GIFT OF Daughters of P'rederick Slate sss A TREATISE ON ELEMENTARY STATICS A TEEATISE ON ELEMENTAKY STATICS BY JOHN GREAVES, M.A., 't FKLLOW AND MATHEMATICAL LECTURER OP CHRIST's COLLEGE, CAMBRIDGE. ^,>^i^ ^' ^ . -> .J O rt J •« r^i Honbon : MACMILLAN AND CO. 1886 [All Rights reserved.] I Cambritigf : PRINTED BY C. J. CLAY, M.A. AND SON, AT Tin: UNIYERKITY PRESS. 4 «^ f < 1 1 « PREFACE. There has existed for some time past a general feeling that the Laws of Motion form the only satisfactory basis on which the science of Statics can be built. So far as I know, all the text-books in use in Cambridge except Prof. Minchin's, treat the subject from quite a different point of view. In this text-book I have endeavoured to supply the wants of students, who are not sufficiently advanced in Pure Mathematics, to read with advantage Professor Minchin's treatise on Analytical Statics. Deducing from the Newtonian definition of force and the parallelogram of velocities, the parallelogram of forces, I obtain the necessary conditions of equilibrium for any material system by means of the third law, without assum- ing the transmissihility of force, or supposing the system to become rioid. From these and certain oeometrical considerations follow the sufficient conditions of equili- brium of a rigid body. This involves the introduction of the conception of the moment of a force about a line, and certain geometrical propositions, which may be regarded as somewhat difficult for a beginner : I am in hopes that these difficulties will not be found insuperable, as it seems to me that there is a distinct gain in clearness and simplicity by this mode of treatment of the subject. The Appendix on indefinitely small quantities has been added 994525 VI PREFACE. to enable the student, who is unacquainted with Newton's Lemvias and the Differ^ential Calculus, to follow the methods used in the chapters on the Centre of Mass and Virtual Work. For the sake of students beginning the subject, easy numerical examples on the preceding propositions have been embodied in the text. The articles, marked with an asterisk, may be reserved for a second reading of the sub- ject. Explanations and illustrations are printed in smaller type than the articles relating to general principles. With the view of making the diagrams more intelligible, the ]bounding lines of physical surfaces are drawn thicker than lines representing forces, and lines diawn merely to obtain a geometrical solution of the problem are dotted. I have referred continually to Thomson and Tait's Natural Philosophy, and have also consulted Jellett's Theory of Friction. Several of the Illustrative Examples are taken from Dr Wolstenholme's Collection. I am much indebted to Mr E. W. Hobson, M. A., Fellow of Christ's College, for many valuable suggestions, and also to Mr J. B. Holt, B.A., Scholar of Christ's College, for his kind criticisms and assistance. JOHN GREAVES. Christ's College, Cambridge, April, 1886. CONTENTS. CHAPTER I. Statics of a Single Particle Parallelogram of velocities Parallelogram of forces Triangle of forces Polygon of forces Conditions of equilibrium . Theorem relating to moments about a point Conditions of equilibrium .... Theorem relating to moments about a line CHAPTER II. Statics of Systems of Particles .... Necessary conditions of equilibrium .... Sufficient conditions of equilibrium for coplanar forces Sufficient conditions of equilibrium for any system . Resultant of two parallel forces .... Equivalence of couples ...... Couples follow parallelogram law .... Poinsot's central axis ...... CHAPTER III. Statics of Constrained Bodies, &c. Conditions of equilibrium Bending moment ....... Tension of string stretched over smooth surface CHAPTER IV Centres of Mass ..... Position of centre of parallel forces . Theorem relating to centres of mass . Centre of mass of a triangle PAGE 1 5 11 14 15 31 34 34 36 54 55 60 63 67 72 75 78 99 100 103 106 123 124 129 132 Vlll CONTENTS. Centi-e of mass of a pyramid . Centre of mass of arc of circle . Centre of mass of zone of sphere Condition of body resting on horizontal plane TAGK 13(; 140 142 149 CHAPTER V. Friction ........ Laws of friction ...... Laws of limiting friction . . . . . Tension of string stretched over rough surface 165 166 167 170 CHAPTER VI. Virtual Work .... Principle for single particle Principle for rigid body . . . Converse of principle for rigid body Positions of stability and instability General theorem relating to potential energy 200 200 202 208 210 217 CHAPTER VII. Machines Lever Wheel and axle Systems of pulleys Inclined plane . Screw Balance . Common Steelyard Danish Steelyard Koberval's balance Differential wheel and axle Differential screw 233 234 238 239 246 249 254 257 258 259 260 202 APPENDIX. Guldin's theorems . . . 269 STATICS. CHAPTER I. STATICS OF A SINGLE PARTICLE. 1. When a point is changing its position relatively to surrounding points, it is said to be in motion relatively to them : if it is not changing its position, it is said to be at rest. If we consider not only the actual change of position, but also the time which the motion occupied, we bring in the idea of ' rate of motion' or ' velocity'. 2. Def. If a point moves so that the distances, measured along its path, between its positions at the ends of equal successive intervals of time, are equal to one another, no matter how short the intervals are, the velocity of the point is said to be 'uniform'. If the distances are not equal, the velocity is ' varying'. For the velocity to be uniform, it is essential that the distances be equal, even when the intervals of time are indefinitely small: for instance, we may imagine a train travelling 30 miles during each of several successive hours, yet we should not describe its motion as uniform, if the distances travelled during the different minutes were not all equal, nor yet, even though the distances travelled during the different minutes were so, provided those travelled during the different seconds were not always the same, and so on indefinitely. G. 1 2 STATICS. dr : i5^f wa wisL to give ^.ny one a clear idea of the mag1aitii"dG 1")^ senile' pbysi'';?! quantity, we describe it as bearing such and such a ratio to some definite arbitrarily chosen amount of that quantity, known to him. The known definite amount is called the ' imit' of the physical quantity generally, while the ratio is called the 'numerical measure', or simply the measure of the particular amount under consideration. If for instance, the area of a certain field be 12| acres, and an acre be chosen as the unit area, the ratio of the area of the field to that of the unit is 12|, which is therefore the numerical measure of the area of the field. We shall suppose then, that we have fixed on some particular length as the unit length, and some particular interval of time as the unit of time. If the velocity of a point be uniform, its numerical measure is the numerical measure of the distance traversed by it during the unit of time. It may happen that the point's velocity, though uniform for a finite interval of time, is not so for the unit of time : in that case, its numerical measure is that of the space the point would traverse during the unit of time, provided it moved throughout with the same velocity as during the finite time. The velocity which we call the unit velocity, or whose numerical measure is one, is the velocity of a point which traverses the unit of length in the unit of time. Def. The mean or average velocity of a point during any interval of time is the velocity with which a point, moving uniformly during that time, would describe the same distance. Its numerical measure is therefore the numerical measure of the distance described, divided by that of the time required. Def. The velocity of a point at any instant, is the limit of the mean velocity of the point during an interval of time including the particular instant, when the interval is diminished indefinitely. STATICS OF A SINGLE PARTICLE. 3 Ex. 1. Compare the velocities of two points which move uniformly, one through 5 feet in half a second, and the other through 100 yards in a minute. Aiis. 2:1. Ex. 2. A railway train travels 160 miles in 6 hours 30 minutes. What is its average velocity in feet per second? Ans. 36*1 nearly. Ex. 3. One point moves uniformly twice round the circumference of a circle, while another moves uniformly along the diameter : compare their velocities. Ans. 27r : 1. Ex. 4. A fly-wheel is 14 feet in diameter, and is observed to go round uniformly fifteen times in a minute : find the velocity of a point in the circumference. Ans. 11 feet per second nearly. Ex. 5. Supposing the earth to rotate about its axis in 23 hours 56 minutes, its equatorial diameter being 7925 miles, find the velocity of a point at the equator relative to the earth's centre, in feet per second. Ans. 1526 nearly. 4. Now a velocity is entirely known, if its direction and magnitude are known. But as a straight line AB can be drawn in any direction, it can be drawn so as to in- dicate fully the direction of a point's velocity, provided we shew Fig.i either by an arrow-head or by the a b order of the letters AB, the sense of the velocity, i.e. whether its direction be from A to B or from B to A. As we can make the line of any length, we can make it so that its length bears the same ratio to some arbitrarily chosen length as the velocity considered bears to the unit of velocity. If this be done, and we know the 'scale', i.e. the length chosen to represent the unit velocity, the line AB represents completely the velocity considered. 5. A point may be moving with several independent velocities at once : for instance, we know that the earth as a whole is describing an orbit about the sun, and that all points on the earth's surface are describing circles about the earth's axis ; if then, a point be moving on the earth's surface, it has relatively to the sun, three independent velocities, viz. its velocity on the earth's surface, the velocity of the point of the earth's surface it occupies at the particular instant, relatively 1—2 4 STATICS. to the centre of the earth, and the velocity of the earth's centre about the sun. Def. When a point has several independent veloci- ties, the single velocity which would alone give the point's motion is called the resultant of the other velocities. Let us consider the case of a point moving in a straight line along the deck of a ship, with uniform velocity relative to the ship, which is sailing with uniform velocity in a straight line along the earth's surface. It is required to find the point's motion relative to the earth's surface, i.e. given its position at one instant, it is required to find its position at the end of a given time. Now since the point's motion on the ship's deck is entirely independent of the ship's motion, if we suppose the point fixed to the deck during the time considered, so that its motion is that of the ship, then the ship to remain stationary while the point moves for an equal time along the deck with its velocity relative to the ship, the final position of the point will be the same as if the two motions had taken place simultaneously, as they really do. The above illustration exemplifies a general axiomatic principle, which may be stated thus : if during a certain time a 'point has several independent motions, its actual position at the end of any portion of that time may he found by imagining that all the motions take place separately during a number of successive periods of time equal to the one considered, instead of supposing that all the motions take place siiyiidtaneously , which is ivhat really takes place. Of course the imaginary motion only gives the same initial and final positions of the point as the real one, aud not in general intermediate ones, although by taking the periods of time very small, but very large in number, the imaginary motion which gives us the real position of the point at the end of each of them, will give us an infinite number of points on the point's actual path. The motions referred to above are not of necessity due to uniform velocities. STATICS OF A SINGLE PARTICLE. 5 6. Prop. If the two independent velocities of a point be represented in magnitude, direction and sense by two straight lines drawn from or to a point, and a paral- lelogram be constructed on them as adjacent sides, the resultant velocity is represented in magnitude, direction and sense by the diagonal drawn from or to the point of intersection of these sides. Let the lines OA, OB represent in magnitude, sense and direction the velocities u, v of the point : complete the parallelogram OAGB, and join 00; then OC shall represent the resultant velocity. If be taken as the Fig.2 initial position of the point, its position at the end of a time t can be found by supposing that it first moves with a velocity OA for a time t, and then with a velo- city OB for the same time. If it moves with a velocity OA or It alone, it will at the end of a time ^ be at a in the line OA, where Oa = ut ; if now it moves with the velocity OB or V alone for a time t, it will arrive at c, where ac is parallel to OB, and ac = vt c then is the position of the point at the end of a time t, when the motions take place simultaneously. But because ac : Oa = AC : OA, c is in 00, i.e. 00 re- presents the direction of the resultant velocity. Also the magnitude of the resultant velocity is to the velocity OA as Oc is to Oa, i.e. as 00 : OA : hence 00 represents the magnitude of the resultant velocity. The sense of the resultant velocity is clearly OC. 6 STATICS. The above proposition which is known as the 'Paral- lelogram of Velocities' holds at any instant, even though the independent velocities be varying velocities : for it is only necessary to suppose that the time t is ultimately indefinitely small, and the above proof holds. Ex. 1. Velocities of 4 feet and 16 feet per second in directions at right angles to each other are simultaneously communicated to a body : determine the resultant velocity. Ans. 16"49 feet per second. Ex. 2. A ship whose head points N.E. is steaming at the rate of 12 knots an hour in a current which flows S.E. at the rate of 5 knots an hour, find the velocity of the ship relative to the sea bottom. Ans. 13 knots an hour. 7. All the objects around us that we can see and touch, and even invisible substances, such as air, are material bodies or composed of matter. The various pro- perties of matter, such as hardness, density, &c., can be in- vestigated, but no definition of matter can be given which would give any idea of it to a being that had had no ex- perience of it. Any limited portion of matter is called a ' material body' or simply a 'body'. When we consider a body whose dimensions are so small that we are only concerned with its motion as a whole, and not with any rotational motion it may have, we describe it as a material particle^ or simply a particle. The term ' mass ' is synonymous with the phrase 'quantity of matter', so that the mass of a body means the quantity of matter in it. We shall learn afterwards (Art. 9) what is meant by saying that the quantity of matter in one body bears a certain ratio to that in another. 8. Statics is the science which treats of the equi- librium of bodies under the action of 'forces'. A defini- tion of the term 'force ' is supplied by Newton's 1st Law, which asserts that 'Every body remains in a state of STATICS OF A SINGLE PARTICLE. 7 rest or of uniform motion in a straight line, except in so far ! as it may he comj^elled by impressed forces to change that state '. Force, then, is that which alters or tends to alter the state of rest or of uniform motion in a straight line, of a body. It is not necessary to suppose that the state of rest or of uniform motion is actually altered by a force, because other forces may be in action which counteract the etfect of the first. If we observe a body moving in any way other than uniformly in a straight line, we infer that it is acted on by force: e.g. when we find that the planets move in nearly elliptic orbits, we know that each is under the action of some force: similarly when we see that a falling body moves with gradually increasing velocity, or that another is stopped, we know that a force has acted on each of them. If a force acts for a time on a body, producing a change in the body's velocity, it is clear that | if it continues to act, it will tend to produce a still further ' change. I 9. The next question that presents itself is ' How is force measured? ' or 'When may this force be said to bear | such and such a ratio to that force?' We know by ex- perience that it requires a greater effort on our part to impart velocity to a large amount of any substance than it does to impart the same velocity to a small amount, but what determines the exact ratio that exists between the two forces? Our own sensations do not give us an accurate scale by which the forces may be measured. The answer is con- tained in Newton's 2nd Law, wdiich asserts that 'Change of ' motion is proportional to the impressed force and takes place in the direction of that force'. ; What does the expression ' Change of Motion mean ? i First, let us suppose that a number of different forces act on the same particle during equal intervals of time, so \ that the only variations in the different cases are the ' 8 STATICS. differences in the mao^nitudes and directions of the forces and in the changes of velocity produced. In this case, ' Change of Motion is understood to be a quantity pro- portional to change of velocity. Hence forces are equal if, when they act for equal times on the same particle, they produce equal changes of velocity, and the ratio between the magnitudes of two forces is the ratio between the respective changes of velocity they produce in the same particle, after acting for equal times. The direction of a force is clearly defined by the latter part of the law as the direction of the change of velocity produced by the force. It is of course to be understood that the change of velocity meant is not necessarily the increase or decrease of the particle's velocity, but that velocity which, com- pounded with the particle's initial velocity, will give the final velocity. Suppose now that a number of equal forces act one on each of a number of particles for the same time, and produce the same changes in their velocities : we express the relation that holds among the particles by saying they are of 'equal mass\ If there be n particles of equal mass, initially all moving side by side with the same velocity, and n equal forces act, one on each, in the same direction, for the same length of time, the particles will finally be found moving side by side with the same velocity. Hence we infer, that to produce in a particle, whose mass is 7i times that of another, the same change of velocity that a given force produces in the latter, and in the same time, a force n times as great as the given force has to be applied to the former. It seems then, that the force varies as the mass of the particle, if the change of velocity produced in a given time is always the same, and we have seen that it varies as the change of velocity produced in a given time, when the mass remains constant. The numerical measure of a force then is proportional to the product of the mass acted upon into the change of velocity produced in a given time. STATICS OF A SINGLE PARTICLE. 9 The expression 'change of motion' means the product of the mass into the change of velocity produced in the given time, or the change of Momentum produced in the given time, if the momentton of a particle be defined as the product of its mass into its velocity. 10. We are all of us familiar with some instances of the manifestation of force. For instance, we may set a body in motion or stop it by pushing it, either directly with the hand, or by means of a rod, or we may pull it by a string attached to it : we may also expose it to the action of the wind or to the pressure of steam. In all these cases the force is exerted by tangible means, but force is often manifested without any tangible means, as in the case of gravity, the name given to the force which causes any body near the earth to move towards it, and the planets to revolve in their orbits about the sun; also in the case of the force which causes small pieces of iron to move towards a magnet held near them. A force ctf this kind is called an attraction. In both theoretical calculations and in actual practice we must fix on some standard force which is to be the unit. In theoretical calculations we take as our unit the * dyne ', which is the force required to generate in one second a velocity of 1 centimetre per second in a mass equal to that of a cubic centimetre of distilled water at 4^C. This is called the 'absolute unit' and the advantage in its use is, that all the terms involved in its definition are the same at all points of the earth's surface and indeed everywhere. It is found that if bodies be allowed to fall towards the earth in a vacuum, so that the air does not resist their motion, the velocities with which they fall are increased every second by an amount always the same at the same point of the earth's surface, and nearly so all over it. The force which produces this change of motion in a body is called its 'weight' : hence the w^eights of different bodies are proportional to their masses, since the change of velocity' produced is the same for all. Assuming for the present, what we shall prove hereafter (Art. 13), that when a body is at rest under the action of two forces, they are equal in magnitude and opposite in direction, we see that the force required to support a body is equal and opposite to its weight, and would, if it acted 10 STATICS. alone, produce in the body the same change of motion upwards that its weight does downwards. In practice in Statics, forces are generally measured in terms of the weights they would support if they acted upwards : for instance in England that force that is just sufficient to support a certain lump of metal kept at the Mint and called the Imperial Pound, is very often regarded as the unit force, the slight variations in this force at different places being of little consequence for practical purposes. The velocity of a falling body is increased every second by 32 feet per second, approximately. Ex. 1. If a body weighing 60 lbs. be moved by a constant force which generates in it in a second a velocity of 5 feet per second, find what weight the force would statically support. Ans. 9'3 lbs. nearly. Ex. 2. During what time must a constant force equal to the weight of one ton act upon a train of 100 tons to generate in it a velocity of 40 miles an hour ? Ans. 3 min. 3^ sees. Ex. 3. A force which can statically support 25 lbs. acts uniformly for one minute 'on a mass of 400 lbs. : find the velocity acquired by the body. Ans. 120 feet per second. 11. Def. The resultant of a number of forces is that single force whose effect is the same as that of the original forces. It frequently happens that there are several forces acting simultaneously on a body : e.g. a kite in the air is acted on by its weight, by the pressure of the wind and by the tension of the string attached to it. In the case of a particle acted on by several forces, we shall shew that there is a single force which could produce exactly the same effect as the other forces do. The second Law of motion states that the change of motion is proportional to the impressed force and takes place in the direction of that force, so that if there are several impressed forces we infer that the actual motion will be the resultant of the several independent motions which the forces would produce if they acted separatel}^ because the law holds for each force, and therefore these independent motions must each be produced. But this STATICS OF A SINGLE PARTICLE. 11 resultant change of motion might be produced by a single force, which is therefore the resultant of the original forces. 12. When a particle is in equilibrium or moving uniformly in a straight line under the action of a number of forces there is no change of motion, and therefore the resultant force must be zero ; conversely, when the re- sultant force is zero, there is no change of motion, and the particle must be at rest or be moving uniformly in a straight line. The necessary and sufficient condition of a particle's being in equilibrium under the action of a number of forces is that their resultant be zero. Note, strictly speaking, if the resultant force on a particle is zero, it only shews that the particle's velocity is undergoing no change, and not that it is necessarily zero. As however in this subject we always suppose the particle initially at rest, if this condition holds, it will always remain so. 18. We have already inferred from Newton's second Law, that the direction of a force is that of the change of motion it produces, and that its magnitude is propor- tional to that of the change of motion : so that a force is completely defined when the magoitude and direction of the change in velocity it produces in a given time, when acting on a particle of given mass, are given. A force may therefore be represeoted completely by the straight line that represents this change in velocity. We are now in a position to prove a most important pro- position, known as the 'Parallelogram of Forces'. Prop. If two straight lines be drawn from or to a point representing in magnitude, direction and sense, forces acting on a particle, and a parallelogram be constructed having these two lines as adjacent sides, the diagonal drawTi from or to the point mentioned will represent the resultant completely. Let OA, OB be two straight lines representing the magnitude, direction and sense of two forces acting on 12 STATICS. Fig. 3 a particle. Complete the parallelogram OACB, having OA, OB for two adjacent sides, and join OC. Since OA, OB represent the forces completely, they also represent the changes in ve- locity they would separately produce in a certain time in a particle of certain mass. By the parallelogram of ve- locities then, OC represents the resultant change of velocity they would produce in the same time in the same particle, and therefore represents the resultant of the original forces. The following particular case of this proposition is very important : siuce the diagonal of a parallelogram always has a finite length unless the two adjacent sides are equal in length and in opposite directions, the resultant of two forces is never zero, i.e. the forces do not counter- balance one another, unless they are equal in magnitude and opposite in direction. Cor. If three forces not in one plane acting on a particle, be represented in every respect by three lines OA, OB, OC drawn from a point, and a parallelopiped be con- structed on these lines as adjacent edges, the diagonal OGoi the parallelopiped represents the resultant in every respect. Fig.4 For OF is clearly the resultant of OA and OB, and since OFGG is a parallelogram {OC, FG being equal and STATICS OF A SINGLE PARTICLE. 13 parallel), OG is the resultant of OF and OC, i.e. of OA^ OB, OC. Ex. 1. Find the resultant of two forces of 12 lbs. and 35 lbs. respect- ively, which act at right angles on a particle. Ans. 37 lbs. Ex. 2. If two forces acting at right angles to each other be in the proportion of 2 to yJ5, and their resultant be 81 lbs. find the forces. Ans. 6 lbs., 3>/51bs. Ex. 3. The resultant of two forces which act at right angles on a particle is 51 lbs. : one of the components is 24 lbs. : find the other. Ans. 45 lbs. Ex. 4. Two forces act on a particle, and their greatest and least possible resultants are 17 lbs. and 3 lbs. : find the forces. Ans. 7lbs., 10 lbs. Ex. 5. Two forces acting in opposite directions to one another on a particle have a resultant of 28 lbs. : and if they acted at right angles they would have a resultant of 52 lbs. : find the forces. Ans. 48 lbs., 20 lbs. Ex. 6. Two forces, one of which is three times the other, act on a particle, and are such that if 9 lbs. be added to the larger, and the smaller be doubled, the direction of the resultant is unchanged : find the forces. Ans. 9 lbs., 3 lbs. Ex. 7. Shew that if the angle at which two given forces are inclined to each other is increased, their resultant is diminished. Ex. 8. If the resultant of two forces is at right angles to one of the forces, shew that it is less than the other force. Ex. 9. If the resultant of two forces is at right angles to one force and also equal to the other divided by J2, compare the forces. Ans. 1 : J 2. 14. Prop. If a particle be in equilibrium under the action of a number of forces, any one of them is equal and opposite to the resultant of the rest. From the definition of a resultant, all the forces but one can be replaced by their resultant without altering their effect, so that this resultant force and the remainino: ..... . ^ force mamtam equilibrium, which we have seen can only be the case when they are equal and opposite. 14 STATICS. 15. The following proposition known as the 'Triangle of Forces ' is practically another way of stating the ' Paral- leloo^ram of Forces '. If three forces acting on a particle can be represented in magnitude, direction and sense by the sides of a triangle, taken in order, the forces are in equilibrium. By the phrase 'taken in order' is meant, that the arrowheads which indicate the directions of the forces, should all point the same way round the triangle, or that no two should both point to or from the same point. Let ABC be a triangle whose sides AB, BC, GA, taken in order, represent in magnitude, direction and sense three forces acting on F'g.5 a particle — the particle shall D^ - — - --^ be in equilibrium. \ ^^ \ Complete the parallelo- \ jy^ V gram BGAD. Since BD is 1 y^ \ equal and parallel to GA^ it \ y^ \ will represent the force repre- y- ^^^ ^c sented by GA : but the re- sultant of the forces repre- sented by B G, BD is represented by BA, and is therefore counterbalanced by the force represented by AB, so that the three forces produce equilibrium. 16. Conversely, if three forces keep a particle in equi- librium, and a triangle be drawn having its sides parallel to the directions of the forces respectively, the sides are proportional to the forces to whose directions they are respectively parallel. Let BG, BD (fig. 5) represent two of the forces: then, since they are in equilibrium, AB must represent the third. But GA is parallel and equal to BD, therefore the triangle ABG has its sides parallel to the three forces, and also proportional to them respectively. Any triangle then, that has its sides parallel to the three forces respec- tively, must have them parallel to the sides of the triangle STATICS OF A SINGLE PARTICLE. 15 ABC, and must therefore be equiangular to this triangle: equiangular triangles are similar ones, so that the forces are proportional to the sides, to which they are respectively parallel, of any triangle drawn in the way described. This proposition may be extended thus : if three forces keep a particle in equilibrium, and a triangle be drawn with its sides making a constant angle measured in the same direction, with the directions of the forces respectively, the sides of the triangle are respectively proportional to the forces with whose directions they make the constant angle. For if the triangle be turned in its own plane through an angle equal to the constant angle, but in the direction opposite to that in which the angle is measured, each of its sides becomes parallel to the direction with which it previously made the constant angle, and the proposition becomes identical with the previous one. 17. The ' triangle of forces ' can be easily extended to the ' polygon of forces ', which is : If a particle be under the action of a number of forces, which can be represented by the sides of a polygon taken in order, the particle will be in equilibrium. Let the sides AB, BC, CD, DE, EF, FG, GA of the polygon ABCDEFG, taken in order, represent a number of forces acting on a particle. Join AC, AD, AE, AF. By the ' triangle of forces ', the forces represented by AB, BC can be counter- balanced by CA, therefore A C represents their resul- tant ; similar the resultant of AG, CD is represented by AD, that of AD, DE by AE, that of AE, EF by AF, and that of AF, FG hj AG\ therefore the re- sultant of forces represented hy AB, BC, CD, DE, EF, FG is represented hy AG', but forces represented by A G, GA counterbalance one another, so that the original forces are in equilibrium. Fig.6 16 STATICS. Note. The forces are not necessarily in one plane. Cor. To obtain geometrically the resultant of a num- ber of forces acting on a particle. Draw a series of straight lines, end to end, AB, BG, CD, BE, FF, FG to represent completely the forces, whose resultant is required, then join A G, it represents completely the resultant. The following particular case of the polygon of forces may be noticed: the resultant of a number of forces on a particle, and in the same straight line, is their algebraical sum, the forces being estimated positive in one direction and negative in the other. The converse of the polygon of forces does not hold, because equi- angular polygons are not necessarily similar. 18. The following theorem, enunciated by Lami, is the parallelogram of forces in another form. Prop. If three forces acting on a particle, keep it in equilibrium, each is proportional to the sine of the angle between the other two. Let OA, OB, DC represent three forces P, Q, R which, acting on a particle, keep it in equilibrium. With OA, OB as adjacent sides complete the parallelo- gram OADB : join OD. ng.7 „ D B STATICS OF A SINGLE PARTICLE. 17 OD, 00 must be equal and opposite, since OD represents the resultant of P and Q. P : Q : E=OA : OB : 0C= OA : AD : OD = sin ODA : sin AOD : sin O^D = sin DOB : sin AOD : sin ^05 = sin BOG : sin AOC : sin AOB = sin Q, R : sinP, R : sin P, Q. 19. The magnitude of the resultant R, of two forces P and Q, which act on a particle, and whose directions make an angle with one another, may be easily found. Let OA, OB represent the forces P, Q respectiveh'. Complete the parallelogram OBCA, and join 00 \ the latter represents R. F'g.8 But 00^ = OB' + BC - 20B . BC. cos OBC, BC=AO,sind OBO = 180°-A0B = 180'-6>, .-. R' = P'-^ Q'- + 2PQ cose. Ex. 1. If forces of 3 lbs. and 4 lbs. have a resultant of 5 lbs.,- at what angle do they act ? Ans. 90'^. Ex. 2. If one of two forces acting on a particle is 5 lbs., and the re- sultant is also 5 lbs., and at right angles to the known force, find the magnitude and direction of the other force. Ans. Osj2 lbs., making an angle of ISo*^ with the other force. G. 2 18 STATICS. Ex. 3. At what angle must forces P and 2P act on a particle in order that their resultant may be at right angles to one of them ? Ans. 120*^. Ex. 4. If three forces, whose magnitudes are expressed by the numbers 3, 6, 9, act on a particle, and keep it at rest, shew that they must all act in the same straight line. Ex. 5. If the three forces in Ex. 4 act in directions represented by the sides of an equilateral triangle, taken in order : determine their re- sultant. Ans. A force 3^3, acting at right angles to the force 6. Ex, 6. Three forces acting on a particle keep it in equilibrium : the greatest force is 5 lbs., and the least is 3 lbs., and the angle between two of the forces is a right angle: find the other force. Ans. 4 lbs. Ex. 7. Two equal forces act at a certain angle on a particle, and have a certain resultant: also if the direction of one of the forces be reversed and its magnitude be doubled, the resultant is of the same magnitude as before : shew that the two equal forces are inclined at an angle of 60". Ex. 8. Determine the resultant of four forces of 5, 6, 9, 10 lbs. acting on a particle and represented in direction by OA, OB, 00, OD, re- spectively, where is the point of intersection of the diagonals of a square ABOD. Ans. 4.^2 lbs., in the direction bisecting the angle OOD. Ex. 9. Forces P, P^3, and 2P act on a particle : find the angles be- tween their respective directions that there may be equilibrium. A71S. Between P and P^3, 900; between P and 2P, 120"; between Pjs and 2P, 1500. Ex. 10. Five equal forces act on a particle, in directions parallel to five consecutive sides of a regular hexagon taken in order; find the mag- nitude and direction of their resultant. Am. The direction is parallel to the third force, and the magnitude equal that of any one force. 20. The following proposition is sometimes useful. If two forces acting on a particle be represented by m times the line OA, and n times the line OB, respec- tively, their resultant is represented by {7n + n) times the line OG, where G is the point between A and B, such that mAG = nBG. STATICS OF A SINGLE PARTICLE. 19 By the ' triangle of forces ' mOA is equivalent to mGA and mOG, and the force nOB to nGB and nOG. But since mAG= nBG, and they are opposite, these two forces counterbalance one another, so that we are left with (m + n) OG only. Fig. 9 21*. Def. Let A^, A^, A^... A,, be a series of points; join A^A^, and take B^ between them, so that AB^ = BA -2 ' join B^A^, and take B^ between them, so that 25,5, = 5,^3; join B^A^ and take B^ between them, so that 35 B, = B,A,, 2 3 3 — 4 = and so on until we arrive at B^_^: this point is called the ' centroid ' oi A^, A^, A^ ... A^^. The centroid of the n points A-^^, Jo...^„is sometimes defined as the point whose distance from any plane is one ?i*'* the sum of the distances of A^, A^...A^ from that plane. We can easily shew that the definition of the centroid we have already given leads to this definition also. Draw A^M^, A.JU &c. J5jA\, B.,N., &c. perpendicular to any given plane. Draw Ajn^m.2 parallel to il/jiYjil/.,, and B^n.^rn^ parallel to N-^N^M.^. JjJ/i + A.2M.2 = 7ijN^ + A.,m,2 + m.2N^ = 271-^N^ + 2BiWi = 2B^N^, A^M^ + A.^M^ + A^M^ — 2B^N^ + m^M^ + A.^m.^ = Sn^N^ + 3i?o«2 = 3i>\iY2 2—2 20 STATICS. This proves the statement for two and three points, and by the method of induction the proof can easily be extended to any number of points. Fig.lO Note. The distances from the plane must be considered positive when they are on one side of it, negative when they are on the other. We may suppose that any number of the points become coincident : for instance, if A<^ and A^ coincide with A-^, B^ and B^ will also coincide with A-^, and B.^ will be in the line A-^A^, and such that B.^^A^ — hB^A-^. We may extend the idea of the centroid by supposing that some of the points are negative, in which case the process of finding the centroid will be some- what modified : for instance, if A^ be a negative point, B.2 will be in AJB-^, but beyond B^ not between B^ and A.^, and such that i?o^3=2Bo£i : as B.^ is the centroid of two positive and one negative point B^ will divide the line B^A^ equally. Also the distance of a negative point from a plane must be taken of opposite sign to what it would be if the point were a positive one, and in estimating the number of points, we must take the difference between the numbers of positive and negative points. 22*. Prop. lfOA^,OA^,OA^kc 0^, represent a number of forces acting on a particle, their resultant STATICS OF A SINGLE PARTICLE. 21 will be represented by r times the line 0B^_^, where B^_^ is the ' centroid ' of A^, A^... A^. For, by the last proposition, putting m = w = l, the resultant of OA^ and OA^ is 20 B^ ; putting iii= 1, ?i = 2, that of OA^ and 20B^ is SOB^, and so on, until we obtain rOB , as the final resultant. After reading Chap. IV. it will be obvious that the centroid of a number of points is the Centre of Mass of equal particles situate one at each point. As a direct result of this proposition we see, that the resultant attraction or repulsion on a particle of any mass of which each particle attracts or repels with a force varying as its distance and its mass con- jointly, is the same as the attraction or repulsion of the whole mass collected at its Centre of Mass. Ex. 1. Find a point such that, if it be acted on by forces represented by the lines joining it to the vertices of a triangle, it will be in equi- librium. The required point must be the centroid of the three points, i.e. (Art. 21) the point of intersection of the lines drawn from the vertices to the middle points of the opposite sides. Ex. 2. is any point in the plane of a triangle ABC, and D, E, F are the middle points of the sides. Shew that the system of forces OA, OB, OC is equivalent to the system OD, OE, OF. It can be shewn that the centroid of the points A, B, (7 is also that of D, E, F. V. Ex. 1. Ex. 3. The circumference of a circle is divided into a given number of equal parts, and forces acting on a particle are represented by straight lines drawn from any point to the points of division : shew that their re- sultant passes through the centre of the circle, and that its magnitude varies as the distance of the point from the centre. The centre of the circle is clearly the centroid of the points. Ex. 4. AOB and COD are chords of an ellipse parallel to conjugate diameters: forces are represented in magnitude and direction by OA, OB, OC, OD: shew that their resultant is represented in direction by the straight line which joins to the centre of the ellipse, and in magnitude by twice this line. The centroid of the points A, B, C, D is midway between and the centre. 22 STATICS. Ex. 5. Straight lines are drawn from any point parallel to the four sides of a parallelogram : find the magnitude and direction of the resultant of the forces represented by these four straight lines. Ans. The direction is along the line joining the point with the centre of the parallelogram, and the magnitude is represented by twice this line. 23. Def. The components of a force, in two or in three given directions, are the forces which acting in those direc- tions, will have the given force for resultant. As it is frequently desirable to replace two or more forces by one (their resultant), so also is it to replace one force by tico (its components), in two given directions in the same plane w^ith it, and sometimes by thi-ee in three given directions, which are not all in one plane and no two of which are in the same plane as the single force. For instance, imagine a particle, free to move in a straight groove, to be pulled by a string making an angle with the groove : it is clear that the tendency of the force is twofold, viz. to make the particle move along the groove and to press it against the groove. Also it is clear that the one effect might be produced by a force along the groove and the other by a force at right angles to the groove : these two separate forces will be the components in the corresponding directions of the force exerted by the string. We have seen that the mechanical problem of compounding two forces into one is the same as the geometrical one of constructing the diagonal of a parallelogram, having given two adjacent sides: so also to resolve one force into its two components in two given directions in its plane, we have to construct the parallelogram, having given one diagonal and lines to which the sides are respectively parallel. 24. To find the components of a given force in two given directions in its plane. Let 00 represent the force. Draw OA, OB parallel to the line giving one direction, and OB, OA parallel to the line giving the other. By the parallelogram of forces, the force OC is the resultant of OA and OB, which, being in the given directions, are the components required. We can easily express the magnitudes of these compo- nents in terms of 00 and the angles the given directions make with 00. STATICS OF A SINGLE PARTICLE. 23 Let P be the force represented by 00 and let the angles 00 A, OOB be a, /3. Then OA : 0^ = sin 00^ : sin OAO = sin OOB : sin AOB = sin /3 : sin (a + /3) ; .'. the component in direction OA = P - — t -^ . ^ sm (a + p) Fig.ll 5, _ n Similarly that in direction OB = P sm a sin (a + y5) * Since we can construct any number of parallelograms having a given diagonal, the number of ways in which we can resolve a single force into two is infinite. The most important case is when the two directions along which the resolution takes place are at right angles to one another. 25. Pef. When the directions of the two compo- nents of a force are at right angles to one another, each component is called the 'resolved part' of the force in the corresponding direction. When we speak then of the resolved part of a force in any direction, it is understood that the force is resolved into two components, one in the specified direction, and the other in the direction at right angles to it, and in the plane containing this direction and that of the original force. Let Ox be a given straight line, and let OA^, OA^, OA^ represent a number of forces P^, P^, P^ , whose directions, which are not necessarily in one plane, make angles 6^, 0^, 6^, &c. with Ox. 24 STATICS. Produce xO backwards to x\ and draw A^M^, A^M^, AJ^I^, &c. perpendicular to xOx. Then OM^, OM^, OM^, &c. Fig. 12 represent the resolved parts of P^, P^, P^, &c. respectively along Ox, and M^A^, M^A^, M^A^ the resolved parts per- pendicular to Ox. It is found convenient to adopt the convention that forces in direction Ox, from left to right, be considered positive, while those in the opposite direction are considered negative. In the above figure it will be seen that with this con- vention the resolved parts along Ox of P^, P^ and P^ are positive, while those of P^ and P^ are negative. OM^ = OA^ cos (9j, A^M^ = OA^ sin <9^, OM^ = OA^ cos (9,, &c. Hence the numerical values of the resolved parts of the forces along Ox are P^ cos 6^, P^ cos 6^, &c. and those per- pendicular to it are P^ sin 0^, P^ sin 0^, &c. It is easily seen that these values also give the alge- braical values of the resolved parts, the signs being deter- mined in accordance with the above convention. Note. When the forces are in one plane, their resolved parts at right angles to Ox are in the same straight line, but not otherwise. STATICS OF A SINGLE PARTICLE. 25 If X, Y are the resolved parts of a force P, in two TT 2 directions, making angles 6 and ^ — respectively, with P, we have seen that .Y= P cos e, and Y= P sin (9, .-. P = JX^TT' and tan 6 = Y/X. 26*. To find the components of a force in three given directions, which are not all in the same plane, and no two of which are in the same plane as the original force. Let the line AB represent the given force. Through both A and B draw three planes parallel to / N / '^^^>^ .....X—^- --^D I AY ' '' H^ 7— -X- -,-.«.'^ ^v ''C A ^ each pair of the given directions. These six planes will form the faces of a parallelopiped of which AB\^ the dia- gonal, and each edge of which will be parallel to one of the given directions. By Art. 13 the edges AE, AC, and AH will repre- sent forces of which AB 'y^ the resultant, and which are therefore the required components. The only case which is of much interest is when the given directions are mutually at right angles to one another : the components are then termed the resolved parts in the corresponding directions. Let P be the given force, X, Y, Z the resolved parts in the directions AG, AE, J. irrespectively, which make angles 26 STATICS. a, P, 7 with AB. But AC = AB cos a, J.^ = AB cos yS, and ^jy = AB cos 7, and AB'=BF' + AF'= AW + AG^ + AH\ .', X =Pcosa, Y= P cos/3, Z—F cosy, and P' = X'+ Y' + Z\ Hence cos^a + cos^/3 + cos^7 = 1. Ex. 1. Shew how to resolve a given force into two others, of given magnitude. When is this impossible ? Ex. 2. Find the components of a force P, when they both make angles of 30° with it. Ans. Each is P/^/s. Ex. 3. Find the components of a force P in two directions, making angles of 60*^ and 45° with P on opposite sides. Am. 2P/(1 + VS) and P ^6/(1 + x/3). Ex. 4. Three forces of 5, 2, and 7 lbs. respectively act on a particle in directions mutually at right angles : determine the magnitude of their resultant. Ans. ^78 lbs. Ex. 5. Three forces, represented by three diagonals of three adjacent faces of a cube which meet, act at a point : shew that their resultant is equal to twice the diagonal of the cube. Each of the forces may be resolved into two components, represented by those edges of the corresponding face, which meet in the point: the three forces are equivalent then to the three forces represented by twice the edges of the cube, which meet in the point, i.e. to twice the diagonal of the cube. A similar result holds for any parallelopiped. The purelij geometrical propositions of the next three Articles are extremely useful. 27. Def. If perpendiculars be dropped from the ends of a given finite straight line on any other given straight line, the length intercepted between the feet of these perpendiculars is called the orthogonal projectio7i of the first line on the second. (The two lines are not neces- sarily in one plane.) Let AB be the given finite line, PQ the line on which it is to be projected. STATICS OF A SINGLE PARTICLE. 27 Draw A a, Bh perpendicular to PQ, then ah is the orthogonal projection oi AB on PQ. Fig.l4 We shall make a convention here, similar to that we have already made about the resolved parts of forces : viz. ii AB be regarded as drawn from A to B, its pro- jection is ah, measured from a to h, whereas if BA be measured from jB to ^, its projection is ha, measured from h to a. The projections are considered positive when measured from left to right as ah is, negative when mea- sured in the opposite direction as ha. These signs apply to figure 14: they are reversed for figure 15. Fig.15 f£' STATICS OF A SINGLE PARTICLE. 47 Since the resultant of P, Q, R passes through 0', its moment about O' is zero. .'. the algebraical sum of the moments of P, Q, R about O'is zero (Art. 37). .•. P.DD'+Q.EE'-R.FF'=0, .'. P(c cos i?-|j+Q(a cos C--)-P (acosP- k)=0; .*. P(ccosP-&cos C) + Q (acosC — c cos A) + R{h cos A -acosB) = 0. In the above figure we see that the forces P and Q tend to move a particle situate at in the opposite way round 0' to that in which R would move it : their moments therefore are of the opposite sign to that of P. The student may verify for himself that the same result would be obtained were the figure different. He should specially notice in the above example that the required result was obtained by expressing that the algebraical sum of the moments about 0' was zero, 0' being on the line of action of the resultant. Ex. 11. A particle of weight IF is supported on a smooth inclined plane by means of two strings of given lengths, which are attached to the particle C and to fixed points A, B in a, horizontal line in the plane and at a given distance apart. It is required to find the tensions of the strings. The sides of the triangle ABC being known, the angles which AC, BC make with the horizontal line AB are known : let 6, 6' be their comple- ments. Let a be the inclination of the plane to the horizon. Draw CD perpendicular to AB. Fig, 29 48 STATICS. The particle is in equilibrium under the action of four forces, its weight ir which acts vertically downwards, the tension T of the string AC, T, that of BC, and the pressure of the inclined plane R, which acts at right angles to the plane. We shall apply the conditions of equilibrium obtained in Art. 33. Since the algebraical sum of the resolved parts of the forces in any direction is zero, those in the directions AB and CD must be zero. .-. ^'sin^'-r8in^ = (i), Tcos^ + T'cos^'-irsino^O (ii). jR occurs in neither equation, because its direction is perpendicular to all lines in the inclined plane, and W does not occur in the first, because its direction is perpendicular to AB. The inclination of CD to the vertical is the same as that of the plane, and is therefore ^ - a, so At that the resolved part of W along CD is - IF sin a. From (i) and (ii) we obtain R can be obtained by equating to zero the sum of the resolved parts in the direction perpendicular to the plane, we have then E-TFcosa = 0, or R=W cos a. The advantages derived from resolving in the particular directions chosen above, are obvious. Ex. 12. Two equal particles are connected by a fine string, the par- ticles and string being in a fine smooth elliptic tube, whose semi-cir- cumference is equal to the length of the string. The particles are acted on by constant repulsive forces from one focus : prove that, if these forces are equal, the particles will be in equilibrium in any position in which the string is tight, and if they are unequal, in only one such position. In this example we shall assume what will be proved hereafter (Art. 81), that the tension of a string, in equilibrium under the action of forces at each end and the pressures of a smooth surface, is everywhere the same : the force at each end must of course be equal to the tension. Let S be the focus from which the repulsive forces act ; P, P' the positions of the j)article8 when the string is tight ; PP' is therefore a dia- STATICS OF A SINGLE PARTICLE. 49 meter of the ellipse. Join P, P' with the foci S, II. SPHP' is obviously a parallelogram. Fig.30. Let F, F' be the repulsive force on P, P' respectively : let the string lie along the semi-circumference PAP', and let it be fixed at A. Each particle is now in equilibrium : let T be the tension of AP, and T' that of AP'. The forces acting on the particle P are F along SP, T along the tangent at P, and the action of the tube along the normal at P, since the tube is smooth (Art. 43). Since P is in equilibrium, resolving along the tangent at P, we have (Art. 33) F ^in SPG -T^O. Similarly F' sin SP'G' -T' = 0, or F' sin SPG -T' = 0, .'. if F=F', T=T', i.e. it is not necessary to suppose the string fixed at A to insure equilibrium, as (Art. 43) the tension of the string wdll adapt itself to preserve equilibrium, if possible. If, however, F and F' are unequal, T and T' will be unequal also, unless sin SPG vanishes, i. e. unless the j)articles are situate at A and A'. But the tension of P^-1 cannot be different from that oi P'A, unless the string is fixed at ^, so that there is not equilibrium generally, when F and F' are unequal, and the string is free to move along the tube. G. 50 STATICS. Examples on Chapter I. 1. ABC is a triangle : D, E,F are the middle points of the sides BC, CA, AB respectively: shew that forces acting on a particle and represented by the straight lines AD, BE, CF will maintain equilibrium. 2. A, B, C are three points on the circumference of a cu'cle: forces act along AB and BC inversely proportional to these straight lines in magnitude; shew that their resultant acts along the tangent at B. 3. Two forces P and Q have a resultant R which makes an angle a with P: if P be increased by R while Q remains unchanged, shew that the new resultant makes an angle ^ with P. 4. The resultant of two forces P, Q, acting at an angle 6 is equal to TT (2m + 1) \/{T?^ + Q^) : when they act at an angle -x-Q, it is equal to (2w - 1) V(^^ + Q^) : shew that tan B = '^^^ . 5. Compare in terms of the sides of a triangle ABC the forces which acting from 0, the centre of the inscribed circle, along OA, OB, OC will balance. 6. Two forces P and P ^2 act on a particle lying on a smooth hori- zontal plane. If P makes an angle of 45^ with the horizon, find the direction of P ^2 in order that the particle may be in equilibrium. 7. Find a point within a quadrilateral, such, that if it be acted on by forces represented by the lines joining it to the angular points of the quadrilateral, it will be in equilibrium. 8. ABC is a, triangle, P any point in BC. If PQ represent the resultant of the forces represented by AP, PB, PC, the locus of Q is a straight line parallel to BC. 9. A heavy particle is attached to one end of a string, the other end of which is fixed. Find the force making an angle of SO** with the hori- zontal which must be applied to the particle in order that the string may deviate by an angle of 45^ from the vertical, and find also the tension of the string. STATICS OF A SINGLE PARTICLE. 51 10. Two forces P, Q act at a point along two straight lines making an angle a with each other, and have a resultant II : two other forces P', Q' acting along the same two lines have a resultant R' ; shew that the directions of R and R' will also include an angle a if FP' + QQ' + 2PQ' cos a = 0, or PP' + QQ' + 2P'Q cos a =^0. 11. If from 0, the centre of the circle inscribed in the triangle ABC, A B forces X cos -^ , Xcos ^ act along OB, OA, prove that the magnitude of the necessary force towards C, in order that the resultant may pass through (J the middle point of AB, is X cot — . 12. A small ring slides on a smooth arc of a circle and rests in equi- librium under the repulsion of three forces P, Q, R, directed from points dividing the circumference in three equal parts : if its position of equili- brium lie on the smaller arc between the points from which the forces Q, R are directed, shew that the pressure exerted by the circle is {P2 + (32 + i^2 -QR + RP + PQ] K 13. Two particles of weights P and Q respectively, are connected by a string which Hes on a smooth circle fixed in a vertical plane : shew that IT if — be the angle subtended at the centre by the string, the inclination of the chord joining P, Q to the horizontal in the position of equiUbrium is 14. OA, OB, OC... are any number of fixed straight lines drawn from a point 0, and spheres are desciibed on OA, OB, OC... as diameters. Any straight line OA' is drawn thi'ough O and a point P taken on it so that OP is equal to the algebraical sum of the lengths intercepted on OX by the spheres. Find the locus of P. 15. Two constant equal forces act at the centre of an ellipse parallel to the directions SP and PH, where S and H are the foci and P is any point on the curve. Shew that the extremity of the line which represents their resultant hes on a circle. IG. Forces are represented by the perpendiculars from the angles of a triangle ABC on the opposite sides: shew that if theu' resultant passes through the centre of the nine-point circle a2 {b^ - c-i) -f ¥ {c^ - a^) -f c- (a-* - ¥) = 0. 4—2 52 STATICS. 17. Three equal forces act at the orthocentre of a triangle ABC, each perpendicular to the opposite side : prove that if the magnitude of each force be represented by the radius of the circle ABC, the magnitude of the resultant will be represented by the distance between the centres of the inscribed and circumscribed circles. 18. The resultant R of any number of forces Pj, P^, F.^, &c. is deter- mined in magnitude by the equation J^^ = 2 (P^) + 2SP^P, cos (P^.PJ, where (PrPg) denotes the angles between the directions of P^, P^. 19. ABCDEF is a regular hexagon, and at A forces act represented in magnitude and direction by AB, 2AC, SAD, 4:AE, 5AF; shew that the length of the line representing their resultant is ^JS51 . AB. 20. Two small smooth rings of weights W and W, connected by a string, slide upon two fixed wires, the former of which is vertical, and the other inclined at an angle a to the horizon. A weight P is tied to the string, prove that in the position of equilibrium cot e : cot : cot a= W : P+W : P+ W'+ W, where 6, (p are the angles which the two portions of the string make with the vertical. 21. ABCD, A'B'C'D' are two parallelograms; prove that forces acting at a point proportional to and in the same direction as AA', B'B, CC, I)'D, will be in equilibrium. 22. A particle is acted upon by a number of centres of force, some of which attract and some repel, the force being in all cases proportional to the distance, and the intensities for different centres being different : shew that the resultant force passes through a fixed point for all positions of the particle, and examine the one apparent exception. 23. From any point within a regular polygon perpendiculars are drawn on all the sides : shew that the direction of the resultant of all the forces represented by these perpendiculars passes through the centre of the polygon, and find its magnitude. 24. Two heavy rings slide on a wire in the shape of an ellipse whose major axis is vertical, and are connected by a string which passes over a smooth peg at the upper focus : shew that if the weights are equal and the length of the string is equal to that of the axis major, there are an infinite number of positions of equilibrium. STATICS OF A SIXGLE PARTICLE. 53 25. Four particles A, B, C, D are attached to the ends of strings whose other ends are tied in a knot at 0. Any two particles repel one another with a force which varies directly as the distance and the product of their masses. Shew that when the system is in equilibrium, the volumes of the tetrahedra OBCD, OCDA, ODAB, OABC are proportional to the masses of ^, Z>, C, D respectively. 26. In an ellipse a polygon PQHS, &g. is described so that the triangles formed with a side as base and the centre of the ellipse as vertex are of equal area. If O be any point in the plane of the ellipse, prove that the line of action of the resultant of the forces represented by OF, OQ, OR, &c. passes through the centre of the ellipse. 27. Two small heavy rings slide on a smooth wire, in the shape of a parabola, w'hose axis is horizontal: they are connected by a light string which passes over a smooth peg at the focus : shew that in the position of equilibrium, their depihs below the axis are proportional to their weights. 28. Forces P, Q, R act in the lines DA, DB, DC and their resultant meets the plane ABC in G, shew that ju m> ' -m. ^'""' ^^«-^ ^-^^^^ If their resultant be parallel to the plane ABC, then P . DB . DC -\- Q . DC . DA + R . DA . DB ^0. 29. is any point on the circle circumscribing a tiiangle ABC, and OL, 031, ON are the perijendiculars from on the sides. The line LMN meets the perpendiculars from A, B, C on the opposite sides in P, Q, R respectively. Prove that if forces act at O represented by OL, OM, ON, OP, OQ, OR their resultant is represented by BOK, where K is the orthocentre. 30. ABC is a triangle and O^O.^O.^ are the centres of the three escribed circles opposite to ^, B, G respectively. At any point P, forces act along PO^^, PO.^, PO.^ represented in magnitude by PO^.BC, POo-CA, PO^.AB, respectively. Shew that if their resultant is of constant magnitude, the locus of P is a circle concentric with the circle circumscribing the triangle Oj^O.^O^^. CHAPTER 11. STATICS OF SYSTEMS OF PARTICLES. 44. When a body composed of a number of particles is in equilibrium, each of these particles is in equilibrium also, and the forces which act on it must therefore satisfy the conditions of equilibrium. But among the forces acting on a particle must be included, not only what are called ' external ' forces, such as the force of gravity, the pressure and tensions due to other bodies, but also 'in- ternal ' forces, i. e. the forces of attraction and repulsion that exist among the different particles composing the body. These forces are by no means always the same in the same body: for example, it is plain that if we try to stretch a rod, the forces that the different particles composing the rod, exert one on another, are different from what they are when we try to compress it. In the former case, the external forces tend to separate particles arranged along a line parallel to the rod's length, in the latter they tend to move them nearer together. To resist these quite opposite tendencies, different internal forces must be called into play. Concerning these in- ternal forces we have Newton's Third Law which asserts that * action and reaction are ahuays equal and opposite ' : i.e., if the particle A exerts on the particle B a force R, (the action) in a certain direction, it is itself acted on by a force R, (the reaction) in the exactly opposite direction, and also in the same straight line, so that the line of STATICS OF SYSTEMS OF PARTICLES. 55 action of each of these forces must be the line joining A and B. 45. Without any further assumption about the in- ternal forces that are exerted when any body is in equili- brium, we can determine conditions which must be satisfied by the external forces in such a case. Since the algebraical sum of the resolved parts in any direction of the forces, which act on each particle of a body in equilibrium, is zero, that of the resolved parts in any direction of all the forces, external and internal, acting on all the particles, is zero also. But as the re- solved part of any action is numerically equal, but of opposite sign, to that of the corresponding reaction, the algebraical sum of the resolved parts in any direction of all the internal forces vanishes separately, for the internal forces consist entirely of pairs of forces, equal and opposite to one another. Hence the algebraical sum of the re- solved parts of the remaining forces, the external ones, is zero. Cor. A system of forces keeping a number of particles in equilibrium will, if applied to a single particle, keep it in equilibrium, since the conditions of Art. 33 are satisfied. 46. In a similar w^ay we can shew that the alge- braical sum of the moments about any line, of the external forces acting on a body in equilibrium, is zero. We have only to substitute ^moments about any line' fov ^ resolved parts in any direction ' and the above proof holds. We may state then, that if any body be in eq^uilibrium under the action of external and internal forces, the alge- braical sums both of the resolved parts in any direction, and of the moments about any line, of the external forces, are zero. If the lines of action of the external forces be in one plane, the algebraical sum of their moments about any 5{) STATICS. l^oint in that plane is zero, being equal to the algebraical sum of their moments about a line through the point in question, and perpendicular to the plane. 47. It is to be noticed that what we have called internal forces are only so relatively — the force which is ex- erted on the particle A by the particle B is an internal one, when we are considering a body or system of bodies con- taining both particles, whereas if B is not contained in the system, the force is an external one. It is then very necessary, in applying the above conditions of equilibrium to a system of particles, to know which forces are external and which internal. The force which is an internal one when we are considering the whole body may become an external one, when only a portion of the body is under consideration. Ex. 1. A jjicture weighing 10 lbs. is supported by a string which passes over a smooth peg, and has its two ends fastened to the picture : if the pension of the string be 10 lbs. , shew that each string makes an angle of GC with the vertical. Apply Art. 45, choosing the vertical and horizontal as the directions along which to resolve. Ex. 2. A rod is supported by means of two strings which are attached to a fixed point, and one to each end of the rod. Assuming that the weight of the rod acts at its middle point, prove that the tensions of the strings are proportional to their lengths. Apply Art. 46, taking moments about the middle point of the rod. Ex. 3. A rod of weight W, is supported at an angle of 60** with the horizon by means of strings attached to its ends, the one attached to the upper end maldng an angle of 60** with the horizon, but in an opposite direction to the rod : find the tensions of the two strings and the inclina- tion of the second to the horizon, assuming that the weight of the rod acts at its middle point. Alls, --f—i 'St^ ^f'. the latter acting at an angle tan~^3Ay3, to the horizon. Apply Arts. 45, 46, choosing an end of the rod as the point about which to take moments, and the horizontal and vertical as the directions in which to resolve. STATICS OF SYSTEMS OF PARTICLES. 57 Ex. 4. A square ABCD, is in equilibrium under the action of four forces, one of 3 lbs. acting along AB, one of 2 lbs. along BC, and one of 3 lbs. along CD ; find the magnitude and line of action of the remaining force. Ans. A force of 2 lbs., acting in direction CB, at a distance equal to BI2.AB from BC. Apply Arts. 45, 46, resolving along AB, BC respective!}', and taking moments about B. Ex. 5. AB is a straight weightless rod, 15 feet long ; 4 lbs. is hung at A, lib. at a jDoint 3 feet from A, and a force of 11 lbs. acts vertically upwards at a jooint 8 feet from B ; find what weight must be attached to the rod to maintain equilibrium and where. Ans. Gibs., 2 ft. 8 inches from B. Ex. 6. Three forces acting at the corners of a triangle, each per- pendicular to the ojDposite side, keep the tiiangle in equilibrium : prove that each force is proportional to the side to which it is perpendicular. Take moments about two of the angular points. Ex. 7. If three forces P, Q, li, acting along the bisectors of the angles of a triangle, at the angular points A, B, C, respectively, keep the triangle in equilibrium : prove that P : Q : R = cos - : cos— : cos — . 2 2 2 Take moments about two of the angular points. 48. We have then found necessary conditions of equilibrium for any body or system of bodies whatsoever, inckiding liquids, flexible strings, &c. We shall hereafter find sufficient conditions of equili- brium for ^ rigicV bodies. Def. When the particles which compose a body, always form the same configuration, or in other words, when the body always retains the same shape and size, whatever forces be applied to it, the body is said to be rigid. We have no experience of bodies, which answer this description perfectly, but we know of many substances 58 STATICS. which answer it more or less approximately : i.e. we know of many substances, which will submit to the action of considerable forces without undergoing any appreciable change in shape or size. The results which we shall prove absolutely true for perfectly rigid bodies, will be so approxi- mately for bodies that are approximately rigid. 49. We have already seen that any system of forces acting on a particle is equivalent to a single force, i.e. there is a single force, such that its effect on the particle could not be distinguished from that of the combined forces. The question now presents itself, whether this is so or not when the forces do not all act on a single particle, but on different particles of a system. If the particles form a rigid body, we shall see that under certain circumstances there exists a force, which together with the given forces would keep the body in equilibrium, so that the effect of this force reversed on the body as a whole, is the same as that of the original forces. But it must be remembered that it is only on the body as a ivhole that the effects are the same necessarily : the internal forces called into play by the single force, are not necessarily the same as those called into play by the system of forces, in fact are generally very different. When we cannot find a single force whose effect on the body as a whole is the same as that of the system of forces, we can always find a different sj^stem of forces whose effect will be the same, though they will not generally give rise to the same internal forces. It is usual to speak of the single force, when such a one exists, as the resultant of the original forces, and the second set of forces as equiva- lent to the first, though it must always be understood that they are so, strictly speaking, only in one sense. Even when the body is not rigid, a single force, or set of forces, which would, if the body were rigid, be equivalent in the above sense is said to be, one the resultant of, the other equivalent to, the original system of forces. 50. Prop. Any rigid body, under the action of any system of forces, can be fixed by applying single forces at STATICS OF SYSTEMS OF PARTICLES. 59 each of any three given points of the body not in the same straight line, the direction of the force at one point being at right angles to the plane containing the three points, and that of the force at a second point at right angles to the line joining it to the third. Let A, B, and G be any three points of the body. The body can be fixed by the following constraints : imagine a very small spherical socket to be made in the body at A, and a ball just smaller than the socket to be placed in it, and the ball to be fixed. Now imagine a very small hoop with its plane perpendicular to AB, to be fixed round J5, and also some obstacle to be placed to prevent C from moving at right angles to the plane ABC. The first constraint prevents the body from moving in any way except by turning about A, and exerts a single force through A as the ball and socket touch in only one point : the second prevents B from turning about A, and there- fore from moving at all, so that the body can now only turn about AB; the second force acts at ri^ht ang^les to AB. The third constraint prevents G from moving round AB, and therefore from moving at all, and exerts a single force through G at right angles to the plane ABC. As G cannot turn about AB, it is clear that the body is now fixed. 51. This proposition can be extended to the case in which any or all of the points A, B, and G are not situate in the body. For we may imagine them made so in effect, without introducing any forces external to the whole sys- tem, by arranging a system of rigid rods, without weight, rigidly connecting them with the body. Co7\ If the lines of action of the external forces all lie in one plane, the body can be fixed by the application of single forces at any two given points A, B in that plane, each force being in the plane, and the direction of one being at right angles to the line AB. For by the last proposition, if a third point C be taken in the plane but not in AB, the body can be fixed by the application of suitable forces P, Q, R at A, B, C respec- GO STATICS. tively, tlie direction of B being perpendicular to the plane, and that of Q perpendicular to AB. The body is now in equilibrium under the action of the internal forces, the original external forces, and the forces of constraint P, Q, a. Hence the algebraical sum of the moments about AB of all the external forces, including P, Q, and R, is zero : but each of these momeuts except that of P is zero, since each of the corresponding forces either intersects AB or is parallel to it, Art. (46). The moment of P must there- fore be zero, i.e. P itself is zero, since P neither meets AB nor is parallel to it. Similarly we may shew that the moment of Q about every line through A in the plane is zero, i.e. Q is either zero, or it lies in the plane in question. Also by taking moments about lines through B in the plane, except AB, we may shew that either Pis zero, or it lies in the plane. 52. Prop. A number of forces acting on a rigid body, their lines of action all being in the same plane, will keep it in equilibrium, provided any of the following sets of con- ditions hold : (1) if the algebraical sum of their moments about each of three given points in the plane, but not in the same straight line, be zero ; (2) if the algebraical sum of their moments about one given point in the plane, and of their resolved parts in any two given directions in the plane, be zero ; (3) if the algebraical sum of their moments about two given points in the plane, and of their resolved parts in any given direction in the plane, not at right angles to the line joining the two points, be zero. (1) Let A, B, (7 be the three given points : if the body is not in equilibrium, it can be fixed by applying forces of constraint P and Q Sit A and B respectively, both in the plane of the forces and ^perpendicular to AB. The whole system of forces including P and Q must satisfy the necessary conditions of equilibrium : therefore the alge- braical sum of their moments about A is zero : but the algebraical sum of the moments of the original forces about A is zero, and the moment of P about A is zero STATICS OF SYSTEMS OF PARTICLES. 61 also ; hence the moment of the remaining force Q is zero, i.e. Q itself is zero, as it does not pass through A. Simi- larly we can shew that the moments of P about both B and C are zero ; hence either P is zero, or it passes through both B and G. As ^4, B, and C are not in a straight line, Pis zero. Hence the body is in equilibrium without any constraint. (2) Let J. be the given point, and B any other point in the plane of the forces : apply forces of constraint P and Q at A and B respective!}' as in (1). Then we shew as before that Q is zero. The forces including P must satisfy the necessary conditions of equilibrium : therefore the algebraical sum of their resolved parts in each of the two given directions is zero ; but the algebraical sum of the resolved parts of the forces excluding P, in each of these two directions is zero, i.e. the resolved part of P in each of these directions is zero. But as the resolved part of a force is only zero, in a direction perpendicular to the force, P itself must be zero. Hence the body is in equilibrium without constraint. Case (3) can be proved in a similar way. 53. Prop. Two equal forces acting in opposite direc- tions along the same straight line on a rigid body, but not necessarily on the same particle, keep it in equilibrium. This is obvious as the two forces clearly satisf}'- the sufficient conditions of ecpiilibrium given in the last article. This proposition is essentially the same as the prin- ciple known as the transviissibility of force, which is generally assumed as an experimental fact, but which we prefer to deduce as above from the Laws of Motion. The formal statement of that principle is as follows : luhen a force acts on a rigid body, it is indifferent on luhat particle in the line of action it acts, provided that particle is part of the body, or rigidly connected tuith it. This follows directly from the proposition just proved. For let A, B 62 . STATICS. be any two particles in the line of action of the force, and rigidly connected with the body. We have just proved that a force equal and opposite to the given force, would counterbalance it, so long as the former acted at a point in AB rigidly connected with the body : hence the given force counteracts a certain other force, whether the given force acts at A or at B. As regards its effect on the body as a whole, we may say then, that it is indifferent at which point we apply the force. It is however in this sense only, that it is indifferent ; if we take into con- sideration the internal forces brought into play in the two cases, they will probably be very different. Imagine, for instance, a sphere resting on a smooth horizontal plane; a force of a certain magnitude, and in a certain direction will give the sjDhere the same change of motion, whether the force take the shape of a push behind or a pull in front, yet the internal forces in the sphere will be different in the two eases, as in the first case the tendency of the external force is to compress the sphere, whereas it has the opposite tendency in the second case. The proof of the converse principle, viz. that if it is indifferent at which of two points a force is applied, the line of action of the force must be the line joining them, is obvious from what has gone before. Ex. 1. A square lamina ABCD is acted upon by a force of 3 lbs. along AB, 2 lbs. along CB, 1 lb. along CD, 2 lbs. along AD, V^lbs. along CA, and Ay 2 lbs. along BD : prove that it is in equilibrium. Ex. 2. A weightless rod AB, 10 feet long, has weights of 7 lbs. hung at each end, and one of 11 lbs. at its middle point : a string is attached to a point 2 feet from A and after passing over a smooth peg vertically above the point of the rod to which it is attached, supports a weight of 10 lbs, : another string attached to a point 4 feet from B supports in a similar way a weight of 15 lbs. Prove that the rod is in equilibrium. Ex. 3. A rigid rod AB, 20 inches long, is acted upon by the follow- ing forces: 3 lbs. at A along BA, ^73 lbs. at right angles to AB, at a point 5 inches from A, 6 lbs. at a point 5 inches from B, and making an angle of GO'' with the part of the rod on the same side as A, and 4 ^73 lbs. STATICS OF SYSTEMS OF PARTICLES. 63 at B making an angle of 30" with AB produced. Prove that there will be equilibrium, provided all the forces are in one plane, and the 3rd force acts on the opposite side of the rod to the 2nd and 4th. Ex, 4. ABCDEF is a regular hexagonal lamina : prove that it is kept in equilibrium by the following seven forces : 2 lbs. along AB, CD, DE, FA, and AD, 3 lbs. along CB and 1 lb. along FE. 54.* Prop. A rigid body under the action of any system offerees, is in equilibrium, provided the algebraical sum of their moments about each edge of any given tetrahedron be zero. Let A BCD be the given tetrahedron, such that the D I \ * N -— ; - -"'--^c A'- B Fig.3l algebraical sum of the moments of the forces about each edge is zero. If the body is not in equilibrium under the action of the system of forces in question, it can be fixed (Arts. 50, 51) by applying suitable forces of constraint F, Q, B a.t A, B, and G respectively. Also Q may be taken perpendicular to AB, and R perpendicular to the plane ABC. Since the bod37- is in equilibrium, under the action of the original forces, together with P, Q, B, these forces must satisfy the necessary conditions of equilibrium. There- fore the algebraical sum of their moments about AB is zero ; but the algebraical sum of the moments of the original forces alone about AB is zero, and the moments 64 STATICS. of both P and Q about AB are clearly zero, so that the moment of the remaining force R must be zero. R being perpendicular to the plane ABC, can neither intersect ^5 nor be parallel to it, so that its moment about AB can only vanish by R itself vanishing (Art. 38). Similarly by taking moments about AC and AD, we see that the moment of Q about each of these lines is zero: hence Q must either be zero, or its line of action must lie in each of the planes BAC, BAD, i.e. be the line AB; the latter alternative is out of the question, because Q is perpendicular to AB : Q must therefore be zero. Again, by taking moments about BC, DB^ and DC, \NQ obtain that the moment of P about each of these lines is zero, i.e. that if P is not zero, its line of action lies in each of the planes BAC, BAD, DAC, which is impossible. P must therefore be zero. All the forces of constraint being zero, we see that the body is in equili- brium under the action of the original forces only. 55.* Pqv]). a rigid body under the action of any system of forces is in equilibrium, provided the algebraical sum of their moments about each of any three given straight lines intersecting in a point, but not in one plane, be zero, and the algebraical sum of their resolved parts alonsr each of these lines be zero also. Let OA, OB, OC he the straight lines, such that the algebraical sum of the moments of the forces, about each of them is zero, and that of their resolved parts along each is zero also. As in the last proposition, if the body is not in equi- librium, it may be fixed by applying suitable forces of constraint P, Q, R at 0, A and B respectively; R may be taken perpendicular to the plane OAB, and Q perpen- dicular to the line OA. Then as before, by taking moments about OA , R is found to be zero ; and Q also by taking moments about OB and OC. But the algebraical sum of STATICS OF SYSTEMS OF PARTICLES. 65 the resolved parts of the original forces together with P, along each of the lines OA, OB, 00 must be zero; hence the resolved part of P along each of these lines must also be zero, i.e. if P is not zero, it is perpendicular to each of the lines OA, OB, OC, which is impossible as they do not lie in one plane. P must therefore be zero. The body is therefore in equilibrium under the action of the original forces alone. The sufficient conditions of equilibrium of any system of forces acting on a rigid body can be expressed in many ways, other than the two given above. 56. We have seen that if two systems of forces are equivalent, either of them reversed will counteract the other; hence it is sufficient for equivalence when both systems are in the same plane, if any one of the following sets of conditions holds. (1) If the algebraical sum of the moments about each of three points in the plane but not in the same straight line, of one system, be equal respec- tively to the corresponding sum of the other. (2) If the algebraical sums of the moments about one point in the plane, and of the resolved parts in two directions in it, of one system be equal respectively to the corresponding sums of the other. (3) If the algebraical sums of the G. 5 GQ STATICS. moments about each of two points in the plane, and of the resolved parts in one direction in the plane, not perpendicular to the line joining the two points, of one system, be equal respectively to the corresponding sums of the other. Analogous conditions of equivalence can be obtained from Arts. 54, 55, for systems of forces which are not in one plane. 57. To find the resultant action on a body of a weightless string stretched round it. Let PABCDQ be a string stretched over a body, A and D being the points where the string leaves the body. The forces acting on the part ABCD of the string are the force due to the part PA, or the tension at A along AP, the tension at D along DQ, and the innumerable actions of the body at every point of ABCD. Since this portion of the string is in equilibrium, the two tensions counteract all these actions along ABCD, i. e. they just balance the resultant of all these actions. But by Newton's Third Law, the resultant action of the string on the body is equal to, opposite to, and in the same straight line as, that of the body on the string. The two tensions counteract the latter of these resultants, i.e. they are equivalent to the former. We may therefore, in considering the equilibrium of the body, suppose that it is acted on directly by the ten- sions at A and B, instead of supposing, what is really the case, that these tensions act on the string ABCD, and so cause it to exert on the body the innumerable small forces, to which the tensions are equivalent. STATICS OF SYSTEMS OF PARTICLES. 67 We arrive at the same conclusion by regarding the body and the por- tion ABCI) of the string as one system of particles: in that case the tensions at A and B are forces external to the system, while the innu- merable actions and reactions between the string and the body are internal forces. Ex. 1. A smooth pulley is supported by a string which passes under- neath it: find the weight of the pulley, if the tension of the string is 10 lbs. and the two parts not in contact with the pulley make angles of 30*' with the vertical. Ans. 10/^3^ lbs. Ex. 2. A rope is passed several times round a fixed rough post, the tensions exerted at the ends of the two parts of the rope not in contact with the post, are 3 lbs. and 2 ^2 lbs. respectively, and these two parts make an angle of Ao^ with one another. Find the resultant action of the rope on the post. Ans. ^yiiy lbs. Ex. 3. A circular cylinder {W) is placed with its axis horizontal on a smooth inclined plane : a weightless string is attached to a point in the plane and after passing over the cylinder supports a weight P, the straight portions of the string being respectively horizontal and vertical : shew that if there is equilibrium, the inclination of the jjlane to the horizon is tan-i {P/(P-MF)}. 58. We have seen how to obtain the resultant of two forces acting on the same particle; if now we have two forces acting on a rigid body, but not on the same particle, we can find a single force equivalent to them provided their lines of action either meet or are parallel, except in the case in which the forces are equal and opposite, but not in the same straight line. If their lines of action meet in a point, we may by the principle of the trans- missibility of force, suppose each force to act at this point, and then their resultant is just w^hat it would be if the forces really acted on a particle, situate there and rigidly connected with the body. 59. When the forces are parallel, their resultant is a force in the same plane, whose resolved part in each of two directions in that plane equals the algebraical sum 5—2 68 STATICS. of the resolved parts of the two forces in the same direc- tion, and whose moment about some point in the plane equals the algebraical sum of the moments of the two about that point. Let A, B he two points where two parallel forces, P, Q respectively act. Then the first two conditions are B F/g.34 satisfied, provided this force acts in the same direction as P and Q, and is equal to their algebraical sum. It must then be parallel to the other two, and at such distances from them that their moments about any point in it are equal in magnitude but opposite in sign. It must then be between them if the signs of P and Q are the same, but not otherwise : its distances from them must be in- versely proportional to their magnitudes. Hence if C be the point where its line of action meets AB, B.AC = Q.BG. When P and Q act in opposite directions, the greater force will clearly lie between the less and the resultant. Cor. The position of C is independent of the direction of the forces, so long as they remain parallel. If the forces P and Q are equal in magnitude and opposite in sign, the preceding solution fails, and we can find no single force, whose effect is equal to that of the two together. Two such forces constitute a couple. STATICS OF SYSTEMS OF PARTICLES. 69 Ex. 1. Four forces, P, 2P, 3P, and 4P act along the sides taken in order of a square : find their resultant, Ans. 2P fj2, acting parallel to the diagonal joining the corner where 2P, and 3P, meet with the opposite corner, and at a distance from it f ^2 times a side of the square. Ex. 2. A uniform beam 4 ft. long is supported in a horizontal position by two props which are 3 feet apart, so that the beam projects one foot beyond one of the props : shew that the pressure on one prop is double the pressure on the other. Ex. 3. If a bicycle and its rider weigh 60 lbs. and 10 stone respec- tively, find how the pressure on the ground is divided between the two wheels, whose points of contact with the ground are 3 ft. 6 inches apart, while the points through which the weights of the bicycle and rider act, are distant horizontally 7 in. and 6 in. respectively from the centre of the driving wheel. Ans. 170 lbs. and 30 lbs, GO. Since a rigid body under the action of any system of coplanar forces, can be fixed by two forces of constraint acting in that plane at two arbitrarily chosen points in it, the system must be equivalent to the forces of constraint reversed : but two forces in one plane can be replaced by a single one, unless they form a couple. Hence any system of forces in one plane is equivalent to a single force or a couple. 61. Prop. If three forces maintain equilibrium, their lines of action must be in one plane, and either all meet in one point or be all parallel. Let P, ft R be the three forces, Aa, Bh, Cc, their respective lines of action. Since the algebraical sum of the moments of a system of forces in equilibrium about any line is zero, that of the moments of F, Q, R about AB vanishes: but as P and Q both intersect this line, each of their moments about it is zero, hence that of R about it must also be zero, i.e. Cc meets AB or is parallel to it. Similarly we can shew that Cc meets, or is parallel to, each of the lines obtained by 70 STATICS. joining any point in Aa, with any point in Bb, which is impossible unless Aa, Bb lie in one plane. Hence all three forces are in one plane. \B Fig.35 If the forces are not all parallel, two of them meet and can be replaced by a single force, which is counterbalanced by the third force, and is therefore in the same straight line with it, i.e. the third force passes through the point of intersection of the other two. Cor. Two forces, whose lines of action are not in one plane, cannot be equivalent to a single force. 62, Def. The moment of a couple is the algebraical sum of the moments of the two forces which form it, about any point in their plane. This moment can easily be shewn to be independent of the position of the point and to be equal to the product of either force into the arm, i.e. the perpendicular distance between the lines of action of the forces. For let P acting at A, and P acting in the opposite direction at B, form the couple. Then the algebraical sum of the moments of the two forces about is STATICS OF SYSTEMS OF PARTICLES. 71 in Fig. (36), P {Oa + Oh)=P. ah, Fig.36 in Fig. (37j, P {Oct -Oh) = P. ah, Al ..Ts_ „ Pig.37 in Fig. (38), P {Oh - Oa) = P,ah, 0'-„ •'.'a 12::=:^. Fig. 38 72 STATICS. where Oa, Oh are the perpendiculars from on the lines of action of the forces. If the body on which the couple acted were only free to turn round 0, the tendency of the couple in all the above figures is to turn the body in the direction in which the hands of a watch move ; the couples are said therefore to have moments of the same sign, or to be like) Avere the tendency of one of them to turn the body in the opposite direction, its moment would be of the opposite sign, and it w^ould be unlike the other two. 63. Prop. Two like couples of equal moment, in the same or parallel planes, are equivalent to one another. (i) When the couples are in the same plane. In this case the two couples form two systems of forces in one plane, such that the algebraical sums of their moments about any point whatsoever in the plane are the same; therefore the systems are equivalent to one another (Art. 56). (ii) When the couples are in parallel planes. Let Pj, Pg be the two equal forces forming one of the couples, acting at the points A, B respectively. In the plane in which the other couple acts, draw CD equal and parallel to AB. Then the effect of Pj, P^ Avill STATICS OF SYSTEMS OF PARTICLES. 73 not be altered by introducing at G, P^, P^, two forces in opposite directions, each equal and parallel to P^, and also two similar forces P^, P^ at D. Join AD, BC, intersecting in 0; then bisects both AD and BC. P3 and Py are equivalent to 2P^ at 0, in the same direction as Pg and P^ and P^ are equivalent to 2P, at in the opposite direction to F^\ 2P^ and 2P^ at will counteract one another, so that we are left with P^ at (J and Pg at D, as equivalent to the original couple. But these two forces constitute a couple like to the original one, equal to it and in the given plane parallel to it: therefore as the original couple is equal to one couple in the parallel plane, it is by (i) equal to any like couple of the same moment in that plane. 64.* The latter part of the last proposition might have been proved in a manner analogous to that adopted for the former, as follows. Let A and B be the two couples : we shall prove that A and B reversed satisfy the sufficient conditions of equi- librium of Art. 5o. Take three straight lines, intersecting in a point, one perpendicular to the plane of each couple, and the other two in the plane of B. It is obvious that the algebraical sum of the resolved parts of the four forces in each of these directions is zero : also the moments of A and B reversed, about the line perpendicular to their planes, are numerically equal but of opposite sign. Hence the algebraical sum of the moments of the four forces forming them about this line is zero. The moment of each of the forces forming B reversed about any line in their plane is zero, and the moments of the two forces forming A, about any line in the plane of B, are equal numerically but of opposite sign ; the alge- braical sum of the moments of all four forces about every straight line in the plane of B is therefore zero. The six sufficient conditions of equilibrium of Art. 55 are therefore satisfied, and the couples A and B reversed 74 STATICS. balance one another; in other words A and B are equiva- lent. Ex. 1. Like parallel forces, each equal to P, act at three of the corners of a rhombus, perpendicular to its plane : at the other corner such a force acts that the four forces are equivalent to a couple: find the moment of the couple, provided the angle at which the last force acts be 60^. Ans. 2 sjs . Pa, where a is a side of the rhombus. Ex. 2. ABCDEF is a regular hexagon : equal forces act along AB, BC, DE, EF, and two other forces, each double any one of the former forces, act along DC and AF: prove that they maintain equilibrium. 65.* Let us consider what we require to know to determine the effect of a couple on a rigid body. It is unnecessary to know the actual position of the plane in which the couple acts, but we must know the direction of the plane, i.e. the direction of a line to which it is perpen- dicular. We do not require to know the magnitude or direction of the forces which compose the couple, but we must know the magnitude of its moment and its sign, i.e. the direction in which it would tend to turn the body round a line perpendicular to its plane, the line being fixed and the body rigidly connected with it. Now a straight line at right angles to the plane of the couple, and of length proportional to the magnitude of its moment, will represent the couple in the first two respects : also, if it be understood that the line is drawn in that direction in which the axis of a right-handed screw moves, when it rotates in the same way as the couple tends to turn the body, the sig7i of the couple will also be re- presented. Fig.40 STATICS OF SYSTEMS OF PAETICLES. To In fig. 40, if the arrowhead on the circle indicates the direction in which the couple would tend to turn the body about AB, supposing the latter fixed and the body rigidly connected with it, the sign of the couple in accordance with the above convention would be represented by AB and not by BA. The line which thus completely represents the couple is termed the axis of the couple. 66.^ We shall now prove that couples follow the Par^allelogi^am Law, in other words, that if from a point the axes representing two couples be drawn, and a parallel- ogram be constructed on these two axes as adjacent sides, the diagonal passing through the above-mentioned point is the axis of a couple equivalent to the two, i.e. their result- ant couple. We may suppose the couples to consist of forces acting at the ends of a common arm, in which case the 76 STATICS. moments of the couples will be respectively proportional to the forces composing them. Let Aahe the common arm, and let AB, ah represent the two equal and parallel forces forming the first couple, AC, ac those forming the second. Draw AB' perpendicular to Aa and AB, equal to AB, and in the direction which by the convention of Art. 65 represents the sign of the first couple : similarly draw A C perpendicular to Aa and AG^ equal to AG and in the proper direction. Then AB' and AG' are the axes of the two couples. Complete the three parallelograms, ABGD, abed, AB'G'U, and join AD, ad, AD'. These parallelograms are clearly equal in every respect, so that AD = ad = AD'. Also AD, ad are parallel, and AD' is perpendicular to AD. But the two forces AB, A G are equivalent to AD, and the two ah, ac to ad, so that the two couples are equivalent to AD, ad, which form a couple of which AD' is the axis. Hence the couples whose axes are AB', AG' are equivalent to a resultant couple of which AD' is the axis. Cor. Hence we may deduce propositions relating to the composition and resolution of couples, analogous to those obtained in Arts. 19 — 26, 30 — 32, relating to the composition and resolution of forces. 67.* Prop. Any system of forces acting on a rigid body can be reduced to a single force acting at any arbitrarily chosen point and a couple. Let A be the arbitrarily chosen point, P any one of the forces. Fig.42 STATICS OF SYSTEMS OF PARTICLES. 77 We shall not alter the effect of the forces by applying at A two forces P^, P^ each equal and parallel to P, and in opposite directions to one another. P^ which is opposite to P, forms with P a couple. Hence P is equivalent to P^ at A, and a couple. The couple vanishes in the case in which A lies in P's line of action. Similarly we may replace each of the other forces by a force at A, equal to it and in the same direction, and a couple. The whole system thus reduces to a series of forces at A, respectively equal to and in the same direction as the several original forces, and a series of couples. But the forces at A are equivalent to a single resultant at A, and the couples to a single resultant couple. Co7\ The magnitude and direction of the single resultant is the same wherever A is, and the resultant couple is the same for all positions of A in a line parallel to the single resultant force. 68.* Prop. Any system of forces acting on a rigid body is equivalent to a single force and a couple whose plane is perpendicular to the direction of the single force. Hcos4)'' Hsin^ R* Fig.43 By the last proposition, the system is equivalent to a single force R acting at any given point A, and a couple H. If the axis of H make an angle (/> with the direction of R, it may be resolved into ^cos in the direction of R and //sin ^ at right angles to that direction. 78 STATICS. Draw AB perpendicular both to R and the axis of H, and make AB equal to (H sin (j))/B, then applying at B two forces equal and parallel to R, but in opposite directions to one another, the system is equivalent to B at B in its original direction, the couples H cos cf), H sin , and the two forces R at A, and B in the opposite direc- tion at B. But the last two forces are equivalent to a couple whose axis is at right angles to both B and AB, i.e. is in the same straight line as the axis of the couple H^in<^: its moment is R . AB or Hsincj). If AB be drawn as in fig. 43, the axes of these two couples are in opposite directions by the convention of Art. 65 : the two couples therefore counteract one another, and we are left with R at B and the couple H cos (p whose axis is along R's direction. Such a force and couple together form what is called a wrench. R's line of action through B is termed Poinsofs Central xlOCIS. The algebraical sum of the moments of the system of forces about the axis of H through A is H, about a line through A, making an angle 6 with AH, the sum of their moments is H cos 6. Hence AH is called the axis of prmcipal moment at A, as the sum of the moments of the forces about it is greater than that about any other line through A. 69.* Prop. The algebraical sum of the moments of the forces about Poinsot's Central Axis is less than that about any other axis of principal moment. For (Art. 68) (fig. 43) the sum of the moments about the central axis is H cos <^, whereas the sum of the moments about the axis of principal moment at A is H. For this reason the Central Axis is sometimes termed the axis of least principal moment. 70.* Prop. Any system of forces acting on a rigid body can be reduced to two equal forces equally inclined to the Central Axis. STATICS OF SYSTEMS OF PARTICLES. 79 For let 00 be the central axis, R being the single resultant force, and H the moment of the resultant couple whose axis is OG. iff Fig.44 Through draw A OB perpendicular to OG, and make OA = OB. We can replace R by jR/2 at A, and i^/2 at B, each in the same direction as R ; we can replace the couple H by a force P dX A, and a force P at B, each perpen- dicular to the plane GOA, but in opposite directions, TT provided P = -jri* The resultant of Pand P/2 at A, and that of P and P/2 at P, will clearly be equal to one another and will make equal angles with P/2, i.e. with the Central Axis. 71. Recapitulation. Regarding any body at rest what- soever, as a collection of particles each of which is at rest, we can assert that the algebraical sum of the resolved parts in an}'' direction, of all the forces, internal as well as external, acting on the body is zero : also that the alge- braical sum of their moments about any line is zero. But as by Newton's Third Law the internal forces consist entirely of pairs, which are equal, opposite and in the same straight line as one another, the algebraical sums of the resolved parts and of the moments of the internal forces are both zero. Any system of external forces which, together with internal ones, maintain a body in equili- brium, must therefore be such that the algebraical sum of their resolved parts in any direction is zero and that of their moments about any line is zero also. 80 STATICS. Next, considering rigid bodies only, we shew that a body under the action of any system of external forces whatsoever can be fixed by the application of suitable forces at three arbitrarily chosen points, and that the direction of one of these forces may be taken perpendicu- lar to the plane containing the three points, and that of another perpendicular to the line joining its point of ap- plication to the third point. When the forces are coplanar, the body can be fixed by applying suitable forces at any two points in the plane of the forces, the directions of both forces being in the plane and that of one perpendicular to ^ the line joining the two points. From this proposition follow the sufficient conditions of equilibrium of a system of coplanar forces acting on a rigid body. These conditions may be given in three different forms, and each form is expressed algebraically by three equations. When the forces are not in one plane the sufficient conditions of equilibrium can be put in many different forms, and each form requires for its algebraical expression six equations. Defining two forces as equivalent, when either counter- acts the other reversed, we deduce the principle known as the * Transmissibility of Force.' The resultant of two parallel forces is obtained by finding from the sufficient conditions of equilibrium, a force which will counteract them, and then reversing it. It is then shewn that if three forces maintain equili- brium, they must be coplanar and either concurrent or parallel. Then we shew that two couples are equivalent when their moments are equal and their planes coincident or parallel ; hence that couples can be represented by straight lines, and that they can be compounded and resolved like forces by the Parallelogram Law. It was shewn that any system of forces in one plane is equiva- lent to a single force or a couple, it can now be shewn that any system of forces whatsoever, acting on a rigid body, is STATICS OF SYSTEMS OF PARTICLES. 81 equivalent to a single force and a couple acting in a plane perpendicular to the force, or to two equal forces, equally inclined to the Central Axis. Illustrative Examples. Ex. 1. If four forces acting aloug the sides of a (luadrilateral are in equilibrium, prove that the quadrilateral is a plane one, and also, that if the quadrilateral can he inscribed in a circle, each force must be propor- tional to the length of the opposite side. Let ABCD be the quadrilateral. The forces along AB, BC have a resultant through B and in the plane ylZ>C; similarly those along AD, DC Fig.45 have a resultant through D and in the plane ADC. But as the four forces are in equilibrium, these two resultants must be in the same straight line, BD, i.e. BD is in each of the planes ABC, ADC and the quadrilateral is a plane one. When ABCD can be inscribed in a circle let P, Q, R, S be the forces along AB, CB, CD, AD, respectively. Since the forces are in equilibrium, the algebraical sum of their moments about A is zero : .'. Q . ABsinB-R . ADsmD = 0; AD = 2B = '''^'^'''^y^CD BC' G. G 82 STATICS. Ex. 2. ABCB is a quadrilateral, and two points P, Q are taken in AB, BG such that AP : PD = CQ : QB. From P, Q, straight lines PP', QQ' are drawn parallel to, equal to, and in the same directions as BG and DA respectively. Shew that forces represented by AB, CD, PP', QQ' are in equilibrium. Fig.46^~'"- The force PP' can he replaced hy two forces parallel to it, at A and D : thefoYcesit A: PP' = PD : AD = BQ : BG; .•. force at A = BQ ; similarly, that at D = QC. The two forces AB, BQ, acting at A, are, by the triangle of forces, equivalent to ^Q; and the two QC, GD, at D, to QD. Hence the four original forces are equivalent to AQ, QD, and QQ', all acting through Q, and represented by the sides of the triangle AQD, taken in order. They are therefore in equilibrium. Ex. 3. A system of forces represented by the sides of a plane polygon, taken in order, is equivalent to a couple, whose moment is represented by twice the area of the polygon. Let the forces be represented by the sides AB, BG, GD, DE, EF, FA, of the polygon ABGDEF. We know that if the forces are not in equilibrium, they are equivalent to a single resultant or a couple (Art. 60). STATICS OF SYSTEMS OF PARTICLES. 83 But as the algebraical sum of their resolved parts in any direction is F/g.47 zero, their resultant is zero, i.e. they are in equilibrium, if they are not equivalent to a couple. Take any jjoint 0, and join OA, OB, OC, &c.: then the moment of AB about is measured numerically by twice the area of the triangle OAB, since the area of OAB is equal to ^AB into the perpen- dicular from on AB: and similarly for the other moments. Hence the algebraical sum of the moments of AB, BC, &g. about is measured by twice the area of the polygon, i.e. is not zero. The system then must be equivalent to a couple, and the moment of this couple is represented by twice the area of the pol^'gon. Ex. 4. On the sides of a right-angled triangle ABC squares are described, the square BCDE on the hypotenuse on the same side oi BC as A, and the squares CAFG, ABHK on CA, AB on the opposite side of each to the triangle: prove that the forces represented by the straight lines AB, BC, CA, BH, HK, KA, CD, BE, EB, AF, EG, GC will form a system in equilibrium. Fig.48. 6—2 84 STATICS. The twelve forces are represented, four by the sides taken in order, of CAFG, four by those of ABHK, and four by those of BCDE. The algebraical sum of their resolved parts in any direction is zero: also by the last example the algebraical sum of the moments about any point, of the first four, is represented by twice the area of CAFG, that of the second four by twice ABHK, and that of the third four by twice BCDE. If the sum of the moments of the first four be considered negative, that of the second four is negative and that of the third four is positive. Hence, since the area of BCDE is equal to the sum of the areas of ABHK, CAFG, the algebraical sum of the moments about any point, of the twelve forces, is zero, i.e. the forces are in equilibrium. Ex. 5. A uniform rod hangs by two strings of lengths I, V, fastened to its ends and to two points in the same horizontal line, distant a apart, the strings crossing one another. Find the position of equilibrium, and shew that if a, a be the angles that I, V make with the horizontal sin (a + a') {V cos a' - I cos a) = a sin (a - a'). Let B be the angle which the rod AB makes with the vertical : let be the point where the strings cross one another. Since the rod is in equi- librium under the action of three forces, two of wliich, the tensions of the strings, meet in 0, the third, the weight of the rod, passes through O. But the weight acts vertically through the middle point of the rod, which STATICS OF SYSTEMS OF PARTICLES. 85 point G, must tlierefore be in a vertical line with O : hence the perpendi- culars AM, BN on OG must be equal. .-. {I- OC) cos a = (r - OD) cos a. OC OD a But -. , = —■ = -. — 7 — ; — K J sm a sm a sm (a + a ) ( , a sin a ) \„ a sin a ) .•.]l- . -7 r cosa= U'-^-— -— ,r cosa, ( sm (a + a)) I sm (a + a ) ^ or sin (a + a) (Z cos a - r cos a') = a sin (a' - a) (1). If b be the length of AB, we have since the algebraical sums of the vertical and horizontal projections of AB, BD, DC, CA are both zero, I sin a-b sin d -t sin a = 0, I cos a-b cos 6 -T I' cos a - a — 0'. These equations with (1) enable us to obtain a, a, and 6, which de- termine the position of equilibrium. The above is an example of a geometrico-statical problem, in which the position of equilibrium, which must clearly exist, is required, and is obtained from geometrical considerations. If the weight of the rod be given, the other unkno^\^^ quantities, the tensions of the strings, can be obtained by using two more conditions of equilibrium, since there are three, and one only has been used. As there are five unknown quantities, and only three sufficient conditions of equi- librium, we must have two geometrical conditions in order to completely solve the problem. Ex, 6. A uniform heavy rod of length a is placed across a smooth horizontal rail and rests with one end against a smooth vertical wall, the distance of which from the rail is h: shew that the angle the rod makes 1 with the horizon is cos"~i (/'/«) '^. Let 6 be the inclination of the rod to the horizon, in the position of equilibrium. The forces acting on the rod AB are its weight vertically downwards through G, its middle point, the reaction of the wall, hori- zontally through .4, and that of the rail C, at right angles to AB. These forces must therefore meet in a point D. Since ADG, ACD are right angles, AD^ = AC . AG, a^cos- 6 = a . h sec d, .'. COS^^=:-. a 86 STATICS. Or, we might have proceeded thus : let R be the reaction of the wall, S that of the peg, and W the weight of the rod. Resolving vertically, we have JF-^Scos^^O (1). Taking moments about A, S . AC=W, AD, .-. S . 7i see ^= TF . a cos d h (2). From (1) and (2) Eesolving horizontally, cos^ 6 — a R-S sin 6 = Hence R and S can be obtained. (3) The advantage of resolving vertically and taking moments about A is that in neither case does the force R come into the corresponding equation . Ex. 7. Shew that the greatest inclination to the horizon at which a uniform rod can rest, partly within and partly without a fixed smooth hemispherical bowl, is sin"^ (l/\/3). Let ADEC be the circular section of the complete sphere, made by the vertical plane containing the rod AB, which rests against the edge of the bowl at C. COD is the horizontal diameter of the sphere through C. The rod is kept in equilibrium by its, weight through G, its middle point, the reaction of the bowl at A, along the normal AO, and that at C perpendicular to AB, and therefore meeting AO on the sphere at E. Since these three forces pass through one point, GE must be a vertical line. Let AGO = 6, r = radius of the bowl. STATICS OF SYSTEMS OF PARTICLES. 87 Then since AC=AE cos EAC=2r cos 8 ,^ ,^ sin AEG ^ cos 20 AG = AE . --. =r^ , =:2r sin EGA COS0 AAEG = ^-E0F=^-2d. (1). (2), Fig.51. Since ^G is half the rod, (2) determines the position of equilibrium. AG Let m = in = AC cos 20 2cos20-l cos2 = ; cos- 6 1 cos2 2-m' clearly has its greatest value when m has its least value, i.e. when m = ^, since AG cannot be less than half ^C Hence the greatest value of 6 is given by 1 2 COS-0: 2-i 3' or sin 6 — —7^ . ^3 88 STATICS. Ex. 8. Four equal spheres rest in contact at the bottom of a smooth spherical bowl, their centres being in a horizontal plane. Shew that, if another equal sphere be placed upon them, the lower spheres will separate if the radius of the bowl be greater than (2 J IS + 1) times the radius of a sphere. Let A, B, C, D be the centres of the four spheres respectively, that of the upper sphere, 0' that of the spherical bowl. Then AB, BC, CD, Oi Fig.52 DA, OA, OB, OC, OD are each equal to the diameter (2r) of any one of the spheres. O and O' are clearly in the vertical line through H, the intersection of the diagonals of the square ABCD. 1 Then co^OAH=^^ = ,^AB AH J2 1 OA OA OAH=45^. jr When the lower spheres are just on the point of separating, there is no pressure between any two of them, so that each of them is in equi- librium under the action of its weight, the pressure of the upper sphere, and that of the hollow sphere. Let W be the weight of each sphere, E the reaction between the upper and any of the lower spheres. From the equilibrium of the upper sphere, resolving vertically, W~4R R = W 2^2* STATICS OF SYSTEMS OF PARTICLES. 89 The resultant of IF acting vertically, and W 2^2 along OA, on the sphere W whose centre is A, makes with the vertical the angle tan ^ 2 x/2 V2 W+ W 1 2 J2 ' J 2 i. e. tan~i I. But this resultant is equal and opposite to the pressure of the bowl which acts along AO'. Therefore tan ^O'H^i, i 1 _o "^26* OA Ji + 1 ^6 .'. 0'A^J'Z6 . AH =2 JlSr. But the radius of the hollow sphere is equal to O'A together with r, therefore radius of the bowl = (2 JVS + 1) r. If the bowl is any larger, 0' will be further from H, and for the pressure of the bowl to counteract the resultant of the other forces on the sphere (centre A), we shall have to suppose that the actions of the two adjacent lower spheres on it are towards their respective centres instead of away from them. But as the spheres are incapable of exerting such forces, equilibrium is not possible, i. e. the spheres will separate. Ex. 9. A heavy har, AB, is suspended by two equal strings of length I, which are originally parallel : find the couple which must be applied to the bar to keep it at rest after it has been twisted through an angle 6 in a horizontal plane. Let C, D be the fixed ends of the strings; CA', DD' the original vertical positions of the strings. Fig. 53 90 STATICS. Draw Aa, Bb at right angles to CA', DB' respectively. Join ab cut- ting AB in its middle point G. Let 2a be the length of AB, and = angle aCA or bDB. Then CA sin aC A = aA^ 2 AB sin ^^ , a .'. Zsin = 2a sin- (1). Let T be the tension of either string : they will from symmetry be the same. Let P be the magnitude of the force which applied horizontally in opposite directions at A and i>, at right angles to AB^ will keep the rod in equilibrium. Besolving vertically, we have [r-2rcos . acos- = 0. aTFsin^ . cos - Hence 2Pa = (2). cos (p ^ (1) and (2) enable us to determine 2Pa in terms of a, I, W and 6. In this example we have assumed as obvious that a couple only is required to maintain equilibrium: it can be shewn however, that the values we have obtained for P and T will satisfy the six conditions of equilibrium of Art. 55. Examples. 1. Four points A, B, C, D lie on a circle and forces act along the chords AB, BG, GB, DA, each force being inversely proportional to the corresponding chord: prove that the resultant passes through common points of (1) AD, BG ; (2) AB, DC ; (3) tangents at B, D, and (4) tan- gents at A and G. 2. If six forces acting on a body be completely represented, three by the sides of a triangle taken in order, and three by the sides of the triangle formed by joining the middle points of the sides of the original triangle, prove that they will be in equilibrium if the parallel forces act STATICS OF SYSTEMS OF PARTICLES. 91 in the same direction, and the scale on which the first three forces are represented be four times as large as that on which the last three are represented. 3. Forces P, Q, R act along the sides of a triangle ABC, and their resultant passes through the centres of the inscribed and circumscribed circles: prove that P ^ Q ^ J? cos B - cos C cos C - cos A cos A - cos B ' 4. Prove that a uniform rod cannot rest entirely within a smooth hemispherical bowl, except in a horizontal position. 5. If a uniform heavy rod be supported by a string fastened at its ends, and passing over a smooth peg; prove that it can only rest in a horizontal or vertical position. 6. A heavy equilateral triangle hung upon a smooth peg by a string, the ends of which are attached to two of its angular points, rests with one of its sides vertical ; shew that the length of the string is double the altitude of the triangle. 7. A fine string ACBD tied to the end ^ of a uniform rod AB of weight W, passes through a fixed ring at C, and also through a ring at the end B of the rod, the free end of the string supporting a weight P; if the system be in equilibrium prove that AC : BC :: 2F + W : W. 8. A horizontal rod, the ends of which are on two inclined planes, is in equilibrium: if a, ^ be the inclinations of the planes, prove that the centre of gravity of the rod divides it into two parts in the ratio of tan a to tan /3. 9. A uniform heavy rod AB has the end A in contact with a smooth vertical wall, and one end of a string is fastened to the rod at a point C such that AC = IAB, and the other end of the string is fastened to the wall ; find the length of the string if the rod is in equilibrium in a position inclined ta,the vertical. 10. A cylindrical ruler whose radius is 2a, and length 2h rests on a horizontal rail with one end pressing against a smooth vertical wall, to which the rail is parallel. Shew that the angle the axis of the ruler makes with the vertical is given by [h sin d + a cos 6) sin^ 6 + 2a cos d = b, where b is the distance of the rail from the wall. 11. Two equal heavy spheres of one inch radius are in equilibrium within a smooth spherical cup of three inches radius. Shew that the 92 STATICS. pressure between the cup and one of the spheres is double the pressure between the two spheres. 12. Along each side taken in order of a polygon inscribed in a circle, acts a force whose magnitude is proportional to the sum of the lengths of the two adjacent sides: prove that the system of forces is equivalent to a system of forces acting along the tangents at the corners of the poly- gon, each such force being proportional to the length of the chord joining the two adjacent points. 13. ABCD is a quadrilateral: forces act along the sides AB, BC, CD, DA measured by a, /3, y, 5 times those sides respectively. Shew that if there is equilibrium ay — §5. Shew also that A ABDjls ABC = a{y- p)l8 {^-a). 14. Into the top of a fixed smooth sphere of radius a is fitted firmly a fine smooth vertical rod. A bar of length 26 has at one end a ring which slides on the rod ; and the bar rests on the sphere. Shew that in equili- brium the angle (a) the bar makes with the horizontal is given by asina = &cos^a. 15. Forces P, Q, R act along the sides BG, CA, AB of a triangle ; shew that their resultant will act along the hne joining the centre of the circumscribing circle to the intersection of perpendiculars if cos B cos C cos C cos A cos A cos B P • O • B •• * • : . ^ ' '* cosC cos S * cos ^ cosC cosi>' cos^ 16. A kite (weight P) having a tail (weight Q) is stationary, with a normal to its face, the direction of the wind, which is horizontal, and the string in the same vertical plane. The tail is attached at a distance a below the kite's centre of gravity, the string at a distance h above. Shew that, neglecting the action of the wind on the tail, the inclination of the kite to the horizon is given by the equation W) . sin2 ^ = {P6 + Q . (a + 6)} cos 6, where IT is the pressure on the kite, when placed perpendicular to the wind's direction. 17. Forces act at the middle points of the sides of a rigid polygon in the plane of the polygon ; the forces act at right angles to the sides, and are respectively proportional to the sides in magnitude : shew that the forces will be in equilibrium if they all act inwards or all act outwards. STATICS OF SYSTEMS OF PARTICLES. 93 18. Shew that it is impossible to arrange six forces along the edges of a tetrahedron so as to form a system in equilibrium. 19. An uniform rod of weight W is supported in equilibrium by a string of length 21 attached to its ends and passing over a smooth peg. If a weight W be now attached to one end of the rod, prove that a IW length — — — of the string will slip over the peg. 20. If four parallel forces balance each other, let their lines of action be intersected by a plane, and let the four points of intersection be joined by six straight lines so as to form four triangles ; then prove that each force is proj)ortional to the area of the triangle whose angles are in the lines of action of the other three. 21. Two rings of weight P and Q respectively, slide on a string, whose ends are fastened to the extremities of a straight rod inclined at an angle 6 to the horizon : on the rod slides a light ring through which the string passes so that the heavy rings are on different sides of the light ring. Prove that in the position of equilibrium the inclination of those parts of the string next the weightless ring, to the rod, is giyen by the equation tan 0/tan ^ = (P+ Q)/(P~ Q). 22. An elastic string passes round three equal right-circular cylinders so that when each cylinder touches the other two along a generating line, the string is just not stretched : shew that if the system be placed on a smooth horizontal plane, the inclination (6) of the plane con- taining the axis of the upper cylinder, and that of either of the lower ones to the horizontal, in the position of equilibrium, is given by the equation (tt+S) Tr=2\(2cos ^- 1) tan^. (IF is the weight of the upper cylinder, and X is the modulus of elasticity.) 23. Two equal circular discs, of radius r, with smooth edges are placed on their fiat sides in the corner between two smooth vertical planes inclined at an angle 2a and touch each other in the line bisecting the angle ; the radius of the least disc w'hich may be pressed between them without causing them to separate =r (1 - cosa)/cos a. 24. A rectangular lamina ABCD is supported with its plane vertical and one edge AB in contact with a smooth vertical wall, by an endless string which passes through smooth rings, one fixed to the wall at A, and two others P, Q fixed in the sides AB, CD of the lamina respectively 94 STATICS. so that VQ is parallel to AB. Prove that the string has the least tension consistent with equilibrium when the position of Q is such that BGl2AD = ia.ulAQD. 25. Forces act through the angular points of a tetrahedron per- pendicular to the opposite faces and proportional to them. Prove that they are in equilibrium if they all act either inwards or outwards. 26. AC, BD are two non-intersecting straight lines of constant length; prove that the effect of forces rej)resented in every respect by AB, BC, CD, DA is the same, so long as AC, BD remain j)arallel to the same plane, and their projections on that plane are inclined at a con- stant angle to one another. 27. A flat semicircular board wdth its plane vertical and curved edge upwards rests on a smooth horizontal plane, and is pressed at two given points of its circumference by two beams w^hich slide in smooth vertical tubes : find the ratio of the weights of the beams to one another when the board is in equilibrium. 28. An endless string is placed round two equal cylinders and the system is suspended from a peg so that the line joining the centres of the cylinders is horizontal. If the pressure between the cylinders be equal to twice the weight of either of them ; prove that the length of the string : the radius of either cylinder : : 4 (2 + tan"^ 2) : 1. 29. A homogeneous circular cylinder rests on two smooth planes inclined to the horizon at angles a and /3 in opposite directions, so that its axis is at right angles to the line of intersection of the planes. Prove that the inclination 6 of the base to the vertical in the position of equilibrium is given by tang= ^s"^(«-/^) r sin {a + ^)+ 2a sin a sin /3 ' where r is the radius of the base and 2a the length of the cylinder. 30. In a triangular lamina ABC, AD, BE, CF are the perj)endiculars on the sides, and forces represented by the lines BD, CD, CE, AE, AF, BF are applied to the lamina ; prove that their resultant will pass through the centre of the circle described about the triangle. 31. An elliptic lamina rests against an inclined plane (a) being sup- ported by a string attached to the extremity of its minor axis, so that its major axis is vertical and the plane of the ellipse is perpendicular to the inclined plane. Shew that the inclination of the string to the vertical is tan~i&^(a--hZ^-tan-a)/(a-- fc-). STATICS OF SYSTEMS OF PAKTICLES. 95 32. A uniform bar of length a rests suspended by two strings of lengths I and V fastened to the ends of the bar and to two fixed points in the same horizontal line at a distance c apart. If the directions of the strings, being produced, meet at right angles, prove that the ratio of their tensions is al + cV : al' + cL 33. Two weights P, P are attached to the ends of two strings which pass over the same smooth peg and have their other extremities attached to the ends of a beam AB, the weight of which is W; shew that the incli- nation of the beam to the horizon = tan~^ ( =- tan a] ; a, h being the ^ ( tan a ) \a + h J distances of the centre of gravity of the beam from its ends, and sina=Tr/2P. 34. A string 9 feet long has one end attachcid to the extremity of a smooth uniform heavy rod two feet in length, and at the other end carries a ring which slides upon the rod. The rod is suspended by means of the string from a smooth peg : prove that if d be the angle which the rod makes with the horizon, then tan ^^S"-"* _ 3~3. 35. A triangle formed of three smooth rods is fixed horizontally, and a homogeneous sphere rests on it. Prove that the pressure on each rod is proportional to its length. 36. A sphere rests on three smooth pegs, which lie in a horizontal plane, and are at distances a, 6, c from one another, prove that the pressures on the pegs are in the ratios a-{h- + c"-a-) : h- {c" + or - b"-) : c^ {a? + }f'-c''). 37. ABC, A'B'C are two triangles inscribed in the same circle ; and forces proportional to the sides of the triangle act along them, but in op- posite dhections round the two triangles. Prove that, if a, /3, y be the angles subtended at the centre of the circle by the sides of the one triangle, and a', j3', y' those subtended by the sides of the other, the forces .,, , . .,., . .„ . a . jS . 7 . a' . ^' . y' will be in equilibrium if sin - sin \ sm ^ = sin - sm ^ sm — . ^ 2 2 2 2 2 2 38. A, B, C, D are four points in space: four forces represented by AB, AD, CB, and CD act along these lines : prove that they have a single resultant, the line of action of which is perpendicular to the shortest distance between the lines AB, DC, and also to that between AD, BC. 96 STATICS. 39. Three equal spheres are placed in contact on a smooth hori- zontal table, and a fourth equal sphere is placed upon them, and then a cone of semi-vertical angle a is placed over the pile of spheres. Prove that the cone will he lifted if its weight is less than —p= tan a of the weight of a sphere. 40. A cylindrical shell, without a bottom, stands on a horizontal plane, and two smooth spheres are placed within it, whose diameters are each less whilst their sum is greater than that of tlie interior surface of the shell : shew that the cylinder will not upset if the ratio of its weight to the weight of the upper sphere be greater than 2c- a-b : c, where a, b, c are the radii of the spheres and cylinder. 41. Three spheres of radius c are placed on a smooth horizontal table so that their points of contact with it are at the angular points of an equilateral triangle. A fourth sphere of radius a and weight W touches the table and each of the other spheres. An elastic string of natural length 27rc and modulus of elasticity /x is placed symmetrically round the first three spheres. If the fourth sphere is just on the point of ascending, shew that ^ircW —21 jx {a - c). 42. A uniform rod, length c and weight W is suspended from a fixed point by two equal elastic strings, the natural length of each being c and the modulus tv. A particle of weight W is placed on the rod at a distance X from its middle point, and when the system is in equilibrium the rod makes an angle a with the vertical. If 6, are the angles the strings make with the vertical, prove that •T sin (6 ~ cp) -2 cot a . sin 6 sin STATICS OF SYSTEMS OF PARTICLES. 97 from each of the given points. If the plane in which either string lies makes an angle a with the horizon, prove that the tension of each = WajSr sin a ; a being the distance between the points. 45. A smooth semi-circular tube is just filled with 2n equal smooth beads that just fit the tube, and the whole is at rest in a vertical plane with the bounding diameter highest. If Ii„^ be the pressure between the with and {m + l)th. beads from the top, then where W is the weight of a bead. Hence deduce that when the beads are diminished indefinitely in size, the pressure between any two is proportional to their depth below the top one. 46. A smooth rod passes through a smooth ring at the focus of an ellipse whose major axis is horizontal and rests w^ith its lower end on the quadrant of the curve which is furthest removed from the focus. Shew that its length must be at least ^a + ^a ^y(l + 8e-), where a is the semi- major axis and e the eccentricity. » 47. A rigid bar without weight is suspended in a horizontal posi- tion by means of three equal, vertical, and slightly elastic rods to the lower ends of which are attached small rings A, B, and C through which the bar passes. A weight is then attached to the bar at any point G. Shew that, on the assumption that the extension or com- pression of an elastic rod is proportional to the force applied to stretch or compress it, and provided the rods remain vertical, then the rod at B will be compressed, if G lie in the direction of the longer of the A p2 4- R C~ two arms AB, BC, and be at a greater distance from B than ^—- — -— - . Ah ~hC 48. A number 7i of equal smooth spheres of weight IF and radius 7* is placed within a hollow vertical cylinder of radius a, less than 2r, open at both ends and resting on a horizontal plane. Prove that the least value of the weight JV of the cylinder in order that it may not be upset by the balls is given by aW'= {n - 1) (a - r) W or a^V' = n(a- r) W, according as n is odd or even. 49. Four equal smooth spherical balls of radius a are piled up within a hollow sphere which is the largest which can retain them in mutual contact, shew that its radius is a (1-1-2 \/li). G. 7 98 STATICS. • 50. Four equal weights IF are tied by strings to four equi-distant points of a loop of a string of length I, which is then placed sym- metrically on a smooth sphere of radius r, all the weights hanging freely down; shew that in the position of equilibrium, the tension of each string is equal to ^^|(l+COsi;)/cOsi;|. 51. A quadrilateral ABCB has the sides I) A, AB, BG equal and the angles DAB, ABC right angles, but AB and CD are not in the same plane. If forces acting along the four sides can be reduced to a couple, its axis will make with A B an angle :COS" .J I C D--AB" V CD--Y'6AB- CHAPTER III. STATICS OF CONSTRAINED BODIES, ETC. 72. The conditions of equilibrium which we have proved in the last Chapter apply to any rigid bodies what- soever. If however the body considered be a constrained one, i.e. one that is not free to move in every way, as for instance one that can only turn about a fixed axis, we can obtain conditions of equilibrium which do not involve the forces of constraint. 73. Prop. If a rigid body under the action of a system of coplanar forces, have one point in the plane of the forces fixed, it is a necessary and sufficient condition of equi- librium that the algebraical sum of the moments about the fixed point of the forces, excluding the force of con- straint, be zero. For the force of constraint acts through the fixed point A, and therefore when there is equilibrium, the resultant of the remaining forces must act through A. But the algebraical sum of the moments of these remaining forces about any point is equal to the moment of their resultant, and therefore that about A vanishes. The condition is therefore a necessary one. It is also sufficient. For if it hold, it can be shewn as in Art. 52 that A being a fixed point, the body is in equi- librium. 7—2 100 STATICS, Ex, 1. A uniform rod which is 12 feet long and which weighs 17 lbs, can turn freely about a point in it, and the rod is in equilibrium when a weight of 7 lbs, is hung at one end. How far from that end is the point about which it can turn ? Ans. 4 ft. 3 in. Ex. 2. ABCD is a square : a force of 1 lb. acts from A to B, one of 4 lbs. from B to C, and one of 15 lbs. from D to C : if the centre of the square is fixed, find the force which, acting along DA^ will maintain equilibrium. Ans. 10 lbs. Ex. 3. ABCD is a square, of which the point A is fixed : a force of 2 lbs. acts along AB, one of 6 lbs. along AD, one of 10 lbs. along JBD, and one of 3 lbs. along BC, find the force along DC which will maintain equilibrium. Ans. (5 \/ 2 + 3) lbs. Ex. 4. A lever ABC, with a fulcrum B, one-third of its length from ^4, is divided into equal parts in D, E, and F. At C, D, and F, forces of 12 lbs., 8 lbs., and 6 lbs. respectively act vertically downwards, and at E a force of 16 lbs. acts vertically upwards. "What force applied to A will cause equilibrium ? Ans. 21^ lbs. Ex. 5. A weightless lamina in the shape of a regular hexagon ABCDEF, is suspended from the middle point of ^jB : shew that it will be in equilibrium with the side AB horizontal, if weights of 3 lbs., 7 lbs., 3 lbs. and 5 lbs. are hung at C, D, E, and F respectively. 74. Pi'op. If two points of a rigid body be fixed, so that it can only turn about the line joining them, it is a necessary and sufficient condition of equilibrium that the algebraical sum of the moments of the forces, excluding those of constraint, about the fixed line, be zero. If there is equilibrium, the algebraical sum of the moments of all the forces about any line is zero, and the moment of the force of constraint at each of the fixed points about the line joining them is zero : therefore the sum of the moments of the remaining forces, excluding those of constraint, about this line, is zero. It is therefore a necessary condition. It can be proved as in Art. 54 that when the alge- braical sum of the moments about any line is zero, there is equilibrium provided two points in the line be fixed. The condition is therefore sufficient. STATICS OF CONSTRAINED BOl?IES, ETC. 101 75. Prop, li one point of a i^^girl-oorly be 'fixed, t'Le necessary and sufficient conditions of equilibrium are, that the algebraical sum of the moments of the forces about each of three lines through the fixed point, but not in the same plane, be zero. It can be shewn, as in the last proposition, that the conditions are necessary. It can be shewn, as in Art. 55, that they are sufficient. 76. To obtain the forces of constraint at the fixed points in any of the cases considered in the last three propositions, we have only to apply the remaining con- ditions of equilibrium found in Chapter II. 77. As we shall often have to consider the case of bodies, such as rods, which are connected by means of hinges ov joints, it will be well to consider what a hinge is. We shall consider smooth hinges onh". The connection may be supposed to be made in several ways. A point of one body may be connected with one of the other body by a very short string. Or one body may end in a very small hall ox pivot, which works in a corresponding small socket or ring in the other body, so that there is contact at only one point. Or we may suppose each body to end in a small ball, which works in a corresponding socket of a small separate body. In each of these cases there is no restriction on either body, excej^t that the two ends must be in contact ; the action on each at the common point must pass through this point, but will adapt itself in magnitude and direction so as to maintain equilibrium, if possible. If three or more bodies are connected by one joint, we may suppose the connection to be made by each having a very short string attached to it, and the strings to be knotted together. Or we may suppose each to end in a small smooth ball, which works in a corresponding socket in a small separate body. 78. In the construction of materials it is often desirable to ascertain the internal forces between one portion of a body and the adjacent portion. When all these are known, we are able to adapt the strength of each part to the force it has to sustain. For instance, if w^e know that the tension at one point of a chain is 102 STATICS. always -half .that, at ^mother, the thickness of the chain at the former point need only be half that at the latter ; a savinof in material and in weiGfht is thus effected. We have learnt that when a body is in equilibrium, the forces exerted on any portion of it by the adjacent portions counteract the remaining forces acting on the portion in question. As, however, there is an infinite number of systems of forces, each of w'hich counteracts a given system, we cannot as a rule determine which system is the one actually exerted, without going beyond the limits of Elementary Statics. If, for instance, a rope composed of several fibres be taut, though we may know the tension of the rope itself, i.e. the sum of the tensions of the different fibres, we cannot say how it is distributed among them. This can only be ascertained w^hen the elasticity of each fibre is known. When a beam is merely stretched, i.e. when the ex- ternal forces all act along it, the only internal forces called into play will be between particles arranged in lines along the beam. If then the beam be supposed to consist of two parts A and B, the action of B on A will be the sum of the forces exerted by particles of B on the adjacent particles of A , all such forces being in the same direction along the beam. This action is equal to the resultant of the forces acting on the portion A, and which are also external to the beam. It is clear that the greater this action becomes, the more likely is the beam to be pulled asunder at the point of junction of A and B ; the action therefore measures the tendency of the beam to break at that point. 79. When the external forces on the beam are not all along it, the action of one portion on another is not so simple as in the above case. Take the following case. Let ABGD be a rectang^ular beam which is firmlv fixed at the end AB in a vice ; along DC let a force 8 be applied : it will of course be perpendicular to the beam. Consider STATICS OF CONSTEAIXED BODIES, ETC. lOo the equilibrium of the portion CDPQ, where PQ is an imaginary section perpendicuhu" to the beam's length. The forces in action are *S' and the innumerable forces due to ABQP, acting at every point of the section PQ B Q Fig 54 ", S Let the latter be resolved along PQ and at right angles to it : the sum of the former components must be equal and opposite to >S', and will with it form a couple. The com- ponents perpendicular to PQ must therefore be equivalent to a couple, equal and opposite in sign to the former. This shews that the forces near P must be in the direction PA, and those near Q in the opposite direction : and therefore that the tendency of >S' is to stretch the fibres near P and crush those near Q. It must follow too, that the magnitudes of the comjionents perpendicular to QP depend on the moment of S about Q, and. not on the magnitude of >S' simply. Hence the greater the moment of S about Q the more likely are the fibres along PQ to give way and the rod to bend at PQ. Since PQ is supposed small compared with QC, the numerical sum of the forces along PQ must be very much greater than S, i.e. a force is far more likely to bend a rod, when applied at right angles to it, than to pull it asunder when apj)lied along it. Similar reasoning wdll apply to abeam under the action of any system of forces. We can shew that the tendency to bend at any point is measured by the algebraical sum of the moments about that point, of the forces external to the rod and acting on one of the parts into which the beam is divided by the point. This tendency to bend is also termed the bendiny moment. 104 STATICS. Ex. 1. A light beam is supported in a horizontal position at its ends, and a weight w is hung from its middle point. Find the bending moment at a point distant x from one end. Aiis. -^ . Ex. 2. If a heavy uniform rod be supported at its middle point, shew that the bending moment at any point varies as the square of its distance from the nearer end. Ex. 3. A uniform rod AB of weight iv and length a is supported in a horizontal position at A and B ; from a point distant x from A a weight w' hangs : find the bending moment at a point distant y from A . y (a - ?/) {a-x)y Ans. ID. 10 . 2a y [a - y) + w' + w a , if ?/ is < .r, , if y is x. 2a a Ex. 4. A uniform rod of weight w and length a, can turn freely about a hinge at one end, and rests with its other end against a smooth vertical wall, distant b from the hinge. Prove that the bending moment at a point whose distances from the two ends are x, y, respectively, is IV . xyh 80. When each of the bodies forming a system in equilibrium is acted on by forces that reduce to three, the problem of finding the position of each of the bodies, Fig 55 fi' or of ascertaining tho different forces, can often be easily solved by constructing a series of triangles each of which STATICS OF CONSTRAINED BODIES, ETC. 105 is the triangle of forces corresponding to one of the bodies. For instance, let us consider the case of a number of par- ticles of equal weight fastened at intervals along a weight- less string, the ends of which are attached to fixed points. Let A, B, Cy D &c. be the positions of the particles, when in equilibrium. Any particle, B for instance, is kept in equilibrium by three forces, its weight vertically down- Avards, and the tensions of the strings BA, BC. Draw a triangle obc, having its sides ba, ao, oh respectively parallel to the lines of action of these forces : then bv the triangle Ffg 56 of forces these lines arc proportional to the forces, to whose directions they are parallel: i.e. weight of ^1 : tension of AB : tension of BC=ab : ao : ob. Produce ab down- wards, and mark off be, cd, de &c., each equal to ah ; join Oc, Od, Oe &c. Then Ob, be represent in every way the tension of BC, on C, and the weight of C respective^, so that CO must represent the tension of CD. Similarly do represents the tension of DE, eo that of EF, and so on. Draw OM perpendicular to abc : then the tangents of the angles that ao, bo, co &c. make with the horizon are aM bM ^ _^ Oil/' OM' OM' OM' hence the tangents of the angles which the strings make with the horizon form an arithmetic series. Also the 106 STATICS. horizontal resolved part of the tension of each string is represented by Oil/, and is therefore the same for all. Such a figure which is drawn to enable us to solve the problem is called a Force-Diagram.. The above results can be obtained very easily by equat- ing to zero the A . S's of the resolved parts in a horizontal and vertical direction of the forces that act on each par- ticle separately. 81. This 'Graphic' method can be applied to prove the following important proposition. Prop. If a weightless string be stretched across a smooth surface, the tension is everywhere the same. Let ABCD &c. be the string : then any small portion Fig 57 of it AB \^ kept in equilibrium by the tensions at its ends, and the resultant of the pressures of the surface on it : as the pressures along AB all act along the normals to the surface at the corresponding points, their resultant's STATICS OF CONSTRAINED BODIES, ETC. 107 direction must lie somewhere between the normals at A and B. Draw Oa, Ob, Oc, Od, Oe &c. parallel to the tensions at A, B, C, D &c. respectively : and also a6, he, cd, &c. parallel to the resultants of the pressures on AB, BC, CD &c. Then by the triangle of forces, each line represents the magnitude of the force to whose direction it is parallel. Since the resultant pressure on AB has a direction be- tween the normals at A and B, and these ultimately, when AB is taken indefinitely small, make indefinitely small angles with one another, ab makes with the normals at A and B very small angles, i.e. makes with Oa, Ob, which are parallel to the tangents at A, B, angles ultimately equal to right angles. Hence the difference bet\veen Oa, Ob must be of the second order of small quantities, similarly those betw^een Ob and Oc, Oc and Od &c. are of the second order, i.e. Oa, Ob, Oc, &c. and the tensions they represent, are all equal. ILLUSTRATIVE EXAMPLES. Ex. 1. OA, AB axe two uniform beams loosely jointed at A, the former being moveable about a hinge at 0. A string attached to B passes over a fixed smooth pully and supports a weight P. If in the position of equili- brium the beams are equally inclined to the vertical, the string will make an angle cos~^ ( -— — j with the vertical, where W, W are the weights of the beams. 108 STATICS. Let a be the inclination of either of the beams to the vertical, and & that of the string. Ffg 59 liesolve the tension of the string (P) at B into two forces Pcosff vertically, and P sin 6 horizontally. Let 2a, 2b be the respective lengths of OA, AB. From the equihbrium of both rods together by taking moments about 0, we have JV . a sin a + W (2a sin a + Z> sin a)-P cos (2a sin a + 2h sin a) + P sin 6 {2h cos a - 2a cos a) = 0. . .(1). Taking moments about A for the equilibrium of A B 7F' . & sin a + P sin ^ . 26 cos a - P cos 6* . 2& sin a ^ (2). Subtracting Tr.asina + 2IF'. asina-2Pasin (^ + a) = 0, or (TF+2ir')sina = 2Psin(^ + a) (3), from (2) Tf'sina = 2P.sin(a-6') (4). Adding equations (3) and (4) we have ( >F + 3 [F') sin a = 4P sin a cos ^, "f^+3Tr'' ^-cos -'(- 4P If the stresses at and A be resolved horizontally and vertically, as shewn in the figure, we can determine them as follows : STATICS OF CONSTRAINED BODIES, ETC. 109 Resolving horizontally and vertically for the eriuilibrium of OA A"+Z = (5). Y'+Y-W = (6). Resolving horizontally and vertically for AB, X+ Pain 6^0 (7), Y+W- Pcosd=^0 (8). Equations (5), (6), (7) and (8) completel}' determine A', }', X' and Y'. Ex. 2. xVn equilateral pentagon consisting of five freely-jointed rods is hung up with one side horizontal; shew that the inclination [6) of either of the upper rods to the vertical is given by the equation sin ^ + 6 sin- ^ + 8 sin^ ^ - 8 sin^ d = ^. ;r„-* Fig. 60 Let AB be the fixed rod. Let be the inclination of CD and DE to the vertical. (The rods are drawn separate to make the figure clearer.) Let W be the weight of each rod, 2a its length. Let the stresses on the different rods at the joints be resolved horizon- tally and vertically : the magnitude of these stresses are indicated in the figure. The stress at D is entirely horizontal) as the rods CD, DE are symmetrical with respect to the vertical. \ 110 STATICS. From the equilibrium of AE, resolving horizontally we have Xj^Xo (1). Taking moments about A Tr.asin6'+ Y.2.2asmd-X.^. 2a cos ^ = (2). From the equilibrium of ED, resolving horizontally' ^. = ^3 (3). Kesolving vertically Y^=W (-1). Taking moments about E W . a sin 0- A'g . 2rtcos 0==O (5). Substituting from (1), (8) and (4) in (2) and (5), 2A'i.cos^ = 31T^'.sin^ • (6). Tf .sin0 = 2AiCos0 ....(7). . • . cot ^ = 3 cot ~ (8) . Since the sum of the horizontal projections of AE, ED, DC, CB is equal io AB, Aa sin 6 + Aa sin = 2a, .-. sin 6* + sin = 1 (9). By eliminating (0) between (8) and (9) we obtained the required result. By substituting the value of 6 just obtained in (6), we determine X^ and by resolving verticall}' for the equilibrium of AE, we obtain another equation which determines 1\. The stresses at the angular points are thus completely determined. Ex. 3. Six equal heavy rods freely jointed at the ends form a regular hexagon ABCDEF, which when hung up by the point A is kept from altering its shape by two light rods BF, CE. Prove that the thrusts of the rods BF, CE are as 5 to 1, and find their magnitudes. We shall suppose that there is a light pivot at B, to which the three rods AB, BF, and BC are attached; and that a similar arrangement is made at C. Let W be the weight of each rod, 2a its length. Since the rods BF, CE are only acted on by the stresses at their ends, these stresses must be along the rods, i.e. horizontal, let them be S and 2' respectively. Since the rod BC is acted on by its weight along its length and the stresses at B and C, these latter forces must also act along BC (Art. 61), i.e. verticallv. STATICS OF CONSTRAINED BODIES, ETC. From symmetry the stress at D on CD is horizontal. in 3 2 'w Fig 61 Let the stresses resolved horizontally and vertically at A, B, C and D be those shewn by the figure. From the equilibrium of AB, by taking moments about A , .(1). From the equilibrium of the pivot B, resolving horizontally and verticallv, S=X, (2). and ^2=73 (3) 112 STATICS. From the equilibrium of J5C, Y^-W-Y, = .(4). From the equilibrium of the pivot C, T^X, (5). i'4=i'5 (6). From the equilibrium of CD, X, = X, (7). Y, = W (8). By taking moments about C, TF.a. Y-^"^'c-2y3. 28. Six equal and uniform heavy rods are hinged together so as to form a hexagon : it is placed with one side on a horizontal plane and is kept in the shape of a regular hexagon by means of a string fastened to the middle points of the two sides adjacent to the base: find the tension of the string and the stresses at the hinges. 29. A parallelogram formed of four rods of uniform material and thickness, jointed at their ends, is suspended from one point, which is connected with the opposite point by a string of such a length that the figure is rectangular : prove that the tension of the string is half the weight of the four rods, and that the direction of the stress between the rods at either of the joints not connected by the string bisects the angle between them. 30. A heavy uniform rod of length 2a turns freely on a pivot at a point in it, and suspended by a string of length I fastened to the ends of the rod hangs a bead of equal weight which slides on the string. Prove that the rod cannot rest in an inclined position unless the distance of the pivot from the middle point of the rod be less than a-]l. 31. A number of equal weightless rods are freely jointed and assume the form of a. regular polygon when subjected to a system of stresses at each joint, all emanating from a point on the circumscribing circle. Shew that, if from a point radii be drawn to represent in magnitude and direc- tion the stresses in the rods, and a polygon constructed so that its sides taken in order represent the system of applied stresses, then the polygon will be equiangular and described about a parabola, and further the angular points of the polygon will all lie on a hyperbola. 32. Two equal beams AB, AC, connected by a hinge at A, are placed in a vertical plane with their extremities B, C resting on a horizontal plane; they are kept from falling by strings connecting B and G with the middle points of the opposite sides ; shew that the ratio of the tension of each string to the weight of each beam is \\J{Q cot"^ + cosec^^), where d is the inclination of either beam to the horizon. 120 STATICS. 33. A trapezium ABCD is formed of four rods joined by hinges at their extremities: BC, AD are equal, and the framework is suspended by a string attached to the middle point of AB. Determine completely the stresses at A and D. If AB = AB = BC = ICD, stress at A : stress at D= >/i9 : Jl. 34. A number of light rigid rods are loosely jointed together at their extremities so as to form a closed polygon, and a force applied to each side perpendicular and proportional to it, their lines of action meeting in a point; prove that, if equilibrium be maintained, the polygon will be inscribable in a circle, and if S be the point through which the forces act, the centre of the circumscribed circle, and SO be produced to S' so that SS' is bisected in 0, the stress at any angular point of the polygon will be perpendicular and proportional to the distance of the point from S'. 35. n equal uniform rods, each of weight W and length Z, are jointed so as to form symmetrical generators of a cone whose semi-vertical angle is a, the joint being at the vertex of the cone. The rods are placed with their other ends in contact with the interior of a sphere whose radius is r, so that the axis of the cone is vertical, and a weight W is hung on it at the joint. Shew that cos a = , , Z V3n-T7'2 + 4nfr'IF and find the action at the joint on each rod. 36. A fire-screen holder with any number of unequal weightless arms projects horizontally from a chimney-piece. Shew that, if the ends of the arms all lie on a circle, the axes of the couples at the hinges all pass through one point. 37. A regular octahedron is formed of 12 uniform rods jointed to- gether at the ends. Along the three diagonals are stretched strings whose tensions are T-^T^T^. Shew that the thrusts along the rods, joining the ends of diagonals the tensions along which are Tj, T„, are Prove also that, if the four diagonals of a cube be treated in a similar way, equilibrium is not possible unless the tensions are all equal. STATICS OF CONSTRAINED BODIES, ETC. 121 38. Three uniform heavy rods of the same material (lengths 2a, 2b, 2c, respectively) hinged together at B and C rest on a vertical circle of radius r, the whole system being in one vertical plane, and such that BC is horizontal. Find the stresses at the hinges, and prove that cot i ^ . (a2 cos2 6 - c- cos2 (^ + b' + 2bc) = cot 1 (c2 cos2 - a~ cos2 9 + ¥ + 2ah) = {a + h+c)r, where 6 and are the acute angles made with the horizon by AB and CD respectively. 39. Three equal heavy rods, in the position of the three edges of an inverted triangular pyramid, are in equilibrium with their lower ends attached to a joint about which each rod can turn freely, and their upper ends connected by strings each of length equal to half that of a rod. Prove that the tension of a string is to the weight of a rod as 1 : ^Jll. 40. A rhombus is formed of four rods of length a, hinged together. Two opposite rods are supported in a vertical plane by two smooth pegs which are separated by a horizontal distance h and vertical distance k. Shew that the product of the horizontal distances of either jDeg from the ends of the nearer unsupported rod is |(^•2 — 2a/i + K-), and that there is no bending moment round a point in either suj^ported rod, whose distance from its supporting peg is three times the shorter of the distances of that peg from an unsupported rod. 41. Four equal uniform rods are jointed freely together so as to form a rhombus: this is suspended by one of the angular points, and a sphere of weight equal to twice that of the rhombus is balanced inside it so as to prevent it from collapsing ; shew that, if the radius of the sphere be to the length of a rod in the ratio 5 : 8 V3, the rods will, in equilibrium, make each an angle tt/G with the vertical. 42. ABCD is a quadrilateral formed by four uniform rods of equal weight loosely jointed together. If the system be in equilibrium in a vertical plane with the rod AB supported in a horizontal position, prove that 2 tan ^ = tau a ~tan/3, where a, /3 are the angles at A and B, and 6 is the inclination of CD to the horizon : also find the stresses at C and D, and prove that their directions are inclined to the horizon at the angles tan"^ h (tan ^ - tan d) and tan~i h (tan a + tan 6) respectively. 122 STATICS. 43. Four equal rods are joined together so as to form a rhombus ABCD, lying upon a smooth horizontal plane, and elastic strings AG, BD of the same substance are stretched along the diagonals : if a be the length of a side of the rhombus, and if the natural lengths of the strings be ^ and >— , find the angles of the rhombus when there 2 2(1+V2) is equilibrium. 44. Seven rods are freely jointed together to form a regular heptagon ABCDEFG. Two equal strings connect G with D and B with E, and the whole system is suspended by the point A. Find the tension of the strings. 45. Three beams AB, BC, CA are joined together at A, B, G ; B being an obtuse angle, and are placed with AB vertical, and A fixed to the ground, so as to form the framework of a crane. There is a pulley at G, and the rope is fastened to AB near B, and passes along BG, and over the pulley. If it support a weight W large in comparison with the weights of the framework and rope, find the couples which tend to break the crane at A and at B. 46. A door is moveable about its line of hinges which is inclined at an angle a to the vertical ; shew that the couple necessary to keep it in a position inclined at an angle ^ to its position of equilibrium is propor- tional to sin a sin /3. 47. Three equal heavy rods AB,BG, GD are jointed to each other at B and G and to fixed points at A and D, where AD is horizontal and equal to the length of a rod. Shew that the horizontal couple required to turn the rod BG through an angle 6 is BG . W . sin ^6, where W is the weight of each rod. 48. The lid ABGD of a cubical box, moveable about hinges at A and jB, is held at a given angle to the horizon by a horizontal string connecting C with a point vertically over A : find the pressure on each hinge. CHAPTER lY. CENTRES OF MASS. 82. We have seen (Art. 59) that the resultant of two parallel forces F, Q, acting at fixed points A, B respectively, is equal to their algebraical sum, and acts along a line parallel to the line of action of either : also that its line of action cuts AB at a fixed point, whose jDosition depends solely on the relative magnitude of P and Q and not on their direction. So too, if we have a niunher of j^arallel forces acting at fixed points, their resultant is equal to their algebraical sum, and its line of action is parallel to that of any of them, and passes through a point whose position depends solely on the positions of the fixed points and the relative magnitude of the forces. For two of the forces are equivalent to their sum acting parallel to them and through a fixed point : this resultant and a third force of the system are also equivalent to the algebraical sum of the three acting parallel to them through a fixed point ; in this way we can go on reducing the number of the forces until we arrive at the final resultant acting through a fixed point. We shall necessarily arrive at the same fixed point, what- ever be the order in which we compound the forces : for if by compounding them in different orders we obtain two points, at either of which the resultant may act, its line of action must be the line joining the two points (Art. 5.S), which is inconsistent with its being always parallel to the directions of the original forces. 124 STATICS. It is assumed above that the algebraical sum of the forces is not zero, otherwise, if they are not in equi- librium, they will not reduce to a single force, but to a couple. 83. Def. The centre of a number of jmrallel forces acting at fixed points, is the point at which their resultant always acts, however their direction alters, so long as their relative magnitudes remain the same. If the points at which the parallel forces act lie in one plane, we can find an expression for the distance of the centre of the forces from any straight line in the plane. Let A^y A^, A^, &c. be the points of application of the parallel forces, P^, P^, P^, &c., and let C be their centre. -^p. X'.. M: A - M '^(P) Mq Mq ■^P2 Fig. 65 Let X'X be any straight line in the plane containing the points of application. Draw A^M^, ^2^^2' ^^-^ (7 J/ perpen- dicular to X'X. Let x^, x^ X be the lengths of these perpendiculars, which are reckoned positive or negative according to the side of the line on which the corresponding point of application lies. As the position of G is independent of the direction of the forces, it will not be affected by supposing P^P^ ... to act parallel to X' X: since the algebraical sum of P^, P^, &c. acting at C, is the resultant of these parallel forces, the algebraical sum of the moments of P^, P^, &c. about CENTRES OF MASS. 125 any point in X' X is equal to the moment of their alge- braical sum at G, about the same point in X' X. ' 7? — 1^ 1 "^ 2^2 + • • • _ '^{^^) ' P, + P,+ ... t{P)' As we can find in this way the distance of the centre of a number of parallel forces acting at fixed points in one plane, from two intersecting straight lines in that plane, its position is completely determined. 84*. When the points of application of the parallel forces are not in one plane, we can find an expression for the distance of the centre from any given idane. Let A^, A^, &c. be the points at which the forces P^, P^, &c. respectively act : let C be the centre. Draw A^2I^,A^M^, Y ^' ^(P) y^l mi /I ^l' ... ^' X, M2 % Fig. 66 '&c., CM perpendicular to the given plane; let x^, x^...x be these respective distances, which are reckoned positive or negative according to which side of the plane the corre- sponding points lie. Let XX be any straight line in the plane. Since the position of G is independent of the direction of the forces, it will not be affected b}^ sujDposing this 126 STATICS. direction to be parallel to the plane and at right angles to X'X. As the resultant of P^, P^, &c. is their algebraical sum acting at G, the algebraical sum of the moments of P^, P^, &c. about X'X is equal to the moment of their algebraical sum acting at G about X'X. .: P, . A^M^ + P^.A^M^ + kc. = (P^ + P^+ ...) GM -, :. Pj^i + P^x^ + &c. = (Pj + P^ + ...) ;^ ; . P,^, + P,^,+ ... _ S(Pt) P, + P, + ... ~ S(P) • When we have found the distance of G from three planes which have only one point in common, its position is completely determined. Ex. 1. is the intersection of the diagonals of a square ABCD, whose side is 1 foot long : find the position of the centre of like parallel forces acting at A, B, C, D and O, respectively proportional to 4, 3, 4, 6 and 9. Ans. In DB, distant 5^% inches from AD. Ex. 2. At the angular points A, B, C of an equilateral triangle, forces of 1, 2, and 3 lbs. respectively act : find the distance of their centre from C. Ans. I sjl .AB. Ex. 3. At four of the angles of a regular hexagon taken in order, parallel forces proportional to 3,-2,7 and - 5 act : find the magnitude of the forces that must act at the remaining angles, in order that the centre of the six parallel forces may be the centre of the hexagon. Ans. 6,-1. 85. Def. Let A^, A^, A^, &c. be a number of particles of masses m^, 7n^, m^, &c. respectively; then if a ]3oint G^ be taken in A^A^, so that m^. G^A^ =m^.G^A^, this point is called the centre of mass or the centre of inertia of the particles A^ and A^. The centre of mass of A^ and a particle of mass {7n^ + m,^) situate at G^ is the centre of mass o^ A^, A^ and A^. That of A^ and a particle of mass {m^ + m^ + ni^ situate at the centre of mass of CENTRES OF MASS. 127 4' A^, A^ and A^ is the centre of mass of A^, A^, A^ and A Continuing this process we obtain the centre of mass of any number of particles. From this definition of the centre of mass of a number of particles it is clear that its position is the same as that of the centre of a number of like parallel forces acting one on each of the particles, each force being proportional to the mass of the particle on which it acts. Hence if the particles of masses m^, iii^.,. be at distances ^,, x^... re- spectively from a given plane, the distance of their centre of mass from that plane is %(mx)/^(m) ; or, in other words, the distance from a given plane of the centre of mass of a number of particles is obtained by multiplying the mass of each by its distance from the plane, and dividing the alge- braical sum of the products by that of the masses. If the jDarticles are not fixed in position, but move so that the configuration formed by them is unaltered in shape, their centre of mass will be a point moving with the configuration, but occupying a position fixed relatively to it. Def. The product of a mass into the distance of its centre of mass from any plane or line is termed its moment about that plane or line. We see from the above, that the algebraical sum of the moments of a number of masses about any plane, and if they are coplanar, about any line, is equal to the moment of the whole mass collected at the centre of mass about the same plane or line. 86. Let us suppose that the above system is acted on by a number of like parallel forces, one on each particle, their magnitudes being proportional to the masses of the par- ticles on which they respectively act : now, no matter how much the direction of the forces varies, or to what extent the particles move, so long as the configuration formed by them remains the same, the resultant of these forces will always pass through the centre of mass, wdiich is fixed re- 128 STATICS. latively to the configuration. Since the magnitude of a particle's weight is proportional to its mass and its direc- tion is towards the earth's centre, the weights of a system •of particles which are not far from one another in com- 13arison with their distance from the earth's centre, are forces approximately parallel, and also proportional to the masses of the particles on which they act. The line of action of their resultant then will approximately always pass through a point fixed relatively to the configuration formed by the particles, if that configuration does not alter, though it move as a whole. This point, which we have called the centre of mass of the system, is on this account often called its centre of gravity/. We may define the centre of gravity thus : the centre of gravity of a body is the point, fixed relatively to the body and through which the resultant of the weights of the particles composing it always acts, however the body move, provided it always moves as if it were rigid. Strictly speaking, there is no such point of necessity for every body, because the weights of the particles com- posing the body are not accurately parallel, but they are so nearly so that their resultant will pass very close to the centre of mass, if it does not pass through it. It is not assumed in the definition of the centre of gravity that the body is a rigid one : any body whatsoever, a flexible string for instance, or a mass of liquid, will have a centre of gravity corresponding to every definite shape of the body, though its position in the body will generally alter with an alteration of the body's shape. If a body be such that the action of gravity on it can always be reduced to a single force passing through a point fixed relatively to the body, whatever be its position re- latively to the earth, the body is termed a centrobaric body, and the point its centre of gravity, in a stricter sense than is usually attached to the term. 87. Def When a substance is such that the mass of any volume of it is proportional to that volume, it CENTRES OF MASS. 129 is said to be homogeneous, or of uniform density : when this is not the case, it is said to be heterogeneous, or of variable density. When a substance is homogeneous its density is mea- sured by the numerical measure of the mass in a unit of volume. When the density of a substance varies, the average density of any volume is measured by the ratio of the numerical measure of its mass to that of its volume. The density at any point is measured by the limit of the average density of an indefinitely small volume containing the point in question. 88. Prop. Having given the centres of mass of a body and of one part of it, to find that of the remaining part. Let m^, m^ be the masses of the two parts forming the body, (7^, Og their respective centres of mass. Join Gfi^, c7 b' c^ Fig 67 and take C between C^ and C^ such that m^^ . CC\ = m^ . CC[^ : then G is the centre of mass of the whole. Since C\ , C, G^ are connected in this way, it is perfectly clear that if G^ and G are given, 0^ is the point obtained by producing Gfi to a distance = (nijm^ . GG^. Gor. In a similar way we can obtain the centre of one part of a system of parallel forces when the centres of the whole system and of the remaining part are known. 89. Projf. If the mass of each of a series of particles be multiplied by the square of its distance from any given point, the sum of the products so obtained is equal to the sum of the products obtained by multiplying the mass of each particle by the square of its distance from the centre of mass of all the particles, together with the product of the whole mass into the square of the distance of the given point from the centre of mass. G. 9 130 STATICS. Let A^, A^, &c., A^ be n particles of mass m^, m^. . . m,^ ; let G be their centre of mass, and any point what- soever. Join GO, and draw AJ\I^, ^J^h^ ^^- perpendicular to GO. Then AO''=^AG''+ OG' -20G. GM. G M Fig 68 The — sign in this equation refers to the above figure where M and are on the same side of G, but if we agree that GM shall be reckoned positive when M is on the same side of G as 0, and negative when on the other, the equation holds for all figures. Hence t (m . AG') = S (m.AG') + t (m . OG') - 2X (m . OG. GM) = t (m ,AG') + OG' . S (m) - 20G . X (m . G3I), But since G is the centre of mass of the n particles, S (m. GM) is zero (Art. 85), and we have S (m .AO') = ^(m.A G') + OG\t (m). (In the above proof, it is not assumed that A^, A^, &c. are in one plane.) Cor. If the mass of each of a series of particles be multiplied by the square of its distance from any given point, the product so obtained is least when the given 23oint is the centre of mass of the system of particles. 90. We will now investigate the positions of the centres of mass of some of the simpler geometric figures. CENTRES OF MASS. 131 Prop. If a body consists entirely of pairs of particles, such that those forming each pair are of equal mass and at equal distances from, but on opposite sides of, a certain point, that point is the centre of mass of the body. For this point is clearly the centre of mass of each pair, and therefore of all the pairs, i.e. of the whole body. Hence the centre of mass of a thin rod, uniform in density and sectional area, is its middle point : that of a lamina, uniform in thickness and density, and in shape, a circle, ellipse, or parallelogram, is its centre of figure. Also the centre of figure of a homogeneous sphere, ellip- soid, or parallelepiped is its centre of mass. The centres of mass of many other figures can be thus determined. 91. Prop. If a body consists entirely of pairs of par- ticles, those forming each pair being of equal mass and such that the middle point of the line joining them is on a certain straight line or plane, the centre of mass of the body lies in that straight line or plane. For this straight line or plane contains the centre of mass of every pair of particles and therefore that of the whole body. Hence any straight line or plane which divides a homogeneous body symmetrically, contains its centre of mass. For instance the centre of mass of the volume of surface of a rio^ht circular cone, with its base at riorht angles to its axis, lies in the axis : that of a segment of an ellipse or parabola lies in the diameter conjugate to the chord cutting^ off" the seo'ment. When we speak of the centre of mass of a surface or plane figure, we suppose the figure to be of very small uniform thickness. Similarly a line or curve is supposed to be of very small uniform sectional area. 92. To find the centre of mass of a plane triangle. Let ABC be the triangle. Bisect BG in D, and join AD. Draw hdc parallel to BG, meeting AD vn d. 9—2 132 STATICS. Then hd ; BD = Ad : AD = dc : DC; .'. hd = dc. Similarly it may be shewn that AD bisects any other line parallel to BG. Hence the triangle consists entirely of Fig 69 pairs of particles, those forming each pair being of equal mass, and such that AD bisects the line joining them: the centre of mass of the triangle is therefore in AD. (Art. 91.) Bisect AC in U, and join BE meeting AD in G. Then we can prove as before that the centre of mass of the triangle lies in BE, as well as in AD: it must therefore be (J, their point of intersection. Join DE : since D, E are the middle points of BG, AG respectively, DE is parallel to AB, and DE=iAB; .: AG : GD = AB : DE=2 : 1; .'. AG = ^AD. Hence the centre of mass of a triangle is obtained by joining the vertex with the middle point of the base and taking the point two-thirds the way down this line from the vertex. CENTRES OF MASS. 133 The centre of mass of the triangle ABC coincides with that of three equal particles placed at A, B, and C. For the centre of mass of those at J5 and C is at D, half-way between them : and that of all three will be in AB at a point (/, such that Bg : gA = l : 2, or Dg = ^BA. Hence G and g are the same point. Cor. By drawing an indefinitely large number of lines parallel to BC and at equal distances from one an- other, the triangle ABC may be divided into an infinitely large number of indefinitely narrow strips, of equal breadth, and having their centres of mass in AD. Now the marss of each strip is proportional to its area, i.e. to its length, and therefore to its distance from A measured along AB : also for the purpose of finding the centre of mass of tlie whole we may suppose the mass of each portion collected at its centre of mass. The problem therefore of finding the centre of mass of all these strips, i.e. of the triangle, is the same as that of finding the centre of mass of an infinite number of masses arranged along AB at equal, but indefinitely small distances, each mass being proportional to its distance from A. The centre of mass in the latter case, then, is at a distance from A equal to two-thirds of AB. Hence the centre of mass of a thin rod, of uniform sectional area, but such that the density at any point varies as its distance from one end, is distant from that end two-thirds the length of the rod. Also the centre of mass of the portion of a paraboloid of revolution, cut off by a plane perpendicular to the axis, is at a distance from the vertex equal to two-thirds the lensfth of the axis cut off. For let the paraboloid be divided into an indefinitely large number of thin slices of equal thickness by planes perpendicular to the axis. Then the volume of any slice is proportional ultimately to the square of its radius, i.e. to 13-:t STATICS. its distance from the vertex, whence the result given above follows at once by the preceding corollarj^ Ex. 1. Weights of 1, 4, 2, 3 lbs. are placed at the corners taken in order of a parallelogram ABCD; a weight of 10 lbs. is also placed at 0, the intersection of diagonals ; find the jDosition of their centre of mass. Ans. If E be the middle point of BC, the point required is in OE, at a distance from equal to one-tenth of OE. Ex. 2. A line AB is bisected in C^ , C^B in Co , C.,B in C.^ , and so on P P ad infinitum, and weights equal to P, — , ^ , &c. are placed at the points Oj, Cg, Cg, &c. Prove that the distance of the centre of mass of the whole system from B is equal to one-third of AB. Ex. 3. Find the centre of mass of seven equal particles "placed at the angular points of a regular octagon. Ans. If A be the unoccupied angular point and the centre of the octagon, the required point is in ^0 produced, at a distance from equal tof^O. Ex. 4. A square ABCD is divided into four equal triangles, by its diagonals, which intersect in : if the triangle OAB be removed, find G, the centre of mass of the remaining three. Prove that if E be the middle point of CD, G is in OE, and OG = ^AB. Ex. 5. The sides of a square ABCD are bisected, and the points of bisection of the opposite sides joined. If the small square, having the angle A, be removed, find G the centre of mass of the remaining three. Ans. G iQ in AC &nACG — ^^ AC. Ex. 6. Out of a circular lamina of radius r is cut a circle, whose diameter coincides with a radius of the lamina : find the position of the centre of mass of the remainder. Ans. The c, m. is at a distance from the centre of the lamina equal to r/6. Ex. 7. A figure consists of a square and an isosceles triangle, whose base is one of the sides of the square : if the side of the square be C inches, and the height of the triangle be 6 inches, find the centre of mass of the figure. Ans. Within the square, |- of an inch from the base of the triangle. CENTRES OF MASS. 135 Ex. 8. A uniform rod, 18 inches long, is bent so that the two parts, 8 and 10 inches long respectively, are at right angles to one another. Find the distance between the centres of mass of the new shape and the original. Ans. vV^ inches. Ex. 9. Equal weights are placed at n-2 of the corners of a regular ?t-sided polygon : find their centre of mass. Ans. If A, B be the unoccupied corners, C the middle point of AB, and the centre of the polygon, the centre of mass is in CO produced at 2 a distance from O equal to OA. n-2 Ex. 10. Having given the position of the centre of mass of two par- ticles A and B, and also that of A and C, find that of B and C. Ans. Join B with E, the c. m. of A and C, and C with D, the c. m. of A and B; let these two lines meet in G. The point where AG meets BC is the c. M. of B and C. Ex. 11. Assuming that the pressure on an indefinitely small area below the surface of a liquid is perpendicular to the area and varies as the area and its depth below the surface conjointly: find where the re- sultant pressure on a parallelogram, one of whose sides is in the surface of the liquid, acts. Ans. At a point whose depth below the surface is two-thirds that of the lowest side, Ex. 12. With the same assumption as in the last example, shew that the resultant pressure on any plane area below the surface of a liquid is proportional to the area and the depth of its centre of mass below the surface conjointly. Ex. 13. Find the centre of mass of a quadrilateral, two of whose sides are parallel to one another, and respectively 6 inches and 14 inches long, while the other sides are each 8 inches long. Ans. In the line joining the middle points of the two parallel sides, 26v/3 at a distance of ^>— inches from the longer side. 15 Ex. 14. Find also the centre of mass of the perimeter of the above quadrilateral. Ajis. In the line joining the middle points of the parallel -sides, at a distance from the greater equal to Vv^ inches. 136 STATICS. 93. To find the centre of mass of a triangular pyramid. Let ABCD be the pyramid. Bisect BG in E, join AE, and take H in it, so that AH=^IAE. Fig 70 B Let dhc be a section of the pyramid, made by a plane parallel to ABC, and let ae be its intersection with the plane ADE. Since the planes ABC, abc are parallel, he, BG are also parallel ; .-. he : ec = BE : EG; .'. he = ec, and e is the middle point of he. Similarly ae, AE are parallel, and ah : ae = AH : AE = 2:3, i.e. h is the centre of mass of the triangle ahc. Hence, if we suppose the pyramid divided into an in- finitely large number of indefinitely thin triangular slices made by planes parallel to ABC, the centre of mass of each slice will lie in the line DH, which must therefore contain the centre of mass of the pyramid. Join BE, and take K in it so that BK = ^DE; join AK intersecting BH in G. Then, as before, we can shew that the centre of CENTRES OF MASS. 137 mass of the pyramid lies in AK, as well as in BII ; the point of intersection G of these two lines must be the required centre of mass, then. Join HK. AH=IAE, and DK^^DE- .'. HK is parallel to AD, and DG: GH = AD : HK = AE :HE=S:l .-. DG = IDH. Hence the centre of mass of the pyramid is in the line drawn from any vertex to the centre of mass of the op- posite face, and is such that its distance from the former point is three times its distance from the latter. Cor. The centre of mass of a triangular pyramid coincides with that of four particles of equal mass placed at its angular points. For the centre of mass of the particles at A, B and C is H, and therefore that of the four is in HD, and at a distance from D equal to three times its distance from H; it is therefore G, the centre of mass of the pyramid. 94. If the above pyramid be divided into an indefi- nitely large number of indefinitely thin slices, such as ahc, of the same thickness, we may suppose the mass of each slice to be collected at its centre of mass //, which lies in DH : also the mass of any slice ahc is proportional to its area, since they are of equal thickness, and there- fore to the square on Dh. Hence finding the centre of mass of a triangular pyramid is the same problem as finding that of an indefinitely large number of masses arranged at equal but indefinitely small intervals along a straight line, each mass being proportional to the square of its distance from one end of the line. We infer then that the centre of mass in the latter case is at a distance from this end equal to three-quarters the length of the line. For instance, the centre of mass of a thin rod of uniform thickness, but whose density varies as the square of the distance from one end, is the point whose distance from this end is three-quarters the length of the rod. 138 STATICS. 95*. To find the centre of mass of a pyramid having any given rectilinear plane figure for its base. Let V be the vertex of the pyramid, ABODE the perimeter of its base. Let abcde be a section of the pyramid made by a plane parallel to the base. Let FQR be any straight Fig 71 line in the plane of the base; join VP, VQ, VR, cutting the plane ahcd in p, q, r respectively; p, q, r may be said to be the corresponding points to P, Q, R respectively. Since pqr, PQR are the sections of parallel planes made by the plane PVQ, they are parallel ; .-. PQ '.pq=VQ : Vq = QR:qr, .', p)q : qr = PQ : QR. Hence, if Q be the centre of mass of two given particles at P and R, q will be that of particles at p and r, provided the masses of the latter particles have the same ratio to one another as those of the former have. Similarly, if we have any number of particles at different points of the base and also another set of particles at the corresponding CENTRES OF MASS. 139 points of the parallel section, the mass of each particle of one set bearing a constant ratio to that of the correspond- ing particle of the other set, we could shew in the same way as we have done for two, that the centres of mass of the two sets are corresponding points, i.e. that they both lie in a straight line passing through the vertex. But we may suppose the two sections A BCD, abed to be made up each of a number of equal particles, the positions of the particles forming one set corresponding to the positions of those forming the other set. Hence, if H be the centre of mass of the base, the point h, where VH cuts the section abed, is the centre of mass of the latter. Dividing- then the pyramid up into an infinitely large number of indefinitely thin slices cut off by planes parallel to the base, we see that the centre of mass of each slice and therefore that of the whole pyramid lies in Vlf. But the pyramid may be divided into a number of triangular pyramids VHAB, VHBC, &c. and the centre of mass of each of these will lie in a plane parallel to the base and at a distance from it one quarter the distance of the vertex from it. The centre of mass of the pyramid must therefore be at G, the point where this plane cuts VH ] i.e. the centre of mass is found by joining the vertex with the centre of mass of the base, and taking a point in the joining line at a distance from the former point three times its distance from the latter. Cor. Since a cone or pyramid with a curvilinear base may be regarded as the limiting case of a pyramid with a rectilinear base, when the number of sides is indefi- nitely large, we can find the centre of mass of a cone in exactly the same way as we find that of a j)yramid with a rectilinear base. 9G*. To find the centre of mass of the surface of a pyramid with a rectilinear base. If the pyramid be the one in fig. 71, its surface may be divided into a number of triangles having the common vertex V : the centre of mass of each triangle and there- 140 STATICS. fore that of the whole surface will lie in a plane parallel to the base and at a distance from the vertex two-thirds that of the base. Also, as in the case of the solid pyramid, the centre of mass of the surface may be shewn to lie in the line joining the vertex with the centre of mass of the perimeter of the base. Hence the point where this line meets the plane mentioned above, is the centre of mass required. The centre of mass then is in the line joining the vertex of the pyramid with the centre of mass of the perimeter of the base, and its distance from the latter point is half its distance from the former. Cor. The centre of mass of the surface of a cone may be found by the same rule, as a cone is the limiting case of a pyramid, when the number of sides of the base is indefinitely increased. 97*. To find the centre of mass of an arc of a circle. Let ABC be the arc, subtending an angle 2a at the centre 0. Draw OB bisecting the angle AOG\ it is clear from the principle of symmetry of Art. 91 that the centre o<[ Fig 72 of mass is in OB. Construct a regular polygon circum- CENTRES OF MASS. 141 scribing the arc ; let PQ be one side of it, touching the circle at R. Draw Aa, Pp, Qq, Cc perpendicular to the tangent aBc at B, RM perpendicular to OB, and QS to Pp. The right-angled triangles PSQ, ORM are similar, since QS, PQ are respectively perpendicular to OM, OR. .-. PQ : QS = OR : OM; .'. PQ.OM=OR.QS = OB.pq; .-. X(PQ. OM) =^{0B. pq) = OB.t (pq) ; .'. OG . perimeter of polygon = OB . ac = OB . chord AC (Art. 85) ; where G is the centre of mass of the polygon. Also, when the sides of the polygon are taken in- definitely small, the limit of the perimeter of the polygon is that of the arc, and their centres of mass are also coincident. Hence the distance of the centre of mass of the arc ABC from is radius OB . chord AC _r sin a arc ABC ~" ~~a 98*. To find the centre of mass of the sector of a circle. Let ABC be the sector : from 0, the centre of the circle, draw OB bisecting the angle AOC. Then (Art. 91) the centre of mass of the sector is clearly in OB. In the arc ABC inscribe a regular polygon ; let PQ be one of its sides, R the middle point of PQ : join OR, and take r in OR such that Or is equal to | OR. Then the centre of mass of the triangle OPQ is r, and the centres of mass of all such triangles are arranged at equal angular intervals along the arc of a circle of radius 0?% and whose anoxic is AOC. But when the number of the sides of the pol3'gon is increased indefinitely, Or becomes equal to 1 05 ultimately, the sum of the triangles of which OPQ is a type becomes the sector AOC, and the centre of 142 STATICS. mass of the latter is that of an infinite number of equal masses arranged at equal angular intervals along an arc Fig 73 \r of a circle of radius \0B, and whose angle is equal to AOC. But the centre of mass of the masses arranged along this arc is that of the arc itself. Therefore the distance of the centre of mass of the sector from _2 0^ X chord ^0 ~3' arc ^(7 Cor. As the segment of a circle is the difference be- tween a sector and a triangle, its centre of mass can be found by the method of Art. 88. The centre of mass of the portion of a circle cut off by two parallel lines can also be obtained, since the figure consists of the difference of two segments. D9. To find the centre of mass of the belt of a sphere cut off by two parallel planes. Let ABh^ the arc of a circle, which by revolving about OE generates the belt ABGD of a sphere in question. Then (Art. 91) the centre of mass of the belt lies in OE. CENTRES OF MASS. 143 Let PQ be the side of a regular polygon, circum- scribing the arc AB) let R be the middle point of PQ, Fig 74 where it touches the circle. Produce PQ to meet OE in T, and draw PM, RK, QN perpendicular to OE, and QL perpendicular to PM. The area of the frustum of the conO; generated by the revolution of PQ about OE = PT.7rP3I^QT.7rQF = ir.{PR + RT)PM-7r{RT-RQ).QN' = 7r.RT.PL + 2irPR . RK = 2'jr.PQ. RK, by similar triangles PQL, RKT = 217 OR. MN by similar triangles PQL, ORK = the area of the belt cut off by the planes PM, QN from the cylinder circumscribing the sphere and having its axis along OE. 144 STATICS. Hence the sum of the areas of any number of frusta of cones, of which the one considered is a type, is equal to the sum of the areas of the corresponding belts of the circumscribing cylinder. But ultimately, when the number of the sides of the circumscribing polygon is taken in- definitely large, the sum of the areas of the frusta of the cones becomes the area of the belt of the sphere. Hence the area of the belt of a sphere, cut off by parallel planes, is equal to that of the coaxial circumscribing cylinder cut off by the same planes. Let G be the centre of mass of the belt of A BCD of the sphere, G' of the corresponding belt of the cylinder. Then OG . area of A BCD = moment of ABCD about the plane through perpendicular to OE = X (27r . 0^ . MN . OK) (Art. 85) = moment about the same plane of the belt cor- responding to ABCD of the cylinder = OG' . area of the belt of the cylinder ; .-. 0G= OG'. Therefore G, G' are coincident, i.e. G is in OE, half- way between the planes which cut off the belt. (Art. 90.) 100. We can easily deduce from the above the position of the centre of mass of the volume of a sector of a sphere, the figure generated by the revolution of a circular sector about one of the bounding radii. Let OAC be the spherical sector generated by the revolution of the circular sector AOB about OB. The centre of mass is in OB (Art. 91). Imagine the sector to be divided into an infinite number of indefinitely small pyramids having the common vertex 0. The centre of mass of each of these pyramids will lie on a spherical cap ahc, generated by the revolution of ah, the arc of a circle, of radius three-quarters that of CENTRES OF MASS. 145 ABC, and the same vertical angle AOB. Supposing the Fig 75 mass of each pyramid to be collected at its centre of mass, the centre of mass of the sector A 00 is clearly the same as that of the spherical cap ahc : its distance from there- fore is equal to \{0m + Oh) or f (Oi/+ OB). If the sector be a hemisphere, OM vanishes, and the distance from the centre of the centre of mass of the volume of the hemi- sphere is f of the radius. Cor. As a spherical segment, the solid figure cut off a sphere by a plane, is the difference between a spherical sector and a right circular cone, its centre of mass can be found by the method of Art. 88. The centre of mass of the solid cut off a sphere by two parallel planes can also be obtained, since the figure is the difference of two spherical segments. Ex. 1. With the same assumption as in Ex. 11, p. 135, find where the resultant pressure acts on a triangle whose vertex is in the surface of a liquid and whose base is parallel to the surface, but below it. Ans. At a point whose depth below the surface is three-quarters that of the base. Ex. 2. Find the centre of mass of a segment of a circle. Ans. It is in the diameter bisecting the segment, at a distance from A y Sill a the centre, -.^ — -. — r— , where r is the radius, and 2a the angle the 3 2a-sin2a segment subtends at the centre. G. 10 1 46 STATICS. Ex. 3. If a figure consist of a cone and a hemisphere on the same base, find the height of the cone in order that the centre of mass of the whole may be the centre of the hemisphere. Ans. ^d times the radius of the hemisphere. Ex. 4. Find the position of the centre of mass of a frustum of a cone, when the radii of the faces are 4 inches and 8 inches respectively, and the distance between them 7 inches, Ans. In the line joining the centres of the faces at a distance of 4| inches from that of the smaller face. Ex. 5. Find also the position of the centre of mass of the surface of the above figure. Ans. At a distance of 3f inches from the centre of the smaller face. Ex. 6. From a cube is cut a tetrahedron, three of whose edges are the edges of the cube which meet in one of the corners. Find the centre of mass of the remainder. Arts. The cm. is in the diagonal of the cube through that corner from which the tetrahedron is cut off, and at a distance from that corner equal to ^ of the diagonal. Ex. 7. Find the centre of mass of a segment of a sphere. Ans. It is in the diameter of the sphere at right angles to the base of the segment and at a distance from the centre of the sphere equal to - . — :/- , where r is the radius of the sphere and h is the distance from 4 2r+h the centre of the sphere of the plane cutting off the segment. 101. The centre of mass of a segment of a parabola. Let BAB' be the segment, AC being the diameter conjugate to the base BB\ Fig 76 CENTRES OF MASS. 147 Divide AC into an infinite number n of indefinitely small equal parts of which MN is a typical one, the rth. Draw PMP', QNQ' chords parallel to BB\ Let >Sf be the focus ; then PM' = 4^AS.AM. The centre of mass of the strip PP'Q'Q is in MN. (Art. 91.) The area PP'Q'Q lies between PP'.^IN sin BOA, and QQ'.MN. sinBGA, and the distance of its CM. from A lies between AM and AN\ therefore the sum of the moments of all the strips about a line through A perpen- dicular to AG lies between t(PP'.MN.AMsmBGA), and ^{QQ' .MN.AN.smBGA), i. e. between ^AS^ . sin BGA . t {AM^ . MN), and 4^A8^ . sin BGA . 2 (AN^ . MN). But AM = '' A G, AN = "^ A C, and MN = — ; ??^ n n therefore the moment of the mass of the parabola lies between ■^AS\AG\ sin BGA. 6 and 4^AS-' . AG\ sin BGA . ^ ^ •" ?i^ 4.4>S^^C-.sin5a^ ,, I.e., = r . (Appendix.) 2 Similarly it can be shewn that the area of the segment _4' ASK ACK sin BGA 3 •J 10—2 148 STATICS. Hence the distance of the CM. of the parabola from A 8AS^ . AG^ . sin BCA 8AS^\ AG' . sin BGA „ , .. Also the CM. is in AC. (Art. 91.) 102. The centre of mass of a rod of uniform thickness, and whose density varies as the mth power of the distance from one end. Let AB he the rod, such that the density at any point P varies at {APy\ PQ 8 Fig 77 Divide the rod into an infinite number n of indefinitely small equal parts, of which PQ is a typical one, the rth. The mass of PQ lies between K.PQ.AP'-' and K.PQ.AQ^ where « is a constant ; therefore the moment of the whole rod about A lies between 1.{kPQ.AP'\AP) and 1. [kPQ , AQ\ AQ) , i.e. between k{AB) n'"^'' and kA 5"^^^ . "^^ rn^" n kAB"""-' I.e. = m + 2 Similarly it can be shewn that the mass of the rod -^ — . Hence the distance from A of the CM. m-f 1 kAB'"-"' kAR"'' m + 1 m + 2 ' m + 1 m + 2 .AB, CENTRES OF MASS. 149 The centres of mass of a triangle, pyramid and para- boloid of revolution might have been obtained by methods similar to those employed in the last two articles. Ex. 1. Find the cm. of a tetrahedron ABCD, which is such that the density at all points in a plane parallel to BCD is the same and propor- tional to the distance of the plane from A. Ans. If G is the cm. of BCD, the required point is in AG, at a distance from A = ^AG. Ex. 2. Find the cm. of a triangular lamina ABC, when the density at any point varies as its distance from BC. Ans. The middle point of AD, where D bisects BG. Ex. 3. Find the c m. of a tetrahedron ABCD, when the density at any point is proportional to its distance from the face BCD. Ans. In ^G, at a distance from A equal to § .AG, when G is the cm. of the face ABC. Ex. 4. The density of a conical shell standing on a plane horizontal base varies as the depth below the vertex : find the dejDth of the centre of mass. Ans. f the height of the cone. 103. Prop. When a body or any system of bodies is in equilibrium under the action of gravity, mutual actions, and the action of one external supporting point, the centre of mass of the whole system, and the supporting point lie in a vertical line. For considering the equilibrium of the whole system, the only external forces acting on it are, its weight acting vertically at its centre of mass and the action of the supporting point : but these two forces cannot maintain equilibrium, unless their lines of action are the same, which will not be the case, unless the centre of mass and the supporting point are in a vertical line. 104. Prop. If a rigid body be placed in contact with a smooth horizontal plane, it will be in equilibrium or not, accordino' as the vertical line drawn throuo-h its centre of mass meets the horizontal plane within the base or not. 150 STATICS. (N.B. By the base is meant the polygon, without re-entering angles, formed by joining the extreme points of the body in contact with the plane.) Let ABODE be the base, the point where the vertical through G, the centre of mass, meets the plane. 6 w ^E B Fig 78 (i) When lies within the base. It is obvious that the direction, in which the weight of the body acting along GO tends to turn it about the side AB of the base, is such that the points 0, B, E, &c. would move downwards if the plane were not there to resist such motion. As the plane is there such motion is prevented. The same remark applies to motion about every other side of the base. Hence the weight will not produce any motion : and the resistances of the plane on the base are passive forces which can only resist motion and not produce it. The body is therefore in equilibrium. (ii) When lies outside the base. In this case, the base and the point must lie on opposite sides of one or more sides of the base. CENTRES OF MASS. 151 Let AB be such a side of the base. Now the reaction exerted by the plane on any point of the body touching it can only be vertically upwards, D w cdJ.- Fig 79 and its moment about AB is therefore of the same sign as that of the weight. The als^ebraical sum of the moments of all the forces about AB cannot therefore be zero, and equilibrium is therefore impossible. If a curvilinear base be regarded as the limit of a polygonal one, with an infinite number of sides, the above reasoning applies to it. In a similar way it can be shewn that a body placed on an inclined plane, sufficiently rough to prevent sliding, will be in equilibrium, provided the vertical through the centre of mass passes through the base, and that if it does not, the body will topple over. It will be seen hereafter that these propositions are merely particular cases of more general propositions. (Art. 123.) Ex. 1. A right-angled triangle ABC, whose sides AB, BC are respect- ively 5 and 6 feet long, is hung from the point A. Find the inclination of BC to the horizon. Ans. tan~^ (f). 152 STATICS. Ex. 2. A plane triangle is hung with its plane horizontal by three vertical chains from the middle points of its edges. How heavy must it be that a 12-stone man may walk anywhere over it without tilting it? Ans. 36 st. Ex. 3. A circular table of weight 20 lbs, rests on three legs, which are on the circumference, and at the corners of an equilateral triangle. Eind the greatest weight that can be placed on any part of the table without upsetting it. Ans. 20 lbs. Ex. 4. A metal lamina, composed of a semicircle and an isosceles triangle (vertical angle 2a) on the same base, is placed in a vertical plane with its curved rim resting on a horizontal plane ; prove that the lamina will rest in any position, provided tan a = l/sj2. ILLUSTEATIVE EXAMPLES. Ex. 1. If the three diagonals of an octahedron intersect in a point 0, the centre of inertia of the octahedron coincides with that of seven particles, one at and one at each of the angular points : the mass of the particle at being unity, and of that at each angular point the ratio of its distance from to the diagonal through the point. Let ABCD be the plane containing two diagonals AOC, BOD : let EOF Fig 80 be the other diagonal. Let us find the distance of the centre of inertia of the octahedron from the plane ABCD. Let Ji^ be the height of the CENTRES OF MASS. 153 pyramid, having base ABCD and vertex F, and let h^ be the height of the pyramid having the same base and vertex E. .'. the volume of first pyramid : volume of second = /?i : h^=OF : OE. The distance of the c, i, of the octahedron from the plane ABCD ~ h-^ + ho ~ 4 (Here, distances from the plane ABCD have been estimated positive when towards F, and negative when in the opposite direction.) The distance from ABCD of the c.i. of the seven particles OF OE ^^^' EF~ '' EF' , OF+OE OA + OG OB + OD EF AC BD ) \ . K /ij + 7*2 h. h^ Hence the distances of the centres of inertia of both octahedron and the seven particles from the plane ABCD are the same: and it could be shewn in a similar manner that their distances from the planes BEDF, ECFA are also the same. The two points are therefore coin- cident. Ex. 2. Find the centre of mass of a segment cut off an ellipse by a straight line. Let DPE be the segment cutting the ellipse, whose semiaxes are CA, b d Fis El CB. Let hpA be the auxiliary circle. Draw the ordinates DD', EE', and let them be produced to meet the circle in d, e, respectively. Join de, 154 STATICS. and produce it to meet DE in T. T is in CA produced. Draw two ordinates PP', QQ' of the ellipse indefinitely near to one another, and let them meet DE in 31, N respectively. Produce them to meet de in m, n, and the circle in p, q respectively. « PP' : pP'=CB : CA^MP' : mP'; .-. MP : mp = CB : CA; .-. the area PQN3I : a,ie& pqnm=CB : CA. Both elliptic and circular segments may be divided up into the same infinite number of strips, of which PQN3I, pqnm are types. Let G be the CM. of DPE, and g that of dpe. The distance of G from CB _ moment of DPE about CB ~ mass of DPE CB ' _ ^{PQmi.cp') -cA^iv or < AN, AM. .^ Oif I.e. „ — IS > or < A0 + -^, AM. ,^ AM -AG I.e. „ -^ IS > or < A0-] ^ , i.e. ,, .41/ is > or < SAO. EXAMPLES. 1. ABG is a triangle, D, E, F are the middle points of its sides, shew that the centre of gravity of the perimeter of ABG coincides with the centre of the circle inscribed in DEF. 2. ABGD is any plane quadrilateral figure, and a, h, c, d are respec- tively the centres of gravity of the triangles BGD, GDA, DAB, ABG; shew that the quadrilateral abed is similar to ABGD. 3. Prove that the centre of gravity of a wedge, bounded by two simi- lar, equal, and parallel triangular faces and three rectangular faces, coin- cides with that of six equal particles placed at its angular points. 4. A thin uniform wire is bent into the form of a triangle ABG, and heavy particles of weight P, Q, R are placed at the angular points: prove that if the centre of mass of the particles coincides with that of the wire P : Q : R = h + c : c + a : a + h. CENTRES OF MASS. 159 5. The perpendiculars from the angles A, B, C meet the sides of a triangle in P, Q, R : prove that the centre of gravity of six particles pro- portional respectively to sin^^, sin-i?, sin'-'C, cos^A, cos-B, cos-C, placed at A, B, G, P, Q, R, coincides with that of the triangle PQR. 6. A plane quadrilateral ABCD is bisected by the diagonal AC, and the other diagonal divides AC into two parts in the ratio p '. q; shew that the centre of gravity of the quadrilateral lies in A C and divides it into two parts in the ratio 2p + q : }) + 2q. 7. A heavy elliptical ring, whose eccentricity is f, is suspended with its plane horizontal by three vertical strings, one of which is attached to the end of the minor axis, one to the end of the major axis, and one to the end of a latus rectum. Prove that the tensions respectively are :J, ^, and T% of the weight of the ring. 8. A triangular table is supported by three legs at the middle jDoints of its sides. A given weight is placed upon it in any position. If weights P, Q, R placed in succession at its angular points will just upset it, prove that P+Q + R is constant. 9. A uniform wire is bent into the form of a circular arc and its two bounding radii, the arc being greater than a semicircle. Shew that if the acute angle between these bounding radii be tan~^ ^, the centre of gravity of the whole wire is at the centre. 10. A triangular lamina is supported at its three angular points and a weight equal to that of the triangle is placed upon it ; find the position of the weight if the pressures on the points of support are proportional to 4a + b + c, a + 4:b + c, a + h + 4:C, where a, b, c are the lengths of the sides of the triangle, 11. Particles are placed at the corners of a tetrahedron respectively proportional to the opposite faces : prove that their centre of gravity is at the centre of the sphere inscribed in the tetrahedi-on. 12. ABCD is a quadrilateral whose diagonals intersect in 0. Parallel forces act at the middle points of AB, BC, CD, DA respectively propor- tional to the areas AOB, BOC, COD, DOA. Prove that the centre of parallel forces is at the fourth angular point of the parallelogram de- scribed on OE, OF as adjacent sides, where E, F are the middle points of the diagonals of the quadrilateral. 160 STATICS. 13. A solid, consisting of a hemisphere and a right circular cone on opposite sides of the same circular base, is in equilibrium, when placed with any point of the hemisphere on a horizontal plane. If the whole solid can just be included in the sphere of which the hemisphere in question is half, prove that the density of the cone is three times that of the hemisphere. 14. If three uniform rods of the same material but of different thick- nesses be formed into a triangle ABC, and if their centre of gravity be at the orthocentre of this triangle, prove that their thicknesses must be proportional to cos{B-0)-Sco&A, cos{C-A)-3cosB, cos{A-B) -ScosC. 15. The corners of a pyramid are cut off by planes parallel to the opposite sides: if the pieces cut off be of equal weight, prove that the centre of gravity of the remainder will coincide with that of the pyramid. 16. Two uniform heavy rods, AB, BC are rigidly united at B, the rods are then hung up by the end A ; shew that BC will be horizontal if sin C=\j2 .sin|2>. 17. A uniform triangular lamina of weight W is suspended from a fixed point by means of strings attached to its angular points : shew that, unless its plane be vertical, the tensions of the strings are W.l^ and similar expressions ; Z^, U, Zg being the lengths of the strings, and a, b, c the sides of the triangle. 18. Find the centre of gravity of a solid sector of a sphere, in which the density at any point varies as the cube of its distance from the centre. 19. A horizontal rod, the ends of which are on two inclined planes, is in equilibrium : if a, j8 be the inclinations of the planes, prove that the centre of gravity of the rod divides it into two parts in the ratio of tan a to tan /3. 20. Find the centre of mass of the segment of a spheroid cut off by a plane perj^endicular to the axis. 21. Shew how to determine the position of the centre of gravity of the area contained between two concentric, similar, and similarly situated ellipses and two straight lines drawn from the common centre. CENTRES OF MASS. IGl 22. If from a triangle ABC three equal triangles ARQ, BPR, CQP be cut off, the centres of inertia of the triangles ABC, PQR will be coincident. 23. From a uniform circular disc, radius a, are cut two circular holes, radii b and c, and centres at distances /3, y from that of the disc, and distance 8 from one another. Find where to cut the hole of radius s/bc, so that the centre of mass of the remainder may be the centre of the disc : if the distance of the centre of this hole from that of the disc be r, shew that r^ + S^^ = {b^ + c^)(^^^+'fj. 24. A rectangular sheet of stiff pajDer, whose length is to its breadth as \/2 is to 1, lies on a horizontal table with its longer sides perpendicular to the edge and projecting over it. The corners on the table are then doubled over symmetrically so that the creases pass through the middle point of the side joining the corners and make angles of 45*^ with it. The paper is then on the point of falling over ; shew that it had originally If of its length on the table. 25. ABC is a triangle; APD, BPE, CPF the perpendiculars from it on opposite sides. Prove that the resultant of six equal parallel forces, acting at the middle points of the sides of the triangle and of lines PA, PB, PC, passes through the centre of the circle which goes through all of these middle points. 26. The inscribed circle of a triangle ABC touches the sides in D, E, F. Prove that the centre of gravity of weights proportional to BC, CA, AB, placed at A, B, C respectively, coincides with the centre of gravity of the same weights placed at D, E, F respectively. 27. Find the centre of gravity of that part of the circumscribing cu'cle of a triangle which lies outside the nine-points circle ; and shew that its distance from the centre of the circumscribing circle is nine times that of the centre of gravity of the triangle. 28. A, B, C, D, E, F are six equal particles at the angles of any plane hexagon, and a, b, c, d, e, f are the centres of gravity respectively of ABC, BCD, CDE, DEF, EFA, and FAB. Shew that the opposite sides and angles of the hexagon abcdef are equal, and that the lines joining opposite angles pass through one point which is the centre of gravity of the particles A, B, C, D, E, F. G. 11 162 STATICS. 29. Find the centre of mass of a solid hemisphere whose density varies inversely as the distance from the centre. 30. A circle whose diameter is equal to the latus rectum of a parabola has double contact with it. Find the position of the centre of mass of the area bounded by the two curves. 31. A triangular lamina ABC hangs at rest from the point A : if AB = c, AG — h, and S represent the area of the lamina, prove that the tangent of the inclination of BG to the vertical is equal to 4S/(&2^c^). 32. A smooth solid hemisphere rests with its flat base against a vertical wall and is supported by a string, one end of which is fastened to the vertex of the hemisphere and the other to a point in the wall. Prove that the inclination of the string to the vertical lies between tan~i (ff) and cos~i(f). 33. Find the centre of gravity of the surface of the octant of a sphere. 34. From considerations of symmetry and from the fact that the centre of gravity of the whole lies in the line joining those of any two parts, deduce the position of the centre of gravity of a circular arc. 35. If the opposite edges of a tetrahedron are equal, prove that the centre of gravity of its six edges and the centre of gravity of its four faces both coincide with the centre of gravity of its volume. 36. If A, B, C be three fixed points, and P any point on a circle whose centre is 0, shew that AP^ . ABOG + BP^ . AGO A + GP^ . A^ OB = constant. 37. From an external point an enveloping cone is drawn to a sphere ; prove that the centre of gravity of a uniform solid bounded by the sphere and cone is at a distance AN^I4:GN from the centre of the sphere, where GA is the radius of the sphere from the centre G drawn towards the outer point and cutting the plane of contact in N. 38. The centre of gravity of a solid hexahedron whose faces are triangles is the same as that of five equal weights placed at the corners, and of an equal negative weight placed at the point where the line forming the two trihedral angles cuts the plane of the other three angles. CENTRES OF MASS. 163 39. A pack of cards is laid on a table and each projects in direction of the length of the pack beyond the one below it : if each projects as far as possible, prove that the distance between the extremities of suc- cessive cards will form a harmonical progression. 40. Prove that the sum of the squares of the sides of the triangle, formed by joining the feet of the perpendiculars, let fall from a point inside a given triangle on the sides, has its least possible value, when the point is the centre of mass of three particles, at the angles of the given triangle, whose masses are proportional to the squares of the opposite sides. 41. A uniform circular disc of weight nW has a heavy particle of weight W attached to a point on its rim. If the disc be suspended from a point A on its rim, B is the lowest point : and if suspended from B, A is the lowest point. Shew that the angle subtended by AB at the centre is 2 sec~i 2 (71 + 1). 42. A thin shell is bounded by two similar surfaces; any closed curve being drawn on the surface, prove that the centre of inertia of the included portion of the shell, and the centre of inertia of the solid formed by drawing lines to the boundary from the centre of similitude, are in a line with the centre of similitude and at distances from it which are in the ratio 4:3. 43. A frustum is cut from a right cone by a plane bisecting the axis and parallel to the base. Shew that it will rest with its slant side on a horizontal table if the height of the cone bear to the diameter of the base a greater ratio than ^y7 : \/17. 44. Four weights are placed at four fixed points in space, the sum of two of the weights being given and also the sum of the other two ; prove that their centre of mass lies on a fixed plane, and within a certain parallelogram in that plane. 45. A sphere, radius (?'), rests on three points at equal distances (a) on a horizontal plane. If one of those points be depressed so that the plane containing the three points is inclined at an angle {6) to the horizon, the sphere will roll off if d exceed sin~i(a/r\/3), but if the point be raised the sphere will roll off if d exceed sin"^ {a/\/3 (4?-- - a-)}. 46. A hemispherical bowl of radius r rests on a smooth horizontal table, and partly inside it rests a rod of length 21, of weight equal that of the bowl. Shew that the position of equihbrium is given by ^sin (a + /S)=r sina= - 2?* cos (a +2/3), 11—2 164 ' . STATICS. where a is the inclination of the base of the hemisphere to the horizon, and 2]3 is the angle subtended at the centre by the part of the rod within the bowl. 47. Two equal segments are cut from a hollow sphere, and are hung up from a point by two equal strings attached to their rims, so that their convexities are outwards. Prove that, if the lengths of the strings be equal to the diameter of either rim, they are inclined to each other at an angle —2 tan"^ (^ tan |a), where 2a is the angle subtended by either segment at the centre of the sphere. 48. A cone of vertical angle 2a is supported by a string passing over two smooth puUies in the same horizontal line, the string being attached to the vertex and to a point in the circumference of the base. Prove that in the position of equilibrium sin (a + ^ + 0)=| cos a sin ^ cos 0, where 6 is the inclination of either portion of the string to the horizon, and is the angle the base of the cone makes with the vertical. 49. The top of a right cone, semi-vertical angle a, cut oS by a plane making an angle /3 with the axis, is placed on a perfectly rough inclined plane with the major axis of the base along a line of greatest slope of the plane; in this position the cone is on the point of toppling over: prove that the tangent of the inclination of the plane to the horizon has one of the values 4 sin 2a ± sin 2/3 cos2a-cos2j8 50. A ring is made up of three arcs, BC, CA, AB, of uniform section, but of different metals: uniform rods OA, OB, OC, made of the same metals as BC, CA, AB respectively, but with sectional area double that of the arcs, connect the points A, B, C with the centre 0. Find the angles a, jS, 7 which BC, CA, AB subtend at 0, in order that the centre of gravity of the whole may be at 0, and shew that, if w^ , Wo, Wg be the weights per unit length of BC, CA, and AB respectively, ( Wj - Wo) tan - tan ^ + (wq - ^3) tan ^ tan ^ + (W3 - Wg) tan ^ tan „ = 0. CHAPTER V. Fkiction. 105. We have hitherto supposed, that the action exerted by one surface in contact with another is neces- sarily along the common normal at the point of contact, in other words that the surfaces are perfectly smootJi. We have however no experience of bodies except such as do, in certain cases, exert on other bodies forces inclined to the common normal at the point of contact, in other words all bodies we are acquainted with are more or less rough. Suppose the following experiment to be made. Take a mass of some material having a plane surface, and fix it so that this surface is horizontal : on it place a portion of some solid material. Now it will be found that, whatever be the materials used, and however highly their surfaces in contact may be polished and lubricated, it is always possible to turn the horizontal surface through a finite angle without the upper bod}^ slippi^ig, though it may topple over. Let W be the weight of the upper body, a the inclina- tion to the horizon of the plane on which it rests : then Fi^.SS 166 STATICS. resolving W into W cos a perpendicular to the plane and TTsin a along the plane, we infer that as W is counteracted by the action of the plane on the body, this action must consist of two components Il{= W cos a) along the common normal, and F (= TTsina) along the plane. The latter force is called the friction. We see also that F/R = tan a. When the body is just about to slide the friction exerted is said to be the limiting friction. 106. The laws relating to statical friction are : (i) Friction always acts in the direction opposite to that, in which the point of the surface acted upon, would move, relatively to the other surface, if there were no friction. (ii) The magnitude is always the least possible required for preserving equilibrium, provided this amount does not exceed the limiting friction. These laws are axiomatic and are particular cases of the general axiom that a Passive force, being entirely due to the tendency to inotion caused by Active forces, only resists such tendency : its direction therefore is always directly opposite to the motion resisted and its magnitude never exceeds the minimum required for preserving equi- librium, and is if possible equal to this minimum. 107. Let us now make another experiment. As before take a plane surface of some material or other, and on it place blocks of different weights, shapes and sizes, but all made of the same material. If now the plane be gradually inclined in any direction more and more to the horizon, it will be found that each and every block, no matter what face it has in contact with the plane, begins to slide as soon as a certain inclination of the plane to the horizon is exceeded, but not before ; also, that when it does slide, the increase per second in its velocity is constant. This angle though constant for the same pair of materials varies con- siderably for different pairs. Any block may topple over before the others slide. FRICTION. 167 Let us see what inferences can be drawn from this experiment. Let a be the inclination of the plane to the horizon, when all the blocks are just about to slide: the friction exerted is in each case limiting, and since FjR = tan a , (Art. 105), the ratio of the limiting friction to the normal pressure is the same for all the blocks. Also since the weights and therefore the normal pressures differ, this ratio is independent of the normal pressure. Since a is the same whatever face of a block rests on the plane, the ratio F:R is independent of the area of the surfaces in contact. Since the increase per second in the velocity of a block is constant, the force on it is constant, i.e. the friction is independent of the velocity. The above experiment confirms the so-called Laws of limiting and dynamical friction. These laws are (i) So long as the substances in contact are unaltered, the ratio of the limiting or dynamical friction to the normal pressure is independent of the magnitude of the latter. (ii) So long as the substances in contact are unaltered, the friction is independent of the area of the surfaces in contact. (iii) When motion takes place, the dynamical friction is independent of the relative velocity of the points in contact. 108. These laws must not be regarded as rigorously true in all cir- cumstances, but only as more or less approximate expressions of the results obtained from the experiments of Coulomb and Morin, who enun- ciated them. More recent investigations would seem to shew, that in certain circumstances they are very far indeed from expressing the amount of friction exerted. According to a report, read before the Institution of Mechanical En- gineers by Captain Douglas Galton, on experiments made by him on the application of brakes to locomotive-wheels, the friction diminishes as the velocity increases beyond a certain limit, and is also less after it has been exerted for some time than when first applied. In the experiments of 168 STATICS. Morin and Galton the surfaces in contact were not lubricated in any- way. Before the same Institution in 1883, Mr Beauchamp Tower read a report on some experiments made by himself on a thoroughly lubricated journal revolving in bearings. These experiments shewed that in certain circumstances the friction per square inch was nearly independent of the normal pressure and that it increased with the velocity of revolution. A rise in temperature was accompanied by a reduction in the friction, though this might be caused by the lubricant becoming more efficient. Professor Thurston states that from his own experiments, he inferred that the friction at first diminished as the velocity increased and then increased again. As however we are only concerned with statical friction we may take laws (i) and (ii) as giving fairly accurately the friction in the cases which we shall have to consider. 109. Def. The ratio of the limitiDg friction to the normal pressure, which ratio we see by laws (i) and (ii) is constant, is called the coefficient of friction for the pair of materials in contact. The angle the total action makes with the common normal at the point of contact is termed the angle of friction, provided the limiting friction is exerted. Hence (Art. 105) the coefficient of friction is equal to the tangent of the angle of friction. The coefficient of dynamical frictio7i is the ratio of the friction to the normal pressure when motion is actually taking place. It is found by experiment to be less than the corresponding coefficient of statical friction, in other words, there is more resistance when motion is just about to take place than when it is actually taking place. Ex. 1. If the smallest force which will move a given block weighing 3 lbs. along a given horizontal plane be \/31bs. ; find the greatest angle at which the plane may be inclined to the horizon without the block sliding. A71S. 30^. Ex. 2. If a weight of 14 lbs., when placed on a rough plane inclined at an angle of 60^ to the horizon, slides down, unless a force of at least 7 lbs. acts on it up the plane, what is the coefficient of friction? Ans. -73. FRICTION. 169 Ex. 3. If a weight of 4 lbs. is just on the point of slipping down a rough plane, inclined at an angle of 45** to the horizon, when a force of 2 lbs. acts up the plane, find the least force which will move the weight up the plane, when the inclination is 30" to the horizon. Ans. 3*01 lbs. Ex. 4. Weights of 4 and 5 lbs. respectively, connected by a hght rigid rod, are placed on a rough inclined plane, with the rod parallel to a line of greatest slope. If the coefficient of friction between the 4 lb. weight and the plane be -6 and that between the other weight and the plane '42, find the greatest inclination of the plane to the horizon, consistent with equilibrium. Ans. tan~^ 'o. Ex. 5. Find the greatest angle at which a plane may be inclined to the horizon so that three equal weights whose coefficients of friction are •5, "6, '7, respectively, may when connected by strings rest on it without sliding. The weights are supposed placed along a line of greatest slope so that each is rougher than the one next below it. Ans. tan"^ {'&). Ex. 6. A uniform ladder rests in limiting equilibrium, with its lower end in contact with a rough horizontal plane and its upper end with a smooth vertical wall. If X be the angle of friction and a the angle the ladder makes with the vertical, prove that tan a = 2 tan X. Ex. 7. If everything is as in Ex. 6, except that the wall is as rough as the ground, prove that a = 2X. Ex. 8. Two particles of equal weight and connected by a light string rest in limiting equilibrum on the arc of a rough vertical circle; prove that the angle the line joining them makes with the horizontal is equal to the angle of friction. Ex. 9. A body is resting on a rough inclined plane of inclination a, the angle of friction being which is greater than a. Shew that the ratio of the least force which will drag the body up the plane to the least force which will drag it down is sin (0 + a) : sin (0 - a). Ex. 10. One end of a uniform rod is on a rough inclined plane to which the rod is perpendicular : at the other end is aj)pHed a force parallel to the plane : if the rod be in equilibrium, prove that the coefficient of friction cannot be less than half the tangent of the plane's inclination. 110. Def. Let a cone be described, having its vertex at the point of contact of the two surfaces, the common 170 STATICS. normal for axis, and the angle of friction as semi-vertical angle. This cone is called the Cone of Friction. The Laws of statical friction, given in Arts. 106, 7, are all included in the following statement. If all the other forces, external and internal, acting on the point of contact he compounded into a single resultant R, the action of the surface in contact will he equal and opposite to R, whatever he the latter's magnitude or direction, provided its line of action does not lie without the cone of friction. Hence a body in contact with rough surfaces will be in equilibrium, provided that to insure its being so, it is not necessary to assume that the total action at any point of contact lies outside the corresponding cone of friction. It should be noticed that the cone of friction is always drawn so that its concavity is towards the body, the action on which we are considering. 111.* To find the relation between the tensions at the ends of a light string stretched over a rough surface, and on the point of slipping. Let APQRZ be the string, which is on the point of slipping from Z to A. FRICTION. 171 Let \ be the angle of friction, fju the coefficient of friction. Let the points A,B,...P, Q,R,...Y, Z he taken on the string so that the ultimately indefinitely small angles be- tween the tangents at consecutive points are each equal to 0. Let us consider a small portion, PQ, of the string. It is kept in equilibrium by the tensions at P and Q and the resultant action of the surface. As in Art. 81, construct a force-diagram Oab...pqr...yz, such that Oa, Oh, ...Op, Oq, Or,...Oij, Oz represent the Fig.87 tensions at A, B,...P, Q, R,...Y, Z respectively. Join ah, bc,...pq, qr,...yz. These last will represent the re- sultant actions of the surface on AB, BC,...PQ, QR,...ZY respectively. As the portion PQ is on the point of slipping from Q to P, the resultant action on it makes with the normal at either P or Q an angle differing from X by an in- definitely small quantity, and on that side of the normal by which it will most assist the tension at Q. Hence in each of the triangles Oah, Opq, &c. the angles at are equal, and the angles Oah, Opq, &c. are each equal to 7r/2 — X ultimately ; the triangles therefore are all similar to one another. Let n be the infinitely large number of portions AB, &c. of the string. Let n6 = a, a finite angle. 172 STATICS. Then Oa = Ob.'''^^-^\ cosX cos (X — 6) op = oq ^— - — ' , ■'■ cos X, &c. &c. .-. Oa = Oz / ^Qs(^-^) ^^^ 0^ ^cos ^ + sin ^ tan XY ; V COS X / .'. log Oa = log 0^ + ^ log (1 — sin'' 6} -}- n log (1+ fJb tan ^) = log 0<2^ — I ( ?^ sin^ ^ + « sin^ ^ + &c. J ^ yLt^?^ tan^ 6 „ + /Lt?i tan 6 — H &c. = log Oz + ycta ultimately ; .-. Oa=Oz . €«'" ; .'. tension at ^4. = e'"'^ . tension at Z. If the string be in one plane, the curve ahcd...z will be a plane one, and as the tangent at every point makes a constant angle with the line joining the point with 0, the curve is an equiangular spiral. The ratio of Oa to Oz might therefore be obtained from the known properties of that curve. If the string be not in one plane, the curve ahc... will not be a plane one ; it can however be made so, without altering the distance of any point from 0, by turning each of the triangles Oah, Ohc, &c. about a side terminating in 0, until they are all in one plane, ^^'hen the curve becomes an equiangular spiral, and the ratio of Oa to Oz can be obtained as suggested above. 112.* We shall sometimes be required to solve problems of the following kind. A system of bodies is in equilibrium FRICTION. 17:- under certain conditions : a gradual change occurs in one or more of these conditions — e.g. the coefficient of friction at one or more points of contact of the bodies is gradually diminished, some external force is gradually altered in magnitude or direction, or the position of one of the bodies is gradually altered. When this gradual change reaches a certain stage equilibrium is no longer possible, and it is required to ascertain the way in which equilibrium is broken, in other words, the nature of the initial motion of the different bodies. The actual way in which equilibrium is broken must satisfy the following conditions. The various forces acting on the different bodies, when such a motion is about to take place, must be able to adapt themselves so as to satisfy the necessary conditions of equilibrium, without in any way violating the laws re- lating to passive forces : they must also be incapable of satisfying the necessary conditions of equilibrium, if the change in the initial conditions increase still further. We shall generally proceed by considering the different ways in which it is conceivable equilibrium might be broken, without violating the geometrical conditions. If only one of these satisfies the above conditions, it is the way required ; if more than one satisfy it, it is beyond the limits of this treatise to obtain a solution of the problem. The following rule will often enable us to solve such problems. If it is inconceivable that equilibrium can be broken, except hy one of the bodies either turning about or sliding past a point of contact luith another body, the former motion luill actually take place, provided it does not involve the assumption that the total action at the point in question lies outside the cone of friction. This rule is a deduction from the axiomatic law re- lating to passive forces (Art. lOG). For if we suppose the body connected with the other body at the point of con- tact by a smooth joint, it can only turn about that point. If now motion be on the point of taking place, the first body will be about to turn about the joint, which will 174 STATICS. exert some action on it. If this action does not necessarily lie outside the cone of friction, it could be exerted at the point of contact if no joint existed, i.e. the motion is the same without the joint as with. On the other hand, if the action at the joint be outside the cone of friction there, it could not be exerted without the joint, i.e. equilibrium is about to be broken by the body sliding past the point of contact in question. When it is necessary to assume that the action at the joint actually lies along a generator of the friction-cone, the question cannot be solved by the above rule, as it shews that slipping is about to occur at the point at the same instant as rolling. Exs. 5 — 8, 10 — 12 are illustrative of this principle and should be studied attentively, ILLUSTEATIVE EXAMPLES. Ex. 1. A uniform rod MN rests with its ends in two fixed straight grooves OA, OB, in the same vertical plane, and making angles a, /3 with the horizon: prove that, when the end M is on the point of slipping down AO, the tangent of the inclination of MN to the horizon is sin (a - j3 - 2e) 2 sin (jS + e) sin (a - e) * Let 6 be the inclination of MN to the horizon, when 31 is on the point of slipping down AO. Draw Mm, Nn, normals to OA, OB, respectively. Since the point 31 is on the point of moving down A 0, the limiting friction is exerted at 31 in the direction 31 A, and the direction of the total action of OA on the rod makes the angle e with Ifm, on the side towards A. FRICTION. 175 Similarly, because N is on the point of slipping np OB, the total action of OB on the rod makes the angle e with Nn on the side to- wards 0. Let the lines of action of the forces on MN at M and N meet in H then, (Art. 61), H is vertically above G, the middle point of the rod. Join HG. BIG : G^= sin 3IHG : sin HMG = sin (a - e) : sin f'^ + d-a + eY, also NG : GH= sin NHG : sin GNH =sm (^ + e) : sin ^ - - /3 + eV, .-. sin(a-e) : cos (a- e-^) = sin (/3 + e) : cos(/3 + e + ^). Hence we obtain tan 6 —- z--. — y—^ — —. — — r . 2 sm [a — e). sm (j3 + e) Ex. 2. A glass rod is balanced partly in and partly out of a cylindrical tumbler with the lower end resting against the vertical side of the tumbler. If a and j8 are the greatest and least angles which the rod can make with the vertical, prove that the angle of friction, X, is sin^ a - sin^ /3 Atan-i sin2 a cos a + sin^ j3 cos j3 * ^ Fig.89 Let AB be the rod, G its centre of mass. Let G be the point of the edge of the tumbler on which AB rests. Draw AD normal to the tumbler at A and CE perpendicular to the rod at C. 176 STATICS. (i) When AB makes the smallest possible angle with the vertical, and is therefore on the point of slipping into the tumbler. Since A is on the point of slipping down, the action there on AB is in the direction AH, which makes the angle X with AD on the side towards C. Similarly, the action at G on the, rod is in the direction CK, which makes the angle \ with CE, on the side away from A. Let KC and AH meet in H, which must therefore be vertically below G, Join GH. Let a be the diameter of the tumbler, and let AG = c. AG : AH=smAHG : sin^GiJ=cosX : sin^S, and AH :AG = 8inAGH: smAHG = cos\ : sin(2X + j3); .-. AG :AC=cos^\ : sin/3 . sin (,8+2X), . •. c: a cosec j8 = cos^ X : sin j3 . sin (j3 + 2X). (ii) When AB makes the greatest possible angle with the vertical and is therefore on the point of slipping out of the tumbler. By reasoning as before, we should have AH and CK on the sides of AD and GE respectively, opposite to those they were on in the first case, and we should arrive at the result obtained there, except that for /3 we must write a, and for X, - X. The result would therefore be c : a cosec a — cos^ X : sin a . sin (a - 2X) ; .*. eliminating c and a, we have sin2 ^ sin (/3 + 2X) = sin2 a sin (a - 2X) ; sin^ a - sin^ B . • . tan 2X = -r-i, . -s ' ^ . sm-^ a cos a + sm^ ^ cos p )► Ex. 3. A uniform rectangular board ABCD rests with the corner A against a rough vertical wall and its side BG on a, smooth peg, the plane of the board being vertical and perpendicular to that of the wall. Shew that, without disturbing the equilibrium, the peg may be moved through a space fi cos a {a cos a + b sin a) along the side with which it is in contact, provided /x do not exceed a certain value : a being the angle BG makes with the wall, and a, b the lengths of AB, BG respectively. Let G be the intersection of diagonals, i. e. the centre of mass of the board. Let P be a position of the peg when there is equilibrium. FRICTION. 1/ / The forces acting on the board are, its weight vertically downwards Fig.90 through G, the reaction of the peg through P and at right angles to BC, and the reaction of the wall through A . The necessary and sufficient condition of equilibrium is that these three forces should meet in a point, as the magnitudes of the reactions at P and A will adapt themselves to secure equilibrium, if the above con- dition holds. Let the first two forces meet in A'; join AK, which is therefore the direction of the reaction of the wall. But AK is not a possible direction of the reaction at ^, if it makes with the normal to the wall an angle greater than tan"^ /i. Draw AE and AF making with the normal on either side of it the angle tan~i fi, and meeting GK in E and F. Draw EM, FN perpendicular to BC. The condition of equilibrium is then that K should lie between E and F, i. e. that P should lie between 31 and jV. We may therefore, without disturbing the equilibrium of the board, move the peg through the space MN along BC. And 2IN=EF cos a = 2fx cos a x horizontal distance of G from wall — ficosa {a cos a + b sin a) . We have assumed above that M and N are both between B and C: if either lies beyond J5 or C, as the peg cannot be moved oft' the board with- out disturbing the equilibrium, it can only be moved along that part of MN which lies between B and C. It is obvious that if /ul be greater than a certain value, either 31 or N will not lie between B and C. G. 12 178 STATICS. ^ Ex. 4. If one cord of a sash window breaks, find the coefficient of friction of the sash in order that the other weight may still support the window. Let ABCD be the window, TFits weight acting at G its centre of mass. Fig.9l We assume that the window fits loosely in the sash, so that there will be contact at only one point on each side; these will be A and G respectively. The unbroken cord at B supplies a force |TF vertically upwards; the resultant of this and W is .|TF vertically downwards at A. Hence in order that equilibrium may be possible, the action at G must be along CA, i. e. the coefficient of friction at G must be not less than tan^CD. Ex.. 5.* A right circular cone, vertical angle 2a, rests with its base on a rough horizontal plane : a string is attached to the vertex and pulled in a horizontal direction with a gradually increasing force : determine how the equilibrium will be broken. Let VAB be a vertical section of the cone containing the direction of FRICTION. 179 the string. Let T be the tension of the string when equilibrium is about to be broken, and IV the weight of the cone. The different ways in which it is conceivable equilibrium may be about to be broken are (1) the cone being lifted bodily from the plane, (2) the cone tilting, with one point of the base resting on the plane, (3) the cone sliding along the plane. (1) is impossible, as in that case the cone would be in equilibrium under the action of W and T. If (2) take place, the cone is in equiUbrium under the action of W, T, and the reaction of the plane at the point of contact, which must there- fore be A ; also the action at A must pass through V, i. e. along A V. This is only possible when the angle ^F makes with the vertical, i.e. a, is less than the angle of friction (X). If, therefore, a be < X, (2) takes place, (Art. 112) : if a be > X, (3) occurs. Ex. 6.* A uniform beam AB lies horizontally upon two others at points A and C ; prove that the least horizontal force applied at J5, in a direction perpendicular to BA which is able to move the beam, is the less of the two forces fxlV and - - , where AB = 2a, AC = h, Tr= weight 2a -b 2 of beam, and ju = coefficient of friction. The vertical pressures at A and C are W . — ^ and W . - respectively. B Fig.93 The maximum frictions that can be exerted at A and C are therefore fiW . and fiW . - respectively. The frictions act horizontally and the maximum friction at either point is exerted, only when motion is about to occur at the corresponding point. The horizontal force ai)j)lied at B is gradually increased until equilibrium is just about to be broken : it is required to find its maximum value, P. 12 2 180 STATICS. Let us suppose that this maximum force P is exerted, and let us first see whether we obtain consistent results by supposing the rod just about to turn round A. In this case as C is about to move at right angles to AB in P's direction, the friction there is the maximum and acts in the opposite direction to P, so that the friction at A must be in the same direction as P. Let F be the friction exerted at A. Taking moments about A and C for the equilibrium of the rod, we have the equations P.2a = iut.W.^.b, P.{2a-b) = F.b; {2a - h) .'. F=fjiW 2b We have shewn (Art. 112) that the rod will turn about A, provided it is not necessary to assume the friction there greater than the maximum ; therefore the rod turns about A or not, according as F or fxW — — is b-a . ^^^ b-a .>ixW 2a -b > ^^ • ~r~ ' i-6- ^^ /*^^ o:: — i. ^^ If then aW. „ — % is > ^^ the rod will turn about A as soon as P ex- 2a -b 2 eeeds -— . If iiW . 7 is < '^ , the rod will slip at A instead of turning about it. 2a- b 2 We can shew in a similar way in this case that it will turn about C, when P = iiW . . . 2a- b b-a ixW If uW . ^ r= ^ , the rod is about to slip at both A and B snnul- 2a -b 2 taneously when P — \ixW: the investigation of the point about which the rod will turn in this case is beyond our present scope. It is easily seen that the point lies between A and C. Ex. 7.* A heavy straight rod, whose sectional area varies as the distance from one end, rests on a rough horizontal jDlane. At the other end, perpendicularly to its length and in the horizontal plane, a force is applied of gradually increasing magnitude : prove that the distance of the FRICTION. 181 point about which the rod begins to turn, from the end first mentioned, is given by the equation 4:X^-6lx^' + P = 0, where I is the length of the rod. Let ^B be the rod, B the end at which the force is appHed. Let us investigate whether or no there is a point in AB about which the rod may be on the point of turning. If there is such a point, let C be it, and let x be its distance from the end A. J i ^\ p Q C Fig.94 y Let S be the force at B, when the motion is just about to take place. The friction then on any point of the rod between C and B acts in the opposite direction to .S', and that on any point in CA in the same direc- tion as S. The friction on any small portion PQ is ^u x weight of PQ, and there- fore the total friction on ^iC is /x x weight of AC, and acts at the c. ii. of AG. Also the total friction on CB is /ix weight of CB, and acts at the c. M. of CB. By Art. 102, the weight of AC = kx', and that of AB is kZ^. also the distance from A of the c. ii. of AC is fx, that of the c. m. of AB is |Z. The weight of the rema,inder CB is therefore k[1--x-), and the distance of its c. M. from ^ is 2 (/,3 _ x^)l {3 (Z^ - x-)} . .-. taking moments about .-1 for the equilibrium of the rod, we have (1). 2 iV^ -x^^ fJLK [t - X-) . .3-7-5 37^ - flKX' .fX = Sl 3 (/2 - X-) Also resolving at right angles to the rod, we have fxK {l^ - x~) - fiK . x- — S (2); .-. eliminating S from equations (1) and (2) 2{P-2x^)^Sl{r--2x-); Since the expression on the left hand is positive when .r = 0, and negative when x — 1, the equation has a root between and I, i.e. it 182 STATICS. determines a real point on the rod. This must be the point, round which the rod is about to turn, since the rod will turn about it rather than slide past it (Art. 112). Ex. 8.* A square lamina is supported in a horizontal position by means of four rough pegs on which its angles A, B, C and D rest. A horizontal force is applied at C at right angles to AG and gradually increased until it moves the lamina. Shew that, if the pressures on the pegs be equal, the lamina will begin to turn about the angle A. Let P be the applied force which will cause the lamina to be just on the point of motion. We know (Art. 112) that the square will be on the point of turning about A, provided that all the necessary equations of equilibrium can be satisfied on such an assumption, without requiring the friction exerted at A to be the maximum. Let be the point of intersection of the diagonals of the square. Let Figr.95 Q be the maximum friction that can be exerted at any of the corners of the square — then if rotation isabout to take place about J, the force at D will be Q along DC, that at C, Q opposite to P, and that at B, Q along CB. Taking moments about A, we have P.AC=Q{AD + AG + AB), • '. P^Q{l+j2). The friction at A must be equal to the resultant of the other four forces, its magnitude is therefore ^{{P-Q-QJ2T'+{Q|^y^-Q|^/2y-\, i.e. zero. A is the point therefore about which the square will begin to turn. FRICTION. 183 Ex. 9. A heavy particle is placed on a rough inclined jDlane whose inclination is equal to the angle of friction : a thread is attached to the particle and passed through a hole in the plane which is lower than the particle, but not in the line of greatest slope : shew that if the thread be very slowly drawn through the hole the particle will describe a straight line and a semi-circle in succession. Let be the hole; OA the horizontal line in the inclined plane through O. Let P be the particle^ W the resolved part of its weight in the plane. The maximum friction that can be exerted on it is JV therefore. Let d be the angle the string PO makes with a line of greatest slope. Let be the angle the direction of motion at any instant, and therefore the friction, makes with a Une of greatest slope. Let T be the tension of the string. (1) When P is above OA. The resolved parts of forces down the line of greatest slope ^W+Tcosd-Wcos(f>. Those perpendicular to the same line=: T sin 6 - Wsin . Let be the centre of the sphere, G the middle point of the beam. Fig.99 Considering the equilibrium of the sphere and beam together, since the horizontal forces acting on them are the frictions at A and D respec- tively, they must be equal. Also from the equilibrium of the sj)hei-e, by taking moments about O, we deduce that the friction at C' = that at Z) = tliat at A. Let us suppose that the sphere is slowly moved away from A, until equilibrium is about to -be broken ; what will be the nature of the motion which is about to happen? Of the three forces acting on the sphere, two, the weight and the reaction at I) act through D, therefore the third, the action at C, is along CD, and the action at D is within the angle CDO. Hence slipping cannot be about to occur at D, as then the angle CDO, and therefore the angle OCD, would be greater than the angle of friction, which is impossible, as it is the angle the action at C makes with the normal. Let DC produced meet the vertical through G in If : join AH. AH is the direction of the action on AB at A. Hence either the angle AH makes with the vertical, or OCD must be the angle of friction, as slipping must occur at either A or C. 186 STATICS. The slipping occurs at J or C according as L AHG is > or < z OCD, i.e. z ODC, i.e. z GHC, according as G is nearer A or D, according as A and B are on the same or opposite sides of the vertical through D. Ex. 11.* A heavy cube with its vertical face smooth, is placed on a rough horizontal plane, and a ladder is placed with the upper end leaning against it, the vertical plane containing the ladder also containing the centre of mass of the cube. A man now ascends the ladder and when he reaches a certain height the equilibrium ceases. Examine the character of the ensuing disturbance. Let AB be the ladder, CDEF the section of the cube made by the F L ^ B K £ I ^ \ \ \ 6 \ / r y N J f \ D Fig.lOO vertical plane through the ladder. Let G' be the centre of mass of the ladder and man when motion is about to take place, W their weight. Let W be the w^eight of the cube, G its centre of mass. Let X be the angle of friction at the different points of contact with the ground. The ways in which we can imagine the equilibrium to be broken are (1) the ladder turning about A and slipping at B, and the cube turning about D, (2) the ladder turning about A and slipping at B, the cube sliding, (3) the ladder slipping at A and J5, the cube remaining stationary, (4) the ladder slipping at A and B, and the cube sliding along the plane, (5) the ladder slipping at A and B, and the cube turning about D, (6) the ladder slipping at A but not at U, and the cube sliding, (7) the ladder slipping at A but not at B, and the cube turning about D. FKICTION. 187 Since the vertical sides of the cube are smooth, the action and reac- tion (E) at B are horizontal. Draw LBK horizontally to meet the vertical lines through G' and G in L and irrespectively. Join AL, BK. The action at A must be along AL. By Art. 112, slipping will not occur at A if the angle AL makes with the vertical, i.e. the angle ALG', be less than X, but will otherwise. (a) Let Z ALG' be tan-i(i?/jr), i.e. if TFtanA^Di^ be >Tr'tanX, (3) or (■4) must occur. (3) occurs if tan X be >RIW, i.e. if W be < W, and (4) otherwise. Ex. 12.* A block in the shape of a rectangular parallelopiped of weight W rests with one edge horizontal on a rough inclined plane; against the block rests a rough sphere ( W) whose radius is less than the tliickness of the block. The inclination of the plane is gradually increased 188 STATICS.' until equilibrium is no longer possible: shew that if the block tilt, the sphere will slide or roll along the i^lane according as the limiting inclina- tion (6) of the plane to the horizon is > or < 7r/-i ; and shew also that if the block slide, the si:)here will slide or roll according as X (the angle of friction) is > or < 7r/4 , and that in the last case, 6 is given by the equa- tion TFsin(\- ^) = TF'sin ^ (cos \-sin X), X being supposed the same everywhere. Let be the centre of the sphere, A and B the points where it touches the plane and block respectively : C the point of the block nearest to A. Let the inclination of the jalane {6) be such that equilibrium is just on the point of being broken. There are onl}- two motions of the block conceivable, (1) turning about its lowest edge, (2) sliding down the plane. Whichever of these two ways the block moves, the sphere will either (a) slip at B and roll at A, or (jS) roll at B and slip at A. The actions at A and B on the sphere must meet in the vertical through 0, in the point H say: join AH, BH, these will be the directions of the respective reactions. As we have seen either (a) or (jS) must occur, one of the angles OBH, OAH must equal X, and as the other angle must be less than X, it is the greater angle of the two that is equal to X. We shall prove that / OBH is > or < Z OA H, according as 6 is < or > 7r/4. If 6* is < 7r/4, z AOH is > Z BOH, .-. AH is > BH, and as BO = OA, and OH is common to the two triangles OBH, OHA, L OBH is > z OAH. Similarly it can be shewn that if ^ is >7r/4, z OAH is > z OBH. Hence whether (1) or (2) happen to the block the sjjhere will roll or slide at A, according as ^ is < or > -rrj^. If (2) and (jS) happen, ^ = X, and X is therefore >7r/4. To find 6 when (2) and (a) happen. Let R be the normal reaction at B, then taking moments about A for the equilibrium of the sphere, Wa sme = R{l + tan X) a. FRICTION. . 189 Resolving along the plane for the equilibrium of the block irsin e + R = {W COS 6 + R tan X) tan X. .-. ir(sin ^ - cos ^ tan X) (1 + tan X) = W sin d . (tan^ X - 1), .-. Trsin(X-^) = TF'sin^. (cosX-sinX). Since 6 cannot be gi'eater than X, this equation shews that if X be > 7r/4 , (2) and (a) cannot happen. EXAMPLES. 1. A body is supported on a rough inclined plane by a force acting along it. If the least magnitude of the force, when the plane is inclined at an angle a to the horizon, be equal to the greatest magnitude when the plane is inclined at an angle /3, shew that the angle of friction is ^{a- ^). 2. Two equal particles on two inclined planes are connected by a string which lies wholly in a vertical plane perpendicular to the line of junction of the planes, and passes over a smooth peg vertically above this line of junction. If, when the particles are on the point of motion, the portions of the string make equal angles with the vertical, shew that the difference between the inclinations of the planes must be twice the angle of friction. 3. A uniform rod is resting on a rough inclined plane, and is moveable on the plane about one end which is fixed : shew that when it is about to slip it makes with the line of greatest slope the angle sin~i {/j. cot a). 4. Spheres whose weights are W, W rest on different and differently inclined planes. The highest points of the spheres are connected by a horizontal string perpendicular to the common horizontal edge of the two planes above it. If jx, p! the coefficients of friction are such that each sphere is on the point of slipping down, ixW=ii'W'. 5. Two equal particles rest upon two equally rough inclined planes, being connected by a string passing over a smooth pulley at the common vertex, the vertical plane which contains the string being at right angles to each inclined plane. If the weight of one particle be increased by a certain amount the system is on the point of motion, and if instead the weight of the other particle be decreased by the same amount the system is again on the point of motion in the same direction as before. Prove that the difference of the inclinations of the two planes is double the angle of friction. 190 STATICS. 6. A lamina is suspended by three strings from a point: if the lamina be rough, and the coefficient of friction between it and a particle placed upon it be constant, shew that the boundary of possible positions of equilibrium of the particle on the lamina is a circle. 7. A uniform heavy rod is in equilibrium in a rough spherical cup ; and the length of the rod subtends a right angle at the centre of the sphere; find the greatest angle the rod can make with the horizon in terms of the angle of friction. 8. Two fixed pegs are in a line inclined at a given angle a to the horizon. A rough thin rod rests on the higher and passes under the lower, the higher peg being lower than the centre of gravity of the rod. The distance of that jDoint from the pegs being a and h respectively, shew that when the rod is on the point of motion {h^-a) ij. = {h-a) tan a. 9. Prove that the direction of the least force required to draw a carriage is inclined at an angle 6 to the ground, where a sin ^ = 6 sin ^, a being the radius of the wheels, h of the axles, and tan (f> the coefficient of friction of the axles. 10. A light string is placed over a rough vertical circle, and a uni- form heavy rod, whose length is equal to the diameter of the circle, has one end attached to each end of the string, and rests in a horizontal position. Find within what points on the rod a given mass may be placed, without disturbing the equilibrium of the system : and shew that the given mass may be placed anywhere on the rod, provided the ratio of its weight to that of the rod does not exceed |(e'^'^-l), where /t is the coefficient of friction between the string and the circle. 11. Two particles of unequal mass are tied by fine inextensible strings to a third particle. They lie on a rough horizontal plane with the strings stretched at a given angle to each other. Find the magnitude and direc- tion of the least horizontal force which, applied to the third particle, will move all three. 12. An equilateral triangle, of uniform material, rests with one end of its base on a rough horizontal plane and the other against a smooth vertical wall : shew that the least angle its base can make with the horizontal plane is given by the equation cot d = 2/^ + 1/^3, fx being the coefficient of friction. 13. Two weights P, Q are connected by a string and rest one on each face of a double inclined plane, the string passing over the cominon vertex, which is smooth : at first P is about to slip downwards and when FRICTION. 191 the weights are interchanged, it is found that P is still just about to slip downwards : shew that if X, X' are the angles of friction for the two planes and a, /3 the angles they respectively make with the horizon, then cos a . cos \' = cos /3 . cos X. 14. Two rough spheres, the larger of which is fixed, rest on a rough horizontal plane, and a uniform board rests symmetrically upon the top of them, its centre of gravity being midway between the points of con- tact : shew that, if tan X' and tan X be the coefficients of friction between the board and the larger and smaller spheres respectively, and motion be about to take place at both points of contact, tan {X'-X) = sin-X. 15. Two rings, each of weight w, shde upon a vertical semi-circular wire, diameter horizontal and convexity upwards. They are connected by a light string of length 21 (supposed less than the diameter 2a) on which is slipped a ring of weight W. Shew that when the two rings are as far apart as possible, the angle 2a subtended by them at the centre is given by (TF-f 2j(;)- tan-(a-{-e) (Z^-a^sin-a)^ JF-Vsin^a, e being the angle of friction. 16. An isosceles triangular prism is placed with its edge horizontal and its base on a rough inclined plane, the inclination of which is gradually increased : shew that the prism will tumble or slide according as ^ is > or <3c/2/t. c is the base of a section perpendicular to the edge and h the height. *c>^ 17. Two hemispheres, of radii a and b, have their bases fixed to a horizontal plane, and a plank rests symmetrically upon them. If fx be the coefficient of friction between the plank and either hemisphere, the other being smooth, prove that, when the plank is on the point of slipping, the distance of its centre from its point of contact with the smooth hemisphere is equal to (a ~b) j /x. 18. A disc in the shape of a sector of a circle lies on a rough table {fj) and is fastened at the centre by a peg. Shew that the least force applied along any tangent to the sector necessary to turn it round is to the weight of the disc as 2/x : 3. 19. A rod rests partly within and partly without a box in the shape of a rectangular parallelepiped, and presses with one end against the rough vertical side of the box and rests in contact with the opposite smooth edge. The weight of the box being four times that of the rod, shew that if the rod be about to slip and the box about to tumble at the 192 STATICS. same instant, the angle the rod makes with the vertical is ^X + ^ cos~i(^cosX), where X is the angle of friction. 20. Three equal heavy rough cylinders are placed in contact along generating lines, lying on a horizontal plane : and two other such cylinders are similarly placed upon them : find the frictions and reactions at the instant when the system is bordering on motion. 21. A sjDhere (radius a) whose centre of gravity is distant c from its centre, rests in limiting equilibrium on a rough plane, which is inclined at an angle a to the horizon : shew that the sphere may be turned through the angle 2 cos"^ ( ~ | and still be in limiting equilibrium. 22. Assuming that the limiting friction consists of two parts, one proportional to the pressure, and the other to the surface in contact, shew that if the least force which can support a rectangular parallelepiped, whose edges are a, b, and c on a given inclined plane be P, Q, R, when the faces in contact are he, ca, ah respectively, then {Q-Ii)hc+{R-P)ac + {F-Q)ah^Q. 23. A rough rod rests over a rough sphere, one end of the rod press- ing on a rough horizontal plane, on which the sphere rests. Shew that there will be limiting equilibrium for the whole system when the rod makes an angle 2X2 with the plane, if the weight of the sphere is to the weight of the rod in the ratio sin {\-\-^ : sin(Xo + Xi), where Xj is the angle of friction between the rod or sphere and the plane, and Xo the angle of friction between the rod and sphere. 24. A rectangular lamina rests in a vertical plane with the middle point of one side in contact with a rough peg, the coefficient of friction being 2, and a point in the opposite side in contact with a smooth peg. If the line joining the pegs make an angle a with the vertical, and the sides in contact with the pegs an angle 6, when the lamina is just about to slip, shew that tan [d-a) — \-2 tan d. 25. A heavy rod PQ is in equilibrium with its ends on a rough parabola whose axis is vertical and vertex downwards : shew that the line joining the intersection of the tangents to the parabola at P, Q to the intersection of the normals makes with the vertical an angle not > the angle of friction. FRICTION. 193 26. A pair of equal rods AB, AC are hinged together at A and have rings at jB, C : these rings are free to slide along fine rough (//,) straight wires OB', OC in the same vertical plane equally inclined at an angle a to the vertical. Shew that in the limiting positions of equilibrium the angle between the rods is either 2 tan-i 2 Lt^fl^ or 2 tan'i 2 lr_^J^^^ . tan a- im, tan a + // 27. A right-angled isosceles triangular lamina rests with its base angles on the arc of a rough circular wire whose plane is vertical and radius equal to either of the equal sides of the triangle. If the equal sides be horizontal and vertical in the limiting position of equilibrium the coefficient of friction is | { ,^17 - 3}. 28. Two uniform rods of equal weight, but different lengths, are jointed together and placed in a vertical plane over two rough pegs in the same horizontal line : if a, /3 be the inclinations of the rods to the horizon, 6 that of the reaction at the hinge, prove that when the rods are on the point of slipping, 2 tan^ = cot (/3-t-X) -cot (a -\), where \ is the angle of friction. 29. An ellipse is placed with its plane vertical and major axis hori- zontal so that one of its vertices A rests against a rough vertical wall. P is a point on the wall vertically above A, and a string of length 21 which has its extremities fastened at the foci S, H passes through P. Find the least value of the coefficient of friction consistent mth the equilibrium of the ellipse. 30. A uniform ladder (length 2a) rests at an angle a to the vertical against a smooth horizontal rail at a height h from the ground. If \ be the angle of friction, between the ground and the ladder, shew that a man of weight n times that of the ladder may ascend a distance along the ladder, {2 (n + 1) h sin X . sec (a - X) cosec 2a-a} I n, without the ladder slipping, 31. A uniform rod AB rests with its ends on a rough circular wire in a vertical plane and the equilibrium is limiting ; shew that the vertical through the centre of the rod meets the circle through AB and the centre of the wire in two points, in one of which the directions of the resultant actions at A and B meet. 32. A uniform rod of mass 31 rests in a horizontal position with its ends on the circumference of a rough vertical circle and subtends an angle 2a at the centre. An insect of mass m starts from the middle point of the G. 13 194 STATICS. rod and crawls gently towards one end. Prove that if the angle e of friction be less than 45*^ it will be able to reach the end of the rod without disturbing the equilibrium provided sin 2e>msin 2aj{M + m). Examine the case when e>45°. 33. A rod resting with one end on a rough horizontal plane, leans against a rough cylinder, which rests on the plane, with its axis at right angles to the rod. Determine how the equilibrium is broken, when the plane is gradually more and more inclined. 34. A uniform rod, length 2a sin a, is placed within a rough vertical circle, radius a, and is on the point of motion, the coefficients of friction at its upper and lower ends are tan \', tan X : prove that if 6 be the in- clination to the vertical of the line joining the centre of the circle to the centre of the rod . . sin (X+V) tan 6 = f, TT ^; 777 r . 2 COS (X + a) cos (X - a) Examine the case when a-f X = 7r/2. 35. One end of a heavy rod AB can slide along a rough horizontal rod ^C to which it is attached by a ring ; B and C are joined by a string : if ABC be a right angle when the rod is just on the point of slipj)ing, fi the coefficient of friction and a the angle between AB and the vertical, shew that _ sin a cos a ^~T+cos2a* 36. A circular lamina, whose centre of gravity is at an excentric point, rests in a vertical plane supported by the loop of a rough string which is attached to two fixed points. If the lamina be on the point of slipping and the radius containing its centre of gravity be inclined at right angles to the radius bisecting the portion of the string in contact with the circle, the angle of contact 0, is given by 1 _ e^^0 2 c ' a being the radius of the circle and c the distance of its centre from its centre of gravity. 37. A rough circular disc of radius a has its centre of gravity at a dis- tance h from the centre, and rests in a vertical plane on two pegs placed at a distance apart < 2^(a--Z>-) and > 2& in a horizontal line: shew that equilibrium is possible for all positions of the centre of gravity provided the angle of friction be not less than sin~i {hja). FRICTION. 195 38. Two equal heavy rods AB, BC, each of length 2a, joined together at B, hang with AB resting on a rough peg P. If /x be the coefficient of friction, and 2a the angle between the rods, shew that AB will slip on the peg if PB < a cos a (cos a- fi sin a) or > a cos a (cos a + fx sin a) . 39. A uniform isosceles triangular lamina rests in a limiting position of equilibrium in a vertical plane between two rough pegs in the same horizontal line : prove that 3c cos X cos {d + a) = '2p sin 2a . sin [26 + 2a — X), where 6 is the inclination of one side to the horizon, X the angle of friction, 2a the vertical angle of the triangle, p the perpendicular from the vertex on the base, and c the distance between the pegs. 40. Three rough jjarticles of masses m^, m.^, m^ are rigidly connected by light smooth wires meeting in a point 0, such that the particles are at the vertices of an equilateral triangle whose centre is 0. The system is placed on an inclined plane of slope a, to which it is attached by a pivot through ; prove that it will rest in any position if the coefficient of friction for none of the particles be less than tana / o o o vi 41. A cylindrical rod with hemisiDherical ends rests in a vertical plane against two equally rough planes, one horizontal, the other vertical : determine the limiting position of equilibrium, and shew that if the co- efficient of friction be not less than the ratio of the length of the straight part of the rod to the total length, it will rest in any position. 42. A uniform heavy rod of given length rests perpendicularly and horizontally across two rough parallel horizontal rails which support the rod at a quarter of its length from each end. One end of the rod is pulled perpendicularly by a string in a downward direction making an angle 6 with the vertical : shew that the rod will move at both points of support at the same time when 6 — t&n~^2iJi.; and in this case find the tension of the string. 43. To the ends of a heavy rod are attached rings which slide on the circumference of a rough vertical cu'cle. Find the force perpendicular to its direction, acting at a given point of it which will just move the rod when in any position : and prove that for all positions it will be greatest when the rod is inchned to the horizon at an angle tan~^(cot 2e +cos 2a /sin 2e), where 2a is the angle subtended by the rod at the centre and tan e the co- efficient of friction. 13—2 196 STATICS. 44. A straight uniform rod of length 2c is placed in a horizontal position as high as possible within a hollow rough sphere of radius a. Prove that the line joining the middle point of the rod to the centre of the sphere makes with the vertical an angle io.TT'^ fxaj \J{a^-e^). 45. A semi-circular arch, composed of an odd number of equal and similar smooth blocks, is constructed upon a rough horizontal plane: prove that the number of blocks must be 3 : and that the coefficient of friction must be not < 1/^3. Also prove that the ratio of the internal to the external arch must not be > the positive root of the equation 2 V^ {x^ -t-.T + 1) + TT (2.r2 - a; - 3) = 0. If the blocks, except the key stone, be rough, and if their number be n, greater than 3, prove that the angle of friction at the ^jth joint from the base must be not < cot~^ {(^i-^i?) tan 7r/2?i} -pirjn. 46. Two particles of equal weight lo connected by a rod without weight rest on a rough plane inclined to the horizontal at an angle a : the coefficient of friction p tan a for one particle is less, and that for the other p tan a greater, than tan a. Prove that, when both are on the point of moving, if in the plane a triangle ABP be constructed whose sides AB, BP, PA are 2, p , p, and be the middle point of AB which is drawn in a line of greatest slope, then OP is the direction and OP . lo sin a is the tension of the rod. 47. An elliptical cylinder placed in contact with a vertical wall and a horizontal plane is just on the point of motion when its major axis is inclined at an angle a to the horizon. Determine the relation between the coefficients of friction of the wall and plane : and shew from your result that if the wall be smooth, and a be equal to 45*', the coefficient of friction between the plane and cylinder will be equal to ^c^, where e is the eccentricity of the transverse section of the cylinder. 48. Two equal spheres rest on a rough horizontal plane, the distance between their centres being c : and a third sphere rests on them : prove that the normal pressure between the two spheres is equal to half the weight of the upper sphere, and that the necessary and sufficient condi- tion of equilibrium is a-\-h>\c . cosec 2e, where e is the angle of friction, and a, h the radii of the spheres. If this condition is not fulfilled how will the lower spheres begin to move? 49. An elliptic lamina of eccentricity e rests upon a perfectly rough equal and similar lamina, the two bodies being symmetrically situated FKICTION. 197 with respect to their common tangent at the point of contact. If a be the incHnation of the major axis of the fixed ellipse to the horizon, and 6 be the inclination, measured in the same direction, of the major axis of the moving ellipse in a position of equilibrium, then sin^ {6 + a)=e'^sin 6 cos ^{6 -a). 50. A chain is formed by 2n rods, equal in length and weight, smoothly jointed together. The two extremities can move by rings on a rough horizontal rod, coefficient fx. Shew that in the limiting position of equilibrium the inclination of either of the upper rods to the vertical is 51. A rough elliptic cylinder rests with its axis horizontal upon the ground and against a vertical wall, the ground and the wall being equally rough ; shew that the cylinder will be on the point of slipping when its major axis plane is inclined at an angle of irj-i to the vertical if the eccentricity of its principal section be a^ { 2 sin X (sin X + cos X) } , where X is the angle of friction. 52. An elliptic lamina moveable about its focus in a vertical plane rests against a smooth inclined plane, the major axis of the ellipse being horizontal. The lower surface of the plane is rough and rests just on the point of moving on a horizontal table. If a, b be the semi-axes of the ellipse, and p the perpendicular from the centre on the inclined plane, shew that the coefficient of friction is \/{{p- - ^^)/(«^- P")} • 53. A circular ring of weight W hangs in a vertical plane over a rough peg, and to the lowest point of the ring a string is fastened. It is kept always horizontal in the plane of the ring, and its tension is gradually increased from zero. Prove that the ring will slip on the peg when the tension of the string reaches the value W tan | { sin~^ (3 sin e) - e} , € being the angle of friction ; and explain what happens if 3 sin e > 1 . If the tension be still further increased to a given value T, find the position of equilibrium. 54. A ring of diameter a is fixed with its plane making an angle a with the vertical, and a rough uniform cylinder is supported by being slipped through the ring: prove that the length of the cylinder must be not less than cosf^rta-X) acos^ . . ,: — , . ' sm (^±a) smX wiiere X is the angle of friction, and d is the inclination to the axis of the 198 STATICS. cylinder of a plane section whose major axis is equal to a. (The sign to be taken in the above expression depends on whether the cylinder and ring make angles with the vertical on the same or opposite sides.) 55. A cylinder is laid on a rough horizontal plane, and is in contact with a rough vertical wall, the coefficients of friction being equal ; a string, coiled round it at right angles to the axis, passes over a fixed pulley and sustains a weight which is gradually increased until equilibrium is broken. Determine the nature of the initial motion. (Jellett's Friction.) 56. Two uniform beams, of the same material and thickness but of different lengths, rest each with one end on a rough horizontal plane, and their other ends connected by a smooth joint. If equilibrium be about to be broken shew in what way it will happen. 57. Two weights, P, Q, whose coefficients of friction are fx-,^ ^lo, each less than tan a, on a rough inclined plane of angle a, are connected by a string which passes through a fixed pulley A in the plane. Prove that if the angle PAQ be the greatest possible the squares of the weights of P, Q, are to one another as 1 - /x.^" cot^ a is to 1 - /x^" cot- a. 58. A rough rod is laid on a horizontal table and is acted on by a horizontal force perpendicular to its length. Find about what point the rod will begin to turn, the point of application of the force trisecting the rod. 59. A cubical uniform block is placed on a rough inclined plane and has two of its faces vertical : it is attached by a string parallel to a line of greatest slope of the plane passing from the middle point of its upper horizontal edge to the middle point of the nearest horizontal edge of another equal similarly situated cube. If /j. (less than unity) be the coeffi- cient of friction for the lower block, the equilibrium will be broken when the inclination of the plane to the horizon is given by 2/j. = Ssmd- cos 6^ by the higher cube tumbling over, provided the friction coefficient for the higher block be great enough. 60. A heavy rod, of length 21, rests horizontally on the inside rough surface of a hollow circular cone, the axis of which is vertical and the vertex downwards. If 2a is the vertical angle of the cone, and if the coefficient of friction is less than cot a, prove that the greatest height of the rod, when in equilibrium, above the vertex of the cone is I cot a 1 + cos2 a + sin2 a ^(sin^ a + 4/U-)| 4 FRICTIOX. 199 61. A cubical block, and a cylinder whose diameter is equal to a side of the cube, are laid upon a rough plane, and are attached to each other by a cord coiled round the middle of the cylinder, and fixed to the middle point of one of the edges of the cube which is parallel to the axis of the cylinder. If the plane be then slowly raised (the cube being uppermost) until equilibrium is broken, what will be the nature of the initial motion? (Jellett's Friction.) 62. Two particles A and B of weight W are connected by a thin weightless rod and placed on a rough inclined plane at an inclination to the line of greatest slope, the coefficient of friction for each particle being /u. A force F is applied to A the lower particle in the direction BA and its direction gradually turned through an angle 6 in the plane. Find the nature of the initial motion of the system. If the particles be placed along a line of greatest slope, prove that both will slip when j'2 + 4jf^2gij2a(sina-ucosa) POO, H :^ 2FTr(/acosa-2sina) ' and find the limits between which F must lie when a W J -J, cos (a + tt / 4) sin (a + e), where e - 7r/4 is the angle of friction, and a is the angle CAB. 64. A cylindrical rod with hemispherical ends and another cylinder are in contact on a rough i^lane, the axis of the former is vertical, that of the latter horizontal. The radius of the horizontal cylinder is such that the other touches it at a point in the rim of its upper hemispherical end. The horizontal cylinder is gradually moved along the plane in a direction perpendicular to its axis, the two remaining in contact : shew that equi- librium is no longer possible unless X be >7r/4, and if X be >7r/4, equi- librium is impossible when the rod makes an angle > 6 with the vertical, where 6 is given by the equation h cos 6 = h + (h + a) cos 2X. Explain how the equilibrium is broken in this case. a = the radius of the hemispherical ends, 2h = the length of the generating lines of the rod, and X = the angle of friction, supposed the same everywhere. CHAPTER VI. VIRTUAL WORK. 113. Def. If the point A at which a force P is acting be displaced to any point B, the distance ^i^ is called the displacement of the point. If from B, BN be drawn perpendicular to P's line of action, the product P. AN is called the work done by B Fig.102 N the force P during the displacement. If N falls on that side of A towards which P acts, the work is said to be positive, if on the other side, negative. We may say then that the product of the force into the projection of the displacement, along the direction of the force, gives the algebraical as well as the numerical value of the work done during the displacement. If the displacement does not really take place but is only imagined to do so, it is said to be a vii^tual displace- ment, and the work which would be done during such a displacement is called the vii^tual work. 114. Prop. If a particle acted on by any system of forces receive any virtual displacement whatever, the alge- braical sum of the virtual work done by the different forces during the displacement is equal to the virtual work done by the resultant. VIRTUAL WORK. 201 Let represent the actual position of the particle, 0' the position to which it is supposed displaced ; let P^, P^, Fig, 103 0/ Pg, &c., be the forces acting on the particle ; 6^, 6^, 6^ &c., the angles their directions make with 00'. The A. s. of the virtual work done by P^, P^, Pg, &c. = P, .00' cos 6', + P,. 00' cos ^, + Pg. 00' cos ^3+.. . = 00' . (P, cos e^ + P^cos 6>, -f ...) = 00' X A. s. of the resolved parts of the forces in direc- tion 00' = 00' X resolved part of the resultant in direction 00' = projection of 00' along the direction of the resultant X the resultant = the virtual work done bv the resultant. It should be observed that in the above proposition the displacement the particle receives is virtual, and entirely unrestricted both as regards magnitude and direction. Co7\ If a particle in equilibrium under the action of any system of forces receive any virtual displacement whatever, the A. s. of the virtual work done by the dif- ferent forces is zero. 202 STATICS. 115. If a system of particles be in equilibrium under the action of external and internal forces, and any number of particles of the system receive any virtual displacements whatever, we have seen that the A. s. of the virtual work done by the forces on each particle is zero, and therefore the A. s. of the virtual work done by all the forces, external and internal, is zero. Prop, If a system of particles in equilibrium under the action of any system of external forces together with internal forces, receive any indefinitely small virtual dis- placement whatever, which does not alter the configuration formed by the particles, the A. S. of the virtual work done by the external forces alone is zero, or more strictly speak- ing, is of an order higher than that of the virtual displace- ment. In this proposition the displacements which the particles receive are much more restricted than in the corresponding theorem for a single particle : here the displacement must be indefinitely small, there it was unlimited in extent : the displacements, too, of the different particles are ako so connected, that if the particles formed a rigid body, these displacements would not involve any alteration in its shape or size, but only an alteration of its position as a whole. We shall first prove that if the displacements be of this character, the virtual work done by any internal force (the action exerted by B) on the particle A is equal and opposite in sign to that done by the reaction exerted by A on the particle B. Let R be the action exerted by B on the particle A, along AB in the direction indicated, then Fig.l04 b' a' - I r ! : ! AMR R^ Jti the reaction exerted by ^ on j8 is in the opposite direction. Let A' , B' be the points to which A, B are supposed dis- VIRTUAL WOllK. 203 placed, then by the conditions relating to the nature of the displacements, the angle (6) between A'B' and AB is small, and the length A'B' = AB. Draw A'M, B'N perpendicular to AB. Then MN= A'B' cos 6 = A'F ultimately, = AB .-. AM=BN. The virtual work done by action R on A = R . AM : that done by reaction R on B = — R . BN = — R . AM. Hence the A. S. of the virtual work done by any action and the corresponding reaction is zero. But the internal forces consist entirely of pairs, each pair being made up of an action and the corresponding reaction : therefore the A.S. of the virtual work done by all the internal forces is zero, since that of each pair is so. We have seen too that the A. s. of the virtual work done by all the forces, both external and internal, is zero, so that that of the virtual work done by the external forces alone must be zero also. In obtaining this result we have neglected quantities depending on the powers higher than the first of the dis- placement, so that strictly sjoeaking the A. s. of the virtual work done by the external forces is not zero, but of an order higher than the first power of the displacement. Cor. If in any system of forces in equilibrium there are two forces equal to one another, and acting in opposite directions along the straight line joining the particles on which they respectively act, the two forces will not enter into the equation of virtual work, provided the virtual displacements of the two jDarticles produce no alteration in the length of the line joining them, or at any rate one of the second order only. Hence, if we have two bodies in contact, and the virtual displacement does not alter the points in contact, the action and reaction between the two bodies will not appear in the equation for the two bodies together. Also, if two particles are connected by an in- extensible string or rod, and they receive displacements which do not involve breakino' or bendino' the strino- O O O 204 STATICS. or rod, the tension of the string, or in the case of the rod, the tension or thrust, whichever it exerts, will not enter into the equation of virtual work for both particles. This may easily be extended to the case of two particles con- nected by an inextensible string which passes round a smooth fixed body : for the distance between them measured along the string is constant, so long as the string neither slackens nor breaks. 116. In applying the above proposition to the case of a rigid body, we may suppose the displacement any slight displacement of the body as a whole not involving any change of shape or size. If we wish to ascertain the inter- nal forces between one portion of a body and another, we may suppose that the first portion is displaced as a whole, without any displacement of the remainder, in which case the actions of this last portion on the first will enter into the equation of virtual work. In solving problems by the principle of virtual work, it is often convenient to make such a displacement that a force, whose magnitude we do not wish to ascertain, may not enter into the equation of virtual work. In that case the virtual work done by that particular force must be zero, or at any rate of a higher order in small quantities than the displacement. For that to be the case the particle on which the force acts must be virtually displaced in a direction making with the force, either a right angle or an angle differing from a right angle by an in- definitely small quantity. In the following propositions, it is understood that the displacements are indefinitely small. 117. Proj). The work done by the tension of an in- extensible string or rod, when one end is fixed and the other attached to a particle which is displaced so that the string or rod is neither broken nor bent, is ultimately of an order higher than the first. It is obvious that the particle can only move in a direction which is ultimately at right angles to the rod or string, i.e. at right angles to the tension. VIRTUAL WORK. 205 Prop. If a rigid body resting in contact with any smooth curve or surface, receive a displacement by sliding along the curve or surface, the work done by the reaction of the curve or surface on the body is ultimately of an order higher than the first. In this case the particle situate at the point of the body touching the curve or surface is the one on which the reaction acts, and this point moves along a tangent to the surface or curve, i.e. at right angles to the normal along which the reaction acts. Prop. If a rigid body resting in contact with any surface, not necessarily smooth, receive a displacement by rolling along the surface, the work done by the reaction of the surface is ultimately of an order higher than the first. Let A be the common point of the rigid body and the surface : let the body be rolled so that the point 'b' originally at A comes to A\ and B becomes the point of contact. Then the arcs AB, A'B are small, and therefore the corresponding chords ; also the angle between these chords is small, so that the base AA', which is the displacement of ^, is of a higher order than AB. Cor. If the rigid body partly roll and partly slide along a smooth surface, it is clear that the displacement is compounded of the two displacements of rolling and slid- ing, and is therefore of an order higher than the first, since each of the latter is so. 206 STATICS. We have already seen that if two bodies in contact receive such virtual displacements that their points of contact remain the same, the action and reaction between the two do not appear in the equation of virtual work for the two bodies : neither will they if in addition to these dis- placements one rolls along the other, or if the bodies be smooth, one slides or partly slides and partly rolls along the other. For either set of displacements alone will not bring these forces into the equation, therefore a combina- tion will not do so. 118. As an illustration of the application of the principle of virtual work, we will by means of it prove the theorem proved in Art. 81, viz., that the tensions at the ends of a weightless string stretched over a smooth surface are equal. Let A, B be the points where the string leaves the surface, and let T, T be the tensions at the ends P, Q respectively. We shall not interfere with the equilibrium of the string if we suppose it to lie in a groove cut in the surface, so that when pulled at one end, it must move along the groove. Let the virtual displacement which is given to the string be produced by pulling the end P in the direction AP to P', so that the end Q must move along QB to a point Q' such that QQ' = PP'. As each portion of the string in contact with the surface moves at right angles to the action of the surface on it, no work is done by the actions of the surface on the string, and the algebraical sum of the virtual work done is T.PP'-T' . QQ', which is therefore zero ; 1. e. T=T', since PP'=QQ'. Apply the principle of virtual work to the solution of the following examples : Ex. 1. The algebraical sum of the moments about any point in their plane of a number of coplanar forces in equilibrium is zero. VIRTUAL WORK. 207 Ex. 2. Two small rings of equal weight slide on a smooth wire in the shape of a parabola, whose axis is vertical and vertex upwards; they will be in equilibrium if connected by an inextensible string which passes over a smooth peg placed at the focus. Ex. 3. Two equal uniform rods freely jointed at their ends rest, one on each, of two smooth pegs which are in a horizontal Hne. Shew that the inclination (6) of either rod to the vertical is given by the equation a sin^ 6 = c, where a is the length of each rod, and c the distance between the pegs. Ex. 4. Shew that in Ex. 27, page 94, the weight of each beam is proportional to the tangent of the angle, which the Hne joining the centre of the semicircle with the corresponding point of contact of the beam makes with the horizontal. Ex, 5. Two equal uniform rods of the same material and thickness have two ends connected by a smooth hinge, and their other ends are attached to small rings which slide on a smooth horizontal wire. Find the position of equilibrium when a circular disc whose weight equals that of either rod, is placed between them so that each rod touches its circum- ference. Ans. Each rod makes with the vertical the angle {6) given by the equation 2a sin^ ^=r cos ^, where a is the length of a rod and r the radius of the disc. Ex. 6. A trij)od formed of three equal uniform rods, three ends being connected by a common joint, and the other three connected, each with the other two, by equal strings, rests with the joint uppermost on a smooth horizontal plane. Shew that the tension of each string is TI'c/3//, W being the weight of a rod, c the length of each string, and h the height of the joint above the plane. Ex. 7. A heavy elastic string rests in the shape of a necklace round a smooth right circular cone whose axis is vertical : shew that its radius is a + ica?l\ tan a, where X is the modulus of elasticity, ^ira the length of the string when un stretched, w its weight per unit length, and a the semi- vertical angle of the cone. 119. We will now prove the converse of the princii^le of virtual work for a single particle, i.e. if the algebraical sum of the virtual work done by a system of forces acting on a particle, be zero for every displacement whatever, the particle is in equilibrium. 208 , STATICS. For let (fig. 103) the forces be P^, P^ &c., the par- ticle, 00' any virtual displacement, 6^, 0^ &c., the angles PjjPgj&c. make with 00\ Then, since the algebraical sum of the virtual work done by the forces = 0, P^ . 00' cos 6*^ + P, . 00' cos 6>, + ... = 0, .-. 00' (P, cos ^^ + P^ cos ^,+ ...) = 0, .*. P^ cos 6^ + Pg cos 6^-\- ... = 0, i.e. the algebraical sum of the resolved parts of the forces in any direction is zero, and the particle is therefore in equilibrium. 120.* The material systems, to which the following propositions refer, are either single rigid bodies, or systems of rigid bodies, connected in such a manner by means of inextensible strings, smooth joints, &c., that the motion of one of them determines that of all of them. Prop. A material system as above, under the action of a system of external and internal forces, such that for every indefinitely small virtual displacement whatever, which does not violate the geometrical conditions, the algebraical sum of the virtual work done by the external forces is of an order higher than the displacement, is in equilibrium. For if the system be not in equilibrium, its motion is a definite one, which does not break the geometrical conditions : now we can conceive a number of smooth surfaces or inextensible strings so arranged, that they do not interfere with the actual motion of the system, but yet render it the only motion possible. If this be done, the whole system can be fixed by fixing one of the moving points in it, and this can be done by applying to it a force (P) of sufficient magnitude, in the direction opposite to the point's motion, since that is the only direction in which the particle can move. The system is now in equilibrium under the action of the original forces, the new forces of constraint and the VIRTUAL WORK. 209 force F. If then any virtual displacement be given to it, the algebraical sum of the virtual work done by these forces is zero : let the displacement be the one which actually takes place, when F is not applied. In this case the new forces of constraint do no work, so that the algebraical sum of the virtual work done by the original forces and F is zero. But we know that the algebraical sum of the virtual work done by the original forces alone is zero ; hence the virtual work done by F is also zero. But as the point on which F acts is one of the moving points of the system, its displacement is of the first order, so that F must be zero, i.e. the system is in equilibrium without F, and without the new forces of constraint. 121.* Projy. A material system as above, under the action of external and internal forces, will, if held in any position, and then let go, move at first so that the alge- braical sum of the work done by the external forces is positive, and of the first order of the actual displacement, provided the position is not one of equilibrium. If the system is not in equilibrium, we can, as before, arrange a system of smooth surfaces, so that the actual motion is the only one possible, and this again can be entirely prevented by applying at one of the moving par- ticles, A, say, of the system, a force, of sufficient mag-nitude F, in a direction opposite to that of A's motion. The system being now in equilibrium we see as before, choosing the virtual displacement that which actually takes place, that the algebraical sum of the virtual work done by the I original forces and F is zero. 1 But it is obvious, since A moves in the direction op- posite to that of the force F, the work done by F is negative and of the first order ; and therefore the alge- braical sum of the virtual work done by the original forces is positive and of the first order of the displacement. The algebraical sum of the work actually done by the i G. 14 210 STATICS. original forces is therefore at first positive and of the first order of the actual displacement. 122. Bef. The equilibrium of a body is said to be stable, when, on moving it slightly from its position of equi- librium, it returns to it ; if it moves still further away from this position, its equilibrium is unstable. If the body remain in equilibrium, the equilibrium is neutral. If a small ring slide on a smooth circular wire placed in a vertical plane, it will be found by experiment that there are two positions of equilibrium, one, the stable one, at the lowest point of the wire, and towards which it will readily return if moved away from it : the other, the unstable one, at the highest point, away from which it will move if disturbed ever so slightly, and in which it is found practically almost impossible to keep it. A uniform sphere resting on a smooth horizontal plane is an instance of a body in neutral equilibrium ; for if it be rolled out of its position along the plane, it will neither return, nor, unless a velocity be given to it, move further away. The positions of stable and unstable equilibrium of a body succeed one another alternately, i.e. there cannot be two positions of stable equilibrium without one of unstable equilibrium between them, and vice versa. For a position of stable equilibrium is one to which the body tends to move when placed near it, so that if there are two such positions of a body, there must be a position between them such that, if the body be placed on one side of it, it will tend to move towards one of the above positions, and if placed on the other side, towards the other: i.e. there is a position of unstable equilibrium between them. Similarly we can shew, that there is a position of stable equilibrium between every two positions of unstable equilibrium. 123.* Prop. When the only external forces acting on the material system of the last two propositions are the weights of the different particles which compose it, and the forces due to the geometrical constraints, such as the reactions of smooth surfaces and the tensions of in exten- sible strings, the system is in a position of stable or unstable VIKTUAL WORK. 211 equilibrium, according as its centre of mass is at a maximum or minimum depth consistent with the geometrical con- ditions of constraint. For the work done by gravity during any displacement is the algebraical sum of the products of the weight of each particle into its vertical displacement (the positive sign being given to the displacement when it is down- wards, the negative when upwards): and this again is equal to the product of the weight of the whole system into the vertical displacement of its centre of mass. Also, during any displacement of the system, consistent with the geometrical conditions, gravity is the only force which does work. We have seen (Art. 121), that when not in equilibrium the system moves so that the work done by the forces is initially positive, i.e. in this case, so that the centre of mass moves downwards. Hence the system always tends to move initially, so that its centre of mass moves towards the adjacent position at a maximum depth and away from the adjacent position at a minimum depth; these positions succeed one another alternately, and it is clear that the former are positions of stable, and the latter of unstable equilibrium. 124.* The cases considered above divide themselves into two classes : one, in which the centre of mass of the system is constrained to move along a certain curve, so that in any position, it is onl}^ free to move in tiuo direc- tions, opposite to one another; the other, in which the centre of mass is constrained to move on a certain surface, so that in any position, it is free to move in any direction in a certain plane, the tangent plane to the surtace at the point. A rod with its ends compelled to move along fixed wires is an illustration of the first class, one placed inside a bowl is an illustration of the second class. If there is a position of the system, such that for all possible small displacements from it, the depth of the centre of mass is diminished , that position will be a position of absolutely stable equilibrium; on the other hand, a 14—2 212 STATICS. position, such that for all possible small displacements from it, the depth of the centre of mass is increased, is one of absolutely unstable equilibrium. A point then on the locus of the centre of mass, where the tangent line or plane is horizontal and below the adjacent points of the curve or surface, corresponds to a position of the system of abso- lutely stable equilibrium, when it is above the adjacent points, to one of absolutely unstable equilibrium. If the locus of the centre of mass be a curve, it may be that there is a point on it, such that the tangent at it is horizontal, and cuts the curve there, i.e. the adjacent part of the curve on one side is above the tangent and that on the other side below it : in other words there may be a point of inflexion at which the tangent is horizontal. Such a position of the centre of mass corresponds to a position of equilibrium of the system, a position from which a dis- placement in one direction will bring about a tendency to return to it, in the other direction, a tendency to recede still further from it. Again, when the locus of the centre of mass is a surface, the shape of the latter may be that of a saddle, or that of the ground at the top of a pass between two mountains; in this case a tangent plane to the ground at the top of the pass is horizontal, and has part of the surface above it and part below it. This position of the centre of mass cor- responds to a position of equilibrium of the system, which is unstable for displacements of the centre of mass in the plane containing the tangent to the path over the pass, and stable for displacements in the plane at right angles to it. 125.* A body BAG rests on a rough fixed body DAE, the surfaces near the point of contact A being spherical : it is required to determine whether, for displacements made by rolling only, BAG is in stable or unstable equi- librium. Let 0, be the centres of the spherical surfaces : we suppose that the common normal oAO is vertical. G the centre of mass of BAG will be situate in Ao. VIRTUAL WORK. 213 het BAGhe displaced by rolling through a small angle so that it comes into the position B'A'C, G' and o' being a ,' Fig.107 the new positions of G and o, P the point of contact of the two surfaces. Let oA=r,OA=R,AG = h, ^AOP= a, and A'o'P-=^, -.' the arc AP = the arc A'P, Rol = r/3. Now the equilibrium is stable or unstable, according as G was originally at a maximum or minimum depth, i.e. according as G' is vertically above or below G, according as Go' cos a — G'G cos (a + ^) is OG^, > {R + r) cos a — {r — h) cos (a + /8) is i^ + li, < < h {1 — cos (a + /3)} is r {cos a — cos (a + ^)] > — R(l — cos a), 214 STATICS. according as h is 7^ . sm - . sm a + — — it sm - < 2 V 2/ 2 k is — ^^ — 7 ;=^ — '' — as a and are small, > (^^ + ^) . < tR (2r + i^) - Rr" ^'''> {R-^rf 1. >R + r 1 1 ris -^— or ^ + -. This result is also easily obtained from the consideration, that BA C will return to its original position or not, accord- ino- as G' lies nearer to Oo than P or further from it. If the concavity of either surface be turned the other way we shall obtain the same result as before, except that the sign of the corresponding radius will be changed. If either surface be plane, its radius is of course infinity. Cor. The above results hold for any curved surfaces, if R and r represent the radii of curvature of the sections made by the plane of displacement. If Y = - + ^ , the equilibrium is said to be critical, h r li and we must proceed to a higher degree of approximation in order to determine whether the equilibrium is really stable or unstable. Ex. 1. A body made uj) of a cone and a hemisphere having a common base, rests with the axis vertical on a rough horizontal table : determine the greatest height of the cone in order that the equilibrium may be stable. Ans. Height of cone = ;^3 . radius of base. Ex. 2. A prolate spheroid rests with its axis horizontal on a rough horizontal plane ; shew that for rolling displacements in its equatorial plane the equilibrium is neutral, and for displacements in the vertical plane containing the axis, it is stable. VIRTUAL WORK. 215 Ex. 3. A right circular cylinder of radius r rests with its axis hori- zontal on a fixed rough sphere (radius R > r): shew that the equilibrium is stable or unstable, according as the plane in which the displacement takes place makes with the vertical one containing the axis of the cylinder an angle cos~^^(r/jR). Ex. 4. A prolate hemispheroid rests with its vertex on a rough hori- zontal plane, prove that the equilibrium is stable or unstable according as the eccentricity of the generating elhpse is less or greater than ^{Sj8). 126.* The material systems for which we have proved the preceding proj^ositioDS have been either single rigid bodies, or rigid bodies connected in such a way that the position of one determined the positions of the others. We caii however easily extend them to include the case of a system of rigid bodies so connected, that it is necessary to know the positions of a number of the bodies in order to know those of all. Prop. If the algebraical sum of the virtual work done by the external forces be zero for all possible small virtual displacements consistent with the geometrical conditions, the above material system is in equilibrium. For if it is not, it will have a definite motion consistent with the geometrical conditions, and without interfering with the actual motions of the bodies we can so arrange a number of smooth surfaces or inextensible strings, that these actual motions are the only ones possible : they need not however all take place, i.e. several of the bodies may move without all the others doing so, and fixing one of them will not of necessity fix all. In this case we can re- duce the whole system to rest by fixing one moving point in each of the bodies, and this can be done by a^^plying forces P, Q, R, &c. of sufficient magnitude in the directions opposite to the actual motions of these points respectively. Now the whole system is in equilibrium under the action of the original forces, the new forces introduced by the smooth fixed surfaces &c., and the forces P, Q, R, &c.: therefore, if the virtual displacements chosen be the actual 216 STATICS. ones, the algebraical sum of the virtual work done by the original forces and P, Q, R, &c. will be zero, because the virtual work done by the reactions of the smooth surfaces is zero. But it is obvious that the work done by each of the forces P, Q, R, &c. is negative, since the particle on which it acts moves in the direction opposite to that of the force : the algebraical sum of the work done by the original forces is therefore positive, which is inconsistent with its being zero, as it is by supposition. Hence each of the forces P, Q, R, &c. is zero, and the system is in equilibrium; and as the smooth surfaces or inextensible strings do not interfere with the actual motion in any way, their removal will not upset the equilibrium of the system. Cor. We see from the foregoing, that when such a material system as the above is not in equilibrium in a particular position, under the action of given external forces, it will, if placed in that position and then released, move so that at first the algebraical sum of the work done by the external forces is positive. By reasoning as in Art. 123 the proposition there proved can be extended to the case of the material systems we have just been considering. Among these systems we can include a mass of liquid, or a heavy inextensible flexible string. In fact the propositions of Arts. 120 — 123 apply to all systems of bodies the internal forces among which can do no work, so long as the geometrical conditions are not violated : they will not however apply to those systems in which the internal forces are capable of doing work ; for instance, systems in which the pressures of compressible fluids, the tensions of elastic strings, and the actions of rough surfaces are included among the internal forces. On the other hand there is no restriction on the nature of the external forces : they may consist of frictions, or the tensions of elastic strings, without atfectiug the validity of these propositions. 127.* In a precisely similar way we could prove the much more general proposition still, that if any material VIRTUAL WORK. 217 system whatsoever, under the action of any system of forces, be placed in any position and then released, it will, if not in equilibrium, move at first so that the work done by all the forces, internal as well as external, is positive. 128.* Def. When the forces, internal as well as ex- ternal, acting on a material system are such, that the algebraical sum of the work done by them, as the con- figuration of the system changes, depends only on the initial and final configurations and not on the paths the different bodies take, they are said to form a conservative system of forces. Def. If any material system is acted on by a con- servative system of forces, the algebraical sum of the work done by these forces, as the configuration of the system changes from any other to some standard configuration, is termed the Potential Energy of the system correspond- ing to the former configuration. It is generally convenient to take the standard configuration such, that the potential energy for every other configuration which is practically considered is positive. 129*. Prop. When any material system is acted on by a conservative system of forces, it is in a position of stable equilibrium when its potential energy has a mini- mum value, and in a position of unstable equilibrium when its potential energy has a maximum value. We have seen (Art. 127), that when the system is placed in any position, except one of equilibrium, and then released, it will move so that the algebraical sum of the work done by the forces is initially positive, i.e. it will move so as to diminish its potential energy. Hence it will move towards a position of minimum, and away from one of maximum, potential energy. The joositions of maximum potential energy then are positions of unstable equilibrium and those of minimum potential energy of stable equilibrium. 218 STATICS. The proposition proved in Art. 123 is a particular case of this theorem. 1*30. Recapitulation. We began by shewing that if a particle be in equilibrium, the total virtual work done by the forces acting on it, during any virtual displacement whatever, is zero. The same theorem is therefore true for any system of particles, when the internal as well as the external forces are taken into consideration ; but if the virtual displacement is a small quantity of the first order, and the system of particles form a rigid body, and also in certain other cases, it was shewn that the total virtual work done by the external forces alone is a small quantity of the second order. The converse theorem was then shewn to hold for single rigid bodies, and also for a system of rigid bodies, connected in certain ways. Also such a system will, if placed in any position, and then released, move so that the total virtual work done by the external forces during the initial small displacement is positive, unless the posi- tion is one of equilibrium. Hence followed the principle, that such a material system, when gravity is the only active force, is in stable or unstable equilibrium, according as its centre of mass is at a maximum or minimum depth consistent with the geometrical conditions. Similarly followed the more general theorem, that for any material system under the action of any conservative system of forces, stable positions are positions of minimum, and un- stable positions of maximum, potential energy. ILLUSTEATIVE EXAMPLES. Ex. 1. Find the amount of work done in stretching an elastic string. Let a be the natural length of the string, \ its modulus of elasticity; let X be the extension of the string. Let the extension x be divided into n, an indefinitely large number, equal parts. Wlien the length of the string is a-\ , the tension is VIRTUAL WORK. 219 T X 7* -1- 1 X . - . - , and therefore the work done in stretching? it to a-\ x lies n a 11 between r X X r+1 X X A . - . - . - , and A . . - . - nan nan i.e. the total work done in stretching the string to the length x _ Xx2 jt l + 2 + 3+...(w-l) _Xa;2 ^ = 00 w 2a ' Hence the work done in increasing the extension from y to x is X(a;2-7/)/2rt. Ex. 2. Shew that the power necessary to move a cylinder of radius r and weight W up a plane inclined at angle a to the horizon by a crowbar of length I inclined at an angle /S to the horizon is sm a Wr I * l + cos(a + /3) ' Fig. 103. Let be the point where the axis of the cylinder intersects the verti- cal plane containing the crowbar AB ; C the point where the same plane meets the generating line in contact with the inclined plane. Let P be the force, which applied at B at right angles to AB will maintain equilibrium. Let the virtual displacement be for AB to turn through a small angle 6, so that its inclination to the horizon becomes /3 + ^. By Art. 117, the actions between the cylinder and crowbar and between each and the plane do not enter into the equation of virtual work. The vertical height of above A {■& AC sin a + OC cos a, i.e. = r {sin a . tan ^ (a + /3) + cos a]~r cos ^ (a - /3)/cos \[oi-\- j8). 220 STATICS. neglecting the weight of the crowbar, the equation of virtual work is cos- cos P.l.d-W .r\ cos a + ^ + d cos a + /3 :0. P = Wr sin a sin cos a + jS cos Wr I Wr sm a cos 2 2 cos' , a + /3 2 cos a + ^ + d 2 sin; 2 sin a I *l + cos(a + j3)' since 6"^ and higher powers are neglected. "s/.. Ex. 3. A straight uniform rod has smooth small rings attached to its extremities, one of which slides on a fixed vertical straight wire, and the other on a fixed wire in the shape of a parabola whose latus rectum equals twice the length of the rod, and whose axis coincides with the straight wire: prove that in the position of equilibrium (stable when the vertex is upwards) the rod will be inclined at an angle of 60'^ to the vertical. "Which is the position of stable equilibrium when the vertex is downwards ? Let PQ be the rod, P being the point on the parabola. Let d be its inclination to the vertical, 2a its length, and G its middle point. Draw PN perpendicular to the axis AN. VIRTUAL WORK. 221 (1) When the axis is upwards, the depth of G below A = AN + PG cos d PN^ ^ 4a2sin2^ = — h a cos 6 = — — ; h a cos d Aa 4a = a(l + cos^-cos2^). The positions of equilibrium are given by the maximum and minimum values of this expression, 1 + cos ^ - cos2^ = f - (A - cos df, i.e. is a maximum when cos^:=^ or when ^ = 60*^. It is clearly a minimum when ^ = 0. Hence ^ = 60^ corresponds to a position of stable equilibrium, and ^ = to one of unstable equilibrium. (2) WTien the vertex is downwards ^ = corresponds to the position of stable, and ^=60*^ to the position of unstable equilibrium. Ex. 4. Two smooth rods which intersect at an angle 2a are placed so that they are equally inclined to the vertical, and the line bisecting the angle between them is inclined at an angle /3 to the vertical. Prove that a spherical ball of radius a will be in a position of unstable equilibrium, if the distance of its points of contact with the rods from the intersection of the rods be a cot a cos ^ \/{l - cos- a sin- /3) ' Let be the centre of the sphere, 0' of the circle in which it is inter- F/g.iio 222 STATICS. sected by the plane of the rods. Let B, C be the pomts where the rods AB, AC touch the sphere : 6 the angle GO' or BO' subtends at 0. Then 00' = acos^, CO' = a sin ^, ^0' = a sin^ . cosec a. The vertical height (h) of above A = A0' cos p+ 00' sin 13, = a (sin ^ cosec a . cos j3 + cos ^ . sin j9). Let cos/3 coseca = rcos 0, and sinjS=rsin0: then h=ar (cos sin ^ + cos 6 sin 0) = ar sin (0 + 0). Now the sphere is in a position of unstable equilibrium when h is a maximum, i.e. when + = 7r/2, i.e. when 0=cot~i (sin a tan ]3). But AB = O'B cot a = a cot a sin 6, a cot a a cot a . cos/S AB = ^(1 + sin2 a tan^' /3) ^/(l - cos^ a sin^ j3) * Ex. 5. Three equal heavy rods are connected by two hinges, a string is attached to the free ends and hung over a peg so that the middle rod is horizontal ; a is the angle which either of the other rods, and /3 the angle either part of the string, makes with the vertical. Prove that 2 tan a = 3 tan j3. Let be the peg, AB, BC, CD the rods. Let 21 be the length of the string, and a that of each rod. The depth of the centre of mass of the whole below is ? cos /3 + |a cos a. VIRTUAL WORK. 223 Since the algebraical sum of the horizontal projections of the sides of OABCD is zero, 2Z sin /3 + 2rt sin a = 2a. If the virtual displacement be such that the rod BC is moved vertically through a small distance, without interfering with any of the connections, no work is done by any of the forces except gravity, and therefore none is done by gravity ; in other words, the displacement of the centre of mass is of the second order. Let, in consequence of this displacement, a become a + d, and /3, /3 + 0, then I cos p + ^a cos a = l cos (iS + 0) +^acos(a + ^), and Zsin/3 + asina = a=Zsin (j3 + 0) + a sin (a+ 6), .-. Isin (^ + ^) sin| + fasin (« + g) siHg^^' / 0\ . / d\ . d ^ and Zoos ( /3 + 2 1 sm- + acos ( a + 2 I sm- = 0; .-. tan^^+|)=|tan^a + ^^, .*. 3 tan/3 = 2tano, since 6 and

, = , -or /(-?'+ Tf' + 2PTr cos (9)... (1). If P could only just balance W, or, in other words, were W on the point of moving P, the relation would be Pp= Wq- r ^m\ . s/[P' + W^ + 2PW . cos 6) . 138. To find the efficiency {E) of the rough Lever. Let Pq be the power just required to move W, when the fulcrum is perfectly smooth. Then P,p = Wq. P Wq But the efficiency, E — ^ = -^ . Therefore from (1) we have r sin\ . W I i + ^—ir- + 2i ^ 9. 1=E + p ■l+i^+2fcos^); .-. pq (1-E)= r sin \.^/{q'-\- p'E' + 2pq.E. cos 6), which gives us E. 139. The Wheel and Axle. This machine consists of a cylinder a (the wheel) with a groove cut round the cir- cumference, and a cylinder b of smaller radius (the axle). MACHINES. 239 The two form one rigid body and have a common hori- zontal axis cc ^ at the ends of which are two pivots c and c', resting in fixed sockets so that the whole can turn about this axis. The power P is applied tangentially at the circum- ference of the wheel, generally by means of a rope, while the weight is suspended by a rope which is wound round the axle so that it tends to turn the machine in the op- posite direction to the power. The apparatus for drawing a bucket of water out of a well is frequently a machine of this kind, the power being applied by means of a handle attached to the wheel instead of by a rope. A windlass for hauling up an anchor on board ship is a modification of the wheel and axle, in which the common axis is vertical, and the power is applied at the end of poles which project from the wheel so as to form radii produced. The wheel and axle, as before stated, is a kind of lever, and we can shew as in Art. 74 that the condition of equilibrium is that the moment of P about the axis should be equal and opposite to the moment of TF, i.e. that P X the radius of the wheel = TF x radius of the axle. Ex. 1. Four sailors, each exerting a force of 112 lbs., can just raise an anchor by means of a capstan whose radius is 1 ft. 2 in. and whose spokes are 8 ft. long (measured from the axis). Find the weight of the anchor. Ans. \\% tons. Ex. 2. If the length of each of a pair of sculls be 8 ft. 6 in., and the distance from the hand to the rowlock be 2 ft. 3 in., find the resultant force on the boat when the sculler pulls each scull with a force of 25 lbs., assuming that the blade does not move through the water. Ans. 18 lbs. Ex. 3. A fly-wheel 10 ft, in radius weighs 15 tons, its axle is 6 in. in radius and revolves in bearings between which and it the coefficient of friction is "2 : find the smallest weight which, hung from a band round the circumference of the wheel, will just turn it. Ans. 333 lbs. nearly. 140. The Pulley. A pulley -block consists of two plates, connected by an axis about which a circular disc, with a groove cut in its circumference, can turn. Ris^idlv connected with the axis is a hook to which a string can 240 STATICS. be attached so as to support the pulley, or by means of which the pulley can support a weight. Sometimes there are several discs, either turning about the same axis or placed one below another ; they then form double, treble, &c. blocks. When the block is fixed, the pulley is said to he fixed; otherwise it is called a moveable pulley. A rope passes along the groove in the circumference of the disc, and, as the latter is supposed smooth, the tension of the rope will be the same on both sides the pulley. If a fixed pulley be used to enable us to overcome resistance, the only object gained by the use of the pulley is that the force applied is enabled to counteract a force in a different direction, though of no greater magnitude. When a single moveable pulley is used, the weight W is attached to the block, and the power P is applied at Fig.118 one end of a rope which passes under the disc of the pulley, the other end of the rope being fastened to a fixed point. It is obvious, when the strings are parallel, that W = 2P, and when they are not parallel, but each makes an angle 6 with the vertical, that W = 2P cos 6. MACHINES. 241 141. There are three systems of pulleys usually de- scribed in text-books. In the first system, the weight is attached to the lowest pulley, which is supported by a rope, one end of which is ?^f 4 M w tt'. T Fig. 1 19 attached to a fixed beam, and the other end to the next pulley above, Avhich is in turn supported in a similar way: the power is applied at the end of the rope supporting the highest pulley. The portions of the ropes not in contact with the pulleys are vertical. Assuming the ropes to be without weight, we can easily investigate the condition of equilibrium. Let there be n pulleys, whose weights in order from the lowest are tu^, tu^, ..AUn, and let the tensions of the ropes supporting them be T^, T^,...T^^. Then from the equilibrium of the different pulleys, we have 2T^= W + w^ (1), i'^h 2T^=T^-^w.^ G. 16 242 STATICS. 2T^^T^+w, (3), 2T„ = T^_, + ^,^ • 00, also P=T^ (n+1). Multiplying equations (2), (3)...(?^+ 1) by 2, 2"^ 2^...2" respectively, and adding, we have 2« . P = Tf + w, + 2^2 + 2X + . . . T-'w^^. If the pulleys be without weight, this equation reduces to2\P=Tf. We can deduce the same equation by the principle of virtual work. Let the virtual displacement be the one which would actually be produced by moving the end of the rope to which P is applied through a small distance x in P's direction. By this, the uppermost pulley would be raised through a height - , the next lower pulley through a height ^2) ^^d so on, the lowest pulley and weight being . X raised through a height — . During this displacement, the actions of the fixed points to which the ends of the different ropes are at- tached do no work, nor is any done by the internal forces of the system (Art. 117). The equation of virtual work then is P .x-w^^.^- w^_^ . 22 - ••• - ['W+^o^) ^, = 0, i.e. 2"P = F + -m;, + ^w^ + 2X + • • • 2"'y,. 142. In the second system there are two pulley- blocks, the upper of which is fixed, and the lower move- able : a rope passes over one of the discs of the upper block and under one of the lower block alternately, the radii of the different discs being such that the portions MACHINES. 243 of the rope not in contact with a pulley are vertical, or nearly so. One end of the rope is attached to one of the two blocks, and at the other end the power is applied. The weight is attached to the lower block. Let W be the weight to be raised, including that of the lower block : let P be the power which just raises it : then the tension of the rope is P throughout, and if there be ?i strings coming from the lower block, the total force exerted by them is nP, and w^e must have W=nP. 143. In the third system, the uppermost pulley is fixed : each pulley has a roj^e passing over it, with one end attached to the weight and the other to the pulley 16—2 244 STATICS. next below. The power is applied at the end of the string passing over the lowest pulley. A TJi I& t m Ulo A To ^2 (n l ff il f n l ^ >^^/ ^ Fig.l2J In investigating the relation between the weight W and the power P which will support it, we shall suppose the portions of the ropes not in contact with the pulleys vertical, and the ropes to be without weight. Let there be 71 pulleys, including the fixed one, and let w^, w^, Wg . . . w^_j be the weights of the moveable ones beginning with the lowest ; let T^,T^...T^he the tensions of the ropes passing over them. Since each pulley is in equilibrium, we have T,=rf, + w^ (1), T,==2T^+w, (2), MACHINES. 245 from the equilibrium of the weight T^+T^+T^-\-... 1\=W {n), also T, = P. n-2 Multiplying equations (1), {2)...{n-l) by 2""\ 2 2^*"^ ... 2 respectively, and adding, we have 2T„ = 7/;^ . 2"-^ + ^t'^ . 2""'-' + . . . 2io^_^ + P . 2". Adding equations (1), (2) ... (?z — 1), and employing equation (n), we have ir-P=2 (Tf- rj + z/;, +^, -h ... i/;„_,. Eliminating T^, we have Tf = P (2'^ - 1 ) + ^, {r" -1)^-10^ (2«-^ - 1) + . . . w„_, (2-1). To deduce the relation between W and P from the principle of virtual work. Let the weight W be supposed moved vertically down- wards through a small distance x. Then the highest moveable pulley, w^_^, will be raised through a height x : the next pulley below will be raised twice the height through which the highest is raised, together with the distance through which the weight descends, i.e. through a height 3.^^^. Similarly we can see that any pulley will rise through a height x, together with twice the distance through which the next pulley above rises. The distances, therefore through which the weights w^^_^,w^^...iu^, are respectively raised are x, (2^ — 1)^', (2^ — 1) ^, . . . (2"~^ — 1) x. Also the point of application of P will be moved vertically upwards through a distance (2"* — 1) x. Hence the equation of virtual work is W .X- w^_^ .x-{r-l) w^_.2 .x-{2^ -1) w^_^ . X - (2"-^ -l}w^.x-{2"-l)P .x=0; :. W= {T - 1) P + (2"-'- l)iu^ + (2"-'- l)ii\^ + ...(2-'-l)^^V2+^^'n-i- 246 STATICS. Ex. 1. If there are three moveable pulleys arranged as in the first system, their weights beginning from the lowest being 9, 3, and 1 lbs. re- spectively, find what power will support a weight of 69 lbs. Ans. 11 lbs. Ex. 2. If in the second system there are altogether nine pulleys and each pulley weigh one pound, what force will be required to support a weight of 86 lbs. ? Ans. 10 lbs. Ex. 3. If the weight supported in the third system be 56 lbs., and each moveable pulley, of which there are 3, weigh 1 lb., find the horizon- tal distance of the centre of mass of the weight from the centre of the fixed pulley, supposing the diameters of all the pulleys equal. Ans. 9/28 the radius of any pulley. 144. The Inclined Plane. A line in the plane per- pendicular to its intersection with the horizontal plane is called the line of greatest slope, and the vertical plane containing this line the principal plane. To find the condition of equilibrium on an inclined plane, where W is the weight and P the power. (i) When the plane is smooth. Let BAG be a section of the inclined plane made by a principal plane, BA being the line of greatest slope and A C being horizontal. Let a = the angle of inclination BAG. The reaction R of the plane acts at right angles to the plane and therefore parallel to the plane BAG: the weight W acts parallel to this plane also, so that the power P must also act parallel to this plane. Let 6 be the angfle which P's line of action makes with AB mea- MACHINES. 247 sured up the plane, 6 being positive when P's direction is above the plane. By Art. 18, P \ W \ R = ^m (TT, R) : sin [R, P) : sin (P, W) = sin a : cos 6 : cos (a + 6). P clearlv has its least value for a given value of W when (ii) When the plane is rough, and the direction of P is in the principal plane. The total reaction R of the plane will act in the principal plane, since W and P do ; its direction cannot Fig.123 make with the normal an angle greater than X, the angle of friction, but may make any smaller angle. Let \' be the angle which R makes with the normal, k! being measured towards the lower part of the plane. We have then the equations R^ = P' + W + 2PTr . COS ('^ + I + a) = p2+ Tr-2PFsin((9 + a), 248 STATICS. and P Bing-V + f-g) ,i„(,^,.) to determine M and X'. When P is just on the point of moving W so that the latter is just about to sHp up the plane, the total reaction will make with the normal an angle X on the side towards the lower part of the plane : in that case * cos (6 — \) Also j^^y cos(. + g) COS {u — X) The value of which will give the least value of P for a given value of IT is X ; i.e. for P to be most effective it should make with the plane an angle equal to the angle of friction. By changing the sign of X in the preceding investiga- tion we can obtain the value of P which will just prevent W from slipping down the plane, when the reaction B will make an angle X on the other side of the normal. This gives us P=W ^^~^~^^ ' cos ((9 + X) ' and ^^^ cos(« + ^ ) cos (6 + X) (iii) When the plane is rough and P's direction does, not lie in the principal plane. Let P make Avith the plane an angle 6, and let its resolved part along the plane make an angle with the line of greatest slope drawn up the plane. Let the re- action R of the plane be resolved into R cos X' along the MACHIXES. 249 normal, and R sin X' along the plane, the latter making an angle /3 with the line of greatest sIojdo. Resolving the Fig. 124 forces at right angles to the plane, along the line of greatest slope, and in the plane at right angles to the line of greatest slope, we have P sin S + Rco^X' — W cos a = 0, P cos 6 cos (j)-\- R sin V cos ^ — W sin a = 0, P cos 6 sin (f)-h R sin V sin fi = 0. These equations are sufficient to determine R, V and /3, when the other quantities are known. If P be on the point of moving W, the reaction R makes with the normal an angle \, so that writing A, for V in the above equations, they enable us to determine P, R and /S, when the other quantities are known. 145. The Screiu. A screw may be supposed con- structed as follows : — Let aa'cl'd be a solid right circular cylinder, and let- AA'DD be a rectangle, whose breadth AA' is equal to the circumference of the cylinder. Draw BB' , CC, DD' &c. parallel to A A' and at equal distances from one another: join AB' , BC\ CD'. Now let the rectangle AD' be supposed wrapped round the cylinder, so that the sides ABCD, A'BC'I)' coincide wdth the generator ad, the points A, A' coinciding in a, B, B' in h, C, C in c, and D, U 250 STATICS. in d. The lines AB', BC, CD' will now form a con- c' B' t Fig.l25 tinuous line going round the cylinder, called a helix. It is clear that in wrapping the rectangle round the cylinder, we have not altered the inclination to A'B' of any of the lines AB', BC, CD', so that the helix everywhere makes the angle B'AA' with the base of the cylinder. This angle is called the pitch of the helix ; and it is equal to _i AB tan" AA / J or tan _i distance between two consecutive coils {ah) circumference of the cylinder Now imagine a solid figure generated by a small rect- angle ahcd, which moves so that one side ad always coincides with a generatins^ line, while a corner a describes the helix, and its plane always contains the axis of the cylinder. Each point in ah will describe a helix, the pitch of the helix being smaller the further the point is from a : the distance between two consecutive coils will be the same for all, but the circumference of the cylinder round which any particular helix would wrap being greater the further the generating point is from a. This thread is called a square one : an angidar thread is sometimes generated by an isosceles triangle ahc, whose plane always contains the axis of the cylinder, and whose MACHINES. 251 base ah moves exactly as the side ad of the rectangle abed which generates the square thread. The solid cylinder, together with the solid figure above described, form a solid screw, which works in a hollow cylinder of the same diameter as the solid one, and with a groove cut in it, which just fits the thread of the solid screw. The hollow screw is generally fixed in a support. The screw is generally used as follows: — The solid screw has at one end an arm at riofht angles to the axis: the power P is applied at the end of this arm / rV u r Fig.I27 252 STATICS. and perpendicular to it, so as to tend to turn the screw round and so move it in the direction of its axis, and thus produce pressure on an}^ body situate at the end of the axis : the pressure which is thus overcome is called the weight (W). 146. To find the condition of equilibrium in a screw with a square thread. Let a be the leng^th of the arm, at the end of which the power P acts : h the distance between two consecu- tive threads. The surface of the groove of the hollow screw will exert pressures perpendicular to the surface of the thread at a very large number of points. Let R be the resistance at one of these points Q, which is at a distance r from the axis of the screw : we have seen that the pitch (a) of the helix, which passes through this point, and has as axis the axis of the screw, is tan~^ hj^irr. The direction of R is normal to the surface of the thread at Q, and therefore to any line in that surface, passing through Q. From the way in which the surface of the thread has been generated, R must be at right angles to the line from Q perpendicular to the axis of the screw ; it must also be at right angles to the tangent to the helix through Q, i.e. it makes an angle a with the axis of the screw. We ma}^ resolve R into two components, one along the axis of the screw, R cos a, and the other R sin a, perpendicular to the axis, and at a distance r from it. Similarly all the resistances, such as R, can be resolved in the same way. Resolving along the axis, we have W= t (R cos a). Taking moments about the axis, Pa = S (Rr sin a). But 27rr sin a = /i cos a ; -r, ^ fRh cos a\ h ^ ^T^ X Wh .-. Pa = t[ — = — S (E cos a) = ^— , MACHINES. 253 W 2'7ra " F h circumference of circle traced out by end of P's arm distance between two consecutive threads We can easily deduce the same relation for a screw with any smooth thread, square or angular, from the principle of virtual work, by a method similar to that used in Art. 155. Ex. A smooth screw makes three revolutions while it advances half an inch, find the power which must be applied at the extremity of an arm one foot long in order to produce a pressure of 144 lbs. Ans. '32 lbs. 147. When the screw thread is not smooth, we can find the condition of equilibrium, if we assume that the breadth of the thread, i.e. the side ah of the rectangfle which is supposed to generate it, is very small. The pitch of the screw will be the same then at every point of the thread : let it be a. Let us suppose that the power (P) is just on the point of moving the weight W, then the limiting friction is called into play at every point of contact of the thread with the groove, and acts in the direction in which it can most efficiently oppose P, i.e. directly opposite to that in which the point of the thread is about to move : it makes then an angle ^73 — "jl, with the axis of the screw, and is perpendicular to the line drawn from its point of applica- tion at right angles to the axis. Let \ be the angle of friction, R^, P.^, Pg ... the normal reactions at the different points of contact ; r the distance of each of them from the axis of the screw. Resolving along the axis of the screw, we have Tf - 2 (P cos a) + 2 (P tan \ sin y.) = 0, taking moments about the axis, Pa — % (Rr sin c<) — S (Rr tan \ cos a) = 0. 254 STATICS. _, . h COS a But r sm a — 27r ' W = (cos a - sin a tan X) 2 {R) = ^^ii^L+2^) 2 (R), cosX -^ //i cos a h tan X cos^ ^\ k* /-n\ and Pa = - ..^ 4- ^ ^ S (P) V 27r 27r sm a / ^ ^ _ /?, cot a sin (a + \) -o / p\ 27r " cos X ' P _ h tan(a-f X) W 27ra * tan a When W is on the point of overcoming P, the relation 1 P h tan (o — X) becomes .^^ = - — . ^ . W 27ra tan a Since the power (P^) which would just move W, when the screw is smooth is Wh/27ra, the efficiency of the rough screw is = tan a/tan (a + X). 148. The wedge, which is a solid prism, whose section is an isosceles triangle, and which is used to split wood, &c., by being driven in by blows of a hammer, is so essentially dynamical in principle, that we shall not discuss it here. 149. Besides its use as an instrument for multiplying Force, the Lever is employed for weighing purposes : in one form it is known as the Common Balance. This in its simplest form consists of a straight uniform beam AB, from the two ends of which scale-pans hang. The lever turns about a fulcrum C, which is situated above it in a short beam CD, which projects at right angles to AB, from its middle point P. The substance to be weighed is placed in one scale- pan, and such weights in the other, that the beam is horizontal when in equilibrium. MACHINES. 255 In well-constructed balances for accurate weighing the fulcrum is formed by the edge of a triangular prism of hardened steel (a knife-edge), ^Q'S ip-^s which rests on a plate of smooth agate. Hence (Art. 137) the effect of friction is rendered very small. Fig. 128, hke the other figures of the machines, is not intended as a realistic representation. It is assumed that the student is familiar with the actual forms of the simple machines. A good Balance should have the following 150. requisites. (1) It should be true, i.e. when loaded with equal weights, the beam should be horizontal. This requisite is obtained by making the scale-pans of equal weight, and the two arms exactly the same in weight, length and sectional area. We can easily test the Truth of a Balance by inter- changing the weights, which kee23 the beam in equili- brium, when horizontal : if the beam settles again into a horizontal position, the weights are equal and the balance true, but not otherwise. (2) A Balance should be sensible, i.e. when the weights differ by a small quantity the deviation of the beam from the horizontal should be easily perceptible. To ascertain how to secure this requisite, we must find the position of equilibrium when the balance is loaded with weights P and Q. 256 STATICS. Let G be the centre of mass of the lever, not includ- ing the scale-pans, W its weight. Let AB = 2a, CD = h, CG = k. Let >Sf be the weight of each scale-pan acting through A, B respectively. Then if 6 be the angle which AB makes with the horizontal when P is placed in the scale-pan hanging from A and Q in the other, we have, by taking moments about C for the equilibrium of the beam, (P -f 8) (a cos O-h sin ^) - (Q + S) (a cos 6 + h sin 6) -Wksmd = 0, •• ^^'^^ {PVQ + 2S)h+Wk' For a given value of P — Q, the sensibility will be the greater, the greater tan 6 is, and for a given value of 0, the sensibility is the greater, the smaller P — Q is, so tan fi that we may take p — ^ as a measure of the sensibility. Hence the second requisite is best obtained by making y^ — ^ — —^^—j ^j^ very large, i.e. by making a large in {P+Q+2S)h-\-Wk ^ ^ -^ * * comparison with h and k. (3) A good balance should be stable, i.e. it should readily return to its position of equilibrium, w^hen moved from it, i.e. its time of oscillation about its position of equilibrium should be small. It is shewn in works on Rimd Dvnamics that the time of oscillation is small when the arm a is small compared with h and k, so that the conditions of sensibility and stability are at variance one with another. In making a balance, however, consideration is paid to the sort of weighing it is required for. In scientific measurements, where the greatest accuracy is desired, the third requisite is sacrificed to obtain the second ; but for ordinary commercial purposes, where it is more necessary to save time than to be very accurate, the reverse is the case. MACHINES. 257 The stability is often measured by the sum of the moments of the forces which tend to bring back the beam into its position of equilibrium, but it is obvious that the time required to do this, and therefore the stability, will depend on the mass to be moved and on its shape, as well, 151. The Common Steelyard. This is a lever used as a balance, in which the necessity of keeping a number YC Yp Fig.129 of weisfhts is obviated. It consists of a straisjht beam AB, which is free to turn about a fulcrum C. The weight to be ascertained is placed in a scale-pan, which hangs from the end A. A fixed moveable weight slides along the beam, which is graduated so that the graduation at which the moveable weight is situate, when the beam rests in a horizontal position, gives the required weight. To shew how the graduations are obtained. Let P be the moveable weight, Q that of the beam and scale-pan, G the point of the beam through which Q acts. Let K be the position of the graduation n, i.e. the position P occupies when there is a weight nP in the scale-pan, and the beam balances in a horizontal position. Taking moments about C, we have nP.AC-Q.CG-P.GK = 0. Putting n — 0, in this equation, we get the position of the zero of the scale, CO = — ^ . CG, or is on the other side of C to G, and at a distance -yj . CG from it. G. 17 258 STATICS. Hence or nP.AC = P.OK, OK=nAG. The graduations are obtained then by marking off dis- tances from 0, equal to AC, 2 AC, SAC, &c. By giving n fractional values we can obtain intermediate graduations. 152. The Danish Bteelyard. This steelyard consists of a beam AB, terminating in a ball B ; from the end A hangs the scale-pan in Avhich the body to be weighed is placed. The fulcrum C is moved until the weight placed in the scale-pan is counterbalanced by that of the steelyard. The beam is graduated so that the position of C, when the beam balances, gives the corresponding weight in A. To obtain the graduations. Let P be the weight of the steelyard and scale-pan, 54 ^' Fig.130 <£) W acting through the point G of the steelyard. It is obvious that the zero graduation is at G, since the fulcrum must be at G, when the beam balances without any weight in the scale-pan. Let C be the position of the graduation n, i.e. the point where the fidcrum is when there is a weight nP in the scale-pan, and the beam balances. Taking moments about C, we have nP.AC=P.CG = P{AG- AG), AG AC = n + 1 MACHINES. 259 Hence the graduations are at a distance from A equal to AG AG AG 3 ' 4 &c. Ex. 1. If the beam of a balance be horizontal, when there are no weights in the scale-pans, shew that if the balance be a false one, the actual weight of a body is the geometric mean of its apparent weights when weighed first in one scale-pan, and then in the other. Ex. 2. If the arms of a false balance be without weight and one arm longer than the other by Jth part of the shorter arm, and if in using it the substance to be weighed is put as often into one scale as the other, shew that the seller loses I per cent, on his transactions. Ex. 3. If the bar of the common steelyard be 18 inches long, weigh 8 lbs. and be suspended at a point 3 inches from one extremity, what is the greatest weight which can be measured by a moveable weight of 2 lbs.? Ans. 16 lbs. Ex. 4. A common steelyard is 12 inches long, and with the scale-pan weighs 1 lb., the centre of gravity of the two being 2 inches from the end to which the scale-pan is attached ; find the position of the fulcrum when the moveable weight is 1 lb. and the greatest weight that can be ascer- tained by means of the steelyard is 12 lbs. Ans. 1 in. from scale-pan. Ex. 5. The moveable weight of a common steelyard is 6 oz. A tradesman diminishes its weight by half an ounce: of how much is a person defrauded who buys what appears to weigh 6 lbs. by this steelyard ? Ans. ^ oz. Ex. 6. Find the length of a Danish steelyard, weighing 1 lb., when the distance between the graduations 4 lbs. and 5 lbs. is 1 inch. Ans. 30 in. 153. RobervaVs Balance. This consists of four uniform rods, AB, BD, DC, CA, freely jointed at their extremities and forming a parallelogram. The rods AB, CD can turn about pivots at their middle points E, F, which are fixed in a vertical support. The rods AB, CD are similar in every respect, as are the rods AC, BD. Equal scale-pans are rigidly connected with AC and BD. The peculiar advantage of this balance is that it does not matter whereabouts the scale-pans the weights to be compared are placed. 17—2 260 STATICS. Let the weight P, when placed in the scale-pan at- tached to AG, counterbalance the weight Q placed in the other scale-pan. If now the system be supposed displaced by the beams AB, CD turning through a small angle, it is clear that the centres of mass of AB, CD suffer no displacement, while that of BD and its scale-pan is raised or lowered through a vertical distance p, say, and the centre of mass oi AC and its scale-pan is lowered or raised through the same distance. The virtual work done by the weight of BD will be equal to, but of opposite sign to, that done by the weight of AC. Also the algebraical sum of the virtual work done by the internal forces of the system is zero. The equation of virtual work is there- fore Pp — Qp = 0, since P, Q move through the same vertical distance as A C, and BD viz. p ; therefore P= Q. This result holds wherever P and Q are placed in their respective scale-pans, i.e. whatever be their distances from the vertical support. 154. The Differential Wheel and Axle. In order to raise a very large Aveight by means of a comparatively small power, with the help of the ordinary 'wheel and axle', it would be necessary to make either the radius of the wheel inconveniently large, or else that of the axle so small that it would be unable to bear the strain put upon it. This difficulty is got over in the 'Differ- MACHINES. 261 ential Wheel and Axle'. This consists of two axles B and C, of different radii, rigidly connected together and El h, nT 1^ D Fig. 132 iw turning about their common axis AE, which is horizontal and turns in fixed sockets. The power P is applied at right angles to the axis, and at the end of an arm AD, the 'wheel'; the weight W is attached to a pulley sup- ported by a rope which is wrapped one way round B, and the other way round C: P and the rope round the thicker axle B tend to turn the machine in opposite di- rections. To find the conditions of equilibrium. Let a, 6, c be the radii of AD, B, C respectively, and T the tension of the rope supporting the pulley. Since the pulley is in equilibrium 2T=W. Since the machine is in equilibrium, taking moments about the axis AE, we have Pa-Th + Tc = 0, /. Pa = T{h-c) = ^l(b-c\ or P: W=b-c: 2a. 262 STATICS. Hence by making the radii of J5 and G as nearly equal as we please, the weight which a given power P can raise, may be increased to any extent. The principle of work also enables us to obtain this result very easily. 155. Hunter s Differential Screiv. This consists of a screw AD which works in a fixed nut C0\ AD is hollow and has a thread cut inside it, in which a solid screw DE G 7 B D^ ~w G Fig.I33 works. DE is prevented from turning round by some means, for instance, by means of a rod FEE' rigidly con- nected with it, and whose ends work in smooth grooves, so that the screw DE can only move in a direction parallel to its axis. The weight W is the resistance exerted by any sub- stance placed between E and the base GG' of the frame- Avork CGG'C . The ])ower P is applied at the extremity of the arm AB which is at right angles to and rigidly connected with the screw AD. Let a be the length of AB, h, E the distances between consecutive threads of ^jD, DE respectively. Let us see the effect of the arm AB making a com- plete revolution. AD will clearly descend through a MACHINES. 263 distance li: DE cannot turn with AD, and therefore will move upwards relatively to J. i) through a space //, i.e. will actually descend through a space h — K : this is therefore the distance throuo^h which the wei c S (6) ; .'. 2 (a) = kl (6), since cS (6) is indefinitely small compared with k S (6). Cor. Hence two infinite series of indefinitely small quantities of the first order, such that each term of the one differs from the corresponding term of the other by a quantity of the second order, are equal. This explains why in Arts. 97, 99, 101 and 102 we have neglected one infinite series and retained another: APPENDIX. 269 this is done when the first series is of a higher order than the second. 3. As an illustration of these principles we will give a proof of Guldin's theorems. One theorem is, that the volume, generated by the com- plete revolution of a plane area about any straight line in its plane and not cutting it, is equal to that of a right cylinder whose section is the plane area and height the length of the path described by the centre of mass of the area. Draw a number of straight lines at right angles to the line AB, about which the revolution takes place, dividing the area >Si into n strips of equal breadth. Let Pp, Qq be two consecutive lines of this system, typical of the rest, M, N the points where they meet AB. Draw PR, pr perpendicular to Qq. The volume, generated by the revolution of PRrp about AB, differs from that generated by PQqp by the volumes generated by the two curvilinear triangles PQB, pqr. But when n is increased indefinitely, the breadth only of the rectangle is diminished indefinitely, whereas both length and breadth of each triangle is diminished in- definitely ; the volumes generated by the latter are there- fore of a higher order than that generated by the former. Hence the total volume generated by the area equals the sum of the volumes generated by the rectangles of which Pr is a type, i.e. = 2 {irPM' . MN- irpM^ . MN) = irt [{PM-pM) (PM + pM) MN] = irt [Pp . MN {PM + pM)]. Also the sum of the areas of the rectangles is the area of the figure 8\ and since they differ by the sum of the areas of the triangles PQR, pqr, &c., which are of a higher order than the rectano-les, the centres of mass of the sum of the rectangles and the figure S must be coincident. 270 STATICS. Therefore x^ the distance of the c. M. of >S' from AB _ ^{Pp. MN . I (PM + pM)] t{Fp. MN) _ J vol. generated by S TT X area S :. volume generated hy S = S . Iirx. The second theorem is, that the area of the surface, generated by the revolution of a curve about any straight line in its plane and not cutting the curve, is equal to the rectang^le, whose lensfth is the length of the curve and breadth the distance of the curve's centre of mass from the straight line. Let PQ be a side of a polygon, either inscribed within or circumscribed about the curve : let R be the middle point of PQ, and therefore its centre of mass. Draw RK perpen- dicular to the line AB, about which the curve revolves. As in Art. 99, it can be shewn that the area of the surface generated by the revolution of PQ about AB '\% ^ttFQ . RK. Therefore the total surface generated by the revolution of the polygon about AB = S (27rPQ . RK) = 27r2 (PQ . RK) = 2ttx X perimeter of polygon, where x is the distance of its centre of mass from AB. When the lengths of the sides of the inscribed and circumscribed polygons are diminished indefinitely and their number increased indefinitely, their perimeters differ by indefinitely small quantities, and their centres of mass become coincident. The surfaces generated by each are therefore equal. It is assumed as axiomatic, that as the perimeter of the curve lies in position between the two polygons which ultimately coincide, it is equal to the perimeter of either APPENDIX. 271 polygon, its centre of mass coincides with that of either polygon, and the surface generated by it is equal to that generated by either polygon. Hence the surface generated by the curve is equal to the product of the length of the curve into the length of the path traced out by its centre of mass. Each of Guldin's theorems can easily be extended to the case in which the revolution is not a complete one. There is no limitation in either as to the number of times, in which a straight line at right angles to ^i^ cuts the generating curve. Ex. Find the volume and surface of an anchor-ring, the figure generated by the revolution of a circle about a line in its plane, and not intersecting it. Alls. Vol. = 27r-a-c, surface = 47r'ac, where a is the radius of the circle, and c the distance of its centre from the line. 4. To prove that the limit of F+ 2^ + 3^+... {n-iy _ 1 ^' ~p+l ' where p is any positive quantity, and n is increased in- definitely. Let S^ denote 1^ + 2^ + 8" + ... {n - ly. j,p-^ _ („, _ ly-^^ = ^p + 1) (^n _ 1)^ + (£±i]i^(,, _l)p-i+ &c. (n - ly^' - {n - 2y^' = [p + 1) {n - 2)^ .•. by addition {p+\)p {p + l) p( p-l) ^ 3! 272 APPENDIX. p+i ' ..p+i + p + 1 (p+1) 7f^' n £ 1 S^_, p(p-l) 1 ^2,^^ But — Jj is obviously < -^ -^ — , i.e. is < 1. Similarly ^.^^.^ is < 1, if jt? is > r. it Hence, when p is integral, P + 1 rf^^ n n^ n^ '" 'nF^'^ ' where A^, A^y A^, &c. are all finite quantities; •*• ^1 = ^ J when n is increased indefinitely. If p be fractional S -^ = ultimately, when ^ is > 7\ it whenp is ^o MAY 1 19 63 16Niaf J^"" •" MAR a ^953. ISJul'SSJL r-* '«-*?-?« f-s 4yt 4r-t ^9it l^C- ''V\ r^ 1 t;l.