546 Elementary mechanics orsolids and fluid 3 1924 001 099 518 H Cornell University B Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924001099518 ELEMENTARY MECHANICS OP SOLIDS AND FLUIDS SELBY JSonoon HENRY FROWDE Oxford University Press Warehouse Amen Corner, E.C. (Jtew 2M MACMILLAN & CO., 112 FOURTH AVENUE ELEMENTARY MECHANICS OF SOLIDS AND FLUIDS A. L. SELBY, M.A. FELLOW OP MERTON COLLEGE AT THE CLAKENDON PRESS 1893 ©xfotb PRINTED AT THE CLARENDON PRESS BY HORACE HART, PRINTER TO THE UNIVERSITY PKEFACE I hope that this book may prove useful to students, who without possessing much knowledge of Mathematics desire to read Mechanics as an introduction to Physics. Those who are acquainted with the elements of Algebra and Geometry will, I think, meet with no serious mathe- matical difficulty ; they may, however, find it convenient to read first the geometrical theorems at the end of Chapter I. Some propositions on the geometry of the ellipse are given in Chapter VI, in order to render the discussion of the law of gravitation more complete. My thanks are due to Mr. J. Walker, M.A., of Christ Church, for his kindness in reading the proof-sheets, and to Professor J. V. Jones, M.A., for many valuable suggestions. A. L. S. CONTENTS CHAPTEK I. KINEMATICS. Measurement of Time and Distance Definition and measurement of displacement Composition and resolution of displacements Displacement of an extended body in one plane Definition and measurement of velocity. Diagrams Composition and resolution of velocities Examples ...... Angular motion ..... Definition and measurement of acceleration The Hodograph. Uniform circular motion Rectilinear motion with uniform acceleration Examples Motion with uniform acceleration Composition and resolution of accelerations Simple harmonic motion Examples Appendix. Elementary Results in Geometry 2 3 6 IO, 12 19 21 22 24 25 27 28 3° 38 40 41 42 CHAPTEK II. THE LAWS OF MOTION. Law I. — Property of inertia and measurement of time Law II. — Discussion of the law Definition of mass. Units of mass and force Definitions of momentum and impulse . Parallelogram of forces 50 52 54 55 59 viii Contents. PAGE Law III. — Consideration of force as stress 63 Centre of mass of two particles 6 4 Centripetal force 67 Illustrations. — Watt's Governor. Time of vibration of simple pendulum 69 Atwood's machine 7 2 Illustrations and examples 74 CHAPTEK III. WOKK AND ENERGY. Work. Power 82 Investigation of relation between kinetic energy acquired and work done by (as) constant, (6) variable forces ... 85 Work done by an impulse 93 Work done by the mutual actions of a system of particles . . 93 Conservation of energy ........ 97 Impact of spheres ... 104 Illustrations and examples 108 CHAPTEE IV. MOTIONS OF AN EXTENDED BODY. Properties of the centre of mass of two or more particles . .112 Moments of inertia 118 Rotation of a body 122 Resultant of parallel forces .124 Couples 126 Centres of mass of a triangle and tetrahedron . . . .129 The compound pendulum 132 Atwood's machine 135 The ballistic pendulum 136 Examples 138 Contents. IX CHAPTEE V. MACHINES. PROBLEMS IN STATICS. Machines treated by the principle of work The lever .... The balance .... Pullies .... The wheel and axle The inclined plane . The screw .... Principle of virtual work Stable and unstable equilibrium Friction Examples .... PAGE I40 141 142 149 155 156 157 159 l6l I63 171 CHAPTEE VI. GRAVITATION. Elementary geometry of the ellipse 178 Kepler's laws 185 The law of gravitation 186 Correction of Kepler's third law 189 Comparison of the masses of the planets 191 Work done by gravitational force 193 Equilibrium of a particle within a thin attracting spherical shell . 195 Potential due to a sphere at an external point . . . .196 Attraction of a solid uniform sphere 198 Mutual attraction of two spheres . 200 Terrestrial gravitation 201 Variation of g at the Earth's surface 202 Variation of g within the Earth . 203 Jolly's experiment 204 Contents. CHAPTER VII. ELASTICITY. PAGE Internal stresses of a body which is subject to external force . 207 Plane strain. Uniform extension and shearing strain . . 208 Strain of a solid body ... 218 Shearing stress and hydrostatic stress 222 Hooke's law 224 Young's modulus ... 225 Torsion ... 227 The torsion balance 229 Flexure ... 231 Work done by strain 234 CHAPTER VIII. HYDEOSTATICS. Nature of pressure in a fluid 237 Variation of pressure in a heavy fluid at rest . Equilibrium of liquids in a TJ tube. The siphon . . . . Proof of Archimedes' principle. Experimental verification of it . Measurement of density of solids and liquids by various methods Determination of the total pressure and resultant pressure on plane surfaces immersed in a liquid The centre of pressure. Examples Transmission of pressure in fluids Bramah's press .... Gases. Torricelli's experiment Fortin's barometer Boyle's law. Closed manometers . Construction and mode of action of pumps . Examples 238 240 241 243 246 247 251 253 256 258 259 262 267 Contents. xi CHAPTEE IX. CAPILLARITY. PAGE Surface energy of fluids ........ 273 Connection between the curvature and the difference of the pres- sures on the surfaces of a film. Application to mercury drops and soap bubbles 275 Conditions of contact of three fluids . . ... 277 Elevation of liquid in tubes 278 Elevation of liquid between two plates 280 Illustrations and Examples .... . . 281 CHAPTEE X. UNITS AND THEIR DIMENSIONS. Dimensions of units 284 Application of the principle of dynamical similarity to the solution of problems . . . ... 289 Answers to Examples ... . ... 295 ELEMENTARY MECHANICS. CHAPTER I. Kinematics. § 1. The Science of Mechanics investigates the con- ditions which govern the motion of bodies. It comprises two parts, Kinematics and Kinetics. The object of Kinematics is to describe and classify the changes in the motions of moving bodies. It is the province of Kinetics to assign the causes of these changes. A body may remain at rest, or in equilibrium, either from the absence of causes which tend to move it, or because such causes balance one another. The part of Mechanics which treats of the conditions under which bodies remain at rest, is a branch of Kinetics and is called Statics. As we cannot hope to explain the phenomena of motion before we know how to classify them, we must first occupy ourselves with Kinematics. We shall therefore discuss in this chapter the simple phenomena of motion without reference to their causes. The form and size of moving bodies will be supposed to remain unchanged during motion. § 2. Measurement of Time and Length. Times and lengths (or distances) are expressed as mul- tiples of convenient units. B 2 Kinematics. [cb. i. The unit of time is generally the second. The unit of length is the centimetre or the foot. A time 5 means 5 seconds ; a length 10 is 10 centimetres or 10 feet ac- cording as the centimetre or foot is the unit. The centimetre (0-3937079 inch) is the International Scientific Unit ; the foot is only used in Britain. Other units might be chosen, as the inch and the minute. Expressed in terms of these units, 5 seconds and 10 feet would be denoted by T V and 120 respectively. Derived Units. Square Measure. Cubic Measure. The most convenient unit of area is the area of the square on the unit of length, for the area of an oblong of length a and breadth b is denoted by ab, when measured in terms of this unit. Similarly the unit of volume is the volume of a cube, each edge of which is of unit length. As the units of area and volume depend on the unit of length, they are called derived units. The centimetre, square centimetre, and cubic centimetre, are denoted by cm., sq.cm., and c.cm. § 3. Displacement. , , When a body is dis- A E c B D pi ace( i so that a point of lg ' 1- it is moved from A to B, the change of position of the point is described by saying that it has a displacement AB. The displacement is only completely described when the magnitude and direction of AB are given. The magnitude of the displacement is measured by the number of units of length in AB. If the direction of the displacement is not also given, B may be anywhere on a sphere with centre A and radius AB. Quantities, such as displacement and velocity, which are Ch. i.] Composition of Displacements. 3 only completely defined when their magnitude and direction are both known, are called Vectors. A vector whose magnitude is AB, and whose direction is that of the line from A to B, can be represented by this line, and is called the vector AB. Equal vectors are those which are represented by equal and parallel straight lines drawn towards the same parts. If a point has successive displacements in the same direc- tion the total displacement is the sum of these displacements. If the displacements are all along the same straight line, as AB, BC, CD, BB, some in one direction some in the other, displacements in opposite directions must be given opposite signs, and the algebraic sum of all is the total displacement in the positive direction. For example, displacements of 3 cm. and 4 cm. to the right, combined with a displacement of 8 cm. to the left, make a displacement of — 1 to the right, or 1 to the left. A body is said to have a displacement of translation, when every point of it has the same displacement. § 4. Composition of displacements. Let successive displacements AB, BC be given to a body ; then since a point originally at A is- finally at C, the total displacement is AC. The final position of the body displaced depends only on the mag- nitude and direction of the displace- ments AB, BC, and not on the order in which they take place. Indeed we may break up each displacement into parts, as AB, FF, FB, and BG, GH, HC, and communicate them to the body in any order we please ; the result is still to produce the same displacement AC. AC is called the resultant displacement, and AB, BC are b a 4 Kinematics. [Ch. i. called its components. The rale for finding 1 the resultant displacement when the components are given is called the Parallelogram Law, and is as follows. Let a parallelogram be constructed whose adjacent sides, AB, AD, represent the magnitudes and directions of the component displacements. The diagonal AC, which is drawn from the angle in which these sides meet, represents the magnitude and direction of the resultant displacement. Conversely, a displacement AC may be replaced by dis- placements AB, BC or AB, AD, where AB is any straight line through A. AC is then said to be resolved into its components AB, BC or AB, AD. The most useful case of resolution is that in which ADC is a right angle. In this case if AC = R and Z GAD = a, the component displacements x, y, along AD, DC respectively, are B cos a, B sin a; and x 2 +y 2 = B 2 ; y = x tan a. (i) Polygon of Displacements . If a body has several successive displacements AB, BC, CD, DE, not necessarily all in the same plane, the resultant displacement is AE ; and if the lines representing the dis- placements form a closed figure, the resultant displacement is zero. This proposition is called K A the Polygon of displacements. The magnitude and direc- tion of the resultant of several displacements, which are all in the same plane, is best found as follows. Ch. i.] Relative Displacement. 5 Take two convenient lines OX, OY in the plane perpen- dicular to one another, and resolve each displacement into its components parallel to these. Then if x be the sum of the components parallel to OX, and y the sum of the components parallel to OY, the mag- nitude of the resultant displacement is Vse^+y* and the angle a which it makes with OX is given by tan o = Relative displacement. First, let two points A and P have equal displace- ments AP, PQ. Then by Euclid I. 33, PQ is equal and parallel to AP, and since there is no change in the length or direction of the line joining the points, each is said to have no displacement relatively to the other. Next, let the displacements be AP, PR. Resolve PR into PQ and QR, of which PQ is equal to AP. QR is called the displacement of P relatively to A ; it is compounded of QP, PR — that is, of the displacement of P, and the reversed displacement of A. Also QR and PQ compound into PR. Therefore the displacement of P relatively to A, and the original vector from A to P, compound into the vector from the final position of A to that of P. The displacement of A relatively to P is equal in magni- tude but opposite in direction to the displacement of P relatively to A. If the displacements AP, PR are along the same straight line, the displacement of P relatively to A is the algebraic 6 Kinematics. [ch. i. difference of the two displacements ; it is also the algebraic sum of the displacement of B and the reversed displacement of A. If we call the resultant of two vectors their vec- torial sum, we may appropriately use the term vectorial difference to denote the resultant obtained when one vector is reversed. When the two displacements are in the same straight line, the algebraic difference is the vectorial difference. Hence in every case the displacement of B relatively to A is the vectorial difference of the displacements of B and A. § 5. Displacement of an extended body. The displacements of an extended body may be of a very complex kind ; there may be no simple relation between the displacements of its points. The two simplest cases of displacement are (i) Displacement of translation ; here all points of the body have the same displacement both in magnitude and direction, and the displacement of any point may be considered as that of the body. (a) Displacement of rotation ; here all points on one straight line (either in the body or fixed relatively to it) * remain fixed, and the displacement can be performed by rotation about this line as axis. Kg- 5. If the body is a plane figure which rotates in its plane, * A point is said to be fixed relatively to a body when its distances from the several points of the body do not change. The straight line joining two such points is said to be fixed relatively to the body. ch. i.] Displacement in a Plane. 7 one point in the plane remains fixed, and the displacement may be described as rotation round this point. Any displacement of a ■plane figure of invariable form in its own plane can he made either by a motion of translation or by a motion of rotation round an axis perpendicular to the plane. Let A, B be the positions of two points of the figure before displacement; C,B their displaced positions: then the final position of any other point of the figure can be found. Join AC and BB and bisect them at -Sand F. Draw BO and FO at right angles to AC and BB re- spectively. They will meet at unless they are parallel, i. e. unless AC and BB are parallel. But if AC and BB are parallel, the displacement can be made by a motion parallel to AC. If AC and BB are not parallel, join OA, OB, OC, OB. Then since AB = EC and BO is perpendicular to AC, OA = OC. Similarly OB = OB. Also in the triangles AOB, COB, the two sides AO, OB are equal to the two sides CO, OB. And AB = CB. Therefore the angle AOB is equal to the angle COB. And taking away the angle COB, the angles AOC, BOB are equal. Therefore by rotation about through an angle AOC, the point A is displaced to C, and B to B. Since all lines in the plane are turned through the angle AOC, AOC is called the angular displacement. Note. — It is supposed that in the displacement, the figure never quits the plane. A displacement such as that which a leaf of a book undergoes when it is turned over is excluded. 8 Kinematics. [Ch. i. Two equal successive angular displacements of a body in opposite directions about parallel axes are equivalent to a displacement of translation. Let the axes be perpendicular to the paper, meeting it in A and B. If 8 be the angular displacement, any line perpen- dicular to the axes is first A ""----^_ J turned through an angle 6 and then through an angle B-"-- — 0. Hence it is parallel to its original position. The displacement of translation is equal to BC the 6 displacement of B. If AB = I, BC = 0,1 sin - • This result may also be stated thus : An angular displacement 6 about any axis can be re- placed by an equal angular displacement about a parallel axis, at distance I, together with a displacement of transla- a tion 2 1 sin - equally inclined to the initial and final positions of the shortest distance between the axes. Examples. 1. Find the resultant of the following displacements, 3 N, 5 E, 6S, and9"W. 3 N and 6 S are equivalent to 3 S, and 5 E and 9 W are equivalent to 4 W. Therefore if B be the resultant displacement, its direction lies between S and W and the magnitude of B isv' 3 2 + 4 2 or 5. } = tan a where a is the angle which B makes with the W. An angle a whose tangent is x is often denoted by tan -1 a;. Thus here a = tan -1 f. The resultant can also be constructed on a diagram by drawing AB, BC, CD, DE to represent the magnitudes and directions of the given displacements. ^!i?then represents the resultant. Ch. i.] Examples of Displacement. g 2. Find the magnitude and direction of the resultant of the displacements 16 SW, 4 V 2 N, and 9 SE. The most convenient directions in which to resolve are SW, and SE. 4 V 2 N resolves into 4 NW and 4 NE, or into -4 SE and -4SW. Therefore the resultant displacement E is compounded of 16-4 or 12 SW, and 9-4 or 5 SE. Therefore E = V 12 2 + 5 2 = 13. The direction of E lies between SW and SE, making an angle tan- 1 A with the SW. E may also he constructed on a diagram as in Ex. 1. 3. Find the resultant of the displacements 30 N, 6V 2 SE, and 1 W. 4. A point has successive displacements 5, 10, 10 parallel to the sides AB, BC, CA. of an equilateral triangle ABC taken in order ; find the direction and magnitude of the resultant displacement. Let AD be perpendicular to BC, and resolve along DA and DB. r^/3 S along AB resolves into - — - along AD and - along DB. 10 */ ^ 10 along CA resolves into along DA and 5 along CD. Therefore the components are 2 2 2 5 And - + 5 - 10= - * , along DB or ~ , along DC. The resultant is V (-) + (^rV or 5- And it makes an angle tan -1 V3 or 60° with DC. Therefore the resultant displacement is 5 parallel to BA. io Kinematics. [Ch. i. We can also find the resultant as follows ; displacements io, io, io have a resultant zero ; and therefore the given displacements have a resultant — 5 parallel to AB or 5 parallel to BA. 5. ABCD is a square, and is the intersection of its diagonals. Find the resultant of displacements V2, 2, 4, 4, 5 V2 respectively parallel to OA, AB, BC, CD, DO. 6. Find the resultant of displacements 1, 8, 3, 4, 5, 6 parallel to the sides of a regular hexagon taken in order. 7. The resultant of two equal displacements is equally inclined to each of them. 8. ABC is an isosceles triangle right-angled at C. Find the resul- tant of three equal displacements a respectively parallel (1) to AB, BC, CA ; (2) to AB, BC, AC. 9. D, E,F a.re the middle points of the sides BC, CA, AB of any triangle ABC. Prove that the resultant of displacements equal and parallel to AD, BE, CF is zero. 10. From a point within a triangle ABC, straight lines OD, OE, OF are drawn, which are perpendicular in direction and proportional in length to BC, CA, AB respectively. Show that the resultant of displacements represented by OD, OE, OF is zero. § 6. Velocity. The displacement of a point from AtoB can only take place by motion along a continuous path between A and B, and must occupy time. The time occupied along the path will be longer or shorter according as the moving point travels more or less quickly ; hence the rate at which a displacement proceeds is an important matter. To take the simplest case, let the path be a straight line along which the moving point travels equal distances in equal times, however small. The rate of displacement of the point is called its Velocity, and is measured by the distance travelled in unit time. Ch. i.] Velocity. n If * be the distance travelled in time t, - is the distance travelled in unit time. Therefore if v he the velocity, s v = r It has been noticed that * means * units of length ; similarly v means v times a particular velocity which is called the unit. Making s = i, and t = i, we have v = i. Therefore the unit of velocity is the velocity of a point which moves through unit distance in unit time. Since this unit is determined when the units of length and time are known, it is a derived unit. A velocity, like a displacement, is only fully represented when its direction as well as its magnitude is known. It is therefore a vector. The formula s = vt applies to motion with velocity of constant magnitude, whether the path is straight or curved, provided that * is measured along the path. Examples. 1. Express a velocity of 30 miles an hour in feet per second. 30 miles = 30 x 5280 feet. 1 tour = 3600 seconds. Therefore v = - — -A = 44 feet per second. 3600 2. A train travels 32J miles an hour. How far does it go in 50 seconds ? 3. A cyclist rides at a rate of 25 feet per second. How long does he take to cover a mile ? When a body moves over unequal distances in equal times, the magnitude of its velocity varies. 12 Kinematics. [Ch. i. The conception of quantities which vary with the time is very important in Physics. It is first necessary to explain the term ' instant of time.' An instant of time is analogous to a point of space ; as a point has no magnitude, so an instant has no duration ; e. g. if a circular disc be divided along a diameter into semi- circles coloured black and white respectively, and a pointer of the form of a sector of the disc revolve continuously round its centre, there is a definite instant at which the forward edge of the pointer passes from the black to the white part of the disc, and a subsequent instant at which the hinder edge does the same. These instants are separ- ated by an interval of time to which we must attach the idea of magnitude. No such idea is attached to an instant. § 7. Diagrams. Let there be some physical quantity, as the height of Q. x' y' Pig. 8. Or M N a barometer or the temperature of a room, which at any Ch. i.] Diagrammatic Representation. 13 instant has a definite magnitude, but may vary from time to time ; the law of variation of such a quantity may be indicated on a diagram. Draw two perpendicular lines x'Ox, y'Oy ; and from any point P draw PM parallel to Oy. OM is called the abscissa, PM the ordinate of P. If a point lies above x'Ox, its ordinate is considered positive; if below x'Ox, the ordinate is negative. The abscissa of a point is positive or negative, according as the point is to the right or to the left of y'Oy. The position of any point is known when the magnitude and sign of the ordinate and abscissa are given. For let the abscissa and ordinate be a, — b, respectively. Along Ox take ON equal to a, and draw NQ 2 downwards equal to b. Then Q 2 is the point required. If Qi 0,2 $3 Qt ^ e a rectangle with its sides parallel to Ox, Oy, and its diagonals pass through 0, Q 1 N= NQ 2 and Q 2 L = BQ S . A point of which the ordinate and abscissa are x, y, is often denoted by (x, y). The points Q L , Q 3 , Q 4 , are respectively (a,b),(-a,-b),(-a,b). Let there be two quantities A and B, which are mutually dependent, so that A varies when B does, and A can be found when B is known. Take any number of points a,b,c,... along Ox ; Oa, Ob, Oc, ... being possible values of A. Through a, b, c ... draw ap, bq, cr ... parallel to Oy, and equal respectively to the values which B has when A has the values Oa, Ob, Oc, .... Then there is a curve passing through p, q, r, ... such that 14 fLinemancs. l^h. i. the abscissa of any point of it gives the value of A, and the ordinate the corresponding value of B ; and when this curve is drawn the relation between A and B is completely expressed. A familiar instance of such a curve is given by a x' /U. a b c y Fig. 9. barometer chart ; the abscissae measured in divisions on the paper represent the number of hours or days that have elapsed since a given instant, and the ordinates represent on a convenient scale the corresponding heights of the barometer. If A is negative, the curve will lie on the left of yOy', above Oaf when B is positive, below Ox' where B is negative. If A is positive and B negative, the curve lies within the angle xOy 1 . § 8. Application to velocity. Curve of Positions. Let a diagram be drawn on which the abscissae are the times reckoned from a certain instant which is called the Epoch, and the corresponding ordinates are the distances traversed by a moving point in these times. Thus in fig. 10 let PM, QN denote the distances traversed Ch. I.] Curve of Positions. 15 in the times OM, ON, by a point. The curve BPQ re- presenting the motion of this point is called the curve of Positions. rig. 10. Draw PL parallel to Ox, and produce QP to meet Ox in E. Draw PF to touch the curve at P. QL MN Then QL is the distance traversed in time MN, and is the average velocity during this time, QL _QL_PM_ PM NM~ LP ~~ ME ~ MF-EF' Now let Q move along the curve towards P, and ulti- mately coincide with it ; then QP tends more and more to coincide with the tangent at P. EF But as Q approaches P, , , diminishes indefinitely, and vanishes when Q coincides with P. Thus when the time MN is diminished indefinitely, the average velocity approaches and ultimately has the value PM M p , for it differs from this by less than any assigned quantity. 16 Kinematics. [ch. i. PM -^tfjp is therefore said to be the velocity of the moving point at the instant M, or at a time OM reckoned from the Epoch. Measurement of Variable Velocity. Thus the magnitude and direction of a variable velocity at any instant, are the magnitude and direction of the average velocity during a time t immediately succeeding that instant, when t is in- definitely diminished. The speed of an express train when passing a point P could not be estimated by the average speed between two stopping-places. If the train were timed over the next mile beyond P, the speed at P would sometimes be determined with fair ac- curacy, sometimes not. But if the time taken to traverse, a yard or foot from P could be accurately found, this would give the velocity at P with great accuracy, for in so short a time the velocity would not vary much. Our system of measuring variable velocity is devised on this plan, but the time of the measured motion is made ex- ceedingly small, and the accuracy of measurement can be increased as much as we please by diminishing this time. The curve of positions has been drawn as a continuous curve with no sharp points ; this covers all cases that we need consider. The tangent to the curve of positions is only parallel to Ox for instants when the point is at rest. When the velocity is uniform and equal to v, the curve of positions is a straight line inclined to Ox, at an angle tan _1 ». § 9. Curve ofVelocity. Let us now take another diagram on which the abscissae represent times as before, but the ordinates are the corres- ponding velocities of a moving point. Any given motion Ch. I.] Curve of Velocity. 17 Pig. 11. of a point will be represented by a curve which we shall call the curve of velocity. If the velocity is uniform and equal to v, the curve is a straight line PQ parallel to Ox at a distance v from it. Let^f and B denote instants of time separated by an in- terval t. Then, since the velocity is uniform, the dis- tance travelled in time t is vt or AP. AB, that is, the area PABQ contained between the axis of*, the curve of velocity, and the ordinates at A and B. The same statement holds when the velocity varies in any manner. For let MN be the curve of velocity, and let AB (fig. 1 a) be divided into any number of parts such as Aa, ab, be, cB, each equal to cl. Through A, a, b, c, B draw ordinates meeting the curve in M, m, n, p, N. Complete the rectangles Mm, mn, np, pN, by parallels to Ox produced to meet BNinf,g,h,L The rectangles vf, Mm are equal, for they are on equal bases and between the same parallels. For the same rea- son the rectangles mn, Iff are equal, and so are pn, kl. Therefore the rectangle fq whose base is d and altitude y p 7 f k > h / n I S f / m V M c ) / V 1 I 7 c ■ I i * Fig. 12. i8 Kinematics. LUH. BN—AM is equal to the sum of the rectangles Mm, mn, np, and pN, whose diagonals are the successive chords between M and N. This statement holds, whatever be the number of parts into which AB is divided. Now during the time Aa, the velocity is less than am and greater than AM, and therefore the distance traversed is less than Am and greater than aM. Similarly in the second interval ah, the distance traversed lies between bm and an. So for all succeeding intervals. Therefore the distance traversed in the whole time AB is less than the sum of the areas Am, an, bp, cN, and greater than the sum of aM, bm, en, Bp. And the area bounded by the curve MN, the axis Ox, and the ordinates AM, BN lies within the same limits. Therefore the distance traversed and the area AMNBA lie within limits which differ by fq or by d (BN— AM). Now by increasing the number of equal intervals be- tween A and B, d can be diminished indefinitely, and d (BN— AM) can be made less than any assigned quantity. Therefore the distance traversed and the area described both lie between limits which can be made to differ by less than any assigned quantity. Therefore the area AMNBA re- presents the distance traversed. Uniformly increasing Velocity. Let the velocity increase uni- formly from v to v in a time t. T A x Make OQ and OA equal to v and Fig. 13- t respectively, and through A draw AMN parallel to Oy, and make AN = v. Join QN, and from any point P on QN dra.w PDL parallel to Oy. Draw OM parallel to QN Ch. i.] Composition of Velocities. ig Let OL=co. Then PL = PD + DL = OQ + ^AM = v + % (v - vX t t Now - (v — v ) is the velocity acquired in a time as and V v is the initial velocity. Therefore PL represents the velocity after a time on, and P is any point on QN. Therefore QN is the velocity- curve. The area OQNA = \OA (OQ + NA) = \ (v + v) t. Therefore * = \ (v + v) t. Whence * is the distance travelled in time t. If the body starts from rest v = o, and * = i vt, where v is the final velocity. § 10. Composition and Resolution of Velocities. Let PQ represent the velocity of a moving point. BrsmPB, QB parallel to two lines Ox, Oy, and meeting at B. Then PB, BQ are the components parallel to One, Oy of the displace- ment in unit time. They are therefore called the components of the velo- city along One and Oy ; Fig. 14. and if PB, BQ are known, the actual or resultant velocity can be obtained by compounding them according to the parallelogram law (§ 4). If iv Oy is a right angle, R the magnitude of the velocity c a 20 Kinematics. [Ch. i. PQ, a the angle between PQ and Ox, X and Y the com- ponents along Ox and Oy, X = R cos a, Y= R sin a, R 2 = X 2 + Y 2 , tan a = J- Relative Velocity. If two points J and 5 are moving with the same velocity, the line AB remains unchanged in magnitude and direction ; B is then said to have no velocity relatively to A. Next, let the velocities be different, AP and BR (fig. 15) representing the displacements in unit time. Then QR is the displacement of B relatively to A in unit ' time. This is called the velocity of B relatively to A, and is obtained by compounding the velocity of B with that of A reversed. The velocity of B is obtained by compounding the velo- ,R city of B relatively to A with the velocity of A. Similarly, if there are three moving points A, B, C, the velocity of C is ~B compounded of the velo- Fig. 15- city of C relatively to B, the velocity of B relatively to A, and the absolute velocity of A. Thus velocities are compounded according to the same law as displacements. The idea of relative motion is of great importance in Mechanics, for we do not know any fixed point in space, or the absolute velocity of any point. The true velocity of a falling stone when it begins its course is not merely that which the thrower imparts to it; with this there Ch. i.] Composition of Velocities. 21 are compounded the velocity of the point of projection round the earth's axis and the velocity of the earth's centre as it travels through space. The motions of any number of points relatively to one another are not affected if their actual velocities are com- pounded with an additional velocity which is the same for all points. A body is said to have a velocity of translation when the velocities of all points of it are the same ; the velocity of any point of it is then the velocity of the body. Examples. 1. A steamer goes 7^2" miles an hour N. in a S.E. wind blowing 6 miles an hour. Prom what direction and with what velocity does the wind appear to blow to a passenger on the steamer. Impress a velocity of 7 -/2 miles an hour S. on both the wind and the steamer. The relative motion is unaltered, the steamer is brought to rest, and we have to compound velocities 6 N.W. and 7 V2 S. supposing the mile and hour are units. 6 N.W. = 3 Vz N. and 3 V2 W. Therefore the relative velocity is compounded of (7 - 3) V~2 S. and 3 «/2 W. Therefore the velocity required is 5 V~2, directed along a line OM in the S.W. quadrant which makes an angle tan -1 f with OW. The wind then appears to blow almost from the N.E. 2. A who walks at a rate of 3 miles an hour starts 13 minutes before B, who walks on the same road at a rate of 4 miles an hour. When will B overtake A ? Impress a speed of — 3 miles per hour on both A and B. Their relative motion is not affected. Also in 13 minutes A walked £§ mile. Hence the question becomes — How long will B take to walk £§ mile at the rate of one mile an hour ? 3. The extremities P and Q of the hour and minute hands of a clock are distant 2 and 4 inches respectively from the centre of 22 Kinematics. [Ch. i. the face. Find the velocity of Q relatively to P at 12, and at 3 o'clock, in foot-second units. Let x and y be the velocities of P and Q. 27T The point Q traverses a distance — in 3600 seconds. Therefore y = — = '000146. 7 x 3 x 3600 , , 22 x 2 1 And x = -: — = — y. 7x6x12x3600 24 The velocity of Q relatively to P is found by compounding y and —x. Therefore the required velocity at 12 o'clock is ff y. At 3 o'clock x is directed vertically downwards. Hence - x is vertically upward. And if B be the required velocity at 3 o'clock, a the angle it makes with the horizon, T, r .. / ! -000146 . R = -000146 V 1 + — , = =- V 577, 24 2 24 ■"" tan a = — . 24 4. In the last example, find the velocity of Q relatively to P at 1.30 and at 10.30. 5. A boat is rowed at the rate of 4 miles an hour across a river, a quarter of a mile broad, being steered in a direction at right angles to the current. If the velocity of the stream is 3 miles an hour, find the resultant velocity of the boat, and the distance below the starting point at which it reaches the opposite side. Angular Motion. Defini- tion of a Particle. Let a body, free to turn £ about an axis through per- pendicular to the plane of the paper, be displaced so that a straight line in it Fi s- l6 - which initially coincides with OA finally coincides with OB. Ch. I.] Angular Motion. Let LAOB- - d, AO = r, arc AB = Then <* = -■ r 2 3 If the body turns through equal angles in equal times, it is said to have uniform angular velocity. If t is the time occupied in the displacement d, - t is the angular displacement in unit time, and is called the angular velocity. Denoting this by o> we have d = tat. . , rd s And no = — = -. & Therefore if v is the velocity of a point A, distant r from the axis, V = TO). When the angular velocity is variable, the displacement d, by which it is measured, must be made very small. If the angular velocity increases uniformly from m to to in a time t, the velocity of A increases uniformly from «o to ra>, and (§ 9) the point A traverses a distance \ (ra> + ru>) t. Therefore the angular displacement d in a time t is given by d = £(u Therefore the acceleration at P is v 2 . I cf 1 IP — > and it is perpendicular to cp, that is, it is directed along PC. § 12. Rectilinear motion with uni- form acceleration. If v be the initial velocity, v the velocity acquired in a time t by a point which has uniform acceleration a, we have by definition v — v n Fig. 21. or v = v + at. (1) Again, it has been shown that if s be the distance traversed, s = i(v +v)t. (2) 28 Kinematics. [ch. i. Substituting in (2) the value of v given by (1) we have s = v t + i at 2 . (3) But we can also write (l) in the form v = v — at, and substituting this value of v in (2) we have s = vt-\at 2 . (4) Again, by (1) t = -> and substituting this value of t in (2) we have 2as = v 2 — v 2 . (5) Each of the equations (1), (2), (3), (4), (5) contains a relation between four of the five quantities s, v Q , v, a,t. When any three of these quantities are given, the other two can be deduced from two of the equations. Thus if s, a, t are given, v can be found directly from (3) and v from (4) ; and if v, v Q , and * are given, a can be found from (5) and t from (2). The velocity and acceleration must be measured in terms of the units which are derived from the chosen units of length and time. A velocity and an acceleration, which are denoted by opposite signs, have opposite directions. Examples on uniformly accelerated motion. 1. The velocity of a point increases uniformly in io seconds from 150 to 200 centimetres per second. To find the acceleration and the distance traversed in the given time. The initial velocity = 1 50 = » . The final velocity = 200 = v. The time of motion = 10 = t. Let a be the acceleration, s the distance traversed. mi '' _ v o 200 - 1 50 Then a = = = = 5. t 10 J And s = h (»j + f) t = 5 (150 + 200) = 1750 centimetres. Ch. i.] Uniformly accelerated motion. 29 2. In 5 seconds a body travels 125 cm. with acceleration 15. Find the initial velocity. Here the distance s, acceleration a, and time t are given. Equation (3) above gives the initial velocity v , 125 = 5« + £x 15x25, whence v — - ^. The velocity is therefore initially in the direction opposite to that of the displacement (125 cm.) and of the acceleration. It is continually diminished and ultimately reversed. To find when the velocity is zero, we write in equation (1) r = o, v = - ~, a = 1 5, whence t = £ sec. The distance traversed in this time is given by The negative sign indicates that this distance is in the direction opposite to that of the acceleration. The body then retraces its path. It reaches its initial position at a time t given by writing s = o in equation (3) This gives t = = f sec. Hence the path of 125 cm. is really described as follows. In £ sec. the body moves 5-^ cm. in the direction opposite to the acceleration. In an equal succeeding interval it retraces the same path backwards, and then in 3 J seconds it travels 125 centimetres with a constantly increasing velocity. 3. Find the initial velocity of a body which moving with acceleration 10 acquires in 5 seconds a velocity of 150. 4. Find the initial velocity of a body which moving with acceleration 15 acquires in 5 seconds a velocity 10. 5. Find the final velocity when the initial velocity is 100, the acceleration 18, and the time is (1) 4 seconds, (2) 6 seconds, (3) 8 seconds. 6. Substituting an acceleration - 18 for 18, solve question 5. 7. A body after moving from rest with uniform acceleration for I minute acquires a velocity of 10 metres per second. Find its acceleration. 8. A train moving at a speed of 60 kilometres per hour is pulled up in J minute. Find its acceleration. 30 Kinematics. [Ch. i. 9. If the velocity of a moving body increase in 10 seconds at a uniform rate from 15 to 25, find the distance traversed during the given time. 10. If the distance traversed in 30 seconds is 1 5 metres and the final velocity is 35 cm. per sec, find the initial velocity. 11. In questions (3) and (4) find the distance traversed. 12. A body starts from rest, and in 5 seconds travels 100 cm. Find the acceleration. 13. If the acceleration of a body be 5, and in 5 seconds it travels 125 centimetres, find the initial velocity. 14. A body moves with uniform acceleration through 100 centi- metres in s seconds and then comes to rest, find the acceleration, the initial velocity, and the average velocity during each second. 15. A body moves from rest with acceleration 32 foot-second units, find the distance traversed and the velocities acquired in 1, 2, 3 seconds. Find also the average velocity during each second, and the distance travelled during the 10th second. 16. The velocity of a moving point increases during motion through 12 centimetres from 18 to 21 centimetres per second. Find the acceleration, and the time of motion. 17. In moving through 10 feet with acceleration 32 foot-second units a moving point has acquired a velocity of 26 feet per second. Find its initial velocity, and the time during which it has been moving. 18 A mass whose acceleration is uniform moves over 483 feet in the fifth second from rest. Find its acceleration, and its velocity at the beginning of the fifth second. § 13. Motion with Uniform Acceleration. A particle starts from a point with initial velocity u along Ox, and has acceleration g along Oy, a perpendicular to Ox. To find its motion. Ch. I.] Motion of a projectile. 3i The component of velocity along Ox is constant, and if OM = MN = NP = u, OM, ON, OP, are the horizontal distances travelled in 1, 2, 3 seconds respectively. Also if Om = \g, On=\g-i?,Op = \g.f; Om, On, Op are the vertical distances travelled in 1, 3, 3 seconds. Completing the rectangles Fi S- S2 - mM, nN, and pP, the actual positions of the particle after 1, 2, 3 seconds are respectively 1 2 O s ; and the particle describes a continuous curve passing through these points. Let F be the position of the particle after a time t, FK perpendicular to Oy. Then OK = \gt % , and FK = ut. M a N P X \ L K \ \ P \ y °3 Therefore FK? = 9 OK. This relation enables us to find all points on the curve when u and g are known. The curve thus determined is called a Parabola, and the quantity — is called its Parameter. If the particle were pro- jected from O v with a velocity compounded of u and#, it would describe the same parabola. Fi g- 2 3- Next, let the initial velocity be inclined to Ox (as OQ), having components h,—k along Ox, Oy (fig. 23). 32 Kinematics. [Ch. i. Let P be the position of the particle at a time t after leaving 0, and draw PM perpendicular to OM. Then OM=M, (l) PM = -kt+igt 2 . (2) And at a time t the velocity along- Oy is —k+gt. k This is negative till t = - , and afterwards positive. Therefore the body moves upwards until t = -', at this instant its velocity is horizontal, and afterwards the motion is downwards, the path being a parabola with para- meter — • 9 The point A is called the vertex of the parabola, and the vertical line through A is the axis. This problem derives its interest from the fact that a falling body has uniform acceleration downwards, when the resistance of the air can be neglected. A falling body projected at any angle to the vertical is often called a Projectile, and its path a Trajectory, y k The Projectile attains its greatest height when t = - ;. substituting this value of t in (2) we find \ H k 2 . that — is the greatest height attained. Therefore the height attained by the parti- cle depends only on the vertical component ■ m of the initial velocity. The Iloclograph of uniformly accelerated motion. Since the acceleration is constant in mag- Fig. 24. nitude and direction, the hodograph is a straight line along which the tracing point moves with velocity g. Ch. i.] Projectiles. 33 Let OH (fig. 24) represent the initial velocity in mag- nitude and direction, and HP = gt ; then OP is the velocity at time t. Let OP' and OP be equally inclined to the horizontal OM. Then OP = OP and MP = MP' = 4 Pi". Hence, considering two points of the path at which the tangents are equally inclined to the vertical, we see that the velocities at these points are equal in magnitude, and the time occupied in passing from one to the other is twice that occupied in passing from one of them to the vertex. Hence since the horizontal velocity is constant, the two points are equally distant from the axis of the parabola. Also, since the vertical velocities at the two points are the same, they are in the same horizontal line. Hence the portions of the path on each side of the vertex A are precisely similar. Examples on the motion of a Projectile. In the following examples it may be assumed that the accelera- tion of a projectile is 980 centimetre-second units or 32 foot- second units. 1. A body is thrown vertically upwards with a velocity of 2058 centimetres per second. How high will it rise, and when will it strike the ground ? The initial velocity % is 2058 ; the acceleration a is - 980 ; the velocity v at the highest point is o. Therefore e 2 + 2 as = o. . •, V 2058x2058 , And s = — 2 — = — 5-^- = 2160.9 cm. -20 2 x 980 Also if t be the time to the highest point s 2160.9 t = — = ? =2.1 sees. i(v + v) 1029 This is also the time of fall from the highest point. Therefore 4-2 seconds is the time of flight. D 34 Kinematics. [Ch. i. 2. A projectile is fired at an inclination 6o° above the horizon, with a velocity of 588 centimetres per second. Find the hori- zontal and vertical components of velocity after 3 seconds. The horizontal component of velocity is initially 588 x | or 294. Since there is no horizontal acceleration, this is the horizontal component of velocity at any later time. The vertical component of velocity v is initially 294^/3, and the vertical component after 3 seconds is v -gt or 294^3 -3x980 upwards; i.e. 294 (10— V3) downwards. 3. A body is projected in a horizontal direction, with a velocity of 14 metres per second, from the top of a tower 22.5 metres high. (a) Find the distance from the foot of the tower at which the body will strike the ground. Let t be the time of fall. The vertical component of velocity is initially zero, and the vertical distance fallen is 2250 centimetres. Therefore 2250 = £ x 980 1\ V225 15 or t = — =? = — • V49 7 The horizontal velocity is initially 1400, and is uniform. Therefore the required distance is 1400 x ^ or 3000 centimetres. (&) Find the velocity of the body when it touches the ground'. The horizontal component is 1400. Since the vertical component is initially zero, its final value is given by v = gt = ^ x 980 = 2100. Therefore the resultant velocity is ■/(l4oo) 2 +(2ioo) 2 or 700V 1 3. The direction is inclined below the horizon at an angle a, such that tan a = § . 4. Three seconds after a stone has begun to fall down the shaft of a mine, a second stone is thrown down the shaft with a velocity of 4410 centimetres per second, and both stones strike the bottom simultaneously. Find the time of fall of the first stone and the depth of the shaft. 5. A ball projected vertically upwards rises 102.9 metres in Ch. i.] Motion of Projectiles. 35 3 seconds. Find, (i) how much, longer, (2) how much higher the ball will rise, (3) its velocity at the instant named. 6. Two bodies are simultaneously projected upwards in directions which make angles 30° and 6o° respectively with the vertical. Prove that if they both rise to the same height, they strike the ground at the same instant, but one will be three times as far from the point of projection as the other. 7. A body is thrown vertically downwards with an initial velocity of 48 feet per second. Find the distance traversed in the fourth second. 8. Two particles are simultaneously thrown vertically upwards from the same point. One has an initial velocity of 144 feet per second, and the other a velocity of 202 feet per second. Find the height of the latter when the former reaches the ground again. 9. A cricketer throws a ball 48 yards. If it rises 36 feet in the air, find its initial velocity and the time of flight. 10. A ball is thrown upwards at an angle 60° with the horizon. If its initial velocity is 48 feet per second, find when it hits the ground. 11. Show that if a ball is projected with an initial velocity of given magnitude, its range on a horizontal plane will be greatest when the direction of the initial velocity is inclined at 45° to the horizon. 12. A ball is projected horizontally from a railway carriage in a direction perpendicular to that in which the train is travelling. If the speed of the train be 30 miles an hour, the initial velocity of the ball 33 feet per second relatively to the thrower, and the height of the point of projection above the line 9 feet, find the velocity of the ball when it hits the ground, and the horizontal distance which it travels. § 14. The following discussion of the motion of a pro- jectile may be useful. It is assumed that the reader is acquainted with the principal properties of the parabola. Let v be the initial velocity, a the angle of projection, i.e. the angle between v and the horizontal. d a 36 Kinematics. [Ch. i. Then v cos a, v sin a are the horizontal and vertical components of the initial velocity, the axis of y being drawn upwards from 0. And if x, y are the horizontal and vertical components of displacement in time t, x = vt cos a, y = vtsina — igt 2 . Eliminating o, x 2 + (y + lgt 2 ) 2 = i> 2 t 2 . Therefore if several particles are projected simultaneously in the same plane from a point, with velocities of equal magnitude, they all lie at a time t on a circle with radius vt, and centre (o, —\ gt 2 ). That is, the centre of the circle descends with accelera- tion g, and the radius increases with velocity v. Again, eliminate t. Then y = x tan a— \ V 2 COS 2 a' or \ gx? (i+tan 2 a)— v 2 xi%Q.a + v i y = o. (1) It has already been shown (§ i$) that this is a para- bola, with its axis vertical and its latus rectum equal to 9 If in § 13 Ox be inclined to the horizon, and FK be drawn parallel to Ox, it can be easily shown, by compound- ing the uniformly increasing displacement along Ox with the uniformly accelerated displacement downward, that FK 2 = — OK, v being the velocity of projection. 9 v 2 Hence the focus of the parabola is distant from the point of projection, and since the tangent at is equally inclined to the axis and to the focal distance, the posi- Ch. i.] Motion of Projectiles. 37 tion of the focus is completely known ; for it is the point ,v 2 . v 2 x ' I — sm %a, cos a a ) • v uy iff > v 2 Since — does not involve a, the foci of all trajectories described from with a given initial velocity lie on a v z circle with centre and radius — • In equation (1) let 00 and y be known ; then the equa- tion is a quadratic in tan a. Hence there are generally two paths by which a projectile starting with an initial velocity v can reach a given point, and if a v a 2 are the values of a which satisfy the equation (1), tan a, + tan a, = • • 2 goo The roots of the equation are imaginary if v 2 < a gy, and this obviously is the case, for the given point cannot be reached. To find t/te range of a projectile on an inclined plane through the $oint of 'projection, when the direction of projec- tion and a normal to the plane lie in the same vertical plane. Let 8 be the inclination of the plane. Then it is required to satisfy equation (1) when y = x tan 8. Substituting in (1) and dividing by as, i ff x tan 8 = tan a — $ -x =— . ircos-'a' 2 v 2 sin (a — 8) cos a or x = — — * ' ; g cos/3 and the range (i.e. the distance travelled on the plane) is a « 2 sin (a — /3)cos a g cos^ a v 2 [sin (a a — 8) — sin /3] or g cos^ 8 38 Kinematics. [Ch. i. Now j8 is fixed. Therefore the range is greatest when sin (2a — 13) = i, that is, when 2 a-/3 = -> i.e. when the direction of projection bisects the angle between the verti- cal and the line of greatest slope on the plane. v 2 The greatest range is —, ^—rr , and is greater on a 6 s ^-(i+sin/3) horizontal plane than on any other, for in this case /3 = o. § 15. Angular Acceleration, about a fixed axis. The angular acceleration of a body is the rate of increase of its angular velocity. "We cannot here consider the general case when the axis of rotation changes during motion. We shall only take the simple case in which the axis is fixed and the rate of rotation alters. If ai and to be the values of the angular velocity at the beginning and end of the time t, ~ ° is the angular velocity gained in unit time. Therefore, a being the angular acceleration, to — o) = at. (1) Also we have already shown in §10 that W -±^t = d. (2) 2 v ' Multiplying (1) and (a) together, o) 2 — u> 2 = 2ad. § 16. Composition and Resolution of Accelerations. Let PQ represent the acceleration of a moving point, OP its velocity at a given instant ; then OQ is the velocity after unit time. Ck. I.J Lomposition of Acceleration. 39 Fig. 25. Take any two lines Ox, Oy through in the plane *OPQ, and resolve OP and OQ into their components OH, HP and OF, FQ along these lines. Draw PM parallel to EF. Then EF or PM is the gain of velocity parallel to One in unit time, and M Q is the gain of velocity parallel to Oy in unit time, and by compounding these we get the acceleration PQ. Therefore the resultant acceleration is obtained by com- pounding the rates of increase of the component velocities parallel to Oso and Oy; these rates of increase are called components of acceleration. The most useful case is that in which mOy is a right angle. Conversely, an acceleration can be resolved into its com- ponents along any two lines co-planar with it. Relative Acceleration, Let OP, OQ denote the initial velocities of two moving points A and B ; PH, QK their accelera- tions. Then OH, OK are the velocities after unit time, and the velocity of B relatively to A is initially PQ and finally HK. Therefore the relative acceleration, or change of relative velocity, is found by compounding HK and QP. And since the resultant of HK, KQ, QP, PH is zero, QP and HK are equivalent to QK and HP. Therefore the acceleration of B relatively to A is obtained Fig. 26. Fig. 27. 40 j\inemum,s. L^ n - *• by compounding the acceleration of B and the reversed acceleration of A, i.e. it is B _ _ the vectorial difference of the accelerations of B and A. § 17. Simple Harmonic Motion. Let APA' be a circle of radius r, A OA', BOB' per- pendicular diameters, P any point on the circle. Then if P moves round the circle in the direction of the arrow with uniform angular velocity a>, its velocity is rw directed along NP the tangent at P, and its acceleration is ra> 2 directed along PO. The time occupied by the moving point in going round the circle is — • Let Q be the foot of the perpendicular from P on AOA'. As P moves round the circle, Q moves to and fro along A'OA with a velocity and acceleration which are the com- ponents along A'OA of the velocity and acceleration of P. The motion of Q is called a simple harmonic motion ; the path described by Q when P goes once round the circle starting from any point is called a complete oscillation ; the time — occupied in describing this path is called the time of a complete oscillation or the periodic time ; the distance OA is called the amplitude of vibration. Let OQ = x. Then PQ 2 = r 2 -x 2 . ON The component of the velocity of P along A A' is ru-wTr> pn ?■" this is equal to no -^pp or w Vr 2 —se 2 . Ch. i.] Simple Harmonic Motion. 41 The acceleration of Q is of magnitude rw 2 - , or u> 2 so ; and it is always directed towards : it is therefore com- pletely represented by — m 2 os. Thus in a simple harmonic motion the acceleration is directed to a fixed point in the line of motion, and is proportional to the distance x from the point. Since the time of vibration is — , it depends only on the law of acceleration, i.e. on the constant co, and not on the amplitude. Phase. The fraction of the time of a vibration which has elapsed in the passage of the moving point from A to Q is called the phase of the motion at Q. a It is — , where is the angle AOP. Examples. 1. Two particles are projected upwards with the same velocity from two points A and B in the same vertical, B being at a height h above A. If the second particle be projected t seconds after the first, and if u be the velocity of the first particle when it meets the second, then h = ut + $gt*. 2. A ball falling from the top of a tower has descended a feet when another ball is let fall from a distance 6 below the top. Show that if they reach the ground together, the height of the , . (a + bY . . '" tower is - 1 feet. 3. A bird wishes to reach a point due N. when the wind is blowing from the S.W. at 20 miles an hour. Given the velocity with which the bird can fly, find the point of the compass for which it must aim. Show that if it goes less than 14. 1 miles per hour, it cannot reach the point. 42 Kinematics. [Ch. i. 4. A railway train starts from the top of a straight incline i-i miles long with a velocity of 22 miles per hour, and on reaching the bottom has acquired a velocity of 32 miles per hour. Find the acceleration in foot-second units. 5. An engine-driver suddenly puts on his break and shuts off the steam when the train is running at full speed. In the first second afterwards the train travels 87 feet, and in the next 85 feet. Find the original speed of the train, the time which it will take to come to rest, and the distance that it will travel in the interval, assuming the break to cause a constant retardation. Find also the time which the train would take, if it were 96 yards long, to pass a spectator standing 484 yards ahead of the train at the moment when the break was applied. 6. A body moves in 2 seconds through a distance of 100 feet from rest, its acceleration being constant. Find the distance traversed in the next second. 7. A bullet is fired with a velocity of 1920 feet per second in a direction equally inclined to the horizontal and the vertical ; find the height to which it will rise, and the velocity at the highest point. 8. Assuming that the earth is a sphere of 4000 miles radius turning on its axis in 86164 seconds, find its angular velocity, and the acceleration of a point on its surface — (1) at the equator, (2) in latitude 45°, the foot and second being taken as units. 9. The hour and minute-hands of a clock are respectively 2 inches and 3 inches long. Find the accelerations of their extremities in foot-second units. Find also the acceleration of the extremity of the hour-hand relatively to that of the minute-hand at 12.0, at 6.0, at 3.0, and at 4.30. Appendix. Proofs of some simple geometrical results are appended for the convenience of beginners. (a) The circular measure of an angle. If r is the radius of a circle, the circumference is Zitr, Ch. I.] Geometrical Theorems. 43 ■7! being the incommensurable number 3-14159 ..., or 3^ approximately. Let be the centre of a circle ABC (fig. a 8). Since angles at the centre are proportional to the arcs on which they stand, angle AOB arc AB four right angles whole circumference. But the circumference is lirr. m . „ angle AOB arc AB , . Therefore „ ?- T - z— = . (1) tour right angles aire v ' Let the unit angle be chosen so that 2tt is the measure of four right angles. This unit is or Kj'-IQW'lQ.... 3-14159- This is called the Radian, and the measure of an angle in terms of radians is called its circular measure. By (1), the angle AOB is — radians or simply — , if it is understood that the radian is the unit. The circular measure of an angle at is therefore the Fig. 28. arc which it subtends on a circle of unit radius with as centre. A right angle is - > i.e. it contains - Radians. 44 Geometrical Theorems. [Ch. i. (b) Definitions of the sine, cosine, and, tangent of an angle. Let AOB be any angle less than- • From M on OA draw MN perpendicular to OB. Then in the triangle MON, MN opposite to the angle AOB is called the perpendicular, ON adjacent to AOB is called the base, and OM is the hypotenuse. rn, +• perpendicular MN . . 1 he ratio A= or -pr^-is called th e sine ot A OB, hypotenuse OM base ON . or j-jjy is the cosine or AOB, hypotenuse OM perpendicular MN. ■ ■ . . r or -jYjTj-is the tangent of AOB. For brevity these ratios are denoted by sin AOB, cos AOB, tan AOB. These ratios and their reciprocals are called the trigono- metrical ratios of the angle AOB. Their magnitudes do not depend on the position of M on OA. For if M ' be any other point on OA, and M'N perpen- dicular to OB, the triangles OM'N', OMN have their angles equal and their sides proportional, so that MN _ M'N ' ON ON' MN M'N' OM '' ~~ ~0M' ' OM ~ OM'' ON = OW Again, take m on OB, and draw mn perpendicular to OA. The triangle mOn, MON have their angles equal, each to each, and ££=?• £?= - ~~~ OM Om OM Om' ON ~ On Thus the values of the trigonometrical ratios deduced from the triangles Omn, OMN are the same, and M may Ch. i.] Geometrical Theorems. 45 be taken on either of the two lines which form the angle at any distance from ; the trigonometrical ratios of the angle A OB will be the same for all positions of M. The ratios are different for different angles ; and when the ratios of an angle are known, the angle can be found. (c) By diminishing the angle AOB, its cosine may be made to differ from 1 by less than any assignable quantity. Let ABB be a circle with centre 0, and diameter BB. Draw AM perpendicular to BB, and join AB, AB. rru Ann 0M 0M Then 1 — cos ^0.2 = 1 — 7^=1— -7^ OA OB MB _ MB BA ~ 7 "BB~°" BA 'BB' But the triangles AMB, BAB have their angles equal, each to each. MB BA Therefore BA ~ BB' BA 2 Therefore 1 —cos A OB = a . -jrjr, . BB* Now by revolving OA round towards OB the arc AB can be made smaller than any given fraction of the diameter BB. BA BA 2 Therefore -^ and a fortiori Rn2 can be made less than :any assignable quantity by diminishing the angle AOB. Therefore cos AOB may be made to differ from 1 by a quantity less than any assignable quantity. BA Since ^r = sin ABB = sin \ AOB, we have BB 1 -cos AOB = % (sin k AOBf, or 3 sin 2 \ AOB. A tangent to a curve is defined as follows — Let P and Q be two points on a curve, PQ the chord 46 Geometrical Theorems. [Ch. i. joining' them. Then if P remains fixed, and Q moves along the curve towards P, the line along which PQ lies when Q coincides with P is called the tangent at P. (d) By diminishing the arc BC of a curve, it may be made to differ from its chord by less than any fraction of the chord, however small. Draw the tangents at B and C to the arc and let them intersect in B. Then the arc is greater than its chord ; it is also less B m c ^an *he sum of the distances BB, BC. For at a point E in BC, draw a D v tangent FEG. Kg- 30. Then FG < FB + BO. And BF + FG+ GC< BB + BC. At H between E and C draw a tangent KHL. Then EK + KL + LC < EG + GC. And adding BF + FE to each, we have BF + FK + KL + LC < BF + FG + GC ^ c ° s arc PC; 1 -nn j ^ COS a -n^, or < 7 arc PC, and > — — arc PC ; for cos a < cos 6, except at the extremity of the oscillation, when cos = cos a. Therefore the time of vibration lies between the periodic times of two harmonic vibrations — a -n„ -, a cos a t,- a = - arc PC, and a = J — — arc PC. Therefore the time of vibration of a pendulum which oscillates through an angle a from the vertical lies between 1 it . I - and i it A / \/ g V g cos a We have already seen that cos a approaches very nearly to 1 when a is moderately small (Chap. I, p. 45). A more accurate value of the time of vibration through an angle a can be found by using the Calculus. 72 The Laws of Motion. [ch. ii. § 13. Atwood's Machine. The essential part of Atwood's Machine consists of a pulley mounted so as to revolve freely on an horizontal axis, and a light string passing over the pulley. Prom the ends of the string hang two equal pans in which masses can be placed. In order to diminish the effect of friction on the motion of the pulley, each end of the axle is mounted on two friction wheels. These revolve freely in fixed bearings mounted on a stand, and as their motion is slow compared with that of the axle, the resistance from friction is insignificant, and the weights of the suspended masses are the only external forces which need be considered. The stand which carries the friction wheels is mounted on the top of a pillar about seven feet high, to which a vertical scale graduated in inches or centimetres is attached. We shall suppose that the string is inextensible, and that the tension in it is the same throughout. It will appear afterwards that the latter assumption is not quite justifiable. Let m, m' be the suspended masses, m being the greater, and let T be the tension in the string, a the acceleration of m downwards. Since the string is inextensible, m' has the same acceleration upwards. By the Second Law of Motion, ma = mg—T, m'a = T—m'g. Therefore a = ,q, T _ imm'g m+ m Ch. II.] Atwood's Machine. 73 Thus the greater mass descends with increasing velo- city, and the acceleration may be dimin- ished by increasing the masses in each pan, keeping their difference about the same. Verification of the Laws of uniformly Accelerated Motion, by Atwood's Machine. At the zero of the vertical scale is placed a stop A, the removal of which permits the mass m to fall. Another stop B which arrests the fall can be placed at any other point, and the distance fallen is measured by the reading on the scale at B. The time of fall can be determined by a stop-watch or water-clock. In order to verify the formula s = \ at 2 , we suspend masses m and m from the strings, and observe the times of fall from A to different positions of B. Then, if * be any one of these distances and t the cor- Q Fig. 42. responding time, — = \ a. t Now a is the same throughout. the same for all values of AB. Therefore — - should be Since •■ + '• g can be found by this experiment, but the method is not satisfactory, even when a correction to be noticed in Chap. IV is introduced. Resistances due to friction are appreciable, and it is best to find g by timing the oscillations of a given pendulum (see p. 71). 74 The Laws of Motion. [Ch. ii. We may also vary the value of a by transferring masses from one pan to the other, and verify the formula by ob- serving the times required to fall through a given distance. Since s is constant, we have at 2 = constant. Again, we can verify the formula v 2 = 2 as. Put equal masses in the pans, and place on the top of the descending pan a small bar of mass n, the ends of which project beyond the sides of the pan. At a distance from A equal to the chosen value of s attach to the scale a slider C carrying a horizontal ring, which is wide enough to allow the descending pan to pass through it, but is too small to admit the bar fx. As the pan passes C the bar is removed, the masses in the pans become equal, and the pans move with uniform velocity unless there is sensible resistance from friction. Place the stop B a convenient distance below C, start the time-keeper when p is removed, and note the time of fall from C to B. Then the distance BC being known, v is known. If the experiment is made for different values of AC, the acceleration is constant, and v 2 is proportional to * ; thus, if AC has the values 1, 4, 9 feet, the corresponding values of v are as 1, a, 3. Examples. 1. A mass M rests on a smooth horizontal table ; a string attached to M passes over a small pulley at the edge of the table and supports a mass m hanging freely. To determine the acceler- ation and the tension in the string. Let a be the acceleration, T the tension. Since the tension in the string is the only horizontal force acting on M, T=Ma. Also, since the acceleration of m is vertically downwards, and Ch - n -] Examples. nc the vertical forces acting on m are its own weight mg, and the tension of the string, mg-T=ma. Therefore a = JOS- T _ ^HHL . m + M' m + M 2. Masses of 3, 4, and 5 lbs. respectively move along the sides of an equilateral triangle ABC taken in order, with velocities 25, 20, and 18 feet per second. To find the resultant momentum of the whole system and the force which would generate this momentum in 10 seconds. The momenta are 75 along AB, 80 along BC, 90 along CA ; these reduce to 75 - 80 along AB. and 90 - 80 along CA, or to 5 along BA, and 10 along CA. 10 along CA can be resolved into 5 along BA and s\/3~ per- pendicular to BA. Therefore the resultant momentum is ' v ^(S + S) 2 +(S-v / 3)'' J or V 175 or 5-/7, and the required force is — — — or \Vy poundals. 3. A mass m describes the perimeter of an equilateral triangle ABC with a velocity x of uniform magnitude. Find the impulse of the blow which m receives at each vertex of the triangle. Let the particle pass at B from AB to BC. "Its momentum is mx, initially along AB, finally along BC. The change of momentum is obtained by reversing the initial momentum, and then compounding with the final. Now the resultant of mx along BA, and mx along BC, is mx V3 along the internal bisector of the angle CBA. Therefore the impulse of the blow is mx V3, and it bisects the angle CBA. 4. A ball of mass 10 lbs. projected upwards with a velocity of 30 feet per second, strikes a ceiling 9 feet above the point of projection and penetrates 4 inches into it. Determine its mo- mentum at the time of striking, the energy of the blow, and the mean resistance to penetration, (g = 32 foot-second units.) The velocity v at the time of striking is given by v* = 900 - 2 x 32 x 9 = 324, or v = 18. The momentum is therefore 180. And the energy of the blow is J. 10 v 1 = 1620. 76 The Laws of Motion. [ch. ii. The mean resistance to penetration being F, and s the distance 4 inches, we have (§ 7), Fs = 1620, and s = $ ; .■. F = 4860 poundals, or 1 5 if pounds' weight. 5. A particle of mass 6 oz. is attached to one end of an in- extensible string, which passes over a smooth pulley and has attached to its other end a mass of 3 oz. To the latter mass is attached a second inextensible string which hangs downwards and terminates in a mass of 2 oz. Neglecting the masses of the strings, to find the acceleration of the system and the tensions in the strings. Let T 1 , T 2 be the tensions, a the acceleration. The first string has a maBS of 6 oz. hanging from one end, and 3 + 2 or 5 oz. hanging from the other. Therefore the motion of the mass of 6 oz. or | lb., gives and the motion of the mass of 5 oz. gives .*. a = ; — - g = £? foot-second units, 6 + s and 7 1 ! = %g{i- <&) = if g ^poundals. To find r 2 . The mass of 2 0z. hanging from the second string moves up- wards with acceleration ftg. •'• T i = H I + ii)fl , = A? poundals. 6. A train whose mass is 112 tons is travelling at a rate of 25 mileB per hour on a level track, and the resistance due to air, friction, &c. is 16 lbs. weight per ton. A. carriage of mass 1 2 tons becomes detached. Assuming that the force exerted by the engine is the same throughout, find how much the train will have gained on the detached part after 50 seconds, and the velocity of the train when the detached part comes to rest, given that = 32. After the carriage is detached, the resultant force on the train ch. ii. Examples. 77 is 16 x 12 lbs. weight, and the force on the carriage is equal and opposite to this. Therefore the acceleration of the train is — or — — 100 x 2240 3500 foot-second units ; similarly the acceleration of the carriage is - ^j . Therefore the acceleration of the train relative to the carriage is *&g, and the distance gained is *x(S°) 2 x^= 320 feet. Let x be the speed of the train in miles per hour when the detached carriage comes to rest. The momentum is then 100 x 2240 x x. The momentum before parting is 112x2240x25, and these momenta are equal, since the forces exerted on the train and detached carriage are equal and opposite. Therefore loox = 112 x 25, 01 x = 28 miles per hour. 7. The time of fall from rest down all chords of a vertical circle to the lowest point is the same. Let^Bbe a vertical diameter, Pany point on pig ' 43 the circle, a the radius of the circle, PAB = 8. The acceleration of a heavy particle falling along AP is g cos 6. Also AP = 2 acos 6. Hence (Chap. I, § 12), if t be the time of fall from P to A, ^4^J = 2 ^. ^ gcosB ^ g Thus the time of fall is the same for all positions of P on the circle APB. 8. A mass of 10 lbs. is whirled round 5 times in a second on a smooth horizontal table at the end of a fine string of length 5 feet. Calculate the tension in the string. The velocity is 2ir x 5 x 5 = 5°' r - The radius of the circle is 5. Therefore the acceleration towards the centreis £5^ or Su'. The total momentum is the same before and after impact. Therefore mu + m'u' = mv + m'v'. ( i ) As soon as the balls come into contact A exerts a pressure on B tending to drive it forward, and B exerts an equal and opposite pres- • sure on A. Now all bodies are really more or less deformed by the application of pressure. As soon therefore as the balls come into contact with each other they begin to compress one another and continue to do so till the velocities of the balls are equal ; at this instant the kinetic energy of the system is diminished by the work done in deforming the balls. But after this instant, the centres of the balls begin to separate again, and the mutual pressure does work on them and increases the kinetic energy. Let us suppose that the kinetic energy finally returns to its initial value. Then \ mu 2 +% m'u' 2 = \ mv 2 + J m'v" 1 . (a) And by (i) and (a), , , . , J v ' v " u — u = v — v. (3) Ch. in.] > Impact. 105 Whence the relative velocity of the balls after impact is the same as before in magnitude but is of the opposite sign. If the ball B is at rest initially, u = o. T ., . m—m' in this case v = ,•%. m + m Therefore if A impinges directly on a ball B at rest, A continues to move in the same direction if its mass is greater than that of B, while its motion is reversed if its mass is less than that of B. If the two balls are of equal mass, A is brought to rest and B moves on with velocity u. The conditions implied above are never realised in nature, for the work done in the first part of the collision is expended in setting the balls in vibration (producing sound, heat, &c.) as well as in compressing the balls. Though the latter part of the energy may be recovered in the later part of the collision, the former is wasted and is converted directly or ultimately into heat. Newton proved experimentally that the formula (3) may be replaced by v >_ v = e{u ^>), (4) where e < 1 for all substances. e is called the Coefficient of Restitution ; it is approxi- mately the same for all velocities of the colliding balls, and depends only on the materials of the balls. Combining (1) and (4) we have (m + m')% — m'v'(i + —) + v(m ) , (m + m')u' = mv{\ +-) + v'(m' ) • Therefore mu 2 + m'u' 2 = mv 2 + m'v' 2 + — — 7 (e- 2 -i)(v-v') 2 . m + m v 106 Work and Energy. [Ch. hi. Therefore the kinetic energy lost in the collision is "' -(e-*-i)(v-v')\ 2,(m + m') or —. k(i— e 2 )(u — u') 2 . The Impact is here said to be direct. When e = i, the balls are sometimes said to be perfectly elastic. Oblique Impact of two smooth latts. Let A and B be the positions of the centres of the balls m and m' at the moment of collision, and let the paths of the centres before collision he in the same plane with AB and make angles a and a with it. The impact is then said to be oblique. Let u, %' be the initial velocities, Flg ' 53 ' v, v' the final velocities, /3, /3' the angles which the paths after collision make with AB. Since the balls are smooth the stress between them during collision is along AB, and the components of velocity perpendicular to AB are not affected by the collision. Therefore u sin a = v sin /3, (5) and %' sin a'= v' sin J3'. (6) Again, the total momentum along AB remains unaltered. Therefore mu cos a + m'u' cos a = mv cos fi + m'v cos /3'. (7) Also the relative velocity of the balls along AB is affected in the same way as before. Therefore v' cos j3'—v cos /3 = e (u cos a—u' cos a'). (8) The equations (7) and (8) are of the same form as (1) and (4) with % cos a, w'cos a,... written for u, %',... Ch - "'•] Impact. IQ7 Therefore mu 2 cos 2 a + m'u' 2 cosV = ircv 2 cos 2 fi + m'v' 2 cos 2 /3' + ^T^"'^ 1 ~ 6 ^ ( w cos a-M ' eos a ') 2 - Therefore by (5) and (6), mu 2 + m'u' 2 = w z + ». v' z + — - -(i-e 2 )(«cos a -a' cos a') 2 . And the kinetic energy lost is mm' . —, — ■ — r\(i— e 2 )lu cos a — u' cos a) 2 . 2,(m + my JK 1 Impact on a fixed smooth plane. Let a small ball moving with velocity u along BA im- pinge on a fixed plane at B. Then if o is the angle between BA and BN, the per- pendicular to the plane, the com- ponents of velocity along and perpendicular to NB are u cos a, u sin a. Let BO be the path of the particle after impact. If x, y are the components of n its velocity along and perpen- dicular to BN, x = eu cos a, y = u sin a. The total momentum does not Fig. 54. remain unchanged since the plane is not free to move. The resultant velocity after impact is u«/cos 2 a + e 2 sin 2 a, and tan NBC = - = - tana. x e 108 Work and Energy. [Ch. hi. If e — i, the angles NBA, NBC are equal. The impulse on the particle is given by its total change of momentum, and is u cos a (i+e) parallel to BN. Examples. 1 . To find the H. P. of an engine which will raise 20 gallons of water per minute from a depth of i8£ feet, and discharge it with a velocity of 100 feet per second. The gallon of water weighs 10 lbs. and g = 32. Therefore the work done per second in raising water is 20 x 10 x 32 37 5 „ . . , . 7 — x — = - x 1 1 84 foot-poundals. And the kinetic energy generated is 1 20 x 10 . ., 5 — x (ioo) 2 = - x 10000. 2 60 3 Therefore the whole work done is -(1184 + 10000) = 18640 foot-poundals. But 1 H. P. = 550 x 32 foot-poundals per sec. Therefore the engine is — — — = i^fo H. P. 550x32 2. To explain the principle of the centrifugal railway, and to find the leastheight from which the car- riage may start. In the centrifugal railway, a carriage A runs down the in- clined track CB and thence round the in- terior of the circle BED. It is required to start the carriage with such speed that it can travel round the circle without dropping off. Let v be the velocity of the carriage at D, r the radius of the circle. Then — is the acceleration of the carriage, and is directed r vertically downwards. Fig. 55- Ch. in.] Examples. 109 And the forces acting on the carriage are its own weight, and the resistance of the rail at D. The acceleration due to these cannot be less than g. Therefore — must not be less than g, for otherwise the particle would fall from the circle. Hence the velocity at D must be at least Vrg. At other points on the circle the velocity is greater than at D, and the radius is inclined to the vertical. Therefore if the carriage does not fall at D, it passes all other points safely. Now assuming the railway to be smooth, the velocity at D is equal to that at C on the same level as D. And the velocity acquired in falling through a height h is V2gh. Therefore h, the vertical distance through which the carriage descends from rest to C, cannot have a smaller value than that g™nby V^^VTgh, or h = £ r ; i. e. the carriage must descend from a point whose height above the highest point of the circle is at least \r. 3. A simple pendulum of length I is held horizontally and then let go. To find the velocity when the string is vertical and its tension T in this position. The velocity v of the bob is given by « 2 = 2 gl. And at the lowest point the acceleration towards the centre is » 2 j or 2 g. Therefore by the Second Law of Motion, ^P= T-mg, or T = zmg. 4. A bullet weighing \ oz. moving with a velocity of 1200 feet per sec. strikes a bank of earth, and penetrates it to a depth of 4 feet ; find the work done and the mean resistance to penetration. 5. A man of weight 160 pounds is in a swing whose ropes are 18 feet long; if he rises through an angle of 60°, what will no Work and Energy. [Ch. hi. be the stress on the ropes when they are vertical ? Give the answer in lbs. weight. 6. A smooth tube is in the form of the arc of a quadrant, of which one extreme radius is horizontal and the other turned vertically downwards. A thread passes up the tube and over its upper edge which is smooth. To the ends of the thread are fastened heavy particles one of which, m, is just within the lower end of the tube, while the other, p, hangs freely. Find the ratio of p to m, that p may draw m just up to the upper end of the tube. 7. How long must a 10 H. P. engine work to impress on a mass of 1 2000 tons initially at rest an ultimate velocity of 3300 yards an hour ? 8. A smooth circular tube in a vertical plane contains a heavy particle. Find the least velocity with which the particle must be projected from the lowest point, in order that it may travel continually round the tube. 9. Find the H. P. of an engine which can start a train of 200 tons, and raise its speed to 30 miles an hour in ij minutes. 10. A railway truck, the mass of which is 6400 lbs., stands on a horizontal railway, the total friction when it is in motion being g'jth of the weight. A man begins to push it with a force of 120 lbs. weight. Find the acceleration ; find also the velocity and the rate of working after t seconds. Show that if his highest rate is ^ H. P., he cannot go on pushing with the initial force for more than about 4.6 sees. 11. A ball is projected with a velocity 120 feet per sec, at an elevation 60°, against a vertical smooth wall, 120 feet distant : where will it impinge ? If the coefficient of elasticity be \, when and where will it strike the ground ? 12. A 6 oz. ball moving with velocity 20 meets and strikes a 4 oz. ball moving with velocity 40. If the coefficient of restitution be \, find the subsequent motion of each ball. 13. A fire-engine pump is provided with a nozzle the sectional area of which is 1 sq. inch and the water is projected through the Ch. hi.] Examples. 1 1 1 nozzle with a velocity of 130 feet per second. Find the H. P. of the engine required to drive the pump, irrespective of the loss by resistance of the working parts. A cubic foot of water weighs 62 \ lbs. 14. "What is the H. P. of an engine which draws a train at a uniform rate of 45 miles per hour, against a resistance equal to 900 lbs. weight ? 15. A shaft 560 feet deep and 5 feet in diameter is full of water ; how many foot-pounds of work are done in emptying it, and how long would it take an engine of 3J H. P. to do the work ? 16. A mass of M pounds is drawn from rest up a smooth inclined plane of height h and length I by means of a string passing over the top of the plane and supporting a mass of m pounds hanging freely. Prove that M will just reach the top of the plane if m is detached after it has descended a distance — . ■ = — - • m h + l 17. A particle is suspended by a string, the upper end of which is attached to a smooth vertical wall : the string is pulled aside through an angle 60° in a plane perpendicular to the wall ; it is then let go, the particle impinges on the wall, and rebounds through 45°. Find the coefficient of restitution. CHAPTEE IV. Motion of an Extended Body. § 1. Centre of Mass of a System of Particles. Let A, B, C, D be four points moving with, any velocities Then if a divides AB so that P.Aa = Q.aB, we have shown (Chap. II, § io) that the velocity of a is compounded of a/_ Kg- 56. P+Q , T and Pv i and _. ^>- are the velocities of the centre of AB mass at the beginning and end of the time t, M is the acceleration of the centre of mass. Since the external forces acting at a point would communicate this acceleration to a particle of mass M, the second property of the centre of mass may be stated thus : — The acceleration of the centre of mass of the system is that which the external forces acting at a point would impress on a particle of mass M. Hence we may state the First and second Laws of Motion as follows : — (1) The centre of mass of a system of particles remains at rest or in uniform motion in a straight line, except in so far as it is made to change that state by force acting on the system from without. Ch. iv.] The Centre of Mass. 115 (a) The change of momentum of the system during any time is measured by the resultant of the impulses of the external forces during that time. A system of particles can only have one centre of mass, for there cannot be two points (determined by the method used in § 1), whose motion is the same in all displacements from rest. Hence in a symmetrical body, the centre of symmetry is the centre of mass : for otherwise there would be at least two points, one on each side of the centre, which have equally good claims to be regarded as the centre of mass. Therefore the middle point of a thin straight rod, and the centre of a circle or sphere, are the centres of mass of the corresponding figures. The centre of mass of a parallelogram or cube is the intersection of the diagonals which join opposite angles. Let A, B be the centres of mass of two parts of a body, of masses P and Q. If M and N be the resultant , momenta of the parts, the mo- mentum of the whole body is compounded of M and N. Divide AB in B so that P . AB = Q . EB. M N The velocities of A and B being -p > -~ , the velocity of W ■ A A f M P * N Q P B is compounded of -^ • -=5 — ^ and 7= • 5 — „ ; or ot M , N P P + Q Q P+Q PTQ and PTQ- Therefore B is the centre of mass of the whole body. Conversely, if the centres of mass of the whole body and a given part be known, the centre of mass of the remainder can be determined. 1 3 E Fig. 58. n6 Motion of an Extended Body. [Ch. iv. For let E, A be the centres of mass of the whole and the given part. Then if P is the mass of the latter, R the mass of the whole body, R — P is the mass of the remainder, and the required centre of mass is situated at B on AE produced, P where BE = r- — - . AE. To find the Centre of Mass of a System of Particles. Let Oy be a given straight line, and let particles of mass m 1 ,m 2 , m 3 start from simultaneously with velocities y 1 ,y 2 ,y i , ■ • . along Oy, not necessarily all positive. The velocity of their centre of mass is m 1 + m 2 + m s + ... Hence by considering the positions of the particles after unit time, we find that the centre of mass of particles m 1} m 2 , m 3 , . . . at distances y lt y 2 , y 3 , . . . from along Oy, is at a distance y from 0, where y = «iri + W-V.+ - . 17 * m 1 + m 2 + m 3 + . . . Draw Ox perpendicular to Oy, and let the particles start from the positions y 1 ,y 2 ,-.- simultaneously with velocities x x , x 2 , x 3 , . . . parallel to Ox ; then, if the velocity of the centre of mass is x, m 1 x 1 + m 2 x 2 + m 3 x s + .. . m 1 + m 2 + m 3 + ... Hence when the particles are at distances x lt x 2 , x 3 ... from Oy and y lt y 2 , y % ... from Ox, their centre of mass is distant x from Oy, and y from Ox. The particles having attained this position, let them move perpendicularly to the plane x Oy with velocities z x , z 2 , z 3 , . . ., velocities being considered positive or negative according Ch. iv.] The Centre of Mass. 117 as they are towards or away from the reader's eye. Then, if the centre of mass then moves with velocity z, ._ m 1 z 1 + m 2 z 2 + m 3 z 3 + ... rn 1 + m 2 + m 3 + .. . Hence when the particles are distant z 1 , z 2> z 3 , . . . respec- tively from the plane xOy, the distance of their centre of mass from the plane is z. Centre of Mass of a Circular arc. If- a closed figure, formed of straight uniform rods, is moving so that each rod has the same velocity per- pendicular to its length, the resultant momentum is zero, for the polygon of momenta is similar to the given figure. By increasing the number and diminishing the length of the sides we can extend the result to a figure bounded by any closed curve. Hence we can determine the centre of mass of a circular arc ABB of radius r, subtending an angle 8 at the centre C. Let particles leave the centre with Velocity r, so that after unit time they are uniformly distributed over the circle, a mass m lying a on unit length of the arc, and let AB = I = ar sin - • The considerations just given show that the resultant momentum is that of a mass Im moving with velocity r perpendicular to AB, i.e. it is a mr 2 sin- • Since the moving mass is mr 0, the velocity of the centre of mass is . Q 2r sin- 2 e Therefore when the particles are at distance r from the 1 18 Motion of an Extended Body. [Ch. iv. centre, their centre of mass is on the perpendicular from C %t to AB at a distance — sin - from C. 6 % For a semi-circle this distance is ' — • 77 Examples. 1. Masses of 2, 4, 6, 7 lbs. lie on a square table ; their distances from one edge are 2, 3, 4, 6 feet, and from an adjacent edge 3, 5, 1, 2 feet respectively. Find the positions of their centre of mass. Here m 1 = 2, m 2 = 4, m 3 = 6, m 4 = 7 ; #2 = 2, # 2 = 3> x 5 = 4j ^4 == 6 ; #1 = 3. 2/2 = 5. Vs = 1. 2k = 2. Therefore _ 4+12 + 24 + 42 82 19 _ i9' 6 + 20 + 6+14 46 y = ■ — = — • 19 19 2. Three vertices of a tetrahedron are 7, 8, 9 feet above, and the fourth is 5 feet below a fixed horizontal plane. Masses of 7, 5, 12 and 15 lbs. are placed one at each vertex. Find the distance of the centre of mass from the fixed plane. m i "= 7. »% = S» m s = 12, m 4 = 15 ; % = 7. z 2 = 8, z 8 = 9, 2 4 = - 5. Therefore (7 + 5 + 12 + 15)5 = 49 + 40+108-75 = 122. A a _ 122 And z = — • 39 The centre of mass is 3^ feet above the plane. § 3. Moments of Inertia. If a particle of mass m moves with angular velocitj' 2 . If there are several particles m lt m 2 , . . . m n , not neces- sarily in the same plane, revolving with angular velocity a> about a fixed axis at distances r x , r 2 , . . . r n from it, the kinetic energy of the system of particles is \ (a*,/, 2 + m 2 r 2 2 + ...+ m n r n 2 ) a, 2 . Ch. iv.] Moments of Inertia. ng The term in brackets is called the Moment of Inertia of the system about the axis. If a particle of mass m 1 + m. 2 + ... +m n revolve round the same axis at a distance h from it, its kinetic energy is equal to that of the given system, if (m 1 + m 2 + ...+m„)k 2 = m^ 2 + m 2 r 2 2 + m 3 r 3 2 + ... + m n r, 2 . The quantity k determined from this relation is called the Radius of Gyration of the system of particles. Since a rigid body revolving round an axis may be con- sidered as a system of particles similar to that which we have considered, its moment of inertia about an axis can always be determined by analysis, and in a few cases it can be determined very simply. Moment of inertia of a t/iin ring about an axis through its centre perpendicular to its plane. Let n particles each of mass m be distributed uniformly round the circumference of a ring of radius a. The moment of inertia required is nma 2 or Ma 2 if M is the total mass of the particles. Let the number of particles be indefinitely increased, the total mass remaining unchanged ; then ultimately the mass is uniformly distributed round the ring, and the moment of inertia is still Ma 2 . Moment of inertia of a uniform thin rod AB about an axis perpendicular to its length through its extremity A. Let I be the length of the rod, and divide it into n equal parts of length h, so that nh = I. Let equal particles, each of mass m, be placed on the rod at distances h, %h, ... nh from A. nm is their total mass, which we shall denote by M. 120 Motion of an Extended Body. [Ch. iv. Their moment of inertia is m \h 2 + (a h) 2 + . . . + (nh) 2 } mh 2 , , , > or -z- »(«+ i)(2rc+i), or — (1+-U1 + — )• 3 v n' x in' Now increasing « indefinitely, we approach the case of a uniform thin rod of mass M, and the moment of inertia Ml 2 approaches indefinitely to the value Moment of inertia of a circular disc of radius a about an axis through its centre perpendicular to its plane. Divide the disc into concentric rings by circles of radii h, ih, <$h, ... (n—i)h where nh = a. The surface of the ring contained between the circles whose radii are (r— i)h and rh is nh? {r 2 — (r— i) 2 } which lies between i-nh 2 {r— 1) and %-nh 2 r. The moment of inertia of this ring lies between 1-nh 2 (r— i).(r — i) 2 h 2 about an axis X at distance h from the centre of mass can be replaced by an equal angular velocity about a parallel axis T through the centre of mass, together with a velocity of translation Aa>, this being the actual velocity of the centre of mass. If K, k are the radii of gyration about X and T, and M is the mass of the body, the kinetic energy is %MK 2 a> 2 ; but by the proposition just proved it is also \Mh 2 v i + \Mk 2 2 ) is the kinetic energy gained in 124 Motion of an Extended Body. [Ch. iv. the displacement 6, and since this is equal to the work done P. ON.0 = lMP( u > 2 -to 2 ), if P is the only force acting. But a> 2 — o) 2 = 2,0.6, where o is the angular acceleration. Therefore P. 0N= M k % a. Mk 2 a is called the moment of angular acceleration. Thus the moment of a force round an axis is the measure of its tendency to produce rotation round the axis, and if several forces act on the body the algebraic sum of their moments is equal to Mh 2 a, Since the conditions of equilibrium under forces in one plane are satisfied, when the centre of mass is at rest, and the angular acceleration round any axis vanishes, they may be stated as follows — (i) The external forces acting on the body are parallel and proportional to the sides of a closed polygon taken in order. (a) The algebraic sum of the moments of all the forces round any point in their plane is zero. § 5. Resultant of Parallel Forces. Let P, Q be two parallel forces applied to a body at the points A and B. It is required to find a single force R which has the same effect on the motion of the centre of mass, and the same moment round any point in the plane as P and Q to- gether. The first condition shows that R is parallel to P and Q, and equal to P+Q, P and Q having opposite signs if they are oppositely directed. Ch. iv.] Resultant of Parallel Forces. 125 or or Then From any point in the plane draw Omn perpendicular to P and Q, and let x be the point in mn through which B acts. Then 0m.P+0n.Q = 0x.(P+Q), Q {On-Ox) = P(Ox- Om), Q . xn = P . xm. Draw xC parallel to mA meeting AB in C. P _nx _BG Q~ mx~ AC The algebraic sum of the moments of P and Q round x is P. xm— Q . xn, which has been shown to be zero. If the forces P and Q are oppositely directed, let P, Q denote their numerical magnitudes, P being greater than Q. Then the resultant is P— Q in the same direction as P, and the point C through which it acts divides AB externally, so that CA_ Q CB~ P" The ratio in which C divides AB does not depend on the position of 0, and therefore the mo- R ment of the force B round any other point in the plane is equal to the sum of the moments of P and q round the same point. Neither does the ratio depend on the direction of P and Q. Fig. 63. C is the same if this direction is altered. 126 Motion of an Extended Body. [Ch. iv. § 6. Couples. In the case when the oppositely directed parallel forces are equal, our construction of the resultant fails. This combination of forces is called a couple ; it does not affect the motion of the centre of mass. Let equal forces P be directed along Am and JBn. The work done by them in a small rotation round is P(On-Om)d or P.mn.d. The distance mn between the lines of action of the forces is called the arm of the couple, and P. mn is called the moment of the couple. If there are two couples whose arms are a and b and moments Pa and Q6, the condition that they should maintain equilibrium is Pa+ Qb = o. For when this is satisfied there is no angular accelera- tion, and the couples do not affect the motion of the centre of mass. The sign of the moment is changed if the directions of the forces of the couple are reversed. This also reverses the direction of the rotation which the couple tends to produce. Two couples therefore balance one another when they tend to produce rotations in opposite directions and have numerically equal moments ; and one couple may be re- placed by another with the same moment, in the same or in a parallel plane. § 7. Let us now consider a body which rotates round an axis under forces which are perpendicular to the axis but not all in the same plane. Then it is still true, as may be seen from the proof in Ch. IV.] Couples. 127 Fig. 64. Then P. AL+Q, But § 4, that the moment of the forces is proportional to the angular acceleration. Let BE be the axis, and let two parallel forces P and Q n act through A and B, perpen- dicular to the plane of the paper. Divide AB in C so that P.AC=Q.CB, and draw AL, CN, BM perpendicular to BE. Then P. AL and Q . BM are the moments of P and Q round BE. Through C draw GCF parallel to BE. BM=P (GL-AG) + Q (BF+FM) = (P+Q) CN-P . AG+ Q . BF. P_ _ CB_ BF q~ AC~ AG Therefore P. AG = Q . BF. And P.AL+Q.BM=(P+Q)CK Therefore the resultant of P and Q, previously found, has the same moment round BE that P and Q together have. Extended definition of the moment of a force. If the force P makes an angle a with the axis, resolve it into components, P cos a parallel to the axis, and P sin a perpendicular to the axis. The former does no work in a rotation round the axis, and the latter does work Pp sin a . 8 where p is the per- pendicular distance of the component P sin a from the axis. Hence generally the moment of a force round an axis is obtained by resolving the force into components at its point of application parallel and perpendicular to the axis and mul- tiplying the latter component by its distance from the axis. We can now show that round any axis the moment of T28 Motion of an Extended Body. [Ch. iv. two parallel forces is equal to the moment of the force which we have already obtained as their resultant. For if two parallel forces P and Q acting- at A and B are inclined at an angle a to the axis, their components perpen- dicular to the axis are P sin a, Q sin a, and the resultant of these has been already found to be (P + Q) sin a acting at G ; but this is the moment of a force P + Q at C, acting parallel to P and Q. § 8. Centre of Parallel Forces. Centre of Gravity. Let parallel forces P, Q, R act at points A, B, G. By § 5 the forces P and Q are equivalent to a parallel force P+Q acting at F, where P.AF=Q.FB. Join CF and divide it in G so that (P+Q)GF= R. GC. Then the forces P, Q, R can be replaced by a single force P+ Q + R acting at G, and the position of G does not depend on the direction of the forces. In like manner it can be shown that when several par- allel forces P, Q, R... act at given points A, B, C,.,, their resultant passes through a cer- g ' 5 ' tain point whose position does not depend on the direction of the forces; this point is called the Centre of Parallel Forces. Now let P = m 1 ff, Q i =m 2 g,R = m 3 g. Then the point G is the centre of mass of particles m u m 2 , m 3 placed at A, B, and C respectively, and the forces are the weights of these particles. Therefore the resultant of the weights of the particles passes through their centre of mass, and the process of finding the point through which the resultant of the weights of any number of particles acts is precisely the same as that of finding the centre of mass. Ch. IV.] The Centre of Gravity. 129 Hence the action of gravity on any body reduces to a single force, termed the weight of the body, the line of action of which always passes through the centre of mass. From this property the centre of mass is sometimes termed the centre of gravity. If a body is supported at its centre of mass G it has no angular acceleration, for the moment of its weight round any axis through G is zero. Hence the centre of gravity of a body is sometimes defined as the point about which the body, if supported there, will balance in all positions. The forces of gravity acting on different parts of a body cannot be considered parallel when the dimensions of the body are comparable with its distance from the Earth's centre. Here two cases arise according as the forces reduce to a single force acting through a fixed point in the body, or, not. In the former case the body is called eentrolaric ; in the latter, it cannot be said to have a centre of gravity. § 9. To find the centre of mass of a triangle ABC. Divide the triangle into thin laminae by straight lines parallel to BC, such as ab, cd. Bisect BCia. B, and join AB, meeting ab and cd in e and/. Then in the similar triangles aeA, BBA BB _ ae BA~e~I' K 130 Motion of an Extended Body. [Ch. iv. . ., , CD be similarly bI = e~A But BD=CD; therefore ae = eb. Similarly tf=fd. Now ef being very small the lamina abdc is a uniform thin rod, and its centre of mass is at the middle point oief. Similarly the centre of mass of every lamina formed by parallels to BC lies on AD. Therefore the centre of mass of the triangle lies on AD. But if BA is bisected in F\ it can be similarly shewn that the centre of mass lies on CF; and therefore G, the intersection of AD and CF, is the centre of mass. Join FD ; then BF= FA and BD = DC. Therefore FD is parallel to AC, and DFG, ACG are similar triangles. __ . DG DF BF 1 Iherelore -^-r = —r^ = -^-r = - • GA AC BA % And G divides AB so that DG - i AD. The centre of mass of a body is frequently called its centroid. § 10. To find the centre of mass of a pyramid on a triangular base. Let OABC be a pyramid on a triangular base ABC. Divide the pyramid into thin triangular laminae by planes parallel to ABC, such as abc. Bisect BC in D. Join AD and take G in AD such that DG=iDA. G is the centroid of the triangle ABC. Join OD, meeting be in d, and OG, meeting ad in g. Ch. iv.] The Centre of Gravity. 131 Then because dga, BGA are parallel, dg_ Og__ ga_^ DG 0G~ GA' but BG=\ GA. Therefore dg = \ga. Also because be, BG are parallel, bd _Od _ do BB~OlD~BC' Therefore bd = do. Therefore g is the centroid of the triangle dbc. O D Fig. 67. Hence when the laminae into which the pyramid is divided are very thin, their centroids all lie on OG. Therefore the centroid of the pyramid lies on OG. Again, regard OAB as the base of the pyramid and C as its vertex, and let b now represent the middle point of OB. Then bB=\ OG. Take H, the centroid of the triangle OAB, and join CH. It can be proved, as above, that the centroid of the pyra- mid lies on CH. And AH= § Ab, and AG = f GD. K % 132 Motion of an Extended Body. [ch. iv. Therefore HG, bJD are parallel, and HG = f IB = £ 00. But since IB is parallel to 00, HG and 00 are parallel, and CH and OG intersect in K, which must be the centroid of the pyramid. And since K = S ' GK = * K0 = * 0G - Mix OLi Thus the centroid of the pyramid is found as follows : — Find the centroid G of the base, join it to the vertex 0, and in Off take a point K such that 0K= % OG. Kis the centroid required. The same construction applies to a pyramid on a poly- gonal base of any number of sides ; and if the number of the sides be increased and their lengths diminished indefi- nitely, the pyramid becomes a cone on any curve as base, and its centroid can be determined by the same con- struction. § 11. The Compound Pendulum. Consider an extended body of mass M oscil- lating on a fixed horizontal axis through perpendicular to the plane of the paper. Let G be the highest position of the centre of mass of the body during the oscillation, G' any other position. Draw OLN, the vertical through 0, and GL, G'N perpendicular to OLN. Let GOL = , G'OL = d, OG = L Then the work done by gravity in the fall of the pendulum through the angle GOG' is ;v , ; Mg.LN=Mg{ON-OL) = Mgh (cos 6 — cos cj>). The resistance of the air is neglected. The only other Ch. iv.] The Pendulum. 133 force acting 1 on the pendulum is the resistance at the axis, and no work is done by this force, for it is applied at a fixed point. The kinetic energy generated in the fall is \ MK 2 co 2 , where K is the radius of gyration round the axis through 0, and co is the angular velocity when the centre of mass passes through G'. Therefore I MK Z co 2 = Mgh (cos 0-cos <£). (1) Now let a simple pendulum of length I oscillate through the same angle as the compound pendulum. Its radius of gyration is I, and therefore by (1) its angular velocity co', when it makes an angle d with the vertical, is given by the equation \ I co' 2 = g (cos 6 — cos c/>). K 2 Hence co'=a> if -7- = I, and the motion of the compound pendulum is precisely the same as that of a simple pen- E 2 dulum of length -7- oscillating through the same angle, provided that the resistance of the air can be neglected. Therefore the time of a small oscillation of the compound VK 2 K 2 -j- , and -j- is called the length of the equivalent simple pendulum. If k be the radius of gyration about a parallel axis through the centre of mass, K 2 =h 2 + k 2 . Hence the time of vibration of the compound pendulum about the axis through is /A 2 + k 2 and it is the same about all parallel axes equidistant from the centre of mass. 134 Motion of an Extended Body. [Ch. iv. Axes of Suspension and Oscillation. We shall now show that there are two different values of h, for which the period of oscillation is the same. If I = — - — , I is the length of the equivalent simple pendulum, and, when this is given, h is determined by the equation h 2 -Ih + k 2 ^ o. If h v li % be the roots of this equation, h^ + h 2 = I, and h x h 2 = k 2 . Therefore the times of vibration about parallel axes, at distances A v h % from the centre of mass, are equal. The axis on which the pendulum swings is called the Axis of Suspension, and the parallel axis distant ^ x + ^ 2 from it, about which the time of vibration is the same, is called the Axis of Oscillation. The time of vibration about either of these axes is equal to that of the simple pendulum, whose length is the distance between them. This furnishes the most satisfactory method of deter- mining g. Eater's Pendulum. This was designed by Captain Kater in 1818. As used in the Clarendon Laboratory at Oxford it con- sists of a brass rod, about 40 inches long, f inch broad, and \ inch thick, terminating at one end in a heavy metal bob. Each of the two axes on which the pendulum can be swung is a knife edge, formed by the intersecting faces of a triangular steel prism. One prism is fixed to the bob of the pendulum, the other is mounted on a small slider, which can be clamped anywhere on the rod. The rod also carries another slider, which can be moved Ch. iv.] The Pendulum. 135 along the rod and fixed at any desired point. By the motion of this slider the times of vibration about the knife edges can be slightly altered. The moveable knife edge is adjusted till the times of vibration about both knife edges are nearly equal ; then the slider is adjusted to make the equality still more perfect. The time of vibration t, and the distance I between the knife edges, being accurately determined, g can be found from the formula g = 2 . t § 12. Atwood's Machine. We shall now apply the principle of energy to correct the formula previously obtained for Atwood's Machine. Let m, m' be the masses {m > m'), r the radius of the pulley, h its radius of gyration, and ju, its mass. Then if v is the velocity of the descending mass after a fall h, v is also the velocity of the rim of the pulley, and - is the angular velocity of the pulley. k 2 v 2 Since /x g- is the energy of the pulley, and | (m + m') v 2 is the energy of the masses m, m', the total energy is k 2 v 2 i (m + m') v 2 + ij. -^ . The work done in the fall h is (m-m')gA. 2 (m — m?) gh Therefore v 2 = . uP . m + m ^ rr r~ (m — m!) g And the acceleration is , w^ 2 • m + m + —r 136 Motion of an Extended Body. [ch. iv. — =- can be calculated when the mass form and dimen- sions of the pulley are known. The effect of the inertia of the pulley can be represented by supposing a mass ^-—^ added to each pan. 2 T If T and T' are the tensions in the strings from which m and m! are suspended, we have T'=m' (g + a); T=m{g-a). And T-T= (m'-m)g + a(m + m')= /*"*,{*"*?■ ■ § 13. Impulse. It has been shown that if a force F acts on a body which is free to turn round a fixed axis perpendicular to F, pF=Mk*a, j> being the distance of F from the axis, M , ^ the mass and the radius of gyration of the body, a the angular acceleration. If the force .Facts during a time t, Ft = l, the impulse of-f; and at = w—w , u> and a> being the angular ve- locities at the beginning and end of the time t. Therefore pl = p Ft = M k 2 at = M P ( m(v-p a> )=I. .". (Mk 2 + mp 2 )a — mpv. It is supposed that the shot comes to rest in the pendulum, before the pendulum has sensibly moved from its position of rest. By making M very large this can be secured, for then = pi. Since the initial velocity of the centre of mass is horizontal and is hat, Mha is the resultant impulse on the pendulum, and is horizontal; and the impulse exerted by the axis on the pendulum is Mha> - 1. Hence the impulse of the pendulum on the axis is I- Mha or I (-«)• There is no impulse on the axis if p = -5- • The point whose vertical distance from the axis is -^- is called the Centre of Percussion. 138 Motion of an Extended Body. [Ch. iv. When two bodies rest in contact along a smooth circle, the action of one on the other reduces to a single force, for the pressures at the circumference of the circle are all perpendicular to it, and their lines of action pass through the centre. Hence when a hody turns on a smooth hinge, the action of the hinge on the body can be represented by a single force. Examples. 1. A horizontal rod of mass 6 lbs. hinged at A is kept in position by a string attached at B, and making an angle 30° with BA. Find the tension in the string. 2. Forces act along the sides of a triangle ABC in the directions AB, BC, and AC, and are proportional to the sides. Find a point in AB about which the moment of the forces vanishes. 3. Forces act on a body which are directed along, and are proportional to, the sides of a polygon. Show that they reduce to a couple, whose moment is proportional to the area of the polygon. 4. Prove that the moment of inertia of a circular disc about a diameter is half the moment of inertia about an axis perpen- dicular to the disc through its centre. 5. Three equal heavy particles lie at the vertices of a triangle. Show that their centroid is the centroid of the triangle. 6. Three equal triangles are cut off, one at each corner, from a given triangle. Show that the centroid of the remaining figure is the centroid of the triangle. 7. From a uniform circular disc the portion bounded by a circle described on a radius as diameter is cut away. Find the radii of gyration of the remainder about axes perpendicular to its plane, passing (1) through the original centre of the disc, (2) the centre of gravity. 8. A rod of length I and mass m hangs vertically, being free to turn about its upper extremity, which is fixed. Find the impulse which being applied at the lower end will start the rod so that it will just reach the horizontal position. ch. iv.] Examples. 139 9. A uniform heavy bar AB is free to turn about a fixed hinge at B, and the end A ia attached by a string to a point C vertically above B, and such that BC — AB, the bar being horizontal. A weight equal to that of the bar is suspended from A. Find the tension of the string, and the magnitude and direction of the reaction at the hinge. 10. A circular disc of cast-iron, 10 inches in diameter and 1 inch thick, acts as a pulley for a cord carrying 10 lbs. at one end and 5 lbs. at the other. Find the angular velocity of the pulley and the linear velocity of the weights 50 seconds after starting from rest. 11. A cylinder 5 cm. long, 5 cm. diameter, and density 8"5, is set rotating by a couple with axis parallel to the axis of the cylinder. The moment of the couple being 10000 cm. dynes, find the angular velocity after 10 seconds. 12. Find the centroid of a conical shell on a plane base bounded by right circular cones on the same axis with equal vertical angles. Deduce the position of the centroid of the surface of a cone. [The volume of a cone is £ that of a cylinder with the same altitude and base.] 13. A rectangular lamina, whose shorter edges are 4 feet long, turns round one of its longer edges 50 times a minute. It weighs 441 lbs. ; find its kinetic energy (a) in foot-poundals, (6) in foot-pounds. 14. A cylinder of uniform density whose radius is 2 feet and weight a ton revolves on its axis 150 times in a minute. Find its kinetic energy. If its motion were retarded by a tangential force of 60 poundals, how many turns would it make before coming to rest ? 15. A circular lamina of uniform density revolves on an axis through its centre perpendicular to its plane. If its mass is 100 lbs., its diameter 3 J feet, and it turns round the axis 150 times a minute, find its kinetic energy in foot-pounds. 16. A mass of 10 lbs. moves round the perimeter of a regular hexagon with a constant velocity of 100 feet a second. Find the magnitude and direction of the impulse on the mass when it passes a corner of the hexagon. CHAPTER V. Simple Machines. Statics. § 1. It has been shown that if forces F lt F 2 , ...I„ in the same plane are applied to points A x , A 2 , ... A n of a material system, the work done by the forces in any displacement is equal to the kinetic energy generated. If the different parts of the system are in motion with uniform velocity either of translation or rotation, no kinetic energy is generated, and no work is done by the forces. Let the point A x be displaced in a very short time t through a distance the component of which, in the direction of F x , is * r The work done by F 1 in the displacement is F^. Let * 2 , * 3 ... *„, be corresponding component displace- ments of A 2> A 3 , ... A„, in the same time. If the motion is uniform the total work done is zero. Therefore F 1 s 1 + F 2 s 2 + ... + F n s n = o. Let * x = v x t, s 2 = v 2 t, ... s n = v„t. Then F lVl + F 2 v 2 ...+F n v n = o. F 8 Now F x v x is the power exerted at A, for -—^ is the work V done in unit time. Hence the total power exerted is zero. If 2^ , F 2 be the only forces whose points of application are displaced, F 1 v 1 + F 2 v 2 = o. (i) ch. v.] The Lever. 141 The system may then be used as a machine, i.e. as an arrangement for producing uniform motion against a given force, without necessarily exerting actively an equal and opposite force. Let motion take place against F 2 so that F 2 v 2 is neg- ative ; F 2 may be called the Resistance. The only con- dition for uniform motion is laid down in equation (1). Thus, if Vi > v 2 we may maintain uniform motion by a force F x less than F 2 . We shall call F x the Driving Force. It has generally been known as ' The Power,' but it is best to use ' Power ' only to denote the rate at which work can be done. If F 1 >F 2 , v 2 >v 1 . This result is sometimes applied to obtain very rapid motion against small resistance, where it is inconvenient to apply force at a rapidly moving point. The Machines that we shall examine are the Lever, the Pulley, the Wheel and Axle, the Inclined Plane, and the Screw. § 2. The Lever. The Lever is a rod, straight or curved, constrained to turn round a fixed axis called the Fulcrum. Let ACB represent the Lever, C the point where the axis meets the plane of the forces, W the Resistance, P the Driving Force. Draw Cm, Cn perpendicular to Am, Bn, the lines of action of P and W. If the lever is pivoted on a smooth hinge at C, no work is done by the resistance at C during displacement. Let the weight of the lever be negligible in comparison with P and W. The condition of uniform motion, or of equilibrium, is obtained by taking moments round C, and is P.Cm=W.Cn. 142 Simple Machines. [Ch. V. Thus a Driving Force P at A can balance a Resistance W at B, and by suitably choosing the magnitudes of Cm W and Cn, the ratio -5 can be made as great or as small as we please. W p- is called the Mechanical Advantage. Levers have been divided into three classes in which the Fulcrum, the Eesistance, and the Driving Force respectively occupy the middle position. The crowbar or spade is a lever of the first kind ; a pair of nutcrackers, a wheel-barrow, and an oar are levers of the second kind ; a pair of tongs, used for lifting coal, is a lever of the third kind. The relation Pv 1 + Wv% = o, shows that, the greater the v efficiency, the smaller is the ratio — • That is, the greater the efficiency, the slower is the motion of B when the speed of A is constant. This may be expressed by the statement — ' what is gained in efficiency is lost in time.' In many cases, the directions of P and W are parallel, and ACB is straight. Then ■ 7r = 7m , and 7^, is the On LH Li si Mechanical Advantage. § 3. The Balance. The Balance is a form of lever, which is employed to determine the mass of a body, by comparison of its weight with the weight of one or more known masses. As the Kg. 69. Ch. V.] The Balance. 143 masses of bodies are proportional to their weights, this is equivalent to a direct comparison of masses. The form generally chosen for an accurate physical balance is roughly shown in vertical section in the figure. The lever AB which can turn very freely on its support is Fig. 70. called the beam, and equal pans are hung from A and B in which the masses to be compared are placed. In a sensitive balance the beam should be both strong and light ; these advantages are obtained when it has the latticed form shown in the figure. In order that the beam may turn freely it is pivoted on the column C by a hori- zontal knife edge, which is the edge of a steel prism attached to the beam. The pans of the balance are suspended from the beam by knife edges similar to the central knife edge. A long pointer, which is vertical when the beam is horizontal, is attached to the beam, and its lower end moves, as the balance swings, over a graduated scale. Thus a very small displacement of the beam can be detected. i 4 4 Simple Machines. [Ch. V. An arrangement called the Arrestment is employed for taking the weight of the beam and pans off the knife edges when the balance is not in use. For further experimental details the reader may consult Walker's ' Theory and Use of the Physical Balance.' When the balance is at rest, each pan is in equilibrium, under the weight of itself and its contents and the force exerted by the beam at the knife edge. Hence this force is vertical and is equal to the weight of the pan and its contents, whatever be the position of the masses in the pan. And the action of the pan on the beam is equal and opposite to that of the beam on the pan. Let A, F, B be the knife edges, E the middle point of AB, G the centre of mass of the beam, and let the beam be symmetrical so that GE bisects AB at right angles. It is required to find the position of equilibrium of the beam when masses, P and Q respectively, are placed in the pans. Let N be the mass of the beam, M the mass of each pan. Through F draw FG perpendicular to AB, and aFlb horizontal, meeting the verticals through A, G, B in a, I, b respectively. Let AE = a, CE = c, FC = h, GE = k ; LEGL = 6, so that is the inclination of the beam to the horizon. The forces which act on the beam are (P + M)g at A, (Q+M)ff at B, Ng at G, and the resistance at the knife edge F. Taking moments round F, and dropping the factor g aF(P + M)-6F(Q + M)-Fl.N=o. ch. v.] The Balance. 145 Also aF=(a — e) cos + h sin 0, IF = (a + c) cos 6 — hsmO, al = AE cos 6 — EL = a cos — <£ sin 0, jf? = al—aF = c cos 0— (£ + £) sin 0. Substituting' the values of a.F, IF, Fl, (P + M) |(«-c)cos0 + ^sin0} = (Q + M) {(a + e) cos - h sin 0} + N {ccos 6 - (h + i) sin 0}. Whence tan 0= ( Q + ^ + ^ " ( ^ ^ ~,1 + ^ " The conditions to be satisfied by a balance are the following- : — (1) The beam should be horizontal when the pans are unloaded, i.e. tan should be zero if P = o and Q = o. This requires that should vanish ; i.e. the arms of the balance should be of equal length. If c = o, we have for any load tanfl _ (Q-*> h(P+Q + zM) + N{k + k)' Since tan = o when Q = P, the beam is also horizontal when the masses in the pans are equal. (2) The balance should be sensitive, that is, a small difference in the masses P and Q should cause a con- siderable deflection of the beam. The ratio -= — „ is a convenient measure of the sensi- ng — " tiveness. Since it is equal to 7-7^ — ^ ^r — j^ — =r , it appears that the sensitiveness diminishes as the load increases unless A = o, i.e. unless all three knife edges are in the same plane. In this case, the sensitiveness becomes tt; • 146 The Balance. [Ch. v. The sensitiveness is therefore increased by increasing the length of the beam, diminishing its mass, and bringing its centre of mass very near to the plane of the knife edges. Equilibrium is altogether impossible if the centre of mass of the beam is above the plane of the knife edges. (3) The oscillations of the beam about its position of equilibrium should be fairly rapid, for otherwise weighing is a slow process. In determining the period of vibration of the beam, we shall suppose that h = o, = o, P = Q. Let a, a> be the angular acceleration and velocity of the beam ; the acceleration of the extremity of the beam is then compounded of a a perpendicular to AB, and au> 2 along AB. Thus if the beam only oscillates through a small angle, the acceleration of each extremity is practically a a, upward at A and downward at B. If the pans have no lateral swing, the forces which they exert at A and B axe respectively T, T where T=(P + M)(# + aa), T'=(P + M) (ff-aa), since P = Q. If is the inclination of the beam to the horizontal, cos is sensibly 1 since is small. The moment of the forces which tend to increase is - 2 (P + M)a 2 a-fflc sin

= o. Hence the beam oscillates like a simple pendulum of length Nr< i +2(P + M)a* Nk Ch. v.] The Balance. 147 In order that the oscillations may be rapid this length should be small. This condition is to some extent at variance with those for sensitiveness. The maximum sensitiveness can then only be gained at some sacrifice of rapidity in weighing. Accurate Weighing. When the arms of the balance are of equal length, the three knife edges in the same plane, and the beam horizontal, it may be inferred that the pans are equally loaded ; but it is not easy to ensure the fulfilment of the two first conditions. There are two methods of using a balance which remove this difficulty. I. Let a, b be the distances of the outer knife edges from the central knife edge when the pans are unloaded. Place the mass Q which is to be determined in one pan and load the other with known masses P, till the reading of the pointer attached to the beam is the same as when , the pans are unloaded. The forces thus introduced are the weights of P and Q and an additional resistance at the central knife edge. Since the position of equilibrium is unchanged, the sum of the moments of these forces round the knife edge vanishes, and Pa = Qb. (1) Remove the unknown mass Q from its pan and place it in the other. Balance against it known masses P' till the position of rest is the same as before. Then Fb = Qa. (a) By (1) and (a) Q 2 = PP', and since P is very nearly equal to P', we may write Q = \{P + P'). L % 148 The Balance. [Ch. v. II. In the left-hand pan of the balance, place a mass greater than Q. In the right-hand pan place Q together with known masses m, which bring the beam to a position of equilibrium in which the pointer falls on the scale. Remove the contents of the right-hand pan and replace them by a mass P, which brings the beam to the same position of equilibrium. Then Q + m and P have, when placed in the same pan, the same effect on the beam. Therefore they are equal, and Q = P — m. The steel-yard. Consider the case when the masses are hung directly from the beam, and one standard mass only is employed in weighing, its effect being adjusted by placing it at different points on the beam. This form of bal- ance is called the Steel- yard. Let F be the fulcrum, G the centre of mass of the beam, P the standard mass suspended at E, Q the unknown mass suspended at B, the extremity of the beam. Draw FC perpendicular to AB. Let CB = a, CG = c, EC=x, N = mass of the beam. The beam will rest in a horizontal position if Px = Nc+ Qa. Take a point E in CF such that Ne=P. CH. Then Qa = Px-Nc = P(CE- CH) = P. HE. We can therefore graduate the steel-yard by dividing HCd 'Q Pig. 72. Ch. V.] The Steel-yard. 149 HA into parts each equal to — , m being any integer, and marking the scale at each division. If the beam is horizontal when P is at the #th division from H, or Q = — • p_ Qa _ Qm HE~ x ! Thus Q is determined at once by reading the graduations at the point where P is suspended. § 4. The Pulley. The Pulley is a circular disc mounted on an axle at its centre, which revolves in a framework called the Block, and the edge of the Pulley is grooved so that a cord wound on it cannot slip off. The Pulley is said to be Fixed or Moveable according as its Block is fixed or moveable. In the Fixed Pulley, the Driving Force P is applied at one end of the cord and the resistance W at the other. The cord between the points A and B is acted on by the forces P and W &dA the resistance of the pulley. The work expended by these forces is employed in bending the cord and giving it kinetic energy. We shall suppose that the cord is very light and perfectly flexible, so that it requires no work to bend it. Under these conditions no sensible work is spent on the cord, and the Fig. 73. 150 Simple Machines. [Ch. v. moments of P and W axe equal and opposite to the moment of the total pressure of the pulley on the cord. Hence the moment of the total pressure of the cord on the pulley is a(P— W), where a is a radius of the pulley. The pulley revolves under this pressure, its own weight, and the resistances at the axle. If the bearings are smooth the resistances reduce to a single force which is directed towards the axis of motion and has no moment round it. The centre of mass of the pulley is on the axis. Hence the moment of the forces which act on the pulley is a(P — W). If the pulley revolves with uniform velocity no kinetic energy is generated, and P = W. If the pulley revolves with angular acceleration a and has a moment of inertia K, a(P— W) = Ka. Hence the tension of the string is not the same through- out, unless the velocity is uniform or zero. The Fixed Pulley can only be used to change the di- rection of a force; its Mechanical Advantage is 1. The Single Moveable Pulley. AFGC is a cord, with the end A fixed, passing under a moveable pulley from the block of which a mass, of weight Q, is hung. It is required to raise the mass with uni- form velocity by applying the driving force P as a tension to the string GC. Let w be the weight of the pulley. If v v v 2 are the vertical velocities of the block, and of a point on the string GC, Pv 1 — (Q,+w)v 2 = o. And since v, = 2 ■ Q + w = a P. Ch. V.] Pulleys. 151 If the weight of the pulley can be neglected, -=• = 2 ; this is the Mechanical Advantage of a single moveable pulley when the strings are parallel. The tension of the string FA = P= \ (Q + w). Thus the pulley is half supported by the ceiling and half by the Driving Force. § 5. First System of Pulleys. A greater Mechanical Advantage can be obtained by combining several pulleys. In the system shown in the figure there are two blocks, the upper fixed, the lower moveable, and several pulleys (two in the figure) may be attached to each block. A string is attached to the upper fixed block, and passes round every pulley ; the driving force P is applied to the free end of the string. The resistance is generally the weight of a body which is raised by the lower block. Let Q be its measure in absolute units of force, and let the weight of the pulley and block be negligible. If the body is raised with uniform velocity v 2 , and the point C on the string, at which P is applied, has velocity v x , Pv 1= Qv 2 . In the figure there are four strings at the lower block, and if the lower block rises through a distance v 2 , C falls through 4 v 2 . Therefore ^ = 4 „ 2> an & 4 P = Q. The Mechanical Advantage here is 4. When there are n strings at the lower block the 152 Simple Machines. [ch. v. Mechanical Advantage is n, if the weight of the lower pulley is neglected. If the weight of the pulley is w, Pv 1 = (Q + w)v 2 , and w the Mechanical Advantage is « — -^ . The pressure on the upper block is P + Q + w. If any forces P and Q are applied to the Pulleys, it is easy to calculate the acceleration of C and of the lower block, when the masses of the pulleys can be neglected. For when the lower block ascends through a height h, G descends through 4 h, and the work done by P and Q is {4P-Q)L This is equal to the kinetic energy generated. If v , v are the initial and final velocities of the lower block, 4 v , 4 V will be the initial and final velocities of C. Hence the kinetic energy generated is ~(» z -V)(« + i6P). And the acceleration of the lower block is y^- or 2 h Q + 16P ' In the general case the acceleration is —= %■ • Q + n*P § 6- Second System of Pulleys. The combination of moveable pulleys shown in the figure may also be used to obtain Mechanical Advantage. Each pulley is supported by a string, with one end attached to the ceiling or some fixed support, and the other to the block of the neighbouring pulley above it. The Driving Force P is applied at D, and the Resistance is the weight of a body hung from the lowest block. Ch. V.] Pulleys. 153 If D rises through a height v 1; the block K and the end of the string YC rise through 4 v 1 . The end of XB rises through 4 • \ v v that of TA through 4 • \ • 4 v i ; and the lowest block Arises through 4 ■ 4 • 4 • 4^1 1 or -j v-,. a 4 1 Hence if v x and v 2 are the velocities of the point ID and the block F, the condition of uniform motion is Pv 1 —Qv 2 = o, assuming that the weights of the pulleys can be neglected. And since v^ = 2 i v 2 , Q = 2*P. Hence the Mechanical Ad- vantage is a 4 when there are four pulleys. If there are n pulleys the Mechanical Ad- vantage is 2 "- We may consider the block K as rising uniformly under the Force P and the tension of the cord YC. The case of the single moveable pulley shows that this tension is a P. In like manner the block H may be considered as rising under this tension and the tension of XB. In this way we find that the tensions of the cords ZB, YC, XB, TA are P, a P, 4 P, 8 P respectively. Pig. 76. 154 Simple Machines. [Ch. v. Hence the total pull on the support is (a*-i)P. The reaction to this, together with P, balances Q. When there are n pulleys the pull on the support is (a B -i)P. There is no difficulty in showing that when the weights of the pulleys are neglected the acceleration of the lowest block for any values of P and Q is 2 n P-Q 2 2n P+Q' ff ' Let us now suppose that the pulleys F, G, H, K have weights w lt w 2 , w 3 , w 4 . It has been shown that their velocities upward are v 2 , 2v 2 , 4V 2 , 8v 2 . Hence the condition for uniform motion is (§ 1), Pvy—Q v 2 — w 1 v 2 —2w 2 v 2 — 4w 3 v 2 — 8 w i v 2 = o, or 2 i P = Q + w 1 + 2iv 2 + 4to 3 + Sto i . The reader will find it a useful exercise to determine the tension of each string in this case. § 7. Third System of Pulleys. If the page is turned upside down the second system of pulleys is converted into that shown in the annexed figure ; in this system each pulley hangs from the block above it, and all the strings are parallel and attached to the weight. We have interchanged the positions of the support and the resistance. Hence when there are n pulleys, Q = (2 M — i)P, and the pull on the support is 2 n P, provided that the weights of the pulleys can be neglected. The reader should have no difficulty in finding the tensions of the cords in terms of P, and thence deducing these results. It can be easily shown that the acceleration of the Ch. V.] Pulleys. 155 hanging body is ~—^ ^-p — ^g for any values of P and Q. Let w lt w 2 , w 3 , w 4 be the weights of the four pulleys F,G,H,K. When Q rises through a height v 2 , F remains at rest, G falls through v 2 , H through 2v 2 + v 2 or $v 2 , K through a x 3 v 2 + v 2 or 7 v 2 , and the point of application of P through 2xyv 2 + v 2 or 15 v a . • Hence by § 1, (a 4 -i)P + (a 3 -i)«> 4 + (a 2 -i)w 3 + w 2 = Q. § 8. The "Wheel and Axle. This consists of two pulleys of different radii revolving in one piece on the same horizontal axis. Tbe Resistance Q is applied to a cord coiled round the Axle, the Driving Force P to a cord wound in the opposite direction on the wheel. If c and b are the radii of the Wheel and the Axle, the condition for motion with uniform velocity is (by taking moments) Pc=Qb. The Mechanical Advantage is j ; it may be increased indefinitely by diminishing the radius of the Axle, but this tends to weaken the machine, so that it may not bear the forces applied to it. Fig. 77. i56 Simple Machines. [Ch. V. Fig. 78. To overcome this difficulty the Differential Axle has been used. It consists of two co-axial circular discs ABX, abx, (Fig. 79) of radii b, b', with grooves cut in their edges, forming the same solid piece with the Wheel. One end of a cord is secured to a point on ABX; the cord quits the groove at X, passes under the moveable pulley AH, and thence passes into the groove of abx at x, the end of the string being fastened in this groove. The machine is now acted on by the Force P, the resistance at the axis, and the tensions T in the strings HX, Ax. The condition for uniform motion is obtained by equating the sum of the moments round the axis to zero. Therefore Pc = T(b~ b'). Also supposing that the pulley HA has no weight, Q = a T cos a ; where a is the angle which each string Ax, Hx makes with the vertical. Therefore P = 0=0. q. 2 Since GHZ, GKH are similar tri- angles. Fi s- 8o - The Mechanical Advantage is therefore the ratio of the length to the height of the plane. Next, let the Driving Force be horizontal. In this case P is perpendicular to GH, Q to KH, and R to GK. Therefore — — — - — • ineretore GH~ HK~ GK And the Mechanical Advantage is the ratio of the base to the height of the plane. § 10. The Screw. Draw two parallel lines AB, CD (Fig. 81) ; on AB take points a, b, c, d, ..., such that Aa = ab = bo = cd = ... = h. And draw am, in, op, dq, ... perpendicular to AB. Join Am, an, bp, cq, .... If the portion of paper ABCB be cut out and bent to form a cylinder with AC as the perimeter of the base, the lines AM, an, bp, cq, ... form a continuous curve on the cylinder known as the Helix. 158 Simple Machines. [Ch. v. Fig. 81. Let a plane figure F, one side of which is the line rs ( < h), be applied to the cylinder, so that * is at a point of the helix, rs parallel to AB S and the plane of F perpendicular to the plane which touches the cylinder along rs. Then, if s moves along the Helix so that these conditions are satisfied, the area F traces out a surface resembling that of a corkscrew. The cylinder and the surface when imitated in metal or wood form a screw, and the surface is called the Thread of the screw ; the angle mAG is called the Pitch of the Screw. We shall only consider the case when F is a rectangle. The screw then formed is shown in the accompanying figure. When used as a Mechanical Power the screw revolves in a fixed hollow cylinder of the same radius with its inner surface grooved to form a screw of the same pitch. The Screw is used to obtain great pressure in the direction of its axis. The Driving Force is applied at the end of an arm perpendicular to the axis of the screw, and is itself per- pendicular to the arm and to the axis. Therefore if P is the Driving Force, and a is the length of the arm, 2-naP is the work done by P in one revolu- tion of the arm. During this revolution the end of the screw advances by the distance h. Therefore if Wis the resistance Wh = 2 it a P. Ch. v.] The Screw. 159 Hence the Mechanical Advantage is — ; — Since -7 A h is the number of threads per unit length (inch or centimetre) of the screw, the Mechanical Advantage is obtained by- multiplying the number of threads per unit length of the screw by the circumference of the arm. § 11. Principle of Virtual Work. In each Machine that has been considered the con- ditions for uniform motion are, by the First Law of Motion, the same as the conditions for rest. Thus in the three systems of Pulleys, a weight P applied as the Driving Force can balance weights nP, 2 n P, or (2 n — 1) P, if the weights of the Pulleys are neglected. Hence the conditions under which a material system can remain at rest in a given position may be obtained by supposing the system to be in motion through this position, and equating to zero the powers of the acting forces. Since the motion is not real but hypothetical, this method is called the Method of Virtual Velocities. Example. — A smooth triangular prism ABC forms a double inclined plane, on which particles m, m' rest, connected by a light string which passes over a pulley at A. To find the condition of equilibrium. Let BC be horizontal, and denote the angles at B and C by B and C. The external forces acting ^J^ on the system composed of the particles and string are the resistances of the planes at P and Q, the weights mg, m'g, and the resistance of the pulley at A. Pig 83. The internal force is the tension T in the string. If m moves with velocity v up AB, m! has the same velocity 160 Statics. [Ch. v. down AC. The tension in the string does work at the rate Tv per unit time in the displacement of m, and at the rate - To in the displacement of m'. Therefore on the whole it does no work. The resistances of the pulley and planes also do no work, for the motion at their points of application is perpendicular to their direction. Therefore mg, m'g are the only forces which do work, and their power is (m'g sin C— mg sin B) v. Hence the condition of equilibrium is m' sin C = m sin B. The tension in the string can be found by considering a virtual displacement of P only. We then have Tv = mgv sin B, or T= mg sin B. In many questions it is more convenient to consider the work done in small hypothetical displacement of the points at which forces act. It has been shown in Chap. IV, § 4, that internal forces acting on a rigid body do no work in a displacement of rotation, and it also follows from Chap. Ill, § a, that they do no work in a displacement of translation. Example. — AC is the base and CB the height of a smooth inclined plane AB; D is a given point in CB produced, and two particles of mass P and R are connected by a string passing over a smooth peg atD. R is placed on AB at Q, and P hangs freely. To find the position of equilibrium. Let DQB = 6, BAC = iu The acting forces are the weights of the par- ticles, the resistance of the plane, and the ten- sion of the string. Let P be displaced Fig. 84. " downward through a dis- tance h to p. Then, the string being unstretched, R will be displaced up the plane to q. The tension does no work since the string Ch. v.] Virtual Work. 161 remains unstretehed, and the resistance of the plane does no work since it is perpendicular to the displacement at Q. The work done by gravity on P is Pg . h. Also DQ - Dq = h, and if qe is perpendicular to DQ, De = Dq (within a negligible quantity) when h is very small. Therefore , „ n „„„ a h = Qe = Qq. cos 8. The vertical component of Qq is Qq sin a or -r- ■ r coad Therefore the work done on E is — Eg . -r- , and cosfl _ , , sin a y y cosfl ' , JB . or cos 8 = — sm a. This method of finding the conditions of equilibrium is called the Method of Virtual Work. § 12. Stable and Unstable Equilibrium. If a body is slightly displaced from a position of equilibrium X to another position T the forces are generally slightly altered, in respect either of their magni- tude or of the direction and position of their lines of action. The body therefore is generally no longer in equilibrium. The forces which act on the body in its position T may tend either to displace the body still further or to restore it to the position X. In the former case the equilibrium is said to be un- stable, since a slight displacement of the body from its position of rest calls into play forces which increase the displacement and prevent the body from coming to rest. If, however, the forces tend to restore the body to the position X, they generate in the displacement from Yto X a certain kinetic energy which carries the body beyond X M 162 Statics. [Ch. v. In the new displacement forces arise which either increase or check the displacement beyond X. In the former case the equilibrium is unstable, in the latter it is stable. If in the position J" the forces which act on the body- maintain it in equilibrium, the equilibrium at X is said to be neutral. Instances of bodies in stable equilibrium are afforded by all ordinary cases of equilibrium, e.g. a stool resting on three legs, or the bob of a pendulum at rest. Cases of unstable equilibrium are only theoretical. A cone can conceivably, but not practically, rest on its vertex, or a chair on two legs, or an egg on its end. An instance of neutral equilibrium is the case of a sphere resting on a smooth horizontal plane. In examining the stability of equilibrium it is useful to notice that the centre of mass of a heavy body tends to assume as low a position as it can consistently with the other constraints to which the body is subject. This position of equilibrium is stable. Examples. 1. A cylinder with elliptic cross section rests on a smooth horizontal plane, to find the positions of equilibrium. These are two in number, the major axis being vertical in the one, the minor axis in the other. The first position is unstable, for gravity can do work in a displacement of the centre of mass, and the moment of the forces when the body is displaced tends to displace the body still further. The second position is stable, since the centre of mass is in its lowest possible position, and the moment of the forces called into play by a displacement tends to restore equilibrium. 2. A prism whose section is a triangle ABC with each angle at the base AB equal to 30° rests on a rectangular face with Ch. v.] Stable and Unstable Equilibrium. 163 BC horizontal ; to find the mass that can be added at A without upsetting the prism. If the prism is slightly displaced round C the acting forces are the weight W of the prism, the weight w of the added mass at A, and the resistance at the point C. Let Q be the centroid of ABC; pro- duce AG to D, and draw OL, AM per- pendicular to BC. Take moments round C. The equilibrium becomes unstable wh en w.CM>W. CL. Now DL = §DM, and Z)C = \DM. Therefore CL = \DM = $ CM. W Therefore the prism will be upset when w > — 3 § 13. Friction. We have till now assumed that when two bodies rest in contact along a plane surface, the stress between the bodies is perpendicular to this plane. Thus a body A resting on a horizontal plane could be set in motion with constantly increasing velocity by any horizontal force, however small. Experience shows that this is not the case ; small hori- zontal forces can be applied to A without causing it to move ; equal horizontal forces opposing these must there- fore exist, and they are attributed to friction between A and the surface on which it rests. Laws of Friction. The body A exerts a vertical force B, equal to its weight, on the plane S. If a gradually increasing horizontal force be applied to A, A will begin to move when this force attains a certain magnitude so. M 2, 164 Statics. [Ch. v. Now let a body B of equal mass be placed on the top of A ; the pressure on the plane is now iR, and experiment shows that the force required to just set A in motion is zx. Similarly, if A exerts a vertical pressure nR on the plane, the force required to just move A is nsc. Hence the First Law of Friction follows. With the same surfaces in contact, the minimum tangential force required to generate motion is proportional to the pressure perpendicular to the surfaces. If the position of A is changed so that any other part of its surface is in contact with the plane S, A exerts the same force on the plane as before, and experiment shows that the same force is required to produce motion. Therefore the horizontal force required to generate motion is independent of the area of the surfaces in contact, the total pressure remaining the same. This is the Second Law of Friction. Thus if R be the force exerted by A perpendicular to S, there is generally a tangential force between A and S, which has a magnitude pR when A is on the point of motion. The Laws just enunciated imply that, when the substances of the bodies which are in contact are given, ju does not depend on their weights or dimensions, jn is a constant, and is called the coefficient of friction between these substances. If a body slides on a rough surface under known constant forces its acceleration when small is uniform. Therefore the force of friction is constant, and the Third Law of Friction follows, viz. : — The force of friction is independent of the velocity, when the velocity is small, and its direction is opposite to that of motion. Friction during motion is generally somewhat less than just before motion. Ch. v.] Friction. 165 Rankine supposed friction to be due to the fact that even apparently plane surfaces consist of small depressions and elevations, and that when two such surfaces are in con- tact their inequalities lock into one another, thus requiring a certain exertion of force to move one body over the other; and, the greater the pressure between the two surfaces, the more deeply do they interlock. When the force of friction has attained its greatest possible magnitude, the body A is said to be in limiting equili- brium. Angle of Friction. Cone of Friction. If a body rests on a rough plane surface, the action of the surface on the body at any point may have any direc- tion lying within a certain cone. For if ON be the normal to the surface at 0, make ON = B, the normal component of the action at 0, and draw NE perpendicular to ON and equal to pR. Then OF represents in magnitude and direction a possible limiting value of the action of the surface on the body. ji ig 86 The angle NOF is called the angle of friction, and the cone formed by the revolution of the triangle NOF round ON is called the cone of friction. All straight lines drawn from within this cone repre- sent possible values of the action of the surface on the body, and there are no other possible values. When a body H rests in contact with two rough surfaces P and Q, the resistance at each surface is gene- rally indeterminate. For let the cones of friction at P and Q be represented i66 Statics. [Ch. V. in section by the angles APJB, CQD. Then if ,any point K of the vertical KG, through the centre of mass of H, lies within both cones, forces along PK, QK, and-STGcanbe found which maintain H in equi- librium, and the values of these forces depend on the position of K. If there is limiting equi- librium at either P or Q, the problem of finding the resist- ances at P and Q is determin- ate, since then the direction of either PK or QK is known. If no point on KG lies within both cones, the position of H is not a possible position of equilibrium. A heavy rough particle P rests in limiting equilibrium on an inclined plane. To find the inclination of the plane. Let W be the weight of the particle P, R the action on P perpendicular to the plane ; then fxR is the action parallel to the plane, and is directed up the plane along the line of greatest slope. Since the particle is in equilibrium, txR:AC:R:CB, Pig. 88. or M = AC CB = tan a. Therefore a body rests in limiting equilibrium on the Ch. v.] Friction. 167 inclined plane, when the angle of friction is equal to the inclination of the plane ; and if the inclination of the plane is adjustable, the angle of friction can be found by tilting the plane till motion is about to take place. A better way of finding the coefficient of friction is the following : — C is a horizontal plane forming the upper surface of one body ; the other body A, of mass N, rests on this by a plane face. A horizontal string passes from A over a pulley E, mounted on p^-| E friction wheels, and from the free end c of the string is hung a box in which masses can be placed. These masses are gradually increased till A begins to move. F - g If If be the suspended mass when motion begins, Mg is the horizontal stress between A and C, and M. -^ is the coefficient of friction. § 14. Initial Motion under Friction. The results just obtained apply only to particles. The equilibrium of an extended body resting on a rough sur- face may be broken in two ways — the body may slip along the surface or roll on some point in its base. If the surface is rough enough, the body will roll before it slides ; an example will make this clearer. A prism, whose section by a plane perpendicular to its edges is a right-angled triangle ABC, rests on a horizontal plane (Fig. 90). To the middle point of the upper edge a horizontal cord is attached, which passes over a smooth pulley and carries a weight W. To find the initial motion of the prism if W is .gradually increased. If there is equilibrium, the forces acting on the prism are i68 Statics. [Ch. v. its weight G, the horizontal tension W, and the resistance of the plane BC which has components, vertical and hori- zontal, equal to G and W. Therefore sliding- begins if W>y. G. But the equilibrium may be broken by motion round C. Let the resultant upward force exerted by the plane on the prism act through a point in BC, distant x from C, and let AC = b, BC = a. Taking moments Fig. 90. round C we have as the condition of equilibrium bW+xG = iaG. Therefore as W increases * diminishes, and when the total resistance acts at C, W=±G. 3 b For any greater value of W there cannot be equilibrium ; the prism begins to turn on its edge through C. Therefore, as W increases, equilibrium is broken by roll- ing if [i. > — 7, and by sliding if ju < —r Friction, on the Inclined Plane. A particle is on the point of gliding up a rough inclined plane under a force P making an angle a with the line of greatest upward slope. To find the condition of equilibrium. Let W be the weight of the particle ; E, i*B the normal and tangential pressures of the plane on the particle, the .' inclination of the plane to the horizon. Ch. v.] Friction. 169 Resolving along the plane, P cos a — T^sin 6 — jx R = o. And resolving at right angles to the plane, P sin a — ^Tcos 6 + B = o. Hence P (cos a + fj, sin a) = W(sm 6 + ix cos 8). This is also the condition for motion with uniform velo- city, provided that the coefficient of friction under motion is substituted for jx. If the particle is on the point of sliding down the plane, P (cos a — ju sin a) = W (sin d— /x cos 0). Friction on the Screw. Let a be the. pitch and h the radius of the screw, and let the driving force P be on the point of over- coming the resistance W. At each point H where the thread is in contact with the sister screw, it is subject to a force r perpendicular to its surface, and to a force l*.r tangential to its surface, opposing the motion, and perpendicular to the plane through H and the axis of the screw. Let R be the sum of all the resistances r at the point of contact of the screws. Resolving parallel to the axis of the screw, W—jR cos a + /j. It sin a = o. (1) To obtain another relation, suppose that the screw is turned through one revolution, and let * be the displace- ment of a point on the thread measured along the path by which the point travels. Then * cos a = 2, it 6, if the thread is very narrow ; * sin a — h, where 7- is the number of threads per unit length. 170 Statics. [Ch. V. If a is the radius of the arm at the end of which P is applied,' the work done in the displacement is Zira.P— Wh—fx R s. Equating this to zero, and substituting for s and h, we have a P cos a = b (Wsin a + pR). (2) Substituting in (a) the value of R obtained from (1), (cos a— pi sin a) aPcosa=(cosa— ^sino) b Wsina + b p W = b Wcos a (sin a + /x cos a) ; _, b W sin a + u cos a or P= ; a cos a — ix sin a Action at a Rough Hinge. It has been shown that if two smooth coaxial cylinders press on each other, the pressures reduce to forces which all meet the axis of the cylinders and can be replaced by a single force. If the hinge is rough, the pressures are not necessarily directed towards the axis, and the action does not necessarily reduce to a single force. Action at a Joint. An illustration will best explain this. ABC is a bar, jointed at B, and suspended by a smooth hinge at A. When in equilibrium it sets itself so that G, the centroid, is vertically below A. Let E be the centroid of BC. BC is in equilibrium under its own weight, equivalent to a force W sA, E, and the action of the joint. Hence the action of this joint must be equivalent to an Fig. 91. Ch. v.] Action at a Joint. 171 upward force W at B and a couple of moment W. BE cos 0, where is the inclination of BC to the horizon. The complexity of the force at the joint is due to one part being under a pressure and the other under tension.. With reference to this, consult chapter VII, § 13. Problems on the equilibrium of systems of bars can be very elegantly treated by the methods of Graphical Statics, but space does not permit us to discuss them here. Examples. When the forces are all in one plane it is necessary to express the following conditions : — (1) The sum of the component forces in any direction vanishes ; this condition is satisfied when the sum of the components vanishes for each of two perpendicular directions. (2) The sum of the moments of the forces round any point vanishes. These may be replaced by the following conditions : — The sum of the moments of the forces round any three non- collinear points in the plane vanishes. For let A, B, C be the three points, F, G, H, etc. the forces. The moment of F round A is equal to the moment of F round B together with the moment round A of a force F atB. Applying this result to each force we have M A = M B + B A when M A , M B are the sums of the moments of the forces round A and B respectively, and B A is the moment round A of the resultant of F, G, H,. . . supposed to act at B. Similarly M a = M B -^B c . Now if M A , M B , M c are all zero, B A = o and B c = o. But B is not on the straight line AC, and therefore the resultant of F, G, H . . . at B cannot pass through both A and C. Hence, as its moment round both points is zero, it must vanish, and the con- ditions (1) and (2) are satisfied. 172 Examples. [Ch. v. 1. A circular cylinder rests with its horizontal axis on a rough inclined plane, and is supported by a string coiled round its middle section and supporting a weight. To find the condition of equilibrium. Let W be the weight of the cylinder, P the weight suspended from the cord. Since these forces are vertical, the resistance at E must be vertical also. Let u be the inclination of the plane. Taking moments round E IT sin a = P(l -sin a). This relation determines the weight P. Fig. 9 2 - 2. A perfectly flexible uniform string hangs from points A and B. To find the conditions of equilibrium. The tension at P reduces to a single force directed along the tangent at P. Let T v T 2 be the tensions at P and Q, 6 X and 6 % the angles which the tangents at P and Q make with the horizon. The string between P and Q is at rest under its own weight and the tensions ; let the arcs AP, AQ be equal to s x and s 2 respectively, and let p be the mass of unit length of the string. Then Tj cos 6 1 = T % cos 6 it T x sin 6 X — T 2 sin # 2 = p (s 2 — Sj) g. Hence the horizontal component of the tension is everywhere the same and is equal to t, the tension at the lowest point. Therefore Tj sin 8 1 — T% sin 6 1 = t (tan X — tan 6 2 ). And pgs 1 + tta,TiB 1 = pgs 2 + tta,nd 2 . The curve is therefore such that s + — tan 6 = constant, where P9 * is measured from A. Also the centre of gravity of the arc PQ lies vertically above the inteisection of the tangents at P and Q, and since the equilibrium Ch. v.] Examples. 173 is stable it lies lower than the centre of gravity of any other arc of equal length joining P and Q. 3. AB and CD are equal rods that can turn freely round a rivet passing through their middle points ; the ends A and C are joined by a thread of given length I, the plane of the rods is vertical, and the system rests with A and C on a horizontal plane. Given weights P and Q are hung at D and B ; find the tension of the thread and the stress on the rivet, the weights of the rods being neglected. Let X and Y be the forces exerted by the plane at A and C. The system formed by the two rods is at rest under the forces X, Y P, and Q. Resolving them vertically, X+Y = P+ Q. Taking moments round 0, — (X— P) = — (Y— Q), whence X-Y=P-Q. Therefore X = P, Y = Q. Let T be the tension in the string. The rod AB is at rest under the forces X (or P), Q, T and the force exerted on it by B at the rivet. If the components of this latter force are x and y, we have by resolving the forces x+T=o, y + P-Q = o. Taking moments round A, we have, if AD = a, a- I tss -x + -y = lQ. We have then three equations from which x, y, and T can be found. 4. A table of weight W is supported on three legs, which meet the horizontal plane through the centre of mass in A, B, C. The distances of the centre of mass from BC, CA and AB being p, q, r, it is required to find the pressure on each leg. Let a, b, c denote the sides BC, CA, AB of the triangle ABC; A, B, C the angles opposite to a, b, c. Let R, 8, T be the pressures on A, B, C. Taking moments round the axis BC we have R . c sin B = p . W. 174 Examples. [Ch. V. Similarly by taking moments round CA and AB, S.asinC = q.W, T.bamA = r. W. Hence R, 8 and T are known. Since R + S+ T = W, we have ap + bq + cr — 2 A ABC. Equilibrium is impossible if the centre of mass of the table lies without the triangle. 5. A cube is placed with one edge on a rough horizontal plane with coefficient of friction P, and a parallel edge on a smooth plane inclined at an angle 45° to the ver- tical ; if 6 is the inclination of the base of the cube to the horizon when on the point of sliding down, show that (1 + 3^) tan 6= i-n. Let the plane of the figure be a vertical plane through the centre of mass of the cube, perpendicular to the inclined plane. AB is the section of the lowest face of the cube, G the centroid of the cube, and if A.B = ia, AG = aVz. The resistances at the edges of the cube reduce to a single force R at A perpendicular to AE, a vertical force S at B, and a horizontal force /i S at B parallel to BE. The weight of the cube is W. Draw the vertical lines AL, BN, OM, and the horizontal line AMN through A. Then AM = AG cos (45 -6) = a\ / 2cos(45— 8) = a (sin 6 + cos 6), AN =2a cos 6. Fig. 94- Ch. v.] Examples. 175 Resolving R into its vertical and horizontal components, and equating the vertical and horizontal forces separately to zero, V2 V2 Therefore W=bi + i)S. ( 1 ) Also the sum of the moments round A vanishes. Therefore AM. W-AN.8+AL. pS =o, or aW (sin 8 + cos 6) — 2 a cos 8 . 8+ 2 a sin 6 . n S = o. Substituting from (1) and dividing by aS, (3 p + 1) Bin. 8 + (n— 1) cos 8 = 0. Which is the relation required. 6. A horizontal bar of length 10 feet is suspended by vertical strings attached to its extremities. If the mass of the bar is 10 lbs. and a mass of 4 lbs. is attached to it at a distance from 4 ft. from one end, find the tensions in the strings. 7. A ladder of length 30 feet rests in limiting equilibrium on rough ground, with its upper end against a smooth vertical wall 24 feet above the ground. If the weight of the ladder is 80 lbs., find the pressure on the ground and the coefficient of friction. 8. A mass of 6 lbs. is supported on a smooth inclined plane, rising 5 feet in 18, by means of a string inclined at an angle 30° to the plane. Find the tension in the string. 9. Two masses P and W balance on the Wheel and Axle. Show that if they are interchanged, W will after one second be de- aW(W-P) scendmg with velocity yri-WP+P 2 ' 10. A mass of 12 lbs. rests on a rough plane inclined 30 to the horizon. What force must be applied to it at an angle 30° to the vertical in order that it may be on the point of moving up the plane, the coefficient of friction being \ ? 11. A barrel weighing 5 cwt. is lowered into a cellar down a smooth slide inclined at 45° to the vertical. It is lowered by two ropes passing under the barrel, one end of each rope being fixed, while two men pay out the other ends of the ropes. What pull in 176 Examples. [Ch. v. pounds weight must each man exert that the barrel may descend with uniform velocity ? 12. In example 2, if the length of the string is I and a is the angle which each upper end makes with the horizon, t tan a = \g pi. 13. A chain hangs between two points. Prove that the square of the tension at any point P is equal to the sum of the squares of the tension at the lowest point and of the weight of chain between P and the lowest point. 14. A right cone, diameter of base a, height &, is placed on a plane with coefficient of friction 71. Find the greatest value p may have that the cone may slide without turning over when the inclination of the plane is gradually increased. 15. A uniform rod rests with one end on a rough horizontal plane and the other end on a smooth plane inclined at 30° to the horizon. Prove that if the rod is on the point of sliding down, its inclination 8 to the horizon is given by 2 fitund = 1 -11V3, l* being the coefficient of friction. 16. Prove that any force can be reduced to another force and a couple in the same plane with it ; and hence that any system of coplanar forces may be reduced to a single force acting at a specified point together with a couple. 17. A particle of mass 50 lbs. hangs from two strings, re- spectively 7 and 8 feet long, the upper ends of which are fastened to pegs 9 feet apart in the same horizontal line. Find the tension in each string. 18. An uniform beam 12 feet long rests with one end against the base of a wall 20 feet high. The beam is supported by a rope 13 feet long attached to the top of the beam and the summit of the wall. Find the tension in the rope, neglecting its weight and assuming the mass of the beam to be 100 lbs. 19. A beam AB of length I leans against a rough vertical wall, the end A being prevented from sliding along the smooth hori- zontal plane AD by a string AD attached to the wall. Find the tension in the string, if p is the coefficient of friction and a the inclination of the beam to the vertical. Ch. v.] Examples. 177 20. A step-ladder in the form of the letter A, with both legs inclined at an angle u to the vertical, is placed on a horizontal floor and is held up by a cord connecting the middle points of its legs, there being no friction anywhere ; prove that when a weight W is placed on the top of the ladder the tension in the cord is increased by TFtan a. Also if W is placed on a step at a height from the floor equal to — of the height of the ladder, the increase of tension is — W tan a. n n 21. A uniform triangular plate with one vertex A resting on a smooth horizontal plane is supported in a vertical plane by a cord attached to the vertices B and C, and passing over a fixed pulley at D. If BC is horizontal, show that the pressure on the plane is one-third of the weight of the plate. 22. A cylinder of radius r whose axis is fixed horizontally touches a vertical wall along a generating line. A flat beam of uniform material of length 2 1 and weight W rests with its ex- tremities in contact with the wall and cylinder, making an angle 45° with the vertical. Prove that in the absence of friction r Via' that the pressure on the wall is £ W, and on the cylinder is W 5 CHAPTEE VI Gravitation. § 1. The Law of gravitation asserts that any two par- ticles of matter attract one another with a force which is proportional directly to the product of their masses, and inversely to the square of their distance apart. The evidence in favour of this law is mainly derived from the consideration of the elliptic orbits described by the planets round the Sun. A few properties of the ellipse that will be useful are proved in §§ 3-4. § 2. An ellipse may be most simply obtained in the fol- lowing way. Let ABA' (Fig. 95) be a circle with centre C. From a point Q on the circle draw an ordinate QN per- pendicular to a diameter AC A', and on QN take a point P such that PN : QNis a ratio < 1. If different positions of Q on the circle are taken, and PN . the fraction -pr^. is the same for all, the point P lies on an ellipse. Since the point Q may lie above or below the line AC A', the ellipse consists of two precisely similar portions, one above, the other below AC A'. C is called the centre of the ellipse, any chord through C Ch. VI.] The Ellipse. 179 is called a diameter, and AC A', being obviously the longest diameter, is called the major axis. If BOB' be the diameter of the circle perpendicular to Fig. 95- ACA', b, b' the points where it meets the ellipse, bCb" is called the minor axis of the ellipse. It divides the ellipse into two similar parts. § 3. Let QN\ qn be two ordinates of the circle ABA', P andjs corresponding points on the ellipse. If T be the point in which Qq meets AA', 7F - = = — • , by the definition of the ellipse. In, qn pn J r Therefore Pp passes through T. Now suppose that q moves along the circle till it coincides with Q ; then p moves along the ellipse till it coincides with P, and the chords Qq, Pp approach and ultimately coincide with the tangents at Q and P to the circle and ellipse respectively. Hence if QN, QT be the ordinate and tangent at Q (Fig. 96), TP is the tangent at P to the ellipse through P whose major axis is AA'. i8o Gravitation. [Ch. VI. Let TP meet the circle in T and Y ; and draw YS, YS' perpendicular to TP, meeting AC A' in S and S'. We shall show that S and S' are fixed points in ACA' whatever be the position of P on the ellipse. Fig. 96. Produce YS to meet the circle in Z, and join CQ. Then, since ZYY' is a right angle, ZY' passes through C (Euclid III. 31). In the triangles CSZ, CS'Y', the angles CSZ, SCZ are equal to CS'Y, 8CY, each to each, and CY = CZ. Therefore (Euclid I. 26) CS = CS' and SZ = S"Y. Now the triangles PNT, SYT are similar. FN SY Therefore (Euclid VI. 4) m = j^ • Similarly the triangles PNT, S'Y'T are similar, and PN _ S'Y' NT ~ Y'T ' PN 2 SY.S'Y' SY.SZ Therefore jyy2 — fY, TY' ~ TY. TY' Ch. VI.] But TY.TY' = TQ 2 . The Ellipse. 181 Therefore PN 2 8T.SZ FN 2 or NT 2 TQf NT 2 ~ TQ 2 ~ SY.SZ Again, in the similar triangles CQT, QNT, NT^ QN _ QN _PN TQ~ CQ~ CA ~ Cb ' „. e FN 2 NT 2 FN 2 Therefore _ = _ = ^__. AndSY.SZ=Cb 2 . Now SY.SZ=AS.SA' (Euclid III. 35) = CA 2 -CS 2 . Therefore CA 2 -CS 2 = Cb 2 , or GS 2 = GA 2 -Cb 2 . Therefore S is a fixed point, and so is S' for CS = CS'. S and S' are called the foci of the ellipse. § 4. The following property of the circle is useful. If QN be the or- dinate of a point Q, QT the tangent at Q, meeting AC A' in T, TYY' any chord through T, meeting the ordinate QN in P. TY FY XT~ PY'' For CN:CQ::CQ: CT, since CQN, CTQ are similar triangles. And CQ = CY= CY'. Therefore CN : CY::CY: CT. Fig. 97. 182 Gravitation. [Ch. vi. And the triangles CNY, CYT are similar (Euclid VI. 6). Therefore LCNY = LCYT. And LYNT= LCYY'. In the same way it can be shown that CNY', CY'T are similar triangles* And that I Y'NC = L CY'Y. But the angles GYY', CZT are equal. Therefore NT bisects the angle YNY' externally, and NQ, being at right angles to NT, bisects YNY' internally. PY NY TY Therefore (Euclid VI. 3 and A) -^ = -^p = ^p ■ Let TP be the tangent at a point P on the ellipse. Then PY'' Therefore (Euclid VI. 2) SP is parallel to Y'CZ. Similarly S'P is parallel to GY. Therefore ISPY = LCY'Y= LCYY' = IS'PY', And SYP, S'Y'P are right angles. Therefore the triangles SPY, S'PY' are similar, SP S'P and sy = WY'' W a result which we shall require later. , Ch - vi -3 The Ellipse. T 8 3 Again, let SP and C I intersect in E. Because SC = CS' and EC, S'P are parallel, EC = i S'P, and PE = \ SP. Therefore SP + S'P = a (PE+EC). But PE = ET, for the angles EYP, EPFare equal. Therefore SP + S'P= 2 (CE+ET) = AA'. (2) The area of the ellipse is to the area of the circle as PN is to QN ox as Cb is to AC. Therefore the area of the ellipse is -nAC % -—- or n AC Cb AC ' ' § 5. Let us now consider a point P moving on a curved path so that the radius vector SP, drawn to P from a fixed point S, sweeps over equal areas in equal times. Let P, Q be positions of the moving point at the beginning and end of a time t. Join PQ, and draw SM per- pendicular to PQ. Fig. 99. When t is made indefinitely small, the chord PQ becomes the tangent at P to the path. Denote the length SM by p, and let v be the velocity at P. Since the area SPQ = J SM . PQ, the area described in unit time is pQ iSM.-? ox \po. t Thus vp = k, where A is a constant for all points of the path. Hence the moment of the velocity at P round S remains the same throughout the motion. 184 Gravitation. [Ch. vi. Now the moment of a velocity, like that of a force, is equal to the sum of the moments of its components, and the velocity at Q is compounded of the velocity at P, and the velocity gained in passing from P to Q. Therefore since the moments of the velocities at Q and at P are equal, the moment of the velocity gained between P and Q is zero. That is, either no velocity is acquired, or the acceleration is directed towards S. But there must be acceleration, as the path is curved, and therefore the acceleration of the moving point is directed towards S. The result, vp = A, may be expressed in another way. For if PN be perpendicular to SQ, the area SPQ is iPF.SQ. If PSQ be a very small angle whose circular measure is 6, PN is sensibly an are of a circle with centre S and radius SP, and SQ differs insensibly from SP. Therefore if SP=r, the area of the triangle SPQ is % r 2 d, a and the area described in unit time is \r 2 - > or \r 2 u, where a> is the angular velocity round S. Therefore r 2 a> = h, a constant. § 6. Let the path of the moving point be an ellipse, the radius vector from one focus S describing equal areas in equal times. Denote the lengths of CA and Cb by a and b. The velocity at P is^or -ST, since ST. S'Y'= b\ Hence if unit velocity is represented by a line of length b 2 -j- 3 the hodograph is obtained by turning the circle ABA' through a right angle round &, and the acceleration at P is parallel to Y'C, that is, it is along PS. Ch. vi.] Planetary Motion. 185 Again, since CY' is parallel to SP, the angular velocity of Y' round C is or (§5)^2- § 7. Kepler's Laws of Planetary Motion. The planets describe orbits round the sun in accordance with the following laws, which were first enunciated by Kepler in 1609. Law I. The areas swept out by a line drawn from the sun to a planet are proportional to the times taken in describing them. Law II. The orbit of a planet is an ellipse with the sun in one of the foci. Law III. The squares of the periodic times of different planets are proportional to the cubes of the major axes of their orbits. Since the sun and planets are extended bodies, it must be understood that the line mentioned in Law I is that which joins the centres of the sun and planets, and that the centre of the sun is the focus of the ellipse described by the centre of the planet. But as the dimensions of the planets are very small compared with their distances from the sun, we shall treat each planet as a particle moving along the path of the centre of the planet. The first law shows that the acceleration of a planet is towards the sun, and the second law shows that the ac- celeration is inversely proportional to the square of the distance of the planet from the sun. This statement is true for each planet, but we have no mean's of comparing the action of the sun on different planets till we interpret the third law. 186 Gravitation. [ch. vi. If \ k be the area described round the sun in unit time, the area of the ellipse irab is described in time —r— ■ This is called the periodic time of the planet which describes the ellipse, and is denoted by T. Hence 3 Or Now -fp., is the same for all orbits described round the h 2 a . sun, by Kepler's third law. Therefore -75- is the same for all the planets. Hence the accelerations of different planets are inversely proportional to the squares of their distances from the sun. § 8. The Law of Gravitation. If a planet describes a circular orbit with uniform velocity its centripetal acceleration can be found, but no inference can be drawn as to the dependence of the ac- celeration on the distance between the planet and the body round which it revolves, for the distance always remains the same. But if several planets describe circular orbits uniformly round a common centre, and the squares of the periodic times are as the cubes of the radii of the orbits, it is clear that the accelerations of the several planets are inversely proportional to the squares of their distances from the centre. For in a circle of radius A described with velocity v the lit A acceleration a is —. > but v = A Therefore a = 4ti 2 A 4tt A 3 T t ~ J_2 T 2. Ch. vi.] The Law of Gravitation. 187 A z But y% * s constant for all orbits round the common centre. Therefore a or A 2 This leads us to consider some new phenomena. Almost at the same time that Kepler announced his laws of planetary motion, Galileo by his application of the telescope to astronomy was enabled to detect the four satellites of Jupiter. These bodies describe approximately circular orbits round Jupiter, they form on a small scale a system resembling the planetary system, and it will be seen from the following table that they satisfy Kepler's third law very closely. Satellite. T A* A s f hrs. min. fi I 42 27-6 604853 io -3 x 3-4096 2 85 14-6 9-62347 3-4070 3 171 42-6 15-35024 3-4076 4 400 318 26-99835 3-4075 It is probable then that the influence of Jupiter on these satellites is of the same kind as that which the sun exerts on the planets. Saturn has eight satellites, and Uranus five, whose mo- tions round their respective primaries obey the same law. Our conception of force as stress leads us to regard these motions as due to the action of the body round which they take place ; thus we say that the sun attracts the planets, * The equatorial radius of Jupiter is taken as i. •)■ T having been reduced to minutes. 188 Gravitation. [Ch. vi. and the planets their satellites, and both these attractions are as the inverse square of the distance. Now it appears (§ 7) that if two planets, e.g. Mars and Venus', could be brought to equal distances from the sun, they would be equally accelerated towards his centre. But it is unlikely that the masses of all the planets are equal, considering their known inequalities in size. Therefore the force exerted by the sun on a planet is proportional to the mass of the planet, other things being equal. Similarly the force exerted by Jupiter on a satellite is proportional to the mass of the satellite, if the law of the inverse square is true ; and the force exerted by Jupiter on the sun (equal and opposite to the force exerted by the sun on Jupiter) is proportional to the mass of the sun. Therefore if M is the mass of the sun, m the mass of Jupiter or any other planet at a distance r from the sun, the attraction between the sun and planet is proportional to — g- and may be denoted by — j- , where h does not depend on either the masses or distances of the bodies. There is at present no evidence that h depends on the physical condition of the attracting masses, or on the medium between them. If, as is generally done, we regard k as constant, we may advantageously employ a new unit of mass, called the astronomical unit and defined as follows : — The mass of each of two equal particles which im- part to one another an acceleration of one centimetre per-second per-second when placed 1 centimetre apart is the astronomical unit of mass, in the centimetre-second system of units. Ch. vi.] The Law of Gravitation. 189 When this unit is used, k becomes equal to 1 and —=- r 2 is the attraction between two masses m and M at a dis- tance r apart. Since the planets attract their satellites, they must attract other planets and cause them to deviate from their calculated paths. Such deviations exist, proving the mutual attraction of the planets, but they cannot be discussed in an elementary work. It may, however, be stated that the inconsistency of the calculated and observed motions of the planet Uranus led M. Leverrier and Professor Adams independently to cal- culate, from the law of gravitation, the orbit of an unknown planet which was presumed to be the cause of the dis- turbance. Both astronomers predicted the magnitude, orbit, and position of this planet with considerable ac- curacy; and guided by their instructions observers had little difficulty in discovering Neptune, which is the dis- turbing body. The discovery of this planet was a striking confirmation of the theory of gravitation. § 9. Correction of Kepler's Third Law. The sun, the planets, and their satellites form a sys- tem of bodies which move freely under their mutual attraction. The centre of mass of this system can be regarded as fixed, and since the mass of the sun is very large, compared with that of the planets, the centre of mass is never very far from the sun's centre. Let G be the centre of mass of the sun S and a planet P. The motion of G is only affected by the action of other planets, and, neglecting this action, we may regard G as at rest. 190 Gravitation. [ch. vi. If * and p are the masses of the sun and planet, SG_ p SP~s + ~p' And if c is the acceleration of P relatively to S, —^— is J s+P the acceleration of G relatively to S, and c or is the accele- s+p s+p ration of P relatively to G. v Now 2a, 2b being the major and s „. minor axis of the orbit round the Fig. too. sun, and T the periodic time, it has been shown that the acceleration relatively to the sun is ah 2 . 7 2 w all 3^2' wnere h = -y- • ATT 2 a 3 Therefore the acceleration relatively to the sun is „, 2 2 , , ~ . 4 -n 2 a 3 s and the acceleration relatively to Cr is . . „,„ „ • (s+pJT^r 2 Since this is the true acceleration the force exerted by 4 -a 2 spa 3 the sun on the planet is (*+p)?V But by the law of gravitation it is -=- - T Q 3 1 Therefore y^r-. . = — 5 , a constant. T 2 (S +p) 4 l! 2 Therefore Kepler's Third Law requires amendment and must run as follows. The cubes of the major axes of the orbits are proportional to the squares of the periodic times and the sum of the masses of the sun and planet. It is found experimentally that Kepler's Third Law is ch. vi.] Kepler's Third Law. 191 not quite accurate, and requires the amendment indicated here. The correction is small, and, having indicated its existence, we shall for the future neglect it. It becomes however very important in the investigation of double stars, since if the periodic time and the mean distance of the orbit described by one star about the other can be found, the total mass of the two stars can be deter- mined in astronomical units. § 10. We can now compare the masses of the members of the solar system which possess satellites. Let «, a' be the mean distances of two satellites from their primaries, T, T the periodic times of the satellites, s, / the masses of the primaries. m, a 3 _ s a' z _ / men T*~ 4v*> T'*~ 4* 2 ' Therefore iy>2 ! ^12 Let us compare roughly the masses of the earth and sun. The earth itself is a satellite of the sun, performing its orbit in 365-26 mean solar days, with a mean distance of 93,000,000 miles from the sun. The moon describes its orbit round the earth in 27^ days, at a mean distance of 238,000 miles. It thus appears that the mass of the sun is about 350,000 times that of the earth. The following table, extracted from Herschel's As- tronomy, supplies sufficient data for the comparison of the masses of the principal planets, and for the verification of Kepler's third law. i9 2 Gravitation. [Ch. VI Satellite. Radius of i Primary. Mean Distance of Satellite from Primary.l Time of Revolution. days hrs. min. The moon 395° 60-27 27 7 43 Inner Satellite of Mars 2273 2.551 7 39 14 Outer „ „ „ 6.423 3° 17 54 Fourth Satellite of Jupiter 45367 27. 16 16 32 / Rhea 9-55 4 12 25 Satellites of \ Titan 3839S 22.14 IS 22 41 Saturn 1 Hyperion 2678 21 7 8 Japetus 64-36 79 7 54 , Ariel 7-4 2 12 29 Satellites of \ Umbriel Uranus j Titania 17653 10-31 16-92 4 8 3 28 16 57 ^ Oberon 22.56 13 11 7 Satellite of Neptune 20000 ii- S 21 43 The only planets concerning which we have obtained no data are Venus, Mercury, and the Asteroids. The masses of Venus and Mercury have been estimated by determining the disturbing effect of Venus on the Earth's motion, and that of Mercury on the motion of Venus. The satellites of Mars were only discovered in 1877, and before then the mass of Mars was determined by its effect on the Earth's motion. The following illustration, given by Sir John Herschel, conveys a good idea of the relative magnitudes and dis- tances of the members of the solar system. On a level field place a globe two feet in diameter ; this will represent the Sun ; Mercury will be represented by a grain of mustard seed, moving on the circumference of a circle 164 feet in diameter; Venus by a pea on a circle 284 feet in diameter ; the Earth also by a pea on a circle 1 The distances are given as multiples of the equatorial radius of the Primary. Ch. vij Work done by Gravitational Force. 193 430 feet across ; Mars by a rather large pin's head on a circle 654 feet across ; the Asteroids by grains of sand in orbits of from 1000 to 1300 feet ; Jupiter by a moderate- sized orange on a circle nearly half-a-mile across ; Saturn by a small orange on a circle £ mile across ; Uranus by a full-sized cherry on a circle more than if miles across ; Neptune by a good-sized plum on a circle about %\ miles across. On the same scale the nearest visible fixed star would be represented by a globe distant 8000 miles from the Sun. § 11. Work done by Gravitational Force. Let us again consider the motion of a particle in an elliptic orbit. It has been shown in § 6 that if v is the velocity at P (%• 98), , , Multiplying the two values of v together, and denoting ah. 2 b~ 2 by ju, + = !L.*L = £.** r §4(l )]. a 8Y a SP L9 4li;j- And since gP = o,a-SP [§4 (a)], v * = »(ip-l)- If V is the velocity at any other point Q on the ellipse, Hence, if m is the mass of the moving particle, im{v 2 -F 2 ) = ixm(^-^)- Q 194 Gravitation. [Ch. vi. Since the term on the left-hand side is the gain of kinetic energy, (iw(sp — n^) is the work done by the attraction to S in the displacement from Q to P. By § 6 the attraction at P is ~rr. • SP* Let us choose a very large ellipse, so that Q can be taken at a great distance ; then -=-^ is very small, and can be made smaller than any assigned fraction by taking Q far enough off. If Q is at an infinite distance, ^7; = o, and ^ is the o(^ or work done on m in bringing it to P from infinity by any path, since the attraction is a conservative force. In the following sections masses will be expressed in terms of the astronomical unit. § 12. We have till now regarded the planets and their satellites as moving particles. Observation, however, shows that they are extended bodies of approximately spherical form, but that nevertheless the law of the inverse square of the distance expresses their mutual action with great accuracy. Assuming then that the law of the inverse square is true for the mutual actions of two particles, we are led to ex- pect that the same law will hold for the mutual action of two spheres, at all events in certain cases. We shall prove this to be true. The motion of each body in the heavens does not depend on its mass, but only on the masses and positions of neigh- bouring bodies. The motion of each body is therefore the same as that of a particle of unit mass, subjected to the attraction of the same bodies, and moving with the same initial velocity, as the given body. ch. vi.] Work done by Gravitational Force. 195 It appears from §11 that if r is the distance of a point M . P from a fixed particle of mass M, — is the work that would he done by the attraction of M in bringing a particle of mass unity from infinity to P M . — is called the gravitation potential at P due to M. If there are several particles of mass M x , M 2 , ... M n re- spectively, at distances r 1 , r 2 , ...r n from P, the gravitation potential at P is Mi + M 2 + . + Ei. If the force is zero in any region the potential there is constant, for the same work is done in passing from infinity to all points of the region. § 13. If to all points of a spherical surface equal centripetal forces tend which vary as the inverse square of the distance, a particle within the sphere is in equilibrium under these forces. Let EKL be the spherical surface, P a point within it. "With P as vertex describe a cone of small aperture cut- ting the plane of the paper in the lines HPL, IPX. Then the areas intercepted on the sphere by the cone about IE and LK are to one another in the ratio IH 2 :LK 2 , or in the ratio PH* : PL 2 . % ig6 Gravitation. [Ch. VI. But if these areas are A, B, the forces due to them at P are in the ratio And A PB Z B B PL' pm ~ PL 2 Hence tne forces are equal and their directions are opposite. Therefore the resultant force at P due to them is zero. In like manner the whole sphere can be divided into pairs of opposite elements whose actions at P neutralize one another, and therefore the force at P due to the whole sphere is zero. Hence the potential everywhere within the sphere is the same and has the same value as at the centre. And if a be the radius of the sphere, M the total mass distributed over its surface in an indefinitely thin layer, the . M potential at the centre is — ■ a This is therefore the potential at any point within the sphere. § 14. To find the potential at a point P outside the sphere due to the shell of mass M. Let C be the cen- tre of the sphere, A'CA the diameter throughP, Q a point such that CQ.CP= CA 2 , L any point on the sphere. Join ZQ, LA and LP. Ch. vi.] Potential due to a Spherical Shell. 197 Then CQ-.CL:: CL : CP, and the angle LCQ is common to the triangles CLQ, CPL. Therefore (Euclid VI, 6) Z CLQ = Z CPL. But Z CLA = Z CAL = Z CPL + Z ALP. Therefore Z QLA = Z ALP. Therefore AL bisects the angle QLP, and QL-.LP ::QA:AP. But CQ:CA::CA:CP. . • . QA:CA::AP:CP. . • . QL:LP::CA:CP. Now let w be the potential at Q due to an element of the sphere at L, v the potential at P due to the same element. Then 11 : 1 1 V:: LQ : LP' or V u LQ CA ~~ LP~CP' CA Since -^ does not depend on the position of L, the ratio V - is the same for all elements of the spherical surface. Therefore if U is the potential at Q, V the potential at P, due to the whole sphere, V _ CA U~ CP' M M And since U= ^ (§ 13), ^=gjp- Hence the potential of a thin uniform spherical shell of mass M is the same as that of a particle of mass M placed at the centre of the sphere. 198 Gravitation. [Ch. vi. The work done by the attraction of the shell in the displacement of a particle m from P to R is and the same work would be done by a particle of mass M at G in the same displacement of m. Therefore the force exerted by the shell on m is the same as that which a particle of mass M placed at C would exert on m, and if Pbe the position of a particle m, the force exerted on m by the shell is p™ • The substance of the sphere being supposed to lie on its surface, the quantity of matter per unit surface is said to measure the surface-density. M If M be the total mass, „-,-„ is the surface-density, ' 4 it . Ca* and is generally denoted by where N is the mass of this sphere. 2oo Gravitation. [Ch. vi. But N=§np. CP S , where p is the density of the sphere. Therefore the force exerted on the particle is $wp . GP. Or, the force exerted hy a solid uniform sphere on an internal particle is an attraction proportional to the dis- tance from the centre. If the density is not uniform, but is the same at equal distances from the centre, p must be taken as the average density of the nucleus PGR. § 16. Mutual attraction of Two Spheres. Let X and ¥ be two spheres, in each of which the density is either uniform or the same at the same distance from the centre. Let A and B be the centres of the spheres, M and m their masses, P any particle of the sphere X The action of the mass M on Y is unaltered if we sup- pose M to be concentrated at A, and it is equal and opposite Fig. 104. to the force exerted by Y on a mass M at A. But this is equal to the force exerted by a mass m at B on a mass M at A, and is therefore AB 2 ' We thus see that the same simple law suffices to explain the motions of the planets (and of the satellites round the Ch. vi.] Attraction by a Solid Sphere. 201 planets), whether they are large or small, provided that the density of a planet is the same at equal distances from its centre. The density of a planet is not uniform, but the condi- tions which determine the density (such as pressure) are probably the same at the same distance from the centre, and we may therefore reasonably regard the planets as satisfying the condition given. § 17. Terrestrial Gravitation. All bodies at a given place on the Earth's surface fall to the ground with the same acceleration 2 7? cos 2 A. Therefore the true acceleration along PC is a — a> 2 R cos 2 X. The effect of the Earth's rotation is therefore to diminish the value of g by an amount which is greatest at the equator and least at the poles. Taking R = 4000 miles = 31,120,000 feet, and we find that at the equator g is diminished by about ^|^th of its value, in consequence of the Earth's rotation. Value of g within the Earth. If the density of the Earth were uniform, the value of g within the Earth would be proportional to the distance from the Earth's centre. At the bottom of a coal shaft, \ mile deep, the value of g would be Fjnnsth P a1 "^ ^ ess * nan it is at the surface. Experiments with the pendulum, conducted by Sir George Airy, have however shown that g is greater at the bottom of the shaft than at the surface. We can account for this by supposing that the density of the Earth increases with the depth below the surface, and this hypothesis is justified, since the mean density of the Earth — determined by a method to be described later — much exceeds the mean density of the rocks, &c, at the surface. 204 Gravitation. [ch. vi. Let M be the mass of the Earth, p its average density, g the acceleration of gravity at the surface, / the accelera- tion at a depth h, p' the density of the nucleus of radius R-h. Then g = ^=^pR. And /= Hp'(R-A). Therefore / > g if p' (R - A) > pR. And ^ being less than x , t/> a if- > . R s 8000 '* " p 7999 § 19. It follows from the law of gravitation that bodies on the Earth attract each other, and experiments arranged on a suitable scale show this. The following method was devised by Jolly. I , Q are the pans of a delicate balance. A glass globe filled with mercury is placed in Q, and balanced by a mass m in the pan I . A sphere of lead, 1 metre in diameter, is now placed vertically below the middle of the pan Q. Under its attraction the pan Q descends and the beam can only be brought to its former position of rest by the addition of a mass <) to m. The globe of mercury in the pan Q is then so much the larger portion of the mass there that t)g may be taken as the force exerted by the sphere on the globe of mercury. By this experiment the mass of the Earth may be deter- mined in grams. Let M, m, p. be the masses (measured in grams) of the Earth, the lead sphere, and the globe of mercury, R the Ch. vi.] Terrestrial Gravitation. 205 radius of the Earth, r the distance between the centre of the glohe and sphere. Then since g = ^- , and 7>g = -^ , ,. mu, R 2 In the experiment m = 57-7 kilos, fx = 5 kilos, d = -589 mg. jffi = 6366 kiloms. r = -5686 m. M can be determined from these data, and thence the M Earth's density, which is -j — =3, or approximately 57. Hence (§ 10) the masses of the sun and the planets can be found, and thence the densities of these bodies can be deduced, since their radii are known. CHAPTER VII. Elasticity. § 1. Stress and Strain. Let a body be maintained in equilibrium by forces P, Q, B applied at points A, B, G of its surface. Since a portion of the body, EGFAE, containing the point A, but not B and C, is also in equi- librium, other forces must act on this portion, and their resultant must be equal and opposite to P. These forces can only be supplied by the action of the other part of the body, EBCFE ; since if this part were removed EGFAE would not remain in equilibrium. Similarly, if we take any other surface within the body, the two portions of matter sepa- rated by the surface will generally exert force on one another. The forces at different points of the surface are not necessarily the same either in magnitude or in direction. These forces, regarded as mutual actions of the parts of a body, are called Stresses. If S be the area of a small closed curve on a plane X within the body, P a point within the curve S, and F the total force which the matter on one side of S exerts across S Fig. 106. Elasticity. 207 on the remaining matter, then F is the Total Stress across S, F. and ~ is the Stress at P across the plane on which S lies. The Stress is called a Tension or Pressure according as the force F is a pull or thrust, and it is uniform if it has the same value at different points of the plane. If the Stress across the plane X is not uniform, the area S enclosing the point P must he made very small ; the F limit to which ■= approaches when S is indefinitely small is then the Stress at P. Examples. 1. A wire of I mm. radius is stretched by the weight of 2 kilo- grams. To find the stress across any cross-section, neglecting the weight of the wire. Here £= ttx (-i) 2 P = 2000 x 981. This is the Total Stress across a section. Therefore the stress is -= or — O IT 2. A cylinder of radius r, and density p, is suspended with its axis vertical, its upper end being fixed. Find the stress across a horizontal section of the rod, distant d from the lower end. Here P is the weight of the cylinder of length d. .'. P — udr 2 pg, and 8 = irr t . . p The stress is -^ or pdg. o If the external forces which maintain a body in equili- brium are doubled, the equilibrium is not disturbed ; it thus appears that the stresses in the body are doubled too. 208 Elasticity. [Ch. vn. It is found that alterations in the internal stresses are accompanied by small changes in the form and size of the body, which are called Strains. The branch of Mechanics which treats of the relation between stresses and strains is called Elasticity. Before investigating this relation, we must find some way of classifying and measuring strains. § 2. Plane Strain. Let us consider the simplest changes in the form of the parts of a plane figure which is strained so that it remains plane. Let M and m be two precisely similar plane membranes, stretched by similar forces applied at their boundaries, S any curve drawn on the membrane M, s the curve on m which coincides with S, when m is superposed on M so that the membranes coincide. If the tensions at the boundary of in are altered, m will assume another form \x with a new boundary, and s will assume a form which we may call (OM 2 + MP 2 ) (i + e 2 ) 2 since e 2 < e t . Therefore Op > OP (i + e 2 ), and similarly Op is the strained position of P. The cubical dilatation is (i+ej) (i+e 2 ) (i+e s )— I or e i + e 2 + e 3> when products of elongations are neglected. § 10. Superposition of Strains. Let the strained figure be further strained with elongations (e/, e/, €3'). Then if Q be the strained position of p, and QJkT, MG be parallel to pn, nf respectively, 0G = 0f(i + O = OF(i + O (1 +. €,), Jf = «/(l + e/) = NF(i + «/) (1 + «„), QJf = ^ (I + e,0 = P2V(i + */) (1 + e,). Neglecting ej e/, e 2 e 2 ', e 3 e 3 ', 00= Qyfi+^ + e/). MG = NF(i+e 2 + e 2 '), QM=PN(i+ e3 + e s '). And Q is the position which P would assume after a strain whose principal elongations are e i + e l'> e 2 + e 2> e 3 + e 3- Shearimg Strain. Let e 3 = o and e t = — e 2 = e. Then if OX, OZare the external and internal bisectors of the angle OAB in the unstrained figure, the lines OX, OY Ch. VII.] Superposition of Strains. 221 are unaltered by strain^ but the angle YOX is altered by % e, CO remaining perpendicular to OX and OY. Since YOX is the angle between the planes COX, COY, the change in this angle is called the shear of the planes. Since e 1 + e 2 + e 3 = o, there is no cubical dilata- tion in this strain. Uniform Dilatation. If e x = e. 2 = e 3 = «, all elon- gations are equal ; this strain is a uniform cubical dilatation of magnitude 3 e. Reduction of Strain. Any strain can be reduced to uniform cubical dilatations and shears. Denote a strain whose principal elongations are e v e 2 , e 3 by Oi, f 2 = € sl Then a simple elongation e 1 is [e ls o, o], and by the principle of superposition of strains it is equal to [!i,li. f j]+p,_Ii 3 ] + [li, 0) _f!]. L 3 3 3 J L 3 3 J L 3 3 J But the first term on the right-hand represents a cubical dilatation and the two others represent shears. Hence a simple elongation is equivalent to a uniform cubical dilatation and two shears, and the most general strain considered is made up of three elongations at right angles to one another, and can therefore be reduced to cubical dilatations and shears. § 11. By the action of stress the position and area of a surface across which stress is exerted are changed, and in 222 Elasticity. [Ch. VII. estimating stress as a force per unit area strict accuracy would be obtained by referring to the strained state of the body. It has, however, been already mentioned that our work is only approximate, and it thence appears that we may refer stress to the unstrained state of the body. Thus, when a wire is stretched by a weight P, the cross-section S of the wire is diminished to S (i — a e), 2e being the areal contraction of the cross-section; we, P P however, take ~- as the stress, and not -^-. x . iS 5(i-2e) This approximation permits us to assume that the effect of two stresses applied simultaneously is the sum of their effects when applied separately, so long as the resulting strain is small. If the strain of the solid is everywhere the same, the stress is the same at all points. £ Shearing Stress. If equal pressures T are applied perpendicularly to the faces he, off of the cube oadbfcga, oh and the edges parallel to it will contract, and all lines perpendicular to oh will be .^ T equally elongated. Let a be the elongation of oa and oc, — /3 the elongation of oh. If equal tensions T are ap- plied perpendicularly to the faces of and ae, —a is the elongation of oh and oc, and /3 is the elongation of oa, due to the tensions T. Kg. 119- Ch. VII.] Shearing Stress. 223 Superposing the two stresses, we find that oc has no elongation, oa and ob have elongations a + /3, — (a + /3). Hence the stress produces shear only, and is called a Shearing Stress. Hydrostatic Stress. If equal tensions T are applied perpendicularly to the six faces of the cube, the resulting strain is the sum of the strains due to each pair of tensions; therefore each edge of the cube is elongated by /3 — 2 a, and the angles of the figure are unaltered by strain. Hence the elongations are principal elongations, and the strain is a cubical dilatation without change of form. The corre- sponding stress may be called a Hydrostatic Stress. Transformation of Shearing Stress. Let the cube oadbfcga be cut in two by the plane agfb, forming a prism on a triangular base fcg ; then if no ex*- ternal forces (as gravity) act on the substance of the prism, its equilibrium is due to the stresses on its faces. Let the triangular faces be free from stress, and let normal stresses T,—Tbe ap- plied to the faces ca, ch. Since the stress on each face is uni- formly distributed, it can be replaced by the total stress acting at the centroid of the face. If oa = I, the prism is in equilibrium under forces l 2 T parallel to ao, l 2 T parallel to ob, and the total stress on the plane abfg. This latter force must therefore be 224 Elasticity. [Ch. vii. Tl 2 >/%, parallel to ba, and it is distributed over the area /•■/a. Therefore the stress on the plane alfg is T, and it is parallel to la. If a plane gah is drawn through ag perpendicular to afg, the stress on this plane is tangential to it, and is of magnitude T. Relation between Stress and Strain. Hooke's Law. The relation between Stresses and the corresponding small Strains was first enunciated by Hooke in 1676, and may be stated as follows: — Strains are proportional to the Stresses which pro- duce them. The principal stresses that we have to consider are hydrostatic stress and shearing stress. Let P be a hydrostatic stress, producing a uniform P . dilatation 8. By Hooke's law — - is the same for all values of P, and if P = kb, h is called the Elasticity of bulk, or the Coefficient of Resistance to Compression. It does not depend on the magnitude of the pressure so long as this is moderate, but it is determined by the nature and physical condition of the strained body. It has been shown that a shearing stress S produces o a shear 2 e. By Hooke's law — is a constant, and if 2e S = 2 n e, n is a coefficient, which depends only on the nature and physical condition of the body which undergoes shear. n is called the Coefficient of Rigidity of the body, or the Elasticity of Form. ck.vii.] Youngs Modulus. 225 § 12. Seduction of any stress to hydrostatic and shearing stress. — Young's Modulus. In general the stress within a body may alter both the form and size of the parts of the body. But any strain Can be reduced to cubical dilatation 8 and shearing strain s, and therefore the stress consists of a hydrostatic stress &b superposed on a shearing stress 2 ns. Hence any stress reduces to a combination of shearing and hydrostatic stresses. An example will make this point clearer. Let a wire hanging vertically with its upper end fixed be 'stretched by a weight W attached to its lower end. Neglect the weight of the wire, and let I, S be its length and the area of its cross-section before stretching, I + A. the length after stretching. A. . . W . Then since -j is the elongation e, and -~- is the stress, W A -—= ixy, where ju, does not depend on W, I, or S. O v The coefficient [i. is the measure of the resistance which the wire offers to stretching, and is called Young's Modulus. Take a small cube in the wire with its faces of, o'c horizontal. W The stress across these faces is -^ , Fig. iai. o which we denote by T, and there is no stress across the other faces. T This stress is equivalent to tensions — across all T T six faces, tensions — and — — across of, oe respectively, 3 3 Q < / / /• / a 226 Elasticity. [Ch. vii. T T and the opposite faces, and tensions — and — - across of, og and the opposite faces. The equal tensions across all faces of the cube form a T T hydrostatic stress - and the corresponding- strain is —= , or 3 3* T a linear dilatation — , ■ gk T T •The tensions-, across of, oe and the opposite faces form a shearing- stress. The corresponding principal T T elongations are along 1 oc and oa and are y— and — ^- • s 6 6n 6n • The tensions across of, og and the opposite faces give similar elongations. Hence on the whole the elongation of the wire is given by €=T ± + ±l ( 9* 3«) And since T = ue, u. — 3/c + n Again, the lateral linear contraction of the wire is {6n 96) and as the wire cannot expand laterally by stretching, k cannot be less than \ n. Modulus of elasticity for simple longitudinal strain X. This strain is equivalent to a cubical dilatation A. and two shears. The hydrostatic stress is a tension k\ on all planes, and if oc be the direction of the elongation, the stress and it is accompanied by equal lateral stresses (k jX €h. vii.] Torsion. 227 corresponding to the first shear is a tension across n m \ of, and parallel planes, and across oe, and parallel planes. The stress corresponding to the second shear is similarly formed. Therefore ( k H ) A. is the longitudinal stress (across of, 2 n T across all planes parallel to the direction of elongation. § 13. Torsion. A straight uniform wire is suspended vertically from a fixed support, and carries a horizontal bar attached to its lower end. The bar is deflected from its position of rest by equal and opposite forces P applied to its ends perpendicular to its length, and comes to rest when it makes a certain angle \j/ with its former position. The lower end of the wire is thus twisted through an angle \jr, and the wire is said to have a twist, or to undergo Torsion. If «S is any horizontal section of the wire, the portion of the wire below S is in equilibrium under the stress across S and the deflecting couple of moment M. Hence the stress across any horizontal section reduces to a couple M with its axis along the wire. Therefore the strain is everywhere the same, and the wire is twisted uniformly throughout its length. The total twist of the wire is measured by the angle \jr, and if I is the length of the wire, -j is called the twist. Q 2 228 Elasticity. [Ch. vii. If the length of the wire is halved, the total twist is evidently halved too, and by Hooke's law if the couple is doubled the twist is doubled. Hence the total twist is proportional to the deflecting' couple and to the length of the wire. Torsion is equivalent to shearing strain, for if the couple M produced any contraction, an opposite couple would produce expansion ; the two couples however clearly produce similar strains, and therefore each can only produce shearing strain. It can be proved mathematically that if the cross-section of the wire is circular, it undergoes no deformation by torsion; but a square or elliptic section is distorted by twist and ceases to be plane. It will be shown in Chapter X that the total twist pro- duced by a given couple in a wire of given length and material is inversely proportional to the fourth power of the radius. Therefore if M is the twisting couple, I, r the length and radius of the wire, \jr the total twist, a E Fig. 122 M = I, where C is a constant depending on the material of the wire, which is called the coefficient of torsion. It can be, shown that C = — • a We may determine C (or n) from the oscillations of a wire as follows. Suspend from the wire a cylinder, the axis of which is the axis of the wire produced. Let H be the mass of the cylinder, K its radius of gyration about the axis. CHiVii.] The Tdrsion Balance. 229 The couple exerted on the cylinder by the wire, when the angular displacement is \j/, is — j-±- , and HK 2 a is the moment of the angular acceleration. Therefore c Hence the vibration is simply harmonic, and its period T Therefore C = * ^ . The Torsion Balance. When the rigidity, radius and length of the wire are known, the moment of any couple applied to the wire can be determined from observation of the twist produced ; but the magnitudes of different couples may be compared by simple comparison of the corresponding twists. A wire used in this manner, provided with proper arrangements for adjusting and measuring the twist, is called a Torsion Balance. As an illustration of its use, we shall describe Caven- dish's method of detecting and measuring the attraction (due to gravitation) between two bodies. AB is a light metal rod of length a a, suspended by a long fine metal wire ; at its ends it carries two small equal spheres A and B. P and Q are two large spheres of lead with their centres in the same horizontal plane as those of A and JB; they are attached to the ends of a lever and their centres are equidistant from the middle point of JB. 230 Elasticity. [ch. yn. By a motion of the lever the centres of the lead spheres can he moved into the positions M and N, or M' and N f , the lines AM, AM', being perpendicular to AB. Let m, m' be the masses of the spheres P and A ; when they are at a distance so apart, the force between them 13 km. m' O e O Figs. 123, 124. ©When the lead spheres are at P and Q, they exert on the whole no couple on the spheres A and B ; when they are at M and N they exert a couple 2aF, where F is the force exerted by M on A. Therefore if C is the coefficient of torsion, the angle through which the wire is twisted when the lead spheres are displaced from P, Q, to M, M, is given by 2aF=—j—- If the spheres are displaced from P, Q, to WW, the de- flection is the same, but is in the opposite direction. Therefore if we observe the. angle (j> through which AB turns when the spheres are moved from M N to M, A'', c And since /, a, r can all be measured, F can be found. Ch. vnj Flexure. 231 To deduce the mean density d of the Earth. Fx 2 Since k = — - , k can be found by this experiment, and ■mm J r ' if M and B are the Earth's mass and radius respectively, kM and d = z — m = : ^— f ■ Thus M and d can both be found. The best experiments made by this method give 5-48 as the Earth's mean density. § 14. Flexure. The complete investigation of the behaviour of a rod under flexure requires advanced mathematical treatment ; its principal results, however, are easy to express. Consider a bar of oblong cross-section fixed at one end and bent by the application of a force F at the other, the force being parallel to one side of the cross-section. The portion of the bar between the free end and a cross- section S is in equilibrium under the force F and the stress across S, the weight being neglected. The stress across S therefore reduces to a force F and a couple of moment Fx, where x is the distance of S from the free end of the bar. Hence when a bar is bent by a single force F the stress is not the same across all sections. The Bending Moment at any section S is the moment, round a point in S, of all the forces between S' and the end of the bar. Uniform Flexure. Let the bar rest symmetrically on 232 Elasticity. [Ch. vii, supports A and B, and be bent by equal weights P applied to its ends C, D. The stress across the section of the rod Fig. 125- at the middle point must be entirely hori- zontal, for an upward force on one half of the rod would imply the existence of a downward force on the other, and considerations of symmetry render this impossible. Hence the conditions for equilibrium of either half of the rod show that the resistances at A and B are equal to P. If AC = BB = I, the Bending Moment at any point of AB is PI, and since the bending moment is everywhere the same, the strain must be the same. Hence the bar when bent assumes a circular form. The complete investigation gives the following results : — The mean line, i.e. the line through the centres of the cross-sections, of the wire is bent into a circular form HK, and is not altered in length by strain. All planes, initially perpendicular to the mean line, become by strain planes through the point C (the centre of HK) perpendicular to HCK. Lines parallel to the mean bine become by strain circles with their centres on the perpendicular from C to the plane HCK. If the whole rod is imagined to be divided parallel to its length into filaments of exceedingly small thickness, each fila- ment shrinks or swells laterally with the same freedom as if it were separated from the rest of the substance. Ch. vii.] Flexure. 233 Let hk be the axis of any filament in the plane HKC, y its initial distance from HK, regarded as positive when above HK, CH- R. hk HK hk-HK Then C/ i =CH=~AH—' m , „ hk — HK AH y .„y 2 . , , , Therefore gg = 'CH = li' if ^2 is neglected. Therefore the stress across a section at distance y from HK is a tension —' At points below HK the stress is a pressure. Let efgh be a cross-section of the bar, E its centroid, ef— a, fg = b. Imagine the bar to be divided into thin filaments by horizontal planes parallel to ef the cross-sections of the filaments being of area S 1} S 2 ,...S n respectively. Let y x , y 2 , . . . y„ Fi §- I2 7- be the distances of the axes of the filaments from E before strain. ^3 The total stress on S 1 is ^w^i, and is a pressure or Jti m R tension according as y x is positive or negative. Its moment round the axis cd parallel to ef is —^- S 1 • The total stress on efgh is made up of the stresses on S 1 , S 2 , ... S n , and reduces to a couple of moment ■2£(tf8 l +rt8 a + ...+tf8.). But y/S x +y 2 % + ... +y n *S n = K* . ab, where K is the radius of gyration of efgh about cd. 234 Elasticity. [Ch. viij Therefore if M is the bending moment M = -~- , where / is the moment of inertia of the cross-section about the horizontal line through its centroid. /n/is called the flexural rigidity. By the method of § 3, Chapter IV, it can be shown that ~ 12 If the rod is turned through a right angle so as to bend about an axis parallel to fg, the flexural rigidity is txd b - and the flexural rigidities in the two cases are as 12 ° b 2 : a 2 . If a bending moment round any other axis is applied to the rod its forces must be resolved into components, one tending to produce flexure round cd, the other, flexure round a perpendicular to cd. As these flexures are not in the same proportion as the forces which produce them, the relation of the bending couple to the resultant flexure is in this case somewhat more complex. Work done by Stress. Hydrostatic Stress. Let the volume of a body be V when free from stress (or subject to negligible stress), V— v when subject ,to hydrostatic pressure P. In estimating the work done we suppose that no kinetic energy is generated, or that the stress is increased very gradually ; we also assume that Hooke's law is satisfied. The strain consists only in change of volume, and no change of shape occurs ; hence the result obtained is the same as for a body of no rigidity, and we may give the body any form we please. Imagine it to be contained in a cylinder of section S, Ch. VII.] Expansion of a Gas. 235 fitted with a piston which moves in through a distance d, when the pressure is increased to P. When the volume is V—nv, the pressure is nP, n being a fraction, and the distance through which the piston has been displaced is nil. Since the total pressure on the piston is nP x S, the force on the piston is proportional to the displacement, and (Chapter III, § 3) the work done in the displacement d is i PS. d. Now Sd = v. Therefore the work done is \ Pv. If the pressure on the piston is originally P , the pressure when the volume has been diminished by nv will be P + nP, and the total work done is P v + \ Pv. If the initial and final pressures are P , P x the work doneis i(P + P 1 )v. Similarly if a wire or string increases in length by I under a stretching force W, the work done in strain is \Wl. Expansion of a gas. The volume of a gas varies considerably, not only with the pressure, but also with the tem- perature ; Hooke's law is not satisfied except for very small changes of volume. Let a gas contained in a cylinder expand from volume v to v, its temperature being so regulated that the pressure is constant and equal to p. The work done by the gas in Fig. 128. 236 Elasticity. [Ch. vii. expanding is p (v — v ), and if 0A = v , OB = v, AM =p, the rectangle AMNB represents the work done by the gas in expanding. Now let the pressure and volume of a given mass of gas vary in any manner, and let the abscissa and ordinate of a point represent the volume and pressure of the gas at a given instant. Then a continuous curve PQB represents the states through which the gas passes in expanding, and the argument of Chapter III, § 3, shows that the work done in expansion from v to v is represented by the area APQR CHAPTER VIII. Hydeostatics, § 1. In a fluid at rest the only possible stress is hydro- static, and, except under unusual conditions, this stress is a pressure. Fluids are divided into two classes, liquids and gases. Every liquid can be brought into the gaseous state by a sufficient increase of its temperature, and every gas can be liquefied by cooling it and increasing to a greater or a less extent the pressure to which it is subjected. By suitable processes a liquid can be continuously and imperceptibly changed into a gas, but in the case of all known substances, under ordinary atmospheric pressure and at ordinary temperatures, there is a marked distinction between liquids and gases. A liquid is only very slightly compressible, and may for our purposes be considered as incompressible. The density of a liquid will therefore be regarded as constant. A gas on the other hand is easily compressed. In the case of air and other gases which are not readily brought to the liquid state, the density is proportional to the pressure, when the temperature is kept constant. Thus if external pressure is removed the gas expands indefinitely. 238 Hydrostatics. [Ch. viii. § 2. In a heavy fluid at rest, the pressure is the same at all points which are in the same horizontal plane. Consider a cylindrical portion of fluid with horizontal ,, axis, terminated by parallel vertical V^ I planes, and let A be the area of Fig. iag. either plane end of the cylinder, p the pressure at one end, p' the pressure at the other end. The cylinder is in equilibrium under its own weight, the resultant of the pressures on the curved surfaces, and the total pressures pA, p'A applied to the ends of the cylinder. The only forces in the direction of the axis of the cylinder are pA, p'A. Therefore these are equal and opposite or p = p'. Cor. Since the cylinder has no vertical motion, the resultant of the pressures on the curved surface of the cylinder passes through the centre of mass of the cylinder, is directed vertically upwards, and is equal in magnitude to the weight of the cylinder. \p § 3. To compare the pressures at different depths in the fluid. Let the axis of the fluid cylinder be vertical, the pressures on the curved surface are then hori- zontal, and the only vertical forces are pA and W (the weight of the cylinder), downwards, and p'A upwards. Therefore p'A =pA+W. If 'the density of the fluid is uniform and equal to p, and h is the height of the cylinder, W= pghA and p' = p + pgh. If it is the pressure of the atmosphere the pressure at a depth h in the liquid is it + pg%. Ch. viii.] Pressure within a Fluid. 239 The common surface of two liquids of different density which do not mix is a horizontal plane. For let AB, PQ be horizontal lines in the same vertical plane AQ, AB lying in the upper liquid, and PQ in the lower. Then it has been proved that the pressures at A and B are equal, and that the pressures at P and Q are equal. Let p, p' be the pressures at A and , P; E, F the points where BQ, AP meet the common surface of the liquids, p, p the densities of the upper and lower liquid respectively. Then if BQ = h, BE = so, AF = y, p'-p = g{px + P '(h-x)} = g{py + p(A-y)}. Therefore x = y unless p = p', and therefore E and F are in the same horizontal plane. Also the equilibrium is not stable (i. e. cannot exist) unless the lower liquid is the denser. Hence also the free surface of a liquid is horizontal, for p may denote the mean density of air in the neighbourhood of the free surface. In the above propositions it is assumed that g is the same in magnitude and direction at all points of the liquid. This ceases to be justifiable when a large liquid surface, as that of a lake or ocean, is considered. Cor. Let a vessel containing liquid move vertically upward with acceleration a. Considering as before the motion of a portion of liquid bounded by a cylindrical surface with vertical axis, we have A(p'—p—pgh) as the resulting force upwards and p hA is the moving mass. .-. pha=p'-jp~pgh, or / =p+pA(g + a). 240 Hydrostatics. [CH..VHT. Q When liquid is contained in a curved vessel of any form, the laws of distribution of pressure that have been laid down are still true, for if two points in the vessel cannot be joined directly by a vertical or horizontal line they can be joined by a series of such lines, all lying in the liquid. § 4. Equilibrium of two liquids in a XT-tube. Let two liquids such as mercury and water, which do not mix, be placed in a TJ-tube, and let P, Q be the free surfaces of the liquids, S their common surface of inter- section. Let h, h' be the vertical heights of P and Q above 8, p the density of the liquid between P and S, p' the density of the other liquid, it the atmospheric pressure. Then ir + pgh is the pressure at iS due to the liquid PS, and v + p'gh' is the pressure due to the liquid QS. Therefore it + pgh = it + p'gh'. Kg. 132. And ph-p'h'. Thus the heights to which the liquids rise above S are inversely as their den- -" sities. The Siphon. The si- phon is a bent tube ACS employed to transfer liquid from one vessel to another at a lower level. Let the arm CA be immersed in a liquid of density />, which fills the portion ACE of the tube. Then if the vertical depth of E below the free surface in Kg. 133. Ch. viii.] Archimedes' Principle. 241 is r, the pressure of the liquid at E is -a + pgr, it being the atmospheric pressure. And the opposing pressure at E is only it. Hence the liquid flows out at E, at a rate which in- creases with r. In order that the siphon may work, it is necessary that it should first be filled with so much liquid that the free surface in the arm BC (at E) lies below the free surface in the vessel from which water is drawn. § 5. Archimedes' principle. Consider any portion A of a fluid in equilibrium, under its own weight and the pressures of the surrounding fluid on its surface. The resultant of the pressures is a single force acting upwards through the centre of mass and equal to the weight of the fluid in A. If a solid body occupies the space A, the pressures at its surface are the same as when A was occupied by fluid. Therefore the resultant pressure of a fluid on a body is equal to the weight of the fluid which occupies the same volume as the immersed portion of the body (called for brevity the fluid displaced) ; and the resultant acts verti- cally upwards through the centre of mass of the displaced fluid. This is Archimedes' principle. Let the body be a homogeneous solid completely im- mersed in a homogeneous liquid. Then the centres of mass of the solid and the displaced liquid coincide. And if V, p be the volume and density of the body, o- the density of the liquid, V(p — a) g is the resultant downward force on the body. Thus a body when wholly immersed sinks or rises ac- cording as its density is greater or less than that of the liquid. 242 Hydrostatics. {Cn. vm. Let the body be only partially immersed, displacing a volume of liquid v. Then (Vp — vB Nicholson's hydrometer is represented in the Jl figure ; it is made of brass and loaded at the bottom, so that it floats in stable equilibrium, with its axis upwards. B is a platform on which weights can be placed, which sink the hydrometer till the mark A, on the wire supporting B, is in the surface of the liquid. Let P be the mass of the hydrometer, M the mass on the platform B when the hydrometer is Pig. 135. immersed in water to the mark A, N the mass when the hydrometer is similarly immersed in some other liquid of density and this is transmitted * GH to the receiver F. The horizontal pressures on the piston K have no effect, and the vertical total pressure on the piston is P , — • jpfy , which is applied to the compression of the body resting on LB. Making — and -^-^ large, a moderate force exerted at H brings a very large force into play at LB. Hence Bramah's Press is a machine of very great mechanical advantage. The following difficulty may perhaps present itself to the reader. A is the base of a conical beaker of area A, filled with liquid to a depth h. The total pressure on the base is pghA. But the base obviously supports the whole mass of liquid and the weight of this is greater than jjghA. This is explained as follows — Imagine a vertical cylinder described in the liquid on the base A. 256 Gases. [Ch. VIII. The liquid outside this cylinder is kept in equilibrium by the horizontal pressures due to the cylin- der, its own weight, W, and the pressures on the sides of the beaker. To these latter forces there is an equal and opposite action of the liquid on the sides of the beaker, and to balance this the base must exert an action on the sides whose vertical com- ponent is equal to the weight W, and is directed upwards. § 10. Gases. Gases are remarkable for their low density, their high compressibility, and their great expansion when heated. All gases have weight ; this can be demonstrated for air by exhausting a flask, weighing it, admitting air and then weighing again ; the increase in weight is the weight of the air in the flask. If the volume of the flask is known, the density of the air can be determined in this way. The density of dry air at 0° C under the pressure of a column of mercury 76 cm. high is -001293. ^ can b® demonstrated in like manner that other gases have weight and their densities can be found. Atmospheric pressure. The following experiment was first performed by Torricelli. A glass tube, rather more than 30 inches long and closed at one end, is filled with mercury, the other end temporarily closed with the thumb, and the tube then inverted in a mercury bath. It will be found, on removing the thumb from the sub- merged end, that the mercury does not all eseape from the ch. viii.] The Atmosphere. 257 tube, but that a column about 30 inches high remains in the tube. Since the pressure at all points of the same liquid in the same horizontal plane is the same, the mercury in the free surface of the bath must be subject to a pressure equal to that exerted hj this column. This pressure was attributed by Pascal to the weight of the atmosphere. To test this idea, he carried the apparatus up the mountain Puy-de-Dome in 1648, and found that the height of the column steadily diminished as greater alti- tudes were attained. This result is now so well known, that the heights of mountains can be approximately estimated from the heights of the barometer column at their summits. Measurement of atmospheric pressure. If a Torricellian tube inverted in a mercury cistern be fitted with a vertical scale, by means of which the height of the column can be read, it forms a barometer, i.e. an instru- ment for measuring the pressure of the atmosphere ; for if h be the height of the barometer column, and p the density of mercury, the pressure is pgh. The standard pressure (commonly called an atmosphere) is the pressure when the height of the barometer is 76 centimetres, and the temperature 0° C. As the density of mercury under these conditions is 13-596, and g = 981, this is rather more than io 6 dynes per sq. cm. A pressure of io 6 dynes per sq. cm. might very well be made the scientific unit of pressure. As the height of the barometer varies, mercury passes be- tween the tube and the cistern, and the level of mercury in the cistern varies also. Hence to determine the length of the barometer column it is necessary to know the position 258 Gases. [ch. viii. of the free surface in the cistern as well as in the tube. This difficulty is best overcome in Fortin's barometer. Here the bottom of the mercury cistern is a bag of chamois leather, and a vertical screw working in a fixed nut presses against the bottom of the bag, so that, by raising or lowering the screw, the level of the mercury in the cistern is also raised or lowered. The tip of a fixed ivory pin pointing downwards indicates the zero of the barometer scale, and if the mercury surface is clean it can, by turning the screw, be adjusted very accurately to touch the tip of the pin. Since the density of the air under standard conditions is ■001293, ^he height of the atmosphere if homogeneous would be — x 76 cm., or about Z miles. •001293 Law of decrease of the pressure of the atmosphere. The pressure of the atmosphere diminishes as the height above sea-level increases. Assuming that the atmosphere is at rest, and that its temperature is constant, we can find the law of variation of the pressure. Consider a portion of the atmosphere contained between two horizontal planes at a small distance t apart ; then if p is the pressure at the lower plane, p 1 the pressure at the upper plane, and p the average density of the air between the planes, p = p 1 + pgt. Since t is very small, p x and p are very nearly equal, and, applying Boyle's Law (§ 11), we may write p = kp. Therefore p x =jp (1 —kgt), and — = 1 — kgt = c, a con- stant if variations in g are neglected. Therefore taking a layer A of any thickness ce, and ch. viii.] Boyle's Law. 259 dividing it into m layers of very small thickness t, we find that if p n is the pressure above the » tlx layer, Pn-l Pn-2 And if P and j» are the pressures above and below the layer A, p = Pc m ; and if A, D be the corresponding densities, d = Dc m . Therefore in an atmosphere at uniform temperature and at rest, the density and pressure diminish in geometric pro- gression when the height increases in arithmetic pro- gression. §11. Boyle's Law. When the temperature of 'a given mass of gas remains constant, the volume varies inversely as the pressure. This law may be verified for air by the following experi- ments. I. To verify Boyle's Law for pressures less than that of the atmosphere. A straight glass tube, closed at one end, is gauged by weighing the quantities of mercury which are required to fill it up to marks, etched on the side of the tube. It is then filled with mercury and inverted in a deep cistern, and a small quantity of air is passed up the tube. The mercury column in the tube falls, and can be adjusted (by raising or depressing the tube in the cistern) till its top coincides with one of the marks. Let h 1 be the height of the mercury column in the tube, H x the height of a barometer column, obtained by reading an adjacent barometer. Then if p 1 be the pressure of the air in the tube, Pi+Pffh = Pff s i> or i>i = P9( H i-h)- s a 260 Gases. [ck. viii. Also Fj the volume of the air is known. Now lot the tube be depressed in the cistern till the air occupies a volume V 2 , and let ft 2 be the height of the column in the tube, R 2 the' height of the barometer, p 2 the pressure of the air. Then p i = pg{H i -h^j. It is found that if the temperature of the gas has re- mained the same in both operations, p l F L = p 2 V % . Therefore the product of the pressure and volume of a given mass of air remains the same for all pressures less than that of the atmosphere if the temperature is con- stant. II. To verify Boyle's Law for pressures above atmospheric pressure. Mercury is poured into a U-tube the longer arm of which is open, and by inclining the tube the air in the closed arm is brought into free communication with the atmosphere, thus taking the same pressure. More mercury is now poured into the long arm, com- pressing the confined air, and causing the columns of mercury in the two tubes to assume differents heights. If h is the height of the free surface in the open tube above the other surface and ir the pressure of the atmo- sphere, n + pgh is the pressure of the confined air; the volume must be obtained by gauging the tube. The volumes occupied by the air under different pressures having been observed, it is found that Boyle's Law is satisfied to a high degree of accuracy. For air, hydrogen, nitrogen, and other gases which are not readily liquefied the discrepancies from this law at moderate pressures are ' so slight that they can only be detected by most careful experiments. For gases which,, Ch. vin.] Boyle's Law. 261 are more easily liquefied, as carbonic acid, the law at ordinary temperatures is a rough approximation to the truth. Manometers, or instruments for direct measurement of pressure, are of two kinds, open and closed. The open manometer consists of an open U-tube con- taining mercury, one arm of the tube being connected to the enclosure within which the pressure is to be measured, and the other arm opening into the air. If the free surface of mercury in the open arm is at a height h above that in the closed arm, the pressure is ■n + pgh, where it is the pressure of the atmosphere. The closed manometer differs from the above in having one arm of the tube closed at the top, instead of opening into the air. The tube is gauged and the volume occupied by the contained air at standard pressure is noted. Thence the pressure of the air when occupying a different volume can be deduced from Boyle's Law, and if p be this pressure, h the same difference of heights as before, p+pgh is the pressure to be measured. Law of Charles. IS the temperature of a mass of gas which obeys Boyle's Law varies, it is found that _pv = B(i + at), where p is the pressure, v the volume, and t the tem- perature. The coefficient a is approximately the same for all such gases, and is about ^\^ when temperature is measured on the centigrade scale. Absolute temperature ^on the air thermometer is defined 262 Gases. [ch. viii. by the relation T=t+2J 3, t being the temperature on the centigrade scale. Thus the zero of absolute temperature is — 273° C, and the laws of Boyle and Charles are expressed by the relation pv = BT, where R is a constant which depends on the nature of the gas. Law of the mixture of gases. Two portions of gas at the same temperature, hut at different pressures, are mixed. To show that if p lt p 2 are the initial pressures of the gases, v 1 and v 2 the initial volumes, V the ■ volume of the mixture under a pressure P, PV = p 1 v 1 +p 2 v 2 . If the pressure of the second gas were p v its volume would be v', where p x v' = p 2 v 2 . Now, when two gases at the same pressure p are mixed, the pressure of the mixture is found to be p, if the total volume of the gases is un- changed. Hence, under a pressure p 1 , the volume of the mixture is v 1 + v' ; and since the mixture obeys Boyle's Law, the volume V under a pressure P is given by the relation PV=p l (0 1 + »') =p 1 v 1 +p 2 v 2 . § 12. Pumps. The suction pump (Kg. 145) consists of a vertical cylin- drical barrel opening below into a smaller cylindrical tube called the suction tube. A piston rod C worked by the pump handle moves up and down in the barrel, and the suction tube dips below the surface of the water that is to be pumped up. At the junction of the pump barrel with the suction tube there is a valve D opening upwards, and an aperture in the Ch. VIII.] Pumps. 263 piston is covered by another valve E, which also opens upwards. The following is the mode of action of the pump. Initially the valves are closed, the air in the barrel and suction tube is at atmo- spheric pressure, and the water surface is at the same level inside and outside the suction tube. Pumping 1 begins by raising the piston from the bottom of the barrel. This in- creases the volume of air in the barrel and lowers its pressure ; consequently the lower valve opens and air passes from the suction tube into the barrel, and the diminution of pressure in the suction tube causes the water to rise into it, until the upward stroke has been completed. When the piston descends the pressure below it increases, the upper valve opens, the lower valve shuts, air is exhausted from the barrel, and the level of the water in the suction tube remains stationary. In the second upward stroke of the piston, the water rises to a greater height, and after a certain num- ber of strokes reaches the pump barrel. The length of the suction tube must not exceed the height of the water ba- rometer (about 34 feet), since the pressure at the foot of the column must be that of the atmosphere. When the water has risen into the barrel, the action changes ; as the piston descends, water as well as air escapes through the upper valve, and at the end of the stroke all the water in the pump ban-el is above the piston. When the piston begins to rise the valve closes, and the water above the piston is c »/ E Fig. 145- 264 Gases. [Ch. VIII. carried up and discharged by the mouth of the pump at the end of the stroke. If the course of the piston carries it more than 34 feet above the water in the well the pump barrel will not com- pletely fill with water during 1 the upward stroke, and the piston will not work to full advantage. If the piston does not descend to the bottom of the barrel the height to which the water can rise is diminished, since when the piston is in its lowest position the air below it is at atmospheric pressure is, and when the piston rises the pressure of this air cannot fall below -=r > H being the height of the barrel, h the height un traversed by the piston. The maximum possible height to which such a pump can raise water by suction is tth 34- E feet. The Force Pump (Fig. Pi g- 14 6 J 46). This consists of a pump barrel, dipping into water, with a valve at the bottom opening inwards and a valve at the side opening outwards. A piston works in the pump barrel and, when it descends, closes the bottom valve and forces water out of the barrel through the side valve. When the piston rises, the side valve closes and the lower valve opens, admitting water from the reservoir into the barrel. Air Pump. The principle of the Air Pump is precisely that of the CH.VIII.] Air Ptimps. 265 f~\ _v A B Fig. 147. Suction Pump ; the receiver A, from which air is pumped, corresponds to the well; the tube B corre- sponds to the suction tube, and the arrangement of the barrel and piston is the same as in the suction pump. When the piston descends, the lower valve closes and the upper one opens, air escaping through it from the barrel. When the piston ascends the upper valve closes, the lower one opens, and air passes from the re- ceiver into the barrel. We can calculate the theoretical rate of exhaustion. Let V be the volume of the barrel, V that of the re- ceiver and the tube leading to it, P the pressure in the receiver when the piston is in its lowest position. When the piston has been raised the air which originally occupied a volume V occupies a volume F+ V, and by PV Boyle's law its pressure is ~p — =7-. The descent of the piston does not alter the pres- sure in the receiver. Therefore each complete stroke diminishes the pressure in the receiver in the ratio V r,j and after n strokes the pressure in the receiver r+r is P ( r+r' In the pump here described, if P be the pressure of the atmosphere, S the area of the piston, p the pressure in the receiver, the total force required to raise the piston is 266 Gases. [Ch. VIII. (P— p)S, and when exhaustion has proceeded pretty far, this force is inconveniently great. This difficulty may be overcome by having two barrels to the pump, the pistons in which are worked by a rack and pinion motion, which depresses one piston while raising the other. At the beginning of a stroke the resultant pressure downwards on one piston is equal to the resultant pressure upwards on the other, and though this equality does not continue throughout the stroke, the arrangement eases the action of the pnmp. This pump also produces more rapid exhaustion, for air is always passing from the receiver into one barrel or the other. Another way of obviating the difficulty of a single barrelled pump is to place a valve opening upwards in the top of the barrel. This re- lieves the piston from atmospheric pressure in its ascent. The advantages of a double barrelled pump can be obtained with a single barrel as follows : — The piston rod is hollow, and the piston valve opens into it ; a valve at the top of the barrel opens upwards. Two openings in the barrel at e and/ communicate through the tubes ghe, gf with the receiver, and are closed by stoppers at the ends of a metal rod which passes through the piston and is carried by friction with it in its vertical movement. When the piston ascends, the opening e and the valve in the piston rod are closed, the opening / and the valve Jc are open. Thus air flows in from the receiver below the / Fig. 148. Ch. viii.] Examples. ■ 267 piston and passes out of the barrel above into the room. When the piston descends, air from the receiver enters above the piston, and below the piston is expelled into the room. Compressing Pump. The pump can also be used for the compression of air ; it is only necessary that the valves should be reversed, so that in the ascent of the piston atmospheric air is admitted into the barrel, and in the descent the air is pushed into the receiver. If ^is the volume of the receiver of a single-barrelled pump, V that of the barrel, and P the atmospheric pres- sure, the pressure in the receiver after one stroke is pjj+ry V ' After n strokes, a mass of air occupying a volume nV at atmospheric pressure has been passed into the receiver, in addition to the volume V which it originally contained. Therefore the pressure after n strokes is „ • Examples. 1. The apparent weight of a piece of platinum when immersed in water is 20-6 gr. When immersed in mercury it is 8 gr. The density of mercury being 13.6, find the volume and density of the platinum. If V is the volume and p the density, V(p~i)g is the apparent weight in water and V{p-iy6)g is the apparent weight in mercury. Therefore F(p-i) = 2o.6, and F(p - 13.6) = 8. Therefore V=i and p = 2i-6. ' 2. 330 ccm. of cork float in a liquid of density 1 .2 with yj ccm. 268 Examples. [ch. vjin immersed. Find the density of the cork and the added -weight that would totally submerge it. The mass of the cork is equal to the mass of liquid displaced, that is to 77 x 1.2 gr. = 92.4 gr. And the density of the cork is — - = — = — • 33° 3° 2 5 3. The barometer stands at 30 inches. Find in tons' weight the pressure per linear horizontal foot on the sides of a rectangular tank 10 feet deep filled with water, the specific gravity of mercury (i. e. its density relative' to water) being 13.6, and the mass of a cubic foot of water being 1000 oz. The pressure of the atmosphere is equal to that of a column of water 3 x 3 ' = 34 feet high. The pressure at the centroid of a vertical face is that due to 34 + S or 39 feet of water. And by § 8 the total pressure per lineal horizontal foot is 39 x 10 x 1000 oz. weight = iofJ| tons' weight. 4. A tube closed at the upper end and containing air dips into a deep cistern of mercury, the mercury standing at the same level inside and outside. If the height of the portion of the tube above the mercury is equal to the height of the barometer (30 inches), find how much the tube must be raised that the contained air may occupy 40 inches of the tube. The original volume of the air is § of the final volume. Hence the final pressure is f x 30 inches = 22-J inches. Therefore the tube must be raised through 40 — 22J = 1 "j\ inches. 5. A barometer which has a little air in it reads 296 inches, the end of the tube being 6 inches above the top of the mercury, when a standard barometer reads 30. What is its reading when the standard registers 29 ? The pressure of the air above the mercury iB 30—29.6 when its volume is 6 (arbitrary units). Let x be the reading required. Then 29— a; is the pressure of the air above the mercury, and 35.6 — a; is its volume. Ch. viii.] ' Examples. 269 Therefore by Boyle's Law (35-6 -x) (29 -a:) = 6x-4, or x i -6^.6x+ 1030 = o. x = 32.3 + 3.65 approximately. The upper sign gives the air a negative volume and pressure, which is impossible. Therefore x = 28.65. 6. Taking a cubic foot of water as iooo oz., find the total pressure of the water arising from its weight on a side of a cistern 7 feet wide and 8 feet deep, if the cistern is filled with water. What horizontal line would divide the side into two parts so that the total pressure on each would be the same ? 7. A body consists of an alloy of two metals of specific gravities Sj , s 2 respectively ; its weight in vacuo is w and in water is w'. Show that the proportion of the two metals (by volume) is s 2 «>' - (s 2 - 1) w : (sj— l)w- s^v/. 8. A sphere whose internal radius is I foot contains mercury which covers f of the vertical diameter; find the resultant pressure on a horizontal plane through the centre of the sphere, assuming that a cubic inch of mercury weighs 3400 grains. 9. If in 10 strokes the mercurial gauge of an air-pump fell from 30 inches to 1 5 inches, what would have been its fall after the first 5 strokes ? 10. A square whose side is 8 ft. long has its plane vertical and its upper edge on the surface of the water in which it is immersed ; find the resultant pressure on one face of it. If the square is fixed and the surface of the water raised a foot, what would now be the magnitude of the resultant pressure ? Find also the centre of pressure in each case. 11. A spherical shell made of material of density p floats half immersed in water. Find the ratio of the internal to the external radius. 12. A cubical vessel is filled with two liquids of density 1 and 6-8. They do not mix and their volumes are equal. Find the ratio of the resultant pressure on the upper to that on the lower half of one of the vertical faces of the cube. 270 Examples. [ch. viii. 13. 1000 cubic inches of air under a pressure of 20 lbs. per square inch are mixed with 800 cubic inches of air under a pressure of 15 lbs. per square inch. Find the pressure of the mixture when it has a volume of 1 500 cubic inches, the temperature being the same in all cases. 14. A hydrometer floating in a liquid of density 1.1 has 5 cm. of the stem above the surface ; when it floats in liquid of density 1.2 it has 6 cm. of the stem above the surface. How much of the stem will be above the surface in a liquid of density 1.3 ? 15. 3 litres of liquid of density 0.8 are mixed with 5 litres of another liquid of density 1.04. Find the density of the mixture, assuming (1) that there is no contraction, (2) that there is a contraction of 5 p. c. of the joint volumes. 16. A wooden vessel 6 in. square and 6 in. in height with a neck 2 in. square and 6 in. in height is filled with water. Find the centre of mass of the water, and the total pressure on the base of the vessel. 17. A piece of silver and a piece of copper, fastened to the ends of a string passing over a pulley, hang in equilibrium when suspended in a liquid of density 1.15. Determine the relative volumes of the masses, the densities of silver and copper being 10.47 an d 8.89 respectively. 18. A piece of cork floats in a beaker of liquid. If the beaker is placed under the receiver of an air-pump and the air above it exhausted, will the cork rise or sink in the liquid ? 19. A vessel shaped like a portion of a cone is filled with water. It is 1 inch in diameter at the top, and 8 inches at the bottom, and is 12 inches high. Find the pressure (in pounds per square inch) at the centre of the base and the total pressure on the base. (A cubic foot of water weighs 1000 oz.) 20. A forcing pump the diameter of whose piston is 6 inches is employed to raise water from a well to a tank. If the bottom of the piston is 20 feet above the surface of the water in the well and 100 feet below the surface in the tank, find the least force which will (1) raise, (2) depress the piston ; friction and the weights of the valves being neglected. Ch. viii.] Examples. 271 21. A Nicholson's Hydrometer weighs 8 oz. The addition of 2 oz. to the upper pan causes it to be sunk in one liquid to the mark, while 5 oz. are required to produce a like effect in another liquid. Compare the densities of the liquids. If the density of the weights employed is 8, what weight must be placed in the lower pan in water to produce the same effect as 2 oz. in the upper pan ? 22. A Nicholson's Hydrometer when loaded with 200 grains in the upper pan sinks to the marked point in water ; a stone is placed in the upper pan, and the weight required to sink it to the same point is 80 ; the stone is then placed in the lower pan and the weight required is 128 grains. Find the density of the stone. 23. A piece of lead and a piece of sulphur are suspended by fine strings from the extremities of a balance beam and just balance each other in water. Compare their volumes, their densities being respectively 11-4 and 2. Which will appear the lighter in air and what weight must be added to it to restore equilibrium ? 24. A siphon is filled with water and inverted into a vessel of liquid of density 1-6. What is the condition that the liquid may flow through the siphon ? 25. If in Bramah's Press a (total) pressure of 1 ton is produced by a force of 5 lbs., and the diameter of the pistons are as 8 : 1, find the ratio of the arms of the lever employed to work the piston. 26. A piece of silver and a piece of gold are suspended from the ends of a balance beam which is in equilibrium when the silver is immersed in alcohol and the gold in nitric acid. The densities of gold, silver, nitric acid, and alcohol being 19.3, 10.5, 1.5 and 085 respectively, compare the masses of gold and silver. 27. A piece of wood floats partly immersed in water, and oil is poured on the water till the wood is completely covered. Explain whether this makes any change (and if so whether there is an increase or decrease) in the quantity of wood immersed in water. 28. If the height of the barometer changes from 29.55 *° 30.33 inches, what is the change in the weight of 1000 cubic 272 Examples. [Ch. viii. inches of air, assuming that ioo cubic inches of air weigh 31 grains at the former pressure, the temperature being con- stantly at 0° C ? 29. 10 ccm. of air are measured at atmospheric pressure, when introduced into a barometer vacuum they depress the mercury which previously stood at 76 cm., and occupy a volume of 15 ccm. By how much has the mercurial column been depressed ? 30. A faulty barometer tube 33 inches long contains air at the top and consequently reads 28 inches when the true pressure of the atmosphere is 29. Find its reading when the pressure of the atmosphere is 28 inches. 31. In an air-pump with one barrel the volume of the receiver is 10 times that of the barrel, and the lower j"jy of the barrel is not cleared by the piston. Find (1) the pressure after two complete strokes, (2) the lowest pressure that can be obtained with the pump. 32. A piece of lead weighing 1 7 grams and a piece of sulphur have equal apparent weights when suspended from the pans of a balance and immersed in water. When the water is replaced by alcohol of density 0.9, 1.4 grams must be added to the pan from which the lead is suspended to restore equilibrium. Find the weight and volume of the sulphur, the density of lead being 1 \\. 33. Two dock gates close a channel 12 feet wide, the depth of water on one side of the gate is 3 feet and on the other 15 feet. Find in tons' weight the force that must act on either gate to prevent them from opening. CHAPTER IX. Capillarity. § 1. When two particles of liquid are separated by less than a very small distance a (called the range of molecular action) they exert a certain force on each other. Thus if m be a particle of liquid round which as centre a sphere P is described with radius a, the particles which exert force on m lie within the sphere P. The potential energy of the liquid is equal to the work that its forces would do in separating its particles to such an extent that the distance between any two of them is greater than a. Let the external forces acting on the particles of the liquid be insignificant in comparison with the molecular or internal forces ; there may however be a uniform pressure of any magnitude on the surface of the liquid. The in^ ternal forces are supposed to form a conservative system. In considering the work done in removing m from the action of neighbouring particles two cases arise, according as the sphere of action of P is entirely within the surface, or meets it. In the first case the potential energy of each particle is a constant U; in the second case the eneigy differs from U by a quantity which depends on the distance of m from the surface. 274 Capillarity. [ch. ix. Thus the potential energy of all particles at a given dis- tance from the surface is the same, and if the distance is very small their number is proportional to the area of the surface. Therefore the potential energy of a liquid containing 11 particles is proportional to n U+ kS, where S is the area of the surface, and h is a constant which depends on the nature of the liquid but not on the form of the surface. A system of particles tends to have as little potential energy as possible, and since AS is the only term in the energy which can vary, it will become as small as possible. Therefore a liquid when free from external forces other than a uniform pressure over its surface, assumes such a form that its surface is either of maximum or of minimum area, according as k is negative or positive. Thus a small drop of mercury assumes a spherical form, for the surface of a sphere is less than the surface of any other solid of the same volume. Pure water in contact with glass tends to assume a maximum surface, spreading out into a thin film. The energy which a liquid possesses in virtue of its surface is called its Superficial Energy. Different liquids possess, for equal surfaces, different amounts of superficial energy, and the superficial energy per unit area is called the Constant of Capillarity. Since the surface tends to be as small as possible, it offers resistance to extension, just as a stretched membrane does. Phenomena of superficial energy are called capillary (from capilla), because they are most conspicuous in very fine tubes. Let ABCD be a rectangular framework, having the bar BC moveable along AB and CD, and containing a liquid film whose superficial energy is T per unit area. Ch. ix.] Superficial Energy. 275 The loss of potential energy in moving BC through a distance h towards AD is %liBC.T, a ABC being the decrease of surface. A tension 2T.BC applied to BC would maintain equi- librium, since the work done against it in this displace- ment is equal to that supplied by the film. This would be a tension T per unit length of BC on each face of the film. We may therefore consider the • surface of a liquid as a membrane stretched by a tension T; that is, across any line of length I on the surface a force Tl is exerted, uni- formly distributed along the line. Hence T is often called the Surface Tension of the liquid. Examples of capillary phenomena. If a wire ring is dipped into a soap solution and withdrawn, it carries with it a film of liquid whose surfaces are plane. If a loop of thread is laid on the film, and the film broken within the loop, the remainder of the film contracts until the loop forms a circle. As the circle has a larger area than any other curve of the same perimeter, the area of the film is the smallest possible. A needle laid gently on water will sometimes float on it, being supported by the tension of the surface. If a mixture of alcohol and water is made to have the density of oil, and by means of a pipette a drop of oil is placed inside the mixture, the drop assumes a spherical form. Here the fluid pressure balances the force of gravity, and the shape of the drop is determined by surface tension only. § 2. Pressure within a spherical drop of liquid. If a thin membrane with a fixed boundary is stretched, and unequal pressures are applied to its faces, the mem- t 2 276 Capillarity. [Ch. ix. brane becomes curved, and the curvature increases with the difference of the pressures. Hence the pressure within a drop of liquid exceeds the external pressure by an amount which depends on the curvature of the drop. Let p be the pressure within a spherical drop of radius r, T the surface tension, P the atmospheric pressure. Consider the conditions of equilibrium of half the drop, bounded by a plane through the centre perpendicular to a radius On. The hemisphere is in equilibrium under the external atmospheric pressure, the pressure across its plane circular boundary, the tensions at the perimeter of this boundary, and its weight. The resultant of the tensions is 2-nrT parallel to nO. The resultant of the pressures on the plane boundary is ■nr 2 p, parallel to On,, and that of the pressures on the curved boundary is ttt z P, parallel to »0(Chap. VIII. § 7). The weight is ■§ itr s pg and its component along nO can be neglected when r is small. Therefore the condition of equilibrium is Trr'*]3 = 2TrrT+Ttr 2 P, or p =P H r Pressure within a soap bubble. The bubble is a liquid film each of whose surfaces has a tension T, and as the bubble is very thin we may take the radii of the faces as equal, and neglect the weight of the bubble. %T Hence the pressure within the film is P H , and the pressure within the bubble is P -1 Ch. ix.] Boundary Conditions. 277 § 3. Conditions of contact of three fluids, e.g. water, mercury, and air. If a drop of water rests on mercury in air, we have three surfaces, mercury-water, water-air, and air-mercury, inter- secting in a line of which a very small part L, of length k, can he considered as straight. Let the line be perpendicular to the plane of the paper, and let OA, OB, OG be tangents in the plane of the paper to the surfaces. The surfaces can be regarded as stretched membranes in which the tensions are T A , T B , T a respectively. Kg . I49 . A cylinder of liquid of height h, which has L for its axis, is in equilibrium under the pressure of the surrounding fluid and the forces kT A , kT B , hT B . But by diminishing the base of the cylinder we can make the resultant pressure on its surface diminish indefinitely, and therefore the forces hT A , JiT B , hT , acting at 0, must satisfy the condition of equilibrium. Hence T A : T s : T c : : sin-50C : sin GO A : sin AOJB, and the angles AOB, BOG, CO A can be found when T A , T B , and T are known. If one of the tensions is greater than the sum of the other two, the fluids cannot rest in contact along a line ; the spreading of oil on water is a consequence of this. The above argument also holds when two fluids are in contact with a solid. A solid in contact with a liquid in air is generally wetted, the liquid-aii- and solid-air surfaces being in contact ; if OG is in the solid-air surface, OA and OB coincide and are in the same straight line with OC; hence T = T A + T B . In some cases, as when mercury is in contact with glass, the liquid does not wet the solid, but the solid-air and liquid-air surfaces intersect at an angle. 278 Capillarity. [Ch. ix. Then if OA is in the liquid-air surface, the conditions of equilibrium give the relation T -T B = T A sin AOB. Hence A OB, or the angle of contact of the solid and liquid is known when T A , T B , T are known. Hence under conditions for which T A , T s , T are constant, the angle AOB is a fixed angle. C o % _ It follows that when a liquid is in Fig. 150. . . ^ contact with a solid its surface makes a definite angle with that of the solid, and the potential energy of the liquid is as small as it can be consistently with this condition. Equilibrium of a heavy drop. In this case the drop still tends to possess a minimum surface, but its centroid also tends to lie as low as possible, and by the Principle of Virtual Work the form of the drop is such that the work done by gravity in a very small change of form is equal and opposite to that done by the surface-tension. Thus, while very small drops of mercury are approximately spherical, a large drop, lying on a hori- zontal plane, assumes a flattened form. § 4. Elevation of Liquids in capillary tubes. When a tube is immersed vertically in water, the free surface rises higher in the tube than outside it, and the narrower the tube, the greater is the elevation of the water. Let r be the radius of the tube, h the difference of the levels of the free surfaces within and without the tube, p and T the density and surface tension of the liquid. When the tube is very narrow we may take the curved surface, or meniscus, to be a sphere. Ch. ix.] Boundary Conditions. 279 If Pis the atmospheric pressure, P is the pressure just within the meniscus, and P j- pgh is the pressure at the level of the free surface outside the tube ; but this is equal to P. „, Therefore — = pgh, or T=\ pgrh. Hence the elevation of liquid in a tube is proportional to the surface tension, and inversely proportional to the radius of the tube and to the density of the liquid. This affords the simplest method of measuring the sur- face tension of such a liquid as water. The radius r of the tube may be determined by measuring the length I of a column of mercury in the tube, and deter- mining the mass m of the mercury. Taking the density of mercury p as about 13-55 a t ordinary temperatures we have irr 2 lp = m, from which r can be found. Depression of mercury in a tube. Mercury in contact with glass and air assumes as small a surface as possible, and the pressure within its surface exceeds that of the air. HeDce the surface of mercury is depressed in the tube and has its convexity upwards. T If a is the angle of contact, is the radius of the cos a surface, and the pressure within the meniscus is ■D ^T F -\ cos o. r Hence the meniscus is at a depth ft below the free surface in the cistern, where 3 T cos a = pgrh. 280 Capillarity. [Ch. ix. Thus the pressure of the atmosphere determined by a mercury barometer is less than the true pressure, owing 1 to the surface tension of the mercury. § 5. Elevation of liquid between two parallel plates. It is first necessary to find the tension T of a cylindrical membrane of radius r, filled with liquid at pressure p and subjected to an external pressure P. Bisect the cylinder by a plane L through its axis and consider the portion intercepted between this plane, the curved surface, and two planes perpendicular to the axis at a distance / apart. Resolve the acting forces along a perpendicular to L. The resultant of the tensions in the membrane is a Tl. The resultant of the internal pressures is 2 rip, and of the external pressures 2rlP (Chap. VIII. § 7). T Therefore p — P + — • r "When two parallel plates are immersed vertically, at a small distance ar apart, in a liquid which wets them, the liquid surface, except near the edges of the plates, is approximately a horizontal cylinder of radius r. Hence if the liquid wets the plates the pressure within T the meniscus is P , and at a depth h below the T meniscus it is P p ok. r This is equal to P if T = prgh. Hence the elevation of the liquid in this case is half of the elevation in a tube of radius r. Apparent attraction of the plates. The external pressure on the free portion of either plate Ch. ixj Elevation of Liquid. 281 is P, and the internal pressure at a height x above the surface in the cistern is P—pgx. Hence at a height x there is an excess of external pressure pgx. If the plates are rectangular with one pair of edges horizontal, and if h is the height to which the water rises, b the width of the plate, the area immersed is bh, and the resultant pressure is \ bh 2 pg. Thus the pressures urge the plates towards one another, and the plates apparently attract one another. The resultant pressure may also be written 1 bT 2 2 pgr* ' The attraction between the plates thus increases rapidly as the distance between them diminishes ; two wet plates fairly pressed together require considerable force to separate them. When the liquid is depressed between the plates, there is also an apparent attraction. Illustrations. 1. A thin flexible string forming a loop of length 2 wr is laid on a plane film, and the film within the loop is broken ; find the tension X of the string. Let T be the surface tension. The string becomes a circle of radius r, each portion of it being pulled outwards by a tension 2 T (for the film has two surfaces) per unit length. A part of the string which forms a semi-circle is in equilibrium under these forces and the tensions X at its extremities, and by Chap. VIII. § 7 the liquid tensions are equivalent to a single force \rT per- pendicular to the (Jiameter which bounds the semi-circle. Therefore 2 X = 4 rT, or X=2rT. 2. A liquid film OAB of surface tension T is bounded by two wires OA and OB, each of length I, and a light inextensible string AB of length — . The wire OA is fixed horizontally and OB can 282 Capillarity. [Ch. ix. turn freely round 0. Prove that if OB rests at an angle 60° to the horizon its weight is 8 TV. Complete the parallelogram OADB. BA subtends an angle 60° ( = — ) at and at D. Also BA = — V 3/ 3 and BD = DA = I. Therefore D is the centre of the circle of -which AB is an arc. The forces which act on OB are its own weight W, the tension 2 27 per- pendicular to OB, the resistance at 0, and the tension t in the string at B. By the last example, t = 2TI. Hence taking moments round 0, W = 8 27. 3. Two soap bubbles of radius *-,, r 2 rest in contact along a circle. Show that the film between the bubbles is a sphere of radius '1'2 Let P be the atmospheric pressure. P+ — is the pressure within the first bubble ; A T P-\ is the pressure within the second. The difference of pressures is4T( — ~— J • Therefore the radius r of the bounding surface is given by I — ! l r r x ~ r 2 At a point common to the three spheres, the tensions are equal and in equilibrium. Hence the spheres cut at an angle of 120°. If r 1 = r i , the bounding surface becomes a plane. Examples. 1 . In a tube whose internal diameter is 1 mm. a certain solution which wets the side of the tube rises to a height of 10 mm. Calculate the pressure within a bubble of this solution of 1 cm. ch. ix.] Examples. 283 radius, the atmospheric pressure being 76 cm. of mercury, the density of the mercury 13.6, and the density of the solution 1.1. 2. A soap bubble of radius r rests within a bubble of radius ir. If T is the surface tension, d the density of air under atmospheric pressure, find the total mass of air inside the larger bubble. 3. Two vertical tubes of radii r lt r 2 , are connected below by a pipe and contain liquid of density p and surface tension T. Prove that the difference of the heights of the liquid columns is pg \ ri V 4. A hydrometer consisting of a glass cylinder of radius r, loaded below, floats in a liquid of density p and surface-tension T. Prove that if the liquid were free from surface-tension, the hydrometer would rise through a height 5. A vertical framework is formed by two horizontal wires of length I, and two light flexible strings of length V, attached to the end of the wires. The upper wire is fixed and the framework encloses a soap film. Calculate the weight of the hanging wire if the strings at their lower extremities make an angle 30 with the vertical. CHAPTEE X. Units and their Dimensions. § 1. Units. Every measurable quantity is expressed by two factors, one a pure number or numeric, the other the unit in terms of which the quantity is measured. Thus in the expression iolb., 10 is the number and the lb. is the unit. It has been already seen that physical quantities are most easily dealt with when they are measured as multiples of units which are derived from the units of mass, length, and time. A system of units connected in this way is called an Absolute System of Units, and any one of the units is called an absolute unit. The units of mass, length, and time are perfectly arbitrary, and if they are altered the other units alter too. The question that we shall now consider is — In what way do the derived units alter, when the fundamental units are changed ? If (I) is the unit of length, e. g. the foot or centimetre, 1(1) is the full expression of a length containing I units. Similarly, if (t) is the unit of time and (m) the unit of mass,- a, time and a mass hitherto denoted by t and m are fully described by the symbols t(t) and m(m). If a new unit of length (L) is chosen, and the length 1(1) contains L of the new units, L(L) = l(l). (i) Ch. x.] Dimensions of Units. 285 Let the units of length, time, and mass be changed from (I) to x(l), from t to y(t), and from (m) to z(m), x, y, z being any numbers or fractions. We shall determine the changes in the derived units. Units of Area and Volume. If (*) is the unit of area when (I) is the UDit of length, the rectangle contained by the lines A (I), k(l) is hk(s). When the unit of length is changed to x(l)or (L), the h k . lines become - (E), - (E). And the rectangle becomes OS OS hk —% (S), where (S) is the new unit of area. OS Ilk Therefore -£ (8) = hk (»), and (S) = x z (s). Hence the unit of area is proportional to the square of the unit length. Similarly the unit of volume is propor- tional to the cube of the Unit length. Unit of Velocity. Let (») be the unit of velocity in the system (l), (m), (t), i.e. the system derived from (I), (m), (t) as fundamental units. If a body traverses a distance 1(1) in a time i(t) its velocity is v(v), where v — -. Now let the units be changed to (L) and (T), and let (V) be the new unit of velocity. Then 1(1) = —(E) and t(t) = —(T), and the new measure x y of the velocity is V, where V — — j= —v. SO v OS f Also V(V) = v(v). Therefore (V) = - (v). Thus the unit of velocity is directly proportional to the 286 Dimensions of Units. [ch. x. unit of length and inversely proportional to the unit of time. Unit of Acceleration. Let v (v) be the velocity acquired in time t {£) by a body. Its acceleration is a (a), where (a) is the unit of acceleration, and a = —. v In the new system the acceleration is A (A), and A = - 7j , = 9 --="-a. T x t x Therefore since A (A) — a(a), (A) = -^(a). Unit of Force. Let a force/(/) be required to communicate an accelera- tion a («) to a mass m (m). Then f = ma. When the units are changed, let the same force be denoted by F(F). Then F=MA = £.tna=£(f). zx xz v ' And since F(F) = f(f),(F) = °^ f. Unit of Density. Let a volume u (u) contain a mass m (m) ; the density is d Id), where d = -. v ' u Changing the units, let U(U) be the new measure of the volume, D(D) the density. Then B=% =^ = ^-.d. U zu z And since D(D) = d(d), (D) = ~ (d). 3G Ch. x.] Dimensions of Units. 287 The numbers which specify a strain and an angle are the same whatever system of units we adopt. When the alteration of a fundamental unit [A) in the ratio 1 : p alters a derived unit (B) in the ratio 1 : p m , (B) is said to be of dimensions m in (A). Thus the unit of velocity is of dimensions 1 in length and — 1 in time. Let (B) and (C) be two units which are of dimensions m and n respectively in [A), and let them become (b) and (c) when [A) is changed in the ratio 1 :p. Then (b) = p m (B), and (c)=p n (C); and by multipli- cation (b) (c) = p m + « (B) (C). Thus the product of two units has dimensions, which may be obtained by adding the dimensions of the units. In the remainder of this chapter we shall generally use small letters to denote numerics, and capitals to denote the corresponding unit. The units already found can be denoted as follows : — Length L. Time T. Mass M. Velocity LT-\ or V. Acceleration LT- 2 , or A. Force MLT~\ or F. Density ML' 9 , or L. Angle Constant. Strain Constant. The dimensions of the products of the units follow from this table. Thus AL and V 2 are both of dimensions L 2 T~ 2 - Each term of any physical equation, e.g. s = vt + J at 2 , 288 Dimensions of Units. [Ch. x. is of certain dimensions in the fundamental units. But the equation is true whatever these units may be, and therefore a change of units changes all terms in the equation in the same ratio. This can only be true if the equation is homogeneous in the fundamental units. Hence all physical equations are homogeneous, and this result may be usefully employed to verify equations, for it is a necessary (but not sufficient) condition of accuracy. It may also be employed to determine the dimensions of physical units. Other fundamental units might be taken, e. g. V, A, F. We then have T= VA~\ M = FA~\ L= Y 2 A~\ whence the relation between any other unit and V, A, F can be found. Astronomical Units. If we regard gravitation as a fundamental property of matter, the unit of mass may be made to depend on the units of length and time. The astronomical unit of mass is defined as that which communicates to a mass at unit distance from it the unit of acceleration. According to this definition and the law of gravitation, a mass m exerts a force mm' d~ z or a mass m' at distance d. Therefore, if a is the acceleration of m!, md~ z = a ; since this relation is homogeneous, mass is of dimensions L Z T~ 2 in this system of units. Substituting L S T~ 2 for M in the formulae above, we obtain the astronomical dimensions of other units in L and T. The unit of density is T~ 2 , i. e. it does not depend on the unit of length. Dimensions of Stress. Coefficients of Elasticity. Iff is the force exerted across an area s, the stress on s \sp, where/ = jew. Ch. x.] Dimensions of Units. 289 Now/" is of dimensions MLT~ 2 in the C. G. S. system, and s is of dimensions L 2 . Therefore p is of dimensions ML~ X T~ 2 , since f = ps is a homogeneous relation. If t is a strain (e. g. dilatation or shear) corresponding to the stress p, h the corresponding coefficient of elasticity (e. g. resistance to compression or distortion), p = kt. Since t is of no dimensions in mass, length, or time, k is of the same dimensions as a stress. Similarly, energy (potential or kinetic) is of dimensions ML 2 T~ 2 , and superficial energy being energy per unit surface is of dimensions MT~ 2 . A couple is of dimensions ML 2 T~ 2 . § 2. Dynamically similar systems. Two systems of bodies are said to be dynamically similar when the numerical data which define one system can by transformation of the fundamental units be transformed into corresponding data defining the other system. Thus they are similar systems constructed on different scales. The consideration of the conditions of dynamical simi- larity often leads to important results by simple methods. The following applications illustrate this : — (1) The simple pendulum. Let I be the length of a pendulum, g the acceleration of gravity. By altering the unit of length in the ratio 1:1' and the unit of acceleration in the ratio g : g\ the length and acceleration become I', g' respectively. Since T — Z%A~%, the unit of time is altered in the ratio V r : V / Hence if the time of vibration was formerly denoted v 290 Dimensions of Units. [ch. x. by t, it is now t\ where t:t' ':: I L : /^. But t' is the time of vibration of a pendulum of length V where the acceleration of gravity is g', oscillating through the same angle as the given pendulum. Hence the time of vibration of a pendulum of length I is proportional to / — ,ff being the acceleration of gravity. V 9 The method fails to give the constant factor aw in the complete formula, and it gives no indication that the time of vibration depends on the amplitude of oscillation. (2) Resistance of a wire to torsion. Let I, r be the length ,and radius of a wire of rigidity n, to which a total twist 8 is imparted by a couple w. Let the unit of length be decreased to — th of its given x value. The length and radius of the wire become xl and xr ; the rigidity becomes x~ x n, and the couple becomes x 2 w. Therefore to twist a wire of length xl and radius xr through an angle 6 requires a couple x 2 w if the rigidity is xr x n\ it follows from Hooke's law that a couple aPw is required if the rigidity is n. Q The total twist of a length I of the wire is — . To produce a total twist 8 in this length requires a couple x i w. Hence a wire of length I, radius xr, and rigidity n requires a couple x i w to give it a total twist 8. Therefore the resistance to twist varies as the fourth power of the radius (Chap. VII, § 13). (3) Kepler's Third Law. A planet of mass m describes an orbit round a luminary of mass m'. If the unit of length is changed from L to ch. x.] Dimensions of Units. 291 - • L, and the unit of mass from M to - ■ M, the astro- x z nomical unit of time changes from T to T%~% z%, since If d is originally any diameter of the orbit, xd denotes the same diameter after the change of units ; and, if t originally denotes the time of describing the orbit, tafcz~^ denotes this time after the units have been changed. Also the periodic time does not depend on the mass of the planet, for the acceleration is independent of this mass. Hence if the planet describes an orbit whose principal diameter is d round a luminary of mass m' in time t, it will describe a similar orbit with principal diameter xd round a luminary of mass zm! in a time os* z~*t. Hence for similar orbits the periodic time is proportional to I —1 m being the mass of the luminary, d the prin- \l m' cipal diameter of the orbit. Thus we have proved Kepler's Third Law for the case of similar orbits. (4) Consider a mass of liquid at rest under its own gravitation. It will assume the form of a sphere, and if the surface is deformed in any way and then released the liquid will be set in vibration. It is required to find the relation of the period of the vibration to the size and density of the sphere. Let t be the time of vibration, d the density of the sphere. As the force in question is that of gravitation, we use the astronomical system of units, and in this system D = T~ 2 '- Hence if we alter the unit of length only, the TJ 2 292 Dimensions of Units. [ch. x. density and time of vibration are denoted by the same numbers as before. Therefore in similarly deformed liquid spheres of the same material the time of yibration is the same whatever may be the radius of the vibrating sphere. In the case of spheres of different materials we must suppose the unit of density, and therefore the unit of time, to vary. If the density denoted initially by d is finally denoted by d', a time denoted initially by t is finally denoted by t /_, , or f. J \] d' Hence for similarly deformed spheres of different densities tX —y= J Vd t being the time of vibration, and d the density of a sphere. If g is the acceleration of a falling body at the surface of the sphere, and a is the radius, and tx N 9 q = — irda? ■+■ a 2 = — it da, 3 3 The reader will find it a useful exercise to construct a system of units analogous to the astronomical system, assuming that the law of force between two masses m, m' at a distance r apart is — — > and thence to compare the periodic times of similar orbits described by m about m! under this law of force. Making n= — i, the isochronism of simple harmonic motions may be verified. Example. — A particle oscillates along a straight line AB under an attraction to A which is proportional to the distance from A. Ch. x.] Dimensions of Units. 293 Show that the oscillations are isochronous when the motion is resisted by a force proportional to the velocity. The above method may also be applied more briefly by noticing that all physical equations are homogeneous. Thus if the time of vibration t of a pendulum is required, it is assumed that it depends only on I the length of the pendulum, and on m and mg the mass and weight of the bob. It can therefore be expressed by the formula Al x m v f, when A, x, y, z are numerical constants. Now this expression is of dimensions T. Therefore x + z = o, V = °, *= -1. and -V) In order that this method may be trustworthy we must be sure that we have enumerated all the quantities on which t (or any other quantity required) depends. Example. — To compare the times of vibration of different liquid spheres held together by their surface-tensions, other forces being neglected. The time of vibration depends on the surface-tension S of dimensions MT~* and the mass M of the sphere. Hence * = VM", and 7.x = - 1, x + y = o. /w Therefore t = A V— , where A is a numerical constant. It must be understood that the vibrations of the spheres are similar, i. e. that the spheres are similarly deformed before being left to vibrate. u 3 2g4 Dimensions of Units. A good illustration of the care needed in applying the Principle of Dynamical Similitude is afforded by an inven- tion which was brought before a recent Commission on Mines. It was an arrangement for preventing a cage from falling to the bottom of a shaft if by any acci- dent the rope supporting the cage broke. A spring, compressed by a weight, would on being freed from pres- sure release a clutch, which after release (in the model) held the guide ropes in the shaft, and so prevented the cage from descending. If by any chance the rope broke, the cage would begin to descend with nearly the accelera- tion of gravity, the pressure of the spring would become very small, the clutch would be released, and the cage stopped. In applying conclusions derived from this model to the cage in an actual shaft, it is necessary to bear in mind that the moving mass m and its velocity v are very considerably greater than the corresponding quantities in the model, and that work J mv 2 must be spent in bringing the cage to rest. The strain on the guides is thus very considerable indeed compared with that in the model, and correct in- ferences could only be drawn if the guides in the model were made correspondingly weak. ANSWEES TO EXAMPLES CHAPTER I. Page 10. Ex. 5. 2 V2 along DB. 6. o. 8. (1) a(v / 2-i) parallel to BA. (2) a\^3, making an angle tan _ 'V2 with AB. Page 11. 2. 33 ft. per second. Page 22. 4. - ooo 1 4 ... inches per second. -000156. .. 5. 5 miles an hour, 330 yards. Page 29. 3. 100. 4. —65. 5. 172, 208, 244. 6. 28, -8, -44. 7. 33J. 8. -ssf 9. 200. 10. 65. 11. 625, —137-5. 12. 8. 13. 12-5. 14. -8, 40; 36, 28, 20, 12, 4. 15. 16, 64, 144. The distance travelled in the 10th second is 304 ft. 16. 4|. T s s sec. 17. 6. I sec. 18. 107^, 429^. Page 34. 4. 6 sees. ; 176-4 metres. 5. (1) 5 seconds ; (2) 122-5 metres; (3) 7840 centimetres per second. 296 Answers to Examples. 7. 160. 8. 522. 9. 48 V 2 at an angle 45 from the horizon, 3 sees. 12. 6008 ... feet per second. Page 42. 4. -oi. 5. 60 miles an hour, 612 yards. 8 seconds. 6. -125. 7. 28800 ft. 960^2. 8. -00007295... 9. -893... xio -9 , .1931... xio -6 . Page 78. 9. CHAPTER II. 7 5 X5 4 7 8 X5 2 4 2 10. 545 cm. per sec, f sec, 13352J. 13. (1) 468 grams weight, 195 ; ( 2 ) 432.75 ; (3) 180. 15. 12 P. 17. i5oV , 2 — 100 along AC. 18. 300 poundals. 19. 3 */%, 3. 20. 8,-32,96,24. 21. 5^ ft. 22. 240,960. 23- ig, 3| lbs. weight, f sec. 24. 214-8. ., 25. 984-94. 26. 20 ft. per sec. 27-29. See §11. 30. -A;- * 4^ 2 31. 16 ft., iof. 32. 8 ft. 33. (i) 9 fin.; (2) 9 f lbs. CHAPTER III. Page 83. 1. 2,060,100 ergs. (3) 40-24. Page 109. 4. 22500 foot-poundals. 5625 poundals. Answers to Examples. 297 2 5. 160 lbs. weight in each rope. 6. -• 7. 96} min. 8. See Ex. 2. 9. 273! H. P. 10. T-V. — • 11. At a height 120^/3 — 64 ft. After 2 sec. 60 ft. from the wall. 12. The subsequent velocities are —16 and 14. 13. 24.1... 14. 108. 15. 3,080,000, 26f min. 17. « 2 = 2 — V2, CHAPTER IV. Page 138. 1. 6 lbs. weight. 7iq (2) ' ifa a , J/ being the mass, and a the v ; 1600 ' ° »■ »Vf. radius. 9. If W is the weight of the bar, the tension is —7=, and the force at the hinge is • V2 2 , , 20 48 , „ 9800 ... , . 11. — — • 13. - jr 2 ft. poundals. 1777 3 14. 56000 7T 2 , 233! 7T. 15 . 30625^ 16> I000> 512 298 Answers to Examples. CHAPTER V. Page 175. 6. 7-4; 6-6. 7. 10 V^; f. 8. -^- lbs. weight. 10. 6 ^ — - lbs. weight. 3 v 3 2 V 3 + 1 11. —V cwt. 14. -j- • 4V2 6 17. .l£°_ . _55°_ 18 _ 32 ^ lbs- weight. 9 -A 9 A 19. ^ (cot a + ii) = iW,W being the weight of the beam. CHAPTER VIII. Page 269. 6. 14000 lbs. weight; 4^/2 ft. from the surface. 8. — lbs. 9. To 15-/ 2 inches. 35 10. 16000 lbs. weight; 20000; 5^; 5 r \ ft. below the upper edge of the square. 11. . 3 /zp-i 12. 4 : 13- 2/> 13. 21 J lbs. per sq. inch. 14. 6£J cm. 15. (1) 0-95; (2) 1. 16. 8-4 in. below the top, 15I lbs. weight. 17. 774 : 932. 19. ^|| , 1 ^f lb. weight. 20 .87^,3"S_*. 21 . I0 . 22> 16 o 23. volume of lead : volume of sulphur : : 1 : 10-4. 25. 7 : 1. Answers to Examples. 299 26. Mass of silver : mass of gold : : 965 x 193 : 17-8 x 10-5 28. ^-5- gr. 29. sof cm. 591 s ° * 30. Beading 27-15 cm. 31. 13 1 (1) — • (2) — atmosphere. v ' 40 v ' 10 32. 2025 31 grams, ifi-fi ccm. 33. ■ 00 v II2 THE END.