LIBRARY OF THE UNIVERSITY OF CALIFORNIA sr\\ **- UIFORNIA LIBRARY OF THE UNIVERSITY OF CALIFORNIA y-v tfw ERSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CA _ RSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CAi / / cr s MECHANICS AN ELEMENTARY TEXT-BOOK THEORETICAL AND PRACTICAL. Sonfcon: C. J. CLAY AND SONS, CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, AVE MARIA LANE, AND H. K. LEWIS, 136, GOWER STREET, W.C. ILeipjtg: F. A. BROCKHAUS. lorfe : MACMILLAN AND CO. GEORGE BELL AND SONS. CAMBRIDGE NATURAL SCIENCE MANUALS PHYSICAL SERIES. MECHANICS AN ELEMENTAEY TEXT-BOOK THEORETICAL AND PRACTICAL FOR COLLEGES AND SCHOOLS. BY R. T. GLAZEBROOK, MA., F.R.S. ASSISTANT DIRECTOR OF THE CAVENDISH LABORATORY, FELLOW OF TRINITY COLLEGE, CAMBRIDGE. CAMBRIDGE: AT THE UNIVERSITY PRESS. 1895 [All rights reserved.] PHYSICS /4 rt t. I o PHYSICS DEPT. CambrtUgt : PRINTED BY J. AND C. F. CLAY, AT THE UNIVERSITY PRESS. PREFACE., rhas now come to be generally recognized that the most satisfactory method of teaching the Natural Sciences is by experiments which can be performed by the learners themselves. In consequence many teachers have arranged for their pupils courses of practical instruc- tion designed to illustrate the fundamental principles of the subject they teach. The portions of the following book designated EXPERIMENTS have for the most part been in use for some time as a Practical Course for Medical Students at the Cavendish Laboratory. The rest of the book contains the explanation of the theory of those experiments, and an account of the deduc- tions from them. This part has grown out of my lectures to the same class. It has been my object in the lectures to avoid elaborate apparatus and to make the whole as simple as possible. Most of the lecture experiments are performed with the apparatus which is afterwards used by the class, and whenever it can be done the theoretical 665385 VI PREFACE. consequences are deduced from the results of these experiments. In order to deal with classes of considerable size it is necessary to multiply the apparatus to a large extent. The students usually work in pairs and each pair has a separate table. On this table are placed all the apparatus for the experiments which are to be performed. Thus for a class of 20 there would be 10 tables and 10 specimens of each of the pieces of apparatus. With some of the more elabo- rate experiments this plan is not possible. For them the class is taken in groups of five or six, the demonstrator in charge performs the necessary operations and makes the observations, the class work out the results for themselves. It is with the hope of extending some such system as this in Colleges and Schools that I have undertaken the publication of the present book and others of the Series. My own experience has shewn the advantages of such a plan, and I know that that experience is shared by other teachers. The practical work interests the student. The apparatus required is simple ; much of it might be made with a little assistance by the pupils themselves. Any good-sized room will serve as the Laboratory. Gas should be laid on to each table, and there should be a convenient water supply accessible ; no other special pre- paration is necessary. The plan of the book will, I hope, be sufficiently clear: the subject-matter of the various Sections is indicated by the headings in Clarendon type ; the Experiments to be performed by the pupils are shewn thus : EXPERIMENT (1). To explain the use of a Vernier and to determine the number of centimetres in half a yard. PREFACE. Vll These are numbered consecutively. Occasionally an account of additional experiments, to be performed with the same apparatus, is added in small type. Besides this the small-type articles contain some numerical exam- ples worked out, and, in many cases, a notice of the principal sources of error in the experiments, with indica- tions of the method of making the necessary corrections. These latter portions may often with advantage be omitted on first reading. Articles or Chapters of a more advanced character, which may also at first be omitted, are marked with an asterisk. I have found it convenient when arranging my own classes to begin with a few simple measurements of length, surface, volume and the like. These are given in Chapter I. The two following chapters deal with Kinematics and treat the subject in the usual method. When questions dealing with Momentum, Force, and Energy come to be considered two courses at least are open to the teacher. It is possible to make the whole subject purely deductive ; we may start with some de- finitions and axioms laws of motion, either as Newton gave them, or in some modern dress and from these laws may deduce the behaviour of bodies under various circumstances. Another and more instructive method, it seems to me, is to attempt to follow the track of the founders of Mechanics, to examine the circumstances of the motion of bodies in certain simple cases in the endeavour to dis- cover the laws to which they are subject. This method has been followed in Chapters IV. and v. I have made viii PREFACE. free use of a piece of apparatus the ballistic balance devised by Professor Hicks of Sheffield, and by its aid the student is led to realize the importance of mo- mentum in dynamics and to study the transference of this quantity from one body to another. The rate at which momentum is transferred is then considered (Chap- ter v.) and a name Force is given to the rate of transference. It is shewn that in many cases the rate of change of momentum is constant; while others are re- ferred to in which the rate of change of momentum depends only on the position of the body. Experiments are described to prove that in a given locality all bodies fall with the same uniform acceleration. It is then shewn that with Atwood's machine, when the rider is on, the weights move with uniform accelera- tion ; and hence the kinematical formulae obtained earlier in the book relating to the motion of a particle moving with uniform acceleration are verified by experiment ; the connexion between the mass moved, the acceleration and the weight of the rider is also investigated. Some idea of the laws of motion in a simple case having been thus obtained from observation and experi- ment, Newton's Laws of Motion are enunciated in Chapter VI. and their consequences are deduced in the ordinary way. Some portions of the preceding chapters are of necessity repeated by this method of procedure, which may have other disadvantages as well. These I hope are counter-balanced by the gain resulting from a more intelligent appreciation of the subject on the part of the learner. Mechanics is too often taught as a branch of pure PREFACE. IX mathematics. If the student can be led up to see in its fundamental principles a development of the conse- quences of measurements he has made himself, his interest in his work is at once aroused, he is taught to think about the physical meaning of the various steps he takes and not merely to employ certain rules and formulae in order to solve a problem. Chapter VIII. deals with the third law of motion and the principle of energy ; while in the succeeding chapters other problems are discussed. The Second Part of the book deals with Statics. I believe it to be desirable that a student should commence the study of Mechanics with Kinematics and Kinetics, and have therefore arranged the book on this plan. At the same time it will, I hope, be found that the Statics is in great measure independent of the other part of the subject, though at the cost of some repetition. It will be possible therefore for a teacher to take it before the Kinematics. In the Third Part some of the simple experimental laws of Hydrostatics are discussed and explained. The book has grown considerably beyond the limits of my lectures, though it is by no means a complete treatise on Elementary Mechanics; still I hope it may prove useful as an introduction to the subject. I have to thank many friends for help. Mr Wilber- force and Mr Fitzpatrick have assisted in arranging and devising many of the experiments. Mr Fitzpatrick has also read all the proofs. Dr Ward's suggestions in many parts of Chapters IV., V., VI. and vil. have been of the highest value. My pupil, Mr G. G. Schott of Trinity X PREFACE. College, collected for me many of the Examples, while Mr Green of Sidney College has most kindly worked through all the Examples and furnished me with the answers. The illustrations have for the most part been drawn by Mr Hayles from the apparatus used in the class. R. T. GLAZEBROOK. CAVENDISH LABORATORY. January, 1895. CONTENTS. DYNAMICS. CHAP. PAGE I. FUNDAMENTAL QUANTITIES 1 Units of Measurement, Methods of Measuring Length, Area and Volume, Terms used in Mechanics. II. KINEMATICS. VELOCITY 20 Motion, Rate of Change, Displacement, Velocity, Speed, Composition and Resolution of Displacements, Parallelogram of Velocities, Graphical Methods of Solution. III. KINEMATICS. ACCELERATION . 56 Change of Velocity, Acceleration, Motion of a body with uniform acceleration, Falling Bodies, Composi- tion of Accelerations. IV. MOMENTUM 77 Mass, The Comparison of Mass, The Ballistic Balance, Impact, Momentum, Impulse. V. RATE OF CHANGE OF MOMENTUM. FORCE . . 92 Measurement of Force, Laws of Falling Bodies, Weight, Atwood's Machine, Experiments on Uni- formly Accelerated Motion. VI. NEWTON'S LAWS OF MOTION 112 Definition of Force, The First Law, The Second Law, Measurement of Force, Comparison of Masses, Units of Force, Dyne, Poundal. Xll CONTENTS. CHAP. PAGE VII. FORCE AND MOTION 125 Action of Force, Law of Gravitation, Equilibrium, Composition of Forces, Parallelogram of Forces, Determination of g. VIII. THE THIRD LAW OF MOTION. ENERGY . . .148 Action and Reaction, Illustrations of the Third Law, Meaning of Action, Conservation of Momentum, "Work, Energy, Unit of Work, Conservation of Energy. IX. CURVILINEAR MOTION UNDER GRAVITY . . .188 Projectiles, The Parabola, Motion on a Smooth Curve, The Pendulum, Value of g. X. COLLISION 211 Impact of two balls, Energy and Impact, Oblique Impact, Impact on a fixed surface. XL MOTION IN A CIRCLE 223 The Hodograph, Circular Motion, Simple Har- monic Motion, The Pendulum. STATICS. I. FORCES ACTING AT A POINT 3 Definition of Force, Composition of Forces, Re- sultant Force, Parallelogram and Triangle of Forces, Conditions of Equilibrium. II. PARALLEL FORCES .35 Resultant of Parallel Forces, Experiments on Parallel Forces, Moments. III. COUPLES 59 Theorems about Couples, Conditions of Equilibrium. CONTENTS. Xlll CHAP. PAGE IV. WORK. EQUILIBRIUM 67 Work done by a system of Forces, Virtual Velocities, Problems on Forces in equilibrium. V. CENTRE OP GRAVITY 86 Centre of Mass, Centre of Gravity, Experiments on Centre of Gravity, Determination of Centre of Gravity, Properties of Centre of Gravity, Stable and Unstable Equilibrium. VI. MACHINES 113 Simple Machines, Lever, Wheel and Axle, Pulleys, Inclined Plane, Wedge, Screw, Combinations of Machines, The Balance. VII. FRICTION 166 Laws of Friction, Limiting Friction, Coefficient of Friction, Angle of Friction. HYDROSTATICS. I. STATES OF MATTER 1 Solids and Fluids, Elasticity, Liquids and Gases, Density and Specific Gravity. II. FLUID PRESSURE 20 Stress, Thrust and Tension, Shearing Stress, Pressure, Average Pressure, Units of Pressure, Trans- missibility of Pressure, Illustrations of Fluid Pressure. III. PROPOSITIONS ON FLUID PRESSURE .... 46 Pressure at points at the same level, Pressure at various points, Level Surface, Effective Surface, Levels, Units of Pressure, Resultant Thrust, Manometers, Safety Valve, Experiments on Fluid Pressure, Surfaces of Equal Density, Equilibrium of two Fluids. XIV CONTENTS. CHAP. PAGE IV. FLUID-THRUST CENTRE OF PRESSURE ... 79 Thrust on a Horizontal Surface, Thrust on a Vertical Surface, Centre of Pressure, Thrust on the Base of a Vessel, Thrust on a Curved Surface. V. FLOATING BODIES 95 Resultant Vertical Thrust, Archimedes' Principle, Floating Bodies, Buoyancy, Stability of Floating Bodies, Metacentre. VI. MEASUREMENT OF SPECIFIC GRAVITIES . . . 118 The Hydrostatic Balance, Hydrometers, Nicholson's Hydrometer, Jolly's Balance, Specific Gravity Bottle, The U Tube Method. VII. THE PRESSURE OF THE ATMOSPHERE . . '. 141 Otto Guericke's Experiment, Pressure of the Air, The Barometer, Corrections to the Barometer Reading, Standard Barometer, Height of Homogeneous Atmo- sphere, Pressure of a Gas, Boyle's Law, Dilatation of Gases, Mixture of Gases. VIII. HYDROSTATIC MACHINES 170 The Pipette, The Siphon, Pumps, Bramah's Press, Air- Pumps, Mercury Air- Pumps, The Condenser, The Diving Bell, The Volumenometer. Answers to Examples i Indexes . xvi DYNAMICS MECHANICS. \ ^ CHAPTER I. FUNDAMENTAL QUANTITIES. METHODS OF MEASUREMENT. 1. Mechanics and its signification. Mechanics is defined by Kirchhoff as the Science of Motion. Its object is to describe the kinds of motion which occur in Nature completely and in the simplest manner. Motion is change of position ; that which moves is known as " Matter." For the complete apprehension of Mechanics the ideas of Space, Time and Mass are necessary and sufficient. To these fundamental notions the idea of Force depend- ing on the mutual action of bodies is subsidiary. It will be our first step to consider in some detail the various quantities with which we have to deal and the methods we employ to measure them. 2. Units of Measurement. Any Quantity is of necessity measured in terms of a unit of its own kind, thus we measure the distance between two points in miles or feet, centimetres or inches; the area of a field in square yards or square metres, the mass of a lump of stone in tons or kilogrammes, the time between two events in hours or seconds. We shall thus have to consider, firstly, what are the units in terms of which the various quantities which occur in G. M. 1 2 MECHANICS. [CH. I Mechanics ,are ,to be measured, and, secondly, how we shall the\ quantities^ with these units. : ; T'aug, , Jbr ^ example/ when it is stated that the distance between ^twd ^ fixed,' points is three feet, it is implied that a certain unit of length called a "foot" has been adopted and that three of these placed end to end exactly cover the distance. 3. Fundamental Quantities in Mechanics. The three fundamental quantities in Mechanics in terms of which other quantities which may occur can be measured are, Length, Time and Mass. 4. The Unit of Length. The unit of length generally used in England is the Yard. Other measures of length, the inch, foot, fathom, mile, etc. are submultiples or multiples of the yard, and can be expressed in terms of it. The Yard is denned by Act of Parliament l as follows : "The straight line or distance between the centres of the transverse lines on the two gold plugs in the bronze bar de- posited in the office of the Exchequer shall be the genuine standard yard at 62 Fahrenheit, and if lost it shall be re- placed by its copies." In accordance with the Weights and Measures Act of 1878 the British Standards are now kept at the Standards' Office of the Board of Trade at Westminster. The copies referred to above are those preserved at the Boyal Mint, the Boyal Society, the Boyal Observatory, and the Houses of Parliament. Another unit of length which is authorized by Act of Parliament in England is the Metre. This was denned by a law of the French Republic in 1795 to be the distance between the ends of a rod of platinum made by Borda, the temperature of the rod being that of melting ice. At this date the distance between the pole and the equator along a certain meridian arc of the Earth's surface 1 18 and 19 Viet. cap. 72, July 30, 1855. 2-5] METHODS OF MEASUREMENT. 3 had recently been measured by Delambre, and it was supposed that Borda's platinum rod represented one ten-millionth of this distance. Further research has shewn that this is not exactly the case, and thus the metric standard of length is not the ter- restrial globe but Borda's platinum rod. The divisions of the metre are decimal. Thus 10 Decimetres = 1 Metre. 10 Centimetres 1 Decimetre. 10 Millimetres = 1 Centimetre. It is this fact and not the actual length which gives the metric system its value for scientific measurements. In such measurements the unit of length is now almost invariably the Centimetre, that is to say, it is one-hundredth part of the length of Borda's platinum rod when at the temperature of melting ice. The relation between these two standards, the yard and the metre, has been the subject of very careful investigation. According to the most recent measurements it has been found that 1 metre- 1-09362 yards = 39-37079 inches. Hence 1 inch = 0-0253995 metre = 2-53995 centimetres. In this book we shall adopt the Centimetre as the unit of length. 5. Methods of measuring Lengths. The measure- ment of a length consists in the determination of the number of centimetres and fractions of a centimetre which are con- tained in it, and the method to be adopted in making the measurement will depend to some extent on the magnitude of the length ; different methods would be required to measure a fraction of a centimetre or many kilometres. Some of the methods used in measuring small lengths will be given later. 12 4 MECHANICS. [CH. I 6. Measurement of Area and of Volume. The units of area and of volume depend directly on that of length; they are respectively a square whose side is one centimetre, and a cube whose edge is one centimetre: in measuring an area we determine the number of square centimetres it con- tains, in measuring a volume we find the number of cubic centimetres in it. The volume of 1000 cubic centimetres is called a Litre, and is often employed as a unit of volume. 7. Experiments on Measurement of Length, Area and Volume. EXPERIMENT (1). To explain the use of a vernier and de- termine the number of centimetres in half a yard. You are given a rod half a yard long and a metre scale divided to centimetres. The scale has a vernier attached. This is another short loose scale which has ten divisions marked on it. On laying this along the metre scale it will be found (as in Fig. 1) that these ten divisions occupy the i i i-i Vri 45 44- 46 47 48 49 52 55 54 45 50 55 Fig. 1. same length as nine divisions of the scale. Each division is therefore y 9 ^ of a division of the scale, so that one division of the vernier is less than one of the scale by -=$ of a scale division, that is, in this case, by 1 millimetre. Place the rod so that one end coincides exactly with the end division of the scale, and place the vernier along the scale so that its end division, marked by an arrow, coincides with the other end of the rod. This division will probably not come opposite to a division of the scale. Suppose it falls between two, say 46 and 47. The rod is between 46 and 47 centimetres long. The vernier enables us to measure the exact length more 6-7] METHODS OF MEASUREMENT. 5 nearly ; for on looking along the vernier it will be seen that its divisions and those of the scale get more and more nearly coincident until some division of the vernier, say the eighth, coincides almost exactly with one of the scale. Let us count back from this to the arrow-head or division of the vernier, remembering that a vernier division is 1 mm. less than a scale division. The distance between 7 of the vernier and the corresponding scale division is 1 mm., between 6 and the scale division 2 mm., and so on, so that the distance between and the scale division, which we have supposed to be 46, is 8 mm. The rod is therefore 46 -8 cm. long. Had the coinci- dence of vernier and scale been at 5 or 6 the rod would have been 46-5 or 46-6 cm. We have merely to note the division of the vernier which coincides with a scale division and re- member that in this case, when 10 vernier divisions coincide with 9 scale divisions, the divisions of the vernier enable us to read to tenths of the scale divisions. Other examples of the vernier should be studied, such as one in which twenty divisions correspond to nineteen of the scale, which therefore reads to twentieths of a scale division. EXPEEIMENT (2). To find the circumference of a circular disc and so to verify the formula Circumference = 2?r x Radius where TT stands for 3^ approximately. Measure the diameter of the disc by laying it on a finely divided scale, or better by the use of the calipers (Fig. 2) a Fig. 2. pair of calipers for the purpose can be constructed out of a draughtsman's T-square and set square. A scale is marked along the straight edge of the T-square and a vernier on the set square. Lay the T-square on the table and place the 6 MECHANICS. [CH. I disc so as to touch both the straight edge and the cross piece of the square. Place the set square against the straight edge and slide it along until it touches the disc. Since now both the cross piece of the T-square and one edge of the set square are at right angles to the straight edge of the former the distance between these two as measured along the scale and vernier will give the diameter of the disc. To find the circumference; make a mark with a pencil, or otherwise, on the edge of the disc, and place this in contact with the zero of a finely divided scale, then roll the disc along the scale, taking care that it does not slide, until the mark again comes in contact with a division of the scale. Note this division. The distance between this division and the zero of the scale is equal to the circumference of the disc. Repeat the observations to secure accuracy. It will be found that the ratio of the circumference to the radius is approximately equal to 2 x - 2 T 2 -. This ratio is usually denoted by 2?r. Thus TT - 2 T 2 - approximately. EXPERIMENT (3). To find the area of a circle and to verify the formula Area = Trr 2 , where r denotes the radius of the circle. You are given a sheet of paper divided by two series of parallel lines at right angles to each other into a number of small squares 1 . The distance between any two consecutive lines is ^ inch, so that each square has an area of '01 sq. inch. Draw on this a circle of some 3 or 4 inches diameter and measure the diameter. The circle will enclose a large number of complete squares. Count these up and reckon their area. The circumference will also intersect a number of squares. Estimate for such intersected squares the total area which lies within the circle. Thus the area of the circle can be found approximately and the formula verified. This method can be applied to any other figure. EXPERIMENT (4). To find the volume of a sphere and to verify tlie formula Volume = Trr 3 , where r is the radius of the sphere. 1 Such paper can be obtained from Messrs Waterlow & Co., London Wall. METHODS OF MEASUREMENT. Measure the diameter of the sphere by the calipers and hence find its radius. Place the sphere in a test tube or small beaker, Fig. 3, which has a mark made on its out- side by means of a file or by gumming on a piece of paper. Fill a burette with water up to a known volume, and let the water run from the bu- rette into the beaker until the latter is filled up to the mark, and note at what level the water in the burette now stands. Find hence the volume of water which has been placed in the beaker. Remove the sphere and the water from the beake*r, and by again letting water run in from the burette find the volume of the beaker up to the mark. The difference be- tween these two volumes is clearly the volume of the sphere, and the formula can -p ig 3 be verified. The volume of any other solid which sinks in water can be found in the same way. Care must be taken to remove all air bubbles from the solid. EXPERIMENT (5). To find the thickness of a glass cover-slip by the screw. The instrument, Fig. 4, consists of a platform with three feet, whose extremities form an equilateral triangle. Through the centre of the platform passes a fourth foot, which can be raised or lowered by means of a screw. The pitch of the screw used is 20 threads to the inch, so that if the platform in which the screw works be held firm and the screw turned once round, its end advances or recedes -^ of 8 MECHANICS. [CH. I Fig. 4. an inch. A disc is fixed on to the screw-head and its edge divided into 100 parts, and a vertical scale divided into -fa inch is attached to the platform. If the disc be turned so as to bring the edge of this scale from any one division to the next, the end of the screw moves one-hundredth of one-twen- tieth of an inch, or ^^(T i ncn j hence, by noting the number of whole turns and parts of a turn made by the disc, we can measure the distance moved over by the screw-point. The whole number of turns are given by the readings of the vertical scale, for the disc moves over one division for each turn. Place the instru- ment on a flat sheet of glass, and turn the disc until the screw- point is in contact with the glass, read the screw-head ; place five or six cover-slips one on the top of the other on the glass and raise the screw until they will just pass underneath it. Read the position of the disc when the point of the screw is just in contact with the top cover-slip, having noted the number of whole turns made by it. From this whole number of turns and the two readings of the disc calculate the total thickness of all the cover-slips, and then, by dividing this by the number of slips, the average thickness of each one. EXPERIMENT (6). To use the screw gauge to measure the diameter of a wire. The Screw Gauge is shewn in Fig. 5. It consists of a metal arm ABC ; through one end of this passes a steel plug D with a planed face, and through the other a screw EF. The pitch of the screw is half a millimetre, and the end E is planed so as to be parallel to the Fig. 5. 7-8] METHODS OF MEASUREMENT. 9 face D and perpendicular to the axis of the screw. The screw can be turned by means of the milled head G until its end E comes in contact with D ; each complete turn of the screw separates the planed faces by half a millimetre. A scale of half millimetres is engraved on the frame of the instrument parallel to the axis of the screw and the milled head G carries a cap H with a bevelled edge. The circumference of this edge is divided into a scale of fifty parts, and when the end of the screw is in contact with D the zero of this scale and the zero of the scale on the frame should coincide. On making a complete turn of the screw the cap is moved back half a millimetre, and the zero mark on the bevelled edge is brought opposite the first division of the linear scale. Thus the divisions of this scale which are exposed register the number of half millimetres between D and E. Since the bevelled edge is divided into fifty parts a rotation through a single part corresponds to a separation of the plane ends by ~ of J of a millimetre or by T ^-Q of a millimetre. Thus, if a division (say 24) of the bevel- led edge coincides with the linear scale, the distance between the plane faces is a whole number of half millimetres, which is given by the number of divisions of the linear scale exposed, together with T 2 ^ of a millimetre. Thus, to measure the distance between the plane ends, read the number of half milli- metres exposed on the linear scale and add to this the number of hundredths of a millimetre given by the reading of the scale on the bevelled edge. When using the instrument to measure the diameter of a wire first test the zero reading ; then hold the wire between D and E and turn the screw-head G until the wire is gently clipped between the two plane faces. In this case the distance between these faces is the diameter of the wire. 8. Other Instruments for measuring lengths. (a) Scales and Compasses. A pair of compasses and a finely divided scale are often the most convenient apparatus for measuring lengths. The compasses are ad- justed until the distance between their points is exactly the length to be measured ; they are then applied to the scale and the length is read off. Instead of an ordinary scale a diagonal scale may be used. This is shewn in Fig. 6. There are eleven equidistant parallel lines running the whole length of the scale dividing it into ten spaces. The scale is 10 MECHANICS. [CH. I divided into inches by lines running across it at right angles to this series of parallel lines. These are numbered 0, 1 , 2,3, starting from D as zero. The first inch, AB and CD, along each of the top and bottom lines Fig. 6. is divided into tenths. The alternate divisions, starting from D to the left, are numbered 2, 4, 6 etc., the alternate vertical divisions starting from G upwards are also numbered. Lines are drawn obliquely across the first rectangular division ABDC of the scale as shewn in the figure ; thus A is joined to division 9 of CD, division 9 of BA to division 8 of CD and so on ; these oblique lines enable us to measure to one-tenth of the small divi- sions of the scale. For consider the distance between any vertical line, say that through the 1 inch division, and an oblique line such as that through the small division 4. When measured along the lowest horizontal line this distance is 1-4 inches, when measured along the top line it is 1'5 inches. Thus on passing from the bottom to the top of the scale it increases by -1 inch, but it increases by an equal amount for each vertical space passed over, and here are ten of these spaces, hence the increase for each vertical space is -01 inch. Thus the distance along the fifth line from the bottom and between the vertical line through the 1 inch mark and the oblique line through the '4 inch mark is 1 inch + 4 tenths' + 5 hundredths or 1-45 inches. Thus to measure on the scale a distance with the compasses place the right leg on one of the vertical divisions at the point where it crosses the bottom horizontal line, say at division 2 inch. Let the left leg of the compass fall between the points in which the fourth and fifth oblique lines cut the bottom horizontal line. Then the distance is between 2-4 and 2-5 inches. Slide the compasses upwards, keeping the right-hand leg in the vertical division through 2, and the line joining the two legs parallel to the horizontal lines on the scale, until the left-hand leg falls on a horizontal line as close as possible to the point in which it is intersected by one of the oblique lines ; let this occur on the fifth horizontal line ; the distance between the legs is greater than 2-4 inches by 5 hundredths of an inch ; thus it is 2-45 inches. 8-9] METHODS OF MEASUREMENT. 11 (&} Caliper-Compasses are made in special forms for measuring the dimensions of curved bodies. Thus, Fig. 7 sA B. shews a pair of Calipers of simpler construction than the slide calipers described in EXPERI- MENT 2. The outside calipers AB can be set so that the points A, B just come in contact with two points on the outside of a cylindrical or convex surface, the distance between which is required, while by means of the inside calipers CD the distance between points on the inside of a cavity within which the instrument can be introduced can be measured ; in either case the distance is found by adjusting the calipers and then laying off the length between the points on a scale. (c) The Beam Compass. This instrument is shewn in Figure 8. A sliding piece (7, fitted with a vernier and a clamping screw, is attached D E to a long straight scale AB. A point D is attached to one end A of the scale and the sliding piece C carries a similar point E. The instrument is adjusted so that when the points D and E are in contact the vernier is at zero on the scale ; the reading then of the scale and vernier in any position gives the distance between the points D and E. The instrument is set so that D and E coincide respectively with the two points the distance between which is required, and this distance can then be read off directly 1 . 9. Time. The next fundamental Physical Quantity which we have to consider is Time. "The idea of Time," says Max- well, " in its most primitive form is probably the recognition of an order of sequence in our states of consciousness." We can 1 For further particulars as to the method of using such measuring instruments, see Grlazebrook and Shaw, Practical Physics, Chapter iv. 12 MECHANICS. [CH. I associate certain sense expressions in a group and separate them off from other groups which we perceive simultaneously. Thus we gain the idea of space, but we have also the power of perceiving things in succession, we recognize a group of sense impressions as like another distinct group, the impress of which is stored in our memory \ we perceive events which follow each other as well as others which have a simultaneous existence. Our measure of time is derived from the apparent motion of the stars ; this apparent motion is a consequence of the mo- tion of the earth round its axis, and we feel that this motion, in the interval from noon to noon, marks off series of like sequences of events ; we recognize that the time occupied in one such complete rotation is approximately constant. Owing to the motion of the earth round the sun the interval between two successive passages of the sun across the meridian of any place differs slightly from day to day. The average of such intervals during the year is the mean solar day. A mean solar day contains 86400 seconds and the funda- mental unit of time is the Mean Solar Second. 1O. Mass. Our third fundamental quantity is called Mass. If we consider the bodies with which we have to deal as composed of Matter, then any body will consist of a definite quantity of matter. This quantity is usually called its Mass. We shall find however in the sequel that we can give a definite meaning to the term Mass as used in Mechanics without attempting to define the term Matter. We have means for comparing with great accuracy the masses of dif- ferent bodies ; we can therefore measure the Mass of any body in terms of some standard Mass. For the present then we look upon Mass as a property of bodies which we recognize by experiment and which we can only define when we have considered those experiments. We do not know what matter is, it may be that it has no pheno- menal existence apart from our conception of it, but it is beyond our province to discuss this here. If we assume that there is a substratum of something we call matter in a body, then the quantity of tbat matter is measured by the mass of the body, and the masses of bodies can be compared in an exact manner. 9-11] METHODS OF MEASUREMENT. 13 There is one case in which there is no difficulty in compar- ing the quantity of matter in two bodies. For consider two cubes of some homogeneous 1 substance such as platinum, each one centimetre in edge, they are alike in all respects ; if they are composed of matter the quantities they contain are ob- viously equal. The two cubes combined will have double the volume of either singly; there is double the quantity of matter in the two that there was in either. The quantities of matter in two portions of the same homogeneous substance are propor- tional to the volumes of the two. We cannot apply this argu- ment to portions of different substances; equal volumes of iron and lead we shall see have different masses, they are said to contain different "quantities of matter"; it is only when we have considered the laws of motion that we can state exactly what is meant when we assert that the "quantities of matter" in two given bodies are equal, and how it is possible to compare the "quantity of matter" in a lump of iron with that in a heap of feathers. 11. Measurement of Mass. Masses are measured in terms of a unit of mass. DEFINITIONS. The mass of a certain lump of Platinum mar kedPS 1844 1 Ib.j deposited in the Standards department of the Board of Trade at Westminster, is the English Unit of Mass and is called the Pound Avoirdupois. The mass of a certain lump of platinum made by Borda in 1795 and kept at Paris is the Unit of Mass on the metrical system and is called the Kilogramme. It was intended that Borda's kilogramme should be equal to the mass of 1 litre or 1000 cubic centimetres of distilled water at 4 C. 2 The exact determination of the mass of such a volume of water is difficult and its value probably differs 1 A homogeneous substance is one which has identical properties at all points. Water or any other liquid, glass, brass, iron are examples of homogeneous substances. Substances, such as a piece of conglomerate rock, which have different properties at different points, are called hetero- geneous. 2 The mass of water which can be contained in 1000 c.c. is greater at this temperature than at any other, hence this temperature was chosen as the standard. See Glazebrook, Cambridge Natural Science Manuals, Heat, p. 88. 14 MECHANICS. [CH. I slightly from that of Borda's platinum mass. Hence the metri- cal standard of mass is really the mass of a lump of platinum and not, as was intended, the mass of a definite number of units of volume of water. Still the statement that the mass of 1 cubic decimetre (1000 c.c.) of distilled water is 1 kilogramme is sufficiently nearly true for most purposes, and enables us to introduce great simplification into many numerical calculations. The unit of mass which is now usually adopted for scien- tific purposes is the Gramme. One Gramme contains one- thousandth part of the mass of Borda's kilogramme Standard. Since a kilogramme (1000 grammes) is very approximately the mass of 1000 c.c. of distilled water, a gramme is very approximately the mass of 1 c.c. of distilled water at 4 C. The divisions of the gramme are decimal : 10 Decigrammes = 1 Gramme, 10 Centigrammes = 1 Decigramme, 10 Milligrammes = 1 Centigramme. Careful experiment has shewn that the kilogramme contains 2-2046 pounds. 12. Density. For a given substance, the mass of a body depends on its volume, while, for bodies of given volume, the mass depends on the substance of which the bodies consist and on its physical state. A large lump of iron has a greater mass than a small lump of iron, but a small lump of iron may be of greater mass than a large lump of cork. It is useful to have some term to denote the mass of a definite volume of any body, say 1 cubic centimetre. DEFINITION OP DENSITY. The Density of any homogeneous substance is the mass of unit volume of that substance. It follows from this definition that to determine the density of a body we must find the number of units of mass in the unit of volume, we require therefore to know the unit of mass and the unit of volume, if these be the gramme and the cubic centimetre respectively, we may say that the density is so 11-13] METHODS OF MEASUREMENT. 15 many grammes per cubic centimetre. Thus in these units the density of water is 1 gramme per c.c., that of iron 7*76 grammes per c.c. In any other units the numerical measures of the densities of these substances would differ from the above. Thus a cubic foot of water contains 998*8 oz. or 62-321 Ibs. ; hence the density of water is 998*8 oz. per cubic foot or 62*321 Ibs. per cubic foot ; iron is 7*76 times "as dense as water, hence its density is 7'76 x 62-321 Ibs. per cubic foot. From the above definition of density we can find a relation between the mass, the volume and the density of a body. PROPOSITION 1. To shew that if the mass of a homogeneous body be M grammes, its density p grammes per cubic centimetre and its volume Y cubic centimetres, then M = Y/a. For by the definition, the mass of 1 c.c. = p grammes, therefore the mass of 2 c.c. = 2p grammes, and the mass of 3 c.c. 3p grammes, hence the mass of V c.c. = Vp grammes. Therefore M= Vp. We may write this as M and thus we have the result that the density of a homogeneous substance is the ratio of its mass to its volume. A result similar to the above holds for any other consistent system of units. Yarious methods of determining by experiment the density of a body will be given later \ 13. The Comparison of Masses. A balance is the instrument usually employed in the comparison of masses. The theory of the balance is discussed in the Statics, and the method of measuring mass is there considered. 1 See Hydrostatics. 16 MECHANICS. [CH. ] 14. The C.G.S. system of measurement. We shall find that the other physical quantities with which we have to deal in mechanics can be expressed in terms of the units of length, time and mass or of some of these units. When we take the Centimetre, the Gramme and the Second as fundamental units we are said to employ the C.G.S. system. This is now generally used for scientific purposes; when a quantity has been measured in terms of these fundamenta] units, it is said to have been determined in absolute measure. If we know the relation between the C.G.S. system and some other system of units it is easy to change from one to the other in our cal- culations. Thus if we wish to change to the Foot Pound Second System we have approximately 1 cm. = -03281 feet. 1 gramme = -002205 pound. Examples. (1). Find the number of cubic feet in a litre. 1 litre = 1000 c.cm. = 1000x-(03281) 3 c. feet. = -03532 c. feet. (2). The density of a piece of glass is 2-5 grammes per c.cm., find it in Ibs. per c. foot. We have lc.cm. = (03281) 3 c. feet = -00003532 c. feet 2-5 grammes = 2-5 x -002205 Ibs. Hence a volume of -00003532 c. feet contains 2-5 x 002205 Ibs. Thus density required '5*<5x -002205 ., ^00003532- lbS ' perC - f 0t - And this reduces to 2-5 x 62-43 or 156-08 Ibs. per c. foot. 15. Terms used in Mechanics. Mechanics is the Science of Motion. In studying motion we shall generally require to know both the Displacement or change in position of the body, and also the time during which that displacement has occurred. 14-15] METHODS OF MEASUREMENT. 17 This branch of the subject is called Kinematics. It may be described as the Geometry of Motion; Geometry deals with Space only, Kinematics has for its subject space and time. When we come to consider the mutual relations between moving bodies the science of motion is called Kinetics ; while in Dynamics we pay special attention to the con- nexion between Force and Motion. In Statics we consider the conditions which must exist among a set of Forces im- pressed on a body which remains at rest. Statics and Dynamics are usually applied to the Mechanics of Solid bodies. The Sciences which deal with the equili- brium and motion of Fluid bodies are respectively Hydro- statics and Hydrodynamics. We are concerned in nature with Material Bodies. A Body is a portion of "matter" bounded in every direction. We shall consider a Body as composed of a number of material Particles. A Material Particle is a portion of "matter" so small that for the purposes of our investigations the distances be- tween its different parts may be neglected. In Dynamics we deal first with the motion of one or more isolated particles, or of a body which we can treat as a particle ; we can afterwards proceed to consider the motion of a body of finite size. We must remember that it will depend on circumstances whether we can treat a body as a particle or not. Thus, apart from a small effect due to the resistance of the air, the shape of a falling stone does not affect the rate at which it falls to the earth ; we may solve a problem relating to a falling stone correctly on the supposition that the whole of the stone is concentrated into one point and that the stone behaves as a particle ; the same would be true of a cricket ball so long as it is in the air, but the motion of the cricket ball, on striking the ground or being hit by the player, depends on its shape and on the amount of spin given to it by the bowler; these must be considered before we can state how it will move immediately after it is struck, in solving this part of the problem we can- not treat the ball as merely a particle. The words " for the purposes of our investigations" in the above definition are of importance; in the present book when considering Dynamics we deal with the motion of particles. G. M. 2 18 MECHANICS. [CH. I EXAMPLES. MEN SITUATION. [Take the value of IT as 22/7.] 1. Eeduce to centimetres (1) 1 ft. 2 in., (2) 2 yds., (3) 5 ft., (4) 1 furlong. 2. Keduce to kilometres (1) 1 mile, (2) 4000 miles, (3) 600 yds., (4) 1 metre, (5) 25 millimetres. 3. Find in centimetres the circumference of circles whose radii are (1) 1 ft., (2) 10 yds., (3) 4000 miles, (4) 750 metres. 4. Find in square centimetres the areas of the circles whose radii are given in Question 3. 5. A circle of radius 5 inches is cut out from a circular disc of radius 9 inches ; find the area of the remainder. 6. The circumference of a circle is 1 mile; find its area. 7. The area of a circle is equal to that of a rectangle whose sides are 44 and 126 feet ; find its radius. 8. A circle and a square have the same perimeter ; determine which has the greater area. 9. An equilateral triangle is described on one side of a square of which the side is 10 feet ; find the area of the figure thus formed. 10. A circle of 20 centimetres radius is divided into three parts of equal area by two concentric circles ; find the radii of the circles. 11. A circle is 20 centimetres in radius ; find the area of a square which can be inscribed in it. 12. A sphere when placed in a beaker as in Exp. 4 displaces 38-786 cubic centimetres of water ; find its radius and its surface. 13. Ten cover-slips are placed under the spherometer as described in Experiment 5, the pitch of the screw being \ a millimetre, and the point is raised 8'56 turns ; find the average thickness of a cover-slip. 14. The density of copper is 8 - 95 grammes per c. cm. The diameter of a piece of copper wire is 1-25 mm. and its length 1025 cm. ; find its mass. 15. Find the density of a cylinder 1 foot in height and 6 inches in radius whose mass is 60 Ibs. CH. I] METHODS OF MEASUREMENT. 19 16. Find the density of a sphere 10 cm. in radius and 5 kilogrammes in mass. 17. Determine the density of the cylinder described in Question 15 in grammes per c. cm. 18. Find the density of a pyramid on a triangular base each side of which is 10 cm. and which has an altitude of 30 cm., the mass of the pyramid being 8 kilogrammes. 19. The density of mercury is 13-59 grammes per c. cm.; find it in grains per cubic inch. c 20. Compare the densities of a sphere 5 cm. in radius, 5 kilos, in mass, and a cylinder 1 foot in height, 6 inches in radius and 60 Ibs. in mass. 22 CHAPTER II. KINEMATICS. VELOCITY. 16. Motion. A Body is said to move when it is in different positions at different times. Thus, in order to de- termine the motion of a body, we have to determine its posi- tion at different times and investigate whether the position changes or not. We may notice first that the motion with which alone we can deal is relative motion. Two passengers seated in a railway carriage are at rest relatively to each other and to the carriage ; they are however in motion relatively to objects by the side of the line along which the train is moving. The planets are in motion relatively to the sun ; the whole solar system, sun, planets and satellites, is in motion relatively to the stars. In all problems of motion we must have some point which so far as that motion is con- cerned we treat as fixed and from which we regard the motion. We may investigate the motion of a cricket ball thrown upwards from the earth, but in the investigation we should usually suppose the point from which the ball is thrown to be at rest ; as a fact of course this is not true, that point and the ball with it partake of the motion of the earth round its centre, of the motion of the earth's centre round the sun, and so on, but for our purposes this is immaterial. DEFINITION. Motion is change of position. 17. Measurement of Position of a Particle. The position of one particle relatively to another is determined 16-18] KINEMATICS. VELOCITY. 21 if we know the length and the direction of the line joining the two. We may say then that one /A particle, A Fig. 9, is in motion rela- tively to a second particle B when- ever the length or direction of the line AB joining the two varies. When a particle is moved from / """"* one position A to another position A' it is said to be displaced. The Displacement of the particle is measured by the length and direction of the line A A'. We are however in general concerned with the rapidity with which the change in position occurs ; moveover the motion may be Uniform or Variable. DEFINITION. The Motion of a particle is Uniform if the particle passes over equal spaces in equal times. The motion of a particle is Variable if the particle passes over unequal spaces in equal times. Since an interval of time is measured by the angle which the earth turns through about its axis in that interval, equal times are those in which the earth turns through equal angles. If then in a series of intervals in which the earth turns through equal angles a moving particle passes over equal distances the motion of the particle is uniform ; if on the other hand the distances traversed by the particle are unequal, while the angles turned through by the earth are equal, then the motion is variable. 18. Rate of Change of a Quantity. The phrase, Rate of, is one which will often occur and which it is desirable to consider. By rate of change is meant generally the change in a quantity which takes place during some given interval of time adopted for convenience as the unit of time or more exactly the ratio which that change bears to the interval of time during which it has occurred. Thus the statement that the Rate of Interest is 3 per cent, per annum means that in one year a sum of ,100 increases by 3. If this rate continues uniform, then in 10 years the. increase on .100 will be 3 x 10 or .30; the total increase is obtained by multiplying the rate of interest by the time during 22 MECHANICS. [CH. II which interest has accrued. Again, if we know that in 5 years .100 has increased by 15 and that the rate of interest has been uniform, we infer that that rate is 15/5 or 3 per cent, per year. We might speak in the same way of the rate of growth of the population of a town, meaning the increase in popula- tion per week or per day as the case may be, or of the daily rate of progress of a building, meaning the amount built in a day. Thus we see (1) Any quantity varies uniformly when it increases or decreases by equal amounts in equal times, and (2) the rate of change of a quantity, which varies uniformly, is the ratio of the change in that quantity to the interval of time during which it has occurred; it is measured by the change which takes place in the unit of time. 19. Average Rate of Change. But Quantities do not always change uniformly. The amount of interest obtain- able for a given sum may vary from day to day. The daily rate of growth of the population of a town will not be the same throughout the year ; more children are born on some days than on others. In such cases we are often concerned with the Average 1 Rate of Change. This average rate of change is found by calculating the actual change during any time and dividing that by the time. Thus the statement that during the year the average daily rate of increase in the population of London was 340, does not mean that 340 children were born on each day, but that during the year 340x365or 124100 were born, so that the total increase during the year is the same as though 340 were born on each day. Or again, consider the case of a railway train which performs a journey of 42 miles in an hour. We should say that its average rate of motion is 42 miles an hour, but this does not imply that it is moving uniformly at the rate which would enable it to traverse the distance in this time ; at times it moves more quickly, at the stations it is at rest for some few minutes, and on starting again its rate of motion is less than 42 miles per hour. We must distinguish 1 The word average as used here has reference to time. 18-21] KINEMATICS. VELOCITY. 23 between its average rate of motion and its rate of motion at any instant of the hour. 20. Graphical Representation of Rate of change. Suppose now we represent on a diagram the two cases of uniform and variable motion thus. Draw a horizontal line, say 30 cms. long to represent an hour ; divide it into half centi- metres so that each 5 mm. represents 1 minute, and at the end of each division erect a vertical line to represent the distance traversed up to the end of that minute. Then, in the case of uniform motion, if we represent 1 mile by 1 cm., since 42 miles per hour is the same as 42/60 or -7 miles per 1 minute, the first vertical line will be e^x 7 cm. or 7 mm. long, the second will be twice this, the third three times, and so on ; the ver- tical lines will increase uniform- ly in height, each will be 7 mm. higher than the preceding ; a line joining these ends will be straight and will be uniformly inclined to the horizon. The figure obtained will be similar to that shewn in Fig. 10. 21. Variable Rate of Change. Consider now the case in which the rate of motion is not uniform ; let us suppose for the present that we may treat it as uniform during each minute, but that it varies from one minute to the next; the increments in the lengths of the various vertical lines will be different. The train starts slowly, the first line will therefore be less than 7 mm. long, the second less than 14 mm. After a time however the train must exceed its average rate of motion, each successive increment will be greater than 7 mm. until an- other station is approached, when they will again decrease ; during the three or four minutes for which the train stops the corresponding vertical lines will all be of the same length, the Flg ' 1L 24 MECHANICS. [CH. II line joining their ends will be horizontal. The diagram ob- tained will resemble Fig. 11. The rate of change of the train's position during different minutes is variable ; unequal distances are traversed in equal times ; the rate of change however for each minute is obtained by finding the alteration during that minute ; moreover, if we multiply the change so obtained by 60, the number of minutes in the hour, we can get the rate of motion in miles per hour. Now the figure just obtained does not represent accurately the motion of the train, for it does not move uniformly for 1 minute, then change its rate of motion and so on; we should get a nearer representation to the truth if we divided each 5 mm. into 60 parts, each representing a second, and supposed the rate of motion to be uniform for each second but to change at the end of every second; this would not be exact, but by proceeding thus and dividing each second into a very large number of very small fractions and supposing the rate of motion uniform during each fraction, but variable from fraction to fraction, we may get as close a representation to the truth as we please. In this manner we come to see that the Rate of Mo- tion at any moment is found by dividing the distance traversed during an indefinitely short interval of time by the number of seconds in that interval. If we divide each minute as shewn in Figs. 10 or 11 into a very large number of parts we shall obtain, instead of the series of straight lines, such as those shewn in Fig. 11, a very much larger number of such straight lines. Each of these will be very short, and it will be impossible to dis- tinguish the many-sided polygon thus formed and a figure bounded by a regular smooth curve as in Fig. 1 2. The one however represents the series of discontinuous changes at very brief intervals, the other the regular continuous change in the rate of motion which actually occurs with a train. We may compare the two pro- cesses with those of mounting .p. a hill (a) by a series of stairs 21-22] KINEMATICS. VELOCITY. 25 or steps, (b) by a gradual slope. If the number of steps in the stair be sufficiently large and the size of each sufficiently small the two processes are indistinguishable, the steps merge into the continuous slope. We arrive then at the definition of the term Rate of Change applicable to all cases. DEFINITION OF RATE OF CHANGE. The Rate of Change of any quantity is the ratio of the change in that quantity to the interval of time during which it has occurred when that interval is sufficiently small 1 . The rate of change therefore is found by dividing the change which has taken place during an interval of time by that interval when the interval is made sufficiently small. Let us now suppose that at a given moment the rate of change ceases to vary and that the quantity concerned con- tinues to change at the same rate as it did at that moment. From this time on the rate of change is uniform, and is equal therefore to the ratio of the change in any interval not necessarily a very small interval to the interval during which that change has taken place. If we take the interval to be one second the unit of time we may in this case say that the rate of change is the change which has taken place in one second ; or the change per second. We may thus sum up with the statement that the Rate of change of any quantity ', when uniform, is the change in that quantity per second, and, when variable, is the ratio of the change which would take place in that quantity to the interval of time during which that change has occurred when that interval is made very small. In measuring the rate of change of any quantity we adopt the process indicated in the definition above and divide the change occurring in any interval by the number of seconds in the interval when this number is suffi- ciently small. 22. Velocity and its measurement. We proceed now to apply the idea of rate of change to various dynamical quantities. 1 This definition it will be seen is the same as the statement on page 21, except that the last clause is additional. This clause is not necessary if the rate of change be uniform. 26 MECHANICS. [CH. II DEFINITION. Velocity is Hate of Change of Position. We have already defined Motion as change of position, we may therefore state that Velocity is rate of motion. Velocity may be either Uniform or Variable. Uniform Velocity is measured by the change in position which occurs per second. Variable Velocity is measured by the ratio of the change in position in a given interval to the number of seconds in that interval when that number is sufficiently small, that is by the change which would occur per second if during the second the velocity remained uniform. A particle moving with uniform velocity describes equal spaces in equal times. A particle moving with variable velocity describes unequal spaces in equal times. To measure velocity we need to know the change in position, or displacement, per second; this change is deter- mined if we know (1) the distance the particle moves through, (2) the direction of motion. The word Speed is employed to denote the distance traversed per second without reference to the direction. DEFINITION. The Speed of a particle is the rate at which it describes its path. A particle moves with uniform Speed if it travels over equal distances in equal intervals of time ; if the Velocity be uniform, not merely will the distances be equal but their directions will be the same. In strictness therefore the velocity of a particle can be uniform only when the particle is moving in a straight line and passes over equal distances in equal times. The term uniform velocity is however often applied where uniform speed would be more accurate ; thus the hand of a clock is said to move with uniform velocity, but since the direction of motion changes the velocity is not strictly uniform though the speed is. 23. Motion of a particle with uniform speed. PROPOSITION 2. To find the distance s cm. traversed in t seconds by a particle moving with a uniform speed of v cm. per second. Since the speed is uniform, the particle passes over v cm. in each second. 22-26] KINEMATICS. VELOCITY. 27 Hence distance traversed in 1 second = v cm., distance traversed in 2 seconds = 2v cm, distance traversed in 3 seconds = 3v cm., distance traversed in t seconds = tv cm. Therefore s = vt. Hence also v = - . t This last result gives a formal proof of the statement that speed when uniform is measured either by the distance traversed per second or by dividing the distance traversed in any interval of time by that interval. 24. Average speed. The Average speed of a particle moving over a given distance in a given time may be defined as the speed with which a second particle, moving uniformly, would describe the given distance in the given time. It is found therefore by dividing the distance traversed by the number of units of time taken to traverse it. Thus, consider two trains, one of which is moving uni- formly at the rate of 50 miles an hour, while the second starts from one station and arrives, after an interval of an hour, at a second 50 miles distant. The average speed of the second train is the speed of the first train. 25. Variable speed. The actual speed of the second train at each moment is not 50 miles an hour. To find its value we should require to determine the distance traversed by the train in some very short interval, and divide that distance by the interval ; probably in the case of a train the speed would not vary much during a single second, and if we could measure accurately the distance traversed in one second we should determine the speed during that second. 26. Units of speed. Speed is measured by the number of units of length traversed per unit of time. The numerical measure therefore of the speed or of the velocity of a particle will depend on the unit of length 28 MECHANICS. [CH. II and on the unit of time. The same speed may be expressed by very different numbers. Thus, since a mile contains 5280 feet, a speed of 60 miles per hour is the same as one of60x5280 feet per hour ; and since an hour contains 3600 seconds this velocity is the same as one of 60 x 5280/3600, or 88, feet per second. In order then to specify a speed or a velocity we must state the unit of length and the unit of time. A velocity of 88 means one which is 88 times as great as the unit of velocity, and conveys no definite information unless we state clearly what that unit is. It may be a velocity of 88 feet per second, or 88 miles per hour, or any- thing else. DEFINITION. A particle has Unit Velocity when it traverses unit distance in unit time. In the c. G. s. system, a particle has unit velocity when it traverses 1 centimetre per second. In stating, then, that the velocity of a particle is v, it is necessary to specify the unit distance and the unit time, and to write, if the c. G. s. system be used, a velocity of v cm. per sec. Examples. (1). Which train has the greater speed, one moving at the rate of 60 miles an hour or one which travels 100 yards in three seconds ? The first train in 60 x 60 seconds moves over 60x1 760 yards; , . 1760 x 60 /. in 1 second it moves -^ yards ; bO x bO /. the speed is 29^ yards per second. 100 The second train in 1 second moves -=- yards ; o /. the speed is 33^ yards per second. Thus the second train is moving 4 yards per second the faster. 26-28] KINEMATICS. VELOCITY. (2). Find in feet per second the speed of the earth round the sun assuming it to describe in 365 days a circle of 92000000 miles radius. Taking the value of IT as 22/7 the circumference of the earth's orbit is 2 x 22 x 92000000/7 miles, 2 x 22 x 1760 x 3 x 92000000/7 feet. Again, 365 days contain 365 x 24 x 3600 seconds. Hence in 365 x 24 x 3600 seconds the earth moves 2 x 22 x 1760 x 3 x 92000000 . = feet ; or the speed is 2 x 22 x 1760 x 3 x 92000000 feet per second. 7 x 365 x 24 x 3600 And this reduces to 97691 feet per second. (3). Find the average velocity of a train which takes 7 minutes to traverse the first three miles after leaving a station, then moves for half an hour at the rate of 40 miles an hour, and finally comes to rest, taking 5 minutes to traverse the last 2 miles. The whole distance travelled is 3 + 20 + 2 or 25 miles. The time taken is 7 + 30 + 5 or 42 minutes. Thus the average speed is 25/42 or about -595 miles per minute. 27. Graphical representation of Velocity. PROPOSITION 3. To skew that a velocity can be represented by a straight line. To determine the value of a velocity we require to know its amount and its direction; these two quantities can be represented by a straight line containing as many units of length as the velocity contains units of velocity and drawn in the direction of motion. Thus velocities may be completely represented by straight lines. 28. The Composition of Displacements. Consider a man in a railway carriage. Let him move diagonally across the carriage from A to , Fig. 13. Then if the carriage is at rest his displace- ment is All, but suppose the carriage to be in motion and let A A 'represent the displace- ment of any point of the Fig. 13. 30 MECHANICS. [CH. II carriage during the interval in which the man has moved from A to B. Draw EC equal and parallel to AA' ; then relatively to the carriage the man moves from A ,to B, but relatively to the rails B has moved from B to C, thus the man is at (7; his actual displacement 1 is AC, and it is made up of the displace- ment AB relative to the carriage and BC relatively to the lines. In this case AC is said to be the Resultant of the two displacements AB and A A, and these displacements are spoken of as the Components of AC. Moreover we may continue this process. The rails are not at rest, they are in motion round the axis of the earth. Let A A", Fig. 14, represent the displacement of any point on the rails. Draw CD equal and parallel to A A"', then both man, carriage and rails have been displaced through a distance represent- ed by CD] the man there- fore will be at D. His dis- placement is AD, and this is the resultant of AB, BC and A _. - . CD while these displacements are the components of AD. DEFINITION. The single displacement which is equivalent to two or more displacements impressed on a particle is called the Resultant of those displacements. Each of a number of individual displacements, the combined effect of which is equivalent to a single displacement, is spoken of as a Component of that single displacement. PROPOSITION 4. To find the resultant of a number of dis- 1 This does not at all imply that the man has moved along the straight line AC, but merely that lie was at A and is at C. 28] KINEMATICS. VELOCITY. 31 Consider in the first place two displacements OA and OA', Fig. 15. Draw AB equal and parallel to OA'. In conse- quence of the displacement OA alone, the particle would be at A, in consequence of the second displacement OA' the point A is brought to B. Thus the particle is brought to B and its resultant displacement is OB. Now let there be three displacements OA, OA', OA", construct the figure as above -pig. 15. and draw EG equal and paral- lel to OA". In consequence of the displacement AA", B is brought to C, thus the particle is at C and its displacement is OC. The general rule, therefore, is obvious. From any point draw OA to represent the first displacement from A, the extremity of OA, draw AB to represent the second, from B draw BC to represent the third, and so on. Thus if P be the last point thus found OP is the Resultant displacement. We notice that the various displacements and their resultant form a closed polygon ; if it should happen that the point P should coincide with it is clear that the re- sultant displacement is zero ; the particle will remain at rest. Thus if a series of displacements can be represented by the sides of a closed polygon taken in order the particle remains at rest. Moreover it is immaterial in what order the displace- ments are made ; we can prove this graphically by drawing the figure in various ways, starting with OA ' or OA" instead of OA, then drawing from A', A'B' to represent OA, and so on; it will be found that we always arrive in the end at the same point P. In general then, if the several displacements be represented by all but one of the sides of a polygon taken in order their resultant is represented by that side taken in the opposite 32 MECHANICS. [CH. II direction. This proposition is called the polygon of dis- placements. 29. Special cases of the composition of displace- ments. PROPOSITION 5. When the component displacements are all in the same straight line the resultant is their algebraical sum. For consider two such displacements. Draw OA, Fig. 16, to represent the first, AB to repre- sent the second ; then AB is in the 6 * ^ * & same straight line as OA and if OA and AB are drawn in the same direction, Fig. 16, then OB = OA + AB. While if OA and AB are drawn in opposite directions, Fig. 17, then ^ OA=OA -AB. 6 * B * A In either case OB is the alge- braical sum of OA and AB. Fig - 17 ' The proposition may clearly be extended to three or more displacements. 30. The Parallelogram of Displacements. PROPOSITION 6. If two displacements represented in direction and magnitude by two straight lines OA, OB meeting at a point be impressed on a particle, the resultant is OC, the diagonal through O of the parallelogram which has OA, OB for adjacent sides. For from A draw AC, Fig. 18, equal and parallel to OB. Join OC and BC. Then OACB is a parallelogram and OC is the diagonal through 0. In consequence of the dis- placement OA the particle is moved to A \ in consequence of the displacement OB, A is moved o"~ to C. Thus OC is the resultant Fig. 18. 2930] KINEMATICS. VELOCITY. 33 displacement and it is the diagonal through of the par- allelogram AOBC. This proposition may be put into a slightly different form, thus : If two sides OA, AC of a triangle OAC, Fig. 19, represent displacements impressed on a particle, then the third side OC represents the resultant dis- placement. In this form it is known as the Triangle of Displacements. p. ^ PROPOSITION 7. To find an expression for the resultant of two displacements at right angles. Let OA, OB, Fig. 20, represent two displacements, P, Q respectively, at right angles to each other. Complete the rectangle A OBC. Let R be the resultant of P and Q, then R is represented by OC. Since the angle OAC is a right angle we have 'p Fig. 20. Hence * PROPOSITION 8. To find an expression for the resultant of two displacements inclined to each other at any angle. Let OA, OB represent respectively two displacements P, Q inclined to each other at an angle y. Complete the parallelogram AOBC. Then OC represents R the resultant of P and Q. Draw CD perpendicular to OA meeting OA produced, Fig. 20 (a), or OA, Fig. 20 (b), in D. Then AOB = y; in Fig. 20 (a) the angle y is less than a right angle ; in Fig. 20 (b) it is greater. G. M. 3 MECHANICS. [CH. II Now in Fig. 20 (a), OD = OA + AD = OA + AC cos DAG Q cos y, CD = AC sin CAD = OBsm = Q sin y. 'P A Fig. 20 (a). Fig. 20 (&). In Fig. 20 (6), OD = OA - AD = OA - A C cos DAG = OA-OB cos ( 1 80 - y) = OA + OB cos y = P + Q cos y, CD = AC sin CAD = OB sin (180 -y) = BC= 12-5 miles, for BC If we take BO to represent a velo- city of 20 miles an hour, OC will re- present one of 15 miles per hour in a direction opposite to that in which A is moving. If then we superpose on the ships a velocity represented by OC, the ship A will be reduced to rest while B will move in the direction BC with Fi a velocity represented on the same scale by BC ; this is a velocity of 25 miles an hour. Thus the relative motion of the two ships is represented by a velocity of 25 miles an hour in the direction BC. The two ships will be nearest apart when B has arrived at D, and this least distance is given by AD. Now AD OB AT ^_ 12-5 BC = 6 miles. Thus the least distance apart of the ships will be 6 miles. CD CO Moreover AD = OB' whence CD 4-5 miles, .-. BD = 17 miles. 36-38] KINEMATICS. VELOCITY. 49 Hence since the relative velocity of B along BD is 25 miles per hour, the ship B will reach D in -1 hour from the time at which it was at B, and at this time the two ships will be the closest together. *37. Angular Velocity. Let P, Fig. 35, be a point which is moving along a plane curve APB, and let be any fixed point in the plane of the curve and OA a fixed line through 0. As P moves the angle PDA varies; the rate of change of this angle is called the angular velocity of the point P about 0, and is measured in general by the ratio of the change in the angle to the interval of time during which that change has occurred when that interval is made sufficiently small. When the angular velocity is uniform it is measured by the ratio of the angle 0, described in the interval of time t seconds, to the time, so that in this case if to be the uniform angular velocity we have 6 *38. Motion with uniform speed in a circle. If the curve described be a circle, with 0, Fig. 36, as centre, we can find a relation between the uniform angular velocity about and v the uniform speed of the particle in the circle. For if s be the arc described in time t measured from A we have, since the speed is constant, But if a be the radius of the circle, Fig ' 36 ' and the circular measure of the angle AOP, then G. M. 50 MECHANICS. [CH. II But (jt>t } vt .. to = . a Hence v = ao>. Hence the speed in the circle is found by multiplying the angular velocity by the radius of the circle. Example. Assuming the earth to be a sphere ivhose radius is 6-436 x 10 6 metres, find in metres per second the velocity of a point on the equator. The earth rotates uniformly through an angle whose circular measure is 2ir (44/7) in 24 hours. its anular velocit is 4 7 x24x3600 ' Hence the velocity of a point on the equator is 44 x 6436 x 1Q3 7x24x3600 ' and this reduces to 468 metres per second. A relation identical with the above holds between the speed, the radius of the circle and the angular velocity about the centre even when the two are not uniform, provided that v and w stand for the values of the speed and the angular velocity at the same moment of time. 39. Graphical Representation of Space passed over by a particle. Draw a horizontal line OX, Fig. 37, divide it into a number of equal parts in the points N- , N. 2 , N s etc. and let each such part represent a small interval of time. From each point draw lines P 1 N IJ P^NZ, etc. at right angles to OX to represent the velocity of the particle at the end of the corresponding interval ; join the points PjP 2 . ... If the intervals be sufficiently small the line joining these points will be a continuous curve. Such a curve is called a velocity curve ; it is denned by the property that if a perpendicular PN be drawn from any point on it to meet the time line OX in N then PN is the velocity of the particle at the time represented by ON. 38-39] KINEMATICS. VELOCITY. 51 Let us now suppose the velocity to be constant ; the lines A P?,N Z , etc. will all be of the same length, the ve- locity curve is a straight line such as PP' parallel to OX, Fig. 38. Let P, P' be two points on the curve and PN, P'N' perpendicular to OX, let t be the time ON Tr ~j represented by NN' and Fi 8- 38 - let v be the constant velocity, s the distance traversed. Then Now = PN v = s vt, = area PNN'P'. Thus in this case of uniform velocity the area -between the velocity curve, the line OX and two lines perpendicular to the line OX represents the space traversed by the particle. Some further consideration shews us that this proposition is always true whether the velocity be uniform or variable. For we have seen that we may approach the case of a continuously varying velocity by dividing the time up into a large number of small intervals and supposing the velocity to remain constant during each interval but to change suddenly at the end of every interval. Draw the velocity curve for such a case supposing for the present the intervals during which the velocity is constant to be seconds. It will consist, as shewn in Fig. 39, of the series of horizontal and vertical straight lines alternately parallel and perpendicular to OX. During the time N 4 Fig. 39. 42 52 MECHANICS. [CH. II the velocity is constant and equa^to P^i, the space traversed is the area P^^N^R^ at the time JV a the velocity changes to P 2 N 2 increasing by P^R^ ; for the next second N^N S the velocity is constant and equal to P 2 ff 2 , the space is represented by P^N^N^R^ and so on. Thus in this case the whole space traversed is the area between the velocity curve, the line OX and the two bounding lines P^N-i and P n N n . This result will be equally true if we divide each second into a large number of parts and suppose the velocity to change at the end of each part. Instead of a single step between P^ and P 2 we obtain a large number of steps ; instead of a single parallelogram such as P^N^N Z R^ we have a large number ; the sum of the areas of these parallelograms is still the space traversed. If now we make the number of parts sufficient each individual step will be indefinitely small, the broken line will merge into the continuous velocity curve, and the sum of the parallelograms will become the area of that curve. Thus the space traversed during a given time is given by the area bounded by the velocity curve, the line OX and two lines perpendicular to X drawn from points on X which represent the beginning and the end of the time. Whenever then we can calculate the area of this curve, we can find the space traversed by the particle. In the important case considered in the next chapter the velocity increases uniformly with the time and the curve is a straight line. The area is bounded by straight lines and can therefore be easily calculated. By drawing a diagram to scale on squared paper and then determining the area by the method given in Experiment 3, we can find the space traversed in many cases in which we are given the relation between the velocity and the time. In solving such a question it is necessary to be careful as to the units in which the lines in the diagram are measured. Suppose, for example, 1 cm. along the time line represents an interval of 1 second, and 1 cm. at right angles to this a velocity of 1 cm. per second ; then an area of 1 sq. cm. represents the distance traversed in 1 second by a particle moving with a velocity of 1 cm. per second ; that is, it re- presents a line 1 cm. in length ; if however we had taken a length of 1cm. at right angles to the time line to represent a velocity of vena. per second, then an area of 1 sq. centimetre would represent a length of v centimetres. 39] KINEMATICS. VELOCITY. 53 Examples. (1). The velocities at the ends of 1, 2... 10 seconds are 5, 7, 9... 28 cm. per second, find by a diagram the space traversed in 10 seconds. (2). The velocities at the ends of 1, 2. ..10 seconds are I 2 , 2 2 ...10 2 cm. per second, find by a diagram the space traversed in 10 seconds. These examples are left for the student to solve with the aid of a ruler and squared paper. EXAMPLES. UNIFOBM SPEED. 1. Find in feet per second the following velocities : (1) 10 miles per hour ; (2) a quarter of a mile in 44 seconds ; (3) 92000000 miles in 8 minutes ; (4) 25000 miles in 24 hours. 2. Find in centimetres per second the velocity of a body which traverses (1) a cm. in b seconds ; (2) a circle of 10 cm. radius in 1 second ; (3) 76 cm. in 10 minutes ; (4) the perimeter of a square 1 foot in edge in 1 minute. 3. The speed of a steamer is 22 knots, reduce this to cm. per second. 4. A particle has a velocity of 30 miles per hour, how many feet does it traverse (1) in 1 minute, (2) in a day, (3) in a year? 5. A man walks a mile in 10 minutes, a second mile in 12 and a third in 15 ; he runs a fourth mile in 5 minutes ; find his average speed (1) in feet per second, (2) in miles per hour. 6. A and B start to walk towards each other from two places 6 miles apart. A walks twice as fast as B. Where will they meet ? The meet- ing takes place 50 minutes after the start, find the speed of each. 7. A starts along a road at a speed of 3 miles an hour, after 40 minutes B follows at a speed of 5 miles an hour, how far must B go before overtaking A ? 8. The velocity of sound is 1100 feet per second, a man in front of a cliff claps his hands and hears an echo after 5 seconds, how far is he from the cliff ? 9. A man climbs a hill inclined at 30 to the horizon, if he rises vertically 1000 feet in an hour find his speed in feet per second. 10. The radius of the Earth's Orbit is 92 million miles and the radius of the Earth 4000 miles, compare the velocities of a point on the equator at midday and at midnight. 54 MECHANICS. [CH. II 11. Find the resultants of the following pairs of velocities in direc- tions at right angles to each other; the velocities are all expressed in centimetres per second : (1) 3 and 4; (2) 6 and 8; (3) 12 and 15; (4) v and i? 2 , where 12. Find by a graphical construction and by the formulee the re- sultants of the following velocities : (1) 3 and 4 at 60; (2) 6 and 8 at 45; (3) 1 and 2 at 30; (4) 1 and 2 at 60. 13. A boat is rowed across a river at the rate of 3 miles per hour, the river is flowing at the rate of 4 miles per hour ; find the velocity of the boat. 14. A ship is sailing at the rate of 10 miles an hour and a sailor climbs the mast 200 feet high in 30 seconds. Find his velocity relative to the Earth. 15. The paths of two ships steaming North and East respectively, with velocities of 12 and 16 miles per hour, meet. The two ships are each 12 miles distant from the point of intersection. Determine after what time they will be closest together and what that closest distance will be. 16. Two equal velocities have a resultant equal to either, shew that they are inclined to each other at 120. 17. The resultant of two velocities u and v is equal to u, and its direction is at right angles to that of u. Shew that v is equal to u ^/2. 18. Find by a graphical construction or otherwise the resultant of the following velocities in the directions of the sides of a square taken in order : (1) 1, 2, 2, 1 ; (2) 3, 4, 5, 6 ; (3) 2, 5, 6, 3 ; (4) 7, 8, 4, 5. 19. Find by a graphical construction or otherwise the resultant of the following velocities in the directions of the sides of an equilateral triangle taken in order : (1) 3, 3, 3 ; (2) 4, 5, 6 ; (3) 5, 8, 10 ; (4) 6, 9, 12. 20. Find the horizontal and vertical components of the following velocities : (1) 1000ft. per second in directions inclined respectively at 30, 45 and 60 to the horizon. (2) 25 miles per hour at 60 to the vertical. 21. Kesolve a velocity of 1000 feet per second into two equal velo- cities inclined at 60 to each other. 39] KINEMATICS. VELOCITY. 55 22. A velocity of 500 feet per second is resolved into two at right angles, one of these is 250 feet per second, find the other. 23. A velocity of 5 miles an hour to the East is changed into one of 5 miles an hour to the North ; find the change in velocity. 24. A velocity represented by one side AB of an equilateral triangle ABC becomes changed into one represented by the side AC; find the change in velocity. 25. Find by a graphical construction or otherwise the resultant of the following velocities, which are given in centimetres per second : (a) 15 to the North, 20 to the East, 20^2 to the North-west, 35 to the West. (&) 1, 2, 3, 4, 5, 6 parallel respectively to the sides of a regular hexagon. 26. One of the rectangular components of a velocity of 60 miles per hour is a velocity of 30 miles per hour ; find the other component. 27. A body moves during each of 5 consecutive minutes with velo- cities respectively of 1, 2, 3, 4, 5 feet per second ; find the space traversed. 28. The spaces traversed up to the end of 1, 2, 3 and 4 minutes by a body moving with constant velocity during each minute are 2, 8, 18 and 32 feet respectively. Shew on a diagram the velocity during each minute. 29. The velocity of a body starting from rest increases by 1 foot per second at the end of every second of its motion. Determine by means of a diagram or otherwise the space passed over in t seconds. 30. The components in two directions of a velocity of 30 miles per hour are velocities of 15 and 25 miles per hour, determine their directions. 31. Two velocities u and v have a resultant U which makes an angle a with the direction of u ; if u be increased by U while v is unchanged shew that the new resultant makes an angle ^ with the direction of u. 32. Two particles are projected simultaneously with equal velocities from the points A and B, one from A towards B, and the other in a direction at right angles to AB ; find how far the former will have travelled towards B when the two particles are nearest to one another. 33. If a point begins to move with velocity u, and at equal intervals of time r, a velocity v is communicated to it; find the space described in n such intervals. 34. Compare the velocities of two trains, one travelling with a velocity of 50 miles per hour and the other with a velocity of 55 feet per second. CHAPTER III. KINEMATICS. ACCELEKATION. 4O. Change of Velocity. The velocity of a particle may change either in magnitude or in direction or in both these respects. Let OA represent the velocity of a partide at a given instant; if the velocity remain uniform, OA will continue to represent it; suppose however that the velocity changes and that after an interval it is represented by OB. If the change occur in the magnitude only, the particle will continue to move in the same direction as before. OAB will be a straight line, and AB, Fig. 40, will repre- sent the velocity which must be added to the original velocity OA ^ ^ j ^ g to give OB. If the change in velocity take place in direction as well as in magnitude, OA and OB will be inclined to each other, Fig. 41, but AB will still represent the change in the velocity, for by the parallelo- gram of velocities A B is the velocity which must be compounded with OA to give OB. o Thus AB represents in direction Fig. 41. and magnitude the velocity which must be superposed on OA to change it into OB. 40-41] KINEMATICS. ACCELERATION. 57 If then we represent by two straight lines drawn from a point the initial and final velocities of a particle, the line joining the extremities of these two lines represents the Change of velocity of the particle. Examples. (1). A particle moving North-east with a velocity of 1 foot per second is observed after a time to be moving East icith a velocity of *J2 feet per second, find the change in velocity. Here, Fig. 42, OA = 1 , OB = >J2 and A OB = 45. Draw AC normal to OA meeting OB in C; then ACB=45 and AC=AO = 1. Hence OC 2 = 2 = OB 2 , thus C and B coincide, and AB the velocity added is 1 ft. per second in the South-east direc- ' 'V 2 tion. -pig. 42. (2). A velocity of 10 feet per second is changed into one of 10 feet per second inclined at 60 to the former, find the change in velocity. In this case, OA = OB = W and 4 OB = 60, /. L OAB = L OB A = \ (180 - 60) = 60. Thus ^LB = (L4 = 10. The additional velocity is one of 10 feet per second inclined at 60 to OA. 10 10 Fig. 43. 41. Acceleration. When the velocity of a particle is variable it is said to have acceleration. DEFINITION. The Acceleration of a particle is its rate of change of velocity. Acceleration may be Uniform or Variable. Uniform acceleration is measured by the ratio of the change of velocity to the interval of time during which that change has occurred, i.e. by the change of velocity in one second ; variable accelera- 58 MECHANICS. [CH. Ill tion is measured by the same ratio when the interval is sufficiently small, that is, by the change in velocity which would take place in one second if during that second the velocity changed uniformly. The numerical measure of an acceleration is the number of units of velocity added per second. Now velocity is measured by the number of units of space traversed per second. When, then, we state that the acceleration of a particle moving with uniform acceleration is a, we mean that in each second an additional velocity of a cm. per second is given to the particle. To define then an acceleration we must know the number of units of space per unit time which are added to the velocity, and further we must remember that this additional velocity is conferred in the unit of time. Just then as when considering a velocity, we speak of so many centimetres per second, so when dealing with accele- ration we speak of so many centimetres per second per second. DEFINITION OF UNIT ACCELERATION. A particle has Unit Acceleration when its velocity increases in each second by 1 centimetre per second. If the units of space or time be changed the numerical measure of a given acceleration is changed also. The method of calculating these changes is shewn in the following. Example. A particle has an acceleration of 32 '2 feet per second per second, find its value (a) in cm. per sec. per sec., (b) in yds. per min. per min. For (a) we have I ft. = 30-48 cm. Now in 1 sec. a vel. of 32-2 ft. per sec. is added, .-. in 1 sec. a vel. of 32-2 x 30-48 cm. per sec. is added. .-. the new measure is 32-2 x 30-48 cm. per sec. per sec. This reduces to 981-5 cm. per sec. per sec. For (6), lfl.=-Jyd., 1 min. = 60 sec. 41-42] KINEMATICS. ACCELERATION. 59 In 1 sec. a vel. of 32-2 ft. per sec. is added, 32 '2 /. in 1 sec. a vel. of yds. per sec. is added, O , 32-2x60 , /. in 1 sec. a vel. of 5 yds. per mm. is added, o , 32-2x60x60 , .-. in 1 mm. a vel. of - yds. per mm. is added, o . 32-2x60x60 , Thus the new measure is ^ yds. per mm. per mm. o This reduces to 38640 yds. per min. per min. It will be noticed that in (b) the change in the unit of time comes in twice. The reason for this is clear, the unit of time affects the measure of the velocity and affects also the time during which, when calculating the acceleration, the change in the velocity is to be reckoned. 42. Uniform acceleration in the direction of motion. The change in the velocity of a body may be a change in magnitude, in direction or in both. For the present we deal only with the case of a body moving in a straight line with uniform acceleration. The change of velocity will be one of magnitude only, and that change will be a uniform one, the speed will vary but not the direction of motion. The velocity may either increase or decrease ; in the first case the acceleration is positive, in the second negative. PROPOSITION 13. To determine the velocity of a body moving in a straight line with uniform acceleration in terms of the initial velocity, the acceleration and the time of motion. Let the initial velocity be u, the velocity after t seconds v, and the acceleration a. In 1 second a velocity of a centimetres per second is added and the acceleration is uniform. Hence in 2" the velocity added is 2a, in 3" the velocity added is 3a, and in t" the velocity added is at. 60 MECHANICS. [CH. Ill Thus at the end of t seconds the velocity is u + at. Hence v = u + at. If the velocity decreases with the time, the acceleration is negative and we have v u at. The proposition can be put rather differently thus. The change in velocity in t" is v-u. Therefore the change per second is (v - u)jt. But the acceleration is the change of velocity per second. Hence /. v-u = at, or v = u + at. PROPOSITION 14. To draw 1 the velocity curve for a particle moving with uniform acceleration. Draw a horizontal line OX, Fig. 44, to represent time and a vertical line Y to re- present velocity. Choose a convenient length to repre- sent the unit of time, and also a convenient length to represent the unit of ve- locity. Mark off along 07 a length OA to represent the initial velocity u. Through A draw AM parallel to the time line and commencing from A divide AM in J/ 1? M 2 , M 3 , etc. into equal parts, each of which shall represent 1 second. At M lt M 2 , etc. draw P^M^, P 2 M 2 , P 3 M 3 , etc. perpendicular to AM, and pro- duce these to meet the time line OX in JV 15 N z , etc. Make P^M 1 equal to a, P 2 M 2 equal to 2a, P S M 3 equal to 3a, etc. Then these 1 For this and similar purposes squared paper such as is used in Experiment 3 is convenient. Fig. 44. 42] KINEMATICS. ACCELERATION. 61 various lines represent the increments of the velocity up to the end of the first, second, third, etc. second, and the lines PI^I, /yV 2 , PS^S, etc. represent the actual velocity at the end of the first, second, third, etc. second. Thus the line AP : P 2 . . . represents the velocity curve and the construction shews that it is a straight line. Examples. (1). A particle moving from rest with uniform acceleration has a velocity of WO ft. per second after 5 seconds, find its acceleration. In each second a velocity of 160/5 feet per second is produced. Hence the acceleration = i4p- = 32 ft. per sec. per sec. (2). A particle moving under a negative acceleration of 32 feet per second per second is projected with a velocity of 160 fekt per second. Find ichen it ivill come to rest and what ivill be the velocity after 10 seconds. In each second a velocity of 32 feet per second is destroyed. Therefore the initial velocity of 160 feet per second will be destroyed in 160/32 seconds. Thus the particle is instantaneously at rest after 5 seconds. The acceleration now produces in each second a velocity in the oppo- site direction of 32 feet per second. Therefore after 5 seconds more, i.e. at the end of 10 seconds, the velocity will be - 5 x 32 or - 160 feet per second. Aliter. Let v be the velocity after t seconds. Then v = 160-32.f. If t : represent the time at which the particle is at rest, at which therefore v is zero, we have = 160-32^; ' t l = 1 ^- = 5 seconds. Again after 10 seconds, v = 160 - 32 x 10 = - 160 ft. per sec. (3). Draw the velocity curve in the case of (2). Draw the time and velocity lines OX and OF, Fig. 45. In OF take OA to represent a velocity of 160 ft. per second. Draw a line from A parallel to OX and in it take Mj so that AM l may represent 1 second ; from J/j draw M 1 P l vertically down to represent a velocity of 32 feet per second. Join AP^ and pro- duce it, AP l is the required velo- city curve. It meets the line OX in N, where from the figure, ON=5AM lt Hence ON represents 5 seconds. 62 MECHANICS. [CH. Ill 43. Acceleration, space traversed and time of motion. PROPOSITION 15. To find the space passed over in a given time by a body starting from rest and moving with uniform acceleration. The space passed over is given by the area of the velocity curve which in this case will be a straight line passing through the point from which the time and velocity lines are drawn. / Let ON, Fig. 46, represent the time t and NP perpendicular to ON the velocity at the end of the time interval. Then PN=at. Fig. 46. Join OP; the velocity curve is the line OP and the space s required is the area of the tri- angle OPN. Now the area of a triangle is half the product of the base and the altitude; .-. s = area OPN= J PN . 0$ = \at.t = \ at 2 . Hence . s = J at 2 . Thus the space passed over in the first second is J a while the velocity at the end of that second is a. The space tra- versed is found by multiplying half the acceleration by the square of the time. PROPOSITION 16. To find the space passed over by a particle moving with uniform acceleration when the particle starts with an initial velocity. The space required will be the area of the velocity curve. In OT, Fig. 47, take OA equal to the initial velocity u\ let ON represent the time t, and NP perpendicular to OX, the velocity after t seconds, so that PN=u + at. 43] KINEMATICS. ACCELERATION. 63 Join A P. Then AP will be the velocity curve and the space required is the area OAPN. Draw AM parallel to OX to Y meet PN in M. Then P M = at, MN = OA = u. Hence s-area OAPN = parallelogram OAMN + triangle APM = OA x ON + J PM > and AM=ON = t. N X Fig. 47. Hence = ut + \ at . t = ut + at 2 . Thus the space actually traversed is found by adding together the spaces the particle would traverse (1) if it moved with the constant velocity u and (2) if it started from rest with the constant acceleration a. PROPOSITION 17. To find the average velocity of a particle moving with uniform, acceleration. We can put the last formula in a different form thus : let v be the final velocity of the particle, then v = PN. Join AN, then the figure OAPN is made up of the two triangles OA N and PAN; the bases of these triangles are OA and PN and their altitude is ON. Thus s = area OA PN = triangle OAN+ triangle APN = \OA. ON+\PN.ON = J (u + v) t. Now we know, 24, that when a particle moves with variable speed the space traversed is found by multiplying the average speed and the time. 64 MECHANICS. [CH. Ill In this case, therefore, the average speed so defined is J (u + v). Thus, the average speed of a particle moving with uniform acceleration is half the sum of the initial and final speeds. The above formula may be applied to the case in which a is negative, we have then s = ut at 2 for v = u- at. In a later chapter ( 68) experiments will be given by which the student can verify for himself the truth of the formulas proved in this section. Those who have a difficulty in following the steps of the proof may adopt the experimental proof. The argument just given may be made clearer to some by giving the following algebraical proof which sums up in symbols its results. Let us suppose the whole time t divided up into a series of n equal small intervals each equal to T, so that nr-t. At the beginning of each interval the velocity will have the values respectively M, w + ar, u+ 2ar,...u + n - 1 ar, and at the end of each interval it will have the values u + ar, u + 2ar, u + 3ar, . . . u + nar. The space traversed will be greater than that which would be traversed, if during each interval the particle moved with the velocity which it has at the beginning of the interval, and less than that which would be traversed if during each interval the particle moved with the velocity it has at the end of the interval. In the first case the space would be s x , where we have s l = UT + (u + ar) T + (u + 2ar) T+ ...... + (u + n-la)r; ...n-l) n(n-l) ' Now T = - ; n 43-44] KINEMATICS. ACCELERATION. 65 and in the second case the space would be 2 , where = unr + ar 2 (1 + 2 + 3 + . . . + ri) But s lies between s l and s 2 and these two quantities can both be made as nearly equal as we please to ut + ^at 2 by making n very large, for then 1/n vanishes. Hence s 44. Acceleration, Velocity and Space traversed. PROPOSITION 18. To find a relation between the velocity, the acceleration, and the space traversed for a particle moving with uniform acceleration. Let u be the initial velocity, v the velocity after t seconds during which the particle has traversed a distance s and a the acceleration. We have proved that v u = at, s = ut + lr at 2 . We wish to eliminate t from these equations. The first gives us V - U a u (v - u) . (v - uf Hence s = + Ja ^ 2 '- , a a 2 .'. 2as 2uv 2u 2 + v 1 + u 2 2uv = v*-u z . Hence v 2 = u 2 + '2as If the acceleration be negative we start with . v-u= -at, s = ut-^at 2 , and find v 2 - u?= - 2as. G. M. MECHANICS. [CH. Ill We can put the proof otherwise, thus v-u we have Hence on multiplication, as = % (v - u) (v +w) = i (v* - w 2 ), or v z u 2 + 2as. 45. Formulae connected with uniform accelera- tion. We have thus proved the following formulae in which the symbols have the meanings attached to them in the preceding sections. (i), 2 ........................ (ii), -y 2 = u 2 + 2as ........................ (iii). We may also write (ii) as s = J (v + u) t ..................... (iv). If the particle start from rest u is zero, and the equations become v =at ........................... (i) a, v 2 = 2as ........................ (iii) a, s =\vt ........................ (iv) a. Examples. (1). A particle starts with a velocity of 3 cm. per sec. and an acceleration of 2 cm. per sec. per sec., find its velocity after 10 sec. and the distance traversed in 10 sec. Let v be the velocity after 10 seconds, s the distance traversed. Then v = 3 + 2 x 10 = 23 cm. per sec. (2). How far must the particle, moving as in Example (1), move in order that its velocity may become 5 cm. per second. To solve this we need equation iii, giving a relation between the velocity and the space. If the space required be s cm., we have = 5 2 -3 2 =16; 44-45] KINEMATICS. ACCELERATION. 67 (3). A particle has a velocity of 20 cm. per second and an acceleration of -5 cm. per sec. per sec., how far will it move before coming to rest ? If v be the velocity after it has traversed s cm. we have v 2 =20 2 -2x5xs. If the particle is at rest for a moment, we have v zero, and hence /. s = 40cm. Thus the particle is brought to rest after moving 40cm.; if the acceleration continue to act it will only remain at rest for an instant, and then commence to retrace its path passing through the starting point with its initial velocity. (4). A particle starts ivith a velocity u and an acceleration - a ; shew that it comes to rest after an interval u/a seconds and passes through the starting point again after an interval 2u/a seconds. The velocity after t seconds is u-at\ when th particle is at rest this is zero, .. . a The distance of the particle from the starting point at t" is ut-^at 2 . When the particle is at the starting point this distance is zero. Then or u-^at = 0; 2u whence t = . a Thus the particle is at the starting point initially and reaches it again after an interval 2u/a. During half this interval the particle is moving from the starting point, during the second half it is moving to it. (5). A particle has an initial velocity of 125 cm. per sec. and an ac- celeration of (a) 10 cm. per sec. per sec., (b) - 10 cm. per sec. per sec. How long will it take in either case to move over 420 cm. ? We know the initial velocity, the distance traversed and the accelera- tion and require to find the time. This is given us by equation ii. Let it be t seconds, then for (a), 52 68 MECHANICS. [CH. Ill solving the quadratic _ -25=^625 or -28. From the solution t = 3 we see that 3 seconds after starting the particle will be at a distance of 420 cm. from the starting point. Now as to the solution t = - 28", we infer from this that it is possible to start the particle from a position 420 cm. from the starting point with such a velocity that after 28 seconds it is at the starting point and is moving with a velocity of 125 cm. per second. Let 0, Fig. 48, be the original starting point, A a t point 420 cm. to the right of A ; then B O A if the particle is projected towards A -p. 4g with a velocity of 125 cm. per second it will arrive at A in 3 seconds, this is the first solution. But it is also possible to start the particle from A towards with such a velocity that it passes through O to B, comes to rest for an instant at B and then returns to O, arriving at with a velocity of 125 cm. per second, 29 seconds after it has left A ; if this be possible then we may say that the particle was at A - 28 seconds before leaving 0. And this is clearly possible. In the first case, the particle arrives at O with a velocity of 155 cm. per second, viz. its original velocity of 125 cm. per sec. and the velocity of 30 cm. per sec. generated in 3" by the acceleration. Suppose now it be projected from A towards with this velocity. It will arrive at after 3 seconds and have a velocity of 125 cm. per second ; it will then continue to move towards B for 12-5 seconds, in which time the velocity of 125 cm. per sec. will be destroyed. Thus it arrives at B 15 '5 sees, after leaving A. It will then return from B to and will arrive at after another 12-5 sees, with the velocity of 125 cm. per second. Thus the interval between the start at A and the time at which the particle reaches with a velocity of 125 cm. per sec. towards A is 15 '5 + 12-5 or 28 seconds. Thus under the given conditions of acceleration and velocity the particle might be at A, 420 cm. from O, either 3 seconds after passing 0, or 28 seconds before passing 0. (b) Taking now the other case in which the acceleration is - 10, we have 25=^(625-336) ~2~ J^ 7 = 4 or 21. The reason for the double value of t is clear as before. The particle starts from and arrives at A after 4 seconds, during which time its 45-46] KINEMATICS. ACCELERATION. 69 velocity has been reduced to 125 - 40 or 85 cm. per second. It moves on with decreasing velocity until it is brought to rest at B after a further interval 85/10 or 8'5 seconds. Thus it takes 12-5 seconds to reach B from O. It now returns towards under the acceleration 10 starting from rest at B and after a further interval of 8-5 seconds again passes through A. Thus it reaches A the second time 12-5 + 8-5 or 21 seconds from the start. 46. Falling Bodies. When a body is allowed to fall to the earth's surface from a point above it, it is found (1) that the acceleration is uniform 1 , (2) that the acceleration is the same for all bodies. The Experiments on which these statements are based will be described later. See 65, 139. This uniform acceleration of all bodies when falling from a given point is spoken of as the acceleration of gravity, or better, the acceleration due to gravity. It is usually denoted by the symbol g. Again, Experiment shews us that the acceleration of a falling body differs slightly at different places on the earth ; it is greatest at the poles and least at the equator. A body falls from a given height more rapidly at the pole than at the equator. The value at the pole is 983-11 cm. per sec. per sec. and at the equator 9 78 '10 cm. per sec. per sec. At Greenwich the value is 981 '17 cm. per sec. per sec. Since 1 foot contains 3048 cm. the value of g at Greenwich is 981-17/30-48 or 32-191 feet per sec. per sec. 2 . Hence we see that the formulae of Section 45 are all applic- able to the case of a falling body. 1 This statement is only true for distances above the surface which are small compared with the radius of the Earth (4000 miles). It may be applied therefore without error to the experiments described in this book. 2 See Example 1, p. 58. In working numerical examples we may use the values 980 cm. per sec. per sec. or 32 feet per sec. per sec. 70 MECHANICS. [CH. Ill When the body is projected downwards and starts with velocity u the acceleration g is in the direction of motion, and v = u + gt ........................... (i), s = ut + %gt* ......................... (ii), v 2 = u z + 2gs ..................... (iii). If the body be "dropped" it starts from rest so that u = 0. If the body is projected vertically upwards and starts with the velocity u the acceleration g is opposite to the direction of motion and we have v = u gt ........................ (i) a, (iii) a. 47. Problems on falling bodies. (1) To find the space passed over by a falling body in the nth second of its motion. Let ! be the space up to the beginning, s 2 up to the end of the nth second, then s 2 s : is the space required. Now, assuming the body to have been dropped, (2) A particle is projected upwards with velocity u. a. Find the height to which it will rise. Let this be H, the particle moves up till it reaches the height ff, then it is instantaneously at rest and finally falls. Hence at a height H the velocity is zero /. 0=u*-2gH, ft. Find the time of rising. Let this be T l , then at time T l the velocity is zero, 46-48] KINEMATICS. ACCELERATION. 71 . y. Find the time of falling. Let this be T z , then T 2 is the time of falling a height H. T- U -T 2 ~n). 44. Find the change of momentum of a body of mass m which is initially moving with velocity u, and then has its velocity deflected through an angle a, but without change of magnitude. Two particles A and B of the same mass m are connected by a light string and rest with the string just tight on a horizontal table whose co- efficient of friction is p. A uniform force P (greater than 2fmig) begins to act on the particle A in the direction BA. Find the acceleration of either particle and the tension of the string at any time. G. M. 10 146 MECHANICS. [CH. VII f 45. A mass of 10 grammes resting on a smooth table is connected by a string passing over a pulley with an equal mass which hangs vertically. The first mass is 1 metre from the edge ; find the velocity of the system at the moment at which it is dragged over. 46. A mass of 200 grammes can just be dragged up an inclined plane by a mass of 100 grammes which is connected to it by a string passing over a pulley and hangs vertically from the top of the plane. Find the inclination of the plane ; find also the velocity produced after the system has moved through 1 metre, if a mass of 5 grammes be placed on the smaller mass. 47. A mass of 1 Ib. slides down an inclined plane whose height is half its length and draws a mass of 5 Ibs. along a smooth horizontal table level with the top of the plane. Find the acceleration ; find also the mo- mentum of the whole after the masses have moved 1 yard. 48. Two masses of 5 and 10 Ibs. respectively are placed on two inclined planes of the same height and the angle 30 ; the masses are con- nected by a fine string passing over a smooth pulley at the top of the two planes. Find the acceleration, the tension of the string, and the distance traversed in 5 seconds. 49. An engine draws a train whose mass is 75 tons up a slope of 1 in 50. If the resistance of the rails be -^ of the weight, find the force exerted by the engine in order that the speed may be constant. The speed at the bottom of the slope is 25 miles an hour : if the engine stops working how far will the train run ? 50. A mass of 10 kilogrammes slides down a rough plane which rises 1 in 10: find the resistance if the speed remains constant. 51. What is meant by the equation W=Mg ? A lift is rising with an acceleration of 8 feet per second per second ; what pressure would a man scaling 16 stone exert on the floor of the lift? 52. . A particle, 'starting from rest, slides down a smooth plane in- clined at an angle of 45 to the horizon in 6 seconds ; find the length of the inclined plane. 53. Two equal masses are at rest side by side. One moves from rest under a constant force F while at the same instant the other receives an 21 impulse 7. Shew that they will again be side by side after a time . 1 F 54. A particle whose mass is 10 Ibs. is moving with a velocity of 30 feet per second. If it is brought to rest in 100 feet by applying a constant resistance, find the magnitude of the resistance. 98] FORCE AND MOTION. 147 55. A particle whose mass is 12 Ibs. is moving with a velocity of 25 feet per second. If a constant resistance equal to a weight of 15 oz. is applied to stop it, find how far it will travel before it comes to rest. 56. A weight of 7 Ibs. is placed on a smooth horizontal board and connected by strings passing over smooth pulleys at each end of the board with weights of 5 Ibs. and 4 Ibs. respectively. Find the acceleration of the weights and the tensions of the strings. 57. If two masses m, m' are connected by a string, whose mass can be neglected, passing over a smooth fixed pulley, find the tension of the string. Two bodies, of mass 2 Ibs. and 30 Ibs. respectively, lie on a smooth horizontal table whose height above the floor is 27 inches. The bodies are ^connected by an inextensible string, whose length is not less than 27 inches, and, when the' string is taut, the smaller mass is dropped through a hole in the table. *Find when it reaches the ground. 102 CHAPTER VIII. THE THIRD LAW OF MOTION. ENERGY. 99. Action and Reaction. In the third Law of Motion Newton stated that To every action there is always an equal and opposite reaction, or the mutual actions of two bodies are always equal and opposite. The Law is based on observation and experiment, in considering it however we are at once met by the question What is meant by Action ? We can learn from Newton's own illustrations what he understood by the term. " If a man presses a stone with his finger," he says, "his finger also is pressed by the stone. If a horse draws a stone by means of a rope the horse is drawn equally towards the stone, for the rope stretched between the two will urge the horse towards the stone and the stone towards the horse ; and this will impede the progress of the one as much as it helps that of the other." " If a body impinging on a second body changes the motion of that body in any manner by the force it exerts, it will itself undergo the same change in its own motion in the opposite direction through the force exerted by the second body (because of the equality of the mutual pressure)." " It is the amount of momentum, not of velocity, interchanged in these actions which is equal, at least in bodies which are otherwise unimpeded. For the changes of velocity which take place also in opposite directions are reciprocally pro- portional to the [masses of the] bodies, since the change of 99-100] THE THIRD LAW OF MOTION. ENERGY. 149 momentum is the same for the two. The law is also true for the case of attractive forces." 10O. Illustrations of the third law. These illustra- tions of Newton's make it clear that the action and reaction contemplated by him was the Interchange of Momentum between two bodies. We shall see later that, in a scholium attached to the law, he interprets the terms in another sense in which they are equally true. If then we are to mean by Action Gain of Momentum, the experiments on impact described in Sections 51 61 afford a verification of the law. Moreover, when two bodies are unimpeded in their action on each other, not only is the whole gain of momentum of the one equal to the whole loss of the other, but, during the time of transfer, the rate at which the one gains momentum is equal to the rate at which the other loses it. The impressed force on the one is equal and opposite to that on the other. The moon has an acceleration towards the earth and the earth towards the moon, astronomical observations shew that these accelerations are inversely proportional to the masses of the moon and the earth respectively. For moon and earth the equation ma = m'a f holds, m and m' being the respective masses, a and a' the respective accelerations. Each gains momentum at the same rate. An experiment, also due to Newton, may illustrate this point. Float a magnet on a block of wood in a large 1 vessel of water; float a piece of soft iron on a second block of wood. Place the iron at a little distance from the magnet and in such a position that the magnet may point directly towards it. Hold the two at rest until the water is quite still, then release them simultaneously. The magnet will draw the iron to itself, the two blocks of wood will impinge and then come to rest 2 on the water. The two blocks gain momentum in opposite directions when free to move, this gain however is the same for each, the force 1 Large because otherwise capillary action at the sides of the vessel may affect the result. 2 It is desirable in performing this experiment to arrange that the impact may be direct, otherwise the blocks of wood may be made to spin round in the water. 150 MECHANICS. [CH. VIII on the one is equal and opposite to that on the other ; after impact the system remains without momentum. The following also affords us an illustration of this law. When a gun is fired it kicks, unless held close to the shoulder it bruises it. If the gun be held firm, it and the man constitute, so far as the action is concerned, but one body, the mass of which is much greater than that of the bullet ; the velocity acquired by this body will be small compared with that of the bullet, but the momentum of the bullet and of the gun and man combined will be equal and opposite. We can shew by the following experiment that this is so. The gun is attached with its barrel in a horizontal position to a massive block of wood which can swing about a horizontal axis above the gun and at right angles to the barrel ; the whole constitutes a pendu- lum; as with the simple pendulum if the gun be displaced from its equilibrium position its velocity can be calculated by observing the arc through which it swings, the velocity of the bullet as it leaves the barrel can also be found. Suppose the gun loaded and fired when at rest at the lowest point of its swing ; observe the velocities of the gun and pendulum and of the bullet and calculate the momentum of each: it will be found that they are equal. A simpler experiment of the same character is as follows. EXPERIMENT 22. To illustrate by experiment the third law of motion. Suspend two balls, Fig. 64, of different mass, so that their centres may be at the same level and equally distant from their points of support ; compress a fairly strong spiral spring and tie it with thread so that the spires are as close together as possible. Place it be- tween the balls ; release the spring by cutting or burning the thread, the balls will move in opposite directions ; determine their velocities by observing the distances through which each ball swings, it will be found that the velocities are in- versely proportional to the masses of the balls. Each ball acquires the same amount of momentum. Hicks' ballistic balance already described may be used for this experiment, with this apparatus the velocities are easily measured. 100-101] THE THIRD LAW OF MOTION. ENERGY. 151 One case mentioned by Newton needs a little further consideration ; when a horse drags a load or tows a boat how is it that the horse is pulled back as much as the load is pulled forward? The horse acquires momentum by the action between his feet and the road. If he is moving with uniform speed and we neglect the relative motion of the various parts of his body the whole of the momentum so acquired is transferred by the rope to the boat. The passage of the boat through the water is resisted and the boat loses momentum to the water, this loss is equal to the gain it acquired from the horse; thus the momentum gained by the horse appears in the water wave which follows the boat. Now the earth has lost momentum estimated in the direction in which the horse is moving equal to that gained by the horse: by the actions just described the water forming part of the earth has gained the same amount in the direction of the horse's motion, thus on the whole there is neither loss nor gain. If the boat is gaining speed the horse acquires from the ground rather more momentum than is transmitted by the rope to the boat. 1O1*. Attraction and the third law. This equality of action and reaction holds also in the case of the Attractions between the various portions of the same body. For consider a body like the earth, imagine it free from ex- ternal action, and suppose it divided into two parts A, B, Fig. 65, by a plane CD; if it be possible let A gain momen- tum from E faster than B gains it from A, then the whole sys- tem is continually acquiring momentum in the direction from B to A and will move off with continually increasing velocity into space in the direction BA ; this is contrary to the first law for the body is free from external action. Example. As an example of the third law consider the case of a shell exploding in the air. The fragments are projected in various directions but the total gain of momentum of the various parts is zero; the momentum gained in one direction by some parts is equal to that gained in the opposite direction by others. Thus if the shell burst into two equal pieces, if its original velocity be 100 feet per second, and the velocity after fracture of one part in the same direction be 150 feet per second we have, calling m the mass of the whole shell and v the velocity of the other part, m x 100=| x 150 +|v. Fig. 65. 152 MECHANICS. [CH. VIII Hence v 50, the one half gains 25m units of momentum, the other loses the same amount and continues to move forward with a velocity of 100 - 50 or 50 feet per second. We see by these and similar examples how to interpret the third law of motion if we give to action the meaning of change or rate of change of momentum. 102. Conservation of Momentum. In all these cases then Action is Transference of Momentum. When- ever a transference of momentum takes place between two bodies the loss of the one body is equal to the gain of the other. From this point of view the third law expresses the Conservation of Momentum. We can observe the facts that the one body gains momentum while the other loses it; we call the rate at which this transference takes place Force, and when the transference is going on we say that Force acts between the two bodies, its action on the two being equal and opposite. 103. Action. Energy. But the third law as Newton pointed out contains more than this. We shall find that, when bodies act on each other, another quantity (Energy) besides Momentum is transferred, and that in this case too there is neither gain nor loss on the whole transaction. In the Scholium to the laws of motion, Newton calls attention to the importance of the quantity obtained by multiplying force and the displacement of its point of applica- tion. He writes after giving some examples " By these I wished to shew how wide-spread and how certain is the third law of motion. For if the action of an agent be measured by the product of its force and displacement 1 , and similarly the reaction of the resisting body be measured by the sum of the products of the resistances and their several displacements, then whether the resistances arise from friction, cohesion, weight or acceleration, in all cases action and reaction 1 The word actually employed is velocity, but the velocities concerned are those which occur during the same moment, they are measured there- fore by the displacements. 101-104] THE THIRD LAW OF MOTION. ENERGY. 153 in every kind of machine will be equal." Action then in the third law may be measured by the product of a force and the displacement of the point at which it is applied; this displace-* ment however as Newton points out is to be estimated in the direction of the force. In other words, when momentum is being transferred to a body at a given point, the product of the rate at which the body is gaining momentum and the displacement of the point at which the transference is taking place may measure the Action which is considered in some aspects of the third law. The meaning and importance of this statement will how- ever be better appreciated after some preliminary explanations and definitions. 104. Work and Energy. When momentum is being transferred to a body and the point at which the transference is taking place is in motion, we say in general that Work is being done. Now let F be the amount of momentum transferred per second the force; let A, Fig. 66, be the point at which the transference is taking place the point of action of the force. Let AB ~ be the direction in which the momentum is being transferred the direction of the force. Let A be displaced to A' and draw A'A l perpendicular to AB to meet it in A lt Then the work done is measured by the product F x AA^ The actual displacement of the point A is AA' } this can be resolved into AA l in the direction of the force and A^A! at right angles to the direction of the force, when this is done AA l} the component of the displacement in the direction of the force, is spoken of as the displacement in the direction of the force and the work done is the product of the force into the displacement in the direction of the force. 154 MECHANICS. [CH. VIII 105. Measurement of Work. DEFINITION. When Momentum is being transferred to a body at a point which is in motion the Work done is measured by the product of the rate at which the body is gaining momentum at that point multiplied by the component of the displacement in the direction of the momentum. Or, in other words. The product of a Force into the component in its own direction of the displacement of its point of application measures the Work done. Thus let F be the force, let the point of application of the force be displaced a distance s in the direction of the force; the work done U is given by the equation U=Fs. When the actual displacement s is not in the direction of the force we resolve it, as in Figure 66 above, into s l in that direction and s 2 at right angles to the direction, in this case the work done is given by Fs^ If the angle A^AA* be called 6, then since A'A^A is a right angle we have from Figure 66, A-iA =AA' cos#. Hence 8 l = s cos 6. Therefore U=Fs cos 6. Now in this expression F cos is the component of the force in the direction of displacement. Thus we see that the work is found by multiplying together either the force and the component of the displacement in the direction of the force or the displacement and the component of the force in the direc- tion of the displacement. Hence Work = Force x component of displacement in the direction of the force. Displacement x component of the force in the direc- tion of the displacement = FscosO. 105] THE THIRD LAW OF MOTION. ENERGY. 155 If, as in Fig. 67, the displacement A A' be at right angles to the force then the displace- , ment in the direction of the force is zero, thus the work done is zero. Hence, when the motion of the point of application is at right angles to the direction p. of the force, no work is done. Again, the component of the displacement with which we are concerned may be as AA^ in A' Fig. 68 in the same direction as that in which the force acts or it may be as AA l in Fig. 69 x in the opposite direction to that A of the force. In the first case, Fi 8- 68 - Fig. 68, work is done on the body by the force, in the second, Fig. 69, work is done by the , body against the force. Thus when a body is gaining momentum at any 1 >- ^ point and when the direc- 1 tion of motion of the point Fig ' 69> is the same as that of the gain of momentum work is done on the body ; when the direction of motion of the point is oppo- site to that of the gain of momentum work is done by the body; when the direction of motion is at right angles to that of the momentum no work is done. Hence, if a body, acted on by gravity only, move in a horizontal direction with uniform speed no work is done, if the body be raised work is done on the body against its weight by the agent raising it, if the body be allowed to fall from a height work is done by its weight. When a body of mass ra is raised a height h the upward impressed force is mg and the work done is mgh. In making this statement it is supposed that the body is raised very slowly. The upward force is really greater than mg, otherwise the body would not move, but a force which is just in excess of mg will raise it ; if the excess be extremely small the motion will be exceedingly slow, and 156 MECHANICS. [CH. VIII the work done will differ from mgh by a very small quantity which we may neglect. If the body is drawn up with a run more work will be necessary. See Section 112. It should- be noticed that two factors are necessary to measure work. It depends on the product of the force or rate of transference of momentum and of the displacement. If we know the amount of work done, we cannot calculate the force unless we also know the displacement : a small force working through a large distance may do as much work as a large force working through a small distance, the work is the product of the two. Again, the work done does not depend on the time in which it is done. An engine which can raise a ton a foot in the course of a year will then have done as much work as one which raises a ton the same distance in a second ; the amount of work done depends solely on the product of the force and the displacement, and is quite independent of the time taken to do it. 106. Unit of Work. Consider now a body which is gaming F units of momentum per second; if the body be displaced a distance s in the direction of this momentum, then the work done is U, where U = F.s. If then the force or rate of gain of momentum be unity and the displacement also be unity the work done is unity. Thus the Unit of Work is done when a particle on which unit force is acting is displaced unit distance in the direction of the force. The value then of the unit of work depends on the unit of force and on the unit of length. 107. The C. G. S. Unit Work. On the c. G. s. system the unit force is a Dyne and the unit distance a Centimetre ; the c. G. s. unit of work then is done when a particle on which 1 Dyne is acting is displaced 1 Centimetre in the direction of the force, this unit of work is called an Erg. 105-109] THE THIRD LAW OF MOTION. ENERGY. 157 DEFINITION. One Erg is the work done when the point of application of a force of 1 dyne is moved 1 centimetre in the direction of the force. The Erg is a very small unit of work for a dyne is a very small unit of force, being rather less than the weight of 1 cubic millimetre of water. Hence, the erg is rather less than the work done in raising a cubic millimetre of water 1 centimetre. For this reason ten million ergs are taken as the practical c. G. s. unit of work and are called a Joule. Thus we have 1 Joule - 10,000,000 - 10 7 Ergs. 108. The P.P. S. Unit Work. On the F.P.S. system the unit force is one poundal. Unit work then is done on this system when the point of application of a force of 1 poundal is moved through 1 foot. This unit is called the Foot- poundal. DEFINITION. One Foot-poundal is the work done when the point of application of a force of 1 poundal is moved 1 foot in the direction of the force. 109. Gravitational units of work. Another unit of force used is the weight of a body whose mass is 1 gramme. When the point of application of such a force is moved 1 centimetre, I centimetre -gramme unit of work is done. Since the weight of a gramme contains g dynes we see that 1 centimetre-gramme unit of work contains g ergs. Moreover since g depends on locality the centimetre-gramme unit of work is different at different points of the Earth, the erg is the same everywhere. DEFINITION. One centimetre-gramme unit of work is done when the point of application of a force equal to the weight of 1 gramme is moved 1 centimetre in the direction of the force. 158 MECHANICS. [CH. VIII Hence, 1 centimetre-gramme unit of work = g ergs = 981 ergs, if we take g as 981 centimetres per second per second. Again, the weight of a mass of 1 pound is taken sometimes as the unit of force. When this is done the corresponding unit of work is the foot-pound. DEFINITION. One foot-pound unit is the work done when the point of application of a force equal to the weight of 1 pound is moved 1 foot in the direction of the force. Since the weight of 1 pound contains g poundals we see that one foot-pound is equal to g foot-poundals, and since in feet per second per second g is 32-2, we see that 1 foot-pound = g foot-poundals = 32*2 foot-poundals. A foot-pound of work is done when a mass of 1 pound is raised 1 foot. The work done in raising a kilogramme through one metre is 1000 x 100 or 100,000 centimetre-gramme units. Since each of these contains 981 ergs the work is 98,100,000 ergs. 11O. Rate of Working. The rate at which work is done is called Power. DEFINITION. Power is measured, when uniform, by the work done per second, when variable by the ratio of the work done in an interval of time to that interval when it is suffi- ciently small 1 . Thus the powers of the two engines mentioned in Section 105 are very different. The second has a power of 1 foot-ton or 2240 foot-pounds per second, while the first, since there are 31536000 seconds in the year, has a power of 2240/31536000 foot-pounds per second. A Horse-Power is the name given to a unit of power in common use. DEFINITION. When 550 footr-pounds of work are being done per second the rate of working is 1 Horse-Power. 1 See Section 22. 109-112] THE THIRD LAW OF MOTION. ENERGY. 159 Thus the second engine is one of rather over 4 horse-power (really 224 1 '55 horse-power). The rate of working in common use on the c.G.s. system is The Watt. DEFINITION. When 1 Joule (10 7 ergs of work) is being done per second the Rate of Working is I Watt. Thus 1 Watt is 10 7 ergs done per second. We can shew 1 that a Joule is about '737 of a foot-pound, so that a Watt is 737 foot-pounds per second. Thus a Horse-power is 746 Watts. 111. Measurement of Power. Since work is mea- sured by force multiplied by displacement, if the force be con- stant the rate of working is measured by force multiplied by rate of displacement. Now rate of displacement is velocity. Hence Power is measured by the product of a force into the velocity of its point of application measured in the direction of the force. In other words, the rate at which work is being done on a particle is the product of its rate of gain of momen- tum and the component of its velocity measured in the direction in which it is gaining momentum. Thus if F be the force impressed on a particle and v 1 the component of its velocity in the direction of F. Then Rate of Working = Fv l = Fv cos 0, if v be the velocity and the angle between the directions of v and F. 112. Expressions for Work and Power. The expressions which have been found for Work and Power may be put into various forms. Thus PROPOSITION 27. To shew that if a body of mass m acquire a velocity v in moving with constant acceleration in a straight line from rest through a space s the work done is ^mv 2 . Let F be the impressed force, a the acceleration, U the work. Then we have F - ma and v 2 = 2as. Hence U Fs = mas = ^mv 2 . Again, if the initial velocity be u and not zero, we have v 2 u 2 = 2as. Hence U = Fs = mas = L See Section 112, Example 4. 160 MECHANICS. [CH. VIII The units in which this result is given will depend on those in which 77i and s are measured, for instance, on the C.G.S. system ra is in grammes and s in centimetres; since we apply the equation F=ma the force is in dynes, hence the work Fs is in centimetre-dynes, or ergs. Thus the work required to give to a mass of m grammes a velocity of v centimetres per second is %mv 2 ergs. If the mass be in pounds, the space in feet, then the work is in foot-poundals. If we wish to use gravitation measure we must remember that a dyne is 1/g of the weight of one gramme. Hence ^mv 2 ergs is %mv 2 /g centimetre- grammes of work and \mv i foot-poundals is %mv 2 /g foot-pounds where g is 981 cm. per sec. per sec. in C.G.S. units or 32-2 feet per sec. per sec. in F.P.S. units. PROPOSITION 28. To find expressions for the rate at which work is being done on a particle of mass m, moving in a straight line with constant acceleration a. Let v be the velocity of the particle at any moment, F the force, t the time from rest, and s the space traversed. Then we have Power = Rate of Working -Fv = mav mv z 2mas /^ = = ma 2 t = = ma v 2as. t t Examples. (1) Find in the various units the work done on a mass of 1 cwt. when lifted through 100 feet. Since 1 cwt. = 112 pounds, work = 112 x 100 foot-pounds, = 112 x 100 x 32-2 foot-poundals. Also 1 lb.= 453-6 grammes, 1 foot = 30-48 cms. Thus work=112 x 453-6 x 100 x 30-48 centimetre-grammes = 112 x 453-6 x 100 x 30-48 x 981 ergs, and this reduces to about 1-519 x 10 11 ergs or 15190 Joules. (2) This same mass is allowed to fall from a height of WO feet. Cal- culate the work done by gravity (a) after it has fatten 50 feet, (b) when it has reached the ground, and determine in each case the rate at which work is being done. The rate at which the mass is gaining momentum or the force is 112 x 32-2 poundals. Thus the work which has been done in 50 feet is 112 x 32-2 x 50 foot- poundals, and in 100 feet it is twice as much, or 112x32-2x100 foot- 112] THE THIRD LAW OF MOTION. ENERGY. poundals. This last expression is the same as the work done in lifting the body to the height of 100 feet. The rate of working is the product of the velocity and the rate at which momentum is being communicated. After falling 50 feet the velocity is J% x 32-2 x 50 feet per second, and the force is 112x32-2 poundals, thus the rate of working is 112 x 32 '2 x ^/2 x 32-2 x 50 foot-poundals per second. This reduces to 2-033 x 10 5 foot-poundals per second ; if we wish to work in foot-pounds per second we have for the power 112 x ^/2 x 32-2 x 50 or 6349 foot-pounds per second. Dividing this by 550 we find for the horse-power 11*54. Thus when a body of 1 cwt. in mass has fallen freely through 50 feet, work is being done on it at the rate of 11 "54 horse-power. When the body has fallen 100 feet we shall have to substitute 100 for 50 in the above formulae, the velocity will be A/2 x 32-2 x 100 feet per second, and we find for the rate of work 8964 foot-pounds per second or 16-17 horse-power. (3) Tioo bodies 1-5 kilos and 1 kilo in mass respectively, suspended by a fine string over a pulley are free to move. Find the work done 5 seconds after motion has commenced, and the rate at which it is then being done. The acceleration is given by 1-5-1 a = f5Tl* Thus a=g/5 cm. per sec. per sec. The rate at which the system gains momentum in the downward direction is (1500 - 1000) g dynes, and this reduces to 5000 dynes. In 5 seconds the velocity (at) is 5 x g/5, or g cm. per second, and the space traversed |at 2 or 250/10 cm. Thus the work done is This reduces to 120-3 x 10 7 ergs or 120-3 Joules. The rate of working being the product of the force and the velocity is 500^ x g ergs per second, or 48-12 x 10 7 ergs per second. This is 48-12 watts. G. M. 11 162 MECHANICS. [CH. VIII (4) Find the value of an erg in foot-poundals and of a horse-power in watts. 1 erg = - centimetre-gramme unit, 981 1 centimetre = -03281 feet, 1 gramme = -002205 Ibs. 03281 x -002205 1 erg= - pr^j - foot-pounds. yoi This reduces to -00000007374 foot-pound. Hence 1 Joule is -7374 foot-pounds. Thus 1 foot-pound = -^^ Joules * = 1-356 Joules. 1 horse-power = 550 x 1-356 Joules per second = 745-8 watts = approximately f of a kilowatt. 113. Measurement of Work. In the above ex- amples we have shewn how to calculate the work done in various cases in which a particle or a body which we may treat as a particle is displaced in the line of action of the force ; we will now consider some cases in which the displace- ment is oblique to the force. Suppose then that a body of mass m is moved from a point A to a point J5, Fig. 70, where B is not vertically above A. Let the motion take place along the line AB and let us calculate the work done against gravity. The weight of the body is mg and its direction is vertical. If then we draw AC vertical and EG hori- zontal meeting in C, the displace- ment may be resolved into AC Fi 70 vertical in a direction opposite to that of the weight and EC horizontal in a direction at right angles to that of the weight. By the definition of work then the work done against the weight is mg x AC or mgh, if h is the vertical distance between A and B. 112-113] THE THIRD LAW OF MOTION. ENERGY. 163 For the purpose of calculating the work, we may look upon the displacement as a vertical one A C (= A) in which mgh units of work are done, and a horizontal one CB in which no work is done. In raising a, body from one point to another the work done against its weight depends only on the difference of level between the two points. Moreover this result is independent of the path described by the body : if it be first moved vertically downwards its weight will do work but this will be cancelled by the work done against the weight in raising the body to its original position. If the path from A to B be a broken one as shewn in Fig. 7 1 Fig. 71. Fig. 72. or a curved one as in Fig. 72, the work done is still the same, the vertical component of the whole displacement is in all cases AC or h and the work therefore is mgh. Thus when the weight of the body is the only force con- sidered, the work done in moving the body from one position to another against its weight depends only on the relative posi- tion of the two points being proportional to their vertical distance apart and not at all on the path of the particle between the points. Now it can be shewn that a similar statement is true for a very large number of actually observed cases of motion. The work done on a body, which is gaining momentum by many of the processes which occur in nature, can be shewn to depend only on the initial and final positions of the body relative to such of its surroundings as influence its motion, and not at all on the path by which it has moved from one position to the other, or on the speed with which it has traversed this path. 112 164 MECHANICS. [CH. VIII There are some cases of motion for which this statement is not true. It is the fact however of its truth in many cases which gives to Work its great importance in mechanics. Action then as used in the third law may mean the V^ork done on a body, reaction will be the work which a body can do in consequence of this, and the law states that these two, when all the forms of reaction are taken into consideration, are equal. In the Statics we shall have numerous examples of this principle and shall describe experiments arranged for its verifi- cation. 114. Motion of a body down a plane. Work. We have already found the value of the work done on a body by its weight when it is allowed to fall freely. If m be the mass of the body, h the height through which it falls, and v the velocity it acquires, then U the work done is given by U=mgh = ^mv z . Let us now consider the work done by its weight on a body which is allowed to slide down a smooth inclined plane of height h. We know that the work done against its weight when the body is lifted from the bottom to the top of the plane is mgh. PROPOSITION 29. To find the work done on a body of mass m in sliding down a smooth inclined plane of height h. Let BA, Fig. 73, be the plane making an angle a with the horizon. Draw BC vertical and AC horizontal, let R be the force between the plane and the body; the weight of the body is mg. Then since the plane is smooth the direction of R is at right angles to it, the displacement of the body is along BA, at right angles to R, hence no work is done by the force R: the only force which does work is the weight mg, the displacement of ing cos a 113-116] THE THIRD LAW OF MOTION. ENERGY. 165 the body in its direction is h ; hence the work done is nigh ; thus an amount of work mgh is done on a body which slides down a smooth plane of height h. Now let v be the velocity with which the body reaches the bottom, the acceleration parallel to the plane is g sin a : if I be the length of the plane we have a velocity v acquired in mov- ing a distance I with acceleration g sin a. Hence v 2 = 2gl sin a = 2gh, for h = I sin a. Thus mgh Jra-y 2 . Hence the work done in sliding down the smooth plane which is equal to mgh is also given by ^mv 2 . Moreover if the body be projected up the plane with velocity v it will just reach the top, and the work done against gravity will be mgh. 115. Work due to Gravity. We have just shewn that if a body be allowed to slide down a smooth inclined plane the velocity with which it reaches the bottom depends only on the height of the plane, and also that if the body starts up a second inclined plane with this same velocity it will rise to exactly the same height as that from which it fell. Work is done by gravity on the body in sliding down; the body will rise again until this same amount of work is done against gravity. Galileo discovered this relation between the velocity and height of fall and verified it by experiments which we will describe shortly. We have deduced the above results on the supposition that the body slides down a smooth flat surface, a plane, so that its path is a straight line; it can be shewn that they are true if the surface be not flat but curved so that the path is a curve, not a straight line ; for we may con- sider the curve as made up of a large number of very short straight lines inclined to each other at very small angles, the proposition is true for each of these lines ; it is also true in the limit when they become a curve, though the proof of this would require some further consideration. *116. Motion down a curve. It is difficult to make observations oil a particle sliding on a smooth curve; 166 MECHANICS. [CH. VIII no curve is perfectly smooth, and the corrections introduced by the friction are considerable. We can however easily investigate motion in which the conditions are the same as on a curve. Thus, if a heavy body be suspended by a string and allowed to swing in a vertical plane it will move in a circle, the conditions of motion will be exactly the same as though it were sliding down the circular arc; the tension of the string acting at right angles to the path takes the place of the resistance of the curve. No work is done by this tension: if h be the vertical distance through which the body rises the work done is always mgh. Consider now a body moving in such a circle ; if by any means we can fix a point in the string, the body will continue to move in a circle but the radius of the circle will be less than before. We can attain this result by allowing the motion to take place in front of a vertical wall or board ; fix a nail or peg into the wall in such a way that the string may strike the nail, which will thus become the centre of the circle in which the body will commence to move. Such an arrange- ment is shewn in Fig. 74; or again, if as in Fig. 75, we allow the string to unwrap itself off or to wrap itself on a curved surface such as FK t the path of the body will not be a THE THIRD LAW OF MOTION. ENERGY. 167 116] circle but will depend on the form of this curve. By properly adjusting this we can make the body to describe any path Fig. 75. which we like and can thus investigate the motion of a body sliding down any smooth curve. In this way Galileo shewed that the Height to which a body when moving under gravity on a smooth curve will rise depends only on the vertical height of its starting point above the lowest point of its path. EXPERIMENT 23. To shew that a body moving under gravity in an arc of a vertical circle will ascend to the same height above the lowest point of the circle as that from which it started. Suspend a heavy body an iron or lead sphere, some 6 or 8 cm. in diameter by a long flexible cord such as a piece of waterproofed fishing-line about 2 metres long. Allow it to swing in front of a vertical wall or drawing-board about a point JS, Fig. 74, and note the position C from which it starts. Observe the position D to which it rises at the end of its first swing and let A be the position it would occupy at rest. Join CD cutting BA in E, then it will be found that CD is horizontal 1 ; the ball thus rises to the same height above A as the point from which it started. 1 This statement is not quite strictly true, D will be a very little lower than G ; the difference in height being due to the friction of the air for 168 MECHANICS. [CH. VIII Now drive a nail or peg into the wall as shewn at F a point in the vertical line AB, and again start the ball from C: when the string becomes vertical the portion BF is brought to rest, the ball proceeds to move in a circle AG with F as centre and rises to the position G before its motion stops. Observation will shew that G is in the horizontal line CD, the ball though now moving in a smaller circle than previously still attains the same height. Repeat the experiment again, driving the nail in however at H a point between A and E, nearer to A than to E, Then the ball after the string has become vertical will describe a circle of radius HA about H. But this circle will not cut the horizontal line CD, the ball cannot rise to the same height as previously, it will be found that the ball completes the whole semicircle HK ; its motion after passing through the point K will depend on the position of H and the radius of the circle. The ball may continue to describe the circle winding the string upon the nail, or the string may become slack for a time and the path of the ball alter. Again, by reversing the direction of motion we may allow the ball to describe first the smaller circle with F as centre, then, when the string becomes vertical contact with F ceases, and the ball proceeds to move about A in the larger circle; it will in this case be found to rise to (7, the same height as previously, and this will be the case for all positions of F which will permit the ball to start from some point in the horizontal line CD and describe an arc of a circle about F. Thus in all these cases the velocity with which the ball starts up the circle AC must be the same. This velocity is acquired by sliding down the various circles corresponding to the different positions of F. Thus the velocity acquired by sliding down any of these circles from points in the horizontal line CD is the same. Again, take a piece of wood 1 , cut into the form of a curve which no allowance has been made. It is possible to prove this and to make an allowance if required by experimenting with balls of the same size but of different material. With the apparatus as described, however, the correction will be very small. 1 Instead of using wood the curve may be made out of a strip of sheet metal bent to the required form. 116-117] THE THIRD LAW OF MOTION. ENERGY. 169 as shewn at FK, Fig. 75, and place it so that the string after passing the vertical comes into contact with the curve. The ball will no longer describe a circle but some curve, as shewn in Fig. 75, depending on the shape of the wood. It will be found however that it still comes to rest at the point G in which its path cuts the horizontal line through G the starting point, or that conversely, if it be started from G it will rise to C. Thus the velocity acquired in sliding from rest under gravity a given vertical height down any curve is the same. *117. Velocity on a curve. The foregoing experiments enable us to find an expression for the velocity acquired in sliding down a curve ; for we have seen that this is the same whatever be the form of the curve provided only that the height through which the body moves is constant. Now in Section 116 it has been shewn that in sliding down a smooth plane the velocity v acquired is given by the equation v 2 = 2gh. Thus in sliding from rest through a vertical height h down any curve a particle acquires a velocity v given by the equation The work done in this case is mgh and this is equal to If the particle do not start from rest the corresponding equation may be found thus. Let u be the initial velocity and let it be acquired by sliding through a vertical height h'. Then the velocity v is acquired by sliding through a height h + Jt'. Hence we have v 2 = 2g ( h + h') , u? = 2gh'. Thus subtracting t>-w*=V- While for the work done we still have U= mgh = fynv 2 - %mu z . This result corresponds exactly to those found in Section 112 for a body falling freely. It can be shewn that it is true for many cases of motion which are actually observed in nature. We may state then as a result of very wide application that when the velocity of a body changes from u to v work has been done on the body and The work done is equal to %mv z - mw 2 . 170 MECHANICS. [CH. VHI Hence, Work must be done in order to increase the velocity of a body, and a body in having its velocity decreased can do work. Work is also necessary in many cases to change the position of a body, while in changing its position the body can do work. 118. Energy. Hence bodies, as we can observe them, have in some circumstances a capacity for doing work on other bodies ; by their action momentum is communicated to those other bodies, which are thereby set in motion: work is done. This capacity for doing work is called Energy. DEFINITION. The Energy of a body is its capacity for doing work, and is measured by the work which the body can do in changing to some standard state as regards its position and velocity. It is sometimes more convenient to measure the energy of a body by the work which must be done on it to bring it to its actual state from some standard condition. Thus a stone at the edge of a precipice has energy, a touch will send it over the edge to the ground below and in its fall it can do work ; we can imagine it connected with another stone just lighter than itself by a fine string passing over a pulley; as it falls it can draw this other stone up. There are of course numberless other ways in which it could do work. A body then has energy when raised above the earth. For such a body it is usual to take as the standard state referred to in the definition that in which the body is at rest on the ground. A body resting on the ground is said to have no energy. When at a height h it has energy measured by the work done to lift it to that height; this, if the mass be m, is mgh. DEFINITION. The energy of the body which depends on its position and not on its velocity is called Potential Energy. Hence the potential energy of a mass m at a height h is mgh. 117-118] THE THIRD LAW OF MOTION. ENERGY. 171 A moving body can do work in being stopped. It has energy in consequence of its motion. This form of energy is called Kinetic Energy. DEFINITION. The energy of a body which depends on its motion is called Kinetic Energy. The kinetic energy of a body is measured by the work which it can do in being brought to rest. PROPOSITION 30. To find the kinetic energy of a body which is moving with uniform velocity and is brought to rest by a uniform force. Let ??i be the mass of the body, v its velocity, F the force, s the distance the body will move before being stopped. Then the work done in stopping the body is Fs. Now we have Fma, and as = ^v z . Thus Work =Fs = mas = Hence the kinetic energy of the body is We may shew that this is a proper measure for the kinetic energy of the body in any case, and not merely when the retardation is constant. Thus we have Kinetic Energy = ^mv 2 = \mv x v = | the product of the momentum and the velocity. To prove this for a variable force we treat the force as uniform for very short intervals of time, but variable at the end of each interval. Let .Fj, F%... be the values of the force and let s 1 , s 2 ... be the short dis- tances traversed while the force has the values F lt F 2 ... etc. respectively. Then when ! etc. are very short the work actually done is Now let v i , v 2 be the velocities at the beginnings of the spaces s 1 , s 2 etc. Then during each of these spaces we may treat the retardation as uniform. 172 MECHANICS. [CH. VIII Hence Fs = mv if the body comes to rest at the end of the nth space s n . Thus adding these terms together F 1 s 1 + F 2 s z + ...F n s n =^mv 1 2 . Hence writing v for ^ as the velocity of the particle we see that the work done in stopping the body is %mv 2 . Thus ^rau 2 is the kinetic energy. 119. Change of Form of Energy. The energy of a body may change its form from potential to kinetic and vice versa. Thus a stone at the top of a cliff is at rest but has potential energy mgh', just before it strikes the ground it is moving with velocity v and has kinetic energy Jra-y 2 , but in this position it has no potential energy; we notice however that since the velocity v has been acquired in falling a distance h, we have mgh \mv 2 . Hence, The kinetic energy at the bottom is equal to the potential energy at the top. We shall find this result to be of the greatest importance : for the present let us consider some other transformations of energy. A bullet shot upwards from a gun starts with kinetic energy but with no potential energy; as it rises its kinetic energy decreases, for its velocity diminishes, but its potential energy increases, for its height becomes greater ; when at the top of its flight it is instan- taneously at rest ; its kinetic energy is zero. The height to which it rises is found by dividing the energy with which it starts by its weight, for if v be the velocity of projection, K the kinetic energy at start, and h the height the bullet reaches, then mgh = Tj'mv 2 = K. 7 K kinetic energy Hence h = = =-7- , ; mg weight or bullet as the bullet falls the potential energy becomes again trans-- formed into kinetic energy. A pendulum 1 bob when at the extremity of its swing has potential energy : if we take the equilibrium position as the standard one from which to measure, and if h be the height of the starting point above this position, then the potential energy is mgh. As the bob moves down to its equilibrium position its potential energy is diminished, its kinetic energy 1 A simple pendulum is a ball at the end of a string. 118-119] THE THIRD LAW OF MOTION. ENERGY. 173 increased until when at the lowest point the energy is all kinetic and is Jm-y 2 ; moreover the kinetic energy in this position is equal to the potential energy at starting, for ^mv* = mgh. As the pendulum passes this equilibrium position and rises again, the transformation of energy takes place in the other direction, the kinetic energy becomes potential; we see moreover that it remains unchanged in amount since the pen- dulum rises to the height from which it started. Many other examples of the transformation of energy might be given; we should find in all the same law, potential energy can be transformed into kinetic or kinetic into potential, but the gain of one is equal to the loss of the other. We will give a formal proof of this statement for one or two cases. PROPOSITION 31. To shew that the energy of a body falling freely remains unchanged during the fall. Let a body of mass m fall from a point A, Fig. 76, at a height h above the ground. Let v be its velocity when at the point P at a depth z below A, and E its total energy in this position. Then since PB is h z the height of the body is h z. Hence its potential energy is mg (h z). Its velocity is v ; hence its kinetic energy is Jra-y 2 . i p Thus E ^mv 2 + mg (h z). But the velocity v is acquired by a fall through the distance z. hz Therefore v 2 = 2z, and \mv i mgz. Hence E = mgz + mg (h z) Fig. 76. Thus the energy in any position is mgh, which is equal to the potential energy at the start : the energy remains unchanged in amount during the motion. 174 MECHANICS. [CH. VIII We may put the proof slightly differently thus. In falling a depth z the body loses potential energy mgz, it gains kinetic energy fynv' 2 , and since v z =2gz these two are equal; thus the total energy does not change. The same result follows if the body be projected down with a velocity M, instead of being dropped. After it has dropped a depth z its total energy E is given as before by But v* = u? Thus E=%m and this is the sum of the kinetic and potential energies at starting. The same result is true when a body slides down a smooth curve; for in this case the same formulae hold, h and z being the vertical distances between the various positions of the body. *120. Mutual Energy. We have thus arrived at the result that in a large number of cases of motion the energy of the moving body remains constant though it alters in form. In the cases with which we have been dealing the energy depends on the position of the body relatively to the earth. We have determined the energy on the assumption that the earth is at rest so far as the motion of the falling body is con- cerned and that the body falls to it ; strictly of course this is not true, the earth moves towards the body and the body towards the earth, their accelerations being inversely as their masses ; if we allow for this we find that it is the sum of the energies of the earth and the body which remain constant. We ought not to speak of the potential energy of the body but of the mutual potential energy of the body and the earth ; in the fall some of this energy becomes transformed into the kinetic energies of the body and the earth ; the sum of these two is equal to the loss of mutual potential energy. We are thus to look upon Energy as a quantity which we can measure and which in such cases of motion as we have been considering remains unchanged during the motion. 121. Forms of Energy. There are however other cases of motion in which energy apparently disappears. A falling stone just before reaching the ground has energy ^mv 2 , after striking the ground it is reduced to rest and has neither kinetic nor potential energy. Two masses which impinge 119-123] THE THIRD LAW OF MOTION. ENERGY. 175 directly with equal momenta and adhere have kinetic energy before impact, they are reduced to rest and apparently lose this kinetic energy by the impact. A body which is allowed to slide down a rough surface has potential energy at starting but it is soon brought to rest in a lower position; a railway train in motion has a large supply of kinetic energy; when the brakes are applied and the train stopped this kinetic energy has been dissipated. Now it can be shewn that in all these cases the energy has merely changed its form. Heat has been produced and it is found that the heat produced is proportional to the energy which has disappeared. The experiments of Joule and others have proved this; the visible kinetic energy of the moving bodies has been changed into the invisible energy of the molecules of those bodies. When the stone falls and strikes the ground, the total energy of the earth and stone remains un- changed ; when the two bodies impinge and are brought to rest, they are heated by the impact and the heat is energy equal in amount to the kinetic energy of the masses. Energy may take other forms besides the potential and kinetic energy of bodies sufficiently large for us to see. The total amount of energy existing in two or more bodies cannot be altered by any mutual action between those bodies 1 . 122. Conservation of Energy. It was said above that in many cases of motion the sum of the kinetic and potential energies of the system considered remains the same ; we have now been led to a wider generalization as the result of observation and experiment. We may in Maxwell's words state it thus. PRINCIPLE OF THE CONSERVATION OF ENERGY. The total energy of any material system is a quantity which can neither be increased nor diminished by any action between the parts of the system though it may be transformed into any of the forms of which energy is susceptible. 123. Conservation of Energy in Mechanics. When stated as above the principle of the conservation of 1 For an account of some of the experiments necessary to prove the statements made above, see Glazehrook, Heat, Chaps. I. and XIII. 176 MECHANICS. [CH. VIII energy is too wide for effective use in Mechanics, it includes the whole of Physical Science. We can however limit it and put it into a form which will be of assistance to us. Now we have seen that there are some cases of motion in which there is no change in the sum of the kinetic and potential energies, while in other cases energy is dissipated as heat or in some other form. In the first case the system considered is said to be a Conservative System; when dealing in mechanics with a conservative system the two forms of energy with which we are concerned are kinetic and potential ; the sum of these two forms is always the same. The gain of kinetic energy in any change is equal to the loss of potential energy during the same change, and vice versa. DEFINITION. In Mechanics a system is said to be Conser- vative when the amount of work necessary to bring it from any one condition to any other is always the same and does not depend upon the steps by which that change is carried out. For example, the same amount of work is necessary to raise a body from one given position to another given position, by whatever path the body be raised, provided that we are concerned only with the mutual action between the earth and the body, and the constraints introduced by smooth surfaces. This system then is a conservative system. It can in fact be shewn that if the impressed force or rate of change of momentum of each part of the system depends only on the position of that part relative to the other parts and not on its velocity, then the system is conservative. A body sliding down a rough surface is losing momentum owing to friction at a rate which depends partly on its speed and on the direction in which it is going ; when sliding down, it gains momentum from the action of the earth but loses it owing to friction ; when sliding up, both actions contribute to the loss of momentum ; this system is not conservative. In order that the principle of the conservation of energy may be of use to us in solving mechanical problems in which we deal only with the kinetic and potential energies of visible bodies it is necessary that the system considered should be a conservative one. 123-124] THE THIRD LAW OF MOTION ENERGY. 177 In a Conservative system the sum of the kinetic and potential energies of the system can only be changed by action exercised on the system from without. Assuming then this principle as established by reasoning of a general character from the fundamental laws and definitions, it may be applied to the solution of individual problems. Thus the system consisting of the earth and a heavy body moving on a smooth surface is a conservative one, the potential energy depends only on the height z above the surface of the earth ; the kinetic energy is ^mv 2 ; if the body start from rest at a height h we have %mv 2 + mgz = mgh, or v*=2g(h-z). Again, the mutual potential energy of two masses m, w^, for which Newton's law of gravitation holds, can be proved to be mm^r, where r is their distance apart. Thus if v, v^ be their velocities, their total energy is If the bodies move to a distance r' apart at which they have velocities v', v/, then the energy is and these two expressions for the energy are equal. Thus %m (v 2 - v' 2 ) + ^ (vf - v/ 2 ) 124. Unit of Energy. Since energy is measured as work the unit of work is the Unit of Energy, its actual value then will depend on the units we use for length, time and mass. On the c. G. s. absolute system the unit of energy is the Erg ; if we are measuring force in grammes' weight, the unit of energy is the Centimetre-gramme. If again we are working in feet and pounds we have on the absolute system as unit of energy the Foot-poundal, and in gravitation units the Foot-pound. When however it is stated that the kinetic energy of a moving mass is Jrav 2 , the truth of the relation F = ma has G. M. 12 178 MECHANICS. [CH. VIII been assumed ; this implies that on the c. G. s. system the force is measured in Dynes, the energy therefore is in Centimetre- Dynes or Ergs, while on the F. p. s. system the energy is in Foot-poundals. In the various statements made in the preced- ing sections, it is of course assumed that a consistent system of units is employed throughout. Thus on the C.G.S. system the statement that the kinetic energy is %mv 2 means that it is Jwu 2 ergs ; if we wish to express it in centimetre- grammes we must remember that 1 dyne is l/g (or -^ T ) of the weight of a gramme. Hence in centimetre-grammes the kinetic energy is ^mv^jg or Jm^/981. Again the kinetic energy of a mass of ra pounds moving with a velocity of v feet per second is ^mv 2 foot-poundals or ^mv 2 /g or Jmv*/32*2 foot-pounds. 125. Energy, Momentum and Force. We have thus been led to deal with two quantities, Energy and Momentum, depending on the motion of bodies: each of these is unchanged in total amount by mutual action between the bodies which make up the system, each can be transferred from one body of the system to another. Energy may exist in various forms, it may change from one form to the other in the course of the motion, but all these forms can be measured in terms of one common unit, and when so measured their sum total remains the same. Momentum we only know in the form of the product of the mass of a body and its velocity. Momentum and kinetic energy are closely connected ; to measure the kinetic energy of a particle we multiply its momen- tum by its velocity and divide by 2. Force stands in a different category to these two quantities, its amount does not in general remain unchanged during the motion; we find however that the conception of force is useful to connect together momentum and energy, and to express certain observed facts. Thus suppose a body of mass m is observed to move from rest with a uniform acceleration a, and to describe a distance s in time t. Then we find that the following three quantities mv/t, ma and ^mv 2 /s each of which we can determine by observation are equal. 124-126] THE THIRD LAW OF MOTION. ENERGY. 179 Each of these may be defined to be the impressed force: calling it F we have mv F = ma = . t s If the initial velocity be u the formulse become _, m (v - u) irav 2 - iwu 2 Jb = ma = - ' . t s We also deal with the products Ft which measures the whole change of momentum or the Impulse, and Fs which measures the Work or whole increase of kinetic energy. We find moreover that in a large number of cases these quantities depend only on the position of the body with refer- ence to surrounding objects and are quite independent of its velocity or direction of motion. To each of these quantities the" name of Force is given. These results are obtained from the observation of certain simple cases of motion. They are then generalized and by their aid the motion of bodies under very complex circumstances can be calculated. Newton founded the Science of Dynamics on the first two of the above relations. Force he defines as rate of change of momentum. In the Corollary to the third law he draws attention to the importance of the last relation and emphasizes some of its principles. In his view it follows as a mathematical consequence of the first relation ; it might of course have been taken as the starting point of the subject, basing it as has been done in the preceding pages on the fact that a body moving along a smooth curve will rise to the same height as that from which it started; this indeed had been done by Galileo, and it was by this method that Huyghens had obtained his results about the motion of pendulums. *126. Graphical construction for Work. When the impressed force is uniform the work done is Fs; when it is variable we suppose it to be uniform while the body moves over a number of very short spaces s l} s 2 etc., but to change at the end of each of these spaces. We can express this by a graphical construction identical with that used in Section 41 to determine the space traversed in terms of the time. PROPOSITION 32. To determine graphically the work done by a force. 122 180 MECHANICS. [CH. VIII Draw a line OX, Fig. 77, to represent space and let NN' be the distance traversed un- der a force F. Draw NP and N'P' at right angles to NN' to represent the force and join PP. Then the area PNN'P' is equal to Fs. But Fs is the work done on the body when moving the - Fig. 77. distance s; hence the area PNN'P' represents the work. If the force be not constant but change at the ends of the spaces s lt s z ... etc., from F to F 2) F 2 to F 3 , etc., respectively, the work done will be represented by the area consisting of a number of rectangles such as N-f^N^ NJP^N^ etc., Fig. 78. N! N 2 N, N 4 N, Fig. 78. By diminishing the spaces N^N^ N 2 N^ etc., during which we deal with the force as uniform, we get the case of a varying force; the broken line P^P^... of Fig. 78 becomes a 126] THE THIRD LAW OF MOTION. ENERGY. 181 continuous curve P^PP^ Fig. 79, and the area P^N^N^P^ repre- sents the work done by traversing the distance N^N^. Thus if we construct a figure on squared paper in which the hori- zontal divisions represent space, and the vertical divisions force, by drawing a curve Pf such that the line PN perpen- dicular to the space line may represent the force when the body has traversed a space Njy, the area P^N^'P' represents the work done during the motion. In such a diagram if the horizontal divisions represent centimetres and the vertical divisions be dynes, then the work in Ergs is given by the number of squares contained within the area. The following is an example of the method. PROPOSITION 33. To calculate the work done in stretching a spiral spring. Let ON-u Fig. 80, be the original length of the spring and suppose it stretched so that its length is ON. Then N^N" represents the extension. Now we have seen, Section 91, that the force required to extend N a spring is proportional to F* 80 the extension. Thus if PN drawn vertical at N represent the force which will just hold the spring extended to the length ON, we see that PN is always proportional to N-jN'. if we wish to extend the spring by twice N^N the force necessary to hold it in this position will be twice PN. Thus the curve corresponding, Fig. 79, which gives the force in terms of the displacement, is a straight line through the points P and Jf lt and the area which deter- mines the work is a triangle. Thus the work done in extending the spring from N^ to N is the area of the triangle Hence if we call F the force, and 8 the final extension produced under this force, we see that Work done = Area 182 MECHANICS. [CH. VIII Examples. (1) Find the energy of a mass of I cwt. while falling from a height of WO feet. The total energy at any point of the fall is equal to the potential energy at the starting-point, and this is equal to the work done in raising the body from the ground. This work has been shewn Example 1, p. 160, to be 151-9 Joules. (2) Compare (a) the momenta, (b) the kinetic energies of a bullet whose mass is 100 grammes moving with a speed of 400 metres per second, and a cannon-ball ivhose mass is 50 kilogrammes moving with a speed of 10 metres per second. Eeduce the speeds to centimetres per second and the masses to grammes, then we have (a) Momentum of the bullet = 100 x 400 x 100 = 4 x 10 6 C.G.S. units of momentum. Momentum of cannon-ball =50 x 1000 x 10 x 100 = 5 x 10 7 C.G. s. units of momentum. (b) Energy of bullet = 100 x 16 x 10 8 = 8xl0 10 ergs. Energy of cannon-ball = J50 x 1000 x 1 x 10 6 = 2-5 xlO 10 ergs. Thus the cannon-ball has the greater momentum while the bullet has the greater energy. (3) The bullet and the cannon-ball are each brought to rest with uniform retardation in 1 second. Determine the impressed forces and the distance each moves. In the case of the bullet 4 x 10 6 units of momentum are destroyed in one second, thus the impressed force is 4 x 10 6 dynes and the retardation is 4 x 10 6 /100 or 4000 cm. per sec. per sec. The distance the bullet will travel while being stopped in one second is thus \ (4000) or 2000 centimetres. For the cannon-ball 5 x 10 7 units of momentum are destroyed in one second, thus the impressed force is 5 x 10 7 dynes, and since the mass is 50000 grammes the retardation is 5 x 10 7 /5 x 10 4 or 1000 cm. per sec. per sec. The distance the cannon-ball moves while being stopped in 1 second is thus 500 centimetres. Thus if both bodies are stopped in the same time and both lose their momentum at uniform rates, the rate of loss for the cannon-ball is -\ or 12-5 times as great as that for the bullet, but the bullet moves *$ or 4 times as far as the cannon-ball. 126] THE THIRD LAW OF MOTION. ENERGY. 183 (4) A bullet 100 grammes in mass is fired from a gun the barrel of which is 75 cm. long and leaves it with a velocity of 400 metres per second ; assuming the pressure due to the powder to be uniform, find the impressed force on the bullet and the time it takes to traverse the barrel. Let F dynes be the impressed force. The kinetic energy of the bullet is 4 100 x (40000) 2 , or 8 x 10 10 ergs. It acquires this energy while moving through 75 centimetres. Hence .Fx75 = 8xl0 10 . 8x4 xlO 10 32 The momentum of the bullet is 4 x 10 6 c. G. s. units. The time during which the bullet is in the barrel is given by dividing this by the impressed force ; let it be t seconds. 4 x 10 6 x 3 3 Then ' = S2^T or three eight-hundredths of a second. (5) An engine developes 5000 horse-power while driving a ship at the rate of 25 miles an hour. Find the resistance offered to the motion. The energy supplied by the engine is employed in doing work against the resistance. The velocity of the ship is 36f feet per second, the rate at which work is being done is 550 x 5000 foot-pounds per second, and this is equal to the resistance multiplied by the velocity. Hence if R repre- sent the resistance in Ibs. weight Rx 36| = 550x5000, D 550 x 5000 x 3 __ A _ . ,, R = -- ^ - =75000 Ibs. weight. (6) A simple pendulum, the mass of ivhich is 1 kilogramme, is started from its loivest point with a velocity of 120 cm. per second ; the pendulum makes one oscillation per second and loses energy from friction and other causes at the rate of I centimetre-gramme unit per second. Determine for how long it will continue to move. The kinetic energy of the pendulum is i x 1000 x 14400 ergs, 1000 x 14400 or | x - -f - - centimetre-gramme units. yyi This reduces to 7339 centimetre-gramme units of energy. Since 1 of these units is lost each second, the pendulum will continue to move for 7339 seconds or for 2 hours 2 minutes 19 seconds. 184 MECHANICS. [CH. VIII EXAMPLES. WOEK AND ENEEGY. 1. Define Energy, and explain how to measure (a) the energy of a bullet as it leaves the muzzle of a gun, (6) the energy of a clock pendu- lum at the highest and lowest points of its swing. 2. State the principle of the Conservation of Energy as employed in mechanics, and illustrate it by some examples. By the use of this principle shew that the velocity acquired by a body falling from rest down a smooth inclined plane depends only upon the vertical height of the plane and not upon its length. 3. Distinguish between work and power. A watt is equivalent to 10 7 ergs per second ; the acceleration of gravity is 981 cm. per sec. per sec. ; find how long a kilogramme has been falling from rest when gravity is doing work upon it at the rate of one watt. 4. Calculate the momentum and the energy of (1) a bullet weighing \ an oz. moving at the rate of 1200 feet per second, (2) a mass of ^ a ton moving at the rate of 6 inches per second. Find the forces required to stop the two in T V second and the work which each would do in being 5. If a body be moving under the action of a constant force, shew that the horse-power developed by the force is proportional to the force and to the velocity of the body. 6. Distinguish between kinetic and potential energy. A pendulum consisting of a ten-gramme bob at the end of a string thirty centimetres long oscillates through a semi- circle; find its kinetic energy when the string makes an angle of 45 with the vertical. Specify the units in which your answer is expressed. 7. Shew how the second law of motion enables us to measure force and mass. What do you understand by "action" in the statement of the third law ? Illustrate your answer by some applications of the law. 8. A mass of 50 grammes moving with a velocity of 12 cm. per second overtakes and adheres to a mass of 30 grammes moving with a velocity of 4 cm. per second. Find the common velocity and calculate the total kinetic energy before and after the impact. 9. A mass of 1 cwt. is moving with a velocity of 1 foot per second. Determine the velocity of a bullet whose mass is 1 oz. when it has (1) the same momentum, and (2) the same kinetic energy as the larger mass. 10. A force equal to the weight of 10 Ibs. acts for a minute on a mass of 1 cwt. Find the momentum and energy of the mass. What is the work done by the force? 126] THE THIRD LAW OF MOTION. ENERGY. 185 11. A bullet whose mass is 1 oz. leaves the muzzle of a gun with the velocity of 1000 feet per second, find its energy in foot-poundals ; and if the length of the barrel of the gun is 3 feet, find the mean pressure ex- erted by the powder on the bullet. 12. A bullet whose mass is an ounce moving with a velocity of 2400 feet per second strikes a block of wood at rest and remains imbedded in the wood: if the resulting velocity of the block and bullet together be 16 feet per second, calculate the mass of the wood. Also calculate the energy lost when the bullet penetrates the wood, and express the result in foot-tons. 13. A man whose mass is 150 Ibs. walks up a hill of 1 in 6 at the rate of 4 miles an hour ; what fraction of a horse-power is he doing ? 14. Find the amount of work done in pushing a mass of 10 Ibs. through 5 feet up an incline of 1 in 10, neglecting friction. 15. Find the amount of horse-power transmitted by a rope passing over a wheel 10 feet in diameter which makes 1 revolution per second, the tension in the rope being 100 Ibs. 16. Point out the transformations of energy that take place during the swinging of a pendulum. State at what point of its swing a pendulum must be if its energy is half potential and half kinetic. 17. How much energy has a mass of 1 cwt. when moving at the rate of 100 yds. per sec. ? In what units is your answer expressed ? 18. A man can bicycle 12 miles an hour on a smooth road. He exerts a downward pressure of 20 Ibs. with each foot during the down- stroke, and the length of down-stroke is 12 inches. His driving wheel is 12 feet in circumference. Find the work he does per minute. 19. A shot whose mass is half a ton is fired with a velocity of 2000 feet a second from a gun whose mass is 50 tons ; neglecting the weight of the powder, find the velocity with which the gun will recoil, if mounted so that it moves without friction along a level tramway. Compare the work done on the gun with that done on the shot. 20. A cannon weighs 35 tons, and the shot half a ton. The velocity of the shot on leaving the muzzle is 1200 ft. per second ; find the velocity of the recoil of the cannon. Neglect the inertia of the gases formed by the burning of the powder. Will the effect of these gases be to reduce or to increase the recoil ? Give your reasons. 21. What is the horse-power required to fill in 3 hours a tank 9 feet deep, 20 feet long, and 10 feet wide, placed on the top of a building 60 feet in height, from a well in which the surface of the water remains con- stantly 24 feet below the ground level ? Give the answer correct to two places of decimals. (Mass of cubic foot of water =62^ Ibs.) 186 MECHANICS. [CH. VIII 22. Assuming that the resistance of a train moving along a hori- zontal railway at 35 miles per hour is equivalent to an incline of 1 in 280 find the horse-power required to take a train of 100 tons along a horizontal railway at a rate of 35 miles per hour. 23. The mass of a ship is 3000 tons ; assuming that the resistance to its motion varies as the square of the speed and that the force required to give it a speed of 1 foot per second is equal to yWinnr tne weight of the ship, find the horse-power requisite to propel it at the rate of 30 feet per second. 24. Define the kinetic energy and the potential energy of a system. A fine string passes through two small fixed rings, A and B in the same horizontal plane, and carries equal weights at its ends, hanging freely from A and B. If a third equal weight is attached to the middle point of the portion AB of the string, and is let go, prove that it will descend to a depth equal to two-thirds of the length AB below AB, and will then ascend again. 25. A mass of 4 cwt. falls from a height of 10 feet on to an inelastic pile of 12 cwt. Supposing the mean resistance to the penetration of the pile to be 1| tons weight, determine the distance it is driven at each blow. 26. A smooth wedge of mass M and angle a rests upon a horizontal plane ; another mass m is placed upon its slant surface, and the system begins to move. Write down (1) the equation of energy, (2) the equation of linear mo- mentum. 27. I strike an anvil with a given hammer, and its velocity on reaching the anvil is the same as that with which it reaches a piece of red-hot iron on the anvil. Will the impulse be the same in the two cases ? What quantities must be known in order to compare impulses ? 28. A bullet weighing 1 oz. leaves the mouth of a rifle whose barrel is 4 feet long with the velocity of 1000 ft. per second. Find the mean force on the bullet, neglecting the friction against the sides of the barrel. 29. A cannon-ball whose mass is 1 cwt. moving with a velocity of 25 yds. per sec. penetrates to a depth of 10 feet into a sandbank. Find the average pressure on the sand. 30. The erg and foot-pound are both units of work: a horse-power is 33,000 foot-pounds per minute; how many ergs per hour would this be? [1 inch = 2-54 centimetres. 1 Ib. = 453-6 grammes.] 126] THE THIRD LAW OF MOTION. ENERGY. 187 31. In a system of distributing power by means of water at a high pressure, the pressure of water is 2000 Ibs. per square inch. How many cubic feet must be used per hour to supply 10 H.-P. (1 H.-P. =33,000 foot- pounds of work per minute) assuming no power to be lost ? 32. A mass of ra pounds is raised up a plane inclined at an angle of 30 to the horizon, and of length I, and reaches the top with a velocity v. Shew that the work done is IK) foot-pounds. *CHAPTER IX. CURVILINEAR MOTION UNDER GRAVITY. 127. Projectiles. When a ball is thrown in the air it does not move in a straight line \ a very little observa- tion is sufficient to shew this; moreover its velocity is con- tinually changing. We can deduce the form of the path from the laws of motion, thus Let us suppose, in order to simplify the problem, that the body is projected in a horizontal direction with a velocity u\ the only accelera- tion which it acquires is g in a vertical direction due to the action of the earth, and it is not difficult to determine the path of the body under these circumstances. We will however in the first instance investigate the motion by the aid of experiment. We have seen that two bodies whatever be their masses when dropped together fall at the same rate. We wish now to shew that, if one of the bodies be projected in a horizontal direction with any velocity while the other is allowed to drop simultaneously from the same height, the two will fall at the same rate and reach the ground together. We may shew this roughly by rolling a ball rapidly along a table ; on leaving the table it will describe a curve in the air ; if now at the moment the first ball leaves the table a second be dropped from the same height the two will reach the ground together. The same fact is better shewn by the aid of an arrangement of apparatus devised by Sir Robert Ball 1 . 1 Experimental Mechanics, Section 511. 127] CURVILINEAR MOTION UNDER GRAVITY. 189 EXPERIMENT 24. To shew that the time of fall of a body is independent of its horizontal velocity. In Fig. 81, ABC is a piece of wood about 2 '5 cm. thick the upper edge of which is curved as shewn. A strip of thin Fig. 81. brass is screwed to each face of the board, the edges of the brass strips being also curved : care must be taken that there shall be no metallic connection between the brass strips. Thus the upper edge of the wooden board forms a kind of groove with brass sides. The brass strips are connected to binding screws. On resting a braes ball on the strips as shewn in the figure, an electric current can pass from one strip to the other through the ball. One binding screw is connected to a battery, the other to the electromagnet described in Section 63 \ a wire also 190 MECHANICS. [CH. IX passes from the electromagnet to the second pole of the battery. When the ball is on the groove the circuit is complete and the magnet is made ; thus it can support a small iron ball. When the brass ball rolls off the groove the circuit is broken between the strips and at the same moment the iron ball drops. If the groove be fixed in such a position that its direction at B is horizontal the brass ball is projected in a horizontal direction. If also the electromagnet D be fixed at the same height as the point of projection J3, then the iron ball is dropped and the brass ball projected horizontally at the same moment from the same height. The velocity with which the brass ball starts can be varied by allowing it to roll 1 down the groove AB from various positions, or by "placing a spring in the groove and projecting the ball forward by its aid. If this plan be adopted a straight horizontal groove will do as well as the curved one shewn. Arrange the apparatus as described and, starting the brass ball from various points in the groove, observe the times at which the two balls reach the floor. It will be found that whatever be the height of the starting point and whatever be the velocity of projection the two always strike the floor simultaneously. Thus the downward acceleration of the two balls is the same; the brass ball although it starts with a horizontal velocity, depending on the distance it has rolled down the groove, gains in each second the same vertical velocity as the iron ball ; in the first second it will fall through \g centimetres, in the second through g centimetres, and in t seconds \g& cen- timetres. The vertical velocity is independent of the horizontal and is the same as that of a body allowed to drop freely. 1 If the ball is not quite spherical then in rolling down it may happen that contact between the ball and the strip is broken, and the iron ball is allowed to drop too soon. This may be avoided by using instead of a ball a cylindrical piece of brass or zinc which slides down on the brass strips ; the friction however in this case is greater and the ball does not start with so great a velocity. 127] CURVILINEAR MOTION UNDER GRAVITY. 191 EXPERIMENT 25. To describe a parabola. Take a straight lath or scale about a metre long ; divide it into equal distances each 10cm. in length and from each point of division suspend by a piece of fine string or thread a small bullet or weight. Adjust the first piece of string to some convenient length, say 3 centimetres, make the second 3 x 2 2 cm., the third 3 x 3 2 cm., the fourth 3 x 4 2 and so on, so that the lengths of any two strings are proportional to the squares of their distances from the end of the rod. Fig. 82. Thus, in Fig. 82, A B is the lath, M 19 M 2 , M 3 , etc. the points of division, P lt P z etc. the bullets, and we have P^M^ = 3 cm., and so on. Hold the lath as shewn in Fig. 82 against a vertical black board or sheet of drawing-paper, mark the positions P lt P 2 etc. 192 MECHANICS. [CH. IX of the bullets. The points so found lie on a curve called a parabola. If we suppose that each of the 10-centimetre divisions is subdivided into (say) 10 parts, and that threads with bullets are hung from these in such a way that the lengths of the threads may follow the same law as before, the bullets will be practically in contact and the curve which passes through them will be a parabola. The curve however can be constructed with sufficient accuracy for our purposes by drawing a free- hand curve through the points Pf z - . - Again, draw a vertical line AN as in Fig. 83 from A, the end of the lath from which the divisions are reckoned, and from P l draw P 1 A r 1 parallel to the lath to meet AN in N^. Then and by construction PM l is proportional to AM 2 . Hence P^ 2 is proportional to AN lf In the figure the lath is horizontal, this however is not necessary ; in whatever direction the lath be held the balls still lie on a parabola ; the size of the curve will depend on the inclination of the lath. EXPERIMENT 26. To determine the path of a body projected in a horizontal direction and to shew that it is a parabola. Arrange the grooved board described in Experiment 24 in front of a vertical blackboard as shewn in Fig. 83 in such a way that the ball after sliding down the groove may start from A in a horizontal direction and fall in a vertical plane parallel to the board. Make a mark C on the groove and in all the experiments start the ball from this mark. Allow the ball to roll down the groove and watch its path. Fix to the board with drawing-pins a number of paper hoops so that the ball in its path passes through each of them ; the proper position of the uppermost hoop is first found by trial, then that of the next below, and so on, the ball being started in each case from the same point C. Mark on the board the positions of the centres of the hoops, remove them and draw with a free hand a curve starting from A and passing through the various marks P lt P 2 ____ Draw a horizontal line from A, and vertical lines P^M^ P 2 M 2 ... from the marks 127] CURVILINEAR MOTION UNDER GRAVITY. 193 to meet it in M lt M s . Measure the horizontal distances AM V AM Z ..., and the vertical distances P l M l etc. Write down the squares of AM l} AM 2 etc., and obtain the quotients, given by dividing each square such as AM* t by Fig. 83. the corresponding vertical distance ; it will be found that these quotients are all approximately equal. But the curve which has this property is the parabola. Thus the path is a parabola. Again, measure the vertical height above A of the point C from which the ball starts ; let it be a. Then it will be found that the constant ratio of A M 2 to PM is equal to 4c&. Thus we have G. M. 13 194 MECHANICS. [CH. IX But if u be the horizontal velocity with which the ball leaves the groove we have seen that u is acquired by falling down a height a, hence u 2 = 2ga. Therefore a = u?/2g, and AM 2 = PM. 9 The curve described by P is a parabola, the point A is called the vertex of the parabola, and the quantity 2u 2 /g is its Latus Rectum. Thus, when a body is projected in a horizontal direction its path is a Parabola. Again, if we suppose the motion of the body at any point to be re- versed, it will proceed to describe the same path in the reverse direction ; thus when projected obliquely its path will still be a parabola. This may be verified by construction in a similar way by arranging the grooved board so that its direction is not horizontal. The path of the drops of water in a water-jet is the same as that of a body projected under gravity, each little particle of water follows the same course as that which would be taken by the body if projected with the velocity with which it starts. By placing a lamp at some distance from such a jet its shadow can be thrown on a white screen placed behind it, the path of the jet can thus be traced and measurements made on it as on the curve drawn as described above. For further details see Glazebrook and Shaw, Practical Physics, Section C. 128. Motion of a Particle projected under Gravity. We have shewn by the result of Experiment 24 that the vertical motion of a falling body is independent of its horizontal velocity. A body which has initially no vertical velocity will in t seconds fall a distance ^gt 2 , whether it start from rest or be projected horizontally. If the body be projected obliquely, its velocity has a vertical as well as a horizontal component; let v be the upward vertical component, u the horizontal component. Then during t seconds the velocity v will have carried the body a distance vt upwards, while owing to the vertical acceleration g the displacement will be \gf downwards. Thus h, the actual height above the point of projection, is given by h = vt- 127-128] CURVILINEAR MOTION UNDER GRAVITY. 195 In the same time the horizontal velocity u will have carried it a distance ut, and it has no horizontal acceleration, hence k the horizontal displacement is given by k = ut. If the actual velocity of projection be U and the direction of motion be inclined at an angle a to the horizon, then we have u = U cos a, v = U sin a. Hence h = Ut sin a - J^ 2 , k=Ut cos a. We will now shew how to prove that the path is a parabola so long as the resistance of the air is neglected. PROPOSITION 34. A particle is projected with a given velocity and at a given inclination to the horizon ; to shew that its path is a parabola. Let P y Fig. 84, be the point of projection and PT, making an angle a with the horizontal line Px, the direction of projection. Let U be the velocity of projection along PT. Fig. 84. 132 196 MECHANICS. [CH. IX The only acceleration which the particle has is vertical and is equal to g. The motion will therefore take place in the vertical plane through PT. To find the position of the particle after t seconds, make PT equal to Ut and from T draw TQ vertically downwards and equal to \gt*. Draw PR vertical and equal to TQ and join QR. Then if the particle had no acceleration it would move uniformly along PT and at the end of t seconds would be at T. Again, if it had initially no velocity it would in t seconds fall vertically and reach the point R. In the actual circumstances the displacement in each of these two directions is, in accordance with the second law, independent of that in the other, the particle is displaced in the direction PT just as far as it would be if it had no vertical acceleration; it is displaced verti- cally just as far as it would be if it had initially no velocity along PT. Thus, at the end of t seconds, PT still represents the displacement due to the initial velocity, while PR or TQ represent the displacement due to the acceleration. Hence the particle is at Q. Now QT=$gf, PT= Ut. f)OT PT Hence t* = ^L, t = , ,, 9 77 2 Therefore PT*=. QT. 9 Now U and g are both constant, therefore the ratio of PT 2 to QT is a constant, and this (Experiment 25) is the funda- mental property of a parabola. Thus the point Q always lies on a parabola. 129. Properties of the path of a Projectile. We will now proceed to investigate various properties of the motion of a projectile. (1) To find the directrix of the parabola. Draw PK vertical and equal to U 2 /2g. Then the velocity at P would be acquired by falling through a height KP. A horizontal line KX drawn 128-129] CURVILINEAR MOTION UNDER GRAVITY. 197 through K is called the directrix. The velocity at any point of the parabola is that due to the fall from the directrix. For if PK is equal to J5T, and if h be the height above P of the particle when at Q, since the energy remains constant we have, if V denote the velocity at Q, Hence V*=2g (H-) = 2gQL if the vertical QT meet the directrix in L. (2) To find the vertical and horizontal components of the velocity at any time. The horizontal velocity initially is f/eosa, and since there is no horizontal acceleration it remains unchanged. The vertical velocity initially is U sin a, and in t seconds under the downward vertical acceleration g an additional vertical velocity -gt is acquired. Hence if u, v represent the components of the velocity at any time u= 7 cos a, v= Ueina-gt. (3) To find the direction of motion at any time. If at any time t the particle be moving with velocity V in a direction making an angle 6 with the horizon, we have = w= 17cosa, Fsin 6 = v= Usin a gt. Hence F 2 U sin a - tan0 = t/cosg (4) To find the position of the particle at any time. Let h be the distance of the particle above P, k its horizontal distance from P. Then h= Utsin a-^gt 2 , k=Ut cos a. (5) To find the time to the vertex. At the vertex A (Fig. 84) the motion is horizontal, the vertical velocity therefore is zero. Hence if t t is the time to the vertex 7 sin a gt i = t _ U sing i~ 7, 198 MECHANICS. [CH. IX (6) To find the height of the vertex. If the height of the vertex AN^ (Fig. 84) be h lt then h is the height of the particle at time t when the vertical velocity is zero. Hence h Ut^ sin a - ^gt-f. But Therefore 9 I/ 2 sin 2 a 9 (7) To find the distance of the vertex from the directrix. At the vertex the velocity is horizontal and is equal to u cos a. Hence if AX be perpendicular from A on the vertex, we have I7 2 cos 2 a In a parabola four times the distance between the vertex and the directrix is the latus rectum. Hence the latus rectum is 2U 2 cos 2 afg. In Section 125 we found the value 2w 2 /# for the latus rectum ; it must be remembered that u is the constant horizontal velocity which is equal to U cos a. Thus the two formulas are the same. A line through the vertex at right angles to the directrix is called the axis of the parabola. A point S on this line at a distance from the vertex equal to AX is called the focus. (8) To find the horizontal distance between the vertex and the point of projection. Let the distance PN (Fig, 84) be ^ . Then fcj represents the horizontal distance which the particle has moved in time ^ . J7 2 cos a sin a U sin a Hence ft, = U cos at-, = = u- , 9 9 writing u for the constant horizontal velocity U cos a. U* sin 2 a Moreover h t = . Hence fc, 2 =- Thus PN* =4AX. AN, and the ratio of PN 2 to AN is the same for all points. (Cf. Section 125.) 129] CURVILINEAR MOTION UNDER GRAVITY. 199 (9) To find the time at which the particle reaches the ground again. Let the time be t z . Then at time < 2 the height h of the particle is zero. Hence = Ut% sin o - \g t 2 2 . Therefore * 2 = which gives the starting-point, or _2*7sina_ 2t 2 9 which gives the time to the point P'. Thus the particle takes as long to descend from A to P' as to rise from P to A. The value of 2 is known as the time of flight. (10) To find the range on the horizontal plane. The range PP' is the horizontal distance which the particle moves in time t%. f 2U 2 sin a cos a Hence Range = Ut. 2 cos a = = . U sin a = 9 9 Hence for a given velocity of projection the range is greatest if sin 2a is greatest, that is if 2a = 90, a = 45. A number of other propositions on parabolic motion might be given ; for these th'b reader is referred to Loney, Elementary Dynamics. Examples. (1) Shew by finding an expression for the velocity of a projectile that it is equal to that due to a fall from the directrix. If H be the distance between the directrix and the point of projection, h the height of the particle at time t above the point of projection. We have U 2 = 2gH, h=Utsina- %gt*. F 2 = U 2 cos 2 a + ( U sin a - gt) 2 U 2 -2g(Ut sin a - U 2 -2gh Thus V is the velocity due to a fall from the directrix. (2) Find the time at which a particle projected with velocity U in direction a will strike a plane through the point of projection inclined at an angle j3. We have with the same notation k=Ut cos a, 200 MECHANICS. [CH. IX Hence U8ina.-%gt=Ucosatau |8. f)TT Therefore t = (sin a - cos a tan /3) _2t7sin(a-/3) g cos/3 (3) jPmd fte angle at which a particle must be projected so as to hit a given point if the velocity of projection be given. _ Let h, k be the vertical and horizontal distances of the point from the point of projection and t the time to the point. i. Then t This gives us a quadratic equation to find a, and corresponding to the two roots we have two directions of projection. (4) A stone is thrown from a cliff 112 feet high with a velocity of 192 feet per second in a direction making an angle o/30 with the horizon; find where it strikes the ground. The vertical velocity is 192 sin 30 or 96 feet per second, the horizontal velocity is 96 */3 feet per second. "When projected up with a velocity of 96 feet per second it will be at a distance of 112 feet beloiv its point of projection after a time T given by = 96T-16T 2 , T 2 -6!T-7 = 0. Hence T=7, orT=-l. Thus the stone will strike the ground 7 seconds after it started ; we also see that the stone might have been projected from the ground with proper velocity 1 second before it started from the cliff ; it would then have passed the edge of the cliff at the moment of starting with a velocity of 192 feet per second at an inclination of 30 to the horizon. Since the horizontal velocity of the stone is 96 >/3 feet per second, the horizontal distance from the cliff of the point at which it strikes the ground after 7 seconds will be 7 x 96^/3 feet. 129-130] CURVILINEAR MOTION UNDER GRAVITY. 201 (5) A man can just throw a stone 392 feet. Find the velocity with which he throws it; find also how high it ivill rise and determine the time of flight. The range is greatest when the angle of projection is 45, and then the range is U 2 lg. U* Hence = 392 feet. 9 Hence U 2 = 32x392, whence U=ll2 feet per second. The greatest height to which it rises is |U" 2 sin 2 a/#, and sin 2 a = |. Hence greatest height is 32 x 392/4 x 32, or 98 feet. The time of flight (2*7 sin a/0) is 112 N /2/32, or 7/>/2 seconds. ISO. The Simple Pendulum. A heavy particle sus- pended by a fine flexible string constitutes a simple pendulum. In practice we cannot of course arrange that the string should be perfectly flexible or that the body which is suspended should be a particle. For most purposes however a spherical ball of wood or metal suspended by a piece of fine string such as a waterproofed fishing-line will serve as a simple pendulum \ the ball may be conveniently from 5 to 7 cm. in diameter. If such a pendulum be drawn aside and then let go, it commences to oscillate backwards and forwards in a vertical plane. The bob of the pendulum moves in an arc of a circle. The distance from the point of suspension to the centre of the suspended sphere is called the length of the pendulum, and we Fig. 85. 202 MECHANICS. [CH. IX treat the motion as though the bob were a heavy particle con- centrated at its centre. Such a pendulum is shewn in Fig. 85. The length of the arc A C through which the pendulum swings, measured from its lowest position, is known as the Amplitude of the vibration. Such a pendulum when once started loses its energy very slowly and will continue to swing for a long time ; its amplitude gradually grows less but the decrease is very slow. Now Galileo shewed that, if the amplitude of oscillation of a pendulum be not large, its time of swing is constant; thus if at starting it takes the pendulum 1 second to move from C through A to D and back to (7, it will continue throughout its motion to take 1 second for each such oscilla- tion. The uniform rate of a clock depends on this property of a pendulum. In order to verify it completely we should need to start a long heavy pendulum and count the number of oscillations in the interval between some two astronomical occurrences, which are always separated by a constant interval of time, such as the transit across the meridian of two known stars. If we find that during any such interval the pendulum makes a number of oscillations which is proportional to the length of the interval, we infer that the duration of each oscillation is a constant number of seconds. We thus arrive at the result that in a given locality the time of swing of a given pendulum is independent of the amplitude. But we can shew more than this, for we find also that the time of swing does not depend on the mass of the bob. The bob may be of any material, provided only the length of the pendulum remains unchanged, and the conditions such that we may treat the motion as that of a simple pendulum, the time of swing is the same. Newton called attention to this and made numerous experiments to verify the fact. EXPERIMENT 27. To shew that the time of swing of a simple pendulum is independent of the mass of the bob. Take a number of spheres of about the same size, but of dif- ferent materials, and suspend them all side by side from some steady support, such as a horizontal bar, by strings of the same length (say 1 metre). This is most easily done by having an eye screwed into each sphere through which the string can 130-131] CURVILINEAR MOTION UNDER GRAVITY. 203 pass. The string is passed through a small hole in a small wooden block, it is then threaded through the eye of the bob and the end is secured to another hole in the same small block. The length of the string can then be adjusted by sliding the block up and down in the same way as the stay-ropes of a tent are tightened, and the friction will hold the block in any position. Adjust the strings of each pendulum carefully till all are of the same length, then start the pendulums swinging simultaneously. To do this, place a board against the spheres and push them all aside to the same extent. On withdrawing the board, the pendulums all start together and will continue if their lengths have been carefully adjusted to keep time for a large number of oscillations ; thus the time of swing is inde- pendent of the mass of the bob. If this experiment be continued for some time it will be found that the lighter spheres begin to lag behind. This, as Newton shewed, is due to the resistance of the air; if the experiment were performed in a vacuum no such effect would be noticed, the effect of the resistance of the air may be allowed for by observing the decrease that takes place in the amplitude of successive swings. When this is done it is found that the mass of the bob does not affect the time of swing. 131. Relation between Weight and Mass. This result affords a more accurate verification of the law that the weight of a body in a given locality is proportional to its mass than can be obtained from observations on a falling body a pendulum is practically a falling body whose motion we can observe for a long period of time. Consider now two of the pendulums. At any moment their velocities and accelerations are respectively the same. Their masses however are different, the force acting in each case is the same definite fraction of the weight of either pendulum, a fraction which depends on the inclination of the pendulum string to the vertical at that moment. Since the acceleration is the same the ratio of the force acting on each pendulum to the mass of the pendulum is the same for the two ; hence the ratio of the weight of the pendulum to its mass is the same for all the pendulums, but this ratio measures = 24-84 cm. 4?H A "seconds pendulum," as the phrase is generally employed, means one which passes through its equilibrium position once a second. Its time of swing therefore is 2 seconds and its length is four times the above. Hence Length of a seconds pendulum under the above conditions = 99 -36 cm. The actual value of g in London is 981-17, and the length of the seconds pendulum is 99 '413 cm. 208 MECHANICS. [CH. IX (2) The value of g at the equator is 978-1, in London 981-17, and at the pole 983-11; the length of the seconds pendulum is 99*413 cm. in London ; find its value at the pole and the equator. Since the time of swing is to be constant, the lengths are proportional to the values of g. Hence Q78"l Length at Equator =^x 99-413 =99-103 cm. yoi*i / qoo.li Length at Pole=^ x 99-413 = 99-610 cm. yoi-17 (3) A pendulum beats seconds at London; find with the above data its time of swing at the equator. The length remains the same; thus the periods are inversely pro- portional to the square roots of the value of g. Hence Period at Equator = = 1-00156 seconds. EXAMPLES. PKOJECTILES. 1. Determine the velocity with which a stone must be projected at an angle of 45 to the horizon in order that the range may be 100 yards. 2. A stone is projected with a velocity of 50 feet per second in a direction making an angle with the horizon, where tan = f . Find the greatest height it attains. 3. If v is the vertical component of the velocity of projection of a particle, prove that the greatest height it attains above the horizontal plane through the starting-point is . 4. A stone is projected with a velocity of 60 feet per second in a direction making an angle with the horizon, where tan # = f. Find its range on a horizontal plane through the starting-point, and the time of flight. 5. A cannon-ball is observed to strike the surface of the sea, to rebound, and to strike the surface again 2000 yards further on after 6 seconds. Find the horizontal velocity of the shot during the rebound and the greatest height the shot attains. 134] CURVILINEAR MOTION UNDER GRAVITY. 209 6. From the top of a vertical tower, whose height above the hori- zontal plane on which it stands is *-g feet, a heavy particle is projected with a velocity whose upward vertical and horizontal components are Qg and Sg feet per second respectively ; find the time of flight and the distance from the base of the tower at which the particle will strike the ground, g being the acceleration due to gravity. 7. A particle is projected with a velocity V in a direction making an angle a with the horizon, under the action of gravity ; find the highest point to which the particle will rise and the range on the horizontal plane passing through the point. If the velocity of projection be 880 feet per second, find in miles the greatest range on the plane, supposing the acceleration due to gravity to be 32 feet per second. 8. A shot is fired horizontally from the top of a tower with a velocity equal to the vertical component of its velocity when it reaches the ground ; shew that it reaches the ground at a distance from the foot of the tower equal to twice the height of the tower. 9. A particle is projected with a velocity u in a direction making an angle 6 with the horizontal. Find the range on the horizontal plane through the point of projection. If the range is 300 feet, and the time of flight 5 seconds, find the velocity of projection. 10. A particle is projected with a velocity u in a direction making an angle 6 with the vertical. Find the greatest height to which it will rise above the horizontal plane through the point of projection. If the greatest height is 49 feet and the velocity at the highest point is 42 feet per second, find the velocity of projection. 11. If a particle be projected horizontally, with a given velocity u, along the surface of a smooth plane, inclined at an angle a to the hori- zontal, what will be the latus rectum of its path ? 12. A bullet is fired from a gun at an elevation of 45, with an initial velocity of 840 feet per second. Find the range on a horizontal plane through the point of projection. If the bullet strikes a bird which rose vertically with uniform velocity from a point on the ground 500 yards distant from the gun at the instant when the shot was fired, find the velocity of the bird. 13. Prove that the range of a projectile on a horizontal plane through the point of projection is 2uv/g, where u and v are the horizontal and vertical components of the velocity of projection, and g is the acceleration due to gravity. 14. A particle is projected with velocity V at an angle a with the liorizon ; find the height of the focus of the path. G. M. 14 210 MECHANICS. [CH. IX 15. A ball rolls from rest at the top of the roof of a house and finally strikes the ground at a distance from the house equal to its breadth. If the roof on both sides of the house makes an angle of 30 with the horizon, prove that the height of the house to the eaves of the roof is three times the sloping distance from the eaves to the top of the roof. [The top of the roof is in the middle of the house.] 16. Shew that if lines be drawn from a point to represent the velocities of a projectile at different instants their other extremities will lie on a vertical line. 17. A vertical line is divided into a number of equal parts A^A Z , A%AZ > -^3-^4 e ^ c - Shew that if a particle be projected from in the ver- tical plane through the line, OA 1J OA 2 , OA% etc. will meet its path in points such that the times of flight from each to the next are all the same. 18. Find the range of a projectile on a horizontal plane passing through the point of projection ; and prove that when the velocity of pro- jection has a given value u there are two possible directions of projection such that the range has a given value R ; provided R is less than a certain distance. 19. Prove that in this case the difference of the greatest heights: ITJA attained in the two paths is \ .1 - 2 - JR 2 . *CHAPTER X. COLLISION. 135. Impact. We have seen already that experi- ment proves that when two bodies impinge there is neither loss nor gain of momentum, and it has been pointed out that in order to calculate the motion when the two impinging bodies do not adhere some further experimental result is necessary. * Newton's experiments already referred to afford the necessary information. He proved by measuring with the aid of the apparatus shewn in Fig. 51, Section 50, that when two spheres impinge directly their relative velocity after impact always bears a fixed ratio to that before impact; thus, if u, u are the velocities of the spheres before impact and v, v' after impact, all estimated in the same direction, the relative velocities are u-u' and v v respectively. Now with Newton's apparatus the velocities can be measured and it is shewn as the result of experiments that the ratio of v - v to u u is always the same for balls of the same two materials; it does not depend on the masses of the balls but only on the substances of which they are composed. This constant ratio is found to be a negative fraction, that is to say if u is greater than u', so that u - u' is positive, then v is less than v' or v v is negative, the ball which is struck has the greater velocity after impact ; the ratio is never greater than unity ; in some cases the balls separate with the same relative velocity as that with which they impinged though the 142 212 MECHANICS. [CH. X direction of this velocity is changed ; in general however the relative velocity is reduced by the impact. If we denote the ratio of v - v to u u' by - e, then the quantity e is never greater than unity : it is called the Coefficient of Restitu- tion. For certain pairs of substances the coefficient of resti- tution is unity. In general we have v - v' , = -e. u-u We can now make use of these two results of experiment to solve some problems on impact. PROPOSITION 35. Two balls impinge directly; to find their velocities after impact in terms of the velocities before impact, their masses, and the coefficient of restitution. Let m, m be the masses of the two balls A, B, Fig. 87, Fig. 87. u, u' their velocities before impact, v, v their velocities after impact, and e the coefficient of restitution. We suppose that initially the two balls are moving in the same direction AB, the velocity (u) of A being greater than that of B. We know that the momentum is unchanged by the impact and that the ratio of the relative velocity after impact to that before impact is e. Thus we have mv + m'v' = mu + m'u, v - v' = e (u - u') - - eu + eu'. 135] COLLISION. 213 Hence, solving these equations (m + m') v = (m - em') u + m' (1 + e) u'> (m + m')v = m (1 + e) u + (m' - em) u. These two equations give the values of v and v' when u and u' are known ; they are simplified if e is unity, in which case the balls are said to be perfectly elastic, or if e is zero, in which case the balls adhere and move with the common velocity (mu + m'u')/(m + m'). In the latter case with which we have already dealt at length the balls are said to be inelastic. (See Section 58.) PROPOSITION 36. To find the Impulse of an Impact. Let the Impulse of the motion be 7. Then we have / = mv - mu = (m'v m'u'). Now subtracting (m + m')u from both sides of the equation which gives v we have (m + m') (v-u) = (m- em) u-(m + m') u + m' (1 + e) u. m' (1 + e) . , x Hence v u = s f (u u). Therefore /="*' + , e \u' -u}. m + m PROPOSITION 37. To find the velocity of rebound after direct impact on a fixed surface. We may deal with this problem by supposing the mass of the second ball to be very large and its initial velocity to be zero. It will remain at rest : we must put m infinite and u' zero in the equations and we get v eu, v' 0. Or we may obtain this from Newton's second experimental result, the relative velocity before impact is u, after impact it is v, and we have v = - eu. We cannot use the first experimental result, for though u' is zero, m' is very large, and we do not know what the value of m'u' is. 214 MECHANICS. [CH. X Examples. In working examples it is much the best plan always to have recourse to the two fundamental principles and not to quote the results for v and v'. Thus (1) A mass of 1 kilogramme moving with a velocity of 50 cm. per second impinges directly on a mass o/lO kilogrammes at rest; the coefficient of restitution is % . Find the velocities after impact. Let the velocities after impact be v and v' respectively in centimetres per second. The momentum before impact is 50000 gramme-centimetre units, and the relative velocity 50 cm. per sec. Hence lOOOz; + lOOOOi/ = 50000, v-v'= -50= -25. Hence IQv' + v = 50, v'=|~f =6 T 9 T cm. per sec., v= - (25 - 6^) = - 18^ cm. per sec. Thus the 1 kilogramme ball returns with a velocity of 18 T 2 T cm. per sec., the 10 kilogramme ball moves forward with a velocity of 6 T 9 T cm. per sec. (2) A ball whose mass is 1 Ib. moving with a velocity of 10 feet per second impinges directly on another ball moving in the opposite direction with a velocity of 5 feet per second. The first ball is observed after impact to continue to move on with a velocity of 5 feet per second, and the coefficient of restitution is f . Find the mass and the velocity after impact of the second ball. Let the mass be m r Ibs. and the velocity estimated in the direction in which the 1 Ib. ball moves v' ft. per sec. The momentum before impact in this direction is 10 - 5m'. The relative velocity before impact is 10 + 5 or 15. Hence 5 + m'v' = 10 - 5m', 5-t>'=-|15=-10. Hence 5 + 15m' = 10 - 5m', or 20wi' = 5, Thus the mass of the second ball is J of a Ib. and it moves after impact in the direction opposite to that of its initial motion with a velocity of 15 feet per second. 135-136] COLLISION. 215 (3) A ball drops from a height of 25 feet on to a horizontal surface and rebounds, the coefficient of restitution is f ; jind how high the ball will rise after striking the surface three times. Let the velocity with which it strikes the surface be u ft. per sec. Then since the velocity is acquired by falling through 25 feet u= ^2g.25 = A/64 x 25 =40 feet per sec. After 1 rebound this becomes f of 40, or 30 feet per second. The ball rises and then falls again, striking the ground with a velocity of 30 feet per second. After this second impact the velocity becomes f of 30, and after the third impact it becomes | x | x 30 or | x 15. The height to which the ball will rise then is Thus the height required is equal to [ , or about 4*448 feet. V 64 y 136. Energy and Impact. PROPOSITION 38. To find the work done when two bodies impinge directly. The kinetic energy of each ball is changed by the impact ; thus work is done, and the work done on either ball is equal to its gain of kinetic energy. Hence the work done on the ball A = \m (-y u) (v + u) =i/ (*+*), if / is the impulse or whole change of momentum. The work done on the ball B = %m'v'*-%m'u' 2 = \m! (v f u') (v + u') = for m' v' u'= mv-u 216 MECHANICS. [CH. X PROPOSITION 39. To find the whole change of kinetic energy on Impact. We have just seen that the ball A gains an amount \I (v + u) of energy, while the ball B loses an amount Hence the total loss of kinetic energy is \I (v + u' - v - u}. Now v v = e (u - u'). Thus the loss of kinetic energy is On substituting the value of / from Prop. 38, we find for the loss of kinetic energy 1 mm' , , , ; (1 - e 2 ) (u' - uf. 2 ra + ra v Now (u' uf being a square is always positive and e 2 is not greater than unity, hence 1 e 2 is positive unless e = 1, when it is zero. Thus the loss of kinetic energy on impact is always positive unless the coefficient of restitution is unity, when there is no loss. In this case kinetic energy is transferred from one ball to the other but its amount is unchanged. In general however kinetic energy disappears. Joule's experiments, already referred to, lead us to believe that the total energy is un- changed, the energy apparently lost in the kinetic form is transformed into heat, the balls are raised in temperature, and the heat needed for this is measured by the loss of kinetic energy. 137. Oblique Impact. In the cases of impact which have been considered it has been supposed that the two balls were moving either in the same or in exactly opposite directions at the point of impact. This is not always the case: consider the two balls A, , Fig. 88, let their directions of motion make angles a, a' before impact with the line joining their centres, which is known as the line of impact, and /?, /3' with the same line after impact, their velocities and masses being as in the previous sections. In considering the problem we have to deal 136-137] COLLISION. 217 with the motion along the line AS and also with that at right angles to it. Now the whole momentum is unchanged, and Fig. 88. this statement is true for the components of the momentum along and perpendicular to the line of impact. If the balls be smooth there is no force between them at right angles to the line of impact. Thus the velocity of each ball in this direction remains unchanged. Newton's experimental result as to the relative velocity before and after impact applies to the velocities along the line of impact. PROPOSITION 40. To determine the motion after impact of two balls which impinge obliquely. The first principle above stated gives us the following results : mv cos (3 + m'v' cos /8' = mu cos a + m'u' cos a' . .. (i), mv sin (3 + m'v' sin /?' = mu sin a + m'u' sin a! (ii). From the second we have v sin /3 = u sin a (iii), v' sin /?' = u' sin a' (iv), while from the third we find that v cos /? v' cos /3' = e (u cos a u' cos a') (v). Of these five equations (ii) is included in (iii) and (iv). Hence we have four independent equations (i), (iii), (iv) and (v) from which we can find v, v, ft and /?'. 218 MECHANICS. [CH. X We may shew as above that kinetic energy is lost by the impact and that the amount so lost is 1 mm . , pr r (l e*) (u cos a u cos a r. 2 m + m v If one of the bodies be fixed we proceed as follows : Example. A billiard-ball moving with velocity u strikes the cushion at an angle a, the coefficient of restitution being e ; find the direction and velocity of rebound. Let v be the velocity of rebound and let the direction of motion after impact make an angle /3 with the normal to the cushion at the point of impact. Then the velocity along the cushion is unchanged, that perpendicular to the cushion is reversed and reduced in the ratio e to 1. Hence if the velocities be estimated as in Fig. 89, we have v sin /3 = u sin a, v cos fi = eu cos a. Thus v 2 = u 2 (sin 2 a + e z cos 2 a) , cot fi = e cot a. If the coefficient of restitution be unity, we have = 1, then v = u, /3 = a. The ball rebounds with its velocity unchanged in amount, and its direction of motion inclined to the cushion at the same angle as before, but on the opposite side of the normal. 138. Action during Impact. The term Action has been used in our discussion of the third law either for the rate 137-139] COLLISION. 219 at which momentum is transferred, or the rate at which work is done. Now in a case of impact the momentum of each ball is changed by a finite amount in a very brief period, the rate of change of momentum is very great, too great for observa- tion ; we do not deal with the rate of change of momentum but with the whole change which occurs during the impact. This whole change is what is meant by the Impulse of the motion, so that when we speak of the Action taking place between two bodies at impact we refer to the whole amount of momentum which is transferred. 139. Moment of greatest compression. It is con- venient in some cases to divide the whole change of momentum into two parts. Consider what takes place at the point of impact : the bodies are deformed, the velocity of the one A is being reduced, that of the other B is being increased, there will be a moment during the very brief interval in which the two are in contact at which the deformation of each ball has reached its maximum amount and the two balls are moving together with the same velocity. Let us then divide the duration of the impact into two parts, the first lasting up to this instant of greatest compression, the second, during which the balls are again separating, from this instant up to the time at which contact ceases. Let 7 X be the momentum transferred in the first part, 7 2 in the second, then we have. PROPOSITION 41. To shew that the impulse after the moment of greatest compression bears a constant ratio to that before, and that this ratio is the coefficient of restitution. By hypothesis the two balls have a common velocity at the instant at which an amount I I of momentum has been trans- ferred. Let this velocity be V ; then Tr mu + m'u' Hence V= 7 - m + m Thus I^ m + m 220 MECHANICS. [CH. X ,. mm' (1 + e) . , . But /= -- - ~-(u-u). m + m Hence I=(l+e)I^ But I = IiI z . Therefore / 2 el-^. Hence the impulse during the second period of the impact bears a constant ratio to that during the first part, and this ratio is measured by the coefficient of restitution. Thus when a ball impinges directly on a flat surface and rebounds, during the first part of the impact all its momentum is transferred to the surface and the ball is reduced to rest, during the second period an amount of momentum e times as great as that which it has lost is acquired by the ball in the opposite direction, it rebounds therefore with a velocity e times as great as that with which it struck the surface. Example. A ball whose mass is 1 Ib. moving with a velocity of 10 feet per second overtakes another ball whose mass is J Ib. moving in the same direction with a velocity of 5 feet per second. The coefficient of restitution is f ; find the impulse up to the moment of greatest compression and the lohole impulse. Let V be the common velocity at the moment of greatest compression, Then since the momentum is unchanged --,. 4 ~ V ' F=9 feet per second. Thus the Impulse up to this time is 1 (9 - 10) or - 1. The ball loses 1 Ib.-foot unit of momentum. The total impulse therefore is 1 (1 + |) or \ Ib.-foot units of momentum. These are lost by the first ball, gained by the second. Hence if v' be its final velocity we have Thus after impact the smaller ball has a velocity of 12 feet per second. 139] COLLISION. 221 EXAMPLES. IMPACT. 1. An inelastic ball impinges directly on another of half its mass at rest, find the new velocity of the two in terms of that of the impinging ball. 2. The velocities of two balls before impact are 10 and 6 feet per second respectively, after impact they are 5 and 8 feet per second respectively ; compare the masses of the two balls and find the coefficient of restitution. 3. Two bodies of unequal mass moving in opposite directions with equal momenta impinge directly. Shew that their momenta are equal after impact. 4. A body whose mass is 3 Ibs. impinges directly on one whose mass is lib., the coefficient of restitution is J. After impact the momenta of the two balls is the same and the smaller has a velocity of 15 feet per second ; find the original velocities of the two. 5. Find the condition that two balls may interchange velocities on direct impact. 6. A body is dropped from a height of 64 feet on to a horizontal floor. If the coefficient of restitution be J, find the height to which the body rises after 3 rebounds. 7. A ball strikes a cushion at an angle of 45. If the coefficient of restitution be f , find the angle of rebound. 8. A ball impinges with a velocity u on an equal ball at rest, the direction of motion making an angle of 30 with the line of the centres ; determine the motion of the two balls afterwards, the coefficient of resti- tution being unity. 9. A ball impinges obliquely on an equal ball at rest, find the direction of impact if the two move afterwards with equal velocities, the coefficient of restitution being unity. 10. Two balls of masses ?/i, ra', whose coefficient of elasticity is e, impinge directly on each other with velocities u, v respectively (u>v). Shew that the impulse between the balls is 11. Of three equal balls A, B, C, placed in this order in one straight line, A moves with a given initial velocity towards J9, while B and C are at rest. Find .B's velocity after it has struck C, assuming the coefficients of elasticity to be the same for the two pairs of balls A, B and B, C. 222 MECHANICS. [CH. X 12. Find the velocities after impact of two smooth spheres which impinge directly, in terms of their masses, their velocities before impact, and the coefficient of restitution. Shew that the velocity of their centre of inertia is unaltered hy the impact, and that the velocity of each body relative to the centre of inertia is reversed in direction and diminished in the ratio of 1 : e, where e is the coefficient of restitution. 13. Two smooth spherical masses ra and mf moving with given velocities u and u' in the same direction collide. Shew that the loss of kinetic energy due to the collision is 2 (m + m') where e is the coefficient of restitution. (-') (!-*), *CHAPTER XL MOTION IN A CIRCLE. MISCELLANEOUS. 14O. The Hodograph. The velocity of a moving particle may be represented at any moment by a straight line drawn from some fixed point, the length of the line represents the magnitude while its direction represents the direction of the velocity. Thus if the particle move with constant velocity the straight line is fixed in magnitude and direction. If the particle move with uniform acceleration in a straight line the direction of the line representing the velocity is fixed; its length however increases uniformly with the time. Thus if OQ represent the velocity at any moment, and if OQi represent it after an interval of 1 second, then QQ 1 is the increase of velocity in 1 second, and is therefore equal to the acceleration. Consider now a particle moving in a curve A, Fig. 90, let P lt TV . . be its positions at different times and drawP^, P 2 T 2 . . . 224 MECHANICS. [CH. XI to represent the velocities F 15 F 2 ... of the particle in those positions. From a fixed point draw OQ : equal and parallel to Fj, OQ 2 equal and parallel to F 2 , and so on for the other points P 3 etc. We thus get a second series of points Q 19 Q 2 , Q 3 ... which have the property that the lines drawn to these points from the fixed point represent the velocities of the particle in the corresponding positions P lt P 2 etc. If a similar construction be made for all points on the curve AB we shall get a second curve CD, which has the property that the lines drawn to it from represent in direc- tion and magnitude the velocity of the particle at the corre- sponding points of AB. This second curve CD is called the hodograph of AB. 141. The Hodograph and the Measurement of Acceleration. Again let Q lt Q 2 in Fig. 90 be two points on the hodograph, P lt P 2 the corresponding points on the path; in moving from P 1 to P 2 the velocity changes from OQ 1 to OQ 2) join QiQri then the change in velocity is represented in direc- tion and magnitude by Q^Q^ for $i(?2 represents a velocity which when compounded with OQ l will give OQ 2 . PROPOSITION 42. To shew that in the case of uniform acceleration the hodograph is a straight line. If P 15 P 2 , Fig. 91, are two positions which the particle occupies after an interval of 1 second, then Q^Q Z is the change in velocity in 1 second; if we know that the accelera- tion is constant then it is measured by the change in velocity in 1 second. Hence in this case the line Q : Q 2 re- presents the acceleration. Now let P s be the position of the particle after a further interval of 1 second and Q 3 the corresponding point on the hodograph. Then in the same way QzQs represents the acceleration, but since the acceleration is 140-141] MOTION IN A CIRCLE. MISCELLANEOUS. 225 constant in direction and magnitude Q 2 Q 3 must be equal to and in the same straight line as QiQ*. Thus the hodograph QiQzQs--- must in this case be a straight line. Hence if a particle move with uniform acceleration the hodograph is a straight line, and the arc of the hodograph traced out in 1 second represents the acceleration. This proposition can be generalised thus. PROPOSITION 43. If P be a particle describing any curve and Q the point on the hodograph which corresponds to P, then the velocity of Q in the hodograph measures the acceleration of P. For suppose that after a short time T, P, Fig. 92, has moved to P', and Q in consequence toQ'. Then the velocity of Q is given by the ratio QQ'/r, when T is made very small. For QQ' is the space tra- versed by Q in time T, and the ratio of the space tra- versed to the small time of traversing it measures the Fig. 92. velocity of Q. But when r is very small, we may treat QQ' as a straight line : moreover we may consider the acceleration of P as constant for the interval T, if that interval be made sufficiently small. Now OQ is the original velocity of P and OQ' is its velocity after the interval r. Hence QQ' is the change in velocity during the interval and the ratio of this change to the interval in which it occurs measures the acceleration of P. Thus the ratio QQ' jr measures the acceleration of P as well as the velocity of Q. Hence the acceleration of P is equal to the velocity of Q. Whenever then we know the hodograph of a path and the velocity with which it is described, we can find the acceleration in the original path. Thus when the hodograph is a straight line described with uniform velocity, the acceleration is constant both in direction and magnitude; G.M. 15 226 MECHANICS. [CH. XI when the holograph is a straight line but the velocity in it is not uniform the acceleration is constant in direction but variable in amount. We proceed to give some other examples. PROPOSITION 44. A particle describes a circle with uniform speed V; to find the hodograph and the acceleration. Let the path of the particle be a circle P^P Z with centre C, Fig. 93. Let P 1 be the position of the particle in the original path, P 2 its position after one second. Fig. 93. ^ P 2 T 2 to represent the velocity: since = P 2 T 2 = V. At P lt P 2 draw the speed is constant Again, since the path is described with uniform speed and P^Pz is the distance traversed in a unit of time we have P^ 2 equal to V. Now let OQi equal and parallel to P^T^ represent in direc- tion and magnitude the velocity at P lt and let OQ 2 represent it at P 2 . Then OQ 2 is parallel to P 2 T 2 . Hence OQ 1 and OQ 2 are respectively perpendicular to CP 1 and CP 2 . Hence the angle Q-fiQi is equal to the angle P^CP 2 . Since the speed is constant the length of the radius vector from is constant. Thus the hodograph is a circle and this circle is described by the point Q with uniform speed. Now Q : Q 2 represents the space traversed in 1 second by the point in the hodograph when moving with uniform velocity. It is therefore equal to the velocity in the hodograph. Thus it 141] MOTION IN A CIRCLE. MISCELLANEOUS. 227 represents the acceleration completely. Also since the velocity at Q l is perpendicular to OQ ly the acceleration at P is perpen- dicular to P^T^ that is, it is along P^C. Hence the accelera- tion at P l is represented in amount by Q^Q^ the arc of the hodograph described in one second, and is directed along Hence if a be the acceleration we have Let r be the radius of the circle Then since the angles QiOQ 3 and P^CP Z are equal their circular measures are the same. a 7= ? ' 72 Thus a = . r Hence when a particle moves with uniform speed V in a circle of radius r, it has at each point an acceleration directed F 2 to the centre of the circle and equal in amount to . r Thus the force towards the centre is mF 2 if m is the mass of the particle. By expressing these results in terms of the angular velocity of the particle we can put them in slightly different form, for if ft be the angular velocity we have (Section 38) F 2 Hence a = = ft 2 r, r Thus when we observe a body moving in a circle with uniform speed we know that it has the acceleration given above. 152 MECHANICS. [CH. XI Li a stone is tied to a string and swung round in a horizontal circle, Me toward the centre is exerted by the string ; this force measures the *? presented by OB. Thus the resultant force R is represented by OB, and since OB = A + AB, Similarly if the forces be impressed in opposite directions, we have Fig. 2, OB = OA - AB. Thus R = P-Q. 6 - * - - - A If the lines of action of the -p. 2 forces P and Q are not in the same straight line we find the resultant by the parallelogram law. 5. Parallelogram of Forces. PROPOSITION 2. If two forces represented in direction and magnitude by two straight lines OA, OB be impressed on a 3-6] FORCES AT A POINT. 5 particle their resultant is represented by OC the diagonal through of the parallelogram which has OA and OB for adjacent sides. Let OA, OB, Fig. 3, represent two forces P and Q im- pressed on a particle. Complete the parallelogram AOBC and draw the diagonal OC. Then 00 shall represent the resultant force R. For OA, OB represent also the accelerations of a particle of w ^ A ., , . , *, ,. -big. o. unit mass on which the forces P, Q are each separately impressed ; and by the parallelogram of accelerations OC represents the resultant acceleration of such a particle; but the resultant acceleration measures the re- sultant impressed force. Hence OC the diagonal of the parallelogram represents R the resultant impressed force. Thus forces are compounded and resolved according to the parallelogram law and Propositions 9 to 13 of the Kinematics, relating to displacements, apply equally to forces. We may notice that the resultant force does not depend on the mass of the particle. If in Fig. 3 above, the mass of the particle be m, and the accelerations corresponding to P, Q, and R be p, q, r; then since P is equal to mp, Q to mq, and E to mr, the lines OA, OB represent mp and mq respectively. Thus OC represents mr on the same scale, hence it represents the resultant force jR. The proof of the parallelogram of forces depends therefore on that of the parallelogram of accelerations. 6. Experiments on the Parallelogram of Forces. The parallelogram of forces can be verified in various ways by direct experiment; we shall describe two such experiments. A student who has difficulty in following the dynamical proof may base his acceptance of the proposition on the direct results of the experiments. A statical proof by Duchayla is often given ; there are many reasons however why its use should be avoided and we shall not include it. 6 STATICS. [CH. I In many statical experiments the impressed force is measured by the weight of a body suspended by a string from some part of the apparatus, and it is often necessary to vary this weight. A convenient arrangement is shewn in Fig. 4. An iron rod about 15 cm. in length has a hook at the upper end ; to the lower end a flat circular disc is ri vetted and the whole is adjusted to have some definite mass such as 1 Ib. or, if c.G.s. units are being employed, ^ a kilogramme. The weights take the form of flat circular discs of iron or brass ; each disc has a slot cut out as shewn in the figure, the slot reaches to just beyond the centre and is wide enough to admit the vertical rod which supports the scale-pan. Two sizes of " weights," say pounds and ^ pounds or ^ and ^ kilos., will be found convenient; when in use they rest one on the top of the other on the scale-pan forming a pile through the centre of which the supporting rod runs ; thus a force equal to the weight of a definite number of half-pounds is easily applied. Fig. 4. Fig. 5. GJ FORCES AT A POINT. 7 In other experiments the force is most easily applied and measured by means of a spring balance. A useful form of pulley is illustrated in Fig. 5. EXPERIMENT 1. To verify by experiment the parallelogram of forces. (a) The apparatus required for this is illustrated in Fig. 6. ] . 6. Two pulleys F, G are attached to a horizontal support. A 1 In the figure the apparatus is shewn supported by a Willis frame work. Such a framework consists of a number of bars which can be secured together in various positions by suitable screw bolts, and is very useful for various statical experiments. In a laboratory where a large number of sets of apparatus for the same experiment are required it is 8 STATICS. [CH. I string AOB passes over these and carries two of the scale-pans just described : these are shewn at L and M. A second string OD knotted at to the first carries a third pan N. Some convenient number of weights is put on each scale- pan and the whole system is allowed to come into equilibrium. Let us suppose the total weights supported at L, M, N re- spectively including the weight of the scale-pans to be P, Q and R pounds-weight, these weights measure the tensions of the respective strings. Thus forces of P, Q and R Ibs. weight act along OA, OB and OD respectively. Since there is equilibrium R is clearly equal and opposite to the resultant of the other two forces P and Q. Now adopt some convenient length to represent the unit of force, e.g. represent a force of 1 Ib. weight by a length of 10 centimetres. Draw on the board lines OA, OB, OD, parallel to the strings and measure off along OA and OB lengths OA and OB to represent the forces P and Q ; complete the parallelogram AOBC and join OC. Measure the length of OC. It will be found to represent in magnitude the force R on the same scale as OA and OB represent P and Q. Place a straight-edge against OC, it will be found that CO when prolonged is in the same straight line as OD the line of action of R. Thus OC repre- sents R in magnitude but is opposite to it in direction, and since the three forces P, Q, R are in equilibrium, R must be opposite to the resultant of P and Q; hence OC represents the resultant of P and Q represented by OA and OB respectively, and OC is the diagonal of the parallelogram of which OA and OB are adjacent sides. Thus the parallelogram of forces is verified. By taking along OD a length OD to represent the force R, and constructing a parallelogram with OA and OD as sides, we could shew that Q is represented by the diagonal of this paral- lelogram. convenient to have a board fastened to the walls, but projecting some little distance from them, and running round the room at some suitable height. Apparatus such as pulleys, etc. can be secured to this and the weights conveniently suspended from them without coming in contact with the walls. Arrangements should be made for supporting a drawing-board behind the strings which carry the weights, in order to solve questions easily by graphical construction. FORCES AT A POINT. In the figure as drawn P, Q, R are forces of 3, 4 and 5 Ibs. weight. It will be noticed that in this case the angle AOB is a right angle and that since 5 2 = 3 2 +4 2 , we have R* = F* + Q\ (b) Knot three strings together at 0, Fig. 7, and attach N Fig. 7. their ends to three spring balances L, M, N. Fix the balances to hooks on the edges of a drawing-board in such a way that the strings may be all drawn tight and the balances stretched. The readings of the balances will give us the forces acting along the strings, let them be P, Q, R respectively. Mark off along OL, OM and ON lengths OA, OB, OC to represent the forces P, Q, R thus if the balances read in Ibs. we might take a length of 1 inch to represent 1 Ib. weight, a force of P Ibs. weight is then represented by a line P inches in length. Draw lines OA, OB, OC on the paper under the string to represent these forces ; then by constructing a parallelogram as before with OA 10 STATICS. [CH. I and OB as sides, we can shew that OC is equal and opposite to the diagonal of this parallelogram, the diagonal represents the resultant of P and Q. We may use this construction to verify another important formula. Measure with a protractor the angles JBOC, CO A and AOB opposite respectively to the forces P, Q and R. Look out in a trigonometrical table the sines of these angles and then calculate the values of the fractions P Q R and - siii BOG' sinCOA sin AOB' the ratio that is of each force to the sine of the angle between the other two. It will be found that the ratios are all equal. We thus have the equations P Q It sin BOG ~~ sin CO A ~ sin AOB ' Thus in the case shewn in Figure 7 in which P=3, Q=4: and 72 = 5, we find JBOC = 143, GO A = 127 and AOB = 90. Hence sin BOG = sin 143 - sin (180 - 143) = sin 37 = '6 approximately. Also sin CO A = -8 ; sin AOB = 1. Hence c sin CO A R - 5 -5 T~ v siuAOB 1 Thus the relation is verified. 7. Further experiments on the equilibrium of forces. The spring balance may be employed in various other experiments to measure force and verify the results of theory. 6-7] FORCES AT A POINT. 11 Thus in Fig. 8 a weight is supported by two strings each of which is attached to a spring balance. The balances are fastened to two points A and B, thus their readings give the tensions in the strings; if a parallelogram be constructed with its sides representing these tensions the diagonal will be vertical and will represent the weight. Moreover we should find in this case that each of the tensions is greater than the weight. In a similar manner we may support a weight as in Fig. 9 by three or more strings, each of which is attached to a spring balance : the readings of the balances give the tensions, and if, 12 STATICS. [CH. 1 starting from any point, we construct a polygon whose sides represent the tensions in direction and magnitude, it will be found that the line joining the starting point to the extremity of the last line so drawn is vertical and represents the weight supported. 8. Composition and Resolution of Forces. Since forces like displacements are combined according to the parallel- ogram law, the various propositions which have been given for the composition and resolution of displacements apply to forces ; for the sake of completeness in this part of the subject we repeat them here. We will first put the parallelogram of forces into a slightly different form. In order to find the resultant of two forces P, Q acting at a point we draw OA, OB, as in Fig. 3 above, to represent the forces and complete the parallelogram AOBC. We have seen that OC is the resultant. Now the position of C can be found somewhat more simply : in the figure AC is equal and parallel to OB, hence AC will represent the force Q in magni- tude and direction though not in point of application, for both forces P and Q act at 0. We may then clearly find the point corresponding in any given case to C thus. From 0, Fig. 10, the point of action of the forces draw OA to represent the force P, from A the extremity of this line draw AC to represent the force Q in magnitude and direction. Join OC, then OC represents the re- sultant of P and Q in point of action, magnitude and direc- tion. For by completing the Fi 10 parallelogram by drawing a line from equal and parallel to AC and another line through C parallel to AO, it is clear that OC is the diagonal of a parallel- ogram whose two sides meeting at represent the forces ; hence OC represents the resultant of P and Q. This construction can be generalized thus. 7-8] FORCES AT A POINT. 13 PROPOSITION 3. To find by a graphical construction the resultant of a number of forces impressed on a particle. Let OA, OA', OA", etc. Fig. 1 1, represent the forces P, P, P", etc. From A draw AB equal and parallel to OA' to repre- sent Q in magnitude and di- rection. Then OB is the re- sultant of P and P'. From^ draw BC equal and parallel to OA" to represent P" in magnitude and direction : then OC is the resultant of forces represented by OB and BC, and OB represents / \ A ,v the resultant of P and P' ; ,^ v hence OC represents the re- sultant of P, P' and P". Proceeding in this way we find the resultant of any number of forces acting at a point ; for if L is the last point found, then the resultant is OL. Corollary. If L coincide with the resultant is zero and the forces are in equilibrium. In this case the forces are represented in direction and magnitude by the sides of a closed polygon taken in order and we have the result that : If a number of forces impressed on a particle be represented in direction and magnitude by the sides of a closed polygon taken in order, the forces are in equilibrium. This proposition is called the Polygon of forces. A special case of this is the Triangle of forces, of this on account of its importance we give a formal proof. PROPOSITION 4. If three forces impressed on a particle be represented in direction and magnitude by the sides of a triangle taken in order, the particle is in equilibrium. For let the sides OA, AC, CO, Fig. 12, taken in order, represent in direction and magnitude three forces P, Q, R 14 STATICS. [CH. I acting on a particle at 0. By completing the parallelogram OACB we see that OB is equal and parallel to AC and therefore represents Q completely. Hence the resultant of P and Q is re- presented by OC; this resultant is therefore equal and opposite to R, hence the particle is at rest. Fig. 12. It should be noticed that the sides are to be taken in the same direction round the triangle. Thus forces represented by OA, AC, and OC are not in equilibrium. The converse of the above proposition is also true. PROPOSITION 5. If three forces impressed on a particle are in equilibrium they can be represented in direction and magni- tude by the sides of any triangle drawn so as to have its sides parallel to the forces. Let P, Q, R be three forces impressed on a particle at which are in equilibrium. In Fig. 13 take OA to represent the force P, from A draw AC to represent Q in direction and magnitude and join OC. Then OC represents the resultant of P and Q, and since P, Q and R are in equilibrium, R must be equal and opposite to the resultant of P and Q, thus CO must represent R ; hence the forces P, Q, R are represented by OA, AC and CO respectively. Again, any convenient length along OA may be taken to represent P, hence any triangle with its sides parallel to P, Q and R will represent the forces. The converse of the polygon of forces is not true. All triangles whose sides are parallel to the forces are similar and have their corresponding sides proportional, hence any one of them may be taken to represent the forces : this is not the case for polygons. A number of polygons can be found whose sides represent the forces, all these polygons are similar; but any polygon with its sides parallel to the forces will not represent them. 8] FORCES AT A POINT. 15 The following examples illustrate this graphic method. Examples. (1) Find the resultant of forces of 2 to the North, 3 to the East, 3 to the South, and 4 to the West, impressed on a particle. Draw a vertical line OA (Fig. 14) upwards, 2cm. in length, to represent the first force. Draw AB to the right, 3 cm. in length ; BC downwards, 3 cm. in length ; CD hori- A j : zontal to the left, 4 cm. in length; then OD is the required resultant. Also if OL be perpendicular on CD, it is clear thai OL is 1 cm. and LD is also 1 cm. Hence OL is ^/2 cm. Thus the resultant force -^ is y^/2 to the South-West. This example is the same as Example 1 on page 44. . j^g. 14. (2) Six forces of 1, 9, 2, 7, 3, and 8 Ibs. iveight respectively are im- pressed on a particle in directions parallel to the sides of a regular hexagon taken in order. Shew that the particle is in equilibrium. The adjacent sides of a regular hexagon make angles of 120 with each other. Take a line of 1 cm. to represent a force of 1 Ib. wt., and draw AB 1 cm. in length to represent the first force, BC inclined at 120 to it 9 cm. in length to represent the second, CD 2 cm. in length to represent the third, and so on. If the figure be carefully drawn, the end of the sixth line representing the force of 3 Ibs. wt. will be found to coincide with A. The hexagon is a closed one, and the particle is in equilibrium. The student should construct this figure for himself to scale. Aliter. The forces of 1 and 7 Ibs. weight acting in opposite directions are equivalent to a force of 6 Ibs.wt. acting in the same direction as the 7 Ibs. wt., the forces of 9 and 3 Ibs. wt. are equivalent to a force of 6 Ibs.wt. acting in the direction of the 9 Ibs. wt. , the forces of 2 and 8 Ibs. wt. are equivalent to a force of 6 Ibs. acting parallel to the 8 Ibs. ; thus we have three equal forces of 6 Ibs. acting away from the particle in directions inclined to each other at 120, and these form a system in equilibrium. (3) The ends of two strings are secured to two fixed points L and M, and are knotted together at 0; a 5 kilogramme weight is suspended from O; find by a graphical construction the tensions in the strings. Fix a drawing-board in a vertical plane behind the strings, and trace on the board their directions. Take some length (say 5 cm.) to represent a force equal to the weight of 1 kilogramme. From draw OA (Fig. 15) vertically upwards 25 cm. in length, then OA represents the suspended weight. From A draw AB parallel to OM to meet the string OL in B. Let Tj , 2 2 be tne tensions in OL and OM. 16 STATICS. [CH. I Then the three forces 1\ , T 2 sides of the triangle OBA taken in order. They are therefore represent- ed by its three sides. Measure the lengths of OB and BA in cm. Then we have Hence \ = - kilos' wt., = kilos' wt. In the figure drawn it is clear that ^ = 6 kilos' wt., T 2 = 4 kilos' wt. (4) The bob of a pendulum weighing 5 kilos is putted aside by a horizontal string until its thread is inclined at 30 to the vertical. Find the impressed force in the string and the tension of the pendulum thread. Let (Fig. 16) be the point of suspension, A the pendulum when dis- placed, OC vertical. Draw AC horizontal meeting OC in C. The impressed forces o are 5 kilos' weight vertical, parallel therefore to OC, the tension T' of the horizontal string parallel to CA, and the tension T of the pen- dulum thread parallel to A 0. The impressed forces therefore are proportional to the sides of the triangle AOC. Also since the angle at is 30 we have AC = ^AO: OC=%j3AO, and JL .?* J_ OC~CA~ AO' '' T=5 ^ = ^ kil s' weight, oc~ v ^ (5) Weights W lt TF 2 , W 3 are attached to three points A l , A 2 , A 3 in a string the ends of which are secured to two fixed points A, B. The whole hangs in a vertical plane and the form taken by the string is drawn to scale. W l is known. Find the weights of JF 2 and W 3 and the tensions of the parts of the string. Take a vertical line X^X 2 (Fig. 17) to represent W-^. From X^ draw 8] FORCES AT A POINT. 17 X^O parallel to AA^ , and from Z 2 draw X 2 parallel to A^A 9 . The three TV O< -W., Fig. 17. forces at A l are parallel to the sides of the triangle therefore we have Thus T l and T 2 can be found by measuring OX l and OX 2 . Draw OX 3 parallel to A 2 A 3 meeting X^^ produced in X 3 ; the three forces T 2 , T 3 and W 2 at A 2 are parallel to the sides of the triangle X 2 OX 3 , they are therefore represented by these sides, thus 7yv~ f\ Y y Y ' UA 2 UA 3 A 3 A 2 Similarly draw OX. parallel to A 3 B and produce X 2 X 3 to meet OX. in X 4 . Then T 3 , T. and W 3 are parallel to X 3 0, OX. and Hence -I* - OZ 4 ~Z 4 X 3 - Thus T. and JF 3 can be found. Hence if W-^ be known, the values of W 2 and W 3 together with the tensions in the different parts of the string are given graphically. In the diagram as drawn it will be found that Hence if W l= =l kilo, then ^ a = 2 kilos, JT 8 =3 kilos. Also T 1= :4-2 kilos, T 2 = 3-7 kilos, T 3 =3-3 kilos, and T 4 = 4-6 kilos. Hence the tensions and two of the weights are found. G. S. 18 STATICS. [CH. r When two or more forces impressed on a particle are given in direction and magnitude it is possible to find expressions for their resultant. We have in the case of two forces to deter- mine the diagonal of a parallelogram two of whose sides are given, while the case of more than two forces involves an ex- tension of the same process. Thus the propositions given on pp. 33 38 of the Dynamics with regard to displacements apply to forces. For the sake of completeness in this part of the book they are repeated here. PROPOSITION 6. To find an expression for the resultant of two forces at right angles. Let OA, OS, Fig. 18, represent two forces P, Q respectively at right angles to each other. Complete the rectangle AOBC. Let R be the resultant of P and Q, then It is represented by OC. Since the angle OA C is a right- angle we have = OA* + Fig. 18. Hence PROPOSITION 7. To find an expression for the resultant of two forces inclined to each other at any angle. Let OA, OB represent respectively two forces P, Q inclined to each other at an angle y. Complete the parallelogram A OBC. OC represents R the resultant of P and Q. Draw CD perpendicular to OA meeting OA produced, Fig. 19 (a), or OA, Fig. 19 (b), in D. Then AOB = y ', in Fig. 19 (a) the angle y is less than a right angle ; in Fig. 19 (b) it is greater. FORCES AT A POINT. Now in Fig. 19 (a), OD = OA + AD = OA + ACcosDAC = OA+OBco$AOB = P+Q cos y, CD = AC sin DAC = OB sin y = Q sin y. 19 In Fig. 19 (6), O.Z> = OA - AD = OA - AB cos DAC = P + $ cos y, CD = AC sin DAC = OB sin (180 -y) = Q siny. Hence in either case we have There are many special cases of this last proposition which can be solved by Geometry without reference to Trigonometry. Thus, suppose the angle between the two forces to be 45. Hence, constructing Fig. 20 as above, we have 22 20 STATICS. CH. I Also R 2 = C 2 = OD 2 + D C 2 And Hence Or again, if 7=60, we have, fig. 21, AD = AC = IQ, CD = ^". These are both given by the general formula by putting 7=45, cos 7 = fj\ and 7 = 60, 0087 = ^. If the two forces be equal the resultant bisects the angle between them ; for, Fig. 22, if OA=AC, B A ~~ - -^C then LAOC-^ACO = L BOG. Join AB, cutting OC in D, then AB bisects OC at right angles. And Fig. 22. 9] FORCES AT A POINT. 21 9. The Resolution of Forces. Just as we can com- bine or compound two or more forces and find their resultant, so conversely we can resolve a single force into a number of others, called its components, which are equivalent to it. PROPOSITION 8. To find, by a graphical construction, the components of a force in any two directions. Let OC, Fig. 23, be the given force, and LM, LN the two given directions. Through draw OA parallel to LM and through C draw AC parallel to LN. These two forces OA, AC acting at have OC for their resultant, hence OA, AC are components of OC and they are parallel respectively to LM and LN, that is, they are drawn in the given directions. PROPOSITION 9. To find an expression for the components of a force in two given directions. Let OC, Fig. 23, represent R the given force, and let OA, OB be the components in directions making angles a, ft, respec- tively with OC. Then AOC = a, Hence Now in the triangle OA C the sides are proportional to the sines of the opposite angles. 22 STATICS. [CH. I 0(7 OA AC Hence sin OA C sin A CO sin A 0(7 ' P P sin (a + ft) sin ft sin a ' Moreover from the figure a + ft = y. D T> sin Hence P = P -7 , smy sin a sin y PROPOSITION 10. 3T0 find the components of a force in two directions at right angles. Let 0(7, Fig. 24, represent the force E, OA, OB two direc- tions at right angles in which the components are required. Let AOC = a. Draw CA, CB perpendicular on the two directions. Then OA, OB represent the components P, OA Also Yffi = cos AOC - cos a, Fig. 24. .*. OA = 0(7 cos a. Hence P = R cos a. Again -^ = cos BOG = sin AOG = sin a, /. OJ5=0(7sina. Hence Q = R sin a. If we put 50(7 = ft we have clearly OB=OC cos ft And in this case a + fi = 90. Thus, when a force is resolved into two others mutually at FORCES AT A POINT. right angles, the component in each direction is found by multiplying the original force by the cosine of the angle between it and the direction of the component. It must be remembered that this result is only true when the two components are at right angles. Thus, let OA, OB (Fig. 25) be two com- _ _, ponents of OC at right angles. Draw OB' making an angle y with OA' and through C draw CA' parallel to OB'. If now OC be resolved into two forces in directions OA and OB' inclined at an angle 7, the component in the direction OA is no longer OA but OA'. A force represented by OA is R cos a, that represented by OA' has not this value. A' Fig. 25. PROPOSITION 11. To shew that if the components of a force be resolved in any direction, then the sum of these resolved parts is equal to the component of the original force resolved in this same direction. Take the case of a force R represented by OC, Fig. 26, which is resolved into two forces P, Q represented by OA and OB respectively. Draw Ox, Oy two lines at right angles through 0, and draw AL, EM and CN perpen- dicular to Ox. Then if we sup- pose all the forces resolved in the directions Ox and Oy it is clear that OL represents the component of P, OM of Q and ON of R. Draw A K parallel to Ox to meet CN in K. Then LN= AK. And in the triangles BOM and CAK the sides are respectively parallel and OB is equal to AC. Hence the triangles are equal. Therefore OM AK= LN. Thus LN represents the component of Q in the direction of Ox. But from the figure ON= OL + LN=. OL + OM. L N Fig. 26. STATICS. [CH. I Hence Component of R in the direction Ox = Component of P + Component of Q. This proposition can readily be extended to the case of any number of forces. PROPOSITION 12. To find an expression for the resultant of a number of forces impressed on a particle in given directions lying in one plane. Let JPj, P z ... be the forces and let them make angles ttj, a 2 ... with a fixed line Ox, Fig. 27, drawn through the point of action. Let Oy be perpen- dicular to Ox. Let R be the resultant force and 6 the angle its direction makes with Ox. Resolve all the forces and the resultant in the two directions Ox and Oy. Then since the resultant is equiva- lent in its effect to the forces the components of the result- ant in each of these two directions are respectively Fig. 27. equal to the sum of the components of the forces in these two directions. The component of the resultant along Ox is R cos 0, the components of the forces are P l cos 04 . . . , P 2 cos a a . . . respectively. Hence R cos = P 1 cos aj -f Po cos a 2 + . . . = 2 {P cos a}, where 2 {P cos a} means the sum of a number of quantities like P cos a. Again resolving parallel to Oy R sin 6 P l sin a x -f P 2 sin a?+ ... = 2 {P sin a}. Thus remembering that sin 2 -f cos 2 = 1, we have by squaring and adding 9-10] FORCES AT A POINT. 25 ^ = [2 {P cos a}] 2 + p {P sin a}] 2 ^2{/ 32 } while by dividing we find sn a tan - cos a 10. Equilibrium of Forces impressed on a par- ticle. We have already found, Prop. 3, the conditions of equilibrium of a set of forces impressed on a particle. If a polygon be drawn whose sides represent the forces in direction and magnitude it will be closed. The same result can be expressed in symbols by the aid of the last proposition thus. If the particle be in equilibrium the resultant of the forces is zero. Thus the components of the resultant in any two directions must also be zero. Hence the sum of the components of the forces in any two directions at right angles must be zero. Hence P 1 cos ttj 4- P 2 cos a 2 += or 2 (P cos a) 0, and P l sin 04 + P 2 sin a 2 + . . . = or 2 (P sin a) = 0. This result is of course applicable to the case of three forces, but in this case we know in addition that the forces are represented by the sides of any triangle drawn parallel to their directions ; hence for three forces we have the following theorem known as Land's Theorem. PROPOSITION 13. When three forces impressed on a particle are in equilibrium each is proportional to the sine of the angle between the other two. Let the three forces be P, Q, R acting in directions OL, OM, ON respectively, Fig. 28. Let ABC be a triangle whose sides are parallel to the forces, BC being parallel to OL, CA to OM and AB to ON. Then from the figure 26 STATICS. [CH. I Hence sin CAB= sin MON, sin ABC = sin NOL, sin EGA = sin J/. R R Fig. 28. But the forces P, Q, R are proportional to the sides BC r CA and AB of the triangle .4-5(7 respectively. Moreover the sides of a triangle are proportional to the sines of the opposite angles. Thus the forces are proportional to the sines of the angles of the triangle which are opposite to- them. T> r\ T> Hence 1 sin A sin B sin C ' Thus -4. _*_ sin MON sin JWZ " sinLOM' This theorem has already been verified by experiment (Exp. We may conclude then that in dealing with questions on the equilibrium of forces impressed on a particle : (a) The direct method of solution is to resolve the forces in any two convenient directions at right angles and equate to zero each set of com- ponents. (6) If the forces be only three in number a graphical solu- tion based on the triangle of forces can easily be obtained. 10] FORCES AT A POINT. 27 The following examples will illustrate these various methods. Examples. (1) Find the resultant of two forces of 3 and 4 kilos' weight respectively impressed on a particle at right angles. Let E be the resultant ; then since the forces are at right angles 2 = 3 2 + 4 2 = 25, R = 5 kilos' weight. This is the result which we verified in Experiment 1. (2) Find the resultant of two forces of 10 and 5 kilos' weight acting at an angle of 60. Let R be the resultant. Then substituting in the formula R* we have R = 5 N /7 kilos' weight. (3) A force of 10 kilos' weight is resolved into two equal forces mutually at right angles ; find these forces. Since the components are equal, they are equally inclined to the resultant. Hence the angle between each of them and the resultant is 45. Thus if P and Q be their values P= 10 cos 45= ^ kilos' wt., >/2 Q = 10 sin 45= - _ kilos' wt. v 2 (4) A force of 15 kilos' weight is resolved into two at right angles, the value of one of these is twice that of the other; find the forces. Let them be P and Q kilos' weight. Then P=2Q, e?2 (1+4) = 152, Q = 3^5 kilos' weight. Hence P= 6^/5 kilos' weight. 28 STATICS. [CH. I (5) The resultant of two forces o/3 kilos' weight and 5 kilos' weight is a force of 1 kilos' weight ; find the angle between the two. Let 7 be the angle, then if R be the resultant of two forces P and Q inclined at an angle 7, we know that x5x3cos 7 , 30 0087 = 49-25-9 = 15, cos 7 = |, 7 = 60. Thus the angle required is 60. (6) Forces equal to the weights of 6, 7, and 8 Ibs. are impressed on a particle in directions inclined to each other at 120. Find their resultant. Three equal forces at angles of 120 are in equilibrium. The given forces are equivalent to forces respectively of 6 Ibs., (6 + 1) Ibs., and (6 + 2) Ibs. The three forces of 6 Ibs. are in equilibrium, and may therefore be removed from consideration, and there are left forces of 1 and 2 Ibs. weight at an angle of 120. Their resultant may be found graphically, or thus .R 2 =l 2 + 2 2 + 2 x 1 x 2 x cos 120 = 5-2 = 3. Thus 12=^/3 Ibs. wt. Or again. Let R be the resultant force and let it make an angle with the direction of the force of 8 Ibs. weight. Eesolve all the forces parallel and perpendicular to the direction of the 8 Ib. force. R cos 6 = 8 cos + 7 cos 120 + 6 cos 120 = 8. R sin 0= 8 sin + 7 sin 120 - 6 sin 120 5/8 2 ' Hence JR= v/3 Ibs. weight. /. = 30. Thus the resultant is a force of */3 Ibs. weight inclined at 30 to the force of 8 Ibs. weight. 10] FORCES AT A POINT. 29 (7) A body weighing 5 Ibs. is supported by two strings, the tension in one string is 8 Ibs. weight and its direction is inclined at 30 to the horizon. Find the direction of the other string and the tension in it. (i) Graphically. Draw AB (Fig. 29) vertically down, 5 cm. in length, to represent the weight, draw BC 8 cm. in length and at 30 to the horizon to represent the tension in the first string, A and join GA. Then CA represents the direction of the second string, and the number of cm. in CA measures in Ibs. weight the tension in that string. (ii) By resolution of forces. Let T be the tension of the string OC, the angle it makes with the vertical. Let OD be the first string and OA the direction of the 5 Ibs. weight. Kesolving vertically 5 = T cos + 8 cos 60. Resolving horizontally T sin = 8 sin 60. From the first equation Tcos0=l. From the second T sin = 4/3. Fig. 29. Hence T 2 =:49, T= 7 Ibs. weight, and tan = (8) The resultant E of two forces P, Q impressed on a particle is equal to P and at right angles to it. Find the force Q. Let AB (Fig. 31) represent P, BG equal to AB and at right angles to it will represent E. Join CA. Then BC is the resultant of forces represented in magnitude and direction by AB and AC acting at B. A Thus a line through B parallel and equal to AC will represent Q. But AC* Therefore Hence B 'R c Fig. 81. STATICS. [CH. I (9) Two forces impressed on a particle are represented respectively by \OA and (tOB, A and B being fixed points and X and fi constants. C is a point in AB such that \AC is equal to pBC. Shew that the re- sultant of the forces is (X + ^t) OC. By the triangle of forces a force \OA along OA (Fig. 32) is equivalent to \OG along OC and \CA acting at parallel to CA. Again fj.OB is equivalent to HOC along OG and ^GB at parallel to GB. Thus the two given forces are equivalent to (X + ju) OG along OC together with \CA and fj.BC in opposite directions parallel to AB. These last two are equal, they therefore balance and may be removed. Hence the resultant is (X + u) OC. Fig. 32. (10) Explain the action of the wind in propelling a ship. Let AB (Fig. 33) represent the direction of the ship's keel; CD the direction of the sail which we suppose to be flat. Let the pressure of the wind be equivalent to a force P acting on the sail in the direction indicated. Kesolve this force into two components, one E at right angles to the sail, the other T along j the sail. This last component pro- duces little or no effect, and we may neglect it; it is only the component perpendicular to the sail which we need to consider. This force E acts on the ship through the mast. We Fig. 33. may resolve E into two components, the one X parallel to the keel, the other Y at right angles to the keel. The resistance offered by the water to motion in a direction at right angles to the keel is so great that the component Y is almost balanced by it ; the ship is built so that the water may offer a small resistance to motion parallel to the keel, and the ship moves in this direction under the impressed force X. The effect of the force in tending to turn the ship round can only be considered later. CH. I] FORCES AT A POINT. 31 EXAMPLES. TRIANGLE AND PARALLELOGRAM OF FORCES. 1. Forces represented by the weights of 10 Ibs. and 15 Ibs. respec- tively act at a point in northerly and easterly directions. Find the magnitude and direction of their resultant. 2. Find the magnitude of the resultant of two forces 12P and 5P when they act at a point and in directions at right angles to one another. 3. ABC is a triangle, D, E, and F the middle points of BC, CA, AB respectively. Forces acting at a point are represented in direction and magnitude by the lines AB, A C, BE ; shew that their resultant will be similarly represented by 3.FD. 4. The sum of two forces is 36 Ibs. and the resultant which is at right angles to the smaller of the two is 24 Ibs. Find the magnitude of the forces. 5. The difference of two forces is 8 Ibs. and the resultant, which is at right angles to the smaller of the two, is 12 Ibs. Find the magnitude of the forces. 6. Can three forces which are in the proportion of 7, 10 and 17 keep a point at rest? 7. Forces represented in magnitude and direction by the diagonals of a parallelogram act at one of the corners, what single force will counteract them? 8. Shew by means of a diagram how to find the part of the pressure of the wind which is available for urging on a ship when the wind blows very nearly from the direction in which the ship is going. 9. Explain how the law of composition of forces may be deduced from the second law of motion. 10. Two forces P and Q have a resultant E equal to P. Draw a diagram representing such a system and shew by means of it that the resultant of two forces equal and parallel to P and E respectively would act at right angles to Q. f 11. Shew how to find the resultant of two forces acting at the same point and explain how to verify the result by experiment. 12. The wind is blowing from the north-east. Explain with a diagram its action in propelling a ship towards the north. 13. Resolve a force of 15 Ibs. into two forces each making with it an angle of 30. 14. Shew how to place three forces which are in the ratio of 3, 4 and 5, so that they may keep a particle at rest. 15. Describe an experiment to prove the parallelogram of forces. 32 STATICS. [CH. I 16. A pendulum consisting of a bob weighing 1 kilogramme at the end of a string 1 metre long is drawn aside until the bob is 25 cm. from the vertical through the point of support, and is held in this position by a horizontal string. Find the forces on the bob (1) when in this position, (2) just after the horizontal string is cut, (3) as the bob swings through its lowest position. 17. Five forces each equal to P act along radii of a circle which are at angular distances 30, 60, 90, 120 and 150 from a fixed radius; determine the resultant. 18. If a heavy body is supported by two strings one of which is vertical, prove that the other must be vertical also. 19. Shew that a vessel may sail due east against a south-east wind. 20. A straight line is drawn parallel to the base BC of a triangle ABC, cutting AB in a point D such that AD is twice BD. If P be any point on this line, prove that the resultant of forces completely repre- sented by AP, BP, CP is parallel to BC. 21. Find the resultant of two forces represented by the side of an equilateral triangle and the perpendicular on this side from the opposite angle. 22. If the magnitude of one of two forces acting at a point be double that of the other, shew that the angle between its direction and that of their resultant is not greater than thirty degrees. 23. If a uniform heavy bar is supported in a horizontal position by a string slung over a peg and attached to both ends of the bar, prove that the tension of the string will be diminished if its length be increased. 24. A weight hangs at the end of a string attached to a peg. If the weight is held aside by a horizontal force so that the string makes an angle of 30 with the vertical, shew that the tension of the string is double the horizontal force. 25. ABCD is a rhombus. Shew, without assuming the Parallelogram of Forces that forces represented by AB, CB, CD and AD are in equi- librium. 26. State how three equal forces must act so as to produce equi- librium and hence find the resultant of two equal forces inclined at 120 to each other. 27. A weight is suspended by means of two equal strings attached to points in the same horizontal line. Shew that if the lengths of the strings are increased, their tension is diminished. 28. A weight hangs at the end of a string attached to a peg. If the weight is held aside by a horizontal force, so that the string makes an angle 60 with the vertical, shew that the tension of the string is double the weight. OH. I] FORCES AT A POINT. 38 29. Two forces P and Q act at the same point and their directions are inclined to each other at an angle of 45. Find an expression for the magnitude of their resultant. Find approximately the magnitude of the resultant if the component forces be respectively equal to weights of 3 Ibs. and 4 Ibs. 30. If a heavy uniform bar is supported in a horizontal position by a string slung over a peg and attached to both ends of the bar, prove that the tension of the string will be diminished if its length be in- creased. 31. A weight is suspended by means of two equal strings attached to points in the same horizontal line. Shew that if the lengths of the strings are increased, their tension is diminished. 32. Find the resultant of two forces of 5 and 10 Ibs. weight respectively acting at an angle of 60. 33. A body is acted upon by two forces, one of 500 dynes due north, and one of 250 dynes north-east ; find the resultant force. 34. -A. mass of 1 Ib. is supported by strings of lengths 3 and 4 feet respectively attached to two points in the same horizontal plane 5 feet apart. What is the tension of each string ? 35. Prove that when a kite is being flown, the position which the string will take up cannot be at right angles to the body of the kite, but will be less steep. 36. Shew that three forces of 5, 6, and 12 Ibs. wt. can never be in equilibrium. 37. A. body weighing 4 Ibs. at rest on a smooth table is acted upon by forces of 3 and 4 Ibs. weight in directions oblique to the table and at right angles to each other. Shew by a diagram how to find the directions of the forces. 38. Three forces keep a particle in equilibrium, one acts towards the east, another towards the north-west, and the third towards the south ; if the first be 5, find the other two. 39. ABCD is a parallelogram, and three forces acting at a point are represented in magnitude and direction by AC, BD and 2DA. Shew that the three forces are in equilibrium. 40. DC and AB are diameters of a circle. Three forces acting at a point are represented in magnitude and direction by AB, DC and 2BD ; shew that they are in equilibrium. 41. Three forces of 5, 12 and 13 Ibs. wt. are in equilibrium. Shew that two of them are at right angles and find the sines of the angles which the remaining force makes with these two. G. S. 3 34 STATICS. [CH. I 42. A weight of 10 Ibs. is suspended from a fixed point by a string 25 inches in length. The weight is drawn aside until its vertical distance below a horizontal line drawn through the fixed point is 20 inches. Shew that the smallest force which will keep the weight in this position will just support a weight of 6 Ibs. hanging freely. 43. A force of 7 Ibs. wt. acts on a particle due north, one of 8 Ibs. wt. due east, one of 6 Ibs. wt. N.N.W. Find by a careful drawing the direction and the magnitude of the force which will balance these. 44. If two forces each equal to 1 Ib. wt. act at a point and their direc- tions make with each other an angle of 60, find to the nearest oz. the magnitude of their resultant. 45. Shew that forces of 99 Ibs. wt. and 5 Ibs. wt. acting at right angles to each other have a resultant which is 17 times as great as the resultant of 5 Ibs. wt. and 3 Ibs. wt. acting at right angles to each other. 46. ABC is an equilateral triangle, AB the perpendicular on BC. Forces each equal to P act along BA, AD and AC respectively, in the directions indicated by the letters. Find the magnitude of their resultant, and shew that it is inclined at an angle 75 to the line AB. 47. Three lines AB, AC and AD in the same plane make the angles BAG, CAD each equal to 30. Forces, equal to P, act along BA, AC and AD in the directions indicated by the letters. Find the magnitude of their resultant, and shew that it is inclined at an angle 75 to the line AB. 48. ABCD is a rectangle, AD=15 inches and ^ = 20 inches; find the magnitude of the resultant of two forces of 16 Ibs. wt. and 25 Ibs. wt. acting along AB and AC respectively. 49. To two points A, B, 5 feet apart on a horizontal beam, the ends of a string ACB are attached, AC being 4 feet and BC 3 feet long. From C a weight of 10 Ibs. is hung. Find the tensions in the strings AC andBC. 50. Four weights of 2, 3, 4 and 5 Ibs. are hung on a string 5 feet long at points 1 foot apart. The ends of the string are attached to two points 3 feet apart in the same horizontal line, and the form assumed by the string is drawn to scale on a sheet of paper. Shew how to find from the figure the tensions in each part of the string. CHAPTER II. PARALLEL FORCES. 11. Rigid Bodies. So far we have dealt only with forces impressed on a particle, or on some body which for our purpose could be treated as a particle. We are to consider now some of the effects of forces impressed on a body, the volume of which cannot be treated as very small. The bodies with which we shall deal are called Rigid Bodies. By this it is meant that they do not alter in shape when force is im- pressed. No body is perfectly rigid, but many substances will resist the application of force and will change in shape by an amount which is practically infinitesimal, when force is applied. Iron, glass and wood have all rigidity, and, though the shape of a body of any of these and other similar materials may vary slightly under the application of a force, the variation will not concern us. Bodies which have rigidity are called Solids. Other bodies which we consider in Hydrostatics are Fluids. The distinction between Solids and Fluids will best be con- sidered later 1 . 12. Superposition of Forces. Now, when a force is impressed on a rigid body, it is found by experiment that any point of the body, which lies in the line of action of the force, may be considered as the point of application of the force. Thus if a body be in equilibrium under two forces P and Q applied at A and B, Fig. 34, the two forces P and Q must be equal and act in oppo- P Q site directions along the line AB. /& ir Now let a force P act on the body at , Fig- 34. 1 See Hydrostatics. 3 2 36 STATICS. [CH. II A in direction AB, Fig. 35. At B introduce two equal and opposite forces each equal to P. These forces are in equilibrium and will _^ -? therefore not disturb the effect of P. A Then P at A and P at B directed Fig. 35. along BA balance ; they therefore pro- duce no effect and may be removed from consideration, and we are then left with P at B in the direction AB producing the same effect as P at A impressed in that same direction. Thus P may be impressed at any point of the body in its line of action without altering its effect. This is called the principle of the transmissibility of force. In proving this principle, as well as in some of the Ex- amples solved in Chapter I., we have made use of another prin- ciple which is of general application. We have introduced two equal forces in opposite directions and have assumed that this does not affect the equilibrium. The truth of the assumption is obvious. Thus we may, without altering the conditions of any problem, superpose upon, or remove from, any system of forces any second system which is itself in equilibrium. 13. Parallel forces. When the lines of action of two or more forces impressed at two or more points in a body meet, we may suppose the forces to be applied at the point of intersection of their lines of action, and find their resultant by the rules established in the last chapter. In general, however, the lines of action of forces impressed on a body do not all meet in a point \ the problem is more complicated. The simplest case of this occurs when the forces are parallel. DEFINITION. Two parallel forces are said to be Like when they act in the same direction, they are Unlike when they act iti opposite directions. PROPOSITION 13 A. To find the resultant of two parallel forces impressed at two points of a rigid body. (i) When the forces are like. 12-13] PARALLEL FORCES. 37 Let the two forces be P, Q acting at the points A, B, Fig. 36, Fig. 36. in directions AL and EM. Take AL and BMto represent the forces. Join AJ3, and at A and B introduce two equal and opposite forces S, represented respectively by AD and BE acting in the directions AD and BE. Complete the parallelograms ADFL and BEGM. The resultant of P and S at A is represented by AF and acts along AF, let it be P l and replace P and S by their result- ant. The resultant of Q and S at j5 is expressed by .#6?, let it be $ x acting along J96r. Then the two forces P and Q are equivalent to P l and ^ t impressed at A and 2? and represented by AF and j56r respec- tively. The lines of action Pj and Q l when produced will meet. Let them be produced and meet at 0. Draw OC parallel to the original direction of the forces P and Q to meet AB in C. Transfer the points of application of P and $1 from A and B respectively to 0. At resolve P into two components parallel to OC and CA, respectively, these components must be equal to P and S. 38 STATICS. [CH. II Again resolve Q 1 at into two components parallel to 00 and CBj these two components will be Q and S. Thus we now have, acting at 0, P + Q along OC and two equal forces S parallel to CA and CB. These forces are equal and opposite, they may therefore be removed and we are left with a single force R equal to P + Q acting along OC parallel to the original direction of P and Q. Transfer the point of application of R to C. Then the resultant of P and Q is R acting at C parallel to the original direction of P and Q. We have now to determine the position of the point C. By the construction, the triangles ADF and ACO are similar. . OC FD P Hence 7T - = 7TT = -= , CA. DA iS or P.AC = S.OC. The triangles BEG and BCO are similar. Thus W_GE Q CB~ EB~ S' or Q.BC = S.OC. Thus Q.BC = P.AC, AC Q Again add unity to each side AC , Q R -p-- P> or P.AB = R.BC. Similarly Q.AB^R.AC. =-: 13] PARALLEL FORCES. 39 (ii) When the forces are unlike. Let the forces be P and Q acting in the opposite directions AL, BM, Fig. 37, at the points A and -6; let P be the greater force. Fig. 37. Take AL and BM to represent the forces. Join AB, and at A and B introduce two equal and opposite forces S represented respectively by AD and BE impressed in the directions AD and BE. Complete the parallelograms ADFL and BMGE as before. Replace P and S, acting at A, by their resultant Pj along the diagonal AF, replace Q and S at B by their resultant $1 along BG. Since Q is less than P the two diagonals AF and GB must meet. Let them meet at ; draw OC parallel to the original directions of the forces to meet BA produced in C. Remove the points of application of P 1 and Q 1 from A and B respec- tively to 0. At resolve P^ into its two components P and S parallel to AL and AD respectively, and resolve Q 1 into its two com- ponents Q and S parallel to BM and BE respectively. The two components S balance and may be removed, and we have left a force R equal to P - Q, in the direction CO pro- 40 STATICS. [CH. II duced, which is equivalent to the two forces P and Q at A and B. Transfer the point of application of R to C. Then the resultant required is R equal to P Q acting at C. To determine the position of C we have, as before, since the triangles AD F and AGO are similar, Thus P.AC = S.OC. Also, since BEG and BCO are similar, Thus Q.BC = S.OC. Hence P.AC = Q.C, J?___ P ~Q A BC~ CA~ BC-CA~ AB' Thus for like forces we have = P + , for unlike forces Jf=P-Q, while in either case the position of the point C is given by P_ _Q__R__ BC~ CA " AB' 14. Couples. We should notice that there is one case of the last proposition (Proposition 13, ii.) in which the con- struction fails; if P is equal to Q the lines AF and GB of Fig. 37 will be parallel; we cannot find their point of inter- section, it is at an infinite distance away and there is no single force which will replace the two ; such a system of two equal unlike parallel forces is called a couple. We shall consider such a system later. 15. Resultant of Parallel forces. When a number of parallel forces are impressed on a body we can find their 13-16] PARALLEL FORCES. 41 resultant by taking two and finding the magnitude and line of action of their resultant then combine with this resultant a third force and so on, in this way the resultant of all the forces can be obtained. It is clearly a force R given by where P 1} P z ... are the individual forces all supposed to act in the same direction ; if one or more of the forces acts in the opposite direction its sign, in the expression for 72, must be changed. 16. Experiments on Parallel forces. For experi- ments on parallel forces a rectangular bar of wood about a metre long with square section, each side of the square being some 2 to 3 cm. in length, is convenient. One face of the bar should be graduated in centimetres or inches as in Fig. 38. Three rings of brass wire A, B, C are made to fit the bar Fig. 38. and can slide along it. These rings have small hoops attached as shewn in the figure. The bar is suspended in a horizontal position by vertical strings attached to A and B while various weights can be supported from C. The ends of the strings attached to A and E are secured to two spring balances D and E. For this experiment Salter's circular balances are the most convenient. In a spiral balance the scale pan drops consider- 42 STATICS. [CH. II ably when loaded, owing to the stretching of the spring ; the drop is much less in the circular form shewn in the figure. The small motion of the end of the spring is magnified by the pointer. The spring balances are supported in some convenient way and the bar suspended in a horizontal position. If the bar be uniform and the rings A, B be equidistant from either end respectively, it will be found that with the central ring C unloaded each balance is equally stretched. This extension is due to the weight of the bar; in making- experiments the readings of the balances thus obtained should be subtracted from those observed when the bar is loaded. EXPERIMENT 2. To determine by experiment the position and magnitude of the resultant of two parallel forces. Suspend the bar as just described from the two balances D and E by the rings A, B. Adjust the length of the strings so that the bar may be horizontal and the rings A, B equidistant from its centre. Take the readings of the spring balances, these should be the same. Suppose them to be '25 kilo. Sus- pend a carrier or scale pan of known weight from (7, and place known weights on it. Let the total weight suspended from C be R kilos weight, this is supported by the tensions in the two strings at A and B. Read the spring balances and subtract from each the reading observed before R was suspended. The differences give the values of the forces P, Q which acting at A and B respectively support R. Note the positions on the scale of the points A, B and C and thus measure the distances AC and EC. Then it will be found that R=P+Q, and also that The resultant of P and Q is clearly a force through C equal and opposite to R, thus the formula which gives the magnitude and line of action of the resultant of two parallel forces is verified. 16-17] PARALLEL FORCES. 43 Repeat the experiment ; shifting the position of C it will be found that as C is moved P and Q both change, but their sum remains constant while they always satisfy the relation P.AC = Q.BC. Moreover if the value of R be altered by changing the suspended weights it will be found that the values of P and Q are also changed but in such a way that the equation P + Q = R is always true. Again if be a point on the bar on the side of A removed from , and the distances OA, OB and OC be measured, it will be found that R.OC = P. OA + Q.OB. See Section 21. 17. Motion about an axis. We will now consider the equilibrium of a body which can turn round a fixed axis, and on which forces are impressed in a plane perpendicular to the axis. Unless some relation exists between these forces rota- tion will take place, we wish to determine the relation which must exist if there is to be equilibrium. This condition is easily found; the forces, which we have to consider, are the impressed forces and the reaction at the axis ; this reaction is a force which necessarily passes through the axis, and it must balance the resultant of the impressed forces. Hence for equilibrium the resultant of the impressed forces must pass through the axis. Suppose now that there are only two impressed forces. We can put the conditions into symbols thus. PROPOSITION 14. To find the condition of equilibrium of a body which can turn about a fixed axis when acted on by two forces in a plane perpendicular to the axis. It is clear from what has been said that the resultant of the forces must pass through the axis. Let the forces P, Q act in the plane of the paper and let the axis, at right angles to that plane, cut it in C. Let a line ABC through the axis cut the lines of action of the forces in A and B and transfer the points of application of the forces to A and B ; then the resultant of two forces P and Q acting at A and B respectively passes through C. 44 STATICS. [CH. II (i) Let the forces be parallel, Fig. 39. Then their resultant R, which is equal to P + Q, passes R Q through C and we have P. AC = Q. BC. Draw DCE perpendicular to the forces to meet their lines of action in D and E and let CD =p, CE=q. The triangles ACD, BCE are simi- lar. p CD CA Q Hence = -=?=, = ^^ = CE~ CB~ P' Fig. 39. Therefore Now p and q are the perpendiculars from the axis on the lines of action of P and Q. Thus in this case the condition of equilibrium is that P x perpendicular from axis = Q x perpendicular from axis. (ii) Let the forces P, Q, Fig. 40, be not parallel but let their lines of action meet at 0. Their resultant R acts through hence R acts along OC. Take OC to represent the resultant and from C draw OF, GG paral- lel to BO and AO respectively to meet OA and OB in F and G. Draw CD and CE perpendicular to the lines of action of P and Q, and let their lengths be p and q. Then OFCG is a parallelogram and OC represents the resultant of the forces P, Q along OF and OG respectively. Thus OF represents P and OG represents Q. Now the diagonal of a parallelogram bisects it. Hence area of triangle OFC = area of triangle OGC, and area OFC = %OF. CD, also area, OGC = OG . CE. Fig. 40. 17-18] PARALLEL FORCES. 45 Thus OF.CD=OG.CE. p CD 00 Q rT PT1PP q~ CE~ OF~ P' Therefore P . p = Q . q. Thus the condition of equilibrium is the same as before, P x perpendicular from axis = Q x perpendicular from axis. It may be shewn that, when three or more forces act, the condition of equilibrium is the same in form; the quantities involved are the strength of each force multiplied by the length of the perpendicular from the point C on the line of action of the force. 18. Moment of a force. The quantity which we have thus been led to consider has been given a name ; it is called the moment of the force about the point. DEFINITION. The Moment of a force about a given point is the product of the force and the perpendicular drawn from the point on to the line of action of the force. Thus the condition of equilibrium just found may be ex- pressed by the statement that The moment of P round C is equal to the moment of Q round C. Again if we consider the force P only it will turn the body in one direction round the axis, while Q, alone, would turn it in the opposite direction. Under the action of P the body would rotate in a direction opposite to that of the hands of a watch, placed face uppermost on the page, under the action of Q it would rotate in the same direction as the hands of the watch. This fact is generally expressed by the statement that the moments about C of P and Q are opposite in sign. The moment of P is said to be positive, that of Q is nega- tive. Thus when the bar is in equilibrium the moments of the forces round C are equal in magnitude and opposite in sign, or in other words, having regard to the difference in sign, we may state that the sum of the moments about C of all the forces is zero. 46 STATICS. [CH. II 19. Experiments on Moments. We shall now describe some experiments to verify this result. EXPERIMENT 3. To prove that, when a body which can turn about a fixed point is in equilibrium under two forces, the moments of these forces about the point are equal and opposite. For this experiment the bar employed in Experiment 2 is again used. It is however suspended by an axis through C, Fig. 41, so that it can rotate in a vertical plane. a I I | I I I no I I I I TT Fig. 41. This is attained either by boring a hole through the centre of the bar and passing a piece of steel wire through the hole, the ends of the wire pass through two screw-eyes fixed into some convenient support, the wire thus forms the axis ; or if more convenient, a screw-eye may be fixed into the bar instead of boring a hole through it. A pin run through these eyes forms the axis. By suspending the bar very close to its centre, the effect of its own weight in tending to turn it will be very small, and may be omitted from consideration. [See 35, 37 Centre of Gravity.] (i) When the forces are parallel. Suspend by means of the wire rings carriers from two points A, B of the bar. Let C be the fixed point or fulcrum. Place weights on the carriers until the bar balances in a horizontal position. Let P be the total weight, including that of the carrier, suspended from A, Q that suspended from B. Determine the weights P, Q for various positions of the carriers A, B and measure in each case the distances AC and BC. Form a table giving in four columns corresponding values of P, Q, AC and C, then calculate the products P.AC and Q.BC. 19] PARALLEL FORCES. 47 It will be found in all cases that they are equal, moreover A and B are on opposite sides of C, hence the moments are opposite in sign. Thus the proposition is verified for two parallel forces. It can be verified for more than two such forces by placing several carriers on the bar and loading until equilibrium is secured. In all cases it will be found that the sum of the moments about (7, each taken with its proper sign, is zero. (ii) When the forces P, Q are not parallel. The bar is supported as before and two spring balances are suspended from a point as shewn in Fig. 42. The hooks of A C B Fig. 42. the balances are connected by strings to the points A and B and the lengths of the strings are adjusted until the bar is horizontal; the readings of the balances give the tensions P and Q ; the perpendicular distances p and q of C from the lines AO and BO are measured, and it will be found as before that P.p=Q.q. In this experiment some small error may be produced by the weights of the balances themselves, but it will not be large if the balances are stretched so that the tension in the strings may be considerable. Another arrangement of this apparatus is shewn in Fig. 43. The bar is used as before, the strings from A and B pass over pulleys and carry weights. Thus the forces P, Q impressed at STATICS. [CH. II A and B are measured by the weights suspended at the ends of the string, the pulleys which should move easily merely serve Fig. 43. to alter the direction in which the forces are impressed. In all these experiments it is necessary that the bar should turn easily on its axis. 20. Principle of the Lever. We have thus verified the law that : When forces in one plane are impressed on a body, which can turn about an axis perpendicular to that plane, and maintain it in equilibrium, the resultant passes through the axis and the sum of the moments of the forces, each taken with its proper sign, about the axis is zero. This law when applied to the case of two forces is often spoken of as the Principle of the Lever. 21. Moments. The moment of a force can be repre- sented geometrically and the representation will be found useful. Thus let AB, Fig. 44, represent a force P, and let C be a point, about which the moment of P is required. Draw CD perpendicular to AJB, Fig. 44 (a), or to A B produced, Fig. 44 (b), and join AC and BC. Then the moment of P about C is equal to P. CD. But P is represented by AB. 19-21] PARALLEL FORCES. 49 Hence P . CD is represented by AB . CD. And AB . CD = 2 area triangle ABC. Thus the moment of a force is twice the area of a A P Fig. 44 (a). Fig. 44 (&). triangle whose base is the line representing the force com- pletely, and vertex the point round which the moment is being taken. It should be noticed, that in making use of this proposition, the line AB must represent the force completely. It must therefore pass through the point of application of the force. Suppose OA, OB be two lines representing forces P, Q impressed on a particle at 0; complete the parallelogram A OB C, then for some purposes, e.g. in order to find the resultant, we may treat AC, which is equal and parallel to OB as representing Q, when so doing we bear in mind all the time that it represents it only in magnitude and direction, NOT in point of application, we cannot calculate the moment of Q about some point such as D, by finding the area of B the triangle DC A we must find the area of DOB. Fig. 45. We proceed now to some theorems about moments which are of great importance. PROPOSITION 15. To prove that the algebraic sum of the moments of two forces about a point in their plane is equal to the moment of their resultant about that point. (i) When the lines of action of the forces meet. Let the forces be P, Q impressed on a particle at in directions OA, OB respectively. G. S. 4 50 STATICS. [CH. II Let H, Fig. 46, be the point about which the moments are OP A Fig. 46. required and let OC be the direction of R, the resultant of P and Q. Draw HC y parallel to OA, to meet OC in C, and OS in B, and from C draw CA parallel to BO to meet OA in A. Then AOBC is a parallelogram having OC for its diagonal. Take OC to represent the resultant JK, then OA and OB must represent P and Q the components of R in the directions (9-4 and OB respectively. Thus the forces P, Q and R are represented respectively by OA, OB and OC. Again the moment of P about H is represented by twice the triangle HO A, that of Q by twice the triangle HOB, and that of R by twice the triangle HOC. Two cases will now arise. (a) If H \$ outside the angle AOB, Fig. 46, the moments of both forces P and Q about H are the same in sign. (b) If H is within the angle A OB, Fig. 47, the moments of the two forces have opposite signs. Fig. 47. 21] PARALLEL FORCES. 51 In (a) Moment of R = 2 A HOC. Now AHOC= AHOB + ACOB = AHOB + ACOA = AHOB+ AHOA, for HC is parallel to OA. Thus the moment of R = moment of P + moment of Q. Again (b) Moment of R = 2 A HOC. Now A HOC= A COB- A HOB = ACOA- AHOB = AHOA- AHOB, for BC is parallel to OA. Moment of R = moment of P moment of Q. In either case The moment of the resultant of two forces is the algebraic sum of the moments of the forces. (ii) When the lines of action of the forces are parallel. Let P, Q be the forces and R their resultant. From H, Fig. 48, the point about which the moments are taken, draw H HACB perpendicular to the lines of action of the forces / ?--. H, meeting them in A, B and C respectively. In the figure the moments of P and Q about H have the same Fig. 48. Moreover we know that since R is the resultant of P and Q, and P.AC^Q.BC. 42 52 STATICS. [CH. II Now the moment of R about H = P(HA + AC} + Q (HB-BC) = P.HA + Q.HB+P.AC-Q.BC Since P.AC = Q.BC. Thus the moment of R is equal to the moment of P together with the moment of Q. If the point about which the moments are taken between the lines of action of the forces as If 1 then the two moments have opposite signs. Also the moment of R = P (R,A -AC)+Q(BC- H^ The proofs which have just been given can be extended to the case of three or more forces. Thus we obtain the result that the sum of the moments of a system of forces in one plane about any point in that plane is equal to the moment of their resultant about that point. 22. Resultant of a number of Parallel Forces. We may notice (1) that this theorem enables us to find the line of action of the resultant of a number of parallel forces in a plane. For let P lt P 2 ... be the forces, R their resultant. Let be any given point in the plane and let OA^... cut the lines of action of the forces at right angles in A lt A 2) ..., and of the re- sultant in G. Then we wish to determine G. Now we have 21-22] PARALLEL FORCES. 53 Also taking moments about R . OG = P 1 . OA^ + P 2 . OA 2 + ... Hence 0^^' Thus (i) the position of G is found ; while (ii) The theorem includes that proved in Experiment 3, viz. that the algebraic sum of the moments of a system of forces about a point in the line of action of their resultant is zero. For, if be in the line of action of the resultant, then OG is zero and the sum of the moments vanishes. Examples. (1) Find the resultant of two parallel forces of 10 and 12 kilos' iveight acting at two points A and B 50 centimetres apart. Let the line of action of the resultant R cut AB in (7. Then E = 10 + 12 = 22 kilos' weight. Also taking moments about A, #.^(7=10x0 + 12x50 = 600. Thus AC=^= 27-27 cm. mm (2) The line of action of the resultant of two forces of 10 and 15 kilos' weight is 20 cm. from that of tlie smaller force. Find the distance between the lines of action of the two forces. Let AB perpendicular to the lines of action of the two forces cut the line of action of the resultant in G. Then AC =20 cm. and it is required to find AB. Take moments about A (W + 15)AC=15.AB, (3) Two men carry a pole 24 feet long supporting it at each end; from the middle point of the pole a mass weighing 3 cwt. is suspended. The weight of the pole is 1 cwt., and may be supposed to act at a distance of 8 ft. from one end. Find the weight carried by each man. Let AB be the pole, P and Q the upward pressures at A and B, C the middle point and G the point 8 ft. from A, at which the weight may be supposed to act. Thus P and Q will be the pressures on the men's shoulders at A and B respectively, and their resultant must just balance that of the 3 cwt. and 1 cwt. 54 STATICS. [CH. II Thus Also taking moments about A Q Q - If cwt. wt. Hence P=2cwt. wt. (4) Masses of 2, 4, 8 and 16 Ibs. are placed at a series of points in a line and at distances of 4, 3, 2 and 1 feet from the edge of a table. Find the magnitude and point of application of the resultant force. Let the resultant force be R Ibs. wt. and let it act at a distance of x feet from the edge of the table. Then = 2 + 4 + 8 + 16 = 30 Ibs. wt. Also taking moments round the edge Rx=2x 4 + 4x3 + 8x2 + 16x1, 52 30 ~30 = lttfeet. (5) A series of forces acting at a point are represented in direction and magnitude by the sides of a polygon taken in order. Shew that the sum of their moments about any point in the plane of the polygon is zero. The forces are in equilibrium ; they have therefore no resultant ; the moment therefore of the resultant is zero ; hence the sum of the moment of the forces is zero. (6) Forces act at the angles A, B, C of a closed polygon, each force being represented completely by one side of the polygon, and all the forces acting in the same direction round the polygon. Prove that the sum of the moments of the forces about any point in the plane of the polygon is con- stant. Let the point (Fig. 49) be within the polygon ; the forces P, Q, R etc. are completely represented by AB, BC etc. Thus moment ofP=2AOAB, moment of Q= 2 A OB C, etc. / \ "^>D Hence sum of moments of forces F< = 2 area polygon, and this is the same for all the positions of within the polygon. Now let be outside the polygon. The moment of P is again represented by 2 A OAB and so on; but in this case some of the mo- ments are positive, some are negative. If the sum of the areas of the triangles which give Fig. 49. 22-23] PARALLEL FORCES. 55 negative moments be subtracted from the sum of the areas giving positive moments, it will be found in any case that the difference is the area of the polygon. Hence in this case also the sum of the moment is twice the area of the polygon. 23. Moment about an axis. So far we have dealt only with forces in one plane. Suppose we have any system of parallel forces and consider a line at right angles to the direction of all these forces. Each of these forces is said to have a moment round this axis ; this is found by drawing a line perpendicular both to the direction of the force and to the axis, and finding the product of the length of this line and the force. This common perpendicular is the shortest distance between the direction of the force and the axis ; hence the moment of a force about an axis, perpendicular to the direction of the force, is the product of the force and the shortest distance between the axis and the direction of the force. We may put this otherwise thus. Pass a plane perpendicular to the axis through the direction of the force, then the moment of the force round the axis is the same as its moment round the point in which the axis cuts this plane. If the forces be not all at right angles to the axis each force can be resolved into two components, one at right angles to the axis the other in a plane through the axis. The product of the component at right angles to the axis into the shortest distance from the axis of its line of action is, in this case, called the moment of the force about the axis. PROPOSITION 16. To prove that for any system of parallel forces the sum of the moments of the forces about an axis perpen- dicular to their directions is equal to the moment of the resultant about the same axis. Let two parallel forces, P, Q be impressed on two particles A, B, Fig. 50, in directions perpendicular to the plane of the paper. Let Ox be a line in the paper about which moments are required, and let AL and EM be perpendicular on Ox. Join AB and divide it in C so that P. AC = Q. BC. The resultant (P + Q) of P and Q acts at C. 56 STATICS. [CH. 11 Draw CN perpendicular to Ox and HCK parallel to Ox, to meet AL and MB produced in H and K. Then Fig. 50. EG _ BK TC~~AH' Therefore P . A H = Q . BK. Now the moment of the resultant = R.CN=(P+Q}CN=P.HL + Q = P(AL- AH} + Q (BM+BK} = P.AL + Q.BM-P.AH+Q.BK = Sum of moments of P and Q. In the same way the proposition can be proved for any number of parallel forces. The proof may also be extended in a similar way to any system of forces; for our purposes it is sufficient to have established it for a system of parallel forces and to have shewn that the sum of the moments of any system of parallel forces, about an axis perpendicular to the direction of the forces, is equal to the moment of their resultant about the same axis. For examples of the use of this Proposition, see Section 38. CH. II] PARALLEL FORCES. 57 EXAMPLES. PARALLEL FORCES. 1. Four equal and like parallel forces act at the angular points of a quadrilateral ABCD. E is the middle point of AB and F of CD. Prove that the centre of the forces is the middle point of EF. Deduce that the lines joining the middle points of the opposite sides of a quadrilateral bisect one another. 2. A uniform rod ABCD moveable about a fulcrum, and thirty feet in length, has weights P, 3P, 5P, 7P attached to the rod at A, B, C and D, which are at equal distances apart. If the rod be in equilibrium, find the distance of the fulcrum from A. 3. A uniform bar of length 2 ft. Sin. and weight 5 Ibs. is supported on a smooth peg at one end and by a vertical string distant 4 inches from the other end. Find the tension of the string. 4. Two light rods AB, BC are rigidly connected at B and meet at right angles. Weights W and W are attached at A and C. If the system can turn about B shew that the tangent of the angle which AB makes with the horizon is . W AB ~ 5. A straight uniform heavy rod of length 6 feet has weights of 15 and 22 Ibs. attached to its ends, and rests in equilibrium when placed across a fulcrum distant 2^ feet from the 22 Ib. weight. Find the weight of the rod. 6. A straight uniform rod of length 6 feet and weight 11 Ibs. is placed across a fulcrum distant 2^ feet from one end to which a weight of 26 Ibs. is attached. What weight must be attached to the other end so that there may be equilibrium ? 7. A uniform bar of length 3 ft. 6 in. and weight 9 Ibs. is supported on a smooth peg at one end and by a vertical string distant 6 inches from the other end. Find the tension of the string. 8. A straight light rod 2 feet long rests in a horizontal position between two fixed pegs, placed at a distance of 3 inches apart, one of the pegs being at one end of the rod ; a weight of 5 Ibs. is suspended at the other end ; find the pressure on each of the pegs. 9. Let P and Q represent two like parallel forces acting at the points A and B respectively of a body. Let C be the point in the straight line AB through which their resultant R acts. If R = 14 Ibs. , Q = 8 ozs., and A B = 58 inches ; find A C and BC. 10. A uniform beam weighing 10 cwt. and lying horizontally between supports 50 ft. apart carries additional weights of 3 cwt., 5 cwt. and 8 cwt., at distances of 10 ft., 20 ft., and 35 ft. respectively from one support. Find the proportion of the whole weight born by each support. 58 STATICS. [CH. II H. ABCD is a square; E, the middle point of AB, is joined to C ; BD is joined. Forces of 4 Ibs. and 6 Ibs. act in AB, BC respectively; of 3 Ibs. and 2 Ibs. in AD, DC respectively; a force of ^/2 Ibs. in BD, and one of 5 ^/5 Ibs. in CE ; shew, by using moments alone, that the system is in equilibrium. 12. A rod AB moveable about a hinge A, has a weight of 20 Ibs. hung on to B ; B is tied by a string to a point C vertically above A and such that GB is 6 times AC: find the tension in the string BO. 13. A dog-cart loaded wi.th 4 cwt. exerts a pressure on the horse's back of 10 Ibs. ; find the position of the centre of gravity of the load, the distance between the pad and axle being 6 feet. 14. Two forces act on a body which can move round a fixed point and the body remains at rest. Shew that the moments of the forces round the point are equal. 15. What is meant by the moment of a force about a point ? A man and a boy carry a weight of 55 pounds between them by means of a pole 5^ feet long, weighing 20 pounds. Where must the weight be placed so that the man may bear twice as much of the whole weight as the boy ? 16. Find the magnitude and position of the resultant of four forces P, 2P, 3P, and 4P acting along the sides of a square taken in order. 17. The sides of a square are 15 inches long, at the ends of one side are two weights of 3 Ibs. each, at the ends of the opposite side two weights of 5 Ibs. each ; where does the resultant of the 4 weights act ? 18. A thin board in the form of an equilateral triangle, and weighing 1 Ib. has one quarter of its base resting on the end of a horizontal table, and is kept from falling over by a string attached to its vertex and to a point on the table in the same vertical plane as the triangle. If the length of the string be double the height of the vertex of the triangle above the base, find its tension. *CHAPTER III. COUPLES. 24. Theorems about Couples. DEFINITIONS, (i) Two equal and opposite parallel forces constitute a Couple. (ii) The line drawn at right angles to the directions of the two forces is called the Arm of the couple. (iii) The product of either force into the perpendicular distance between the lines of action of the two forces the arm of the couple is called the Moment of the Couple. (iv) A line drawn at right angles to the plane containing the two forces and proportional to the moment of the couple is called the Axis of the couple. (v) The couple is said to be positive when, to an observer looking along the axis from the point of application of one of the forces, the forces tend to turn the body on which they are im- pressed in the same direction as the hands of a clock appear to move. We are now prepared for some propositions about couples. PROPOSITION 17. The algebraic sum, of the moments of the forces which constitute a couple about any point in the plane of the couple is constant and is equal to the moment of the couple. Let each force of the couple be P and let 0, Fig. 51, be any point in the plane of the couple. Draw OAB perpendicular to the lines of action of the two forces. 60 STATICS. [CH. Ill Then the algebraic sum of the moments of the forces is P.OB-P.OA. Fig. 51. Thus the sum of the moments = P(OB-OA) = P.AB = moment of couple. PROPOSITION 18. Two couples impressed on a rigid body in one plane balance if their moments are equal and opposite. Let the forces of the one couple be P, P and its arm p, the forces of the other couple Q, Q and its arm q. Then we have that the moment P . p is equal to the moment Q . q. (i) If the lines of action of P and Q meet. Let two of the forces P, Q meet in 0, Fig. 52, and the other two P, Q in 0'. Draw O'M, O'N perpendicular to the directions of P and Q respectively, then O'M = p Thus we have the result that P.O'M=Q.O'N, or the moment of P about is equal and opposite to that of Q about 0. Hence by Section 22, 0' is on the line of action of the resultant of P and Q which act at 0. Fig. 52. 24], COUPLES. 61 Similarly, by drawing perpendiculars from on the lines of action of the two forces P, Q which act at 0', we can prove that is on the line of action of the resultant of these two forces. These two resultants are equal in amount and the one acts at 0' along O'O the other at along 00'. Hence they balance each other. Thus the two couples are in equilibrium. (ii) If the lines of action oj the two forces do not meet but are parallel. Let ACBD perpendicular to the common direction meet them as shewn in Fig. 53. We know that P . AB = Q . CD. Fig. 53. Then the resultant of P at B and Q at C is P + Q acting upwards at a point L where = Q.AC+Q.CD Again the resultant of P at A and Q at D is (P + Q) acting downwards at L' where = Q.AD. Thus (P + Q)AL = (P+Q)AL' J or AL = AL'. Hence L and L' coincide, thus we have P + Q acting upwards and P + Q acting downwards at the same point. Hence the system is in equilibrium. 62 STATICS. [CH. Ill It follows therefore from this proposition that we may alter the direction of the forces of a couple, keeping the two parallel, without modifying its effect. We may also alter the forces, so long as we alter in the inverse ratio the distance between them and thus keep the moment constant. Further we may alter the points at which we consider the forces impressed, if only the moment remains unchanged. Thus the forces of a couple may have any value, and one of them may be supposed to be impressed at any point in the plane of the couple, so long as the moment of the couple remains unchanged in magnitude and direction. PROPOSITION 19. Any system of couples impressed on a rigid body in one plane is equivalent to a single resultant couple whose moment is the algebraic sum of the moments of the in- dividual couples. Since two couples of equal moment, opposite in direction, balance, any two couples of equal moment, the same in direc- tion, produce equivalent effects. Thus we may replace any couple by another couple of the same moment with one of its forces passing through any given point. Suppose now that P lt P 2 ... be the forces of the various couples and p lt p z ... their arms. Replace the couples by another set of couples having the same forces P 1} P 2 ... and the same arms but such that all the forces are parallel, and that one force of each couple passes through a fixed point 0. Let OA^A Z ... Fig. 54, be perpendicular on the lines of actions of the forces and meet the second force of each couple respectively in A lt A 2 .... Then we have a system of parallel forces P lt P 2 ... etc. at 0, together with a second system of forces in directions opposite to these, viz. P l at A lt P 2 at A 2 etc. Moreover OA l = p l , OA 2 = pz etc. yp, +P, + p. Now the first system of forces has a resultant R acting at 0. Also R = P l + P 2 + . . .. Fig. 54. 24] COUPLES. 63 The second system has a resultant also equal to R at some point C in OA^A Z ... the direction of this force R being opposite to that of R at 0. The position of C is found by taking moments about and is given by = sum of moments of original forces. Now 7? at and R at C, acting in opposite directions form a couple whose moment is R . OC, this we have just seen is the sum of the moments of the original couples. Thus any system, of couples in a plane is equivalent to a resultant couple, whose moment is the sum of the moments each with its proper sign of the original couples. PROPOSITION 20. A force and a couple impressed on a rigid body cannot maintain it in equilibrium. Let 0, Fig. 55, be any point in the line of action of the force P. Change the arm of the couple, keeping its moment constant, until the forces of the couple are also P. Place the couple so that one of its forces acts at in a direction opposite to the im- pressed force P. Then we have two equal and opposite forces P impressed at and a third force P in a direction parallel to these but at a distance p from them. The two forces at ba- Fig. 55. lance, and we are thus left with an un- balanced force acting at A parallel to the impressed force P but at a distance p from it. Such a force cannot maintain equili- brium. PROPOSITION 21. A force impressed on a rigid body at any point is equivalent to an equal and parallel force, impressed at any other point, together with a couple whose moment is the moment of the force about that point. 64 STATICS. [CH. Ill Let a force P be impressed on a body at any point A, Fig. 56. Let be any other point of the body and draw ON perpendicular to the direction of P. At introduce two equal and opposite forces P lt P 2 equal and parallel to P. This will not affect the equilibrium. Then P at A and P z at constitute a couple whose moment is P . ON and there is left a single force P 1 at 0, equal and parallel to P. Thus the force PatA is equivalent to an equal and parallel force P at together with a couple of moment P . ON about 0. PROPOSITION 22. Any system of forces acting at different points of a rigid body in one plane is equivalent to a single re- sultant force acting at any given point of the plane together with a couple, whose moment is equal to the sum of the moments of the forces about that point. For, let there be a number of forces P, Q, R... acting at points A, j&, C. Let be any given point in the plane. Then by the last Proposition each of the given forces is equivalent to an equal and parallel force through together with a couple whose moment is the moment of the force about 0. The forces P, Q, R... impressed at have in general a single resultant. The couples have in general a single result- ant couple whose moment is equal to the sum of the moments of the couples, and therefore to the sum of the moments of the forces about 0. PROPOSITION 23. To find the conditions of equilibrium of a system of forces in one plane impressed on a rigid body. Such a system is, we have seen, equivalent to a single resultant force and a single resultant couple. Since a single force cannot balance a couple it is necessary and sufficient for equilibrium (i) that the resultant force should be zero, (ii) that the resultant couple should be zero. In order that the resultant force may be zero the sum of its components in any two directions at right angles must be zero. 24] COUPLES. 65 Since the moment of the resultant couple about any point is the sum of the moments of the forces about that point, it is necessary in order that the resultant couple may vanish that the sum of the moments of the forces about any point in the plane should vanish. If the resultant force is zero we are left with a resultant couple ; now the moment of a couple is constant, hence in this case the sum of the moments of the forces is constant about any point in the plane; and therefore if the sum of the moments of the forces vanishes about one point, it will vanish about all. The necessary and sufficient conditions for equilibrium of a system of forces in one plane therefore are (i) The sum of the resolved parts of the forces in each of two directions at right angles is zero. (ii) The sum of the moments of the forces about some one point in the plane is zero. Conditions equivalent to these hold for the ease of forces not acting in one plane. The following Examples illustrate the foregoing propositions. (1) Forces of 4, 5, 6, 7 Ibs. weight act at the angular points A, B, ' . = the work done by the resultant. If one of the forces act in a direction such as AB, Fig. 63, the result is still the same, for the work done by P 2 is in this case negative since the displacement takes place in a direction opposite to the force and M is to the left of L. Thus the work done by forces - ( OL - L M + MN ) a = N~ . a = the work done by the resultant. If the forces form a system in equilibrium, then the point in the diagram, corresponding to C, which forms one extremity of the line representing the last force, coincides with 0, the sum of the components of the forces -in any direction is zero, and the work done is zero. Thus, if a system of forces impressed on a particle be in equi- librium, no work is done in any displacement of the particle, provided that the forces are not altered by the displacement. If we make the displacement very small, we may, even in cases in which the forces do depend on the position of the Fig. 63. 70 STATICS. [CH. IV particle, assume without error in calculating the work that the forces remain unchanged. Thus the conditions of equilibrium, for a system of forces impressed on a particle, express the fact that the particle can be slightly displaced without expenditure of energy. Starting from this proposition as an axiom we could estab- lish the various laws already obtained as to the composition and resolution of forces. *28. Work done on a rigid body. We proceed now to shew that the conditions of equilibrium for a body under a system of forces in one plane also express the fact that if the body be in equilibrium no expenditure of energy is necessary to give it a slight displacement. PROPOSITION 25. To find the work done by a force when a body on which it is impressed receives a slight rotation about an axis at right angles to the direction of the force. Let AB, Fig. 64, be the direction of the force P impressed in the plane of the paper on a body. Let the body be turned about the point through a small angle so that OA is brought into the posi- tion OA ; then AA' is the displace- ment of A and if the angle be very small we may treat A A either as a small straight line or as an arc of a circle. M Moreover A A is ultimately, Fig. 64. when the angle is very small, at right angles to OA. Draw A'A 1 perpendicular to AB the direction of the force, and ON perpendicular from on the same line AB. Then since OAA' is a right angle AjAA' = 90 -OAN = NO A, and the angles at A l and N are both right angles, thus the triangles NO A, A^AA are similar. 27-28] WORK. EQUILIBRIUM. 71 A,A ON Hence -J-T-, = ^-r. AA OA A A' thus AiA = ON.~.. OA AA' Now -j^- is the circular measure of the angle 0. OA Hence A^A = ON x 6. Now the work done by the force P is equal to P . AA. Thus Work done = P . AA^ = P . ON x = x moment of P about 0. Hence, when a body is displaced through an angle 6 about an axis at right angles to the impressed force, the work done by the force is found by multiplying the moment of the force about the axis by the angular displacement. Now the circular measure of an angle is merely a number, the ratio of two lines, we see therefore how it is that the moment of a force and work are both measured in the same units, foot-pounds, foot-poundals or centimetre-dynes as the case may be. PROPOSITION 26. To find the work done, by a system of forces impressed on a body in one plane, when the body receives a small rotation in that plane about some point. If a number of forces are impressed on the body then 0, the angle through which the body is rotated, is the same for them all. Hence, if be the point about which rotation takes place, then the work done by each force is found by multiplying its moment about by the angle 0. Hence the whole work done = 6 {sum of moments of forces about 0\. Hence if the sum of the moments about any point is zero no work is done during any small rotation about that point. 72 STATICS. [CH. IV *29. Equilibrium of a rigid body. Now we have seen that one of the conditions of equilibrium of a body under a system of forces in one plane is, that the sum of the moments of the forces about any point in the plane should be zero. If this condition be satisfied no work is done by turning the body through a small angle about any point in the plane. Again any system of forces is equivalent to a resultant force acting at and a resultant couple about 0. The moment of this couple is the sum of the moments of the forces. If remains fixed, work is done only by the resultant couple, none is done by the resultant force, and the work done by the couple is found by multiplying its moment by the circular measure of the angle turned through. Again if all points on the body are displaced the same amount, no work will be done by the resultant couple. For the work done by one force of the couple is equal to that done against the other. Now any motion of a rigid body in one plane is made up of a translation, in which all points of the body receive equal parallel displacements, and by which some one point of the body is brought to its new position, together with a rotation of the body as a whole about this new position. For the position of the body will be known if we know the position of two points ; suppose then that the two points A, B B Fig. 65. (Fig. 65) come to the position A', B', so that A 'B' is equal to AB. We can bring AB to the position A'B' in two steps. First keep it parallel to itself and move A to A'. Then AB will come in the position A'B l parallel to AB. Secondly turn A'B^ about A till B l comes to B', then AB has been brought to A'B'. 29-30] WORK. EQUILIBRIUM. 73 If a system of forces be impressed on the body the system is equivalent to a resultant R acting at A and a couple about A ; let G be the moment of the couple. Owing to the first displacement the couple does no work, the whole work done is the product of R into the component of the displacement A A' in the direction of R. Owing to the second displacement since A 1 remains fixed the force R does no work. All the work done is done by the couple G and is equal to the product of G and the angle E^A'E'. Thus, if e denote the angle between the direction of R and AA', and the angle B^A'B', the whole work done is given by .AA'cose+Gx 0. In order then that no work should be done in any small displacement by which the body is brought from one position to another it is necessary and sufficient that both the resultant force and the resultant couple should vanish. Now the general form of the condition of equilibrium is that both the resultant force and the resultant couple should be zero. Thus in this case the condition of equilibrium expresses the fact that when a body is in equilibrium no expenditure of energy is needed to produce a small displace- ment. Work is done by some forces and against others, if there be equilibrium these two amounts of work are equal ; in Newton's phrase the action is equal and opposite to the reaction. 30. Virtual Velocities. The principle which we have just proved for the case of forces in one plane is often spoken of as the Principle of Virtual Velocities. It may be enunciated thus : Suppose each point of a rigid body to which a system of forces is applied to receive any small displacement consistent with its geometrical relations 1 . Multiply each force of the 1 By this phrase we mean that the displacements of each point must be such as can take place without altering the form or size of the body. 74 STATICS. [CH. IV system by the component of the displacement of its point of application estimated in the direction of the force ; then form the sum of the products so obtained. If the forces be in equi- librium this sum, is zero, and conversely, if the sum be zero for all possible displacements, the system of forces is in equi- librium. We may state this otherwise thus : Calculate the total work done in any small possible . dis- placement. Then (i) if the forces are in equilibrium this work is zero, (ii) if the work is zero for all such displacements then the forces are in equilibrium. The word Virtual is used in describing the Principle because the displacements considered are not actually made ;. they are hypothetical and the work is calculated on the suppo- sition that they are made. The displacements are spoken of as velocities because it is usual to suppose them to take place in the same time. The Principle is also called the Principle of Virtual Work. A number of problems can most conveniently be solved by the direct application of this Principle. Examples are given in Section 33. For the present we are concerned with shewing that the conditions of equilibrium merely express the fact, that no expenditure of energy is required in order to displace slightly a body under a system of impressed forces in equilibrium. 31. Conditions of Equilibrium. It has been shewn that a system of forces impressed in one plane on a rigid body is equivalent to a single force and a couple, and that for equilibrium both the force and the couple must be zero. It has also been pointed out that this condition expresses the fact that no expenditure of energy is needed to give the body a slight displacement. Now these conditions can be put into various forms each of which may be useful in special cases. We proceed then to 30-31] WORK. EQUILIBRIUM. 75 illustrate them by some problems and to state various useful theorems. In solving problems the two following rules will express the conditions of equilibrium in the most straightforward manner. (i) Equate to zero the sum of the components of the forces in two convenient directions at right angles. (ii) Equate to zero the sum of the moments of the forces about some convenient point. In this way three equations are obtained and the problem, if determinate, can be solved. Examples. (1) A uniform ladder of Jcnoivn length I and weight W rests on rough ground against a smooth vertical wall at a known angle a with the horizon. Find the reaction at the points of contact of the wall and the ground, assuming the iceight of the ladder to be equivalent to a single force W acting at its middle point. Let AB (Fig. 66) be the ladder, BC the wall. Let R be the reaction at the wall. Since the wall is smooth, the direction of R is horizontal. Let P be the reaction at the ground and let it make an angle 6 with the horizon. Since R acts horizontally and W vertically, the horizontal and vertical directions will be two "convenient" directions at right angles in which to resolve. The vertical components of the forces are W downwards, and P sin upwards. The horizontal components are R outwards from the wall and P cos 6 to the wall. Thus : Fig. 6 6. Kesolving vertically, W=Psme .................. ............... (1). Resolving horizontally, R=Pcos9 ................................. (2). For the third condition we may take moments about any point, but since the direction of P passes through A, by taking them about A we eliminate two of our unknowns, P and ; it will clearly be " convenient" to do this. Take moments about A, R.BC=W.1.. ...(3). 76 STATICS. [CH. IV But by geometry, Hence from (3), From (1) and (2), BC = AB sin a = Zsin a, AC=AB cosa = Zcos a. and W tan0 = =2 tan a. XI Hence R, P and are determined. We might have solved the problem by resolving in two other directions or taking moments about some other point, but less conveniently. For a different solution see page 78. (2) A uniform rod AB of length I and weight W is hinged at A to a vertical ivall; the end B is connected by a horizontal string to the wall, and the rod is inclined to the wall at an angle a; a weight W is suspended from B. Determine the tension in the string and the direction and magni- tude of the pressure on the hinge. Let T be the tension in the string, X and Y, in the directions indicated, the horizontal and vertical components of the force at the hinge. Resolving vertically, c Y= W+ W. Resolving horizontally, T=X. Take moments about A, . I sin a ' A Tlcosa=W'lsma+W W Fig. 67. Thus the forces are all determined. 32. Problems on Forces in one plane. The con- ditions of equilibrium can be put into different forms. Thus : 31-32] WORK. EQUILIBRIUM. 77 PROPOSITION 27. To prove that a system of forces is in equilibrium if the sum of the moments of the forces about each of three points not in the same straight line is zero. For let the sum of the moments about A be zero, then the system is either in equilibrium or has a resultant force through A. If the sum of the moments about B is also zero, this result- ant must pass through B. Thus it must act in the line AB. But similarly since the sum of the moments about C is also zero the resultant acts along AC', and since A, B, C are not in a straight line this is impossible, hence the resultant is zero, and the system is in equilibrium. In the case in which the forces acting on the body reduce to three, the following proposition is useful. PROPOSITION 28. When three forces in one plane maintain a body in equilibrium their lines of action meet in a point or are parallel*. (i) Let us suppose the directions of two of the forces meet in a point 0. Replace these two by their resultant acting through 0. Then the resultant and the third force maintain equilibrium, hence the line of action of the third force passes through 0. (ii) Suppose that the directions of two of the forces are parallel, replace these by their resultant which will also be parallel to the two forces. Then this resultant must balance the third force. Hence the three forces act in parallel directions. Corollary. By combining this with the triangle of forces we see that, if three forces whose directions are not parallel maintain a body in equilibrium, they can be represented by the sides of a triangle drawn parallel to their lines of action respec- tively. 1 It may be shewn that if three forces maintain a body in equilibrium their lines of action must be in one plane. 78 STATICS. [CH. IV In applying this proposition to problems it frequently happens that the lines of action of two of the forces are known. By determining the point of intersection of these two lines of action, the direction of the third force can be found, and then by an application of the triangle of forces the values of the forces can be obtained. We will apply this to the problem of the ladder already solved, and to some other cases. Examples. (1) A uniform ladder of known length I and weight W rests on rough ground against a smooth vertical wall at a known angle a with the horizon. Find the reaction at the points of contact of the wall and the ground, assuming the iveight of the ladder to be equivalent to a single force W acting at its middle point. Let G (Fig. 68) be the middle point of the ladder. The weight W acts vertically through G, the pressure of the wall R acts horizontally through B. Let the lines of action of these two forces meet at 0, then the pressure of the ground must act along AO. Let OG meet the ground in D, then the three forces W, JR, and P are parallel respectively to OD, DA, and AO, and act at 0, hence they are proportional to these lines. Thus IL- = ~ = W_ OD DA P_ AO' And lsina, A = Hence = I jsin 2 a + CC J- P= while if 6 is the angle the direction of P makes with the ground, ^? DA. (2) Two equal weightless rods AC, CB are connected by a smooth joint at C, the rod AC is free to turn about a smooth point at A, and the end B of the rod CB is free to move along a smooth groove AB. Forces P, Q in the plane ABC act at the middle points of and perpendicularly to the rods AC, CB respectively, both forces being directed towards the inside of the triangle ACB. Find the position of equilibrium and shew that if P=2Q the triangle is equilateral. Since CA = CB (Fig. 69) the triangle CAB is isosceles. Let the angle CBA-8. 32] WORK. EQUILIBRIUM. Fig. 69. The forces on the rod CB are the pressure at B normal to AB, let this be R ; the force P acting at G, the middle point of CB perpendicular to the rod, and the resistance of the hinge at (7, let this be S, and let the lines of action of R and P meet in M. Then since there are three forces on the rod, their lines of action meet in a point. Hence S acts along CM. Moreover tGMB=GMC=0, and G is the middle point of BC. Thus M G, the direction of P, bisects the angle CM B. Hence the two forces R and S are equal, and resolving the forces at M along M G we have P = 2Scos0=2Ecos0. Now the action of BC on CA at the hinge must be equal and opposite to that of CA on BC. Hence if MC be produced to N there is a force S acting on AC at C along CN. Let H be the middle point of AC and draw HN perpendicular to AC to meet CN in N. The force Q acts along NH. Thus the lines of action of two of the forces on AC meet in N. The third force acting on AC is T, the pressure at the hinge A. This force must therefore act along AN, and for the rod AC we have forces S, T and Q acting at N. Moreover L HNA = L HNC t and therefore S = T. Thus resolving along NH, Q = 2S cos HNC. Again in the figure Therefore, and Hence Thus Therefore, Hence we have = L lACB=18Q-28. L HCN= 180 - (180 - 20) - (90 - 0} = 30-90. L HNC = 90 - L HCN= 180 - 30. Q = 2S cos HNC = 2S cos (180 - M = -2S cos 30. P = 2Scos0, Q=- 2S cos 30. 80 STATICS. [CH. IV Therefore 6 is given by the equation _cos3 p cose =4 sin 2 0-1. If the triangle be equilateral = 60, sin0 . Hence p=3-l = 2. Q = 2P. 33. Virtual Velocities. A number of problems may readily be solved by the principle of work. In applying this principle we have to suppose the system displaced and to calculate the work done in the displacement by each force. We notice (i) that if any part of the system moves over a smooth surface no work is done by the pressure of the surface, for the displacement is at right angles to the direction of the force ; (ii) that if any part of the system consists of two rods connected by a smooth joint, no work is done by the mutual forces acting on the rods at the joints, for the displacement is the same for both rods at the joint, but the force which is impressed on the one rod is equal and opposite to that on the other, hence the work done is zero. The follow- ing examples illustrate the principle. Examples. (1) A weight P is tied to each end of a string ivhich hang* over two smooth pegs AB in the same horizontal line. At a point in the string midway between A and B a weight W is suspended. Find the angle which the string makes with the vertical when there is equilibrium. Let the angle be 0. Let the weight W be displaced a small vertical distance x so that O (Fig. 70) may come to 0', and in conse- quence let the weights P each rise a distance y bringing them to P'. Then the work done by W is W.x, P'f X>v ' + p ' that done against each of the weights P iBP.y. P^ Hence W.x = 2P.y. From draw OL perpendicular to AGf. Then since 00' is very small AL is equal to AO. 33] WORK. EQUILIBRIUM. 81 Now the length of the string from P to is the same as that from P' to 0'. PP' + P'A+AO Hence Therefore LO'=PF=y. And in the triangle L 00' the angle at L is a right angle, and the angle LO'O is 6. Hence L0' = 00' cos 0, Therefore JF 2P' This result could of course have been more readily obtained by re- solving the forces at in a vertical direction, this at once gives JF=2Pcos0. (2) Apply the principle of work to find the position of equilibrium for the rods in Example 2, p. 78. The reactions at A, B, C all disappear from the equation of work for the reasons given above ; the only forces which we have to consider are P and Q. Suppose now that the rod AC is brought into the position AC' by being turned through a small angle about A, and in consequence let CB come to the position G'B'. Let H', G' be the new positions of H and G. And let HH', which may be treated either as a small arc of a circle about A or a short straight line perpendicular to AC, be equal to x. Then CC' = 2x. Draw CL and C'M perpendicular on AB, and draw C'N parallel to BA to meet CL in N. Now since G and G' are respectively the middle points of BC and B'C', the component in any direction of the displacement GG' is half the sum of the components in the same direction of the displacements GG' and BB'. Now L ACG' is a right angle and / ACE is 180 - 20. Hence Fig. 71. G.S. 82 STATICS. [CH. IV BB'=2LM=2C'N = 2CC'sinC'CN and the projection of CC' in the direction of P is CC' sin C'CB which is equal to 2x cos 26. Again the projection of BB' in the same direction is, since P acts inward, - BB' sin ABC. Moreover AL Hence Thus the projection of BB' on the line of action of P is - 4# sin 2 6. Hence Projection of GG' on the line of action of P = {2x cos 26 - 4x sin 2 6} = x {1-4 sin" 6>}. Thus the work done by Q is Qx, that done by P is Px {1-4 sin 2 e\. Hence Qx + Px {l-4sin 2 0} =0, =4 sin 2 0-1, as before. (3) The opposite angular points A, <7, B, D of a rhombus ABCD are connected respectively by two elastic strings. When the whole is in equi- librium the tension in AC is P, that in BD is Q; find the angle of the rhombus. Let the diagonals intersect in O (Fig. 72). Let the rhombus be displaced so that it becomes A'B'C'D', so that A moves a distance AA' along OA and B a distance BB' along BO. Let OA = a, AA'=a, OB = b, BB' = p. Then a and /3 are small and the A principle of work gives -P.AA' + Q. BB' = Q. Hence = ?. Q a Let c be the side of the rhombus, then AB=c=A'B', 33] WORK. EQUILIBRIUM. 83 and a 2 + 6 2 = A B 2 = c 2 = A'B'* = (a + a)* + (b - ) 2 = a 2 + 2aa + a 2 + 6 2 - 2bp + p?. And if a and /3 are very small we may neglect a 2 and /3 2 . Hence 2aa- 26/3 = 0, -=? a 6 Hence 7! = ?= tan 2L4C. Q 6 Thus the angle between the sides of the rhombus is given in terms of the tension. EXAMPLES. 1. A uniform rod of weight W is supported from a point by two strings. One of these makes an angle of 60, the other an angle of 30, with the rod. Find the tensions in the strings. 2. Forces P, Q, and R act along lines OA, OB, and OC which are met by a line through O' in A, B, and C. Shew that if the resultant of the forces passes through 0', then ^O'A O'B ^O'C P OA + Q OB +R UC= ' 3. A rod whose length is 10 feet, and which is thicker at one end than at the other, balances about its centre when 10 Ibs. is hung from one end and 20 from the other ; while if 40 Ibs. instead of 20 is hung from the second end the fulcrum is at 4 feet from that end. Find the weight of the rod, and the position of its centre of gravity. 4. A heavy uniform plank 9 feet long can turn about a point 3 feet from one end. A man whose weight is double that of the plank stands upon it, with one of his feet midway between the centre of the plank and the point of support, and the other foot 2 feet from the end. Find the pressures which each of the man's feet must exert on the plank in order to preserve equilibrium. 5. A heavy rod equal in length to the radius lies in a smooth hemi- spherical cup, the centre of gravity of the rod being one-third of its length from one end. Shew that if 6 be the angle made by the rod with the vertical tan = 6. ABC is an equilateral triangle, and forces P, P and 2P act along BA, AC and CB respectively. Find the magnitude of the resultant and the point at which it cuts the line AC, produced if necessary. 6-2 84 STATICS. [CH. IV 7. A straight rod has its ends moveable on the arc of a smooth fixed curved wire in a horizontal plane ; if a string is fastened to the centre of the rod, find by a geometrical construction the direction in which it can be pulled without moving the rod. 8. 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 G is horizontal, and determine, by geometrical construction, the stresses at A and B. 9. If a weight equal to the weight of either rod be attached to the centre of the lower rod, prove 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 B = 3 tan a ; and that this stress is to the weight of either rod in the ratio *l + 9 tan 2 a : 4. 10. Two equal heavy uniform beams AB, BC, each of weight W, jointed at B so as to make an angle a with one another, rest in a vertical plane with the ends A, C on a smooth horizontal plane, and A C is joined by an inextensible string. Determine the tension of the string. 11. AB is a uniform rod of weight W attached by two light strings AC, BD to two points C, D in the same horizontal line. Assuming AC to be the shorter string, shew that it is possible to choose a weight P and place it on the rod so as to maintain it in a horizontal position, when P is at a distance from the middle-point of the rod equal to W + P sin (0-0) AB P sin (0 + 0) ' ~2* ' where and are the acute angles made with CD by A G and BD re- spectively. 12. A small bead P is free to move on a given smooth circular wire and is acted on by two forces represented by PA and PB, where A andB are fixed points in the plane of the ring. Find the positions of equilibrium. 13. The sides of a triangular framework are 13, 20, and 21 inches: the longest side rests on a horizontal smooth table and a weight of 63 Ibs. is suspended from the opposite angle. Find the tension in the side on the table. 14. A gate is hung in the usual manner by two hinges on a gate-post. Indicate the forces acting on the gate when it hangs open and in equili- brium, and shew that it may happen that the reaction of one of the hinges is wholly horizontal. Give a verbal explanation as well as a diagram. CH. IV] WORK. EQUILIBRIUM. 85 15. A curtain-ring can slide along a horizontal pole, which is at a height of 4 feet above my hand: if a string 10 feet long is attached to the ring, shew by a diagram in what direction I must pull the string so as to move the ring with most effect, and at what point of the string I must take hold. 16. Two small heavy rings of weights W and W connected by a light string slide on two wires in the same vertical plane making equal angles a with the horizon. If the string makes an angle with the horizon shew that (W+ W) tan $ = ( W- W) cot a. 17. A ladder rests against a smooth wall, the ground being also smooth. Compare the horizontal forces which must be applied to the bottom of the ladder to preserve equilibrium, when a weight equal to the weight of the ladder is placed on the ladder at the top and bottom respectively. CHAPTER V. CENTRE OF GRAVITY. 34. Centre of Mass. Consider two particles A, B, Fig, 73, of mass m, in' respectively, let them be connected by an in- extensible rod whose mass we may neglect, and suppose two parallel forces act on them, the force on each particle being pro- portional respectively to the mass of that particle. Let the two forces then be ma and m'a, a being some constant. These two forces have a resultant which acts at a point C in the line AB, dividing AB so that ma . AC = m'a . BC, Fig. 73. or AC EG m The point C is called the centre of mass of the two particles. Whatever be the position of the line AB the point C is fixed in that line and the resultant force ma + m'a always acts through it. 34] CENTRE OF GRAVITY. 87 Again if we have three masses m ly m z , m 3 at A lt A%, A s Fig. 74, and forces m^a, m 2 a, m 3 a are impressed on these, the centre of mass of ra x and m 2 will be a A, point C in A-^A^ such that we may consider the forces and m z a to be replaced by their resultant (m-^ + m 2 ) a impressed at C. Then this force at C and Fig. 74. m s a at A 3 will have a resultant acting always at G a point in A 3 C such that CG = This resultant will be (ra x + ra 2 + m 3 ) a. The point # is the centre of mass of the three particles. In general, since the resultant of any number of parallel forces impressed on a rigid body at various points is a force equal to the sum of the individual forces, whose line of action always passes through a fixed point in the body, we see that if parallel forces m^, m 2 a..., be impressed respectively on each of the particles m l) m.,..., of a rigid body these forces will have a resultant equal to (m-^ + m^..) a whose line of action passes through a fixed point of the body. This point is called the centre of mass of the body. DEFINITION. The Centre of Mass of a body is the point of action of the resultant of a system of parallel forces impressed on each of the particles of the body, each force being proportional to the mass of the particle on which it is impressed. This point is fixed in the body. The position of the centre of mass does not depend on the direction of the parallel forces but only on their amounts and on the points at which they are impressed, thus if the body be turned in any way, the forces still remaining parallel, their resultant still acts through the same point in the body. 88 STATICS. [CH. V Again there is only one centre of mass, for if there be two, ff l and H 2 turn the body if necessary until the line H-JEL^ is at right angles to the forces. Then the resultant force acts in one line through // x and in a second parallel line through N 2t which is impossible. Hence there is only one centre of mass. 35. Centre of gravity. The weight of a body is the resultant of the weights of the particles of which the body is composed, the weight of each particle is proportional to its mass and, if the body is small compared to the earth, the lines joining the particles to the centre of the earth are all parallel, so that the weights of the particles form a system of parallel forces, each being proportional to the mass of the particle on which it is impressed. The resultant of these forces passes through a fixed point in the body the centre of mass or as it is more often called when considered in connexion with the weight of the body the centre of gravity. DEFINITION. The weights of the various particles of which a body is composed form a system of parallel forces ; these forces have a resultant equal to their sum. This resultant passes through a point which is fixed in the body however it be placed. This point is called the Centre of gravity of the body. Thus we may treat the weight of a body of finite size as a single vertical force impressed on the body at a definite point ; this point is its centre of gravity. If we allow for the fact that the weights of the various particles are not strictly parallel forces, it does not follow that their resultant passes in all cases through a fixed point in the body. The body has no centre of gravity though it has a centre of mass. It is clear that if the only impressed force be the weight of the body and the centre of gravity be supported the body will balance in any position in which it may be placed. 34-35] CENTRE OF GRAVITY. 89 We shall describe first an experimental method which can sometimes be applied in order to find the centre of gravity of a body, and then shew how to determine by calculation the position of the centre of gravity for each of certain bodies. The following proposition will be needed. PROPOSITION 29. To prove that, if a body is suspended freely from one point, the centre of gravity is either vertically above or vertically below the point of suspension. Two forces only act on the body, viz. its weight and the force exerted at the point of support. These two forces must be equal and their lines of action must coincide. Now the weight acts vertically through the centre of gravity. Hence the point of support must be in the same vertical as the centre of gravity, and the direction of the force at the point of support must be vertical. Two cases of this proposition arise. In the one, Fig. 75, the centre of gravity G is vertically below the point of support ; if the body in this position be slightly displaced it will tend to return to it ; the equilibrium is said to be stable. Fig. 75. Fig. 76. In the other case, Fig. 76, the centre of gravity G is verti- cally above the point of support, the body, if displaced, will not return to its original position ; the equilibrium is said to be un- stable. 90 STATICS. [CH. V 36. Experiments on Centre of gravity. EXPERIMENT 4. To find the centre of gravity of a plane lamina 1 . Attach a string to any point A, Fig. 77, of the body and suspend it by the string. Draw, by the aid of a plumb line hanging from the same support as the string, on one face of the lamina a vertical line AG through^!. The centre of gravity lies in this line, pro- vided the lamina be very thin. Suspend the lamina by a string attached to another point , and draw on the same face a vertical line EG through , intersecting AG in G. The centre of gravity lies in this vertical line. Hence the cen- tre of gravity must be G, the point where these two lines Fig. 77. intersect. To verify this, sus- pend the lamina from a third point (7, the vertical line through C will also be found to pass through G. Determine in this way the position of the centre of gravity for a triangular lamina and for a square with a corner cut off; and shew that they agree with their theoretical positions. Sections 37, 39. EXPERIMENT 5. To find the centre of gravity of a frame- work. On one of the bars of the frame-work a light wire is fixed, this has a ring at one end. Suspend the frame- work from any 1 A lamina is a thin flat sheet of any material, such as a sheet of paper or cardboard or of very thin metal; for the experiment a thin wooden board will serve. The centre of gravity will be in the interior midway between the surfaces of the board. 36-37] CENTRE OF GRAVITY. 91 point A, Fig. 78, and bend the wire so that a vertical string through A passes through the ring. Sus- pend the framework from a second point B, and keeping the string from A through the ring determine where the vertical through B cuts this string, this point will be the centre of gravity of the frame-work. Bend the wire until the ring just comes into this position, so that when suspended from A or B the vertical from the point of sus- pension passes through the ring. It will be found that when suspended from any other point the vertical through that point passes through the ring. The centre of gravity of a solid body cannot be found in this manner, because of the impossibility of reaching the point in the interior of the body which is always vertically below the point of support. 37. Centre of gravity found by Calculation. The position of the centre of gravity can be found in some cases from consideration of symmetry. PROPOSITION 30. To find the centre of gravity of a uniform straight rod. The centre of gravity is clearly the middle point of the rod, for let AB, Fig. 79, be the rod. Bisect AB in G. Let P and Q be p two equal particles of the rod equi- ^ 1 1 \ distant from G. Then the result- G ant of the weights of these two particles passes through G. And the Fig. 78. Fig. 79. whole rod can be divided into a number of such pairs of equal particles, the centre of gravity of each pair is G, hence the centre of gravity of the rod is G. PROPOSITION 31. To find the centre of gravity of a lamina in the form of a parallelogram. Let ABCD, Fig. 80, be a parallelogram composed of some material of uniform thickness and density. 92 STATICS. [CH. V Bisect the sides AB and CD in E and F and join EF ; bisect <7 and DA in # and K and join ffiT intersecting EF in . Let PQ meet EF in R. Then since D~~ ~~f~ EF is parallel to BC and bi- Fi 8Q sects AB and CD it also bi- sects PQ. Thus R is the middle point of PQ. Now we may consider the strip PQ as a uniform straight line ; its centre of gravity therefore is at R. The whole parallelogram may be considered as made up of a series of strips such as PQ, the centre of gravity of each of these is on the line EF, thus the centre of gravity of the whole is on the line EF. In a similar way the parallelogram may be divided up into a series of narrow strips parallel to A D ; the centre of gravity of each of these will be at its middle point, that is on the line HK. Hence the centre of gravity of the whole is on the line HK. Thus the centre of gravity of the whole parallelogram is G, the point of intersection of HK and EF. PROPOSITION 32. To find the centre, of gravity of a uniform triangular lamina. Let ABC, Fig. 81, be a uniform triangular lamina. Bisect the sides BC, CA, AB in D, E and F, and join AD and BE intersecting in G. Then G shall be the centre of gravity required. Divide the triangle into a series of narrow strips, such as PQ, by lines drawn parallel to BC. Let AD meet PQ in R. Then since PQ and BC are paral- lel, and are met by the three lines AB, AD and AC we have PR^BD B D QR CD Fig. si. 37] CENTRE OF GRAVITY. 93 Thus R is the middle point of PQ f Hence R is the centre of gravity of the strip PQ. Hence the centre of gravity of all strips such as PQ lies on AD. Thus the centre of gravity of the triangle is in AD. Similarly, by dividing the triangle into strips parallel to CA, we can prove that the centre of gravity of the triangle is in BE. Therefore the centre of gravity of the triangle must be G, the point where AD and BE intersect. Again since D and E are the middle points of CB and CA respectively, DE is parallel to BA and equal to \BA. Moreover, since DE and AB are parallel and are met by AD and BE, LADE = LDAB, and L BED = L EBA. Hence the triangles GDE and GAB are similar. DG DE Thus AG = TB = ^' Hence DG Similarly E and if CG be joined it will when produced pass through F the middle point of AB and FG = Thus the centre of gravity of a triangle is found by joining any angular point to the middle point of the opposite side, and taking a point on this line at a distance from the angle equal to ^ds of the whole length. PROPOSITION 33. To shew that the centre of gravity of a triangle is the same as that of three equal masses at its angles. It is clear that the point G thus found will be the centre of gravity of three equal masses placed at the angular points of the triangle. For consider two equal mases m at B and G, their centre of gravity is D midway between them, and we may 94 STATICS. [CH. V replace them by 2m at D. Now since AG = 2GZ> we have Thus G, which is the centre of gravity of the triangle, is also the centre of gravity of m at A and 2m at D. It is therefore the centre of gravity of the three masses, m at A, B and C. PROPOSITION 34. To find the centre of gravity of three uniform rods, BC, CA and AH forming a triangle. Let a, b, c be the lengths of the rods respectively, their masses are proportional to a, b and c. Bisect the rods in D, E, F, Fig. 82, then the points D, E, F are the centres of gravity of the rods and we have to find the centre of gravity of masses a, 6, c at the points D, E, F respectively. Let H, a point in FE, be the centre of gravity of masses 6 at E and c at F, K of masses c at F and a at D, L of masses a at D and b at E. _b_AC F.D EH~c~AB~~ z lED' 1 nen -r^ - ^ 3 D Therefore DH bisects the angle FDE and we have to find the centre of gravity of b + c at If and a at D. It is clearly a point in DH. But by considering first c at F and a at D we can shew similarly that the centre of gravity required is a point in EK. Hence the centre of gravity of the three rods is G, the point in which DH and EK coincide. This point is clearly the centre of the circle inscribed in the triangle DEF. 38. Formulae connected with Centre of Gravity. Formulae can be found which enable us in many cases to obtain by calculation the position of the centre of gravity of a body. Thus: 37-38] CENTRE OF GRAVITY. 95 PROPOSITION 35. To find the position of the centre of gravity of a number of particles in a straight line. Let A^A 2 , Fig. 83, be the positions of the particles their masses. Take any point in the line and let X-L, x 2 ... be the distances of j- 1 j-^ | j- the particles from 0. "* i 4 Let G be the centre of Fig. 83. gravity. Then G is the point of application of the resultant of a series of parallel forces proportional to ra^ m 2 , etc., acting at A lt A zt etc., respectively. And by Proposition 1 5 the moment of the resultant of these forces about is equal to the sum of the moments of the forces. Hence, taking moments about 0, tn., 1 . Hence OG ' 2(m)' where as before ^,(mx) stands for the sum of a series of quantities like mx. PROPOSITION 36. To find the centre of gravity of a number of particles in a, plane. Let m 1 , m 2 be the masses of the various particles placed at points A l} A 2 etc. in a plane. Let 0, Fig. 84, be any fixed point in the plane, and Ox, Oy two lines at right angles meeting in 0. Let A 1 M l , A 2 M Z , etc., be perpendicular on Ox, and A-Jj^ A Z L^ etc., perpendicular on Oy. Let A-iL-L^x-^ A 2 L 2 = x 2 , etc. A-iMi = y 19 A Z M Z = 2/0, etc. 96 STATICS. [CH. V Let G be the centre of gravity and let GL, perpendicular on Oy, x, and GM, perpen- dicular on Ox, y. We require to find the point of application of the re- sultant of forces proportional to m lt in z , etc., impressed on particles at A u A 2 etc. The point of application of the resultant is the same whatever be the direction of the forces so long only as they all remain parallel. Let us sup- pose then that they act per- Fig. 84. pendicularly to the plane of the paper. They are then at right angles to Ox and Oy and by Prop. 15 the moment of the resultant about Ox and Oy is equal to the sum of the moments of the forces about these lines. Hence, taking moments about Oy, A 3 A 2 A< G 1 M MI M 2 GL = ra x . A i L l Thus x = GL = m = ?<*) , 2 (ra) while, taking moments about Ox, .) GM= m^ . A 1 M 1 mj + m z + . . . 2 (m) By these two equations the position of G is determined. A similar method can be applied to a series of particles in space. In this case we shall have to consider three axes at right angles and obtain the formulae. We will illustrate the results by some examples. Examples. (1) Find the centre of gravity of masses of 10, 20, 30, 40, and 50 grammes arranged in a straight line at intervals of 10 cm. apart. Let be the position of the 10 gramme mass. The distances of the various masses from are therefore 0, 10, 20, 30 and 40 cm. respectively. Let G be the centre of gravity, 38] CENTRE OF GRAVITY. 97 then, taking moments about 0, OG (10 + 20 + 30 + 40 + 50) = 10-0 + 20-10 + 30-20 + 40-30 + 50-40, (2) Masses of 10, 20, 30 and 40 grammes are placed at the four angles of a square each side of which is 20 cm. in length, and a mass of 50 grammes at the centre; jind the centre of gravity of the whole. Let ABCD (Fig. 85) be the square, E the centre, and the masses be placed as shewn in the figure. (i) By application of the formula. Draw lines Ex, Ey through E parallel to the sides of the square. Let G be the centre of gravity, x and y its distances from Ey and Ex respectively, x being measured to the right, and y upwards. The distances of the angles of the square from Ex and Ey are each 10 cm., but in taking mo- _ _ ments about Ex it must be noted that the A.10 B.20 moments of the 30 and 40 grammes are of opposite sign to those of the 10 and 20 Fig. 85. grammes. And similarly, when taking mo- ments round Ey, the moments of the 10 and 40 grammes are negative, those of the 20 and 30 grammes positive. Hence, taking moments about Ey, (10 + 20 + 30 + 40 + 50) = 50-0 + 40-10 + 30-10 + 20-10 - 10-10 =500-500 = 0. Hence x = and G lies in Ey. This result is obvious from the symmetry. Taking moments about Ex, (10 + 20 + 30 + 40 + 50) = 50-0 + 40-10 + 30-10 - 20-10 - 10-10 = 400. 400 Therefore y = ^ = 2%cm. Thus the centre of gravity is on Ey at a distance of 2| cm. above E. G. S. 7 98 STATICS. [CH. V (ii) By geometrical construction. Divide DA (Fig. 86) into 5 parts and let H be the first division from D so that DH_l 40 F 3(^ ~' E|50 10 20 H is then the centre of gravity of 40 grammes H at D and 10 grammes at A, and it is at a distance of 4 cm. from D. Similarly, if K be a point on CB at a distance of 8 cm. from C, then K is the centre of gravity of 30 grammes at C and 20 at B. A We have now to find the centre of gravity of 50 grammes at H, 50 grammes at K and 50 Fig. 86. grammes at E. Join HK, cutting in L a line EF through E parallel to the side DA. Let EF meet DC in . Then HL = LK, and L is the centre of gravity of 50 grammes at each of the points H and K. Also, since Dff=4 cm. and CK=8 cm., .FZ = 6cm., and therefore EL = 4 cm. We have now to find the centre of gravity of 50 grammes at E and 100 grammes at L. This will be a point G in EL such that This is of course the same point as was found by the previous method. (3) Where must a mass of 100 grammes be placed in order that the centre of gravity of the system and the 5 masses described in the previous question may be at the centre of the square ? The centre of gravity of the four particles at the angles of the square has been shewn to be at L (Fig. 86) in the line EF at a distance of 4 cm. from E. Produce LE to L' making L'E = EL = cm., and place a mass of 100 grammes at L'. The centre of gravity of 100 grammes at L and 100 grammes at L' is at E half way between them ; the 50 grammes is already at E. Hence in order that the centre of gravity of the whole may be at E, the 100 grammes must be placed at L'. (4) A uniform wire ABC is bent at B so that the angle ABC is 60, and suspended from the point A. The part AB is a cm. long. Find the length of BC in order that when the whole is in equilibrium BC may be hori- zontal. Let BC=x cm. 38] CENTRE OF GRAVITY. 99 Let H and K (Fig. 87) be the middle points of AB and EG. Draw AD vertical and HL horizontal. Then AB = a, hence BD = | , and = |; also BK=l , hence DK= X = - % . & 4 Now H is the centre of gravity of AB, and the mass of AB is pro- Fig. 87. portional to its length a. Again K is the centre of gravity of BC, and the mass of EC is proportional to its length x. Thus, taking moments about A, Hence =\x(x-a). Thus 2. 9L 39-40] CENTKE OF GRAVITY. 103 Hence GG l = ^a-%h = Take moments round G. Then %ah . Thus GG 9 =- -h GG 1 2a-fc 6(2a-/i) Thus the position of G 2 is found. The results of these last three examples may be verified by experiment by cutting out in stiff card or sheet metal lamina of the shape indicated, and determining the positions of their centres of gravity by the method indicated in Experiment 4. 4O. Equilibrium of a body resting on a horizon- tal surface. When a heavy body rests on a flat horizontal surface it is in equilibrium under its weight, which acts verti- cally downwards through its centre of gravity, and the upward pressures at the points of contact with the surface. These upward pressures have a vertical resultant, and this resultant must balance the weight; it must therefore act vertically upwards in a line which passes through the centre of gravity of the body. If the form and position of the body is such that this is impossible the body cannot be in equilibrium. Thus suppose the body is a vertical lamina, a sheet of card or metal having the shape shewn in Fig. 92, and that it is in contact with the table at two points A and B, The pressures of the table at A and B are both up- ward vertical forces. The line of action of their resultant must lie between A and B. Hence for equilibrium the position of the centre of gravity must be such that a vertical drawn through it must fall between A and B. If the centre of gravity be in a position such as 6 ? , Fig. Fig. 92. 92, the lamina can remain in equilibrium, if it have a position such as G in Fig. 93 equilibrium is impossible. 104 STATICS. CH. V In the same way if the body rest on three points, like the legs of a three-legged table, the resultant of the upward pres- sures balances the weight. If a triangle be formed by joining the points in which the three legs rest on the floor, the result- ant upward pressure must act within this triangle, therefore the vertical through the centre of gravity must fall within the ^ig. 93. triangle. Or suppose again that the body rests in contact with the table at more points than three. Imagine a string drawn tightly round the body so as to include all these points of con- tact, thus forming a closed polygon whose sides are either straight or convex outwards. The pressures at the various points of support all act vertically upwards at points within the area thus denned ; the resultant pressure therefore acts vertically upwards at some point within this area, and this resultant pressure, since it balances the weight, must pass through the centre of gravity. Thus if equilibrium is possible the vertical through the centre of gravity must fall within the area thus denned. This area is sometimes spoken of as the base of the body, and the proposition is expressed in the statement that, in order that a body, resting on a plane under gravity, may be in equilibrium, it is necessary that the vertical through the centre of gravity should fall within the convex polygon formed by joining the extreme points of contact of the body and the plane. This polygon must have no reentrant angles. Thus if ABODE (Fig. 94) be points of contact having a reentrant angle at D, the boundary of the base is ABCEA. A string stretched round the points of support would pass from C to E. The resultant pressure must lie within tbis polygon, though it might quite well lie without the polygon ABCDE. Examples. (I) A right-angled tri- angle ABC having angles at B and C of 30 and 60 respectively, rests in a vertical E plane on a horizontal table, the side AC being vertical and A being the right angle. Fig. 94. 40] CENTRE OF GRAVITY. 105 The point C is joined to a point D in AB and the triangle DAG is removed. Find the largest triangle lohich can thus be removed without disturbing the equilibrium of the rest. Bisect EC (Fig. 95) in L and join DL. triangle CBD lies in DL. If the angle BDL is acute, a vertical from the centre of gravity must fall within the "base" BD, if it be obtuse the vertical must fall outside the "base," the limiting position of D then will be found by drawing LD vertical, and in this case LD is parallel to CA. Thus since L is the middle point of BC, D is the middle point of BA. Hence the area of the triangle DBC = ihe area of the triangle DAG = % area of the triangle ABC, and half the original triangle may be removed without disturbing the equilibrium. The centre of gravity of the B Fig. 95. (2) A' brick 8x3x4 inches in size rests with its smallest face on an inclined plane, the 3-inch side being horizontal; the brick is prevented by the friction from slipping down. Find the greatest angle to which the plane can be raised without causing the brick to fall over. Let ABCD (Fig. 96) be a section of the brick by a vertical plane, AB being the inclined plane. Let G be the centre of gravity of the brick. Then so long as the vertical through G falls between A and B, equilibrium is possible. In the limiting position GA is vertical. Thus the angle GAD is equal to the angle of the plane. Now AG passes through C. CD Thus tan GAD= tan CAD=j- Fig. 96. =f 4- Thus the plane can be raised until it makes with the horizon an angle whose tangent is . (3) A circular table rests on three legs attached to three points in the circumference at equal distances apart. A weight is placed on the table, determine in what position the weight is most likely to upset the table, and find the least value of the loeight which when placed in that position will upset the table. If the table is upset by placing a weight on it, it will at first turn round an axis passing through the lower ends of two of the legs. A given weight therefore will be most effective in turning the table over when its 106 STATICS. [CH. V moment round such an axis is greatest. This will be the case when the weight is as close to the edge of the table as possible, and at a point D midway between two of the legs A and B. Again let ABC (Fig. 97) represent the top of the table and G its centre of gravity, A, B, G being the points at which the legs are attached, and D the middle points of the arc AB. Join GD cutting the line AB in K. Then if W be the weight of the table, W that of the weight which is placed on it, the table will not upset so long as the mo- ment of W about AB is not greater than that of W. Hence in the limiting condition we must have W . DK = W . GK. But since the angle AGB is 120 and DA is equal to DB, the triangles DGA, DGB are equilateral and the figure DAGB is a rhombus, thus DG is bisected in K. Hence f>K=KG. B Thus the table will not upset so long as the weight supported at D is less than that of the table. 41. Stability of Equilibrium. Consider any body supported at one point, such as a lamina, which can turn round a horizontal axis through a point 0, Fig. 98. We have seen that the condition for equilibrium is that the centre of gravity should be in the vertical through 0. Three cases however may occur; the centre of gravity may be below 0, or above 0, or it may coincide with 0. Consider the first case and sup- pose the body to be slightly displaced ; so that the centre of gravity G is brought to G'. Then the weight of the body W acting through G' has a moment about which tends to bring the body back to its original position, the equilibrium is said to be stable. Fig. 98. If however G be above as in Fig. 99, and the body be displaced so that G may come to G' the moment of the weight 40-41] CENTRE OF GRAVITY. 107 about tends still further to in- crease the displacement, the equi- librium is unstable. And thirdly if and G coin- cide the body will balance in any position however it may be turned about 0, the equilibrium is said to be neutral. The above illustra- tion affords an example of what is meant by the terms stable, un- stable and neutral, which are applicable generally to bodies in a position of equilibrium. Fig. 99. DEFINITION. Consider a body which has been slightly dis- placed from a position of equilibrium. If the body tends to return to that position, its equilibrium is stable. Thus a weight suspended by a string from a fixed point is in stable equilibrium, so is an egg resting with its shortest diameter vertical, or a sphere which has been loaded at one point and rests on the table with the loaded part downwards. DEFINITION. A body at rest which, after receiving a small displacement, tends to move further away from its equilibrium position is in unstable equilibrium. Thus it is possible to make an egg rest on its point, or to balance a stick on its lower end, but the very slightest disturb- ance upsets the equilibrium ; again a loaded sphere may rest with the load uppermost, but if ever so little displaced it will turn until the load comes to the bottom. These are all cases of unstable equilibrium. A wheel which has a load attached at one point can rest with this load either below or above the axle. In the second position the equilibrium is unstable, if the wheel be disturbed the load will move until it settles itself in the lowest position. DEFINITION. A body is in neutral equilibrium when after receiving a small displacement it will rest in its new position. A truly balanced wheel or a uniform sphere or cylinder resting on a flat surface are all in neutral equilibrium. 108 STATICS. [CH. V iNow in these various cases, in which the weight of the body is the only impressed force in addition to the reaction of the supports ; we notice, that for equilibrium the centre of gravity is either as high as possible or as low as possible. The potential energy of the body depends on the position of its centre of gravity ; in the first case the potential energy has a maximum value, in the second case it has a minimum value. In either case, if the body be very slightly displaced, the height of the centre of gravity, and therefore the value of the potential energy, is at first altered by a quantity which is itself very small compared with the change in the position of the body. Thus let the body be turned through a very small angle 6, so that OG (Fig. 100) becomes OG', and L GOG' = 0. Draw G'K perpendicular to OG, then the centre of gravity is raised a distance GK. Now if / GOG' is very small, then OG'G is very nearly a right angle. Hence the triangles KGG' and G'GO are similar. Thus Hence KG__ GG'~ KG = GG' GO' GG' 2 GO Fig. 100. And if GG' is small GG^/GO is very small indeed. Compared with its horizontal displacement, the change in height of the centre of gravity is very small. This proposition is found always to be true, in an equili- brium position the potential energy has always a maximum or a minimum value, the change in potential energy consequent on a small displacement is very small when compared with the displacement, it depends on the square of the displacement. Again, in all the cases of unstable equilibrium, the centre of gravity is as high as possible, the potential energy has a maxi- mum value, the change produced by a small displacement is very small but, such as it is, it tends to reduce the potential energy of the body, the energy tends to take the kinetic form, the potential energy tends to decrease. When the equilibrium is stable the centre of gravity is as low as possible, the potential energy has, in the equilibrium position, a minimum value, the change due to the displacement, though very small, is an increase; to produce it, work must be 41] CENTRE OF GRAVITY. 109 done against the impressed forces, the body must gain energy ; this cannot take place without a supply of energy from with- out, hence a position of rest in which the potential energy is a minimum is a stable position, for to displace the body work must be done on it from without. In the unstable position some of the potential energy can be transformed into kinetic, and this is a change which will go on of itself if once started by a small displacement, the position therefore is unstable, the body can do work on external bodies if properly connected with them and when once started will do that work. If the body in the unstable position be quite free from all impressed force except its weight, and the (frictionless) reaction at the point of support, it acquires kinetic energy in falling as fast as it loses potential energy, it therefore passes through the position of stable equilibrium with an amount of kinetic energy equal to the potential energy it has lost and this, if there were no friction and no air resistance, would carry it up on the other side to the former unstable position ; in practice, however, some of the energy is dissipated as heat and in other ways ; the body does not rise to the unstable position but comes instantaneously to rest before reaching it. It then falls back through the stable position and continues to oscillate about this until the kinetic energy it has acquired in the fall is all dissipated, when it comes finally to rest in this position with a minimum of potential energy. Thus a position of minimum potential energy is one of stable equilibrium because work must be done to displace the body from this position, and to do this work needs a supply of energy from without. A position of maximum potential energy is one of unstable equilibrium because it is possible for some of the potential energy to be transformed into kinetic energy without an ex ternar supply, and this is a change which in nature can take place of itself. We do not know why there is this tendency for the trans- formation of energy or by what process it goes on, all that we observe is that motion which can take place without gain of 110 STATICS. [OH. V energy to the system, and which merely involves transforma- tion of energy will occur, while motion which involves a gain of energy will not occur unless energy be communicated from without. EXAMPLES. 1. A cylinder, 50 ft. long, balances on a log put under it at 30 ft. from one end, and it also balances on the log put under its centre, when a weight of 50 Ibs. is placed at one end and 120 Ibs. at the other ; find the weight of the cylinder. 2. Find by a diagram the centre of gravity of 3 cylindrical rods, of unequal lengths but small uniform thickness, so placed as to form a triangular figure. Where is the centre of gravity, in this case, geometri- cally situated ? 3. A circular table weighing 20 Ibs. is supported by vertical legs attached to 4 points of the rim forming a square ; find from what parts of the rim a hundredweight can be hung without overturning the table. 4. Find the centre of gravity of a lamina formed by a square having a part cut off by means of two cuts reaching from the centre to the two adjacent corners. 5. Twelve equal heavy particles are placed round the circumference of a circle at equal distances from each other. Two of the particles which have three particles between them are now removed; find the centre of gravity of the remaining ten particles. 6. If from a uniform lamina in the form of an equilateral triangle of side a the triangular portion formed by joining the middle points of two of its sides is cut away ; find the distance of the centre of gravity of the remaining piece from the centre of gravity of the whole triangle. 7. From a body whose centre of gravity is known a portion whose centre of gravity is known is cut away. Find the centre of gravity of the remaining piece. If the body is a uniform lamina in the form of a rectangle, and the triangle formed by joining its centre of gravity G to the ends of one of the sides is cut away ; find the distance of the centre of gravity of the remaining part from the point G, where a is the length of the adjacent side. 8. A parallelogram ABCD weighs 3 Ibs. and is divided by its diagonal BD into two parts, one of which, viz. the triangle BCD, is twice as heavy as the other. If a weight of 1 Ib. is placed at the corner A of the paral- iebgram, find the centre of gravity of the system. CH. V] CENTRE OF GRAVITY. Ill 9. A regular hexagon is inscribed in a circle, and weights of 1 Ib. each are placed at 5 of the angular points of the hexagon, and 3 Ibs. at the centre of the circle. Find the centre of gravity of the system. 10. If three equal triangles are cut off a triangle by lines respectively parallel to the three sides, shew that the centre of mass of the remaining figure coincides with that of the original triangle. 11. From a square piece of paper ABGD a portion is cut out in the form of an isosceles triangle whose base is AB and altitude equal to one third of AB. Find the centre of gravity of the remaining portion. 12. A table with a heavy square top ABGD rests upon four equal and heavy legs, placed at A, B, E and F, where E and F are the middle points of BC and CD. Shew that the table will be upset by a weight upon it at C, just greater than the weight of the whole table ; and find the greatest weight which may be placed at D without upsetting the table. 13. A circle of radius r touches internally at a fixed point, a fixed circle of radius R ; find the centre of gravity of the area between them, and its ultimate position when r increases and becomes ultimately equal toR. 14. The middle points of two adjacent sides of a uniform rect- angular lamina are joined and the lamina is cut in two along the joining line. Find the centre of gravity of the larger portion. 15. From a body, weight W, a piece of weight w is cut and moved a distance x; shew that the centre of gravity of the whole moves a distance xio-r-W in the same direction. ABGD is a trapezium, the angles at B and G being right angles. Shew that the distance of its centre of gravity from B C is 3 (AB + CD) 16. A triangle is cut off from a uniform square plate by a section along a line joining the middle points of two adjacent sides. Will it be possible to balance the remainder in a vertical position with one of the sides that has been cut in contact with a horizontal plane? 17. ABC is a triangle; forces represented by SAB and 4AC act along the sides AB and AC. Prove that their resultant cuts BC at a point G, such that BG = jBC. 18. A rod whose length is 10 feet, and which is thicker at one end than at the other, balances about its centre when 10 Ibs. is hung from one end and 20 from the other ; while if 40 Ibs. instead of 20 is hung from the second end the fulcrum is at 4 feet from that end. Find the weight of the rod and the position of its centre of gravity. 19. A uniform rod of weight W is supported from a point by two strings. One of these makes an angle of 60, the other an angle of 30 with the rod. Find the tensions in the strings. 112 STATICS. [CH. V 20. A thin square board whose weight is 1 Ib. has one quarter of one edge resting on the end of a horizontal table, and is kept from falling over by a string attached to an upper corner of the board and to a point on the table in the same vertical plane as the board. If the length of the string be double that of the edge of the board, find its tension. 21. A beam 12 feet long rests on two supports, distant 2 feet from each end. The beam weighs 1 cwt. Find the greatest weight which can be supported from one end without overbalancing the beam. Find also the pressure on each support when this weight is suspended. 22. A circular hole 1 foot in radius is cut out of a circular disc 3 feet in radius. If the centre of the hole be 18 inches from that of the disc, find the centre of gravity of the remainder. 23. Where must a circular hole of 1 ft. radius be punched out of a circular disc of 3 ft. radius, so that the centre of gravity of the remainder may be 2 inches from the centre of the disc ? 24. Two isosceles triangles are on the same base but on opposite sides of it, and the altitude of one is 6 inches and of the other 2 inches. Find the distance from the common base of the centre of gravity of the whole figure. 25. A cylinder of wood 12 inches long is 4 inches in diameter for 8 inches of its length, and 3 inches in diameter for the remaining 4 inches. Determine the position of its centre of gravity. 26. ABC is a triangle, whose sides AB, BC, CA are 6, 10 and 8 inches long: at A, B and C respectively, are weights of 7, 8 and 9 Ibs. Shew that the centre of gravity of the weights coincides with that of the perimeter of the triangle. 27. The middle points of two adjacent sides of a uniform tri- angular lamina are joined and the lamina is cut in two along the joining line. Find the centre of gravity of the larger portion. 28. How, practically, may the centre of gravity of a heavy beam be found of which one end is heavier than the other? If it be made up of two uniform cylinders whose lengths are as 3 : 5, and weights as 3 : 1, where is the centre of gravity ? 29. A uniform bar 4 yards long weighing 12 Ibs. has three rings each weighing 6 Ibs. upon it at distances 1 foot, 5 feet and 7 feet from one end. At what point will it balance ? 30. One corner of a square is cut off by a straight line passing through the middle points of two adjacent sides. Find the position of the centre of gravity of the remainder. 31. A uniform triangular plate hangs from one angle with the base horizontal ; shew that the triangle is isosceles. CHAPTER VI. MACHINES. 42. Simple Machines. There are various contrivances by which the amount or the direction of a force impressed on a body can be modified. By impressing a small force at one point of a body we may be able to give rise to a large force acting, it may be, in a different direction at some other point ; or vice versa, some small force may be produced through the action of a larger force impressed elsewhere. A contrivance for either of these purposes is called a machine. DEFINITION. An apparatus for making a force impressed on a body at a given point and in a given direction available at some other point or in some other direction is called a Machine. We should notice at the outset that through the action of a machine the force exerted may be greater than that applied, yet it follows from the principle of energy, that no more work is done by the machine than is done on it. Energy supplied to the machine at one point is transmitted by it to some other point ; the amount of such energy, except for frictional loss, remains unchanged. The words " except for frictional loss " are of course important, for in nature there is very considerable loss in any machine. More work must be done on the machine than it can do. Now any machine such as a pump, a steam-engine or a crane, consists of a number of simple parts, these we shall find it desirable to classify and deal with separately. Each of these parts is spoken of as a Simple Machine. G.S. 8 114 STATICS. [CH. VI We shall suppose further that the machine is in equilibrium, and that a single force impressed at one point just balances another force impressed in general at some other point. We look upon the first force as exerted by some other body on the machine, it is often called the Power, though the name is not a good one, for power means rate of doing work. The second force we consider as a force which the machine exerts on some other body, this is ordinarily described as the Weight. In addition to these we have the reactions at the points of support of the machine. The names Power and Weight come from the fact that the simplest machines were no doubt originally devices to enable men to raise weights. By means of certain contrivances a man is able, though he can exert but little force, to raise a heavy weight ; the force he can exert measures the power, the weight he raises is the weight. The simple machines consist in all cases of bodies which are constrained so as to be capable of motion in some definite manner, two forces applied to such a body balance and the problem is to find the relation between them. Now we suppose the machines to be frictionless, and the fundamental principle which will apply to all is that if the machine be supposed to receive any very slight possible dis- placement, the work done by the one force just balances that done against the other. If then we measure the distance which the point of appli- cation of each force is displaced in the direction of the force, and multiply that displacement by the corresponding force the two products will be numerically equal. Work is done by one force and against the other, the amount of work in each case being the same. If one force be large and the other small, they may balance, but in this case the displacement of the point of application of the first is small, that of the second is large. Work is measured by the product of two factors, Force and Displacement. In many cases it is convenient to change it from the form of a small force multiplied by a large displace- ment into the form of a large force multiplied by a small displacement. A machine enables this to be done. 42-43] MACHINES. 115 If P, Q are the two forces which balance on a machine, p, q the corresponding displacements of the points at which the forces are impressed measured parallel to the lines of action of P and Q, then P.p=Q.q. Hence -=-i, or the displacements are inversely as the corresponding forces. This result is sometimes expressed by the statement that "what is gained in Power is lost in Speed." If the "Power" P be small and the "Weight" Q large, then p is large and q is small, so that, in order to raise a large "Weight" by the aid of a small "Power," the point at which the "Power" is applied must move through a large distance compared with that traversed by the weight. In most machines the relation between the " Power " and the "Weight" can be found most simply by making use of this principle ; the problems which occur will however be solved in this way and also by the direct application of the conditions of equilibrium of a system of forces. We shall thus obtain verifi- cations of the principle for the simple machines. DEFINITION. When two forces, a "Power" and a " Weight," impressed on a machine maintain it in equilibrium, the ratio of the iveight to the power is called the Mechanical Advantage of the Machine. The reason for the name is clear; the object of most machines is to balance a large "Weight" with a small "Power." When this can be done mechanical advantage is gained by the use of the machine. The Simple Machines may be classified as : (i) The Lever, including the Wheel and Axle. (ii) The Pulley. (iii) The Inclined Plane, including the Wedge, (iv) The Screw. 43. The Lever. The Lever is a rod or bar which may be either straight or curved and which can move only about a fixed point. 82 116 STATICS. [CH. VI This point is called the fulcrum, and is denoted by C in the figures. A force P applied at one point A balances another force W applied at a second point B we wish to find the relation between the two. The lines of action of the two forces and of the reaction at the fulcrum must lie in one plane and meet in a point or be parallel ; we will take this plane as the plane of the paper. The conditions of equilibrium are obtained from the same principle in all cases. The resultant of the forces P and Q impressed at A and H, Fig. 101, must pass through C. Hence the moment of P round C must be equal to that of Q about C. Thus if CL, CM be perpendicular from C on the lines of action of the We will now consider a little more in detail the various cases which arise, and in the first place Fig. 101. we will deal with a straight lever and suppose the forces P and Q to be parallel and at right angles to the length of the lever. PROPOSITION 39. To find the mechanical advantage of the straight lever. Levers of this kind are usually divided into three classes. CLASS I. The points A and B at which the Power and the Weight are applied, Fig. 102 (a), are at opposite ends of the lever and the fulcrum C is be- tween them. Levers of this class are a crowbar as ordinarily ^^^^^^,,^^^ employed to raise a weight, the beam of a balance, a pair of p j, scissors, or the handle of an ordi- Fi 102 , ^ nary pump. For the conditions of equilibrium we have if R be the pressure on the fulcrum and a, b the lengths of the arms CA, 43] MACHINES. 117 CB respectively, R = P + W, P.CA = W.CB. W CA a Hence J=CI! = b' Thus the mechanical advantage may be greater or less than unity according as the Power acts at the end of the longer or the shorter arm. CLASS II. The Power and the Weight act on the same side of the fulcrum C but in opposite directions, the Power being applied at a greater distance from the fulcrum than the Weight. Among levers of this class are an oar and a pair of nut- crackers. In the case of the oar, the portion of the blade in the water is the fulcrum, the power is applied by the oarsman, the pressure of the rowlock corresponds to the weight, the fulcrum is of course not absolutely fixed. For this class then we have Fig. 102 (b) R= W-P, 1>.CA=W.CB. p AR W CA a B Hence = ^^ = T- . A \^^^M^MMM^^^^Ar. P CB b And since a is greater than b \v the mechanical advantage is al- Fig. 102 (b). ways greater than unity. CLASS III. The Power and the Weight act in opposite directions as in Class II., but the Power is nearer the fulcrum than the Weight. As examples we have some forms of the treadle of a lathe or sewing machine, or a pair of spring shears, the blades of which are held open by a spring and are closed by the pressure of the hand applied at a point between the spring and the blade. Another important example is the bone of the forearm, the fulcrum is the elbow joint, the power is applied by a muscle STATICS. [CH. VI attached to the arm not far from the joint, the weight being held in the hand. Tn this case we have, Fig. 102 (c), R=P- W t P.CA=W.CB. A p Therefore A T P^CB^b' y w Fig. 102 (c). and since a is less than b the mechanical advantage is less than unity, a small weight is raised by a large power, but the point of application of the power moves over a small distance while the weight is con- siderably displaced. 44. Bent Levers. If the lever be not straight, or the forces P and W be not parallel, we still find their ratio by taking moments round the fulcrum. Again let the directions of Pand W, Fig. 103, meet at } then R is the result- ant of P and W at and it passes through (7, thus if y be the angle between OA and OB, R acts along OC while its value is given 1P = P 2 + W 2 + 2PW cos y. Fig. 103. We may obtain an equation to find the direction of R thus, supposing AGE to be straight. Let OAB = a, OB A = ft, and let OCB = 6. Then R acts along OC, P and W along OA and OB respectively. Kesolve the forces perpendicular to AB. R sin 6 = P sin a + W sin ft. Kesolve parallel to AB. - JFcos/3. 43-45] MACHINES. 119 Squaring and adding we have E 2 = P 2 + TF 2 - 2P W cos (a + ft) . Dividing tan0 = Psina+ JFsin/3 Pcosa- JFcosjS* In the above equations we have not taken into account the weight of the lever ; this can if necessary be done. Assuming the forces all to be vertical, we have to add the weight of the lever to the pressure on the fulcrum and include the moment of the weight applied at the centre of gravity in the equation of moments. 45. Application of the Principle of Work. We can readily obtain the relation between the power and the weight for the lever by an application of the principle of work. This has already been done in the general case in Section 22, for it was proved there that when a body can turn about an axis the work done by any force is found by multiplying the moment of the force by the circular measure of the small angle turned through. Hence if the work done is zero the sum of the moments of the forces is zero and this applied to two forces gives us the principle of the Lever. The ratio of the two forces is equal to the inverse ratio of the arms at which they act. We will however apply the principle of work to the case of a straight lever on which two parallel forces are impressed at right angles to the arms. Let ACE, Fig. 104, be the lever, ACE being a straight line, Jfk A M A, Fig. 104. 120 STATICS. [CH. VI let P, W be the forces impressed at the points A and B re- spectively in directions perpendicular to the lever. Let CA = 0, CB = b. Let the lever be turned about C through a small angle 9 into the position A'GB'. Draw A'A lt B'B l , perpendicular to the directions of P and W respectively, and A'M, B'N perpen- dicular on ACB. Then A,A=A'M, and B,B = B'N. Work done by P = P . AA { = P . A'M. Work done against W= W.BB^ = W.B'N. Hence, since these two amounts of work are equal, we have P.A'M= W.B'N. P B'N A'C W=A^f=WC> for the triangles A' CM, B'CN are similar. Also A'C = AC = a, B'C = BC = b. W a Hence p = 6 ' which is the result required. We may also obtain this result by direct experiment. The bar employed has already been described, Section 18, and is shewn in Figure 105. EXPERIMENT 6. To find by experiment the relation between the Power and the Weight in a lever and to verify the law that the work done by the Power is equal to that done on the Weight. You are given a straight graduated bar ACB, Fig. 105, moveable about C as a fulcrum. This point is very approxi- mately coincident with the centre of gravity of the bar, which will therefore balance about C. The weight of the bar is thus 45] MACHINES. 121 directly supported by pressure at the fulcrum and need not be further considered. Rings A and B slide on the bar and Fig. 105. from these rings respectively weights which we will call P and W are supported. Place A with its weight P in any con- venient position on the bar and adjust either the weight W or the position of the ring B until the bar rests in equilibrium in a horizontal position. Measure the distances AC and BC. Then it will be found that P x AC = W x BC. Again, measure the heights of A and B above the floor or some other con- venient horizontal plane. Then lower the end A and fix the rod in an inclined position A'CB'. Measure the heights of A' and B' and hence determine the distances a and b say, through which the Power has been lowered and the Weight raised. It will be found that P x a = W x b, or the work done by the power -is equal to that done on the weight. Various forms of balances exemplify in their action the principle of the lever. These will be dealt with in a separate section. (See 59.) Examples. (1) Weights of 10 and 15 Ibs. are suspended from the ends of a lever 12 feet in length ; find the point at which they balance. Let AB (Fig. 106) be the lever, C the fulcrum, the upward pressure at C is 25 Ibs. Let the 10 Ib. weight be at A. Take moments about A . 25. AC = 15. 12, 15 . 12 3 . 12 , ~25~ = 5 ~ 7 ft I AC= Fig. 106. 122 STATICS. [CH. VI (2) A straight rod is loaded so that its centre of gravity is ^ of its length from one end. When iceights of 5 and 10 Ibs. are supported from the ends, the rod balances about its middle point ; find the iveight of the rod. Let the length of the rod be I feet and let it weigh W Ibs. The centre of gravity is I from the centre and the 5 Ib. weight clearly hangs on the same side of the centre as the centre of gravity. Hence, taking moments about the middle point, Thus JT=151bs. (3) Two weights P and Q are suspended from points A and B in a straight rod of weight W. The rod can move about a fulcrum C. If A , C, B and the centre of gravity G be in a straight line and the rod be in equi- librium when inclined at an angle 6 to the horizon, shew that it will be in equilibrium in any other position. Let the direction of P, W and Q meet a horizontal line through C in M , L and N (Fig. 107) respectively. Fig. 107. Take moments about C. Then P . CM = W .CL + Q. CN. Now CL = CGcos0, CM=CAcosd, CN= Hence P . GA cos 6= W . CG cos 8 + Q . CB cos 6. Thus dividing out by cos 6 we have Since this relation does not involve the angle 6 it will be true for all values of 6. It should be noticed however that cos 6 must not be zero, if it were we should not be justified in dividing out by it. Thus in the initial position the rod must not be vertical; clearly if it were vertical it would be in equilibrium whatever the weights might be. 45-46] MACHINES. 123 46. The Wheel and Axle. The apparatus is shewn in Fig. 108. It consists of a wheel or drum of considerable Fig. 108. diameter round which a rope can be coiled and which can turn about an axis through its centre. A string coiled on this wheel carries the power P. A drum of smaller diameter the axle is mounted on the same axis. Round this a rope is coiled in the opposite direction to the first and carries the weight W. Thus when P is lowered the rope round the wheel is uncoiled, that round the axle is coiled up and W is raised. PROPOSITION 40. To find the mechanical advantage of the Wheel and Axle. Fig. 109 represents a plan of the wheel and axle, perpendicular to the axis round which it can ro- tate. It is clear that the machine acts like a lever. C the centre is the fulcrum, CA, CB radii of the wheel and axle respectively are the arms. Let CA-^a, CB=b. Then for the conditions of equi- librium we have either Fig. 109. 124 STATICS. [CH. VI (i) By taking moments about G, P.CA=W. CB. W CA a ThUS 7- 65= ft' or (ii) By the principle of work. Let the apparatus be turned through a small angle so that the line A'CB' may become horizontal. Then clearly the power P drops a vertical distance A A', an amount A A' of rope is uncoiled, while W is raised a distance BE'. An amount of rope BB' is coiled up. Hence W.BB' = P.AA'. Therefore -J and since ACB and A'CB' are straight lines AA_ AG a BB'~~BC~ b' Hence as before W_a 7> ~l' In the above we have treated the rope as though it were a mathematical line of no thickness. In experiments it may quite well happen that the thickness of the rope is comparable with the radius of the axle, if this is so we may suppose the power and the weight to act respectively at the centre of the rope, and we have then to add to the radii, both of the wheel and of the axle, half the thickness of the rope. Since the mechanical advantage of the wheel and axle is given by the ratio a/6, we could, by making a large and 6 small, raise by means of a small power a very large weight, were it not for the fact that if 6 be too small the axle will not be sufficiently strong to carry the weight. "We cannot reduce b beyond a certain limit without endangering the machine. This difficulty is avoided in the differential wheel and axle. See Section 55. The mechanical advantage of the wheel and axle can be determined by experiment by finding the weight which a given power can support. Friction will however probably prevent any very close agreement between experiment and theory. 46-48] MACHINES. 125 47. The Pulley. The Pulley is a small circular disc or wheel with a groove cut in its outer edge round which a string can pass. The wheel can turn on an axis through its centre, the ends of this axis are carried by the block within which the pulley turns. When the block is fixed as in Fig. 110, the pulley is said to be fixed, in other cases it is move- able. The weight is attached to one end of a string which passes over the groove of the pulley ; the power in the case of a fixed pulley can be applied at the other end of the string. If, as we shall suppose, the supports of the pulley are smooth the tension at all points of the string must be the same throughout, and the power will then be equal to the weight. For consider the two points A, J3, where the string leaves the pulley, and let C be the centre, P the power applied at A , W the weight suspended from B. Then, taking moments about (7, we have P.CA=W.CB. But CA = CB. Hence P= W. The fixed pulley is useful only in changing the direction of a force. 48. The single move able pulley. In this instru- ment the weight W is suspended from the pulley block ; the string passes round the pulley, one end of the string is secured as at G to a fixed support, the power P is applied upwards as at A. In this case as in others the strings may either be parallel or inclined to each other. 126 STATICS. [CH. VI PROPOSITION 41. To find the relation between the Power and the Weight in a single moveable pulley. (i) When the strings are parallel. The forces acting are the tensions of the two parallel strings AD and BC, Fig. Ill, and the weight; the _ weight acts vertically, hence the two strings are vertical. Moreover the tensions D are equal and each is equal to P. Thus we have '2P upwards balancing W, which acts downwards. Hence Thus 2P = W. or a given power can raise twice its own weight. (ii) When the strings are not parallel. Since the tensions in the two strings are equal and are balanced by the weight, their resultant is equal and opposite to the weight, but the resultant of two equal forces bisects the angle between the forces. Thus the two strings are equally inclined to the vertical, let 0, Fig. 112, be the angle between either string and the vertical. The tension in each string is equal to P, hence resolving vertically = W. Thus w p If the weight of the pulley be w and it be sufficient to be considered then the downward vertical force is W+w, and the last equation be- comes W+w = 2 cos 0. 48-49] MACHINES. 127 If the " Weight " be not vertical but be applied in some other direction by means of a string or rope attached to the pulley block, then its direction still bisects the angle between the strings as in Fig. 113, and a similar equation holds. PROPOSITION 42. To apply the Principle of Work to a single move- able pulley with parallel strings. Suppose the pulley raised a dis- pj g> n$ tance x, so that its centre may move from to 0'. The simplest way of doing this, Fig. 114, is to suppose both ends of the string to be raised an equal dis- tance x from C and D to C' and D' respec- tively, the work done, since the tension in each string is P, is 2 Px. Now suppose that C' is lowered to C again, the pulley being kept fixed, D' will rise an equal distance x to D", but no work will be done by this, for the work done at one end of the string just balances that done in the other. Thus the end D at which the power P is applied is raised 2x and the total work done is as above equal to 2P . x. Hence P.2x=W.x. or If the strings are not parallel we have two equal forces at a point balanced by a third and the problem is the same as that solved in Section 33. D' c ' Fig. 114. 49. Systems of Pulleys. Various combinations x>f Pulleys are in common use. Some of these will be described. PROPOSITION 43. To find the mechanical advantage of the first system of pulleys. 128 STATICS. [CH. VI The first system of pulleys consists of a number of pulle} r s, each of which is suspended by a separate string. One end of each string is attached to a fixed support, the other end of the string after passing round a pulley is fastened as shewn in Fig. 115, to the block of the next pulley. The weight is hung Fig. 115. from the lowest pulley, the string from the highest moveable pulley passes over a fixed pulley and to it the power is applied. Thus each pulley after the lowest is acted on downwards by its weight and by the tension of the string which connects it to the pulley next below, and upwards by the two tensions in the parts of the string by which it is supported. Thus the tension in the string round any pulley is half the sum of the weight of that pulley and the tension of the string below it. Thus let w ly w 2 ,w 3 ... be the weights of the moveable pulleys A lt A 2 ,... etc., $!, 2 > Is--) the tensions of the strings round A lt A 2 , etc., and suppose there are n moveable pulleys, then 49] MACHINES. 129 = -(t, + w a ) = - and so on. Moreover t n = P. Hence p= <1 ,=i(r +Wl ) + 4 1 , 2 H or multiplying up by 2 71 , 2 n P=W+w l + 2w 2 If the weights of the pulleys be neglected the expressions become simpler though the principle is just the same. Thus, if there be four moveable pulleys, the tension in the W W W first string is -^-, in the next ^ , and in the fourth -^ . Hence -^ = P. Therefore p-=2 4 =16. Thus with this system a given " Power " P could support a " Weight " of 2 n . P where n is the number of moveable pulleys. PROPOSITION 44. To apply the Principle of Work to the first system of Pulleys. Let the weight and the first pulley rise a distance x. The end of the string round this pulley rises 2x ; thus the second pulley rises 2x, the next pulley rises twice this or 2 2 x. Thus if there be n pulleys as before the "Power" P moves a distance G. S. 9 130 STATICS. [CH. VI Hence P. 2 n x = w . '2 n ~ l x. Therefore 2 n P=(W + w 1 ) + 2m, + 2X + ... 2 n ~ 1 w n , which is the same result as previously obtained. PROPOSITION 45. To find the mechanical advantage of the second system of pulleys. In the second system there are two sheaves of pulleys in separate blocks. The string is attached to one of the blocks, Fig. 116, in the figure it is the upper and passes round the pulleys in turn first under one in the lower block, then over one 49] MACHINES. 131 in the upper and so on. The pulleys are sometimes arranged with a common axis, sometimes the various pulleys in a block are placed one below the other as in the figure. In either case the tension of the string is equal to the power; let there be n strings at the lower block. The upward force will be found by multiplying the tension by n, the down- ward force is the weight supported JF, together with the weight of the lower block w lf Hence nP- W+w L . We can apply the principle of work thus. If the lower block be raised a height X, a length x of each string will be left slack. Hence the end of the string can move a distance nx. Thus, P.nx=(W+w l )x, nP=W+w l . PROPOSITION 46. To find the mechanical advantage of the third system of Pulleys. In this system, Fig. 117, one end of each string is attached w to a bar which carries the weight. The uppermost pulley is fixed ; a string passes over it and supports the next pulley, another string passes over this and supports the third, and 92 132 STATICS. [CH. VI so on, the last string passes over the last moveable pulley and to it the Power is applied. Now if t , &> . . t n be the tension in the strings beginning from that over the topmost pulley w it w 2 ... the weights of the pulleys. Then t n = P, , = 2P + w n , Also Now we know that l + 2 + 2 2 +... + 2'"- 1 =2''-l. Thus W = P (2* - 1) + w w (2"- 1 - 1) + w 2 (2~ 2 - 1) + Wl . If the weights of the pulleys be neglected the expression is simplified, for the tension in each string, beginning from the power, is clearly twice that in the string before. t n = 2"- 1 P. Thus = P{2 n -l}. We may apply the Principle of Work thus. Let the weight and the bar carrying it rise a distance x. If all the pulleys retained their positions fixed there would be a length x slack in each string. 49-50] MACHINES. 133 In consequence of this alone each pulley after the first could be lowered a distance x-, the first pulley is fixed, hence the second pulley drops a distance x\ in consequence of this drop the third pulley will be lowered through twice this distance or 2x ; to this must be added the direct drop x due to the rise of the weight, which would have taken place even if the second pulley had been fixed. Thus the actual drop of the third pulley is 2x + x, or 3#, and we may write this (2 2 -l)#. In consequence of this the fourth pulley drops twice as far or 2 (2 2 - 1) x, and to this we must add the fall x for the direct rise of the weight. Thus the drop of the fourth pulley is or *(2 3 -2+l), or #(2 3 -l). The law is now clear, the nth pulley drops x (2 n ~ 1 - 1), and the Power *(2-l). Thus the equation of work gives Wx = Px (2 W - 1) + w n x (2"- 1 - 1) + ... + .. . + wtfc (2 2 - Therefore W= P (2 M - 1) + w n (2"- 1 - 1) + ... 50. Experiments -with Pulleys. It is not possible to verify by direct experiment these results. In any system of pulleys the friction is considerable, a smaller power than that given by the equations is sufficient to support a given weight, a larger power than is given is necessary just to raise the weight. We can however verify by direct measurement the result that when a pulley rises a distance x, and one end of the string round it is held fixed, then the other end rises a distance 2x. Thus consider the first system of pulleys. Support a Weight W from the lowest pulley, and let the Power P be another weight supported in a suitable scale-pan. Adjust two vertical scales as shewn in Fig. 115 above, by the side of the power and the weight respectively. Displace the system by 134 STATICS. [CH. VI raising the weight W a measured distance a on its scale, and observe the distance through which P descends, it will be found to be equal to 2 n P where n is the number of moveable pulleys. Similar observations can be made for the other systems of pulleys. In the third system of pulleys the bar to which the weight is attached will not remain horizontal unless the point of attachment of the weight is the point of action of the resultant of the tensions. Thus if K be the point of attachment, A the end of the bar to which the string carrying the weight is attached, and 2a the distance between the strings, that is the diameter of each pulley, then taking moments about K By substituting the values of t lt f 2 , etc., found in Section 49, the posi- tion of K can be found. 51. The Inclined Plane. Consider a plane inclined to the horteon at an angle a. A weight could be raised to the top of such a plane if we neglect friction or take means to make it small with the application of a smaller force by causing it to slide up the plane, than by lifting it directly. Hence the inclined plane is one of the mechanical powers. The solution of the problem depends in part on the direc- tion in which the force is impressed. PROPOSITION 47. To find the mechanical advantage of an inclined plane when the Power acts parallel to the plane, Let BAG, Fig. 118, be an inclined plane, AC being hori- zontal and EC vertical. Let the angle BAG be equal to a, and con- sider a body of weight W at rest on the plane, which is supposed to be smooth. Let AJB, the length of the plane be equal to I, and let the height BC be h. The forces acting are the weight W vertical, the power P parallel to the plane, and the resistance R at ri - 118 - right angles to the plane. The rela- tion between these quantities can be found in various ways. 50-51] MACHINES. 135 (i) By the resolution of forces. The direction along the plane, in which P acts, and a line at right angles to this, along which R acts, will clearly be " con- venient " directions in which to resolve the forces. The angle between the normal to the plane the direction of R and the direction of W is clearly a. Hence the com- ponent of W, perpendicular to the plane, is TFcos a, and along the plane it is W sin a. Hence resolving along the plane P= JFsina, and perpendicular to the plane 72 = JFcosa. Thus P and R are found in terms of W and a. BC h Moreover, sin a = ^ - j . A.JJ I -p j Therefore ^ = sin a = j , or P.l = W.h. (ii) By an application of the principle of work. The work done by P in moving the body a small distance s along the plane is P . s, if at the same time the body rise a height z t the work done against z is W . z. No work is done by the resistance R, since the motion is everywhere at right angles to the direction of R. Hence P . s = W . z, H4 from the figure. Hence W.h = P. I, as before. We can obtain the result directly by considering the work done in moving from A to B. 136 STATICS. [CH. VI (iii) By the triangle of forces. Since the forces P, W and R, maintain the body in equili- brium, they must be proportional to the sides of any triangle drawn parallel to them. Let G, Fig. 119, be the particle. Let GK, vertical, meet AC in K, and GL at right angles to the plane meet AC in L. Draw KM parallel to the plane to meet GL in M. Then P, R and W are respec- tively parallel to KM, MG and GK. Hence P_ B^ W^ KM~ MG~ GK' Again the triangles CBA^MKG are similar. KM EG K L Fig. 119. Hence Thus GK AB ~ sin a, MG GK AC -Jg-0080. ___^_ W~GK~AB~l R MG AC PROPOSITION 48. To find the mechanical advantage of an inclined plane when the power acts horizontally. Let AB, Fig. 120, be the plane, BC being vertical and AC horizontal. Let P be the power acting hori- zontally, and let a be the angle of the plane. Let BC h and Let W be the weight and R the resistance of the plane. We can find the mechanical advantage in various ways as follows. Fig. 120. 51] MACHINES. 137 (i) By the resolution of forces. Resolve horizontally and vertically. The components of R are R cos a vertical, and R sin a horizontal. P is horizontal and IF" vertical. Hence, resolving vertically, W = R cos a. Resolving horizontally Thus P fisma -fP = -^ -- W R cos a = tan a. (ii) By the principle of work. In moving the body from A to B, since the displacement in the direction of P is AC, while that opposite to the direc- tion of W is CB, the work done by P is P . AC, and that done against W is W.BC. Thus P. AC = W.BC. P BC Hence = = tana ' (iii) By the triangle of forces. Let G, Fig. 121, be the body, and let GK vertical meet AC in K, and GK normal to the plane meet it in L. Then P, R and W are parallel respectively to KL, LG and GK. P R _W KL~TG~GK' But the triangles LKG and BCA are similar. P KL BC Hence _ ~~ R LG AB Fig 138 STATICS. [CH. VI If the force be inclined as shewn in Fig. 122 at an angle e to the plane, it is usually simplest to resolve parallel and perpendicular to the plane. Since three forces which maintain a body in equilibrium are in one plane and R and W lie in a vertical plane through the particle, the direction of P must also be on this vertical plane. Eesolving parallel to the plane we have P cos e = W sin a. Kesolving perpendicular to the plane we have Fig. 122. Some of these results can be verified by experiment. EXPERIMENT 7. To prove that on an inclined plane, when the Power acts parallel to the plane, the ratio of the Power to the Weight is equal to that of the height of the plane to its length, and to verify the Principle of Work. The plane is a wooden board shewn at AB, Fig. 123, to which a sheet of glass is attached. The board is hinged at A to a second board, which can be clamped to the table. At B, which is at some convenient distance (say 10 inches from A), Fig. 123. 51-52] MACHINES. 139 there is a thumb-screw. By means of this there can be clamped to the plane a vertical rod with a slot parallel to its length. The rod is graduated in such a way that the height of B above the lowest point A can be read off directly. Thus the height h can be measured, and the length is a known constant. A pulley is fixed at the top of the plane. The "Weight" W consists of a heavy brass roller mounted in a frame so as to turn with very little friction, a string attached to the frame passes over the pulley and supports a scale-pan, into which various weights can be placed. The " Power" P is the weight of this scale-pan and weights. The frame and pulley are arranged so that the string between them when tight is parallel to the plane. Thus the Power P acts on the Weight W parallel to the plane. Set the plane so that h may have some convenient value, say 5 inches. Observe the value of P required to support W for this value of h. To do this accurately find the value, P l , which will just drag W up the plane and the value, P 2 , which will only just let it roll down. The proper value of P, that is, the value which it would have if there were no friction, is the mean of P l and P 9 . When the observations are made it will be found that W. h = P . I, or that P : W = h : L Again, if a is the angle which the plane makes with the horizon h/l = sin a. Hence P = TFsin a. In our case since the length I is 10 inches we have to divide the height by 10 to get sin a, thus if h - 5 inches, sin a = -5 = J, and a = 30. Again, suppose the weight is allowed to move down the plane, it will have fallen a vertical height of h inches. Thus the work done by the weight will be W . h units of work ; and clearly the power P, since it is attached to W by the string, must have been raised I inches and the work done on it will be Pl units ; but by what we have seen P . I = W . k, which proves the principle of work for the inclined plane in this case. Take a series of values of P for various values of h and thus shew that in all cases P . I = W . h. 52. The Wedge. This is a sort of double inclined plane or prism, Fig. 124, made of iron or steel or some such 140 STATICS. [CH. VI hard material and used for splitting wood or other like purposes. Thus if AC, Fig. 125, be a wedge of angle a driven into a piece of wood by a weight W applied downwards, and we Fig. 124. suppose R, R', the pressures which the two faces of the wedge exert on the obstacle, to act normally to its faces AB and BC, and further that these two faces are equally inclined to the vertical, at angles therefore of - , we can find the relation between R, R' and W, thus, supposing the wedge to be smooth. The vertical components of R and R' balance W, the hori- zontal components of R and R' are in equilibrium. Hence resolving vertically R COS n~ ft' cos 9 Hence and 2' 52-53] MACHINES. 141 In reality the friction involved in the use of the wedge is enormous. We ought to consider two other forces F, F' acting parallel to the faces of the wedge. If we suppose the machine to be symmetrical with regard to the ver- tical line bisecting the angle a, we have and unless we know the relation between E and F we cannot carry the solution any further. 53. The Screw. Consider a sheet of paper cut into the form of a right-angled triangle BAG, Fig. 126 (a), and wrap it round a cylinder so that the base AC of the triangle Fig. 126 (a). Fig. 126 (6). may be at right angles to the axis of the cylinder. The hypotenuse AB will form a spiral curve round the cylinder, Fig. 126 (b). Now imagine a projecting thread to be fixed to the outside of this cylinder so as to coincide with the spiral curve thus drawn ; the cylinder then becomes a " screw." Sup- pose now that the paper is wound inside a hollow cylinder of the same radius as the solid cylinder just described and that a hollow groove is cut in the surface of this hollow cylinder, the groove being of such a form that the projecting thread just fits it. The hollow cylinder constitutes a " nut " in which the 142 STATICS. [CH. VI screw can turn, Fig. 127. Let the nut be held fixed and the end of the screw inserted in it. Turn the screw round its axis by means of a lever ; as it is turned its end moves outwards through the nut parallel to the axis. If the axis be verti- cal, a weight W can be raised by the applica- tion of a power P to the end of the lever. In any case if the nut be held fixed, force can be exerted by the end of the screw. The angle a which the thread of the screw makes with a plane at right angles to the axis is called the pitch of the screw. Fig 127 The distance measured parallel to the axis between two consecutive turns of the thread is constant and depends on the pitch and on the radius of the cylinder on which the screw is cut. For let K, Fig. 126 (a), be a point on the paper triangle which when it is rolled on to the cylinder comes directly over the point A. Draw KL perpendicular to the base AC, parallel that is to the axis, when the paper is rolled on the cylinder, L will coincide with A, and KL is the distance between two threads. Hence KL -- h. Again AL is clearly the circumference of the cylinder so that if b be its radius we have AL = circumference of a circle of radius b = 2irb. KL But -r-- = tan KA L = tan a. AL Thus KL = AL tan a, or h = 2?r6 tan a. We notice further that when the screw makes one complete revolution, a point such as K is brought into the position previously occupied by A, the end of the screw advances a distance h parallel to the axis. 53-54] MACHINES. 143 It is impossible to obtain a screw in which there is no friction. We must therefore suppose that at each point of the thread there is acting a normal force at right angles to the thread, and a tangential force parallel to it. Suppose these forces to be uniformly distributed, and let R and F be their values for each unit of length of the screw. Let I be the whole length of the screw. R may be resolved into a force R cos a parallel to the axis, and R sin a at right angles to it i ; and F has for its components jFsina parallel to the axis, jPcosa at right angles to it. The forces at right angles to the axis have moments round the axis, the others have not. Let us suppose also that we are trying to raise the weight, then the frictional force helps the action of the weight and opposes that of P. Suppose further that P acts at the end of an arm a in a direction at right angles to the axis. Then resolving vertically Taking moments about the axis Pa = Rlsiu a .b + FlcoBa.b. If W and P are known these equations will give us R and F; we can- not use them to find the mechanical advantage unless we have some relation between F and R. If we suppose F is zero, which in practice is never the case, then W=Rlcosa, Hence p = - cot a. PROPOSITION 49. To find the mechanical advantage of the Screw. We can obtain the result most easily by the Principle of "Work. For while W is being raised a distance A, the point of application of P moves once round a circle of radius a, hence its displacement is 2-n-a ; moreover the direction of P is tan- gential to this circle. Thus the work done is P . 2ira. Hence P . 2va = Wh = W . 2irb tan a. W 2-rra a Thus -= = ,7-7 - = r cot a. P 2-n-b tan a b 54. Combinations of Simple Machines. A com- plex machine is usually made up of a number of Simple Machines. In these the "Weight" of the first becomes the 144 STATICS. [CH. VI " Power " of the next and so on. In such a case the mechani- cal advantage of the whole is found by multiplying together those of all the simple machines. For let P lf P z , P 3 , etc., be the powers, W l} TF 2 , TF 3 , etc., the weights, Then m^...m n the mechanical advantages. Hence r, P ' * n-1 Thus multiplying all together VYl i . TTig ^3 Wl n W = -=^ = mechanical advantage of the whole. Some special forms of machines are described below. 55. The differential Wheel and Axle. In this apparatus shewn in Fig. 128, the axle consists of two drums or cylinders of different radii, b and c. A rope, the ends of which are coiled in opposite directions round these two drums, passes under a single moveable pulley from which the weight W is supported. As the axle is turned the rope is coiled up on one drum and uncoiled from the other; the motion of the weight depends on the differ- ence of these two effects. The power P is usually applied at the circumference of a wheel of radius a. W Fig. 128. 54-55] MACHINES. 145 In Fig. 129, let AC OB represent the machine as viewed from a point on the axis. Sup- pose it to be turned through a small angle so that A'C'OB' may become horizontal. The power falls a distance aO, the end C of the string round the pulley falls a distance cO, while B rises a distance bO; hence the loop of string carrying the pulley is shortened by (b c) 0, and therefore the pulley and weight rise J (b - c) 0. Thus the Principle of Work gives or W_ 2a P~b^~c' w Hence by making b and c nearly equal, W can be made Fig. 129. very large compared with P t with- out unduly reducing the strength of the machine. For a given motion of the Power the distance traversed by the Weight can be made very small, hence the ratio of the " Weight " to the " Power " can be made very large. We can solve the problem without using the Principle of Work thus : Let T be the tension in the string carrying the pulley, then taking moments about 0, Hence Pa =T(b- c). But from the equilibrium of the pulley 2T = W. Thus P G. S. a 10 146 STATICS. [CH. VI 56. The Differential Screw. This machine consists of two screws of slightly different pitch h and k. The axes of the two screws coincide, the second screw works inside the cylinder on which the first is cut. Thus if H, Fig. 130, be the outer screw, and K the inner screw, then on giving the outer screw one complete turn its point will move down- wards through a distance A, and if the inner screw did not turn in the outer, it too would be displaced this same distance ; but the inner screw does turn relatively to the outer, its point is in conse- quence raised relatively to the outer screw a distance k, thus the point of the inner screw actually descends a dis- tance (h - k). Hence if P be the "Power" impressed at Fig. 130. one end of an arm a, we have Thus W 2 57. Cog Wheels. A train of cog wheels is virtually a combination of Wheels and Axles. Consider two such wheels, Fig. 131. Let A and B be their centres and let them be in Fig. 131. 56-58] MACHINES. 147 contact at C in the line A B. Let R be the force between the wheels at (7, and let A C a, EC = b. Draw AL, BM perpen- dicular on the direction of JR. Let the " Power " be a force P acting at an arm p, the " Weight " a force Q acting at an arm q. Then for the equilibrium of A, P.p = R.AL. For the equilibrium of B> Also the triangles A CL, BCM are similar. Thus P-P AL AC a Hence Q = p b P a' q' ff\ Now - is the mechanical advantage of the wheel A con- sidered as a bent lever ; while - is that of the other wheel. q We have thus found the mechanical advantage of the whole, and see that it is the product of those of the two parts. 58. The Spanish Barton. This forms a useful com- bination of Pulleys shewn in Fig. 132. A and B are two moveable pulleys which are suspended by a string over a fixed pulley C: the "Weight" W is attached to A. A string passes from a fixed point D under A and over B, and the " Power " P is attached to this. Let w 1} w 2 be the weights of the pulleys A and B. Suppose the weight and the Pulley A raised a distance x. In consequence of this, if the pulley B were fixed, a length 2x of the string would be left slack. Thus P would fall a distance 2x ; but since A and B are connected by a string, when A 148 STATICS. [CH. VI is raised a distance x, B falls the same distance. In con- sequence of this the "Power" falls a further distance 2x : thus on the whole the "Power" falls 4#. Hence W and A each rise x, P falls 4:X and B falls x. Therefore ( W + w^) x = wjx + P . 4#, 4P. In practice w 1 and w z would usually be equal. Hence the mechanical advantage (W/P) is 4. We can also find the tension of the strings and solve the problem thus. Let T be the tension in the string over the fixed pulley, that in the string round the moveable pulleys is P. Hence for the equilibrium of A and for that of B, Therefore 59. The Balance. The balance, Fig. 133, in its sim- plest form consists of a lever which can turn about a fulcrum : it is used for comparing the masses of two bodies, or rather for Fig. 133. 58-59] MACHINES. 149 determining in terms of the standard mass the mass of any other body. This is done by comparing the weights of the bodies. From the arms of the lever two equal and similar scale-pans are suspended, the standard mass is placed in one of these, the body whose mass is required in the other, and, if the balance be in adjustment, the two masses are equal when the arms are horizontal. To determine when this is the case with accuracy, a vertical pointer is attached to the beam near the fulcrum, the lower end of this pointer moves over a horizontal scale, being adjusted so as to rest at the centre of the scale when the beam is horizontal. The arms of the beam ought, as we shall see, to be equal and similar if the balance is to be accurate. The requisites of a good balance are : (i) Truth. (ii) Sensitiveness. (iii) Stability. A balance is said to be True if the beam be horizontal whenever equal masses are placed in the scale-pans. A balance is Sensitive when the beam deviates appreciably from its horizontal position for a very small difference P Q, in the two masses P and Q. A balance is Stable when the beam if disturbed from its equilibrium position readily comes back to it. We shall consider how to secure these conditions separately. The points from which the scale-pans are suspended are A and I>, Fig. 134, these in a good balance are steel knife-edges, YP+S Q-fS Fig. 134. secured to the beam in such a way that their edges are at right angles to the length of the beam. The scale-pans are attached to small flat plates of steel, or better of agate, which 150 STATICS. [CH. VI rest on these knife-edges when the beam is in use and hang with their centres of gravity below the respective knife-edges. The weights thus act vertically through A and B. The fulcrum C is also a similar knife-edge resting on a steel or agate plate. In good balances there is an arrangement by which the plates are lifted off the knife-edges when the balance is not in use. It is desirable 1 in a balance that the fulcrum C and the points of support of the scale-pans A and B should be in one straight line, and we shall assume this condition satisfied. When the balance is loaded, the forces acting are the weight of the beam, let this be TF, the weights of the scale- pans S, S' respectively and the weights P, Q of the masses in the scale-pans. The weight W acts at the centre of gravity of the beam; if the beam remains horizontal when the scale- pans are removed, this point must be vertically below the fulcrum, let it be 6f, then CG is at right angles to ACB. Let CG = k, AC=a, C = b, PROPOSITION 50. To find the condition that a balance may be true. The balance is true provided the beam be horizontal what- ever equal masses are placed in the scale-pans ; if the beam be horizontal the centre of gravity G is vertically below the fulcrum, thus the weight of the beam W has in this case no moment about the fulcrum. Suppose now the scale-pans are empty, and the beam hori- zontal ; the only forces which have a moment about the fulcrum are the weights S, S' of the scale-pans; these act vertically through A and B respectively. Hence taking moments about the fulcrum, S . a = S' . b. 1 It can be shewn that if this condition be not true the sensitiveness of the balance will vary with the load ; the condition cannot be always accurately satisfied, for as the load increases the beam bends and the points A and B are brought down below the fulcrum. 59] MACHINES. 151 Suppose now that two equal masses P, P are placed one in either scale-pan : if the balance be true the beam will still be horizontal, thus taking moments again or P.a + S.a = P.b+S' . b. But S.a = S'.b. Hence subtracting P . a = P . b. Thus a = b, and since S . a = S' . b, we must have also 8& Therefore if a balance is to be true the arms must be equal in length and the scale-pans equal in weight. PROPOSITION 51. To find the condition that a balance may be sensitive. In finding this condition we assume that the balance is true. The condition required is that, when the weights P and Q differ slightly, the beam should be sensibly inclined to the horizon. Let the centre of gravity of the beam be at G, a distance h below the fulcrum. Then if the beam is displaced, its weight W will have a moment about the fulcrum tending to restore it ; in order that the displacement for a small difference (P Q) may be considerable the moment of the weight about C must be small ; this moment will depend on the value of W . h, and can be made small by making W small or h small or both, thus for sensibility the beam must be light and its centre of gravity near the fulcrum. But the displacement will also be great if the moment due to the difference of the weights is large. This moment, since the balance is true, is (P Q) a, and for a given difference P Q can be made large by making a large. Thus the sensitiveness is increased by lengthening the arms. The sensitiveness therefore may be made considerable either by (a) increasing the length of the arms, or (ft) reducing the weight of the beam, or (y) bringing the centre of gravity of the beam near the fulcrum. 152 .STATICS. [CH. VI In order to secure this last condition there is usually, in good balances, a short vertical wire attached to the beam above the fulcrum. This has a screw thread cut on it and a metal sphere can be screwed up or down on this wire; by raising the sphere the centre of gravity of the beam is raised and the sensitiveness increased. We may put the results just obtained more briefly thus. Let the beam be displaced through an angle 6 and let a horizontal line through C meet in L, M and K, Fig. 134, the verticals through A, B and G, which are the lines of action of P + S, Q + S and W respectively. Then the angles ACL and CGK are both equal to 6. And CM = CL = GAcos0= a cos 6, CK= CG sin 6 = h sine. Thus taking moments about C, (P + S)CL = (Q + S)CM+W.CK. Hence (P-Q)a cos B-Wk sin 8, tan0 a P^Q=W7h- Now t&n6/(P-Q) is a measure of the sensitiveness. The balance is sensitive if this fraction be large when P - Q is small. Thus for sensi- tiveness a\Wh is large. Hence a must be large or Wh small, or both these conditions must hold. PROPOSITION 52. To find the condition that a balance may be stable. For this it is necessary that, when the pans are equally loaded, the beam after displacement should return rapidly to its position of equilibrium. Now if the pans be equally loaded and the balance be displaced the moments of the loads about the fulcrum balance, and the only moment tending to restore equilibrium is that of the weight of the beam. This depends on the product W.h', hence for stability this moment must tend to bring the beam back, thus G must be below the fulcrum, and for rapid action the product W . h must be con- siderable. It will not however be sufficient to make W large, for if this only is done the mass to be moved is increased as well as the impressed force and the acceleration is not changed. For great 59-60] MACHINES. 153 stability then h must be considerable, the centre of gravity must be well below the fulcrum. It will be noticed that this condition is antagonistic to one of those obtained for sensitiveness ; fair sensitiveness and reasonable stability can be secured by making h not very small and giving the balance long arms. The relative importance of the two conditions depends on the purposes for which the balance is required. Stability and rapid action would be the main desideratum in a balance employed for weighing coals; a man conducting a chemical or physical research would attach greater importance to sensitiveness, though in this case also rapid action is impor- tant. A complete discussion of the question would take us beyond the limits of this book. GO. Use of a balance. It is important to be able to test a balance for accuracy and to determine any errors which it may possess. Now in using a balance we have always to bring the beam back to the horizontal position. This may be done by placing weights in one or other of the pans until the pointer comes back to zero at the middle of the scale ; instead of wait- ing however till the pointer is actually at the middle of the scale, we may notice the distance it moves on either side of the zero mark, and adjust the weights until these oscillations are equal, we then know that when the beam is at rest the pointer will be at zero 1 . We now proceed to describe some experiments with a balance. EXPERIMENT 8. To determine the ratio of the arms of a balance, and to weigh a body correctly in a balance with unequal arms. Let the balance come to rest with its pans unloaded, let S and S' be the weights of the scale-pans, a and b the lengths of the arms. 1 A more exact method of " weighing by oscillation" is given in Glaze- brook and Shaw's Practical Physics, Section 12, p. 107. 154 STATICS. [CH. VI If the pointer does not come to zero, it may be because the pans are unequal in mass, or the arms in length, or it may be that the balance case is not level so that the stem which carries the beam is not vertical. Test for this by a spirit level, assuming the maker has set the stem at right angles to the bottom of the case so that the knife-edges and the plate on which the fulcrum rests will be horizontal when the bottom of the case is. If the pointer does not come to zero 1 , load one of the scale- pans with some shot or bits of lead foil until it does. Then, since the beam is horizontal, we have if S and S f now denote the weights of the scale-pans and loads used to bring the beam horizontal, Sa = S'b. Let W be the weight of the body whose accurate weight is required. Place it in the left-hand scale-pan and place weights P in the right-hand scale-pan until the beam is again horizontal. Then (W+S)a = (P + S')b. Hence Wa = Pb. Now interchange W and P : if the beam remain horizontal the arms are equal, if not, let Q be the weights in the left-hand scale-pan which are required to balance W when in the right- hand scale-pan. Then (Q + S)a=(W + S')b. But Sa = S'b. Hence Qa= Wb (i), And we have already seen that Wa=Pb (ii). 1 It is not important that the pointer should read zero exactly, pro- vided its position with the balance unloaded is noted, and that it is always brought back to this position when a body is being weighed. To secure this it is usually simplest to adjust the balance so that the unloaded reading may be at the middle of the scale or very close to it. 60] MACHINES. 155 Hence Therefore 1 Also dividing (i) by (ii) W = ~P' Hence TP = PQ, Thus P and Q are known weights : hence the ratio of the arms and the true value of W is found. EXPERIMENT 9. To determine the difference between the weights of the scale-pans. Level the balance as in Experiment 8, and note the position of the pointer on the scale. Interchange the scale-pans so that S may now hang from the arm b, S* from the arm a. If the pointer remains in the same position, the pans S and S f are equal in weight. If not, we can find their difference thus. Let S be suspended from the arm a, and suppose that a small weight w must be added to S to make the beam horizontal. Then (S + w) a = S'b. . suppose n S'a=(S + Interchange S and S' and suppose now that w' in the pan S is required for equilibrium Thus Hence S-S' + w=(S'-S)--w'-. 'a a 156 STATICS. [CH. VI Thus Therefore S'-S= Hence if b/a is known we can find S' - S. If w and w' be both very small and 6/a very nearly unity, we shall not alter the value of this quantity much by putting bja equal to 1, and then If 6 be accurately equal to a then clearly w is equal to w', but by pro- ceeding as above and interchanging the pans we take into account the only important part of the effect due to a difference between 6 and a. 61. The common or Roman Steelyard. This balance consists of a lever A, Fig. 135, supported on a knife- Fig. 135. edge at C. A hook at A carries the pan in which the object to be weighed is supported. The arm BC has a number of divisions marked on it, and from these a standard weight P of constant magnitude can be suspended. The weight Q is determined by finding the division at which B must be sus- 60-61] MACHINES. 157 pended in order that the beam may be horizontal. Practically then the steelyard is a balance such that the length of one arm can be adjusted : we weigh by adjusting this length, not by altering the weight. If the centre of gravity of the steelyard itself coincided with C the fulcrum, the weight W would be directly proportional to the distance from C of the point from which P is suspended, the beam could be graduated by dividing it into equal parts from C, each of these parts being equal to GA. In practice this is not the case : we have always to take into account the weight of the steelyard. This is done in the following way. Suppose G is the centre of gravity and W the weight of the steelyard. Let the scale-pan be unloaded and adjust P until the beam is horizontal. Let the position of P so found be 0. The point G is usually between C and A, while is between C and E. Then if a weight P were always kept at it would just balance W at G. If then a weight Q be put into the scale- pan and another weight equal to P supported at some point along AC, the weight Q will be measured by the distance of this point from C. But the effect of the two weights P is the same as that of a single weight P placed at a distance beyond the second weight equal to CO. The weight Q is then measured by the distance of this weight from 0, in other words the steelyard is graduated from 0, not from C. This may be put more briefly using symbols thus. Let P suspended at B balance Q at A. Then P.CB= W.CG+Q.CA. Now W . CG = P . CO by experiment. Hence P. CB = P . CO + Q . CA. Thus Q.CA=P.(CB-CO) = P.OB. OR Hence Q = P - 158 STATICS. [CH. VI Hence if we take points 1, 2... etc., along AC such that their distances from are AC, 2 AC... respectively, the weight Q is equal to P, 2P, 3P, etc., according as P is at 1, 2, 3, etc. Thus the steelyard is graduated from 0. 62. The Danish Steelyard. This consists of a bar AB, Fig. 136, terminating in a ball at B, the weight of which Fig. 136. constitutes the power, the bar is graduated and the fulcrum is moveable. The body to be weighed is suspended from A, and the fulcrum C is shifted until the steelyard is horizontal. Then if P be the weight of the bar and G its centre of gravity, the moment about C of P acting at G is equal to that of W at A. Thus to graduate the bar we have or Thus (P+ W)AC=P.AG, P.AG P + W Thus by making W equal successively to P, 2P, 3P, etc., the successive graduations can be found. 63. The Letter Balance. (Roberval's Balance.) A ommon form of letter balance is shewn in Fig. 137. The beam ACB turns about a fulcrum at C. The scale- pans are supported above the beam on knife-edges at A and B. Two equal vertical rods AD, BE are attached below the scale-pans, and the lower ends D, E of these rods are connected 61-63] MACHINES. 159 by joints to a horizontal bar DE parallel and equal to the Fig. 137. beam. This bar can turn about its middle point F, which is vertically below the fulcrum. By this arrangement, as the balance swings, the rods AD, BE always remain vertical and the scale-pans horizontal. Moreover it follows readily from the principle of work that if the arms of the beam are equal the weights may be placed anywhere on the scale-pans. For if the beam be slightly displaced, every point of the one scale-pan rises the same vertical height h, say, while every point of the other falls an equal distance h. Hence if P and Q be the weights at any point of either scale-pan respectively, and there be equilibrium with the beam horizontal, we must have P=Q. In an ordinary balance, if the weights be put on at any point of the scale-pan, the pan swings about its point of support until the centre of gravity of the weights is vertically under this point. The weights therefore always act vertically through the ends of the beam. If the pan be not free so to swing, then in general the arm at which the weights act, the horizontal distance that is between the fulcrum and the vertical line through their centre of gravity, will depend on the position of the weights in the scales and the balance if not 160 STATICS. [CH. VI actually useless would be very troublesome to use. In the balance just described this difficulty is avoided by securing that each point of the pan rises or falls an equal amount. EXAMPLES. 1. A balance has unequal arms. A piece of lead placed in the left pan weighs apparently 580 grams ; when it is placed in the right pan its weight is apparently 560 grams. Calculate the ratio of the lengths of the arms of the balance. 2. A uniform rod ABCD moveable about a fulcrum and twenty feet in length has weights P, 2P, 3P and 4P attached to the rod at A, B, G and D which are at equal distances apart. If the rod be in equilibrium find the distance of the fulcrum from A. 3. In a given Koman steelyard where must the fulcrum be if the smallest weight that can be measured is half the moveable weight, as- suming that the beam weighs 4 times the moveable weight ? 4. Discuss the effect of an increase in the value of the acceleration of gravity (1) upon the sensitiveness, (2) upon the stability, of a balance. 5. Describe a balance and explain the conditions upon which the sensitiveness of a balance depends. 6. Why is it that the sensitiveness of a balance depends upon the sharpness of its knife-edges ? 7. Describe and explain the various precautions which are necessary for the accurate determination of the mass of a body by means of a balance. 8. What is meant by stable, neutral, and unstable equilibrium? Give examples of each of these. 9. A false balance rests with the beam horizontal when unloaded, but the arms are not of equal lengths ; a weight W, when hung- at the end of the shorter arm 6, appears to balance a weight P, and when hung at the end of the longer arm a it appears to balance a third weight Q ; shew that Can you suggest another way of ascertaining correctly the weight of Tf? 10. How would you compare the "stability" and the "sensibility" of two balances ? 11. How may an Inspector with one standard pound test a trades- man's scales and weights ? CH. Vl] MACHINES. 161 12. An object is placed in one scale-pan of an ordinary balance and it is balanced by 20 Ibs. The object is then put into the other scale-pan, and now it takes 21 Ibs. to balance it. When both scale-pans are empty the scales balance. What is the matter with the balance, and what is the true weight of the object ? 13. The arms of a false balance are in the ratio of 20 to 21. What will be the loss to a tradesman who places articles to be weighed at the end of the shorter arm if he is asked for 4 Ibs. of goods priced at 3s. per lb.? 14. The arms of a balance are 2 ft. long. One of the scale-pans is a circular disc, whose diameter is 6 inches, and which is fixed to the end of an arm of the balance by a rod passing through the centre of the pan and rigidly attached to the pan at right angles to its plane. Shew that a 1 lb. mass placed in such a pan may balance any mass between 18 and 14 ounces in the other pan. 15. A weight W is supported on a smooth inclined plane at an angle of 30 to the horizon, by a string attached to a point in the plane; find the tension of the string. If in the preceding case the pressure on the plane is E, and in the case in which the string is inclined at an angle of 60 to the horizon [the in- clination of the plane remaining the same] the pressure is E', prove that 2E = 3E'. 16. Find the horizontal force that would support a weight W on a smooth plane inclined at an angle of 60 to the horizon. If on the same inclined plane the weight W is supported by two equal forces one acting horizontally and the other acting along the plane upwards, and the pressure on the plane is E, prove that E = W. 17. The height of an inclined plane is 4 feet and it requires a power P acting along the plane to support a weight W. If the height is altered, the length of the plane remaining the same, and 3P, acting along the plane, is now necessary to support the weight W, find the new height. 18. A weight W is supported on a smooth plane inclined at an angle a to the horizon by means of a force inclined at an angle 6 to the plane. Find the magnitude of the force, and the pressure on the plane. If there is no pressure on the plane, in what direction does the force act? 19. A body of weight W is supported on a smooth plane inclined at an angle a to the horizon by means of a force inclined at an angle a + p to the horizon. Find the magnitude of this force. If the force is vertical find the pressure on the plane. 20. A weight rests on a smooth inclined plane. Shew that the smallest force which will keep it in equilibrium must act along the plane. If the weight be the weight of a ton, and the inclination of the plane be 45, what is the power ? G. S. 11 162 STATICS. [CH. VI 21. Two inclined planes of equal heights are so placed that they have a common vertex; a weight lies on each of the planes, and the weights are connected by a string which passes over the common vertex ; in this position there is equilibrium; the lengths of the planes are respectively 6 and 3 feet ; the weight which rests on the shorter plane is 10 Ibs. Find the other weight in one of the following ways, neglecting all friction : (1) by means of the "triangle of forces," (2) by means of the " principle of work." 22. Find the ratio of the power to the weight on a smooth inclined plane when the former acts horizontally. If the weight be the weight of a ton, and the inclination of the plane be 30, what is the power ? 23. A. weight W is supported on a smooth plane inclined at an angle 30 to the horizon, by a string inclined at 60 to the horizon ; find the tension of the string. 24. Find the horizontal force that would support a weight W on a smooth plane inclined at an angle of 45 to the horizon. If on the same inclined plane the weight W is supported by two equal forces, one acting horizontally and the other acting along the plane upwards, find the pressure on the plane. 25. Find the ratio of the power to the weight when a body is kept in equilibrium on a smooth plane, inclined at an angle of 30 to the horizon, by a horizontal force. A given force is applied to support a weight on an inclined plane. Will the greater weight be supported, when the force acts horizontally, and the plane is inclined at an angle of 30 to the horizon ; or when the force acts parallel to the plane, and the plane is inclined to the horizon at an angle of 60 ? 26. An inclined plane 14 feet long has one end 8 feet and the other end 10 feet above the level of the floor. If 294 ft.-lbs. of work are done in dragging a mass of 1 cwt. up the plane, find the friction. 27. Describe the system of pulleys in which each string is attached to a bar from which the weight is suspended, and find an expression for its mechanical advantage, the weights of the pulleys being neglected. If there are 3 strings attached to the bar, what power will support a weight of 35 Ibs.? CH. Vl] MACHINES. 163 28. Describe the system of pulleys in which each pulley hangs in the loop of a separate string, and find an expression for its mechanical ad- vantage, the weights of the pulleys being neglected. If there are four moveable pulleys, what power will support a weight of 50 Ibs. ? 29. Determine the relation of the power to the weight in a system of 5 pulleys, of which the weight may be neglected, one end of each string being fixed to a beam. Find also how far the weight is raised when the power moves through 12 ft. 30. Find the mechanical advantage of the system of weightless pulleys in which each pulley hangs in the loop of a separate string, one end of which is fastened to a fixed beam ; all the strings being parallel. If there are 7 pulleys and the weight is 8 cwt. find the power in Ibs. 31. Find the ratio of the power to the weight in the system of pulleys in which all the strings are parallel and are attached to the weight. If there are 6 pulleys and the weight is 9 cwt. find the power in Ibs. 32. In a certain system of pulleys it is found that the power descends 1 ft., while the weight rises 1 inch. What power will be required to raise a weight of 1 cwt. ? 33. Describe the system of pulleys in which each string is attached to a bar from which the weight is suspended, and find an expression for its mechanical advantage, the weights of the pulleys being supposed equal. If there are three strings attached to the bar, and the weight of each pulley is 1 Ibs. , what power will support a weight of 42| Ibs. ? 34. Describe the system of pulleys in which each pulley hangs in the loop of a separate string, and find an expression for its mechanical ad- vantage, the weights of the pulleys being supposed equal. If there are 4 moveable pulleys, each weighing f of a lb., what power will support a weight of 56f Ibs. ? 35. In the system of three equal pulleys, one fixed and two moveable, in which each string is attached to a bar supporting the weight W, prove that, neglecting the weights of the pulleys and the bar, if P is the power, 7P = W, and find the point on the bar from which W must be suspended. If, when there is equilibrium, one-third of the weight falls off, prove that the acceleration of the remainder of the weight will then be one twenty-third of the acceleration due to gravity. 36. Find the ratio of the power to the weight in that system of pulleys in which there is only one string. In such a system a power P supports a weight W; if P and W are in- terchanged prove that the weight to be added to P to produce equilibrium 164 STATICS. [CH. VI 37. Draw carefully a system of pulleys in which each pulley hangs by a separate string and the ratio of the power to the weight is 1 to 32. 38. Find the conditions of equilibrium in the wheel and axle. 3t on rough weight W is b (1 + sin X) Shew that if the axle rest on rough bearings, the least power (acting downwards) that will raise a weight W is a - b sin X where a, b are the radii of the wheel and axle and X the angle of friction. 39. If in order to raise a weight of 144 Ibs. through an inch by means of 'a wheel and axle' I must move my hand through a distance of one foot, what power must I exert? 40. -A- balance is apparently in adjustment when no weights are in the scale-pans. A certain mass is put into the right pan and requires weights of 45 '63 grams in the left to maintain equilibrium ; on putting the mass into the left pan it is found that 45*81 grams are needed in the right. Find the mass and ratio of the arms. 41. In a wheel and axle the radius of the wheel is 3 feet. The axle is of square section, the side of the square being 6 inches long. Find (i) the greatest, (ii) the least vertical power that must be exerted to slowly lift a weight of 252 Ibs. in the usual manner. 42. A straight uniform lever AB, 12 feet long, balances about a point in it 5 feet from B, when weights 9 Ibs. and 13 Ibs. are suspended at A and B. Find the weight of the lever. 43. A straight uniform lever whose weight is 16 Ibs. balances about a point one foot from its middle point when weights 6 Ibs. and 10 Ibs. are suspended from its ends. Find the length of the lever. 44. Find the power required to support a weight W in a system of 4 pulleys in which each string is attached to the weight and the pulleys are supposed weightless. If the weights of the pulleys are taken into account and each weighs 1 lb., find what power will support a weight of 78 Ibs. 45. Find the power required to support a weight W in a system of 4 pulleys in which each pulley is supported by a separate string, one end of which is fastened to a fixed beam, and the pulleys are supposed weightless. If the weights of the pulleys are taken into account and each weighs 1 lb., find what power will support a weight of 65 Ibs. CH. Vl] MACHINES. 165 46. Find the condition of equilibrium in the system of pulleys in which the same string goes round all the pulleys. If a weight of 6 Ibs. just supports a weight of 28 Ibs., and a weight of 8 Ibs. just supports a weight of 42 Ibs. ; find the number of pulleys, and the weight of the lower block. 47. Find the condition of equilibrium in the system of pulleys in which each string is attached to the weight. If there are 5 moveable pulleys each weighing half-a-pound, and the weight is 35 Ibs., what is the power ? CHAPTER VII. FRICTION. 64. Friction. The term "friction" has been used occasionally in the last chapter: we have seen that in many cases when a body rests on a surface the force between them is riot wholly at right angles to the surface, but has a component along the surface. The surface is then said to be rough and the component of the force along the surface is called friction. It remains now to consider the nature of friction a little more fully. DEFINITION. When a body rests on a rough surface and the impressed force has a component along the surface a force is called into play tending to balance this component and prevent motion. This force is called Friction. The Direction of friction is opposite to the component of the other forces resolved parallel to the surface, opposite that is to the direction in which motion would take place if there were no friction. The Amount of friction up to a certain limit is always just sufficient to prevent motion, but only a limiting amount of friction can be called into play. Thus if a body rest on a horizontal table the pressure of the table balances the weight, these forces are both vertical, there is no component in the direction of the surface and no friction is called into play. Apply a small force parallel to the surface, the body does not move, sufficient 64] FRICTION. 167 friction is exerted just to stop the motion ; increase the force still further, until the force parallel to the surface reaches a certain limit depending on the normal pressure and on the nature of the surface, the body does not move : when however this limit is exceeded, motion takes place. Thus consider a ladder resting as in Example (1), p. 75, against a smooth wall on rough ground ; when the foot of the ladder is near the wall the friction needed to maintain it in position is small ; as the foot is withdrawn from the wall and the slope increases more friction is required and is called into play, until at last there comes a position in which the ladder begins to slip, the friction which needs to be exerted is greater than the ground can exert, the limiting position has been past. We may put this in symbols thus : Let E be the normal pressure of the smooth wall, W the weight of the ladder acting vertically at its centre of gravity, Y the vertical stress at the foot and X the horizontal component of the action the friction. Then, as in Example (1), p. 75, Eesolving vertically Y=W. Eesolving horizontally X = R. Taking moments about the lower end Hence X=%Wcota. Now as the angle a decreases this increases : when it becomes greater than the maximum friction which the ground can exert the ladder will slip. Let us call F this maximum friction, then X must be not greater than F. Hence ^ IF cot a is not greater than F, and cot a must riot be greater than 2*7 JF. Many other observations shew us tbat there is a limit to the maximum amount of friction which can be called into play. The following experiments will help to shew on what the limiting amount depends. EXPERIMENT 10. To investigate the Laws of Limiting Friction. Thus take a well-planed board of hard wood and a small piece of wood, say six inches by nine in size ; rub one surface of this small piece of wood as smooth as possible with sand- 168 STATICS. [CH. VII paper and then cut the wood in two so as to have two pieces, one of about twice the size of the other, with surfaces as nearly alike as possible. Fix a screw eye into one end of each of these pieces of wood and attach a string to each. Fix a pulley at one end of the horizontal board over which the string may pass and support a scale-pan and weights. Secure some pieces of lead to the smaller piece of wood and thus make the weight of the two the same. Place the larger piece of wood on the board and put a considerable weight, 5 or 6 kilos, on it. Pass the string from the wood over the pulley and suspend the scale-pan as in Fig. 138. Load the scale-pan until the wood just begins to Fig. 138. slip on the board. The weight required for this can be found fairly closely by gently tapping the board. Note the total weight suspended ; this measures the maxi- mum friction which can be exerted between the board and the wood. Note also the total weight supported by the board ; in- cluding in this the weight of the wood itself l . This measures the normal stress between the two. Place more weights on the wood so as to increase the force between it and the board ; it will be necessary to place more weights in the scale-pan in order to start the motion; the limiting friction is increased. Determine in this way the limiting friction for a number of loads and form a table in which the first column is the Limiting Friction, the second column the Normal Force be- tween the surfaces. 1 It is convenient to make this up to some fraction of a kilogram such as ^ or by securing some pieces of lead to its upper side. 64-65] FRICTION. 169 Divide the numbers in the first column by the correspond- ing numbers in the second; it will be found that the quotients are the same. Thus when the surfaces in contact remain the same the ratio of the limiting friction to the normal force is a constant. Now repeat the experiment, using the second or smaller piece of wood, it will be found that for the same normal stresses as previously the limiting friction is also practically the same. The areas of the surfaces in contact have been altered but not the nature of those surfaces or the state of their polish; the ratio of the limiting friction to the normal stress is unchanged. Replace the wooden slide-piece by another of different material; the law, that the ratio of the limiting friction to the normal force is constant, still holds; but the value of this constant ratio differs from that found in the previous ex- periment. 65. Laws of Limiting' Friction. We are thus led to the following Laws of limiting friction. (i) The ratio of the limiting friction to the normal force between any two given surfaces is a constant. (ii) This constant ratio depends on the material of the surfaces in contact and the state of their polish, but not on their area or shape. These two laws of Limiting Friction together with the definition and statements given in Section 64 are sometimes enunciated together as the Laws of Friction. These laws, though they express fairly well the result of experiments, are probably not rigorously true; they are how- ever usually employed in considering problems involving friction. It is customary to give another law which however does not concern us in Statics. When sliding motion takes place the ratio of the friction to the normal force is still found to be constant for any two given 170 STATICS. [CH. VII surfaces and independent of the velocity ; this constant however is slightly less than the constant ratio of the limiting friction to the normal force. Thus it requires rather greater force to start a body moving on a rough surface against friction than to maintain it in motion with uniform speed when once started. 66. Coefficient of Friction. The constant ratio of the limiting friction to the normal stress for two surfaces of given material in a definite state of polish is called the Coefficient of Friction. Thus if F is the limiting friction, R the normal force and /A the coefficient of friction, we have F or =p. The experiments described in Experiment 10 give one method of determining /A. The following method is generally more convenient. Consider a body lying on a rough horizontal surface under its weight and the reaction of the surface, the body is in equilibrium and there is no friction. Tilt the horizontal surface, the weight will have a component down the surface. Friction is called into play to balance this component. Con- tinue to tilt the surface until the body just begins to slide down ; when this is the case the limiting amount of friction has been reached, and this limiting amount of friction is just equal to the component of the weight down the surface when tilted at the angle at which sliding just begins. The following Experiment will enable us to verify the laws of friction and to find the coefficient of friction by this method. EXPERIMENT 11. To prove that on a rough surface the limiting amount of friction is proportional to the normal force, and to find the coefficient of friction. The apparatus consists of a mahogany board, Fig. 139, some 12 or 15 inches long and 3 or 4 inches in width. This is 65-66] FRICTION. 171 hinged at one end to another similar board which can be clamped to the table. At the end of the second board remote Fig. 139. from the hinge a vertical support is fixed and the hinged board can be raised and secured in any position by means of a string passing through a small screw eye at the top of this support. A graduated vertical rod is also screwed as shewn at EC to the base board and the height of the plane at B can be easily measured on this rod. The point G is at some convenient distance (say 10 inches) from the hinge A: by dividing the height in inches by 10 we get at once the tangent of the inclination of the plane. Let the angle BAG be a. One or more small boards of various sizes and materials can be placed on the inclined plane and weights can be placed on these small boards to vary the force between them and the inclined plane. In the apparatus shewn each small board has one or more thin brass rods screwed to it, the weights used in the other experiments can be piled on it so that each weight when the plane is tilted is prevented from slipping by the rod which passes through its centre. 172 STATICS. [CH. VII Place one of the small boards with some convenient weight on it on the plane. Raise the plane, gently tapping it from time to time, until the small board begins to slide down. When this takes place note the height BC and thus find the tangent of the inclination of the plane to the horizon. Now when the board just begins to slide the maximum friction has been reached and this just balances the component of the weight down the board. Thus if R be the normal force, F the friction in any position and W the weight, we have resolving along the plane, ^= JFsina, and resolving perpendicular to the plane R= JFcosa. Hence F W sin a R IT cos a = tan a. Now experiment shews that, so long as the material and state of polish of the surfaces remain the same, the slipping just begins at the same angle. Thus replace the weights by others and repeat the experiment, the slipping takes place at the same angle as before. Replace the small board by another of the same material and polish but of different area ; it will just begin to slip at the same angle as previously. Hence if F now stand for the maximum amount of friction and a for the angle at which slipping takes place we see by the experiment that a is constant so long as the material and polish remain the same. Hence since the ratio FjR is equal to tan a we see that FjR is constant under these same conditions. This ratio is defined to be ^ the coefficient of friction. F Hence u = -= = tan a. -ft Thus the coefficient of friction is found by reading BC the height of the plane and dividing it by AC the base. 66-67] FRICTION. 173 67. Angle of Friction. DEFINITION. The angle a whose tangent gives the coefficient of friction is called the Angle of Friction. We can give a more general meaning to this angle thus. Consider any body resting on a rough surface. Let the friction be F and the normal force R. The resultant of these two will be a force P, which combined with the other forces must maintain equilibrium. Now let P act at an angle 6 with the normal force R. Then resolving P along and perpendicular to R we have Hence F tan 6 = -yr . Thus in any case the ratio FjR measures the tangent of the angle which the resultant force between the surface and the body makes with the normal to the surface. Thus, when F reaches its limiting value, reaches its greatest value and becomes a the angle of friction. Hence the angle of friction is the angle which the resultant force makes with the normal when the friction has reached its greatest value ; or in other words it is the greatest angle which the resultant force can make with the normal. So long then as the resultant force is inclined to the normal at an angle less than the angle of friction, equilibrium is possible; when this angle becomes greater than the angle of friction motion must take place. In the case of a body on an inclined plane the resultant force due to the plane must be always vertical, for the only other force is the weight ; and the resultant of the normal force and the friction must just balance the weight. Again, the angle which the normal to the plane makes with the vertical is the angle of the plane. Thus the angle between the direction of the resultant force and the normal to the plane is the angle of the plane ; so long as the angle of the plane is less than the angle of friction equilibrium is possible; when the angle of the plane is just greater than the angle of friction slipping begins. 174 STATICS. [CH. VII The following is a table of approximate values for the coefficient of friction. Wood upon wood ... ... ... -5 lubricated -2 Wood upon polished metal -6 lubricated ... -12 Metal upon metal ... ... ... -18 ,, lubricated ... ... -12 EXAMPLES. FEICTION. 1. Explain what is meant by the coefficient of friction. A cubical block rests on a rough plane, one end of which is gradually raised. Find the greatest value of the coefficient of friction which will just permit the block to slide down the plane before falling over. 2. State the laws of friction ; and assuming that the friction is the same when a body is moving as when it is at rest, find the time taken by a body to fall down a rough inclined plane from rest. 3. A brick whose dimensions are 8x4x3 inches rests on a rough plane in such a way that it cannot slip, and the plane is gradually tilted about a line parallel to one edge of the brick; shew that the angle through which the plane can be raised without upsetting the brick depends on which face of the brick is on the plane ; find also the greatest and least angles for which the brick will just not upset. 4. What is meant by the angle of friction ? The lower half of an inclined plane is rough, the upper half being smooth. A particle is allowed to slide from the top and is brought to rest by friction just as it reaches the bottom. Find the ratio of the friction to the weight of the particle, assuming it to be independent of the velocity. 5. A block W l rests on an inclined plane and is supported partly by friction and partly by the tension of a cord which passes over a pulley H. VIl] FRICTION. 175 at the top of the plane and carries a weight JP 2 . Shew how to find (graphically or otherwise) the values of W 2 which will (1) just prevent W : from slipping down, and (2) just make W^ begin to slip up, when the co- efficient of friction and the inclination of the plane are known. 6. A block of iron weighing 10 Ibs. rests on a level surface plate. A string attached to the block passes over a pulley so placed above the sur- face plate that the string makes an angle of 45 with the vertical. After passing over the pulley the string supports a weight. Find the least value of this weight which will make the block slip, the coefficient of friction being |th. 7. A heavy body is to be drawn up a rough inclined plane. If the force is the least possible prove that its inclination to the plane must equal the angle of friction. 8. Describe an experimental method for finding the coefficient of friction between two substances. A semicircular disc rests in a vertical plane, with one part of its curved surface touching the ground and another touching a vertical wall. Shew that it will rest in any position in which its straight-edge makes an angle with the vertical greater than where /j. is the coefficient of friction between the disc and the ground and between the disc and the wall. [The centre of gravity of a semi- circular disc is at a distance from the centre equal to (4/3?r) times the radius.] 9. A heavy body rests on a rough plane inclined at the angle 30 to f) the horizontal, the coefficient of friction being p . v 3 A horizontal force along the plane is applied to the body, and is gradually increased until the body begins to move ; find the direction in which the body begins to move, and the magnitude of the horizontal force. 10. A uniform rod AB of length 2 a rests with the end B in contact with a rough vertical wall, and is supported by a smooth peg fixed at a distance 6 from the wall, and the end B is on the point of slipping up- wards. Shew that the inclination of the rod to the vertical is then given by the equation sin 2 6 (sin 6 - /* cos 6) = - , fi being the coefficient of friction between the rod and the wall. 11. How would you place a brick whose length is double its breadth, and breadth double its thickness, on a rough inclined plane, so as to be least likely to tumble over? Would it be less likely to slide down with one face in contact than another? Give reasons for your answers. 176 STATICS. [CH. VII 12. A weight W rests in equilibrium on a rough inclined plane, being just on the point of slipping down. On applying a force W parallel to the plane, the weight is just on the point of moving up. Find the angle of the plane and the coefficient of friction. 13. What is the coefficient of friction when a body weighing 50 Ibs. just rests on a plane inclined at 30 to the horizon? If the plane were horizontal what horizontal force would be required to move the body? 14. Find what horizontal force will be required to support a weight of 3 cwt. upon a smooth inclined plane, whose height is f of its length. 15. A force P acting up an inclined plane supports a weight Won it. If E be the reaction of the plane, prove that P : W : R :: height of plane : length : base. 16. A mass of 1 cwt. rests on a rough inclined plane of angle 30. If the coefficient of friction be lj*J& find the greatest and least forces which, acting parallel to the plane in both cases, can just maintain the mass in equilibrium. HYDEOSTATICS CHAPTER I. STATES OF MATTER. 1. Solids and Fluids. In Dynamics and Statics we have dealt with some parts of the Mechanics of Solids. A Solid Body such as a lump of iron or a lump of wood has a definite Volume. It also has a definite Shape. If force be applied to it, both the shape and the volume are generally changed, though the change in many cases is very small. The force required to produce a given change of shape or size differs for different substances. Great force must be applied to a lump of iron to produce an appreciable alteration in shape or volume; a solid india- rubber ball can be squeezed from its spherical form by a much smaller force than is necessary in order to change to an equal extent the shape of a similar sphere of iron. Again, if the force applied be not too large, both the iron and the india-rubber will regain their original shape and volume when it is removed. Iron and india-rubber are both Elastic Substances. DEFINITION. An Elastic Substance is one which has a definite shape and volume when free from the action of external forces ; when such forces are applied, the shape, or the volume, or both are changed ; when the forces are removed provided they have not been too great the substance recovers its shape and its volume. G. HYD. 1 2 HYDROSTATICS. [CH. I If the forces are too great the body may be strained beyond recovery, the limits of its elasticity may be passed ; when the forces are removed the body does not regain its former shape and volume. A solid body then can offer resistance (a) to forces tending to change its volume, (b) to forces tending to change its shape. In consequence of the former it is said to have Volume Elasticity ; in consequence of the latter it is said to have Elasticity of Form or, as it is called, Rigidity. A perfectly rigid body is one in which no change of shape is produced by the action of a finite force ; no known body is perfectly rigid, though the rigidity of most solids is so great that for many purposes we may treat it as perfect. Solids possess these two kinds of elasticity in very different degrees. Thus the shape of a piece of india-rubber is easily altered ; experiment however shews that it requires consider- able force to change its volume. A piece of cork can by the application of moderate force be squeezed into a much smaller volume ; its shape however is not necessarily greatly changed in the process. In comparison with its volume elasticity the rigidity of cork is considerable. Any body which has Elasticity of Form or Rigidity is called a Solid. In consequence of its rigidity the body preserves its shape. It is a matter of everyday experience that there are numbers of bodies which exhibit little or no tendency to retain their shape. Ice is a solid, on melting it becomes water, the particles of the water slide freely over each other and the shape assumed depends on that of the vessel in which it is contained. Water, alcohol and numerous other substances can be poured from one vessel to another. Such substances have practically no rigidity; they offer practically no resistance to forces tending to produce sliding motion of their particles and thus to change their shape. They are called Fluids. We distinguish then between Solids and Fluids thus. DEFINITIONS. A Solid is a body which can offer perma- nent resistance to forces tending to change its shape. 1-2] STATES OF MATTER. 3 A Fluid is a body which can offer no permanent resistance to forces tending to change its shape. In other words, a Solid has rigidity, a Fluid 1 has no rigidity. 2. Fluids. Water and alcohol have been instanced as examples of fluid bodies, we can readily pour them from one vessel to another, and these fluids adapt themselves almost instantaneously to the shape of the vessel into which they are poured ; they offer practically no resistance to forces tending to change their shape ; moreover the change of shape follows very rapidly on the application of the slightest force. There are other substances, however, in the case of which, time is necessary, before a change of shape will occur under the action of a force. Honey or treacle can be poured from one vessel to another, but they pour slowly. We can make a heap of treacle in the middle of a dish or plate, but on leaving it, the heap is gradually flattened out and the plate covered. The treacle has no permanent rigidity, it can offer no resistance to a small force, such as its weight, if that force acts for a sufficient time ; treacle like water is a fluid, but it has the property of Viscosity and in consequence yields slowly to forces tending to change its shape. A Perfect Fluid is one which yields instantaneously to such forces. No fluid in nature is perfect, water and alcohol have some slight viscosity, but the amount is so slight that they may be treated as perfect in comparison with fluids such as treacle, honey or glycerine, which are called Viscous Fluids. It is sometimes difficult to draw the line between a solid and a fluid or between a viscous fluid and one which is practically perfect. Thus consider a stiff jelly made by melting gelatine in water and allowing it to solidify ; the jelly retains its shape when cold and recovers it again after being slightly squeezed, it is a solid on mixing it with more water we obtain a sticky, viscous liquid, if the quantity of water be con- siderably increased the viscosity can be made very small indeed and the fluid is practically perfect. 1 It will not however be sufficient to give this last statement as a definition of a fluid unless at the same time a definition of rigidity be given. 12 4 HYDROSTATICS. [CH. I Again, a piece of pitch or of cobblers' wax has a definite shape, but it can only retain it for a short time, if placed on a flat surface it will gradually flow over the whole ; pitch then must be classed as a fluid, but as a very viscous one. We may shew this by placing some pitch in a wide-necked funnel and leaving it in a fairly warm place, the pitch will gradually flow through the funnel. The same fact is illustrated by supporting a rod of sealing-wax at two points near its ends respectively, in time the rod is seen to bend, sinking in the middle ; a small force produces change of shape but time is necessary in order that the effect may take place. We must also distinguish between viscous fluids and plastic solids. Beeswax and paraffin wax are both solids, they have a definite shape and will retain it indefinitely, the application of quite a small force however is sufficient to mould a piece of beeswax into a new form ; the rigidity of such a substance is extremely small, the limits within which it will recover its form are very narrow ; it is said to be Plastic. If a paraffin candle be substituted for the sealing-wax in the experiment just described, it will not sag as the wax did, it can support its weight without continuous yielding and does not gradually change in shape under so small a force. The paraffin is a soft solid. We thus see the importance of the word permanent in the above definition of a fluid ; in the experiments to be described it will generally be assumed that the fluids employed are not viscous, though we shall see that in dealing with the equi- librium of fluids Hydrostatics we need not consider viscosity. See Section 15. 3. Liquids and Gases. Fluids then differ from Solids in that they have no Rigidity, they have however Volume Elasticity. A fluid, like a solid, will resist a force tending to reduce its volume, and will recover that volume when the force is with- drawn. But Fluids can be divided into two main classes possessing this property in very different degrees. Water and air are both fluids ; very great force is needed to produce even a small change in the volume of a mass of water, that of a mass of air (Section 79) can be changed easily. Some fluids are practically Incompressible, the change of volume produced by the application of even a very large force is extremely small. Such fluids are called Liquids. 2-3] STATES OF MATTER. 5 Water, Oil, Alcohol, Vinegar are liquids. Other fluids are very easily compressible ; these are called Gases. Such are Air, Oxygen, Hydrogen, Carbonic Acid. DEFINITIONS. A Liquid is a substance which can offer no permanent resistance to forces tending to change its shape, but which offers very great resistance to forces tending to diminish its volume. A Gas is a substance which can offer no permanent re- sistance to forces tending to change its shape, and which offers only a small resistance to forces tending to diminish its volume. To illustrate the difference between a Liquid and a Gas let us imagine a cylinder closed with a tightly-fitting piston and suppose the area of the piston to be 100 square centimetres. Let there be a litre (1000 c. cm.) of water in the cylinder, then the depth of the water will be 10 cm. Now suppose a weight of 100 kilogrammes is placed on the piston so that each square centimetre of the piston has to carry an additional weight of 1 kilogramme, then it has been shewn that supposing the piston to move without friction it would sink by about one two-thousandth (^^oir) ^ a centi- metre; the change of volume of the whole litre produced by this pressure would be ^ of a cubic centimetre, the change in each cubic centimetre therefore would be -a^tRre f a cubic centi- metre. Thus the volume of a given mass of a liquid is very nearly constant and is only changed very slightly by the application of considerable forces. We may treat a liquid as a fluid of invariable density. See Section 5. Now let us suppose that the water is removed from the cylinder and replaced by an equal volume of air at atmospheric pressure 1 . Then, on repeating the experiment, it would be found that the piston if it were perfectly frictionless would sink about five centimetres, the volume of the air would be about halved. Each cubic centimetre would now occupy half a cubic centimetre. Air is about 10,000 times as compress- ible as water. 1 See Section 67. 6 HYDROSTATICS. [CH. I The density of the air is hereby doubled. Gases then are fluids the density of which depends on the pressure to which they are subjected. The experiment could not be carried out in this simple form because of the friction of the piston against the sides of the cylinder, but this difficulty can be avoided by means of a suitable modification of the apparatus. See Section 79. 4. Free Surface of Liquids. There is moreover another distinction between a liquid and a gas. Imagine that the walls of the cylinder just described are continued some distance above the piston, on raising the piston the level of the upper surface of the water still remains at about 10 centimetres from the bottom, above it there will be an almost empty space containing a little water vapour ; the water will have a free surface separating it from the empty space above. If, however, the cylinder contain a gas this will no longer be the case, the gas will expand as the piston rises, the whole space below the piston will be occupied by the gas, its density and the pressure it exerts on the sides of the cylinder will diminish ; there will be no free surface. Thus we may say that an incompressible fluid a Liquid- is a substance which can offer no permanent resistance to forces tending to change its shape, but which has a definite density ; it will therefore not increase indefinitely in volume if the force on its surface be diminished ; when placed in any vessel it will occupy the lower portion of the vessel completely and will have a free surface. A Gas is a substance which can offer no permanent resist- ance to forces tending to change its shape ; its density however depends on the forces impressed on its surface, it will increase indefinitely in volume if these forces be sufficiently diminished, and, if placed in an empty closed vessel of any size, will fill it completely and have no free surface. For the purposes of this book we may treat liquids as incompressible. 3-5] STATES OF MATTER. 7 5. Density. Equal volumes of different substances differ in mass and therefore also in weight. We have already (Dynamics, Section 12) given a definition of the term Density which has been used in the last Section, and deduced some results from it. The definition is as follows : DEFINITION. The Density of any homogeneous substance is the mass of unit volume of that substance. It follows from this definition that to determine the density of a body we must find the number of units of mass in the unit of volume, we require therefore to know the unit of mass and the unit of volume ; if these be the gramme and the cubic centimetre respectively we may say that the density is so many grammes per cubic centimetre. Thus in these units the density of water is 1 gramme per c.cm., that of iron 7*76 grammes per c.cm. In any other units the numerical measures of the densities of these substances would generally be different. Thus a cubic foot of water contains 998-8 oz. or 62-321 Ibs.; hence the density of water is 998-8 oz. per cubic foot or 62-321 Ibs. per cubic foot; iron is 7 -7 6 times as dense as water, hence its density is 7-76 x 62-321 Ibs. per cubic foot. From the above definition of density we can find a relation between the Mass, Volume, and Density of a body. PROPOSITION 1. To shew that if the mass of a homogeneous body be M grammes, its density p grammes per. cubic centimetre and its volume V cubic centimetres, then M= Vp. For by the definition, the mass of 1 c.cm. = p grammes, therefore the mass of 2 c.cm. = 2p grammes, the mass of 3 c.cm. 3p grammes, hence the mass of V c.cm. = Vp grammes. Therefore M= Vp. We may write this as M 8 HYDROSTATICS. [CH. I and thus we have the result that the density of a homogeneous substance is the ratio of its mass to its volume. A result similar to the above holds for any other consistent system of units. To determine then the density of a piece of homogeneous material we require to know its mass, which is obtained in terms of a standard mass by weighing (Statics, Sections 59, 60), and its volume, which may be found in some_cases by direct measurement, in others by the displacement method (Dynamics, Experiment 4), in others again by one or other of the methods described in the following pages, Experiments 15 etc. In any case we should notice that the measure of the density will depend on the units adopted for the measurement of the mass and the volume; these units must be known in order to determine the density completely. 6. Specific Gravity. In many cases, however, we are only concerned with the relative masses or the relative weights of equal volumes of two substances ; it may be sufficient for us to know that a lump of iron is 7 '7 6 times as heavy as an equal volume of water, or that the weight of a piece of pine wood is about -56 of that of an equal volume of water. DEFINITION. The Specific Gravity of a substance is a number which expresses the ratio between the weight of the substance and that of an equal volume of some standard sub- stance, usually water. Thus, if W be the weight of this substance, W the weight of an equal volume of some standard substance, and , where o> represents the weight of unit of volume of the standard substance. Let W be the weight of an equal volume V of the standard substance, then, since w is the weight of each unit of volume of the standard, the weight of V units of volume is Tco. Hence W is equal to Fw. W W . Thus if we know the volume of a body, its specific gravity referred to some standard substance, and the weight of unit of volume of that standard, we can calculate the weight of the body. The calculation is simplified if we take water as the standard substance, the volume of 1 cubic centimetre as the unit of volume and the weight of 1 gramme as the unit of weight, for, since the weight of 1 cubic centimetre of water is 1 gramme weight, we have in this case the weight of the unit of volume as the unit of weight ; thus w is unity and W V the weight of a unit of volume of the standard substance, we have, assuming no chemical action to occur, F= F 1+ F 2 +...+etc. Fora) = weight of whole = sum of weights of components = Fj + F 2 11] STATES OF MATTER. 17 Examples, (l) If a volume of 10 c.cm. of a liquid of density 8 grammes per c.cm. be mixed with 15 c.cm. of a liquid of density 7 grammes per c.cm., find the density of the mixture. The volume of the mixture is 10 + 15 or 25 cubic centimetres. The mass of the mixture is 10 x -8 + 15 x -7 or 18-5 grammes. Hence its density is 18 '5/25 or -74 grammes per c.cm. (2) An alloy of zinc (sp. gr. 7'2) and copper (sp. gr. 8-95) has a mass of 467 grammes. Its volume is 60 c.cm. Find the volume of each component. Let v l c.cm. be the volume of the zinc and v 2 c.cm. that of the copper. Then the mass of the zinc is 7'2 x Vj grammes, that of the copper is 8-95 x v 2 grammes. The sum of these two is the total mass 467 grammes, the sum of the two volumes is the total volume 60 cubic centimetres. Thus ^ + ^2 = 60, 7-2 t^ + 8-95 8 = 467. Hence, solving these equations, 1-75^ = 537-467 = 70, 1-75 v 2 = 467 -432 = 35. Thus Vj = 40 c.cm. v 2 =20 c.cm. (3) A mixture is made of 14 cubic centimetres of sulphuric acid (specific gravity 1'85) and 6 cubic centimetres of water. The specific gravity of the mixture is found to be 1-615. Determine the amount ef con- traction which has taken place. If there were no contraction the volume of the mixture would be 14 + 6 or 20 c.cm.; let the actual volume be V c.cm. Then, since the density is 1-615 grammes per c.cm., the mass is Fx 1-615 grammes. The masses of the components are 14 x 1-85 or 25*9 grammes and 6 grammes respectively. The mass of the mixture is the sum of the masses of its components. Hence Fx 1-615 = 25-9 + 6 = 31'9. Therefore F = 19*75 c.cm. Hence the contraction required is 20 - 19*75 or '25 c.cm. G. HYD. 18 HYDROSTATICS. [CH. I EXAMPLES. 1. What is meant by the density of a substance ? How would you find the density of water ? 2. The density of a substance being defined as the mass of a unit of volume of the substance, shew precisely how the density of a liquid may be experimentally determined. 3. Describe the experiments you would make in order to determine the mass of a cubic centimetre of water. 4. Explain clearly the distinction between specific gravity and density and shew how the numerical value of these quantities depends on the choice of fundamental units. 5. Find the density and specific gravity of the following body : A rectangular pillar having a square base each side of which is one foot in length, the height of the pillar being 10 feet and its weight half a ton. (The weight of a cubic foot of water may be taken as 1000 ounces.) 6. The density of copper is 8'95 grammes per c.cm. The diameter of a piece of copper wire is 1*25 mm. and its length 1025 cm. ; find its mass. 7. Find the density of a cylinder 1 foot in height and 6 inches in radius whose mass is 60 Ibs. 8. Find the density of a sphere 10 cm. in radius and 5 kilogrammes in mass. 9. Determine the density of the cylinder described in Question 7 in grammes per c.cm. 10. Find the density of a pyramid on a triangular base each side of which is 10 cm. and which has an altitude of 30 cm., the mass of the pyramid being 8 kilogrammes. 11. The density of mercury is 13-59 grammes per c.cm. ; find it in grains per cubic inch. 12. Compare the densities of a sphere 5 cm. in radius, 5 kilos, in mass, and of a cylinder 1 foot in height, 6 inches in radius and 60 Ibs. in mass. 13. Find the mass in Ibs. of a cube of gold each side of which is 4 inches. 14. An iceberg is 30 fathoms high, 40 fathoms wide and 30 fathoms thick ; find its mass in tons. CH. l] STATES OF MATTER. 19 15. A carboy of sulphuric acid has a mass of 98 kilogrammes ; find its volume. 16. It is desired to float a piece of slate a metre square by 5 centimetres thick by means of cork floats. What volume of cork is required ? 17. Find the specific gravity of a mixture of glycerine and alcohol (i) in equal parts by weight, (ii) in equal parts by volume. 18. A piece of brass is made from 2 Ibs. of copper and 3 Ibs. of zinc, the volume of the copper being 6 cubic inches, and that of the zinc 13 \ cubic inches ; find the specific gravity of the brass. 19. A cylindrical tube, 16 cms. long, holds when full 1 gramme of mercury, sp. gr. 13-6 ; find the sectional area of the tube. 20. The specific gravity of a faulty iron casting which weighs 3 Ibs. is found to be 5-8. If the normal specific gravity of cast iron be 7'2 ; find what volume of the faulty iron is unoccupied by iron. (A cubic inch of water weighs 0-57 oz.) 21. If the specific gravity of a mixture of glycerine and water be 1-094, find the relative weights of glycerine and water in the mixture, the specific gravity of pure glycerine being 1-26. 22. The mass of a piece of brass is 25 grammes and the density of brass is 8-4 grammes per c.cm., find the volume of the brass. 22 CHAPTER II. FLUID PRESSURE. 12. General considerations on Stress. Consider a block of wood lying on a smooth horizontal table, let a weight be placed on the wood, as in Fig. 1. The wood is in equilibrium, the downward force exerted by the weight is balanced by the upward force between the table and the wood. Imagine the block divided into two parts by a horizontal plane CDj the upper part is acted on by the weight which presses it downwards ; since there is equilibrium the weight must be balanced by a force exerted upwards by the lower part of the block on the upper ; if now we consider the lower part of the block a downward force is exerted on its upper surface equal to the upward force which this surface exerts on the upper portion. This downward force is balanced by the upward force exerted by the table ; the two portions of the block are squeezed together across the section CD ; the block is said to be under Stress. The two equal forces acting in opposite directions on the two portions of the block across the Fig. 1. 12-13] FLUID PRESSURE. 21 section CD constitute a Stress. In the case considered the forces are at right angles to the surfaces on which they act ; the external forces, the weight and the pressure of the table, are such as to bring the two portions of the body, separated by the plane CD, more close together than they could other- wise be ; each portion of the body " thrusts " or pushes the other. We speak of the force which each part of the body exerts on the other across the plane CD as a normal Thrust or more simply as a " Thrust." 13. Thrust and Tension. The Stress which we have just been considering consists of a simple Thrust acting in opposite directions on the two sides of any horizontal plane such as CD, by which we imagine the body to be divided. It should of course be noticed that the division is imaginary, the two parts of the body on either side of any horizontal plane thus act on each other ; it is not necessary actually to cut or divide the body to give rise to the action. There are, of course, many other ways in which we can apply a simple thrust to a surface; thus when we push a body with a long pole, directing the push along the axis of the pole, any section of the pole at right angles to the axis is subject to a Thrust, the portion of the pole on one side of the section thrusts and is thrust by that on the other, the stress across the section is a simple thrust. But now suppose one end of the pole is attached to some body and that we pull at the other; if we now consider a section of the pole at right angles to its length, the portion of the pole on one side of the section is pulled by that on the other; the pole throughout its length is subject to a Pull or Tension instead of a Thrust. The Stress across each section at right angles to the length is a simple Tension acting along the pole. In both these cases however the stress is at right angles to the surface to which it is applied. 22 HYDROSTATICS. [CH. II 14. Shearing Stress. Let us return now to the block of wood on the table and imagine it divided, as in Fig. 2, by a plane CD inclined to the horizon. The upper part is acted on by the weight which presses it vertically down; since there is equi- librium this vertical force must be balanced by the force which the lower part of the block exerts on the upper, this latter force then must be ver- tical. But the plane CD is inclined to the horizon, hence in this case the force which the lower portion of the block exerts on the upper is inclined lg ' ' to the plane across which it acts. The force therefore may be resolved into two com- ponents, one, at right angles to the plane, constituting a normal thrust on the upper part, the other, parallel to the plane, preventing the upper part from slipping down; this second component constitutes a " Shearing Force." It is balanced by the equal and opposite shearing force exerted on the portion of the block below the plane CD and these two forces constitute a Shearing Stress. The components of the stress across the plane CD are a normal thrust which tends to compress the solid into a smaller space and a shearing stress tending to make it slide parallel to the plane CD and thus to change its shape. 15. Stress in Solids and Fluids. Hence, if we imagine any plane drawn in a solid body, the Stress across the plane due to the action which the portion of the solid on one side of the plane exerts on that on the other may be either a Thrust, a Tension or a Shearing Stress. In consequence of its Rigidity a solid can withstand shearing stress, it yields slightly until the impressed force is 14-15] FLUID PRESSURE. 23 just balanced by the elastic forces called into play by the yielding and then retains its new form so long as the force is impressed. A fluid however cannot permanently withstand shearing stress. Consider a solid body ABC, Fig. 3, resting on a table and let DE be a plane inclined to the horizon, dividing the body into two parts. Then the weight of the portion DEE is balanced by the force across this plane. If this force were a normal thrust perpendicular to DE the portion DEE would slide down, its weight has a component parallel to the plane; this component however is ba- lanced by the shearing force exerted across the plane. Now, suppose the portion Fig- 3 - DEE to become fluid; it runs away, the fluid has no rigidity and in consequence can exert no shearing force across any plane, there is therefore no force to balance the component of the weight parallel to DE, and equilibrium can no longer be maintained; the fluid yields to the impressed shearing force. We have thus arrived at the following results. If we imagine a body to be divided into two parts, each part exerts a force on the other across the dividing surface. These two forces are equal and opposite and are spoken of together as a Stress. In general the forces can be resolved into components respectively at right angles to, and parallel to the surface across which they act. And we can state the following Definitions. DEFINITION. The components of a Stress at right angles to the surface to which it is applied constitute a Thrust or a 24 HYDROSTATICS. [CH. II Tension. The components parallel to the surface constitute a Shearing Stress. A Solid is a body which offers permanent resistance to any form of stress, so long at least as the stress is not too great. If the stress be too great the solid may yield or break. A Fluid is a body which offers no permanent resistance to continued shearing stress, however small. It follows from this that any shearing stress, however small, will in time produce motion in a fluid if the fluid be very viscous it will be a long time before flow takes place under a small shearing stress: if, on the other hand, the viscosity be small, flow follows rapidly. If there were no viscosity no shearing stress could ever be exerted whether the fluid were at rest or in motion. DEFINITION. A Perfect Fluid is a body which, whether at rest or in motion, can never offer resistance . to a shearing stress, however small. There are no fluids known which satisfy this definition. When however a fluid is in equilibrium, whether it be viscous or not, there can be no shearing stress, for since a fluid can offer no permanent resistance to shearing stress even a small shear, if it existed, would in time disturb the equilibrium. We thus arrive at the result, that In any fluid in equilibrium there is no shearing stress. 16. Fundamental Property of a Fluid. The following Proposition then expresses the Fundamental Property of a fluid. PROPOSITION 5. To prove tliat, when a fluid is in equi- librium, the force, which it exerts on any surface with which it is in contact, is at right angles to that surface. For, let ACS, Fig. 4, be a portion of the surface and, if it 15-17] FLUID PRESSURE. 25 be possible, let the force P which the fluid exerts on a small portion of the surface near C be inclined to it in the di- rection DC. Then the force which the surface exerts on the fluid is P, acting in di- rection CD. This force can be resolved into a force R at right angles to the surface, constituting a normal thrust on the fluid, and a tangential force T parallel to the surface. C Fig. 4. This constitutes a shearing force and tends to make the fluid particles slide over the surface; since the fluid has no rigidity it cannot resist this force and motion will take place, which is contrary to the supposition that the fluid is at rest. Hence there can be no tangential force such as T, therefore the whole force is R, at right angles to the surface. Thus the force exerted by a fluid on any surface with which it may be in contact, or by one portion of a fluid on any other portion, across any surface separating the two, is a normal thrust at right angles to the surface. Experiments have shewn that it is possible for a fluid under certain circumstances to sustain a tension or pull ; these circumstances however occur very rarely, for our present purposes we may suppose that the only stress which can exist in a fluid is a thrust. 17. Stress distributed over a Surface. Imagine now that we have a piece of a stiff board resting on a table; on placing weights on the upper side of the board, force is exerted on the table, and this is balanced by the force * which the table exerts on the weights. Suppose the board is divided into a number of squares each 1 centimetre in edge and therefore 1 square centimetre in area; there is in general a force between each of these squares and the table, though the forces acting on each of the squares need not be equal. Sup- pose, however, that the same weight, 50 grammes say, is placed 1 Part of this force is due to the weight of the board, we suppose this to be small compared with the weights it carries and neglect its effect. 26 HYDROSTATICS. [CH. II on each square ; then the upward force on each square will be 50 grammes' weight ; the resultant upward force will be found by multiplying by 50 the number of square centimetres con- tained in the area of the board. The force in this case is Uniformly Distributed over the surface, and the thrust or resultant upward force is found by multiplying the force per unit area of the surface by the number of square centimetres in the area. DEFINITION. The Thrust on a surface is said to be Uniformly Distributed over the surface when it is the same on every equal area of the surface. In the example given above the thrust on each square centimetre is 50 grammes' weight, it is uniformly distributed ; but suppose that, instead of placing 50 grammes on each square centimetre, the weights were irregularly placed, so that on some squares there were more than 50 grammes, on others less, while the total weight carried remained the same; the total thrust would remain the same, but its distribution would be variable. In the illustration it is of course possible that the thrust of 50 grammes' weight which acts across each square centimetre may not be uniformly distributed over that square centimetre ; if we suppose the square centimetre to be divided into a large number of very small equal areas and if the thrust on each of these areas be the same, then the distribution over the square centimetre is uniform, this case is included in the definition by the introduction of the word " every." 18. Pressure at a Point. It is found convenient to give a name to the thrust per unit area of any surface. DEFINITION. When a thrust is uniformly distributed over a surface, the thrust on each unit of area is called the Pressure at each Point of the surface. Now let P be a thrust which is uniformly distributed over a surface, let p be the pressure at each point of the surface, and let a square centimetres be the area of the surface. 17-19] FLUID PRESSURE. 27 Since on each square centimetre there is a thrust p, on a square centimetres the thrust is pa, but the total thrust is P. Hence P = pa, P and p - . a. Thus The Pressure at each Point of a surface eocposed to a uniform thrust is found by dividing the thrust on the whole surface by the number of units of area it contains. Now, even when the thrust is variable over the surface, we may treat it as uniform over a small area a square centi- metres if that area be sufficiently small ; and in this case, if P be the total thrust on the area, the ratio P/a gives the pressure at each point of the area. DEFINITION. When the thrust over any surface is not uniformly distributed the Pressure at each Point of the surface is the ratio of the thrust on a small portion of the surface which includes the point to the area of that portion when that area is sufficiently small. Thus, if P be the thrust on a surface of area a under variable pressure, p the pressure at each point of that surface, then P when a is taken so small that the thrust over the portion of surface considered may be treated as uniform. 19. Average Pressure. The ratio of the total thrust on any plane surface to the area of that surface is known as the Average Pressure at each point of the surface : if the total thrust be P, and the area of the surface a, then the average pressure is P/a. If the thrust be uniformly distributed the average pressure and the pressure at each point are the same ; if the thrust be variable, the pressure at any point is the average pressure on a portion of the surface containing the point, and so small that the distribution of thrust over that portion may be treated as uniform. 28 HYDROSTATICS. [CH. II If a, the area of the surface, be unity, we see that the average pressure is equal to the thrust on the surface. Thus The Average Pressure is the thrust per unit of area. 2O. Examples of Uniform and Variable Thrust. If we consider a horizontal surface above which a mass of sand is piled, there will be a thrust on the surface, arising from the weight of the sand : if the sand be piled to a uniform depth all over, the thrust will be uniform, and the pressure at each point of the surface the same ; if, on the other hand, the upper surface of the sand be uneven, the thrust will usually be variable and the pressure will differ from point to point. Or again, consider a rectangular vessel filled with water having vertical sides and a horizontal bottom; the vertical forces acting are the weight of the water and the upward thrust of the bottom, these two are equal ; moreover the thrust is uniformly distributed and the pressure is the same at each point of the base. The same would be true if the vessel contained a solid which just filled it, but the solid would exert no force on the sides of the vessel ; when, however, it contains fluid each side is subject to a horizontal thrust the amount of which can be calculated. This thrust, it can be shewn, is not uniformly distributed, the force on any small portion of the surface near the top of the liquid is less than that on an equal portion near the bottom, the pressure is variable from point to point. Thus the pressure at the bottom of a dock of uniform depth is uniform, that on the dock gates is variable. * 21. Units of Pressure. The pressure at a point is found we have seen by dividing the thrust or normal force impressed on a definite surface by the area of that surface ; we therefore speak of a pressure of so many units of force per unit of area ; the numerical measure of a pressure depends on the unit of force and on the unit of area. We may express it in dynes or in grammes- weight per square centimetre, or in poundals per square foot. A common unit of pressure adopted in England is pounds-weight per square inch. 19-21] FLUID PRESSURE, 29 Suppose, for example, it is found that the force acting on a surface 100 square centimetres in area is 50 kilogrammes' weight, and that the pressure is uniform ; the pressure at each point of the surface is 50/100 or '5 kilogrammes' weight per square centimetre. Similarly, if the force on a square 1/100 of a square inch in area be 2 Ibs. weight, then the pressure is or 200 Ibs. weight per square inch. It must be clearly remembered that pressure is not force. In order to determine the thrust or force, impressed normally on a given plane surface by fluid pressure uniformly dis- tributed, we must multiply the pressure at each point by the area of the surface. If the pressure is not uniform, the problem of finding the total thrust is more complex ; if the average pressure at each point be known, the thrust is found by multiplying the average pressure by the area. Examples, (l) A surface is subject to a pressure of 15 Ibs. weight to the square inch; determine it in grammes 1 weight per square centimetre, and also in dynes per square centimetre. 1 square inch contains (2-54) 2 or 6'45 sq. cm., 1 Ib. contains 453*6 grammes. Thus the thrust on an area of 6'45 sq. cm. is 15x453*6 grammes' weight. Hence the pressure is 15 x 453 '6/6 '45 grammes' weight per square centimetre or 1055 grammes' weight per square centimetre. Now the weight of 1 gramme contains 981 dynes. Hence the required pressure is 1055 x 981 or 1 '033 x 10 6 dynes per sq. cm. Thus the pressure in question, which we shall see is about that exerted by the atmosphere, is equivalent approximately to a weight of 1 kilogramme per square centimetre; we might produce it by erecting a vertical tube 10 metres (1000 cm.) high and one square centimetre in area; if such a tube were filled with water the thrust on the base 1 sq. cm. in area would be the weight of 1000 c.cm. of water or 1 kilogramme. (2) The total thrust on a surface 5 square feet in area is found to be the weight of 1 ton, find the average pressure in Ibs. weight per square inch. The surface contains 5 x 144 or 720 square inches, one ton is 2240 Ibs. Thus the pressure is 2240/720 or 3 Ibs. weight per square inch. 30 , HYDROSTATICS. [CH. II (3) Taking the atmospheric pressure at 15 Ibs. weight per square inch, Jind the weight supported by a square mile of the earth's surface. One square mile is 4,014,489,600 square inches; the thrust in pounds' weight is found by multiplying this by 15. It is therefore 60,217,334,000 Ibs. weight. 22. Graphical Solutions. The following graphical method of representing the pres- sure at any point of a plane surface is sometimes convenient. We have seen that pressure is measured by the force impressed per unit of area : taking a square centimetre as the unit of area, the pressure may be given as so many grammes weight per square centimetre : now imagine the surface to be horizontal and erect on each square centimetre a vertical column of some homogeneous substance, of such a height that its weight may be equal to the force exerted by the fluid on the square centimetre which supports the column. Since the weight of a cubic centimetre of water is 1 gramme weight if the column be of water, h cm. in height, its weight will be h grammes; this is supported by the portion of the surface, 1 square centimetre, on which it rests, and if the weight of this column is to represent a pressure of p grammes weight per square centimetre, we must take h equal to p. Hence the pressure may be represented by the height of a column of water 1 sq. centimetre in area. The number of centimetres in the height of this column will be equal to the number of units of pressure in the pressure it represents. The height of this column of liquid is sometimes spoken of as the " head " of liquid which gives the pressure. Thus we might speak of the pressure of the atmosphere, which is about 1 kilogramme weight per sq. cm., as due to a " head " of water 10 metres in height. There is no need to select water as the substance by the aid of which the pressure is measured ; it is, however, in most cases convenient to do so. 23. Pressure within a Fluid. So far we have dealt with the thrusts which a fluid exerts on a surface with which it is in contact ; we may compare these 21-23] FLUID PRESSURE. 31 to the force between a solid block such as that shewn in Fig. 1 above, and the table on which it rests. But we have seen (Section 15) that we must, in the case of the block, suppose stresses to exist throughout its substance. For a horizontal plane such as CD, Fig. 1, we have a normal thrust exerted in opposite directions on the two sides of the plane ; if the plane be oblique as in Fig. 2, the normal thrust is accompanied by two tangential forces constituting a shearing stress. In the same way stresses exist throughout the substance of any fluid. Consider a mass of fluid in a vessel and suppose, for simplicity, that the sides of the vessel are vertical, imagine a horizontal plane CD, Fig. 5, drawn in the fluid. The forces acting on the fluid above this plane are its weight and the downward thrust of the atmosphere on its upper surface these act in a ver- tical direction together with the thrusts of the sides ; the directions of these last forces are horizontal and they are in equilibrium among themselves. In order then that equilibrium may be maintained Fig. 5. there must be an upward vertical thrust across the horizontal plane CD, equal to the weight of the fluid above the plane, together with the downward thrust due to the atmosphere. Or again, if the plane CD be inclined to the horizon, as in Fig. 6, there will, as in the solid, be a stress across it; this stress however will differ from that in the solid in that the forces which compose it are at right angles to CD. In the fluid there can be no stress along CD such as exists in the solid. In the fluid there is there- fore a thrust across CD, this thrust a vertical component which has balances the weight of the fluid above, together with the downward Fig. 6. 32 HYDROSTATICS. [CH. II vertical thrust of the atmosphere; it also has a horizontal component and this balances the resultant horizontal thrust on the vertical sides of the vessel. In the solid the resultant force across CD, Fig. 2, is vertical ; there is no force on the vertical sides and hence no horizontal component to the force across CD. There is, in consequence, a shearing stress in the solid across CD. The fluid cannot support such a stress, the wedge of fluid above CD, Fig. 6, would change in form and slide down were it not for the thrusts impressed on it by the sides of the vessel. 24. Pressure at a Point within a Fluid. Consider now a small plane surface of area a immersed in a fluid ; the fluid on either side of the surface exerts a thrust on the surface ; if the fluid on one side could be removed it would be necessary to exert a force on that side in order to balance the fluid thrust on the other. Let the magnitude of this force be P. Then P measures the thrust in the fluid across the surface of area a. The ratio Pja is defined as the Average Pressure at each point of the surface. If the thrust over the surface be uniformly distributed then the ratio P/a is the Pressure at each Point of the surface. If the thrust over the surface be not uniformly distributed then, in order to find the pressure at any point, it is necessary to reduce the area of the surface until it is so small that the distribution of thrust over it may be treated as uniform; when this is the case the ratio P/a measures the pressure at any point of the surface. DEFINITION. In order to find the pressure at a point of a fluid, imagine a small plane surface of area a. immersed in the fluid so as to contain the point. The Pressure at the Point is measured by the ratio of the thrust on one side of the surface to the area of the surface, when that area is so small that the distribution of thrust over it may be treated as uniform. 23-25] FLUID PRESSURE. 33 The meaning of the term pressure at a point may perhaps be made clearer from the following. Let A, Fig. 7, be a point in a fluid at which the pressure is required. Imagine a small plane surface placed at A and a tube inserted in the fluid in such a way that the surface may form a piston in the tube; suppose further that it is possible for the piston to move without friction in the tube. Now let all the fluid be removed from the tube on one side of the piston : the thrust on the other side will drive the piston down the tube unless force be applied to it ; suppose that a force P applied at right angles to the """".' _ piston holds it in its place, then P measures > '* the thrust on the piston and if a be the area of its surface P/a is the average pressure at each point of its surface. If the area of the piston be so small that the thrust may be taken as uniformly distributed P/a will be the pressure at each point. The arrangement described above does not constitute a practical means of measuring the pressure ; it is not an experiment to illustrate the measurement of pressure; it is merely an illustration, impossible to realize in practice, of what is meant by fluid pressure. Practical means of measuring the pressure at a point will be given in Sections 36 40. 25. Pressure in different directions. At any point in a solid imagine a small surface drawn, con- taining the point, and consider the thrust across this surface; it will of course depend on the forces which act on the solid ; it will also, in general, depend on the direction within the solid in which the surface is drawn, thus, if the weight of the substance be the only force, there will be a normal thrust on a horizontal surface but no normal thrust on a vertical surface; the thrust depends on the direction of the surface. This is not the case in a fluid ; the thrust on a surface of very small area placed at a given point is the same in whatever direction the surface be placed. If, when the surface be horizontal, there be a thrust of P grammes weight upon it, then it can be shewn that there is the same thrust upon it when it is vertical, or in any other position, so long as it is sufficiently small and passes through the given point. In fact it follows from the fundamental definitions that, The pressure at a point in a fluid is the same in all directions about the point. G. HYD. 3 34 HYDROSTATICS. [CH. II The direction therefore of the small surface placed in the fluid so as to form a piston in the illustration given in the last section is immaterial, the thrust upon it is not changed by turning it about any point in itself. This fundamental property can be deduced from the fact that there is no shearing stress in a fluid. The proof is given below (Proposition 6). A direct experimental proof would require somewhat complicated apparatus, the law however is involved in many of the experiments which will be described, and the student, who finds a difficulty in following the mathe- matical proof, may believe it because the results of experiment bear out theoretical deductions from the law. The proof that the pressure is the same in all directions about a point is based on the following considerations. Suppose that all the particles of a fluid are acted on by some force, such as their weight, and consider the matter which lies within some surface drawn in the fluid; the resultant impressed force acting on this matter will be proportional to the volume of fluid within the surface, and this force is balanced by the thrusts on the surface arising from the fluid pressure. Thus if, to make ideas definite, we consider a small cube in the fluid and suppose the weight of the fluid to be the only impressed force, the forces acting on the cube are the weight of the fluid which it contains and the six thrusts, one on each of the six faces of the cube; the resultant of these thrusts therefore must balance the weight. Now the thrust on each face is proportional to the area of the face, while the weight is proportional to the volume of the cube. Thus we have the resultant of six forces, which depend on the area of the faces, balancing a force which depends on the volume of the cube. Suppose now that the cube is reduced in size so that each edge becomes say x^-th of its previous length, the faces will then become yj^th of what they were while the volume of the tube will be y^^th of its previous value; the forces then which depend on the surface will be reduced 100-fold, those which are proportional to the volume will be reduced 1000- fold, or ten times as much ; if the edge be again reduced to a 25] FLUID PRESSURE. 35 tenth, the forces proportional to the volume will again be reduced ten times as much as those which depend on the area of the surface. Thus, proceeding in this manner, we see that we can make the forces which depend on the volume as small as we please when compared with those which depend on the surface ; in the end, then, when the cube has become very small its weight may be neglected in comparison with the forces which arise from fluid pressure, and these forces form a system in equilibrium among themselves. This statement is true whatever be the shape of the small portion of the fluid which we consider ; let us apply it to a small triangular prism in the fluid. ^PROPOSITION 6. To prove that the pressure at a point in a fluid is the same in all directions about the point. Let ABC, Fig. 8, be a small triangle in the fluid, draw lines A A', BB', CC' each I centimetres in length at right- angles to AEG; join A'B', B'C' and C'A' and consider the portion of fluid within the prism thus formed. Let a, 6, c be the lengths of the sides of the triangle BCA, p 1} p 2 and p s the average fluid pressures on the faces BCC'B', CAA'C', ABB' A' re- spectively. The areas of these faces are respectively la, Ib and Ic square centimetres, hence the normal thrusts are Iap 19 lbp%, and lcp s . Now we have seen that when the prism is made very small, these three forces form a system in equilibrium among themselves ; but, when three forces are in equilibrium, they can be represented by the sides of a triangle to which they are parallel. The three forces in question are at right angles to the sides of the triangle ABC, hence, if this triangle were turned through a right angle in its own plane, its sides would be parallel to the directions of the forces ; thus the three forces 32 36 HYDROSTATICS. [CH. II are respectively proportional to the sides a, b, c of the triangle BCA. The ratio then of each force to the corresponding side must be the same for all the forces. Now these ratios are Pilaja, p 2 lb/b and p s lc/c. Hence p^l = p^lp^ or Pl =p 2 =p s . Thus the average pressure on each face is, when the prism is made very small, the same; but, in this case, the average pressure on a face is the pressure at the point, to which the triangle is reduced, estimated in the direction of the normal to that face: thus the pressures at right angles to the faces of any very small prism enclosing the point are ultimately equal, or in other words, The pressure at a point in a fluid is the same in all directions about that point. It should be noticed that the proof depends on the fact that the force on each face is at right angles to that face ; if there were a shearing force parallel to the face as well as the simple thrust the proposition would not be true. We may put the proof in mathematical form thus. Let ABCC'B'A' be a small triangular prism in the fluid. Let the face AGO 'A' be horizontal. Let I be the length of the prism a, &, c, the sides of the triangle BCA. Let Pi,pz,p 3 be the average pressure on the faces, and w the weight of a unit of volume of the fluid. Let d be the perpendicular distance of the vertex B from the base GA. The volume of the prism is %bdl and its weight is %wbdl this force acts vertically and is therefore at right angles to the face ACG'A'. The other forces are^aZ, p 2 bl and p 3 cl at right angles to the faces. Kesolve these vertically p 2 bl = \wbdl +p^l cos C +p 3 cl cos A. Eesolve the forces horizontally p^al sin G=p 3 cl sin A. But we know that a sin C= c sin A. 25-26] FLUID PRESSURE. 37 Thus Pi=P 3 , also b = CA=acos C + ccosA. Hence substituting in the first equation p 2 b = ^wbd +p l (c cos A + a cos C) Thus p 2 -p l = ^o)d. Now when the prism is made very small, so that p l and p 2 become the pressures in two different directions at a point, then d is indefinitely small. The difference therefore between p l and p. 2 can be made as small as we please, or p l is equal ultimately to p 2 . Hence ultimately Pi=Pz=Pa- Now p% is the pressure in a vertical direction, p l , p s pressures in any other two directions. Hence the pressure in a vertical direction is equal to that in any other, thus the pressure is the same in all directions about a point. 26. Transmissibility of Fluid Pressure. If the pressure at any point of a fluid is changed, that at all other points is changed also ; it follows, from the fundamental property of a fluid, that, for a liquid, the change of pressure at all points is the same. A fluid in this respect differs from a solid. Imagine a cylinder fitted with a piston and place in it a portion of a solid which just fits the cylinder loosely. Put weights on the piston, the force thus applied to the top of the solid is transmitted to the base; unless the solid expands laterally under the force, so as to fit the cylinder more tightly, there will be no pressure on the sides; if, however, the substance in the cylinder is a fluid this is no longer the case; the addition of the weights increases the pressure or force per unit of area on the top of the fluid, this increase is transmitted by the fluid in all directions; the pressure at each point is increased and, as we shall shew, for a liquid, the increase of pressure is the same at all points. This may be illustrated by the following arrangement. 38 HYDROSTATICS. [CH. II Fig. 9. Imagine a vessel fitted with a number of openings, each closed by a piston, as shewn in Fig. 9; suppose the whole to be filled with liquid. In order to keep the pis- tons in their place a force must be applied to each. Suppose now that the force on any one piston is in- creased, the other pistons will be driven out, and in order that equilibrium may be maintained it is necessary that additional force should be applied to each of them. If we could secure friction- less pistons, all of the same area, it would be found that the force applied to each piston would be the same ; if, however, the pistons differ in area, the force necessary to maintain any piston in position would be found to be proportional to the area of the piston ; the ratio of the force to the area over which it is applied is the same for all. An increase of fluid pressure applied at one point is trans- mitted equally to all other points. The experiment in this form is impossible; we cannot obtain frictionless pistons. The principle, however, is illus- trated by the action of various pieces of apparatus which will be described shortly, see Sections 27, 28, and by some experiments which will be better understood when we have considered some of the methods for measuring fluid pressure. We proceed now to give a formal proof of the principle. PROPOSITION 7. An increase of pressure, at any point of a liquid at rest, is transmitted without change to every other point. For let A, B, Fig. 10, be two points within the liquid. (i) Suppose first that the line AB lies entirely within the liquid. 26] FLUID PRESSURE. 39 Construct a small cylinder, having the line AB for its axis, and consider the forces acting on the fluid within this cylin- der. They are (1) the thrusts on the ends A and B, parallel to the ^pj^gpW (2) the thrusts on the curved surface at right angles to the axis, Fig. 10. (3) the resultant of the external impressed force. Now the liquid is in equilibrium and the thrusts on the curved surface have no component parallel to the axis. Thus the difference between the thrusts on the two ends must balance the component of the impressed force in the direction of the axis. But this component remains the same even though the pressure be changed. Hence the difference between the thrusts on the ends A and B is a constant ; but the areas of these ends are equal. Thus the difference between the pressures at the two ends is always the same. Hence, if by any means the pressure at the point A is increased, that at B is increased by the same amount, other- wise the difference between the two would change and it has just been proved that this difference is unchanged. (ii) Suppose that the line AB does not lie entirely within the liquid. Join the points A and B by a series of straight lines AP, PQ, QB, Fig. 11, etc. each of which does lie entirely in the liquid. Then the proposition just proved holds for each of the pairs of points A, P . P, Q etc. Hence if the pressure at A be in- creased, that at P is increased equally ; but if the pressure at P is increased that at Q is increased equally, and so for all the points. Hence the increase of pres- sure at B is equal to that at A. Fig. 11. 40 HYDROSTATICS. [CH. II Thus any increase of pressure produced at any point of a liquid in equilibrium is transmitted without change to every other point. The proposition is not true, for a gas or compressible fluid, in any case in which it is necessary to take into account the weight or other force impressed on the gas; for if a gas be subject to increased pressure its volume is diminished and its density is increased. The weight therefore of the gas within the cylinder AB is changed by the change of pressure, and in consequence the difference of pressures between the two ends is changed also. The density of a gas is however usually very small, the weight therefore of a limited portion is generally small compared with the force to which each unit of area of it$ surface is subject; for many purposes we may omit the consideration of the weight of the gas entirely and may suppose that in the case of a gas we are dealing with a fluid acted on throughout its mass by no impressed forces. We may shew that in this case the pressure is the same at every point. PROPOSITION 8. To prove that if a fluid be acted on by no impressed force the pressure is the same at every point. Let A, J3, Fig. 12, be two points in such a fluid. Join AB and suppose the line AB to be entirely within the fluid. Con- struct a small cylinder about AB as axis. The cylinder is in equilibrium under (1) The thrusts on the two ends acting parallel to the axis. (2) The thrusts on the curved surface acting at right angles to the axis. Hence the thrusts on the two ends are equal and opposite; but the areas of the ends are the same. Thus the pressures at the two ends are equal. But A and B are any two points in the fluid. Hence the pressure is the same at any point in the fluid. If the line AB does not lie entirely in the fluid the proof can be extended as in Proposition 7 (ii). 26-27] FLUID PRESSURE. 41 Thus, provided the volume of gas considered is so small that its weight may be neglected, we may take the pressure in a gas to be the same at all points. The above statement would not of course apply to the pressure throughout any large volume of a gas such as the atmosphere. The pressures at the top and bottom of a mountain are very different. Delicate pressure gauges will enable us to detect the difference in pressure between the attics and the basement of a house. 27. Hydrostatic Bellows. This apparatus, designed by Pascal, illustrates the principle of the transmissibility of pressure in a fluid. A stout bladder, such as is used for a football, or a leather bellows, is attached to a piece of tube. The tube is fixed in a vertical position and the bladder rests on the table. A piece of light board is placed on the bladder and a weight rests on the board. Water is then poured down the tube; the water flows into the bladder, causing it to expand and raise the weight; the level of the water in the tube stands, as at (7, Fig. 13, some distance above the level of the board. To explain the action, we notice that the weight is sup- ported by the upward thrust of the water on the underside of the board; this upward thrust depends, partly on the pressure of the water and partly on the area of the surface of the board which is in contact with the bladder, and its value is obtained by finding the product of the two; if, therefore, the area in contact with the board be large, the upward thrust may be considerable, even though the pressure is not large. Suppose now, when the whole is in equilibrium, the level of the water in the tube is at B. Let more water be poured into the tube, and suppose that the weight does not rise. Fig. 13. 42 HYDROSTATICS. [CH. II The downward thrust over the surface of the water at B is increased by the weight of the water poured in, the pressure therefore at B is increased. Hence, if w be the weight of water poured in and a the area of the section of the tube, the increase in the thrust on the area a at B is w ; thus the increase in pressure is w/a. This increase of pressure is transmitted equally to all points, hence the upward pressure at all points of the under surface of the board is increased by w/a. Let the area of this surface be A, then the total upward thrust is increased by Aw /a. In order that the board may not rise the weight upon it must be increased by an amount W equal to A w/a. If the weight be not increased the board will rise and the level of the water in the tube will sink, thus reducing the pressure until equilibrium is again established. In this arrangement we see that a weight w of water can support a weight W placed on the board and that Hence, if A is large compared with a, W will be large compared with w. By making the area of the board con- siderable and that of the tube small, a large weight W can be supported by a small weight w of water. This fact is sometimes described as the hydrostatic paradox: the principle involved in the above experiment is made use of in Bramah's Press (see Section 101). Example. The area of the tube used in an experiment like that described in Section 28 was 10 square millimetres, the area of the board 100 square centimetres. If 10 grammes of water are poured into the tube, find the additional weight which the board can support. We have 10 sq. mm. = *1 sq. cm. Thus the increase of pressure is 10/-1 or 100 grammes weight per square centimetre. The increased upward thrust on the board 100 sq. cm. in area is 100 x 100 or 10000 grammes weight. Thus if the board is not to rise an additional downward force of 10 kilos weight must be applied to it. 27-28] FLUID PRESSURE. 43 28. Illustrations of Fluid Pressure. The apparatus shewn in Fig. 14 again illustrates the transmissibility of fluid pres- sure. In it M and N are two cylinders of different diameters which are filled with water and communicate through a tube at the bottom. They are fitted with pistons; the pressure at any point of the two pistons is the same; the upward thrusts on the pistons are proportional to their areas ; thus, if a down- ward force w be applied to the Fig. 14. smaller piston, a larger force W must act on the larger piston in order to maintain equilibrium. If the pistons be assumed to be frictionless, the relation of W to w is found thus : Let A, a be the areas of the two pistons respectively. Then on the smaller piston there is a downward thrust w, the thrust per unit area of the piston is therefore wja; this then is the pressure in the fluid and it is transmitted to each unit of area of the piston A. The total upward thrust then on this piston is Aw/a and this upward thrust must balance W. Hence w= Aw , a or W_w A ~ a In consequence of the friction, however, the relation of W to w given by experiment would differ from this. Numbers of other illustrations of the effects of fluid pressure can be given. Thus (i) Fill with water, a glass tube 30 or 40 cm. in length, 44 HYDROSTATICS. [CH. II Fig. 15. closed at one end, and place it with its open end downwards, in a vessel of water ; the water remains in the tube, as shewn in Fig. 1 5 ; the pressure of the air on the free surface of the water in the vessel is transmitted to the water in the tube. The upward thrust over the open end of the tube is sufficient to support the contained water. (ii) Repeat the experiment with a tube, one end of which is closed with a piece of thin india-rubber, the india- rubber is stretched and takes the form shewn in Fig. 15, the pressure on its upper surface is greater than that exerted by the water on the lower surface. (iii) Again, take a hollow cylindrical vessel, such as a tin can or bucket, and place it bottom downwards in a vessel of water, the can floats ; to sink it under water force is necessary, the upward thrust arising from the fluid pressure is greater than the downward force, the weight of the vessel, hence it floats; if a small hole be bored in the bottom the water spouts up in a jet. (iv) Place a small beaker or tumbler mouth downwards in water, as in Fig. 15 a, and depress it below the surface; the water rises in the beaker, compressing the air it contains; the greater the depth to which the beaker is lowered, the greater will be the compression ; the pressure in the water increases with the depth. (v) The water supply of a town usually comes from a reservoir at some height above the town, in consequence the water in the pipes is under pressure ; a jet of water allowed to flow from a hose-pipe will rise to a height which depends partly upon this pressure. Fig. 15 a. 28] FLUID PRESSURE. (vi) If a fluid be allowed to escape from a tall vessel through a hole in the side, the velocity with which it flows out depends on the pressure; if holes be made at various depths, as in Fig. 16, in the side of the vessel, the water flows more rapidly from the lower holes than from those above ; the pressure is greater at the greater depth. Fig. 16. (vii) Grind the end of a glass tube flat, cover it with a plate of flat glass and hold the glass against the bottom of the tube with a string, as shewn in Fig. 17. Lower the whole some depth into a vessel of water and release the string, the glass remains in contact with the end of the tube and does not fall off. The upward thrust, due to the pressure of the water on the bottom of the glass, is more than sufficient to counterbalance its weight. In the next chapter we give some fundamental propositions on fluid pressure 'when the only force acting on the fluid is its weight. We then describe some ex- periments on the measurement of fluid pressure and the numerical verification of some of the laws. Fig. 17. CHAPTER III. PROPOSITIONS ON FLUID PRESSURE. 29. Fluid Pressure. We assume in the following propositions 1 that the only forces, impressed on the portion of fluid which we consider, are the thrusts due, either to the action of surrounding fluid, or to solids with which the fluid is in contact, together with the weight of the portion of fluid considered. If then we take any portion of the fluid, say that within some small sphere or cylinder, described in the fluid, the forces on this portion of fluid are the thrusts on its surface and its weight ; these forces must form a system in equilibrium : we can determine from this the relation between the fluid pressure and the weight. Proposition 9 deals with the pressure at points at the same level in a fluid. Two cases of this proposition arise. (i) It may be possible to join the two points by a straight horizontal line or a series of straight horizontal lines which lie entirely in the fluid ; thus any two points in an ordinary bath of water could be joined by a straight line. Suppose, however, that there is a sponge or a piece of soap in the bath, 1 Similar propositions may be proved in a very similar way for a fluid at rest under other forces than its weight. For these the reader is referred to Greaves' Elementary Hydrostatics (Cambridge University Press). 29] PROPOSITIONS ON FLUID PRESSURE. 47 then a point on one side of the soap cannot be joined by one straight horizontal line to a point on the other side without cutting the soap; we can however join the two points by means of two or more such lines. (ii) It may be impossible to join the two points by horizontal lines lying entirely in' the fluid ; thus, if there is a vertical partition stretching completely across the bath and reaching part way down, a point on one side of this cannot be connected with a point on the other side by horizontal lines lying entirely in the fluid; the same is true of two points, one in each leg respectively, in a fluid filling a U tube. Proposition 9 applies to the first case ; the second is dealt with in Section 30. PROPOSITION 9. If a fluid be at rest under the action of gravity, the pressures are equal, at any two points which can be joined by a single straight horizontal line lying wholly within the fluid, or by a series of such lines. In Fig. 18, let A, B be the two points in the same horizontal plane. Fig. 18. (i) Suppose that the straight line AB lies wholly within the fluid. About AB construct a cylinder of very small section with its ends at right angles to AB and consider the forces impressed on the cylinder. 48 HYDROSTATICS. [CH. Ill They are (i) The thrusts on the ends A, B which act in opposite directions along AB. (ii) The thrusts on the curved surface of the cylinder which are everywhere at right angles to AB. (iii) The weight of the fluid within the cylinder, the line of action of which is vertical and therefore at right angles to A B. Thus the only impressed forces in the direction of AB are the thrusts at A and ; these forces then must be equal and opposite; but the area of the end at A is equal to that at B. Hence the fluid pressure at A is equal to that at B. (ii) If AB cannot be joined by a single horizontal straight line lying wholly within the fluid, but by a series of such lines AP, PQ, ... etc.. each of which does lie in the fluid, then the proposition is true for each of these lines. Hence the pressure at A is equal to that at P, the pressure at P is equal to that at Q, and so on; thus the pressures at A and B are equal. The next proposition deals with the pressures at two points in a fluid, one of which is vertically below the other. PROPOSITION 10. To find the difference of pressure between two points in a fluid, one of which is vertically below the other. Let A, B, Fig. 19, be the two points and suppose that the line AB lies wholly in the fluid. Fig. 19. 29] PROPOSITIONS ON FLUID PRESSURE. 49 Consider a small vertical cylinder on AB as axis with its ends perpendicular to AB, let a be the area of either end and let p, p be the pressures at A and B respectively. - The forces on the fluid composing this cylinder are (i) The thrusts on the curved surface; these are hori- zontal, at right angles therefore to AB. (ii) The thrusts on the ends; the values of these are pa and pa. respectively ; they act vertically in opposite directions parallel to AB. (iii) The weight of the fluid contained in the cylinder; the direction of this force is also vertical, and parallel there- fore to AB. Thus the difference between the thrusts on the ends must balance the weight of the fluid in the cylinder. Hence p'apa. = weight of fluid in the cylinder. Suppose now that the fluid is homogeneous, so that its density is the same throughout; let o> be the weight of a unit of volume; let h, li be the depths of the points A and B below some fixed horizontal surface. Then AB = h'-h. Now the volume of the cylinder is AB x a and its weight is o> . AB . a or w (h r h) a. Hence p'a pa = o> (ti h) a. Therefore p p = CD (h' h}. Now w (h' h) is the weight of a column of fluid of unit cross section and of height equal to the vertical distance between the two points. Hence The difference of pressure, between two points in the same vertical line, is equal to the weight of a column of fluid of unit cross section, and of height equal to the distance between the points. Thus in a homogeneous fluid the difference of pressure between two points in the same vertical line is proportional to the distance between the two points. G. HYD. 4 50 HYDROSTATICS. [CH. Ill Corollary. If the fluid be not homogeneous it is still true that the difference of pressure is the weight of the column of fluid of unit cross section which extends from one point to the other. 3O. Pressure at various points in a heavy fluid. By combining the two propositions just proved we can shew (i) that In any homogeneous fluid, the pressures at any two points in the same horizontal plane are equal, and (ii) that The difference of pressure between any two points is the weight of a column of fluid, of unit cross section, whose height is the vertical distance between the points. For suppose that A, B, Fig. 20, be two points in a fluid in the same horizontal plane which, however, because of some barrier cannot be joined by a horizontal line entirely within the fluid. Fig. 20. Draw AC and ED vertical and let and D be two points in the same horizontal plane below the barrier which can be so joined. Join CD. Then, since AB and CD are both horizontal, the distance AC is equal to ED. The pressure at A is less than that at C by the weight of a column of unit cross section and of height AC. The pressure at B is less than that at D by the weight of a column of unit cross section and of height ED. The weights of these two columns are equal, and the 29-30] PROPOSITIONS ON FLUID PRESSURE. 51 pressure at C is by Proposition 9 equal to that at D, for CD is horizontal. Hence the pressure at A is equal to that at B. If it be not possible to pass from A to B, by a single step of the nature indicated, it can always be done by a series of such steps. To prove (ii) let A, B, Fig. 21, be any two points. Draw BC vertical and from A draw AC horizontal to meet BC in C. Fig. 21. Then, by Proposition 10, the pressure at C exceeds that at B by the weight of a column of fluid of unit cross section and of height BC, while by Proposition 9, the pressure at A is equal to that at C. Hence the pressure at A exceeds that at B by the height of a column of fluid of unit cross section and of height equal to the vertical distance between the points. If it be not possible to pass from A to B by a single step of the nature just described, it can always be done by a series of such steps. PROPOSITION 11. The surface of a liquid, subject to con- stant pressure and at rest under gravity, is horizontal. Let A, B, Fig. 22, be two points in the same horizontal plane in a liquid at rest under gravity. Let 7T be the constant pressure to which the surface is subject and the weight of a unit of volume of water, then The pressure at a depth h below the surface of water then is p 7T + ah = U>HQ + o)h to (HQ + A). Suppose now that we consider a horizontal surface at a height H above the surface of the water, then h + H is the depth of the point below this surface ; this imaginary surface is sometimes spoken of as the Effective Surface, and we see that the pressure at a point is proportional to the depth of the point below the effective surface. In this case the height H Q is called the height of the water barometer (see Section 76). We may, it is clear, without affecting the circumstances within the water, suppose that its surface is covered with a. layer of water of sufficient depth to produce over that surface the pressure which actually exists there ; the upper boundary of this layer being free from pressure, and imagine that the thrust over the actual surface is due to the weight of this superposed water and not to the atmosphere. If we are con- sidering the pressure in some other liquid, not water, it will be most convenient to suppose the superincumbent layer to consist of this same liquid. 54 HYDROSTATICS. [CH. Ill PROPOSITION 12. Two liquids which do not mix are placed in a vessel, to find the pressure at a point in the lower liquid. Let A t Fig. 23, be a point in the lower liquid. Let c Fig. 23. ABC drawn vertically from A meet the common surface of the two liquids in B and the upper surface of the upper liquid in C. Let AB h, EC = h'. Let o> and o/ be the weights of unit volume of the lower and upper liquids respec- tively, p } p' the pressures at A and , 7T the pressure at C. Then from the upper liquid we have p = 7T + '. EF= '. EC, or <&'(DE-AB) = vt(BC-EF) (1). Again, the pressure at D is equal to that at A. Hence TT + to' . EF + co . DE = TT + o>' . BC + w . AB. Thus .AB + , E be the levels of the mercury columns in the two limbs. Draw EF horizontal to meet the column DB in F. Read the positions of E and D on the scale and thus find the height DE or DF; let it be h centimetres, let p be the pressure on the surface at JS, 7T the atmospheric pressure, and o) the weight of a unit volume of the liquid in the manometer. Now, since the pressures at two points in a given fluid in the same horizontal plane are equal, the pressures at E and F are equal; but the pressure at E is p, hence that at F is also p. The pressure at F is the atmospheric pressure together with the weight of a column of fluid of height DF and unit cross section. Hence p 7T + (ah. Thus the excess of the pressure at E over the atmospheric pressure is wh. If h be in centimetres, co in grammes weight per cubic centimetre, this difference of pressure will be in grammes weight per square centimetre. If mercury be the liquid used, the height h will measure the " head " in centi- metres of mercury. The choice of a liquid to be used will depend to some extent on the pressure to be measured ; with a dense liquid like mercury a comparatively small head corresponds to a considerable pressure, hence, to measure pressures only slightly in excess of the atmospheric pressure, the head of mercury necessary would be small, and a small error in measuring it would mean a considerable error in the value of the pressure. If a liquid of smaller specific gravity be used, the " head " necessary to measure a given pressure will be increased in the inverse ratio of the specific gravities, a given error in measure- ment will produce a proportionately less error in the result. Sulphuric acid, the specific gravity of which is about 1'842, is often used ; water may be employed in some cases ; it has however the disadvantage that it evaporates rapidly above the column EB, and the pressure due to the water vapour may cause error ; sulphuric acid, on the other hand, absorbs water 64 HYDROSTATICS. [CH. Ill quickly, its density therefore changes and this is a source of error. This form of manometer is most useful to measure the pressure of a gas in a confined space ; this pressure we have seen is the same throughout the mass. If it be used to measure that of a liquid we must remember that the pressure measured by the height of the manometer column is that at the surface E ; if we wish to use it to measure the pressure of a liquid in a vessel connected to the manometer at C we must allow for the weight of the column of liquid between G and E. EXPERIMENT 7. To measure the pressure of the gas in the gas-pipes of the Laboratory. For this purpose connect the end C, Fig. 29, by means of a piece of india-rubber tubing with the gas-pipe ; turn on the gas and read the difference in height between the two columns of liquid ; the result gives the excess of pressure in the gas-pipes, over the atmospheric pressure, measured as a "head" of the liquid in the manometer. If the specific gravity of the liquid be known, the pressure can be reduced to any other units. For this experiment water is a convenient liquid to use. Example. The pressure in a gasometer exceeds the atmospheric pressure by 10 inches of mercury and the barometer 1 stands at 30 inches. If the specific gravity of mercury be 13-6 and that of sulphuric acid 1'84, determine the difference of level in a sulphuric acid gauge attached to the same gasometer and the pressure on the ivalls of the gasometer in Ibs. weight to the square inch. The equivalent head of water is 10 x 13*6 inches and of sulphuric acid it is 10 x 13-6/1-84 or 73-9 inches. A cubic inch of water weighs 62-5/1728 or '03617 Ibs. weight. Thus the pressure in Ibs. weight per square inch is 136 x -03617 or 4-92 Ibs. weight. We see from these results that whereas with a mercury gauge an error of *1 inch in the height would mean an error of 1 per cent, in the pressure; with a water gauge it would mean an error of about 1 in 1300, and with acid gauge of about 1 in 740. (ii) Other forms of Siphon Gauge. In some cases, for measuring high pressures, the end A of 1 The height of the barometer measures the atmospheric pressure. See Section 68. 36] PROPOSITIONS ON FLUID PRESSURE. 65 the gauge is closed as in Fig. 30. When the pressure on the end E of the mercury column increases this end is driven down, and the air in AD is compressed. By measuring the extent of this compression the pressure of the air in AD can be found by Boyle's law (see Section 79), and hence the pressure at E can be obtained ; this pressure is chiefly due to the compressed air ; in most cases the difference in pressure due to the column of liquid DF will be small compared with that due to the compressed air and may be neglected j the pressures at E and F may be treated as the same. B Fig. 30. Fig. 31. For measuring low pressures the tube AB is completely filled with mercury and its end is sealed up as in Fig. 31 ; then when the end C is open the atmospheric pressure forces the mercury up AB to the top of the tube, it stands at a lower level in the open tube ; when the pressure in this second tube is sufficiently reduced the mercury in AB falls. Suppose that, as in Fig. 31, the surface of the mercury is at D, and in the other tube at E ; draw EF horizontal, the pressure above the mercury at D is zero, and the pressure at E is equal to that at F. The pressure at F is measured by the height of the column DF, which thus gives the pressure in the reservoir attached G. HYD. 5 66 HYDROSTATICS. [CH. Ill to C. This form of gauge is commonly used with an air-pump. (See Section 98.) 37. Barometer Tube Gauge. Another gauge for low pressures is shewn in Fig. 32. It consists of a vertical tube AB which dips at A into a vessel of mercury and communicates at the top with the vessel in which the pres- sure is to be measured. As the pressure in this vessel is reduced below that of the atmo- sphere the mercury rises in the tube l . Let its top surface be at D, a height h above the surface of the mercury at A in the reservoir, let E be a point 2 within the tube at the same level as A, 7T the atmospheric pressure, p the pressure at D, and o> the weight of a cubic centimetre of mercury. The points A and E in the mercury are at the same level, hence the pressures at these points are the same. But the pressure at A is the atmospheric pressure 7T. Hence the pressure at E is also 7T; now consider the column of mercury above E in the tube BA, the pressure at its top is p, and its height is h, hence the pressure at E is p + n>h. Thus 7T= + tah. Hence p = 7T It should be noticed in all these cases that the size of the tube need not be taken into account. 38. The Safety-valve. Another form of pressure gauge for high pressures is the safety-valve of a steam-boiler. A spherical or conical plug A, Fig. 33, fits accurately into a circular opening connected with the boiler, the pressure in the boiler tends to raise this plug, it is kept in position by a downward force applied from above. 1 See Section 70. 2 E is not shewn in the figure. 36-39] PROPOSITIONS ON FLUID PRESSURE. 67 This downward force is usually exerted by a lever, the fulcrum of the lever is fixed, as shewn at C ; the arm of the lever Fig. 33. presses on the plug at A, the extremity of the arm either carries a weight TF, or, if the boiler be not steady, is attached to a spiral spring. Let W be the weight at B, or the downward pull registered by the spring. Let P be the upward thrust at A, p the pressure in the boiler, and a the radius of the orifice closed by the plug. The effective area exposed to vertical fluid pressure is ira 2 , hence the upward thrust is p . Tra?. Thus P=p.ira z . But P x horizontal distance between C and A = Wx horizontal distance between C and B. And hence, arm of lever W '__ Tra 2 distance of fulcrum from valve ' If p exceeds this value, the steam just begins to escape; the pressure therefore can be measured by adjusting either the weight W or the length of the arm CB until the steam just begins to blow off. 39. The Bourdon Gauge. This consists of a tube AJ3, Fig. 34, of thin metal whose axis is bent into the form of the arc of a circle. One end of the tube A is closed, the other, , communicates with the vessel in which the pressure is to be measured, the- section of the tube is elliptical. Now when the pressure in the tube increases, this elliptical section tends to become circular. This causes the axis of the tube to uncurl slightly so that 52 68 HYDROSTATICS. [CH. Ill the end A moves upwards ; this slight motion of the tube is communicated to a pointer which moves over a circular scale Fig. 34. by means of a lever working a rack and pinion. The scale is graduated by the application of known pressures and then any pressure applied to the inside of the tube can be measured. 40. Experiments on Fluid Pressure. We have already described in Section 29 observations on some effects of fluid pressure. Thus the column of water remains in the glass tube, Fig. 15, because the atmospheric pressure at each point of its base transmitted through the fluid is greater than the pressure due to the column of water above. The pressure, however, at the top of the column is less than the atmospheric pressure ; hence the stretched india- rubber which covers the top of the tube in the same figure is bulged in. We may use the apparatus shewn in Fig. 17 to verify some of the laws of fluid pressure. For let us suppose that the radius of the glass tube is a centimetres, the area of its cross 39-40] PROPOSITIONS ON FLUID PRESSURE. 69 section is TTO? square centimetres; hence if p be the fluid pressure on the glass covering its base, the upward thrust is PTTO? and if h be the depth to which it is sunk, the value of p is 7T + ah where TT is the atmospheric pressure, h) grammes weight per square centimetre. We have thus obtained from the experiments an expres- sion for the pressure at a point in a fluid under gravity. 1 The interior area at the bottom of the tube is of course less than the exterior by an amount depending on the thickness of the walls, but the atmospheric pressure acts vertically downwards at the top of the tube on an area exactly equal to this difference. 41-42] PROPOSITIONS ON FLUID PRESSURE. 73 *42. Surfaces of Equal Density. There are two theoretical propositions of importance with which we may conclude the chapter. * PROPOSITION 14. To shew that the densities at two points in a fluid at rest under gravity at the same depth are the same. Let A, , Fig. 36, be two points at the same depth in a fluid at rest under gravity. Fig. 36. The pressures at A and B are equal, let each be p. Take a point C a very short distance below A and a point D at the same short distance below J5, so that AC -ED. Then C and D are at the same depth and the pressure at C is equal to that at D ; let the value of this pressure be p. If C is very near to A we may treat the fluid as though its density between A and C were constant, and equal to its mean value between these points, let this mean value be pjj similarly let p 2 be the mean value of the density between B and D. Then we shall shew that p l = p 2 > For, by considering a small cylinder of fluid between A and C, we find p' While, by considering a cylinder of equal height between B and D, we have p p + gp 2 BD. Hence Pl AC = p 2 D. But AC = D. HYDROSTATICS. [CH. Ill Therefore p L = p 2) or the density is the same at two points at the same depth. Corollary. It follows from this that the common surface of two different liquids which do not mix is a horizontal plane ; for, if not, let the surface lie as EF, Fig. 37. Then draw AB horizontal, so that A may be in one liquid, B in the other. The density at A differs from that at B, which is contrary to the proposition just proved; thus EF cannot be as drawn, it must be horizontal. Thus, in a fluid under gravity, surfaces of equal pressure are also surfaces of equal density. * PROPOSITION 15. When two fluids which do not mix are in stable equilibrium, the upper fluid 'must be lighter than the lower. To prove this, imagine a closed tube filled with the two fluids, and having a stop-cock at the bottom by which it can be divided into two parts. If it be possible, let the heavier fluid fill the upper part of the tube. Then there will be equilibrium so long as the surfaces of sepa- ration A, B, Fig. 38, in the two branches of the tube are at the same level. Let the fluids now be displaced so that the surfaces of separation take the positions (7, A and suppose the stop-cock Fig. 38. 42-43] PROPOSITIONS ON FLUID PRESSURE. 75 closed. Draw CE, DF horizontal to meet the fluid again in E and F. Let o^, w 2 be the weights of unit volume of the upper and lower fluids respectively, and let h be the vertical distance between DF and EC. Then, since C and E are points in the same fluid at the same level, the fluid pressures at these two points are equal. Let each be p. Let p l be the pressure at />, p z that at F. Then, since the vertical distances CF and DE are equal, while the density of the fluid in DE is greater than that in FC, and the pressures at C and E are equal, the pressure at D is greater than that at F. For we have p l p + o^/t, j). 2 =p + o> 2 A, and w 1 is greater than w 2 ; hence p 1 is greater than p v Now let the stop-cock be opened, then F and D are points at the same level in the same fluid; hence, for equilibrium, the pressure at these two points must be the same; but we have shewn that the pressure at D is greater than that at F, hence the equilibrium cannot be maintained and the column CE of the fluid will move. Moreover the surface C will ascend and E will descend ; thus, if the two fluids be slightly disturbed from their original position, motion will be set up in such a way that the disturbance goes on increasing ; the heavier fluid comes to the bottom, the original position was unstable. If the lighter fluid had been at the top originally, then the same method of proof would have shewn that when the disturbance took place the pressure at D would be less than that at F, thus the column FD would, when the stop-cock is opened, move back to its original position; the equilibrium would be stable. The proof will apply to a more general case when the two fluids instead of being contained in a tube are placed one above the other in an open vessel. Equilibrium of Two or more Fluids. The proposition may be illustrated in various ways ; if oil arid water, which do not mix, be placed together in a beaker, 76 HYDROSTATICS. [CH. Ill the oil rises to the top and floats on the water; it is lighter than the water, and the stable position of equilibrium is one with the oil on the top. It is possible, however, to arrange that the water should be above the oil. Thus, take two small tumblers or wine-glasses of the same diameter, fill the one with water, the other with oil. Cover the top of the former with a thin card and invert it, holding the card so that no liquid escapes. Place it mouth downwards above the vessel containing the oil, the card separating the two liquids ; on shifting or removing the card, so as to open a communication between the two, the oil will gradually rise into the upper vessel and the water sink into the lower one ; the initial position with the water uppermost is unstable, and hence the transference occurs. EXAMPLES. 1. Assuming the atmospheric pressure to be 1 kilo. \vt. per square centimetre, find the pressure in water at the following depths : 25 cm., 1 metre, 1 mile, 5 kilometres ; and in mercury at the following depths : 1 cm., 1 metre, 25 metres, 1 kilometre. 2. The pressure at a certain point in a vessel of salt water is 35 Ibs. wt. per square inch. Find the depth of the point, assuming the atmospheric pressure to be 15 Ibs. wt. per square inch. 3. Determine the height of the mercury column which would produce the pressure given in Question 2. 4. Compare the pressures at equal depths in alcohol, carbon disulphide and water, neglecting the atmospheric pressure. 5. What head (1) of water, (2) of mercury is equivalent to a pressure of 14 '5 Ib. per square inch ? 6. If the head of water above a point be 100 yards, what is the pressure at the point? 7. Find the heads (1) of water, (2) of mercury corresponding to pressures of 1 kilo wt. per square centimetre ; 30 Ibs. wt. per square inch ; one million dynes per square centimetre. 8. Find the pressure in poundals per square foot due to a head of 30 inches of mercury. CH. Ill] PROPOSITIONS ON FLUID PRESSURE. 77 9. What is the pressure at a depth of 60 fathoms below the surface in sea water ? 10. A cylinder 3 feet in diameter is fitted with a piston and filled with water. A weight of 5 tons weight is placed on the piston, find the pressure in the water. 11. The head of water in a pipe communicating with a cylinder having a piston 2 feet in diameter is 200 feet. Find the force the piston can exert. 12. The pistons of a press are 2 inches and 10 inches in diameter ; what is the pressure in the liquid when the small piston carries a load of 5 cwt. and what force can the large piston exert ? 13. The pressure in a liquid at a depth of 60 inches is 30 Ibs. per square inch ; what is the thrust at a depth of 30 feet (1) on a square foot, (2) on a square yard ? 14. The pressure in a well at a depth of 95 feet is four times that at the surface ; find the pressure of the atmosphere per square inch of the surface. 15. Assuming the atmospheric pressure to be 1 kilo wt. per square centimetre, find the pressures at depths (1) of 76cm., (2) of 380cm. below the surface of mercury. 16. The pressure in a water-pipe at the base of a building is 40 Ibs. wt. per square inch, on the roof it is 20 Ibs. wt. per square inch ; find the height of the roof. 17. A vessel in the form of a cube 1 metre in edge is filled with water ; find the resultant thrust on its base. 18. What volume of mercury must be placed in the vessel to produce the same resultant thrust? 19. Express the pressure of the atmosphere in pounds weight per square foot when the height of the water barometer is 32 feet. 20. A vessel is partly filled with water and then olive oil is poured on until it forms a layer 6 inches deep ; find the pressure per square inch at a point 8'5 inches below the surface of the oil, neglecting the atmospheric pressure. 21. A tube 20 feet long with one end open is filled with water and inverted over a vessel of water ; what is the pressure in the water at the top of the column ? The height of the water barometer is 33 ft. 22. A vertical tube is fixed alongside of a vessel and communicates with its bottom. The vessel contains mercury, water and olive oil, the depth of each being 10 inches. How high is the column of mercury in the tube? 23. Describe how the pressure at a point 12 inches below the surface of the water in a vessel may be measured by experiment, and how the experiment may be varied to shew on what this pressure depends. 78 HYDROSTATICS. [CH. Ill 24. The atmospheric pressure at the surface of a lake is 15 Ibs. per square inch. Find at what depth the pressure is 45 Ibs. per square inch, the weight of a cubic foot of water being taken to be 1000 ounces. 25. Describe an experiment to shew that the difference between the pressures at two points in a fluid at rest under gravity is proportional to the difference in their depths. 26. What will be the thrust on a square board whose side is 1 foot when sunk in water to the depth of 20 feet, the board being horizontal and the height of the barometer at the surface of the water 30 inches ? The specific gravity of mercury is 13-59. 27. What is the pressure in Ibs. per square inch at a point in mercury at a depth of 2 feet, the specific gravity of mercury being 13-59? The pressure of the atmosphere being neglected. 28. A layer of oil 25 cm. in depth and of specific gravity 0-82 floats on a quantity of water at the same depth. Find the difference between the pressure at the top surface of the oil and that at the bottom of the water. 29. 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 communicating with the cylinder, and therefore also containing water. Find the area of the piston so that for each pound placed upon it, the surface of the water in the tube may increase its vertical distance from the piston by 1 inch. [A cubic foot of water weighs 1000 ounces.] 30. A cylindrical barrel, the area of whose bottom inside the barrel is 5 square feet, has its axis vertical. A vertical pipe (area of its internal section 18 sq. inches) is screwed into a hole in the top of the barrel, and water poured in until the barrel is full, and also 7 inches of the pipe above the barrel. The uniform thickness of the top being 1 inch, find (i) the upward thrust of the water on the top of the barrel, (ii) what extra volume of water must be poured in, so that the upward thrust may be doubled ? [A cubic foot of water weighs 1000 ounces.] CHAPTER IV. FLUID THRUST. CENTRE OF PRESSURE. 44. Thrust on a Horizontal Surface. A value has already been found for the resultant thrust on a horizontal surface exposed to fluid pressure. It is, if we omit the pressure on the free surface of the fluid, the weight of a column of the fluid having the horizontal surface for its base and the depth of that surface for its height. This follows from the expression found in Section 35, for, if A be the area and p the pressure at each point of the area, then, since the pressure is uniform, the resultant thrust P is given by the equation P = Ap. But if k is the height of the surface, and o> the weight of unit of volume of the fluid, then p = a>h. Hence P = A5 (20,). Now (Statics, 38) we know that if z is the depth of the centre of gravity of a number of particles a 15 a 2 , ... then _ S (za) _ And in the case in point, 5 (a) = area of surface A. Hence Az = 3 (za). Hence P = o>5 (za) = uAz. 1 If the pressure on the upper surface of the fluid is to be considered must be measured from the effective surface, see Section 32. 45-46] FLUID THRUST. CENTRE OF PRESSURE. 83 Again, Az is a volume of fluid having the area A for its base and z, the depth of the centre of gravity, for its height ; while i body ly weighing it in air and in water. Archimedes' Principle gives us a m^ans of finding the volume of any body which sinks in water. Thus weigh the body in air, let it be W grammes weight. Weigh it in water, let it be W grammes weight. Then the buoyancy or upward thrust is W W grammes weight, and this is the weight of a mass of water equal in volume to the body. But the volume of WW grammes of water is W - W cubic centimetres. Thus the volume of the body is WW cubic centimetres. If we are weighing in pounds we must, in order to find the volume, remember that a cubic foot of water weighs 62 '32 pounds. Hence the volume of 1 Ib. of water is 1/62-32 cubic feet and the volume of W-W pounds is (W- JF')/62-32 cubic feet. This method of finding the volume of a body is made use of in many of the experimental determinations of specific gravity. (See Chapter vi.) 51. Floating Bodies. When a body is floating in a fluid partially immersed, the volume of liquid displaced is less than that of the body ; the upward thrust is the weight of the liquid displaced, and acts vertically through the centre of gravity of the fluid displaced. PROPOSITION 23. To find the conditions of equilibrium of a body floating freely. Let ABC, Fig. 54, be the body and let G be its centre of gravity ; let H be the centre of buoyancy, i.e. the centre of gravity of the fluid displaced. Then the body is in equili- brium under two vertical forces, viz. its own weight W acting 102 HYDROSTATICS. [CH. V downwards at G, and the weight of fluid displaced W, acting upwards at H. It is necessary for equilibrium that these two .-^,-4 K ' L Fig. 54. e eqfual ahcf opposite, for when two forces main- tain a body in equilibrium they must be equal and their lines of action must lie in the same straight line. Hence the conditions of equilibrium are (i) The weight of the body is equal to the weight of fluid displaced. (ii) The centre of gravity of the body is in the same vertical line as that of the fluid displaced. PEOPOSITION 24. A solid of given volume and density floats freely in a fluid of given density, to flnd the volume immersed. Let V, p be the volume and density of the solid, V the volume immersed, p the density of the fluid. Then the weight of the solid is equal to the weight of fluid displaced ; hence the mass of the solid is equal to the mass of fluid displaced. But the mass of the solid is Vp ; the volume of fluid displaced is V, and its density is p', its mass therefore is V'p'. Hence Vp = V'p. Therefore V'=V?-,. P If the fluid be water the ratio p/p is the specific gravity of the solid. Hence, since p/p is equal to V'/V we see that, When a solid floats in water its specific gravity is the ratio of the volume immersed to the whole volume of the solid. 51] FLOATING BODIES. 103 Corollary. If we can easily measure the value of V we can use the result to compare the densities of different fluids. For we have , VP P -'--^' Now suppose the solid allowed to float in a fluid of density p", and let V" be the volume immersed. Then P = r Hence J 7 ' Hence The densities of two fluids in which a given solid can float are inversely as the volumes immersed. We can prove this more briefly thus. When the solid floats, the mass of fluid displaced is equal to that of the solid. Hence the masses of the two fluids displaced are the same, hence their densities are inversely proportional to their volumes, i.e. to the volumes of the solid immersed. This principle is made use of in the common Hydrometer. See Section 58. PROPOSITION 25. To find the conditions of equilibrium of a solid immersed in a fluid and held by a string. (i) When the solid is totally immersed as in Fig. 55 the solid is acted on by three forces ; its own weight, the buoyancy Fig. 55. of the fluid, and the tension of the string. Since the solid is totally immersed, if we suppose it to be homogeneous, the centre of gravity of the fluid displaced coincides with that of the solid. 104 HYDROSTATICS. [CH. V Thus the first two forces are vertical and act at the same point; the tension of the string is therefore vertical and its direction passes through the centre of gravity of the solid. Thus the point of attachment of the string is in the same vertical as the centre of gravity. Let W be the weight of the solid, W that of the fluid displaced, the tension T is the difference between W and W. If W is greater than W the tension acts upwards, supporting the solid, and we have T= W- W. In this case the solid is denser than the fluid. If W is greater than W, or the fluid denser than the solid, then T-W'-W^ and the string helps to keep the solid submerged. # (ii) When the solid is not totally immersed. In this case the centre of gravity G and the centre of buoyancy H do not coincide. Let A, Fig. 56, be the point at which the Fig. 56. string is attached, then the forces W and W, acting at G and H respectively, are parallel, being both vertical in direc- tion ; hence the third force T is also parallel to them, the string is vertical, and, assuming that W is greater than W, T = W- W. Again, let ALM meet in L and M the vertical lines through G and H respectively ; then taking moments about A we have W.AL= W .AM. This equation will determine the weight of the fluid which is displaced, and then the former equation will give T. 51-52] FLOATING BODIES. 105 In case (i) it has been assumed that the solid is homogeneous so that its centre of gravity coincides with that of the fluid displaced ; if this is not true the methods of (ii) must he applied to (i). 52. Buoyancy of the Air. Experiments will be described in Section 66, which prove that air has weight. When therefore a body is weighed in air it is acted on by an upward thrust equal to the weight of the air displaced. Hence the apparent weight of a body, when weighed in air, is not its true weight. The buoyancy of the air is illustrated by the ascent of a balloon. An air-tight bag of silk or some other light material is filled with hydrogen or coal gas or some other gas of less density than air, the weight of air displaced is then greater than the weight of the balloon, which can therefore rise and draw up with it a light car carrying passengers. The necessary condition is that the weight of air displaced should be greater than that of the balloon and its load. In a fire-balloon the gas used is heated air, this is less dense than cold air ; the bag is filled over a piece of sponge or cotton wick soaked in methylated spirits and ignited. EXPERIMENT 15. To illustrate the buoyancy of the air. In Fig. 57, A is a glass bulb with thin walls. A portion of the tube from which the bulb was blown is left attached, as Fig. 57. shewn at J5(7, being closed at the end C. By means of a small metallic counterpoise the whole is balanced with the stem horizontal on a knife edge at B. The weight of air displaced by the bulb is much greater than that displaced by the tube EC and the counterpoise. 106 HYDROSTATICS. [CH. V Since there is equilibrium, we have, taking moments round JB, Moment of bulb and contained air moment of air displaced by bulb = moment of counterpoise and tube moment of air displaced by these. The apparatus is then put inside the receiver of an air- pump (see Section 96), and the air is exhausted. As the exhaustion proceeds the bulb A sinks while the counter- poise rises. By withdrawing the air the upward thrust due to it is diminished ; the buoyancy effect on the bulb is greater than that on the counterpoise, this is shewn by the ascent of the latter. 53. Corrections for weighing in air. The buoyancy of the air affects the result of weighings made in air and a correction is in consequence needed. The correction is very small in most cases, but it is desirable to shew how to introduce it. # PRO POSITION 26. To find the correction to the apparent weight of a body due to the buoyancy of the air. . Let M be the true mass of the body to be weighed, p its density ; let M Q be the mass of the weights, p their density, and let X be the density of the air. The volume of the body is Mjp and the mass of air dis- placed by it is MXjp, the mass of air displaced by the weights is M \/ Po . Now if the balance be true the difference between the weight of the body and the weight of air it displaces is equal to the difference between the weight of the " weights " and the weight of air they displace ; and since the weights of these various bodies are proportional respectively to their masses ' 52-54] FLOATING BODIES. 107 Therefore Now the value of X varies with the pressure and tempera- ture of the air, it is however very small compared with the density of solids or liquids, thus \/p and X/p are very small quantities. If then we divide by the denominator 1 X/p and neglect X 2 //> 2 , and higher terms which will be very small indeed, we find Po For air under standard conditions X is about -0012 grammes per c.cm., while p n for brass weights it is about 8 grammes per c.cm., thus X//o is about -00015. The value of \[p depends on the density of the body weighed: for water p is 1 gramme per c.cm. and \/p is -0012. Thus the true mass M of a volume of water weighed with brass weights is given in terms of the apparent mass M by the formula M= M (1 - -00015 + -0012) = M (1 + -00105). The correction in this case amounts to about -1 per cent. 54. Experiments on floating bodies. The following experiments illustrate some of the laws of floating bodies. (i) The mass of an egg is greater than that of an equal volume of fresh water, it is less than that of an equal volume of a strong solution of salt in water ; thus an egg will sink if placed in fresh water, it will float if placed in a strong solution of salt in water. If a vessel is half filled with salt solution and then fresh water is carefully poured on to the top, the two liquids mix where they come into contact, forming layers of variable density ; the egg placed in the fresh water will sink, but after oscillating up and down for some time will come to rest in a position in which the mass of fluid displaced is equal to that of the egg. 108 HYDROSTATICS. [CH. V (ii) The Cartesian diver. A small glass bulb, Fig. 58, has an opening in its lower side ; to the bulb is attached a small counterpoise, the weight of which is adjusted so that the whole just floats in water with some air in the bulb the counterpoise often takes the form of a glass figure ; hence the name of the apparatus. The water is contained in a tall jar and its top is closed with a piece of india-rubber. On pressing the india-rubber down the pressure of the air above the water is in- creased; this pressure is trans- mitted to the air in the bulb, which contracts in volume. More water enters the bulb, the weight of the bulb with its contents becomes greater than the weight of water which it displaces. Hence the diver sinks. When the pressure at the top of the vessel is relieved, the air in the bulb expands and the diver rises unless the vessel is too deep. If the vessel exceed a certain depth the pressure at the bottom, due to the water, may be so great even when the air-pressure on the surface is relieved, that the air in the bulb cannot expand sufficiently to again allow the diver to rise. By placing the vessel under the receiver of an air-pump and exhausting, it may again be brought to the top. # 55. Stability of equilibrium of a floating body. So far we have dealt only with the conditions of equili- brium ; the problem whether the equilibrium is stable or not remains to be discussed. #(i) When the body is totally immersed and just floats. Fig. 58. FLOATING BODIES. 109 54-55] The centre of buoyancy is the centre of gravity of the fluid displaced; if the body and the fluid be homogeneous it coincides with the centre of gravity of the body. In this case the two forces which act on the body pass through this point, and any position is one of equilibrium. The equilibrium therefore is neutral. But if the centre of gravity and centre of buoyancy do not coincide, because, for example, either the body or the fluid is not homogeneous, various cases arise. Suppose in the first case the body is not homogeneous. It may for example consist of a piece of wood with some lead fastened to it just sufficient in amount to allow it to float, or of a glass bulb and stem with some mercury in the bulb. Then we must distinguish between displacements in which the body moves without rotation for these the equilibrium is neutral and displacements in which the body is turned about some axis. Fig. 59 (i), (ii) represent two cases which may occur ; in each G is the centre of gravity, H the centre of buoyancy, in (i) G is below H, in (ii) it is above, and in each case the body has been displaced from its equilibrium position. The weight of the body acts downwards at G and the buoyancy acts upwards at H\ these forces are equal and thus constitute a couple. In (i) the couple tends to right the body, the equilibrium is stable; in (ii) it tends to increase the displacement, the equilibrium is unstable. For stable equili- brium the centre of gravity must be as low as possible. Or take again the case of the egg floating in a solution of 110 HYDROSTATICS. [CH. V salt and water of variable density ; the lower layers are the denser ; if the egg is depressed it displaces a volume of fluid greater in weight than itself, hence it rises again ; while if it is raised the fluid displaced becomes less in weight than the egg and it sinks back to its original position, thus the equilibrium is stable. With the diver on the other hand the reverse is the case ; when it begins to sink it continues to sink until the air pressure on the surface is reduced, or the bottom is reached. # (ii) When the body is only partially immersed. The circumstances are now somewhat different. In the first place, such a body, if it can float at all, is in stable equi- librium for vertical displacements; if it be pushed down the buoyancy is increased, and it rises again ; if it be raised out of the water the buoyancy is decreased, and the body sinks back. But, though this is the case, the equilibrium for rotational motion may be either stable or unstable. A thin flat board could be made to float with its flat faces vertical and its edge downwards; in this position, however, it would be unstable and would tend to turn over until the flat sides became horizontal and the narrow edges vertical, when its equilibrium would be stable. When the body is displaced, it is clear that in general the centre of buoyancy shifts its position in the body. Let 6r, Fig. 60, represent the centre of gravity of the body, H its centre of buoyancy in the undisturbed equilibrium Fig. 60. 55] FLOATING BODIES. Ill position, H' the new position of the centre of buoyancy when the body is displaced. Let us suppose that the form of the body is such that ff, G and H' lie in a vertical plane, and also that the motion is such that the volume of water displaced remains unaltered, so that the buoyancy is unchanged 1 . Draw H'M vertical through H', it will meet the line HG\ let M be the point of intersection, then the nature of the equilibrium will depend on the position of M. The buoyancy now acts vertically upwards through J/", the weight vertically down- wards through (2, and these two forces constitute a couple. If M be above G the couple tends to restore the body to its original position, if M be below G the couple tends to displace the body further. The position of H' and therefore of M depends on the shape of the body. If the displacement be very small the point M is known as the Metacentre, and the condition of equilibrium for small displacements is that the metacentre should be above the centre of gravity. ^DEFINITION. Imagine a floating body to be displaced through a small angle about a horizontal axis in such a way that the volume of fluid displaced may remain unchanged, and suppose further ', that the displacement is such that the vertical line through the new position of the centre of buoyancy will meet the line joining the centre of gravity to the original position of the centre of buoyancy. The point of intersection of these two lines wJien the displacement is very small is the Metacentre. Thus we see that for small displacements the stability of a boat or other floating body depends on the position of the metacentre relative to the centre of gravity; for a boat of given shape and given weight it is important to keep the metacentre as low as possible. When the dis- placements to be dealt with may, as in the case of a ship, be considerable, other points besides the position of the metacentre must be attended to. Examples, (l) A piece of ivood weighing 24 grammes floats in water with 2/3 of its volume immersed. Find the density and the volume of the wood. (i) Since 2/3 of the volume is immersed and the weight of the wood is equal to that of the water displaced, each cubic centimetre of the 1 This condition is satisfied in the case, for example, of a ship which is made to heel either by the wind or through shifting some of the cargo. 112 HYDROSTATICS. [CH. V wood weighs as much as 2/3 of a cubic centimetre of water or 2/3 of a gramme. Thus the density of the wood is 2/3 of a gramme per cubic centimetre. The volume of the wood is its mass divided by its density or 24/f c.cm. and this comes to 36 c.cm. Thus the volume is 36 c.cm. (ii) Otherwise thus. Let V be the volume in c.cm. Then ^F is the volume of water displaced, and F grammes is the mass of water displaced; this is equal to the mass of the body. Hence |F=24, V 36 c.cm. Also Density = mass/ volume = 24/36 = | grammes per c.cm. (2) A piece of wood of specific gravity '6 is floating in oil of specific gravity '85, what fraction of its volume is immersed? Since the volume of oil displaced is equal in weight to the wood, the two volumes are inversely as the specific gravities. Thus fraction of whole immersed = f = '7. (3) A body whose volume is 30 c.cm. and specific gravity 1-5 is placed in a vessel and just covered with water. What thrust does it exert on the bottom? The weight of the body is 45 grammes, the weight of water displaced is 30 grammes, hence the thrust on the vessel due to the body is 15 grammes weight. (4) The mass of a balloon and its car is 3000 Ibs., the mass of air displaced is 3400 Ibs., with what acceleration does the balloon rise? The resultant upward force is (3400 - 3000) or 400 Ibs. weight. This is equal to 400 x 32 poundals. The mass moved is 3000 pounds. Hence the acceleration is 400 x 32/3000 or about 4-26 feet per second. (5) A vessel containing water is placed on the pan of an upright spring balance and a body suspended from a second spring balance is immersed in the water. Examine how the readings of the two balances are altered. The reading of the second balance is reduced by the buoyancy of the water, acting upwards on the suspended body ; the reading of the first balance is increased by the same amount, for the upward thrust of the water on the body is just equal to the additional downward thrust on the bottom of the vessel, due to the immersion of the body in the water. Thus the sum of the two readings is unchanged, and this clearly must be so for the total mass supported by the two balances is not changed. See Experiments 11 and 12. 55] FLOATING BODIES. 113 (6) The specific gravity of sea water is 1-028 and of ice '918. What fraction of the volume of an iceberg floats out of ivater? The weight of the iceberg is equal to the weight of water displaced and the weight of any body is proportional to the product of its volume and its specific gravity. Hence, Volume of sea water displaced x 1-028 = volume of iceberg x -918. volume immersed _ -918 "whole volume" ~ 1^028 * Therefore volume above water _ 1028 - 918 _ 110 _ whole volume 1028 ~~ T028~ (7) A body weighing 10 Ibs. floats in a liquid with 1/3 of its volume above the surface. What weight must be placed on the body in order just to sink it? The specific gravity of the body is 2/3, hence the weight of water displaced by the body when just immersed is f of 10 Ibs. or 15 Ibs. Hence the body will just float totally immersed if a weight of 5 Ibs. be placed on it. (8) A rectangular block of boxwood 10 cm. in depth and of specific gravity *9, is floating in water with its upper surface horizontal. Oil of specific gravity -6 is poured on to the water, shew that, neglecting the buoyancy of the air, the wood will rise through 1-5 cm. Initially 9/10 of the wood is immersed, thus the height out of the water is 1 cm. ; the weight of the wood is equal to the weight of water displaced. When the oil is poured on, the weight of the wood is equal to the weight of water displaced together with the weight of oil displaced. Thus the weight of water displaced must be less than before, the wood therefore rises; let it rise h cm. and let A square cm. be the area of its upper surface. The volume of the wood is 10 A c.cm., that of the water displaced is (9 - h) A c.cm., and of oil (1 + h) A c.cm. Equating then the weight of the wood to the weights of oil and water we have Hence 4ft = 6 or ft =1-5 cm. Thus the wood rises 1-5 cm., remaining with 7-5 cm. in the water and 2'5 cm. in the oil. (9) A cylindrical rod iceighted at one end floats in water, determine the conditions of stable equilibrium. If the rod can float with half its length immersed and at any angle to the horizon, shew that the weight which is added is equal to that of the rod. Let W be the weight of the rod, 2a its length, W the weight added, and the weight of a unit volume of water, I the length immersed, and a G. HYD. 8 114 HYDROSTATICS. [CH. ^ the area of the cross section, and h the distance from the bottom of the centre of gravity of the rod and weight. Then the weight of water displaced is law, hence W+W^law, also taking moments about the bottom of the rod h(W+W l )=aW. The centre of buoyancy which is at a distance |Z from the bottom must be above the centre of gravity. Hence Z is greater than h. Thus I is greater than 2h, or aw W+ Wi ' (W+ TFj) 2 . or - l > 2aaw. Now 2aaw is the weight of a volume of water equal to the volume of the rod, let this be W, then the condition for stability is that (W+ JFj) 2 is greater than W. W. If the rod floats half immersed then I = a, while if any position is one of equilibrium, the centre of gravity coincides with the centre of buoyancy, hence Hence W+W l = 2W or W=W^ EXAMPLES. FLOATING BODIES. [For a Table of Specific Gravities see p. 15.] 1. A piece of oak 35 c. inches in volume floats in water, what volume of water does it displace ? 2. What is the weight in water of 1 kilogramme of iron ? 3. Ten kilogrammes of cork are totally immersed in water. What is the resultant thrust of the water, and what will be the acceleration with which the cork will rise if let go ? 4. Find the resultant upward thrust on the following bodies when totally immersed : (i) 100 c.c. of iron in water. (ii) 250 grammes of copper in salt water. (iii) 500 c.c. lead in sulphuric acid. 5. A lump of iron floats in mercury. What fraction of its volume is immersed ? CH. V] FLOATING BODIES. 115 6. A lump of iron is supported in water by cork so that ^ of its volume is out of the water. Compare the volumes of the cork and iron. 7. What is the apparent weight of 25 grammes of iron when weighed in alcohol ? 8. Find the weight of water displaced by 56 grammes of copper. 9. An iceberg floats with 2000 c. feet above the surface of salt water ; find its volume. 10. A piece of cork floats in water with 50 c. inches above the water ; find its volume. 11. A body weighing 30 grammes floats in water with of the volume submerged ; find its volume. 12. A cylinder floats, with its axis vertical, totally immersed in water covered with a layer of oil. If % of the cylinder be in the water, find its specific gravity. ' 13. Find the weight of a glass ball 2 inches in diameter (1) in air, (2) in water, (3) in alcohol. [Sp. gr. glass 3-6.] 14. A vessel of water is placed on one pan of a balance and counter- poised. If 35 grammes of lead, supported by a string are immersed in the water, what additional weights are required to restore the balance ? 15. Two bodies whose specific gravities are 2'5 and 7*5 balance when each is weighed under water ; find the ratio of their weights. 16. A man weighing 10 stone floats with 5 cubic inches out of the water ; find his mean specific gravity and his volume. 17. A piece of oak J immersed in water is supported by a string. What portion of the weight is carried by the string ? 18. A lump of iron floats totally immersed partly in mercury and partly in water. What volume is there in each liquid ? 19. Find the specific gravities of bodies which float with the following volumes above and below the surface of water respectively: (1) Wood 21 : 24, Ice 21 : 237, Cork 3:1, Oak 1 : 3. 20. A ship weighing 1000 tons passes from fresh to salt water, if the area of a section at the water line be 15000 square feet and the sides where they cut the water be vertical, how much will she rise ? 21. A piece of iron weighing 275 grammes floats in mercury with f of its volume immersed. Determine the volume and density of the iron. 22. How much water will overflow from the edges of a cup just full of water when a cork 2 cubic inches in volume is gently placed in it so as to float ? 82 116 HYDROSTATICS. [CH. V 23. A cylindrical cork 4 inches long is to be loaded with iron of the same section so as just to float. How long must the iron load be ? 24. A cone of a certain material floats point downwards with f of its axis immersed ; find its specific gravity. 25. A vessel contains water and mercury. A cube of iron, 5 cm. along each edge, is in equilibrium in the liquids, with its faces vertical and horizontal. Determine how much of it is in each liquid, the densities of iron and mercury being 7'7 and 13-6 respectively. 26. A piece of wood is floating on water, and oil is then poured on to the water. How is the volume of wood in the water affected ? 27. A cubic metre of wood floats in water with f ths of its volume immersed. Calculate the depth to which it would sink in a liquid of specific gravity -8. 28. A block of wood 10 Ibs. in weight floats in water with two-thirds of its volume immersed. What force will be required just to sink it? Also what weight of a metal (specific gravity 5) must be placed on it so that both metal and wood may just be immersed? 29. A closed cubical vessel with walls one inch in thickness is to be made of metal whose specific gravity is f. Shew that in order that the vessel may float in water its internal dimensions must be at least 64 cubic inches. 30. A block of hard wood, 5 feet long, 6 inches wide and 4 inches thick weighs 52 Ibs. Determine whether it will float (1) in ordinary water, (2) in sea water, the specific gravity of which is 1-026. If in either case it does float, how much of it will project above the surface? 31. A body whose specific gravity is 3 and whose weight is 6 Ibs. is supported by a string with half its volume immersed in water. What is the tension of the string, the density of water being unity ? 32. A small hole drilled at one end of a thin uniform rod is filled with some much denser material. It is observed that the rod can float in water half immersed and inclined at any angle to the vertical. Shew that the specific gravity of the rod is . 33. A beaker of water is placed on the pan of a balance and counter- poised, a piece of glass suspended from a separate support is immersed in the water and it is found that 20 grammes have to be added to the counterpoise to restore equilibrium. Explain this and calculate the volume of the glass. 34. Account for the ascent of a balloon. How would the lifting power of the balloon be altered if the atmo- spheric pressure be diminished ? 35. Why does a ship rise when it goes out of a fresh- water river into the open ocean ? CH. V] FLOATING BODIES. 117 + 36. The specific gravity of ice is 0-92, that of sea water is 1-025. What depth of water will be required to float a cubical iceberg whose side is 100 feet ? 37. A- piece of iron (specific gravity 7*2) is covered with wax (specific gravity 0'96) and the whole just floats in water; its mass is 36 grammes; find the mass of the iron and the wax respectively. 38. A cubic foot of ice at the freezing point, one of whose edges is one foot long, floats in ice-cold water, but so that it is capable only of rotation about one edge which is hinged in the surface of the water : a weight of 42 ounces placed on the top of the cube of ice 10 inches from its hinged end just immerses the ice in the water. Find the specific gravity of the ice. 39. A large stone is held suspended under water by a rope. Explain why the load on the rope is less under these conditions than if the stone were suspended in air. If an addition of 10 Ibs. to the load on the rope will break the rope, how much of the stone may be raised out of the water before the rope breaks? CHAPTER VI. MEASUREMENT OF SPECIFIC GRAVITIES. 56. General Considerations. In many methods of determining specific gravities the fact that the resultant upward thrust on a body immersed in a liquid is equal to the weight of liquid displaced is made use of, for by this means we obtain the weight of a volume of liquid equal in volume to the body, and the specific gravity has been defined, Section 6, as the ratio of the weight of the body to the weight of an equal volume of some standard substance, usually water. The procedure generally adopted is to weigh the body in air and then in water ; the difference between the two gives the weight of water displaced, that is, the weight of an equal volume of water, and the ratio of the weight in air to this difference, is the specific gravity. The Hydrostatic Balance and various forms of Hydrometers give examples of this method. 57. Hydrostatic Balance. This is an ordinary balance arranged so that a body suspended from one end of the beam may be easily weighed in water. In some forms a hook is attached below one of the scale- pans ; this scale-pan is often attached to the beam by shorter chains than the other and the body to be weighed can be suspended from the hook. 56-57] MEASUREMENT OF SPECIFIC GRAVITIES. 119 In other forms, as shewn in Fig. 53 (a) (p. 98), a wooden stand or bridge rests on the floor of the balance case over one of the scale-pans which can swing freely below it, the supports of the scale-pan passing on either side of the bridge; the body to be weighed is suspended from a hook attached to the knife edge which carries the scale-pan. The vessel of liquid is placed on the bridge and the body can be immersed in it. A spring balance often forms a convenient instrument for use as a hydrostatic balance. Jolly's balance is a special form of spring balance so used. In many of the experiments to be described the solid has to be weighed in water. This introduces several sources of error. If the solid is very light the weight of the wire or thread by which it is supported may need to be considered, moreover a capillary force is exerted on the wire where it cuts the liquid; it is therefore important that the supporting wire should be as fine as is consistent with strength. A horsehair or a piece of fine wire may be used, a piece of thread will serve but it absorbs moisture and so varies in weight as it gets wet. Water again contains dissolved air, this collects in bubbles on the sides of the containing vessel or of any body immersed in the water and the apparent weight of the body is reduced by the buoyancy of these bubbles. The bubbles must be carefully brushed off the solid before weighing; for very accurate work it is desirable to boil the water and allow it to cool previous to use, or it may, if convenient, be placed for a time under the exhausted receiver of an air-pump. These precautions are clearly specially necessary in the case of a small body. We proceed now to consider some experiments with the Hydrostatic balance. EXPERIMENT 16. To find, by the Hydrostatic balance, the specific gravity of a solid body which sinks in water, and to determine its volume. Suspend the body by a piece of fine wire from the pan of the balance and weigh 1 it. Let the weight be W grammes. 1 In all these experiments several observations of the weight must be taken. For exact work the "method of oscillations," Glazebrook and Shaw, Practical Physics, 12, p. 107, should be employed. For precau- tions to be observed in determining specific gravities, see Glazebrook and Shaw, Practical Physics, 15-19. 120 HYDROSTATICS. [CH. VI Immerse the body in water and weigh it again. Let the weight be W grammes ; then W W measures the upward thrust of the water on the body, and this, we know, is equal to the weight of water displaced. This water is clearly equal in volume to the body. Thus W grammes is the weight of the body and W W grammes is the weight of an equal volume of water. Now Specific gravity weight of body weight of equal volume of water The volume of the body which is equal to that of the water displaced 1 is W W cubic centimetres. Thus, in the case of a piece of glass, the following observations were made : Weight in air =76*8 grammes. Weight in water = 46 32 grammes. Weight of water displaced = 30 '48 grammes. 7 A Q Specific gravity = ^^ = 2-52. Also volume of glass = 30*48 cubic centimetres. EXPERIMENT 17. To find, by the Hydrostatic balance, the specific gravity of a liquid. Weigh a body, say a piece of glass, in air ; let the weight be W grammes. Then weigh it in water. Let the weight be TPi grammes ; weigh it in the liquid whose specific gravity is required, let the weight be W z grammes. Then W W z = weight of a quantity of liquid equal in volume to the body, and W- W^ = weight of a quantity of water equal in volume to the body. W W<> Hence the specific gravity of the liquid = -^ 2 1 In this statement the variation in the density of water with tempe- rature is neglected. 57] MEASUREMENT OF SPECIFIC GRAVITIES. 121 Thus, using the same piece of glass as in the last experiment : Weight in air =76 -8 grammes. Weight in water = 46 *32 grammes. Weight in liquid = 43*88 grammes. Weight of water displaced = 30 '48 grammes. Weight of liquid displaced = 32 -9 2 grammes. 32*92 Specific gravity = 3^43 = * * 08 * EXPERIMENT 18. To determine with the Hydrostatic balance the specific gravity of a solid body lighter than water. For this purpose a sinker is attached to the body of such a weight that the combination will sink in water. (i) Weigh the solid in air, let the weight be JF, weigh the sinker in air, let the weight be w, weigh the sinker in water, let the weight be w', weigh the combination in water, let the weight be W. Then Weight of water displaced by sinker = w w. Weight of water displaced by combination = W+ w W. Weight of water displaced by solid = W+w-W'-(w-w') = W- W' + w'. Hence weight of body Specific gravity = r-r-r * ^ ; ; 5 weight of equal volume of water _ W ~ W-W + w" Thus we have the following observations for a piece of wax. Weight of solid = 26*65 grammes. Weight of sinker (copper) =11 *38 grammes. Weight of sinker in water =10*10 grammes. Weight of combination in water = 9*16 grammes. Hence Weight of combination in air = 38*03 grammes. 122 HYDROSTATICS. [CH. VI Weight of water displaced by combination = 28 '87 grammes. Weight of water displaced by sinker = 1-28 grammes. Weight of water displaced by wax = 27*59 grammes. O/ / p* Specific gravity of wax = ~ = -966. '-/ 1 *oy (ii) Otherwise thus. If it is convenient to support the sinker below the solid so that it can be immersed in water while the solid is not, the following method requires fewer weighings than that given above. Weigh the solid in air, let the weight be W. Attach the sinker below and weigh again with the sinker only in water, let the weight be JFj. Raise the vessel containing the water so that the solid is immersed as well as the sinker and weigh the combination in water, let the weight be W. Then, W 1 = weight of solid + weight of sinker weight of water displaced by sinker. W = weight of solid + weight of sinker weight of water displaced by sinker weight of water displaced by solid. W l W = weight of water displaced by solid. W Hence Specific gravity = - ^ . W i rr 58. The Common Hydrometer. The principle of this instrument is most easily understood by considering a hollow cylindrical glass tube loaded at one end, so that it will float vertically in water and some other fluids, and having a graduated scale of millimetres either fixed inside or engraved on the glass ; the scale is adjusted so that its zero may coincide with the bottom of the tube. Observations are made by noting the depth to which the hydrometer is immersed ; this depth will measure the volume of fluid displaced. Since the weight of the hydrometer remains unchanged, the 57-58] MEASUREMENT OF SPECIFIC GRAVITIES. 123 weight of fluid displaced is the same, whatever fluid it be im- mersed in. Hence, if the tube sink to depths d, d' in two different fluids, the densities of these fluids are inversely ' as d to d 1 . If the first fluid be water, the specific gravity of the second is cl/d'. Suppose, for example, that the instrument is so adjusted that it sinks to division 100 in water, and to division 92 in a solution of salt in water. Then the specific gravity of the solution is 100/92 or 1-086. Now an instrument such as this would be far from sensitive. A change of 1 mm. in the position in which it rests would mean an altera- tion of 1 per cent, in the specific gravity, and for accurate work this would be useless. Let us suppose however it were possible to have the tube over a metre long and let the reading in water be 1000, the reading in the salt solution 922, then the specific gravity is 1000/922 or 1-085, an alteration of a millimetre in the reading will now mean an alteration of about one in a thousand, the instrument is more sensitive, but it is too long to be of use. In such an instrument however it is only the upper part of the stem which need be graduated. Thus if the graduations extended down to 800 the specific gravities of fluids from 1 to 1-250 could be measured, the rest of the stem will never rise out of the fluid. There is no need therefore for the lower part of the instrument to take the form of a straight stem at all. It may be made in any convenient form provided that its weight and volume are the same as those of the straight stem it replaces and also that it is so constructed as to float with the stern vertical. In practice then the hydrometer usually takes the form shewn in Fig. 61. A is a hollow glass tube ending below in a bulb C which contains mercury so adjusted as to make the instrument float in a vertical position. The upright stem B is hollow and contains a paper scale. The scale is usually not divided into equal parts as in the simple form of apparatus described above, but in such a way that the readings of the scale may give directly the specific gravity of the fluid in which the instrument is immersed. Let us suppose it is to be used for determining the specific gravities of fluids denser than water, 1 See Section 51, Prop. 24 (Coroll.). I Fig. 61. 124 HYDROSTATICS. [CH. VI then the instrument rises as the fluid in which it is immersed is made more dense. The mercury in the bulb and the position of the scale then are adjusted so that, when floating in water, the top division of the scale which is marked I'OOO is level with the surface. The scale is usually adjusted so that each division is equiva- lent to a change of '001 in the specific gravity of the fluid, and each tenth division is marked ; the divisions of the scale are not equal, but decrease in length as the bottom of the stem is approached. It is not difficult to shew how the points on the scale may be obtained by calculation ; in practice, however, it is simpler to determine a few points by immersing the instrument in turn in fluids of known specific gravities; the distances between these points are then sub- divided by eye. If the instrument is to be used to determine the specific gravity of fluids lighter than water, the mercury is adjusted so that it floats in water immersed just up to the bottom of the stem ; when immersed in fluids of less density it sinks further; thus the graduation 1 '000 is at the bottom, and the graduations run up the tube, be- coming less towards the top. Fig. 63 (i), (ii), shew an hydrometer floating in water and in spirit. In order to get sufficient range without having very long stems, a series of hydrometers is generally em- ployed. These are adjusted in such a way that one instrument may sink just up to the top of the stem in a given fluid, while the next will float in the same fluid with the whole of the stem exposed. Thus the scale runs on from one instrument to the next, and by a (i) Fig- 62. (ii 58-59] MEASUREMENT OF SPECIFIC GRAVITIES. 125 proper choice of the hydrometer the specific gravity of any given fluid can be determined. EXPERIMENT 19. To use the common hydrometer to find the specific gravity of a liquid and to check its graduations. Immerse the hydrometer in the liquid and note the reading. This gives the specific gravity. To check the graduations determine in some other way the specific gravity of the liquid, e.g. by the hydrostatic balance, Section 57, or by the specific gravity bottle; the two results should agree. 59. Nicholson's Hydrometer. This instrument, shewn in Fig. 63, consists of a hollow bulb to which a thin stem is attached, the stem carries a tray or cup, into which weights can be placed. Below the bulb hangs a second tray or basket. This is weighted with mercury, which is ad- justed so that the instrument may float vertically in water. On the stem there is a mark, when the instrument is used it is loaded so that this mark is just in the surface of the liquid in which it is floating. Thus whatever be the liquid the volume dis- placed is always the same. EXPERIMENT 20. To weigh a body and to determine its specific gravity by the use of Nicholson's Hydrometer. lg> Place the hydrometer in a tall vessel of water and take care that it floats freely and that no air-bubbles are attached to it. Weights are to be placed in the upper cup to sink the hydrometer down to the mark. To avoid the inconvenience caused by these weights falling into the water, the top of the vessel is covered with two pieces of glass 1 which fit together and close it. The stem of the instrument rises through a hole which has been drilled in the glass. Place weights on the upper cup till the instrument sinks to 1 The glass is not shewn in the figure. 126 HYDROSTATICS. [CH. VI the mark. It will be found that there is not a perfectly definite position of floatation for given weights, the hydrometer will, within limits, rest in any position. This is due chiefly to the capillary action between the water and the stem, the position will be more definite if the stem is free from grease ; it may be cleaned before the instrument is used by being rubbed with cotton-wool soaked in alcohol. Suppose the instrument is floating with the mark just below the surface. Take off' some small weights till the mark just rises above the surface and note the weight left on ; put on weights until the mark again sinks below and note the weights ; do this several times and take the mean. Let it be w f Now place the solid in the upper cup ; the instrument sinks. Take off weights until the mark is again in the surface. Determine as above the exact weight to be taken off. Let it be JF, then W is clearly the weight of the solid. Place the solid in the lower pan, taking care that no air- bubbles adhere, the weight supported is the same as before, but the solid is now acted on by the buoyancy of the liquid displaced, the instrument therefore rises. Place additional weights on the upper pan until it again sinks to the mark, determining their value as before. Let it be W, then W is clearly the weight of liquid displaced by the solid. Hence the specific gravity is W/ W. Instead of reckoning the weights taken off in the second operation and those added in the third, it may be more convenient to reckon the weights on in each case. Let them be W l when the solid is not in either pan, w when the solid is in the upper cup, w' when it is in the lower cup. Then the weight of the solid is W-^-w and the weight of water displaced 10' w. Hence the specific gravity is (W l - w)l(w'-w). Thus in an experiment with sulphur the hydrometer was in adjustment with 8 '35 grammes in the cup, the weights taken off when a piece of sulphur was placed in the upper pan were 5 "81 grammes. This then was the weight of the sulphur; the weights added to these when the sulphur was transferred to 59-60] MEASUREMENT OF SPECIFIC GRAVITIES. 127 the lower pan were 2*92 grammes, giving the buoyancy of the sulphur. Thus specific gravity of sulphur = 5-81/2-92 = 1-99. EXPERIMENT 21. To find the specific gravity of a liquid with Nicholson's hydrometer. For this purpose we require to know the weight of the hydrometer; let it be W Q . Place the instrument in water and determine as before the weight required to sink it up to the mark. Let it be TFj. Place the instrument in the liquid and let the weight required to sink it be W 2 . The weight of water displaced, since the hydrometer is floating, is W + W^. The weight of liquid dis- placed is W Q -f W 3 . The volumes of these two weights are the same, each being equal to the volume of the hydrometer up to the mark. Thus the specific gravity of the liquid is In an experiment with a solution of salt in water containing 20 grammes of salt in 100 grammes of the solution, the weight of the hydrometer was 11*27 grammes, the weight required to sink it in water was 8*35 grammes. Thus the weight of water displaced was 19*62 grammes. The weight required to sink it in the salt solution was 9 -88 grammes, thus the weight of salt solution displaced was 21*15 grammes. Hence specific gravity of salt solution -21-15/19-62=1*078. 6O. Jolly's Balance. This, as shewn in Fig. 64, consists of a long spiral spring which carries two light scale-pans one below the other. A vertical scale graduated on mirror glass is mounted behind the spring and a white Fig. 64. 128 HYDROSTATICS. [CH. VI bead is attached to the end of the spring. The division of the scale opposite to the top of the bead can be read accurately, by looking at it with one eye closed, from such a position that the bead exactly covers its own image in the mirror. When using the apparatus the lower pan is kept always immersed in a vessel of water. The method of making measurements and the theory of the instrument are much the same as with Nicholson's hydrometer. Weights are placed in the upper pan until the bead comes opposite to some convenient division of the scale. The body to be weighed is then placed in the upper pan and weights removed until the bead is in the same position as before ; this gives the weight of the body. The body is then transferred to the lower pan, thus causing the bead to rise, weights are added until the bead again occupies its sighted position; these weights give the buoyancy, and by dividing the weight of the body by its buoyancy we get the specific gravity of the body 1 . 61. Specific Gravity Balls. In order to obtain a rapid determination of the specific gravity of a liquid of which only a small quantity is available, a set of specific gravity balls is useful. These are small glass bulbs loaded with mercury, and so adjusted that each will just float in a liquid of definite specific gravity. A number of these balls are placed in the liquid to be examined, some of the balls sink, others float ; if one be found which will just float the specific gravity of the liquid is known. If it happens that no one ball just exactly floats, limits can be found by noting the specific gravity of that ball ; among those that sink, which most nearly floats ; and the specific gravity of the ball, among those that float, which most nearly sinks. The specific gravity of the liquid lies between these two. 62. The Specific Gravity Bottle. An experiment involving the use of the specific gravity bottle has already been described (see Section 9). Two forms of bottle are shewn in Fig. 65 (i) and (ii). In (i) the bottle has a narrow neck which is open ; a fine mark is 1 For practical details see Glazebrook and Shaw, Practical Physics. p. 137. 60-62] MEASUREMENT OF SPECIFIC GRAVITIES. 129 made on the neck and in use the bottle is always filled exactly up to this mark. In (ii) the bottle is closed with a ground (i) Fig. 65. (ii) glass stopper, which is perforated with a small hole. The bottle is filled to the top of the neck with water or whatever other liquid is being used, and the stopper is then pushed home. The surplus liquid escapes by the hole in the stopper, leaving the bottle completely filled. The specific gravity bottle is used to find the specific gravity of a liquid or of a solid which is either in the form of a powder or can be broken into small fragments and inserted in the bottle. Bubbles of air easily collect among the fragments or on the sides of the bottle ; care must therefore be taken to free the liquid used from air as far as possible. The following experiments may be performed with the specific gravity bottle. EXPERIMENT 22. To find the specific gravity of a liquid with the specific gravity bottle. Weigh the bottle empty 1 . Let its weight be W grammes ; 1 Before the bottle is weighed empty it should be dried. This is done by connecting a glass tube with a bellows and blowing air into the bottle. The tube is held over a Bunsen burner to heat the air before it enters the bottle. G. HYD. 130 HYDROSTATICS. [CH. VI then fill it with water and again weigh it. Let the weight be W ; finally fill it with the liquid and weigh it. Let the weight be W 2 . In these last operations care must be taken that all traces of air-bubbles are removed from the sides of the bottle and from the stopper. W 2 W is the weight of liquid which fills the bottle, W l - W is the weight of an equal volume of water. W W Hence the specific gravity of the liquid = -==? ^. . \V -i ww In some cases it is convenient, instead of weighing the empty bottle, to counterpoise it with shot and lead foil. Then two weighings only are necessary. The following observations were made. Weight of empty bottle 6 -85 grammes. Weight of bottle filled with water 31 '82 grammes. Weight of bottle filled with liquid 30-43 grammes. Hence Weight of water filling bottle 24-97 grammes. Weight of liquid filling bottle 23-58 grammes. 23-58 Specific gravity of liquid = = -944. EXPERIMENT 23. To find the specific gravity of a solid in small fragments with the specific gravity bottle. Weigh the fragments of the solid ; let the weight be W. Fill the bottle with water and place it with the solid on the pan of the balance. Let the combined weight be W r Place the solid in the bottle and fill it up with water, taking care to get rid of all air-bubbles; let the weight be JF 2 . This weight TF 2 will be less than W because water will have been displaced from the bottle by the solid and W l - W will be the weight of the water displaced. The volume of this water is equal to that of the solid. W Hence specific gravity of solid = _ . W i W 62-63] MEASUREMENT OF SPECIFIC GRAVITIES. 131 The following observations were made. Weight of solid 5*67 grammes. Weight of solid, bottle and water 37*49 grammes. Weight of solid in bottle, bottle and water 34-62 grammes. Hence Weight of water displaced 2-87 grammes. Specific gravity of solid = -^ = 1-97. 63. Solids soluble in water. In the descriptions just given of the various methods of determining specific gravity it has been assumed that the solids used could be immersed in water. This of course is not always the case ; salt, sugar, and many other substances when placed in water dissolve. If the substance whose specific gravity is required is soluble in water some other liquid must be employed in which it will not dissolve ; thus sugar may be weighed in alcohol. The result of the observations give us the ratio of the weight of a given volume of sugar to that of an equal volume of alcohol. Let us call this s a- a . Now the specific gravity of alcohol is the ratio of the weight of a given volume of alcohol to the weight of the same volume of water, let this be a ana tne silver / T . In what proportion were gold and silver mixed in the crown ? From the experiment it follows that the mean specific gravity of the crown was 14, that of the gold *- and of the silver s -. Suppose that in each cubic centimetre of the crown there is v c.cm. of gold, there will be l-v c.cm. of silver. The weight of gold will be proportional to 77v/4 and of silver to 21(l-v)/2; the weight of a c.cm. of the whole is proportional to 14. 14 = Therefore t? = | of a cubic centimetre. Hence the crown was composed of by volume of gold and of silver. 63-64] MEASUREMENT OF SPECIFIC GRAVITIES. 133 64. The U-Tube Method. When two liquids which do not mix are poured into the two vertical limbs of a U-tube, the heights of the columns of liquids in the two tubes above the common surface are, as we have seen, different. The pressures at the common surface are the same in the two liquids ; these pressures are measured respectively by the heights of the two columns, and since the densities of the two are different the heights are different. A method of comparing the densities of two liquids which do not mix is based on this. The apparatus employed is shewn in Fig. 66. The U-tube is mounted on a stand, and scales of millimetres are fixed beside each limb of the tube. The heights of the columns can be read on these scales. EXPERIMENT 24. To compare the densities of two liquids by observations on the heights of balancing cohtmns. (i) Let ABCD, Fig. 66, be the U-tube. Let the one limb AB contain oil, the other water, and let B be the common surface of the two. Draw BC horizontal to meet the water in the other limb in C. Let AB = h, CD = h', let o> be the weight of a unit of volume of the oil, co' of that of the water. Then since B and C are points in the same fluid the water in the same horizontal line, the pressure at B is equal to that at C. Let 7T be the atmospheric pressure. Pressure at B = ir + tah. Pressure at C = 7T + /o>' measures the specific gravity of the fluid in AB. Hence in this case Specific gravity of fluid in AB = h'/h. #(ii) If the liquids mix the apparatus requires modifica- tion; one form which is then useful is shewn in Fig. 67. Two U -tubes ABC, DEF are used, one limb of each being much longer than the other. The two shorter limbs are connected to- gether by a piece of india-rubber tubing as shewn at G. A small quantity of liquid is poured into each tube, thus enclosing some air in the part CGD. Additional quantities of liquid are then poured into the tubes in turn; the air in the space CGD is gradually compressed, and finally the columns stand as in the figure. Draw CB and DE horizontal. Let AB = h, FE = h', and let w, 7i'. 64] MEASUREMENT OF SPECIFIC GRAVITIES. 135 Hence w'ti and as in (i) specific gravity of fluid in A> Another form of the apparatus suitable for two fluids is shewn in Fig. 68. EXPERIMENT 25. To compare the densities of two liquids which will mix by means of the inverted \J-tube. An inverted U-tube A BCD, Fig. 68, dips into two beakers containing the liquids. At the top of the tube there is an opening to which a short length of india- rubber tubing is attached. This can be closed by a clip or by the insertion of a piece of glass rod. Air is sucked out of the tube through this upper opening, and the liquids rise in the two limbs of the tube. The rise of the liquid is caused by the atmospheric pres- sure acting on the surface of the liquids in the beakers (see Section 68). The heights of the columns of liquid in the two tubes will be found to be different. Scales are fixed to the apparatus and by their means the heights of the columns can be measured. Measure the two heights AB, CD, reckoning from the level of the liquid in the beakers in each case, let them be h and h', and let to and a/ be the weights of unit volume of the two liquids. Let 7T be the atmospheric pressure. Then the pressure of the enclosed air at B and C is the same, and the pressures at A and D of the liquids within the tubes are equal to the pressures at the same level of the liquids in the respective beakers. Each is therefore equal to 7T. Fig. 68. 136 HYDROSTATICS. [CH. VI Hence from the column AB 7T pressure at B + wA, and from column CD 7T = pressure at C + w'ti. Therefore 9 x 76 or 1032-8 grammes weight. Again, the weight of 1 gramme in London contains 1 981 dynes or absolute c.G.s. units of force. Thus the pressure of the standard atmosphere is 1032-8 x 981 or 1013177 dynes per square centimetre. This is approximately 1-013 x 10 6 , or rather greater than one million dynes per square centimetre. Now one square inch contains 6-451 square centimetres. Thus the thrust on a square inch is 1032-8 x 6-451 grammes weight. Also 1 Ib. contains 453-6 grammes. Hence the pressure of the standard atmosphere in pounds weight per square inch is 1032-8x6-451 _., Q -46*6- - rl4 ' 69 ' Thus there is a thrust of nearly 15 pounds weight on each square inch of the Earth's surface. In England the standard height of the barometer is usually taken to be 30 inches. Since 1 inch is equal to 2 -54 centimetres, 30 inches is equal to 76-2 centimetres. Thus the standard height of the baro- meter on the metric system differs slightly from that adopted in England. 77. Water Barometer. A barometer might be made with other liquids than mercury; water, for example, might be used, but in this case the tube would be very long ; for, since mercury is 13-59 times as dense as water, the height of a water column which would balance the column of the mercury barometer would be 13-59 times its height or 13-59 x 76 centimetres. This reduces to 1032*8 centi- metres or 10-328 metres. Since the weight of 1 cubic centimetre of water is 1 gramme weight, the height of the water barometer in centimetres is equal to the pressure per square centimetre in grammes weight. 1 Dynamics, Section 85. 154 HYDROSTATICS. [CH. VII Again, 1 foot contains 30-48 centimetres, so that the height of the water barometer in feet is 1032-8/30-48. This comes to about 33-88 feet. Another objection to the water barometer is that the vapour pressure 1 is considerable and increases rapidly with the temperature. In consequence, therefore, the column will be depressed by a considerable amount and this amount will vary with the temperature ; thus the correction for the pres- sure in the closed space above the column becomes considerable and is troublesome to apply. The glycerine barometer is free from this disadvantage and is sometimes used. 78. Height of the Homogeneous Atmosphere. Since the barometer column is balanced by the weight of a column of air extending from the earth's surface upwards as far as there is air, it is possible, if the density of the air be known, to calculate the height of this column. But the density of the air decreases, as we ascend, according to a complex law, and a limit to the height of the atmosphere cannot thus be found. We may, however, calculate how high the air would be if it were homogeneous throughout and of sufficient height to pro- duce the pressure actually observed. Now this pressure is equal to the weight of 1032'8 grammes per square centimetre, and the weight of 1 cubic centimetre of dry air at freezing point and standard pressure is '001293 grammes weight. The air column therefore must be 1032-8/-001293 or about 7-988 x 10 5 centimetres. But 10 5 centimetres is 1 kilometre. Hence the height of the homogeneous atmosphere is 7*98 kilometres. Again, 1 mile is 1 -609 kilometres. Thus the height of the homogeneous atmosphere is 7-98/1-609 or about 4-97 miles. We may say then that the pressure on the Earth's surface is about the same as it would be if the earth were surrounded 1 See Glazebrook, Heat, Section 116. 77-79] THE PRESSURE OF THE ATMOSPHERE. 155 by an ocean of air, 5 miles in depth and of the same density throughout as the air is at the Earth's surface. *79. Measurement of heights by the Barometer. Observations with the air-pump have shewn (Section 69) that the reading of the barometer depends on the pressure of the air, and we have seen that air like other fluids has weight. Now the pressure at any point of a body immersed in a heavy fluid depends on the depth to which that body is immersed ; if the depth be reduced by raising the body, the pressure is reduced also. Thus consider a flexible bag tied on to the end of a glass tube as shewn in Fig. 76. Fill the bag with mercury or some other fluid and immerse it in a vessel of water, keeping the upper end of the tube above the surface of the water. The bag is squeezed by the water pressure and the mercury rises in the glass tube, becoming higher as the bag is depressed. Or again, instead of enclosing the mer- cury in the flexible skin, place it in a beaker and insert in the mercury one end of a long glass tube open at both ends, then immerse the beaker and tube in a vessel of water, keeping the lower end of the tube under the mercury and the upper end above the surface of the water. The pressure of the water on the surface of the mercury drives it up the tube. A column of mercury is supported by the water pressure in just the same way as the barometer column is supported by the air. Clearly also, if the depth of the beaker be altered, there will be a relation between the alteration of depth and the change of height of the column. Alterations in depth could be measured by observing the change in height of the column. When the liquid is water the alteration of depth is given by multiplying the change in the height of the mercury column by the specific gravity of mercury. The same principles apply to the atmosphere. If the rise of the mercury in a barometer tube really be due to the weight Fig. 76. 156 HYDROSTATICS. [CH. VII of the air above the mercury in the cistern, then when the barometer is raised by carrying it up a mountain, the column will fall and there will be a relation between the amount of the fall and the height of the mountain. This was pointed out by Pascal, and his prediction that the column would fall was verified by Clermont, who in 1648 made use of the method for the first time, to measure the height of the Puy-de-D6me. In employing the method to determine the difference in level of two stations the barometer is read at the two stations. From the difference in readings the weight of the column of air between the two can be calculated ; hence, if the density of the air be known, the height of the column can be found, and this height is the difference of level required. It remains therefore to calculate the relation between the weight of the column and its height. Now the density of the air depends on its temperature and pressure. When the pressure is p centimetres of mercury and the temperature t Centigrade, the density, p, is given in terms of p , the density at C. and 76 cm., by the formula 1 273 p p ~ P X W^+t X 76' but p and t both vary as the mountain is ascended and the calculation becomes complex. We may, however, assume, if the difference in level is not very great, that the column of air considered is of the same weight as it would be if the air were throughout at a uniform pressure and temperature equal to the mean of those observed at the two stations. Observe then the pressure and temperature at each of the two stations and calculate from the above formula the value of the density corresponding to their mean, assuming the density at C. and 760 mm. pressure to be -001293 grammes per c.cm. Let H be the required difference in level, 7^ and h 2 the two barometer readings corrected in the manner described in Section 72, then the column of air balances a mercury column of height h z h^ Thus if p be the density of the air pH= 13-59 x^-/^). 1 See Sections 80, 83 ; also Glazebrook, Heat, Section 102. 80] THE PRESSURE OF THE ATMOSPHERE. 157 13-59 Hence H - (A - h{) x . P Example. The barometer at the lower station reads 751-9 mm. and at the upper 633-7 mm., while the temperatures of the air at the two are 13 C. and 7 C. respectively. Find the difference in level. The difference in pressure is 118-2 mm., the mean pressure is 692-8mm. and the mean temperature 10 C. The mass of a column of air 1 square centimetre in section between the stations is 13*59 x 11-82 grammes. The density of air at 692-8 mm. pressure and 10 C. is 001293 x 692-8 x 273 760 x 283 Hence the difference in level is 13-59x11 -82x760x283 001293x692-8x273 C or about 1413 metres. In obtaining the result no allowance has been made for the aqueous vapour in the air. In consequence of its presence the density will be rather greater than the value used above. The difference in level, therefore, should be less. We may obtain a more accurate formula thus : Suppose that at the lower station the height of the barometer is h and that, in going to a height of z metres, it falls to a height kh , k being a proper fraction. Suppose further that the temperature is uniform and that z is so small that we may treat the density of the air as constant throughout each stratum of thickness z, though it changes as we pass from one stratum to the next. On rising through a second distance z the barometer will fall by the same fractional amount as previously, for the fall is proportional to the average density of the stratum through which the barometer is being carried; and this average density in the second stratum bears the same relation to the pressure at its under side or kh as the average density in the first stratum does to the pressure 7t . Hence, at the top of the second stratum, the height of the barometer is On rising through a third stratum the pressure falls to k s h and so on in succession. Thus, for a series of heights in arithmetical progression, the barometer readings form a series in geometrical progression with the common ratio k. Now, let H be the total height through which the barometer is raised, and let the distance H be divided into ?i-layers, each of thickness z, throughout each of which we may treat the density as constant. 158 HYDROSTATICS. [CH. VII Then H=nz, while, if h be the barometer reading at the top, h=k n h Q . Therefore fcn=r r- h Hence, taking logarithms, n log fc = log ( r ) =log h - log fc . Vox TJ- But 7i= . z Hence H =~- fc (log A - log h ). Again, if p be the average density of the air in the first layer of -thickness z at the lower station and a- the density of mercury, then g, and let this tap be closed. Release the clamp F; mercury will run down the tube BC carrying the air before it and will completely till the tube. Close the clamp F. The atmospheric pressure will sustain a column of mercury equal in height to that of the barometer in the tube BC. Above this column there will be a vacuum. Now open D. Air from the receiver expands Fig.92. into this vacuum, the mercury column falls somewhat and the pressure in the receiver is reduced. Close D and open F; the stream of mercury down BC carries the air again out of that tube and, on closing F, the mercury stands at the barometric height. Again open D; more air enters BC from the receiver and this, on closing D and opening F, can again be removed. This description may serve to explain the action of the pump, but, in practice, the tap D is unnecessary and G. HYD. 13 194 HYDROSTATICS. [CH. VIII the clamp F is only used to stop the flow when the receiver is sufficiently exhausted. It is found that if the receiver is permanently connected to the vertical tube BC and the clamp F is opened, the process described goes on in a partial manner continuously. The mercury column descending from E breaks into drops at B\ as the pressure in BC is reduced, through the air being carried down by the mercury, the air from the receiver expands into the tube and is carried down between the drops. There is no need therefore alternately to open and close the tap D; hence it may be removed and the risk of leak which it gives rise to may thus be avoided. The process continues until the degree of exhaustion in the receiver is comparable with that of a Torricellian vacuum. As the mercury from the reservoir E flows away, it is replaced by mercury overflowing from 6?, which is caught in a suitable vessel at one side. In the more modern forms of Sprengel's pump, which are used for exhausting the bulbs of incandescent lamps and other work needing a very high exhaustion, the tube ABC is bent in the manner shewn in Fig. 93. The efficiency of the pump is thereby increased. 103-104] HYDROSTATIC MACHINES. 195 G-eissler's Air-pump. This pump in its simplest form consists of two glass reservoirs of considerable capacity. One of these A, Fig. 94, is fixed, the other, , can be raised or lowered at will; the two are con- nected by a piece of stout india- rubber tubing. The moveable reservoir B is open to the atmosphere, while A can be put into communication with the atmosphere through a tube which enters at its top ; the tube is closed by a tap C. A second tube, con- nected to A, communicates with the receiver to be exhausted through a pipe, which can be closed by a second tap D. The reservoir B when in its lowest position is filled with mer- cury. Close the tap D and open C. Raise B slowly until it is at a slightly higher level than A. The mercury passes from B into A, driving the air out until A is filled with mercury up to the level of the tap (7, all the air being expelled. Close the tap C and Fig. 94. lower B; the mercury passes back into B and stands at the barometric height in the tube between the two reservoirs. Now open the tap D. The air from the receiver expands into A and the pressure in the receiver falls. Close D and raise the reservoir B, the air which now fills A is compressed by the mercury as it rises into A ; when the mercury is nearly at the same level in the two, indicating that the air above the mercury in A is at atmospheric pressure, open the tap C and continue to raise B until the mercury in A rises again to C. Then close C and repeat the process. In this way the air is drawn from the receiver and a high degree of exhaustion is attained. 132 196 HYDROSTATICS. [CH. VIII In the form of pump just described the tap D is a source of difficulty. It is nearly impossible to make the fit so good that there shall be no leak when the tap is turned, and thus the vacuum which can be obtained is impaired. #105. Topler's Air-pump. In Topler and Hagen's modification, Fig. 95, the taps are done away with. Com- munication is made with the receiver through a long in- verted U-tube ADE which enters at the bottom of the reservoir A. The height AD is greater than that of the barometer. To reduce the risk of fracturing the pump by a sudden inrush of air from the receiver, when the mercury falls below the level of the tube DA, a side tube connects C the top of the receiver to a point just above the junction of the reservoir A and the in- verted U-tube. From the top of the reservoir A a tube CFG bent twice at right angles runs downwards and ends, either under mercury in a small vessel, or in a siphon bend as shewn in the figure. The distance of the bottom of this bend below C is greater than the barometric height. As the vessel B is raised the mercury in its ascent closes the mouth of the tube ADE and thus shuts off the air in the receiver from that in the vessel A . The air in A is then compressed, and finally it is expelled through CFG, by raising B until the mercury begins to flow down this tube. The vessel B is then lowered. The Fig. 95. 104-106] HYDROSTATIC MACHINES. 197 atmospheric pressure forces mercury up CF, but since CF is greater than the barometric height the mercury does not reach C and the vessel A is completely shut off from the atmosphere. A vacuum is thus formed in the reservoir A until the level of the mercury it contains falls below the point where the tube from the receiver enters, when this is the case, air enters from the receiver. The process is then repeated. It is necessary that DA should be greater than the barometric height for, as the vessel B is raised, the air- pressure on the surface of the mercury in B forces mercury up the tube DA and, when the exhaustion is considerable, this mercury will rise to very nearly the barometric height above the level of C. In using the pump the vessel B can be raised and lowered by hand, in general however some mechanism for doing this is attached to the stand of the pump. 1O6. The Condenser or Compressing Syringe. This is an air-pump arranged to compress air into a vessel. A piston P with a valve E opening inwards works in a barrel AJ3, Fig. 96 (a). The vessel into which the air is to be Fig. 96 (a). compressed communicates with the end of the barrel through a tube. This tube is closed by a valve F opening outwards from the barrel. There is also usually a stopcock in the tube so that communication between the vessel and the air may be cut off at will. Let the piston be at the end of the barrel near the valve F. On withdrawing it the pressure in the barrel is reduced ; the atmospheric pressure opens the valve E and the barrel is filled with air at atmospheric pressure. The piston is then depressed, the valve E is closed and F is opened ; hence all the air from the barrel is forced into the vessel ; on again withdrawing the piston the process is repeated. At each 198 HYDROSTATICS. [CH. VIII downstroke a barrelful of air at atmospheric pressure is forced into the vessel. A form of condensing pump used for the tyres of bicycle wheels is shewn in Fig. 96 (b). The piston rod DC is hollow Fig. 96(6). and contains a valve at D. At G there is communication with the tyre. When the piston is pressed up against the end of the cylinder at B air can enter the cylinder behind the piston through the aperture B. The cylinder is then pushed forward and the piston moves past the aperture JS, thus cutting off the cylinder from the atmosphere ; as the piston is moved towards A the air in AD is compressed and forced through the valve D into the tyre. We may find the density of the air, after any number of strokes, thus. Let V be the volume of the vessel, v that of the barrel. At each stroke a volume v at atmospheric pressure enters. Thus after n strokes the air in the receiver would, at atmospheric pressure, occupy a volume V + nv, and if p be the density of air at atmospheric pressure, the mass of air in the vessel is p(V+nv). But its actual volume is V and, if p n be its density, its mass is p n V. Thus p n V = p(V+nv\ Hence p n = p ( 1 + n n = p( Again, by Boyle's Law the pressure of air is proportional to its density. Thus if p n be the pressure after n strokes, 7T the initial pressure, Examples. (1) If the volume of the vessel is ten times that of the barrel, how many strokes are required to double the pressure ? We are to have p n = 2v, also v/F=10. 106-107] HYDROSTATIC MACHINES. 199 Thus 2=1+ io~ Hence the pressure is doubled after 10 strokes. (2) When the piston of a condenser is pushed as far down as it will go a volume v' is left beneath it. The valves will open when there is a difference of pressure p between the two sides. Shew that the pressure of the air inside the receiver can never be greater than (IT - p) v/v' - p, where IT is the atmospheric pressure. When the piston is drawn up there is a volume v of air below in the barrel; the pressure of this air is ir-p 1 . As the piston is pushed down this air is compressed ; let us suppose that its pressure at its greatest is just insufficient to open the lower valve, then when the stroke is complete its volume is v' ; its pressure therefore is (ir-p)vlv'. Since the valve just does not open the pressure inside is less than this by p. Hence the greatest pressure inside is 107. The Diving-bell. This is an apparatus for enabling a man to descend to a considerable depth under water, thus a sunken vessel could be examined, or the foundation of a pier laid or repaired, or work of other kinds carried out. Take a beaker and immerse it mouth downwards in water: as the beaker is depressed the air it contains is compressed, but the water does not rise so as to fill the beaker completely, there is always air in the upper part of the beaker, a fly or small animal might live there for some time. The beaker is a diving-bell in miniature. The bell, Fig. 97, consists of a large bell-shaped or cylin- drical vessel closed at the top but open underneath. This can be lowered mouth downwards into the water, its weight is greater than the weight of water which would fill it, hence it sinks in the water. As it sinks the air it contains, like that in the beaker, is compressed, and the water rises in the bell but never fills it. The bell is so constructed that a person can stand within it and thus be lowered into the water without depriving him of air to breathe. The bell is usually filled with air by two tubes leading to the atmosphere above. Through 1 If, as has been assumed in the text above, the valves open with no difference of pressure it would be IT ; it is less than this by the pressure required to open the valve. 200 HYDROSTATICS. [CH. VIII one of these fresh air is forced into the bell, through the other foul air is withdrawn. The pressure in the bell depends on the depth to which it is sunk. The difficulty of working under very great pressure limits of course the depth at which it can be used. Fig. 97. The tension on the chain supporting the bell is equal to the difference between the weight of the bell and the weight of water it displaces : as the bell sinks the water rises inside, the weight of water displaced therefore is reduced and the tension increased. Examples. (1) A conical wine-glass 4 inches in height is lowered mouth downwards into ivater until the level of the water inside is 34 feet below the surface; the height of the water-barometer being 34 feet, what is the height of that part of the cone which is occupied by air ? The pressure in the wine-glass is doubled, thus the volume of air is half what it was. But the volumes of two cones of the same angle are proportional to the cubes of their heights; hence, if z is the height of the conical volume of air in the bell, we have 107] HYDROSTATIC MACHINES. 201 Therefore z = - = 3-175. Hence the height required is 3-175 inches. (2) A cylindrical diving-bell Qfeet in height and 5 feet in diameter is lowered till its top is 45 feet below the surface. What volume of air at atmospheric pressure that due to 34 feet of water must be pumped in to Jill the bell completely ? JTo solve this it is simplest to suppose the bell to be full and then to what will be the volume at atmospheric pressure of the air actually in the bell.] Let V be the volume of the bell. When the bell is full the level of the water inside is 45 + 6 or 51 feet below the surface. The pressure therefore is that due to a head of (34 + 51) feet of water or 2^ atmospheres. If the air then were at at- mospheric pressure its volume would be f of F. The volume of air added is at atmospheric pressure f of F. Now V- j x 25 x 6 c. feet. Hence the volume required is 9 x 25 x 7r/4 or 177-9 c. feet. (3) A piece of ivoodjloats half immersed at the top of the water, Iww much of it will be immersed when floated in water within the bell mentioned in Example 2? Let 2v be the volume of the wood in cubic centimetres, a the density of air a't atmospheric pressure referred to water. Let v + x be the volume of the wood in the air in the bell, v - x then will be the volume in the water. The density of the air in the bell is f Jl ft. per sec. */5 sees. 48. 14 ft. per sec. 24 ft. per sec. per sec. 49. 159 ft. approx. 50. 3888000 miles per hr. per hr. CHAPTER VII. (Page 142.) 1. (i) 100,000 units of momentum ; (ii)20f; (iii) 2,682,240; (iv) 990,000. 2. momentum of second mass = momentum of first mass x 231-5. 3. 27 cm. per sec. per sec. 1,666,666| dynes. 4. impressed force on second mass = impressed force on first mass x -2572. 5. 469-57T x 10 J1 C.G.S. units of momentum. 6. 45-4 cm. 7. impulse = 30 J5 x m, where ?M mass of cricket-ball ; average force = 1500 >/5 x m poundals. 8. 2215 cm. per sec. 9. 256 Ibs. 8192 poundals. H. 62J poundals. 12. 363|f cm. per sec. 181|f cm. 21822f cm. 13. 24 ft. per sec. 14. 10f pounds. 15. 21 T 9 T pounds. 16. 60,500 ft. = !!& miles. 17. - ITT ft. per sec. per sec. ^^ of the weight. 20. forces are equal. 21. 200 poundals. 22. 2133| poundals. 23. a dyne. 25. 17| poundals. 28$ ft. 26. 5 ^30 mm. =27-4 mm. 27. = 59-6ft. per sec. 28. 16 ft. per sec. 29. 255-7 cm. per sec. 30. 38080 poundals. 31. 2| Ib. weight. 10 ft. per sec. per sec. iv ANSWERS TO EXAMPLES. 32. 38'8 ft. per sec. per sec. 33. 14-85 cm. per sec. per sec. 34. /2 yds. 2. 14-06 ft. 4. (1) 108 ft. (2) 3 sees. 5. (1) 1000 ft. per sec. (2) 144 ft. 6. (1) 13 sees. (2) 1040 ft. 7. t5 *~~ miles. 9. 100 ft. per sec. 10. 70ft. per sec. 12. (1) 22,050ft. (2) 553-48 ft. per sec. 14. CHAPTER X. (Page 221.) 1. of that of the impinging ball. 2. 2 to 5. 4. 15 ft. per sec. in opposite directions. 5. Their masses are equal and e is unity. 6. '0878 ft. 7. tan-if- 8. v = \u\ /3 = 90. v' = ^u; /y = 9. cosa = ^-. 11. -^ .. CHAPTER XI. (Page 236.) 2. 1'65 revolutions per second. 3. lengthened by -08 inches. 5. (1) In first position pressure on support = 800 ^/f grins, wt., tension of string = 200 ^/f ,, weight of bob (1000 grms. wt.). (2) In second position pressure on support 250 >Jl5 grins, wt. (3) In third position pressure on support (in ) kilos, wt. \ 2 j/15/ VI ANSWERS TO EXAMPLES. 7. Intensity of gravity at B = intensity at A x 1-00023. 14. '804 ft.-pounds. 15. 24 ft. per sec. 16. 80854 ergs. 23. 40 x/6 ft- per sec. 24. 9 to 8. 25. 1. 27. 28 J2 ft. per sec. 36| ft. 28. I of the way across. 29. 3 times as far as it fell in the first case. 30. 20^ ft. After sec. and 1J sees. 31. 31-6 ft. per sec. 33. (1) 13i ft. per sec. kinetic energy of projectile (2) kinetic energy of gun = |^ = 34. (1) 4f ft. per sec. per sec. (2) 109f poundals. 35. (1) 8 miles per hour. (2) 933,333^ ft.-poundals. 36. (1) if ft- per sec. per sec. (2) 33 poundals. MISCELLANEOUS EXAMPLES. (Page 238.) 3. Increased 12,960 times. 4. - sees. */9 6. lHlb.-wt. 7. . 8. 45 to the direction in which the ball is coming. 9. 400 ft. 10. 7 miles per hour. 12. 1108 ft. per sec. 13. 5 ^13 miles per hour. 14. Increased 3600 times. 15. 6| miles. 16. ^VV ft- per sec. per sec. 17. 15 *J miles per hour. 18. 5 ft. per sec. per sec. 19. | sees. 21. ^ + sin" 1 ^ with the motion of the train, j 22. 4-65 sees. 37. (1) 448 ft.-poundals. (2) 18 ft. 40. 195 T 5 F ft.-lb. 41. 22500 ft.-lb. 42. (1) 22400000 ft.-lb. (2) 678-8 H.P. 43. T \. 44. 9856 ft.-lb. 47. |f ft. per sec. 48. 3^ pounds. 49. (1) 19200 F.P.S. units of momentum. (2) 1648571f ft.-poundals. (3) 1648571f ft.-poundals. 50. (1) -& f t. per sec. per sec. (2) 147 T 3 T miles per hour. 51. 8 /s/6 ft. per sec. 52. 11 ft.-poundals. 54. 38554687-5 ft.-lb. ANSWERS TO EXAMPLES IN STATICS. CHAPTER I. (Page 31.) 1. 5 Vl3 Ib.-wt. 2. 13P. 4. 10 Ib. and 26 Ib. 5. 5 Ib. and 13 Ib, 13. 5 \/3 Ib. each. 14. Place the forces parallel to the sides of a right-angled triangle whose sides are 3, 4 and 5. 16. Weight of bob = 1 kilogramme. Tension of horizontal string = '258 kilogramme. pendulum =1-033 17. P (2 + \/3). 21. -^ times the side of the triangle. 26. A force equal to the given forces bisecting the angle between them. 29. (i) P 2 + <222.P.Q. (ii) 6-48 Ib.-wt. 32. 5 \/7 Ib.-wt. 33. 699-5 dynes. 34. f Ib.-wt. and Ib.-wt. 38. 5 V2 to the North-west, 5 to the South. 41. sin- 1 ( - H) with the force 5. sin- 1 T \ with the force 12. 44. 27-7 oz. 46. */2 . P. 47. \/2~. P. 48. 39 Ib. 49. 6 Ib. and 8 Ib.-wt. CHAPTER II. (Page 57.) 2. 21 ft. 3 in. 3. 3 Ib.-wt. 5. 5 Ib. 6. 17 Ib. 7. 5 Ib.-wt. 8. 35 Ib. and 40 Ib. 9. AC=2 in. = 56 in. 10. |f and f|. Vlll ANSWERS TO EXAMPLES. 12. 100 Ib. 13. -131 ft. from the axle. 15. 1J ft. from the man. 16. 2 x/2 . P at an angle of 45 with the force 4P. 17. In the line joining the middle points of the two sides, and at a distance of 5f in. from the side which supports the two 5 Ib.-wt. 18. flb.-wt. CHAPTER IV. (Page 83.) 1. ^; ~^- 3. 50 Ib. 4 ft. from the thicker end. 4. Pressure on shorter end = f of the man's weight. ,, longer ,, =f ,, 6. P, $AC horn C. 10. iJFtana. 12. The radius to P bisects the line AB. 13. 20 Ib. 15. The end. 18. 3 to 1. CHAPTER V. (Page 110.) 1. 350 Ib. 3. From any point on the rim within a distance of ^ of the whole circumference from any of the 4 legs. 4. At a distance of ^ of the side of the square from the centre of the square ; the straight line joining the centre of gravity with the centre of the square is parallel to a side of the square. 5. At a point on the diameter drawn from the middle one of the 3 particles mentioned and at a distance of -Q5d from that point where d is the length of the diameter. 6. In the perpendicular drawn from the angle which is adjacent to the 2 bisected sides, and at a distance j- below the c. G. of the whole 6\/3 triangle. 7. In the diameter of the rectangle parallel to the side a and at a distance of ^a from G. 8. At the point G in the straight line A C where AG = ^.AC. 9. In the diameter drawn from that angular point on which no weight is placed and at a distance of ^ of the diameter from that point. 11. In the straight line drawn parallel to BC from the middle point of AB and at a distance of f f of the side of the square from this point. 12.. Half the weight of the whole table. ANSWERS TO EXAMPLES. IX 13. In the diameter drawn from the point at which the two circles touch r -2 one another and at a distance of - from the centre of the larger circle. 14. In the diagonal drawn from the angle enclosed between the two bisected sides and at a distance of ff of the diagonal from this point. 18. 50 Ib. 6 ft. from thinner end. 19. -^- W. \W. 20. - 7=lb. 2 cwt., 3 cwt., 0. 1 + W3 22. In the straight line joining the two centres, and at a distance of 1 ft. 8^ in. from the centre of the hole. 23. The centre of the circular hole must be 16 in. from that of the disc. 24. H in - 25. \\ in. from the thicker end. 27. On the line joining the angle removed to the c. of G. of the whole and \ of this distance from the angle removed. 28. rV of total length from the heavier end. 29. 5 ft. from the end. 30. In the diagonal drawn from the angle enclosed between the two bisected sides and at a distance of ^- of the side of the square from this point. CHAPTER VI. (Page 160.) 1. Left arm = right arm x 1-018. 2. The fulcrum is at C, that is, 13| ft. from A. 3. f of distance of centre of gravity from the end at which the weight is hung. 12. True weight = 20-494 Ib. 13. 7i pence. 15. \W. 16. J3W. 17. 12ft. 18. e ' 08 W (cos a sin a tan 0) depending on direction of force. Vertically up. 19 ' cos^* 1 ' Nothin g' 20. ^ton. 21. 20 Ib. 22. j-ton. 23. ^. 24. TF, w. 25. 1 : -T= . When the force acts parallel to the plane. X ANSWERS TO EXAMPLES. 26. 51b.-wt. 27. 51b. 28. 3* lb. 29. (1) 1 : 16 ; (2) 9 in. 30. 7 Ib. 31. 16 Ib. 32. 9 lb. 33. 6jlb. 34. 4^1b. 35. From a point distant f- of the whole length of the bar from the end to which the string attached to the fixed pulley is connected. W_ radius of wheel 38 ' p- radius of axle ' 39 ' 40. (1) 45-72 grams; (2) 1-002 to 1. 41. 21x/21b. 21 lb. 42. 21b. 43. 16ft. 44. (1)^. (2)4ilb. 45. (l)- (2)51b. 46. 7 pulleys. 14 lb. 47. dftr lb. CHAPTEE VII. (Page 174.) 1. . / : ^ . 3. tan" 1 f, tan' 1 f . 6. 2 sin a. \/ gsma-^gcosa 9 \/2 lb. ; tan -1 -j= to the horizon ; - times the weight. 11. With the largest face on the plane and the length up the plane. No. 12 . 30 : 1, 13 . -L ; ^n, 14. 2 cwt. 17. \ cwt. ; 0. ANSWERS TO EXAMPLES IN HYDROSTATICS. CHAPTER I. (Page 18.) 5. Density, 112 Ib. per cubic foot. Specific Gravity, 1-792. 6. 112-58 grms. 7. 76 '39 Ib. per c. ft. 8. 1-1937 grms. per c. cm. 9. 1-2237 grms. per c. cm. 10. 18*476 grms. per c. cm. 11. 3437 grains per c. inch. Density of sphere =7 . 803 . 13 44 . 68 lb . Density oi cylinder 14. 199,200 tons. 15. 53 litres. 16. 72-4 litres. 17. (i) "975. (ii) 1-027. 18. 7 "09. 19. '00459 sq. cm. 20. 2-82 c. inches. Weight of glycerine zl. TIT T. A. f [ = 7lo. I 2i2t. * "o c. cm. Weight of water CHAPTER III. (Page 76.) 1. Pressure in water at depth of (1) 25 cm. is 1025 grms.-\vt. per sq. cm. (2) 1 metre is 1100 ,, (3) 1 mile is 161,900 (4) 5 kilometres is 501,000 ,, Pressure in mercury at depth of (1) 1 cm. is 1013-6 grms.-wt. per sq. cm. (2) 1 metre is 2360 (3) 25 metres is 35,000 (4) 1 kilometre is 1,361,000 2. 44-91 ft. 3. 40-66 inches. 4. '795 p, 1-28 p, p, where p is the pressure in the water. Xll ANSWERS TO EXAMPLES. 5. Head of water = 33-4 ft. Head of mercury = 29 -48 inches. 6. 130-2 lb.-wt. per sq. inch. 7. Heads of water, 10 metres, 829-4 inches, 1019-4 cm. Heads of mercury, 73-53 cm., 60 '98 inches, 74 '95 cm. 8. 68,400 poundals. 9. (II + 160-3) lb.-wt. per sq. in. where 11 = pressure of the atmosphere in lb.-wt. per sq. inch. 10. 1584-5 lb.-wt. per sq. ft. 11. 39270 lb.-wt. 12. Pressure in liquid =178-25 lb.-wt. per sq. inch. Force exerted by piston = 14, 000 lb.-wt. 13. (1) 25,920 lb.-wt. per sq. ft. (2) 233,280 lb.-wt. per sq. yd. 14. 13-74 lb.-wt. per sq. inch. 15. (1) 2033-6 grms.-wt. per sq. cm. (2) 6168 grms.-wt. per sq. cm. 16. 46-08 ft. 17. 1000 kilogrammes weight. 18. 73,500 c. cm. 19. 2000 lb.-wt. per sq. foot. 20. 41-3 lb.-wt. per sq. ft. 21. 5-63 lb.-wt. per sq. inch. 22. 11-41 inches. 24. 69'12 ft. 26. 3373 lb.-wt. 27. 11-76 lb.-wt. per sq. inch. 28. 45-5 grms.-wt. per sq. cm. 29. '192 sq. ft. 30. (i) Upward thrust on top of barrel = 208 J lb.-wt. (ii) volume of water =144 cubic inches. CHAPTER IV. (Page 93.) 2. (a) 4-948 lb.-wt. (b) 61,500 tons-wt. thrust on base with vertex upwards _ 3 vertical thrust on curved surface, vertex downwards ^5 * 5. 1-0167 ft. thrust on base of large cistern _ **' ^ thrust on base of small cistern ~ thrust on vertical sides of large cistern _ * ' thrust on vertical sides of small cistern ~~ 7. 21-38 lb.-wt. 8. 283^ lb.-wt. 9. A force equal to the weight of 562J Ib. applied to the centre of the lower edge of the face. 10. 1875 lb.-wt. 11. 50,625 lb.-wt. thrust on side 12 =A- 13. 351,5624 lb,-wt. thrust on bottom 14. 1,001,9531- lb.-wt. 15. 25-323 ft. 16. 2273J lb.-wt. 17. 46,256 lb.-wt. 20. = Force on small plate ANSWERS TO EXAMPLES. Xlll CHAPTER V. (Page 114.) 1. 25-9 cubic inches. 2. 871-1 Ib. 3. Kesultant thrust of water = 41 kilogrammes- wt. Acceleration = 3106^ cm. per sec. per sec. 4. (1) 100 grms.-wt. (2) 28-66 grms.-wt. (3) 925 grms.-wt. 5. "57 of its volume. 6. Volume of cork 15-8 times the volume of the iron. 7. 22-44 grms. 8. 6-257 grms. Q. 19,000 c. ft. 10. 65-8 c. inches. H. 45 c. cm. 12. 0-97. 13. (1) -5454 Ib. (2) -3939 Ib. (3) -425 Ib. 14. 3-07 grms. 15. Wt. of most dense body 9 times the wt. of least dense body. 16. Mean spec, gravity =0-9987. Volume= 3875-72 c. inches. 17. -55 of the weight. 18. '537 of the volume of the iron is in mercury, and -463 of the vol. of the iron is in water. 19. Spec, gravity of wood = -533. Spec, gravity of cork=-25. Spec. gravity of ice = -919. Spec, gravity of oak=-75. 20. '726 inches. 21. Volume of iron = 36-4 c. cm. Density of iron = 7*56 grms. per c. cm. 22. "48 cubic inches. 23. *45 inches. 24. Sp. gr. = -5. 25. 2 -34 cm. in water. 2-66 cm. in mercury. 27. 93-75 cm. 28. (1) Pressure required = 5 Ib.-wt. (2) Wt. of metal = 6 Ib. 30. (1) Floats in ordinary water with -0064 inches above the surface. (2) sea-water -108 31. 51b.-wt. 33. 20 c. cm. 36. 89 '8 ft. 37. Mass of iron = 1-66 grms. Mass of wax = 34 -34 grms. 38. 0-93. 39. 277-0 c. inches. CHAPTER VI. (Page 137.) 1. (1) 2-637. (2) 11-37. (3) 2-68. (4) 17-57. 2. (1) '240. (2) -8535. (3) -530. (4) -601. 3. (1) 1-0714. (2) 1-200. (3) 1-276. (4) 1-305. 4. Spec, gravity =3 -57. Volume = 7-69 c. cm. 5. 4-04. 6. 1-069. 7. 43-79 grms. 8. 2-476. 9. (1) 1-064. (2) 0-920. (3) 1-032. 10. 2-077. 11. 2-768. 12. Mass of silver = -997 times the mass of the gold. 13. 1- 14. *309 of the volume of the mixture is alcohol. xiv ANSWERS TO EXAMPLES. 15. 82-9 cm. 16. 6-36 inches. 18. 12-288. 19. Specific gravity = 3. Volume = 1 c. cm. 20. '923. 22. 8-88. 23. 1-5. 24. '80. 25. '787. 26. 13-92 grms. 27. 1'44. 28. l'48c. cm. 30. 2-23. 31. Mercury falls 1-99 cm. in one leg, and rises 1-99 cm. in the other. 65-9 cm. of oil are required. 32. 27'2 inches. CHAPTER VII. (Page 166.) 1. (1) 3800 c. cm. (2) 2992 c. cm. (3) 3839 c. cm. (4) 47'1 c. cm. 2. 39-1 litres. 3. 18-12 kilogrammes. 4. 340-5 kilogrammes. 5. 144-16 grms.- wt. per square cm. 6. Add 13-36 mm. 7. 4-05 mm. 8. 75-394 grms. 9. Pressure of hydrogen = 14-4 times the pressure of the air. 10. (1) 749 -54 mm. (2) 27-445 inches. (3) 21-176 inches. 11. (1) 783 c. cm. (2) 306 '1 c. cm. (3) 414-? cubic inches. 12. 26-48 inches. 13. 2-401 litres. 14. 24-73 inches of mercury. 15. 8752 yards. 16. Increases by 10 cubic feet. 17. 1-782 inches in diameter. 18. '001441 cubic inches. 19. Diminished to -^th of its former value. 20. 2-85 cubic ft. 21. fth of the air escapes, or, in other words, the escaped air occupies a volume of 85 cubic inches. 22. 601b.-wt. 23. 8932 yards. 24. '2533 grms. 27. Weight on mountain fth weight at sea-level. 29. 30 inches of mercury, 30. Pressure of the inside air is less than the pressure of the outside air by the pressure of the column of water left in the bottle. 32. 8408 metres. 33. 6181 ft. 35. 30 '5 inches CHAPTER VIII. (Page 204.) 4. (1) 30 inches. (2) 513-2 inches. 6. 64-41b.-wt. 7. 4 strokes. 8. 8-1 Ib.-wt. 9. 47,142f Ib.-wt. 10. 37,500 ft.-lb. per minute. 11. (1) Force required = 13-635 Ib.-wt. (2) Work done = 24, 543 ft.-lb. (3) Wt. of water raised = 981- 72 Ib. 12. 10-2 strokes. ANSWERS TO EXAMPLES. XV 13. 981-75 Ib.-wt. during the back stroke and 3927 Ib.-wt. during the forward stroke. 14. 15,590 ft. -Ib. 15. 304-5 inches. 16. (1) x original pressure. (2) 1 Q4 ' 8 5?6 x original pressure. 14,348,907 X0ngmal 17. th of a stroke. 18. 16 strokes. 19. When the piston in its upward stroke has traversed T ^ of the barrel. 20. ^i^th f the atmospheric density. 21. Volume of receiver = 4 times that of the barrel. 22. 167 strokes. 24. 5 strokes. 26. WUT f tlie atmospheric density. 27. 10 strokes. 28. '2727 cubic feet. 30. 2L& ft. 31. 27-16 ft. 32. Depth of top of bell=3 inches. 33. (1) 1-155 F. (2) -311 F. Where F= volume of the bell. 34. Half way up the cylinder. 35. 300 cubic feet. 36. Until its depth be doubled (neglecting the length of the bell). 37. Top of the bell is o (ti -h)- - ft. below the surface. ( h ) 38. Weight raised = 1000 tons. Number of strokes required = 12,000. 39. ITS sq. feet. 40. ton weight. 41. Weight raised = 2800 Ib. Number of strokes = 240. 42. '426 inches. 43. Load required = 14,137-2 Ib. Head of water = 1152 feet. 44. 191 Ib. per sq. inch. 45. -5 grms. per c. cm. 46. 23^ c. cm. 47. 102-3 Ib.-wt. 48. 60 Ib. per sq. inch. 49. 150 Ib. per sq. inch. 50. (1) Volume raised = 19,161 c. feet. (2) Eeduction of volume^ 2903 c. ft. INDEX TO DYNAMICS. The references are to the pages. Acceleration, unit, 58 ; uniform, 59; space traversed, 62 ; velocity and space traversed, 65; formulae connected with, 66, of a falling body, 94, 101; measurement of, and the Hodograph, 224 Accelerations, composition and resolution of, 71 Action of Force, 125 Action, 148, 152; during impact, 218 Amplitude of vibration, 202 Angular velocity, 49 Area, measurement of, 4 Attraction and the third law of motion, 151 At wood's machine, 103; experi- ments with, 104, 105, 108; rate of change of momentum with, 107 Average rate of change, 22 speed, 27 Ballistic balance, Hicks', 80; ex- periments with, 82, 86 Bodies, falling, 69; problems on, 70; impact of elastic, 88; ac- celeration of falling, 94, 101 Body, motion down a plane, 164; down a curve, 165, 166 Centimetre-gramme, 177 C. G. S. system of measurement, 16 ; unit of force, 118 ; unit of work, 156 Change of momentum, 89 ; rate of, with Atwood's machine, 107 Change, of velocities, 56 ; of form of energy, 172 Circle, motion in a, 229 Circular motion, 230 Comparison, of masses, 14, 83, 121 ; of forces, 120 ; by weighing, 131 Component of displacement, 30 Composition, of displacements, 29; special cases of, 32; of accelerations, 71 ,, of velocities, 38; of forces, 134 Compression, moment of greatest, 219 Conservation, of momentum, 91, 152; of energy, 175 Conservative system, 176 Constrained motion due to gravity, 102 Curve, motion down, 165, 166 ; velocity on, 169 Density, 14 Displacement, 16; resultant and components of, 30 ,, composition of, 29; special cases of, 32 ; parallelogram of, 32 ; resolution of, 36 Dynamics, 17 INDEX. XVll Dyne, 118 ; value of, 122 Earth, shape of the, 232; conse- quences of rotation of, 230 Elastic bodies, impact of, 88 Energy, 152, 170, 179 ; and work, 153; potential, 170; kinetic, 171; change of form of, 172 ; mutual, 174; forms of, 174; conservation of, 175; in mechanics, 175; unit of, 177 ; and impact, 215 Equilibrium, 130 Erg, the, 177 Experiments, on the parallelogram Law, 46; on the measurement of mass, 78; with the ballistic balance, 82, 86; with Atwood's machine, 104, 105, 108; on falling bodies, deductions from, 109 ; on the value of g, 139141; with pendulums, 204 Expressions for work and power, 159 Falling bodies, 69 ; problems on, 70 ; acceleration of, 94, 101; deduc- tions from experiments on, 109 ; and the second law of motion, 123 Foot-pound, 177 Foot-poundal, 177 Force, 92, 152, 179; measurement of, 93, 117; methods of measur- ing, 132; impressed, 101, 107; and impulse, 110 ; Newton's defi- nition of, 112 ; c. G. s. unit of, 118 ; F.P.S. unit, 118; comparison of, 120 ; action of, 125 ; gravitational unit of, 129; representation of, 135 Forces, composition of, 134 ; paral- lelogram of, 135, 136 Forms of energy, 174 Formulae, uniform acceleration, 66 F.P.S. unit force, 118; unit work 157 Fundamental quantities in Me- chanics, 2 0, experiments on the value of, 0. M. 139 141; value of, in different latitudes, 207 Galileo, achievements of, 112 Graphical construction for work, 179, 180 representation, of rate of change, 23; of velocity, 29; of space, 50 Gravitation, law of, 126129 Gravitational unit of force, 129; of work, 157 Gravity, constrained motion due to, 102 ; work due to, 165 ; motion of a particle projected under, 194 Harmonic motion, simple, 233 Hicks' ballistic balance, 80; ex- periments with, 82, 86 Hodograph, the, 223; and measure- ment of acceleration, 224 Horse-power, 158 Hydrodynamics, 17 Hydrostatics, 17 Impact, 211 ; condition of rest after, 86; of elastic bodies, 88; oblique, 216 ; action during, 218 and energy, 215 Impressed forces, 101, 107 Impulse, 89 and force, 110 Instruments for measuring lengths, Kinematics, 17 Kinetic energy, 171 Kinetics, 17 Law of gravitation, 126129 ,, of Motion, first, 113; second, 115 Laws of Motion, Newton's, 113 Length, unit of, 2; methods of measuring, 3, 4 Mass, 12, 77; measurement of, 13, 78; unit of, 84; and quantity of matter, 84 ; relation between weight and, 124, 203 XV 111 INDEX. Masses, comparison of, 15, 83, 121 Material particle, 17 Matter, quantity of, and mass, 84 Measurement, units of, 1 ; of lengths, 3, 4 ; of area and volume, 4; of mass, 13, 78; experiments on, 77; c. G.S. system of, 16; of position of a particle, 20; of velocity, 25; of force, 93, 117; methods of, 132 ; of work, 154 156, 162163; of power. 159; of acceleration and the Hodo- graph, 224 Mechanics, definition of, 1 ; funda- mental quantities in, 2; terms used in, 16, 17; conservation of energy in, 175 Methods of measuring force, 132 Moment of greatest compression, 219 Momentum, 85, 179; change of, 89; transference of, 90 ; conservation of, 91; rate of change of, with Atwood's machine, 107 Motion, definition of, 20; of a particle, 26; with uniform speed, 49; constrained, due to gravity, 102; Newton's Laws of, 113; first law of, 113 ; second law of, 115, and falling bodies, 123; problems on, 136, 139; illustra- tions of the third law, 149; at- traction and, 151; of a body down a plane, 164 ; down a curve, 165, 166; of a particle projected under gravity, 194; in a circle, 229; circular, 230; simple har- monic, 223 Mutual energy, 174 Newton's definition of force, 112 Laws of Motion, 113 Oblique impact, 216 Oscillation of a pendulum, period of, 205 Parallelogram of displacements, 32 ,. of velocities, 41 Parallelogram Law, experiments on, 46 ,, of forces, 135, 136 Particle, measurement of position of, 26 ; motion of, 26 ; motion of projected, under gravity, 194 Pendulum, the simple, 201 ,, period of oscillation of a, 205 Pendulums, experiments with, 204 Plane, motion of a body down, 164, 165 Position of a particle, measurement of, 20 Potential energy, 170 Poundals, 119 ; value of, 123 Power, 158 ; measurement of, 159 ; horse-, 158; expressions for, 159 Problems on Motion, 136139 Projectile, properties of the path of, 196 Projectiles, 188192 Quantity, rate of change of, 21 Quantity of matter and mass, 84 Rate of change, of a quantity, 21 ; average, 22; graphical represen- tation of, 23; variable, 23; of momentum with Atwood's ma- chine, 107 Bate of working, 158 Reaction, 148 Relation between weight and mass, 124, 203 Relative velocity, 47 Representation of a force, 135 Resolution, of displacements, 36, 42; of accelerations, 71 Rest after impact, 86 Rotation, consequences of the Earth's, 230 Simple harmonic motion, 233 Space, graphical representation of, 50 Speed, 26; average, 27; variable, 27; units of, 27; motion with uniform, 49 Statics, 17 INDEX. XIX Stress and force, 111 System, conservative, 176 Terms used in Mechanics, 16, 17 Time, 11 Transference of momentum, 90 Uniform velocity, 26 ,, speed, motion with, 49 ,, acceleration, 59 ; formulae connected with, 66 Unit acceleration, 58 of force, C.G. s., 118; F.P.S., 118; gravitational, 129 ,, of work, C.G.S., 156; F.P.S., 157; gravitational, 157 of energy, 177 Units, of measurement, 1 ; of length, 2 ; of speed, 27 ; of mass, 84 Value, of a Dyne, 122; of a Poundal, 123; of g in different latitudes, 207 Variable rate of change, 23 ,, velocity, 26 Variable speed, 27 Velocities, composition of, 38, 42; parallelogram of, 41; resolution of, 42 Velocity, measurement of, 25 ; uni- form and variable, 26 ; graphical representation of, 29; relative, 47 ; angular, 49 ; change of, 56 ; on a curve, 169 Vibration, amplitude of, 202 Volume, measurement of, 4 Watt, the, 159 Weighing, comparison of masses by, 131 Weight, 101 ; relation between mass and, 124, 203 Work and Energy, 153 Work, measurement of, 154 156, 162, 163; C.G.S. unit of, 156; F.P.S. unit of, 157; gravitational unit of, 157 ; due to gravity, 165 ; graphical construction for, 179, 180 INDEX TO STATICS. The references are to the pages. Amount of Friction, 166 Angle of Friction, 173 Application of the Principle of Work, 119 Axis, motion about an, 43; mo- ment about an, 55 Axle and Wheel, the, 123 Balance, the, 148; use of, 153 the Letter (Eoberval's), 158 Barton, the Spanish, 147 Bodies, rigid, 35 Body, equilibrium of a rigid, 70, 71 ; equilibrium of, resting upon a horizontal surface, 103 Calculation, centre of gravity found by, 91 Centre of Mass, 86, 87 of Gravity, 88, 89 ; experi- ments on, 90 ; found by calcula- tion, 91 ; formulae connected with, 94, 95 ; properties of, 100 Coefficient of Friction, 170 Cog Wheels, 146 Combinations of Simple Machines, 143, 144 Common, or Koman Steelyard, 156 Composition and Resolution of Forces, 12, 18 Danish Steelyard, the, 158 Differential Wheel and Axle, the, 144 Differential Screw, the, 146 Direction of Friction, 166 Equilibrium, 3; of a rigid body, 70, 71 ; of a body resting on a horizontal surface, 103 ; stability of, 106 Equilibrium of Forces, further ex- periments on, 10, 11; impressed on a particle, 25 Experiments, on the parallelograms of forces, 5; on parallel forces, 41 ; on moments, 46 ; with pul- leys, 133 Force, Resultant, 4 ,, representation of a, 3 ; measurement of a, 31; moment of a, 45; work done by a, 67 Forces, parallelogram of, 4; ex- periments on, 5 Forces, equilibrium of, further ex- periments on, 10, 11 ; impressed on a particle, 25 Forces, Composition and Reso- lution of, 12; Resolution of, 21; work done by a system of, 68 ; problems on, in one plane, 76 Forces, parallel, 36 ; Resultant of, 40,52 Formulae connected with centre of gravity, 94, 95 Friction, 166 ; Amount and Direc- tion of, 166; Laws of limiting, INDEX. XXI 169, 170; Coefficient of, 170; Angle of, 173 Gravity, Centre of, 88, 89; experi- ments on, 90 ; found by cal- culation, 91 ; formulae connected with, 94, 95 ; properties of, 100 Horizontal surface, equilibrium of a body resting on, 103 Inclined Plane, the, 134 Laws of Limiting Friction, 169, 170 Letter Balance, the (Eoberval's), 158 Lever, the, 115, 116; principle of, 48 Levers, Bent, 118 Limiting Friction, laws of, 169, 170 Machines, Simple, 113; combina- tions of, 143, 144 Mass, Centre of, 86, 87 Measurement of a Force, 3 Moment about an axis, 55 ,, of a force, 45 Moments, 48; experiments on, 46 Motion about an axis, 43 Moveable Pulley, 125 Parallel Forces, 36; resultant of, 40, 52 ; experiments on, 41 Parallelogram of Forces, 4 ; experi- ments on, 5 Particle, equilibrium of forces im- pressed on a, 25 Plane, the inclined, 134 Power, 114 Principle of the lever, 48 ,, of Work, application of the, 119 Problems on Forces in one plane, 76 Projection, 67 Properties of the centre of gravity, 100 Pulley, the, 125; the single move- able, 125 Pulleys, systems of, 127, 128; ex- periments with, 133 Representation of a Force, 3 Resolution and Composition of Forces, 12, 18 Resolution of Forces, 21 Resultant Force, 4 of Parallel Forces, 40, 52 Rigidbodies,_35 ^ Body, equilibrium of a, 70, 71 Roman or Common Steelyard, 156 Screw, the, 141; the differential, 146 Simple Machines, 113; combina- tions of, 143, 144 Spanish Barton, the, 147 Stability of equilibrium, 106 Steelyard, the Common, or Roman, 156; the Danish, 158 Systems of Pulleys, 127, 128 Theorems about couples, 59 Velocities, virtual, 73, 80 Wedge, the, 139, 140 Weight, 114 Wheel and Axle, the, 123; the differential, 144 Wheels, Cog, 146 Work, application of the Principle of, 119 Work done by a force, 67; by a system of forces, 68 INDEX TO HYDROSTATICS. The references are to the pages. Air, buoyancy of, 105 ; corrections for weighing in, 106; pressure of, 142 ; measure of pressure of, in receiver, 191 Air-Pump, Hawksbee's, 186 ; Smea- ton's, 187; Tate's, 189; general considerations, 190 ; Mercury, 193 ; Sprengel's, 193 ; Geissler's, 195; Topler's, 196 Aneroid Barometer, the, 152 Atmosphere, Height of the Homo- geneous, 154 Average Pressure, 27 Balance, Hydrostatic, 118; Jolly's, 127 Balls, Specific Gravity, 128 Barometer, the, 143; Tube Gauge, 66; Fortin's, 146; forms of, 151; Standard, 151 ; Aneroid, 152 ; Water, 153 ; measurement of heights by, 155 Barometer reading, corrections to, 148 Bell, the Diving-, 199 Bellows, Hydrostatic, 41 Bottle, Specific Gravity, 128 Bourdon Gauge, the, 67 Boyle's Law, 159; deductions from, 162 ; variations from, 163 Bramah's Press, 181 Buoyancy of the air, 105 Calculation of Thrust, 61 Centre of Pressure, 83 Charles and Dalton, law of, 164 Columns in tubes of unequal area, 136 Common Hydrometer, the, 122 Condenser, the, 197 Corrections to barometer reading, 148 Curved Surface, thrust on a, 90 Dalton and Charles, law of, 164 Deductions, from experiments on fluid pressure, 71 ; from Boyle's Law, 162 Density, 7 ; propositions on, 15 ,, Surfaces of equal, 73; of gases, 142 Dilatation of Gases by heat, 164 Diving-bell, the, 199 Double-barrelled pump, 188 Effective Surface, 53 Equilibrium, of fluids, 75; stability of, 108 Experiment, of Otto Guericke, 141 ; Pascal's, 146 Experiments, on fluid pressure, 68 ; deductions from, 71; on floating bodies, 107 Floating Bodies, 101 ; experiments on, 107 ; stability of equilibrium of, 108 Fluid-Pressure, 46; transmissibility INDEX. XX111 of, 37; illustrations of, 43; ex- periments on, 68 Fluids, 1, 3; stress in, 22; funda- mental property of, 24; pressure within, 30; pressure at a point within, 32; pressure at various points in heavy, 50 ; equilibrium of, 75 Fortin's Barometer, 146 Free Surface of Liquids, 6 Gases, 4, 5; density of, 142; dila- tation by heat, 164 ; pressure of a mixture of, 164 Gauge, the Barometer Tube, 66 ; Bourdon, 67 Geissler's Air-Pump, 195 Graphical Solutions, 30 Gravity, Specific, 8; definitions of, 10; measurement of, 11; values of, 14 ; propositions on, 15 Guericke, Otto, experiment of, 141 Hawksbee's Air-Pump, 186 Heat, dilatation of gases by, 164 Height of the Homogeneous Atmo- sphere, 154 Heights, measurement of by baro- meter, 155 Homogeneous Atmosphere, Height of, 154 Horizontal Surface, Thrust on a, 79 Horizontal Thrust, 92 Hydrometer, the Common, 122 ; Nicholson's, 125 Hydrostatic Bellows, 41 Balance, 118 Illustrations of Fluid Pressure, 43 Jolly's Balance, 127 Law, Boyle's, 159 ; deductions from, 162; variations from, 163; of Charles and Dalton, 164 Level Surfaces, 52 Liquid, finds its own level, 55 Liquids, 4 ; Free Surface of, Manometers, 62 Measurement, of Specific Gravity, 11; of heights by barometer, 155; of pressure of air in a receiver, 191 Mercury Air-Pumps, 193 Mixture of gases, pressure of, 164 Nicholson's Hydrometer, 125 Pascal's experiment, 146 Pipette, the, 170 Press, Bramah's, 181; work done in, 182 Pressure, at a Point, 26 ; Average, 27; Units of, 28, 58; within a fluid, 30; at a point within a fluid, 32; of the air, 142; in different directions, 33; trans- missibility of fluid-, 37 ; illustra- tions of fluid-, 43; fluidr, 46; at various points in a heavy fluid, 50; centre of, 83; of a mixture of gases, 164 Pump, the common, 175 ; the Lift-, 178; Force-, 179; continuous action, 180; Hawksbee's Air-, 186; Smeaton's Air-, 187 ; double- barrelled, 188 ; Tate's Air-, 189 ; Sprengel's, 193; Geissler's, 195; Topler's, 196 Eesultant vertical thrust, 90 95 ,, horizontal ,, 92 Safety-valve, the, 66 Shearing Stress, 22 Siphon, the, 170; experiments with, 172 Smeaton's Air-Pump, 187 Solids, 1 ; stress in, 22; soluble in water, 131 Solutions, Graphical, 30 Specific Gravity, 8; definitions of, 10; measurement of, 11; values of, 14; propositions on, 15 Specific Gravity Balls, 128 Bottle, the, 128 Sprengel's Air-Pump, 193 XXIV INDEX. Stability of equilibrium, 108 Standard Barometer, 151 Standard Substance, the, 11 Stress, general considerations on, 20; Shearing, 22; in solids and fluids, 22; distributed over a surface, 25 Surfaces of equal density, 73 Surface, stress distributed over a, 25; Level, 52; Effective, 53; Thrust on a horizontal, 79; on a vertical, 80; on a curved, 90 Syringe, the, 172 Tate's Air-Pump, 189 Tension, 21 Thrust, 21; examples of uniform and variable, 28 ; Calculation of, 61 ; on a horizontal surface, 79 ; on a vertical surface, 80 ; on base of a vessel, 86; on a curved surface, 90; resultant vertical, 95 Topler's Air-Pump, 196 Transmissibility of Fluid-Pressure, 37 Tube Gauge, Barometer, 66 Tubes of unequal area, columns in, 136 U-Tube Method, the, 133 Uniform Thrust, examples of, 28 Units of Pressure, 28, 58 Values of Specific Gravities, 15 Valve, the Safety-, 66 Valves, 174 Variable Thrust, examples of, 28 Vertical Surface, Thrust on a, 80 Vertical Thrust, resultant, 95 Volumenometer, the, 202 Water Barometer, 153 CAMBRIDGE : PRINTED BY J. 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