i 14 ii iilF'^ iiiii:;i ii Pilli ''^^iliilillliilllii ill i liiiiiii iliiiliiliiii Iji I iiiflil liii iiiii liii^^ ill: i iill 1! ■:il I I iii iiiiiPii mm Am •■•I'll'. :iil ill ^iiHi'i! l-i^iii i^H :ii!tp1i!!iii iiiiliiffiiil;ii:i!!li'i':;:;:;!;!;| lit'v''-!' ■; -ii.-i,*! iiiliiiii'liilililiiilinjliiiiiliiitiiK^ Ii 111 ill: 1! ! i i h! isSi i'i i 'it t iii ; :; u i4 :Hi;u ;-.i: i» ■iiJiliftJlMI ®0meII Uromitg J itmg BOUGHT WITH THE INCOME | FROM THE SAGE ENDOWMENT FUND THE GIFT OF Hi^nrQ W. Sage 1891 G|a.|9*. A....&.G.s..y:..^ ^BMiim'fKffiiiSLSRP''^'' "nathematics inci ,. 3 1924 031 260 247 olin,anx Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031260247 George Bell & Sons' CAMBRIDGE MATHEMATICAL SERIES. Crown 8vo. ARITHMETIC. With 8000 Examples. By Charles Pendlebury, M.A., F.R.A.S., Senior Mathematical Master of St. Paul's, late Scholar of St. John's College, Cambridge. Complete. With or without Answers. 6th edition. 4J. 6d. In two Parts, with or without Answers, 2s. 6d. each. Part 2 contains Commercial Arithmetic. (Key to Part 2, "js. 6d. net. ) In use at Winchester ; Wellington ; Marlborough ; Charterhouse ; St. Paul's ; Merchant Taylors' ; Christ's Hospital ; Sherborne ; Shrewsbury ; Bradford ; Bradfield ; Leamington College ; Felsted ; Cheltenham Ladies' College ; Edinburgh, Daniel Stewart's College ; Belfast Academical In- stitution ; King's School, Parramatta ; Royal College, Mauritius ; &c. &c The Examples separately, without Answers, ^th edition. 3J. (Answers free to Masters.) ALGEBRA. Choice and Chance. An Elementary Treatise on Permutations, Combinations, and Probability, with 640 Exercises. By W. A. Whit- worth, M. A., late Fellow of St. John's College, Cambridge, a^h edition, revised, ds. EUCLID. Books I-VI. and part of Book XI. Newly translated from the original Text, with numerous Riders and Miscellaneous Examples in Modern Geometry. By Horace Deighton, M.A., formerly Scholar of Queen's Col- lege, Cambridge ; Head Master of Harrison College, Barbados. New edition, revised, with Symbols. 4J. 6d. A Key to the EXERCISES (for Tutors only), 5^. Or in Parts: Book I., is. Books I. and IL, is. 6d. Books MIL, 2S. 6d. Books III. and IV. , is. 6d. In use at Wellington ; Charterhouse; Bradfield; Glasgow High School ; Portsmouth Grammar School; Preston Grammar School; Eltham R.N. School ; Saltley College ; Harris Academy, Dundee, &c. &c. EXERCISES ON EUCLID and in Modern Geometry, containing Applications of the Principles and Processes of Modern Pure Geometry. By J. McDowell, M.A., F.R.A.S., Pembroke College, Cambridge, and Trinity College, Dublin, ^rd edition, revised. 6s. TRIGONOMETRY, The Elements of. By J. M. Dyer, M.A., and the Rev. R. H. Whitcombe, M.A., Assistant Mathematical Masters, Eton College. 2nd edition, revised. 4J. 6d. TRIGONOMETRY, Introduction to Plane. By the Rev. T. G. Vyvyan, M. A., formerly Fellow of Gonville and Caius College, Senior Mathematical Master of Charterhouse. j,rd edition, revised and corrected, y. 6d. ELEMENTARY ANALYTICAL GEOMETRY. By Rev. T. G. Vyvyan. \In the Press. CONIC SECTIONS, An Elementary Treatise on Geometrical. By H. G. Willis, M.A., Clare College, Cambridge, Assistant Master of Manchester Grammar School. 5^. CONICS, The Elementary Geometry of. By C. Taylor, D.D., Master of St. John's College, Cambridge, "jth edition. Containing a New Treatment OF THE Hyperbola. 4^. 6d. Mathematical Works. CAMBRIDGE MATHEMATICAL SERIES SOLID GEOMETRY, An Elementary Treatise on. By W. Steadman Aldis, M.A., Trinity College, Cambridge; Professor of Mathematics, University College, Auckland, New Zealand, i^h edition, revised. 6s. ROULETTES and GLISSETTES, Notes on. By W. H. Besant, Sc. D., F. R. 8. , late Fellow of St. John's College, Cambridge. 2nd edition, ^s. GEOMETRICAL OPTICS. An Elementary Treatise. By W. Steadman Aldis, M.A., Trinity College, Cambridge, e^h edition, revised. 4^. RIGID DYNAMICS, An Introductory Treatise on. By W. Steadman Aldis, M.A. 4J. ELEMENTARY DYNAMICS, A Treatise on, for the use of Colleges and Schools. By William Garnett, M.A., D.C.L. (late Whitworth Scholar), Fellow of St. John's College, Cambridge ; Principal of the Science College, Newcastle-on-Tyne. yh edition, revised. 6s. DYNAMICS, A Treatise on. By W. H. Besant, Sc.D., F.R.S. 2nd edition, los. 6d. HYDROMECHANICS, A Treatise on. By W. H. Besant, ScD., F.R.S., late Fellow of St. John's College, Cambridge. S^A edition, revised. Part I. Hydrostatics. 5^. ELEMENTARY HYDROSTATICS. By W. H. Besant, Sc.D., F.R.S. lyh edition, rewritten. 4^. 6a!'. HEAT, An Elementary Treatise on. By W. Garnett, M. A. , D. C. L. , Fellow of St. John's College, Cambridge ; Principal of the Science College, New- castle-on-Tyne. 6th edition, revised. 4J. 6d. THE ELEMENTS OF APPLIED MATHEMATICS. Including Kinetics, Statics, and Hydrostatics. By C. M. Jessop, M.A., Clare College Cam- bridge. [/» t)ie Press. MECHANICS, A Collection of Problems in Elementary. By W. Walton, M.A., Fellow and Assistant Tutor of Trinity Hall, Lecturer at Magdalene College. 2nd edition. 6s. PHYSICS, Examples in Elementary. Comprising Statics, Dynamics, Hydrostatics, Heat, Light, Chemistry, Electricity, with Examination Papers. By W. Gallatly, M. A., Pembroke College, Cambridge, Assistant Examiner at London University. 41. MATHEMATICAL EXAMPLES. A Collection of Examples in Arithmetic Algebra, Trigonometry, Mensuration, Theory of Equations, Analytical Geometry, Statics, Dynamics, with Answers, &c. By J. M. Dyer, M. A (Assistant Master, Eton College), and R. Prowde Smith, M.A. 6s. CONIC SECTIONS treated Geometrically. By W. H. Besant, Sc.D., F. R. S. , late Fellow of St. John's College. ?ith edition. 4.?. 6d. Solutions to the Examples, 41. ANALYTICAL GEOMETRY for Schools. By Rev. T. G. Vyvyan, Fellow of Gonville and Caius College, and Senior Mathematical Master of Charter- house. 6th edition. 4?. 6d. CAMBRIDGE MATHEMATICAL SERIES. THE ELEMENTS OF APPLIED MATHEMATICS. PRINTED BY C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS. THE ELEMENTS OF APPLIED MATHEMATICS INCLUDING KINETICS, STATICS, AND HYDROSTATICS. BY C. M. JESSOP, M.A., LATE FELLOW OP CLABE COLLEOE, CAUBBIDOE, LECIUBEB IN. MATHEMATICS IN THE DUEHAM COLLEGE OF SCIENCE, NEWCASTLE-ON-TYNE. LONDON GEORGE BELL AND SONS. CAMBBIDGE: DEIGHTON, BELL AND CO. NEW YOEK: 66, FIFTH AVENUE. 1894 PREFACE. THE basis of the present elementary treatise on Applied Mathematics was formed by a manuscript written by my father; on his death the task devolved upon me of com- pleting the work, and by various additions the original scope of the book has been considerably extended. It is designed mainly for use in Schools and for various public examinations, and I am not without hope that the explanation of the principles of the subject herein contained is sufficiently detailed to render it valuable to private students who are not in a position to obtain much assistance from teachers. I have adopted the arrangement, which is now generally approved, of considering successively kinematics, kinetics, and forces in equilibrium. The chief innovation is the introduction of a separate chapter on Graphical Statics. Throughout, the C.G.S. system has been treated side by side with the foot-lb.-sec. system. A large number of worked-out examples has been added in illustration of every point of importance throughout the book. The most fundamental propositions have been treated in the most elementary manner consistent with giving valid proofs, most of them can be mastered by students whose geometrical knowledge does not extend beyond the first three books of Euclid, but in order to give completeness, trigonometrical methods have been added, usually in smaller type. J. b 11 PREFACE. The works to which I am under chief obligation are Matter and Motion by the late Prof. Clerk Maxwell, and The Properties of Matter by Prof Tait. I have also found Prof Minchin's Hydrostatics and Elementary Hydrokinetics, and Prof Hicks' Elementary Dynamics of assistance. I am under deep obligations to several friends for their assistance, among whom I should like to mention specially E. W. Hobson, Esq., Sc.D., F.R.S., who has read carefully certain portions of the proofs and whose advice I have found of the greatest benefit ; D. Rintoul, Esq., M.A., Assistant Master in Clifton College, who has read the proofs of Part I., and to whom I am much indebted for the detection of errors and for valuable suggestions ; A.. Larmor, Esq., M.A., Mathe- matical Head Master of the Londonderry Academy, who has read over the whole of the manuscript and has given con- stant assistance throughout, many of the methods adopted being due to his suggestion. A number of the figures are due to the kindness of T. H. Easterfield, Esq., M.A., late Scholar of Clare College. The answers to the examples have been verified through- out by an experienced computer, J. W. F. Allnutt, Esq., M.A., and I am sanguine that few errors have escaped detection. To the Cambridge University Local Examinations and Lectures Syndicate I am indebted for permission to use the examples set in their examinations, and to the Oxford and Cambridge Schools' Examination Board for a similar con- cession. C. M. JESSOP. Newcastle-on-Tyne, Oct. 1893. m CONTENTS. PART I. (KINETICS AND STATICS). CHAPTEE I. ABTIOIiES PAOES 1 — 29 Motion in a straight line 1 23 CHAPTER II. 30 — 44 Composition and Besolution of Velocities . 24 — 39 CHAPTEB III. 45—64 The Laws of Motion 40—57 CHAPTEE IV. 65—70 Projectiles 58—64 CHAPTEE V. 71 — 84 Composition and Besolution of Forces . . . 65 — 84 CHAPTEE VI. 85—103 Parallel Forces. Moments 85—103 CHAPTEE Vn. 104—112 Couples 104—111 CHAPTEE VIII. 113—134 Centre of Gravity 112—129 CHAPTER IX. 135 — 143 Forces in equilibrium 130—145 IV CONTENTS. CHAPTEE X. ARTICLES PAQEU 144—160 Work and Energy 146—161 CHAPTEE XI. 161—192 The simple machines 162—196 CHAPTEE Xn. 193—201 Friction 197—211 CHAPTEE Xin. 202—208 Impact 212—218 CHAPTEE XIV. 209—216 Graphical Statics 219—233 CHAPTEE XV. 217—224 Circtdar Motion 234—248 PART II. (HYDROSTATICS). CHAPTEE XVI. 1—15 Fluid Pressure 244—258 CHAPTEE XVn. 16 — 25 Pressure on immersed surfaces .... 259 — 270 CHAPTEE XVIH. 26—37 Specific Gravity 271—285 CHAPTEE XK. 38 — 54 Properties of Gases 286—303 CHAPTEE XX. 55 — 64 Machines depending upon fluid pressure . . . 304—321 Miscellaneous Examples 322 332 Answebs to Examples 333 344 ERRATA. p. 2, Art. 6, /or 'it is measured by' read 'it is represented by.' p. 13, Ex. 1, /or '8 feet per second' read '8 foot-seoB. per second.' p. 16, Examples Y. 4, omit 'and the distance...... seventh second.' ,„,.„„ „ ., , '32-2x3' , '32-2x12' p. 16, Art. 22, smaU prmt, for ^^g^ read .ggg^ . p. 30, Examples VII. 2, for '1^157"' read 'iJlSO.' p. 42, Art. 51, add the words 'The mass of 1000 grammes is called a kilogramme.' p. 69, after 'Bee' add 'Art. 72.' p. 143, Ex. 16, /or 'tan- (^^I:! . Q' r«a^ 'tan- (^^ . Q .' p. 175, Ex. 8, for 'W+w' read 'Tr+2io.' INTRODUCTION. As we shall afterwards have occasion to refer to the projection of a straight line, we shall now define it, and prove the most important proposition relating to the pro- jections of straight lines. I. Definition of the projection of a straight line. N MB Fig. ii. If from the ends of a straight line PQ, perpendiculars PM, QN be let fall upon any straight line AB, the straight line MN thus intercepted is called the projection of PQ upon AB. The following rule with regard to the sign of the pro- jection of a line should be carefully observed, viz. that a projection MN measured from left to right as in fig. (i) is taken to be positive, while a projection measured from right to left, as in fig. (ii), is taken to be negative. This may be also stated thus, the projection of PQ is = — the projec- tion of QP VI INTRODUCTION. II. The projection upon any straight line of one side of a polygon is equal to the algebraic sum of the projections of the other sides. A / Fig. i. Let PQRS be a polygon, draw Pp, Qq, Rr and Ss from its vertices upon any straight line AB. Then we see that ps=pq + qr + rs, or, projection of P)S= proj. PQ +proj. QR + proj. RS. Notice that in fig. (ii), rs, the projection of RS, is nega- tive as stated above. THE ELEMENTS OF APPLIED MATHEMATICS. CHAPTER I. MOTION IN A STRAIGHT LINE. 1. The ideas which form the basis of our subject are those of space and time, which, without attempting to define, we shall regard as primary notions. 2. Another idea which is fundamental is that of matter. Matter may be defined as that which affects our senses so as to produce such ideas as resistance, size, shape, &c. * The first of these, namely resistance, is that which is most closely connected with Mechanics. The quantity of matter contained in any given body is called the mass of the body. The masses of two or more bodies may be compared in various ways such as weighing, or lifting. 3. The Measure of a quantity. In what follows we shall have occasion to use the word measwe. The measure of any quantity such as a length, a time &c., is the number of times it contains that quantity of the same kind which we have agreed to call the unit. Thus the measure of a line 6 inches long is 12 if we agree that half-an-inch shall be the unit of length. If the unit of time is one second then the measure of half-an-hour is 1800, and so on. * ' Whatever can occupy space ' is sometimes given as the definition of matter. Tait, Properties of Matter. J. 1 I THE ELEMENTS OF APPLIED MATHEMATICS. 4 Units of length. There are two units of length in common use, one is the foot, the other, which is used mainly in scientific measure- ments, is the centimetre. A centimetre is the -j-^ part of a metre, a metre being 39"37 inches. It is the length of a certain bar of platinum, kept at Paris, when at the temperature 0° C. 5. Velocity. The velocity of a moving point is its rate of motion, or its rate of change of position. Thus velocity may be of any magnitude and be in any direction. In the present chapter we shall only consider the motion of a point which always moves in the same straight line. When in what follows we speak of the motion of a bodu, the motion considered is one of translation, i.e. one in which the velocity of each particle of the body is the same at the same instant, in other words the body moves without rotation. The velocity of the body is then that of any one of its particles. 6. Uniform velocity. The velocity of a body is said to be uniform when it moves over equal distances in equal times, whatever those times may be. It is measured by the space moved over in the unit of time. Velocity is said to be variable, when distances passed over in equal times are not equal. It is measured at any instant, by the distance which would be passed over by the body in a unit of time if the velocity were to remain the same as at the given instant for a unit of time. The more advanced student will be able to see that this definition of variable velocity may be replaced by the following : if s is the measure of the space described in t seconds after the given instant then, when s and t are very small, the velocity of the body is the limit of the fraction -. 7. Unit of velocity. The unit of velocity is the velocity of a body which moves over a unit of length in a unit of time. A body MOTION IN A STRAIGHT LINE. 3 is said to have a velocity 10 (or generally v), when 10 (or v) units of length are passed over in the unit of time. The units of space and time are generally a foot and a second. These are sometimes called " standard " units. We shall call the velocity of one foot per second one foot-second. A velocity of one centimetre per second is called a kine. 8. Space described in t seconds. The foot and the second being the units of length and time a body which moves uniformly with a velocity v, passes over V feet each second, and therefore vt feet in t seconds. Hence if s is the number of feet described in t seconds s = vt, and v = -. 9. Average velocity. The average velocity of a body whose actual velocity is variable is that velocity which would have carried the body through the distance actually described in the same time at a uniform rate. For instance, suppose a body to pass over 2, 3, 4, 5 feet respectively in 4 consecutive seconds. The entire distance is 14 feet. The average velocity is 3^ feet per second. If the velocity is uniform the average velocity is the same as the actual velocity. Ex. 1. A point moves with h, uniform velocity of 12 feet per second, find the whole space described in 3 minutes. Twelve feet are passed over each second, therefore in three minutes, or 180 seconds, 180 times that distance will be described. The answer is then 12 x 180 feet, or 720 yards. Ex. 2. A point moves for 10 seconds with an average velocity of 5 feet per second ; during the first part of the time the velocity is 3 feet per second, and during the latter part it is 7 feet per second ; what is the length of this part t Let s be the space described in 10 seconds ; thus j!=50 feet. 1—2 4 THE ELEMENTS OF APPLIED MATHEMATICS. Let «i and «2 denote the spaces described in the first and second intervals respectively of t^ and i^ seconds ; then ;r=3, n and r = 7; h hence 3 = 4 + 3x5 = 19, hence the required velocity is 19 feet per second. Ex. 2. A body starting with a velocity of 128 feet per second has an acceleration 32 in the direction opposite to this initial velocity, when will the body be brought to rest 1 After t seconds we have ,. 2) = 128-32*, when V is zero = 128 — 32*, or <=4 seconds. MOTION IN A STRAIGHT LINE. 7 Ex. 3. A train having a uniform acceleration starts from rest and at the end of 3 seconds has a velocity with which it could travel through one mile in the next 5 minutes. Find its acceleration. If a be the acceleration, the velocity at the end of 3 seconds is Za feet per second. With this velocity the train would pass over one mile in 5 minutes, or 1760 x 3 feet in 5 x 60 seconds. _ 1760x3 •■• ^"=T^r60-' hence a = 5'86 foot-second units. EXAMPLES, II. 1. A body starting with a velocity of 90 feet per second after 5 seconds has only a velocity of 50 feet per second; what is its ac- celeration ? 2. A body whose initial velocity is 10 feet per second and final velocity 50 feet per second, has an acceleration of 10 ft. -sec. units ; in what time will it gain this final velocity ? 3. A point starting from rest, after 6 seconds has a velocity of 18 feet per second ; find its acceleration. 4. A point whose initial velocity is 20 feet per second, has an acceleration of 32 ft.-sec. units ; find its velocity after 5 seconds. 5. The initial velocity of a body is 11 feet per second, and the body's velocity is increased each second by a velocity of 7 feet per second ; find when it will be moving at the rate of 60 miles an hour. 6. A body has a velocity of 9 feet per second at the beginning, and of 10 feet per second at the end of a given second ; what will be its velocity after 5 more seconds ] 14. Representation by an area of the space de- scribed. On a given straight line OA measure off a line OP containing t units of length, and on a perpen- dicular line OB measure off a line OQ, containing v units of length. Complete the rectangle OPRQ. The area of this rectangle is OP x OQ rio. i. or vt, hence since vt=s, the area of the rectangle measures the space described by a bodjr moving for t seconds with a uniform velocity v. THE ELEMENTS OF APPLIED MATHEMATICS. 15. Lines representing the velocity of a moving point. • 7 /^ e y^ 5 // y y^ y \^ a O A B C D E K Fig. 2. Let the line OK be of length t and be divided into equal portions OA, AB, BG, CD &c. Draw the vertical line -4 a to represent the velocity of the point at the end of the first portion of time. Join Ool and produce it, drawing verticals through 5, (7, ... to meet it at/3, y Then we easily see that B^^IAa, Gy = 3Aa, &c. Thus the lines Aa, B^, Gj ... represent the velocities of the point at the ends of the first, second, third . . . portions of the time, KL representing its final velocity or at. 16. To prove that s = ^at". In the figure of last Article draw horizontal and vertical lines through a, yS, y ... forming two sets of rectangles, one within and the other without the triangle OKL. Consider the rectangles on the base BG. The smaller one measures the space described by a point in a time re- presented by BG, if moving . with constant velocity B/3, Art. 14 ; or a smaller space than that actually described by the point since its velocity during the time BG is greater than B^. Again the larger rectangle measures the space which would be described if the point moved with constant' ve- MOTION IN A STEAIGHT LINE. 9 locity C7, during the time BG, or a greater space than the actual one since the point's velocity is less than O7 during the time BC. Hence the true space described by the point in the time BG is between these values. During each of the times represented by OA, AB, BG ... a similar result is true. We have therefore that the measure of the sum of the inner rectangles is< space described, outer > space described. Now if we divide OK into a very great number of equal parts, the sum of each set of rectangles will approach the area of the triangle OKL very closely, as we see from the figure on the right. In fact the difference of each sum from the area of OKL is equal to ^KL X OA which becomes indefinitely small as we increase the number of divisions. The measure of the space actually described is therefore the area of the triangle OKL = \0K xKL = ^txat = ^at^ Thus s = ha't^- If the body has initially a velocity u, draw OM perpendicular to OK and of length u, complete the rectangle OMNK. In this case the velocity of the body at each instant is greater than its former value by u, an.d the space described will be measured by the sum of the areas OLK and OMNK, that is s = \af + ut. Fig. 3. 17. Alternative proof for the formula s = ut + Jat^ To find the space passed over in the time i by a body moving with an acceleration a, and possessing an initial velocity u, we may also adopt the following method. 10 THE ELEMENTS OF APPLIED MATHEMATICS. Divide the given time t into n equal intervals, each of length - ; the velocities at the beginnings of the first, second, . . . mth of these intervals are u, u+a-, u+2a- , ... , u + (n—l)a- ; since a velocity a - is added during each interval. Art. 12. Now if the body moved uniformly during each interval with the velocity which it actually has at the beginning of that interval, each space so described would be less than the space actually described in that interval. And the sum of the spaces described in this supposed manner is t ( t\t u - + lu + a- n \ n -+ M+2a- - + ... n \ nj n + (u + (n-l)a~] - Art, 8, = n- + ^(l + 2 + 3+...+¥::a) smce 1 + 2+3+ ... +n.-l=— ^^ — - n n' = ut + ],at^(l-~) . 2 \ n/ Again, if the body moved uniformly throughout each interval with the velocity it has at the end of that interval, each space so described would be greater than the space actually described in that interval. The sum of the spaces described in this manner is (u + a-) - +(u + 2a-] -+ ... + (u + na-]- \ nJ n \ nJ n \ nj n = nu - + -J (1 + 2 + ... + n) MOTION IN A STRAIGHT LINE. 11 ^ aP n(n + l) n^ 2 = ut+^ae(l + -) . The space actually described lies therefore between ut + ^at^ (l - -A and ut + ^at^ (l + -) • By making the intervals small enough, and therefore n large enough, we can make - as small as we please, thus causing these two expressions to continually approach each other in value. The actual space s described is therefore ut+^at^ (iv), as this is the limiting value to which each expression approaches as n increases. 18. If the direction of the acceleration is contrary to that of the initial velocity u, we get in the same way, s = ut — ^at\ N.B. If the body started from rest or with initial ve- locity zero, putting u equal to zero in the formula just proved we obtain s = iat' (v). We could of course have got this value for s by going through the proof of the last Article omitting u throughout. Ex. 1. A body has an initial velocity of 100 feet per second, and moves with an acceleration of 10 ft. sec. units ; what is the distance passed over in 5 seconds ? If g be the required distance s=100x5+ixlOx52 =625 feet. Ex. 2. The initial velocity of a body is 40, and it moves with an acceleration of —2; find when it will be 400 feet from the starting- point. We have iOO=40t-tK Solving this quadratic equation we obtain 20 as the required number of seconds. 12 THE ELEMENTS OF APPLIED MATHEMATICS. Ex. 3. A body starts from rest and moves with a uniform accelera- tion of 18 ft. -sec. units. Find the time required by it to traverse the first, second, and third foot respectively. For the first foot we have l = 9t% .•. t=l of a sec. The time required to traverse the second foot is that taken to traverse 2 feet — (the time taken to traverse 1 foot), hence it is f-i = ^(V2-l)secs. Similarly we find as the time for the third foot, V3' 1 s/2^sl^Z^^^,_ EXAMPLES. III. 1. With an acceleration of 32 ft.-sec. units, how far will a particle move in 10 seconds starting from rest, and what will be its velocity at the end of that time ? 2. A particle moves with uniformly increasing velocity. Show that the whole space is proportional to the square of the whole time. 3. A body moves over 3 ft., 5 ft., 7 ft., 9 ft. respectively, in 4 con- secutive seconds ; find its average velocity. 4. A body has an initial velocity of 20 feet per second, and a positive acceleration of 32 ft.-sec. units; how long will it take to pass over 1800 feet ? 5. Starting with a velocity of 200 centimetres per second, a body has an acceleration of -2 centimetre-second units. Find when its velocity is zero and how far it has gone in the time. 6. In what time will a body acquire a velocity of 60 miles an hour, if it starts with a velocity of 28 feet per second, and move with the foot- sec, unit of acceleration ? 19. To show that v'' = u^ -f- 2as. We have seen that v = u + at, and that s = ut + ^at\ From the first equation v'^ = u^ + iuat + aH', or v^ = w'+2a (ut + ^ at% MOTION IN A STRAIGHT LINE. 13 hence from the second v^ = u' + 2as (vi). Ex. 1. A body has an initial velocity of 6 feet per second and moves over a distance of 4 feet with an acceleration of 8 feet per second ; what is the final velocity ? Here v^=e^+2x8xi, = 36+64=100; .■. ■»= 10 feet per second. Ex. 2. If the velocity of a body increase from 12 feet to 13 feet per second while it moves over a distance of 5 feet, what is the accelera- tion? Applying the formula, 132= 122+ 10a; X32— 122 we have that a = — — — =2-5. Ex. 3. The speed of a railway train increases uniformly for the first three minutes after starting, and during this time it travels one mile. What speed in miles per hour has it now gained, and what space did it describe in the first two minutes ? Since the train passes over one mile in three minutes we have 3 X 1760=1 X (3x60)2; 3x1760x2 44 „^ •••"= (3x60)2 =Y35ft.-sec. units. The velocity gained in passing over a mile is by Art. 19 V2 X ^ X 3 X 1760, =ija feet per second, or 40 miles per hour. The space described in the first two minutes =ixA^x (2x60)2 =is^ feet =f of a mile. EXAMPLES. IV. Ex. 1. A railway train whose mass is 100 tons moving at the rate of a mile a minute, is brought to rest in 10 seconds by the action of a uniform force. Find how far the train runs in the time during which the force is applied. 14 THE ELEMENTS OF APPLIED MATHEMATICS. Ex. 2. A body whose initial velocity is 30 feet per second, passes over a distance of 50 feet and has then a velocity of 60 feet per second, what is the acceleration ? Ex. 3. A body which has an acceleration of — 32 ft.-sec. units has an initial velocity of 32 feet per second ; find how far it will go before coming to rest, and in what time it will do so. Ex. 4. A body moves for 6 seconds with a constant acceleration during which time it describes 81 feet, the acceleration then ceases, and during the next 6 seconds it describes 72 feet ; find its initial velocity and its acceleration. 20. Recapitulation. Collecting the results of the last three Articles we see that if the body starts from rest v = at (1), s=|< (2), s^iaP (3), v''=2as (4). or (1) the velocity gained in t seconds is t times the ve- locity gained in one second, (2) the distance passed over in t seconds is t times the average distance passed over in one second, (3) the space described varies as the square of the time, (4) the space described varies as the square of the velocity. If the body has an initial velocity u v = u + at (5), u+v « = -2~« (6)> s = ut + ^at'' (7), v"- = u^ + 2as (8). MOTION IN A STRAIGHT LINE. 15 21. Space described in a given second. To find the space described in a given second by a body moving with constant acceleration a, when the body starts from rest, we proceed as follows : Space described during the *th second = space described in t seconds — (space descr. in f —1 seconds) ^ a(2t-l) 2 Thus the spaces described in the first, second, third, . . . wth seconds are a 3a ba {2n —\)a 2' Y' Y' ■■■ 2 ■ If the body starts with a velocity u we find the space described in the tth second in the same way, it is equal to (ut + ^at') -{u(t-l) + ^a{t- ly} _ a(2«-l) -u+ 2 . Thus the spaces described in the first, second, third, . . . nth seconds are a 3(1 oa , (2*1— 1) a w+2' ""^Y' ^+"2" '•■•'"■'" 2 ■ Ex. 1. A body starting from rest has an acceleration of 32 ft. -sec. units ; find the space described in the 8th second. = 240 feet. Ex. 2. The initial velocity of a body is 21 feet per second, and it passes over 54' feet in the 6th second ; find the acceleration. .. „, 12-1 54=21 + — g— a; .■. a=6 foot-seconds per second. Ex. 3. A body moving with uniform acceleration passes over 126 feet and 246 feet in the fourth and eighth seconds respectively ; find its initial velocity and acceleration. 126=M-f|a, 246=M+Jg^a, which give a =30 and ie=21. 16 THE ELEMENTS OP APPLIED .MATHEMATICS. EXAMPLES. V. 1. A body moving with uniform acceleration describes 520 feet in the 7th second from rest ; find the acceleration. 2. Starting from rest a body describes 330 feet in the 6th second of its motion ; find the space described in the ninth second and in nine seconds. 3. A body whose initial velocity is zero has an acceleration of 32 ft.-sec. units ; compare the distances passed over in the 6th and 12th seconds. 4. A body moving with uniform acceleration describes in the seventh and twelfth seconds after starting 23 and 33 feet respectively ; find its initial velocity, its accleration, and the distance described from rest before the beginning of the seventh second. 5. A body moving with uniform acceleration describes in the last second of its motion J of the whole distance. It started from rest and described 6 inches in the first second ; find how long it was in motion and the distance it described. 22. Falling bodies. Experiments show that if a body fall it will move with an acceleration which is always the same in the same lati- tude and which is due to the attraction of the Earth. It is called the " acceleration of gravity." Its measure is denoted by the symbol "g!' When a foot and a second are the units the value of ^r is 32'2. 32'2 X 3 32-2 feet are equal to centimetres or 981 centimetres, thus when a centimetre is the unit of length g is 981. Thus from what has gone before we see that, taking g as 32, velocity gained hy a falling body in t seconds = 32t ft. sees., distance fallen through = 16i^feet, velocity gained in falling through a height h = H\Jh ft. sees., time occupied in falling through a height h = ^>^h sees. 23. Motion under gravity. The acceleration, due to gravity, of a body projected verti- cally upwards is opposite in direction to the velocity of projection, and is therefore denoted by — ^. MOTION IN A STRAIGHT LINE. 17 In the formulae already proved put a equal to — .^, we thus get s = ut- \gt\ v = u -gt, ■y" = M^ - 2gs. When the body is projected dotunwards we have the equations s = ut + ^gt% v = u + gt, «" = «" + 2gs. If the body is let fall we have s = ^gt^ u = gt, v' = 2gs. 24. Time to reach a given height. If the given height be h, then writing h for s in the first equation of last Article, we have h = ut — ^gt'. This is a quadratic equation to find t, whose roots if real are both positive. If the roots are imaginary u is not great enough for the body to reach a height h. The smaller root is the time at which the body is at the given height when ascending, the larger root gives the time at which it is at the same height when descending. Ex. 1. A body is projected vertically upwards with a velocity of 80 ft. per sec. ; when will it be at a height of 64 feet ? from which, t=l, or 4. It is therefore at the given height one second or four seconds after the beginning of its motion. Ex. 2. A body is projected vertically upwards with a velocity of 64 feet per second, how high will it rise in 2 seconds ? If X be the required height, ar=64x2-16x4 =64 feet, J. • 2 18 THE ELEMENTS OF APPLIED MATHEMATICS. 25. Velocity at a given height. If A is a given height and m the velocity of projection, we have v'^'^u'- 2gh, or t) = + Ju' - 2gh. Now u and h are the same whether the body is ascending or descending, hence the velocity is the same in magnitude but opposite in direction in these two cases. Ex. 1. A body is projected with a velocity of 120 feet per second; what will be its velocity at the height of 200 feet ? If D be the required velocity «-'=(120)2- 64x200, or D=40 feet per second; Ex, 2. If the upward velocity of projection be 60 feet per second, find when the velocity will be 20 feet per second. If h be the height at which the velocity is 20, (20)2=(60)2-64xA; .-. A=^^5)!__W=50feet. b4 The times at which the body is at this hefght are found from the equation 50=60* -ie«2, from which ?= 1 J or 2J seconds. 26. Greatest height of a projected body. At the highest point the velocity is zero; therefore if A is the height of the highest point the body reaches, = u^-2gh, or h = ~. Hence the velocity with which a body must be projected so as to just reach a height h is ^Igh. Ex. 1. A body is projected vertically upwards with a velocity of 64 feet per second, how high will it rise before beginning to descend ? If A is the required height, A=^'=64 feet. b4 MOTION IN A STRAIGHT LINE. 19 Ex. 2. A ball is allowed to fall to the ground from a certain height, and at the same instant another ball is thrown upwards with just sufficient velocity to carry it to the height from which the other falls. Show where and when the two balls will pass each other. Let the given height be h. The ball thrown upwards with a velocity just sufficient to carry it this height must be thrown with velocity fjzgh. Let the time up to the instant of meeting be t seconds. Let «i be the distance described by the falling ball and i^ *^^* described by the one thrown upwards during the time t, then «, = 16«2 S2=i\ligh.t-\et^; hence adding Si + Si='\/2gh.t. But A = «i + «2) - -^l- and h 27. Time to reach the greatest height. At the highest point the velocity is zero, u :. = u — gt, or t=-. 28. Time of flight. When the body has returned to the starting point the total space described is zero, therefore if t be the whole time, = ut- \gt\ The roots of this quadratic equation are and — . The root corresponds to the time of starting, the root — is = time of ascent + time of descent = whole time of flight. ^ u. But - was found to be the time of ascent, hence - is also 9 ^ 3 the time of descent. 2—2 20 THE ELEMENTS OF APPLIED MATHEMATICS. Ex. 1. A body projected upward returns to the point of projection at the end of six seconds, find the height reached and the velocity of projection. Since the time of flight is 6 seconds, 2u = — , 9 hence taking g as 32, the velocity of projection is 96 feet per second. The greatest height = — = ^-~ = 1 44 feet. Ex. 2. A ball thrown vertically upwards rises 200 feet in 4 seconds, in what tinae will it return to the point of projection ? By Art. 24 we find the initial velocity to be 114 feet per second. Therefore the time of flight is Jj^ or 7J seconds. 29. The velocity gained by a fall from rest through a height h is 8VA. The velocity with which a body must be projected to just reach a height h is by Art. 26 also 8^/h. Hence the velocity gained by a fall through any height from rest is = velocity with which a body must be projected to just reach that height. Ex. 1. A stone is thrown vertically upwards with such a velocity as will just take it to the level of the top of a tower 100 feet high. Two seconds afterwards another stone is thrown up from the same place with the same velocity. Determine when and where the stones will meet. The velocity of projection of the first stone is 80 feet per sec. The- time of reaching the top is ff, or 2^ seconds. It therefore reaches the top half a second after the projection of the second stone. This latter will in half a second reach the height 80 X J- 16 (if feet, or 36 feet. Its velocity is now 80 - 32 x J, or 64 feet per second. . It is therefore moving with this initial velocity at the instant when the first stone is beginning to fall. Thus the spaces described by the two stones together with the space already described by the second stone, or 36 feet, is the height of the tower. .-. 36 + 64t-iefi+l6t^ = 100, from which t is found to be 1 second. The time from when the first stone was thrown up is 3^ seconds. MOTION IN A STRAIGHT LINE. 21 Ex. 2. A stone is thrown upwards with a velocity due to a fall through a height ^ , from a point at a height h above the ground. Find when it will strike the ground. The velocity due to a fall through a height 5 is Jgh. Art. 23. The time of flight is therefore 2 /-. Art. 28. Time of falling through the height h with the initial velocity ^gh is given by h = \fght+\gt\ Art. 23. The whole time before it strikes the ground is therefore EXAMPLES. VI. [g is taken to be 32 unless it is otherwise stated.] 1. How far will a body fall from rest in four seconds? Find with what velocity a body must be thrown upwards to return to the hand in four seconds. 2. A stone after falling for one second strikes a pane of glass, in breaking through which it loses half its velocity. How far will it fall in the next second ? 3. A body moving from rest with a uniform acceleration passes over 10 feet in the first two seconds after starting, how far will it be from the starting-point at the end of the third second ? 4. Through what vertical distance must a heavy body fall from rest to acquire a velocity of 161 feet per second ? If it continue falling for another second after having acquired the above velocity, through what distance will it fall in that second 1 5. A stone projected vertically upwards has risen 120 feet in one second. What was its initial velocity of projection, and how far will it rise during the next second ? 6. If a stone reaches the ground again 6 seconds after it has been projected upwards, what was its height above the ground at the end of the first second ? 7. A heavy particle is dropped from a given point, and after it has fallen for one second another particle is dropped from the same point. What is the distance between the two particles when the first has been moving for 5 seconds ? 22 THE ELEMENTS OF APPLIED MATHEMATICS. 8. A balloon has been ascending vertically at a uniform rate for 4-5 seconds and a stone let fall from it reaches the ground in 7 seconds. Find the velocity of the balloon, and its height when the stone is let faU. [The stone when let go has the velocity of the balloon (upward). If this be u the distance through which the stone falls is -Mx7 + 16x72. But this is the height to which the balloon rose in 4-5 seconds, or « X 4-5. Hence ■mx4-5 = 16x49-7m; .•., M = 68'2 feet per second. And height of balloon =i 68-2 x 4-5 feet = 306-9 feet.] 9. Frdm a balloon which is ascending with a velocity of 32 feet per second a stone" is let drop which reaches the ground in 17 seconds. How high was the balloon when the stone was dropped ? 10. A rifle-bullet is shot vertically downwards from a balloon at the rate of 400 feet per second. How many feet will it pass through in two seconds, and what will be its velocity at the end of that time ? 11. A stone dropped into a well reaches the water with a velocity of 80 feet per second, and the sound of its striking the water is heard 2^^^ seconds after it is let fall. Find from these data the velocity of sound in air. 12. A ball is let fall from a height of 256 feet, and at the same time a ball is thrown upwards to meet it : find this velocity of projection if the balls meet at a height of 112 feet. 13. The space passed over in any given time may be represented by an area. Explain clearly the meaning of this statement and under what conditions it is true. Show how to employ it to determine the space passed over by a body in 10 seconds after it starts from rest, and has its velocity increased by one foot per second at the beginning of each second. 14. Explain by reference to a diagram why a stone only falls 16 feet during the first second while yet the force of gravity generates in that time a velocity of 32 feet per second. 15. Find the distance traversed in 10 minutes by a train which has at first a velocity of 20 miles per hour, and which has its speed in that time diminished at a uniform rate to 5 miles per hour. 16. A train running 60 miles an hour is pulled up by its breaks in 900 yards. At what rate would it be running if it could be pulled up in 100 yards ? MOTION IN A STRAIGHT LINE, 23 17. Prove that if a body is projected vertically upwards with the velocity of 64 feet per second and 3 seconds afterwards a second body is let fall from the point of projection, the first body will overtake the second body 1^ seconds later, 36 feet below the point of projection. 18. Prove that space described from rest in the n?+n +lth second is =the sum of the spaces described in the first n seconds and in the first n+1 seconds. 19. A body moving with uniform acceleration /passes over a space « in a time T, show that its velocity at the beginning of the time was 20. The space described by a body in the 5th second of its fall from rest was to the space described in the last second but three as 9 : 11. For how many seconds did the body fall ? 21. Supposing the moon to be a sphere of 2000 miles in diameter which rotates on its axis in 27 days, and that the circumference of a circle is to its diameter as 22 : 7, compare the velocity of a particle in the moon's equator with that of a railway train which travels 57 miles in an hour and a half. CHAPTER II. COMPOSITION AND RESOLUTION OF VELOCITIES. 30. Composition of Velocities. A body may have at the same time, as shown in the next Article, two or more velocities. It will be there shown that they may be replaced by a single velocity which produces their joint effect. This single velocity is called the resultant velocity, and of this velocity the original velocities are said to be com- ponents. 31. Parallelogram of velocities. If two simultaneous velocities of a body be represented in magnitude and direction by the two sides of a parallelogram drawn from a point, the resultant velocity is represented by the diagonal of the parallelogram drawn from the point. Let OA and OB represent the q — j< two velocities whose magnitudes are Fio. 4. u and v; complete the parallelo- gram, we shall show that OC represents the resultant velocity. Notice that if the body moved for one second with the velocity ti only, it would describe OA ; if it moved for one second with the velocity v only, it would describe OB. COMPOSITION AND RESOLUTION OF VELOCITIES. 25 We can see as follows how it is possible for the body to have the velocities u and v at the same time : let a line O'A', which starts from the position OA , move parallel to itself, while 0' moves along OB with velocity v ; let a point P, starting from 0', move along O'A' with velocity u. The motion of P is thus due to the two simultaneous velocities u and v. We shall now show that P always lies upon 00. For let the figure represent the position of P at the time T, where T is any fraction of a second. Draw PK parallel to OB. ™ PK=00' = vT,\ ,,_ , PK vT V AG whence ?Ti^ = -7jt= - = 7T7 5 OK uT u OA .". OP produced passes through Oj Euc. VL 26, or P always lies upon 00. Again, at the end of a second O'A' will coincide with BG, and P with A', that is with G. Thus since in one second P has moved along OC, it has a velocity represented by OC. That is a velocity represented by OG is the effect of the velocities represented by OA and OB. Observe that P moves uniformly along OC since 0P= T. OG. 32. In the same way if a particle possesses three simul- taneous velocities we can find the resultant of two of them and then the resultant of this and the third velocity ; that is a single velocity equivalent to the three simultaneous velocities. 33. The triangle of velocities. If two simultaneous velocities possessed by a body are represented in magnitude and direction by two sides of a triangle taken in order, the third side of the triangle taken in the opposite direction will represent their resultant. 26 THK ELEMENTS OF APPLIED MATHEMATICS. For in the figure of Article 31 0(7 represents the resultant of velocities represented by OA and OB, i.e. by OA and ALi. 34. The polygon of velocities. Just as a body may possess two simultaneous velocities, it may also possess three or any number of such velocities. The simultaneous velocities of a body being represented in magnitude and direction by the sides of a polygon except one, taken in order, then will the line which closes the polygon, taken in the reverse direction represent the resultant velocity. Take four velocities repre- sented by AB, BC, CD, DE : the resultant velocity will be represented by AE. For by Art. 31 the resultant of AB and BG is AG, AG audi GD is AD, AD&nADEisAE. Notice that if A and E coincide the resultant velocity will be zero. Hence when a body has velocities represented by the sides of a closed polygon taken in order it has no resultant velocity or is at rest. 35. Special cases of the parallelogram law. (i) When the velocities are at right angles to each other. Let OA and OB represent the velocities u and v. Then since OG^ = OA^ + AG\ (resultant)^ = w^ + v^, Fi(x. 6. or if w is the magnitude of the resultant, w = Ju'' + v\ COMPOSITION AND RESOLUTION OF VELOCITIES. 27 (ii) When the contained angle is 30°. Fia. 7. «/.?f^^ ^^ perpendicular to OA. Then since GAD (or BOA), IS 30°, ACD is 60°, and the triangle GAD is half an equilateral triangle; .-. Oi) = ^ = ^ 2 2 ' V 4 2 2' But OG'==OI> + GD% or OC^ = (0^ + ADy + GD\ hence «,= =(„ + ^^^J + ^J. . .•, lU' = u'' + V^ + UVi\/S. (iii) When ^05 is 45°. B Since AOB is 45°, AGD is 45°, and AD = CZ). Hence AG'' = 2 AD', or AD = ~ . 28 THE ELEMENTS OF APPLIED MATHEMATICS. I<2 Or, w = u^ + if + uv »J2. (iv) When ^05 is 60°. A D Fig. 9. Here ACD is 30°; and ACD is half an equilateral tri- angle, hence as above. 00^ = {0A+ADy + 01> w = f M + s 1 + -r- = u" + v^ + uv. (v) When AOB is 120° D A Fig. 10. DAC is 60° and 00^ = 0I> + OD^ = (OA - ADf + GIP ; ■uv. COMPOSITION AND RESOLUTION OF VELOCITIES. 29 (vi) When ^05 is 135°. B C D A Fio. 11. DAG is 45°, and AD = DG = AG V'2' w { vy v'^ (vii) When J. 05 is 150°. ^ C HereZ).4a=.30°, and VS CD = ^AG, AD=^AG. Hence w^={u—^ v\ +-j — u^ + v* — \/3uv. All the preceding cases can be got from the relation 0C^= 0A^+0B^+20A . OB cos a, where a is the angle between OA and OB. This gives vj^=u^+i^+2uv coa a. 30 THE ELEMENTS OF APPLIED MATHEMATICS. EXAMPLES. 1. Find the resultant velocity in the following cases : (i) 11=1, v= 2, included angle 30°. (ii) M=3, v= 3, 45°. (iii) it=5, «=11, 60°. (iv) M=l, t) = 20, 150°. 2. From the top of the interior of a railway carriage a stone is let fall. If the train is moving at the rate of 20 miles an hour show that the velocity of the stone is f Jlbl feet per second when it has fallen one foot. 3. A balloon rising vertically with a velocity of 30 feet per second is also carried by the wind over a horizontal distance of 40 feet in a second. Find its total velocity. 4. Find the resultant of two velocities, of 10 feet and 20 feet per second respectively, incUned at an angle of 120°. 5. A ship sailing westwards with a velocity of 16 knots receives an additional velocity of 16 knots from a current so that its velocity is still 16 knots. What is the direction of the additional velocity? 6. A man walks in 12 seconds across the deck of a ship which is sailing due north at the rate of 4 miles an hour, and finds that he has moved in a direction 30° east of north. How wide is the deck and what is his actual velocity ? 7. If two velocities of 9 feet and 7 feet per second respectively possessed by a body, include an angle whose cosine is J, show that the resultant velocity is 12 feet per second. 36. On the choice of components. A given line may be the diagonal of an infinite number COMPOSITION AND RESOLUTION OF VELOCITIES. 31 of parallelograms. Hence a velocity may be resolved into two velocities in an infinite number of ways. If the directions of the components be given their magni- tudes are found at once. Let OX, OF be the given directions and OG the given velocity. Through G draw GA and GB parallel to OF and OX, then we have seen that OA and OB are the components of OG. An important case is when OX and OY are at right angles ; in that case OA and OB are called the rectangular components of OG. 37. The resolved part of a velocity. c O B X Fig. 14. Let OG be a given velocity and OX a given direction, from G draw GB perpendicular to OX. Then OB and BG are rectangular components of OG. OB is called the resolved part of OG along OX ; observe that it is the projection of 00 on OX. 38. Resultant of any number of velocities. When a body has several simultaneous velocities in different directions its resultant velocity may be found : — (i) By repeatedly using the parallelogram of velocities, viz., by finding the resultant of two velocities and then the resviltant of this and a third velocity, and so on. (ii) By the Polygon of Velocities. (iii) A third method given in Art. 41. It is sometimes useful to remember that since the diagonals of a parallelogram bisect each other, the resultant of two velocities OA and OB is 2 . OD, where D is the middle point oiAB. 32 THE ELEMENTS OF APPLIED MATHEMATICS. 39. Resolution of velocities. Any velocity may be split up into two velocities, i.e. since velocities represented by OA and OB have the same effect as a B LC velocity represented by 00, a velocity 00 may be replaced by the two velocities OA and OB. This is called the resolution of a velocity, and the velocities Fig. 15. OA and OB are called the com- ponents of 00. 40. The resolved resultant equals the sum of the resolved components. By Art. 34, if lines be drawn in order to represent the velocities of a body, the line which closes the polygon will represent their resultant. Also the projection of this last line on any direction is equal to the sum of the projections of all the other lines; (see Introduction). Hence the resolved part of the resultant velocity equals the sum of the resolved parts of the component velocities, in any direction. The sum here meant is of course the algebraic sum. 41. Third method of finding the resultant velocity. M X COMPOSITION AND RESOLUTION OF VELOCITIES. 33 Let a particle have velocities represented by Oa, Oh,.... Through draw two lines OX, F at right angles to each other. Through a, b,... draw a^l, bB,... perpendicular to OX; and aA', bB',... perpendicular to OY. Then OA, OB... are the resolved parts of the velocities along OX, OA', OB"... are the resolved parts of the velocities along Let OR be the resultant velocity, and OM, ON its resolved parts. By the last Article, OM=OA +0B +... ON=OA' + OB' + .... Hence OE = J(OA + OB + ...y + {OA'+OB' + ...y. The resultant is therefore found by resolving the veloci- ties along two lines at right angles and compounding the resolved parts. If a is the inclination of a velocity V to OX, then we have by the preceding ; Fio. 17. resolved part of Falong OX=projection of Fon 0X= Fcosa, Or= Or= Vsina, J. 3 34 THE ELEMENTS OF APPLIED MATHEMATICS. If there are several velocities V, V, ... whose inclinations to OX are a, a', ...; the components of the resultant R are Fcosa+ F'cosa' + =X, Fsin a+ V' sin a'+ = Y. Also R=s/X^+rK If 6 be the angle which R makes with OX, ^ T Fsin g+F' sin a' + . tan e--p - p^^^g ^_^y, ^^g ^,_|_~ Ex. 1. Find the horizontal And vertical components of a velocity of F feet per second when inclined at an angle (i) of 30°, (ii) of 45°, (iii) of 60° to the horizon. Let AC represent the given velocity in the three cases ; draw CD perpen- dicular to AB, AD and BC represent the required components. (i) "When 54 C is 30°. The angle 4 Ci) is 60°, and ABC is half an equilateral triangle, hence AD='^AC, DC=\AC, see (ii), p. 27; and the components are ^ 1^> i ^• (ii) When BAC is 45°. The angle ACD is also 45°, hence the components are AD=^AC, DC=-j^AC, see (iii), p. 27; -If -If (iii) When BAC is 60°. The angle ACB is 30°, and AD=iAC, 1)0='^-^ AC; the components are iV, ^F. COMPOSITION AND RESOLUTION OF VELOCITIES. 35 Ex. 2. Two velocities v^ and v^ in the directions OA and OB include an angle of 30°, find the resolved part of their resultant along a line which makes an angle of 60° with the direction of OA and 30° with the direction of OB. Drawing perpendiculars from A and B on the given line OD we see as before, since the angle between OA and OB is 60°, the resolved part of v^ along OB is §»j ; and since the angle between OB and OB is 30° , the resolved part of »2 along OB is ^ ijj. Hence resolved part of resultant = sum of resolved parts of OA and OB Ex. 3. Velocities of 6, 7, and 8 feet per second are possessed by a body in directions making angles of 120° with each other ; find their resultant. Fig. 19. Fig. 20. Take for the line OX of Art. 41, the direction of the velocity of 6 feet per second. Let OQ and OR represent the other two velocities. Draw QM and RW perpendicular to XO produced. Then since §OX=120°, QOM=eO% ROX =120°, R0N=80\ the components along OX are OP, -^OQ, -iOB. The components perpendicular to OX are ^OQ, -^OR. 3—2 36 THE ELEMENTS OF APPLIED MATHEMATICS. The negative sign denotes as usual that a line is measured either to the left or down/wards. The component of the resultant along OX is therefore OP-\OQ-\OR=e,-i-%=-^. The component of the resultant perpendicular to OX is -f(7-8)=-f. If OS be their resultant OS^=l+l=Z or 0=-18. Ex. 2. Find the change of velocity, when the initial velocity is 5 feet per second east, and the final velocity 5 J2 feet per second north- east. The final velocity is evidently the hypotenuse of an isosceles right- angled triangle. Hence the change of velocity is 5 feet per sec. north. 38 THE ELEMENTS OF APPLIED MATHEMATICS. 43. The Parallelogram of Accelerations. The change of velocity oi a body arising in one second if uniform is called the acceleration. (See Art. 11.) Observe, that as in the last Article, the direction as well as the magnitude of the change has to be taken into account. Two such changes of velocity are of course compounded by the parallelogram of velocities, i.e. tivo accelerations are compounded by the parallelogram lavr. This pro- position is called the Parallelogram of Accelerations. 44. Relative velocity. Let two bodies A and B be moving in different directions, and with different velocities u and v. The velocity of either viewed from the other will appear to be different both in magnitude and direction from either u or V. To determine the motion of B as seen from A. Apply to each body a velocity equal and opposite to that oiA. The relative motion will be un- C . changed, but A will be brought to / \ rest, and B will have a resultant velo- city along BG, i.e. B as seen from ^ A will appear to move with a velocity "* a ~ *" represented in magnitude and direc- Fja. 22. tion by BG. Thus in all cases the relative velocity of a point B, with reference to a point A, is found by combining with the velocity of jB a velocity equal and opposite to that of A. Ex. 1. A ship is steaming due south with a velocity of 10 knots, while another is steaming north-east at the rate of 15 knots. Find the velocity of the second ship.with reference to the first. Let a velocity equal and opposite to that of the first be given to each vessel ; the first is brought to rest and the second has two velocities; viz. a velocity of 10 knots northwards, 15 knots north-eastwards. The included angle is 45°, hence the resultant is \/l00+225 + 150V2 knots, or 23 knots nearly. COMPOSITION AND RESOLUTION OF VELOCITIES. 89 Ex. 2. A ship is sailing due north at the rate of 7 miles an hour; in what apparent direction, as seen from the ship, and with what vdocity must a man run on its deck that his actual direction may be due west and his actual velocity 7 miles an hour ? If the ship be brought to rest, the man will be moving due west with a velocity of 7 miles an hour, he has also a velocity of 7 miles an hour south. His apparent direction is therefore south-west and his relative velocity 7 ij% miles an hour. EXAMPLES. IX. 1. A railway train moving at the rate of 30 miles an hour passes another moving at the rate of 5 miles an hour in the same direction. Find the apparent velocity of the first train from the second train. 2. If the trains in the last question are going in opposite directions find the apparent velocity of either viewed from the other. 3. A train moving at the rate of 60 miles an hour is struck by a stone moving at right angles to the train with a velocity of 33 feet per second. Find the magnitude of the velocity with which the stone appears to strike the train. 4. Two straight lines of railway contain an angle of 60°; two engines run, one on each line, each from the point of intersection of the lines at the rate of 30 miles an hour. Find the magnitude of their relative velocities. 5. A ship is sailing north-east with a velocity of 10 miles an hour, and to a passenger on board the wind appears to blow from the north with a velocity of 10 ^2 miles an hour. Find the true velocity of the wind. 6. The velocity of a ship in a straight course is Sj^j miles per hour, a ball is rolled across the deck perpendicular to the ship's length with a velocity of 3 yards in a second, show that it will pass over 5 yards in one second nearly. 7. A steamer is going due north with a velocity v, the smoke from its funnel points ff" south of east. If the wind is due west find its velocity. 8. A cricket-ball is moving in the line of wickets with a velocity of 30 feet per second and is struck by a blow which had the ball been at rest would have sent it with a velocity of 40 miles an hour at right angles to the line of wickets. In what direction will it go 1 9. A company of soldiers is marching along a road at the rate of 3 miles an hour, the column is 3 yards wide and there is just room for one man between two consecutive ranks. A man crosses over from one side of the column to the other walking at the rate of 5 miles an hour. In what direction does he walk and how long does he take to cross over? CHAPTER III. THE LAWS OF MOTION. 45. Hitherto we have been dealing with the motion of a body without considering what produced the motion. When we begin to think of the causes of the motion of bodies we encounter the ideas oi force and mass. In the science of Dynamics we treat of the forces which act on a body ; it is usually regarded as consisting of two parts (i) Statics, when the forces are themselves in equilibrium ; (ii) Kinetics, when the forces are not in equilibrium. Statics is thus a particular case of Kinetics. 46. Force. Force is that which changes, or tends to change, a body's state of rest or of uniform motion in a straight line. To completely determine a force it is necessary to know, (i) its point of application, (ii) its direction, (iii) its magnitude. A simple instance of a force acting on a body occurs when a body on the ground is pulled by a string. The point of application of the force is the point of attachment of the string to the body, its direction is that of the string, its magnitude is the amount of the pull exerted by the string. 47. Representation of forces by lines. A force may be represented by a straight line, for a line maybe drawn; (i) from any point, and hence from the point of ap- plication of the force, (ii) in any direction, and hence in that of the force, (iii) of any magnitude, and hence, as in the case of velocity, Art. 10, to represent the force on a given scale. THE LAWS OF MOTION. 41 4Si. Mass. The quantity of matter contained in a body is called its mass. Two bodies have equal masses when equal forces applied during equal times produce in them equal changes of velocity, i.e. equal accelerations. For instance if two bodies resting on a smooth table are pulled by exactly equal forces for the same time, then if their masses are equal, equal velocities will be generated. Again, we have seen that it is found by experiment that all falling bodies have the same acceleration, see Art. 22. In the case of any two falling bodies, the forces acting on them are their weights, due to the attraction of the earth; hence if they have equal weights, since their accelerations are equal their masses must also be equal; hence, equal masses have equal weights. Hence, for instance, all portions of matter which at a given place weigh one lb. have equal masses. 49. How to measure the mass of a body. Since equal masses have equal weights, if a body A weigh m times as much as another body B, then the mass of J. is m times the mass of B, for we may suppose A divided into m portions each of which has the same weight as B and hence the same mass as B. If we take the mass of B as the unit of mass, we say that the measure of A's mass is m. Art. 3. Hence to measure the mass of a body we compare its weight with that of the unit of mass. 50. If there are two bodies A and B such that the mass of A is double that of B ; then if a force P is required to produce a velocity F in 5 in a given time, a force 2P will be required to produce it in A in the same time. For A may be divided into two parts, each equal to B,.tQ each of which a force P must be applied. 42 THE ELEMENTS OF APPLIED MATHEMATICS.' Hence we see that the velocity produced in a body by a force depends upon the mass of that body as well as on the time during which the force acts ; which also follows from the definition of equal masses in Art. 48. 51. Unit of mass. There are two imits of mass in ordinary use, of which one is the Imperial pound, and is the mass of a certain piece of platinum kept at the Exchequer Office. Although the mass of a given body is of course invariable, its weight, i.e. the force with which it is attracted by the Earth, varies slightly in different latitudes. This is shown by the fact that the indication of a given body on a spring-balance at different latitudes will not be quite the same. The other unit of mass, called a gramme, is the xcnre*^ part of a certain piece of platinum called a kilogramme kept in Paris. The gramme is very nearly the mass of a cubic centimetre of water at 4° C. IVtomentuin. If the measures of the mass and the velocity of a body are m and v respectively the product mv is said to measure the momentum of the body, thus the body has mv tinies the momentum of a body of unit mass moving with unit velocity, so that the first body is said to have mv units of momentum. 52. We now give the laws of motion as stated by Newton. The relations between force and mass laid down in these laws are accepted as true because the conclusions drawn from them agree with observation and experiment. 53. First Law of Motion. Every body continues in its state of rest or of moving vmiformly in a straight line except in so far as it is made to change that state by external forces. This law states the property of matter known as Inertia, or the inability of a body to change its state of motion. "Matter is as it were the plaything of force; submitting to any change of state that may be impressed upon it, but THE LAWS OF MOTION. 43 rigorously persevering in the state in which it is left until force again acts upon it." (Tait, Properties of Matter.) For instance, a stone projected along the surface of ice is at last brought to rest by the (slight) resistance of the air and the (slight) friction between it and the ice, if these retarding forces were indefinitely diminished the stone would go indefinitely far. 54. Second Law of Motion. Change of motion is proportional to the impressed force and takes place in the direction in which the force acts. By " change of motion " is meant change of momentum in a given time. In other words, if a force acting on a body for a certain time produces a certain momentum, double the force will in the same time produce double the momentum, three times the force will produce three times the momentum, and generally, M times the force will produce M times the momentum. The unit force is taken to be that force which in one second produces one wnit of momentum. Hence, if a force whose measure is F, in one second produces a velocity / in a mass m, it produces mf units of momentum, and hence by this law it is mf times the unit force, i.e. its measure is mf, but we called its measure F, .-. F = mf Also, since f is the velocity produced in one second it is the acceleration, or a, hence P = ma. That is to say the measure of a force is = the measure of the mass x the measure of the acceleration produced in the mass. Thus the second law enables us to measure forces. The most important case is that of gravity, here ^=the body's weight = W, and a = acceleration of gravity =5?; .•. W=mg. 44 THE ELEMENTS OP APPLIED MATHEMATICS. 55. Unit of force. The British units of length and time are a foot and a second, and the unit of mass is the mass of a pound weight, hence the unit of force is, that force which acting for one second on the mass of a lb. produces in it a velocity of one foot-second. But the weight of a lb. produces in a body whose mass is a lb. a velocity of 32-2 foot-seconds per second, hence the unit offeree = ^^^^ of the weight of a lb. = wt. of^ oz. nearly. This unit of force is called a Poundal. The C. Gt. S. system *. In the C.G.s. system the unit of force is that force which acting on the mass of one gramme for one second produces in it a velocity of one centimetre per second. This force is called a Dyne. The weight of a gramme produces in a body whose mass is a gramme an acceleration of 32-2 ft.-secs. per second, that is, an acceleration of — ,^ — kines per second (see Art. 7). The value of the fraction is 981 nearly, hence a Dyne is = -^ of the weight of a gramme. Ex. 1. A body weighing 4 lbs. starts from rest and is acted on by a force equal to the weight of J lb. for 8 seconds, find the distance de- scribed. The body has the mass of 4 lbs., .'. m=4. The force equals weight of 4 oz., 8 poundals, .-. F=&. Hence the equation F=Ma gives M 4 Also s=^ai!2=^.2.82=64. The space described equals 64 feet. Ex. 2. A body whose weight is 1 cwt. is acted on by a force which produces in it an acceleration of 36 ft.-secs. per hour, find the magnitude of the force. * The centimetre-gramme-seoond system. THE LAWS OF MOTION. 46 An acceleration of 36 ft.-secs. per hour is an acceleration of 12~x 3 i^r — ^r- foot-secs. per second, 60 X 60 ^ or j^ ft.-secs. per sec. ; hence a=^^. Thus the measure of the force=112 XiJjj=l"12. The force is 1-12 poundak. Ex. 3. A force equal to the weight of 3 grammes acts on a body for 10 seconds, and causes it to describe 10 centimetres in that time, find the mass of the body. If a is the acceleration, then since 10 cms. are described in 10 seconds we have 10=^a{l0y, or a=i in p.G.s. units. fix The equation F=ma gives us that the force equals — Dynes, but it is also equal to the weight of 3 grammes or 3 x 981 Dynes, .-. |: = 3x981, o m = ax5x981, = 14715, or the mass is 14715 grammes. Ex. 4. A mass of 10 lbs. falls from rest through a distance of 10 feet and is then brought to rest by penetrating mud to the depth of 2 feet : find the resistance of mud (suppcsed uniform). The velocity acquired in falling through 10 feet is 8^10, Art. 22. Also if F measures the force of resistance of the beam and a the resulting acceleration, we have F=^ 10a. The body enters the beam with velocity 8 ;^10 which is reduced to zero in passing through 2 feet, hence since w2 = 2a«, Art. 23, 640=20!. 2, or a = 160. Hence F= lOot = 1600 poundals = 50 lbs. weight, nearly. EXAMPLES. X. 1. Find the force which acting on a cwt. for 1 second gives it a velocity of 5 yards per minute. 2. A mass of 10 lbs. is placed on a smooth horizontal table and is acted on by a force equal to the weight of 10 lbs., find the distance it will describe in 1 second. 3. A mass of 400 tons is acted on by a force of 112000 poundals; how long win it take to acquire a velocity of 20 miles an hour ? 46 THE ELEMENTS OF APPLIED MATHEMATICS. 4. In what distance will a force equal to the weight of 1 ounce be able to stop a mass of 40 lbs. which at the time the force begins to act has a velocity of 60 feet per second? 5. A bullet moving at the rate of 200 feet per second is fired into a thick target, which it penetrates to the extent of 6 inches ; if fired into a target 3 inches thick with equal velocity with what velocity would it emerge, supposing the resistance the same in both cases ? 6. A mass of 10 lbs. falls 100 feet and is then brought to rest by penetrating 1 foot into sand. Find the resistance of the sand. 7. If a force equal to the weight of 10 lbs. act upon a mass of 10 lbs. for 10 seconds what will be the momentum acquired? 8. A certain force acting on a mass of 10 lbs. for 5 seconds produces in it a velocity of 100 feet per second. Find the force and the accelera- tion it would produce in the mass of a ton. 9. A train of 200 tons weight is urged forward with a force equal to the weight of 1 ton, while it is retarded by a force equal to the weight of 10 lbs. per ton. What is its acceleration, and in what time will it acquire a velocity of 10 miles an hour? [If -fis the impelling and F' the retarding force, F—F'=ma.'\ 10. While a train travels half a mile on a level line its speed increases uniformly from 15 miles an hour to 30 miles. Find the ratio of the pull of the engine to the weight of the train. 11. A body resting on a smooth horizontal table is acted on by a horizontal force equal to the weight of 2 ounces, and moves on the table over a distance of 10 feet in 5 seconds starting from rest. What is its mass ? 12. A force equal to the weight of a gramme acts on a body whose mass is 27 grammes for one second. Find the velocity of the body and the space passed over. 56. The independence of forces. . We should observe in the second law of motion that the body on which the supposed force acts may be either at rest or in motion. Also the body is not restricted to be under the action of one force only ; hence we infer that, When several forces act on a body, each force produces the same effect that it would produce in the body at rest. The truth of this is easily seen in particular cases, e.g. if a stone fall from the top of the mast of a ship in motion it will strike the deck at the foot of the mast, that is at the same place at which it would have THE LAWS OF MOTION. 47 fallen, had the vessel been at rest. Now the stone when it begins to fall is moving with the velocity of the ship ; the only change is therefore in its velocity in the vertical direction, that is, in the direction of action of the force influencing it, viz. its weight. 57. The following consequence of the Second Law should be noted. When we have any number of forces P, Q, JR ... acting on a body of mass m, producing accelerations a,b,c..., then since P = ma, Q = mb, R = mc we see that the forces are in the same proportion as the accelerations they produce ; hence if lines be drawn parallel and proportional to the forces, they will also be parallel and proportional to the accelerations produced, or lines which represent the forces will also repre- sent the accelerations they produce. 58. Impulse. If ^be the time during which a constant force F acts on a mass M, we have seen that F=Ma=Ml; hence Ft = Mv, v being the change of velocity. The product Ft is called the impulse of the force, or shortly, the impulse. In certain cases such as that of a sudden blow of a hammer it is difficult to measure either the force or the (very short) time during which it acts, but its effect, i.e. the change of mo- mentum or its impulse, can be measured comparatively easily. Ex. 1. A body whose weight is 12 lbs. is acted on by a force which changes its velocity from 30 to 40 miles an hour. Find the impulse. The given velocities are 44 and 58§ feet per second. . • . /= impulse = m x (change of velocity) = 12(58f-44) = 176 units of impulse. Ex. 2. A body is struck and starts ofif with a velocity of 12 feet per second. The time during which tlje force lasts is j^ part of a second, 48 THE ELEMENTS OF APPLIED MATHEMATICS. the mass of the body is 3 lbs., find the average value of the force. If x is the measure of the force -^=3x12, ;!;=3600 poundaJs. Ex. 3. A cannon-shot of 1000 lbs. weight strikes directly a target with a velocity of 1500 feet per second, and comes to rest. What is the measure of the impulse? If the shot rebounded with a velocity of 200 feet per second what would the impulse be ? (i) /= 1500 X 1000 = 1,500,000 units of impulse, (ii) Here the change in momentum is 1000 (1500+200) = 1,700,000 units of impulse. Ex. 4. A ball falls from a height of 64 feet and rebounds to a height of 25 feet, find the impulse, and the average force between the ground and the ball if the time during which they are in contact is ^ of a second, the mass of the ball being 2 oz. Alls. 13 units of impulse, 208 poundals. Ex. 5. A shot whose weight is 4 cwt. leaves a fixed 80 ton gun, the velocity at the muzzle being 2000 feet per second ; if the shot moves over 20 feet when in the gun find the average force which has acted on the shot. Ans. 44,800,000 poundals. 59. Third Law of Motion. To every action there is always an equal and opposite re- action; that is to say, the actions of two bodies upon each other are always equal ar^d in opposite directions. Many familiar instances may be cited; if a body rest upon a table, the pressure of the body on the table is equal and opposite to the pressure of the table on the body. When a horse is drawing a cart, the force exerted by the cart on the horse is equal and opposite to that exerted by the horse on the cart. A difficulty sometimes occurs here to beginners who ask ' why then do the horse and cart move at all ?' This objection arises from over- looking the fact that there are other forces acting, viz. the resistance of friction to the cart's motion and the friction between the horse's feet and the ground. The cart is acted on by the pull of the horse and the resistance to its motion, which usually cannot exceed a certain amount, if therefore the pull be great enough the cart will move forward. The horse is acted on by the pull of the cart and the action of the ground on his feet ; as he treads he as it were pushes back and is therefore pushed forward by the same amount ; if this forward push be great enough, i.e. if he is strong enough, it will be greater than the pull of the cart, and he will move forward. THE LAWS OF MOTION. 49 The force with which the sun attracts the earth is esqual and opposite to that with which the earth attracts the sun. This mutual action between two bodies is called a stress. A tension is a case of a stress in which the forces act from each other, a pressure in which they act towards each other. 60. Applications of the Laws of Motion. We shall now consider a number of cases illustrating the use of the laws of motion. One such application is the following : A shot of mass m is fired from a gun of mass ilf with a horizontal velocity v, the velocity of recoil of the gun being V; the gun being horizontal. The forces acting on the shot and gun are : (i) their weights and the upward pressure of the hori- zontal plane on the gun, and these forces are vertical, (ii) the explosive action of . the powder, this gives two equal and opposite forces, whose resultant is therefore zero. Hence on the shot and gun taken together there is no horizontal force, therefore by Law ii. the total horizontal momentum is unaltered by the explosion. But the original horizontal mxjmentum was zero, after the explosion it is MV~ mv, hence MV— mv = 0, or F= -^ v. 61. Bodies connected by a string. When two bodies are connected by a stretched string the tension is the same at every point of the string, provided the mass of the string is so small that it may be neglected. For let T and T' be the tensions at the ends of a straight portion of the string and m the mass of the intervening portion of the string, then if a be the acceleration of the string, we have • T-I"+mg=ma. If m is zero this gives T= T J. 4 50 THE ELEMENTS OF APPLIED MATHEMATICS. 62. The motion of two masses connected by a string affords a good illustration of the second law of motion. Let two masses M and m be connected by an inextensible string passing over a smooth peg. The mass of the string is negligible ; hence the tension is the same at every point of it. Let T be the tension of the string, and suppose M to djBScend and m to ascend, since the string does not stretch, the down- ward velocity of M equals the upward velocity of m at every instant, hence their accelerations are numerically equal, let a be this common value. The force on M downwards equals Ma by the second law of motion, but the force on M downwards is its weight— the tension, or Mg — T, hence Ma = Mg- T, nvTI Fig. 23. or a = g- Again, the force on m upwards equals ma, but the force on m upwards is the tension — weight of m, or hence or T-mg, ma = T — mg, T a = a. m ^ Equating these two values of a M~ m 9 5'. and T a= — m rp_ 2mM _M—m ' ^ ~ M+m ^' THE LAWS OF MOTION. 51 W— w If the weights of M and m are W and w, a = „ g, since W— Mg, w=mg. Hence if the masses are equal there is no acceleration, and they will either remain at rest or will move with uniform velocity. The tension will then equal the weight of either mass. Ex. Two masses weighing 2 lbs. and 1 stone are connected by a string passing over a smooth pulley, find the acceleration of either. Alls. 1^. 62 a. Let us consider what would be the effect of suddenly attaching a third mass m' by a string to the ascending mass m at an instant when M and m are moving with a velocity u. The string connecting m and w! becomes suddenly stretched, and the three masses will now be moving with a velocity v. This sudden change in the velocity is brought about by an impulsive tension in the two strings. Let T be the impulse (Art. 58) of the tension in the string joining M and to, and T' the impulse of the tension in the string joining m and m', then we see that in consequence of T the momentum of M is changed from Mu to Mv, T asoA T' m mu to mv, T' to' zero to m'v. We therefore have T=M{u-v), T'-T=m{u-v), T'=m,'v; hence m'v— M {u — v) = m{u — v), from which u {M+ m)=v {M+m+m'). 63. Atwood's Machine. This machine is used to verify the laws of motion and to determine approximately the value of g. It consists essentially of a light string passing over a fixed pulley and having attached to its ends two equal masses each of weight P. On a graduated pillar AB a platform B, and a ring E, can slide up and down, and can be fixed by screws in any required position. One of the weights P can pass through the ring. The axle of the pulley rests upon four friction wheels, only two of which are represented in the figure. This greatly reduces the frictional resistance. 4—2 52 THE ELEMENTS OF APPLIED MATHEMATICS. There is also a bar of weight Q which will, not pass through E. The bar Q is placed upon P at some point G, we have now a weight P + Q on the right and a weight P on the left, hence, by the last Article, P + Q will descend with acceleration 2P+Q^' If the measured distance CE is h feet, the velocity acquired at JS is g] , Art. 16. 2Qh \2P + Q' Q is caught off by the ring, and the system will now move uniformly, with velocity 2Qh J 9 V2P + Q- If the measured dis- tance UD equal k feet, and t seconds be the observed time from E to B, 2Qh H or 2P + Q g] t, Art., 8; 2P + Q gf. Fia. 24. In this equation every- thing has been measured except g, hence g is deter- mined. Iliaccuracies enter into this method of determining THE LAWS OP MOTION. 53 " g" since no account has been taken of the mass of the pulley, the resistance due to friction, or the resistance of the air. The advantage of Atwood's machine is that, by properly choosing P and Q, the acceleration a^ r\ 9 ^^'7 be. made comparatively small. A falling body falls too quickly for us to calculate the time it takes to fall through any moderate height. Ex. 1. The two ends A and 5 of a string passing over the pulley of an Atwood's machine are loaded as follows : A with 16J ounces, B with 15^ ounces. Mnd the tension. The forces acting on A are the tension Tand the weight of 16^ ounces, or 33 poundals, its mass is ff pounds, hence 33— ^ acceleration oi A= , oo 32 7^—31 similarly acceleration oiB— , Ox 32 33- y y-31 33 ~ 31 ' 32 32 ■ „ 33x31 , , or T= — ^r — poundals. Ex. 2. In Atwood's machine one of the two weights is heavier than the other by half an ounce. What must be the weight of each in order that the heavier one may fall through one foot in the first second? Since the weight falls through one foot in the first second its acceleration is 2. Also if w-^ be the weight of the lighter weight, that of the heavier is Wi + 1. Hence by Art. 62, or 2 = — - — 32, °^ 2«;i + l ' Thus the lighter weight is 7^ poundals, 8^ 54 THE ELEMENTS OF APPLIED MATHEMATICS. Ex. 3. A weight § on a smooth table is connected by a string with a weight P which hangs vertically. Find the acceleration of P and Q. Q If/ is the acceleration of either — ° ^f=P-T, ^f=T. Hence by addition (P+§)-^=P, _L. i Pg Fig. 25. '^^^ EXAMPLES. XI. 1. Two masses of 48 and 50 grammes respectively, are attached to the string of an Atwood's machine, and starting from rest the greater mass falls through 10 centimetres in one second. Find the acceleration due to gravity. 2. A mass of 7 lbs. is attached to one end of a string passing over a pulley, and two masses of 3 lbs. and 6 lbs. to the other end. After 4 seconds the smallest mass is detached. How much farther will the 6 lbs. fall, and when will it be brought to rest? 3. Masses of 3 lbs. and 5 lbs. respectively are connected by a string passing over a pulley ; after one second the string breaks, for how long and how far will the 3 lbs. ascend? 4. Two weights of 7 oz. and 9 oz. are connected by a string 40 feet long which is hung over a smooth pulley so that the 7 oz. weight touches the ground. When the system has been in motion for two seconds the string is cut, and both weights reach the ground at the same time. Find the height of the pulley from the ground. 64. Pressure between moving bodies in contact. When a heavy body rests on a horizontal platform the pressure of the platform on the body is equal and opposite to the pressure of the body on the platform, by the third law of motion. Suppose the platform to move vertically downwards with an acceleration a, the force on the body whose mass we sup- pose to be m is ma. ■ THE LAWS OF MOTION. 55 But the force on the body is the difference of its weight and the upward pressure P of the platform upon it, hence mg — P = ma, or, P = mig-a) (i); thus the pressure is diminished. If the platform is ascending, the upward force on the body is P — mg, hence in this case P — mg = ma, or P = m(5r + a) (ii); thus the pressure is increased. Ex. 1. A body whose weight is 10 lbs. is placed on a horizontal plane moving vertically upward ; if the pressure of the body on the plane is equal to the weight of 16 lbs. find the acceleration. Here P=16x32 poundals, m=10, .: 10a=32(16-10), or a=19-2. Ex. 2. A balloon ascends vertically with uniform acceleration, so that a weight of 2 pounds exerts a pressure on the bottom of the car equal to the weight of 32J oz. ; find the height the balloon wUl reach in one minute. The force on the body is 65 - 64 poundals, or one poundal. The mass of the body is 2, the acceleration is therefore ^. Hence the height reached in one minute is i . i (60)2 feet, or 300 yards. EXAMPLES. XII. 1. A body whose weight is 112 lbs. is placed on a lift which moves with a uniform acceleration of 12 ft.-sec. units. Find the pressure on the floor when the lift is descending. 2. The pressure on the bottom of a bucket which is being drawn up the shaft of a mine is equal to the weight of 133 lbs. ; if the contents of the bucket weigh 1 cwt. what is the acceleration ? 3. A body whose weight is 4 stone is placed on a lift moving wi1;h uniform acceleration of 12 ft.-sec. units ; find the pressure on the floor of the lift when it is (i) descending, (ii) ascending. 4. A mass of 40 lbs. rests on a horizontal table which is made to ascend, (i) with a constant velocity of 2 feet per second, (ii) with a constant acceleration of 8 ft.-sec. units; find in each case the pressure on the plane. 56 THE ELEMENTS OF APPLIED MATHEMATICS. 5. A man suddenly jumps oflF a table with a 20 lb. weight in his hand, what is the pressure of the weight on his hand while he is in the air 1 6. A balloon ascends with constant acceleration, so that a mass of 561bs. exerts a pressure of 84 lbs. on the bottom of the car. When will it be 200 feet high and what will be its velocity at that time ? 7. A cord passing over a smooth pulley supports two scale-pans, the weight of each being 3 ounces. If weights of 4 and 6 ounces be placed in the scale-pans find the acceleration and the tension of the string, also the pressure between the masses and the scale-pans. 8. Of two forces one acts on a mass of 5 lbs. and produces in it a velocity of 5 feet per second in ^pj of a second, the other acts on a mass of 625 lbs. and produces in it a velocity of 18 miles per hour in one minute, compare the forces. 9. A ball whose mass is 3 lbs. is falling at the rate of 100 feet per second. What force expressed in lbs. weight will stop it (i) in 2 seconds, (ii) in 2 feet? 10. A cannon-ball weighing 600 lbs. and moving with a velocity of 1000 feet per second penetrates a target to a depth of 15 inches. Find the pressure on the target, supposing it to be uniform. 11. A shot of 1000 lbs. leaves a gun with a velocity'of 1500 feet per second. How long must the shot have been under the action of the powder supposing the average pressure upon it to have been equal to the weight of 1200 tons ? 12. A balloon is moving upwards with a speed which is increased at the rate of 4 feet per second in each second ; find by how much the weight of a body of 10 lbs. as tested by a spring balance in the balloon would differ from its weight under ordinary circumstances. 13. In what time will a weight of 16 lbs. draw another of 12 lbs. over a fixed pulley through 32-2 feet,.and what velocity will the weights have at the end of the time? 14. The two weights in an Atwood's machine are 240 grammes each, and an additional weight of 10 grammes being placed on one of the two it is observed to descend through 10 metres in 10 seconds, show that g'=980 nearly. 15. A man of 12 stone weight and a sack of 10 stone weight are connected by a rope over a smooth pulley. The man pulls himself up by the rope and diminishes his downward acceleration by ^, find the upward acceleration of the sack and show that the acceleration of the man relative to the rope is 3'2. 16. A rope hangs over a smooth pulley and a man of 12 stone lets himself down with acceleration /', while a man of 1 1^ stone pulls himself up with acceleration/. Find/' in order that the rope may remain at rest. THE LAWS OF MOTION. 57 17. Two weights of 5 lbs. and 7 lbs. respectively are fastened to the ends of a cord passing over a frictionless pulley supported by a hook. Show that when they are free to move the pull on the hook is llf lbs. weight. 18. Two equal weights A and B connected by an inelastic thread 3 feet long are laid close together on a smooth horizontal table 3 '5 feet from the nearest edge, and B is also connected by a stretched inelastic thread with an equal weight C hanging over the edge. Determine the velocities of the weights when A begins to move and also when B arrives at the edge of the table. 19. One end of a string is fixed, it then passes under a moveable pulley to which a weight W is attached. The string then passes over a fixed pulley and a (smaller) weight P 'is attached to the other end, all the 3 sections of the string are vertical. Prove that the acceleration of TF— 2P W is -^ — jp g. The weights of the pulleys are neglected. 20. A man hangs by a rope passing over a fixed pulley to the other end of which a weight equal to that of the man is attached. Prove that he cannot raise himself up above the level of the weight as he climbs up the rope. 21. To one end of a string passing over a fixed pulley hangs a weight P, and to the other end a pulley over which passes a string at one end of which hangs a weight § and at the other a weight R resting on a table. The system being allowed free motion find the pressure on the table supposing R so heavy that it is not raised from the table. 22. To one string of an Atwood's machine a mass of P lbs. is attached. To the other string a mass of § lbs. is attached where P>Q. Above the mass of Q lbs. is placed a mass of -■ — j~- lbs. which can be detached in the ordinary manner during the motion. The system starts from rest and moves for t seconds, at the end of which time the mass of — jy-^ lbs. is detached. Show that in t seconds more the system will be for an instant at rest, and that the motion will then be reversed in direction. CHAPTER IV. PROJECTILES. 65. In Chapter i. we discussed the motion of a particle projected vertically upwards or downwards. In the present chapter we shall investigate the motion of a body projected in any direction. Fig. 26. Let the velocity with which the body is projected be V, and let u and u' be the vertical and horizontal components of 7. During its motion the body is, owing to gravity, acted on by a downward acceleration g, its vertical and horizontal velocities after t seconds being therefore respectively u — gt, u'. We see that the horizontal velocity is not altered, since there is no horizontal force, the vertical force, by the second law of motion, only affecting the vertical velocity. The body's vertical velocity will be zero after a time T given by u-gT = 0, or T=-. u Hence after - seconds the body begins to descend. PROJECTILES. 59 As in Chapter i. we see that the time from projection to the greatest height is equal to the time from the greatest height to the ground. We thus get the following formulae ; velocity after t seconds is the resultant of ^ . , . , ' (M—^t vertical, ,, J -r. J • J J ■ { u't horizontal, the space described m t seconds is -^ ^ , . . , [ut — igir vertical, the time to reach the greatest height is - . Inserting this value of t we get as the following ex- pression for the greatest height, g ^^ \9i H And since the time of flight is equal to — , we find as the range, or horizontal distance described, , 2m 2mm' MX — = . g g If a is the angle which the direction makes with the horizon, u = Fsin a, m'= Fcosa, hence the greatest height is — = , 2 F^ sin a cos a the range is • Also the distances described in t seconds are Fcos a X i horizontal, Fsin a X i - \gt^ vertical. EXAMPLES. XIII. 1. Find the horizontal and vertical spaces described in 3 seconds, when the components of the velocity of projection in those directions are 100 and 200 feet per second. 60 THE ELEMENTS OF APPLIED MATHEMATICS. 2. Find the greatest height to which a body will rise and its range, if it is projected with horizontal and vertical velocities of 400 and 800 feet per second. 3. The greatest height to which a body rises is 100 feet, find how far it will rise in 2 seconds. 4. The velocity with which a cannon-ball leaves the gun has for its vertical and horizontal components velocities of 7-^ and 10 miles per minute, find its rang^. 5. Show that the direction of the velocity of the body in t seconds after projection makes an angle 6 with the horizontal such that tan 5 = — ^ 2_ . K cosa 6. A particle is projected at an inclination 6 to the horizon where" cos 5=^, with a velocity of 1200 feet per second. Find the greatest height it attains and its range on a horizontal plane through the start- ing point. 7. A body is projected at an inclination a to the horizon, such that with a velocity of J82 feet per second. Show that after J of a second its direction of motion is inclined at an angle of 45° to the horizon. 66. The Qreatest Bange. The horizontal range we have seen to be . For a given velocity of projection this will have the greatest value when sin2a=l, or a = 45°. Hence the greatest range for a given velocity of projection is got by projecting the body at an angle of 45° with the ground. 67. Range on an inclined plane. A body is projected with a velocity V in a direction a from the foot of an in- clined plane making an angle /3 with the horizon ; it is required to find the range on the inclined plane, or how far up the plane the body will strike it. After * seconds the body will strike the plane at some point P, from P draw the vertical line P^. PROJECTILES. 61 Then OiV= horizontal space described in t seconds = Vcoa at, PiV"=vertioal = Fsina t-l . , - Vsinat-hgt^ . qt .-. tan^= — = f^ = tana-jpf^^ . Fcosai 27cosa This determines t, giving 2FcQsa t=- 9 (tan a- tan j3). Thus /^n /^ilT « FcOSa < aF^COS^a,. 0P= ON sec = = -— (tan a - tan S). coS|3 gi COS p ' Therefore the range is -r- (tan a - tan j3) " g'COS^ ^ ' _ 2 F^ cos g sin (a — ;3) ~ g cos^ j3 Hence if F and /3 are given the range is greatest when 2 cos a sin (a — 0) is greatest, and 2 cos a sin (a - /3) = sin (2a - /3) - sin ^. Thus for the greatest range sin (2a - /3) is greatest or 2a-i3=90°, .=§+«•. 68. The patb of a projectile. So far nothing has been said as to the form of the curve described by a projected body. AM B P^ L K Fig. 28. Let P be the position of the body after * seconds, and ^the highest point of its path. 62 THE ELEMENTS OF APPLIED MATHEMATICS. We have already seen that OJV^V COS at, PJV=raiQat-yi^, Draw the horizontal line PL, then HL=HK- PiV^=I!|i^ -^{^Vg sin « t-gH^) {Vaiaa-gtf ^9 .(i). Also PL= OK- 0N= Foos a t. g ... PL=^^^^^{Vama-gf) (ii). Comparing (i) and (ii) we see that PL-^JJl^S^HL. g Thus the path is a parabola whose vertex is H and latus rectum g 69. Directrix and focus of path. The latus rectum of the parabola is , hence if S is the c cir TJT7 ^^cos^a F^ COS 2a focus SK=HK- —= = s • ^9 ^9 t' COS (X The height of the directrix above H is — =r— — • 2g 70. Velocity due to fall ft'om directrix. Produce ^P to meet the directrix in M, then MP^SL+ ^^^ = 1{ F^cos^„+ ( Fsina -^0^}, _ (velocity of body)' 2g Thus velocity of hoiy='ij2gMP, = velocity gained by falling through a distance MP, Art. 19. PROJECTILES. 63 Hence the velocity of a body at each point of its path is that which it would have gained by falling freely from the point on the directrix vertically above it. EXAMPLES. XIV. 1. Find the direction and velocity with which to project a ball that it may pass horizontally over the top of a wall 50 yards off and 75 feet high. 2. A stone is projected into the air with a velocity of 200 feet per second in a direction inclined at 60° to the horizontal plane. With what velocity must another stone be projected vertically that the two stones may rise to the same height above the horizontal plane ? 3. A boy with a stone aims at a mark 25 yards from him on a level 10 feet below his shoulder, with what velocity must he throw the stone horizontally so as to hit the mark 1 4. A body is thrown from one extremity of the horizontal base of an isosceles triangle so as to pass just over the vertex and fall on the other extremity of the base. If a is the base angle and the angle of projection show that tan /3 = 2 tan a. 5. Two particles projected with the same velocity from pass through the same point P, prove that if a and |3 are the angles of projection where i is the angle which OP makes with the horizon. 6. A shot whose mass is -th of the mass of the gun and carriage is flred at an inclination 6 to the horizontal. If a be the, inclination of the gun prove that tanfl = (l + -j tana. 1. A balloon is moving vertically upwards with the velocity V and a rifle is aimed so as to hit it, the bullet's initial velocity being (v/3 + 1) V. If the angle of elevation of the balloon from the rifleman- be 45°, prove that the elevation of the rifle must be 60°. 8. From a point on a hill of inclination 30° one particle is projected up the hill and the other down with equal velocities, the angle of pro- jection being in each case inclined to the horizontal at 45°. Show that the range of one particle is nearly 3| times that of the other. 9. Prove that 4 times the square of the number of seconds in the time of flight in the range on a horizontal plane is the height in feet of the highest point of the path. 64 THE ELEMENTS OF APPLIED MATHEMATICS. 10. A wet open umbrella is held with the handle upright and made to rotate round that handle at the rate of 14 revolutions in 33 seconds. If the rim of the umbrella be a circle of one yard diameter and its height above the ground 4 feet, prove that the drops shaken off the rim meet the ground in a circle of 5 feet diameter, the. circumference of the rim being -^^ feet and the effect of the air being neglected. 11. Three bodies are projected simultaneously from the same point and in the same vertical plane, one vertically upwards, another at the elevation of 30°, and the third horizontally. If their velocities be in the ratio oi 1 : 1 : y/3 prove that the bodies will always be in the same straight line. 12. If the times taken by a projectile from P to Q and from Q to R are equal, its horizontal velocity being V, then if V^, Fg and Fj are the velocities at P, Q and R respectively, 13. A vertical line is divided into a number of equal parts A^A^, A^A^, A^A^ etc. Show that if a particle be projected from in the vertical plane through the line, OA-^, OA^, &c. will meet its path in points such that the times of flight from each to the next are all the same. 14. A number of bodies are projected simultaneously from the same point in the same vertical plane with different velocities such that if lines be drawn from parallel and proportional to these velocities the extremities of these lines lie in a certain straight line AB. Prove that after a certain interval all the bodies will be situated in a straight line through parallel to AB. 15. A bomb-shell on striking the ground burst, scattering its fragments with velocity V, find the area of ground covered by the fragments assuming that the shell falls on a horizontal plane. CHAPTER V. THE COMPOSITION AND RESOLUTION OF FORCES. 71. When two or more forces act on a particle, that single force which would produce their joint effect is called their resultant. 72. The parallelog^ram of forces. When two forces acting at a point are represented in magnitude and direction by two sides of a parallelogram which are drawn through the point, their resultant will he represented by the diagonal drawn- through the point. Let a particle whose mass is m be acted on by two forces represented by OA and OB. Fia. 29. Now lines which represent forces also represent the accelerations produced by the forces. Art. 57. Hence forces represented by OA, OB, and OG will pro- duce accelerations represented by the same lines. But the resultant of the accelerations OA and OB is the acceleration OG, from the parallelogram of accelerations, Art. 43. Hence the force represented hj OG produces the accelera- tion OG, that is it produces the joint effect of the forces OA and OB and is therefore their- resultant. Observe that since ^C is equal and parallel to OB, the result may be put as follows : . OG represents the resultant of the forces which are repre- sented by OA and AG. J. 5 66 THE ELEMENTS OP APPLIED MATHEMATICS. 73. The following special cases are important : (i) the resultant of two forces acting at a point in the same direction is their sum, (ii) the resultant of two forces acting at a point in opposite directions is their difference. 74. Verification of the Parallelogram of Forces. Take three flexible strings and tie them together. Let two of them pass over smooth pegs at any distance apart, the third hanging freely. Attach weights P, Q, and B to their ends. « Let the system right itself; when it has settled, measure off on the strings lengths OA, OB and 00 proportional to P, Q and R ; then the tensions in the three strings, being equal to P, Q, and R are proportional to OA, OB and 00. Complete the parallelogram AOBD. It will be found by measurement that OD is equal in magnitude and opposite in direction to 00. But the effect of the force represented by 00 is equal and opposite to the joint effect of the forces represented by OA and OB, since the forces balance each other ; hence OD represents the resultant of the forces represented by OA and OB. O R Fia. 30 75. Forces are compounded and resolved like ve- locities. When three or any number of forces act at a point, we find the resultant of two of the forces and then the resultant of this and a third force, and so on. Forces, therefore, are compounded and resolved in the same way as velocities, viz. by the parallelogram-law, hence COMPOSITION AND EESOLUTION OF FORCES. 67 if in Art. 35 we write forces P, Q, and E instead of ve- locities u, V, and w, we are enabled to find the resultant of forces P and Q acting at different angles. Ex. 1. Find the resultant of forces of 3 and 4 lbs. weight acting at an angle of 90°. Ans. Vs + 1 6 = 5 lbs. weight. Ex. 2. Find the resultant of forces P and Q acting at ^n angle of 60°. Ans. 'JF^+Q^+PQ. 76. Components of a force. When forces P and Q have R as resultant, P and Q are called components of R. As was shown in the case of velocities we see that a force may be resolved into components in an infinite number of ways. When P and Q are at right angles they are called the rectangular components of R. As an example of the resolution of a force into components take the case of the action of the wind on the sail of a vessel. First resolve the force exerted by the wind into a component R perpendicular to the sail and another com- ponent along the sail, the latter component will have no effect. Next resolve It into components P and Q in the direction of the length of the vessel and perpendicular to it respectively. The component Q produces merely a slight broadside motion called leeway, the component P causes the vessel to move in Fia. 30 o. the direction of its length. We easily see that by setting the sail so as to be nearly parallel to the direction of the wind the vessel may be m,ade to go in a direction nearly opposite to that of the wind, neglecting the effect of leeway. Referring to Ex. 1, p. 34, we see how to find the hori- zontal and vertical components of a force R when inclined at angles 30°, 45°, or 60° to the horizon. The components are /3 1 — R, 5 P for an inclination of 30° to the horizon, 5—2 68 THE ELEMENTS OF APPLIED MATHEMATICS. -7^R, -j^R for an inclination of 45° to the horizon. I -8. f ^- .60° As in Art. 40 we see that when several forces act at a point we have. The resolved part of their resultant in any direction is equal to the (algebraic) sum of the resolved forces in that direction. Ex. 1. Two forces acting at right angles, have as resultant a force equal to 6 lbs. weight, one of the forces is 4 lbs. weight, find the other force. Let a; be the required force, then we have, since the forces a; and 4 whose resultant is 6 act at right angles, a;='\/3e- 16=^20=2^/5. The required force is one of 2 ^5 lbs. weight. Ex. 2. A particle is acted on by a force whose magnitude is un- known, but whose direction makes an angle of p 60° with the horizon. The horizontal component of the force is known to be 1-35 lbs. Determine the force and its vertical component. Let OP be the force, draw PJV perpendicular to the horizontal line OiV. Then OJV and P# are the horizontal and vertical components. And since POA" is 60° and OPA is 30°, the triangle OPA is half an equilateral triangle, hence OA=iOP, 0P=20A= 2-7. Also PN=sl'OW- Oi\^2=V(2-'7)2- (1-35)2 = 1-35x^3 = 2 -3 lbs. nearly. Ex. 3. The resultant of two forces acting at a point 0, in the directions OA and OB, and represented in magnitude by a. OA, fi.OB is represented by ' (a+^) 00, COMPOSITION AND RESOLUTION OP FORCES. 69 where C is a point in AB such that a.CA^^.BG. For by the parallelogram-law the resultant of the forces u. OC and a. CA is a.OA\ also the resultant of /3 . OC and . C5 is ;3 . OB. ^^o. 32. Thus the forces a. OA and /3 . OB are equivalent to (a -1-/3) OC, since forces represented \ij a.CA and j3 . C5 are equal and opposite. Cor. Putting CA=^CB, or a=^=l, we see that the resultant of OA and 05 is 2 . OC. See EXAMPLES. XV. 1. Find the resultant of the following forces : (i) forces of 5 and 11 lbs. acting at an angle of 30°; (ii) -1 ... v/2 60°; (iii) v/3 ... VS 60°; (iv) 1... 4 150°. 2. Show that the resultant of two equal forces bisects the angle between their directions. 3. Two forces acting at right angles are to each other in the ratio of 1 to ,JT, and their resultant is 8 lbs. ; find the forces. 4. Two forces acting at an angle of 120° have a resultant equal to 2 V3 lbs. weight ; if one of the forces is 4 lbs. weight, find the other. 5. Find the magnitude of two forces such that if they act at right angles their resultant is v'l'^lbs. weight, while when they act at an angle of 120° their resultant is ^13 lbs. weight. 6. A force equal to the weight of 15 lbs. acting vertically- upwards is resolved into two forces, one being horizontal and equal to the •syeight of 10 lbs. ; find the other force. 7. The magnitudes of two forces are as 4 : 5 ; the direction of their resultant is at right angles to the smaller force ; find the ratio of the larger force to the resultant. 8. The resultant of two forces P and Q acting at a certain angle is R, the resultant of P and §' acting at the same angle is R', what would be the resultant of R and a force equal and opposite to i2'? 70 THE ELEMENTS OF APPLIED MATHEMATICS. 9. In a triangle ABC, B, E and F are the middle points of the sides, show that forces acting at a point and represented by '2,AD, ^BE and 2Ci''are in equilibrium. 10. A vertical force of 10 lbs. is resolved into two equal components, one of them making an angle of 30° with the vertical, find the magni- tude and direction of the other. 11. If the directions of two forces be inclined to one another at an angle of 135°, find the ratio of their magnitudes that their resultant may be equal to the smaller force. 12. Find the components in directions due E. and N.W. of a force equal to the weight of 12 lbs. acting N.E. 13. If a given force acting at a given point in a given direction be resolved into two equal forces, prove that the extremities of the lines representing the equal forces always lie on a fixed straight line. 14. The resultant of forces of 5 lbs. and 6 lbs. is 7 lbs., find the cosine of the angle between the forces of 5 and 6 lbs. 15. The direction of a force of 10 lbs. weight makes an angle a with the horizon such that cos a=f, find its horizontal and vertical components. 16. AB, AG represent forces of 33 and 25 lbs. respectively. If CD were drawn perpendicular to AB, AD would represent on the same scale 15 lbs., prove that the resultant of the forces in AB and AC is 52 lbs. 17. A force 8P is resolved into two forces, each of which is equal to 5P, find the sine of the angle between the equal components. 77. The triangle of forces. When forces acting at a point have a zero resultant, they are said to be in equilibrium. We shall show that forces represented in magnitude and direction by OA, AG, GO are in equilibrium. For since the resultant of the forces OA and OB is OG, Art. 72, and since AG is equal and parallel to OB, forces repre- FiG. 33. sented by OA and AG have 00 as resultant. Thus the effect of the three forces OA, AC and GO is that of 00 and GO, which is zero ; thus there is equilibrium. COMPOSITION AND RESOLUTION OF FORCES. 71 Hence, if three forces acting at a point can be represented by the sides of a triangle taken in order, they are in equi- librium. This proposition is known as the triangle of forces. 78. Converse of the Triangle of Forces. The converse of the triangle of forces is also true, viz. that if three forces acting at a point are in equilibrium, any triomgle which has its sides parallel to the forces will have those sides also proportional to the forces. b <: - ^ Fig. 33 a. Let the sides of the triangle aha be parallel to the forces P, §, R which are in equihbrium, it is required to prove they are also propor- tional to P, § and R. Let the sides BC and CA of the triangle ABC represent the forces P and Q in magnitude and direction. Then since forces represented by BC, CA and AB are in equilibrimn by the Triangle of Forces, the side AB must represent R in magnitude and direction. Hence BC : CA : AB=P : Q : R. Also the sides of the triangle ahc are parallel to those of ABC, therefore ahc and ABC are similar triangles, and BC : CA :AB=bc : ca -.ah. Therefore P : Q : R=bo : ca : ah. 78 a. Iiami's Theorem. When three forces are in equilihriuni each force is proportional to the sine of the angle between the directions of the other two. 72 THE ELEMENTS OF APPLIED MATHEMATICS. The forces P, Q and R are in equilibrium, and ABO is any triangle having its sides parallel to their directions. Then by the Converse of the Triangle of Forces P:Q: R=BG : CA : AB. The angles of the triangle ABO are supplementary to those between the directions of P, Q and R, and since BO -.CA : AB=sm ^ : sin 5 : sin 0, we have P : Q : R=sm A : sinB : sin or, P:Q:R=sia QOR : sin ROP : sin POR. ' Ex. 1. Three forces acting at a point are in equilibrium ; if they are equal find the angle between their directions. By the Converse of the Triangle of Forces, the forces can be repre- sented by the sides of an equilateral triangle. They therefore make angles of 120° with one another. A Fig. 34. Ex. 2. A weight of 24 lbs. is suspended by two flexible strings one of which is horizontal, and the other is inclined at an angle of 30° to the vertical ; what is the tension in each string ? AD and AO being the strings, the sides of the triangle AOB are parallel to the forces, which are the tensions of the strings and the suspended weight. The angle OAB is 30°, the angle ABO is 90°, hence BO=iAO, AB=y/3BC. Art. 35. Therefore tension in AO : suspended weight =A0: AB= tension in AC= V3 X 24 lbs. ■■ x/3' 16 s/3 lbs. And tension in ^i> : suspended weight =5C : AB=-j-; or, tension in AD =--x 24 lbs. = 8 x/3 lbs. COMPOSITION AND RESOLUTION OF FORCES. 73 Ex. 3. Three forces whose magnitudes are 3, 6 and 9 lbs., act at a point in the directions of the sides of an equilateral triangle taken in order, find their resultant. By the triangle of forces, the forces 3, 3, 3, in the assigned directions are in equilibrium, they may therefore be removed. There are left forces of 3 and 6 lbs. acting at an angle of 120°. Their resultant is therefore 3^3 lbs.; see Art. 35. EXAMPLES. XVI. 1. Three forces acting at a point are in equilibrium; the greatest force is 5 lbs., the least 3 lbs., and the angle between two of the forces is a right angle. Find the other force. 2. Three forces represented by the nimibers 1, 2, 3, act on a particle in directions parallel to the sides of an equilateral triangle taken in order ; find their resultant. 3. A weight of 10 lbs. hangs fastened to the ends of two strings, the lengths of which are 3 and 4 feet, the other ends of the strings being attached to two points in a horizontal line distant 5 feet from each other, find the tension of each string. 4. Three forces cannot be in equilibrium if the sum of any two is less than the third. 5. At what angle must two equal forces act so that their resultant may equal each of them ? 6. If three forces P, P, Q act at a point in directions such that each force is equally inclined to the directions of the other two, find their resultant. 7. A body is acted on by three forces, one of 2 lbs. due west, one of 4 lbs. north-east, and one of 2 ^2 lbs. due south, find their resultant. 8. When two of three forces in equilibrium are given in magnitude, the third force increases as the angle between the first two forces diminishes. 9. Give a geometrical construction for resolving a force into two others inclined at a given angle, one of which is to be of given magnitude. 10. A weight of 10 lbs. is supported by two forces, one of which is horizontal and the other inclined at 30° to the horizon. Find the forces. 11. Forces of 5 lbs., 6 lbs. and 7 lbs. acting at a point are in equi- librium, find the cosines of the angles between them. 74 THE ELEMENTS OF APPLIED MATHEMATICS. 79. The Polygon of Forces. If any number of forces acting at a point be represented in magnitude and direction by the sides of a polygon taken in order, the forces will be in equilibrium. Let AB, BG, CD, DE, EA, be the sides of a pentagon representing forces on a particle at 0. Join AG, AD. The resultant oi AB, BG is AG, see Art. 72, AG,GD\sAD, AD,DE is AE. Hence the resultant of all the forces is the resultant of AE and EA, hence the resultant vanishes, or the forces are in equilibrium. Ex. Five equal forces act on a particle in directions parallel to five consecutive sides of a regular hexagon taken in order ; find the magni- tude and direction of the resultant. By the Polygon of Forces the resultant is represented by the line closing the figure drawn from the starting point ; hence it is equal to any one force and is parallel to the last side. Notice that the converge of the polygon of forces does not hold good, since two polygons may have their sides parallel and yet not propor- tional. Ex. 1. Three forces acting at a point are represented by adjacent sides of a regular hexagon taken in order, find their resultant. Ans. That diameter of the circum-circle which is parallel to the middle side. Ex. 2. Find the resultant of four forces of 4, 5, 7 and 8 lbs;, acting on a particle, and represented in direction by the successive sides of a square. Ans. 3 x/2 lbs. bisecting the angle between the forces of 7 lbs. and 8 lbs. COMPOSITION AND RESOLUTION OF FOECES. 75 Ex. 3. In the hexagon ABCDEF, the lines AD, BE, EB, EC, CF, represent in magnitude and direction five forces acting at a point, find their resultant. Ans. AF. Ex. 4. Taking an inch to represent in magnitude a force of 1 lb. weight, by means of an ordinary foot-rule and a diagram, find the resultant of the following forces acting at a point : 3J lbs. due E., 4 lbs. due S.E., 1 lb. due N.E., 6J lbs. due N. so. Having given that the direction of the resultant of two forces is that of the diagonal of the parallelogram of which the lines represent- ing the forces form two adjacent sides, we may prove that this diagonal represents the resultant in magnitude also in the following manner. Fig. 38. Let OA and OB represent two forces, then assuming that the direction of their resultant is that of OG we have to show that its magnitude is represented by OG. Produce GO to B so that OB re- presents the magnitude of the resultant. Complete the parallelogram AOBE. Join OE. Then the forces represented by OA, OB and OB are in equilibrium, hence OB is in the same line with the resultant of the forces repre- sented by OA and OB ; but we are given above that OE, the diagonal of the parallelogram AOBE, is the direction of this resultant ; hence B, and E are in the same straight line. Then since the figure AGOE is a parallelogram, OCis equal to AE, and since the figure AOBE is a parallelogram, OB is equal to AE, therefore OB is equal to OG, which was to be proved. 81. Resultant of any number of forces. As in the case of velocities we may find the resultant of any number of forces acting at a point in either of the following ways ; by use of (i) repeated applications of the parallelogram-law, 76 THE ELEMENTS OF APPLIED MATHEMATICS. (ii) the polygon of forces, (iii) resolving the forces along two lines at right angles and then compounding the forces acting in these lines, see Art. 41. The following examples show the application of the third method. Ex. 1. The directions of three forces, acting at 0, of 2, 3, and 5 lbs. make angles of 30°, 45°, and 60° respectively with a line OA, find their resultant. Pia. 39. Let OP, OQ and OR represent the forces, we resolve along and per- pendicular to OA. /3 1 The components of OF along OA and OB are 2 x^, 2 x - , p. 34. OQ OR ^''V2'^''V2' 5x-, 5x n/3 Hence the resolved parts of the forces along OA and OB re- spectively are /34. 3 , 5 1 , 3 , 5V3 Therefore the resultant is A-^'*T^-D'<'-^'* n/2 " 2 J = 2^152 + 40^3+42^/2+76), =9'81bs. nearly. COMPOSITION AND RESOLUTION OF FORCES. 77 Ex. 2. A particle is acted on by three forces of P, 2 V2P, 3 y'2P lbs., the angles between the first and second, and the second and third being 45° and 90° respectively ; find the resultant. It shortens the work if we resolve the forces A along and perpendicular to the directions of one of ) the forces themselves, let this be the force P. Let X be the sum of components of all the forces in the direction of F, and T be the sum of components of all the forces perpendicular to P. Then Z=P+2 V2Px-^-3 V2P x 1 V2' =0, r=2 V2Px-^ + 3V2P X 4 = 5P. Thus the resultant is perpendicular to the force f and equal to the weight of 5P lbs. EXAMPLES. XVII. 1. Three equal forces each of 2 lbs. weight act on a particle. The angle between the directions of the first and second is 30°, between the second and third is 60°. Find their restdtant. [Resolve the second force.] 2. A particle is acted on by three forces, one of 2 lbs. East, one of 2 lbs. North, one of 2 a/2 lbs. North-west. Find the resultant. 3. Three forces act at a point, the angle between the first and second is 90°, and between the second and third is 60°. The second and third forces are each equal to F and the first is ,J3F. Find the resul^nt. 4. Forces of 1, 2 and ^3 poundals act at a point in the directions OA, OB and OC ; the angle AOB is 60°, and the angle AOC is 90°, find the resultant. 5. Find the resultant of the following forces acting on a particle ; 3jlbs. due E., 4 lbs. due S.E., lib. due N.E., e^lbs. due N. 6. Four forces of 12, 10, 6 and 8 lbs. weight act on a particle. The angle between the first and second is 30°, between the second and third 120°, between the third and fourth 90°. Show that the components along and perpendicular to the direction of the first force are 8+2 V3 lbs., 8- 4^3 lbs. 7. In the quadrilateral ABCD the forces represented by AB, BD and DC have the same resultant as those represented by AD, DB and BC, the forces being supposed to act at a point. 78 THE ELEMENTS OF APPLIED MATHEMATICS. 8. Forces are represented by the radii of a circle drawn to the angular points of a regular inscribed polygon, show by the Polygon of Forces that these forces are in equilibrium. 9. ABO, A'B'C are two triangles in one plane, show that a hexagon can be constructed each of whose sides is equal and parallel to one of the sides of the two triangles. 10. Forces acting at a point are represented by the sides AB, BC, CD and the diagonal DB of a square, find the resultant force. 11. ABCD is a parallelogram, show that the resultant of two forces represented by ^C and DB is represented by ^AB. 12. Forces of 2, 5 and 8 lbs. act parallel to consecutive sides of a regular hexagon taken in order; show that the sum of their components parallel to the middle force is one of 10 lbs. 13. ABGD is a square and forces acting at a point are represented in magnitude and direction by AB, 2BC, WD and SDA, what line represents their resultant? 14. Six forces act at a point parallel to the sides of a regular hexagon taken in order, the forces being of 3, 4, 6, 8, 10, and 11 lbs. ; find their components parallel and perpendicular to the first force and show that their resultant is 11 lbs. 15. Forces P, Q, R, S acting at a point are represented in direction by the sides AB, BC, CD, DA of a square taken in order. Find the magnitude of their resultant. 16. Forces P, Q, R act at a point and are parallel to the sides of an equilateral triangle taken in order. Find the magnitude of their resultant. • 17. The angles between the direetions of three forces in equilibrium are 120°, 150°, 90°, find the ratios of the forces. 18. Three forces respectively equal to 10 lbs. wt., 10 lbs. wt. and 10 V2 lbs. wt. are in equilibrium, find the angles between their directions; 19. Forces of 3 and 3 ^3 lbs. wt. act at a point, the angle between them being 150°, find their resultant force and its inclinations to them. 20. Two forces of 3 lbs. and 4 lbs. wt. respectively, act at an angle of 60°. Find the sines of the angles their resultant makes with them. 21. A heavy particle is held at rest by means of two strings attached to it, one of which is horizontal. If the tension of one string is double that of the other, find the inclination to the vertical of the string which is not horizontal. COMPOSITION AND EESOLUTION OF FORCES. 79 82. Motion down an inclined plane. As an example of the resolution of forces take the case of a body which falls down a smooth inclined plane. The forces acting on it are its weight W which is vertical and the reaction of the plane H which is per- pendicular to the plane. TF may be resolved into two components at right angles to each other, TTcos a perpendicular to the plane, TFsin a along the plane. ■^"** ^^" The force Wcosa is balanced by M, hence TF" cos a = R, because there is no motion perpendicular to the plane, there remains a force W sin a or mg sin a along the plane. The acceleration along the plane is therefore — or g sin a, Art. 54. If t) is the velocity gained by falling down the plane a distance s, v^ = 2gsm a.s Art. 19 = 2gh, where h = s. sin a = vertical height fallen through. Hence the velocity gained by falling a vertical height h is always the same, whatever the inclination a of the plane may be. Ex. Two bodies, one on each face of a double inclined plane, are connected by an inextensible string, which passes over the vertex ; to find the motion. «\7 ""^ A' ^H^ FiQ. 41a. Let T be the tension of the string and P and Q the weights of the bodies, /the acceleration of each body along the plane. 80 THE ELEMENTS OF APPLIED MATHEMATICS. Suppose P to be the body which descends and Q that which ascends ; the forces on the bodies along the planes in the direction of motion are Therefore Psina-Tand T-Qsva.^. ~ f=Psma-T, 9 ^ f=T- Q sin fi; hence P+Q f—Psina-Qsin^, ' J. Psina-$sinj3 f~' P+Q ^• 83. Motion down chords of a vertical circle. Let a body whose mass is m slide down a chord AP of a circle in a vertical plane, starting from A the highest point. Let the angle which AP makes with the vertical be d. The force acting on the body in the direction of J. P is mg cos 6. The body's acceleration along AP is therefore g cos 6. Hence if the body starts from rest at A and takes a time t to reach P — AP = y cos e . t\ AP = AB cose, AB cos 9 = ^gcosd . t^, 2AB J , /2AB — , and < = * / • 9 V 9 time taken to reach P does not depend But or f=- Thus the upon 6 and is therefore the same for any other chord drawn through A. We thus see that the time taken by a body to slide down any chord of a vertical circle starting from the highest point is the same, and equal to that taken to fall freely through the distance AB, the diameter of the circle. COMPOSITION AND RESOLUTION OF FORCES. 81 84. Change of Units. We have hitherto usually taken a foot, a second and the mass of a lb. as units of length, time and mass respectively. It is often necessary to change from one system of units to another. The examples which follow will show how the change is effected. Ex. 1. Express a velocity of 10 miles an hour in terms of foot- second units. 10 miles contain 10 x 1760 x 3 feet, or 52800 feet, and one hour contains 3600 seconds, hence a velocity of 10 miles an hour is a velocity of 52800 feet in 3600 sees., or a velocity of ^^^^ feet in one second, that is, the given velocity is one of ^^=^-^ feet per second. Ex. 2. The measure of a velocity is 7 when 4 feet and 6 seconds are the units of length and time, find its measure when one foot and one second are imits. The velocity 7 here means, 7 times the velocity of a body moving 4 feet in 6 seconds, or the velocity of a body which moves 28 feet in 6 seconds, or ^ feet in one second. The required measure is therefore ^, or the given velocity contains ^ foot-sees. Ex. 3. A certain acceleration has as its measure 11, when 4 feet and 6 seconds are taken as units, find its measure in foot -second units. An acceleration 11 here means that a velocity whose measure is 11 is added every 6 seconds. Now a velocity whose measure is 11 when 4 feet and 6 seconds are units has, as the last example shows, 11 x f as its measure in foot- second units, or contains ^ foot-sees. Thus every 6 seconds ^ foot-sees, are added, or, every second ^ foot-sees, are added. Hence ||, or J^, is the required measure. Ex. 4. The measure of a force is 13 when the mass of 10 lbs., 5 feet, and one minute are units, find its measure in foot-lb.-second units. A force 13 here means a force which generates in a mass of 10 lbs., acceleration 13, or in a mass of one lb. an acceleration 130. 82 THE ELEMENTS OF APPLIED MATHEMATICS. By proceeding as in the last example we find that the measure of this acceleration in foot-second units is 130x5 13 or — 3600 ' 72 13 The required measure is therefore =^ . Ex. 5. The measure of a certain velocity is 8 when 5 feet and 13 seconds are taken as units, find its measure when 3 feet and 7 seconds are the units. Let a; be the required measure, by Ex. 2 we find that this velocity contains -=- foot-sees. Also a velocity which is 8, when referred to 5 feet and 13 seconds, contains 1^ , or ^ foot-sees. lo io But this number of foot-sees, must be the same as the last, hence 3a; 40 280 Y = 13' °'-^=39-- EXAMPLES. XVIII. 1. Express in foot-second units a velocity of one mile per hour. 2. A velocity has 5 as its measure when 2 feet and 20 minutes are units, find its measure in "foot-second units. 3. The measure of a velocity being 10 when a yard and 15 seconds are units, find its measure when 2 feet and 3 seconds are units. 4. The measure of an acceleration is 14 when 5 feet and 10 seconds are units ; find its measure in foot-second units. 5. Eind the measure of the acceleration of gravity in centimetre- second units. 6. The measure of a force is unity when 14 lbs., a yard and one minute are units ; find its measure in foot-lb. -sec. units. 7. What is measure of the acceleration of gravity if the imit of length is one yard and the unit of time that of falling from rest down one yard ? 8. If 1 lb. weight is the unit of force, one foot and one second being units of length and time, what is the weight of the body whose mass is the unit of mass 1 COMPOSITION AND RESOLUTION OF FORCES. 83 9. The sides of a quadrilateral are 1, 3, 5, 6. Forces of 2, 6, 9 and 12 lbs. wt. respectively act on a particle parallel to the sides of the quadrilateral taken in order. Find their resultant. 10. If the component of a force P in the direction OA is equal to that of Q in the same direction, and if their components in another direction OB are also equal, prove that P is equal to Q. 11. Three forces acting at one point are in equilibrium, one of them is turned round this point through a given angle, find the direction of the resultant of the three forces. 12. A train weighing 200 tons is running at 40 miles an hour down an incline of 1 in 20, find the resistance necessary to stop it in half a mile. 13. A train ascends a gradient of 1 in 40 by its own momentum for a distance of j mile and then stops, the resistance from friction, etc. being 10 lbs. per ton and the weight of the train 250 tons, find its initial velocity. 14. A stone leaves the top of a tower 320 feet high with the velocity acquired by sliding down an inchned plane (of inclination 30°) for a distance of 32 feet. Show that it strikes the ground about 111 feet from the foot of the tower. 15. Find what force a horse has to exert to prevent a railway truck weighing 5 tons from descending a smooth incline of 1 in 300. 16. If the resultant of two forces PA, PB pass through a point Q the resultant of the forces QA, QB will pass through P. VI. If the greatest possible resultant of two forces P and Q is m times the least possible, their inclination when their resultant is ^ their sum is a, where cos a= - . ^ — =t • 18. A weight is supported by two strings fastened to two points in the same horizontal line, the strings being equally strong, but one rather longer than the other. If more weights be continually added to the first one which string will break first ? 19. The resultant of two forces of 12 lbs. and 5 lbs. weight is 13 lbs. weight, what will the resultant be if the forces receive an increase in magnitvde of 3 lbs. weight ? 20. At what angle must the forces A+B and A-B act, in order that their resultant may be V-4^+3^ ? 21. A body P is attracted towards the points A and C the opposite extremities of the diagonal AG oi the parallelogram ABGD by forces proportional to PA and PG respectively, and is repelled from B and D by forces proportional to PB and PD. Show that it is in equilibrium wherever P be situated. 6—2 84 THE ELEMENTS OF APPLIED MATHEMATICS. 22. A heavy particle of weight W is supported on an inclined plane, whose inclination to the horizon is a, by 3 forces each equal to P tending upwards and acting respectively along the plane and making angles a and 2a with it. If sin a=f show that F is equal to ^ W, and that the pressure on the plane is ^ W. 23. A system of forces acting on a particle is represented by straight lines in magnitude equal and in direction parallel to straight lines drawn from the angles of a quadrilateral to the middle parts of each of the opposite sides. Prove that the forces are in equilibrium. 24. Resolve a force P acting along the diagonal of a given square into two components acting along the straight lines joining one end of that diagonal with the middle points of the opposite sides. 25. A body starting from rest down an inclined plane describes 40 feet in the third second, find the plane's inclination. 26. A weight of W lbs. is drawn from rest up a smooth inclined plane of height h and length I, by means of a string passing over a pulley at the top of the plane and supporting a weight of w lbs. hanging freely. Prove that in order that W may just reach the top of the plane, w must be detached after it has descended a distance W+w hi w h+l ' 27. A and B are two fixed points on a circle, P a point on the circumference, if two constant forces act along PA, PB, prove their resviltant is constant in magnitude and passes through a fixed point. CHAPTER VI. PARALLEL FORCES. MOMENTS. 85. The motion of the bodies we have been considering so far has been merely one of translation, i.e. one in -which all the points of the body move in parallel lines with the same velocity. But when a force acts on a body it usually produces rotation. For instance a billiard-ball on a smooth table if struck at a point begins to rotate as well as to move along the table. A blow applied to a smooth cube lying on a table renders this rotational movement still more obvious. Thus the velocities of the different points of the body are not the same either in direction or magnitude. In the case of a particle or very small body the motion of translation is the only one that need be considered. Take the case of a body to which a single force P is applied. Owing to the cohesion of the particles of the body, each particle is acted upon by forces due to its connexion with the other particles, such forces are usually called internal forces. For each particle the Second Law of Motion holds, viz. that its mass-acceleration in any direction is equal to the force acting upon it in that direction. Hence by addition we see that the total mass- acceleration of the body in any given direction is equal to the sum of all the forces acting on its particles in that direction. ~ Now it is easy to see that the internal forces destroy each other, for if A and B be two particles of the body, the force on A due to B is equal and opposite to the force on B due to A, by the Third Law of Motion, thus the internal forces taken together destroy each other in pairs. We are therefore left with the force P, hence it follows that the nMss-accderation of the body in any direction is equal to the 86 THE ELEMENTS OF APPLIED MATHEMATICS. component of P in that direction. If several forces Q, R, . in addition to P, we have, in like manner, that are applied the mass-acceleration of the body in any direction is equal to the sum of the components of P, §, iJ ... in that direction. When a force acts at a point of a body the line through the point in the direction of the force, produced both ways, is called the line of action of the force. Thus in the figure the force represented by LM acting on the flat body of the shape represented in the figure at M, the line AB is called the line of action of the force LM. A L 4 Fig. 43. We shall show that the force LM produces the same effect at whatever point in its line of action it acts, provided the point is in the body. 86. Transmissibility of Force. The body being acted on by a force P at the point C, let us apply a force at any point B in the line of action of P so as to keep B fixed. That being done, the force P cannot turn the body about B, hence the body will not move, since B is fixed. Thus the body has no acceleration and therefore the force at B must,_by what was shown in the last article, be equal and opposite to the force P. Thus we see that two equal and opposite forces whose lines of action are the same produce exactly opposite effects on a body. Pio. 44. Pig. 45. PARALLEL FORCES. MOMENTS. 87 Now let a force P act at M, then at any point N iii its line of action we can suppose to act two equal and opposite forces each of magnitude P, for two such forces produce no eflfect. By what has just been shown the force at M and one of the forces at N destroy each other, and there remains a force P acting at N. Thus we can replace a force by an equal force acting at any point in its line of action. This fact is called the Principle of the Transmissibility of Force. 87. The Moment of a Force. If a perpendicular be drawn from a point upon the line of action of a force, the product of the length of the perpendicular and the magnitude of the force is called the moment of the force about the point. For instance, let AB be the line of action of a force whose Fig. 46. measure is F, from a point draw OH perpendicular to AB. The moment of F about is Fp, where p is the measure of OH; e.g. if ^ is 4 poundals, and OH is 3 feet the moment is 12. 88. The perpendicular vanishes if the line of action of the force passes through the point about which the moment is taken. Hence the moment of a force about any point on its line of action is zero. Conversely, when the moment of a force about any point is zero, its line of action passes through that point. It will be shown in Art. 93 that when a body under the 88 THE ELEMENTS OP APPLIED MATHEMATICS. action of a force turns round a point the moment of the force with regard to that point measures the power of the force to produce rotation. The unit moment is that of one poundal about a point one foot distant from it, The word moment signifies power or importance, it has nothing to do with momentiim. 89. Sign of the moment of a force. The moment of a force about a point is said to be positive when the force tends to turn the body in the direction opposite to that of the hands of a watch, and negaHve when in the same direction as the hands of a watch. When there are several forces the sum of their moments about any point is, of course, their algebraic sum. Ex. 1. ABCD is a square whose side is 2 feet long ; find the moments about B and C of the following forces ; 4 lbs. along AB, 9 lbs. along CB, 2 lbs. along DA, and 20 lbs. along JDC. The moment about B of the force along The moment about C of the force along fAB is zero, CB is zero, DC is 2 X 20= 40 units of moment, DAia-2x 2= - 4 1AB is -2x4=-8 units of moment, OB is zero, DC is zero, DA is -2x2=-4 Ex. 2. In the above square along the hnes CB, BA, DA, DB forces act respectively equal to 4, 3, 2 and 5 lbs. ; find the algebraic sum of the moments of the forces about C. \CB is zero, The moment about C of the JBA is 6, force along ]DA is -4, [DB is -5^2; hence the algebraic sum is - 5-05 units of moment, nearly. PARALLEL FORCES. MOMENTS. 89 90. Geometrical representation of t)ie moment of a force. We have seen that if LM represents a force and OH is O the perpendicular from on the line of action of LM, then the moment of this force is measured by OH X LM, but this product measures twice the area of the triangle OLM; hence the moment of a force about a point is measured by twice the area of the triangle formed by joining the point to the extremities of the line representing the force. Again, twice the area of the triangle OLM is equal to LO X NM, and NM is the component of LM perpendicular to LO, Art. 76. 91. The moment of resultant force equals sum of moments of component forces. Let AP and AQ be two forces having a resultant AR, we O R O R P have to prove that the sum of the moments oi AP and AQ about any point is equal to the moment of AR about 0. 90 THE ELEMENTS OF APPLIED MATHEMATICS. Join AO^ and draw AMN8 perpendicular to AO, draw also PM, QN, and R8 parallel to AO. The moment oi AP about is = AOx component of AP perpendicular to AO, Art. 90 = AOx AM. Similarly, the moment of J. Q about 0^ = AOxAN, AB =A0xA8, =AOx{AN+AM). See p. 68. =sum of moments of^P and J.Q. Notice that in the second figure the moment of J. P is -AO.AM. 92. It follows in the same manner as the foregoing, Art. 81, (i), that when any number of forces act at a point, the moment of the resultant is equal to the sum of the moments of the component forces. 93. The rotatory power of a force depends on its moment. Take a body moveable about a fixed point 0, and let it be acted on by two forces P and Q whose moments about are equal and opposite. It follows from what we have just seen that the moment of the resultant of P and Q is zero. Hence this resultant passes through 0, Art. 88, and there- fore produces no rotation about 0. Thus the tendencies of P and Q to produce rotation are equal and opposite, and hence forces of equal moment have equal rotatory powers. 94. Moments of any number of forces. When any number of forces in one plane have a resultant, the algebraic sum of their moments about any point is equal to the moment of this resultant about 0. PARALLEL FORCES. MOMENTS. 91 Let the forces be P, Q, R, ... , their final resultant is obtained by replacing, P and Q by their resultant jRj, i^iandi? R^, and so on until only one force is left. Now the moment aboutO oiR^ =algeb.sumof the moments of P and Q, Art. 91, Ri = algeb.sumof the moments of Ri and R, = algeb. sum of the moments of P, Q, and R, and so on. Hence the moment of the final resultant = algeb. sum of the moments of all the forces P, Q, R, ... . 95. Composition and Resolution of parallel forces. We have shown how to find the resultant of any number of forces in one plane which act at a point. If the forces are applied at different points of a body their resultant may be found, as in the last Article, by finding the resultant of any two intersecting forces, then the resultant of this and a third force, and so on. In the case of parallel forces, which meet at an infinite distance, we adopt a special method for finding the re- sultant. Like parallel forces are those which act in the same direction; unlike parallel forces are those which act in opposite directions. 96. Resultant of two like parallel forces. Let P and Q be two parallel forces; we may suppose them to act at A and B respectively ; at A and B introduce two forces F equal and opposite, this will make no difference in the resultant of P and Q, see Art. 86. 92 THE ELEMENTS OF APPLIED MATHEMATICS. The lines of action of the resultants of F and P, and of F and Q will meet at some point 0, and we may suppose these resultants to act at 0. Art. 86. Through draw OG parallel to AP and BQ, cutting AB mG. The force acting at along OA may be replaced by its components, F parallel to GA, and P acting along OG. Similarly the force acting at along OB may be resolved into two : viz. # parallel to GB, and Q acting along OG. The two forces F acting at balance each other and may be removed. We are left with the single force P + Q acting along OG. To determine the position of G. The sides of the triangle OGA are parallel to P, F and their resultant, hence by the converse of the Triangle of Forces, OG P GA~ P- Similarly OG Q GB~F- Hence P.GA = Q.OB. That is. AB is divided in the inverse ratio of the forces. PARALLEL FOECES. MOMENTS. 97. Resultant of two unlike parallel forces. Fia. 51. In this case P and Q are unlike parallel forces of which Q is the greater. Suppose them to act at any points A and B on their lines of action. At A and £ apply equal and opposite forces F. The resultants of F and P and of F and Q will meet in some point which is to the right of Q, since, Q being greater than P, the resultant of Q and F is more bent to- wards Q than the resultant of P and F is bent towards P. Through draw OQ parallel to AP and BQ and meeting AB in G. The force acting at along OA may be replaced by F parallel to GB, and P acting along 0(?. Similarly the force acting at along OB may be replaced by F parallel to BG, and Q acting along GO. The forces F balance each other, and we are left with Q — P acting along GO. To determine the position of G. The sides of the triangle OGA are parallel to P, F and their resultant, heiice by the converse of the Triangle of Forces, OG _P AG~F- «. .,1 OG Q Similarly . ^ = -p • 94 THE ELEMENTS OF APPLIED MATHEMATICS. Hence P.AQ = Q.BG. That is, AB is divided externally in the inverse ratio of the forces. 98. Recapitulation. When two parallel forces P and Q act at points A and J5 of a body ; 1. The magnitude of the resultant is the sum or differ- ence of P and Q according as they are like or unlike. 2. Its direction is that of each force when they are like, that of the greater when they are unlike. 3. It acts at a point G in AB, or in AB produced, such that P.AG = Q.BG. Got. 1. The resultant of any number of parallel forces may be found by finding the resultant of any two of the forces, then the resultant of this and a third force, and so on. The magnitude of this final resultant is the algebraic sum of the forces. Cor. 2. The magnitude of the resultant of two equal and unlike parallel forces is zero, and it acts at an infinite distance. This case will be considered in Chapter vii. Ex. 1. Two like forces of 12 and 20 lbs. weight act at points 4 inches apart, find their resultant. The resiiltant is a force of 32 lbs. weight. Let X be the distance AG, then BG=A - x and we have .'. 32is;=80, or a; =2| inches. Ex. 2. Two unhke forces of 48 and 72 lbs. weight act at points 14 feet apart, find their resultant. The resultant is a force of 24 lbs. weight. Let x feet be the distance BO, then AG=x+ 14, and AS{x: + 14:) = nx, ^=1^8^28 feet. 24 PARALLEL FORCES. MOMENTS. 95 Ex. 3. Kesolve a force of 30 lbs. weight into two like forces 6 feet apart, one of them being 9 inches from the given force. Let the forces be P and Q, then P+Q=m lbs. weight, AG=Q,BG^QZ, also 9P=63§, from which it follows that P=26i lbs., Q=^ lbs. Ex. 4. Four forces F, ^F, ZF, AF act along the sides of a square taken m order, find their resultant. Let a be the length of a side. The resultant of F and 3F is '2.F actine at a pomt E in BC produced such that ZF.CE=F{a+GE), or G^=%- Similarly the resxiltant of 'i.F and ^F is 2i?' acting at a point K in CB pro- duced such that '^F.BK=%F{a+BK), or BK=a. These two forces intersect in L, their resultant is 2 ^%F and acts along LM, the figure LKMN being a square. M K A D -^B JF 2F C Fig. 52. EXAMPLES. XIX. 1. Two parallel forces of 15 and 20 lbs. weight act at points 20 inches apart, find their restiltant when they are (i) like, (ii) unlike. 2. The resultant of two like forces is 12 lbs. and it acts at a distance of 2 inches from the larger component which is 8 lbs., find its distance from the smaller component. 3. The resultant of two unlike forces is 15 lbs. and it acts at distances of 2 feet and 6 feet respectively from the forces, find the forces. 4. Two men carry a weight of 160 lbs. between them on a pole, the weight being three times as far from one man as from the other; find how much weight each supports, the weight of the pole being dis- regarded. 96 THE ELEMENTS OF APPLIED MATHEMATICS. 5. A uniform rod 12 feet long and weighing 18 lbs., can turn freely about a point in its length ; the rod is at rest when a weight of 8 lbs. is hung at one end. How far from the end is the point about which the rod can turn ? [The weight of the rod acts at its middle point.] 6. A bridge girder rests on two stone piers, the weight of the girder being a tons, and one of the piers being only capable of supporting 6 tons, at what distances must each pier be from the centre of the girder in order that the stronger pier may support as small a weight as possible ? 1. A man carries a bundle at the end of a stick which is placed over his shoulder, if the distance between his hand and his shoulder be changed, how does the pressure on his shoulder change 1 8. A uniform rod whose weight is 8 lbs. is placed upon two props which are in the same horizontal line and 6 inches apart. Find the distance to which the ends of the rod extend beyond the props, if the difference of the pressures on the props is 4 lbs., and the length of the rod 3 feet. 99. Moment of the resultant of parallel forces. PRO Fig. 53. From any point in the plane of two like parallel forces P and Q draw OAOB perpendicular to the lines of action of the forces. Now, sum of the moments of P and Q about = P.OA + Q.OB = P (00 -AQ)+ Q(OG + OB) = {P + Q)OG, since P.AG = Q.QB, Arts. 96, 97, = moment of the resultant about 0. PARALLEL FORCES. MOMENTS, 97 If the point about which moments are taken is between the forces, we have sum of moments of P and Q = Q.OB-P.OA = Q{00 + GB)-P{AG-OCf) = {P + Q)00 = moment of resultant. PRO Pia. 54. In precisely similar manner we show that when the parallel forces are unlike, the moment of the resultant about any point equals the sum of the moments of the components. 100. Referring to Art. 94, we see that the algebraic sum of the moments of any number of forces is equal to the moment of their resultant whether the forces meet at a finite distance or are parallel. Ex. 1. Four weights of 5, 14, 6 and 25 lbs. respectively hang at distances 2, 4, 9 and 12 feet from one end of a rod without weight. Find the magnitude and the position of their resultant. B C D E 14 6 Fig. 55. ZS Let AF be the rod and B, C, D, E the points at which the weights are attached. Since the forces are like and parallel their resultant is their sum 50 lbs. Let 3s be the distance of the resultant from A, then since moment of resultant = sum of moments of components, 50. a!=5x 2 + 14x4+6x9+25x12; .-. ^=^ = 8|feet. Ex. 2. A rod 14 feet long without weight has a weight of 4 lbs. suspended from its middle point. The rod can turn about one end. If the rod is to be sustained by a force at one end of 11 lbs. weight, where must an additional weight of 63 lbs. be attached, in order that the rod may remain at rest? J. 7 98 THE ELEMENTS OF APPLIED MATHEMATICS. The system of forces (including the force at the jointed end) must be in equilibrium, or there is no resultant force. Hence the algebraical sum of the moments about any point vanishes. Take moments about the jointed end. Then the force at that point has, of course, no moment and we have, if x is the distance from that end of the force of 63 lbs., 63xi!;+4x7-llxl4=0; .■. x=% feet. Ex. 3. A system of forces in one plane being represented in magni- tude and position by the sides of a closed polygon taken in order, show- that the sum of their moments with regard to any point in the plane= is constant. (i) Let the point be inside the polygon, then denoting the sides by a, 6, c, &c. and the perpendiculars from hy p, q, r ... the sum of the moments is pa + qb+rc+ , which is twice the area of the polygon. (ii) Let the point be outside the polygon. Here one of the forces has a moment about _ opposite in sign to the moments of all the 6 , other forces, and V \ the sum of the moments =pa — qb+rc + .., / \ = twice area of polygon. Fig. 56. EXAMPLES. XX. 1. The resultant of two unlike parallel forces is 6 lbs., and one acts 10 inches from the greater force which is 10 lbs., find the distance between the forces. 2. A uniform bar 12 feet long and weighing 16 lbs. is supported at each end and a weight of 48 lbs. is hung at a point 2 feet from one end ; find the pressure on each of the supports. 3. Four parallel forces 1, 6, 9, 8 act at points 4 inches apart along a weightless rod ; where must the rod be supported that it may remain in equilibrium? 4. A uniform iron rod 6 feet long weighs 9 lbs., and from its extremities weights of 6 lbs. and 12 lbs. respectively are suspended. From what point must the rod be supported in order that it may remain balanced in a horizontal position ? [The weight of the rod acts at its middle point.] PARALLEL FORCES. MOMENTS. 99 5. A heavy uniform beam, whose mass is 50 lbs., is suspended in a horizontal position by two vertical strings each of which can sustain a tension of 35 lbs. without breaking. Where must a mass of 20 lbs. be placed so that one of the strings may just break? 6. A weightless rod has equal weights attached to it, one at 15 inches from one end and the other at 9 inches from the other ; it is supported by two vertical strings attached to its ends, if each string cannot support a tension greater than the weight of 50 lbs., find the greatest magnitude of the equal weights. 101. Centre of parallel forces. When any number of parallel forces act at fixed points of a body, we shall prove that there is a certain point, called the centre of the parallel forces, at which the resultant always acts, however the forces are turned round their points of application, provided they remain parallel to each other and of the same magnitude. Fia. 57. Let the parallel forces P, Q, B ... act at points A, B, C... Join AB and divide it at g^ so that P.Ag, = Q.Bg,. Then g^ is a point at which the resultant of P and Q acts. Art. 96. Now gi has been found independently of the direction of P and Q, hence it is the same whatever their direction. Again join g^ and C and divide it at g^ so that {P + Q)g^.=R.Og,. Then g^ is a point at which the resultant of P + Q and R acts. 7—2 100 THE ELEMENTS OF APPLIED MATHEMATICS. That is a point through which the resultant of P, Q and R passes. Moreover we see that its position does not depend on the direction of the parallel forces. Hence the resultant of P, Q and R passes through g^, however these forces are turned round their points 0/ appli- cation, provided they still remain parallel to each other. Thus g^ is the centre of the parallel forces P, Q and R. The centre of any number of parallel forces may be found by continuing this process. Notice that the parallel forces are not restricted to lie in one plane. 102. Distance of the centre f):om any line. To find the distance of the centre of any number of parallel forces P,Q, R ... from any line LM. Since the position of the centre does not depend on the direction of the forces its position remains unaltered if we suppose the forces turned round so as to become parallel to LM. From A,B, G ... draw perpendiculars p, q, r ... upon LM. B 1' "^ A 9 X P r L V Pio. 58. Let K be the required centre through which the re- sultant P ■\^Q-\-R ... passes, and let x be its distance from LM. Then V being any point in LM, we have moment of P + Q + R ... acting at K about V= sum of moments of P, Q, ... , Art. 100; .-. {P + Q + R+...)x = Pp + Qq + Rr.... PARALLEL FORCES. MOMENTS. 101 Hence, for instance, if there are only three forces _ Pp + Qq + Br • "^ F + Q + E ■ Ex. 1. Equal weights hang from the comers of a weightless triangle, find the point in the triangle at which it must be fastened in order to lie horizontally. The resultant of any two of the forces acts midway between them, denoting each weight by P this resultant is 2P. Then if D is the middle point of BC, the resultant of 2P at I) and P at jl is 3P at O, where, OD being x and AJ) being I, '2,P.x=P{l-!t;), or 30= I Fia. 59.. Ex. 2. A beam, the weight of which is equivalent to a force of 10 lbs. acting at its middle point, is supported on two props at its ends. If the length of the beam be five feet where must a weight of 30 lbs. be placed so that the pressure on the props may be 15 lbs. and 25 lbs. respectively? The 30 lbs. weight is to be the resultant of the 10 lbs. weight and the upward pressures, hence if x be the distance of the 30 lbs. weight fi'om the prop whose pressure is 15 lbs. 15x0 + 25x5-10x1 ,, „ ^ X = — ? = 3 J feet. Ex. 3. Weights of 5, 6, 9 and 7 lbs. hang from the corners of a horizontal square whose side is 27 inches long. Find the point where a single vertical force must be applied to the square to balance the efiects of the forces at the corners. We want to find the centre of the four parallel forces; an upward force of 27 lbs. applied there will main- tain equiUbrium. The position of the centre is not altered if we turn the forces about their points of application till they have the position given in the figure. The resviltant still passes through the centre of parallel forces ; but its direction is the line KU, where n.BE=16.CE, hence since BC=i1 inches, BE=16 inches. D 7 F B.6 C 9 Fia. 60. 102 THE ELEMENTS OF APPLIED MATHEMATICS. Similarly if the forces be turned till two of them lie upon DA and two upon CB their resultant acts along KF, and we find BF to be 1^ inches. Thus K, the centre of parallel forces, is 16 and 12 inches respectively from two sides of the square. 103. Moments about a line. In Art. 87 we called the product of a force and its distance from a point in its plane the moment of the force about the point. It is also the moment of the force about a line through perpendicular to the plane of the paper. In Art. 102 we found the relation Pp-\-qq + Rr+... = {P + Q + B+...)x; where x is the distance of the centre of the parallel forces from the line LM. Now turn all the parallel forces round their points of application till they are all perpendicular to the plane of the paper, then Pp, Qq &c. are the moments of the forces P,Q,.. . about the line LM; hence we see that the sum of the moments of the component forces P, Q ... is equal to the moment of their resultant about LM. EXAMPLES. XXI. 1. A man and a boy have to carry a load of 100 lbs. slung on a pole (whose weight may be neglected) carried horizontally and 10 feet long. Their carrying powers are in the ratio of 8 : 5. Where in the pole should the weight be hung so that it may be fairly divided? 2. A rod of uniform thickness has half its length composed of one metal and the other half of another metal. The rod will balance about a point distant J of its whole length from one extremity. Compare the weights of equal volumes of the two metals. " 3. A uniform rod which is 12 feet long and which weighs 17 lbs. can turn freely about a point in its length, and the rod is in equilibrium when a weight of 7 lbs. is hung at one end. How far from the ends is the point about which it can turn ? 4 Three like parallel forces acting at the angular points A, B, C of a plane triangle are respectively proportional to the opposite sides a, h, c. Find the distance of the centre of parallel forces from the side BC. PARALLEL FORCES. MOMENTS. 103 5. If the sum of the moments of a system of forces about a point A is zero and also about a point B, show that it is also zero about any point in AB. 6. Show that the difference of the moments of a force P about two points A and B in its plane equals the moment about either point of a force equal to P acting at the other point. 7. ABCB is a rectangle, AB, BC adjacent sides are three and four feet long respectively. Along AB, BG, CB taken in order forces of 30, 40, 30 lbs. act respectively, find their resultant. 8. If any three forces act along the sides of a triangle taken in order, prove that their resultant cannot meet the triangle. 9. The lines of action of two forces P and Q and their resultant R are cut by a third line in the points A, B and respectively, and P, Q •are each resolved into two forces, one parallel to AB and the other parallel to R. Prove that the components parallel to R are to each other as BC : AC. 10. A uniform beam 4 feet long is supported in a horizontal position by two props which are three feet apart, so that the beam projects one foot beyond one of the props ; show that the pressure on one prop is double the pressure on the other. 11. The sides BC, CA, AB of a triangle are three, four and five feet long respectively; find the magnitude and direction of a force acting at C whose moments about A and B are 7 and 5 respectively. 12. The magnitude of a force is known and also its moments about two given points A and B. Find by a geometrical construction its line of action. 13. A triangle ABC can turn freely in its own plane about the centre of its inscribed circle which is fixed, and forces proportional to y-z, z-x, x-y act along the sides BC, CA and AB respectively. Show that the triangle remains at rest. 14. Forces P, Q, R act along the sides BC, CA, AB of a triangle. Show that their resultant will act along the line joining the centre of the circumscribing circle to the intersection of perpendiculars if _ Tf_'^^ B cos G cos C cos A cos A _ cos^ ■ ^ ' ~cobG cosB ' coaA cosC ' cosB coaA' 15. Four forces acting along the sides AB, BC, CD and DA of the quadrilateral A BCD are in equilibrium ; having given that the first acts from A towards B, find the directions of each of the other three. CHAPTER VII. COUPLES. 104. We have already seen that if F and Q are two unlike parallel forces their resultant is equal to P ~ Q. When P is equal to Q the two forces are said to form a " couple.' A couple therefore consists of two equal unlike parallel forces. The perpendicular distance between the lines of action of the forces is called the " arm " of the couple. 105. Moment of a couple. o' Pio. 62. The forces P, P, whose arm is a, form a couple; it is required to find the sum of the moments of these forces about any point 0. COUPLES. lO^ Through draw OAB perpendicular to the forces. The sum of the moments equals P . OB — P . OA, the moments having opposite signs, = P{OB-OA) = P.AB = P.a. Again, if we take moments about any point 0', within the forces, the sum of the moments equals P.O'B + P.O'A, = P(0'B+0'A) = P.AB = P.a. Hence in all cases the moment of a couple about any point in its plane is equal to the product of one of the forces, and the arm. 106. Sign of a couple. If a body on which the couple acts were pivoted about- either the point or the point 0', the rotation would be in the direction of the hands of a watch. This is called the Tiegative direction of rotation, see Art. 89, the contra-clock- wise direction being called positive. When the directions of the rotations produced by twa couples are the same the couples are said to be like. 107. Axis of a couple. The axis of a couple is a line drawn through any point perpendicular to the plane of the couple of such magnitude as to indicate the magnitude of the couple. The direction of rotation produced by the couple, or its sign, is indicated by the sign of its axis, which is determined as follows : Place a watch on the plane of the couple face upwards ;, if the direction of rotation is contra-clockwise the axis is drawn upwards and is considered positive, if clockwise the axis is downwards and considered negative. 106 THE ELEMENTS OF APPLIED MATHEMATICS. 108. Couples with equal and parallel axes of the same sign are equivalent. We shall now show that two couples in the same plane having equal moments, and therefore parallel and equal axes, are equivalent, that is, we may replace a couple by any other couple in its plane having the same moment. First, take two couples of equal and opposite moment, the forces being all parallel. A ,P B ' C Q Fig. 63. Draw a line ABGB cutting the lines of action perpen- dicularly; then if AB = a , CD = b, and P, Q the forces of the respective couples, we are given that Pa = Qb. Observe that the moment of the upper couple is positive, that of the lower negative. The resultant of the upper force P and the lower force ^ is P + Q acting at a point which is such that Pia + BG) = Q(b + CG), Art. 96, or P.BG = Q.GO. But the resultant of the two middle forces is P + Q in the opposite direction and acts at the same point G. Hence all the forces are in equilibrium since they can be replaced by two equal and opposite forces. Thus the Q-couple destroys the efifect of the P-couple, hence if the forces of the Q-couple were reversed they would be equivalent to the P-couple. COUPLES. 107 Second, if the forces of the couples are not all parallel, but form a parallelogram. Fig. 64. If a and b are the distances between the pairs of parallels, then since the moments are equal, Pa = Qb, but AB.a = AD . b, since each is the area of the parallelogram A BCD, hence P _AB Q~AD- Thus the sides of the parallelogram represent the forces P and Q in magnitude and direction, hence by Art. 72, the resultant of the forces which meet at J. is represented by GA , G AG, and these being equal and opposite forces the four forces are in equilibrium. Thus if the forces of the Q-couple were reversed they would be equivalent to the P-couple. 109. The forces of a couple may be transferred to a parallel plane without altering their effect, Take CD equal and parallel to AB and draw through it a plane parallel to that of the couple ; at both G and D we may suppose equal and opposite forces of magnitude P to act in this plane. 108 THE ELEMENTS OF APPLIEB MATHEMATICS. Since AB and CD are equal and parallel they are FiQ. 65. opposite sides of a parallelogram, hence AD and BG bisect each other at 0. The upward forces P at 5 and G have a resultant 2P acting upwards at 0, downward forces P at ^ and D have a resultant 2P acting downwards at 0. These forces 2P destroy each other and we are left with a downward force P at C and an upward force P at D, form- ing a couple exactly equal to the original couple. 110. Resultant of couples in the same plane. Let P, Q, B, ... be the forces of any number of couples, their arms being p, q, r, ... respectively. The moments of the respective couples are Pp, Qq, Br, &c. We may, by Art. 108, replace these couples by couples whose forces are Pp Qq Br L ' L' L' •• and whose arm is, in each case, L. We may also, by the same Article, move the couples in their plane till their arms come to coincide, we have then one couple of which the force is L^ L^ L^ ••■' COUPLES. 109 and whose arm is L, that is, its moment is the algebraic sum of the moments of the original couples. Ex. 1. Forces P, 2P, 4P, 2P act along the sides of a square ABCB taken in order, find the magnitude and position of the resultant. The forces along BG and BA form a couple, this couple may be turned through a right angle, we then have forces 3P and 6P acting along AB and CD respectively ; their resultant is SP outside the side CD and distant from- it a side of the square. Ex. 2. Along the sides AB, CD of a square there act equal forces of 3 lbs. weight, along the sides AB and CB there act equal forces of 7 lbs. weight, find the moment of the resultant couple, a side of the square being 4 feet in length. The moment of the couple formed by the last two forces is 28, first -12; hence the moment of the resultant couple is 16. Ex. 3. ABCB is a square whose side is 3 feet, along AB, BC, CB, BA forces act equal to 1, 3, 11, 7 lbs. weight respectively, and along AC, BB forces equal to 7 a/2 and 3 v/2 lbs. weight, find their resultant. At the point B introduce forces of 3 lbs. weight acting along AB and BA. 'Bj the Triangle of Forces, we may remove the forces acting along BB, BC, BA since they are in equilibrium. There are left the forces along AC, CB, BA together with a force of 4 lbs. along AB. At the point A introduce forces of 7 lbs. weight acting along AB and BA. As before, the forces acting along AC, BA, BA may be removed. There are left forces of 11 lbs. weight acting along AB and CB, these form a couple whose moment is 33 units of moment. Fia. 67. 110 THE ELEMENTS OF APPLIED MATHEMATICS. 111. A Force may be replaced by a force and a couple. Take any point 0, then at we may suppose two equal and opposite forces to act which are equal in magnitude to a given force P. This given force together with one of the forces at forms a couple, and there is left a force equal to P passing through 0. Hence we have as equi- valent to the original force P (i) a force equal to P in magni- tude and direction passing through 0, (ii) a couple whose moment is equal to Pa, where a is the perpendicular distance of the original force from 0. Conversely, we may replace a force and a couple by a single force. For if the moment of the couple be Q and the force P, we may take the forces of the couple equal in magnitude to P, the arm being then ^ . Now let the forces of the couple be moved till one of them acts in the same line but opposite direction to the given force P thus balancing it, there is left one force equal to P and distant ^ from the given force. Cor. A single force and a couple cannot produce equili- brium. 112. Any number of forces in one plane reduce to a force or to a couple. Take any point 0, then as in the last Article, any force P may be replaced by an equal force through together with a couple. Doing this for all the given forces we have COUPLES. Ill (i) a number of forces passing through 0, which have a resultant force, (ii) a number of couples which may be replaced by a re- sultant couple, Art. 110. If the resultant in (i) does not vanish we are left with a force (through 0) and a couple which as we have seen reduces to a single force. If the resultant in (i) vanishes we are left with a couple. CHAPTER VIII. CENTRE OF GRAVITY. 113. Every material body consists of an infinite number of particles, each particle being acted upon by a force, called its weight, directed to the centre of the Earth, due to the Earth's attraction. If the body be small compared with the Earth, these forces are practically parallel. By the theory of like parallel forces, Art. 96, they have a resultant parallel to them and equal to the sum of the weights of the particles. This resultant passes through the centre of the parallel forces, Art. 101, however the body be placed ; in the present case this point is called the Centre of Gravity. The centre of gravity of a body is therefore that point through which the line of action of the weight always passes, in whatever position the body may be. 114. Every body has only one centre of gravity. If possible let a body have two centres of gravity, A and B. Then we have seen that the line of action of the weight passes through both A and B for every position of the body. But this line of action is vertical and hence cannot pass through both A and B when the line AB is itself not vertical. Hence there can only be one centre of gravity. CENTRE OF GEAVITY. 113 115. Position of the centre of gravity. It has been shown that if there are any number of parallel forces P, Q, R ... acting at points whose distances from a given line in their plane are p,q,r ... , the distance of the centre of parallel forces from that line is equal to Pp + Qq+Rr+ ... P + Q + B+... ■ Now let there be any number of particles in a plane whose masses are m^, m^, m^ and whose distances from a given line in that plane are z,,, z^, z%, ... . We have therefore a series of parallel forces in^, m^g, .... The required distance is therefore TOi^f + m^ + ... or, TJh^l + Ifn^i + ■ mi + m2+ ... The distance of the centre of gravity from any other line is obtained by a similar expression. 116. Since the C.G. is the point at which the body's weight may be supposed to act, the body if fixed at that point will balance about it in every position. 117. Determination of the position of the C.Cr. hy experiment. Fig. 69. Suspend the body from any fixed point A in it. The forces acting are its weight and the force at the point of suspension. Since the body is at rest these forces must be equal and opposite. Therefore the vertical line through the C.G. must pass through A. 114 THE ELEMENTS OF APPLIED MATHEMATICS. Release the body and suspend it from any other point B. Then in this position also the C.G. must lie in BQ. Hence the C.G. will lie at 0, the intersection oi AG and BG. 118. Position of the C.G. found by inspection. If a body has a Centre of Symmetry, that point is its c.G. By a Centre of Symmetry is meant a point such that for every point P of the body we can find a point P' in PO produced so that OP = OP' ; in the case of a circle the centre is evidently a centre of symmetry. For in this case the body may be broken up into pairs , of particles of equal weight on lines passing through the centre of symmetry at equal distances from it. The centre of symmetry is then clearly the C.G. of each pair of particles and therefore of the whole body. Hence it follows that: the C.G. of a uniform rod is its middle point, circular ring or circular area is its centre, sphere is its centre, square or cube is its centre. 119. C.G. of three equal particles placed at the vertices of a triangle. Let three equal particles, each of weight W, be placed at the vertices of a triangle ABC. The C.G. of the weights at B and C is at D the middle point of BG, thus the C.G. of the A three weights must lie in the median line AD and is a point G such that /^w\q "^ 2WxBG=WxAG, OY 2DG = A0. ^f 5 ^<^ Hence the point G divides AD in 'w /^ the ratio of 2 to 1, and ^'''- ^°- AG^%AD. But we see in a precisely similar manner that the C.G. of the three weights must lie in the other two median lines CENTRE OF GRAVITY. 115 BE and GF and divide them in the ratio of 2 to 1. Hence the , lines AD, BE and GF must all intersect in the point G which is the C.G. of the three weights. 120. O.Gr. of a triangular plate. ABG is a thin triangular plate of uniform thickness. Divide the plate into strips, such as PQ, parallel to the side BG. The C.G. of each strip is at its middle point, and the middle points of all these strips lie in the line AD joining A io D the middle point of BC. JHence the C.G. of the plate lies in AD. Similarly we see that the C.G. of the plate lies in BE and GF, hence by Art. 119 it is the same as the c.G. of three equal weights placed at the vertices of the triangle. Hence the C.G. divides each median line in the ratio of 1 to 2. We have therefore proved that, (i) The C.G. of a triangular area is the same as that of 3 equal particles placed at its vertices, (ii) The C.G. of a triangular area is the point of inter- section of the median lines and is distant from each vertex # of the median, or AO = %AD. 121. C.G. of a parallelogram. A Fio. 72. Since the diagonals of a parallelogram bisect each other, we see by the foregoing that the C.G. of the triangles ABG and ADG, that is of the whole figure, lies on BD, similarly it lies on AG. Hence it is at 0, the intersection of the diagonals. 8—2 116 THE ELEMENTS OF APPLIED MATHEMATICS. EXAMPLES. XXII. 1. Show that the c. G. of a uniform rod is the same as that of equal particles at its ends. 2. Prove that a parallelogram has the same o.G. as four equal particles at its vertices. 3. Show, by dividing a parallelogram into strips parallel first to one pair of parallel sides and then to another, that its c. a. is its intersection of diagonals. 4. The c. G. of a triangle is the same as the c. g. of 3 equal particles placed at the middle points of its sides. 5. If the c. G. of a triangle coincides with the centre of the circum- scribed circle, the triangle is equilateral. 6. A triangular board is suspended by a string attached to one comer. What point in the opposite side will be in line with the string ? 7. Show that the c. g. of the circumference of a circle and that of any number of equal particles arranged at equal distances along its circumference are the same. 122. General rule for finding the C. G. Divide the body into portions whose weights and the positions of whose c.G.s are known. The weight of each portion acts at its C.G., Art. 113. The c.G. may then be found. 123. When the body can be divided into two portions of which the weights w^, w^ and the c.G.s are known, we proceed as follows : Let gi, g^ be the given centres, join g^, g^, then by the theory of parallel forces. Art. 96, the required point G divides g^gi, so that w^ . g^G = w^ . g^G. Ex. 1. Find the c.G. of two spheres of 8 oz. and 24 oz. weights, connected by a rigid rod without weight, the distance between their centres being one foot. Let a; be the distance of the c.G. from the centre of the smaller sphere, then 8a;=24(12-a;), .•. a; =9 inches. CENTRE OF GEAVITY. 117 Ex. 2. A uniform rod weighing 7 lbs. is 6 feet long ; if a 2 lb. weight be placed at one end, find the centre of gravity of the whole. Since the rod is uniform its weight acts at its middle point. Let x be the distance of the c.a. from the middle of the rod, then 7.r=2(3-^), .'. x=\ feet, or 8 inches. Ex. 3. On the same base and on opposite sides of it isosceles triangles are described whose vertices are distant 12 and 18 inches respectively from the base. Find the o.a. of the quadrilateral. Let ABCD be the quadrilateral : the line BD is perpendicular to, and bisects, the base AG. Hence g^ and g^, the c.G.'s of the two triangles, are on BD. The distance g^^ is 10 inches, and the weights of the triangles are proportional to their areas. Hence if a: be the distance of the c.G. of the quadrilateral from ^2 J^Cxl8x«=^.4Cxl2(10-a;), or ^=4 inches. 124. To find the C. Gr. of a body when a portion is removed. ff G f ^ / i w W W-7 Fig. 74. Let W be the weight of a body whose c.G. is 0, w that of a portion of it whose CO. is g, required the C.G. of the body got by removing the portion w from the body W. Let X be the distance of its C.G. g' from Q, then since W is made up of the portions w and TT— w, we have {W — w)x = w.gG, or x=gG w W- ■w 118 THE ELEMENTS OF APPLIED MATHEMATICS. Ex. To find the c.g. of the figure ABCD got by cutting the triangle COD out of the square ABCD whose side is | of an inch, being the centre. Let X be the distance of the required c.G. from 0. Then we know that is the c.a. of the whole square, and g that of COD, where Oy=f of half the side of the square = J inch. Also area of square = j^ sq. inches, area of COD=-^^ sq. inches. Hence since the weight of the square is made up of the weights of the fig. ABCOD and of the triangle COD, Pig. 75. (A-A)^=ixA> or ■x=^Ya.ch.. EXAMPLES. XXIII. 1. Find the c.g. of two small bodies whose weights are "01 oz. and •002 oz., their distance apart being 3 feet. 2. Two spheres whose radii are 10 and 11 inches respectively are in contact ; if their weights are 5 and 6 lbs., find the position of their c.g. 3. Two uniform rods are placed so that the line 1 foot 10 inches long joining their centres is perpendicular to each, if their lengths are 10 and 12 inches find the distances of their c.G. from their ends. 4. A rod 12 inches long whose weight is 20 lbs. has a body of weight 2 ounces attached to a point one inch from an end, find the c. G. of the rod and attached weight. 5. Find the c. g.s of the following bodies : (i) A square with a square portion removed, the line joining their centres being perpendicular to a side of each. The sides of the squares are 10 and 3 inches long respectively, the distance between their centres being 2 inches. (ii) A circular plate with a circular portion removed, the weights of the portions being 15 and 4 lbs. and the distance between their centres 15 inches. (iii) A square plate with a circular portion removed, the boundary of this removed portion touching a side of the square and passing through its centre, the line joining their centres being perpendicular to a side of the square, the side of the square being 10 inches long. (iv) A uniform rod with a piece ^ of its length taken out, the centre of the piece being 3 inches from the centre of the rod. CENTRE OF GRAVITY. 119 125. C.G. of weights in the same straight line. Let the body whose C,G. is required be divided into any number of portions whose c.G.s lie in the same straight line, let the weights acting at these points be w^, w^, ... and let their distances from some fixed point on the line be Xi, x^, ... , then the distance of the C.G. required from being x, we have (Wi + Wa + ...)x = WiXi + w^2 + Art. 115. Ex. 1. Two heavy particles weighing ijespectively 3 and 5 ounces are attached to the ends of a straight rod 8 inches long, weighing 2 ounces. Find the c. o. of the system. The sum of the weights is 10 ounces. Taking for the point the end to which the 5 ounces are attached 10x5;=3x8 + 2x4; .*. x = 3'2 inches. Ex. 2. A telescope consists of three tubes each 10 inches in length sliding within one another, and their weights are 8, 7 and 6 oimces. Find the position of the c. g. when the tubes are drawn out to their full length. The sum of the weights is 21 ounces, the weight of the different tubes act at points distant 5, 15 and 25 inches respectively from one end. .-. 21x^=8x5 + 7x15+6x25 = 295; hence x= 14^ inches from the thicker end. EXAMPLES. XXIV. 1. Ten 1 lb. weights are attached to points of a weightless rod distant one inch apart, find their c. g. 2. A rod is pivoted at its middle point and weights of 5 and 6 lbs. are attached to its ends, the rod being 20 inches long, where must a weight of 3 lbs. be attached in order that the rod may rest horizontally ? 3. Four triangles have the same base and the opposite vertices in the same line perpendicular to the base, the distances of these vertices from the base being 5, 6, 7 and 8 inches, find the distance of the c. g. of the four triangles from the base. 4. A uniform rod ^1.8 is 6 feet long and weighs 4 lbs. A lb. weight is attached to the rod at A, 2 lbs. at a point distant one foot from A, 3 lbs. at 2 feet from A, 4 lbs. at 3 feet from A and 5 lbs. at B. Find the distance of the c.G. of the system from A. 120 THE ELEMENTS OF APPLIED MATHEMATICS. 5. A straight rod 6 feet long and heavier towards one end is found to balance about a point 2 feet from the heavier end, but when sup- ported at its middle point it requires a weight of 3 lbs. to be hung at the lighter end in order to keep it level. What is the weight of the rod? 6. A bar of uniform thickness and 5 lbs. weight has a weight of 10 lbs. at one end and. 12 lbs. at the other, it balances about a point 4 inches from the nearer end, find its length. 126. C.Gr. found by takings moments about a line. If there are several bodies of -weights Wj, Wj, ■■• and if Xi, X.2, ... are the distances of their c.G.s from any line Ox in their plane we have, by Art. 102, if x is the distance of their c.G. from this line WiXi + W^l+ ... X — Wi + Wa + W3 + . . . Similarly if Oy be any other line, and y^,y^,... the dis- tances of the C.G.s from it, then y being the distance of the C.G. from this line - ^ Wiyi + Way, -I- Ways + • • ■ Wi-t-Ws+Ws-l-... Ex. 1. Four heavy particles whose masses are 2, 3, 4 and 5 gramme* are placed at the corners A, B, C, D oi a, horizontal square; find the c. G. of the four particles. Taking moments about AD, a being the length of N B a side, 3 G (2 4- 3 -H 4+ 5) (?Jf = (3 -1- 4) a. / Similarly, taking moments about AB, 'S ^ ■ (2 + 3-1-4 + 5) (yiV=(4+5) a, Fig. 76. ^„ . Ex. 2. Five masses of 1, 2, 3, 4 and 5 ounces weight respectively are placed on a square table. Their distances from one edge of the table are 2, 4, 6, 8 and 10 inches, from an adjacent edge 3, 5, 7, 9 and 11 inches. Find the distance of their c.G. from these two edges. One distance x is given by -^ 1x2+2x4+3x6+4x8+5x10 ''~ 1 + 2 + 3 + 4+5 ' =7J inches. Similarly y=8J inches. CENTRE OF GRAVITY. 121 Ex. 3. Weights proportional to 3, 4 and 5 are placed at the corners of an equilateral triangle whose side is of length a, find the distance of their c. g. from the first weight. Take moments about a line through A parallel to BC ; then since the weight at A has no moment about this fine, we get 4x:^a + 5x^a 12 3^/3 Similarly taking moments about AD, the perpendicular to BC, ^^2~^^2 12 a '24' .-. 4(? = V^^2=V(9^!±la=-6a nearly. Ex. 4. To find the c.G. of the perimeter of a triangle formed by three rods of uniform section. Let J), E, F be the middle points of the sides of the triangle ABC, then «, ff, r being the perpendiculars from A, B ana v on the opposite sides, we know that ^ (i) EF, FD and DE are parallel to BC, CA and AB respectively, (ii) the perpendiculars from D, E, F in EF, FD and DE are \p, \q and \r respec- tively. We may suppose the weight of each rod to act at its middle point, hence taking moments about EF, we get as- the distance of the c. G. from EF, BC-^CA+AB- But iyOx5C=area of triangle ABC=S; hence the distance of the c.G. Of from EF equals ^c+CA+AB " D Fia. 77. 122 THE ELEMENTS OF APPLIED MATHEMATICS. In the same way it may be shown that the distances of the c.e. from t;he sides ED and DF may also be shown to equal vQiQ^ij^jg '• hence since its distances from the sides of the triangle I)£F are all ■equal it is the centre of its inscribed circle. EXAMPLES, XXV. 1. Weights of 2, 3, 4 and 5 lbs. respectively are placed at the corners of a square, and weights of 1, 6, 7, 8 lbs. are placed between them, viz. the weight 1 halfway between the weights of 2 and 3 lbs., the weight of 6 lbs. halfway between the weights of 3 and 4 lbs., and so on. Find the c. g. of aU the weights. 2. Find the distance of the c. G. of half a hexagon from its base. 3. Three equal uniform rods are placed so as to form three sides of a square ; find their c. G. 4. Equal particles are placed at 5 of the vertices of a regular hexagon, find the position of their C. G. 5. Weights of 1, 2, 3, 4, 5 and 6 lbs. are placed at the vertices of a regular hexagon, find their c. G. 6. Squares are described on the three sides of an isosceles right- . angled triangle, find the c. G. of the complete figure so formed. 7. Pour books are placed one above another on a table. The lowest projects 4 inches over the edge of the table, the next 2 inches beyond the lowest, the next ^ an inch beyond the second, the upper- most 1 inch beyond the third. Each book is 16 inches in breadth and length and has an edge parallel to the edge of the table. Find the distance of the c. g. of the 4 books from the edge. 8. Find the c. G. of a figure in the shape of a cubical box without a lid, the sides being of small uniform thickness and an edge of the box being 14 inches long. 127. C.G. of the tetrahedron. Divide the tetrahedron into indefinitely thin plates parallel to the base BOB, PQR being one of them. , The c.G. of the triangular plate PQR lies in the line joining P to 8 the middle point of QR, Art. 120. The points such as P for the different plates all lie in the line AB, the points such as S lie in the line AK, where K is the middle point of CI). Art. 120. CENTRE OF GRAVITY. 123 Hence the plane through AB and AK contains the C.G. of all the plates and hence of the whole tetrahedron. In a similar way we may show that the C.G. of the tetra- hedron lies in all the planes which pass through an edge of the tetrahedron and the middle point of the opposite edge; it is therefore the point of intersection of these planes. Again, suppose four equal weights placed at A, B, G and D, the C.G. of the weights at. C and D is at K, hence the C.G. of the four weights lies in the plane through AB and AK. Similarly it is seen to lie in any plane through an edge and the middle point of the opposite edge. Thus the tetrahedron and the four equal weights have the same C.G. But we have seen, Art. 119, that the C.G. of three equal weights at B, C and D is at G' the C.G. of the triangle BGD, hence the C.G. of the four weights, and also of the tetrahe- dron, is at ff, where Ag = SgQ, or it lies f of the way down the line joining a vertex to the C.G. of the opposite face. 128. Equivalent particles for any plate. We have seen that a triangular plate has the same c.g. as three particles each J of its weight placed at its vertices. Hence the weight of the triangle may be replaced by the weights of three particles each J of its weight acting at its vertices. Hence in finding the c.g. of a plate of any form, if we divide it into triangles we may replace the weights of each triangle by three weights each J of it acting at its vertices. The figure represented is a thin plate of weight w bounded by the polygon ABCDEF. Let E be its c. a., join K to the vertices. Then denoting the weights of the plates AKB, BKC, ... hjWi,w^, ... we may replace the weight w^ by 3 particles of weight -J acting at A, B and K respectively. 124 THE ELEMENTS OF APPLIED MATHEMATICS. Similarly for the other triangular plates. We thus have weights i{w^+Wi) at A, ^(w^+w^) at B, J(w2+'<'3) ^^ <^> ^^^ ^° o°' together with J (wi+m'2+ ...+Wg), or ^w, at K. The centre of all these parallel forces, or the c. G. of the plate, is iT; and since one of them ^w acts at K the c. G. of the weights at the ver- tices is also K. 129. CO. of a pyramid standing on any plane polygon. A pyramid whose vertex is stands on the base formed by the polygon ABCDEF. Let K be the c. g. of the base. Join K to and to the vertices of the base. The c. g. of the pyramid is the c. G. of the tetrahedra OKAF, OKBA, ... Also in finding the c. G. the weight of each tetrahedron may be replaced by four particles each equal to J of its weight acting at its vertices. Henxje if W is the weight of the whole pyramid and W,, If j ... the weights of the tetrahedra we thus get weights iW&tO; ^WsLtK; and weights i ( W^+ W^) at 4, i ( Tfi+ Tfa) at B, &c. And since the tetrahedra are all of the same height the weights W-i, W^ ... are proportional to the areas AKB, BKC, ... that is to the weights w^, w^ ... of last article. Hence we have just seen that the weights at the vertices ji,-B, ... have K for their c. G., and their sum is \ W. Hence the weight of the whole pyramid may be replaced by two weights, one j W at and one ^W 3,t K. The c.G. of these is a point Gin AK such that ZQK=^OK, or Oa=iOK. The c. G. of the pyramid is therefore j of the way down the line join- ing the vertex to the c.G. of the base. 130. CO. of any pyramid. We may suppose the number of sides of the polygon in the last article to increase indefinitely, thus giving a curvilinear base. The resiilt then becomes the c.G. of a fiyramAd whose hose is any closed cwrve is f of the way down the line joining the vertex to the c.G. of the hose. CENTRE OF GRAVITY. 131. The arc of a eirde. 125 Fig. 79. Let Q be the o. G. of the arc BAG. Let a be the radius of the circle, and 2a the angle BOG, it is clear that O lies on the line bisecting the angle BOG. Similarly G^j the c.a. of the arc AC lies on the bisector OB of the arc AG ; also OQO-^ is a right angle because OyOO' is perpen- dicular to OA, therefore 0G,=2^. cos^ Thus if G is the c. G. of the whole arc, and G^ the c. G. of half the arc, COS 2 hence it follows that if G^ is the c.g. of the arc GB, OG^ OG 0G,=. Similarly OG, cos -2 cos 2 cos p OG cos 2 cos p cos p 0(?„ — OG a a a S^rCOSxi ... cos =- Now when n is very large OG^ becomes nearly equal to the radius, and sin a cos ^ cos p, cos 5- to 2" a Hence when n is indefinitely large -^ sin a 2a sin a OG=a — -=a- 2aa = radius X chord 126 THE ELEMENTS OF APPLIED MATHEMATICS. 132. Sector of a circle. Let the sector be divided into an indefinitely great number of small equal sectors. Each of these small sectors may be regarded as a triangle, whose c.G. therefore lies f of the way down the radius. The c.G.s of all the small sectors lie on an arc of a circle whose radius is ^OA, and the mass of the whole sector may be supposed uniformly distributed over this arc. Hence if G is the c.G. of the sector _^ „ sin a . OAx chord AB Fig. 80. 133. Zone of a sphere. Let the hemisphere A CB be cut by two planes parallel to its base, they intercept between them a portion called a zone. The planes also intercept between them a zone on a circular cylinder whose base is the same as that of the hemisphere. We shall shew that these two zones have the same area and the same c.G. Fig. 81. For take two planes very near together, they will intersect on the sphere and cylinder two elementary zones or belts. One elementary zone is produced by the revolution oiPQ about 00 and the other by the revolution oiP'Q'. Let R be the middle point of PQ, join OR and draw RS parallel to OA. The arc PQ may be regarded as straight, and since the triangles PKQ and OSR are similar we have P9^0R PK RS ^ '• Now the belt on the sphere may he regarded as a truncated cone, and the area of its surface is therefore equal to the slant side x mean of bounding perimeters, = Pqx%-^R8 = 27r0^xJ'Z, from(i) = %iTOAy.Pq =area of belt on cylinder. CENTRE OF GRAVITY. 127 Hence the areas of the belts on the sphere and cylinder are equal. Thus the corresponding elementary zones on the sphere and cylinder have the same areas, they have also the same c.G., viz. S. To find the o. G. of the entire spherical belt we suppose the weights of the elementary zones to act at their c. &.s in the line 00, and then find the c. &. of these weights. But we have just seen that we get the same weights at the same points, by finding the c. G. of the cylindrical belt. Hence the entire belts on the sphere and cylinder have the same c.G. But the c.G. of the cylindrical belt is halfway between the bounding planes, hence the c. G. of the spherical belt is also halfway between the bounding planes. For a hemisphere, therefore, the c. G. is halfway down the radius. 134. Sector of a spbere. Let the given sector be divided into an indefinitely greater number of equal pyramids, by dividing the spheri- cal surface into small equal areas and join- ing their boundaries to 0. The c. G. of each of these pyramids lies of the way down the corresponding radius, t. 129. Hence the c.G.s all lie on a spherical cap abc and the mass of the sector may be supposed to be uniformly distributed over this cap. But we have seen that the c. G. of such a cap is halfway between m and b, hence if G is the required o. G. 0G= Om+^mb, =iOM+%MB, =1{0M+0B). For a hemisphere OJ/" vanishes, hence 00=%OB. EXAMPLES. XXVI> 1. A square of cardboard is divided into four equal squares, one of the squares being cut out, find the c. G. of the remainder. 2. Four equal heavy particles lie in a straight line ABOD. If their mutual distances are a, ar, ar^ respectively, find r when the c.G. is ate. 128 THE ELEMENTS OF APPLIED MATHEMATICS. 3. At each angle of a square and at the middle points of the sides are placed weights 1 lb., 2 lbs., 3 lbs., 4 lbs., 5 lbs., 6 lbs., 7 lbs. and 8 lbs., beginning at an angular point and going on in order round the peri- meter, and a weight of 8 lbs. is placed at the intersection of the diagonals. Find their c. &. 4. Weights 5, 4, 6, 2, 7, 3 are placed at the corners of a regular hexagon taken in order, find their c. &. 5. ABC is a triangle, find a point in it such that forces repre- sented by OA, OB and OC shall be in equilibrium. 6. If the c. G. of a quadrilateral coincides with that of four equal weights at the vertices of the quadrilateral, show that the quadrilateral is a parallelogram. 7. A uniform rod is broken into two parts of five and seven inches length which are then placed so as to form the letter T, the longer portion being vertical, find the position of the c. G. 8. A square is bisected by a line through its centre making a given angle with the sides ; find the c. G. of either half. 9. If weights 1, 2, 3, 4, 5 and 6 are situated at the angles of a regular hexagon the distance of their c. Q. from the centre of the circumscribing circle is f of the radius. 10. Five masses of 1, 2, 3, 4, 5 ounces weight respectively are placed on a square table. Their distances from one edge of the table are 2, 4, 8, 8, 10 inches and from the adjacent edge 3, 5, 7, 9, 12 inches. Find the distance of their c.g. from the two edges. 11. From a body of weight W a piece of weight w is cut off and moved a distance x; .show that the o.G. of the whole is thereby moved a distance r^ in that direction. W 12. Prove that the c.G. of four equal weights at the angles of a quadrilateral coincides with the c.G. of the parallelogram formed by joining the middle points of the sides. 13. The angle 5 of a triangle ABG is a right angle, AB is 7J inches and BC is 12 inches, at A, B and C are placed particles whose weights are proportional to 4, 5 and 6 respectively, find the distance of their c. G. from B. 14. Three forces PA, PB, PC diverge from the point P and three others AQ, BQ, GQ converge to the point Q ; show that the resultant of the six is represented in magnitude and direction by SPQ. 15. A rod 12 feet long has a weight of 1 lb. suspended from one end and when 15 lbs. are suspended from the other end it balances at a point 3 feet from that end, while if 8 lbs. are suspended there it balances at a point 4 feet from that end. Find the weight of the rod and the position of its c. q. CENTRE OF GPIAVITY. 129 16. A uniform metallic plate is out in the form of a quadrilateral ABCB so that the diagonal AC bisects it and BD cuts it into two parts in the ratio of 2 : 1. Show that its c.g. divides ^C in the ratio of 5 : 4. 17. Two bodies are projected together from a point with different velocities and elevations. Show that their c. G. moves as if it were a heavy particle projected from the same point. 18. A triangular plate ABC, obtuse angled at C, stands on a table in a vertical plane having the side AC va. contact with the table, show that the least weight which suspended from B will overturn it is 19. If one diagonal of a quadrilateral bisects the other and the c. G. be at the middle point of the bisecting diagonal, show that the quadrilateral is a parallelogram. 20. Show that the c.g. of a trapezium ABCB, in which AB is parallel to CB, divides the perpendicular distance between the parallel sides in the ratio AB-\-%CB: 2AB+CB. 21. Two sides of a rectangle are double of the other two, and on the longer side an equilateral triangle is described ; find the c. G. of the figure made up of the rectangle and triangle. 22. A circle of radius r touches internally at a fixed point a fixed circle of radius B; find the c.g. of the area between them, and its iiltimate position when r increases and becomes ultimately equal to R. CHAPTER IX. FORCES IN EQUILIBRIUM. 135. When forces act on a body in such a manner as to produce no motion they are said to be in equilibrium. The simplest case is that of two equal and opposite forces having the same line of action. 136. Equilibrium of three forces in one plane. When three forces in one plane are in equilibrium their lines of action meet in one point. ■Q +R Fig. 83. For the resultant of two of the forces P and Q may be supposed to act at the intersection of their lines of action, but this resultant must have the same line of action as the third force R, since there is equilibrium, hence the line of action of R must pass through 0. In the case when the forces P and Q are parallel their resultant is parallel to them and hence also R must be parallel to P and Q.- FORCES IN EQUILIBRIUM. 131 Ex. 1. A uniform rod AB is suspended with its end in contact with a smooth vertical wall AC, by a string CE. If AE is=^AB, show that AB will be horizontal. Since the rod is uniform the weight acts at its middle point of J 5. The forces acting on the rod are, its weight, the tension of the string, and the reaction of the wall (which is perpendicular to the wall). Two of these forces, viz. the weight and the ten- ^= sion of the string, pass through E, hence also the "" reaction of the wall must pass through E, or AE must be perpendicular to the wall, i. e. horizontal. Fio. 84. Ex. 2. A rod AB can turn round a fixed hinge at A and the end B rests against a smooth vertical wall. Find the reaction of the wall and the reaction of the hinge, the weight of the rod being 10 lbs., its length 18 inches, and the distance of the hinge from the wall 6 inches. The weight of the rod acts vertically through the middle point of AB, the reaction £, of the wall acts along BB perpendicular to the wall, let J) be the intersection of the lines of action of these forces. We see that B', the reaction of the hinge, must pass through B. The sides of the triangle ABE axe proportional to the forces. Art. 78. B_EA R_AB Hence ^ ^^, ^ -^^. EA==^AF=Z; BE=BF=s/AB^-AF^='^l%^-&=lQ) are attached by strings to the corners B, D and hang freely. Prove that in the position of equilibrium the inclination of J -B to the horizon will be where a and h are the lengths of the sides AB, AD of the rectangle. 144 THE ELEMENTS OF APPLIED MATHEMATICS. 17. A hollow vertical cylinder, radius 2a, height 3a, rests upon a horizontal table ; a rod is placed within it with its lower end at the circumference of the base, the rod rests upon the opposite point of the upper rim and projects over. How long must the rod be in order that it may cause the cylinder to topple over, the rod and cylinder being of equal weight ? 18. A heavy uniform rod of length 2a rests partly within and partly without a smooth hemispherical bowl of radius r; if ^ be the inclination to the horizon of the rod Sj-cos 2d=acosd. 19. A heavy triangular lamina rests inside a smooth hemispherical bowl ; prove that the pressures at the three angular points are equal. 20. A uniform beam rests with a smooth end against the junction of the ground and a vertical wall ; it is supported by a string fastened to the other end of the beam and to a staple in the vertical wall. Find the tension of the string, and show that it will be equal to half the weight of the beam if the length of the string is equal to the height of the staple above the ground. 21. A cylindrical tube of mass M stands upright on the ground. Two equal smooth spheres are placed within it, one resting on the ground and the other supported by the cylinder and the other sphere. If the mass of either sphere be m, its radius a, and the radius of the cylinder f a, show that the cylinder is on the point of toppling over provided that 2m=3M. 22. Three equal strings of no sensible weight are knotted together to form the equilateral triangle ABO, and a weight W is suspended from A. If the triangle and weight be supported with BC horizontal by means of two strings, each at the angle of 135° to BC, prove that W the tension in BC is -^ (3 - ,^3). 23. Two equal weights of 112 lbs. are joined by a string passing over two pulleys A and B in the same horizontal line. A weight of 1 lb. is attached to the middle point of the string between A and B ; find the position of equilibrium. 24. Three uniform heavy rods AB, BC and CA of lengths 5, 4 and 3 feet respectively .are hinged together at their extremities to form a triangle. Prove that the whole will balance with AB horizontal about a point distant 11 of an inch from the middle point of AB to- wards A. 25. If W be the total weight of the 3 rods in the last question prove that the vertical components of the action at the hinges A and B when the rod is balanced are J^J W and J^ W. FORCES IN EQUILIBRIUM. 145 26. Show that if the pins of the two hinges of a door are not in the same straight line the door cannot open, and if the straight line of the pins is not vertical the door wiU tend either always to open or always to close. 27. If a system of forces be represented in magnitude and position by all but one of the sides of a closed polygon taken in order, their resultant will be parallel in position and proportional in magnitude to the remaining side, and that its line of action will be at a distance from that side inversely proportional to its length. J. 10 CHAPTER X. WORK AND ENERGY. A Force is said to do work when its point of application moves in a direction not perpendicular to that of the force. 144. Measure of Work. The work done is measured by the product F . s ; where F is the measure of the force and s the displacement of the point of application in the direction of the force. Fio. 100. Thus if a force F acts at a point J. of a body, in the direction AB and A comes to G, the work done by F is FxAC. But if the point of application of the force F comes to A', which is not in AB, draw A'G perpen- dicular to AB. AG is the displacement of the point of application in the direction of the - force, hence the work done by the force i^is F>^AG ^'"^-loi- Work is said to be positive when the point to which the force is applied moves in the direction of the force, and negative when it moves in the opposite direction. In the latter case the force is said to have work done against it. WORK AND ENERGY. 147 No work is done by a force when its point of application moves perpendicular to the force, for here s = 0. For in- stance no work is done by the weight of a body sliding along a horizontal plane. 145. Unit of Work. The unit of work is that done by the unit of force in moving its point of application unit distance in its direction. The unit of work adopted by engineers is the Foot- Found, or the work done in lifting the weight of a pound through one foot vertically. This unit depends on the value of " g " at the particular place, if we wish the corresponding absolute unit, we use the Footrpoundal which is the work done by a poundal when its point of application moves one foot. The unit used in scientific measurements is the Erg. This is the work done by a Dyne when its point of application moves through a centimetre. The erg being too small a unit for practical purposes is generally replaced by the Joule, which contains 10' Ergs. Ex. Find the number of "units of work done against gravity in raising a ton of coals through a distance of 60 yards, i;'=20xll2, « = 60x3; .-. work done=20x 112x60x3=403200 foot-pounds. EXAMPLES. XXVIII. 1. A man weighing 12 stone goes up a staircase of 30 steps so as to rise one foot vertically each step. How many foot-pounds of work must he do before reaching the top ? 2. Show that the work done iu raising 1 cwt. through a height of 10 yards is equal to the work done in raising 1 lb. a height 3360 feet. 3. Find the work done in drawing up a Venetian blind, the number of bars being 50, the distance between each bar three inches, and the weight of each bar four ounces. 4. Find the work done by a force which acts for two seconds on a' body whose mass is m lbs. and gives it a velocity of 10 feet per second. 5. A gun whose weight is one ton is drawn 100 feet along the groimd ; if the resistance due to the roughness of the ground is ^th of the weight of the gun, find the work done against the resistance. 10—2 148 THE ELEMENTS OF APPLIED MATHEMATICS. 146. Work done in falling down an inclined plane. When a body falls down a smooth inclined plane, the resistance of the plane being perpen- dicular to the plane does no work. Art. 144. The work done by gravity is equal to w . h, where w is the weight of the body and h the height of the plane, since h, is the projection of the displacement on the direction of gravity. Hence the work done in descending an inclined plane depends only on the weight of the body and the height of the plane. 147. Rate of Work. It is to be noticed that the work does not depend on the time. The work done being equal to Fs, the rate of work or Power of an agent is equal to Fj, or F.v. V The Power of an agent, therefore, is equal to the work done by it if working uniformly for a unit of time, or the number of foot-pounds per second. 148. A Horse-Power is the power of an agent which can do 33000 foot-pounds per minute, or 550 foot-pounds per second. Hence if H is the Horse-Power Fv = 6^0H, the force F being measured in lbs. weight. In the c. G. s. system the unit of power used is a Joule per second, or 10^ ergs, per second, this is called a Watt. A horse-power contains about 746 Watts. Ex. 1. At what uniform speed can an engine of one horse-power draw a tramcar of 2 tons weight, supposing the resistance to motion to be equal to the weight of 50 lbs. ? Let V be the required velocity. The rate at which work is done against resistance is 50v foot-pounds per second. WORK AND ENERGY. 149 Also Power of engine = 550 foot-pounds per second ; .-. 550= 50i), or D=ll feet per second, or 7 J miles per hour. Ex. 2. A train of 100 tons is pulled by a locomotive on the level at a constant speed of 30 miles per hour, the resistance amounts to 15 lbs. per ton. Find the minimum horse-power of the engine. The resistance to motion is 1500 lbs. ; and a speed of 30 miles per hour is 44 feet per second. Therefore the rate at which the engine works is 1500x44 foot- pounds per second. TT . . , . 1500x44 Hence mimmum horse-power is — ^^ — , ox 120. Ex. 3. If the train described in the last example be moving at a particular instant with a velocity of 15 miles an hour, what is the acceleration at that instant ? The engine is doing 120 x 550 foot-pounds of work per second, while the train is moving at the rate of 22 feet per second. Therefore the force of the engine, or , is here'equal to — ^ pounds' weight, or 3000 lbs. weight. Of this 1500 lbs. weight is needed to overcome the resistance, hence the remainder 1500 lbs. weight is the eflfective force, therefore effective force is 1500 x 32 poundals, 1500x32 1500x32 3 acceleration = mass of train 2240x100 14' EXAMPLES. XXIX. 1. The mass of a complete train is 60 tons, and the resistance to its motion equal to 20 lbs. weight per ton. The locomotive has 240 horse-power; what is the highest speed the train can have ? 2. An engine is required to raise in 4 minutes a weight of 12 cwt. from a pit whose depth is 600 feet. Find the horse-power of the engine. 3. An engine draws a train weighing 96 tons at the rate of 15 miles an hour, the resistance to the motion of the train amovmting to 8 lbs. per ton, find the horse power of the engine. 4. Determine the rate in h. p. at which an engine must be able to work to generate a velocity of 30 miles an hour onthe level in a train of mass 60 tons in 3 minutes after starting, the resistance to motion being 10 lbs. per ton weight. 150 THE ELEMENTS OF APPLIED MATHEMATICS. 149. Energy. The Energy of a body is its capacity for doing work. The body on which work is done will be found in conse- quence to have conferred upon it an increased capacity for doing work, hence a body's energy has its origin in work. Energy is of two kinds, Kinetic and Potential. 150. Kinetic Energy. The Kinetic Energy of a body is that part of its Energy which is due to its motion ; a revolving wheel, a train in motion, a falling body, all possess kinetic energy. The measure of the kinetic energy of a body is the amount of work which can be done by the body before this energy is destroyed. 151. Kinetic Energy is equal to ^mv^ The measure of the K. e. of a body whose mass is m and velocity v is ^mv^ foot-poundals. We shall show that this is true when the body does work against its weight. Let the body be moving upwards with a velocity v, we know it will be brought to rest after rising a height h, where v^=2gh, Art. '24. Now mg is the weight of the body, hence mgh foot- poundals is the work done against its weight before the body is brought to rest, but this amount of work measures the K. E., therefore K. E. of body is = mgh foot-poundals = ^mv^ foot-poundals. 152. Change in Kinetic Energy equals work done. When a body acted on by a constant force F, producing an acceleration a, has its velocity changed from u to v, in going a distances, v'--u^=- 2as ; Art. 24. . ■. ^mif — ^Tnw' = mas = F.s. Hence the change in the K. E. is equal to the work done. WORK AND ENERGY. 151 Ex. 1. A body whose mass is 5 lbs. is thrown vertically upwards with a velocity of 32 feet per second; find its k.e., (i) at the moment of projection, (ii) after half a second, (iii) after one second. (i) At the moment of projection the velocity is 32, hence K;.B.=|mii2=2560. (ii) After half a second the velocity is 16, .-. k.e. =640. (iii) After one second the velocity is 0, .". k.e.=0. Ex. 2. A ball weighing 6 lbs. is rolled on a floor with a velocity of 8 feet per second ; if the resistance of the floor to the motion of the ball is j^th of the pressure on it, how far will the ball roll ? If s be the required distance, the work done against the resistance is 1% X gf X s, and this is equal to the K. b. of the ball, hence j%x32x 5=^x6x82; .-. s= 10 feet. Ex. 3. A cannon-ball whose mass is 60 lbs. falls through a vertical distance of 400 feet ; what is its kinetic energy? The velocity acquired in falling through 400 feet is 160 feet per second .-. K.E.=^xeOx(160)2 =768000 units of kinetic energy. EXAMPLES. XXX. 1. A fourteen-ton gun on being fired recoils and is brought to rest by a uniform resistance equal to the weight of 3 tons. How far does the gun recoil, the velocity of the ball being 1200 feet per second and its mass 112 lbs. ? 2. A ball whose mass is 100 grammes is thrown vertically upwards with a velocity of 980 centimetres per second ; what is the kinetic energy of the body (i) at the moment of propulsion, (ii) after half a second, (iii) after one second ? 3. A shot of 1000 lbs. moving 1600 feet per second strikes a fixed target; how far will the shot penetrate the target which exerts upon it an average pressure equal to the weight of 12,000 tons? 4. A ball whose mass is 10,000 grammes is discharged with a velo- city of 6000 centimetres per second ; find its k. e. in ergs. 5. A ball whose mass is 3 lbs. is moving at the rate of 100 feet per second ; what force will stop it (i) in 2 seconds, (ii) in 2 feet 1 152 THE ELEMENTS OP APPLIED MATHEMATICS. 153. Potential Energy. The potential energy of a system of bodies is the energy which is due to their relative positions. Take the simple case of two bodies one of which is the Earth and the other a body near its surface, if this body be allowed to fall it will do work, thus the relative positions of the Earth and the body afford a capacity for doing work. We may, as an abbreviation, speak of the potential energy of a single heavy body. The measure of the potential energy of a system is the amount of work which is done in changing from the given configuration, or relative arrangement, to some other stand- ard configuration. Returning to the above illustration, when a body whose mass is m falls to the ground from a height h, mgh foot-poundals of work are done, and the configuration of the system is changed from one in which the body and Earth are separated by a distance h, to one in which they are in contact. This last is taken as the standard configu- ration. Using the abbreviation used above we may say that the potential energy of a body of mass m and height h is mgh foot-poundals. A bent spring and compressed air are also instances of systems possessed of potential energy, for a spring does work in straightening and compressed air in expanding. 154. Conservation of Energy. The principle of the Conservation of Energy may be stated thus : If a body, or system of bodies, be under the action of forces which depend only on the position of the body or system of bodies, the sum of the Potential and Kinetic Energies is constant. In the case of gravity this is at once seen to be true ; for if a body fall from a point A, h feet from the ground, to a point P, x feet from the ground, we have v'' = u'' + 2g{h-x\ or w" -I- 2gx = u^ + 2gh, . : ^^ww^ -I- mgx = ^mw' + mgh, or the total energy at A equals total energy at P. WORK AND ENERGY. 153 155. It should be observed that it is incorrect to speak of a single particle or body as possessed of energy. For as force implies the action of one body on another, the energy which is due to the stress between the bodies belongs to the pair. Therefore although it may be convenient to consider one of the bodies, usually the Earth, to be at rest. Kinetic Energy is due to the relative motion of the two bodies and Potential Energy to their relative position. Supposing the bodies to form one system the following definitions sire more accurate. Kinetic energy is the energy which a system possesses by virtue of the relative motion of its parts. Potential energy that which it possesses in virtue of the relative position of the parts. 156. Another expression for virork done. In Art. 144 we saw that the wofk done by a force F was Fx AC, when its point of application moved a distance AC in the direction of F. Let AP represent the force and AB the displacement, draw BC perpendicu- lar to AP and PQ perpendicular to AB produced. Then B, G, P, Q are four points on the same circle. Hence AP xAG = ABxAQ. Euc. iii. 36. But AP X AG, or F x AC, is the work done, also AQis. the component of F in the direction of the displacement, hence work done is = displacement x component of F in the direction of the displacement. 157. Work of resultant equals sum of works of components. By the last Article, when any number of forces act at a point which undergoes a displacement, the total work done is displacement x (sum of components of forces in direction of displacement). But this is = displacement x component of resultant of the forces in direction of displacement = work of resultant of the forces. B Q Fia. 103. 154 THE ELEMENTS OF APPLIED MATHEMATICS. 158. Principal of Virtual Work. If, in the last Article, the resultant is zero, or the forces are in equilibrium, we see that the total work done by the forces is zero. When the displacement is not actual but it is merely supposed that the point of application of the forces receives a displacement, the work done by any of the forces is said to be virtual, and the displacement is called a virtual displacement. The Principal of Virtual "Work states that when a set of forces are in equilibrium the algebraic sum of the virtual work of all the forces is zero, the forces being supposed to remain the same during the displace- ment. To secure that the forces may not sensibly alter during the wtual displacement it is usually taken very small. The displacements are taken so that the work done by the forces which we do not wish to find does not appear, for instance if we take the displacement of a rigid body such that the distances between its particles are not altered, the internal forces (see Art. 85) will do no work. If the displacement consists of a small rotation round any point in the plane of the bodyj it is easy to see that, if the forces are in equilibrium, we thus get the equation of moments round the given point. 159. The work done hy the weights of the system of particles forming a body in any displacement is equal to the work done by their resultant acting at their c. e. Let Wi, W3, W3, ... be the weights of the particles, hi, A3, hs, ... be their distances from any horizontal plane, also let x-i, Xi, X3, ... be the displacements. Then if h be the height of the c. G. before displacement, and h + or. after, we have wJh + wji^ + ... = {Wi + w^ + ...)h, Art. 103 ; Wi {hi+x^ + iVi {h^ + x^+ ... ={Wi+Wi+ ...){h + X), hence subtracting, WiflJi + WaiCa + . . . = (Wi + Wa + . . .) ^ ; which proves the theorem. Ex. 1. Find, the velocity of a particle sliding down a smooth inclined plane. The forces acting on the particle are its weight and the pressure of the plane. This latter force is perpendicular to the plane and hence does no work during the particle's motion. Art. 144. The work done by the weight is mgh, where h is the height of the plane. WORK AND ENERGY. 155 The particle's kinetic energy at the top is zero, at the bottom is ^v^, where v is its velocity at the bottom. ^But change in kinetic energy equals work done, Art. 152. .•. ^mv'^^mgh, or, v^ = 2ffh. If the particle has an initial velocity u, the change in the kinetic energy is ^mv^-^mu% hence in this case ^m,v^ — ^mu^=mgih, or, v^=u^ + 2gh. Ei. 2. Find the height to which a ball will rise whose mass is 2 grammes, and which is projected vertically upwards with a velocity of 20 metres per second before the velocity is reduced to 5 metres per second, supposing the resistance of the air to be 10 dynes per centi- metre. 2 X 2000l ^ The initial kinetic energy of the body is — - ergs. Let k centimetres be the required height, then the work done against gravity is 1962A ergs, the work done by the resistance is lOA ergs, hence the total work done is 1972A ergs. „.„,,.,. . 2 X (500)2 Ihe final kinetic energy is ^ — - ergs. Hence since change in kinetic energy equals work done, we have 2 X (2000)2 2 X (500)2 i . i , -,„„„ ^-= ^ — ^ = 1972A, from which A=1902 cms. nearly. Ex. 3. A pit is sunk whose opening is a square with a side of 6 feet, and which has a depth of 50 feet. Eind the work done in raising the earth, supposing a cubic foot of earth to weigh 100 lbs. The number of cubic feet to be raised is 36 x 50, and the weight of this is 36 X 50 X 100 lbs., or 180000 lbs. The c.G. of the whole mass is 25 feet from the surface, the work done is therefore 180000 x 25 foot-lbs. EXAMPLES. XXXI. 1. A body is projected vertically upwards with a velocity of 80 feet per second, find its velocity when it has reached a point 20 feet higher than the point of projection. 2. Find the work done by an engine of a train of 360 tons weight in going two miles starting from rest ; the motion is on the level and friction is ^^ of the weight, the final speed being 45 miles per hour. 3. A train of 150 tons moving with a velocity of 50 miles an hour has the steam suddenly shut off and the train comes to rest in 440 yards. Find the constant force applied to the train. 4. Find the work done by gravity on a stone whose mass is J a lb. during the sixth second of its fall from rest. 156 THE ELEMENTS OF APPLIED MATHEMATICS. 160. Change of units. In the present Article we investigate the method of finding the measure of a given physical quantity when the units of mass, length and time are changed in any given way, the original measure and units being known : — Velocity. Let V be the measure of a velocity when a feet and t seconds are taken as -the units of length and time, find its measure when a' feet and t' seconds are taken as the units of length and time. With the former units unit velocity is = the velocity of a body moving a ft. in t seconds a. . = - it. m z 1 second = - foot-sees. z But the given velocity is v times this, and therefore is —foot-sees. X Again, if v' is the measure of the given velocity with the latter units, we see in a similar manner that the given ve- locity is v'a' —7- foot-secs. Hence — = —r , since each of these numbers is the number of foot-secs. contained in the given velocity. If the former units are one foot and one second, a = t = l, and the equation to find v' becomes v'a' ' = -f Acceleration. Let / be the measure of a certain acceleration when a feet and t seconds are units, find its measure /' when a' feet and t' seconds are units. WORK AND ENERGY. 157 With the former units, unit acceleration is the addition of unit velocity in t seconds, or, the addition of - foot-sees, in t seconds, or, the addition of - foot-sees, in one second, r But the given acceleration is /times this and hence is an addition of •'— foot-sees, in one second. p With the latter unit we get similarly that the given acceleration is an addition of ^-r- foot-sees, per second. Hence — = ^^ , since each of these numbers is the number of foot-sees, added per second. From this equation we find /' in terms of the given quantities a, t, a', t', f. Ex. Find the measure of the acceleration of gravity when a yard and a minute are the units of length and time respectively. Let us take a = l, t = l, then we know/=32. We are given that as' =3, <'=60, and we have to find /'. But we have seen that / s =/' ^ ■ Therefore 32=/A=_£, or /' = 38400. Force. With a feet, t seconds, and m lbs. as units of length, time and mass, the measure of a force is F, find its measure F' when a' feet, t' seconds, and m' lbs. are taken as units. In the first case unit force produces in m lbs. unit velocity in t seconds, m lbs. - foot-sees, in t seconds, V lib. -r~ foot-sees, in one second. V 158 THE ELEMENTS OF APPLIED MATHEMATICS. And the given force produces F times this, hence it pro- duces F -— foot-sees, in 1 lb. per second. Similarly it produces — -7^ — foot-sees, in 1 lb. per second; Notice that the" unit forces are -^ and --^^ powndah respectively. Momentum. L'et M be the momentum of a body with the former units and M' with the latter. In the former system the unit of momentum is that possessed by m lbs. moving with - foot-sees, and hence — times the momentum of 1 lb. moving with one foot-sec. The given momentum is therefore — - — times the mo- mentum of one lb. moving with one foot-sec. Hence as before Mma _M'm'a' Kinetic Energy. In the former units E is the measure of a given amount of kinetic energy, and E' is the measure in the latter. The first unit of kinetic energy is the kinetic energy possessed by m lbs. moving with - foot-sees, or - —^ x (the energy possessed by one lb. moving with one foot-sec). Hence as before :kE^^W'"'"" t^ 2 ^'2 WOKK AND ENERGY. 159 Work. The kinetic energy is equal to the work done. Art. 152. Hence if W and W be the measures of the work in the two systems Notice that the units of work are -g- and — ^ foot-poundals respectively. Power. Let P and P' be the measures of the power of a given agent in the two systems. In the first system, the unit of power is the power of that agent which does the unit of work in t seconds, or which does - X (unit of work) in one second ; hence it does r x (~;j foot-poundals) in one second. Thus the given power does P —7^ foot-poundals in one second. Hence as before, we find p ma? _ -p/mfa'^ EXAMPLES. XXXII. 1. What is the measure of a velocity of one foot per second when a yard and an hour are the units of length and time respectively? 2. Find the measure of the acceleration of gravity when the imits of length and time are a mile and a minute. 3. If the weight of a lb. is the unit of force, a velocity of one yard per second the unit of velocity, the mass of 4 lbs. the unit of mass, find the units of length and time. 4. The units of length and time being the same as in question 1, find the measure of a poundal, the unit mass being one lb. 160 THE ELEMENTS OF APPLIED MATHEMATICS. 5. Having given that the centimetre is '3937 inches and the gramme weighs '0022046 lbs., find the measures of the dyne and erg in foot-lb.-second units. 6. The unit force being the weight of a ton, the unit acceleration that due to gravity, the unit velocity that of a body which has fallen from rest 5 seconds, find the units of mass, length and time. 7. If the unit of mass is the mass of a ton, the unit of momentum that possessed by one lb. moving at the rate of one mile per hour, find the unit of velocity. 8. If one poundal is the unit of force and one foot-sec. the unit of velocity, show that there are as many lbs. in the unit of mass as there are seconds in the unit of time. 9. The mass of n lbs., n feet and rfi seconds being taken as units, show that the unit force is one poundal. 10. The unit of work is that required to raise a ton weight through a vertical distance of 10 feet, the velocity of 10 miles an hour is the unit of velocity, find the unit of mass. 11. How many foot-pounds of work must be done in order to raise 10,000 gallons of water from the bottom of a mine 110 fathoms deep, and what is the horse-power of an engine which can do this in 5 minutes ? 12. At the bottom of a coal-mine 275 feet deep there is an iron cage containing coal weighing 14 cwt., the cage itself weighing 4 cwt. 109 lbs. and the wire rope that raised it 6 lbs. per yard. Find the work done when the load has been raised to the surface and the h.-p. required to do that amount of work in 40 seconds. 13. Find the h.-p. of an engine able to drive a train of 100 tons on a level line on which the resistance is jj^ of the load at a speed of 30 miles an hour. 14. A cylindrical shaft has to be sunk to a depth of 100 fathoms through chalk, the weight of a cubic foot being 143-75 lbs., the diameter of shaft 10 feet, what h.-p. is required to lift out the material in 12 working days of 8 hours each ? 15. A tank 24 feet long, 12 feet broad, 16 feet deep is to be filled with water from a well the surface of which is always at a depth of 80 feet below the bottom of the tank, find the work done in filling the tank and the h.-p. of an engine that will fill the tank in 4 hours, a cubic foot of water weighing 62-5 lbs. 16. The mass of a fly-wheel is 1200 kilogrammes and the radius of its rim one metre. Supposing the whole mass concentrated in the rim find the energy of the wheel when making 7 turns a second. WORK AND ENERGY. 161 17. A straight rod ACB without weight has two particles of equal weight fastened to it, one at the end B and the other at the middle point C, and the rod can swing about A. If it be held horizontally, and then allowed to swing, prove that the greatest velpcity acquired by the end B will be the same as that of a particle which has fallen freely from rest through a height =| of the length of the rod. 18. A chain weighing 8 lbs. per foot is wound up from a shaft by the expenditure of 4,000,000 units of work, find the length of the chain. 19. Prove that a train of W tons going up an incline of 1 in. m will acquire a velocity (jy---- ^^ gt, and energy * V^ m 2240;' ^'' foot-tons after t seconds from rest, P being the pull of the engine in tons, and R the resistance on the level in lbs. per ton. 20. Find the charge of powder required to send a 32 lb. shot to a range of 2500 yards with an elevation of 30°, supposing the initial velocity is 1600 feet per second when the charge is half the weight of the shot, and that the initial energy of the shot is always proportional to the charge of powder. 21. Having given that an engine of 60 h.-p. is required to drive a steamer 80 feet long at the speed of 9 knots, find the h.-p. of an engine which will drive a similar steamer 240 feet long and similarly im- mersed at 18 knots, assuming that the resistance is proportional to the wetted surface and to the square of the velocity through the water. 22. A fine string passes through two small fixed rings A and B in the same horizontal plane and carries equal weights at its ends. If a third equal weight is attached to the middle portion AB of the string and is let go, prove that it will descend to a depth =§4.3 below AB and then ascend again. 23. Find the h.-p. transmitted by a belt moving with a velocity of 600 feet per minute passing round 2 pulleys, supposing the difierence of tension of the two parts to be 1650 lbs. 24. An inelastic pile of ^ a ton is driven 12 feet into the ground by 30 blows of a hammer of 2 tons falling 30 feet. Prove that it would require 120 tons in addition to drive it down very slowly. 25. A train whose mass is mlbs. moves against a constant re- sistance equal to p times its weight ; it starts from rest and moves with constant acceleration till the steam is shut off and arrives at the next station, distant a from the starting point in t seconds. Show that the greatest horse-power exerted by the train is C '""V 9 "^ where C is a constant depending on the units employed. -"-'2 — 2a J. 11 CHAPTER XI. THE SIMPLE MACHINES. 161. The instrument by which effort is applied to lift a weight or overcome any kind of resistance is called a machine. Part of the effort is spent in overcoming the re- sistance of the machine itself due to friction, imperfect flexi- bility of ropes, &c. Resistances of this kind are called wasteful as distinguished from that which it is the object of the machine to overcome, this latter is called useful re- sistance. 162. Efficiency. The ratio of the useful work done by a machine to the whole work done is called its efficiency. If there were no wasteful resistance the efficiency would be unity. In such a case the machine is said to be perfect, and this we shall assume it to be. 163. Mechanical advantage. In what follows it is supposed that an applied force P, by means of a machine, supports a weight Q, then any force greater than P will move Q. The ratio -^ is called the mechanical advantage of the machine and is usually greater than unity. THE SIMPLE MACHINES. 163 164. Simple IVIachines. The simpler machines are the following : (1) The Lever, (2) The Wheel and Axle, (3) The Pulley, (4) The Inclined Plane, (5) The Screw. The Lever, The lever consists of a rigid bar (straight or curved) which can turn freely abqut a fixed axis called the fulcrivm. By its means an applied force P balances a force Q, a pressure R being produced on the fulcrum. There are three classes of levers; the first is when the fulcrum is between P and Q, _=£__ FiQ. 104. the second, when Q acts between P and B, t=, 1 Fig. 105. the third, when P acts between Q and R. c Fig. 106. Examples of the first kind are, a pair of scales, a crowbar, a poker resting on the bar of a grate ; double levers are a pair of scissors. 11—2 164 THE ELEMENTS OF APPLIED MATHEMATICS. Examples of the second kind are, an oar when the end in contact with the water is at rest, this end is the fulcrum, the force Q acts at the rowlock ; a wheelbarrow, the fulcrum being the point of contact with the ground. An example of the third kind is the limb of an animal; the socket is the fulcrum, the force P is the action of a muscle attached to the bone near the socket, the force Q is the weight of the limb ; a pair of tongs is an example of a double lever of this class. 165. Conditions of equilibrium. For equilibrium R must be the reversed resultant of the parallel forces P and Q, hence in all* cases P.AG = Q.BG. Art. 96. Also, in the first case R = P + Q, in the second case R = Q — P, in the third case R = P— Q. If the lever is bent, draw from C lines CM and CN perpendicular to the directions of P and Q, then taking moments about C, Fig. 107. we have .P . CM= Q . ON. 166. In the foregoing we have neglected the weight of the lever, if it is of weight w and its c. G. at A O Ct B w Q Fig. 108. by taking moments about C as before, we have P.AG+w.OG=Q.BG. THE SIMPLE MACHINES. 165 167. The distances AG and BG are called the "arms " of the lever ; AG and since mechanical advantage = ^= ^-=^, Art. 163, we see that the mechanical advantage is = -p-, . Class I. Mechanical advantage is gained by making P's arm longer than Q's arm. Class II. Mechanical advantage is always gained. Class III. Mechanical advantage is always lost. If the lever gives mechanical advantage we see that the distance moved through by P is greater than that moved through by W. Ex. 1. Two weights balance attached to the ends of a lever of the first class, one weight is 10 lbs. and its arm is 6 inches, the other arm is 15 inches, find the other weight. Let Q be the other weight, here P=10, hence 10x6 = §xl5, or §=4 lbs. Ex. 2. The mechanical advantage of a lever is 4 and the press\ire on the fulcrum is 8 lbs., find the balancing forces, the lever being of the second class. We have given that ^ = 4, Q-P=B. Hence P=| lbs., W=^\hs. Ex. 3. A man who wishes to raise a rock leans with his whole weight on one end of a crowbar five feet long which is propped at the distance of four inches from the end in contact with the rook. The man's weight being 160 lbs., what force does he exert on the rock, and what is the pressure on the prop ? The force he exerts is equal to that of a weight ic, where as is given by the equation 4^=160x56, or ii;=2240lbs. The pressure on the prop is = PF'+P=2400 lbs. Ex. 4. Two forces whose measures are 6 and 8 act at the end of a rod 16 feet long and are inclined to the rod at angles of 30° and 45° respectively ; find the position of the fulcrum when there is equi- librium. 166 THE ELEMENTS OF APPLIED MATHEMATICS. Let AB be the rod and x the distance of the fulcrum from A. B h The perpendicular distance of the force 6 from the fulcrum is ^ . The distance of the fulcrum from 5 is 16-a;. 16 — a; The distance of the force 8 from the fulcrum is „ X „ \%-x \/2 %x hence -^ = 64-4^, or a; = 10'4 feet nearly. EXAMPLES. XXXIII. 1. The arms of a straight lever are 5 and 7 inches long respec- tively, a weight of 2 lbs. is attached to the shorter arm, find the weight to be attached to the longer arm for equihbrium. 2. A weightless rod 7 feet long has weights of 4 lbs. and 10 lbs. hung at its extremities. Find the position of the fulcrum when there is equilibrium. 3. A uniform rod which is 16 feet long and which weighs 17 lbs. can turn freely about a point in itself, the rod is in equilibrium when a weight of 7 lbs. is hung at one end. Find the position of the fulcrum. 4. The two arms of a straight lever are 18 inches and 50 inches long respectively, and its weight is 10 lbs. If a weight of 58 lbs. be applied at the end of the longer arm, what weight must be apphed at the end of the other that there may be equilibrium ? 5. A straight lever whose length is 5 ft. and weight 10 lbs. has its fulcrum at one end. Weights of 3 lbs. and 6 lbs. are fastened to it at distances 1 ft. and 3 ft. from the fulcrum, and it is kept horizontal by a force at the other end. Find the pressure on the fulcrum. 6. If the pressure on the fulcnmi be equal to 10 times the difference of the forces, find the ratio of the arms. 7. A uniform wire is bent so as to form two straight lines incUned at an angle of 120°, one of which is twice as long as the other. The wire is suspended from its angular point. Find the position of equilibrium. THE SIMPLE MACHINES. 167 8. A uniform heavy rod has a weight of 5 lbs. hung from one end and balances on a fulcrum 5 feet from that end. This weight is replaced by a weight of 10 lbs. and then the rod balances when the fulcrum is 4 feet from that end. Find the length and weight of tte rod. 9. A man seated in a boat pulls at the handle of each of a pair of sculls with a force of 25 lbs. weight. If the distance of the rowlock from the end of the blade of each scull be 4 times that of the rowlock from the hand, find the resultant force on the boat, 10. The arms of a bent lever are at right angles to one another and are in the ratio of 5 to 1. The longer arm is inclined to the horizon at an angle of 45°, and carries at its end a weight of 10 lbs. ; the end of the shorter arm presses against a horizontal plane, find the pressure on the plane. 168. Balances. One of the uses of the lever is to determine weights, for this purpose it is used in the forms of the Common Balance, the Roman steelyard, and the Danish steelyard. 169. The Common Balance. Fig. 109. This consists of a beam somewhat in the shape, of a lozenge which turns freely about a fulcrum G, consisting of a wedge-shaped knife-edge attached to the beam and resting on a fixed support. The centre of gravity Q of all the parts of the balance lies vertically below C when the beam is horizontal. To the ends of the beam there are attached knife-edges upon which scale-pans are supported. 168 THE ELEMENTS OF APPLIED MATHEMATICS. 170. If the weights placed in the scale-pans are not equal the beam will not be horizontal, and will take up a position of equilibrium in which the moments of the weights about G will be counterbalanced by the moment of the balance itself about G. 171. A pointer or rod GD (not shown in the figure) is often attached to the centre of the beam. In the position of equilibrium this is vertical. 172. Requisites of a good balance. 1. The balance must be true, that is when loaded with equal weights the beam should be horizontal. This condition is secured if, (i) the arms are equal in length, (ii) the scale-pans are equal in weight, (iii) the C.G. of the beam be vertically below G. 2. The balance should be sensitive, that is for any small difference in the weights the deviation of the beam from its horizontal position should be large enough to be easily ob- served. This will be secured by (i) increasing the length of the beam AB, (ii) decreasing the length of the rod GD, (iii) diminishing the weight of the beam. See Ex. 8, p. 170. 3. The balance should be stable. That is when dis- turbed it should quickly return to its horizontal position. 173. False Balances. The effect of weighing with an untrue balance will be now considered, 1.. The arms of a balance are of unequal length a and h, a body of weight w lbs. weighs w^ lbs. when placed in one scale and w^ lbs. when placed in the other. Here we have wa = wj), w^a= wh. THE SIMPLE MACHINES. 169 °'" W= JwiWs (1). The true weight is the geometric mean of the apparent weights. 2. The arms are of equal length but the C.G. of the balance is at a horizontal distance c from C, the middle point of the beam. Let w' be the weight of the balance, w the true weight of a body and Wi, Wj the weights which have to be used to balance it when it is placed in the two scale-pans in suc- cession. Here if a is the length of either arm, by taking moments about C, wa + w'c = Wja, w^a + w'c = wa. Hence (w — w^) a = {w^ — w) a, or 2w = Wi+W2 (2). The true weight is the arithmetic mean of the apparent weights. 3. If the arms are of unequal length and also the C.G. of the balance not vertically below C, we have by taking mo- ments as before w .a = w'c + wj}, w^a = w'c + wb; .•. w{a + b) = Wib + w^ (3). EXAMPLES. XXXIV. 1. The arms of a balance are respectively 8| and 9 inches long, the goods to be weighed being suspended from the longer arm. Find the real weight of goods which apparently weigh 27 lbs. 2. A body the weight of which is 28 ounces, when placed in one scale of a balance with unequal arms, appears to weigh 14 ounces ; find its weight when placed in the other scale. 3. The apparent weights of a body are 4 and 16 lbs. resjjeotively, when weighed from the two arms of a balance. Find the ratio of the lengths of the arms and the true weight of the body. 170 THE ELEMENTS OF APPLIED MATHEMATICS. 4. If the beam of a false balance is uniform and heavy, show that the arms are proportional to the differences between the true and apparent weights. 5. In a given balance it is found that Sl'OVS grammes in one scale balance 51"362 in the other, and 25-592 balance 25'879; show that the arms are equal, but that the scale-pans differ by '287 grammes. 6. The beam of a balance is 6 feet long and it appears correct when empty, a certain body placed in one scale weighs 120 lbs., when placed in the other, 121 lbs. Show that the fulcrum must be distant about jJg of an inch from the centre of the beam. 7. A dealer has correct weights but one arm of the balance is shorter than the other. If he sells two quantities of a certain drug, each apparently weighing 9J lbs. at 40s. per lb. weighing one in one scale atid one in the other scale, will he gain or lose ? 8. If in Art. 168, the length of the beam is =2a, CD = h, CQ==h, and 8 the angle which the beam makes with the horizon when the two weights P and Q are placed in the pans, show that, if W is the weight of the beam, tang ^-P-^^" 9. If the arms of a false balance be without weight, and one arm longer than the other by J of the shorter arm, and if in using it the substance to be weighed is put as often in the one scale as in the other, show that the seller loses | per cent, in his transactions. 174. The Roman Steelyard. This steelyard is a lever of the first class and consists of a bar moving about a fulcrum G and having a given weight w sliding on the longer arm. To the end of the shorter arm the body vrhose weight is required is attached and the weight w is moved along its arm until there is equilibrium. o D «^?G* j: /S III 1 1 1 T o Fig. 110. □ 1 w 1 [D The longer arm is graduated so that the point at which w is placed indicates the weight of the body. THE SIMPLE MACHINES. 171 To graduate the Roman steelyard. The point from which the graduations start must be where w is placed to balance the weight of the steelyard itself when no weight is attached. Let this point be 0. Then W being the weight of the steelyard w.GO=W' .Ca (i). Again, if a weight W is attached and there is equilibrium when w is moved to P, taking moments about C, we have w.GP=W' .CGJtW.AG (ii). By subtracting (i) from (ii) we get W OP w.OP^W.AG, or -=^ (iii). w AG ^ ' Hence if OB be divided up into portions each equal to AG, starting from 0, the graduation at which w rests gives the number of times W contains w. If w is one lb. the. graduations indicate pounds. The graduations are all at equal distances AG and if we wish to be correct to ounces each graduation must be divided into 16 equal parts. If the centre of gravity Q of the bar be in the longer arm the point will be found to be in the shorter arm. Ex. 1. If the fulcrum be four inches from the point to which the weight is attached, and the centre of gravity five inches from that end, and the weight of the bar equal to the moveable weight, find the position of zero gi'aduation. We have given that TF'=w, also G is one inch from C and in the longer arm, hence is also one inch from C and in the shorter arm. Ex. 2. The moveable weight is originally one pound and a weight of three pounds is substituted, the graduations remaining the same, how is a buyer afiected by the change, the c. G. being in the shorter arm? If the moveable weight were at P the buyer is charged for the OP weight 3 ^7-7- lbs. If the graduations had been constructed for a move- ijA. able weight of 3 lbs., then instead of OP we should have had OP where 0' is such that W .CG=ZCO', and the buyer should be charged for OP 3 ---J lbs. Hence CO is less than CO and therefore OP greater than OP. The buyer therefore gets more for his money than he should do. 172 THE ELEMENTS OF APPLIED MATHEMATICS. Ex. 3. The weight of a steelyard is 10 lbs., the body to be weighed is suspended from a point 4 inches from the fxilcrum and the c. a. of the steelyard is 3 inches on the other side of the fulcrum. Where should the graduation corresponding to one cwt. be situated, the moveable weight being 12 lbs.? The zero of graduation is got in this case from the equation 10 X 3 = 12 X CO by (i). Hence is 2J inches from the fulcrum. Also 112x4=12xOP, or OP=iJ^, from which it follows that the point required is distant from the fulcrum J-J^ - f or 34| inches. EXAMPLES. XXXV. 1. A steelyard 4 feet long has its c. G. 11 inches and its fulcrum 8 inches from A. If the weight of the machine be 4 lbs. and the moveable weight 3 lbs., find how many inches from A is the graduation denoting 15 lbs. weight. 2. In a Roman steelyard the sliding weight is 10 lbs., the gradua- tions for a difference of one stone are 3^ inches apart, find how far from the fulcrum is the point at which the bodies to be weighed are attached. 3. If the moveable weight be equal to the weight of the beam, and if the zero of the graduations bisect the distance between the fulcrum and the point of suspension of the body to be weighed, then the first graduation will coincide with the c. G. of the beam. 4. If a steelyard by use lose -^ of its weight, its c. G. remaining unaltered, show how to correct the graduations of the steelyard. 5. A steelyard is 12 inches long and with the scale-pan weighs 1 lb., the c. G. of the two being 2 inches from the end to which the scale-pan is attached. Find the position of the fulcrum when the moveable weight is 1 lb., and the greatest weight that can be ascertained by the steelyard is 12 lbs. 6. The beam is 32 inches long, the body to be weighed being attached at one end A ; the fulcrum is distant 5 inches and the o. G. of the beam 7 J inches from A. The weight of the beam being 1 lb. and that of the moveable weight 3J lbs., find the heaviest weight that can be weighed by this instrument. 7. A steelyard is damaged, either by rust or by losing a portion of its longer arm, causing it to weigh inaccurately, show that it may be repaired by the attachment of a suitable weight, so that we can use the original graduations. THE SIMPLE MACHINES. 173 175. The Danish Steelyard. This consists of a bar having at one end a heavy lump of metal, to the other end the body to be weighed is attached. We then observe about what point in its length the bar balances. Thus the fulcrum is moveable. AW+w o I J II p p ppp W Fig. 111. To graduate the Danish Steelyard. Let w be the weight of the bar and G its centre of gravity. Then if a weight W be attached, and the system balances about P, the upward force at P is Tr + w, and taking moments about B we get w.B0 = {'W4-w)BP, or BP = .^^BO. W -\-w Now first let W='w, then BP^ = \B0, mark this position of the fulcrum P,; next let W=2w, then BP,^^^BG = \BG, mark this position P^; in the same way taking W to be 3w, 4sw, &c. in succession we get the points Pg, Pi &c., where BPs = iBG, BPi = ^BG,&oc. Thus the bar is graduated, and if for instance w is one lb., the bodies which cause the fulcrum to take the positions Pi, P^, P3 &c. are of the respective weights one lb., two lbs., three lbs. &c. 174 THE ELEMENTS OF APPLIED MATHEMATICS. Observe that the distances BP^, BP^ &c. are in harmonica! progression. By taking IF equal successively to \w, Ji», \w, ... we get the corre- sponding distances of the fulcrum from B as f 5(?, ^BO, ^BQ, ... These points should he marked ^, ^, i) ... Notice that meir distances from O are in harmonical progression. Ex. 1. A Danish steelyard weighs 10 lbs. and the distance of its c. G. from the scale-pan is 4 feet, find the distances of the successive points of graduation from the c. g. By this Article BP= W+IO ^ ^" When 1^=1, jBPi=^2 = 33L feet, or (?Pi= j^ feet,, W=% BPi=^=^i&&t, or ffP2=ifeet, Tf=3, 5P3=fg= 3^5 feet, or (7P3=i§ feet, and so on. Ex. 2. A Danish steelyard has at one end a ball of metal 3 inches in diameter and weighing 8 lbs. The bar weighs 2 lbs. and is 12 inches long. It is graduated from end to end, the spaces being \ an inch apart. What are the greatest and least weights that can be measured by it. Let X be the distance of the c. G. of the bar and ball from the centre of the ball, then 8x=9,(^^—x), or a;=l;^ inches. Thus the o. G. is at the end of the bar. Also by this Article When the weight is greatest BP is least or ^ an inch, .-. 1{10+W) = \W, or Tf=2301bs. When the weight is least BP is greatest or 11|^ inches, .-. 2^(10-1- Tf)= 120, or F=J§lbs. EXAMPLES. XXXVI. 1. In a Danish steelyard the distance between the zero graduation and the end of the instrument is divided into 30 equal parts, and the greatest weight that can be weighed is 5 lbs. 7 ozs. ; find the weight of the instrument. 2. Find the length of the scale of the steelyard whose weight is 1 lb. and in which the distance between the graduations denoting 3 and 4 lbs. is one inch. 3. In a steelyard the fulcrum rests halfway between the second and third graduations, find the ratio of the weight in the scale-pan to the weight of the instrument. THE SIMPLE MACHINES. 175 4. The length of the scale of the steelyard is 24 inches and its weight is 7 lbs. ; find the distance between the positions of the fulcrum in weighing 14 lbs. and 21 lbs. respectively. 5. From a steelyard in which A is the point of the beam from which the scale-pan is suspended, and ff is the c. g. of the steelyard and scale-pan a small particle. of weight w at the middle point of AG has been broken off ; show that the apparent weight of a body determined by the steelyard will be too great by ^(n—l)w, where n is the ratio of the apparent weight of the body to that of the steel- yard and scale-pan. 6. Supposing the steelyard to become coated with rust, what would you do in order to use the steelyard without altering the graduations ? 7. If the weight of a Danish steelyard is 5 lbs. and the fulcrum is at a distance of 3 inches from the end for a weight of 10 lbs., show that in order to balance a weight of 15 lbs. the fulcrum must be moved f of an inch. 8. If a weight W balances with the fulcrum at a distance a from the c. G. of the steelyard, the distance the fulcrum must be moved in order to restore the balance when a weight w is added is equal to 176. The Wheel and Axle. This machine consists of two cylinders of different radii having the same axis and rigidly connected. At their ends are pivots which turn freely in fixed supports. The weight is raised by means of a rope coiled round the smaller cylinder, the force is applied by a rope coiled in the opposite direction round the larger cylinder. Fio. 112. 176 THE ELEMENTS OF APPLIED MATHEMATICS. Conditions of equilibrium. Neglecting the thickness of the ropes, take moments about the axis which gives, since the pressure on the axis has no moment, P.a^W .h (1); where a and h are the radii of the wheel and axle respec- tively, and P is the applied force. 177. Mechanical Advantage. W The mechanical advantage, or -p , here is a _ radius of wheel h radius of axle By making =- very great we can theoretically increase the mechanical advantage to any extent. Practically there are limits to the increase, if a be large the machine is unwieldy, if h be small the axle will be too weak. When a large mechanical advantage is needed, a modified form called the Differential Wheel and Axle, which will be ■ afterwards explained, is used. 178. The Capstan and Windlass are forms of the Wheel and Axle, the force is applied at the end of a spoke lying in a plane perpendicular to the axis. In the Capstan the axis is vertical, in the Windlass horizontal. Ex. 1. If the radii of the wheel and axle be respectively 40 inches and 4 inches, what weight would be supported by a force equal to the weight of 30 lbs. ? find also the pressures on the supports on which the axle rests. Let X be the required weight, then 4iB= 30x40, or ^=300 Ibpi The pressure on the supports = P+ 1^=330 lbs. weight. Ex. 2. A wheel and axle is used to raise .a. bucket from a well. The diameter of the wheel is 15 inches, and while it makes 7 revolutions the bucket which weighs 30 lbs. rises 5J feet. Find what is the smallest force that applied to a point on tiie circumference can turn the wheel. THE SIMPLE MACHINES. 177 Let X be the required force, then by the principle of work, Arts. 157, 158, the total work done is zero since the forces are in equilibrium, or a; X 2jrr X 7 - 30 X 5^=0, or a?x4^x^x7-30x^=0, from which we find a; to be J a lb. weight. Ex. 3. A weight is to be raised by means of a rope passing round a horizontal cylinder 10 inches in diameter turned by a winch with an arm 2^ feet long. Find the greatest weight which a man could so raise without exerting a pressure of more than 50 lbs. on the handle of the winch. Let X be the weight required. The radius of the cylinder is 5 inches. The arm of the winch, which corresponds to the radius of the wheel, is 30 inches. .-. i»x 5 = 50x30, or a;=3001bs. EXAMPLES. XXXVII. 1. If the radius of the wheel and axle be respectively 3 feet and 4 inches, what force must be applied to raise a weight of 40 lbs. ? 2. A man pushing at the end of a pole 4 feet long works a capstan whose diameter is 2 feet ; with what force must he push to overcome a resistance of 600 lbs. weight? 3. If the radius of the wheel be treble that of the axle, and the force and weight are together equal to 48 lbs. weight, find the mag- nitude of each. 4. Four sailors each exerting a force capable of raising 116 lbs. raise an anchor by means of a capstan whose radius is 1 ft. 2 inches and whose spokes are 8 feet long (measured from the axis). Find the weight of the anchor. 5. A weight of 17 lbs. just balances a- weight of 79 lbs. What will be the radius of the axle if that of the wheel is 17 inches ? 6. The radii of the wheel and axle are 17 inches and 4 inches and the weight is 12 lbs., find the applied force. 7. A force of 10 lbs. will raise a weight of § lbs., a force of -^ lbs. will raise a weight of 10 lbs., show that the radius of the wheel is twice the radius of the axle. 8. A man exerting a force of 50 lbs. weight works a capstan. He walks 4 yards round, two feet of rope being pulled in ; what is the weight raised ? J. 12 178 THE ELEMENTS OF APPLIED MATHEMATICS. 9. A cage is suspended by a rope passing over an axle and a man standing in the cage draws himself and the cage up by pulling at a rope passing over the circumference of the wheel. If the joint weight of the man and cage be 14 stone and the radius of the wheel 5 times that of the axle, the rope being pulled into the cage at the uniform rate of 2 feet a second, find the tension exerted by the man and the horse-power at which he works. 179. The Pulley. (i) Fig. 113. The pulley consists of a wheel which can turn freely about an axle, the axle is attached to a framework called the block. In a groove on the circumference of the wheel runs a string acted on by a force at each end. The pulley in fig. (i) is attached to a fixed beam above and can only turn on its axle, the lower pulley of fig. (ii) can move up and down as well as turn. The first is called a fixed pulley, the second a moveabk pulley. The tension of the rope connected with a pulley is assumed to be the same throughout its length, although this is not strictly true owing to Mction. The weight of the string is negligible. 180. The fixed pulley has unity for its mechanical ad- vantage, for W and P being each equal to the tension of the W string are themselves equal, hence -p- is unity. THE SIMPLE MACHINES. 179 The use of a fixed pulley is to change the direction of the applied force. In the second figure a weight W is attached to the block, W is supported by the (equal) tensions of the string on each side, each tension is equal to P, hence W=2P, and the mechanical advantage is 2. The portions of the string are taken to be parallel. 181. System of pulleys. Pulleys are combined to form a system as represented in the figures. The upper pulleys are fixed, the blocks of the lower ones are joined together. The same string passes round all the pulleys, to one end of it a force P is applied, its other end is fastened to the upper block. In the case when there is one more pulley above than below, the end is fastened to the lower block. In practice the two sets of pulleys turn on the same axes. Here the strings cannot be strictly parallel, but the, error is small enough to be disre- garded. The attached weight is W, the weight of the lower block of pulleys is w, and n is the number of strings connected with the lower block. The whole weight W+w is sup- ported by the tensions of n parallel strings, each tension is equal to P being the same' throughout the string, hence W + w = nP. W — nP, and the mechanical ad- 12—2 Fie. 114. If w is negligible vantage is n. 180 THE ELEMENTS OF APPLIED MATHEMATICS. 182. Principle of VTork. If the lower block receive an upward displacement equal to h, each portion of the string is slackened, and P descends through a space h for each portion of string, that is, through a total distance nh, if n is the entire number of strings in the lower block. Hence by the Principle of Work, Art. 157, since the forces are in equilibrium Wh-Pnh=0, or W=nP. Since the space traversed by the lower block is always - th of that traversed by the applied force, the acceleration of the lower block f is - , where / is the acceleration of the point of application of the force P in the case when W is not equal to nP. Ex. 1. What force will be necessary to support a weight of 32 lbs., there being nine pulleys each weighing one lb. ? Here since the number of pulleys is odd, the number of parts of the rope at the lower block is odd and the rope is attached to the lower block. There are thus 4 pulleys in the lower block. The entire weight supported is 36 lbs. .-. 9P=36, or P=41bs. Ex. 2. If a weight of 10 lbs. support a weight of 18 lbs., and a weight of 11 lbs. support a weight of 20 lbs., find the number of strings and the weight of the lower block. Let n be the number of strings and w the weight of the lower block 18+?»=10re, 20+w=11m. Hence m = 2, w=% Ex. 3. Find the smallest weight which can be raised with mechanical advantage if there are 10 pulleys, each pulley weighing 4 lbs. Here there are 5 pulleys on the lower block, the total weight is 1F+20. .-. Tr+20=10P. Por mechanical advantage greater than unity W must be greater than P, 10 TF „ „ greater than PF+ 20, W must be greater than ^ lbs. wt. THE SIMPLE MACHINES. 181 EXAMPLES. XXXVIII. 1. If there are 9 pulleys altogether and each pulley weighs 1 lb., what force is required to support a weight of 104 lbs. ? 2. What weight can be supported if there are 4 pulleys in the lower block and the total number be even, the weight of the lower block being 4 times the force P 1 3. If weights of 5 lbs. and 6 lbs. support weights of 73 lbs. and 88 lbs. respectively, what is the weight of tlae lower block, and how many pulleys are there in it ? 4. If the lower block weigh 24 lbs. and contain 5 puUeys, the string being fastened to the lower block, what weight can be raised by a force equal to 21 lbs. wt. ? 5. Find the pressure on the ground if a man with 8 weightless pulleys sustains a weight J of his own. 6. Ten weights each of 20 lbs. are to be lifted to a height of 8 feet from the ground. Show how a system of puUeys may be arranged to lift them together, the weight of the pulleys being neglected, by exerting a force equal to one of them. 1. A man whose weight is 12 stone raises 3 owt. by means of a system of pulleys, there being 4 pulleys in each block, and the string being attached to the upper block. What is his pressure on the ground? 8. What must be the relation between the radii of the pulleys in the lower block in order that they may be all grooved in the same piece? 183. Two other arrangements are usually described. The figures show arrangements of three pulleys, there being as many strings as pulleys. In I. each string is fastened to the supporting beam, in II. each string is fastened to the beam which supports the weight. Notice that one figure is the inversion of the other. W is the weight supported, the weights of the pulleys are neglected, the force applied is P. 182 THE ELEMENTS OF APPLIED MATHEMATICS. I. II. T. V IP \J \ W Fig. 115. Case I. The tensions of the strings being 2\, T^ and ^'3, we have since the tension of a string is the same throughout the tensions of the strings of the next pulley keep it in equilibrium, hence since they are parallel, by Art. 96 similarly T^ = 2T^ = 2^P, also Tf = 22's = 2=P. Thus W=8Pisthe required relation between Tf andP for equilibrium. The mechanical advantage is 8. If 'there are n pulleys we see similarly that the relation would be Tr = 2»P. THE SIMPLE MACHINES. 183 Case II. The tensions being T^, T^ and T^, we have r,= 2Tx=2P, 2's = 22'j = 2^P. And W = T^+T^ + Ts = P + 2P + 2^P=P(l + 2 + 2=) = P(2'-1). Hence in this case the relation between W and P is Tf=(2=-l)P = '7P. For n pulleys the relation would be ■pr = (2«-l)P. 184. If we take the weights of the pulleys into account, the previous equations are replaced by the following ; Wj, Wj, and Wj being the weights of the pulleys. Case I. ^1=-?, the first pulley is in equilibrium .under the forces T^+w^ downwards and 2?^ upwards, hence T^+'Wi=2Ti, or r2=2P-Wi, similarly T^-irWi=iT^, or T^='i{2P-w.^-w^, =2^F-2wi-w^. Also W+w^=2T3, or W=2{2^P-2wi-Wi)-v)3; .-. ]F=23i'-2%i-2W2-W3. If there are n pulleys we find similarly ir=2»P-2''-iwi-2»-2w2 -Wn. When the weight of each pulley is w^ PF'=2»/'-(2»-l)wi. Case II. 2\=P, T2=2Ti + Wi=2P+Wi, T3=2T2+w^==2{2P+Wj)+Wi, Also W=Ti+T^+T3 =P+2i'+tffi+22p+2wi+Wj, =P(1+2+22)+Wi(1+2) + m;2. =7P+3wi+W2- 184 THE ELEMENTS OF APPLIED MATHEMATICS. If there are m pulleys we find similarly =P(l+2 + ... + 2''-i)+Wi(H-2 + ... + 2»-2) +W2(l + 2+...+2»-s) + ... + (22-l)w„_2+(2-l)w^i. =P(2''-l)+Wi(2»-i-l)+W2(2''-2-l) + ... + (22-l)w„_2+(2-l)«;„_i. If the weight of each pulley is Wj we have Tr=P(2»-l) + Wi{2»-i+2»-H...+2-(»i-l)} =P(2»-l) + Wi(2»->i-l). Notice that the weights of the pulleys in this case assist P. 185. The pull on the fixed beam. The fixed beam undergoes a downward pull which is clearly the resultant of the total weight supported and P. In Case I. if B is this pull, we have R = W-P, in Case II. R = W+P. If the pulleys have weight we must add the sum of their weights to W. Ex. 1. There are 4 moveable pulleys arranged as in Class I., find TT if P is equal to the weight of 10 lbs., neglecting the weights of the pulleys. Here F=2* x 10=160 lbs. weight. Ex. 2. If in the last case the weight of each pulley is 4 ozs., find W. In this case TF=2*xlO-(2*-l)^ = 160-J^ = 156} lbs. Ex. 3. If the pulleys are arranged as in Class II., find W when P is the weight of 10 lbs., neglecting the weights of the pulleys. Here pr=(2*- 1)10=150 lbs. Ex. 4. Taking the weight of each pulley as 4 ozs., and P as 10 lbs., find W. In this case lF=(2*-l)10+(2*-4-l)^ = 150+llx^ = 152f lbs. THE SIMPLE MACHINES. 185 186. motion of tbe Systenu. In the system of pulleys of Art. 181 if P descends a distance h the n strings in connexion with W are shortened a total amount h, hence each is shortened a distance - . n Hence if v is the velocity of P, the velocity of F is - r, therefore if / is the acceleration of P, the acceleration of TF is - /; For the motion of P we have the equation ^f=P-T. (i), and for the motion of W we have the equation -^=nT-W (ii), from (i) and (ii), £fnP + ^=nP-W, In Case I., Art. 184, if the lowest pulley receive an upward dis- placement h, each portion of the string passing round it will be slack- ened. The next pulley must therefore be raised through a distance 2A to tighten the string, similarly, the next must be raised through twice the last distance, or 2%. If there are n pulleys the space moved through by the last will be 2^~^h. If the velocity of the lowest pulley is v, the velocities of the others are 24>, 2^1;, ...2»-i», and therefore if the acceleration of the lowest pulley is /, the accelera- tions of the others are 2/, 2y,...2»-v. For a system of three pulleys we therefore have, |23/=ri-p, if From these equations T^, T^, T^ and/ can be found. 186 THE ELEMENTS OP APPLIED MATHEMATICS. EXAMPLES. XXXIX. Exs. 1 — 8 come under Case I., Exs. 9—16 under Case II. 1. If the weight supported is 1 cwt. find the force applied, using 3 pulleys of Case I., neglecting their weights. 2. If the weight of each pulley in the last question be 2 lbs., find the applied force. 3. There are 4 pulleys, each weighing 2 lbs. What weight can be raised by a force equal to the weight of 20 Ibs.i 4. There are three equal pulleys, and a force equal to the weight of 3J lbs. is required to support a weight of 21 lbs., find the weight of each pulley. 5. If ttere are 3 pulleys which weigh respectively P, JP, JP beginning with the lowest, if a force P be applied, show that it can support a weight 5P. 6. There are 5 pulleys whose respective weights are 5, 4, 3, 2, and 1 lbs., beginning with the lowest, what force will support a weight of 71 lbs.? 7. What is the mechanical advantage when there are n pulleys each as heavy as the weight to be raised? 8. By use of 4 pulleys of equal weight a certain weight can be supported by a force of 7 lbs. weight, but if a fifth similar pulley be used the same weight can be supported by a force of 4 lbs. weight. Find the supported weight and the weight of a pulley. 9. In the arrangement of Case II. find the weight supported by three pulleys, by use of a force of 10 lbs. weight, neglecting the weights of the pulleys. 10. If each pulley in the last question weighs 5 ozs., find the weight supported by a force equal to 10 lbs. weight. 11. If the pulleys are of different weights, show that the most ad- vantageous arrangement is got by placing them in order of magnitude, the greatest being lowest. 12. If four pulleys each weighing 2 lbs. be used, find the force required to support a weight of 238 lbs. 13. Find what weight can be raised by a force of 7 lbs. weight, if there are 3 pulleys arranged in the least advantageous manner whose weights are 5, 4 and 3 lbs. respectively. 14. Find the mechanical advantage when the pulleys are five in number and the weight of each equal to ^th of the applied force. 15. If a force P supports a weight W, show that a force P+w would support a weight W+ii/, where w is the weight of each pulley and w' is equal to (2"- 1) w. 16. In the case where 3 pulleys are used, if the diameter of each pulley be 4 inches, find to what point of the bar the weight should be attached in order that the bar may remain horizontal. THE SIMPLE MACHINES. 187 187. The inclined plane. An inclined plane is a plane making an angle less than a right angle with the horizon. A line in the plane perpendicular to its intersection with the horizon is called a line of greatest slope, and a plane passing through the Vertical and a line of greatest slope is called a principal plane. Fm. 116. In the figure AB, CD and EF are lines of greatest slope. The plane is supposed smooth, and hard enough to sustain any pressure. 188. A body is kept in equilibrium on an inclined plane : by the action of three forces, viz. its weight W, the pressure of the plane R (perpendicular to the plane), and a force P. Since W and R both lie in the same principal plane it is evident that that for equilibrium P must also lie in this plane. We shall now take the two cases when P acts horizontally and along the plane respectively. Case I., P horizontal. 188 THE ELEMENTS OF APPLIED MATHEMATICS. The sides CA, BG, AB of the triangle ABC are respec- tively perpendicular to the forces P, W and R, hence by the converse of the Triangle of Forces, Art. 78 P:W:R = GA:BG:AB, CA BG~AB' This may also be expressed P : W : iJ = height of plane : base : length. Since F= W^ , it follows that P= Tf tan a, or W=Rcosa. Case II., P along the plane. Fio. 118. Make the vertical line A'B' = AB, also make the angles at'%A' and B' equal respectively to those at A and B; the triangles ABG and A'B'G' are then equal in all respects. Euc. I. 26. Also the sides of A'B'G' are parallel to the directions of P, R and W, viz. FA' to W, A'G' to P, G'B' to R, hence by the triangle of forces P:R:W= A'G' : EG' : A'B' = AG:BG:AB. THE SIMPLE MACHINES. 189 Or, P:R: W= height : base : length. AG Since P= Tf-jn, .'. P equals Tf sin a, BC iJ= ^-jDi •'• -^ equals PTcosa. 189. We may apply the method of work to Cases I. and II. as follows : In Case I. let the body be displaced from the top to the bottom of the plane, then R does no work, and the total work done by W and by F must be zero. Art. 157, hence since the forces are in equilibrium WxAC-PxBG=0. In Case II. take the same displacement, then similarly WxAG-PxAB = 0. Ex. 1. Find the weight which a force of 10 lbs. acting horizontally can support on an inclined plane whose height is 2 feet and base 13 feet. ^^^"^^ P=hiliht' or W= 10 X ^ = 65 lbs. weight. Ex. 2. What is the inclination of the plane to the horizon on which a horizontal force of a/3 lbs. can support a weight of 3 lbs. ? hd^^ P^^/3^J_ base - W~ 3 ^3' Thus the required angle is 30°. Ex. 3. An inclined plane rises 2 in 5, what force acting along the plane will sustain a weight of 200 lbs. ? Here ^=200 x f =80 lbs. weight. EXAMPLES. XL. 1. A weight of 8 lbs. is supported on an inclined plane by a horizontal force of 6 lbs. weight, if the length of the plane is 20 feet, what is its height ? 2. A truck whose weight is 3 tons is kept at rest by a rope on a gradient of 1 in 60, find the tension of the rope. 3. What force acting horizontally could keep a weight of 12 lbs. at rest on a smooth inclined plane whose height is 3 feet and base 4 feet, and what is the pressure on the plane ? 190 THE ELEMENTS OF APPLIED MATHEMATICS. 4. What horizontal force will keep in equilibrium a weight of 9 lbs. on an inclined plane and produce a pressure of 15 lbs. weight on the plane ? 5. If the length of the plane is 40 inches and the height 8 inches, what is the mechanical advantage in the two cases ? 6. In the case when the applied force acts horizontally the plane is turned over so that the height becomes the base, show that the force required to support a given weight will be greater if the original base exceeded the height. 7. What force must be applied at the circumference of a wheel 6 feet in diameter in order to drag a ton weight up a smooth inclined plane of 1 in 50 by means of a rope wound round an axle of 9 inches diameter ? 8. If the pressure- on the plane in Case II. be half the weight, what is the angle of the plane ? 9. If the height of a plane be 3 J feet and the length 12 feet, and a string can just bear a weight of 24J lbs. hanging freely, find the greatest weight it can support when fastened to a point on the plane. 10. If A and b are the height and base of an inclined plane, and if a force P can support a weight Tf when acting horizontally and a weight W when acting along the plane, show that W -.W :: b : '/W+b^. 11. If P be the horizontal force which can support a weight W and jP the force along the plane which is sufficient to support W, then F : F' :: length of plane : base of plane. 190. When P makes an angle 6 with the plane we proceed as follows : The sum of the resolved parts of the forces along the plane is zero, p. 68. .-. Pcos5- Tfsina=0 (i). Also the sum of the resolved parts perpendicular to the plane is zero, .-. Psin5+^=Tfcosa (ii), or R=Wcosa-Psia6 ™ IF sin a . . „ ... = Ircosa ^— sm5, from (i) Fig. 119. cos 6 W = ;; (cos a COS ^ - siu a siu 6) : cos d ^ ' ' R= Wcos(a+6) COB 6 THE SIMPLE MACHINES. 191 Ex. 1. Show that the direction of the least force required to support a given weight on an inclined plane is along the plane. Frona (i) if P is the force and 6 its inclination to the plane Pcos5= TTsina, p_ Tf sina ~ cosd ' we see that P is least when cos Q is greatest, or when 6=0. Ex. 2. A body whose weight is 20 lbs. is kept at rest on an inchned plane by a horizontal force of 10 lbs. together with a force of 10 lbs. acting up the plane, find the inclination of the plane to the horizon, and also the pressure on the plane. Kesolving along the plane we have 20 sin a= 10+ 10 cos a, ^v* resolving perpendicular to the plane, iJ=20 cos a+ 10 sin a. -*-io From the first equation ^^'»- ^2"- sin a , siU'^ a 1+COSo ^' {I + COS of , 1 — cos a , hence — — =i, 1 + cosa *' from which cos a = f . It follows that sin a = Vl-^= f , hence R=W lbs. weight. EXAMPLES. XLI. 1. A body whose weight is 20 lbs. rests on an inclined plane whose inclination to the horizon is 60°, find the supporting force, it being supposed to make an angle of 30° with the normal. 2. A body weighing 6 lbs. is placed on a smooth plane which is inclined at an angle of- 30°, find the two directions in which a force equal to the weight of the body may act to maintain equilibrium; 3. A weight 2P is kept in equilibrium on an inclined plane by a horizontal force P and a force P acting parallel to the plane ; find the inchnation of the plane to the horizon and the pressure on the plane. 192 THE ELEMENTS OF APPLIED MATHEMATICS. 4. If the pressure on the plane be an arithmetic mean between the weight and the applied force, and the inclination of this force to the horizon be 2a, where a is the inclination of the plane, sin2a=f. 5. A force P acting at an angle with the plane whose cosine is ^ keeps a weight at rest. If P act at half its former incUnation, find in what direction a force f P must act in order to keep equilibrium. 191. The Screw. The form of the screw is most easily described as follows : Take a right-angled triangle of paper ABO and a cylinder DF, then keeping BG parallel to the axis of the cylinder, wrap the base BA round the cylinder, the hypotenuse AC will form a spiral line on the cylinder. Any line FG on the surface of the cyhnder parallel to the axis will cut this spiral in a series of points. The distance between two successive points is called the step of the screw. A broad groove is cut between the spirals leaving a ridge called the thread. The screw works in a nut whose groove fits the thread. As the screw turns it also rises; the distance risen per unit angle turned through is called the pitch. We see that when the screw has turned through four right angles it has risen a distance equal to a step, hence if ^ is the pitch, a step is equal to p x 27r. If in the figure QB is a step, PR must be equal to the circumference of the cylinder, hence if r is the radius of the cylinder a step=2jirtana, where a is the / CAB. Hence px27r = 27rr tan a, or ^=rtana. THE SIMPLE MACHINES. 193 If a weight W is placed on the screw and the nut is held, the screw wiU descend, to support W a force P is applied at the end of a lever represented in the figure. The condition of equilibrium is got from the principle of work as follows : Let the screw descend through a step, then if P acts at a distance f from the axis of the screw F'xstep=Px27r/, or Wxpx27r=Px27rt'', hence Wp=Pi'. 192. Differential Macbines. The DifferentiaZ Wheel and Axle. W. W. 2 2 Fig. 123." This consists of three unequal cylinders having a common axis. Bound the largest is coiled the rope by means of which the force is applied, round the other two the portions of the rope which support the weight. We see from the figure that the ropes round the largest and smallest cyUnders are coiled in the same manner, the rope round the middle cyUnder in the opposite way to this. J. 13 194 THE ELEMENTS OF APPLIED MATHEMATICS. If IF is the weight attached to the pulley, the tension of each rope is ^. Let c, a, and 6 be the radii of the cylinders, beginning with the largest. Taking moments about the axis, we have Pc+ f- w W 2c • • P~ a-h By making a and 6 nearly equal we can get a large mechanical advantage. Ex. In the differential wheel and axle, if the radius of the wheel be one foot and the radii of the two portions of the axle 5 and 4 inches respectively, what force wiU support a weight of 48 lbs. ? Ans. 2 lbs. weight. The Differential Screw. Fig. 124. CO OC is a strong frame. In CO a groove is cut in which the thread of the screw on AD works, AD is hollow and the solid screw DE works inside it. To the lower screw a board is attached which is guided to work up and down by smooth vertical grooves cut in CG, thus the lower screw can move up or down but cannot rotate. The substance to which pressure is to be applied is placed beneath the board, W is the resistance offered by it. THE SIMPLE MACHINES. 195 Let I be the length of AB, p and p' the pitches of the respective screws. When AD has made a complete rotation it has descended a stepov a distance %tp, in the same time the smaller screw has ascended a distance 2jr»', relatively to the larger one, or descended a distance 2jT{p—p'), and this is the distance through which resistance is over- come, hence if P is the applied force we have by the principle of work P%tI=W%t{:p-p'), W I where I is the distance AB from the axis at which P is applied. The motion of the lower screw, corresponding to a considerable motion of AD, is extremely small. The Differential Pulley. T/\ FiQ. 125. The figure on the right-hand side represents the prin- ciple of this machine, a sketch of which as used is given in the figure on the left. 13—2 196 THE ELEMENTS OF APPLIED MATHEMATICS. An endless chain ABODE passes round the circumfer- ences of two concentric wheels which are supplied with teeth. It is found that for ordinary weights the chain from ^ to ^ may hang freely. Let a and h be the radii of the larger and smaller wheels and T the tension of the chain which supports the weight W, then Also by taking moments about the centre of the wheels, we have Pa + CT = Ta ...P = jTF^^. EXAMPLES. XLII. 1. The arm of a screw-jack is one yard long and the screw has two threads to the inch. What force must be applied to the arm to sustain a weight of half a ton ? 2. If a weight of 3 tons be raised 6 feet by a screw making. 240 revolutions, find the force, the arm being 2 feet long. 3. A man by exerting a force of 12 lbs. with each hand can sustain a weight of 8 cwt. by means of a screw which has a double arm of 4 feet total length, find the pitch of the screw. CHAPTER XII. FRICTION. 193. No surface is perfectly smooth, and when two surfaces are in contact the small ridges on one surface fit into small depressions on the other, so that the roughness of the surfaces retards the motion of one on the other. This retarding force due to roughness is called the force of friction. This force is called into play in using all machines and part of the force applied is spent in over- ooming friction, part only in doing useful work. That friction is in many cases of practical advantage is seen from the fact that it is the friction between our feet and the ground which enables us to walk, and without fric- tion we could not keep hold of objects. Another example of the use of friction is afforded by the locomotive engine. Take the case of an engine weighing 30 tons and suppose that the part of the weight borne by the driving-wheels is 10 tons. We shall see subsequently (Arts. 194, 195) that the force of friction is equal to the pressure x a number called the coefficient of friction, and since the coefficient of friction for wrought iron on wrought iron is '2, the force of friction which acts on the driving-wheels of the engine is 20 X '2 tons, i.e. 2 tons. This friction opposes the rotation of the driving-wheels, i.e. acts in the direction in which the train is moving. It is thus the force which impels the train. It is found that a pull of about 10 lbs. for every ton is sufficient to sustain motion, hence the engine should be able to draw along the level a weight of 4480 -r 10 tons, i.e, 448 tons. 198 THE ELEMENTS OF APPLIED MATHEMATICS. 194. Methods of estimating ft-iction. One method of determining the friction between two materials we shall now describe ; a weighted slab of one material I I is placed on a horizontal planed ^ ^ -l^ surface of the other material, the ^ slab is pulled by a horizontal string with gradually increasing force, more and more friction is p ^^„ thus called into play to counter- act this force, until at length the slab begins to move uni- formly. The friction then exerted is called limiting friction. The gradual increase of force is obtained by attaching a series of weights to the string. The weight required to make the slab move uniformly on the table is equal to the limiting friction, which can there- fore be calculated. As the result of experiment we thus arrive at the follow- ing law : — If F is the limiting friction, and R the pressure of the slab on the plane (equal to the weight of the slab), then F=f,R, where fi is the same for the two given materials whatever the value of R. The quantity /4 is called the coefficient of friction. In the last Article the value of fj. was '2. It is necessary to give the slab a slight motion, since otherwise, owing to its weight, the surfaces become very slightly compressed and a force of coherence is introduced in addition to fnction. 195. The Angle of Friction. Another method of finding the value of fi is that of placing the weighted slab on a plane of the other material, and then tilting the plane until the slab begins to slide with a uniform motion down the plane. As before the slab should be started. Let 6 be the inclination of the plane for which the slab moves with Fig. 127. FRICTION. 199 uniform velocity down the plane, then since the slab has no acceleration, by resolving along and perpendicular to the plane we obtain if F is the limiting friction F-Wsme = 0, iJ-TFcose = 0. Therefore J" = i2 tan e. Thus iM is equal to tan e. The angle e is called the angle of friction. It is found that F=a+iiR, where a is a small quantity independent of F and R gives more accurate results, thus for pine- wood on pine-wood it is found that the values of J" (in lbs.) calculated from the formula i?'=l-44-(-0-2525, give results differing from the actual values of F by only about -3 lbs. The values of R for which this holds lie between 14 lbs. and 112 lbs. (Sir R. Ball's Experimental Mechanics.) The formula F=fi.R is usually sufficiently accurate for most purposes. 196. The Laws of Friction. As a consequence of these experiments the following re- sults have been established. 1. The greatest amount of friction is called into action when motion is about to take place, it is then called the limiting friction. 2. Limiting friction is proportional to the pressure, or F = iiR. For different materials in contact fi has of course different values. 3. Friction is independent (i) of the extent of the area of contact, (ii) of the velocity with which one body moves over the other. Since friction tends to prevent motion, not to originate it, it is often called a passive resistance. The above laws relate to sliding friction, when a body such as a wheel rolls on the ground rolling friction is called into play, this is very much less than sliding friction. 200 THE ELEMENTS OF APPLIED MATHEMATICS. 197. The cone of fiiction. +R Take the case of any body touching a table at the point 0, there is a normal force E, and a force of friction F acting along the table, we have seen that the magnitude of F may be anything between zero and fiR. The resultant of F and JR is called the total resistance of the table. It is easy to see W that p is the tangent of the angle which the total re- sistance makes with the normal, hence if is this angle, tan = -^ . It F But we have seen that -p may have any value between zero and fi, hence tan 9 must lie in value between and fi ; that is between and tan e, where e is the angle of friction, in other words 6 lies between and e. So that the total reaction cannot make with the normal a larger angle than e. If we describe round the normal as axis a cone whose semi- vertical angle is e, this is called the cone of friction, and we see that the direction of the resultant reaction lies within this cone. By increasing R we increase F, so that we may make F of any magnitude we please, but cannot bring its direction outside this cone. This result is clearly true for all surfaces touching at one point. FRICTION. 201 198. Beam resting against a wall. As an application of the preceding we shall take the case of a beam resting in a vertical plane against a rough vertical wall and the ground. Fig. 129. Let the vertical plane through the beam cut the cones of friction at A and B in the lines As, Ar, Bs, Bp. The total reaction at A must lie within the triangle Asr, and the total reaction at B must lie within the triangle Bsp. Hence their intersection is at some point within the area pqrs. For equilibrium the line of action of the weight of the beam must pass through 0, Art. 136. In the case of the first figure there cannot be equilibrium, in the case of the other, where the line of action of the weight does inter- sect the area pqrs, there will be equilibrium. 199. Body falling down a rough inclined plane. Another instance of the effect of friction is afforded by the case of a body falling down a rough plane whose inclination to the horizon is a. Since there is no acceleration perpendicular to the plane, we have jB-Trcosa = 0. The force down the plane is TT sin a - F, where F=fiR. Fig. 130. 202 THE ELEMENTS OF APPLIED MATHEMATICS. Hence the force down the plane is W sin. a — fi Wcos a, and the body's acceleration is g (sin a — /u, cos a). If V is the velocity with which the body reaches the bottom, supposing it to have started from rest at the top, if = 2g (sin a — /i cos a) x AB. The velocity is thus less than it would be if there were no friction by 2giJ, cos a x AB. Part of the work done by the weight of the body in falling has been used to generate its kinetic energy, the other part has been spent in overcoming friction, the energy corresponding to the work thus spent appears in the genera- tion of heat. 200. Examples. As illustrations of the laws of friction the solutions of several simple examples are added. 1. A ladder rests against a wall and the ground, the coefficients of friction between the ladder and the wall and ground respectively are fi and ^j. Find the inclination of the ladder when it is on the point of slipping down, (it is then said to be in limiting equilibrium). "/ ^ B 7 / X A / R. c/^w i/,mr r w Fio. 131. Kesolving horizontally and vertically and taking moments about B we have, I being the length of the ladder and W its weight and 6 its inclination to the horizon, ^R-Ri=0, R+iiiRi= W, Rl cos 6= W- cos 6+fiRl sin S. FBICTION. 203 From the first two equations, W R: hence from the third equation cot^=Ji^=^ R-Z l-^A^l Notice that the direction of the total reactions at A and B intersect on the line of action of W. 2. If in the last case a weight w be placed on a rung of the ladder at C, where BG—nl, find the limiting position of equilibrium. In this case the equations are fiR-Ri=0, R+iiiRj^=W+w, Rlcos6= W - cos d+wnlcos 6+ fiRlsm 6. Hence i?=^,cot^=_^— = ^^(^^+") 1+m R-^-nw Tr(l-,.Mi)-2«'(»ip,.i + l-l) A Observe that if F'(l-;i/ii)=2M;(w/i/xi+l-l), ^ is zero, orthe ladder will rest in any position. Ex. 3. Find the direction and magnitude of the least force required to drag a heavy body up a rough inchned plane. The forces acting on the body are its weight W, the normal pressure of the plane R, the friction jiR, and the re- quired force P. Under the action of these forces it is just on the point of moving up the plane. Hence these forces are just in equilibrium. Resolv- ing along and parallel to the plane we have Pcosd= Prsino-)-/xi?, Psin e= TFcos a- iJ. ^'<*- ^^'^■ Multiplying the last equation by ^ and adding we get P(cos^-|-;isin5)= Tr(sina-l-/iCOSa). Now put /i=tan e, then we have p_ „ sina-t-cosatanf _ ™sin^a+e) ~ cosfl-f-sin^tauf ~ cos(d— t)' In order that P may be as small as possible cos (d- e) must be as large as possible, that is we must have 6=i. Hence the required force P makes with the plane the angle of fric- tion and is of magnitude TTsin (a-t-f). 204 THE ELEMENTS OF APPLIED MATHEMATICS. Ex. 4. A solid cube rests on a rough table, it is pulled by a horizontal string attached to a point in one face, it is required to deter- mine whether the cube will slide or turn about its edge. Let ABCD be the section of the cube which contains the string. We suppose this section to bisect the cube and contain G its centre of gravity. If possible let the motion begin by the cube turning about its edge through A. The forces are the weight of the cube W, the tension of the string and the total reactions of the edge. All these total reactions may be replaced by one force R' which passes through A. It is clear that R' must pass through the intersection K oiW and P, and this it cannot do if the angle OAK is greater than e. OK Hence for turning about the edge yr-j i^ust be less than tane, OK 1 that is, ^r-T < u, or OA > - x half the side of the cube. OA '^ /i Ex. 5. A drawer is to be pulled out by a single force parallel to the •drawer, it is required to find how far from the middle of the end of the drawer the force can be applied so as not to cause the drawer to jam. Let the point N be where P must be applied so that jamming may just begin. Let ON=si:, where is the centre of the end of the drawer whose length is I. The drawer will now press against its sides at B and B, the total resistances at B and D meet on the line of action of P. Since there is equilibrium, by resolving per- pendicularly to the length of the drawer we see that/i-iE'=0, resolving parallel to it we get P=ij,R+iiR' = 2nR, taking moments about Px=Bl, FiQ. 134. hence I If P is applied at a distance from greater than le the drawer will not move however great P may be. If the drawer is very long, or I great compared with AB, the point N may lie beyond A, i.e. the drawer wiU not jam at all. FBICTION. 205 EXAMPLES. XLIII. 1. A weight of 60 lbs. is on the point of motion down a rough inclined plane when supported by a weight of 25J lbs. parallel to the plane, and on the point of motion up the plane when under a force of 32^ lbs. parallel to the plane, find the coeflacient of friction. 2. Show that the difiFerence between the greatest and least forces which acting at an angle « to a plane inclined at an angle u to the horizon sustain a weight TT is 2 Tf ^^ — ^ — -— , where \ is the cos2e+cos2\ ' angle of iWotion. 3. A ladder rests against a vertical wall and the ground, the co- efficients of friction there being /i and /*'. If the ladder is on the point of slipping at both points, then if 6 is the inclination of the ladder to the horizon, 2 tan 6=—,—iu 4. A weight W is placed on a rough horizontal table and moved along by a weight P hanging over the edge and attached to TF by an inextensible string, prove that the acceleration of W is P-yiW where /i is the coefficient of friction. 5. A beam rests against a wall and the ground, the coefficient of friction at both ends being J, find when it is on the point of slipping. 6. A uniform beam standing on a horizontal floor and leaning against a horizontal wall is just on the point of slipping when it is equally inclined to both floor and wall which are equally rough. What is the coefficient of friction, and what is the horizontal resistance of the wall? 7. Find the work done in dragging a load of 6 cwt. up a rough inclined plane whose height is 3 feet and base 20 feet, the coefficient of Motion being f ^- 8. A homogeneous sphere cannot rest upon an inclined plane how- ever rough. 9. A uniform ladder rests at an angle of 45° with the horizon with its upper extremity against a rough vertical wall and its lower ex- tremity on the ground. If /i, p,' be the coefficients of friction between the ladder and the ground and wall respectively, show that the least horizontal force which will move the lower extremity towards the wall is l + 2^-,M' . 2 l-p' 206 THE ELEMENTS OF APPLIED MATHEMATICS. 10. A heavy cube (of weight 100 lbs.) rests on a rough floor on which it cannot slip. Prove that the least force required to begin to raise one edge of it off the floor is about 35J lbs., and find where it must be applied. 11. A man holds the end J of a uniform stick AB in his hand, the other end B being on the ground. If the stick be always kept at the same inclination (30°) to the horizon, and the angle of friction between B and the ground be 15°, the horizontal force required to pvsh B with uniform velocity is to that required to pull it as ^^^3 + 1:2. 12. Prove that a train going at the rate of 45 miles per hour will be brought to rest in about 378 yards by the brakes, supposing them to press with f of the weight on the wheels of the engine and brake-van, which are J of the weight of the train, the coefficient of friction being •18. 13. A sphere of weight W is placed on a rough plane inclined to the horizon at an angle a which is less than the angle of friction, show that a weight W— ; — fastened to the sphere at the upper cos a - sm a end of a diameter parallel to the plane will just prevent the sphere from roUing down the plane. 14. Two rough spheres of equal radii but unequal weights W-^ and W^ rest in a spherica;l bowl, their c. G.s coincide with their centres and the line joining them is horizontal and subtends an angle 2a at the centre of the bowl, prove that the coefiicient of friction between them IS not < ■— — ?7Ftan TTi+TFj (M)- 15. Weights W and W of two different substances (coefficients of friction (X and y!) are supported on a rough double inclined plane (of angles a and a') by means of a string passing over the vertex. If the weight W be on the point of descending, prove that W _ sin a'+fi cos a' W siaa-jioosa ' 201. Effect of fi-iction on the simple machines. We shall now investigate the effect of friction in the case of the simple machines considered in the last chapter. Its effect is negligible in the case of the balance hung on knife-edges, but the friction. at the axis becomes apparent in the case of the puUey and the wheel and axle, in the FRICTION. 207 case of the inclined plane and the screw the friction is con- siderable. The pulley. The axle moves in a socket which it very nearly fits, thus the contact is at one point and the total reaction makes with the common normal to the axle and its socket the angle of friction. Fig. 135. The direction of the total reaction is thus seen to touch a circle whose radius is a sin e, where a is the radius of the axle. To diminish the effect of friction it is usual to make the axle of a pulley as small as is consistent with strength and the radius of the pulley large, so that a comparatively small force applied at the circumference of the pulley may have the same moment as the friction on the axle. The wheel and axle. If c is the radius of the axle on which the machine rests we have seen that R, the total resistance acts at a distance c sin e from its centre, also for equilibrium R = P+Q. Hence taking moments about the centre of the axle, Pb = Qa + (P + Q)csme. If there were no friction we should have Pob = Qa. The efficiency (Art. 162) is found by taking the ratio of P„ to P, hence efficiency Po PJ) _Qa _a 5 — c sin 6 ~ P "^ P6 ~P6~6"a + c sin e' Fig. 136. When € is very small this is nearly equal to l--r(a+b)e. 208 THE ELEMENTS OF APPLIED MATHEMATICS. The inclined plane. Let P be the force acting at the angle with the plane and just large enough to drag a weight TT up a rough inclined plane, then we have R), Pcose=F + Wsina, Psin0 = Wcosa-E, „T, Fig. 137. Whence P (cos 6-\- fimid) = W(sin a + /x cos a), -r, TirSin 0. + u. cos a _. sin (a + e) or P=W a-, ^-Q = ^ 7a \ • COS0+ /J, sua. cos {0 — e) If the inclination of the plane is less than the angle of friction, a force will be required to drag the body down the plane, if P' be this force making an angle with the plane P'cos0 = P-Trsina, P' sin (j>=W cos a— It, also F = fiB ; .•. P' (cos (j) + /i sin (f>) = W (fi cos a — sin a), T>' — W^ cos a — sin a _ ™- sin (e — a.) cos ^ + /i sin (^ cos ( '»'= M- Thus the velocity of each ball is reversed in direction. 206. Kinetic Energy is lost by impact. The fact that kinetic energy is lost in the impact of two spheres may be shown as follows ; let E and E^ be the total kinetic energy before and after impact respectively, then since by elementary algebra (m + to') (mu" + m'u'^)={mu + m'uy—2mm'uu' + mm'{u^+u'% therefore %(m + m')E= (mu + m'u')' + mm' (u - u'f (3). 216 THE ELEMENTS OF APPLIED MATHEMATICS. Similarly 2 (m + m')^i = (mv + rnVf + mm' (v-v'y (4). By subtracting and using (1), we have 2 (m + m') (E- E^) = mm' {{u - u'f -{v- v')% Hence if u — v! is numerically greater than v—v', E is greater than E^ . By use of (2) we see that E-E,= ^."^"l ,, {u - u'y (1 - e% 2 (m + m ) and since e is less than unity ^ — ^i is a positive quantity. The kmetic energy which is apparently lost reappears in the form of vibrations of the molecules of the spheres. 207. When the spheres are not moving at impact in the direction of the line joining their centres, their respective velocities perpendicular to the line of centres are not altered by impact if the spheres are smooth, and their velocities resolved in the direction of the line of centres satisfy equa- tions (1) and (2). 208. Let 7i be the impulse between the spheres up to the instant of greatest compression, and /j the impulse from that time until contact ceases. At the instant of greatest compression the spheres are both moving with the same velocity which we may denote by V. Then since through /j the respective momenta of the bodies are changed from mu and m'u' to mF and m'V, I^ = m{u-V) = m'{V-u'). Similarly /2 = m(F— ■u) = m'(i;' — F). Hence \m mj \m m J therefore -=? = — — ; = e. Ix u—u IMPACT. 217 EXAMPLES. XLV. 1. A sphere impinges directly on another sphere of double its mass moving with half its velocity. Show that if the coefficient of elasticity be ^, the striking sphere will after the impact move with half its original velocity, and find the velocity of the other sphere. 2. A ball of mass M moving with velocity Y impinges directly on a ball of mass m moving with a certain velocity, (i) in the same direc- tion as M, (ii) in the opposite direction ; if the coefficient of elasticity is f and the subsequent velocity of m is four times as great in the former case as in the latter ; show that the original momenta of the balls were in the ratio 1 - 2MjZm. 3. A ball 5 oz. in weight faUs from a height of 20 feet upon a fixed horizontal plane, and on rebounding reaches a height of 11 J feet; find the coefficient of elasticity and the measure of the impulse. 4. A cannon-ball strikes a smooth sea at an inclination of 1 in 100 ■and rebounds from the water at an equal inclination. Compare the blow with which it strikes the water with that required to stop it. 5. An elastic sphere let fall from a height of 16 feet above a fixed horizontal table will come to rest in 8 seconds after describing 65 feet, supposing the sphere to keep rebounding from the table with a co- efficient of elasticity |. 6. A series of equal perfectly elastic balls are arranged in a straight hne, one of them impinges directly on the next, ar)d so on ; prove that if their masses form a g. p. of which r is the common ratio the velocities with which they successively impinge will form a g.p. of which the ■common ratio is = . l+r 7. A ball A of mass m impinges directly on another ball B of mass to' which is at rest. After the impact B impinges directly on a third ball C of mass m" which is also at rest. If G has imparted to it the same velocity as A had at first, and all the balls are perfectly elastic (to -1- to') (to' -)- to") = 4toto'. 8. Two equal marbles A and B lie in a smooth horizontal circular groove at opposite ends of a diameter. A is projected along the grpove and after a time t impinges on B, show that a second impact will occur after a time — . e 9. A series of spherical balls of elasticity e are projected from a point and suffer reflexion at a smooth plane wall ; show that their directions after reflexion pass through the same point. 218 THE ELEMENTS OF APPLIED MATHEMATICS. 10. A ball is projected vertically upwards with a velocity of 160 feet per second, and when it has reached its greatest height it is met in direct impact by another equal ball which has fallen through 64 feet; find the times from the instant of impact to that in which the balls reach the ground, their coef6cient of elasticity being J. 11. A number of balls are dropped simultaneously from the heights m?, v? &c. feet above a perfectly elastic plane, where m, n &c. are whole numbers. Taking ff as 32 prove that they wiU all be in their original positions after ^M seconds, where if is the l.o.m. of m, n, &c. 12. A ball projected with velocity v at an inclination a will keep ricochetting from a smooth horizontal plane for a time —n j and will , v' sm 2a have a range — r^ r . S'(l-e) 13. The velocity of a sphere moving on a smooth horizontal plane is reversed in direction after impinging successively on two fixed smooth vertical planes of the same material at right angles to each other. 14. A ball is projected from a point in a horizontal plane and makes one rebound, show that if the second range is equal to the greatest height which the ball attains, the angle of projection is tan~i 4e. 15. The diameters of each of two equal balls are two inches in length and the balls are moving in opposite directions each with velocity V with their centres on two parallel lines one inch apart. The coeflS- cient of elasticity is | ; find the velocity and direction of motion of each ball after impact. 16. A ball is projected from a point in the floor of a room of height h with velocity v and elevation 6 in a v ertical plane perpendicular to one of the walls so that sin 6= */-^ . After meeting one wall, the ceiling and the opposite wall it returns again to the floor. If the coeflScient of elasticity be e, the distance between the walls a and the distance of the point of projection from the first wall d, prove that the distance from the second wall at which the ball meets the floor is e^{R-d)- ae, where R is the horizontal range of a body projected with velocity v at an angle 6. 17. There are two equal perfectly elastic balls, one is at rest and is struck obliquely by the other, show that after impact their directions of motion are at right angles. CHAPTER XIV. GEAPHICAL STATICS. 209. In the present Chapter we shall investigate some graphical methods for finding the resultant of forces given in position and magnitude, and the stresses in the bars of frameworks. 210. Construction for the line of action of the resultant. Fig. ii. a Fig. 142. Take four forces F, G, H and K, whose lines of action are shown in fig. i, draw a line ah representing F in magni- tude and direction, he representing 0, cd representing jff, and de representing K. Then ae will represent in magnitude and direction the resultant of F, G, H and K, Art. 79. We proceed to find the line of action of this force. Take any point (which is called the pole) in fig. ii, and join it to the points a, h, c, d and e ; from any point L on the line of action of F draw LT and LM parallel to Oa and Ob respectively, MM parallel to Oc, NB parallel to Od and RT parallel to Oe. 220 THE ELEMENTS OF APPLIED MATHEMATICS. We shall prove that jT is a point on the line of action of the resultant force. For we can replace F by forces acting along LM and LT, G ML ... MN, H NM ... NB, K RN ... BT. But from fig. ii, by Art. 78, we see that the components of J* along the lines LT and LM are aO and Ob, G ML ...MN ... bO ... Oc, H NM... NB ... cO ... Od, K BN ... BT ... dO ... Oe. Thus the forces along the lines LM, MN, NB destroy each other, being equal and opposite, leaving the forces represented by aO and Oe acting along LT and BT, hence T their intersection must be a point on the line of action of the resultant. The line of action of the resultant of F, Q, H and K is therefore the line through T parallel to ae. When the forces are all parallel the figures are modified as represented below. K \ ^ ^ \ ^ ■■ F Fio. 143. GEAPHICAL STATICS. 221 Since the forces are parallel the lines ab, he, cd, de are in one straight line, parallel to the direction of the forces, and we find the point T in an exactly similar manner. The above method is quite general and enables us to find the line of action of the resultant of any number of forces. 211. For any given system of forces the figure on the left is called the Funicular polygon and that on the right the Force polygon. If the forces are in equilibrium their resultant ae vanishes and therefore the Force polygon is closed. We have seen that the system of forces reduces to one along LT and one along RT. When there is equilibrium these two forces must be equal and oppositely directed along the same straight line ; thus LT and RT coincide. Hence we see that when there is equilibrium each vertex of the Funicular polygon lies on the line of action of a force. 212. Henrici's notation*. The following notation due to Prof Henrici furnishes a convenient method of indicating forces. d I.R 'Q Fig. 144. Take the case when there are five forces OP, OQ, OR, 08, OT meeting in a point and in equilibrium, and attach, letters to the portions of the plane between each two consecutive forces, then the force OP is indicated by AB, OQ BG, OR CD, and so on. * Sometimes called Bow's notation. 222 THE ELEMENTS OF APPLIED MATHEMATICS. Notice that we go round in the opposite direction to that of the motion of the hands of a clock. The force BA would denote a force equal and opposite to the force AB. On the right is the force polygon. The forces being in equilibrium the force polygon is closed. It is to be observed that the force AB is represented in magnitude and direction by the line ah, the force BC by the line he, &c. 213. Frameworks. When a number of rigid bars are joined together at their ends by smooth pins they are said to form a frame. We shall suppose for the present that the weights of the bars may be neglected. T P_ 'Q. T -< — B i — t-T v-i l a— >■ Fio. 145. The following is an important proposition: The action of a bar on a pin at its end acts along the har. For consider the bar PQ, it is in equilibrium under the actions of the pins at P and Q upon it, hence these actions must be equal and opposite and must therefore each act along PQ, let the magnitude of each of them be T, thus the action of the pin P on the bar PQ is T, and hence the action of PQ on the pin P is a force equal and opposite to T, i.e. a force along the har. These two equal and opposite forces of magnitude T acting on the bar are said to form the stress in the bar. 214. Examples. We shall use the methods which have been described to investigate some simple cases. I. A jointed frame PQR in the form of an equilateral triangle has a weight W attached to the joint P and the ends of its base, which is horizontal, rest upon fixed supports. Find the forces along the bars. Letters are attached to the spaces divided from each other by the different forces, as explained in Art. 211. GRAPHICAL STATICS, 223 Thus the weight F at P is denoted by AB, the upward force at Q which is equal to JTf by BG, the force with which the bar QR acts on the pin Q by CD, and so on. Observe that in the case of any pin we take the forces acting on it in the contra-clockwise direction. The force polygon consists here of a vertical line in which ah is of length W, he and ca each of length ^W. Through a and h draw ad and hd parallel respectively to the bars PR and PQ, join dc. The pin P is in equilibrium under the action of the forces AB, BD and DA, but the sides of the triangle ahd are parallel to these forces and ah is equal to the force AB, hence hd and da are equal respectively to the forces BD and DA. Art. 78. The pin Q is in equilibrium under the forces DB, BG and GD, but the sides dh and he of the triangle dhc represent in magnitude and direction DB and BG, hence the side erf represents the force CD in magnitude and direction. Hence cd must be parallel to the bar QR. We saw that the action of the bar QP on the pin P is represented by hd, hence dh is the action of the pin on the bar, that is the bar is subjected to a compression, such a bar is called a strut. The action of the bar QR on Q is represented by cd, and therefore dc represents the action of the pin on the bar, thus the bar is subjected to a tension, such a bar is called a tie. 224 THE ELEMENTS OF APPLIED MATHEMATICS. II. A frame PQRS has a weight W attached to the pin R and rests on two supports at P and S. Fig. 147. The supporting forces BG and GA at P and S are only given in direction, not in magnitude, we shall now find their magnitudes. Draw ab to represent W on some scale, the forces BC and CA are represented by he and ca but the position of c is not as yet known. Take any pole and join Oa and Ob, we have to find the direction of the line Oc. To do this we construct the funicular polygon, i.e. take any point K on the line of action of BG and draw KL parallel to Ob, through L draw LM parallel to Oa, we shall show that Oc is parallel to KM whose direction has been found. By observing the force polygon we see that W (= ab) may be resolved into a force along KL (==0b) and a force along ML (= a 0), BG (= be) may be resolved into a force along LK (= bO) and a force in direction parallel to Oc, GA (=ca) may be resolved into a force along LM(= Oa) and a force in direction parallel to cO. geAphical statics. 225 The forces W, BG, GA are in equilibrium and their com- ponents along KL and ML destroy each other, hence the forces reduce to two, one through K and one through M, these two forces must be equal and opposite and therefore act along KM, and since we have seen that these forces are each parallel to Oc, therefore Oo is parallel to KM. Hence c is found by drawing through a line parallel to KM, and the forces BG and GA are represented by he and ca. The pin R is in equilibrium under the forces AB, BE and EA. Then by the same reasoning as in Ex. I. if we draw ae and be parallel to the bars jR^S and RQ we find that he and ea represent the forces BE and EA. The pin Q is in equilibrium under the forces EB, BD and DE, but EB being represented by eb through 6 and e draw bd and ed parallel to the bars QR and QS, then as before we see that bd and de represent the forces BD and DE. The forces which keep P in equilibriiim are DB, BG and GD, join cd, then db and he having been shown to represent DB and BG cd must represent GD; thus cd is horizontal. The pin 8 is in equilibrium under th« forces AE, ED, DG and GA which we have seen to be represented by ae, ed, do and ca. The stresses in all the bars have now been found *- Particular attention should be paid to the fact that the force denoted by any two of the large letters, e.g. BD, will be found in the other diagram by looking for the corre- sponding small letters, viz. in this case hd. To find whether any particular bar, say QS, is a tie or a strut we proceed as follows : Take one of its pins, say Q, the force on Q alon^ Q8 is DE, now comparing with the line de in the other diagram we see that the force m the pin Q is from Q towards 8, or the bar pulls the pin in, hence the pin pulls the bar out, thus Q8 is a tie. * That is, we can find their magnitudes by actually measuring the lines ae, he &a. These lines represent the forces in the scale in which ah repre- sents W. 15 226 THE ELEMENTS OF APPLIED MATHEMATICS. III. A framework of five bars PQRS has two opposite sides horizontal and to the pins at P and Q weights W and W are attached, the frame resting upon fixed supports. The lines db and be are drawn to represent W and W respectively. Any pole is taken and joined to a, b and c ; KL, LM and MN are drawn parallel to Oc, Oh and Oa, then as before, a line Od drawn parallel to KN will meet ah in d so that cd and da represent the forces CD and DA, Through a and h, af and hf are drawn parallel to the forces AF and BF, and the forces on the pin P will be repre- sented by ah, hf and fa. We then find the other forces as before. IV. A symmetrical framework of the form shown in the figure is loaded at its highest point with a weight W and rests upon smooth supports. The diagram giving the forces is shown, the student will easily be able to follow the steps of the construction. GRAPHICAL STATICS. 227 V. A system of bars consisting of two horizontal bars joined by cross bars equally inclined to the vertical is called a Warren girder. Equal weights are attached to the lowest pins of such a girder and the system is supported at each end. A ad B 1 C f D I E I F 1. Fig. 150. We find the forces on the pins in the order 1, 2, 3, 4, 5 as indicated in the figure. The forces NP, PC, CD at the pin 5 are represented by np, pc and cd; in order to get the force MN-vfe have to draw through d a line drni parallel to DM and through n a line nm parallel to MN, but these lines intersect in n, showing that n and m coincide, hence the force MN is zero and there is no stress in the correspond- ing bar. 215. Heavy jointed bars. Fio. 151. In the present Article we shall discuss the equilibrium of a system of bars connected by smooth joints or pins ; in the figure, for instance, we have four such jointed bars, on 15—2 228 THE ELEMENTS OP APPLIED MATHEMATICS. the right the bars are drawn separated from the joints to show more clearly the action between the bars and pins. Consider the pin which joins the bars AB and BG; its weight is supposed to be negligible, hence it is in equili- brium under the two forces R and S, the actions on it of the bars AB and BG respectively as shown on the right. It and 8 must therefore be equal and opposite. Next consider the bar BG. The forces acting on it {exclusive of the actions of the pins at its extremities) such as its weight &c., may be re- placed by two forces, one acting at B and the other acting at G. Let these forces be F and F' respectively. The bar BG is in equilibrium under the two forces (i) the resultant of F and 8 acting at B, (ii) ^'and^f' G. The forces (i) and (ii) must therefore be equal and opposite, and therefore act along the bar BG, let the magni- tude of either be T. Similarly if G and G' act on BA at B and A, the result- ant of G and R will be a force T' along BA. Lastly consider all the forces which act at B. Tj.-'- Fig. 152. Since T is the resultant of F and 8 we see that F, 8, and r reversed are in equilibrium, hence the force 8 on the pin B may be replaced by F and T reversed, similarly R may be replaced by G and T' reversed, thus the pin is in equilibrium under the action of (1) the resultant of F and G, (2) T reversed, (3) T' reversed. GRAPHICAL STATICS. A corresponding result holds for each joint. Hence the following rule : We may regard each pin as being in equilibrium under the action of forces along the bars which it joins and the resultant offerees corresponding to F and 0. If the only forces acting on the bars are their weights, F and G are half the respective weights of BG and BA. 216. The result just obtained may be applied to solve such questions as the following : — 1. Five equal bars are connected by smooth pins, the ends are held at points A and F in the same horizontal line. The lowest bar is horizontal and the two lower bars make angles of 60° with the vertical, find the inclination to the vertical of the two upper bars and the actions at the joints. For convenience the bars will be denoted by the letters a, yS, 7, S and e. Here the force on each bar (exclusive of the actions of the pins at its ends) is its weight W, hence the forces at the ends of a bar are each ^W, and each pin is in equili- brium under the action of forces along the bars it connects and a vertical force W. On a vertical line measure off lengths ah, be, cd, and de, each equal to W. Through c and d draw cO and Od respectively parallel to y and S intersecting in 0. Join Oa, Ob, Oe. 230 THE ELEMENTS OF APPLIED MATHEMATICS. The sides of the triangle dcO, taken in order, are pro- portional to the forces acting on the pin D, dc equals* W, hence cO and Od equal the forces along 7 and S respectively. Two of the sides of the triangle edO, viz. ed and dO, equal two of the forces on the pin E, hence Oe equals the force along EF. It follows that Oe is parallel to e. We see similarly that the forces in a and y8 on the pin B are aO and Ob respectively. Since dc = W, cO = dc tan 60° = WjS, hence each component of the stress in 7 is of magnitude Wj3. Also dO = W sec 60" = 2W, hence the components of the stress in S and /8 are each of magnitude 2W. If , ^ ce 2W 2 ^ and Oe = 2Fsec^ = ^^^^ = Tr^/7. To find the reaction on the pins. The reaction on the pin E being the action of either bar which it joins upon it, is the resultant of forces represented by Oe and ^ed, it is thus represented by Op, where p is the middle point of ed. Similarly the reaction at D is repre- sented by Oq, where q is the middle point of de. 2. Any number of equal bars are connected by smooth pins and two extremities fixed, if the system hangs freely the bars make angles with the horizon whose tangents are in Arithmetical Progression. Denote the bars by a, /3, 7, S,... and let F be the lowest pin. * The measure of the line and the measure of the force are equal. GRAPHICAL STATICS. 231 Measure oflf equal lengths ah, be, cd, de,ef,... to represent Fio. 154. W, through / and e draw fO and eO parallel to f and e, join to the points a, b, c, &c. Then as before, the sides of the triangle /eO represent the forces on the pin F, those of the triangle edO the forces on the pin E, and so on. Hence the lines Oe, Od, ... are parallel to the bars e, S . . . . Draw the horizontal line On. The bars a, /8, ... make with the horizon the angles aOn, bOn, . . . whose tangents are an bn en On' Wi' Oil' ••■' W and these form an a.p. whose common difference is ^r ■ 3. A hexagon formed of jointed bars ABGDEF has its lowest side horizontal and rests on a peg Q at the middle point of AF. The middle points of AB and FE are joined by a string and the figure is in the form of a regular hexagon. Find the tension of the string. Let the tension of the string be T. In addition to the weights of the bars we have a force of magnitude T at the middle points oi AB and FE and a force equal to 6 TT acting vertically upwards at 0. These have to be resolved into T forces s- , and 3 TT at the ends of their respective bars. 232 THE ELEMENTS OF APPLIED MATHEMATICS. Denote the bars by a, /8, 7, B, e, ? respectively. F g A Fig. 155. The pin A is in equilibrium under forces along a and /S, T 2 W vertically upwards and — horizontal. £1 T The pin B is acted on by forces in y3 and 7, -^ horizpntal, and W vertically downwards. The pin C is acted on by forces in 7 and S and W down- wards. Hence the force polygon is constructed as follows : — T. Let ah represent the resultant of -^ and 2W a,t A, be ^andFat JS, cd, drawn vertically downwards, represent W at C. Through a and b draw aO and bO parallel to a and 0, they therefore include an angle of 60°. Join Oc and Od. Then as before it follows that Oc and Od are parallel to 7 and S. The part of the force polygon below ad is exactly similar to the part above and refers to the left side of the hexagon. Draw bn perpendicular to ad. The horizontal components of ah and be are each ^T and bn is equal to 2 W. GRAPHICAL STATICS. Thus we have Ore = 2Fcot60° = ^, 73 hence adding Od=Fcot60°=?= 73 ^T = dn = jl, or T=2j3W. CHAPTER XV. CIRCULAR MOTION. 217. Uniform motion in a circle. Fig. 156. Consider the case of a body moving in the circum- ference of a circle with uniform velocity v. The rate at which the radius joining the body to the cefitre turns round is called the body's angular velocity. This we shall denote by 0). Thus if 6 be the angle turned through in a time t, e Hence if T be the whole time taken to describe the cir- cumference — =T. (O Also if an arc s is described in t seconds s e V = - = a- = aw. V V CIRCULAR MOTION. 235 218. Acceleration in circular motion. Let P and Q be successive positions of the body, then since its velocities at P and Q are equal in magnitude and perpendicular to OP and OQ respectively we may represent its velocity at P and Q by OP and OQ. The velocity required to change the velocity at P to that at Q is therefore by the Triangle of Velocities represented by PQ and is pei-pendicular to PQ and ^^'*- ^^''■ directed inwards. Hence change of velocity in passing from P to Q _ PQ _ „ . 9 velocity at P OP ~ "^ 2 = (when is very small). Hence if t denote the (very small) time occupied in going from P to Q, 1 • p _ change of velocity in passing from P to Q T a = - . velocity at P = mv = — = aar. a Hence the measure of the acceleration at P is - and is directed inwards along the radius. The body has thus no acceleration along the tangent at P. 219. We have seen that the acceleration of a particle moving uniformly in a circle of radius a is — , hence if its mass is m the force acting on the particle inwards along the radius is m - . This is the force which is required to keep a the particle in its circular path; if, for instance, the particle is revolving on a smooth table at the end of a string attached 23.6 THE ELEMENTS OF APPLIED MATHEMATICS. to a fixed point, then the tension T of the string is the force which acts on the particle and keeps it in its path, therefore we must have a If the string breaks, the particle will move off along the tangent at the point at which it is when the string breaks, there being now no force to constrain it to move in a circle. 220. The Hodograph. Fig. 158. When a body is describing any curve, from any fixed point let lines be drawn which represent in magnitude and direction the velocities of the body in its successive positions. The extremities of these lines form a curve which is called the Hodograph of the path described by the body. Take as an example the case of uniform circular motion. The lines Op, Oq &c. are drawn to represent in magnitude and direction the velocities of a body describing the circle PQ with uniform velocity v. The points p, q &c. thus them- selves lie on a circle. We notice that. Op being the velocity of the body at P and Oq its velocity at Q, pq will represent the change of velocity of the body in passing from P to Q, hence if P and Q are indefinitely near points, the measure of the velocity of the point which describes the hodograph is equal to the measure efthe acceleration of the body. CIRCULAR MOTION. 237 221. The method of the last Article may be employed to find the acceleration of the body describing the circle, as follows: — The triangles POQ, pOq are similar, therefore pq _0p _v PQ~OP~a- If t be the small time occupied in passing from P to Q, then since the chord PQ may be taken as being equal to its arc, therefore —r- = V, and ^ = acceleration at P, t t pq _ acceleration at P ■■ PQ V ' or from above !i ^ acceleration at P a V ' thus the required acceleration is — , and acts parallel to pq and therefore parallel to PO. 222. Harmonic Motion. Let the point P describe a circle of radius a with uniform velocity v. From P draw PN per- pendicular to OA, a fixed radius of the circle. Then clearly the velocity of the point N is equal to the component of the velocity of P along OA. There- fore the acceleration of N is the component of the acceleration of P parallel to OA. But the latter is aa^ cos PON=(o\ ON, hence writing x for ON, we have found that the acceleration of iV^ is m^ .X acting towards 0. The motion of N, which may be said to be the projection of uniform circular motion, is called Harmonic motion. 238 THE ELEMENTS OF APPLIED MATHEMATICS. This motion is oscillatory, when at A the point N is at its greatest distance from 0, it will then move towards 0, will pass through and reach A', its velocity then changes direction, and it returns through to A again. The whole time taken from leaving A to reaching A again is clearly the 2 same aS the time taken by P, or — . Art. 217. The distance OA is called the amplitude of vibration, the time — the period of vibration. w ^ The velocity of N is the same as that of P resolved parallel to OA and is therefore = aw sin PON = CO PN = CO Ja^ - a;'. In harmonic motion we have therefore, the acceleration = co^. cc, the velocity = o) Va^ — a^, the period = — , m the amplitude = a. We observe that the period does not depend on the amplitude. 223. Particle falling down any smooth curve. We saw in Art. 82 that when a particle has fallen down a smooth inclined plane its increase of velocity is that which it would have acquired if it had fallen freely through the height of the plane. Similarly if the particle slides down a series of planes ABODE inclined at different angles as in the figure, its final velocity will be that which would be due to the entire vertical height if falling freely. Notice that this is a particular case of the theorem in Art. 152, that the CIRCULAR MOTION. 239 change in the kinetic energy is equal to the work done by the forces. Fpr if m is the mass of the particle and h the height fallen through, and v, u the final and initial velocities, then since we have seen that v^ — u'' = 2gh, therefore ^m (ijF — u') = mgh. And mgh is the work done by the weight of the particle ; the reaction does no work since it is at each point perpen- dicular to the direction of motion. When we take the portions of the inclined plane very small and increase their number indefinitely we get the case of a smooth curve. Hence when a particle slides down a smooth curve the velocity acquired is that which is due to the total vertical height fallen through, 224. Motion of a pendulum. We shall now discuss the motion of a particle attached to the end of a string, the other end of which is attached to a fixed point. The particle will oscillate in a ver- tical circle and will, with the string, constitute a simple pendulum. We shall confine ourselves to the case in which the whole arc through which the particle oscillates is very small and shall show that in that case the period of the oscillation is the same for all such arcs. Let one extreme position of the particle be L and P any particular position of it, the angle LOA being equal to a and the angle POA to 0. Draw the radius OQ to bisect the angle 6. It is clear that since OQ always moves over half the arc which P does in the same time that its velocity is half that of P. Draw QM vertical and let K be one extreme position of M, so that when P is at L, M is at K. If V is the velocity of P, since ^ is the velocity of Q, the 240 THE ELEMENTS OF APPLIED MATHEMATICS. velocity of iif is ^ cos „ , which may be taken to be „ , sirwe 6 is very small. The motion of the particle is the same as if it were oscillating on a smooth arc of a circle, the reaction of the arc being equal to the tension of the string, and we have seen that in that case v^ = 2g y. vertical distance between L and P, = 2ga (cos 6 — cos a), = ^1<0K^-0M^). a Hence v = 2 sj\ JOK^-OM\ V Also since the velocity of Jf is ^ , we have that velocity of ilf = a/^ JOK^ - OM". V Cff Therefore by Art. 222 M moves with harmonic motion, of which the period is ^ = 2.y^. And this is also the period of P, hence the period of a simple pendulum for small oscillations is equal to ■'/I seconds. 9 The expression for the period is seen to depend only upon the length of the string and the value of g at the place where the pendulum swings. A method qfjmding the value of g is thus afforded us, viz. by observing the time of oscillation of a simple pendulum whose length is known. CIRCULAR MOTION. 241 The efifect on the period of the pendulum of slightly increasing its length may be found as follows : If the length is increased by a small quantity m, the period becomes if we neglect the square of - . Hence the period is increased by ^ « njag seconds. Seconds Pendulum. A pendulum such that its period is two seconds, i.e. one which occupies one second in its swing, is called a 'seconds' pendulum. The length a of such a pendulum is given by the equation TT A /- = 1, or a = -„ feet. V 9 TT^ Taking the value of g as 32, the length of the seconds pendulum is 3'3 feet nearly. 225. Conical Pendulum. When a particle of mass m, attacljed by a string to a fixed point, moves uni- formly in a horizontal circle whose centre is vertically below the fixed point the string and particle are said to form a conical pendulum. Let be the inclination of the string to the vertical, then since the weight of the particle is supported by the vertical component of the tension T of the string Tcosd-- FiG. 162. ■ mg ...(1). To maintain the uniform circular motion of the particle a force — is required, where v is the velocity of the particle and a the radius of the circle it describes, therefore T sin must be such that Tsia0 = — . 16 242 THE ELEMENTS OF APPLIED MATHEMATICS. If the particle makes n revolutions per second we have V = ^ima, therefore Tsm.6 = miir^n^a = 4iin-n^nH sin 6, where I is the length of the string, or T=4mnrVl (2). Therefore from (1) cose = — ^ 47rVi We have now determined the tension of the string and the inclination of the string to the vertical. EXAMPLES. XLVI. 1. A particle is placed on a rough horizontal plate (/i=§) at a distance of 9 inches from a vertical axis about which the plate can rotate; find the greatest number of revolutions per second the plate can make without moving the particle. 2. At what number of turns per minute must a mass of 10 lbs. revolve horizontally at the end of a string 15 inches long so as to cause the same tension in the string as if one lb. were hanging vertically? 3. A pendulum whose length is I makes m oscillations in 24 hours. When its length is slightly altered it makes m+n oscillations in 24 2w hovirs. Show that the diminution of the length is — I nearly. i. A heavy particle of mass m is moving on a smooth table in a circle being connected by a string, which passes through a hole in the table at the centre of the circle, with a particle of mass 2m which hangs vertically. What must be the velocity of the first particle ? 5. A locomotive engine weighing 9 tons passes round a curve 600 feet in radius with a velocity of 10 miles an hour; what force tending towards the centre of the curve must be exerted by the rails? 6. A clock which gains 15 seconds a day has to be set right, find the alteration in the pendulum which should beat seconds, g being 32. 7. The weight of 29 "905 cubic inches of mercury in London is equal to that of 29 'SOS cubic inches in Manchester. How many seconds will a pendulum clock gain in a year in Manchester if properly regulated for London? CIRCULAR MOTION. 243 8. Taking the values of g at the Equator and at the pole to be respectively 32-09 and 32-25 respectively, find how much a Clock regulated by a pendulum which would beat true seconds at the pole wiU lose in an hour at the Equator. 9. A cannon weighing 12 cwt. hanging horizontally by two vertical suspending ropes at its ends projects a ball weighing 36 lbs. and is raised by the recoil 2-25 feet above its lowest position. Find the momentum and energy of the ball and of the cannon at the instant after discharge. 10. A particle weighing ^oz. rests on a horizontal disc and is attached by two strings 4 feet long to the extremities of a diameter. If the disc be made to revolve 100 times a minute about its centre, find the tension of each string. 11. When a train is travelling in a curve of 242 yards radius at 15 miles per hour, the string by which a hea-srjr particle is attached to the roof of a carriage will be inchned to the vertical at cot"! 48. 12. Show that pieces of mud thrown from the top of a cab- wheel whose diameter is d feet, the cab moving with a velocity of v feet per second, will when they strike the ground, be at a distance Jw^cJfeet in front of the position then occupied by the point of contact of the wheel with the ground. 13. If T is the time of revolution of the bob of a conical pendulum at the bottom of a shaft of a mine of depth I, the pendulum being suspended from the surface of the Earth, the value of g at the bottom of the shaft, is -yj^ (1 — I , where a is length of the Earth's radius. 14. A pendulum performs 21 complete vibrations in 44 seconds, on shortening its length 47-6875 cms. it performs 21 complete vibra- tions in 33 seconds ; from these data find the value of g. 15. Two small bodies of weights 8 and 27 ozs. weight are laid on a smooth table and connected by a string 5 feet long passing through a small fixed ring in the table at the distance of 2 and 3 feet from the bodies and in the straight hne joining them. They are then projected at right angles to the string towards the same side of it with velocities of 3 and 2 ft. -sees, respectively. Prove that each body wiU move in a circle and that if the string breaks after ^ seconds the bodies will meet at the end of the next second. 16. Two pendulums oscillating at two different places lose t and T sees, a day respectively ; if the places at which they oscillate be interchanged they lose t and T' sees. Prove that <-f- T=t!-\- T nearly. 16—2 PART II. CHAPTER XVI. FLUID PRESSURE. 1. A RIGID body is such that it offers an indefinitely great resistance to change of form. Strictly speaking there are no absolutely rigid bodies, for if force be applied to a body it will in general produce in the body a change both of size and shape. Bodies which have no tendency to recover their size or shape when they have undergone a change in these respects are said to be plastic. Of this nature are such bodies as putty, clay, mortar and malleable metals. Bodies which tend to recover their original form when a distorting force is removed are said to be elastic. 2. Fluid Bodies. A perfect fluid is a substance such that the stress between it and any (small) area in contact with it is entirely perpen- dicular to that area, and such that it offers no resistance to change of shape. The area spoken of may be either a portion of the surface of an immersed solid or any area taken within the fluid itself. Such a fluid behaves as if it consisted of very small smooth hard spheres. A fluid is a substance which offers very little resistance to change of shape. The stress between a fluid and any (small) area in contact with it is not quite perpendicular to that area, although nearly so, owing to the existence of a property called viscosity somewhat resembling friction. In the sequel we shall assume the fluids treated of to be perfect fluids. FLUID PRESSURE. 245 3. Liquids and Gases. Fluids are divided into two classes, viz. liquids and gases. The volume of a given mass of liquid cannot be sensibly changed by pressure alone, liquids are therefore said to be incompressible. I If a given mass of liquid be placed in a vessel it will adapt itself to the shape of the vessel and occupy a portion of the vessel equal to its own volume. The surface of the liquid not in contact with the vessel is called the free surface. Compressible fluids are called gases. A gas will entirely fill a closed vessel containing it and therefore has no free surface. Liquids are not absolutely incompressible although nearly so. 4. Fluid Pressure. The fundamental property of a fluid at rest is that already mentioned, viz. that the force exerted by a fluid at rest on any surface in contact with it is at each point perpendicular to that surface. If J. is a portion of the area of a surface of which one side is in contact with a fluid and P the force which would counter- balance the actioh of the fluid on A, the force P reversed is called the pressure of the fluid on A. When equal pressures are exerted by the fluid on equal areas of the surface wherever situated, the pressure is said to be uniform. 5. Intensity of pressure at any point. P . The fraction -j is the pressure per um.it area of .4, it is called the intensity of pressure at any point of A supposing the pressure co be vmform over A. If the pressure is not uniform over the surface we proceed as follows : take a part of the surface a, which is so small that we may regard the pressure over it as uniform, then if P be the whole pressure on a, P . the fraction — is called the intensity of pressure at any point of a and is denoted by p, thus P =pa.. 246 THE ELEMENTS OF APPLIED MATHEMATICS, 6. The method of separate equilibrium. The method which is adopted in order to obtain the pro- perties of fluids at rest is as follows : the equilibrium of some definite portion of the fluid is considered, the forces which act upon this portion are (i) its weight and (ii) the pressures of the remaining portion of the fluid upon its surface, and these forces are in equilibrium. In order to separate mentally the portion of fluid considered, more clearly, from the rest of the fluid, it is sometimes supposed to become iolidijied. But this supposition is quite unnecessary. 7. The intensities of pressure at all points in the same horizontal plane are equal. Fis. 1. Take a horizontal prism of fluid with very small ends. Let one end A BCD be a vertical square whose side is x, the other, LMNR, a rectangle of sides x and y. The forces act- ing on this prism are in equilibrium. The pressure of the rest of the fluid on ABCD is horizontal and in the direction of the length of the prism. First, let LMNR be inclined to the vertical at some angle 0. Then if ^ and^' are the intensities of pressure at the centres of the two ends, the forces on those ends are pxx^ and p'xxy respectively, Art. 5. We shall prove that p=p'- Denote these two forces by P and Q. The forces which tend to move the prism in the direction of its length are (1) P, (2) the component of Q in that direction. And these are the only forces in this direction since the pressures -on the other faces and its weight act perpendicular to the length of the prism. Hence since there is equilibrium we have P = component of Q along the prism. FLUID PRESSURE, 247 Draw MS parallel to AB, the sides of the triangle LMS are respectively perpendicular to Q and its horizontal and vertical components, hence horizontal component of Q : Q = MS : LM = x:y. Therefore P : Q = x:y, that is px^ : p'xy = x:y, .'. p =p'. Y N Q Fig. 2. Secondly, let LMNR be vertical but inclined to ABGD at some angle 0, draw LS parallel to AD, then as before component of Q in the direction of the prism : Q = L8:LM = x:y. Therefore P:Q = x:y, that is pa?:p'xy=.x\y, :. p=p'. Hence whatever the direction of the face LMNR the intensity of pressure at its centre is equal to the intensity of pressure at the centre of ABGD. It follows that the in- tensities of pressure in any direction are the same at all points in the same horizontal plane. Thus we see that at all points of a horizontal area the intensity of pressure is the same, hence the pressures on equal areas are equal, or the pressure is uniform and the; intensity of pressure is the pressure per unit area. 248 THE ELEMENTS OF APPLIED MATHEMATICS. Both cases of the foregoing might have been shortened as follows j P= component of Q along the prism = § cos 5, .•. px'^=p'm/ cos 6, and x=i/ cos 6, .'. p=p' whatever may be the value of 6. Alternative proof that the intensity of pressure at any point is the same for all directions. Consider a very small portion of fluid in the form of a tetrahedron OABC, the faces through being mutually at right angles. Since the tetrahedron is in B equilibrium the sum of the resolved parts of the forces acting upon it in any direction is zero. Take the direction perpendicular to the face BOC. The forces in this direction (i) the pressure on the face BOC, ^_ (ii) the resolved part of the pressure on ABC, C (iii) the resolved part of the weight of the p » tetrahedron. Hence if 6 be the angle between OA and the perpendicular to ABO we have, p being the intensity of pressure over BOG and p' over ABC^ p X area BOC-p' x area ABCy. cos 5+ resolved part of weight of tetrahedron=0 (1). Now the resolved part of the weight of the tetrahedron is, ^ being the angle between OA and the vertical, volume of the tetrahedron x mass of unit volume of the fluid x cos 0, and the volume of the tetrahedron is= J x area J.5Cx perp. from on ABC ; which, since the tetrahedron is very small, is negligible in com- parison with the area ABC ; hence we may omit the last term of (1) as negligible in comparison with the others. Hence p X area BOC=p' x area ABC x cos d. And since the area BOG is the projection of the area ABC, area ABC cos 6 = area BOC •■• P=P'- Hence however the direction of the area ABC be altered the intensity of pressure over it remains the same provided that the area remains indefinitely near 0. By considering the equilibrium of a horizontal right circular cylinder of fluid we see that the intensity of pressure is the same at all points in the same horizontal plane. FLUID PRESSURE. 249 8. The intensity of pressure in a homogeneous heavy liquid varies directly as the depth. To find the intensity of pressure at any point P, consider the equilibrium of a thin vertical cylinder ^^^^ of fluid, whose base ,A is horizontal and r=l- contains P, the upper end being in the sur- r^ face. — - A' The forces, in equilibrium, acting on this cylinder are, (i) the pressures on its curved surface, w- 4 which are horizontal, (ii) the pressure on the base, which is vertical, (iii) the weight of the cylinder acting vertically down- wards. Hence we see that, the upward pressure on the base = weight of the cylinder. Also, A being the area of the base, this upward pressure is pA, and w being the weight of unit volume of the fluid the weight of the cylinder is whA, hence pA = whA, or p = wh. Thus the intensity of pressure is proportional to the depth. Notice that p is the pressure on a horizontal unit of area at the depth h. We have not taken into account the vertical pressure of the atmosphere on the free surface of the fluid, if its intensity is n, we have p = wh + Ii.. The atmospheric pressure on a square inch is 14'7 lbs., this is the weight of a vertical column of water whose height is 34 feet and base one square inch, or of a column of mercnury 30 inches in height. -E^. p^.^ _ : = h ■ A'- , , , , _ A B 250 THE ELEMENTS OF APPLIED MATHEMATICS. 9. The f]ree surface of a heavy liquid at rest is a horizontal plane. Take two points A and B in the same horizontal plane ■within the fluid. The intensities of pressure at A and B are equal, Art. 7, hence if h and h' are the depths of A and B below the free surface, we have by the last Article, wh+Il = wh' + Il, or h = h'. Similarly we show that all points of the free surface are at the same vertical "*' distance from this horizontal plane, hence the free surface is itself a horizontal plane. Liquids maintain their level. The proof of this proposition applies to the case where the free surface consists of several detached portions. If, for instance, we have a number of tubes fitting into the top of a closed vessel and water be poured through one of the tubes, it will, after filling the vessel, rise to the same height in each tube. This fact is sometimes referred to by saying that ' liquids maintain their level.' 10. The surface of separation of two heavy liquids 'Which do not mix is a horizontal plane. When we have two liquids which do not mix, the lighter liquid being therefore uppermost, the intensity of pressure at any point P in the lower liquid is, by reasoning precisely similar to that of Art. 8, wh + w'k + n, where w and w' are the weights of unit volume of the upper and lower liquids, and h, k the respective portions of a ver- tical line through P contained in the two liquids. This intensity of pressure is the same for all points P in the same horizontal plane, Art. 7, and since the free surface of the upper liquid is horizontal, Art. 9, h + k is the same for all points in the same horizontal plane ; hence, for points in such a plane, both h and k are the same, i.e. the depth of the upper liquid is the same. FLUID PRESSURE. 251 11. The height of the free surface of a column of liquid above a horizontal plane is called the "head" of the liquid with regard to that plane. Pressure is often given in "feet" of water or "inches" of mercury, the meaning being that the pressure is that due to a head of the given number of feet of water or inches of mercury. Ex. 1. Find the pressure per square inch at a depth of 100 feet in water. If p is the required pressure p=ioh, Art. 8, and since the weight of a cubic foot of water is 1000 oz., w is :rfr^ oz. ; also A=1200, hence p=1200x^^2gOz. wt. =43'4 lbs. wt. nearly. Ex. 2. Find the pressure per square inch at a depth of 30 inches in mercury. Mercury is 13 '6 times as heavy as water, the weight of a cubic inch of mercury will therefore be yfna ^ ^^'^ oz. Hence the pressure per square inch is ^ x 13'6 x 30 oz. wt. = 14'8 lbs. wt. nearly. Ex. 3. Find the pressure on a horizontal plane 5 sq. feet in area placed 25 feet deep in mercury. The weight of a column of mercury 25 feet in length, the area of , , . „ ^ . 1000x13-6x25,, , ^, -whose base is one sq. foot, is ^3 lo^-j hence the pressure on ,, , . 1000x13-6 X 25x5,, •,n„n-n,i, 4. the plane is r^ lbs. = 106250 lbs. wt. EXAMPLES. I. 1. Find the pressure per square inch due to a head of 34 feet of -water. 2. Find the head of water to which is due a pressure of one lb. per square inch. 3. If the pressure at a depth of 26 feet in a long vessel containing liquid is 3 times the pressure at a depth of 8 feet, find the pressure on & square foot of the free surface. 252 THE ELEMENTS OF APPLIED MATHEMATICS. 4. If the pressure at the bottom of a well is 4 times that at the depth of 2 feet, what is the depth of the well if the pressure of the atmosphere is equivalent to 34 feet of water 1 5. A hole 6 inches square is made in a ship's bottom 20 feet below the water line. What force must be exerted to keep the water out by holding a piece of wood against the hole, assuming that a cubic foot of sea- water weighs 64 lbs.? 6. In two different fluids the intensities of pressure are the same at the depths of 5 and 6 inches. Neglecting atmospheric pressure compare the intensities of pressure at depths of 8 and 10 inches. 7. On the top of a house 47 feet high is placed a cistern 6 feet deep. What force is exerted on a plug in a tap whose section is of area half a square inch when the cistern is half full ; the plug being 2 feet above the ground? 8. The circular cork of a bottle has a diameter of '75 inches. It requires a force of 35 lbs. to push it in. How far must the bottle be sunk in the sea that this may take place, neglecting atmospheric pressure? 12. Transmission of fluid pressure. When pressure is applied to a portion of the surface of a fluid at rest this pressure is transmitted equally to all other portions of the fluid. For if to an area at P in the surface of the fluid a force F is applied, and Q is any other point within the fluid, con- struct a cylinder having the area at P and an equal area at Q as ends. Then since the cylinder is in equilibrium and a force F has been applied at P, there must be an additional pressure equal Pm g and opposite to F on the end Q, since the pressures on the sides of the cylinder are all perpen- dicular to the axis. This is known as Pascal's Principle. The operation of this principle might be shown as follows : — Let a closed cistern be filled with water, the cistern being provided with air- tight pistons passing through holes in its sides. If the areas of the pistons are equal and a force be applied to one of them tending to push it in, an equal forc^ must be applied to each of the other pistons to counteract the effect of the first FLUID PRESSURE. 253 force. And if one piston has an area half that of each of the others, a force applied to it makes necessary the applica- tion of double that force to each of the other pistons. Thus pressure applied to any area of the boundary of the liquid produces an equal pressure on every equal portion of the boundary. Ex. 1. ABCD is a vertical sectiop of a closed vessel filled with water. At two places in the upper surface pistons whose sections are squares are fitted, the sides of the squares being one and two inches respectively. A force of 7 lbs. weight being applied to the first piston find the force that must be applied to the second to maintain equi- librium. Fio. 7. The area of the sections of the pistons are 1 and 4 sq. inches respectively. Hence on the second piston 4 times the force applied to the first piston must act, therefore 28 lbs. must be applied to the second piston. Ex. 2. If the section of the first piston is a square whose side is 6 inches long, and if a weight of 72 lbs. is placed upon it, what weight must be applied to the second piston, if its area is one square inch ? Ans. 2 lbs. 13. Hydraulic Press. The Hydraulic Press consists essentially of two cylinders of different sections connected by a small pipe, the cylinders are filled with water, and in each cylinder an air-tight vertical plunger works. If G and c are the two plungers and A, a the areas of the sections of the plungers, when a force of P lbs. weight is applied to the smaller plunger it produces a pressure of — lbs. weight per unit of area, and hence a total pressure of a p — X J. lbs. is transmitted to the base of the larger plunger. a 254 THE ELEMENTS OF APPLIED MATHEMATICS. If this just supports a weight W we have Tf = Px-. a Fio. 8. On the down stroke of the smaller plunger the valve on the right is displaced, and water is forced into the large cylinder a? the large plunger rises. On again raising the small plimger the valve on the right closes and the large plunger is kept in its position while the valve on the left is raised and water enters the small cylinder from the tank below as explained in the Article on Pumps. In the figure the tap T is turned so as to form a channel for the water from the small to the large cyUnder. The water in the large cylinder can be released by turning the tap through a right angle when the water escapes downwards through the tap. In connexion with the large cylinder there is a gauge O to indicate the water pressure. FLUID PRESSURE. 255 By means of this apparatus a very large pressure may be secured by the application of a comparatively small force, provided that the area A be considerably greater than a. The force P is usually applied by means of a lever, the total mechanical advantage will therefore be the product of the mechanical advantages of the lever and the press. The Hydraulic press remained for a long time compara- tively useless since it was found that the water leaked out round the larger piston when the pressure became consider- able. To remedy this Bramah invented the cupped leather collar, of which a section is represented. A groove aa is cut round the cylinder and in it is placed a leather ring bent so as to have a semicircular section. When water enters this collar it presses one side of it against the groove and the other closely against the piston, so that no water can escape ; as the pressure increases the collar is pressed more and more tightly against the piston. Ex. In a Bramah press the larger plunger has 100 times the area of the smaller, the arms of the lever are 4 and 16 inches long, what pressure will be transmitted to the larger plunger by a force of 40 lbs. weight ? The mechanical advantage of the lever is 4, and of the press is 100, hence the mechanical advantage of the combination is 400, thus the total pressure on the larger plunger is 16000 lbs. weight. 14. Hydrostatic Paradox. Let A and a be the areas of the sections of a wide and a narrow cylinder connected by a pipe. The wide cylinder is fitted with an air- tight lid on which a weight W can be placed. If water be poured in at the top of the narrow cylinder it will raise the weight a distance depend- ing on the amount of water so poured in. If the water in the narrow cylin- der stands at a height h above that ' ' in the wider, the pressure on the area a at P is by Art. 8 wah. Q JWL h ! 1 256 THE ELEMENTS OF APPLIED MATHEMATICS, The pressure transmitted to A is therefore — X wah, or wAh. a If W be the weight supported W=wAh. This result is independent of the section of the narrower cylinder, and by taking the section extremely small we get the result that a very small amount of water can be made to support a very great weight. This fact is sometimes known as the Hydrostatic Paradox. Observe, however, that a con- siderable increase in the length of the longer column of liquid only increases the length of the shorter column very slightly. 15. Liquids in a bent tube. Let BAG be a bent tube containing two liquids which do not mix, say mercury and water. Let a be the area of their common surface a.t A,h the height of the column of water above A, h' of the column of mercury above A, the pressure over a. due to the mercury must be equal to the pressure over a due to the water, hence if w and wf are the weights of the unit volume of water and of mercury respectively wah = w'ah', or h K w w ' Fw. 10. Hence the heights above the common surface are in- versely proportional to the weights of unit volume of the two liquids. Ex. 1. Two liquids that do not mii'are contained in a bent tube; the difference of their levels is 4 inches and the distance of the free surface of the heavier above their common surface is 6 inches, compare the weights of their units of volume. The height of the lighter liquid above the common surface is 10 inches, hence 10 : Q=v/ : w ov w' -5 :3. FLUID PRESSURE. 257 Ex. 2. Two vertical cylindrical vessels A and B are connected at the bottom by a very narrow tube, and stand on a horizontal table. The diameter of 4 is 6 inches, that of 5 is 4 inches. A liquid 1-4 times as heavy as water fiUs the vessels to a height of 6 inches above the base, when an equal volume of water is poured carefully on the top of the liquid in A. Where will be the common surface of the hquids when equilibrium has been restored, assuming that the liquids do not mix? The volume of water poured in is tt x 78 cubic inches. 78 Hence the height of the column of water is -^ inches ; and there- fore by this Article the height of the other liquid above the common surface is - — =-7 inches. 9x1-4 Let X be the distance of the common surface from the base, then since the new volume is twice the original volume Ty + a; j X 97r + ^g-^j^ + /K^ 4jr = 2 (6 X 9,7 + 6 X 4n-), which gives a; =4*1 inches nearly. Ex. 3. Water is poured into a U tube the arms of which are 6 inches long until they are half full. As much oil as possible is then poured into one of the arms. What length of the tube will it occupy if water is half as heavy again as the oil? Let a; be the distance of the common surface above the base of the tube, then if A is the area of the cross-section the weight of oil is that of (6—af)%A cubic inches of water. The height of the water above the common surface is obviously 6 — 2x inches, hence {6-x)%=6-2x, or x=l^, thus the length of tube occupied by the oil is 4^ inches. EXAMPLES. II. 1. If the diameter of the larger plunger of a Bramah press is twenty times that of the smaller one, what pressure is applied to the latter if that on the former is half a ton 1 2. The diameters of the two plungers of a hydraulic press are 2 inches and 20 inches respectively, and the lengths of the arms of the lever are 2 inches and 18 inches. Find the weight which can be suppoi-ted by a force of 40 lbs. weight. 3. If a pressure of one ton is produced by a force of 5 lbs., and the diameters of the plungers are in the ratio of 8 to 1, find the ratio of the lengths of the arms of the lever employed to work the piston plunger. J. 17 258 THE ELEMENTS OF APPLIED MATHEMATICS. 4. A U tube open at both ends is partially filled with mercury and water is then placed in one leg. If the mercury originally stood half way up the leg of the U tube and then one leg is filled up to the top with water, find the height of the mercury in the tube. 5. How many cubic inches of water must be poured into one leg of a U tube to raise the mercury one inch in the other, mercury being 13 '6 times as heavy as water ? 6. The lower ends of two vertical tubes whose cross- sections are 1 sq. inch and "1 sq. inch respectively are connected by a tube. The tubes contain mercury; how much water must be poured into the larger tube to raise the level of the mercury in the smaller tube by one inch ? 7. A U tube whose arms are of equal length has mercury poured in until each surface is 5 inches from the top of the tube. Water is now poured into one branch and alcohol which is 'SS times as heavy as water into the other so as to fill the tube. Find the length of the tube occupied by each liquid, mercury being 13 '6 times as heavy as water. 8. Find the greatest depth in fathoms at which a submarine diver can work, supposing he can bear the pressure due to 5 atmospheres, taking the atmospheric pressure to be 15 lbs. per square inch. 9. A long hollow cylinder 20 sq. inches in section is closed at the bottom by a flat plate without weight and is immersed in water till the base is 21 inches below the surface. What weight must be placed upon the plate to detach it from the cylinder ? [A cubic foot of water weighs 1000 ozs.] 10. The pressure in a water-pipe at the basement of a building is 34 lbs. to the square inch, whereas at the third floor it is 18 lbs. to the square inch, find the height of the third floor. CHAPTER XVII. PRESSURE ON IMMERSED SURFACES. 16. Pressure on the base of a containing vessel. When liquid is contained in a vessel the total pressure on the base varies according as the sides are vertical or inclined. The figures illustrate different cases. (i) Sides vertical. Here the pressure between the liquid and the sides is entirely horizontal, arid the weight of the liquid is therefore entirely counterbalanced by the upward pressure of the base of the vessel. Fig. 11. In this case the pressure on the base is equal to the weight of the entire liquid. (ii) One side inclined inwards as in fig. (ii). Take any two points A and B in the base of the vessel, then since A and B are in the same horizontal plane, intensity of pressure at 5 = intensity of pressure at A = w.AG. 17—2 260 THE ELEMENTS OF APPLIED MATHEMATICS. Thus the pressure on each unit of area of the base is that which would be due to sustaining the weight of a cylinder of liquid ending in the free surface of the liquid. Thus the total pressure on the base is equal to the weight of a cylinder of liquid whose base is the base of the vessel and whose height is that of the free surface. Here the pressure on the base is greater than the weight of the contained fluid, the reason being that at any point in the oblique side there is a downward pressure on the liquid, which is, by Pascal's Principle, transmitted to the base. (iii) One side inclined outwards as in fig. (lii). Here the pressure on the base is less than the weight of the contained liquid. The inclined side supports the weight of a portion of the liquid. In every case the pressure on the base depends only on the area of the base and its depth below the free surface. 17. Floating bodies. We have seen that when a surface is in contact with a liquid every portion of it is sub- jected to a normal pressure by the liquid. The single force which is the resultant of all these normal pres- sures is called the resultant pressure of the liquid. When a body is partly immersed in a liquid we can find the resultant pressure of the liquid upon it as follows : — The only part of the body acted on by the liquid is the immersed portion of its surface; let us imagine the body removed and the space it occupied in the liquid filled up by a portion a of the same liquid (thus a is the amount of liquid which was displaced by the solid), then obviously the action of the liquid on a is the same as was its action on the body, but a is clearly in equilibrium, and is acted on by (i) its weight acting vertically through its c. G., (ii) the action of the rest of the liquid on a ; PRESSURE ON IMMERSED SURFACES, 261 hence we conclude that the resultant pressure of the liquid on the body 1. is equal to the weight of the fluid displaced, 2. acts vertically upwards through the C.G. g of the fluid displaced. If the body is floating freely in the liquid we see that its weight must be equal to the weight of the fluid displaced. And generally when a body is immersed in a liquid it is acted on by an upward force equal to the weight of the liquid it displaces. It is usual to say that the body loses so much of its weight, and the force necessary to support it is called its apparent weight. This is known as the Principle of Archimedes, which states that 'the loss of weight in liquid is equal to the weight of liquid displaced,' If the body floats entirely immersed its weight is equal to that of its own volume of liquid. Ex. 1. A stone attached to a string is just immersed in a water- cistern whose base is 8 inches by 4 inches, if the surface of the water rises half an inch what is the diminution of the weight of the stone? Since the water rises half an inch the volume of water displaced by the stone is 16 cubic inches, the weight of the stone is therefore diminished by the weight of 16 cubic inches of water or 9^ ozs. Ex. 2. If gold is 19'3 times as heavy as water and mercury 13'5 times as heavy, how much will a cubic inch of gold weigh when immersed in mercury? The weight of a cubic inch of gold is 19*3 x ^=55 ozs., 1000 , mercury ... 13 '5 x ozs. Therefore the weight of a cubic inch of gold when immersed in mercury is 5'8 x yj^ ozs. EXAMPLES. III. 1. Find the volume of a body whose weight is that of 460 grammes in air and 60 grammes in water. 2. A body whose volume is 3 cubic feet and which is 1"4 times as heavy as water is placed in a vessel in which there is water enough ot cover it, iind the entire pressure of the body on the base of the vessel. 262 THE ELEMENTS OF APPLIED MATHEMATICS. 3. The edge of a hollow cube of lead is 8 cms. long, and its thick- ness is 2 cms., find its weight in water if lead is 11 '35 times as heavy as water. 4. A piece of metal which is 9 times as heavy as water, when weighed in water has the weight of 56 grammes, find its true weight. 5. A vessel shaped like a portion of a cone is filled with water. It is one inch in diameter at the top and eight at the bottom, its height is 12 inches. Find the intensity of pressure (in lbs. per square inch) at the centre of the base, and the total pressure on the base. 6. A vessel in the shape of a pyramid on a square base, a side of which measures 6 inches, is filled with water. If the pyramid is 18 inches high find the pressure on the base. 7. A body whose weight in air is 0"9 lbs. weighs 0-48 lbs. in water and 0'6 lbs. in a certain liquid. Compare the weights of equal volumes of this Uquid and of water. 8. A vessel is quite full of water, and a piece of wood is put to float on the water; does this alter the pressiffe on the bottom? Answer the question supposing that the vessel had not been overflowing. 9. A piece of wood floats partly immersed in water, and oil is poured upon the water till the wood is completely covered ; explain whether this will make any change in the portion of wood below the surface of the water. 18. Immersed body attached to a string. Fig. 13. Special cases of equilibrium are those of an immersed body either sustained or held down by a string. The forces in equilibrium are PRESSURE ON IMMERSED SURFACES. 263 1. The weight of the body. 2. The resultant pressure of the liquid acting upwards at the C.G. g of the displaced liquid. 3. The tension of the string. Since these parallel forces are in equilibrium we have in the former case weight of the body = weight of liquid displaced + tension of string, in the latter case weight of the body = weight of liquid displaced — tension of string. In the latter case the body is clearly lighter than its own bulk of liquid. Ex. A body six times as heavy as water and whose volume is 36 cubic inches is suspended by a staring so as to be totally immersed in water, find the tension of the string. The weight of water displaced is 36 x r^=^ ozs., and the weight of the body is six times this, hence we have from which, tension =104J ozs. weight. 19. Resultant Pressure. We have defined the resultant pressure on an immersed surface as the force which is the resultant of all the pressures of the liquid on the surface. Art. 17. In the case of a plane area these pressures are all perpendicular to the plane, and hence parallel to each other. So that finding the resultant pressure is the same thing as finding the resultant of a number of like parallel forces. The point in the area at which the resultant pressure acts is called the Centre of Pressure. Magnitude of the Resultant Pressure. To find the mag- nitude of the resultant pressure we proceed as follows : Divide the area A into very small equal areas a, and let the intensity of pressure at one such area be p, then the pressure on this area is pa — wza, where z is its distance from the free surface. 264 THE ELEMENTS OF APPLIED MATHEMATICS. To get the resultant pressure we take the sum of all these pressures giving w^za. (2 indicating " the sum of"). But if z be the depth of the c.G. of A below the free surface, _ Xzoi Hence resultant pressure = wSza. = wAz = Awz = A y. intensity of pressure at the C.G. of A. We thus get the following rule : The resultant pressure on a plane area whose measure is A, is equal to the prodvAst of A and the intensity of pressure at the c.G. of the area. Taking the atmospheric pressure into consideration p has the value wz+n, and the resultant pressure is wAz+AU. Ex. 1. Find the magnitude of the resultant pressure on the side of a cubical vessel filled with water. If a inches is the length of a side, the depth of the c.a. of the side is ia, thus in this case Aiiiz has the value a' x r^rrrr x 7: or ,>.„- oijs. wt. ^ ' 1728 2 1728 This is half the pressure on the base. Ex. 2. Find the magnitude of the resultant pressure on a vertical rectangle 10 inches long and 6 inches broad immersed in water with the longer side horizontal and with the upper side 2 inches below the surface The lower edge is 8 inches below the surface and therefore the c. G. is i (2 + 8) inches, or 5 inches, below the surface. Therefore the resultant pressure is eOxr-s^QxS ozs. wt., that is 173 '61 ozs. weight nearly. Ex. 3. A cubical vessel contains two liquids which do not mix, find the resultant pressure on a side. The figure represents a side of the vessel, the heavier liquid fills it to a depth AB, the lighter liquid to a depth BC. Let a be the length of AD, h of AB, and c of BC; let w and w' be the weights of unit volume B' " of the two liquids. Take a point P below the surface of the A''- lower liquid, the pressure on a small area a is, j, , , Art. 10, {wMN+ ti/FM]a = [wPiV+ {iv' - w) PM] a. N N' Ip- M ti P A PRESSURE ON IMMERSED SURFACES. 265 Hence the resultant pressure on AB' is viSPNa +{v/-w) XPNa. For a point P in the upper liquid, the resultant pressure on a small area a is wPNa, hence the resultant pressure is WS,PNa. Thus the entire resultant pressure is ivSPJVa (taken over the whole area ACC'A') +{v/-w) ^PMa (taken over the area ABB' A'). In other words we find the resultant pressure as follows : Suppose the vpper liquid to extend to the bottom and And the result- ant pressure it produces, then suppose it removed and a liquid the weight of whose unit volume is w' — w to occupy the volume of the lower liquid and find the resultant pressure it produces, and then add the two pressures together. Applying this to the present case we have as the required resultant pressure toxa(b+c) X — ^-f-(w' — *») «6 X Q /,^c\^ ,ab^ The same method applies to the case of any number of liquids which do not mix and which are contained in one vessel ; we add the pressures produced by a number of layers of liquid, each layer escteTuMng to the bottom. Ex. 4. A hollow cone stands with its base on a horizontal table. The area of the base is 100 square inches, and the height 8'64 sq, inches. Its weight is equal to the weight .of water it will contain. When filled with water compare the liquid pressure on the base with the pressure of the base on the table. The volume of a cone is \ area of base x height. Hence the weight of water contained is that of ^^ cubic inches of water. The pressure on the base is the weight of a column of water whose base contains 100 sq. inches and whose height is 8'64 inches. It is therefore equal to the weight of 864 cubic inches of water. The pressure on the table is that arising from the weight of the cone and the water it contains, it is therefore the weight of 2 x s§A cubic inches of water. Hence we have pressure on base : pressure of base on table =864 : | 864=3 : 2. Ex. 5. A hollow cone is filled with water, its base being horizontal and vertex upperrnost. Find the resultant pressure on the curved surface of the cone; The figure represents the cone and the cylinder which stands on the base. 266 THE ELEMENTS OF APPLIED MATHEMATICS. Let the space between the cone and the cylin- der be filled with water and the water inside the cone removed. The intensity of pressure at any point of the curved surface is the same in magnitude as before, but opposite in direction. The resultant pressure is the weight of this portion of fluid, which is twice -pj^^ 15. the weight of the fluid contained by the cone. Hence the resultant pressure on the curved surface of the cone acts vertically upwards and is equal to twice the weight of the water. EXAMPLES. IV. 1. The base of a triangle is one foot in length and its altitude 10 inches. What will be the resultant pressure on the triangle when it is immersed with its vertex in the surface of the water and the middle point of its base 4 inches below the surface, neglecting atmospheric pressure and taking the weight of a cubic foot of water as being 62-5 lbs. ? 2. In the vertical side of a water-tank there is a rectangular plate whose upper edge is horizontal and eight feet below the surface of the water, the depth of the plate is one foot and breadth two feet. Find the resultant pressure on.the plate. 3. A triangular plate is immersed in water with one side in the surface and the opposite angle two decimetres below the surface. If each side of the triangle be 6 decimetres, find the resultant pressure. 4. A vessel whose base is a square the side of which is 3 inches long contains mercury to the depth of one inch, and water is poured upon the mercury to the depth of 10 inches. Find the pressure on the base and on a side of the vessel, taking mercury to be 13"6 times as heavy as water. 5. Find the pressure on one side of a decimetre cube if half full of mercury and half of water. 6. A vessel in the shape of a decimetre cube is filled to one-third of its height with mercury while the rest is filled with water. Find the resultant pressure on a side in kilogrammes weight. 7. A cube whose edge is one foot is suspended in water with its upper face horizontal and at a depth of 2^ feet below the surface. Find the pressure on each face. 8. A square plate whose area is 64 sq. inches is immersed in sea- water, its upper edge which is horizontal being 12 inches below the surface. Find the pressure on the plate when it is inclined at an angle of 45° to the horizon, assuming a cubic foot of sea-water to weigh 63 lbs. PRESSURE ON IMMERSED SURFACES. 267 9. A vessel containing water is placed on a table. The area of the base of the vessel is 10 sq. inches and the depth of the water is 4 inches. If ^ of the water is vertically over the base and the weight of the vessel is 5 ozs., find the intensity of pressure at any point of the base, the resultant pressure on the base, and the pressure on the table. 10. A sphere whose internal radius is one foot contains mercury which covers f of the vertical diameter. Find the resultant pressure on a horizontal plane passing through the centre of the sphere, assum- ing that a cubic inch of mercury weighs 3500 grains. 20. Determinatioii of the Centre df Pressure of a plane area. The Centre of Pressure of a plane area is the point in the area at which the resultant pressure acts. Its position can be easily found by a method which will be now explained. Let ABG be a plane area S ; , through each point of the boundary of S draw vertical lines to meet the free surface. The pressure on each small elementary area of 8 is equal in magnitude to the weight of the vertical column of liquid which stands upon it and is perpendicular to the area. The pressures on the various elements of area form a system of parallel forces perpendicular to the area, and the centre of this system of parallel forces will be unaltered if we turn each of them, about its point of application, into the vertical direction. The forces are then simply the weights of the various columns of liquid standing vertically over the several elements of the area, and the centre of parallel forces is the point P where the vertical through the c. g. of the superincumbent liquid cuts the area. 21. Triangle with vertex in the surface and base parallel to the surfoce. Let ABC be a triangle having its vertex A in the free surface and the side BG parallel to the free surface. The vertical planes through AB, BG and CA meet the free surface in the lines AD, ED and EA. The portion of fluid considered whose weight is sustained by AGB is a pyramid whose vertex is A and base BGED. We have seen that the vertical through the c.G. of this pyramid meets ABG in the required centre of pressure P. But the c. G. is at g, f of the way down AG, where G is the c. G. of the base BGED. Therefore the centre of pressure P is f of the way down AF, where F is the middle point of BC. Fig. 16. 268 THE ELEMENTS OF APPLIED MATHEMATICS. 22, Triangle with its base in the sur£aee. B Fia. 18. ABC is a triangle having the side BC in the surface. Proceeding as before we get a tetrahedron. The c. g. of this is f of the way down AG, where O is the c. G. of the area BCD. Drawing gpP vertically we see that Gp=^DG=lx%DE=^DE, .-. Dp=lDE-lDE=lDE, or Dp=\DE and .-. AP=IAE. 23. Farallelogrsuu with its edge in the sur£ice. The parallelogram is ABCD and the application of the method gives the wedge-shaped figure ABGDEF. This may be divided into thin plates all parallel to BCF. The c. G.s of all these plates lie in a plane which is parallel to CDEF and J of the dis- tance of AB from it. Hence the centre of pressure P is J of the way down the line Gff, which joins the -pm. 19. middle points of AB and CI>. 24. Triangle with its vertex in the surface. Fig. 20. PRESSURE ON IMMERSED SURFACES. 269 ABC is a triangle with its vertex in the surface and its base not parallel to the surface. The depth of the centre of pressure may be found as follows : — Produce BC to meet the surface in D, then if z is the depth of P the centre of pressure, by taking moments about AD, we have by the theory of parallel forces, (resultant pressure on -45i)- resultant pressure on ACD)z =res. press, on -45i)x depth of its cent, press. -res. press, oto. ACDy. depth of its cent, press. Let /3 and y be the depths of B and C respectively below A, then res. press, on ABD~s\^y. ADy.wy.\^, depth of cent, of press, of ABD=^^; ACD=\y,(. AD y.wK.ly, ACD=\y. Hence we have z given by the equation 25. Resultant pressure on a curved 8ur£tee. If a portion of a curved surface S be immersed in a liquid we have seen that the resultant vertical pressure on Sis equal to the weight of the vertical column of liquid which it supports. To U find the resultant pressure on S in any given horizontal direction, take a hori zontal column of liquid terminated by Fie. 21. S and by a plane perpendicular to the given direction. Then since there is equilibrium the pressure on the vertical end 8' is equal to the resolved part of the pressure on S, the remaining forces acting on the cylinder being all perpendicular to its length. In other words, the resultant pressure on S in any given horizontal direction is equal to the resultant pressure on the projection of *S on a plane perpendicular to the given direction. EXAMPLES, y. 1. A cone floats vertex downwards in water with its axis half im- mersed and is just submerged when a weight of one lb. is placed on its horizontal top. What is the weight of the cone ? 2. A uniform cylinder floats in mercury with 5*1432 inches of its axis immersed. Water is poured on the mercury to the depth of one inch and it is found that 5'0697 inches of the axis are below the surface of the mercury. Find the weight of a cubic inch of mercury. 3. The water on one side of a rectangular flood-gate 12 feet high and 10 feet wide rises to the top, and on the other side to half the height. Find the magnitude of the resultant pressure on the gate and the level at which it acts. 270 THE ELEMENTS OF APPLIED MATHEMATICS. 4. The sides of a cistern are vertical and its base is a horizontal square of which the side is J3 feet long. Find its depth if when it is full of water the pressure on each side is the same as on its base. 5. A piece of glass weighs 47 grammes in air, 22 grammes in water and 25 '8 grammes in alcohol. Find the weight of a cubic centimetre of alcohol. 6. Hiero's goldsmith received a mass of m lbs. of gold to make into a crown but constructed one of gold-silver alloy, this Archimedes dis- covered, and found that when placed in a vessel full of water it caused a volume v to overflow, while equal masses of gold and of silver caused the overflow of volumes v^ and v^ respectively. Of what amount did the goldsmith cheat the king, p^ and p^ being the prices of gold and silver per lb. respectively ? 7. A steamer in going from salt into fresh water was observed to sink 2 inches, but after burning 50 tons of coal to rise one inch. Prove that the steamer's displacement was about 6500 tons, supposing the densities of salt and fresh water are as 65 : 64. 8. A right circular cone of density p floats just immersed with its vertex downwards in a vessel containing two liquids of densities the first portion of mercury breaks off, this creates a va- cuum between this portion and that which follows it, and hence the mercury now passing B is on its lower side unsupported by any pressure, hence the air pressure at D cuts it and a portion of air is inserted in the cavity. This air is carried off and emerges at E, and this process is re- peated until the air is re- moved from C. The tube Di: is full of small columns of falling mercury separated by air. When the air has been exhausted these column's of mercury coalesce at the bot- tom of the tube to form a solid column whose height is that of the barometer; as the successive portions of mer- cury fall upon this column a cracking noise is heard which is due to the fact that the impact occurs without the ' damp- ing' effect of interposed air. Of course as portions of mercury fall upon the top of the column other portions leave at the bottom so as to maintain the barometric height. u Fig. 29. 312 THE ELEMENTS OF APPLIED MATHEMATICS. ^ The use of the bulb B is to collect such air as is carried down by the mercury from the cup A, so as to ensure that this air is not carried into the receiver. Notice that the length of DE must be greater than the height of the barometer. 61. Condensing Air-Pump. This instrument consists essentially of a receiver and a barrel to which it can be firmly joined or disconnected at pleasure. The use of the instrument is to fill the receiver with compressed air. In the barrel works a piston provided with a valve opening inwards, there is also a valve, opening in the same direction, at the bottom of the barrel. Communication between the barrel and re- ceiver can be cut ofif by means of a stop-cock. To use the condenser, the barrel is firmly attached to the receiver and the stop-cock opened ; as the piston descends its valve will close and it will force the air contained in the barrel into the receiver. As the piston ascends again its valve is opened by the pres- sure of the external air and the other valve closes, the barrel is thus again filled with air. By each down-stroke of the piston there is forced into the receiver a mass of air whose volume at atmospheric pressure is that of the barrel. Hence if B is the volume of the barrel, and A that of the receiver, and p the density of the external air, after n strokes a Fig. 30. mass of air nBp has been forced into the re- ceiver, and since there was a mass Ap. in the receiver origi- nally the total mass is p(A+nB). If p^ is the density of the compressed air in the receiver the mass of air in the bafrel is Apn. MACHINES DEPENDING UPON FLUID PRESSURE. 313 Hence p^A -p{A+ nB\ or p^ = p{\+n-j^. Ex. If the volume of the barrel of a bondenser is 80 cubic centi- metres and the volume of the receiver 1000 cubic centimetres, how many strokes are necessary to raise the air-pressure in the receiver from one atmosphere to five atmospheres ? We have seen that p»=p ( ^ + ** ^ ) and pressure is proportional to density, therefore by. the question, K 1 , 80 400 ^- '='+'^1000' "'• «=-8-='''- The required number of strokes is 50. EXAMPLES. XVIII. 1. If the volume of the receiver in a Smeaton's Air-Pump is 5 times that of the barrel, find the pressure in the receiver after 3 strokes of the pistop, the barometric height being 30 inches. 2. In a Smeaton's A^r-Pump the volume of the barrel is 24 cubic inches, that of the receiver is 1000 cubic inches, and the piston will not descend quite to the bottom of the barrel but leaves -^ of its length untraversed. Find the density of the air in the receiver after two strokes. 3. In the process of exhausting a common receiver, after 10 strokes of the pump the mercury in the gauge stands at 20 inches, the baro- metric height being 30 inches. At what height will the mercury in the gauge stand after 20 more strokes 1 4. If the receiver holds 2 ounces of air and one-eighth of this.be removed by the first stroke of the piston, how much will be removed by the second stroke ? 5. Show that the upper valve of the barrel of an air-pum]f) will ■ k open when the piston is at a distance x from the bottom where a;=rl, where k is the reading of a barometer placed, in the receiver, h the barometric height and I the length of the barrel. 6. If the volume of the receiver be six times that of the barrel, and if a barometer placed in the receiver stands at 28 inches after one stroke of the piston, where will it stand after two more strokes ? 314 THE ELEMENTS OF APPLIED MATHEMATICS. 7. Find the ratio of the volume of the receiver to that of the barrel in a condenser, if at the end of the fourth stroke the density of the air in the receiver is J of the original density. 8. If the volume of the barrel is J of the volume of the receiver, find the pressure in the latter after 20 strokes. 9. If the volume of the receiver be 7 times that of the barrel, after how many strokes will the density of the air in the receiver be twice that of the external air ? one 62. The Siphon. The Siphon is a bent tube with open ends having branch longer than the other. In order to use it, it is filled with liquid and both ends temporarily closed. It is then placed with the shorter branch dipping into a vessel of liquid whose contents are to be transferred. If the ends are opened the liquid will pass in a continuous* stream through the tube. The explanation of its action Pia, 3i_ is as follows : The pressures at D and E are equal, hence the pressure at E is atmospheric, also pressure at A downwards = pressure a,t E + pressure due to the column AE = atmo. pressure + pressure due to AE. Thus the (downward) pressure of the liquid at A exceeds the (upward) atmospheric pressure at A and the liquid will run out at A. If DG, the height of the highest point of the liquid above D, be less than k, where crk = ph, (a- and p being the densities of the liquid and mercury respec- tively, and h the height of the barometer), the liquid will rise in DC and flow through the tube until DC becomes equal to k. * See however, Cotterill and Slade, Applied Mechanics, p. 482. MACHINES DEPENDING UPON FLUID PRESSUEE. 315 Ex. What would happen if a small hole were made (i) in the shorter, (ii) in the longer arm of a siphon in action ? (i) If the hole were made above the level of the liquid, all the liquid below the hole would descend into the vessel, all the liquid above would ascend and pass through the longer arm. ' If the hole were below the level of the water there would be no effect on the discharge. (ii) If the hole were above the level of the liquid in the vessel, all the liquid below the hole would descend, all above it would flow back into the vessel. If the hole were below the level of the liquid in the vessel, the liquid would issue from this hole. 63. The Balloon. A balloon consists of some light flexible substance (usually- silk) enclosing a gas lighter than air. A balloon when free to move is acted on by two forces, (i) the weight of the envelope, appendage's and en- closed gas, (ii) the upward pressure of the atmosphere. If the weight of the balloon is less than the weight of the displaced air, it will rise. The difference of the forces (i) and (ii) is called the 'lifting force' of the balloon. Before beginning to ascend the balloon is not entirely inflated, for a reason which will be immediately seen; as the balloon ascends the external pressure decreases, hence the gas inside the balloon expands and the volume of dis- placed air increases. If the balloon had been fully inflated at first it would probably burst when the external pressure diminished. If V be the volume of the gas, p the pressure and p the density of the external air, at any instant before the balloon is completely inflated, by Boyle's law, pF is constant and therefore pVis constant. See p. 287. Hence the mass of displaced air remains the same until the balloon is quite inflated; therefore the difference of the forces (i) and (ii) or the Hf ting force remains the same. When the balloon is fully expanded the lifting force rapidly diminishes, but the balloon will rise until the weight of displaced air is equal to the weight of the balloon. 316 THE ELEMENTS OF APPLIED MATHEMATICS. Ex. I. An empty balloon weighs in air 1200 lbs. If a cubic foot of air weighs 1 J ozs., how many cubic feet of gas of sp. gr. '52 (referred to air) must be introducsed before the balloon will begin to ascend ? The weight of a cubic foot of the gas is 13 5 13 ^X^ozs. = 2qOzs., 13 therefore the weight of x cubic feet is g^:;!; ozs., and the weight of a) 5 cubic feet of air is ^ ^ ozs. Hence in order that the balloon may just begin to rise, xi being the req. number of cubic feet, 6 13 -r x= 1200 X le+jr;; X, .-. ^=32000 cubic feet, 4 20 ' Ex. 2. What is the effect of an increase of barometric pressure on the lifting power of a balloon partially filled with gas ? Taking 8 lbs. as the weight of 100 cubic feet of air, find approximately the volume of hydrogen (whose sp. gr. referred to air is ■07) which a baUoon must contain in order that its total lifting power may be equal to the weight of 750 lbs. With an increase of pressure the weight of air displaced by the solid parts of the balloon is greater, hence the lifting power is increased. Let F be the required volume, the weight of the hydrogen is F> Hence the lifting power is ''^ife'^ISo^^'- FxA(i_.o7)=FxjgLibs.wt., and this is to be equal to the weight of 750 lbs., hence F= 10,000 cubic feet, nearly. 64. The Diving-Bell. The Diving-Bell is a large metal vessel open at the bottom and heavier than the weight of water it would dis- place. It is lowered with the open end downward into water. The contained air is compressed and the water rises in the bell. When the surface of the water within the bell is at a depth of about 33 feet below the free surface of the water outside, the ball will be half-filled with water, for the pres- sure of the air within it will be that of two atmospheres. MACHINES DEPENDING UPON FLUID PEESSUEE. 817 Length occupied by air at any given depth. Let h be the length of the bell which is supposed to be cylindrical, d the depth of the top of the bell below the Fio. 32. surface, x the length of the bell occupied by air. Let 11 be the atmospheric pressure, p the pressure of the air within the bell. Then if h is the height of the water barometer n = wh, p = Ii. + w{!}o + d), /I , , 7\ T-th + x + d = w{h + x + d) = Ti h And n6 = px by Boyle's law, therefore X li h h p h + ai + d ' .: a^ + x(h+d)-bh = (i). The positive root of this equation gives the length of the bell occupied by air. If the absolute temperature has changed from T to T', we have p. 293, ^ = ^ , and the equation to determine x ip 318 THE ELEMENTS OF APPLIED MATHEMATICS. Position of equilibrium of diving-bell. The bell will be in equilibrium if disconnected from the chain when the weight of the water displaced is equal to the weight of the bell. Notice that the equilibrium is unstable, for when pushed down slightly the air within, being subjected to greater pressure, displaces less water, hence the bell will gink; if, on the other hand, it is pulled up a little, the air within being subjected to less pressure will expand and displace more water, hence the bell will rise. If X is the length of the bell occupied by air in this position, I the length of the bell, 1^ the weight of the bell, and W the weight of water which the bell would contain, we have -7 ^ '"~ W ' If the weight of the bell is greater than the weight of water which it would contain there is always a tension in the chain supporting the bell. To find the volwme of air at atmospheric pressure which must be introduced into a bell at a given depth so that the water may be driven out. Let V be the volume of the bell, then air which fills the bell when its top is d feet below the surface would occupy a volume V T at atmo. pressure. Hence the amount of air at atmo. pressure which has to be added, is yh_±d±b_y^^b+d ^..^_ Ex. 1. The depth of the surface of water inside a diving-bell is found to be 132 feet. The length of the bell is 5 feet. If the water barometer stands at 33 feet, what height has the water risen iu the beU? Let X be the distance of the water in the bell from the top of the bell, then 132 -^=depth of the top of the bell. Hence by (i), a;2+a;(33 + 132-^) - 165 = 0. Therefore x=\ foot, and the water has risen 4 feet in the bell. MACHINES DEPENDING UPON FLUID PRESSURE. 319 Ex. 2. A diving-bell whose volume is 200 cubic feet rests at the bottom in water 150 feet deep. If the height of the barometer be 29-5 inches, find how many cubic feet of, air at atmospheric pressure must be introduced in order to fill the bell with air. The height of the water barometer is 29-5 X 13-6 . ^ „„ . . , j2 feet =33-4 feet, 150 hence by (ii), 200 x ^^ cubic feet or 898 cubic feet must be introduced. But the bell originally contained 200 cubic feet of air, hence 1098 cubic feet are required. Ex. 3. A cylindrical diving-bell weighs 2 tons and has an internal capacity of 100 cubic feet, while the volume of the material composing it is 20 cubic feet. The bell is made to sink by weights attached to it. At what depth may the weights be removed and the bell just not ascend, the height of the water barometer being 33 feet 1 When the bell is at rest the weight of the bell is equal to the weight of the water displaced. Let y be the number of cubic feet ttien occupied by air, weight of water displaced =(y -1-20) 1000 ozs. wt., .-. (3/-I-2O) 1000=2x2240x16, y=51'68 cubic feet. The pressure of the air in the bell is due to a depth of «-h33 feet, .; . by Boyle's law, (33 -|-^) 51 -68 = 100 x 33, hence «=30"8 feet, nearly. EXAMPLES. XIX. 1. A diving-bell 6 feet long is sunk until its top is 66 feet below the siuface. Find the height to which water will rise inside the bell if the height of the barometer is 34 feet. 2. A cylindrical bell 4 feet long whose volume is 20 cubic feet is lowered into water until its top is 14 feet below the surface of the water and air is forced into it until it is | full. What volume would the entire quantity of air occupy under atmospheric pressure, the water barometer standing at 33 feet f 3. In a diving-bell the surface of the water inside is 17 feet below the surface of the water above. What portion of the bell is filled with water, the barometer standing at 34 feet ? 4. A cylindrical tube one metre in length has one end sealed and contains dry air at the usual pressure and temperature. The tube is dipped vertically with its open end downwards into a tank of mercury 320 THE ELEMENTS OF APPLIED MATHEMATICS. till the air within it is compressed to f of its former volume. Find the distance of the top of the tube from the free surface of the mercury in the tank, the height of the barometer being 75 cms. of mercury. 5. A cylindrical diving-bell whose height is 5 feet is lowered until the depth of its top is 55 feet ; iind the length occupied by air, the barometer standing at 33 feet ; find also how much air must be forced in to completely expel the water. 6. A cylindrical diving-bell whose length is 5 feet is sunk to the bottom of a river on a day when the atmospheric pressure at the surface is that due to 32 feet of water, and the water is found to rise 1 foot in the bell. Find the depth of the river. 7. A diving-bell is lowered into water at a uniform rate, and air supplied by a force pump so as to keep the bell just full of air without allowing any to escape. How must the mass of air supplied be varied as the bell descends i 8. The volume of a balloon when starting is 64000 cubic feet and its mass including that of the enclosed gas is 2 tons ; with what acceleration will it begin to ascend, the mass of a cubic foot of air being 1-24 ozs. ? 9. Assuming that a cubic foot of air weighs IJ ozs., and a cubic foot of water 1000 ozs., a balloon so thin that tbe volume of its substance may be neglected contains 1-5 cubic feet of coal gas, the balloon weighing 1 oz. The balloon floats, without ascending or descending, in the middle of the room. Find the sp. gr. of coal gas (i) referred to air, (ii) referred to water. 10. A balloon which will hold 300 cubic feet of gas is partly inflated with 100 ozs. of a gas whose sp. gr. is J referred to air at atmospheric pressure. Find the height of the barometer at the place where the balloon comes to rest, the sp. gr. of air at sea-level being •0013, and the weight of the envelope 2 ozs. 11. An inverted U-tube whose arms are equal in length just dips into two vessels containing liquids of the sp. grs. s and ^ respectively. The tube is filled with the liquids and the free surfaces of the liquids in the vessels are at the same level. If m and y are the distances of the surface of separation of the liquids in the tube from the highest and lowest points of the arm in which it is, show that where h is the height of the water barometer. 12. Two cylinders with vertical sides are each filled to a depth of 40 feet, one with distilled and the other with sea-water. As much as possible is drawn from each by a siphon, the water in the second vessel then stands ^ of a foot above that in the first vessel. Find the sp. gr. of sea-water, the mercury barometer standing at 30 inches. MACHINES DEPENDING UPON FLUID PBESSUEE. 321 13. Two equal cylinders each containing an equal quantity V of liquids which will not mix and whose sp. grs. are s and s' are connected by a U-tube exhausted of air and of small bore which reaches to the bottom of each cylinder ; find how much liquid will run from one vessel into the other. 14. A small siphon is filled with water and the shorter end is closed ; will, water run out at the other end ? - 15. A cylindrical piece of cork of height k is floating with its axis vertical in a basin of water. If the basin is placed under the receiver of an air-pump and the air pumped out, prove that the cork will sink a distance = (1 —s)h, where o- is the ratio of the densities of air and 1 — O" water, and s that of cork and water. 16. If a vessel to be emptied by a siphon contain water and the siphon itself be filled with some liquid whose sp. gr. is a-, find the minimum length of the longer lim^) when the length of the shorter is 5 feet, in order that the siphon may work. 17. The volume of the receiver of a condenser is m times that of the barrel, and when the piston is in its lowest possible position there is a space between the piston and the valve of the receiver =— of the volume of the barrel ; find the greatest condensation that can be effected by the condenser. 18. An iron bell whose mass is one ton and which can just float in water mouth upwards is immersed mouth downwards in water. What is the tension of a rope supporting the bell if the depth of the surface of the water within the bell below the free surface be equal to the height of the water barometer, the sp. gr. of iron being 7-6 ? 19. What force is required to draw apart a pair of hemispheres 4 inches in diameter and placed together, if the pressure of the external air is 15 lbs. per sq. inch and the hemispheres are exhausted so that the pressure of the air within them is ^ of that outside ? 20. A piece of wood floats in the water inside a diving-bell. If when floating outside the bell it is half immersed, find what fraction of it is not immersed when the surface of the water inside the bell is at a depth of 20 feet, having given that the height of .the water barometer is 32-5 feet and the density of atmospheric air -0013. 21 If h be the range of the piston of an air-pump, a its distance from the top of the barrel in its highest position, /3 its distance from the bottom in its lowest position and p the density of atmospheric air, show that the limiting density of the air in the receiver will be 21 322 THE ELEMENTS OF APPLIED MATHEMATICS. MISCELLANEOUS EXAMPLES. (Dynamics.) 1. Two trains, whose lengths were respectively 130 and 110 feet, moving in opposite directions on parallel rails were observed to be 4 sees, in completely passing each other, the velocity of the longer train being double that of the other. Find the rate of each train. 2. A ship sailing N.E. through a current running 4 miles an hour after 2 hours' sailing has made good 4 miles S.E. Find the velocity of the ship and the direction of the current. 3. If the resistance of the air is always | of the weight of a body in vacuo, find how high a body will go if shot up vertically with a velocity of 900 ft. -sees. Prove that the body wiU again reach the point of pro- jection after 62-5 Sees. 4. A carriage is slipped from a train moving 40 miles per hour. How far will it travel before coming to rest, reckoning the resistance of the rails ^ of the weight? 5. When two unequal weights are connected by a string passing over a rough peg which has the effect of preventing motion until the tension of the string at one end be greater than that at the other by -th part of the latter tension, prove that the effect on the acceleration will be the same as if the peg were smooth and the smaller weight increased by - th part of itself. 6. There are n forces acting at represented by 0.4 j, OA^,, ... OA^- The middle point oi A^A^ is joined to^j, the middle point of OA^, the middle point of the line thus drawn is joined to ^84, a point on OA^, such that 0^4 = ^0.44, and so on. Prove that if P be the middle point of the last of all the Hues thus drawn the resultant of all the forces is repre- sented in magnitude and direction by 2" "^ OP. 7. A number of smooth rods meet in a point A and rings slide down the rods starting from A. Prove that after a time t the rings are all on the surface of a sphere of radius \gfi. 8. A body of mass M hanging vertically draws a body of mass M' up a smooth inclined plane by means of a string passing over a smooth pulley at the top of the plane. If M starts from the top of the plane, which is 14 feet from the ground, determine the velocity of M' just before M strikes the ground. 9. ABCD is any quadrilateral and is the intersection of two straight lines bisecting the opposite sides of the quadrilateral. Prove that forces acting at represented by OA, OB, OC and OD are in equi- librium. MISCELLANEOUS EXAMPLES. DYNAMICS. 323 10. The sides AB, BC, CD and DA of the quadrilateral ABCD are bisected at E, F, O and H respectively. Prove that the resultant of the two forces represented by EG and HF is represented in magnitude and direction by AC. 11. A ball weighing 12 lbs. leaves the mouth of a cannon hori- zontally with a velocity of 1000 feet per second ; the gim and carriage, ■together weighing 12 owt., slide on a smooth plane whose inclination to the horizon is 30°. Find the space of recoil up the plane, having given that the pressure caused by the explosion on Ihe ball and on the end of the bore of the cannon is the same. 12. Four forces P, Q, R and S, no two of which are parallel, act in one plane. The resultant of P and Q meets that of R and S in A, the resultant of P and R meets that of Q and S in B, that of P and S meets that of Q and Rm.C; prove that A, B, C lie in one straight line. 13. In the triangle ABC the line DE is drawn parallel to the side BG and meeting the other sides in D and E, the lengths of DE and BC are b and a respectively. If h be the line drawn from A to bisect BC, prove that the distance of the c. g. of the figure BCED from A is ^ a{fl + b) 14. A heavy triangular plate lies on the ground. If a vertical force applied at the vertex A is just great enough to begin to lift that vertex oflf the ground, show that the same force will sufiBce if applied at B or C, the other vertices. 15. A fly-wheel is brought to rest after n revolutions by a constant- irictional force applied tangentially to the circumference. If k be the kinetic energy of the wheel before the friction is applied and r its radius show that the force is kj^jrur. 16. If equal triangles be cut from the comers of a given triangle by lines parallel to the respective opposite sides, the o. G. of the remain- ing portion will coincide with that of the triangle. 17. Forces P, Q, B, S act in the sides AB, BC, CD and DA of a rectangle ABCD. Find the necessary and sufficient conditions that they should be equivalent to a single force acting at A. 18. ABCD is a square, E the middle point oiAB is joined to C and BD is drawn. Forces of 4 and 6 Ibs.-wt. act respectively in AB and BC, and forces of 3 and 2 lbs.-wt. in AD, DC respectively, forces of ^2 and 5 ijb lbs.-wt. act in BD and CE respectively, prove 6y taking moments ■alone that the system is in equilibrium. 19. ABCDEF is a regular hexagon, and five forces each equal to P act along the straight lines joining A to the other vertices. Show that their resultant is P (2 -1-^3). 20. Two bodies connected by means of a string passing over a smooth peg touch one another at one point, show that the stress between them cannot be horizontal unless their weights are equal. 21—2 324 THE ELEMENTS OF APPLIED MATHEMATICS. 21. A man sits in a chair which is suspended from the axle of a pulley ; this pulley rides on a rope which is fixed to a horizontal beam above and passes over a second pulley fixed to the same beam down again to the man's hand. If the man's weight together with the tackle is 150 lbs. what force must he exert to just support himself, the three portions of the string being parallel 1 22. A quadrilateral ABCD is capable of having a circle inscribed in it, and forces represented by BA, DA, DG, EG act on a rigid body; show that the resultant acts along the straight line passing through the centre of the circle and the middle points of the diagonals. 23. A uniform rod AB rests incUned at an angle a to the horizon with the end A on a, rough horizontal plane and just about to slip, and with the end B supported by a string inclined at an angle /3 to the horizon. Prove that the coefficient of friction is l/(tan /3- 2 tan a). 24. A cannon-ball of mass m is shot from a gun of mass M (which is free to recoil in a horizontal direction) so that its muzzle-velocity relative to the ground is V. Show that its greatest range is — , and is obtained by giving the gun an elevation of tan-* ( 1 + 1? ) • 25. A stone is thrown from a given point with horizontal and ver- tical velocities v, and v respectively; at the instant it reaches the highest point a second stone is thrown from the same point with hori- zontal velocity 3m so as to hit the first. Find what must be the vertical velocity of the second stone. 26. A light string has masses P and Q attached to its ends and is put over two fixed pulleys, the portion between them supporting a moveable pulley of mass R. Find the acceleration of P, all the portions of the string being vertical. 27. Two masses P and § lie on a smooth horizontal table near each other and are connected by a string on which is threaded a ring of mass R. The ring hangs over the edge of the table,, prove that it falls with an acceleration R{P+Q) R{P+Q)+iPQ^- 28. On a smooth wire bent into a circle and placed in a vertical plane slide two rings whose weights are as 1 to ;^3, the rings being con- nected by a light straight rod which subtends a right angle at the centre. Determine the positions of equilibrium. 29. A bullet weighing 1 oz. strikes a block of wood at rest with a velocity of 2400 ft.-secs. and remains imbedded in it, if the resultant velocity of the block and bullet is 16 ft.-secs. find the weight of the wood and the loss of kinetic energy. MISCELLANEOUS EXAMPLES. DYNAMICS. 325 30. A particle is sliding down the smooth face of a wedge whose other smooth face is in contact with a table on which it is free to move. If the angle of the wedge is 60° and its mass 4 times that of the particle show that its acceleration is " '^ . 31. A fly-wheel whose mass of 66 lbs. is practically concentrated round a circle of 6 feet circumference is so mounted as to drive a machine that requires exactly j\y of a h. p. to keep it going. The wheel is started with a speed of 40 revolutions a second. Assuming that the whole energy is absorbed at a uniform rate by the machine, find how long it wiU work. 32. A blow with an inelastic hammer-head of mass ^ lb. drives a nail one and a half inches of its length into a board, the hammer-head being reduced to rest. If the velocity of the hammer-head when it first touched the nail was 15 ft.-secs., find the average resistance of the board to the nail's motion. 33. A 50-ton engine moving at the rate of 10 miles an hour impinges on a truck at rest weighing 10 tons and they both move on together, find their velocity and calculate the loss of kinetic energy. 34. Two equal heavy rods AB, BG are freely jointed at B and have their middle points connected by an inelastic string of half the length of either rod; if the system be suspended by a string attached to the end A, prove that the inclination of AB to the vertical will be tan~i ^ . 3 Show also that the tension of the string will be -j= W, and the stress V ' 2 at B will be equal to -v= W, where W is the weight of a rod. V ' 35. A locomotive is pulling a train of 10 carriages, each weighing 4 tons, up an incline of 1 in 100, the tension of the coupling between the two middle carriages is equal to 1000 lbs. wt., and the resistance due to friction, &c. is 16 lbs. wt. per ton. Find the acceleration of the train and the pull exerted on the first carriage by the engine. 36. The system of pulleys described as Case ii. is modified by making the string which passes over each pulley A, pass round a small pulley attached to the weight, and the string is then fastened to the pulley A, the strings being all parallel. Show that if the weights of the pulleys be neglected and the number of strings be n, the mechanical advantage is 3»-l. 37. From the lower end of the block of a pulley A is hung a weight W. and the upper end is supported by a strmg which after passing over a fixed pulley B supports a pulley C: another weight P is hung by a string which after passing over C and under A is tied to a fixed point. Find the condition of equilibrium, the pulleys being unequally heavy and the strings all parallel. 326 THE ELEMENTS OF APPLIED MATHEMATICS. 38. Two equal masses connected by an inextensible weightless thread that passes over a light pulley hang in equilibrium. Show that the tension of the thread is unaltered when -th of its mass is added to one n and 7:th of its masses is removed from the other. n.+2 39. Two marksmen, using guns with the same muzzle-velocity, fire at the same moment and simultaneously hit the same mark. The one, who fires horizontally, is at the top of a tower of height h, and the other is on the ground. How far off from the tower is the latter, if the horizontal distance of the mark from the tower is c? ? 40. A mass of ^ lb., falls from a height of 18 inches upon dough, which it penetrates to the depth of 3 inches ; what is the average resist- ance exerted by the dough? 41. A weight TT hangs by a string over a pulley. A monkey takes hold of the other end, and at an instant when W is at rest begins to climb and climbs a height hint seconds without disturbing W. Deter- mine his motion and find his weight. If at the end of the t seconds he ceases to climb, how much further wiU he ascend in the next t seconds ? 42. A shot whose mass is i a ton is discharged from a gun whose mass is 110 tons, the gun's backward motion is checked by a constant pressure equal to the weight of 10 tons, and the recoil is observed to be 6^ feet. Show that the muzzle-velocity of the shot is 1320 feet per second. 43. A man starts at right angles to the bank of a river at the uniform rate of IJ miles per hour to swim across; the current for part of the way is flowing uniformly at the rate of 1 mile per hour, ana for the remainder of the way at double that rate. He finds when he has reached the other side that he has drifted down the stream a distance equal to the breadth of the river. At what point did the current change ? 44. A heavy board in the form of a right-angled triangle ABC is suspended at the right angle G. If two equal particles starting simul- taneously from G slide down the sides CA, CB, show that they will reach A, B aX the same moment, and that the motion will not disturb the equilibrium of the triangle. 45. A table with a heavy rectangular top ABGD rests upon 4 equal and equally heavy legs placed a,t A, B, E, P where E and F are the middle points of BG and CB. Show that the table will be upset by a weight placed upon it at G just greater than the weight of the whole table, and find the greatest Weight that may be placed at D without up- setting the table. MISCELLANEOUS EXAMPLES. DYNAMICS. 327 46. An endless string without weight of length I hangs in two loops over two smooth pegs in the same horizontal line at a distance a apart, and on each loop is placed a small smooth ring, one of weight W and the other of weight W'. Find an equation giving the tension of the string. 47. By the principle of work show that the pull of a locomotive engine ia pdH/p lbs., for a mean efifective pressure p lbs. on the square inch, where d is the diameter of each of the cylinders, I the length of the stroke, B the diameter of the driving-wheel of the engiigie. 48. Show that the mechanical advantage of the pulley in the cases I. and II. (p. 182) are reduced to (1 +»!.)" and (l-|-m)"-l, when it is found that the friction of the ropes causes the tension to be reduced to m times its value in passing round a pulley. 49. Two equal rods AG, BG, each of weight W, are hinged together at G and placed to stand with A, B on a, rough horizontal plane. If AB=2a and A=height of G above the plane, prove that if they are just in equilibrium it = ar- Also if u, exceed this value show that a weight = W — — r might bp a — fi/i placed at G without slipping taking place at A or B. 50. An elastic ball of mass m moving on a smooth horizontal plane impinges on a ball of equal size but of different mass m' which is at rest on the plane. If just before impact the line of motion of the impinging ball be inclined at the angle a to the line joining the centres of the balls, prove that the direction of the impinging ball wiU be turned through a right angle if m=m' (e cos^ a — sin^ a). 51. Two elastic spheres equal in all respects are moving towards each other with equal velocities, their centres being on two parallel lines whose distance apart, is d^. Prove that after impact they will move away from each other with equal velocities, so that their centres are on two parallel lines whose distance apart d^ is given by 6?/ {e2rfz+(l -e^) d^} = dPdi\ where d is the diameter of either sphere. 52. Two equal ivory balls are suspended in contact by two equal parallel strings so that the line joining the centres of the ball is hori- zontal and 2 feet below the points of attachment of the threads. Find the coefficient of elasticity when it is found that allowing one ball to start from the position when its thread makes an angle of 60° with the vertical causes the other ball after impact to come to rest in a positioa where its centre is 1 foot 8 inches from its original position of equi- librium. 53. A particle whose mass is 3 ozs. revolves on a smooth horizontal table, being attached to a fixed point by a light string of length 10 feet. If the greatest tension the string can bear is equal to 1 cwt., find the velocity of revolution required to break the string. 328 THE ELEMENTS OF APPLIED MATHEMATICS. 54. If an elastic ball be reflected in succession by each of two smooth vertical planes at right angles in a horizontal plane, show that its directions before the first impact and after the second are the same. 55. If the unit of time is 1 minute, the unit of length 1 mile and the unit of mass the mass of a ton, what unit of force is implied in the equation F=ma% 56. The unit of work is the work done in raising 50 tons through 20 feet, the unit of acceleration is 16 ft. sees, per second, the unit of density that of*a substance of which a cubic yard weighs 2 cwt. Find the units of length, time, mass and force. 57. If an hour be the unit of time and 4000 miles the unit of length, find the value of g. 58. If in a friotionless wheel and axle a force of 10 lbs. wt. supports a load of 120 lbs. wt., find the acceleration of the load when 10 lbs. wt. is taken from it and added to the force. 59. A truck whose mass is m tons is drawn from rest by a horse through a feet and is then moving at the rate of h feet per second. If the resistances are equivalent to c lbs. wt. per ton, show that the work done by the horse is 35mJ2+mac foot-lbs. 60. A ball of mass m^ strikes a ball of mass mj, which is at rest, with velocity u. Both balls are free to move inside a smooth horizontal circular tube, prove that after n impacts the k.e. of the system is JmjM^ (-2 S — j , where e is the coefficient of elasticity. 61. A uniform rod of length 2a sin a rests within a rough vertical circle of radius a, show that the greatest possible inclination of the rod to the horizon is tan~i ( — = — ^„ . „ | . \C0S'* a-i^ svaf' a) 62. A picture-frame rectangular in shape rests against a smooth vertical wall from two points in which it is suspended by parallel strings attached to two points in the upper edge of the back of the frame, the length of each string being = the height of the frame. Show that the picture will rest against the wall at an angle tan""i — , where a is the height of the frame, and 6 its thickness. 63. A smooth wedge of angle a is free to slide along a smooth horizontal table in a direction perpendicular to the edge of the wedge. On the surface of the wedge move two particles of masses m and m! connected by a fine inextensible string which passes round a smooth peg driven into the wedge. The two straight portions of the string lie along lines of greatest slope. If M is the mass of the wedge, prove that the tension of the string is 2»im' {M-\- M. -H m') S' sin a if (m -(- m') -)- 4m7A' -^ sin^ a{m — m!)'^' MISCELLANEOUS EXAMPLES. DYNAMICS. 329 64. If a pendulum fits loosely on a horizontal axis of radius a and is found to make a constant inclination 6 to the horizon when the axle is kept rotating, the angle of friction ^ between the rubbing surfaces is given by sin<^=-sintf, where h is the distance of the c. g. of the eft pendulum from the point of suspension. 65. Show that a railway carriage running round a curve of radius r will upset if the velocity is greater than \/^, where a is the distance between the rails, and h the height of the c. G. of the carriage above the rails. 66. The sides of a triangular framework are 13, 20 and 21 inches long, the longest side rests on a horizontal smooth table and a weight of 63 lbs. is suspended from the opposite angle. Find the stress in the side on the table. 67. If four forces in one plane be in equilibrium and the lines of action of all be given but the magnitude of only one, show how the magnitudes of the other three may be determined by the graphical method. Four heavy rods equal in all respects are freely jointed together at their extremities so as to form the rhombus ABCD. If this rhombus be suspended by two strings attached to the middle points of AB and AD, each string being inclined at an angle 6 to the vertical, the angles of the rhombus will be '2,6 and n- — 25. 68. Two equal heavy rods AC, BC are jointed together at G and have their other extremities A and B jointed to fixed pegs in the same vertical line. Prove that the direction of the stress at C is hori- zontal and determine by a geometrical construction the stresses at A and B. If a weight which is equal to that of either rod be attached to the centre of the lower rod, show that if a is the inclination of each rod to the vertical and 6 the inclination to the vertical of the stress at C tan fl=3 tan a, and that this stress is to the weight of either rod in the ratio \/l+ Stanza : 4. 69. Three equal uniform rods, each of weight W, are joined to form an equilateral triangle. If the system be suspended by the middle W point of one of the rods the stress at the lowest angle is — = , and that ^ 2'v'3 at each of the upper angles i^W^Y- 70. Three uniform rods whose weights are proportional to their lengths are freely jointed together to form a triangle ABG which is placed with its plane vertical and its side BC on a horizontal plane. Show that 6, the inclination of the stress at ^ to the horizon, is given by the equation tan-Bsin (5-fl)=tan Csin (C+5). 330 THE ELEMENTS OF APPLIED MATHEMATICS. MISCELLANEOUS EXAMPLES. (Hydrostatics.) 1. A body whose specific gravity is p floats half-immersed in a fluid but is I immersed in a mixture of equal volumes of that fluid and water. Eina p, neglecting atmospheric pressure. 2. A cone floats with its axis vertical and vertex downwards in two fluids whose specific gravities are s and s+s', if the surface of separation of the liquids bisect the axis of the cone, show that its specific gravity is «+ Js'. 3. A piece of wax weighs 4^ grammes in air. A piece of pla,tinum whose volume is ^ of a cubic centimetre and specific gravity 21 is attached to the wax, and the two together weigh in water 1 J grammes. Find the sp. gr. of the wax. 4. Two vessels contain each 3 pints of fluid, the sp. gr. of one being twice that of the other. Two pint tumblers are filled, one out of each vessel, and then each tumbler is emptied into the vessel from which it was not drawn. Prove that after this process has been gone through 3 times the sp. grs. of the fluids are to each other as 41 : 40. 5. A ship sailing from the sea into a river sinks 6 inches, and after discharging a part of her cargo rises 2 inches. Show that the weight of the cargo discharged is to the weight of the ship as tr — 1 : 3o-, where 0- is the sp. gr. of salt water ; the section of the vessel made by the water line being supposed the same in the three cases. "6. A siphon barometer is so constructed that the long closed tube has an internal sectional area equal to J of a sq. inch, while the short open tube has an internal sectional area of 4 a sq. inch. Find what fall will take place in the long tube of this barometer when the true pressure of the air falls one inch. 7. A common hydrometer weighs 2 oz. in air and is graduated for specific gravities varying from 1 to 1'2. What should be the volumes in cubic inches of the portions of the instrument below the graduations 1, 1"1, 1'2 respectively; it being assumed that a cubic foot of water weighs 1000 ozs. ? 8. If a common hydrometer float in water with one inch of the stem above the surface and in a liquid of sp. gr. 1'2 with 2 inches above the surface, what will be the sp. gr. of a liquid in which it will float with 3 inches above the surface? 9. A cylindrical diving-bell 10 feet high is sunk to a certain depth and the water is observed to have risen 2 feet in the bell. As much air is then pumped in as would fill -££q of the bell, if at atmo. pressure, and the surface of the water in the bell is observed to sink through one foot. Find the depth of the top of the bell and the height of the water barometer. MISCELLANEOUS EXAMPLES. HYDROSTATICS. 331 _ 10. A cylindrical vessel whose external and internal radii are 5 inches and 3 inches respectively, floats in water with the bottom 8 inches below the svu^face. When a liquid, whose sp. gr. is required, is poured into the vessel to a depth of 20 inches it sinks until the bottom is 15 inches below the surface. Find from these data the sp. gr. of the liquid. 11. A cubical box, of one foot external dimensions, made of material of thickness one inch, floats in water immersed to a depth of 3J inches. How much water must be poured in so that the water inside and outside may be at the same level 1 12. A piece of iron whose mass is 26 lbs. is placed on the top of a cubical block of wood floating in water and sinks it so that the upper surface of the wood is level with the water. The iron is then removed. Find the mass of the iron that should now be attached to the bottom of the wood so that the top may as before be in the surface. 13. A small hole drilled at one end of a thin imiform rod is filled .with some much heavier material. It is observed that the rod can float in water half-immersed and inclined at any angle to the vertical. Show that the sp. gr. of the wood is ^. 14. A block of wood, the volume of which is 26 cubic inches, floats in water with f of its volume immersed, find the volume of a piece of metal, the sp. gr. of which is 8 times that of the wood, which when attached to the lower part of the wood causes it to be just immersed. 15. A vertical cylinder is fitted with a smooth piston resting on water contained in the cylinder, from the side of the cylinder close to its base rises a vertical tube commvmioating with the cylinder and also containing water. Find the area of the section of the piston so that for each lb. placed upon it the surface of the water in the tube may increase its height above the piston by one inch. 16. The barrel of a Smeaton's air-pump is of the same capacity as the receiver and connecting tube together ; supposing that the valve at the bottom of the barrel is the first to cease working and that there is no leakage, show that the most complete exhaustion the air-pump can give is at the end of 8 strokes, the area of the valve is -01 of a sq. inch and its weight j^ ozs., the atmo. pressure being 2112 lbs. per sq. foot. 17. A rectangular-shaped box is constructed with its ends (weight- less) hinged to the base and capable of moving without friction between the sides but so as to enable the box to contain water. The tops of these equal ends are connected by a piece of inelastic string so that when water is poured in they are inclined inwards and make equal angles with the vertical. Show that the tension of the string varies as the cube of the depth of the water. 332 THE ELEMENTS OF APPLIED MATHEMATICS. 18. Show that the Centre of Pressure of a parallelogram immersed with one angular point in the surface and one diagonal horizontal lies in the other diagonal and is at a depth =|^ of the depth of the lowest point. 19. A bucket half-full of water is suspended by a string which passes over a pulley small enough to let the other end fall into the bucket. To this end is tied a ball whose sp. gr. o- is greater than 2. Show that if the ball does not touch the bottom of the bucket and no water overflows, equilibrium is possible if the weight of the ball lies between W and ^ W, where W is the weight of the bucket and water. 20. A cylindrical diving-bell of height b is immersed in water with its highest point at a depth z below the surface. If the barometer rises so that the increase of the whole pressure on the top is P, show that the alteration in the tension of the chain is approximately 2 \ '\/{z+hy+4bh/ ' where h is the height of the water barometer. ANSWERS TO THE EXAMPLES. Ex. I. Page 4. 1. 4. 88. 2. 26,400 feet. Jmin. 5. 40. 6. Ex. II. 3. If miles pel 8-15 minutes n( Page 7. ■ minute. jarly. 7. 2 mil 1. 4. a=8. K = 180. 2. 4 sees. 5. 11 sees. Ex. III. 3. G. Page 12. 3. 15 feet per second. 1. 5. 1600 ft., 320 ft. per see. 3, 100 sees., 10,000 cms. 6. 6 ft. per sec. 1 min. 4. 10 sees. Ex. IV. Page 13. 1. 4 440 ft. u=15, a=-4. 2. 27ft.-sec. aniis. 3. Isec, 16 ft. Ex. V. Page 16. 1. 80 ft.-sec. units. 2. 510 ft., 2430 ft. 3. The distances are as 11 : 28. 4. m=10, a=2. 5. 3 sees., 44 ft. Ex. VI. Page 21. 1. 256 ft., 64 ft. per sec. 2. 32 ft. 3. 22i£t. 4'. 405 ft. nearly, 177 ft. 5. 136 ft. per sec, 88 ft. 6. 80 ft. •f] 144 ft. 9. 4080 ft. 10. 864 ft., 464 ft. per sec. ui 1200 ft. per sec. 12. 85Jft.-secs. 15. 11,000 feet. 16. 20 miles per hr. 20. 9 sees. 21, Vel. of train : yel. of particle as 3-9 : 1 nearly. 334 ANSWERS TO THE EXAMPLES. Ex. VII. Page 30. 1, (i) 2-9; (u) 5-5; (iii) 14-2; (iv) 19-1, approximately. 3. 50 ft. per see. 4. 17-3 nearly. 5. 30° E. of S. 6. Vel. = :r^ ft- per sec, width of deck=40'6 ft. 15,^3 Ex. VIII. Page 36. 1. 15j3ft.-seos., 15ft.-secs. 2. 28-3 ft.-seos. nearly. 4. 100 ft. 5. 9, 37-4, nearly. 6. 6'5 ft. per min. nearly. 7. 5 ft.-secg., 5-4 ft.-secs. 8. 138 ft.-seos. ne?irly. Ex. IX. Page 39. 1. 25 miles per hour. 2. 35 miles. 3. 94 ft.-secs. nearly. 4. 30 miles per hour. 5. 10 miles per hour from N.W. 7. " "ot 6. 8. If a is the angle its direction makes with the line of wickets, tan a=||. 9. He walks in a direction making an angle sin-^ f with his rank, in i^ sees. Ex. X. Page 45. 1. 28potindals. 2. 16 ft. 3. 4. 36,000 feet. 5. 141 ft.-seos. 6. 32,000 poundals. 7, 3200 units of momentum. 8. 200 poundals, -^ ft.-sec. units. 9. -017 nearly, 855-5 sees. 10. -0086. H. 5 lbs. 12. 36^ cms. per sec, 18J cms. Ex. XI. Page 54. 1. 980 in o.G.s. units. 2, 52 feet, in 6J sees. 3. J sec, 1 foot. 4. 36 feet. Ex. XII. Page 55. 1, •s'ij of a ton weight. 2. 6 ft.-sec. units. 3. (i) Acwt.; (ii) fjcwt. 4. (i) 40 lbs. wt.; (ii) 50 lbs. wt. 5. Zero. 6. 5 sees., 80 ft.-secs. 7. Acceleration =4 ft.-sec units, tension =wt. of */lbs.; pressures are 9 poundals, lOJ poundals. 8. The forces are equal. 9. (i) 4-7; (ii) 234-4 nearly. lO. 3348 tons wt. nearly. 11. -0174 sees, nearly. 12. True weight 8-9 lbs! 13. 3-7 sees., 17 ft.-secs. 16. 24/-h23/'=^. 18. 6-5 ft.-seos., 7-3 ft.-secs. 21. R - ^^ ■ P + iQ ANSWERS TO THE EXAMPLES. 335 Ex. XIII. Pagk 59. 1, 300 ft., 456 ft, 2. 10,000 ft., 20,000 ft. 3. 96 ft. 4, 12,100 yds. 6. 2700 yds., U,400 yds. Ex. XIV. Page 63. 1. 45°, 40;,y6 ft.-seoa. 2. 173 ft.-seos. nearly. 3. 30 VlO ft.-seos. 15. nT*/g2 Bci. feet. Ex. XV. Page 69. 1. (i) 15-5 lbs. wt. nearly; (u) 1-5 lbs. wt. nearly; (iii) 3 lbs. wt.; (iv) 3-2 lbs. wt. nearly. 3. 1 lb. wt., ^7 lbs. wt. 4. 21bs. weight. 5. 2+;^31bs.wt., 2-^3 lbs. wt. 6. 18 lbs. nearly. 7. 5:3. 8. Q~Q'. 10. 5-8 lbs. making an angle of 30° with the vertical. H. Ji ■ 1. 12. 12^2 lbs., 12 lbs. 14. -\. 15. ^ lbs., ^^ lbs. 17. II . Ex. XVI. Page 73. 1. 4 lbs. 2. A force represented by ,^3. 3. The tensions are equal to the weights of 6 and 8 lbs. 5. 120°. 6. P~Q. 7, 2{v/2-l)lbs.wt. due East. 10. 20 lbs. wt., 10 ^3 lbs. wt. 11. ^ i| , £ . 5. oo 7 Ex. XVII. Page 77. 1. 4-8 lbs. wt. 2. 4 lbs. wt. North. 3. ^'3^. 4. 4 poundals. 5. 8-29 lbs. 10. The resultant is rep. by AB. 13. CA. 15. V-P^ + g2 + JS2 + S2 - 2PR - 2QS. 16. JP^ + Q'+B^-QB-BP-PQ. 17. ^3 : 1 : 2. 18. 135°, 135°, 90°. 19. 3 lbs., 120°, 30°. 20- |#7^*^''^"'*'- 21. 30°. Ex. XVIII. Page 82. 1. 41 ft.-seos. 2. ihi- 3. 3. 4. iV- 5. 981. 6. ,k- 7. 2. 8. 32. 9. llb.wt. 11. If A is the angle through which the force is turned, the resultant makes an angle 90° - -^ with the direction of this force produced. 12. 14-1 tons weight nearly. ,13. 50 feet per sec. nearly. 336 ANSWERS TO THE EXAMPLES. 15. A of a ton weight. 18. The shorter. 19. 17 lbs. weight. 20. 120°. 24. If P is the given force each component is ^^ P. 25. 30°. Ex. XIX. Page 95. 1. (i) 35 lbs. wt. distant 8| inches from the greater force ; (ii) 5 60 2. 4 inches. 3. 22^, 7^ lbs. weight. 4. 40 lbs., 120 lbs. 5. 4^7- 6. The distances are as & : a - 6. 7. If W is the weight of the bundle, and a, x, the distances of the bundle and his hand from his shoulder, the pressure is 1^ 1 1 + - ] . 8. 16^ inches, 13^ inches. Ex. XX. Page 98. 1. 15 inches. 2. 16 lbs., 48 lbs. wt. 3. Where the force 9 acts. 4. 8 inches from the middle point. 5. In the middle. 9. 50 ; — - lbs. wt,, where I is the length of the rod in inches. i + D Ex. XXI. Page 102. 1. 6,?^ feet from the boy. 2. As 1 : 5. 3. 4^, 7| feet. 4. The required distance is equal to the radius of the inscribed circle of the triangle. 7, 40 lbs. weight, parallel to BG and distant 3 feet from BC. 11. It is represented by CD, where D is a point in AB such that AD : DB :: 7 : 5. 15. The other forces are in the directions CB, CD, AD. Ex. XXIII. Page 118. 1. The c.a. is 6. and 30 inches from the respective ends. 2. Itt inches from the point of contact. 3. 13 inches, 2;^34 inches. 4. xf X inches from the centre of the rod. 5. (i) if inches from the centre. (ii) 5^\ inches from the centre of the larger plate, (iii) fl inches from the centre of the square, (iv) J an inch from the centre of the rod. ANSWERS -TO THE EXAMPLES. 337 Ex. XXIV. Page 119. 1. The midpoint of the middle weights. 2. BJ inches from the midpoint. 3. 2A inches. 4. 3^ feet. 5. gibs. 6. 8J|in. Ex. XXV. Page 122. 1. The CO. is ||o, l^a distant from the sides which meet where the weight 2 is placed, a being a side of the square. 2. -yg , if a is a side. 3. - from the middle side. 4. The o.G. lies on the production of the line joining the centre to the vertex at which there is no particle and at a distance ^r, where r is the radius of the circumscribing circle. 5. -^ a. from the first side. 26 6. o- ,„ a from the vertex, where a is a side. ^7 \/^ 7. 2 inches. 8. -Y- inches from the bottom. Ex. XXVI. Page 127. 1, g— ^ from the centre of the square, if a is a side. 2. r=l + J2. 3. The CO. is distant ^ a, ^^a from the sides through the first vertex. 4. The centre of the inscribed circle. 5. is the c. a. of the triangle. 6. 2^^ in. from the angle. 8. If S, y are the distances of the c. q. from the middle point of one side 2a^=Sx'. 10. 7fl in., 8S in., from two edges. 13. S|^ inches. 15. The c. g. bisects the rod, which weighs 12 lbs. 2{2+y2^2Jr 3B 21. On the base of the triangle. 22. 55-; . -s" • jti + V £ Ex. XXVII. Page 142. 1, The diagonal through the point is inchned to the vertical at an angle a such that tan a = J. 2. Tension=101b8.wt., pressure = 6 lbs. wt. 3. Bequiredweight=6'61bs. A 6. Each is a force of 3 lbs. wt. 7. Thrust =iWeot--. 8. To a point in the same horizontal plane as the centre. 9. Pressure = jr. 10. Two ozs. 11. W'^ 2 vZ-B^-r" 13. tan-f^-^i^^ ). 17. 15a. \sin a + 2cos 0/ 23. The angle between the strings is 2 cos-' ^i^. J. 22 338 ANSWERS TO THE EXAMPLES. Ex. XXVIII. Page 147. 1. 5040. 3. 79H foot-PO^tlB. 4. 50m foot-lbs. 5. 22,400 foot-lbs. Ex. XXIX. Page 149. 1. 75 mUes per hour. 2. 6^- 3. 30-72. 4. 130^. Ex. XXX. Page 151. 1. lilfeet. 2. 48,020,000 ergs; 12,005,000 ergs; 0. 3. W feet' 4, 18 X 10'" ergs, 5. 150 poundals, 7500 poundals. Ex. XXXI. Page 155. 1. 71-55 ft.-secs. nearly. 2. 90,367,200 foot-lbs. 3. 1^ tons weight. 4. 88 foot-lbs. Ex. XXXII. Page 159. 1. 1200. 2. W- 3. Unit of length = I feet, unit of time = f sees. 4. 3' X 20*. 5. -00007, -0000000024 nearly. 6. The units are 2240 lbs., 800 feet, 5 seconds. 7, twststs foot-sees. 10. 3332yVT lbs. 11. 66,000,000 ft.-lbs., 2000 h.p. 12. 660,000, 30 H.p. 13. 179-2. 14. 10-7 nearly. 15. H.p. =3-2. 16. 1,161,600 joules. 18. 1000 feet. 20. x/31bs. 21. 4320. 23. 30 H.p. Ex. XXXIII. Page 166. 1, If lbs. wt. 2. 5 ft. and 2 ft. from the ends. 3. Fulcrum is 2J ft. from the centre. 4. 170 lbs. wt. 5. 9# lbs. wt. 6. The arms are as 9 : 11. 7. Shorter wire is inclined to horizon at the angle oot~' 2 ^3. 8. 13J feet, 15 lbs. 9. 62J lbs. wt. 10. 50 lbs. wt., neglecting weight of lever. Ex. XXXIV. Page 169. 1. 26J lbs. 2. 56 ozs. 3. 2 : 1, 8 lbs. 7. He loses £19 x ' ab Ex. XXXV. Page 172. 1, 44 inches. 2. 2J inches. 4. If 0' be the new zero of graduation C(y=:^CO. 5. One inch from one end. 6. 18 lbs. ANSWERS TO THE EXAMPLES. 339 Ex. XXXVI. Page 174. 1. 3 ozs. 2. 20 inches. 3, y. 4. 2 inches. 5. Fix on a weight so that the o.o. and the mass are both the same as before. Ex. XXXVU. Page 177. 1. 4^ lbs. weight. 2. 120 lbs. wt. 3. 12, 36 lbs. wt. 4. 3181flbs. wt. 5. 3^ inches. 6. 2tflbs. wt. 8. 300 lbs. wt. 9. 32f lbs. wt., ^^4ss H.P. Ex. XXXVIII. Page 181. 1. 12 lbs. -wt. 2. 4P. 3. 2 lbs. wt., 7 pnUeys. 4. 207 lbs. wt. 5. li of the man's weight. 7, 9 stone weight, neglecting the weight of the pulleys. 8. The radii of the wheels in the upper block are in the proportion 2:4:6 &o., and those in the lower block in the proportion 1 : 3 : 5 &c. Ex. XXXIX. Page 186. 1. 141bs. wt. 2. 15|lbB. wt. . 3. 290 lbs. wt. 4. 1 lb. wt. 6. 4 lbs. wt. 7. Unity. ,8. 971bs. wt., llb.wt. 9. 701bs. wt. 10. 71J lbs. wt. 12. 14|lbs. wt. 13. 62 1bs. wt. 14. 36-2. 16. The distance from the point of action of Tj is ^ {20)^+10j»i + 8m»j}. Ex. XL. Page 189. 1. 12 feet. 2. 1 owt. 3. 9 lbs. wt., 15 lbs. wt. 4, 121bs. wt. 5. 724,5. 7. Sflbs. wt. 8. 60°. 9, 841bs. wt. Ex. XLI. Page 191. 1. 20^3. 2. 60°, 120° with the plane, 3. ooB-if>2P. 5. 120° with inclined plane. Ex. XLII. Page 196. 1. 2fJlbB. wt. 2. 13Albs. wt, 3, A. 340 ANSWERS TO THE EXAMPLES. Ex. XLIII, Page 205. 1 — 7=::= . 5, It makes the angle tan~i ^- with the horizon. 2^2759 6. V2-1, 2^. 7. 3808ft.-lbs. 10. At the middle point of the edge vertically above the one which moves off the floor. Ex. XLIV. Page 209. 1 j;^^^ sm(o-e) 3,097,600 sin X foot-lbs. '■• I t Bm(o + e) Ex. XLV. Page 217. 1. -T- , where u is the vel. of the striking sphere. 3 35 /5 3. e = J, impulses— ^. 4. As 1:50. 10. 3-7 sees., 45 sees. nearly. 15. The veL of each ball is '^^ v, and makes an angle a with the line o 4 joining their centres where tan a=^—Tx. Ex. XLVI. Page 242. 4. 8 Jr, where r is the radius of the circle. 5. 2,^ cwt. 2 6. Pendulum must be lengthened :pg— ^ inches == -014 inches nearly. 7, 8692 nearly. 8. 9 sees, nearly. 9, Momentum of each is 12' x 112 units of momentum, k.e. of cannon = 123x56, K.E. of baU=1122x288. 10. -107 lbs. wt. 14. 981. ANSWERS TO THE EXAMPLES. 341 HYDEOSTATICS. Ex. I. Page 251. 1. 14-8 lbs. nearly. 2. 2-304 feet. 3. 62-5 lbs. 4. 110 feet. 5. 53J lbs. 6. As 24 : 25. 7. lOiV lbs. 8. 178A feet. Ex. II. Page 257. 1. 2-8 lbs. 2. 36,000 lbs. 3. As 7 : 1. 4. || of height of tube. 5, 27-24 where A is the area of the section of the tube. 6, 14-96 cubic inches nearly. 7, 5-08 inches of water, 4-97 inches of alcohol. 8. ■ 23 fathoms nearly. 9. 243 ozs. wt. nearly. 10. 37 feet nearly. Ex. III. Page 261. 1. 400 cubic ems. 2. 75 lbs. wt. 3. 4572-8 grammes wt. 4. 63 grarmnes wt. 5. 7 ozs. wt. nearly, 349 ozs. wt. nearly. 6. 375 ozs. wt. 7. As 5 : 7. 8, Pressure unaltered in the first case, in the second increased by the weight of the wood. 9, The wood will rise. Ex. IV. ' Page 266. 1. 5-8 lbs. wt. nearly. 2. 1062-5 lbs. wt. 3. 6 ijsx 1000 grammes wt. = 10,392 grammes wt. nearly. 4. 123 and 116 ozs. wt. nearly. 5. 2075 grammes wt. 6. l-2kil.wt. 7. 156J, 187i, 218f lbs. wt. respectively, neglecting atmo. pressure. 8. ^4 (3+;^2) lbs. wt. 9. Pressure on table is 33-9 ozs. wt., pressure on base is 23-1 oz. wt. nearly. 10. 7,524,000. Ex. V. Page 269. 1. The cone weighs f of one lb. 2. i}°9« X wt. of cubic inch of water=13-6 x m% ozs. nearly. 3. 33, 750 lbs. wt. acting at 7J feet from the top. 4. 2^3 feet. 5. -848 grammes. 6. U?i -P2) ^ _^^ ■ Ex. VI. Page 273. 1. #. 2. 4168-8 grammes wt. 3. 8-9. 4. 150 cubic decimetres. 5, 4-76 grammes wt. ; sp. gr. =2-97 nearly. 6, One cubic foot has the mass 8000 ozs. 342 ANSWERS TO THE EXAMPLES. Ex. VII. Page 274. 1. 6-1 inches. 2. ^- 3. 2| cubic feet. 4. 13-5. 5. -082 inches nearly. 6. M owt., 1^ owt. 7. 21. 8. 466 : 387. 9, SW. 10. 348^$ grammes wt. 11. ISjlbs. 12. 8 grammes. 13. -672 inches. 14. 150 gr. W W 15. (i) -o' (") i ,. where W=vi. of body, 5 = sp. gr. of iron. 16. 965 : 1780. 17. 18 inches. 18. As 81 : 46. 19. 48, 52. 20. 3i inches, A of an inch. 21. '965. 22. HJ- 23. 6 inches, 4-8. 24. 4-9 inches nearly. Ex. VIII. Pagk 278. 1. 5f. 2. (i) -95; (ii) 1. 3. \h 4. 6-21 nearly. Ex. IX. Page 279. 1. 7A- 2. Half that of mercury. 3. Hi- 4. 2. 5. -9. Ex. X. Page 281. 1. 2i»j. 2. -^#, 5 grammes wt. 3. 1'20 nearly. 4. '8. 5. iHlb. 6. -8. 7. 7. 8. ^A'^''"'-^- Ex. XI. Page 283. 1, 15/(24, 23, 22, ... 16). 2. |. 3. liV 4. l^^- 5. 6H. Ex. XII. Page 285. 1. 5. . 2. 323^ grs. 3. 1-080. 4. 2f oz. Ex. XIII. Page 289. 1, 1-36 cubic inches. 2. 60 inches. 3. 4 cubic inches. 4. 14-21 lbs. 5. The weights are as 7 : 4. 7. TT ^1- inches. 8. 8-06 grains. Ex. XIV. Page 291. 1. The densities are as 17 : 22. 2. 584-6 cms. nearly. 3. ^'i^xlO*. 4. 29-2 cms. of mercury. 5. 26 grammes. 6. 2Jf feet. ANSWERS TO THE EXAMPLES. 343 Ex. XV. Page 294. 1. 124-2 grammes wt. 2. 58-5 cms. of mercury. 3. "05 grammes nearly. 4. _ 57° c. 5. 96 J , where m is the mass of the gas in lbs. and s its sp. gr. referred to air. Ex. XVI. Page 301. 1. -49 lbs. wt. nearly. 2. 10-8 inches of tube. 3. 22-6 cms. nearly. 4. rk- 5. 32 lbs. per sq. inch. 7. '^-— . 8. 4-4 nearly. 9. 30-5 nearly, IQ. 60 inches. H. 5-037 inches nearly. 13. The piston will rest (i) at a height of 4-8 feet nearly from the base, (ii) at a distance of 5-9 feet nearly from the base. 14, The mercury wiU rise in the air vessel to a height ^ above the base. 20. 3-73 ozs. Ex. XVII. Page 308. 1. 37-7 feet. 2. 5^ feet. 3. 2500 lbs. 4. 280 lbs. wt. nearly. 5. 14 inches nearly. 6. 21-57 lbs. per sq. inch. Ex. XVIII. Page 313. 1. 17-4 inches of mercury nearly. 2. '96/3, whei-e p is the density of atmo. air. 3, S-S inches. 4. ^ oz. 6. 20f inches. 7, As 12 : 1. 8. Six times atmo. pressure. 9. 7. Ex. XIX. Page 319. 1. 4 feet. 2. 22^ cubic feet. 3. J. 4. 61-25 cms. 5. 1-8 feet nearly, 1^ F. 6. 9 feet. 7, If F is the volume of the bell in cubic feet, n feet per sec. the rate at which it is lowered, and h the height of the water barometer, the nV quantity of air at atmo. pressure supplied per see. is — cubic feet. 8. 3f ft.-sec. units. 9. t's, -00058. 10, 8 inches nearly. s — s' k 12, 1-03. 13, 4 F . 16. 5 - - , where k is the depth to which the shorter arm is immersed. 17, Max. density=mx density of atmo. air. 18. ^f of a ton weight. 19. 176 lbs. wt, appUed to each hemisphere. (t=^.) 20. fngT- 344 ANSWERS TO THE EXAMPLES. MISCELLANEOUS EXAMPLES. DYNAMICS. Page 322. 3. 11. 25. 29. 32. 35. 40. 43. 46. 53. 56. 58. 13/x, 27JV miles per hour. 62-5 seconds. 4. tt of a mile. 8. J28 2. 2 x/3 miles per hour ; 15° to W. of S. M- -M'smay T+M' I 2-49 feet. 17. Q .AB + R . AI) = 1. Vertical velocity =«. 26. Acceleration of P= M- 21. 50 lbs. F.B-iQ.R+iP.Q {P+Q)R + 4:PQ 9'3125 lbs., 99i per cent. k.e. is lost. 31. 18 minutes. 450 poundals. 33. 8 J miles per hour; ^ of the k.e. is lost. tVV; 2000 lbs. 37. iP=W+A-C. 38. dr^JW^. Wg hgfi 3J lbs. wt. Three quarters distance across. I 1 1 = + ■ 41. 7 revolutions per second nearly. 224000 X Vl poundals. iw9- ■ 66. 6§lbs. g + 2hlt^' 2gt^+h' 45. iW. 52. I 55. 102g lbs. 57. IOt't- HYDROSTATICS. Page 330. 1. P=H. 3. i-. 7. 2-88, 3-14, 3-46 cubic inches nearly. 9. 3 inches ; 33 feet. 10. U- 26s 12. -^ , where s is specific gravity of the iron, 15. 27-648 sq. inches. 6. § inches. 8. 1-5. 11, 900 cubic inches. 14. 2 cubic inches. CAMBRIDGE: PRINTED BV C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS. October 1893. A CATALOGUE OF EDUCATIONAL W0RK5 Published by GEORGE BELL & SONS. CONTENTS. Grammar School Classics PAGE 2 Cambridge Greek and Latin Texts 3 3 4 4 Cambridge Texts with Notes Public School Series Critical and Annotated Editions Translations, Selections, &c 5 6 Lower Form Series Classical Tables 6 Latin and Greek Class-books 6 Cambridge Mathematical Series 8 Cambridge School and College Text-books ... 9 Bookkeeping, Geometry, and Trigonometry ... 10 Mechanics and Natural Philosophy II Foreign Classics 12 French Class-books 12 Gombert's French Drama 12 German Class-books ■ 13 English Class-books 14 Bell's English Classics 14 Psychology and Ethics, AND Music 15 Geology, Technology, and Agriculture . 16 History ... 17 Dictionaries ■ 17 Divinity . 18 Reading Books, &c • 19 w.c. George Bell and Sons' GRAMMAR-SCHOOL CLASSICS. A Series of Greek and Latin Authors, with English Notes. Fcap. 8vo. Cfflsar : De Bello aallloo. By George Long, M.A. 4s. Books I. -in. For Junior Classes. By G. Long, M.A. Is. 6d. Books IV. and V. Is. 6(J. Books VI. and Vn. , Is. 6(i. Books I., n., and III., with Vocabulary, Is. 6d. each. Catullus, Tibullus, and Propertius. Selected Poems. With Life. By Kav. A. H. WratiBlaw. 28. 6i. Cloero: De Seneotute, De Amioitia, and Select Epistles. By a«orge Long, U.A. Bs. Cornelius Nepos. By Eev. J. ¥. Macmichael. 2s. Homer: Iliad. Books I.-Xn. By F. A. Paley, M.A., LL.D. 48. 6d. Also in 2 parts, 2a. 6d. each. Horace: With Life. By A. J. Maoleane, M.A. 3s. 6d. In 2 parts, 28. eacli. Odes, Book I., vrith Yocabolaiy, Is. 6d. Juvenal : Sixteen Satires. By H. Prior, M.A. 3s 6d. Martial: Select Bpigrame. With Life. By F. A.Paley,M.A.,LL.D. 4i. 6i. Ovid : the Fasti. By F. A. Paley, M.A., LL.D. 3s. 6d. Books I. and n.. Is. ad. Books III. and IT., Is. 6d. Books T. and TI., !<. 6d. Sallust : Oatilina and Jugurtha. With Life. By G. Long, M.A. and J. G-. Frazer, 38. 6d., or separately, 28, each. Tacitus : Germania and Agricola. By Bev. P. Frost. 2s. &d. Virgil: Bucolics, Georgics, and Mneid, Books I.-IV. Abridged from Professor Conington's Edition. 43. 6d. — ^neid. Books T.-2II., 48. 6d, Also in 9 separate Volumes, as follows, Is. 6d. eadi;— Bucolics — Georgics, I. and II.— Georgics, III. and IV.— iBneid, I. and II.— III. and IV.— V. and VI.— VII. and VIII.— IX. and X.- and XI. and XII. .«)neid, Book I., with Vocabalary, Is. 0d. Xenophon : The Anabasis. With Life. By Eev. J. F. Macmichael. 38. 6d. Also in 4 separate Tolumes, l8. 6d. each ; — Book I, (with Life, Introduction, Itinerary, and Three Maps) — Books II. and III. — lY.and V. —VI. and VII. The Cyropsedia. By G. M. Gorham, M.A. 3s. 6d. Books I. and II., Is. 64.— Books V. and VI., Is. 6(1. Memorabilia. By Percival Frost, M.A. A Grammar School Atlas of Classical G-eography, containing Ten selected Maps. Imperial 8vo. 3s. Uniform with the Series. The New Testament, tn Greek. With English Notes, &o. By Hot. J. F. Macmichael. 4s. 6d. The Four Gospels and the Acts, separately. Sewed, 6d. each. Ediicational Works. CAMBRIDGE GREEK AND LATIN TEXTS. Aeschylus. By F. A. Paley, M.A., LL.D. 2». Caesar : De Bello Gallloo. By G. Long, M.A. le. M. Cicero: De Senectute et De Amlcltla, at Eplstolca Selectee. By Gt. Long^M.A. Is. 6iJ. Ciceronis Oratlones. In Verrem. By G. Long, M.A. 2». 6. Newton's Principla, The First Three Sections of, with an Appen- dix ; and the Ninth and Eleventh Sections. By J. H. Evans, M.A. 5th Edition, by P. T. Uajn, M.A. 48. Analytical G-eometry for Schools, ByT.G.Vyvyan. 6th Edit, 4«.6(J. Qreek Testament, Companion to the. By A. C. Barrett, M.A, 5th Edition, revised. Fcap, 8vo, 58, Book of Common Prayer, An Historical and Explanatory Treatise on the. By W. a, Humphry, B.D, 6th Edition, Fcap. Svo. 2s. 6(i. Muaio. Text-book of. By Professor H. C. Banister. 15th Edition, revised. 5s. Concise History of. By Bev. H. G. Bonavia Hunt, Mns. Doc. Dublin. 12th Edition, revised. 3s, 6d, ARITHMETIC, (^*« "'^o *''* *""' foregoing Series.) Elementary Arithmetic. By Charles Pendlebury, M.A., Senior" Mathematical Master, St. Paul's School; and W. S. Beard, F.E.G.S., Assistant Master, Christ's Hospital. With 2S00 Examples, Written and Oral. Crown Svo. Is. 6d. With or without Answers. Arithmetic, Examination Papers in. Consisting of 140 papers, each containinsr 7 questions. 357 more difloult problems follow. A col- lection of recent Public Examination Papers are appended. By 0. Pendlebury, M.A. 2s. 6d. Key, for Masters only, 5s. Graduated Exercises in Addition (Simple and Compound). By W. S. Beard, C. 8. Department Rochester Matbematioal School. Is, For^ Candidates for Commercial Certificates and Civil Service Exama. a2 10 George Bell and Sons' BOOK-KEEPING. Book-keeping Papers, set at various Public Examinations. Collected and WrittoD by J. T. Medhnrst, Lecturer on Book-keeping in the City of London CollfiKe. 2nd Edition. 3s. A Text-Book of the Principles and Practice of Book-keeping. By Professor A. W. Thomson, B.So., Koyal Agricultural College, Cirences- ter. Crown 870. 5s. pouble Entry Elucidated. By B. W. Poster. 14th edition. Fcap. ito. 3s. 6(2. A. Kew Manual of Book-keeping, combining the Theory and Practice, with Specimens of a set of Books. By Phillip Orellin, Account- ant. Crown 8vo. 3s. 6d. Book-keeping for Teachers and PupUs. By Phillip Crellin. Cro«™ 8to. Is. 64. Key, 2s. net. GEOMETRY AND EUCLID. Euclid. Books I.-VI. and part of XL A New Translation. Br H. Deighton. (See p. 8.) ' The Definitions of, with Explanations and Exercises, and an Appendix of Exercises on the Krst Book. By E. Webb. M A. Crown 8yo. Is, «■» j , ..»• Book I. With Notes and Exercises for the use of Pre- paratory SclloolB, Sec. By Braithwajte Amett, M.A. 8vo. 4s. 6d. The First Two Books explained to Beginners. By 0. P. Mason, B.A. 2nd Edition. Pcap. 8yo. 2s. 6*. The EnunoiaUons and Figures to Euclid's Elements. By Bev. J. Brasse, D.D. New Edition. Foap. 8vo. Is. Without the PiKures. 6d. Ereroiaes on Euclid. By J. McDowell, M. A. (See p. 8) Geometrical Conic Sections. By H. G. Willis, M.A. (See p 8 ) Geometrical Conic Sections. By W. H. Besant, So.D. (See p' 9) Elementary Geometry of Conies. By C. Taylor, D.D. (See p. s!) An Introduction to Ancient and Modem Geometry of Conlos ByC. Taylor, D.D., Master of St. John's CoU.,Camb. 8vo. 15s An Introduction to Analytical Plane Geometry. Bv W P Tumbull, M.A. Svo. 12s. j ". i. Problems on the Principles of Plane Co-ordinate Geometrv By W. Walton, M.A. 8yo. 16s. wuioixy. Trillnear Co-ordinates, and Modem Analytical Geometry of Two Dimensions. By W. A. Whitworth, M.A. 8vo. 16a. An Elementary Treatise on SoUd Geometry. By W. S. Aldls, M.A. 4th Edition revised. Cr. 8vo. 6s. Elliptic Functions, Elementary Treatise on. BvA. Cayley, D.So. Demy 8to. [jf e™ Editim Preparing. TRIGONOMETRY. Trigonometry. By Eev. T. G. Vyvyan. 3s. 6d. (See p. 8.) Trigonometry, Elementary. By J. M. Dyer, M.A., and Eev. E. H. Whitcombe, M.A., Asst. Masters, Eton College. 4s. 6d. (See p. 8.) Trigonometry, Examination Papers in. By G. H. Ward,'M A Assistant Master at St. Paul's School. Crown 8to. 2s. 6(J. ' "' Educational Works. 11 MECHANICS & NATURAL PHILOSOPHY. statics, Elementaiy, By H, Goodvrin, D,D. Fcap. 8to, 2nd Edition. St. DynamioB, A Treatise on Elementary. By W. Gamett, M.A., D.C.L. Stli Edition. GrovniSvo. 6s. Dynamics, Bigid. By W. S. Aldis, M.A. 4s. Dynamics, A Treatise on. By W. H. Besant, So.D.,F.B.S. 10». 6d. Elementary Mechanics, Problems in. By W. Walton, M.A. Nev Edition. Crown 8vo. 6s. Theoretical Mechanics, Problems in. By W. Walton, M.A. 3rd Edition. Demy 8to. 16s. Struotxiial Mechanics. By E. M. Parkinson, Assoc. M.I.C.E. Grown 8vo. 4s. 6d. Elementary Mechanics. Stages I. and 11. By J. C. Horobin, B.A. Is. 6d. each. [Stage III. prejiai-wij. Theoretical Mechanics. Division I. (for Science and Art Ex- aminations). By J. 0. Horobin, B.A. Crown 8vo. 2s. 6d. Hydrostatics. By W.H.Besant,Sc.D. Cr.8vo. 16th Edit. 4s. 6d. Hydromechanics, A Treatise on. By W.H. Besant, So.D.,F.E.8. 8to. Sth Edition, revised. Part 1. HydrostatioB. ."is. Hydrodynamics, A Treatise on. Vol. I., 10s. 6d. ; Vol. U., 12s. 6d. A. B. Ba£set, M.A., T.B.S. Hydrodynamics and Sound, An Elementary Treatise on. By A. B. Basset, M.A., P.E.S. Demy 8to. 7s. 6d. Physical Optics, A Treatise on. By A. B. Basset, M.A., P.E.S. Demy 8to. 16s. Optics, Geometrical. By W. S. Aldis, M.A. Crown 8vo. 4th Edition. 4s. ™ , ^ „, „ Double Beflraotion, A Chapter on Fresnel's Theory of. By W. S. Aldis, M.A. Svo. Zs. „ „ , Eoulettes and Glissettes. By W. H. Besant, So.D., P.E.S. 2nd Edition, 5s. Heat, An Elementary Treatise on. By W. Garnett, M.A., D.C.L. Crown 8to. 6tli Edition. 4s. 6d. Elementary Physios, Examples and Examination Papers in. By W. Gallatly, M.A. 4s. Newton's Prinoipia, The First Three Sections of, with an Appen- dix ; and the Ninth and Eleventh Sections. By J. H. Evans, M.A. Sth Edition. Edited by P. T. Main, M.A. 4s. Astronomy, An Introdnction to Plane. By P. T. Main, M.A. Fcap. Svo. cloth. 6th Edition. 48. Mathematical Examples. Pure and Mixed. By J. M. Dyer, M. A. , and K. Prowde Smith, M.A. 6s. Pure Mathematics and Natural Philosophy, A Compendium of Facts and rormnlss in. By Q. B. Smalley. 2nd Edition, revised by J. McDowell, M.A. Fcap. Svo. 2s. „ tx ^ j • tm^ Elementary Course of Mathematics. By H. Goodwm, D.D. 6th Edition. Svo. 16s. _ , , . . .^^ A CoUeoUon of Examples and Problems in Arithmetic, Alffebra. Geometry, Logarithms, Trigonometry, Conio Sections, Mechanics, &of,wah Answers. By Kev. A. Wrigley. 20th Thousand. 8s. 6. 6d. The New Testament for English Readers. By the late H, Alford, D.D. Vol. I. Part I. Srd Edit. 12s. Vol. I. Part II. 2nd Edit. 10s. 6(1. Vdl. 11. Part 1. 2nd Edit. 16». Vol. II. Part II. and Edit. 168. The areek Testament. By the late H. Alford, D.D. Vol. I. 7th Edit. 11. 8». Vol. II. 8th Edit. 11. 48. Vol. III. 10th Edit. 18s. Vol. IV. Part I. 6th Edit. 18s, Vol. IV. Part II. 10th Edit. lis. Vol. IV. 11. 128. Companion to the O-reek Testament. By A, 0, Bairett, M.A. 6th Edition, revised, Fcap. Svo. 6a. Guide to the Textual Criticism of the New Testament, By Rev. E. Miller, M.A. Orown 8vo. 48. The Book of Psalms. A New Translation, with Introdnotions, &e. By the Rt. Rev. J. j. Stewart Perowne, D.D., Bishop ot Worcester. Svo. Vol. I. 8tfa Edition, 18s. Vol. II. 8th Edit. 16s. Abridged for Schools, 7th Edition. Orown Svo. I0«, 6d, History of the Articles of Religion. By 0, H. Hardwick. Srd Edition. Post 8vo. Gs. History of the Creeds. By Bev, Professor Lnmby, D.D. Srd Edition. Orown Svo. 7s. 6*. Pearson on the Creed. Carefnlly printed from an early edition. With Analysis and Index by E. Waif ord, M.A. Post Sro. 5s. Liturgies and 0£9ces of the Church, for the Use of English Readers, in Illastration of the Book of Common Prayer. By the Rev. Edward Bnrbidge, M.A. Orown Svo. Os. An Hlstorloal and Explanatory Treatise on the Book of Oommon Prayer. By Bev. W. G. Humphry, B.O. 6th Edition, enlarged. Small Post Svo. 28. 6d. ; Oheap Edition, Is. A Commentary on the Qospels, Epistles, and Acts of the Apostles. By Rev. W. Denton, A.M. New Edition. 7 vols. Svo. 9s. ea/sh. Notes on the Catechism. By Bt. Bev. Bishop Barry. 9th Edit. Foap. 2s. The Wlnton Church Catechlst. Questions and Answers on the Teaohing of the Ohnroh Oateohism. By the late Rev. J. S. B. Monsell, LL.D. 4th Edition. Oloth, 3s. ; or in Four Farts, sewed. The Church Teacher's Manual of Christian Instruction. By Rev. U, F. Sadler. 43rd Thonsand. Zs, 6ol, Educational Works. 19 BOOKS FOR YOUNG READERS. A Series of Heading Books designed tofaoilitate the acquisition oftfiepower of Reading by very yowng Children. In 10 vols, cloth, M. each. Those with an asterisk Imve a Frontispiece or other Illnstrations. 'The Old Boathouse. BeU and Fan; or, A Cold Dip. *Tot and the Cat. A Bit of Cake. The Jay. The Black Hen's Nest. Tom and THei. Mrs. Bee. I SvAtahle *The Cat and the Hen. Sam and his Dog Bedleg. ) , ^°''» Bob and Tom Lee. A Wreck. | W^ts- *The New-bom Lamb. The BoBewood Box. Poor Fan. Sheep Dog. *The Two Parrots. A Tale ol the Jnbilee. By M. E. i Wintle. 9 Ulnstrations. *The Story of Three Monkeys. *Story of a Cat. Told by Herself. The Blind Boy. The Mute Qlrl. A New Tale of Babes in a Wood. *Queen Bee and Busy Bee. 'O-ulI's Crag. "The Lost Plga. Syllabic Spelling. By 0. Barton. In Two Parts. Infants, 3d. Standard I., 3d. GEOGRAPHICAL READING-BOOKS, By M. J. Babbinqion Wabd, M.A. With numerous Illustrations. The Child's O-eography. For the Use of Schools and for Home Toition. 63. The Map and the Compass. A Beading-Book of Qeography, For Standard I. Kew Edition, revised. Sd. cloth. The Round World. A Beading-Book of Geography. For Standaf d 11. New Edition, revised and enlarged. lOd. About England. A Beading-Book of Qeography for Standard III. With nnmerons Ulnstrations and Coloured Map. Is. 4«J. The Child's Geography of England. With Introductory Exer- cises on the British Isles and Empire, with Qnestions. 2s. Qi. Buitahle for Stmdwrdi I. til. ELEMENTARY MEGHAN ICS. Ey J. C. HoBOBiH, B.A., Principal of Homerton Training College. Stage I. With nwmerovs Ill/it&traMfyns. Is. 6d. Stage II. With, li/wineroua lUual^ralwna, Is. 6d. Stage III. [Preparing. 20 George Bell and Sons' Educational Works. BELL'S READING-BOOKS. VOB SOHOOLS AND FABOOHIAIi LIBBABIB8. Post Svo. Strongly bmmd in cloth, It, each. ^Adventures of a Donkey. *Iilf6 of Columbus. "O-irlmm's O-erman Tales. (Selected.) 'Andersen's Danish Tales, niustrated. (Selected.) •Uncle Tom's Cabin. * a-reat Bngllshmeu. Short Lives for Young Children, Q-reat Bngllshwomen. Short Lives of. O-reat Scotsmen. Short Lives of. Parables ffom Nature. (Selected.) By Mrs. Qatty, Lyrical Poetry. Selected by D. Munro. * Bdgeworth's Tales. (A Selection.) "Scott's Talisman, (Abridged.) •Poor Jaok. By Captain Marryat, B.N. Abgd, •Dickens's Little NeU. Abridged from the ' The Old Curiosity Shop.* •Oliver Twist. By Charles Dickens. (Abridged.) •Masterman Ready. ByOapt. Marryat. Illus. (Abgd.) *a-vilUver's Travels. (Abridged.) •Arabian Nights. (A Selection Bewritten.) •The Vicar of Wakefield. Lamb's Tales trom Shakespeare. (Selected.) •Robinson Crusoe. Illustrated. •SetUers In Canada. By Oapt. Marryat, (Abridged.) •Southey's Life of Nelson. (Abridged.) •Life of the Duke of "Wellington, withMaps andPlans. •Sir Roger de Ooverley and other" Essays from the Tales of the Coast. By J. Eunciman, Spectator. 'These Vohimes are IlhtstrateiJ. Suitable for Standard m. Stanimd ir. standard V. Standards ri.,i VII. Uniform with the Series, in limp cloth, 6d. each. Shakespeare's Plays. Kemble's Beading Edition. With Ex- planatory Notes for School Use. JUIilTIS OMSA.R. THE MBBOHANT OF VENICE. KING JOHN HENET THE FIFTH. MACBETH. AS TOU LIKE IT. ' '•it 31' i'' h