — Consider the direct 
 impact of two bodies weighing IF and IF, lb respectively. 
 
§ 270] IMPACt. 331 
 
 With our usual notation 
 W = energy before impact = Wu^/^g + W^u^/^g ft-pounds. 
 W, = energy after impact = Wv'^/2g + Pr,v,y2^ ft-pounds. 
 
 The change of energy produced by the impact is the differ- 
 ence of these two expressions. Substitute for v, v, their 
 values from Art. 265, and we find after reduction 
 
 W - W. = (1 - «^)-f^^_(» - "J'/Si?. 
 
 When e = 1, or the bodies are perfectly elastic, then w = W, , 
 and there is no change of energy. When e < 1, or e = 0, 
 then w > W,, and the expression indicates a loss of energy 
 produced by the impact. 
 
 But from the principle of the conservation of energy (Art. 
 217) there can be no loss of energy in the system. When 
 energy disappears in one form it reappears in another form. 
 In the present case the energy of impact is broken into two 
 parts, one in producing motion of the impinging bodies, and 
 the other, the so-called loss, in producing sound, heat, etc., 
 and it may be in deforming the bodies. 
 
 270. Whether the change of energy produced by impact is 
 to be regarded as a loss or not depends upon the end to be at- 
 tained. If that is the propulsion of a missile or the driving 
 of a pile, then change of form, heat, etc., are prejudicial, and 
 the energy used in producing them is lost. If, on the other 
 hand, change of form is the main thing, as in molding metal 
 under a hammer or in riveting, this so-called loss becomes the 
 useful energy, and the energy of motion useful in the former 
 case becomes prejudicial in this. 
 
 -- Ex. 1. Two trains weighing 60 tons and 80 tons come into 
 collision with velocities of 60 miles/hour and 45 miles/hour 
 respectively. Find the energy expended in the destruction 
 of the cars, supposing them inelastic. 
 
 Ans. 25,410,000 foot-pounds. 
 2. Two trains of equal weight, moving with velocities of 
 30 miles an hour each and in opposite directions, collide. 
 
332 
 
 ELASTIC SOLIDS. 
 
 [§271 
 
 Show that the loss of energy produced by the impact is the 
 same as in the case of a train moving at 60 miles an hour 
 striking another at rest. 
 
 In the latter case find the velocity with which the debris 
 will be moved along the track. 
 
 Also, show that before impact the total energy in the one 
 case is double that in the other. 
 
 3. Find the loss of kinetic energy if a ball weighing 10 lb 
 falls from a height of 16 ft and rebounds after striking the 
 ground to a height of 4 ft. A7is. 120 ft-pounds. 
 
 4. A ball weighing w lb falls from a height h ft on a fixed 
 plane. Show that the loss of energy from the impact is 
 (1 — e'^)wh ft-pounds, the coefficient of restitution being e. 
 
 271. Ajjplications, — The case of impact that occurs most 
 frequently in practice is when the 
 bodies are inelastic and one is at rest 
 before impact. Here w, = and 
 e = 0. Hence from Art. 265 
 
 v = v^ --^Wu/{W+WJ. 
 
 Consider, for example, the pile- 
 driver. 
 
 It is assumed that the pile is in- 
 elastic, that there is no deformation 
 of the head of the pile, the blow 
 being instantaneous, and that the re- 
 sistance of the ground is uniform. 
 
 Suppose a pile weighing IT, lb 
 is driven vertically s ft into the 
 ground by a ram weighing TTlb fall- 
 ing through a height of 7i ft.^ Let 
 V ft/sec be the velocity of ram and 
 
 pile after the blow is given. The 
 
 velocity of the ram before striking being denoted by u ft/sec, 
 we have 
 
 u' = 2gJh 
 v= Wu/(W + W,). 
 
§ 272] IMPACT. 333 
 
 The kinetic energy of ram and pile causing penetration is 
 
 i( TT + w,y/g = i jruy( W + W,)g 
 
 = WVi/( W + PF.) f t-pounds. 
 
 The work done by the force of gravity on ram and pile is 
 
 (fr.+ TF,)5ft-poiind«. 
 
 If F pounds denotes the average resisting force offered by 
 the grouTid, we have by the principle of work 
 
 Fs = W%/{W-YW,) + (Tf + W,)s, 
 
 and F is found. 
 
 At the last blow, the value of s being small, the second term 
 may be disregarded in comparison with the first, and we have 
 
 Fs=z Wh/(W^W,). 
 
 Still more approximately, by neglecting the weight of the pile 
 in comparison with that of the ram, 
 
 Fs = Wli, 
 
 For example, to find the ultimate load a pile weighing 500 
 lb could carry if the last blow from a height of 25 ft of a 
 one-ton ram sinks the pile one inch. The three formulas give 
 482,500, 480,000, 600,000 pounds, respectively. 
 
 272. The kinetic energy dissipated by the blow is evidently 
 
 Wh - W1i/{W-\-W,) = WWfi/(W-\-W,) ft-pounds, 
 
 and appears principally as sound and heat. 
 
 It is useful to notice that, the energy of the ram before 
 impact being Wh, the loss of energy will be less the more 
 nearly W/( W -\- W^) is equal to W, that the more nearly 
 ir/(Pr + ^^,) is equal to unity, that is, the greater W is in 
 comparison with W^. Hence the ram should be large in weight 
 compared with the pile. 
 
334 ELASTIC SOLIDS. [§ 273 
 
 With the riveting-hammer, steam-hammer, etc., in which 
 change of form is the end to be attained, the useful work 
 done by the hammer depends on WWJi/{ W + ^i)> which is 
 the more nearly equal to Wh the greater W^ is in comparison 
 with W; that is, the heavier the anvil is in comparison with 
 the hammer. 
 
 This was first put in practice by Nasmyth, the inventor of 
 the steam-hammer. "I may mention," he says, "that pile- 
 driving had before been conducted on the cannon-ball prin- 
 ciple. A small mass of iron was drawn slowly up and sud- 
 denly let down on the head of the pile at a high velocity. 
 This was destructive, not nnp2ilsive, action. Sometimes the 
 pile was shivered into splinters without driving it into the 
 soil; in many cases the head of the pile was shattered into 
 matches, nnd this in spite of the hoop of iron about it. On the 
 contrary, I employed great mass and moderate velocity. The 
 fall of the steam-hammer block was only 3 or 4 ft, but it went 
 on at 80 blows the minute, and the soil into which the pile was 
 driven never had time to grip or thrust it up." 
 
 273. In building, the safe load to be carried by the pile is 
 some fractional part of the load P required to drive the pile, 
 say one nth part. 
 
 Then nP = F, and 
 
 nPs = W'h/{W-\-W,), 
 or P = W'li/{W-\-W,)ns. 
 
 The fraction 1/n is to be determined by experience. . An 
 average value is 1/4. 
 
 Ex. 1. A steam-hammer weighing 500 lb has a stroke of 3 ft. 
 If the piston-pressure is 1000 pounds and the blow is verti- 
 cal, find the work delivered in 6 blows. 
 
 A71S. 27,000 foot-pounds. 
 
 2. A pile weighing half a ton is driven 12 ft into the 
 ground by 30 blows of a hammer weighing 2 tons falling 30 
 ft. Prove that it would require 120 tons in addition to the 
 hammer to be superposed on the pile to drive it down slowly, 
 supposing the resistance of the ground uniform, 
 
§273] IMPACT. 335 
 
 3. Find the safe load for a pile weighing 500 lb to carry if 
 the pile sinks 0.1 ft at the last blow under the 5-ft fall of a 
 500-lb ram. A7is. 3125 lbs. 
 
 4. A pile is driven s ft vertically into the ground by n blows 
 of a steam-hammer fastened to the head of the pile. Given 
 p the mean pressure of the steam in pounds/in^, d the di- 
 ameter of the piston in inches, I tlie length of tlie stroke in 
 ft, W the weight in lb of the moving parts of the hammer, 
 and W^ the weight in lb of the pile and fixed parts of the 
 hammer attached to it, and Ji the mean resistance of the 
 ground in pounds, prove 
 
 nW(W-}- 7rpd'/4)l = Es{ W -f W,), 
 
 5. In firing from a rifle of weight W\h a bullet of weight 
 TT, lb with velocity ?> ft/sec, show that the energy of recoil 
 is Wy/2Wg. 
 
 6rt. Prove that the mean resistance of the wood is 204 pounds 
 to a nail weighing 1 oz, supposing a hammer weighing 1 lb 
 striking it with a velocity of 34 ft/sec drives the nail 1 in into 
 a fixed block of wood. 
 
 6^. If the block is free to move and weighs 68 lb, prove 
 that the hammer will drive the nail only 64/65 in. 
 
 6c. Prove that the nail is 0.0052 sec and 0.005128 sec in 
 penetrating the wood in the two cases, during which the 
 block if free will move 0.015 in. 
 
 7. A hammer weighing W lb strikes a nail weighing PT, lb 
 with a velocity of ti ft/sec and drives it s ft into a piece of 
 wood which is fixed in position. If the resistance of the wood 
 is II pounds, find the time of penetration. 
 
 Ans. Wn/Eg sec 
 
 8. If water flowing in a pipe 50 ft long, with velocity 24 
 ft/sec, is shut off in 0.1 sec by a stop-valve, show that the 
 water-pressure in the pipe near the valve is increased by 162.5 
 pounds/in". 
 
 EXAMINATION. 
 
 1. Distinguish stress and strain. In what senses is the 
 term strain used ? 
 
 2. What is meant by the elastic limit ? How is the work- 
 ing stress related to it ? 
 
 3. State Hooke's law. 
 
336 ELASTIC SOLIDS. [§ 273 
 
 4. Define the modulus of elasticity of volume. Of form. 
 
 5. What is meant by Young^s modulus of elasticity ? 
 
 6. State Coulomb's law of torsion. 
 
 7. Explain how to determine moments of inertia by means 
 of the torsion pendulum. 
 
 [Suspend a body whose moment of inertia (/J is required 
 and note the time of oscillation (t^). Replace it by a body 
 whose moment of inertia (/) is known by computation or 
 otherwise, and note the time of oscillation (t). Then 
 
 I^ = it;/t\] 
 
 8. A nut is placed on a table, the forefinger of the left hand 
 placed upon a suture of the nut, and a blow given to the 
 finger by the right fist. The finger will be uninjured, but 
 the nut will almost certainly be cracked. 
 
 Explain the general principle involved. 
 
 9. State the experimental law on which the determination 
 of the motion of- two elastic balls after im23act depends. 
 
 10. Two inelastic spheres are moving in one straight line. 
 Find the result of a collision upon their velocities. 
 
 11. An inelastic sphere impinges on another of twice its 
 weight at rest. Show that the impinging body loses 2/3 of 
 its velocity by the impact. 
 
 12. A ball weighing 2 lb impinges obliquely upon an 8-lb 
 ball at rest. The coefficient of restitution is 0.25. Show 
 that after impact the balls move off at right angles. 
 
 13. Show that an inelastic ball will, after impact with a 
 smooth plane, slide along the plane. 
 
 14. Two perfectly elastic spheres collide directly. Show 
 that their kinetic energy is unaltered by the impact. 
 
 15. A body collides with an equal body at rest. ShcTw that 
 the energy before impact cannot exceed twice the energy 
 after impact. 
 
 16. A block weighing 100 lb falls through 25 ft and is 
 brought to rest in 1/10 sec. Find the average pressure ex- 
 erted on the block, Ans. 1250 pounds. 
 
§ 273] EXAMINATION. 337 
 
 17. A nail is driven 1/10 in into wood by a blow from a 2- 
 Ib hammer having a velocity of 16 ft/sec. Find the average 
 pressure offered by the wood. A7is. 960 pounds. 
 
 18. An 80-ton gun discharges an 800-lb shot with a velocity 
 of 1500 ft/sec. If the recoil is resisted by a constant press- 
 ure of 15 tons, how far will the gun recoil ? Ans. 4.7 ft. 
 
 19. In driving a nail of weight w, lb with a hammer of 
 weight w lb and velocity u ft/sec the energy spent in produc- 
 ing deformation of the head of the nail, sound, etc., is 
 
 ^—u^/1g foot-pounds. 
 
 and the energy spent in driving the nail is 
 y^/lg foot-pounds. 
 
 W + 10^ 
 
 20. A shot weighing w lb is fired from a gun whose weight 
 is H^lb with velocity v ft/sec relative to the gun. Show that 
 the actual velocities of shot and gun are Wv(W ■\- w) and 
 wvl{ W + ?^) ft/sec respectively. 
 
 21. Prove that if a link of a chain which is coiled up 
 loosely near the edge of a table is allowed to hang over the 
 edge the chain will slip over with a uniform acceleration ^/3. 
 
 22. What is the effect of a charge between light and heavy 
 cavalry, the light cavalry having the greater energy and the 
 heavy the greater momentum ? 
 
338 METRIC UNITS. [§ 274 
 
 CHAPTER IX. 
 METRIC UNITS. 
 
 274. In Chapter II it was explained that there are two sys- 
 tems of units in use in Mechanics, the absolute system and 
 the gravitation system. The British gravitation system, 
 being that in most common use in this country, has been em- 
 ployed in this book so far. There is no system of absolute 
 units depending upon British measures that is generally 
 acknowledged or is likely to come into use.* 
 
 The metric system of measures, on the other hand, pro- 
 vides a nomenclature for absolute and gravitation units. 
 The metric gravitation system, like the British gravitation 
 system, is local in character; the absolute system, known as 
 the C.G.S. system, is cosmopolitan, having been devised for 
 international purposes. 
 
 For clear understanding of the units at present in use it 
 will be necessary to go into considerable detail and to present 
 the subject in historical order. 
 
 The Standards of the Archives. 
 
 275. (a) Unit of Length. — In 1790 the French govern- 
 ment, at the suggestion of Talleyrand, asked the Academy of 
 Sciences to propose a scheme tor the reformation of the con- 
 fusion existing among the weights and measures of the coun- 
 try. The report of the committee of the Academy — which 
 included Laplace, Lagrange, and Borda — was adopted in 
 1791. The committee resolved that the 1/10^ part of an arc 
 
 * The system involving Ibe " poundal" as unit of force is quite super- 
 fluous. 
 
§ 276] STANDARDS OF THE ARCHIVES. 339 
 
 of the meridian extending from the equator to the pole 
 should be the basis of the standard of length and be called a 
 meter. 
 
 Accordingly an arc of the meridian was measured and the 
 length of the meter determined. The standard of the meter, 
 which consists of a platinum bar about 1 in wide and 1/7 in 
 tliick, was constructed by Borda, presented to the National 
 Assembly by Laplace, and on the same day deposited among 
 the Archives of France. It is hence called the metre des 
 a?' chives. 
 
 The meter thus fixed upon as the standard of length was 
 defined by a law of the French Republic as the distance at 
 0° C, the temperature of melting ice, between the ends of the 
 Borda platinum bar just described. 
 
 Later determinations have shown that the length of the 
 quadrant is not exactly 10' meters. According to Clarke it 
 has a length of 10,002,015 meters, being 2015 meters [5/4 
 mile] in excess. Other determinations give different values. 
 Hence if the meter were lost or destroyed it would be impos- 
 sible to replace it with any great degree of accuracy by refer- 
 ence to the quadrant of the meridian as natural unit. 
 
 276. It is probable that some other natural unit, as the 
 wave-length of homogeneous light, may be chosen as the ulti- 
 mate standard of length. Michelson found (1893) that the 
 meter and length of a light-wave of red cadmium light might 
 be compared with the same degree of accuracy as is now pos- 
 sible in the comparison of two meter bars.* He says: "If, 
 therefore, the meter and all its copies were lost or destroyed, 
 they could be replaced by new ones which would not differ 
 from the originals more than do these among themselves." 
 
 " The science of the eighteenth century sought to render 
 itself immortal by basing its standard units upon the solid 
 earth; but the science of the nineteenth century soars far 
 
 * The meter = 1,553,168.5 wave-lenglhs (Michelson); =1,553,163.$ 
 wave-lengths (Benoit). 
 
340 METRIC UNITS. [§ 277 
 
 beyond the solar system and connects its units with the ulti- 
 mate atoms which constitute the universe itself." 
 
 277. (b) Unit of Weight. — The standard of weight is the 
 kilogram, which is a certain cylinder of platinum constructed 
 under the direction of the French Academy at the same time 
 as the meter, and with it deposited among the archives of 
 France. It is hence called the kilogramme des archives. 
 
 The kilogram was intended to equipoise in a vacuum a 
 cubic decimeter or liter of water at 3.9° C, the temperature 
 of greatest density of water; that is, it should be of the same 
 weight. But as weight and length have no natural relation, 
 it was impossible that this relation could be exactly realized. 
 The numerical determination changes as methods of com- 
 parison become more perfect, though within small limits. 
 
 Without doubt two weights can be compared at least a 
 thousand times more accurately than either of them can be 
 reproduced by weighing a specified volume of water, and for 
 that reason the kilogram, like the English pound, can now 
 be regarded only as an arbitrary standard of which copies 
 must be taken by direct comparison. (Harkness.) 
 
 The International Standards. 
 
 278. In 1875, by the concurrent action of the princi- 
 pal governments of the world — seventeen in number — an 
 international bureau of weights and measures was estab- 
 lished at the Pavilion de Breteuil, Sevres, near Paris, France. 
 Under the direction of the international committee two 
 ingots were cast of platinum-iridium in the proportion of 
 nine parts of the former to one of the latter metal. From 
 one of these a number of meter-bars were prepared, and 
 from the other a number of kilogram-weights. These 
 standards of length and of weight were intercompared. 
 After the comparison had been made, one of the bars, the 
 length of which at 0° C. was found equal to that of the Metre 
 des Archives, and one of the kilograms, with which the others 
 
§ 280] INTERNATIONAL STANDARDS. 341 
 
 were compared, were selected as international prototype stand- 
 ards. The others were in 1889 distributed by lot to the dif- 
 ferent governments and are called national prototype stand- 
 ards. Those apportioned to the United States — meter 
 number 27, kilogram number 20 — are in the keeping of the 
 Office of Standard Weights and Measures, Washington. The 
 standards allotted to Great Britain are meter number 16 
 and kilogram number 18, deposited with the Standards De- 
 partment of the Board of Trade, London. The ultimate 
 standards are the international prototype standards at BreteuiU 
 which are defined as follows: 
 
 The International Standard Meter is derived from the 
 Metre des Archives, and its length is defined by the distance 
 between two lines at 0° Centigrade on a platinum-iridium bar 
 deposited at the International Bureau of Weights and Meas- 
 ures. This bar is in transverse section nearly of the form of 
 the letter X, known as the Tresca form. 
 
 The International Standard Kilogram is a lump of platinum- 
 iridium deposited at the same place, and its weight in vacuo 
 is the same as that of the Kilogramme des Archives. 
 
 Like the Kilogramme des Archives it is in shape a cylinder, 
 having the same diameter as altitude (39 mm). 
 
 279. Of the two definitions of the liter given in Art. 277 
 the international committee decided that one should be chosen 
 which is most convenient for purposes of measurement. They 
 accordingly in 1880 resolved that the liter (1) should denote 
 the volume of a kilogram of water at the temperature of 
 maximum density. The 1/1000 part of the liter is called the 
 milliliter (ml). 
 
 The liter is therefore not precisely the same as the cubic 
 decimeter, nor the milliliter the same as the cubic centimeter. 
 
 280. The international prototype meter and kilogram are 
 now regarded by the Office of Weights and Measures, Wash- 
 ington, as the standards of length and weight in the United 
 States. In the absence of any material normal standards the 
 units in ordinary use, the yard and the pound, are derived 
 
342 
 
 METRIC UNITS. 
 
 [§281 
 
 from the international units in accordance with the act of 
 July 28, 1866. This act gives the relations: 
 
 1 yard = 3600/3937 meter; 
 1 lb avoir. = 1/2.2046 kilogram. 
 
 The British imperial yard is equal to 36/39.37079 meter, 
 and the imperial pound is equal to 1/2.2046212 kilogram. 
 
 The following are the principal determinations of the value 
 of the meter in terms of the inch : 
 
 Date. 
 
 Authority. 
 
 Inches. 
 
 1818 
 1835 
 1866 
 1885 
 1893 
 
 Kater 
 
 Buily 
 
 39.37079 
 39.869678 
 39.370433 
 39.36985 
 39.370155 
 39.3701 
 (preliminary value) 
 
 Clarke 
 
 Comstock 
 
 Rogers 
 
 International Bureau. . . 
 
 No value of the inch in terms of the meter nor of the pound 
 in terms of the kilogram has as yet (April, 1896) been adopted 
 by international agreement. 
 
 281. The subdivisions and multiples of the meter with 
 their equivalents in British measures are as follows. The 
 abbreviations are those given by the international committee 
 and should be used. . 
 
 Micromillimeter 
 
 Micron 
 
 /* 
 
 mm 
 cm 
 dm 
 
 m 
 
 km 
 
 1/10' m 
 
 1/10« m 
 
 1/1000 m 
 
 1/100 m 
 
 1/10 m 
 
 Im 
 
 10 m 
 
 100 m 
 
 1000 m 
 
 10000 m 
 
 
 Millimeter 
 
 0.03987 in 
 
 Centimeter . . 
 
 
 
 
 Meter 
 
 39.37 in 
 
 Decameter 
 
 
 Hectometer 
 
 
 Kilometer 
 
 Myriameter 
 
 0.6214 mile 
 
 
 
§282] 
 
 APPROXIMATE VALUES. 
 
 343 
 
 The subdivisions and multiples of the gram with their 
 abbreviations and equivalents in British measures are as 
 follows : 
 
 Microgram 
 
 Milligram 
 
 Centigram 
 
 Decigram 
 
 Gram 
 
 Decagram 
 
 Hectogram 
 
 Kilogram 
 
 Myriagram 
 
 Quintal 
 
 Tonneau (Millier) 
 
 1/W g 
 
 1/1000 g 
 
 1/100 g 
 
 1/10 g 
 
 Ig 
 
 10 g 
 
 100 g 
 
 1,000 g 
 
 10.000 g 
 
 100,000 g 
 
 1,000,000 g 
 
 0.01543 grain. 
 15.48235 grains 
 2.2046 lb. 
 
 The following approximate values are for many purposes 
 close enough: 
 
 Length. 
 
 1 millimeter = 1/25 inch 1 inch 
 
 1 centimeter = 2/5 inch 1 foot 
 
 1 meter =1.1 yard 1 yard 
 
 1 kilometer = 5/8 mile 1 mile 
 
 = 2.5 centimeters 
 = 30.5 centimeters 
 = 0.9 meter 
 = 1G09.3 meters 
 
 Weight. 
 
 1 gram = 15.4 grains 
 
 1 kilogram = 2.2 lb 
 1 tonne = 2205 lb 
 
 1 ounce = 28.3 grams 
 1 lb = 453.6 grams 
 1 ton = 10/11 tonne 
 
 282. The metric system is in use among nearly all the 
 nations of Europe, of America, North and South, also in 
 Japan. The principal exceptions among civilized nations 
 are the United States, Great Britain, and Russia. In 1864 
 its use was made legal but not compulsory in Great Britain, 
 and in 1866 in the United States. The act of 1866 reads: 
 
 " It shall be lawful throughout the United States of Amer- 
 ica to employ the weights and measures of the metric sys- 
 
344 METRIC UNITS. [§ 283 
 
 tern; and no contract or dealing or pleading in any court 
 shall be deemed invalid or liable to objection because the 
 weights or measures expressed or referred to therein are 
 weights or measures of the metric system. 
 
 "The Secretary of tlie Treasury is hereby authorized and 
 directed to furnish each State, to be delivered to the Governor 
 thereof, one set of the standard weights and measures of the 
 Metric System for the use of the States respectively." 
 
 This was the first general legislation upon the subject in 
 the United States, and the metric system is thus the first and 
 thus far the only system made legal throughout the country. 
 (See Art. 65.) It is now used in the U. S. Coast and Geo- 
 detic Survey, the U. S. Pharmacopoeia, and to some extent in 
 the Mint, the Geological Survey, the Post Office, and the 
 Weather Bureau. 
 
 Our fractional silver coins represent metric weights. Thus 
 the 50-cent piece weighs 125 decigrams, the 25-cent piece 625 
 centigrams, etc. 
 
 283. The standards being defined, we proceed to explain 
 the two systems of metric units in use in Mechanics. (See 
 Arts. 65-69.) 
 
 (1) The gravitation system, used in commerce and the 
 arts. 
 
 (2) The absolute system, used in theoretical investigations 
 and in laboratory work. 
 
 Gravitation Units. — The fundamental units of the metric 
 gravitation system are the units of length, weight, and time. 
 
 The unit length is the meter [m] or centimeter [cm]. (Art. 
 278.) 
 
 The unit iveight is the kilogram [kg] or gram [g]. (Art. 
 278.) 
 
 The unit time is the second [sec]. (Art. 7.) 
 
 The principal derived units are : 
 
 The unit velocity , which describes unit distance in unit 
 time, is denoted by the symbols 1 cm/sec; 1 m/sec; 1 m/min; 
 etc., according to the units of length and time employed. 
 
§ 283] ORAVITATIOK UKITS. 346 
 
 'i'lie unit acceleration, which changes the velocity one 
 unit in unit time, is denoted by the symbols 1 cm/sec"; 
 1 m/sec'; 1 m/min'; etc. 
 
 The unit force, or the force of gravity on the unit of 
 weiglit [the kilogram or the gram], is called the kilogram or 
 the gram. 
 
 The term kilogram is used in a double sense, corresponding 
 to the use of the term pound in the British system. (Art. 68.) 
 
 The unit impulse is the impulse of unit force [1 kilogram] 
 in unit time [1 second], and is called a second-kilogram. 
 
 The unit momentum is the momentum of 1 kg moving 
 with unit velocity, and is the same as the unit of impulse. 
 
 Thus the momentum of IF kg moving with a velocity of 
 V m/sec is Wv/g second-kilograms, g being equal to 9.81 
 m/sec\ 
 
 The unit work, or the work done by unit force acting 
 through unit distance, is called the kilogrammeter [kgm] or 
 the gram- centimeter [gem]. 
 
 It may also be defined as the work done in lifting unit 
 weight through unit height. 
 
 Thus a force of i^ kilograms acting through s meters does 
 a work of Fs kgm. Or the work done in lifting a weight of 
 W kg through a height of h meters is Wh kgm. 
 
 The unit energy is the same as the unit of work, (Art. 
 214.) Thus the energy of Wkg moving with a velocity of v 
 m/sec is Wvy2g kgm. 
 
 The energy stored in a weight of W kg at a height of h 
 meters above the earth's surface is WJi kgm. 
 
 The unit activity [or power] is the kilogrammeter per 
 second [kgm/sec]. The enlarged unit is the cheval vapeur, 
 which is 75 kgm/sec. 
 
346 METRIC UNITS. [§ 283 
 
 I. 
 
 Ex. 1. A horse trots 12 km in 1 h 40 min. Find his aver- 
 age speed. Ans. 2 m/sec. 
 
 2. Show that an acceleration of 500 cm/sec'^=18 km/min'. 
 
 3. Show that the velocity of the earth round the sun is 
 3 X 10" cm/sec, the mean distance of the earth and the sun 
 being 1.487 X 10"' cm. 
 
 4. A point has velocities of 4, 4, 8 cm/sec inclined at 120° 
 to each other. Find the resultant velocity. 
 
 Ans. 4 cm/sec. 
 
 5. Find the centripetal acceleration of a point which 
 moves in a circular path of 1 m diameter with a velocity of 
 10 cm/sec. A7is. 2 cm/sec^ 
 
 6. A railroad train timed past posts 1 km apart took 2 min 
 and 1 min to pass over two consecutive distances. Find the 
 velocity at the middle post, assuming the acceleration con- 
 stant. Ans. 50 km/hour. 
 
 If t^f t^ were the times, show that the acceleration is 
 
 2(t,-' - t-')/(t,-h t^) km/sec\ 
 
 II. 
 
 7. A force of 5 kilograms acts on a weight of 10 kg. Find 
 the distance described in 10 sec. A71S. 245.25 m. 
 
 8. Show that a brake resistance of 102 kilograms per tonne 
 will bring to rest a train running at 72 km/hour in about 
 20 sec. 
 
 9. If a carriage weighing W kg is placed on a smooth level 
 road and acted on by a pull of P kilograms for t sec, find the 
 velocity acquired. A?is. v = Pgt/W. 
 
 Find the momentum acquired. 
 
 10. Explain why a waterfall li meters high can support a 
 column of water %h meters high. 
 
 III. 
 
 11. Two forces of 4 kilograms and 7 kilograms act at an 
 angle of 55°. Show that their resultant is 9 kilograms 855 
 grams. 
 
 12. The wind is blowing from the southwest. Show that a 
 vessel may be sailed due west. 
 
§ 283] ORAVITATIOK UNITS. 347 
 
 13. Two cords making an angle of 60° support a chandelier 
 weighing 50 kg. Find the pull in each cord. 
 
 A71S. 50/|/3 kilograms. 
 
 14. Three forces of 3, 4, 5 kilograms act at 120°. Find their 
 resultant. 
 
 15. A locomotive weighing 98.1 tonnes runs round a curve 
 of 800 m radius with a velocity of 72 km/hour. Find the 
 centrifugal force. Aiis. 5000 kilograms. 
 
 16. A ball is thrown upwards with a velocity of 20 ni/sec. 
 Find its velocity when half way up. 
 
 17. A marline spike falling from a mast down a hatchway 
 took t sec to fall to the bottom of the hold, a depth of h 
 meters. Find the velocity with which it struck. 
 
 A 7is. li/t + gt/l m/sec. 
 
 18. The wheels of a train running at 36 km/hour on coming 
 to a drop of 5 mm in the rails will go about 32 cm before 
 touching the rails. 
 
 IV. 
 
 19. Weights of 1, 2, 3 g are placed at the angles of an equi- 
 lateral triangle whose sides are 12 cm in length. Find the 
 distance of the C.G. from the angle 1. Ans. 2 4^19 cm. 
 
 20. A body appears to weigh 240 g when placed in one 
 pan of a balance, and 250 g when placed in the other pan. Find 
 its real weight. Ans. 100 1^6 g. 
 
 21. The length of the beam of a false balance is 1 m. A 
 body placed in one scale weighs 8 kg, and in the other 8.1 kg. 
 Find the lengths of the arms of the balance. 
 
 22. A steelyard is 1 m long and weighs 2 kg. It is sus- 
 pended at a point 10 cm from one end. The movable weight 
 is 1 kg. Find the greatest weight that can be weighed. 
 
 Ans. 17 kg. 
 
 V. 
 
 23. A 10-kg weight slides with constant speed down a 
 rough plane which rises 1 in 10. Find the resistance. 
 
 Ans. 1 kilogram. 
 
 24. A 10-kg weight rests on a plane whose inclination is 30°. 
 Find the force of friction called into action. 
 
 Ans. 5 kilograms. 
 
 25. What force will haul a 10-kg weight along a rough 
 
348 METRIC UNITS. [§ 284 
 
 table with a uniform acceleration of 1 m/sec', the coefficient 
 of friction being 0.5 ? Ans. 6,02 kilograms. 
 
 VI. 
 
 26. Find the number of kgm necessary to raise 200 cm' of 
 water to a height of 1 km. A^is. 200 kgm. 
 
 27. The pitch of a screw is 30 mm, and the force exerted by 
 each of two men with levers 2 m long is 15 kilograms. Find 
 the greatest weight that can be raised. Ans, 12.57 tonnes. 
 
 28. In a differential pulley the radii of the upper sheaves 
 are 190 mm and 200 mm. If the pull on the chain is 20 
 kilograms, find the greatest weight that can be raised. 
 
 A71S. 800 kg. 
 
 29. Find the distance in which a boat weighing 50 tonnes 
 and moving at the rate of 4^ km/hour can be brought up by 
 a rope round a post if the greatest pull the rope is capable of 
 sustaining is 1 tonne. A7is. 4 m nearly. 
 
 30. Show that a train running at 36 km/hour can be 
 brought to rest in about 34 m by the brakes, supposing them 
 to press on the wheels with 3/4 of the weight of the train 
 and that the coefficient of friction is 0.2. 
 
 31. A train starts from a station at the foot of an incline of 
 30° and of length I meters. If the train weighs W kg and it 
 reaches the top of the incline with a velocity v m/sec, show 
 that the work done is W{l/2 + vy2g) kgm. 
 
 284. Absolute Units. — As already explained in Chapter II, 
 a system of units is absolute when it does not involve the ac- 
 celeration of gravity at any place. 
 
 The fundamental units are those of length, time, and mass; 
 the term mass being used instead of weight because in an ab- 
 solute system comparisons are made fundamentally by kinetical 
 methods and not by weighing. Comparisons of masses may, 
 however, be made by the beam balance (Art. 63). 
 
 285. C.G.S. Units. — Gauss and Weber, who first developed 
 the absolute system of units, employed the millimeter as unit 
 of length, the milligram as unit of mass, and the second as 
 .unit of time. 
 
 In 1873 a committee of the British Association recom- 
 mended as dynamical and electrical units the centimeter, 
 gram, and second as the fundamental units, and also recom- 
 
§285] C.G.8. UNITS. 349 
 
 mended a system of nomenclature. Additions were made to 
 the nomenclature in 1888. 
 
 This system, involving as fundamental units the centimeter 
 as unit of length, the gram as unit of mass, and the mean 
 solar second as unit of time, is called the C.G.S. — centimeter- 
 gram-second — system. II is growing more and more inter- 
 national in character and is used in theoretical investigations 
 in Astronomy, Physics, Electricity, and Mechanics to the ex- 
 clusion of every other system. 
 
 The principal derived units are as follows : 
 
 The unit velocity which describes unit distance in unit 
 time [1 cm in 1 sec] is called a kine {Kivecj). 
 
 The unit acceleration which changes the velocity 1 unit in 
 time [1 kine in 1 sec] is called a spoud {anovdff). 
 
 The \\x\\t force which produces unit acceleration in unit mass 
 [1 spoud in 1 gram] is called a dyne (6vva^i5). Thus a 
 force of F dynes acting on vi grams produces an accelera- 
 tion of a cm/sec', given by 
 
 F = ma. 
 
 The unit momentum is the momentum of unit mass moving 
 with unit velocity [1 gram with velocity 1 kine], and is called 
 a bole (/?6Ao5^). Thus the momentum of m grams moving 
 with a velocity of v cm/sec is mv boles. 
 
 The unit impulse which generates unit momentum is also 
 called a bole. Thus a force of -F dynes acting for an inter- 
 val of t sec on a mass of m grams generates an impulse of Ft 
 boles, and is equivalent to mv boles if v is the velocity change. 
 
 The unit work^ which is the work done by unit force acting 
 through unit distance [1 dyne through 1 cm], is called an erg 
 (epyov). Thus a force of F dynes acting through s cm 
 does a work of Fs ergs. 
 
 The unit energy is the same as the unit work, the erg. 
 Thus the energy of m grams moving with a velocity of v 
 cm/sec is mv'/^ ergs. 
 
350 METRIC UNITS. [§ 286 
 
 286. M.K,S. Unils. — In practical electricity and in me- 
 chanical pursuits many of the C.G.S. units are found to be 
 inconveniently small. Accordingly what is known as the 
 M.K.S. system, in which the meter, kilogram, and second are 
 the fundamental units, is being adopted as a worki7ig system. 
 It has as yet received only a partial nomenclature. 
 
 In this system 
 
 The unit distance is 1 meter [= 10' centimeters]. 
 
 The unit mass is 1 kilogram [= 10^ grams]. 
 
 The unit time is 1 second. 
 
 The unit force which produces unit acceleration [10' 
 cm/sec''] in unit mass [10^ grams] is consequently 10' dynes. 
 The name gauss * has been suggested for this unit. 
 
 Another unit of force is the megadyne, being one million 
 dynes. 
 
 The unit work is the work done by unit force acting 
 through unit distance [10^ dynes through 10' cm], that is, 
 10' ergs, and is called a joule,! so that 
 
 1 joule = 10' ergs. 
 
 Another unit of work is the megalerg, or a million ergs, 
 being the work done by a megadyne acting through a cm. 
 
 The unit energy is the same as the unit of work, the joule. 
 Thus the energy of m kilograms moving with a velocity of v 
 meters/sec is mv^/2 joules. 
 
 The unit power is the power of an agent which can do a 
 work of one joule per second, and is called a watt,J: so that 
 
 1 watt = 10' ergs/sec. 
 The kilowatt [1000 watts] is the unit of power in electric 
 
 * After Gauss of Gottingen (1777-1855), wlio first published an abso- 
 lute system of units of dynamics. 
 
 f After Joule of Manchester (1818-1889), who first measured accu- 
 rately the mechanical equivalent of heat. 
 
 X After James Watt of Glasgow (1736-1819), who mftde the steam- 
 engine a commercial success. 
 
§ 286] C.G.S. UNITS. 351 
 
 lighting. Roughly, 3 kilowatts are equivalent to 4 horse- 
 powers. 
 
 The kilowatt-hour is analogous to the horse-power-hour 
 (Art. 209), and is a unit of work or energy. 
 
 So also the watt-second is a unit of energy, and is equiva- 
 lent to a joule. 
 
 I. 
 
 Ex. 1. A point moves with a uniform speed of 100 kines. 
 In what time will it pass over 1 km ? Ajis. 1G min 40 sec. 
 
 2. Show that the velocity of a point on the equator arising 
 from the earth's rotation is ahout 46,300 kines, the mean 
 radius of the earth being 2 x 10y;T cm. 
 
 3. A particle starts with a velocity of 100 kines and moves 
 under an acceleration of — 2 spouds. In what time will it 
 come to rest ? A us. 50 sec. 
 
 4. A ball is dropped in an elevator which is rising with an 
 acceleration of 1 spoud. Find its acceleration relative to the 
 floor of the elevator. 
 
 5. A ball is dropped from the top of an elevator 2.45 m 
 high. Find the time in which it will reach the platform 
 when the elevator is descending with an acceleration of 490 
 spouds. A us. 1 sec. 
 
 6. The maximum piston-speed of an engine cross-head is 
 200 kines. P'ind the average speed, neglecting the obliquity 
 of the connecting-rod. Ans. 400/ tt kines. 
 
 7. Find the distance passed over in the fifth second from 
 rest of a body falling at a place where g = 980 spouds. 
 
 Ans. 44.1 m. 
 
 II. 
 
 8. A body of mass 10 g is acted on by a force of 10 dynes. 
 Find the resultant acceleration. A7is. 1 spoud. 
 
 9. Find the momentum of a body of mass 10 g (1) moving 
 with a velocity of 10 kines, (2) acted on by a force of 10 
 dynes for 10 sec. Ans. 100 boles. 
 
 What is the energy in each case ? 
 
 10. A body of mass 10 g moving with a velocity of 10 kines 
 is acted on by a force of 10 dynes for 10 sec. The inclination 
 of the force to the original direction of motion is 60°. Find 
 the final velocity. Ans. 104^3 kines. 
 
 11. A body of mass 10 g produces a certain deflection in a 
 spring-balance at a place where the acceleration of gravity is 
 
352 METRIC UNITS. [§ 286 
 
 981 spends, and at another place it is noted that it reqnires 
 10.01 g to produce the same deflection. Find the valne of 
 the acceleration of gravity at this place. 
 
 Ans. 080.02 spouds. 
 
 12. A body of mass m grams is moving with a velocity of v 
 cm/sec. In what time will a force of F dynes reduce it to 
 rest? 
 
 13. Find the force which acting on 1 kg for 10 sec will 
 produce a velocity of 1 kine. Ans. 100 dynes. 
 
 14. If 1 gram of air occupies 773 cm^ and behaves like 
 a cloud of inelastic particles, prove that the pressure of 
 the wind against a plane perpendicular to its direction is 
 0.00129 y'* barads, if the velocity of the wind is v kines. 
 
 III. 
 
 15. A body slides down a smooth plane inclined at 30° to 
 the horizon. Through how many cm will it fall in the sec- 
 ond second ? A^is, 735f cm. 
 
 16. A body of mass m g revolves uniformly in a circle of 
 radius r cm and makes 7i rev/min. Show that 
 
 the centrifugal force = 7ur(7rn/30y dynes. 
 
 If t sec is the time of revolution, 
 
 the centrifugal force = mr(27r/ty dynes. 
 
 17. Find the length of a seconds-pendulum at a place 
 where g = 981 spouds. Ans. 99.3 cm. 
 
 18. If the speed of a bicycle is 18 km /hour, show that a 
 piece of mud thrown from the highest point of the front 
 wheel 70 cm in diameter will strike the ground 10/i^7 meter in 
 advance of the point of contact of the wheel with the ground. 
 
 19. Forces whose magnitudes are as 3 to 4 act at a point 
 and in directions at right angles. If their resultant is 2 
 dynes, find the forces. Ans. 1.2, 1.6 dynes. 
 
 20. A mass of 6 grams is held up by two threads, one 
 horizontal and the other inclined at 30° to the vertical. 
 Find the pulls in the threads. _ 
 
 A71S. 39241^, 19624/3 dynes. 
 
 21. Kesolve a force of 12 dynes into two equal forces each 
 inclined at 30° to it. Ans. iVd dynes. 
 
 22. A cylinder 10 cm in diameter stands on a rough plane 
 and is just on the point of tumbling over when the inclina- 
 
§ 287] DIMENSIONS. 353 
 
 tion of the plane to the horizontal is 30°. Find the height 
 of the cylinder. ' Aiis. lOV^ cm. 
 
 23. A kg weight is thrown vertically upward with a ve- 
 locity of 981 kines. Find its energy at the instant of pro- 
 pulsion and also after 1 sec. Ans. 481,180,500 ergs; 0. 
 
 24. How much work is done against gravity by a man 
 whose mass is 80 kg in walking a distance of 1 km up a 
 mountain path inclined at 30° to the level ? 
 
 Ans. 392,400 joules. 
 
 25. A 5-kg cannon-ball is discharged with a velocity of 
 500 m/sec. Show that the energy = 625,000 joules. 
 
 26. A kg weight is lifted through 1 m. Find the increase 
 of potential energy. Ans. 9.81 joules. 
 
 27. A gun recoils with a velocity of 327 cm/sec and runs 
 up an incline of 1 in 4 to a distance of 218 cm. Find the 
 value of the acceleration of gravity at that place. 
 
 A71S, 981 cm/sec'. 
 
 28. A particle of mass 1 gram moving with a S.H.M. vi- 
 brates 100 times a second. If the range of vibration is 1 cm. 
 at its mean position, find the energy of the particle. 
 
 29. The energy of a particle of mass m which executes a 
 simple harmonic motion of amplitude a cm and period T sec is 
 
 •Jwa'cj?' sin' cot 
 
 where go = 27r/T, 
 
 30. The kinetic energy of translation of the earth relative 
 to the sun is 10" X 2.69 joules, the radius being 2 X 10'/ ;r 
 cm, and the density 5.6 g/cm." 
 
 31. The electrical power unit adopted at Niagara Falls is 
 5000 electrical horse-power. Show that this is equivalent to 
 3700 kilowatts. 
 
 287. Change of Units — Dimensions. — The numerical value 
 or measure of a quantity is the number of units that must be 
 added together to produce the given quantity; in other words, 
 it is the ratio of the quantity to the unit. If the unit is 
 changed, the numerical value is chans^ed in inverse ratio. 
 Thus a line 100 cm long may be stated as being 1000 mm 
 long or 1 m long, the unit in the first change being 10 times 
 as small and in the second 100 times as large as the original 
 unit. Hence the numerical measure varies inversely as the 
 
354 METRIC UNITS. [§ 289 
 
 unit. Of course the quantity itself is quite independent of 
 the unit employed to measure it — just as the matter of this 
 book is in no way affected by the size of type used by the 
 printer. 
 
 288. In mechanics certain units are assumed as fundamen- 
 tal units. They are called fundamental because a change in 
 any one of them does not imply a change in any of the others. 
 They are arbitrary, being chosen as the result of circumstances 
 or as seems most convenient for the purpose on hand. Those 
 usually but not necessarily taken are the units of length, 
 time, and mass.* They are denoted by the symbols L, T, and 
 M respectively. 
 
 But besides the fundamental units there are other units 
 which are systematically derived from them by definition, and 
 which therefore depend on them. A system of this kind in 
 which a magnitude can be expressed in terms of the funda- 
 mental units is called an absolute system. The C.G.S. system 
 is an absolute system, the derived units being systematically 
 defined in terms of the units of length, time, and mass. 
 
 289. The relation between a derived unit and the funda- 
 mental units from which it is derived may be stated as an 
 algebraic formula. The indices of the fundamental units in 
 the formula of the derived unit are called the dimensions of 
 this unit. The formula itself is a dimensional formula. 
 Knowing the dimensions, we can at once write down the 
 change in the value of a derived unit due to any changes in 
 the fundamental units of that system, or we can convert 
 units of one absolute system into those of another. 
 
 290. The doctrine of dimensions was first given by Fourier 
 in 1821 in his Analytical Theory of Heat, but brought into 
 use by Maxwell. 
 
 The view of dimensional formulas here given is the primary 
 one that the formulas indicate numerical relations between 
 
 * Maxwell's statement that "the whole system of civilized life is 
 symbolized by a foot rule, a clock, and a set of weights " is for the 
 present too sweeping. All electrical and magnetic units have not yet 
 been expressed in terms of these units. 
 
§ 291] DIMENSIONS. 355 
 
 units; in a word, that they are change-ratios. As the tend- 
 ency of modern science is to express everything dynamically, 
 it is probable that dimensional formulas may be so expressed 
 as to indicate the nature of the quantity — the way in which 
 it is built up. At present they do not do this. For example, 
 the table on p. 367 will show that torque and work, though 
 by no means identical, are of the same dimensions. 
 
 For a proposed extension of their meaning along quater- 
 nion lines see Phil. Mag., Sept. 1892. 
 
 291. The principal derived or secondary units are as follows : 
 
 1. Velocity. — From the definition 
 
 unit velocity = unit length/unit time, 
 or V = l/t 
 
 = ltS 
 
 and the unit of velocity is said to be of one dimension with 
 respect to length and minus one dimension with respect to 
 time. 
 If we change to units l, , T, of length and time when 
 
 then unit vel. = L,/t, = tL/t. 
 
 Ex. Find the unit of velocity if the units of length and 
 time are the mile and minute. 
 
 Unit vel. = l./T, = 5280l/60t = 88l/t; 
 
 that is, the unit of velocity is %'^ ft/sec. 
 
 2. Acceleration. — Unit acceleration is rate of change of 
 unit velocity. Hence 
 
 A = V/T = L/t" = LT-', 
 
 and is said to be of dimensions 1 in length and — 2 in time. 
 
 3. Force. — Unit force gives unit acceleration to unit mass. 
 Hence 
 
 F = MA = ML/t' = LMT"'. 
 
356 METRIC UNITS. [§ 291 
 
 4. Impulse, — Unit impulse is given by unit force acting for 
 unit time. Its notation therefore is 
 
 5. Momentum. — Unit momentum is given by unit mass 
 moving with unit velocity. Its notation therefore is 
 
 MV = LMT"', 
 
 or an impulse is of the same dimensions as a momentum. 
 
 This also follows from unit impulse being that which gen- 
 erates or destroys unit momentum. 
 
 6. Work. — Unit work is given by unit force acting through 
 unit distance. Hence 
 
 F X L =^ LMT'' X L = L'MT"'. 
 
 7. Kinetic energy (translation) is of the same dimensions as 
 work. For 
 
 8. Power or Activity. — Unit power is unit work per unit 
 time. Hence 
 
 unit power = w/T 
 
 = L^MT-'/T 
 = L'MT-'. 
 
 9. Density,— "{J mi density is unit mass per unit volume. 
 Hence 
 
 unit density = m/l'. 
 
 10. Pressure. — Unit pressure is unit force per unit area. 
 
 Hence 
 
 unit pressure = v/C 
 
 = LMT-'/L' 
 
 = l-'mt"'. 
 
 11. Torque. — Unit torque is unit force into unit length. 
 
 Hence 
 
 unit torque = f X L 
 
 = LMT-' X L 
 
§ 292] DIMENSIONS. 357 
 
 12. Heat.— In the dynamical theory of heat it is necessary 
 to consider the dimensions of the unit of heat. 
 
 The unit of heat is the calorie, or the quantity of heat re- 
 quired to raise unit mass through unit temperature, or 1°. 
 Hence 
 
 dimensions of unit heat = DM 
 
 when D denotes a degree of temperature on any scale. 
 
 Now to produce a unit of heat requires j units of work 
 when J is known as Joule's equivalent, or 
 
 1 unit heat = j units work. 
 Let, as the result of experiment, 
 
 H units heat = w units work. 
 .-. 1/H = J/W, 
 or J = w/h. 
 
 Hence the dimensions of j are 
 
 L'MTVdM = D-'L'T-% 
 
 which enables us to find the value of j when the units of work 
 and of temperature are changed from one system to another. 
 292. By considering the dimensions of the units involved 
 in a mechanical formula we may check at least its possibility 
 and thus save the labor of going over the demonstration. 
 For example, in Art. 28 it was found that 
 
 V* = ar. 
 
 Considering the dimensions only, we have for the separate 
 terms, r being a length, 
 
 L'T"', LT-' X L, or l'T'*, 
 
 and the expression is therefore homogeneous, each term being 
 of the same dimension. Hence v' = ar is possibly true. 
 
 / 
 
358 METRIC UNITS. [§ 293 
 
 A similar test would show that such an expression as 
 mH + 2FI' = 3m's 
 
 is absurd. 
 
 If, then, in the course of an investigation we had arrived at 
 an equation of this form, we should at once conclude that 
 there was an error somewhere. By applying the principle we 
 may check our work at intervals and save carrying along 
 mistakes. 
 
 293. It is often required to pass from an absolute to a gravi- 
 tation system of units and vice versa. In these cases the rule 
 given in Art. 66 must be observed. 
 
 For example, to find the number of watts in a horse-power: 
 
 Let X = the number of ergs per sec in 1 H.P. 
 
 Now 1 H.P. = 550 foot-pounds per sec. 
 
 Then 
 
 X X L'MT-" = 32.2 X 550 X L/M,T-% 
 
 the second being the unit of time in both systems. 
 But 
 
 100 cm = 39.37/12 ft. 1000 g = 2.2046 lb. 
 
 • .-. L, = 1200/39.37L. /. M, = 1000/2. 2046|V|. 
 
 /1200\' 1000 
 Hence x = 32.2 X 550 X (3^^^) X ^j^^ 
 
 = 746 X 10' ergs/sec 
 = 746 watts. 
 
 1. Show that a velocity of 1 kine = 0.0328 ft/sec. 
 
 2. " " " " '' 1 ft/sec = 30.48 cm/sec. 
 
 3. " " " " " 1 mile/hour = 44.7 cm/sec^ 
 
 4. " " " " " 1 km/hour = 27.78 cm/sec. 
 
 5. " " " " " 1 km/hour = 0.62137 miles/hour. 
 
 6. " " " " " 1 knot = 1853.25 m/hour. 
 
 7. " " an acceleration of 1 spoud = 0.0328 ft/sec'. 
 
 8. " " " " " 1 ft/sec' = 30.48 cm/sec\ 
 
 9. Express in miles per hour the speed of a vessel which 
 steams at the rate of 16.5 knots. Ans. 19 miles/hour. 
 
§ 293] DIMENSIONS. 359 
 
 10. Find the numerical value of a velocity of 10 ft/sec if 
 the fundamental units of length and time are the yard and 
 minute. A7is. 200. 
 
 11. Find theunitof velocity if one meter is the unit of length 
 and one spoud the unit of acceleration. A7is. 10 kines. 
 
 12. Show that the foot per second and mile per hour units 
 of velocity are as 15 : 22. 
 
 13. Find the numerical value of g (981 spouds) if one 
 meter is the unit of length and one minute the unit of time. 
 
 Ans. 36 g, 
 
 14. The unit of velocity is 1 mile per minute and the unit 
 of time 1 minute. Find the unit of acceleration. 
 
 A71S. 22/15 ft/sec». 
 
 15. If the acceleration of gravity is taken as the unit of 
 acceleration, and a velocity of 43 j\ miles an hour the unit of 
 velocity, find the units of time and distance. 
 
 Atis. 2 sec; 128 ft. 
 
 16. Find the unit of time if g is taken as unity, one ft 
 being the unit of length. Ans. 1/4^ sec. 
 
 17. The unit of length is a meter, the unit of mass a kilo- 
 gram, and the unit of time a minute. Find the unit of 
 force. A71S. 250/9 dynes. 
 
 18. If the unit of length is 1 meter, the unit of velocity 100 
 kines, and the unit of energy 100 ergs, find the v.^it of power. 
 
 Ans. 10~* watt. 
 
 19. Taking L, T, W as units of length, time, and energy, 
 find the units of mass, momentum, and force. 
 
 Ans. WL-'T'; WL-'T; WL"'. 
 
 20. Show that 
 
 1 gram [force] = 981 dynes. 
 
 1 grain [force] = 63.57 dynes. 
 
 1 gram-centimeter = 981 ergs. 
 
 1 kilogrammeter = 10** ergs, approx., 
 
 = 9.81 joules. 
 1 pound [force] = 445,000 dynes 
 
 = 2/5 megadyne, roughly. 
 1 foot-pound = 1.356 joules 
 
 = 0.138 kilogrammeter. 
 1 joule = 3/4 foot-pound, approx. 
 
 1 watt = 3/4 foot-pound /sec, approx. 
 
 1 horse-power = 745.9 watts. 
 
 1 kilowatt = 1^ horse-power, approx. 
 
 1 poncelet =981 watts. 
 
360 METRIC UNITS. [§293 
 
 1 watt-second — 10 megalergs. 
 
 3 kilowatt-hours = 4 horse-power-hours. 
 
 1 pound/ft' = 479 barads. 
 
 21. Show by considering its dimensions that 
 
 mvt 4- 2Ft' = 3ms 
 
 is a possible expression. 
 
 22. Show by a consideration of dimensions that the dis- 
 tance described from rest by a body acted on by a force in 
 the direction of its motion varies as the square of the time. 
 
 23. Show that the equation 
 
 Fs = mvy2 
 is homogeneous. 
 
 24. The mean value of Joule's equivalent J" is 4.2x10' ergs 
 when the unit of work is the erg and the unit of temperature 
 1° 0. Express this in foot-pounds when the unit of work is 
 the foot-pound and the unit of temperature 1° F. 
 
 Let X = the number of foot-pounds required. 
 
 Then x X 32.2 X L'^D*' T'' = 4.2 X 10^ X L.'^Dr'T "'. 
 But 100 cm = 39.37/12 ft. 5° C. = 9° F. 
 
 /. L, = 39.37/1200L. .-. D, = 9D/5, 
 , 4.2 X 10' ^ /39.37\' ^ 5 
 
 = 780.1 foot-pounds. 
 
 [Griffith's value (1893) is 779.97. The value in common 
 use is 772, which is certainly too small.] 
 
 25. Find the value of / when the unit of work is the kilo- 
 grammeter and the unit of temperature 1" C. 
 
 Ans. 0.428 kgm. 
 EXAMINATION. 
 
 1. Describe the international standard units of length and 
 weight. 
 
 2. What is the international unit of time ? 
 
 3. How are the inch and lb in the United States defined by 
 the act of 1866 ? 
 
 [In terms of the meter and kilogram by the relations 
 1 m = 39.37 in, 1 kg = 2.2046 lb.] 
 
§ 293] EXAMINATION. 361 
 
 4. What are the legal equivalents in Great Britain accord- 
 ing to the act of 1878 ? 
 
 Ans. 1 m = 39.37079 in; 1 kg = 2.2046212 lb. 
 
 5. Explain the prefixes micro-, milli-, centi-, deci-, decar, 
 hecto-, kilo-, myria-. 
 
 6. How many mm in an inch ? How many cm in a foot ? 
 
 7. Show that 8 kilometers is equal to 5 miles, nearly. 
 
 8. Show that 10 kilograms is equal to 22 lb, nearly. 
 
 9. Should the abbreviation c.c. be used for cubic centi- 
 meter ? 
 
 [No, use cm'.] 
 
 10. What is a metrical tonne ? 
 
 11. How many cm' in a fluid ounce ? 
 
 12. How many ounces in a liter of water? 
 
 13. Is this true : 1 tonne = 1 m' water; 1 milligram 
 = 1 mm' water ? 
 
 14. How many kilograms in a quintal ? 
 
 15. What is a fundamental unit in mechanics ? a derived 
 unit? 
 
 16. State the fundamental C.G.S. units, the fundamental 
 British units, commonly employed. What other units might 
 be chosen ? 
 
 17. Describe the C.G.S. system of units. 
 
 18. What fundamental units does the absolute unit of force 
 involve ? 
 
 19. Define the terms erg, joule, watt, and watt-second. 
 
 20. Express an erg in British units. 
 
 A71S. 7.37 X 10-" foot-pounds. 
 
 21. A force de cheval is 75 kgm/sec. Express it in British 
 units. Alls. 542.48 foot-pounds/sec. 
 
 22. Show that 1 force de cheval — 736 watts. 
 
 23. What is a micron ? a megadyne ? a megalerg ? 
 
 24. A megadyne is roughly equivalent to a kilogram. 
 
 25. Give the adopted abbreviations for micron, meter, 
 cubic centimeter, square millimeter, gram, kilogram, kilo- 
 grammeter, quintal, tonne. 
 
362 METRIC UNITS. [§ 293 
 
 26. " The pressure of the atmosphere upon every cm' of 
 the earth's surface is 1033.3 grams, a little more than a megor 
 dyne." Check the statement in italics. 
 
 27. A watt represents 10 megalergs per second. 
 
 28. Point out the errors in the following: 
 
 (a) " The service of express trains from London to Edin- 
 burgh has received an acceleration of one hour." 
 
 (b) " The couple exerted is evidently 7ic dynes." 
 
 (c) " The momentum of the needle when moving with the 
 angular velocity od is Iqd^/2" 
 
 (c?) " The momentum is all expended in balancing the 
 couple." 
 
 (e) " The O.G.S. unit of work is the gram-centimeter." 
 
 29. How does the length of the meter compare with that of 
 the seconds-pendulum ? 
 
 30. How many watts in a horse-power ? 
 
 31. " A kilowatt is a unit of power and involves the idea of 
 time. A kilowatt-hour is a unit of energy or work, precisely 
 like the term foot-pound, or horse-power-hour, or joule, into 
 which it is directly convertible." Examine this statement. 
 
 32. Is the following rule of thumb true ? — 
 
 "A motor yielding a horse-power for every kilowatt con- 
 sumed has an efficiency of 75 per cent." 
 
 33. Show that 1 joule = 0.00000028 kilowatt-hour. 
 
 34. Show that 1 kilowatt-hour = 1.34 horse-power-hours. 
 
 35. Show that 1 watt-second = 1 joule. 
 
 36. What two functions do dimensional formulas perform 
 in physical investigations ? 
 
 AVhat other function has been suggested ? 
 
 37. Show that the equation 
 
 Ft = mv 
 is homogeneous. 
 
 38. Is the equation 
 
 astv -{■ sv^ = gH* 
 homogeneous ? 
 
§ 293] tXAMlNATlOir. 363 
 
 39. What is meant by a barad ? [A pressure of 1 dyne/cm'.] 
 
 40. Resolve a force so that its component in a given direc- 
 tion shall have a given value. 
 
 41. Show how a stick may be balanced on the finger at a 
 constant inclination to the vertical by making the finger de- 
 scribe a horizontal circle with given velocity. 
 
 42. A projectile is fired with velocity u kines and at an ele- 
 vation d. What are the horizontal and vertical components 
 of its velocity after 2 seconds ? 
 
 43. Determine the height to which the mud-splashes from 
 the hind wheel will reach on the back of a bicyclist. 
 
 44. A mass m grams is moving with velocity u kines. A con- 
 stant force acts on it in the direction of the motion until the 
 velocity is v kines. Prove that the work done by the force is 
 
 mv^fH, — rmi^/^ ergs. 
 
 45. A mass of 1 decigram executes 250 simple harmonic 
 vibrations in 1 sec. If the amplitude of the vibration is 2.5 
 mm^ find the maximum force exerted. 
 
 Ans. 6250;r' dynes. 
 
 46. A simple pendulum is pulled aside until the bob is 
 raised h ft above the horizontal through the lowest point and 
 then let go. Find its velocity as it swings through the lowest 
 point. 
 
 47. A vessel steams directly away from the rock of Gibraltar 
 at a speed of 20 knots. Show that the rock appears to sink 
 with a constant acceleration of 5.5/10* ft/sec'. 
 
 Hence show how a passenger on board could compute the 
 height of the rock. 
 
 48. The actual relation of prototype meter No. 16 is 
 
 prototype 16 = 1 m — 0.6/i ± 0.1// at 0° C, 
 
 so that it may be said that the meter has been verified with a 
 probable accuracy of 0.1 part in a million. 
 Explain the meaning of this last statement. 
 
364 
 
 MECHANICAL UNITS— BRITISH TO METRIC. 
 
 Length. 1 inch = 25.4000 * millimeters (mm) log 1.40483 
 
 1 foot = 0.304801 meter (m) " 9.48402 
 
 lyard= 0.914402 meter " 9.96114 
 
 1 mile = 1.60935* kilometers (km) " 0.20665 
 
 Area. 1 sq inch = 645.16 sq millimeters (mm'^) log 2.80966 
 
 lsqfoot= 0.09290 sq meter (m^) " 8.96802 
 
 lsqyard= 0.83613 sq meter *♦ 9.92227 
 
 1 sq mile = 2.59000 sq kilometers (km«) " 0.41330 
 
 Capacity. 1 U. S. gallon (231 in^) = 3.78543* liters (1) log 0.57812 
 
 1 imperial gallon (277.463 in^) = 4.54683 liters.. . " 0.65771 
 
 Weight. 1 grain = 64.7989* milligrams (mg). .. log 1.81157 
 
 1 ounce (avoir.) = 28.3495 * grams (g) " 1.45255 
 
 lib (avoir.) = 453.5924* grams " 2.65667 
 
 1 ton (2000 lb) = 907. 1849 kilograms (kg) " 2.95770 
 
 Force. 1 grain = 63.57 dynes. 
 
 1 pound = 4.45 X 10* dynes. 
 
 Stress. 1 pound/sq inch = 70.307 grams/sq centimeter., log 1.84700 
 1 pound/sq foot = 4.8824 kilograms/sq meter.. " 0.68863 
 
 Energy. 1 foot-pound = 1.3563 X 10' ergs 
 
 = 0.13825 kilogrammeters (kgm). . log 9.14067 
 1 horse-power-hour = 1,980,000 foot-pounds. 
 
 Activity 1 foot-pound/minute = 0.0226 watt log 8.35411 
 
 (Power). 1 horse-power = 33,000 foot-pounds/minute 
 
 = 746 watts 
 = 76 kilogrammeters/second. 
 
 Velocity. 1 foot/second = 0.3048 meter/second log'9.48402 
 
 1 mile/hour = 0.4470 m/sec " 9-65031 
 
 = 1.6093 km/hr " 0.20665 
 
 * These are the values given by the Office of Weights and Measures, 
 Washington, based on the act of July 28, 1866. 
 
365 
 
 MECHANICAL UNITS-METRIC TO BRITISH. 
 
 Length. 1 millimeter (mm) = 0.03937 inch (iu) log 8.59517 
 
 1 meter (m) = 39.37 * inches '• 1.59517 
 
 = 3.28083* feet (ft) " 0.51598 
 
 1 kilometer (km) = 0.62137* mile " 9.79335 
 
 Area. 1 sq centimeter (cm«) = 0. 1 550 *sq inches (in«). log 9.19033 
 
 Isq meter (m*) = 10.764* sq feet (ft*). .. . " 1.03197 
 
 = 1.196* sq yards " 0.07773 
 
 1 sq kilometer (km«) = 0.3861 sq mile " 9.58670 
 
 Capacity. 1 liter (1) = 61 .023 * cubic inches (iu^) log 1.78549 
 
 = 1.0567* U. S. quarts " 0.02395 
 
 = 0.8797 imperial quart " 9.94433 
 
 Weight. 1 gram (g) = 15.432 * grains log 1 . 18842 
 
 1 kilogram (kg) = 2.2046 * lb avoir " 0.34333 
 
 1 tonne (t) =1. 1023 * tons (2000 lb) " 0.04230 
 
 = 0.9842 ton (2240 lb) " 9.99308 
 
 Force. 1 dyne = 0.01573 grain log 8. 19673 
 
 = 0.00102gram " 7.00860 
 
 1 megadyne = 10' dynes. 
 
 1 gram = 981 dynes log 2.99167 
 
 1 kilogram =9.81 X 10^ dynes. 
 
 Btreie. 1 gram/sq centimeter = 0.01422 pound/sq inch, log 8.15290 
 1 kilogram/sq meter = 0.20482 pound/sq foot.. " 9.31137 
 1 barad = 0.00102 gram/sq cm. . .. " 7.00860 
 
 Energy. 1 erg = 7.37 X 10-8 foot-pounds. 
 
 1 megalerg = 10* ergs. 
 1 joule = 10'' ergs = 1 watt-second. 
 
 1 kilowatt-hour =1.34 horse- power-hours. 
 
 1 gram-centimeter = 981 ergs log 2.99167 
 
 1 kilogrammeter = 9.81 joules 
 
 = 7.2330 foot-pounds •' 0.85932 
 
 Activity. 1 watt = 1 joule/second 
 
 = 0.10194 kilogramraeter/second log 9.00834 
 
 = 44.2385 foot-pounds/minute " 1.64580 
 
 1 kilowatt =1.34 horse-power. 
 
 1 force de cheval = 75 kilogrammeters/second. 
 
 1 poncelet = 100 kilogrammeters/second 
 
 = 981 watts log 2.99167 
 
366 
 
 GRAVITATION UNITS. 
 
 
 
 
 British Units. 
 
 Metric Units. 
 
 Dynamical 
 
 Sym- 
 
 Defining 
 
 
 
 
 
 
 
 Quantities. 
 
 bols. 
 
 Equations. 
 
 
 
 
 
 
 
 
 Names. 
 
 Abbrevia- 
 tions. 
 
 Names. 
 
 Abbre- 
 viations. 
 
 Length .... 
 
 l,s 
 
 
 foot 
 
 ft 
 
 meter 
 
 m 
 
 Time 
 
 t 
 
 — 
 
 second 
 
 sec 
 
 second 
 
 sec 
 
 Weight .... 
 
 W, w 
 
 V 
 
 v=s/t 
 
 pound 
 
 lb 
 
 kilogram 
 
 meter per sec- 
 ond 
 
 kg 
 
 Velocity.. . 
 
 foot per sec 
 
 ft/sec 
 
 m/sec 
 
 Accelera- 
 
 
 
 
 
 
 tion ...... 
 
 a 
 
 az=v/t 
 
 foot per sec 
 per sec 
 
 ft/sec3 
 
 meter per sec 
 per sec 
 
 m/sec« 
 
 Force 
 
 F 
 
 F-wa/g 
 
 pound 
 
 .... 
 
 kilogram 
 
 
 Impulse.... 
 
 1 
 
 \=Ft 
 
 sec'd-poujid 
 
 — 
 
 second-kilogram 
 
 
 Momentum 
 
 1 
 
 \=wv/g 
 
 sec'd-pound 
 
 — 
 
 second-kilogram 
 
 
 Work 
 
 w 
 
 V/=Fs 
 
 foot-pound 
 
 — 
 
 kilogrammeter 
 
 kgm 
 
 Energy .... 
 
 w 
 
 V4=iwvyg 
 
 foot-pound 
 
 .... 
 
 kilogrammeter 
 
 kgm 
 
 Power (ac- 
 
 
 
 
 
 
 
 tivity)... 
 
 F 
 
 p^yN/t 
 
 ft-pound per 
 sec; horse- 
 power 
 
 ft-pound/sec 
 
 kgm per sec 
 
 kgm/sec 
 
 Moment 
 
 
 
 
 
 
 
 (torque) . 
 
 T 
 
 T=Fl 
 
 
 
 
 
 Moment of 
 
 
 
 
 
 
 
 inertia . . . 
 
 I 
 
 I^l.wr^ 
 
 ^ 
 
 
 • 
 
 
ABSOLUTE UNITS. 
 
 367 
 
 
 
 
 
 C.G.8. Units. 
 
 Practical Units. 
 
 Dynamical 
 
 Sym- 
 bols. 
 
 Defining 
 Equa- 
 tions. 
 
 Dimen- 
 sions. 
 
 
 
 
 Quantities. 
 
 Namt 8. 
 
 Abbrevi- 
 ations. 
 
 Names. 
 
 Abbrevi- 
 ations. 
 
 Length 
 
 1,8 
 
 
 L 
 
 centimeter 
 
 cm 
 
 meter 
 
 m 
 
 Time 
 
 t 
 
 .... 
 
 T 
 
 second 
 
 sec 
 
 second 
 
 sec 
 
 Mass 
 
 M, m 
 
 .... 
 
 M 
 
 gram 
 
 g 
 
 kilogram 
 
 kg 
 
 Velocity, 
 linear.. .. 
 
 V 
 
 V=8/t 
 
 LT • 
 
 kine 
 
 
 
 
 Accelera- 
 tion, lin'ar 
 
 a 
 
 a=v/t 
 
 LT" 
 
 s.wud 
 
 
 
 
 Velocity, 
 angular... 
 
 ia 
 
 w=r/r 
 
 T ' 
 
 radian/sec 
 
 
 
 
 Accelt*rat'n, 
 angular... 
 
 \ 
 
 a=u>/t 
 
 T» 
 
 radian/sec» 
 
 
 
 
 Force 
 
 F 
 
 F^ma 
 
 LMT" 
 
 dyne 
 
 
 megadyne 
 
 
 Impulse .... 
 
 1 
 
 \=Ft 
 
 LMT • 
 
 bole 
 
 
 
 
 Momentum. 
 
 1 
 
 |=mv 
 
 LMT ■ 
 
 bole 
 
 
 joule 
 - niega- 
 lerg 
 joule 
 
 
 Work 
 
 w 
 
 V^ = Fs 
 
 L'MT-' 
 
 erg 
 
 
 
 Energy... . 
 
 w 
 
 W = imt;a 
 
 L'MT « 
 
 erg 
 
 
 J 
 
 Power 
 
 p 
 
 P=Yf/t 
 
 L'MT' 
 
 .... 
 
 
 watt; kilo- 
 watt 
 
 kw 
 
 Moment 
 (torque) . . 
 
 T 
 
 T-^Fl 
 
 L«MT-» 
 
 
 
 
 
 Moment of 
 inertia... 
 
 I 
 
 7=2mr9 
 
 L'M 
 
 
 
 
 
 Density. .. 
 
 B 
 
 «=m/Z» 
 
 LM» 
 
 
 
 
 
 Pressure 
 
 P 
 
 p=F/l^ 
 
 L'MT' 
 
 barad 
 
 
 
 
 Modulus of 
 elasticity . 
 
 E 
 
 
 L'MT" 
 
 
 
 
 
INDEX. 
 
 Absolute units, 59, 348, 367 
 Acceleration, 18 
 
 unit of. 19 
 due to gravity, 55, 
 129 
 Accelerations, composition of, 25 
 resolution of, 25, 26 
 Action and reaction, 54 
 Activity, 250 
 Ampere, 3 
 
 Amplitude of a 8. H.M., 30 
 Angular acceleration, 283 
 
 unit of, 283 
 Angular velocity, 278 
 unit of, 279 
 Archimedes. 1, 159. 172 
 Aristotle, 69 
 Arm of couple, 153 
 Atwood machine, 65 
 
 Baily, 342 
 Balance, 172 
 Ballistic balance, 55 
 Ballistic pendulum, 315 
 Barad, 363 
 Bell-crank lever, 173 
 Belts, 219, 255 
 Bitilar suspension, 314 
 Blackburn pendulum, 131 
 Bole, 349 
 Borda, 175, 339 
 
 Capstan, 242 
 
 Carlyle, 43. 68 
 
 Center of gravity. 159 
 
 oscillation, 303 
 parallel forces. 144 
 percussion, 308 
 
 Central forces, 106 
 
 Centrifugal force, 120, 121, 285 
 
 Centripetal force, 120, 123 
 
 Centroid, 159 
 
 C.G.S. units, 348 
 
 Circular motion, 26, 119, 132 
 
 Clarke, 342 
 
 Coefficient of friction, 207 
 
 restitution, 327 
 
 Composition of accelerations, 25 
 couples, 152 
 energies, 261 , 269 
 forces, 74, 81, 141, 
 
 144, 145 
 harmonic motions, 
 
 32 
 velocities, 13, 281 
 
 Compression, 51, 321 
 
 Comstock, 342 
 
 Conical pendulum, 125 
 
 Conservation of energy, 260 
 
 Constrained motion, 90, 112 
 
 Coulomb, 208, 323 
 
 Couple, 149. 154 
 
 D'Alembert. 67 
 D'Alembert's principle, 288 
 De Morgan, 139 
 Descartes, 49 
 
 Differential pulley, 238. 247 
 Dimensions of units, 353 
 Displacement of a S.H.M., 30 
 Dynamics, 3 
 Dyne, 349 
 
 Earth round the Sun ? 39 
 Effective force, 288 
 Efficiency, 245, 253 
 Elasticity, modulus of, 321 
 Elevation of outer rail, 188, 205 
 Energy, 50, 60, 256 
 
 of impact, 330 
 
 369 
 
370 
 
 INDEX. 
 
 Energy, of rotation, 309 
 
 of translation, 256 
 of vibration, 353 
 units of, 258, 345, 349 
 Equations of motion, 48, 287 
 Equilibrium, 85 
 
 three forces, 85, 158 
 any forces, 88, 146, 
 
 156, 185 
 on a rough incline, 
 
 211 
 on a smooth incline, 
 116 
 Erg, 349 
 
 Falling bodies, 92 
 
 Fly-wheel, 311 
 
 Foot, 5 
 
 Footpound, 227, 258 
 
 Force, 45 
 
 centrifugal, 120, 285 
 centripetal, 120, 139 
 elements of, 72 
 representation of, 73 
 transfer of, 156 
 transmissibility of, 73 
 units of, 59, 349, 350 
 
 Force of inertia, 288 
 
 Foircault's pendulum, 282 
 
 Fourier, 354 
 
 Free motion, 90 
 
 Friction, 206 
 
 angle of, 209 
 belt, 219 
 
 coefficient of, 207, 210 
 rolling, 208 
 
 Friction brake, 252 
 
 Galileo, 44, 49, 67, 96, 144 
 Governor, Watt, 127 
 Graphic statics, 180, 192 
 Gravitation units, 57, 344, 366 
 Gravity, center of, 159 
 Greenhill, v, 62, 103 
 Gyration, radius of, 299 
 
 Harkness, 340 
 Harmonic curve, 33 
 Harmonic motion, 29, 107 
 Henrici, 196 
 Hoisting machine, 240 
 Hooke's law, 321 
 Horse-power, 251, 350 
 
 Horse-power hour, 251 
 Huygens, 122, 124, 303 
 
 Ice-boat, 70 
 Impact, 325 
 
 energy of, 330 
 Impulse. 49, 60. 306, 349 
 Inclined plane, motion on, 113 
 equilibrium on, 
 116, 234, 245 
 Independence of forces, 49 
 Indicator, 229 
 Inertia, 44 
 
 force of, 288 
 
 law of, 44 
 
 moment of, 290 
 Instantaneous center, 272 
 Isochronism, 31, 129 
 
 Jointed frame, 193 
 
 Joule, 350 
 
 Joule's equivalent, 357 
 
 Kepler, 122 
 
 Kilogram of the archives, 340 
 
 international, 341 
 Kilogrammeter, 345, 359 
 Kilowatt, 350, 359 
 Kilowatt hour, 350 
 Kine, 348 
 Kinematics, 3 
 Kinematical units, 4 
 Kinetic energy, 258 
 Kinetics, 85 
 Knot, 9 
 
 Lami's theorem, 87 
 J^aws of motion, 44 
 Leibnitz, 67, 258 
 Lever, 172, 236 
 Lewes, 122, 132 
 Liter, 341 
 
 Mach. 47, 125 
 
 Machine, 232 
 
 Mass, 47, 55, 57, 348 
 
 Mass jicceleration, law of, 48 
 
 Maxwell, 61, 260, 354 
 
 Mechanics, 2 
 
 Mechanical units, 364, 365 
 
 Megadyne, 350 
 
 Megalerg, 350 
 
 Mendeuhall, 62 
 
INDEX. 
 
 371 
 
 Meter, of the archives, 338 
 international, 341 
 
 Metric system, use of, 348 
 
 Micron, 342 
 
 Moment of a couple, 153 
 of a force, 149 
 of inertia, 290 
 units of. 150, 293 
 
 Momentum, 49, 60, 349 
 
 Motion in a circle, 118, 132 
 
 under constant force, 91 
 under variable force, 106 
 under impulse, 306 
 on a smooth incline, 113 
 on a rough incline, 216 
 
 Nasmyth, 334 
 Nautical mile, 9 
 Neutral equilibrium, 190 
 Newton, 75, 122, 129 
 
 Oscillation, motion of, 29, 106 
 
 of a pendulum, 129, 
 
 301 
 energy of, 353 
 
 Parallel forces, 142, 180 
 Parallelogram of acceleration, 25 
 forces, 74 
 velocities, 14 
 Pearson, 40 
 Pendulum, Blackburn, 131 
 
 compound, 301 
 
 conical, 125 
 
 seconds, 129 
 
 simple, 128 
 Period of S.H.M., 30,32 
 Phase, 30 
 Piioronomics, 3 
 Pile driving, 332 
 Piston velocity, 275 
 Planetary motion, 112 
 Plumb-line, 124 
 Pole of stress diagram, 146 
 Polygon of forces, 88 
 Poncelet, 359 
 Potential energy, 258 
 Pound, imperial, 57, 61 
 troy, 58 
 United States, 58. 342 
 
 [force], 59 
 Power, 250 
 
 unit of, 261, 345, 350 
 
 Projectiles, 98 
 Pulley, 237, 246 
 
 Radius of gyration, 299 
 Range of a projectile, 101, 103, 105 
 Relative motion, 36 
 Repose, angle of, 210 
 Resolution of accelerations, 25 
 forces, 79 
 velocities, 16, 282 
 Resultant. 74 
 Roberval, 15 
 Rogers, 342 
 Rotation, 271 
 
 energy of, 309 
 Row-boat, 187 
 
 Sail-boat, 80 
 Screw, 240, 248 
 Seconds pendulum, 129 
 Shear, 52 
 
 Smooth surface, 113 
 Speed, 7 
 Spoud, 349 
 Stability, 189. 263 
 Standards of length, 4. 338 
 time, 5, 46 
 weight, 57, 340 
 Statics, 85 
 
 Steam-engine, 181, 274 
 Steam-hammer, 334 
 Steelyard, 177 
 Stevinus, 1, 75 
 Strain, 321 
 Stress, 51 
 
 law of, 52 
 
 diagram, 195 
 Strut, 194 
 
 Tension, 51 
 Tie, 194 
 Time, 5, 46 
 Toggle-joint, 277 
 Torque, 153, 356 
 Torsion, 322 
 
 pendulum, 823 
 Traction, force of, 255 
 Trajectory, 98 
 Translation, 6, 72 
 Triangle of forces, 86 
 Troughton scale, 5 
 Troy pound, 58 
 
372 
 
 INDEX. 
 
 Units, absolute, 39, 348, 367 
 derived, 8, 19, 355 
 fundamental, 4, 354 
 gravitation, 59, 344, 366 
 of acceleration, 18, 349 
 of activity, 250, 350 
 of energy, 258, 345, 350 
 of force, 59, 349, 350 
 of impulse, 349 
 of length, 4, 5, 338, 341 
 of momentum, 60, 349 
 of pressure, 356, 363 
 of time, 5 
 of velocity, 7, 349 
 of weight, 57, 340 
 of work, 227, 349 
 
 Varignon's theorem, 151 
 Velocity, 7 
 
 angular, 278 
 Vis viva, 258 
 
 Watt, 350 
 
 Watt second, 350 
 
 Watt, James, 127, 251 
 
 Watts to horse-power, 358 
 
 Wave-length, 34 
 
 Weight, 57, 61 
 
 Wheel and axle, 236 
 
 Winch, 243 
 
 Woodward, iii 
 
 Work, 50, 60, 224 
 
 against friction, 244 
 principle of, 230 
 of resilience, 325 
 unit of, 227, 349 
 
 Worm wheel, 244 
 
 Yard, imperial, 4, 342 
 
 United States, 5, 342 
 Young, Thomas, 50, 324 
 Young's modulus, 324 
 
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 1^0.31. THE SANITARY CONDITION OF DWELLING- 
 
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 W. Hildenbrand, C.E. 
 
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 TIOJ 
 
 ION AND USE. Compiled by George W. Plympton. Fourth edition 
 
 No. 36 MATTER AND MOTION. By J. Clerk Maxwell. M.A. 
 .Second American edition. 
 
SCIENCE SERIES. 
 
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 Prof. William Cain, A.M., C.E. 
 
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 OF ARTICULATED LINKS. By J. D. C. de Roos. 
 
 No. 48. THEORY OF SOLID AND BRACED ARCHES. By 
 
 William Cain, C.E. 
 
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 Thomas Craig, Ph.D. 
 
 No. 50. DWELLING-HOUSES : THEIR SANITARY CCi;- 
 
 STRUCTION AND ARRANGEMENT.S. By Prof. W. H. Corfield. 
 
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 Thomas Nolan. 
 
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 M. Argand. By Prof. Hardy. 
 
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 Fifth edition. 
 
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 an introduction by Prof. R. H. Thurston. 
 
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 A. de Varona. 
 
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 Description of the Edison System. By L. H. Latimer, to which is 
 added the Design and Operation of Incandescent Stations, by C. J, 
 Field, and the Maximum Efficiency of Incandescent Lamps, by John 
 W. Howell. 
 
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D. VAN NO STRAND COMPANY'S 
 
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 * MENTS. By S. W. Robinson, C.E. 
 
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 By W. P. Gerhard. Sixth edition, revised. 
 
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 No. 65. POCKET LOGARITHMS TO FOUR PLACES OF DECI- 
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 No. 66. DYNAMO-ELECTRIC MACHINERY. By S. P. Thompson. 
 With notes by F. L. Pope. Third edition. 
 
 No. 67. HYDRAULIC TABLES BASED ON " KUTTER'S 
 
 FORMULA." By P. J. Flynn. 
 
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 with additions by A. R. Wolff. 
 
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 edition, revised and enlarged. 
 
 No. 70. EXPLOSIVES AND EXPLOSIVE COMPOUNDS. By 
 
 M. Bertholet. 
 
 No. 71. DYNAMIC ELECTRICITY. By John Hopkinson, J. A. 
 Schoolbred, and R. E. Day. 
 
 No. 72. TOPOGRAPHICAL SURVEYING. By George J. Specht, 
 Prof. A. S. Hardy, John B. McMaster, and II. F. Walling. 
 
 No. 73. SYMBOLIC ALGEBRA; OR, THE ALGEBRA OF 
 
 ALGEBRAIC NUMBERS. By Prof. W. Cain. 
 
 No. 74. TESTING MACHINES : THEIR HISTORY, CON- 
 STRUCTION, AND USE. By Arthur V. Abbott. 
 
 No. 75. RECENT PROGRESS IN DYNAMO-ELECTRIC MA- 
 CHINES, ^eing a Supplement to Dynamo-Electric Machinery. By 
 Prof. Sylvanus P. Thompson. 
 
 No. 76. MODERN REPRODUCTIVE GRAPHIC PROCESSES. 
 
 By Lieut. James S. Pettit, U.S.A. 
 
 No. 77. STADIA SURVEYING. The Theory ot Stadia Measurements. 
 By Arthur Winslow. 
 
 No. 78. THE STEAM-ENGINE INDICATOR, AND ITS USB 
 By W. B. Le Van. 
 
 No. 79. THE FIGURE OF THE EARTH. By Frank C. Roberts, C.E. 
 
 No. 80. HEALTHY FOUNDATIONS FOR HOUSES. By Glenr 
 Brown- 
 
SCIENCE SERIES. 
 
 No. 8i. WATER METERS : COMPARATIVE TESTS OF 
 
 ACCURACY, DELIVP:kY, etc. Distinctive features of the Worth- 
 ington, Kennedy, Siemens, and Hesse meters. By Ross E. Browne. 
 
 No. 82. THE PRESERVATION OF TIMBER BY THE USE 
 
 OF ANTISEPTICS. By Samuel Bagster Boulton, CE. 
 
 No 83. MECHANICAL INTEGRATORS. By Prof. Henry S. H. 
 Shaw, C.E. 
 
 No. 84. FLOW OF WATER IN OPEN CHANNELS, PIPES, 
 
 CONDUITS, SEWERS, ETC. With Tables. By p. J. Flynn, C.E. 
 
 No. 85 THE LUMINIFEROUS ^THER. ByProf.de Volson W^ood. 
 
 No. 86. HAND-BOOK OF MINERALOGY; DETERMINATION 
 
 AND DESCRIPTION OF MINERALS FOUND IN THE UNITED 
 STATES. By Prof. J. C. Foye. 
 
 No. 87. TREATISE ON THE THEORY OF THE CON- 
 STRUCTION OF HELICOIDAL OBLIQUE ARCHES. By John 
 L. Culley, C.E. 
 
 No. 88. BEAMS AND GIRDERS. Practical Formulas for their Re- 
 sistance. By P. H. Phil brick. 
 
 No. 89. MODERN GUN-COTTON : ITS MANUFACTURE, 
 
 PROPERTIES, AND ANALYSIS. By Lieut. John P. Wisser, U.S.A. 
 
 No. 90. ROTARY MOTION, AS APPLIED TO THE GYRO- 
 SCOPE. By Gen. J. G. Barnard. 
 
 No. 91. LEVELING: BAROMETRIC, TRIGONOMETRIC, AND 
 
 SPIRIT. By Prof. I. O. Baker. 
 
 No. 92. PETROLEUM : ITS PRODUCTION AND USE. By 
 
 Boverton Redwood, F.I.C., F.C.S. 
 
 No. 93. RECENT PRACTICE IN THE SANITARY DRAIN- 
 AGE OF BUILDINGS. With Memoranda on the Cost of Plumbing 
 Work. Second edition, revised. By William Paul Gerhard, C. E. 
 
 No. 94. THE TREATMENT OF SEWAGE. By Dr. C. Meymott 
 
 Tidy. 
 
 No. 95. PLATE GIRDER CONSTRUCTION. By Isami Hiroi, C.E, 
 Second edition, revised and enlarged. Plates and Illustrations. 
 
 No. 960 ALTERNATE CURRENT MACHINERY. By Gisbert 
 Kapp, Assoc. M. Inst., C.E, 
 
 No. 97. THE DISPOSAL OF HOUSEHOLD WASTE. By W, 
 Paul Gerhard, Sanitary Engineer. 
 
 No, 98. PRACTICAL DYNAMO-BUILDING FOR AMATEURS. 
 HOW TO WIND FOR ANY OUTPUT. By Frederick Walker. 
 Fully illustrated. 
 
 No. 99.® TRIPLE-EXPANSION ENGINES AND ENGINE 
 TRIALS. By Prof. Osborne Reynolds. Edited, with notes, etc., by 
 F. E. Idell. M. E, 
 
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 29Nlav'5tWK 
 
 «BC« 
 
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 LD 21-100m-9,'47(A5702sl6)476 
 
IW304882 
 
 
 THE UNIVERSITY OF CALIFORNIA UBRARY