M: , I,, TV §0b«vt ^mv^ ^%m»tm ^ mn to 1303 y tomeii unrverslty Ubrary ^I'rfilSmilNfirKi '"''■o''uc*>on to practical mec „ 3 1924 031 304 078" olin,anx The original of tliis bool< is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031304078 PRACTICAL JilECHANICS s E o O asr r> EX>iTi03sr. By the same Author. FIRST LESSONS IN THEORETICAL MECHANICS. Pp. 2S6, with 164 Diagrams. Price %8. 6(2. In writing this volume the Author has had in view the wants of a rather numerous class of readers— those who wish to study the first Sriuciples of mechanics before they ave obtained the knowledge of geometi^, algebra, and trigonome- try, which most elementary books on the subject presuppose. Some knowledge of arithmetic and geo- metry, it is true, must be assumed in discussing the most elementary questions as to forces; but a very large portion of the principles of mechanics admits of exposition and Illustration without demanding of the student a knowledge of more than arithmetic, a few rules in men- suration, enough 'geometry to make accurate diagrams with compasses, scale and protractor, and enough algebra to solve a simple equation. No more than this is needed for the study of these lessons, with the ex- ception of those given in the sixth chapter, on Motion in a Circle, and a few articles and examples, occur- ring, for the most part, towards the end of the book. Throughout the work the Author has endeavoured to explain as clearly as possible the leading ideas of the subject, to illus- trate them by a great number of examples and questions, and to get rid of all difficulties that are not inherent in the subject. ELEMENTAET INTEODUCTION TO PRACTICAL MECHANICS ILLUSTBATED SY NVMEB0U8 EXAMPLES BEING THE SIXTH EDITION OP 'ELEMENTARY EXAMPLES IN PRACTICAL MECHANICS' BY THE EEV. JOHN F; TWISDEN, M.A. Professor of Mathematics in the Staff College; formerly Scholar of Trinity College, Cambridgei and Author of ' First Lessons in Theoretical Mechamcs * LONDON LONGMANS, GEEEN, AND CO. 1880 LONDON : PRINTED BY SPOTTIS-.VOO0E AND CO., NEW- STREET SQTTAHK AND PAELIAMENT STREET PEEFACE TO THE SIXTH EDITIOJ^". In issuing the Sixth Edition of the present work, it may not be improper to mention that the Second Edition, published in 1863, was altered in several respects from the First Edition, published in 1860, and that each sub- sequent Edition has undergone a careful revision. In the present Edition some changes have been made in the technical terms employed, and two chapters have been rewritten : viz. Chapter IX., Part I., on the ' De- flection and Rupture of Beams ; ' and Chapter III., Part II., on ' Force and Motion.' At the end of the latter will be found the three laws of motion, as stated by Newton, together with his illustrations of them, translated from the Introduction to the 'Principia.' All the Examples have been worked through several times, and it may be presumed that the Answers are correct, with few exceptions. The unit of force in which they are expressed — with such exceptions as are apparent from the context — is the gravitation unit. Vl, PREFACE TO THE SIXTH EDITION. the force of one pound, as defined on p. 48. What is meant by the absolute unit of force is explained in Chapter III. of Part II., and the subject is illustrated hy some Examples, On the whole, it is hoped that the later Editions have been considerably improved. J. F. T. October 1, IgSO. PBEFACE TO THE FIRST EDITION, The following Teeatise is designed to be an Introduction to the science of Applied MechanlGS : in this it differs from all the elementary works commonly in use, which are introductory to Rational Meehcmics. How great a difference is caused by this circumstance will appear from an inspection of the Contents ; it may, howeverj be men- tioned that, at the least, one-half of the present work has no counterpart in any Elememtary Treatise that has fallen under the author's notice. That so great a divergence from the usual type should be possible seems sufficient reason for believing that something is wanting in the ordinary works ; but how far the present will supply that want is, of course, another question. It was originally intended to be a book of Examples, and a supplement to others already in exist- ence : it was, however, found that by a few additions it could be made independent, and it was thought that what was gained in point of convenience by completeness, would more than compensate a small increase of size and cost. VIU PREFACE TO The work is intended to comprise two courses : the first is contained in Chapter I., the iirst section of Chapter II., and Chapter III. of Part I., and in Chapter I. of Part II. ; the second forms the remainder of the book. The first course may be read by any one who understands arithmetic, a little algebra, practical geometry, and the rules of men- suration ; in many of the Examples it is intended that a geometrical construction should take the place of calcu- lation : instances of the use of construction are given in Examples l78, 216, &c. In this course the principles of the science are merely stated, their formal demonstra- tion being reserved to the second course ; in other words, the order most convenient for teaching and learning has been followed at some sacrifice of the systematic develop- ment of the subject. The second course presupposes that the reader is acquainted with Euclid, algebra, and trigo- nometry, as commonly taught in schools ; a very few Examples are inserted which require some acquaintance with co-ordinate geometry and the differential calculus ; * the reason for their insertion will generally be obvious from the context in which they occur. Frequent use has been made of simple geometrical limits ; they wiU pro- bably present but little difficulty to the reader: some remarks on the subject of limits will be found in the Appendix. Very many Examples require numerical answers ; it is hoped that but few of the arithmetical operations will prove laborious to any one who possesses a proper facility in manipulating numbers, and it must be remembered * Most of these Examples are contained in Chap. IX., Part I. ; the others are distinguished by an asterisk. THE FIRST EDITION.' IX that few things are more important to a learner in the earlier stages of his progress than that he should be con- tinually referred to the numerical results that follow from the formulse he investigates. Hints and explanations have been freely given in connection with the more difficult Examples, and it is hoped they will be found sufficient to enable the reader to complete the solutions, though many of them are important mechanical theorems, and some of them but rarely to be met with (e.g. Examples 134, 149, 393, 429, 522, 553, 566, &c.). A list is subjoined of the principal works referred to in drawing up the present Treatise ; particular instances of obligation are acknowledged in the footnotes in the course of the work.- A more explicit recognition of assistance is due to the Eev. H. Moseley, Canon of Bristol ; about two hundred of the Examples were given by him to his classes at King's College, London, in the years 1840, 1, 2, 3 ; these he very kindly placed at the author's disposal, and also gave him permission to use freely his excellent Treatise on the ' Mechanical Principles of Engineering ' — a permis- sion of which great use has been made. Staff College : August 1860. WORKS REFERRED TO. M. PoissoN, Trait4 de M4caid(Jile. M. PoNCELET, Introduction a la Micaniqne indnstrielle. M. PoNCELET, Trait^ de Micanique appliqute SMX Machines. M. MoBiN, Aide-M^lnoire de Mecanique pratique. M. MoHiif, Notions fondamentales de Mecanique pratique. Db. T. Yoting, Lectures on Natural Philosophy. Ebv. W. Whbwell, D.D., History of the Inductive Sciences. Bev. H. Moselet, Mechanical Principles of Engineering. . Eet. E. Wili-is, Principles of Mechanism. Db. Baneiiie, Applied Mechanics. Sib W. Thomson and Mb. P. G. Tait, Natural Philosophy, Vol. I. CONTENTS. PART I. CHAPTEB I. PAGE Properties of Materials — Weight — Expansion by Heat — Elongation by Tension — Eesistance to Rupture by Tearing and Crushing . 1 CHAPTER 11. Sect. I. Efficiency of Agents — Modulus of Steam Engines — Water- wheels — Work of Animate Agents — Cost of Labour . . .18 Sect. II. Work done by a Variable Force — Steam Indicator — Work of Raising Weights . 36 CHAPTER III. Statement of the Fundamental Principles of Statics — Parallelogram of Forces — Principle of Moments — Applications, viz. : the Lever — tlie Steel-yard— Equilibrium of Walls— Thrusts of Rafters . 47 CHAPTER IV. The Fundamental Theorems of Statics' 77 CHAPTER V. Centre of Gravity — Geometrical Applications . . . . .112 CHAPTER VL Limiting Angle of Resistance — ^Laws of Friction— Inclined Plane — Wedge — Screw — Friction on a Pivot — Endless Screw . .131 XU CONTENTS. CHAPTEB VII. PAGE Equilibrium of Body resting on an Axle — Rigidity of Hopes — Pulley — Capstan — CaiTiage--wheel 157 CHAPTER VIII. The Stability of Walls — The Line of Resistance — Pressure of Water —Pressure of Earth 173 CHAPTER IX. Bending Moment— Deflection of Beams — Equation of three Moments • — Strength of Beams 184 CHAPTER X. Virtual Velocities — ^Work done by a Eorce — Machines in a State of Uniform Motion — Modulus of Machine — Toothed Wheels — Epicycloid — Form of Teeth — ^Equilrbrium of Toothed Wheels . 202 FART IT. CHAPTER I. Velocity — Motion of Falling Bodies — Motion produced by a given Force— Accumulated Work — ^Motion on a Curve — Centrifugal Force — Oscillation of Simple Pendulum — Centre of Oscillation . 231 CHAPTER II. Unifbrmly accelerated Motion — Composition of Velocities — Motion of Projectiles 262 CHAPTER III. Mass, Momentum — Absolute Unit of Force and of Work— Motion on an Inclined Plane — Newton's Laws of Motion .... 265 CHAPTER IV. Constrained Motion of a Particle — Centrifugal Force — Small Circular Oscillations — Longitudinal Vibrations of a Rod . . . . 279 CONTENTS. Xlil CHAPTER V. PAGE Moment of Inertia — Eadius of Gyration 289 CHAPTEE VI. Angular Velocity — Effective Force — D'Alembert's Principle — Pres- sure on a Fixed Axis of Eotation ' 297 CHAPTER VII. Kinetic Energy of a Rotating Body — Smeatou's Machine — Atwood's Machine — Ply-wheel — M. Morin's Experiments on Friction — Compound Pendulum 30R CHAPTEE VIII. Impulsive Action — Impact of Smooth Balls — Impulse on Axis of Eotation — Centre of Percussion — Axis of Spontaneous Eotation — Ballistic Pendulum 321 APPENDIX. On Limits 335 TABLES. Table of Specific Gravity 2 Expansion produced by Heat 9 Moduli of Elasticity 11 Tenacity 12 Crushing Pressures 14 Moduli of Steam Engines 23 Moduli of Water-wheels 26 Work done by Living Agent? 28 Horizontal Transport of Burdens 30 Cost of Labour 34 Coefficients of Friction 139 Friction of Axles 159 Eigidity of Ropes 164 ModuH of Eupture 200 Acceleration due to Gravity 250 PEACTIOAL MECHANICS. CHAPTEE I. ON SOME OF THE PHYSICAL PROPERTIES OF MATERIALS. 1. Properties of Materials. — The present chapter is intended to serve as an introduction to those that follow. It contains examples illustrative of the more obvious physical properties of the materials commonly used in construction and machinery. These physical properties are (1) Weight; (2) Expansion or Contraction, produced by change of temperature ; (3) Elongation and Compres- sion, produced by Tension or Pressure; (4) Eesistance offered to Eupture by Tension ; (5) Eesistance offered to Eupture by Compression. 2. Weight. — For estimating the weight of bodies with sufficient accuracy it may be assumed that the weight of a cubic foot of water is 1000 oz. This number is easily remembered, and is within a very little of the truth. In every example contained in the following pages wherein the weight of bodies is concerned, it will be assumed that the weight of a cubic foot of water is 1000 oz., unless the contrary is specified. As a matter of fact, a cubic foot of pure water at 39° F. (when its density is greatest) weighs 998-8 pz. It may also be convenient for the reader to remember that a gallon contains 277-274 cubic inches, B 2 PRACTICAL MECHANICS. and that a gallon of water at the standard temperature (62' F.) weighs 10 lbs. Ex. 1. — ^A reservoir is internally 12 ft. long, S ft. -wide, and 3 ft. deep: determine the weight of the water it contains when fall, and the least error produced by considering that each cubic foot weighs 1000 oz. Ans. Weight, 5 tons, cwt. 50 lbs. Error, 13i lbs. Ex. 2. — ^A cylindrical boiler terminated by plane ends, is internally loft, long and 4 ft. in diameter ; through the lower half pass lengthwise 60 fire- tubes, 3 in. in external diameter : determine the volume and weight of the water contained in it when the surface of the water passes through the centres of the ends. Ans. Vol. 57'43 cubic ft. "Weight, 1 ton, 12 cwts. qr. 55 lbs. Ex. 3. — The surface of a pond measures 10 acres; in the course of a period of dry weather the surface falls 1 J in. by evaporation : what is the weight of the water that has been withdrawn ? Ans. 1520 tons, nearly. 3. Specific Gravity. — The specific gravity or specific density of a solid or liquid substance is the proportion which the weight of a certain volume of that substance bears to the weight of an equal volume of water ; thus when it is stated that the specific gravity of cast iron is 7'2070, it means that a cubic foot, or a cubic inch, &c., of cast iron weighs 7*2070 times as much as a cubic foot, cubic inch, &c., of water; consequently a cubic foot of cast iron will weigh 7207 oz., and in general, if s is the specific gravity of a substance, a cubic foot of it will weigh 1000 s oz., at least with sufficient accuracy in almost all cases. The following table gives the specific gravities of some common materials : — Table I. SPECIFIC GRAVITIES. Metals. Platinum (laminated) . 22-0690 Pure Gold (hammered) . 19-3617 Gold 22 carat (do.) . .17-6894 Mercury .... 13-6681 Lead (cast) . . . 11-3523 Pure Silver (hammered) . 10 5107 Copper (cast) . . . 8-7880 Brass (cast) Steel (hard) Iron (east) ,. (wrought) Tin (cast) . Zinc (cast) 8-3958 7-8163 7-2070 7-7880 7-2914 7-1908 SPECIFIC GRAVITY. Stones and Earth. Marble (-white Italian) . 2-638 Portland Stone . . 2-145 Slate (Westmoreland) . 2-791 Coal (Newcastle) . 1.2700 Granite (Aberdeen) . . 2-625 Brick (Eed) . . 2-168 Paving Stone . . 2-4168 Clay. . . . 1-919 Mill Stone . 2-4835 Sand (River) . . 1-886 Grindstone , . 2-1429 Chalk (mean) . . 2-316 Woods (7)ry). Elm. . 0-588 Oak (English) . . 0-934 Fir (Riga) . 0-753 Teak (Indian) . . 0-657 Larch . 0-522 Cork . 0-240O Mahogany (Spanish) . . 0-800 .1 foot length of Hempen rope weighs in lbs. 0-045 x (cire. in inches)^. 1 „ „ Cable weighs in lbs. 027 x (oiro. in inches)^, 1 onbie foot of Brickwork weighs 112 lbs. Note. — The above numbers, where printed to four places of decimals , are taken from Dr. Young's Lectures on Natural Philosophy, vol. ii. p. 503 ; where printed to three places of decimals, from Mr. Moseleys Mechanics of Engineering, 1st ed. p. 622. A definite specific gravity is assigned to each substance to prevent ambiguity in working the following examples. It will be remarked, however, that different specimens of the same substance have different specific gravities : thus of 16 specimens of cast iron the specific gravities have been found to vary from 7-295 to 6-963. The reader must, therefore, bear in mind that the numbers in the text give mean values from which the specific gravity of any specimen of a given substance will not largely vary — though the limits of variation are greater with some sub- stances than with others. A similar remark applies to all quantities determined by experiment. Ex. 4. — What is the weight of a rectangular block of marble 63 ft. long, and in section 12 ft. square? Ans. Weight, 667 tons, 14cwts. 3 qrs, Ex. 5. — The girth of a tree is 3 ft. at top, 3 ft. 9 in. at bottom, it is 14 ft. long. Determine its weight according as it is larch, oak, or mahogany. Also, its value at the following prices : larch, 2s. &d. ; oak, 7s. ; mahogany, 19«. per cubic foot rough. Ans. Vol. 12-74 cubic ft. Weight: Larch, 416 lbs. Oak, 744 lbs. Mah. 637 lbs. Price: „ lllls. lOd. „ il. 9s. 2d. ., 111. 2s. Id. [The volume to be determined as that of the frustum of a cone.] ■ Ex. 6. — Pind the weight of a rectangular mass of oak, 12 ft. long, 4 ft. broad, and 2^ feet thick. Whit would be the weight of a mass of granite .qf the same dimensions? ^n«. Oak, 62 cwts. 2 qrs. 5 lbs. Granite, 175 c-vrts. 3 qrs. 3^ lbs. B 2 '4 PRACTICAL MECHANICS. Ex. 7. — Find the separate weights of a cast-iron ball, 4 in. in radius, and of a copper cylinder 3 ft. long, the diameter of whose base is 1 in. Deter- mine also the diminution in the weight of the ball if a hole were cut through" it which, the cylinder would exactly fit, the axis of the cylinder passing through .the centre of the sphere. Also, find .the error that results from considering the part cut away a perfect cylinder. Ans. Weight of sphere, ni8'09 oz., „ cylinder, 143-8 oz. „ part cut from sphere, 26'204 oz. Error, 0-102 oz. Ex^S. — If a 10-in. shell were of cast iron, and were 2 in. thick, what would be its weight supposing it complete ? If the weight of a 10-in. shell were 86 lbs., what would be its thickness supposing it complete ? Ans. (1) 107 lbs. (2) 1-41 in. Ex. 9.^A hammer consists of a rectangular mass of wrought iron, 6 in. long, and 3 in. by 2 in. in section ; its handle is of oak, and is a cylinder 3 ft. 6 in. long, on a base of 1 in. in radius. Determine its weight. Ans. 12-83 lbs. Ex. 10. — A pendulum consists of a cylindrical rod of steel 40 in. long, on a base whose diameter measures j- in. ; to the end of this is screwed a steel cylinder i in. thick, and 1 i in. in radius, which fits accurately a hollow cylinder of glass, containing mercury 6 in. deep, the glass vessel weighing 3 oz. Determine the weight of the pendulum. Am. 360-8 oz. Ex. 11. — Determine the weight of a leaden cone whose height is 1 ft. and radius of base 6 in. ; determine also the external radius of that hollow cast iron sphere which is 1 in. thick, and equals the cone in weight. ^»s. (1) 185-74 lbs. (2) 8-02 in. Ex. 12. — A rectangular mass of cast iron 6 ft. long, 6 in. wide, and 3 in. deep, has fitted square to its end a cube of the same materials whose edge is li ft. long ; find its weight. Ans. 1858 lbs. Ex. 13, — It is reckoned that a foot length of iron pipe weighs 64-4 lbs. when the diameter of the bore is 4 in. and the thickness of the metal 1 i in. : what does this assume to be the specific gravity of iron ? Atis. 7-197. Ex.:U.—A. cast-iron, column 10 ft. high and 6 in. in diameter will safely support' a weight of 17i tons, whether it bo solid, or hollow and 1 in. thick ; determine :— (1) the weight of a solid column ; (2) the number of equally strong hollow columns that can be made out of 300 solid columns ; (3) the price of 500 solid columns at 10s. per cwt. and of 500 hollow columns at lis. 3d. per cwt. ; (4) the cost of sending the 500 solid and the 500 hollow columns to a given place at the rate of 10a. 6d. per ton. ^«s. (1) 884-4 lbs. (2)900. (3) 1974J. 3«. solid. 1233/. 16«. hoUo*. (4) 1031. 13s. solid. 571. 12s. hollow. BRICKWORK. .6 Ex. 15. — Determine the weight of a hollow leaden cylinder whose leDgth is 3 in., internal radius 1^ in., and thickness IJ in. Am. 26'121 lbs. Ex. 16. — Determine the weight of a grindstone 4 ft. in diameter and 8 in. thick, fitted with a wrought-iron axis of which the part within the stone is 2 in. square, and the projecting parts each 4 in. long with a section 2 in. in diameter. Ans. 1135 lbs. Ex. 17. — Determine the weight of an oak door 7 ft. high, 3 ft. wide, and 1^ in. thick. Ams. 153 J lbs. Ex. 18. — There is a fly wheel of east iron the external radius of whose rim is 5 ft. and internal radius 4 ft. 6 in. ; it is 4 in. thick, and is connected with the centre by 8 spokes 4 in. wide and 1 in. thick, strengthened by a flange on each side 1 in. square (so that their section is a cross 4 in. long aiid 8 in. wide), each spoke is 4 ft. long ; the centre to which they join the rim has the same thickness as the rim, is solid, and (of course) 6 in. in radius : determine the weight of the whole. Ans. 2969 lbs. Ex. 19. — Therearetworoomseachl00ft.longand30ft. wide; theoneis floored with oak planking Ij in. thick ; the other with deal planking (Riga fir) 1 J in. thick. Determine the weights of the floors and their cost, the price of deal being 3s. and oak 7s. per cubic foot. Atis. Deal floor weighs 17,648 lbs. costs 561. 5s. Oak „ 18,242 lbs. „ 1092. 7s. 6d. Ex. 2,0. — ^A cubic foot of copper is drawn into wire ^ of an inch in dia- meter ; what length of wire is made J Am. 46,936 ft, Ex. 21. — It is said that gold can be drawn into wire one millionth part of an inch thick ; what wiU be the length of such a wire that can be made from an ounce of pure gold? Ans. 1,793,448 miles. Ex. 22.— It is said that silver leaf can be made ys^so °^ a° i°'='i t^^ck j , how many ounces of silver would be required to make an acre of such silver leaf? Ans. 254-3 oz, 4. Brickwork. — The measurement and determination of the weight of brickwork depend upon the following data : — (1) A rod of brickwork has a surface of 1 square rod (or 301 square yards) and a thickness of a brick and a half, i.e. of 1 ft. 1^ in., or it contains 306 cubic feet. (2) A rod of brickwork contains about 4500 bricks in mortar, or 5000 bricks laid dry. (3) A rod of brickwork requires 3^ loads (i.e. 3i cubic' yards") of sand and 18 bushels of stone hme. 6 PRACTICAL ItlECHANICS. (4) A brick measures 8| x 4^ x 2| inches, i.e. a quarter of an inch each way less than 9 x 4^ x 3 inches. (5) A bricklayer's hod measures 16 x 9 x 9 inches, and can contain 20 bricks. Labourers, however, commonly .put 10 or 12 bricks into it.* Ex. 23.— Ho w many irods of brickwork are there in a square tower 1 1 7 ft. high and 28 ft. by 7 ft. at its base, externally, and 3 bricks thick ? Deter- inine the number of bricks required to build the tower and their price at 1^. 10s. per thousand. Ans. (1) 62-43 rods. (2) 236,000 bricks. (3) S5il. Ex. 24. — A tower the base of which measures externally 9 ft. square is 50 ft. high and 2 bricks thick ; how many bricks are required to build it, and how many loads of sand and bushels of lime ? Determine also the cost of the materials if the bricks Oost 11. 10s. per thousand, sand 5s. id. per load, and lime Is. Sd. per bushel. Ans. (1) 7-35 rods. (2) 33,000 bricks, 25t loads of sand, 132^ bushels of lime. (3) Cost 671. Ss. 2d. Ex. 25. — How many rods of brickwork are there in a reservoir of a rect- angular form, the internal measurements of which are 20 ft. long, 6 ft. wide, and 12 ft. deep ; the work being 2 bricks thick, viz. both walls and floor ; and the reservoir being open at the top ? Ans. 4-43. Ex. 26. — Find how many rods of brickwork there are in a wall 360 ft. long, 17 ft. high, and 2 bricks thick ; and determine the cost of the material ftom the data in Ex. 24. Ans. (1) 30 rods. (2) 2751. 10s. Ex. 27. — If the wall in the last example had an additional 2 ft. of foun- dation 3 bricks thick, and were supported by 20 square buttresses reaching to the top of the wall 2 bricks thick, on foundations 3 bricks thick, and taeasuring 2 J ft. in a direction perpendicular to the face of the wall ; deter- mine the number of rods of brickwork in the foundations and buttresses. Ans. 10'2 rods. Em. 28.— What would be the cost of the carriage of the bricks in the wall described in the last two examples at Bs. 6d. per thousand ? Ans. 49/!. 15s. Ex. 29.— The following are the actualdimensionsof the brickwork of the outer shell of the chimney of St. KoUox, Glasgow. Commencing from the top, there are five divisions ; the tops of these divisions are respectively 435i, 3501 210^, 114 J, 84J ft. above the ground ; the external diameters at the tops of the divisions are respectively 13 ft. 6 in., 16 ft. 9 in. 24 ft. 30 ft. 6 in., 35 ft. The diameter on the ground is 40 ft. ; the thicknesses of * Weale's Contractor's Price Book for 1859, p. 280. EXPANSION AND CONTRACTION BY HEAT. 7 the divisions are respectively IJ, 2, 2J, 3, and 3 J bricks; below ground the brickwork reaches 14 ft., with a uniform external diameter of 40 ft.; the first 8 feet are 3 ft. thick ; in the remaining 6 feet the thickness gradu- ally increases to 12 ft. Determine the number of rods of brickwork contained in the chimney; the number of bricks employed, their cost at 1/. 11«. 3d. per thousand; also, if the mortar were of sand and stone lime, determine the number of loads of sand and bushels of stone lime required, and their cost at 5s. id. per load, and 1«. 8d. per bushel re- spectively. [The surface of each division of (he chimney may be considered as that of a conic frustum ; the real volume of each division will be the difference between the volumes of two conic frustums. A sufficiently close approxi- mation may be obtained by multiplying the mean surface by the thickness and considering the slant side equal to the height ; the volume of the part below ground is to be determined accurately.] Ans. (1) 218 rods, or 981,000 bricks. (2) Cost of bricks, 15322. 16s. 3d. (3) 763 loads of sand, costing 203Z. 9s. id. (4) 3924 bushels of lime, costing 3272. 5. Expansion and contraction by heat. — It is found that all bodies experience a small change of volume on the application of heat. In general, the change is one of in- crease,* and with sufficient accuracy may be considered to obey the following law within moderate ranges of tempera- ture. If a volume v be increased by A; V when its tem- perature is raised one degree, it will be increased hynxkv when the temperature is raised n degrees, i.e. the in- crease of volume is proportional to the increase of tem- perature. The same rule holds for the expansions in length, which a body experiences from an increase of temperature, In order to fix the conception of a degree of temperature (with sufficient accuracy for our present purpose), it will be proper to mention that when heat is applied to ice the water produced by melting retains a constant temperature imtil the whole of the ice is melted. This temperature serves as one fixed point, and is called the freezing point. Moreover, boiling water in free contact with the air also keeps at a constant temperature (at least when the baro- * Water, near freezing point, is a conspicuous exceptioUk 8 PRACTICAL MECHATJICS. meter stands at a given height). This fact, therefore, supplies a second fixed point, and is called the boiling point, viz., when the barometer stands at thirty inches. These two points being fixed, the graduation is arbitrary. The scale of Fahrenheit's thermometer (which is commonly used in England) is constructed by dividing the space be- tween the freezing and boiling points into 180 equal parts, termed degrees, and by commencing the graduation 32° below freezing point, so that the freezing point is marked 32°, and the boiling point 212°. In the centigrade ther- mometer (now commonly used in scientific investigation) the graduation begins at the freezing point and the in- terval between the freezing and boiling points is divided into 100 equal parts called degrees.* It is easy to see that if at any temperature Fahrenheit's thermometer stood at F° and the centigrade at c°, we should have F°— 32 c° 180 100 Ex. 30. — The density of water is greatest at 3°-3 on the centigrade scale ; what is the same temperature called on Fahrenheit's scale ? Ans. 39°-02 F. Ex. 31. — The standard temperature not nnfreqnently referred to in English experiments is BCF. ; what would the same temperature he called on the centigrade scale? _4^_ 15°'53 C. Ex. 32. — If the centigrade thermometer stood at 5° below zero, or at — 5° C, what would the same temperature he marked on Fahrenheit's scale ? Ans. 23° F, Ex. 33.— What degree on the centigrade scale would be the equivalent to —4° on Fahrenheit's scale? ^,;^_ —20° C The following table gives the fractional part of the whole by which substances expand when heated : f * In Eianmur's thermometer the freezing point is marked zero, and the jiO go nO boiling point 80°: consequently ■„,. =^. ioU oU f From Dr. Young's Natwal Philosophy, vol. ii. p. 390. EXPANSION PRODUCED BY HEAT. Table II. EXPANSION PEODUCED BY HEAT. In length: Glass Tube „ Platinum „ Cast Iron Wroughfl » Iron J „ Steel rods 1, Brass rods „ Lead „ Copper . In volume : Mercury . II .1 in") glass ( ap- K , parent) J Tomporatui'O raised from 82° to 212° P. Tempemture raised 1° P. Authority 0-00077615 0000856 0-0011094 0-001156 0-0011447 0-0018928 0-002867 0-001700 0-00000431 0-00000476 0-00000617 00000642 0-00000636 0-00001052 0-00001592 0-00000944 0-00010415 0-00008696 Eoy Borda Eoy Borda Roy Eoy Smeaton Smeaton Eoy Committee of Eoyal Society &. 34. — The length of the base line of the Ordnance Survey on Hounslow Heath -was found to be 27,404 ft.; this was measured first hy glass tubes, and tlien by steel chains ; if, in correcting the glass tubes for temperature, a uniform error of 1° in excess had been committed, and in correcting the steel chain an error of 1° in defect had been committed, -what would have been the difference between the apparent measurements ? • Am. 3-51 in. Ex. 36.— If the wrought-iron rails on a railway are 10 miles long when at a temperature of 32° below freezing, by how much will they lengthen if their temperature is raised to 88° F. ? Am. 29-83 ft. Ex. 36. — Eamsden's brass yard exceeded Shuckbnrgh's by 0002605 of an inch ; what would be the difference of their temperatures when accu- rately the same length ? Ans. 6°-6 F. Ex. 37. — Two rods, respectively of iron and brass, ab and o d, are fas- tened together in the middle; they are accu- rately the same length, at62°F.; to their ends are fastened by pivots tongues c A E and d b f •' which are perpendi- "^' D cular to the bars, at 62° F. ; in consequence of the unequal expansion or con- tjaotion of the bars the tongues will assume different positions, as shown by the dotted lines ; it is required to determine the lengtli of c e, that the point b may remain unmoved by the expansion or contraction of the bar. The length of A B is 10 ft. and the distance a c is 1-725 in. Ans, c e = 4-426 in. Fis. 1. 10 PRACTICAL MECHAMCS. Ex. 38. — If the expansion in length of a suhstanee is e times the length at a given temperature, show that the expansion in rolume 'vrillbe yery nearly 3 e times the volume at that temperature. Ex. 39.— The volume of a mass of lead being a cubic foot at 60° F., what wiU be its volume at 0° F. ? and what at 88° F. ? Ans. At 0° F. 0-997134 cubic ft. At 88° F. 1-00133728 cubic ft. Ex. 40.— There is half a cubic inch of mercury in a thermometer at 32° F. ; when the temperature is raised to 92° F. the mercury ascends 4 in. ; what is the diameter of the bore of the glass tube ? Ans. 0-0288 in. 6. Elongation prodMced by tension. — The principle on whicli this determination is made is the following : — Suppose the length of a beam or bar to be L feet, the area of its section to be K square inches, then if by the appli- cation of a tension of P lbs. its length becomes L + Z, it appears from experiment that Z : l:: — : E K where E is a constant number depending on the nature of the material, and called the Modulus of Elasticity. It is found that all substances obey this law when the degree of extension does not exceed certain limits; the limits are different in different substances, and in many are very narrow. It appears also that within these limits (i.e. the limits of elasticity) a tension producing a certain degree of extension wiU, if applied in the opposite direc- tion so as to become a pressure, produce an equal degree of compression. It will be observed that - is the tension or pressure per square inch of the section of the beam or bar. It is also plain that if - were equal to E then would I be equal to L, so that the modulus of elasticity is that tension per square inch of the section of a bar which would double its length if its elasticity continued perfect. It is, per- haps, unnecessary to remark that no solid substance has limits of elasticity any way approaching this in extent. ELONGATION PRODUCED BY TENSION. 11 Table III. MODULI OF ELASTICITY.* Material Modulus Material . Modulus Wrought Iron bars Cast Iron . „ Brass . Steel (hard) . Copper wire . 29,000,000 17,000,000 8,930,000 29,000,000 17,000,000 Oak (English) . Larch Fir (Eiga) Elm , 1,450,000 ■ 1,060,000 1,330,000 700,000 Ex. 41. — By how much would a bar of wrought iron J of an inch square and 100 ft. long lengthen under a tension of 2 tons (neglecting the weight of the bar) ? Ana. 0-247 ft. Ex. 42. — Determine the elongation of a steel bar 2 in. square and 40 ft. long when subjected to a tension of 40 tons. What would have been its elongation had it been of cast brass ? .pounds per minute. The con- clusion, as far as regards the efficiency of the animal, is not very correct ; it has, however, fixed the meaning of the term horse-power when applied to a steam engine. Hence Def. — A steam engine works with one horse-power when, it yidds 33,000 foot-pounds per minute. Of course an engine of n horse-power yields n times 33,000 foot-pounds per minute. Ex. 73. — The piston of a steam engine is 15 in. in diameter, its stroke is 2J ft. long ; it makes 40 strokes per minute ; the mean pressure of the steam on it is 15 lbs. per square inch; what number of foot-pounds is done by the steam per minute, and what is the horse-power of the engine ? Ans. 265,072 ft.-pds. 8-03 H.-P. Kv. 7i. — A weight of 1 J tons is to be raised from a depth of 50 fathoms in 1 minute; determine the horse-power of the engine capable of doing the work. Ahs. 30^H.-P. Ex. 75. — The resistance to the motion of a certain body is 440 lbs ; how many foot-pounds must be expended in making this body move over 30 miles in one hour ? What must be the horse-power of an engine that does the same number of foot-pounds in the same time ? Ans. 69,696,000 ft.-pds. 351 H.-P. 13. Application of the foregoing priTidples. — A con- siderable number of practical questions can be answered by means of the principles already laid down, viz. such ques- tions as the horse-power of the engine required to do a certain amount of work, the time in which an engine of a certain power will do a certain amount of work, &c They are all done by following the same method, viz. First, from a consideration of the work to be done, obtain the number of foot-pounds that must be expended in a certain time. Next from a consideration of the power of the agent EXAMPLES OF THE WORK OF STEAM, il obtain the ninnber of foot-pounds yielded in the same time. One of these expressions will contain an unknown quantity, but, since by the terms of the question they are equal, they will form an equation from which the unknown quantity can be readUy determined. Ex. 76. — An engine is required to raise a weight of 13 cwts. from a depth of 140 fathoms in 3 minutes; determine its horse-power. Let X be the required horse-power ; then the number of foot-poimds yielded in 3 minutes will equal 33,000 x a; x 3 ; also the number of foot- pounds required to raise 13 cwts. from a depth of 140 fath. equals 13 x 112 X 140 X 6. And since these two numbers are equal we have 33,000 X 3 X a:= 13 X 112 X 140 X 6. .-. j;= 12-35 H.-P. Ex. 77. — In how many minutes would an engine working at 25 horse- power raise a load of 12 cwts. from a depth of 160 fathoms ? Am. 1'564 min. Ex. 78. — A locomotive engine draws a gross load of 60 tons at the rate of 20 miles an hour ; the resistances are at the rate of 8 lbs. per ton ; what must be the horse power of the engine ? [The reader must bear in mind that the work to be done is to overcome a resistance of 480 lbs. through 20 miles in one hour.] Ans. 25'6 H.-P. Ex, 79. — ^What must be the horse-power of an engine that raises 20 cubic feet of water per minute from a depth of 200 fathoms ? Aiis. 45 ^ H.-P. Ex. 80. — How many cubic feet of water would an engine working at 100 horse-power raise per minute from a depth of 25 fathoms ? Ans. 352. Ex. 81. — How many cubic feet of water will an engine of 250 horse- power raise per minute from a depth of 200 fathoms? Ans. 110 cub. ft. Ex. 82. — It being required to raise 100 cubic feet of water per minute from a depth of 495 ft., what must be the horse-power of the engine ? Ans. 93f H.-P. Ex. 83.-^There is a mine with three shafts which are respectively 300, 450, and 500 ft. deep : it is required to raise from the first 80, from the second 60, from the third 40 cubic feet of water per minute ; what must be the horse-power of the engine ? Ans. 134|i H.-P. Ex. 84. — At what rate per hour vrill a locomotive engine of 30 horse- power draw a train weighing 90 tons gross, the resistances being 8 lbs. per ton ? Ans. 15625 miles. jElc. 85. — What is the gross weight of a train which an engine of 25 horse-power will draw at the rate of 2o miles an hour, resistances being 8 lbs. per ton ? .^w. 46 875 tons. Ex. 86. — A train whose gross weight is 80 tons travels at the rate.^ 20 miles an hour ; if the resistance is S lbs. per ton, what is the horse- power of the engine ? ^»«- 34^ H.-P. 22 PRACTICAL MECHANICS. ^JT. 87. — An engine working with the same power as that in the last example draws a train at the rate of 30 miles an hour ; the resistances being 7 lbs. per ton, what is the gross weight of the train ? Aiu. 60|2 tons. Sx. 88..— What must be the length of the stroke of the piston of an engine, the surface of which is 1500 square inches, which makes 20 strokes per minute, so that with a mean pressure of 12 lbs. on each square inch of the piston, the engine may be of 80 horse-power ? Ans. 71 ft. Ex. 89. — The diameter of the piston of an engine is 80 in., the length of the stroke is 10 ft., it makes 11 strokes per minute, and the mean pressure of the steam on the piston is 12 lbs. per square inch : what is the horse- power? Jns. 20106 S.-P. Ex. 90. — Find the horse-power of an engine that will raise in one minute 100 cubic feet of water from a depth of 600 feet. Ans. IIS^^ H.-P. Ex. 91. — A train weighing 60 tons is drawn along a railway at the rata of 20 miles an hour ; the resistances being 8 lbs. per ton, find the horse- power of the engine. Ans. 211 H.-P. Ex. 92. — The p^linder of a steam engine has an internal diameter of 3 ft. ; the length of the stroke is 6 ft. ; it makes 6 strokes per minute ; under what effective pressure per square inch would it hare to work in order that 75 horse-power may be done on the piston? Ans. 67'S4 lbs. Ex. 93. — What must be the horse-power of a stationary engine that draws a weight of ISO tons along a horizontal road at the rate of 30 miles per hour, friction being 8 lbs. per ton ? Ans. 96 H.-P. 14. Modulus of a Tnaohi/ne. — An agent rarely, if ever, does a considerable amount of useful work directly, but nearly always through the intervention of a machine, by which the motive power of the agent is so applied as to overcome the resistance in the most convenient manner. For instance, when a steam engine raises water out of a shaft, the motive power is the pressure of the steam on the piston, the resistance to be overcome is the weight of the water, the beam, crank, &c., of the engine are the means by which the motive power is applied so as to overcome the resistance. Now it will be remarked that each part of the machine offers more or less resistance to the motion, so that a certain part of the work done by the motive power must be expended in overcoming these resistances, i.e. in reference to the purpose of the MODULI OF STEAM ENGINES. 23 machine, must be expended uselessly. The remainder of the work done by the motive power will be expended usefully in accomplishing that purpose. If the number of foot-pounds done by the agent is represented by u, the number expended in overcoming prejudicial resistances by V^, and the number expended usefully by Ui, all in the same given time, then it admits of proof in the case of a machine moving uniformly, that U = tJ„ + U,. It also appears that in most machines "u, bears to D a constant ratio, so that o,=K u where the letter K denotes some proper fraction, depending on the nature of the machine ; this fraction is called the modulus of the machine ; the following table, taken from General Morin's Aide-Mimoite de M6ccmique Pratique, gives the value of K for different classes of steam engines : — Tabld VI. MODULI OF STEAM ENGINES. Vftlno of K 1 Description of Mndiine Borae-power Bost working Ordinnry do. Watt's low-pressure engine 4 to 8 0-60 0-42 10 „ 20 0'56 0-47 30 „ 100 0-60 0-54 Cornish engines, working up to 80 0-44 085 by expansion and con- densation 30 „ 40 049 0-89 40 ,. 60 0S7 0-46 50 „ 60 0-62 0-60 «0 „ 70 0'66 0-63 70 „ 80 0-82 0-66 80 „ 100 0-70 0-59 High-pressure engines, up to 10 0-50 0-40 working without ex- 10 „ 20 0'55 0-44 pansion or condensation 20 „ 80 „ SO 40 0-60 0'65 0-48 0'S2 above 40 0-70 56 24 PRACTICAL MECHANICS. Ex. 94. — The diameter of the piston Vjf a steam engine is 60 in. ; it makes 11 strokes per minute ; the length of each stroke is 8 ft. ; the mean pressure per square in. 15 lbs. The modulus of the engine being 0-65, determine the number of cubic feet of water jt will raise per hour from a depth of 50 fathoms. [The number of foot-pounds done by steam on piston in one hour equals irx30'x8xloxllx60; this number multiplied by 0-65 will give the number of foot-pounds usefully spent in raising water; hence the number of cubic feet of water is found.] Jns. 7763 cub. ft. Ex. 95.— The diameter of the piston of an engine is 80 in., the mean pressure of the steam is 12 lbs. per square inch, the length of the stroke is 10 ft., the number of strokes made per minute is 11. How many cubic feet of water will it raise per minute from a depth of 260 fathoms, its modulus being 0-6 ? Am. 42-46 cub. ft. Ex. 96. — If the engine in the last example had raised 55 cubic feet of water per minute from a depth of 250 fathoms, what would have been its modulus? Ans. 07771. Ex. 97. — How many strokes per minute must the engine in Ex. 96 make in order to raise 1 5 cubic feet of water per minute from the given depth ? Ans. 4. Ex. 98.— What must be the length of the stroke of an en^ne whose modulus is 0'66, and whose other dimensions and conditions of working are the same as in Ex. 95, if they both do the same quantity of useful work ? Ans. 9-23 ft. Ex. 99.— The diameter of the cylinder of an engine is 80 inches, the piston makes per minute 8 strokes of lOJ ft. under a mean pressure of 15 •lbs. per square inch ; the modulus of the engine is 0-55. How many cubic feet of water will it raise from a depth of 112 ft. in one minute ? Ane. 485-78 cub. ft. Ex. 100.— If in the last example the en|^ne raised a weight of 66,433 lbs. through 90 ft. in one minute, what must be the mean pressure per square inch on the piston ? j^_ 26-37 lbs. Ex. 101.— If the diameter of the piston of the engine in Ex. 99 had been 85 in., what addition in horse-.power would that make to the useful power of the engine ? ^^ ig.gg h.-P. 15. Work of water-wheds. — Hitherto we have con- sidered only one kind of motive power, viz. the pressure of steam. The same principles are appUcable to machines worked by any other motive power, as by the muscular force of animal agents, the pressure of moving air, or of faUing water. The last of these, viz. the power of falling water, is, next to steam, the most conspicuous example of WOKK OF WATER-WHEELS. 25 work done on a large scale by an inanimate agent. We shall therefore consider somewhat particularly the appli- cation of this power by means of water-wheels. It is plain that 1 lb. of water, in descending through 1 foot, must accumulate as much work as would be required to raise it through 1 foot, and hence if P lbs. of water descend through h feet, they will accumulate P h foot-pound of work ; and if, moreover, we suppose this water to descend .^.gainst an obstacle, such as the float boards of a water- wheel, the amount of work so accumulated will be done upon the wheel, and this work may then be applied to any useful purpose after a certain deduction has been made on account of prejudicial resistances. It must be borne in mind that the height of the fall is the difference between the levels of the surface of the water EIG, 2. FlQ. 3. in the reservoir and in the exit canal or tail-race ; in the case of overshot wheels it is supposed that the extreme circumference of the wheel is just in contact with the sur- face of the water in the tail-race. The height is represented by A B in the accompanying figures ; of which fig. 2 repre- sents the ordinary undershot wheel with plane float boards ; fig. 3 the breast wheel, in which the water acts upon the float boards considerably above the level of the tail-race. Fig. 4 represents the overshot wheel. 26 PRACTICAL MECHANICS. Fia. i. The following table exhibits the moduli of various kinds of water-wheels. It is founded on results given in General Morin's Aide-Mimowe. In the table H denotes the length of the line A B in figs. 2, 3, 4, and h denotes the length of B c in fig. 3 : — Table VIL MODULI OF WATER-WHEELS. Description Modulus (1) Undershot wheels, with flat float boards . 0-25 to 0-30 (2) Breast wheels with flat float boards (a) when — = y . (b) „ 1=1. . . . (0 „|=|. . . . (d) ., I=l. . . . (e) „ ^=1. . . . 0-40 to 0-45 0-42 „ 0-49 0-47 0-55 0-65 „ 0-70 (3J Breast wheels with curved float boards (Pon- celet's construction) for H greater than ^ feet 60 to 0-65 (4) Overshot wheels, when the velocity is small and the buckets half filled 0-70 to 0-75 WORK OF WATER-WHEELS. 27 Ex. 102. — The mean section of a stream is 5 ft. by 2 ft. ; its mean velo- city is 36 ft. per minute ; there is a fall of 13 ft. on this stream, at which is erected a water-wheel whose modulus is 0'65 ; determine the horse-power of the ■wheel. Am. 66 H.-P.. Ex. 103.— In how many hours ■would the wheel in the last examplegrind 1000 quarters of wheat, it being assumed that each horse-po^wer will grind 1 'bushel per hour? Atis. 1428 hours. Ex. lot. — How many quarters of wheat ■will the same wheel grind in 72 hours? Ans. 50'41 quarters, JSx. 105,— Suppose the wheel in Ex. 102 to have replaced an undershot ■wheel ■with flat float boards, ■whose modulus was 0'25, determine the num- ber of quarters of wheat each ■wheel will grind in 24 hours. Ans. (1) 6-6. (2) 16-8. Ex. 106. — Ho^w many cubic feet of water must be made to descend the fall per minute in Ex. 102, 3, that the ■wheel may grind at the rate of 3i quarters per hour ? .4ms. 1749'5. Ex. 107. — Given the stream in Ex. 102, 3, what must be the height of the fall to grind Ij quarters per hour; flrst, if the modulus of the wheel is 0-40, next, if it is 0-47, and lastly, if it is 0*65 ? . ■ J»s. (1)37-7 ft. (2) 32 ft. (3) 23-2 ft. Ex. 108. — The mean section of a stream is 8 ft. by 1 ft. ; its mean velo- city is 40 ft. per minute ; it has a fall of 17^ ft. ; it is required to raise water to a height of 300 ft. by means of a water-awheel ■whose modulus is 0'7 ; ho^w many cubic feet will it raise per minute ? Ans. 13'07 cub. ft. Ex. 109. — To what height ■would the wheel in the last example raise 2 J cubic feet of -water per minute ? Ans. 1742| ft. Ex. 110. — The mean section of a stream is Ij ft. by 11 ft. ; its mean velocity is 2| miles per hour ; there is on it a fall of 6 ft. on which is erected a wheel whose modulus is 0'7 ; this -wheel is employed to raise the hammers of a forge, each of which weighs 2 toiis, and has a lift of 1 J ft. ; how many lifts of a hammer will the wheel yield per minute ? Ans. 142 nearly. Ex. 111. — In the last example determine the mean depth of the stream if the wheel yields 135 lifts per minute. Ans. 1'43 ft. Ex. 1 12. — In Ex. 110 how many cubic feet of ■water must descend the fall per minute to yield 97 lifts of the hammer per minute ? Ans. 2483 cub. ft. Ex. 113. — Determine how many quarters of corn the mill in Ex. 110 might be made to grind in six days if it were to ■work for 13 hours daily. Ans. 281-5 quarters. , Ex. 114.— Down a 14-ft. fall 200 cub. ft. of ■water descend every minute, and turn a wheel v?hose modulus is 0-6. The wheel lifts water from the bottom of the fall to a height of 54 ft. ; how many cubic feet ■wiU be thus raised per minute ? If the ■water were raised from the top of the fall to the same point, what would the number of cubic feet then be ? Am. (1) 31-1 cub. ft. (2) 34-7 cub. ft. [Of course in the second case the number of cubic feet of water taken from the top of the fall being x, the number of feet that turn the wheel will be 200 -a.] 28 PRACTICAL MECHANICS. Ex. 115— Water has to be raised from a mine 120 ft. deep, the whole of the water raised forms a stream with a fall of 30 ft., the machinery by which the water is raised is worked by a steam engine of 60 horse-power, and an overshot wheel whose modulus is 0-716 turned by the steam ; determine the whole number of cubic feet raised per minute. Ans. 267-8 cub. ft. Ex. 1 16.— In the last example, if the ground allowed an exit to be made for the water 30 ft. below the mouth of the shaft (by which of course the fall is entirely lost), what must be the horse-power of the engine to raise per minute the same amount of water as before ? Ans. 48-6 H.-P. 16. The work of living agents. — The efficiency of men and animals is estimated in the same manner as that of the inanimate agents already considered, viz., by the number of foot-pounds of -work they are capable of yielding in a given time. The number yielded under given circumstances by any particular agent must of course be determined by expe- riment. The results of experiment on this matter are re- gistered in the tables that follow; they are based on similar tables given in General Morin's Aide-Mimoire. It must be borne in mind that these tables give mean results when the agent works in the best manner. It would be very possible for the agents to work -with greater velocities than those assigned, but were this done they would yield a much smaller daily amount of work — compare the work done by a horse walking with that done by a horse trotting. Table VIII. WORK DONE BY MEN AND ANIMALS. Nature of Labour Daily Dura- tion of ■ft'ork in Hours Eoot-poonds per Day Foot- pounds per Min. Founds raised or Moan Force exerted Velocity Feet per Min. Miles Hour (1) Baising weights ver- tically. Amanmountinga, gentle incline or ladder with- out burden, i.e. raising his own weight. 8-0 2,032,000 4230 146 29 0-33 Labourer raising weights with rope and pulley, the rope returning ■vrithont load 6-0 568,000 1560 40 39 0-44 WORK OF LIVING AGENTS. Table VIII. ( 29 Nature of Labour DaUy Dura- tion of Work in Hours Foot-pounds per Day Foot- pounds per Min. Pounds raised or Mean Force exerted Velocity Feet per Min. Miles per Hour Labourer lifting weights by hand 60 631,000 1480 44 34 0-38 Labourer carrying ■weights on his back ' up a gentle incline or up a ladder and re- , turning unladen 6-0 406,000 1130 146 8 009 Labourer wheeling ma- , teriala in a barrow up an incline of i in 12 and returning with the empty barrow - 100 313,000 520 130 4 0-046 Labourer lifting -earth with a spade to a mean height of 6j- feet 100 281,000 470 6 78 09 (2) Aotion cm Machines. Labourer walking and pushing or pulling horizontally 8-0 1,500,000 3130 27 116 1-32 Labourer turninga winch 8-0 1,260,000 2600 18 144 1-64 Labourer puUing and pushing alternately in a vertical direction 80 1,146,000 2390 XI 216 2-70 Horse yoked to a cart and walking 100 1 15,688,000 26,160 150 175 2-00 Do. to a whim gin 8-0 8,440,000 17,600 100 175 2-00 Do. do. trotting 4-6 7,036,000 26,060 66| 391 4-44 Ox yoked to a whim gin and walking 80 8,127,000 16,930 145 117 1-33 Hul» do. do. 8-0 5,627,000 11,720 66| 176 200 Ass do. do. 80 2,417,000 6030 30 168 1-95 30 PRACTICAL MECHANICS, The following table gives the useful effect of men and animals employed in the horizontal transport of burdens. The second and third columns give the useful effect, viz. the product of the -weight in lbs. and the distance in feet. The reader must not mistake this for the fooi>-pounds done by the agent, the agent being employed not in rais- ing the weight, but in overcoming the passive resistances, friction, &c., which depend on the weight indeed, but are only a fraction of it. Table IX, USEFUL EFFECT OF AGENTS EMPLOYED IN THE HOETZONTAL TEANSPOET OF BUEDENS. Agent Dura- tion of Daily Work Useful Effect Daily TJseful Effect per Minute Pounds Trans- ported.* Velocity ' | Peet per Min. Miles ( perHr. Man walking on a horizontal road without Burden, i.e. transporting his own weight 10-0 26,398,000 42,330 146 292 3-32 Lahourer transport- ing materials in ii truct on two wheels, returning with it empty for a new load 10-0 13,025,000 21,710 22« 99 1-12 Do. do. in a wheel- barrow 10-0 7,815,000 13,030 180 ., 100 1-14 Labourer walking with a weight on his back 7-0 6,470,000 13,030 90 145 1-64 Labourer transport- ing materials on his back and re- turning unburden- ed for a new load 6-0. 5,087,000 14,110 145 97 110 * Exclusive of the weight of the barrow, truck, cart, &c. (Poncelet, JUeo. Ind: p. 247.) WORK OF LIVING AGENTS. 31 Table IX. ^ccmtinwd). Agent Dura- tion of Daily Work Useful Effect DaUy Useful Effect per Minute Pounds Trans- ported Velocity Feet per Min. Miles per Hr. Do. do. on a hand- barrow 10-0 4,298,000 7160 110 65 0-74 Horse trsaisporting' materials in aoart, ■walking, always laden 10-0 200,582,000 334,300 1500 223 2-53 Do. do. trotting 4-5 90,262,000 334,300 750 446 5-06 Do. transporting materials in a cart, returning with the cart empty for a new load 100 109,408,000 182,350 1500 121 1-38 Horse walking with a, weight, on his back 100 , 34,385,000 57,310 270 212 2'41i DOi ds. trotting 7-0 32,092,000 76,410 180 424 . 4-82. Ex. 117. — How many men would be required' to raise by means of a capstan an anchor weighing 1 ton from a depth of 30 fathoms; ini 1 5 minutes ? Jns. 9 nwarlyc Ex. 1 18. — In what time would 20 men raise the anchor in the last example ? Am. 6'4 min. Ex, 119. — Through how great a distance would 30 men raise the' anchor in Ex. 117 in each minute ? Ans. 42 ft. nearly. Ex. 120. — There is a well 150 ft. deep, a labourer raises water from it by a rope and pulley ; how many cubic feet of water will hei raise in a day ? Ana. 60 cub. ft; Ex. 121. — How many cubic feet) of water would a steam' engine of 10 horse-power raise, from this well in 24 hours ? How many labourers would be required to do the same, amount of work if they raised the watei by ■wheel-and-axles, and how many if they raised it by means of capstans.? How many horses woiild do the same amount of work walking in whim gins? ^»ss. (1) 50,688 cubic feet. (2) 380 labourers. (3) 317 labourers. (4) 56 horsesi Ex. 122. — In how many minutes could 20 men. working on a capstan raise an anchor weighing 2 tons from a depth of 200 fathoms ? Ans. 85*88 min. 82 PRACTICAL MECHANICS. Ex. 123. — How many men would in 40 minutes raise the anchor in the last example ? ■^'**- *3 men. Ex. 124.— Through how many fathoms could 15 men raise the anchor of Ex. 122 in 10 minutes ? ^«»- 174 nearly. Ex. 125. — If 1 3 men are required to raise an anchor through 180 fathoms in 20 minutes, what must be the weight of that anchor ? Ans. 753J lbs. Ex. 126. — ^A town is situated 25 miles from the mouth of a coal pit, from which coal is taken to the town by a level railway on which the re- sistance is 10 lbs. per ton; the engine employed is of 15 horse-po-wer and weighs with its tender 10 tons ; each truck weighs 3 tons and contains 7 tons of coal ; on each journey the engine takes 5 full trucks and returns with 5 empty trucks ; supposing no time to be lost at the ends of the journey, how many tons of coals will be taken to the town in 48 hours ? How many horses would be required to conyey the same quantity of coals in the same time ? Ans. (1) 445 tons. (2) 665 horses. 17. Remarks on the work yidded by different agents. — The following remarks upon the preceding tables and examples are worthy of the attention of the reader : — (1) Every agent must be allowed to move at a certain rate in order to do the greatest amount of work it is cap- able of yielding ; thus, a horse walking does considerably more work than a horse trotting, as an inspection of the tables will show. And this is true not of animate agents only, but also of inanimate ; thus the work yielded by the consumption of a given quantity of coal will be larger in, the case of a slow than of a fast engine. (2) Also, in order that an animate agent may do its greatest amount of work, it must not be required to exert more than a certain force. This is also plain from an inspection of the table. (3^ It follows from the above considerations that though two agents may be capable of doing the same work in the same time, it may be in practice impossible or dis- advantageous to substitute the one for the other. Thus an ox and a horse walking in a whim gin do very nearly the same amount of work ; but since the ox moves more slowly, and exerts a greater force than the horse, it would generally be disadvantageous to substitute a horse ON THE COST OF LABOUR. S3 for an ox in a machine requiring a slow heavy pressure. Again, in cases where great speed is a desideratum, it ■would generally be impracticable by any machinery to make the slow agent perform the labour of the rapid agent ; as, for instance, in the case of locomotion. 18. On the cost of labour. — The chief elements in the cost of labour may be enumerated as follows : — (1) In the case of human labour, the whole cost is the wages paid. (2) In the case of a horse, the elements of expense are attendance, keep, and the original cost ; the last is but a small portion of the expense. Thus, if we suppose a horse to cost 20l. and to continue in working order for ten years, and reckon the value of money at four per cent, per annum, the element of cost would be 2*465Z. yearly, or not quite Is. per week. ("3) In the case of a steam engine, the chief elements are the original cost and subsequent repairs, attendance, and fuel. Of these elements the most important is that of fuel ; and accordingly there is a special definition of the power of an engiiae with reference to the consujnption of fuel. The definition is as follows : — Def. — The numbjer of foot-pounds of work yielded by an engine in consequence of the consumption of 1 bushel (i.e. 84 lbs.") of coal, is called the duty of that engine. The extent to which the economy of fuel may be carried is very remarkably illustrated by the engines employed to drain the mines in Cornwall. In 1815, the average duty of these engines was 20 millions; in 1843, by reason of successive improvements, the average duty had become 60 millions, effecting a saving of 85,000Z. per annum ; *■ * Soime on the Steam Engine, p. 171. It may be remarked that this result depends largely on the construction of the boiler ; 1 lb. of coal in the Qornish boiler evaporates 11^ lbs. of Tatar, while in the waggsn-shaped boiler 87 is the masioium.— Faibbairu , ■ Z7a«/«i Information, p. 177. D 84 PRACTICAL MECHANICS. it is stated also, that, in the case of one engine, the duty was raised to 125 millions. The actual cost of 1 ,000,000 foot-pounds of work, when done by dififerent agents, cannot be specified with great precision ; but a sufficiently accurate notion of the relative cost of different agents may perhaps be obtained from the annexed table, which has been calculated upon the follow- ing suppositions : — (1) The wages of a labourer, Ss. a day. (2) Keep of a horse, 2s. a day ; attendance of 6 horses, 33. a day ; cost of each horse, 2d. a day. (3) Steam engine of 50 horse-power, at an annual cost of 51. per horse-power; attendance, 12s. a day; coal, 6d. a bushel.* Table X. COST OF LABOUR. Character of Agent Coat per Million Foot-pounds (1) Labourer carrying weighte up a ladder • 88-67 pence (2) Labourer raising weights by rope and pul'ey . 63-94 „ (3) Labourer turning a -winch .... 28-80 „ (4) Labourer turning a capstan 2400 „ (5) Horse in a whim gin trotting . 4-648 „ (6) Horse in a whim gin walking . 3-791 „ (7) Horse walking in a cart . 2-040 „ (8) Steam engine, duty 20 millions . 0-429 „ (9) Steam engine, duty 90 millions .... 0-196 „ * In Wealds Contractor' t Price Booh for 1859 the prices of various steam engines are estimated to be from 2bl, to 35J. per horse-power, boilers and fittings included ; as the nominal horse-power (which is determined by measurement) is considerably less than the working horse-power, the estimate in the text is very ample ; that estimate assumes 502. to be the cost of a horse-power, and that 10 per cent, will represent interest on capital, repairs, and restitution. It may interest the reader to consider the following statement taken from Mr.R. Stephenson's paper on Railway. Economy which forms an appendix to Mr. Smiles's Life of George Stepheiuoa. In 1854 there were in the United Kingdom 5,000 locomotive engines costing from 20002. to 25002. apiece, and consuming annually 13 million tons of coke, made from 20 million tons of coal. It appears moreover that if a railway ON THE COST OF LABOUE. 35 Ex, 127^ — How many bushels of coal must be expended in a day of 24 hours in raising 150 cubic feet of water per minute from a depth of 100 fathoms ; the duty of the engine being 60 millions ? Ans. 13d bushels. Ex. 128. — ^Determine the number of horses working in whim gins re- quired to do the work of the last example. Determine also the weekly saving effected by employing steam power, supposing the total weekly ex- pense of the engine to be double the price of coals consumed ; the coals costing 10s. a ton ; and each horse 20s. a week, Ans. (1) 960 horses. (2) 92il. 11«. Od. weekly saving. Ex. 129. — There are three distinct levels to be pumped in a mine, the first 100 fathoms deep, the second 120, the third ISO ; 30 cubic feet of water are to come from the first, 40 from the second, and 60 from the third per minute ; the duty of the engine is 70 millions. Determine its working horse-power and the consumption of coal per hour, >ims. (1) 191 H.-P. (2) 5-4 bushels. Ex. 130. — In the last example suppose there is another level of 160 fathoms to be pumped, that the engine does as much work as before for the other levels, and that the utmost power of the engine is 275 H.-P. Find the greatest number of cubic feet of water that can be raised from the fourth level. Ans. 46j cub. ft. Ex. 131. — An engine raises every minute A cubic feet of water from a depth of a fathoms, b cubic feet of water from a depth of b fathoms, and c cubic feet of water from a depth of c fathoms. The diameter of the piston of the steam engine is d in., the length of the stroke I ft., it makes n strokes per minute ; also it consumes o bushels of coal in twenty-four hours, and has a modulus m. Determine (1) the pressure per square inch upon the piston ; (2) the horse-power of the engine (as measured by pressure of steam on piston) ; (3) its duty. . ,.,1500(Aff, + Bi + cc) ,„,Ao + B6-i-ce "*■ ^ ' irdHmn • ' 88m .„. 540,000(Aa + Bft + ce) oiH Ex. 132. — Water is to be raised from three levels of 20, 30, and 40 fathoins respectively; 10 cubic feet of water are to be taken per minute from the first, 20 from the second, and 40 from the third. The engine con- sumes 15 bushels of coal in a day. The diameter of the piston is 4 ft., it makes 10 strokes of 6 ft, each per minute. The modulus of the engine is 0'65. Find the pressure per square inch on the piston, the horse-power (as measured by pressure of steam) and the duty of the engine. Atis. (1) 12-75 lbs. (2) (nearly) 42 H.-P. (3) 133,000,000 duty. company start with lOO new engines, about 20 or 26 will need repair at the end of four years, and after that there will always be about 26 in the workshop. s 2 38 PRACTICAL MECHANICS. Ex. 133. — In Ex. 126 suppose the engine and tracks on the one hand, and tlie horses and carts on the other, to want renewal every ten years ; suppose also that each horse and cart costs iOl., that one man attends to every six horses and is paid 3s. a day, that each horse's keep is 1«. 6d. a day, that there are two tnmpikes on the road at each of which there is a toll of 6d. ; determine the cost of transporting 445 tons of coals. Next suppose the engine and tender to cost 1000^., each truck 1201. (15 trucks are leqnired to prevent loss of time) ; that there are three drivers and three stokers each at 6s. a day ; that money is worth 5 per cent., and that each mile of ruad cost 10,000/. to make, and 365/. a year to keep in repair ; determine in this case the cost of transporting 445 tons of coals. Also if coal cost 3«, a ton at the pit mouth, what will it cost in the town according to each method of transport, neglecting profit ? Ans. (1) 214/. (2) 123/. (3) 12«. 6d. a ton ly cart. (4) 8s. 6d. a ton by rail. [Interest on the cost price of engine, trucks, horses and carts can be neglected.] Section II. 19. On the work done by a variable force. — ITiere are two important questions in the subject of work which we shall treat in the present section : they are, (1) the work done by a variable force, when exerted through a certain distance ; (2) the total amount of work done in Fig. 5. raising a number of weights through differ- ent heights. As an introduction to the theorem which follows, it may be remarked that, if a con- stant force of p lbs. act through a distance of s feet, and if a rectangle a B c D be drawn, of which the base a b represents the s feet on scale, and the perpendicular ad represents the p lbs. on the same scale : then, since the area of a B c d contains p s square units on the same scale, the area will correctly represent the work done by p. WORK DONE BY A VARIABLE FORCE. 37 Proposition 1. If a variable force act through a certavii distance, and if a curve be drawn in such a manner that the ahscissa and correspondi/ng ordinate of any point represent re- spectively the distance through which the force has acted and the magTdtude of the force, then will the area of the curve between any two ordinates represent the work done by the force while acting through a distance represented by the difference between the extreme abscisace. When the force has acted through a distance represented on a certain scale by a n, suppose it to be represented on the same scale by pn ; also, fm. o. when it has acted through a distance a m, suppose it to be represented on the scale by Q M ; let the curve P Q be drawn in such a manner that any ordinate (P3N3) _ represents the force when it ^ T^ -NTTir-N-N.-N. "m" has acted through a distance AN3 ; we have to prove that the area pNmq represents the work done by the force while acting through the distance N M. Divide n m into any number of equal parts in N„ Nj, N,, .... draw the ordinates PiN,, PjNj, PgNj .... and com- plete the rectangles pn,, PjN^, PjN, .... Now, we shall nearly represent the actual case if we suppose the force, while acting successively through the short distances nn,, Jir,N2, Njifg .... to retain unchanged the magnitude it has at the beginning of those distances respectively ; and we shall represent the case more nearly the smaller we make the distances, i.e. the greater the number of parts into which we divide N M : the actual case being the limit con- tinually approached as the number of parts is increased. But if the force acts uniformly through each distance, p PS /I I' PS u p<^^ ^ It 38 PRACTICAL MECHANICS. it will do a number of units of work represented hy the sum of the rectangular areas PN,, y^s^, PjNj . . . ., and this being true whatever be the number of the small distances, the work actually done will be properly repre- sented by the limit of the sum of these rectangles, i.e. by the curvilinear area p N m Q. Cor. — It must be borne in mind that the scale must be the same for lbs, and for feet ; thus, if the scale be in inches, P N must be as many inches long as the force con- tains lbs., and N m must be as many inches long as the distance represented contains feet ; this being so, the area of the curve in square inches will give the number of foot- pounds of work. Ex. 134. — A rope I ft. long and ■weighing w lbs. per foot hangs by one end ; determine the number of foot-pounds of work required to wind up a ft. of the length. Fio. 7. Take a b on scale equal to I, draw A c at right angles to A b and on the same scale equal to w I, join B c ; in A B take any point N, draw P N parallel .to a c, then, ; pir:NB::cA;AB::M;: i. Therefore pn = to n B, i.e. the ordinate pk represents on scale the weight of the rope left hanging when the extremity has been raised through a space a n. Therefore the area ABC represents the number of foot' pounds required to wind up the whole rope, and the area c a p m the number of foot-pounds required to wind up a length A N of the rope. Hence if u is the required number of foot-pounds, Hence also the number of foot-pounds (tt,) required to wind, up the whole rope is given by the formula Ex. 135. — ^A weight of 2 cwt. has to be raised from a depth of. 100 fathoms by a rope 3 in. in circumference ; determine the number of foot- pounds that must be expended in raising it, and the number of minutes in which 4 men would do the work by means of a capstan. Am. (1) 207,300 ft.-pds. (2) 16-5 min. "WORK DONE BY A VARIABLE FORCE. 39 Ex, 136. — How heavy will that anchor he which 13 men will raise by means of a capstan from a depth of 180 fathoms in 40 min., supposing the cable to weigh 1125 lbs. (neglecting the buoyancy of the water) ? Ans! 945 lbs. Ex. 137. — A chain each foot of which weighs 8 lbs. is suspended from the top of a shaft, the depth of which is 50 fathoms ; determine the number of foot-pounds required to wind up each successive 100 ft. of its length ; determine also the length of the chain which will require twice as many foot-pounds to wind it up. Ans. (1) 200,000, 120,000, 40,000 ft.-pds. respectively. (2) 424 ft. Ex. 138. — If a chain 300 ft. long and weighing 8 lbs. per foot is wound up in 4 min., how many men working on a capstan would do it? How many horses walking in a whim gin ? How maiiy steam horses ? How many of each agent would be required if the Weight per foot of the chain were doubled ? And how many if the length of the chain were doubled ? Am. (1) 29 men. (2) 5-1 horses. (3) 2^ hbrse-power. (4) 57 men. 10-2 horses. 5^ horse-power. (5) 115 men. 204 horses. 1012 horse -power. Ex, 139. — A chain is a ft. long ; divide it into ». parts such that the winding up of each may require the same number of foot-pounds. Ans, a a ^(V«- Vw-l), -^(Vw-l- v'»-2), —=.(Vn-2~ .v/»-3)&c. V« Vn • Vn Ex. 140. — Coal is raised from the bottom to the mouth of a pit 150 ft. deep in loads of a quarter of a ton ; the box containing it weighs 1 ewt., the rope by which it is raised is 3 in. in circumference ; determine the number of foot-pounds spent in raising the coal, and the number spent in raising the box and rope. If the lifting engine works with 10-horse power, determine the weight of coals raised in 2 hqurs, supposing the ascent and descent of the box to take equal times. Ans. (1) 84,000 ft.-pds. to raise coal. (2) 21,3561 ft.-pds. to raise • box and rope. (3) 47 tons. Ex. 141. — In the last example suppose machinery to be employed by means of which the same drum winds up the rope of an ascending box and unwinds that of a deseeding box. Determine thenumber of tons raised in 2 hours.* ~ Ans. 118 tons. [Of course the work done by the descending' box and rope will nearly 6qual that expended onnhe aseejiding box and rope — the weight of box and rope can therefore be neglected-.] .- * The primary object of .this mod^ of working was, probably, to save time, the saving of labour being an accidental result ; though that saving is very considerable. 40 PBACTICAL MECHANICS. Ex. 142. — SetermiDe the num'ber of tons raised under the conditions of Ex. 140 and 141, supposing h a minute is expended in filling or emptying the box. ' ^««. (1) ISl^tons. (2) 39| tons. Ex. 1 13.— If 4 cwt. of material are drawn from a depth of 80 fathoms by a rope 5 in. in circumference, how many foot-pounds are expended in raising them, and what horse-power is necessary to raise them in 4§ minutes ? Ans. (1) 344,640 ft.-pds. (2) 2-32 H.-P. Ex. 1 44. — A Tope 3 in. in circumference is strong enough to bear a working tension of 4 cwt. ; how many foot-pounds are wasted in the last example by using a rope 5 in. in circumference? Ans. 82,944 ft.-pds. Ex. 145. — ^A winding engine raises to the surfece a load of 12 cwt. in 6J minutes from a depth of 115 fiithoms ; tlie rope employed is a flat rope composed of 3 ropes each 3 in. in circumference. What is the horse-power of the engine ? Ans. 5-67 H.-P. Ex. 146. — If the engine in the last example have a cylinder 20 in. in diameter, and makes per minute 15 strokes of 2 ft. 10 in., under what mean pressure per square inch of steam does it work if its modulus is 0-55 ? Ans. 25'5 lbs. 20. The steam indicator. — A very instructive appli- cation of Proposition 1 occurs in the steam indicator, which Fig. 8. may be suflBciently described as fol- lows: AB is a small hollow cylinder containing a powerful spring, which can be partly setn through the aper- ture EF; within the indicator is a small piston or plunger (marked in the figure by dotted lines^ which is kept down by the spring, so that if it is forced up, the compression of the spring gives the amount of the com- pressing force, which can be read off on the scale CD by means of the pointer G H, which rises and fells with the plunger. The end H of the pointer carries a pencil, the point of which rests against a sheet of paper wrapped round a cylinder K L ; if this cylinder be stationary, and the pencil move, a vertical straight line will be described ; if the pencil be THE STEAM INDICATOR. 41 stationary, and the cylinder revolve, a horizontal straight line will be described ; but if both the pencil move and the cylinder revolve, a curved line -will be described. To obtain the required curve it is necessary that the cylinder K L should turn in contrary directions during the up and down strokes of the piston. This is efifected by means of a clockspring placed within the cylinder k l. On the up stroke the string m n, which is fastened round the cylin- der, is pulled in the direction M N, causing the cylinder to turn from left to right and winding up the spring. On the down stroke the string tends to slacken, the spring uncoils and turns K L back from right to left. The instrument is used in the following manner : — The end A being screwed into an aperture properly con- structed, the steam in the interior of the cylinder of the steam engine can be admitted into the indicator by opening the cock p ; at first, however, the cock p is shut, so that the pointer remains stationary. The end of the string MN is attached to some part of the engine * in such a manner that the cylinder k l makes one revolution while the piston of the steam engine makes a stroke ; this being done, and the cock kept shut, the pencil will trace on the paper a straight line called the atmospheric line: on the next stroke the cock is opened, and now the steam pressing on the plunger the pencil will rise or fall according as the pressure of the bteam is greater piq. s. or less than that of the atmo- sphere, and will describe a curve that will return into itself at the end of a double stroke (or revolu- tion). The area of the curve thus described will give the amount of work done by the steam during a single stroke. * Generally the radius-shaft. 42 PRACTICAL MECHANICS. To explain this, suppose ABCDKFtobe the curve given by the indicator (which, it may be remarked, is described continuously in the direction abcdefa), AG the atmo- spheric line; draw PNQ any double ordinate, then PN represents the excess of the steam pressure above that of the atmosphere when the ascending piston is at a certain point, and N Q represents the defect of the vacuum pres- sure below that of the atmosphere when the descending piston is at the same point. Now the effective pressure of the steam is the excess of the steam pressure above the vacuum pressure ; but p N = steam pressure — atmospheric pressure, NQ=atmospheric pressure —vacuum pressure, .•.PN + NQ=steam pressure —vacuum pressure ; therefore P Q represents the effective pressure of the steam when the ascending piston is at the point corresponding to N, i.e. assuming the vacuum pressure at any point of one stroke to be the same at the same point of the next stroke. If, then, for the sake of distinctness,* we suppose each inch of the ordinate to denote a pressure of 1 lb. and each inch of the abscissa (i.e. of the atmospheric line) to denote a foot of the stroke, the area of the curve will give the number of foot-pounds of work done during a single stroke by the steam on an area equal to that of the plunger, and if the area of the piston of the steam engine be n times that of the plunger, the work done by the steam during a single stroke will be n times that given by the curve. The area of the curve may be found by Simpson's rule, viz. — Divide A G into any even number of equal parts, and draw the corresponding ordinates ; take the sum of the extreme ordinates, four times the sum of the even ordi- nates, and twice the sum of the odd ordinates (i.e. ex- * In practice the scale would be considerably less than this. Fia. 10 \ ^^ ^ ~- ^ WORK OF LENGTHENING BARS. 43 '06pting the first and last), add them together, and multiply the sum by one of the parts of the abscissa ; the product will be three times the area of the curve.* Ex. 147. — Let the curve shown in the figure be that given by a stroke of 5 ft. ; let A B be divided into 10 equal parts, and let the ordinates 1, 2, 3, 4, . . . .be drawn; suppose themtoirepresent respectively 19, 22, 22, 17'5, 13, 11, 9, 7'5, 6, 6"5, 4 lbs. pressures per square inch. The radius of the piston being 20 in., determine tlie number of foot>-pounds of ■work done per stroke, and the Tnean effective pressure per i«.»45«7b«ii)U square inch on the piston — i.e. the constant pressure that would do the same work. .4»«. (1) 79,000 ft.-pds. (2) 12 6 lbs. Ex. 148. — Determine the number of foot-pounds of work and the mean pressure per square inch on a piston 3^ feet in diameter having a stroke of 5 feet, if the ordinates measured at intervals corresponding to three inches of the stroke give the following pressures— 503, 12-67, 1804,20-73, 21-03, 21-11, 21-26, 20-72, 20-14, 18-63, 15-46, 1324, 10-83, 8-53. 6-49, 4-87,3-99, 3-74, 3-62, 3-25, 2-75. Jm. (1) 87, 600 ft.-pds. (2) 12-65 lbs. per sq. in. 21. Work expended on the elongation of bars. — It is plain that if a rod be lengthened by a gradually in- creasing force, the force exerted at any degree of elongation will be proportional to that elongation; so that if the abscissae represent the degree of elongation, and the ordinates the stretching force, the area which gives the units of work will be a triangle." Hence : Ex. 149. — There is a bar the length of which is r and section k; it is gradually elongated by a length I ; if its modulus of elasticity be b, show that the work expended on its elongation -will be given by the formula l'- TT = — KE. 2l Ex. ) 50. — The pumping apparatus of a mine is connected with the engine by means of a series of wrought-iron rods 200 ft. long ; the section of each rod is f of a square inch ; the tension when greatest is estimated at 6 tons ; how many foot-pounds of work are expended at every stroke upon the elongation of the bars ? ■^■ns. 830 ft.-pds. " The curve given by the indicator is useful in other ways besides that mentioned in the text. — Bourne on the Steam Engine, p. 246. a PRACTICAL MECHANICS. Ex. 151.— A bar of wrought iron 100 ft. long with a section of 2 square inches has its temperature raised from 32° F. to 212° F. ; how many foot- pounds of work has the heat done? An$. 3875 ft.-pds. 22. The work expended in raising weights through various heights. — The questions arising out of this im- portant part of the present subject are solved by means of the following proposition. Proposition 2. When any weights are raised through different heights, the aggregate of the work expended is equal to the work that would be expended in lifting a weight equal to the sum of the weights through the same height as that through which the centre of gravity of the weights has been raised. Letw„W2, Wj be the weights of each separate body ; conceive a horizontal plane to pass below them all ; let h^, Aj, ^3 . . , , be the heights of those bodies above the plane before they are lifted, and let H be the height of their common centre of gravity ; then (Prop. 16) h(w, + Wj + W3 . . . .) = w,A, + WjJj + w,A,+ ... (1) A\.SQ,\et \,k^,k^ .... be the heights of the weights respectively, after they have been lifted, and K the height of their common centre of gravity ; then S(W, + W2 + W, . . . .)=W,A, + W2i2 + W,A,+ ... (2) hence, subtracting (1) from (2), we obtain (k-h) (w, + w,+w,...) = w, (A,-A,) + Wj (Aj-Aj) + w, (A,-J,)... (3) Now, w„ Wjj, Wj, .... are severally raised through the heights k^—h^,k^—h^,k^—h^ . . . .; therefore the right-hand side of equation (3) gives the aggregate work expended in lifting them ; hence that work is equal to (K-fl)(Wi + W2 + W3 ....), RAISING WEIGHTS. 4» i.e. to the work that must be expended in hfting a weight W, + W2 + W3+ . . . through a height k-H. (Q. E. D.). Cor. — In the case of the transport of bodies along any parallel lines, the principle enunciated in the theorem will hold good, since the resistances bear a constant ratio to the weights. Ex, 152. — How many foot-pounds of work must be expended in raising the materials for building a column of brickwork 100 ft. high and 14 ft. square? and in how many hours would an engine of 2 horse-power raise them? ^»s. (1) 109,760,000 ft pdt. (2) 2771 hours. [Since the material has to be raised irora tha ground, the common centre of gravity will hare to be raised from the ground to the centre of gravity of the column, i.e. to its middle point 50 ft. above the ground.] Ex, 153. — A shaft has to be sunk to the depth of 130 fathoms through chalk; the diameter of the shaft is 10 ft. ; how many foot-pounds of work must be expended in raising the materials ? In how long a time could this be done by a horse walking in a whim gin ? How many men working in a capstan would do it in the same time ? Determine the expense of the work supposing the horse to cost 3s. 6d, a day, and the wages of a labourer to be 2s. 6d, a day. ' Ans, (1) 3467 million ft.-pds. (2) 409-6 days. (3) 6-62 men. (4) Cost of horse lU, lis. Cost of men 288?. Ex. 154. — If the work in the last example is to be done in 24 weeks by a steam engine working 8 hours a day, 6 days a week, what must be the horse-power of the engine? Ans. 1'521 H.-P. Ex. 155. — In Ex. 153 suppose the box in which the material is raised to weigh I cwt., the rope to be 3 in. in diameter, and each load to be 4 cwt. of chalk, also suppose the box to take as long in ascending as in descending and that | of a minute is lost in unhooking and hooking at the bottom of the shaft and the same at the top ; when the shaft is 100 ft. deep determine the time that elapses between the starting of one load and the starting of the next ; the engine working at IJ horae-power. Ans, 2'62 min. Ex, 156.— Determine the same as in the last example when the shaft is =" ft- deep. ^„,,. nix + 00io.v"-^ ^.. ^.^ 5500 Ex, 157. — ^Determine the whole time of raising the materials of the shaft in Ex. 153 under the conditions of Ex. 165. Ans, 3331 hours. Ex. 158. — Referring to Ex, 153, 156, suppose the drum of the winding machine to have two ropes wound round it in contrary directions, so that it unwinds one rope while winding up the other, and that consequently an 48 PRACTICAL JklECHANICS. empty tox descends while a full one is being raised (as iu Ex. 141); deter- mine the time that must elapse between two consecutive lifts of i cwt. ■when the shaft is 100 ft. deep. ' ^re«. 1-166 min. Ex. 169. — Obtain a determination similar to that in the last example, when the shaft is X ft. deep. , 448x .n.oi:„;_ Ans ■ + Zo mm. Ex. 160. — Obtain the whole time of lifting the materials from the shaft under the circumstances of Ex. 168. Ans. 1246 hours, Ex. 161. — In how long a time would a 16 horse-power engine empty a shaft full of water, the diameter of the shaft being 8 ft. and the depth 200 fathoms ? If the engine has a duty of 30 millions determine the amount ot coal consumed in emptying the shaft. Ans. (1) 76 hours. (2) 75-4 bushels. Ex. 162. — There is a certain railway 200 miles long ; it may be assumed that in the course of 10 years there will be 50,000 tons of iron railing laid down, and that it will be equally distributed along the line. How many foot-pounds of work must be expended in conveying the rails (neglecting the weight of the trucks), if the depot is at one end of the line ? And how many if the dep6t is in the middle of the lice ? The resistances being reckoned at 8 lbs. per ton. Ans. (1) 211,200 million ft.-pds. (2) 106,600 million ft.-pds. Ex. 1 63. — How many journeys of 200 miles performed by a train weigh- ing 50 tons does the difference of the results in the last example represent? Kesistances 8 lbs. per ton. Ans. 250 journeys. DEFINITION OF MECHANICS, 47 CHAPTEE III. ELEMENTARY STATICS. 23. Mechanics. — The science of Mechanics is that which treats of the motion and rest of bodies as produced by force. The words, ' as produced by force,' are added in order to exclude the science of pure motion or mechanisTn, which treats of the forms of machines, and in which ma- chines are regarded merely as modifiers of Tnotion. Into all questions which are properly mechanical the idea of force must enter. Force may be defined to be any cause which puts a body in motion, or which tends to put a body in motion when its effect is hindered by some other cause. On this definition the following remark is to be made : Suppose a given weight is supported by a string passing over a pulley and fastened to a fixed point at the other end ; next, suppose an equal weight to be supported by a man's hand ; lastly, suppose an equal weight to be supported by the elastic force of a spring. Now, here we have three physical agents, viz, the reaction of the fixed point transmitted through the stringj the muscular power of a man, and the elastic force of a spring, very different in many respects, but agreeing in their common capacity to support a given weight. They may clearly be regarded as equal, when viewed with reference to that capacity. In short, as, in geometry, we regard all bodies as equal which can successively fill the same space, without any regard to their physical qualities, such as weight, colour, &c., so in mechanics we regard all forces as equal which will severally balance by direct opposition a given weight 48 PRACTICAL MECHANICS. irrespectively of their physical origin. By the weight of a body is meant the mutual attraction between the earth and that body ; as this attraction has different amounts when the body is at different places, the weight of a body, when used as a standard of force, must be deter- mined with reference to some assigned place. Thus : — there is kept in the Exchequer Ofi6.ce a piece of platinum called the standard pound (avoirdupois) ; the attraction of the earth on that body at London is a force of 1 lb., and any force which by direct opposition can support that body in London is also a force of 1 lb. If we suppose two forces each of 1 lb. to act in the same direction at a point and to be balanced by a single force, that force is one of 2 lbs. ; and similarly a force of three, four, or more pounds can be defined. 24. Statics and dynamics. — It follows, from the de- finition, that, in Mechanics, we can consider a force either as producing motion, or as concurring with others in producing rest. Accordingly, the science of mechanics is divided into two distinct though closely connected branches^ viz. statics and dynamics. Of these, statics is that science which determines the conditions of the equi- librium of any body or system of bodies under the action of forces. Dynamics is that science which determines the motion, or the change of motion, that ensues in a body or system of bodies subjected, to the action of a force or forces that are not in equilibrium. 25. Determination of a Force. — From what has al- ready been said, it appears that the magnitude of any force is assigned by considering the weight it would just support if applied directly upward; in other words, we arrive at the magnitude of any force by comparing it with the most familiar and measurable of forces, viz. weight. A little consideration will show that the effect of a force in any case depends not only on its amount, but FORCES ACTING ALONG A LINE. 49 Fig. 11, also on its point of a/pplication, and the line along which it acts. We may say, therefore, in general terms, that a force is completely determined when we know (1) its matgnitude, (2) its point of appli- cation, (3) its line of action, and (4) its direction along that Une,* A Ime is frequently said to repre^ sent a force ; when this is the case it must be drawn from the point of application of the force along the line of its action, and must contain as many units of length (say inches) as the force con- tains units of force (say Itis.) It is of great importance that the student should attend to all the conditions which must meet when a line correctly represents a force. Suppose a force of p lbs. (fig. 11) to act on a body at the point A ; if the force is a p'lill, as in the first figure, the line A b containing as many inches as p contains lbs. will repre- sent the force; but if the force is a push, A B must be measured, as in the second figure. 26. Resultant and components. — If we consider any forces that keep a body in equilibrium, it is plain that any one of them balances all the others : thus, if three strings be knotted together at A, and be pulled by forces of p lbs., Q lbs., and e lbs. respectively so adjusted as to bal- ance one another, it is plainly a matter of indifference whether we consider that p balances Q and R, or that Q balances R and Fia. 13. * The student must notice the distinction between the line of a(iti(>n and the dwection of a force : e.g. in fig. 14 (p. 52) f and a act in the sarne direction along difierent lines. E 50 PRACTIOAL MECHANICS. p, or that H balances p and Q. Let us considet that E balances p and Q; now K Would of course balance a force k' exactly equal and opposite to itself: so that if we substitute e' for p and Q, or vice versd, P and Q for R', in either case E is balanced, and the force e' is equivalent to P and Q; under these circumstances, e' is called the resultant of p and Q ; and P and Q are called the com- ponents of b'. Hence we may state generally, Def. — That force which is equivalent to any system of forces is called their resultant. De/.^Those forces which fortn a systpin equivalent to a single force, are called its components. 27. Resultant of forces acting aiong the same straight line. — If' the forces act iii the same direction the re- sultant must be their sum. If some act towards the right and some towards the left, the first set can be formed into a single force (p) acting towards the right, the ; second set can be formed into a single force (q) acting towards the left : the resultant of these two, and therefore of the original set of forces, will be equal to the differ- ence between p and Q and will act in the direction of the greater. If the forces are in equilibrium, the sum of those acting towards the right must equal the sum of those acting towards the left. Ex. 164. — ^If three men pull on a rOpe to tlie right with forces of 31, 20, and 27 lbs. respectively, and are balanced by two men who pull with forces of 40 and p lbs. respectively, find p. Ana. 38 lbs. Ex. 165.— In the last example find the resultant of the 6 forces (1) if p = 30 lbs. ; (2) if p= 40 lbs. Jm. (1) 8 lbs. acting towards the ri^t. (2) 2 lbs. acting towards the left. Ex. 166. — There is a rope A b and men pull along it in the following manner: the iSrst with a force of 60 lbs. towards A; the second with la force of 37 lbs. towards b ; the third with a force of 35 lbs. towards A ; the fourth with a force of 20 lbs. towards a ; the fifth with a force of 64 lbs. towards b ; the sixth with a force of 27 lbs. towards A ; the seventh with a force of 52 lbs. towards b ; the eighth with a force of 30 lbs. towards B. PARALLEL FORCES. 61 Determine fhe single force that must act along a b to balance them, and find whether it acts towards a or b. Ans. 41 lbs. acting towards b. Ex. 16 7. — In the last example suppose the second force to act towards A, find the resultant. Ans, 33 lbs. acting towards a. 28. The terms reaction, thrust, and tension are of frequent occurrence in Mechanics, and it is important that their meaning should be distinctly understood. With regard to the first of them, it must be borne in mind that the only forces mth which we are acquainted are exerted between different portions of matter ; and if, for the sake of distinctness, only two bodies, a and b, are con- sidered, the following statement is found to be universally true : — If a exerts a force ons, then b exerts an equal oppo- site force on A — a fact commonly expressed by saying that to every action there is an equal opposite reaction. Now let AB (fig. 13) be a body urged by a force T against a fixed plane a c, and let the motion which T tends to commu- nicate to the body be prevented by the fixed plane ; that fixed plane must supply a force (k) which exactly balances T; and the body ab is really compressed between two forces R and t, of which the former is the ^^ ^^ Reaction of the fixed plane, and the latter the Thrust along ab. A Thrust and a Eeaction com/press or tend to com- press the body on which they act. If, on the contrary, the body (a b) had been acted on by two equal opposite farces T and E tending to produce elongation, it is said to sustain a tension T. One of the forces pro- ducing a tension may, of course, be a reaction ; thus, if one end of a string is tied to a nail fast in a post, and the other end to a suspended weight of 10 lbs., the string is stretched by two forces each of 10 lbs., viz. the weight and the reaction of the nail, and the string is said to sustain a tension of 10 lbs. s% ■ 52 PEACTICAL MECHANICS. 29. Beaultcmt of two parallel forces.-^F'ucst, let p and Q (fig. 14) be' the two parallel forces acting in the Pig. m. same direction at the points A and B ; join ^ AB and divide it in c in such a manner A that AC ; cb::q : p 1 then the resultant (e') equals P-fQ and acts through c along a line parallel to A P or b Q and in the same direction as p and Q. If c rests on a fixed point P and Q wiU balance about c and the fixed point wiU sustain a pressure e'. Secondly, let P and Q (fig. 15) be the two parallel forces acting at A and B in ^A A^' opposite directions. Suppose Q to be J I the greater. In ab produced take a ^ point c such that AC : cb::q : p T then the resultant (r') equals q— p and acts through along a line parallel to A p and B Q and in the same direc- tion as Q. If c rests on a fixed point p and Q will balance about C and the fixed point will sustain a pressure e'. Ex. 168.— :If weights of 12 lbs. and 8 lbs. are Ining from A and b respee- tirely, the ends of a rod 5 ft. long, and if the weight of the rod is neglected, determine the distance from A of the point round which these forces balance, and the pressore on that point. Ans. (1) 2 ft. (2) 20 lbs. Ex. 169. — Let a b be a rod 12 ft, long (whose weight is neglected), from A a weight of 20 lbs. is hung, 'and an unknown weight (p) from B, it is found that the two balance about a point 3 ft. tzom a ; determine f. Ans.6lVoa. Ex. 170. — If a weight of 16 lbs. is hung from the end a, and 12 lbs. from •the end b of a rod (whose weight is neglected), and if they balance about a Doint c, whose distance from a is 4J ft., what is the length of the rod? Am. 10 i ft. Ex. 171. — Draw a straight line a b, S ft. long ; forces of 5 lbs. and 7 lbs. act at A and b respectively at right angles to ab and in opposite directions. Determine their resultant. 1»ARALLEL FORCES* 63 30. Conditiona of equilihrium of three pwrcdlel forces. — In the last article we saw that the forces p and Q acting severally at A and B are equivalent to the force r' acting at c ; now r' will clearly be balanced by an equal opposite force R ; and therefore p and Q acting at a and B wiU be balanced by the force R acting at c. Hence the following conditions must be fulfilled by three parallel forces that are in equilibrium on a given body : — (a) Two of the forces (p and q) must act in the same direction, and the remaining force (r) in the opposite direction, the line along which the latter acts lying be- tween those along which the former severally act. (6) The sum of the former forces (p and q) must equal the latter force (r). (c) If any line be drawn cutting the lines of action of the forces (p, Q, R, in A, B, c, respectively) the portion of the line between any two forces is proportional to the remaining force, i.e. Bc : ca::p : Q CA : ab::q : R AB : bc':r : p 31. Centre of gravity. — Since each part of a body is heavy, it follows that the weight of a body is distributed throughout it; there exists, however, in every body a certain point called its centre of gravity, through which we may suppose the whole weight of the body to act, whenever that weight is one of the forces to be considered in a mechanical question. It admits of proof that the centre of gravity of any uniform prism or cylinder is the Huddle point of its geometrical axis : and as a uniform rod is merely a thin cylinder, its centre of gravity will be at its middle point. Ex. 172. — Two men, A and B, carry a weight of 3 cwt. slimg on a pole, the endB of which rest on their Bhoulders ; the distancfe of the- weight from A is 6 ft., and bom b is 4 ft. ITind the pressure sustained by each man. 54 PEACTICAL MECHANICS. If p is the pressure sustained ty A and a that sustained by b p + Q=3 owt. and 6 : 4 : : a : P therefore r= 1| cwt. and q = 1| cwt. Ex. 1 73.— There is a beam of oak 30 ft. long and 2 ft. square ; at a dis- tance of 1 ft. from one end is. hung a weight of 1 ton ; how far from that end must the point of support be on which the beam when horizontal wiH rest, and what will be the pressure on that point ? Jm. (1) 11-61 ft. (2) 9245 lbs. Ex. 174.— If a mass of granite 30 ft. long, 1 ft. high, and 3 ft. wide is supported in a horizontal position on two points each 3 inches within the ends (and therefore 29J feet apart), find the pressure on each point of support. ^^' 7383 lbs. Ex. 175. — If in the last example another mass of granite with the same section and half as long is laid lengthwise on the former, their ends being square with each other; determine the single force to which their two' weights are equivalent, and the line along which it acts, and hence the pressure on the two points of support, Ans, (1) Besultant acts 17°5 feet fi:om one end. (2) Pressures on points of support respectively 9197 and 12,950 lbs. Ex. 176. — If in the last case the upper block is shifted round through a right angle in such a manner that the middle point of the upper block is exactly over a point in the axis of the lower, and the end of the lower in the same plane with one face of the latter, determine the pressures on the points of support. " .4ks. 7695 lbs. and 14,452 lbs. Ex. 177. — ^A ladder AB, 50 &.long,weighsl201bs. ; its centre of gravity is 10 ft. from A ; if two men carry it so that its ends rest on their shoulders, determine how much of the weight each must support. If the one of them nearer to the end b is to support the weight of 40 lbs., where must he stand ? Ans. (1) 96 lbs. and 24 lbs. (2) 20 ft. from B. 32. The parcdlelogra/m of forces. — ^When two forces act at a point along different lines, their resultant is determined by the following rule, which is called the principle of the parallelogram of forces : — If two forces act at a point, amd if li/nes be drawn representing those forces, and on them as sides a parallelogram, be con- structed, that diagonal which passes through the point will represent the resultant of the forces. The student, when applying this principle to any particular ease, must bear in mind the meaning of the words a Ime represents a force (Art. 25). Fig. It, page 65. PARALLELOGRAM OF FORCES. 56 Ex. 178.— If at a point a. of a body two ropes ap and a a are fastened and are pulled in directions A p, aq at right angles to each other by forces of 120 and 100 lbs. respectively ; cletermine the magnitude and direction of the resultant pull on the point A. (See fig. a.) Along* A p measure on scale A b containing 120 units of length, and along A Q measure A c containing 100 units of length ; complete tjie rect- angle bc and draw the diagonal ad; this line represents the magnitude and direction of the resultant. In fig. a the scale employed is 1 in. for. 40 lbs. ; the results obtained by construction were the following r= 156-8 lbs. and PAB = 40° 6'; the measurement of the angle was made with a common ivory protractor, so that the number of minutes was determined by judgment : on calculating the parts of the triangle a b D, the results obtained were b = 1 56'2 lbs. and p A b = 39° 48'. It will be observed that when the construction is made on a small scale and with common instru- ments we can obtain by the exercise of moderate care a result that can be trusted to within the one-hundredth part of pjg_ ^g the quantity to be determined. The same remark applies to all the questions that were solved by the constructions from which the figures in the present volume were copied. If in this example the point A were to be pushed along the line A p by a force of 120 lbs., the resultant would of course be determined by the construction shown in the annexed figure. Ex. 179. — Draw ab and Ac two lines at right angles to each other, a force of 50 lbs. acts from A to B, and one of 70 lbs. from A to c. Find their resultant by construction to scale. Ex. 180. — ^Modify the construction of the last example, when the second force is made to act from c to A. Ex. 181. — Draw an isosceles triangle, abc, right-angled at c ; forces of 60 lbs. act from A to b and from B to c respectively. Show by construction that the resultant is a force of about 46 lbs. and that the line representing it bisects the angle between c B and A b produced. Ex. 182. — Draw two lines ab and AC at right angles to each other; a force of 50 lbs. acts along a line A D bisecting the angle B AC. Determine by construction the components of the force along a b and A c. Ex. 183. — Draw A b and A c lines containing an angle of 135° ; within this angle draw ad at right angles to A b ; suppose a force of 100 lbs. to act from A to D. Show by construction that it is equivalent to forces of 100 and 141 -4 lbs. acting respectively from a to b and from a to o. * The examples in the present chapter may be worked by construction ; if solved by calculation, some will be found to lead to very long arithmetical work, e.g. Ex. 184. 56 . PRACTICAL MECHANICS. 33. Resultant of more them two forcea.-^Siace the rules of Arts. 27, 29, and 32 enable us to determine the resultant of any two forces (with one exception, explained in the next chapter) acting in the same plane, it is obvious that the resultant of three forces can be found, by finding the resultant of any two of the forces and then finding the resultant of that resultant and the third force, as shown in the following example. The same method can be extended to four or more forces. Ex. 184.-^Let ABc1}e an isosceles triangle, right-angled at c ; a force of 100 lbs. acta from A to B, a force of 80 lbs. acts from A to c, a force of 80 lbs. acts from b to c. Find the resulta.nt of the three forces. Fig. 17. Draw the triangle A b c and mark by arroir- heads the direction of the forces acting along the lines. On any scale take c d and c b to re- present the forces acting from b to c and A to c respectively. Complete the parallelogram cdfe; the diagonal cf represents the resultant of the two forces of 80 lbs. Produce F c to meet A B in a, take s e eqiial to c f, and g r on the same scale to represent the force acting from A to B ; complete the parallelogram o n k l. Then a k represents the required resultant; which is a force of 151 lbs. and acts along the line q e in the direction s to k. Ex. 1 85. — Let a b c s be the comers of a square taken in order, produce A B to E, make b e equal to a b, and draw e f parallel to b c. If forces of 10 lbs. apiece act from A to B, b to c, and c to d respectively; show that their resultant will be a force of 1 lbs. acting along e f in the direction b to o. Ex. 186.— In the last example if the force along cb has its direction reversed so as to act from c to b ; show the resultant is still a force of 10 lbs. but acts along d a from d to A, Ex. 187.— In Ex. 185 suppose an additional force of 15 lbs. to act from D to A ; show that the resultant of the four forces is determined as follows : — produce b a to o, take a o equal four times a b, the resultant is a force of 5 lbs. acting through o parallel to d A and in the direction d to A. Ex. 188. — Draw an equilateral triangle ab c, let a force of 20 lbs. act from A to b, one of 20 lbs. from b to o, and one of 30 lbs. from a to c ; show that the resultant will be a force of 50 lbs. acting in the direction A to c along a line parallel to A c, drawn through a point p in b c such that b p is three-fifths of b c. Ex. 189. — Determine the resultant in the last case when the direction of the force along a c is rpversed ; the other forces remaining unchanged. Tia. 17a. V PAEALLELOGRAM OF FORCES, 57 34. Condition of equilibrium of three forces. — If three forces, P, Q, and R, whose directions are not parallel, act on a body, it is necessary and sufficient for equi- librium that any one of them (p) be equal and opposite to the re- sultant of the other two (q and k) ; the resultant of Q and R being found by the parallelogram of forces. It is worthy of remark that this condi- tion involves the condition that the three forces act along lines which pass through a common point. Ex. 190. — Three ropes, p a, Q A, a A, are knotted together at the point a ; on each a man pulls ; the angle p a q = 120°, q a e = 132°, and therefore B A p= 108° ; if the man who pulls on a p exerts a force of 24'5 lbs., find ■with what force the other men must piiU that the three may balance each other. [Produce p a to c and measure off on scale a c = 24 J, this line must re- present the resultant of a and b, therefore drawing B c parallel to a o and c d parallel to A b, the forces o, and b will be represented by the lines A d and a b respectively, and can be found by measuring them on scale or by calculating their lengths by trigonometry.] Ans.Ci = Sl-35lha. B = 28-551bs. Ex. \91 . — If in the last example the rope A p were pulled with a force of 28 lbs. ; A a with a force of 36 lbs. ; and A b with a force of 12 lbs., deter- mine the angles p A q, a A b, and bap, .4»«. QAB = 134° 9'. EAP = 63°46', PAa = 162°5'. Ex. 192. — liiaEx. 190 pa is pnlled by a force of 28 lbs., Q A by a force of 40 lbs., and the angle p a a is 135°, determine the magnitude of the force along B a, when they are in equilibrium, and the angles b a «, and bap. Am. aAB=136°34'30". EAP= 89° 25' 30", B= 28-28 lbs. 5S . PEAOTICAL MECHANICS. Fig. 18. . En. 193.— Let A B c D be a rectangle ; A b is 7 ft. long, B c is 3 ft. long ; join E F the middle poinls of a d and b o ; at e act two forces, p and a, in such directions that p e f = 46° and a b F = 30°; the force p = 520 lbs. ; find o (1) when the resultant of P and a acts through B, (2) when it acts through F, (3) when it acts through c. Ms. (1) 421 lbs. (2) 738 lbs, (3) 1420 lbs. Ex. 194. — A boat is dragged along a streani 50 feet wide by men on each bank ; the length of each rope from its point of attachment to the bank is 72 feet ; each rope is pulled by a force of 7 cwt. ; the boat moves straig'ht down the middle of the stream ; determine the re- sultant force in that direction. If, in the next place, one of the ropes is shortened by 10 ft., by how much must the force along it be diminished that the direction of the resultant force on the boat may be unchanged ? What will now be the magnitude of the resultant force? Am. (1) 13-13 cwt. (2) || cwt. (3) 12-08 cwt. 35. Note. — In a large number of questions the solidity of the bodies concerned does not enter the question, ex- cept so far as it afifects the determination of their ■weight ; it being manifest from the conditions of the question that all the forces act in a single plane ; in many such cases a complete enimeiation would be long and troublesome to the reader, while an imperfect enunciation is without any real ambiguity ; wherever this happens the imperfect enunciation will be preferred; thus, in the next example all the forces are supposed to act in a vertical plane passing through the centre of gravity; and the dia- gram ought, strictly speaking, to be that given above, fig. 19, in which the dark lines are all that are shown in the figure which accompanies the example. Pig. 19. P-6- AN PAEALLELOGRAM OF FORCES. 69 p FlQ. 20. D E Vw Ex. 195.— Let ab c d represent a rectangular mass of oak 2^ ft. thick, AB and A D are respectively 4 ft. and 12 ft. long; it is pulled at D by a horiiontal force p, and is prevented from sliding by a small obstacle at a ; find p when the mass of oat is on the point of turning round a. Am. 1050| lbs. [Find a the centre of gravity of a b c D, and through it draw the vertical line ef meeting d o in E, the weight will act along the line e p, and the re- sultant of p and w must pass through a since the body is on the point of turning round A ;— the remainder of the investigation is conducted as before.] Ex. 196.— A BCD represents a block of oak 35 ft. long and 3 ft. square ; the point A is kept from sliding; the body is held by a rope c e 60 ft. long in such a position that the angle d a e is 57° ; determine the direo- ^'°- ^'• tion and amount of the pres- sure on the point a, and the tension of the rope. [Through G, the centre of gravity of the block, draw the vertical line a w, meeting e c in F ; the forces that balance upon the block are the weight w, the tension t of the rope and the resistance of the ground at the point A ; this force must pass through f, and then we have three forces acting in known directions through p ; &c.] Ans. (1) Tension 8453 lbs. (2) Pressure on ground, 23,900 lbs. making with vertical an angle of 17° 39', Ex. 197. — On every foot of the length of a wall of brickwork whose section is a b c b a force acts on the upper angle c, in a direction making an angle of 45° with the inner side b o ; determine this force when the result- ant of it and of the weight of the wall passes through the angle A at the bottom of the wall ; the height of the wall being 20 ft. and its thickness 4 ft. Ana. 1584 lbs. Ex. 198. — ^If in the last example there were a bracket c b on the inside of the wall, c e being in the same line with d o, the top of the wall, and the force (inclined at the same angle as before) were applied at e, 2 ft. from the inside of the wall ; what must be its magnitude if the resultant of it and of the weight of one foot of the length of the wall passes through the point a ? determine also the point in which the resultant would cut A b, the base of. the wall, if the force were the same as in the last example. '.4»&-(l) 1810 lbs. (2)2|in, 60 . 1>EACTICAL MECHANICS. Ex. l99.^If A B are two points in the same horizontal line 10 ft. apart ; A c and B c ropes 10 ft. and 5 ft. long respectively tied by the point o to a' weight w of 3 cwt. ; determine the tension of each rope. Ans. Tension of A c = 86-8 lbs. Tension of e c = 303-6 lbs. [The triangle A b c is, of course, fixed in position, the weight w will act- vertically through and be supported by the reactions of A and B transmitted along the ropes.] 36. Triangle of forces. — The reader will remark on reference to fig. 17a, that if lines be drawn parallel to the directions of p, Q, and R respectively, they will form a triangle ahc similar to a b c, whose sides will therefore have to each other the same ratios as the forces, each side being homologous to that force to whose direction it is parallel. This fact is frequently of great importance. Thus in Ex. 195, if A e be joined, the sides of the triangle A E F are respectively parallel to the forces, so that EF : fa:: w : p and since e f, f A, and w are known, p is at once found. Again, in Ex. 196, if A H be drawn parallel to E c, the sides of the triangle A f h wiU be parallel to the forces, so that FH : ha::w : T and FH : af:: w : r from which T, the tension of the rope, and R, the pressure on the ground (or the reaction of the ground to which it is equal and opposite) are at once found. Hence also can be deduced a very simple construction for finding the re- sultant of any two forces p and Q. Eeferring to fig. 47, p. 83, draw any line 6 c parallel to A p in the direction A to P ; from c draw c a parallel to A Q in the direction A to . Q and of such a length that p : Q::6c;:ca join ah; if r' is the required resultant we shall have b' : p::a6 : 6c REACTION OF SMOOTH SUEFACES. 61 and k' will act at A along a line parallel to ba and in the direction 6 to a. If the force at A had its direction re- versed so as to act as E in fig, 47, viz. in the direction a to b, it will, as we have already seen, balance the forces P and Q. 37. Reaction of smooth surfaces. — We have already seen (Art. 28) that if a body is urged against a second body and thereby kept at rest, the second body reacts against the first. We have now to add that if we suppose the bodies to be perfectly smooth the reaction can only be exerted in the direction of the common perpendicular to the surfaces of contact. The supposition of perfect smoothness is commonly very far from the truth, but by making it we avoid a great deal of complexity in our reasoning and results. So long as both surfaces resist the tendency of the pressures to crush them any needful amount of reaction can be supplied, but, as already stated, only in the direction of the perpendicular, if the surfaces are perfectly smooth. Ex. 200. — A body whose weight is w rests on a smooth plane A b in- clined at a given angle bac to the horizon; determine the force p which acting parallel to ^'®- 22. the plane will just support the body. Find s, the centre of gravity of the body, and through it draw a vertical line a w, cutting in d the direction of p ; through d draw D b at right angles to a b, then b, the reaction of the plane, must act along B D, and we have three forces p, w, and E in equilibrium acting in known direc- tions ; and since the magnitude of w is known, that of B and p can be found by the usual con- struction : viz. take D H to represent w, draw H K parallel to d p, and e l parallel to D H, then D E is proportional to b and D i. represents p. Ex. 201. — In the last example show that p : b : w : : bc : ca : ab, Ex. 202.— In Ex. 200 if A were 43° and w were 1000 lbs., find p and E. Ans. 707 lbs. (each). Ex, 203.— In Ex. 200 if A were 30° and p were 200 lbs. what weight could P support ? Ans. 400 lbs. «2 PRACTICAL MECHANICS. Via.: Ex. 204. — If a cylinder whose weight is w rests between two planes a b and A c inclined at different angles to the horizon (as shown in the figure) ; determine the pressnies on the planes. The weight w will act vertically through o^ and will be supported by the reactions b and b, of the planes A B and A c : as these forces must act at right angles to the planes respectively, their directions will pass throngh o, and their magnitudes can be determined as usual. The pressures on the planes are, of course, equal and opposite to B and b, respectively. Ex. 205. — In the last case if b a s and cab are angles of 30° and w a weight of 112 lbs., determine the pressures. Jms. 64-6 lbs. apiece. Ex. 206. — Explain the modification that Ex. 204 undergoes if both A B and A c are on the same side of the vertical line drawn throngh A ; and deter- mine the pressures when w equals 11 2 lbs. and c a b and b a c are each 30°. AiiS. B- 112 lbs., B, = 194 lbs. 38. Trcmsmiasion of force by vnea/ns of a -perfecSy flexible cord. — If a cord is stretched by two equal forces p and Q, one acting at each end, they will balance each other, and the tension of the cord is equal to either (Art. KQ.24. 28); suppose the cord to pass round a portion A B of a fixed surfece, as shown in the figure, the portions A p and B Q of the cord will be straight, while ab wiU take the form of the surface (which is supposed to be convex), and if p and Q continue in equilibrium they must be exactly equal, provided the surface ab is perfectly smooth and the cord perfectly flexible ; conditions which are supposed to hold good unless the contrary is specified. Hence force is trans- mitted without diminution by means of a perfectly flexible cord which passes over perfectly smooth surfeces. Ex. 207. — Let A and b be two perfectly smooth points in the same iori- zontal line, and let w be a weight of 100 Ibsl tied at <; to cords which- pass over A and b, and let w be supported by weights p and Q tied to the ends of these cords respectivelyi -and suppose the whole to come to rest in such a positiun that bag equals 30° and A c b equals 90° ; find p and a. ' « PEINCIPLE OF MOMENTS. 63 Since the forces r and Q are transmitted without diminution to c, w is supported by a force p acting along c A and by Q along CB. Hence draw c d vertically and such that on scale it represents the vertical force which balances w, and complete the parallelogram acbc, then ca and c i represent the transmitted forces that support w : — hence p equals 50 lbs., and q equals S6'6 lt». Fia.ss. Ex. 208. — ^In the last example show that the pressures on a and b are equal to 86'6 lbs, and 167'3 lbs. and that their directions bisect the angles F A. c and a B c respectively. ^ 39. ITie prmdple of moments. — ^A large class of ques- tions has reference to the equilibrium of a body one point of which is fixed ; in these cases it is frequently sufficient to determine the relation between the forces that tend to turn the body round the point, the actual amount and direction of the pressure on the point not being required ; under these circumstances the relation sought is ^ven at once by a principle called the Principle of Moments. The definition of the moment of a force is as follows : If y represents any force, and A is any point, and an is a perpendicular let fall on p's direction, then if the number of units of force in p is multiplied by the number of units of length in A N, the product is called the moment of the force p with reference to the point A. The principle of moments in its general form will be found in the next chapter; for present purposes the following statement will be sufficient. If am/y nv/mber offoreea actmg in the sam,e pUme keep a body i/n equilibrium found a fixed povnt, and if their mMnents with reference to that poi/rtt be taken, the sum of the mxtyrvents of those forces which tend to turn the body from right to left round thefioeed point, will equal the sum, of the moments of those forces which tend to turn the body from left to right. The following case will exemplify the mode of ;applying 64 PRACTICAL MECHANICS. the principle of moments. In Ex. 196, let it be required only to determine the tension of the rope. Construct the figure to scale (see fig. b) ; determine G, the centre of gravity of the block, draw the vertical line G w, cutting A K in M ; draw A N at right angles to c e ; if T is the ten- sion of the rope, and w the weight of the block which can be found to equal 18,388 lbs., then the moments of T and w are respectively A N X T and A M x 18,388 ; and the prin- ciple of moments assures us that these two are equal. In the construction from which fig. b was drawn, the scale employed was 1 inch to 10 feet; and it was found that AM equals 8*25 ft., and an equals 18"1 ft.; hence was obtained for T a value of 8381 lbs. ; the value of T as determined by calculation is 8453 lbs. The student is recommended, as an exercise, to work by this method aU the previous examples in the present chapter to which it can be readily applied, viz. Ex. 172, 173, 174, 175, 176, 177, 195, 197, 198. 40. The lever. — This is the name given to a rod capable of turning round a fixed point (called the fulcrum) and acted on by the reaction of the fixed point and by two other forces : as most machines are used for the purpose of moving bodies, one of these forces is to be overcome, or opposes motion, and this is called the weight, the other force which produces the motion is called the power. When the lever is in equilibrium the moments of the power and the weight with reference to the fulcrum must be equal ; and, of course, those forces wiU tend to turn the lever in different directions round the fulcrum. Levers are sometimes classified as belonging to the first, second, and third orders respectively ; those of the first order have the fulcrum between the power and the weight, as the beam of a pair of scales, or a poker when used to stir a fire ; levers of the second order have the weight between the power and the fulcrum, as a crowbar E THE LE"VER. 65 •yrhen used to lift a weight one end resting on the ground, or an oar used in rowing, in which ease the water is the fulcrum ; levers of the third order have the power between the weight and the fulcrum, as the limbs of animals, e.g. when a man has a weight in his hand and extends his arm the forearm is a lever of which the elbow is the fulcrum and the power is the contractile force of the large muscle of the upper-arm acting by means of tendons fastened into one of the bones of the forearm — of course in such a case the power must be very much larger than the weight. Many simple instruments consist of two levers fastened together by, and capable of turning found, a common fulcrum ; these are called double levers, and are classified as double levers of the first, second, and third orders respectively ; a pair of scissors and of pincers are of the first order, a pair of nut-crackers of the second order, and a pair of tongs of the third order. Ex. 209. — Let AB be a lever 16 ft. long, movable about a fulcrum d at a distance of 6 ft. from b ; a weight of 28 lbs. is suspended from a and from b a weight of 336 lbs. Find the weight that must be hung at e (which is 7 ft. from d) to balance the lever, Ans. 243 lbs. Ex. 210, — Let AB be a lever 8 ft. long, the end A resting on a fulcrum; a weight of 40 lbs. is hung at c, 3 ft, from a. The lever is held in a horizontal position by a force p, acting vertically upward at B. Find p and the pressure on fulcrum. Ans. F=i IS lbs, Pressure on fulcrum 25 lbs. Ex, 21 1,— Let A B and D E Kg. 26. be levers turning on fiilcrums B and F, connected by a bar D c, loosely jointed at d and c ; x A B and D E are respectively S I and 6 ft. long, A c is 3 ft., and I «» »• F E is 9 in. long ; the weight ' F at B equals 1000 lbs, and is balanced by q acting at a ; find q. 4ni. STy lbs. Ex. 21 2, — A crane B D is sustained in a vertical posi tion by the tension P PRACTICAL MECHANICS. (rfa rope ae; its dimensions are .as follows. — bc, bd, be, and ac respectively 19, 13, 11, and 16 ft. long; the angle c B D equals 108°; a weight p of 7 ewt. is supported by a rope that passes over a pnlley d and is fastened to c. Determine the tension of the rope a b, the weight of the crane and the dimensions of the pnlley being neglected. Ans. 7'329 cwt. Ex. 213. — ^Let bcde repre- sent a block of Portland stone whose dimensions are 5 ft. long, 2 ft. high, and 2J ft. wide; a rope FPd is attached to it, which after passing over a pnlley p, is pulled vertically downward by a force Q, which is just sufficient to raise the block : determine a on the supposition that the di- mensions of the pulley can be neglected, having given that ep equals 6 in. and ba and a p re- spectively 15 and 13 ft., the point A being vertically under p. Jm. 1942 lbs. Ex. 214. — -In the last example determine the amount and direction of the pressure on the ground through the point c. 41. The ateelr^ard. — If a beam A B rests on a fine axis passing through its centre of gravity (g), and on the arm B G is placed a movable weight w, then if a substance equal in weight to w is suspended from A, the beam wiU balance when w is at a dis- " "" tance from G equal to a G ; if the substance equals twice the weight of w, the beam vrili balance when w's distance from G equals twice A G ; and so on in any proportion. Hence, if the beam is made heavy at the end a, so that G. is very near A, the arm b g can be divided into eqiuiZ divisions which shall indicate Fig. 29. B • 7 Ti EQUILIBKIUM OF WALLS. 67 the "weight of a substance suspended at A by means of the position occupied by w when it balances that substance. An instrument constructed on this principle is called a steel-yard, and is used when heavy substances have to be weighed, and extreme accuracy is not required ; the ad- vantage it possesses arises from the fact that the weights employed are much less heavy than the substance to be weighed. A very common application of the principle of the steel-yard can be seen in the weighing machines employed at most railway stations. Mc. 216.^— Show that the graduations of the steel-yard must he equal even if the centre of gravity of the beam do not coincide ■with the axis ; but that the graduations must begin from that point at which the movable ■weight ■would hold the beam in a horizontal position. [Let F be the fulcrum, G the centre of I'lo- 30. gra^vity, and w the ■weight of the beam ; i' p *> t o _A suppose that o is so chosen that ■w at o balances w at G, then 1 rox^w = PGx w Fig. 31. No^w, suppose that a body weighing m ■w is hung at A, and that the beam is kept horizontal by "W at p ; then, measuring moments round F, ■we have FPX^W + FGX«) = FAX»W Therefore, by addition, OPX W=FAXMW Hence, if the weight of the body is w, o p must eq^ual f A ; if twice w, o p must equal twice f a, and so on in any proportion.] 42. The equilibrium of walls. — The question What is the force which, acting in a certain specified manner on a given wall, will be just sufficient to overthrow it ? can be answered by an application of the Principle of Moments ; the general method of con- sidering this important question is as follows :— ^ ^ Let A B c D represent the section of a wall, the base a b being on the level of the ground ; let it be acted on by a F 2 «8 .PRACTICAL MECHANICS, force p along the line p q ; now, it is considered that a wall, to be stable, must be capable of standing irrespec- tively of the adhesion of the mortar ; * hence, if we sup- pose B D to be a continuous mass, and simply to rest on the section A B, and determine the force p which will be on the point of turning the mass round the point A, we shall obtain the greatest force acting in the manner specified that the wall can support ; the force is, of course, deter- mined by the rule that its moment with reference to the point A equals the moment of the weight of the wall with reference to the same point. Ex. 216.— A wall of brickwork 2 ft. thick and 25 ft. high sustains on the inner edge of its sunimit a certain pressure on every foot of its length ; the direction of this pressure is inclined to the horizon at an angle of 60° ; find its amount when it will just not overthrow the wall, (See fig. c.) Draw the section of the wall a b c to scale ; make the angle ban equal to 30°, then the pressure p acts along the line p n ; draw c n perpendicular to p N ; through G, the centre of gravity, draw the vertical line o m, cutting c B in M ; the principle of moments gives us P XCN = W XCM The weight w equals 5600 lbs. ; c m equals 1 ft. ; c N, as obtained by mea- surement, equals 10-8 ft. ; whence p equals 618 lbs. When p is found by calculation it equals 520 lbs. Ex. 217. — In the last example suppose the pressure to be applied by means of a bracket, at a horizontal distance of 3 ft. from the inner edge of the summit ; determine its amount when it will just not overthrow the wall. Am. 685 lbs. 43. The effect of buttresses. — Let fig. 32 represent the elevation of a wall, fig. 33 its plan, and fig. 34 its section made along the line A B ; if now we neglect the * ' Though ordinary mortar sometimes attains in the course of years a tenacity equal to that of limestone, yet, when fresh, its tenacity is too small to be relied on in practice as a means of resisting tension at the joints of the structure, so that a structure of masonry or brickwork, requiring, as it does, to possess stability while the mortar is fresh, ought to be designed on the supposition that the joints have no appreciable tenacity.' — Eankine, Applied ^eohcmios, p. 227. Fig. e, case 68. BUTTRESSES. Fid. 32. weight of the buttresses their efifect in supporting the wall will be understood by inspecting fig. 33 ; for it is manifest that the wall woiild fall by being caused to turn round the line X Y ; but, if the buttresses were removed, by being caused to turn round ' the line xy; so that, in the formal* case, the moments must be measured round M (fig. 34), in the latter round K : in other words, the introduction of buttresses diminishes the moment of p, and increases that of the weight of the wall. Their fib. 33. useful effect is still farther increased by the fact that if the moment of the weight of the buttress is taken into account, it increases the moment of the weight of the wall. It is to be observed that if c d (fig. 32) and E f be drawn at equal distances from a B, and at a distance from each other equal to the distance between the centres of tw6 consecutive buttresses, then we may consider that the total pres- sure on c F is supported by the Weight of the portion of the wall between c D and EF, and by the weight pf the buttress. It must be remembered that the above explanation applies to the case in which the pres- sure is distributed uniformly along the top of the wall ; ^I u "LF Fig. 34. 70 PRACTICAL MECHANICS. which in this case is supposed to be so strong as not to bulge between the buttresses. In many instances, however, particularly in large ecclesiastical buildings, the whole, or nearly the whole, weight of the roof and its lateral thrust act on the buttresses, and not on the portion of the wall between the buttresses ; in such cases the wall serves as a curtain between the buttresses, and not as a support to TTm 55 ^^^ Tooi, and, of course, the moment of the lateral thrust must equal that of the weight of the buttress. Ex. 218. — In the last example, if the wall were sup- ported by buttresses 2 ft. thick,* to what can the pres- sure on each foot of the length of t^e wall be increased without overthrowing it — the weight of the buttresses being neglected ? Ans. 2609 lbs. Ex. 219 — In Ex. 216, suppose the wall to be sup- ported by counterforts reaching to the top of the wall, I ft. thick, 1 ft. wide, and 10 ft. apart from centre to centre, determine the pressure on each foot of the length of the wall that can be supported — (1) when the direc- tion of the pressure is inclined at an angle of 60° to the horizon ; (2) when the direction is inclined at an angle of 30° to the horizon, ^ras. (1) 1145 lbs. (2) 562-8 lbs. Ex. 220. — In each case of the last example deter- mine to what the pressure can be increased if the buttress assumes the form of a Gothic buttress, as indicated in the annexed diagram, where A c and c i: are each a foot square, and c D and A B are respec- tively 20 and 10 ft. high. Ans. (1) 1903 lbs. (2) 876 lbs. 44. The thrust of props. — LetAB represent a beam or prop resting on a fixed support at the end A ; and suppose it to be acted on by certain pressures which are balanced by the reaction of the end A. That part of A the reaietion which acts along the * The thickness of a pier or buttress is supposed to be measured in a dirsQtion perpendicular to the face of the wall. Fig. 3G. PKOPS. 71 Fig. 37. axis of the beam A B is called the thrust of the prop, and is, of course, equal to the thrust produced by the pres- sures on the prop, the two being equal and opposite. If no pressure acts on the beam except at the end B, it is plain that the whole reaction from A must pass along the beam. In the following question, which concerns the thrust of props, it will be assumed that the thickness of the prop can be neglected, except so far as it affects its weight. Ex. 221. — A wall of brickwork, 23 ft. high and 2 ft. thick, sustains on the inner edge of its summit a pressure of 1000 Ihs. on every foot of its length, whose direction is inclined at an angle of 65° to the yertieal ; it is supported at every 5 ft. of its length by a prop 25 ft. long, resting against a point 3 ft. from the top ; determine the thrust on the prop. Ans. 7758 lbs. [If the annexed figure represent a section of the wall and prop, the forces acting are w, the weight of the wall, p, the pressure on the summit, and these are balanced by T, the thrust of the prop, and the reaction of the ground A b : now, unless the prop is wedged up against the wall, there will not be more reaction than is just sufficient to support the wall ; consequently the resultant of p, w, and T must pass through a, at which point it will be balanced by the reaction of the ground ; hence, by measuring moments round A we can find t.] Fig 38. 45. The thrust along rods con/nected by a smooth hinge. — Let ab be a rod cap- able of moving freely round a joint or hinge at A ; if it were acted on by a force it would turn round A, unless the force acted through A. Suppose two such rods, A B and A c, to be connected by a perfectly smooth joint at A, while their ends b and c rest against immovable obstacles, 72 PRACTICAL 5IECHANICS. and let us suppose the rods to -be geometrical lines and without weight ; let a weight w be hung at A, and let it be required to determine the pressures against the fixed obstacles caused by w. Now (Art. 44), the reactions at B and c, which support w, must pass along b a and c A ; hence, if we take A a to represent w and complete the parallelogram Abac, the lines A b and a c will represent the thrusts caused by w along A b and A c, and these are respectively equal to the reactions by which they are balanced (Art. 28). 46. The thrust along a rafter. — The case which we have just explained enables us to determine the thrust produced on the summit of a wall by each rafter of an isosceles roof: letAB, AC, represent two of the principal rafters of such a roof, and let the whole weight sustained by each rafter (including its own weight) be repre- sented by w ; this weight will act at the middle point of the rafter, and therefore can be replaced by weights equal to ^w acting at each end of the rafter ; so that the whole weight sustained by ab and AC may be distributed as shown in the figure, viz. it will be equivalent to w acting at A, -gW at B, and -Jw at c ; then the thrusts along the rafter (t) will be produced by w acting at A, and can be determined as explained above, viz. take Ap to represent w, and complete the parallelogram Arpq, then a r and A q represent the thrusts in question : the total pressure on the wall at b wiU be found by compounding t with ^w. When the determination of the pressure is made for the purpose of ascertaining whether a certain wall will support the roof, it is much better not to compound the pressures T and Jw, but to regard the wall as acted on by the two uncompounded forces. Fig. 39. ■ A r^^ X ■B^ <\ ° 1 Pig. d, page 73. THRUST ALONG EAPTHRS. 73 Er. 222. — There is a roof weighing 25 lbs. per square foot, the pitch of which is 60° ; the distance between the side walls is 30 ft. ; determine the magnitude and diirection of the pressure on the foot of each rafter, the rafters being 6 ft. apart. (See fig. d.) Let ABO represent the roof ; then the weight (w) supported on each rafter equals 3750 lbs. ; hence, when the weight is distributed, we have w at c, — at A, and — at b ; draw c w vertical, and take c d to represent 3750 lbs. ; 2 2 r . draw D E parallel to b c [which is broken in the figure as indicated by the letters a, a and b, i] ; then c e represents the thrust (t) along the rafter. The total pressure on the wall (b) is the resultant of _ and t acting at A ; take A F to represent on scale 1875 lbs. and a h equal to c e ; complete the parallelogram f h ; then A k gives the magnitude and direction of the re- sultant B ; it was found from fig. d that b equals 3885 lbs. and the angle EAF equals 16°; the results given by calculation are that K equals 3903 lbs., and that the angle s A f equals 16° 6'. Ex. 223. — If in the last example the waUs were 20 ft. high, 21 ft. thick, and of Portland stone, -would they support the roof ? Ans. The wall will stand — the excess of the moment of the weight of 5 ft. of its length oyer that of the thrnst being 29,620. Kv. 224. — If in the last example the walls be supported by buttresses 20 ft. apart from centre to centre, 16 ft. high, 2 ft. wide, and 2* ft. thick, would these support the wall if its thickness were reduced to IJ ft. ; and what would be the excess of the moment tending to support 20 ft. of the length of the wall over that which tends to overthrow it ? Kg. «. Jns. (1) Yes. (2) 221,000. Ex. 225. — Show that the total pres- sure on each wall is equivalent to a vertical pressure w, and a horizontal pressure wxBC-i-4AD. (Art. 46.) jE^. 226. In the case of an equi- lateral roof show that the horizontal pressure equals 0-29 w. 47. The equilibrium, of a iriarigular frame. — A tri- angular firame A b c consisting of rods loosely jointed at the angles is in equUibrimn under the action of three forces acting one at each angle ; it is 74 PEACTICAL MECHANICS. required to find the thrust or tension to which eiach rod is subjected. Let the forces p, Q, e act at the angles A, B, c respectively, and suppose their directions to pass through a point 0. In the first place they must be in equilibrium, or otherwise they would make the frame itself move ; draw h h parallel to E, then we know that the forces, P, Q, K are proportional to ho, o k, and k h. Take A a, b 6, c c equal to oh, ok, and k h respectively, and complete the parallelograms A a' a a", B 6' 6 b", Ccf cc"; it will be found that A a" = b 6', B 6" = C c', and c c"=Aa'. We see therefore that each rod is under the action of a pair of equal forces which tend to crush A b and A c, and to stretch B c ; these forces are severally proportional to A a", c c", and B b". These lines there- fore measure the thrusts of a b and A c and the tension of b c. The magnitudes of these forces can be obtained by a more simple construction, thus : — ^Draw 6 c parallel to A o, and containing as many units of length as p con- tains units of force ; draw ca, ah parallel to o B and o c respectively ; draw a d parallel to B c, c? c to A B, and _ join b d (the line bd is parallel to A c, as the student can prove)^ and we shall have ca, ab, ad, dc, bd con- taining as many units of length respectively as the forces Q, E, the tension of b c, and the thrusts of a b and A c contain units of force. This is plain from an inspec- tion of the figure, since a 6 c is the triangle of forces for the three forces in equilibrium at the point o, cd aforthe three forces at the point b, dab for the forces at c, and bed for those at A. Eeferring to Art. 46 ; if the ends B and C of the rafters are connected by a beam b c (fig. 39), called a tie beam, they will constitute a triangular frame like that we have just considered ; it can be easily shown that the tie beam is subject to a tension equal to the horizontal thrust of each rafter, i.e. equal tp w x B c -*-4 AD (Ex. 225). Under these circumstances the roof TRIANGULAR FRAME. 75 will act on the walls merely by its weight, and each wall will, of course, support half the whole weight of the roof. Ex. 227. — If in fig. 40 the point o fall within the triangle, show that all the bars will be compressed or all stretched. Ex. 228. — Two rafters A B and A c are each 20 ft. long, their feet are tied by a wrought-iron rod bc whose length is 35 ft., and a weight of 1 ton is suspended from A ; determine the tension it produces on the tie, the weight of the rafters, &c., being neglected. If the rod have a section of a quarter df a square inch, determine the weight that must be suspended at a to break it. As. (1) 2024 lbs. (2) 18,590 lbs. Ex. 229. — There is a roof whose pitch is 22° 30', the rafters are 40 ft. long; the weight of each square foot of roofing is IS lbs.; determine the jjiameter of the wrought-iron tiefleBBssary to hold the feet of the principal -rafters with safety, supposing them 10 ft. apart. Ans. 1-28 inches. 48. Note. — The foregoing remarks as to the thrusts of the rafters and the tension of the tie beam, apply to the cases in which the joints are perfectly smooth : as this is never the case, the thrusts, &c., may not equal the cal- culated amount ; but it is generally considered that re- liance should never be placed on the resistance offered by a joint to the revolution of a rod round it. It will be instruc- tive, however, to consider the case in which the rods and the joint -at A (fig. 41) are perfectly rigid. Suppose two points, h and c, to be taken near to A, and joined by a rod he; if this rod were inextensible, and if there were no tendency in the materials to give either by crush- ing or tearing at h and c, then would h c act the part of a tie beam, and there would be no horizontal thrust on the wall, which, as before, would merely have to support the weight of the roof. If we suppose the rod bc to be replaced by a metal plate firmly fastened to the beams, as shown \}j ab doix\ 76 rEACTICAL MECHANICS. fig. 42, this would tend to render the attachment of the beams rigid, the horizontal thrust being more or less Fig. 42. neutralised by the resistance by the materials to crushing on the bolts, and to the tearing of the plate across a d. Hence, under all circumstances, the walls have to sustain the whole weight of the roof, and besides this, a horizontal thrust which will more nearly equal w X B ch-4 a d, as the joint is less rigid. 77 CHAPTEE IV. THE FUNDAMENTAL THEOREMS OF STATICS, 49. Axioms. — The following chapter contains demon- strations of the fundamental theorems of statics, so far as forces acting in one and the same plane on a rigid body are concerned. It may be well to invite the reader's attention to the order of proof adopted. In the j&rst place the case of two forces and their resultant is fully discussed, together with the conditions of the equiUbrium of three forces, and the case in which two forces do not have a resultant. In the next place, the results obtained for two forces are extended to any number of forces. Lastly, a peculiar property of parallel forces — the possession of a ' centre ' — is proved. The demonstrations are of a very abstract character, and should be thoroughly mastered. Appli- cations of several of the theorems have been already given in Chapter III., and many more will be found in the succeeding chapters. The demonstrations are based on certain assumed elementary principles or axioms. The assumption of these principles is, of course, not arbitrary, but justified by experience of the action of forces. The axioms are as follow : — Ax, 1. The line which represents the resultant of two forces acting at a point, falls within the angle made by the lines that represent those forces, (See Art. 25.) Ax. 2, If two equal forces act at a point, the line that represents their "resultant bisects the angle between the lines that represent those forces. Ax. 3. If a force acts upon a body, it may be sup- 78 PEACTICAL MECHANICS. posed to act indififerently at any point in the Kne of its action, provided that point is rigidly connected with the body. Ax. 4. It is necessary and sufficient for the equilibrium of any system of forces, that one of them be equal and opposite to the resultant of all the rest. Ax. 5. If a system of forces in equilibrium be imposed on or removed from any system of forces, it will not afifect the equilibrium of that system, if it be in equilibrium, nor its resultant, if it have a resultant. Proposition 3. The principle of the parallelogram of forces (Art. 32) is true of the dArection of the resultant of two equal forces. Fig. 43. Let the equal forces P and Q act at the point A along the lines a p and A Q ; let A B represent the force p, and A c the force Q, then will a b equal AC; complete the parallelo- gram ABCD, and draw the diagonal A D. We are to show that the resultant of P and Q acts along the line A D. Since A c equals A B it equals c D, therefore the angle c A D is equal to the angle A D c, but since c D is parallel to A B, the angle A D c is equal to the angle bad, therefore the angle bad equals the angle CAD, and the line A D bisects the angle p a q ; but the line of .action of the re- sultant of p and Q bisects the angle P A Q (Ax. 2), therefore the resultant acts along ad. Q. E. D. 50. Remark. — The following proposition may be re- garded as the foundation of the science of statics ; the de- monstration generally seems obscure to readers who meet with it for the first time : this results from the somewhat unusual /or^??, of the proof; it may therefore be weU to re- PARALLELOGRAM OF FORCES. 79 mark that the demonstration consists of two parts ; in the first part it is shown that if the principle is true in two cases, viz. with regard to the pair of forces p and P, and the pair p and p.^, it must also hold good in a third case, viz. in regard to the pair of forces p and Pj + Pji this part of the proof is purely hypothetical, as much so as in the case of a demonstration by reduction to an absur- dity ; the second part of the proof takes up the argument, but as a matter of fact the proposition is true in two certain cases ; therefore it must be true in a third case, therefore in a fourth case, and so on. Proposition 4. The pri/ndple of the parallelogram of forces is true of the direction of the resultant of any two coTnmensur- able forces. Let the force P act at the point A along the line A B, and the forces P, and Pj at the point A along the line A c : take A B, A c, c D, fis. 44. respectively proportional to P, Pp and Pj, and complete the parallelograms B c, E D, then is the figure B D a paral- lelogram; draw the diagonals A E, c F, and A F, and suppose the points c, D, E, F to be rigidly connected with a. (a) The lines a b and A c represent the forces P and Pj ; assume that their resultant acts along A E ; then can p and p, be replaced by their resultant acting at A along A E, and, since A and E are rigidly connected, by that resultant acting at E along A E (Ax. 3) ; but this re- sultant acting at E can be replaced by its components acting at E, viz. by P, along B E, and by P along c E ; and these again, since C and F are rigidly connected with E, by P, acting at f along b f, and P acting at c along c E. 80 PBACTICAL MECHANICS, (6) Since a and c are rigidly connected, 7^ may be supposed to act at c along c D ; then c E represents the force p, and c d the force Pj ; assume that their resultant acts along CF, then by reasoning in the same manner as in paragraph (a) it can be shown that the forces p and Pa can be transferred to r. (c) Thus it, follows from our two asswm/ptidns that the forces p, Pj, Pj may be supposed to act indifferently at A or F, therefore each of these must be a point in the direction of their resultant, i.e. their resultant must act along the line A F. Now A b represents the force P and A D the force Pj + Pj ; hence, if the proposition is true of the pair of forces p and p,, and of the pair of forces P and Pj, it must also be true of the pair p and Pj + Pj. (d') But it appears from Prop. 3, that the proposition is true of equal forces, i.e. of any pair p and jp, and of another equal pair 'p and jp, therefore it will be true of the pairja and js+js, i.e. of -p and 2p; again, since the pro- position is true of the pair p and j9, and of the pair p and 2p, it must be true of the pair p and p + 2p, i.e. oi p and 3p ; similarly it is true oi p and 423, otp and 5p, &c., and generally of j3 and rnp. (e) Again, since the proposition is true of the pair of forces mp and p, and of the pair mp and p, it must be true of the pair mp and p+p, i.e. of mp and 239 ; simi- larly it must be true of mp and 3p, of mp and 423, and generally of mp and np, (J) Now, any two coTnmensurable forces p and Q must have a common unit (e.g. a pound, an ounce, &c.), and therefore can be represented by m,p and np ; hence the theo- rem is true of any two commensurable forces, Q. E. D. Exercise. — The above demonstration may be put into a slightly different form, as follows : In the first place, suppose the forces p, p, and Pj to be equ^ ; then the reasoning in § (a) and § (4) of Prop. 4 no longer proceeds from an assumption, but is based directly on Prop. 3 ; and the reasoning PARALLELOGRAM OF FORCES. 81 in § (c) establishes the truth of the proposition in the ease of the two forces p and 2p. The reasoning can be repeated for forces p, 2p and p, and the ease p and 3p -will be estat)lished ; and by a repetition of the reasoning the cases p and 4p, p and 5p, and generally p and «ip are estab- lished, A slight modification of the figure will then enable the reasoning to be extended to the case of »p, p, and p, so that the case nv and 2p will be established, then the cases mp and 3p, »p and 4p, and generally »p and OTP. The student, having first masterec^ Prop, i, will find it a useful exer- cise to write out the proof in this form. Proposition 5. The pri/nciple of the parallelogram of forces is true of the direction of the resultant of any two incorriTnensurable forces. Let p and Q be the two forces represented by the lines A B and A c ; complete the parallelogram a B c D, then will the resvdtant (e) of P and Q act no. 45. along the line joining A and d. For if not suppose E to act along any other line, this line must fall within the angle paq (Ax. 1), and therefore must cut either d D or D B ; let it cut B D in the point E. Now, by continually bisecting A B, a part can be found less than D E ; set off distances equal to this part along AC, and let the last of them terminate at F (it cannot terminate at c, since a b and A c are incommensurable) ; therefore f c is less than this part, and therefore also less than d e ; draw f g parallel to c D, this line will cut B D, in a point G between D and e, join a a. Suppose A f to represent a force q' and F c a force q, then wiU Q equal q' + q; now q' and p are commensurable, therefore their resultant (e') will act along the line A G. But the resultant e of p and Q must equal the resultant of p, q', and q ; i.e. of e' and q ; but e' acts along A G, and q along a c, and therefore (Ax. 1) their resultant E must act within the angle G a Q ; but by the G 82 PEACTICAL JIECHANICS. supposition it acts along ae without the angle gaq; which is absurd. Therefore, &c. Q. E. D. Proposition 6. The principle of the parallelogram of forces is true of the magnitude of the resultant. Let p and Q be the two forces acting at the point a, and let them be represented by the straight lines a b and Pie. 46. A 0, complete the parallelo- gram A B c D, and draw the diagonal A D ; we have to prove that not only does the resultant (h) of p and Q act along the line A D, but also that it is represented in magnitude by that- line. Suppose r' to be the force which balances p and Q, it must act along D a produced. Let A E represent r' ; complete the parallelogram c E, and join A F ; the resultant of Q and e' must act along a f ; but since p balances Q and r', it must act along f a produced ; therefore F A B is one straight line, and is parallel to c D, so that F D is a parallelogram. Hence we have f c equal to A D, but F c equals a e, therefore E A equals a d. But R is equal and opposite to r', which is represented by a e, and therefore R is represented in magnitude by A D. Q. E, D. 51. Application of trigonometry to statics. — It is manifest that the sides of the triangle A c D (Prop. 6) are proportional to the three forces p, Q, e', which are in equilibrium. And hence if any triangle acd he drawn similar to A c d, its sides will be proportional to the forces. Such a triangle will be formed by drawing lines re^pecjtively parallel to the directions of 4;he forces, each force being an homologous term to the side parallel to its direction. The forces a,t A act in the directions d c, PAEALLELOGRAM OF FORCES. 83 ca, ad respectively, as shown by the arrow-heads. A similar remark applies to a triangle formed by drawing lines at right angles to the directions of three forces in equilibrium. The relations between three forces in equilibrium are thus reduced to the relations between the sides of a triangle; and of course all the trigono- metrical relations between the sides and angles of that triangle will be analogous to relations between the forces and the angles between their directions. The two most important of these relations are proved in the following proposition : — Proposition 7. If three forces, v, Q, e, are in equilibrium, and act at a povnt a, to show that the following relations obtain : — (1) p : Q :: sin Q A R : sin R A p Q : R :: sin R A p : sin p a Q (2) r2=p^+q2 + 2 PQ cos P AQ (1) Draw the tri- fio. 47. angle ah c whose p sides bc,ca,ab are respectively paral- lel to the forces ^ p, Q, R. Then it is ^ evident that the angles a, b, c are re- spectively equal to 180°-QAR, 180°— RAP, 180°-PAQ; now be : ca::sin bac : sin c6a::sin qar : sin rap ca : ab:: sin cha : sin acb :: sin rap: sin p aq But by Art. 51— he i ca'.'.v : Q, ca : a6::Q : R. 84 PRACTICAL MECHANICS. therefore p : Q :: sin Q A E : sin H a p and Q : e :: sin e a p : sin p a q These proportions are sometimes expressed by the rule, 'If three forces are in equilibrium, each force is pro- portional to the sine of the angle contained by the other two.' (2) Employing the same figure, we have, by a well- known theorem in trigonometry, + Q we have M*K=OX.R = OX.P + OX.Q = OA.P + AX.P + OB.Q— BX but AX.P = BX.Q (Prop. 8^ therefore m'e=oa.p + ob.q =M*P + M*Q A similar proof will apply to every position of o, and to cases in which p and Q act in contrary directions. Hence, &c. Q. E. D. Ex. 246. — If the point o (Prop. 9) be taken in the direction of the re- sultant, show that the moments of p and q are equal and have opposite signs. Exercise. — ^Prop. 3-10 can be proved by reasoning in the foUoTving manner : — Mrst. Assume as an axiom that the resultant weight of a uni- form rod acts through its middle point ; and bearing in mind the remark in Article 23, that any force can be substituted for an eqiuil force without reference to its physical origin, observe that Ex. 214 gives an independent proof of Prop. 8. Secondly. ObservethatitfoUowsfrom Axiom 2 (Art. 49), that when a body is acted on by two equal forces in the same plane, and has one point in the plane fixed, it will be at rest, provided the forces act at equal perpendicular distances from the point, and tend to turn the body round the point in opposite directions. This observation, combined with Prop. 8, will establish Ex. 246. Thirdly. The principle of the parallelo- gram of forces, so far as the direction of the resultant is concerned, can be easily deduced from Ex. 246. The student who has first mastered Prop. 3-10 will find it a most instructive exercise to write out proofs of the same propositions, adopting the method of proof above indicated. 54. Statical couples. — In Cor. 2 to Prop. 8 it was shown that if p and Q are two parallel ko. 54. forces acting at A and B in opposite directions, then if Q is greater than p their resultant R will be a paral- lel force acting in the same direc- tion as Q through a point x given by the proportion AX ; bx;;q : p AB.P or BX = Q-P ♦i" 92 tEACTICAL MECHANICS, Now, if we suppose q to be gradually diminished, but A B and P to remain unaltered, the magnitude of E (or Q— p) will continually diminish and BX will continually increase, and in the limit when Q becomes equal to P, the magnitude of the resultant is zero, and x is removed to an infinite distance ; in other words, two equal parallel forces acting in opposite directions have no resultant, and therefore cannot be balanced by any single force. Such a pair of forces constitute what is called a statical couple. If, in fig. 54, we suppose P and Q to be equal, and A B to be at right angles to their directions, A B is called the arm of the couple, and A B x p its moment. A little consideration will show that the sum of the moments of the forces with regard to any point in the plane of the couple will equal a B x P ; and moreover, that if the sign of the sum of the moments with reference to one point is positive, it will be positive when taken with reference to any point in the plane of the couple ; and if negative, negative ; e.g. the couple represented in the diagram has a negative moment. Proposition 11. If two couples of equal mam^ents a/nd of opposite signs act i/n the same plane on a rigid body they will bala/nce one another. First. Let the forces which constitute the two couples 1^0. 66. not act along parallel lines, then must the four lines by their in- tersection form a parallelogram* Let A BCD be the parallelogram ^ thus formed, and let the forces ^^ (p, p') of the one couple act along AB and CD, then must the forces (q, q') of the other couple act along A D and c b, since the moments of the couples RESULTANT OF TWO COUPLES. 93 have contrary signs ; draw A m and A to at right angles to c D and c B, then since the moments of the couples are equal Am,XP = A«XQ also A7i,XAD = AmXAB since each product is the area of A B c D ; therefore ADXP=ABXQ or p : q:: AB : ad therefore a b and A d represent (Art. 25) the forces p and Q, and therefore the diagonal A c represents their resultant (r). In like manner p' and q' are repre- sented by c p, and c b respectively, and therefore c a represents their resultant (r'). Hence, the four forces P, Q, p', q' are equivalent to a pair of equal opposite forces R and r', and therefore are in equilibrium. Secondly. Let the four forces act along parallel lines ; draw a straight line cutting ^^ gg those lines at right angles in „ ^ A, b, c, d, respectively ; and let p * » and Q act in the same direction, .1 — -, i — X J and p' aiid q' in the opposite di- | | rection, then the moments of the "' ^ couples wiU have contrary signs ; now r the resultant of p and Q equals p + Q, let it act through the point x, then we have AXXP=CXXQ also since the moments of the couples are equal AB XP = CDX« therefore bxxp = dxxq or bxxp' = dxxq' hence the resultant (r') of p' and q' acts through the point X, and as it equals p' + q', the four forces p, Q, p', q' are equivalent to two equal forces, R and r' acting in 94 PKACTICAL MECHANICS. opposite directions along the same line, and therefore are in equilibrium. Cor. 1. Hence two couples of equal moments and of the same sign and acting in the same plane on a rigid body are equivalent to one another, since either would be balanced by a couple of equal moment and of contrary sign. In other words, there will be no change produced in the effect of a couple by supposing it to act anywhere in its original plane, and by supposing its arm to be lengthened or shortened, provided the forces undergo a corresponding change, so that its moment remains unaltered in sign and magnitude. Cor. 2. Hence, also, if M and N are the moments of two couples acting in the same plane, they will be equiva- lent to a single couple whose moment is their algebraical sum M + N. For let both couples be reduced to equivalent couples having arms of the same length a, then if p and p' are the forces of the one, and Q and Q' of the other, we shall have a p or a p' equal to M, and a Q or a q' equal Pig. 67. to N ; now place the couples so that their arms coincide, then if both moments are positive, the couples will He as shown in the figure, i.e. they are equivalent to a , p, pair of parallel forces, p+q and p' + a' \ e an angle of 31^ 45' (b and A on opposite sides of bc), the resultant is a force of 7'6 lbs. acting from B to b. 96 PRACTICAL MECHANICS. Ex. 254. — ^When fonr forces acting in the same plane at a point are in equilibrium, ehov that a quadrilateral figure can be drawn, the sides of which are related to them in the same manner that the sides of the triangle in Art. SI are related to three forces in equilibrium. Ex. 255. — From the above example show that the resultant of three forces acting in the same plane at a point can be represented by a side of a quadrilateral, and state exactly how the quadrilateral must be drawn, Ex. 256. — Extend the results in Ex. 264 and 256 to any number of forces (v. Art, 59). 56. The resultant of any number of forces acting along the same straight Ivne. — Since the resultant of two such forces is their (algebraical) sum, the resultant of those two and a third force must be the (algebraical) sum of the three, and the same wiU be true of any number of forces ; hence, if any nurriber of forces act along the same straight line thei/r resultant wiU equal their algebraical sum,. If their algebraical sum is zero, the forces will be in equilibrium. In the following general theorems the term 'sum' means 'algebraical sum.' 57. The resultant of any number of couples acting in the same plan£.~— Since the moment of the resultant of two such couples is the sum of the moments of the two couples ("Prop. 11, Cor. 2), that of the resultant of those two and a third will be the sum of the moments of the three, and the same wiU be true of any number of couples; hence, if amy number of couples act in the same plane, the moment of thei/r resultant equals the sum of thei/r several mAyments. If the sum of the moments is zero, the couples will be in equilibrium ; for if aU the couples are reduced to equivalent couples with equal arms, and these arms are superimposed on each other, it is plain that the moment of the resultant couple can only become zero by each force of the couple becoming zero ; i.e. the whole reduces to two systems of forces which are severally in equilibrium. PEINCEPLE OF MOMENTS. ^7 58. Extension of the principle of moments to any of forces. — ^Let Pj, P,, P3J . . . . P„ be any sys- tem of forces acting in one plane on a rigid body ; let R, be the resultant of p, and P^, Ej of Ej and P3, and so on, and E the resultant of £„_, and P„. Now, if the moments are taken with respect to "any one point in the plane, we shall have m'E, = ?n,*p, + m'Pj m*Ej=m'E, + m'Pj m*R = m*Ei,_2 + m'Pn therefore, by addition, m'R=m'Pi + m'P2+w!,*P3 + . . . + m*p„ Hence, if amy forces act vn a plane, the sum of their moments wiiJi respect to any povnt in that plane, will equal the mom,ent of their resultant with respect to that povnt. A little consideration will show that if the forces reduce to a couple, the moment of the couple will equal the sum of the moments of the several forces. Of course, if the point is taken in the direction of the resultant, its moment, and therefore the algebraical sum of the moments of the forces, will equal zero. Now, if a body acted on by any forces be kept at rest round a fixed point, the resultant must pass through that point ; and therefore in this case the algebraical sum of the moments of the forces round that point will equal zero ; a statement which coincides with that already given CArt. 39). It is plain that in this case the forces cannot be reduced to a couple ; for if they could be so reduced they could not be balanced by the reaction of the fixed point. 98 PRAOnOAL MECHANICS. Proposition 12. To determine the resultant of any system of forces acting along paralld H/nes vn one plane. Let Pi, Pj, Pj, .... be the forces ; take any point o, and draw o a at right angles to the direction of the PIS. 58, forces, and cutting the lines ,., along which they act in Np N2>N3, .... let ONi = ; Pi, o s-=p^, Ng=:p3 ; " ' S3 ^' also let E, the resultant of the forces, act along a line cutting A in m, and let o M=r ; we have to find the magnitudes of K and r. Now the re- sultant of any two parallel forces equals their sum, there- fore the resultant- of those two and a third force will equal the sum of three, aiid so on for any number of forces, therefore their resultant must equal their sum, or R=Pl+P2+Pg+. . . . again, the moment of E round o must equal the sum of the moments of the separate forces, therefore Rr=pj9, + P2r>, + PjP3 + . . . . The former equation gives E and the latter r. Cor. I. Let the resultant of Pg, Pg, .... be e', and let its direction cut o A at a distance from o equal to r' ; then it will be necessary and sufficient for the equilibriuni of p„ Pj, Pj, • . . . that Pj be equal and opposite to r', i.e. that r' equal pi, and that Pj +e' equal zero ; but e'=P2 + P3 + . . . . and ^'''^='^iP2 + ^nP3 + ' ■ • • Therefore it is necessary and sufficient for the equilibrium of the system of forces that RESULTANT OF ANY NUMBER OF FORCES. 99 P1+P2 + PS + . . « . =0 and ^iPi + PiPi + ^BPs + - . . ■ =0 By the words ' necessary and sufficient for equilibrium' is meant that on the one hand if the forces are in equi- librium the above equations will be satisfied, and on the other hand if the above equations are satisfied the forces will be in equilibrium. Cor, 2. If the equations when formed lead to the fol- lowing result, and ^iPi + ^tPii + ^3Pa + ' • • • = a finite quantity, the system of forces reduces to a couple. Ex. 257. — A Aniform rod is 3 ft. long and •vreighs 2 lbs. ; weights of 1 lb., 3 lbs., 5 lbs., and 6 lbs. are suspended on it in order at distancei; of 1 ft. apart. Determine completely the resultant of the forces, Ans. 17 lbs. acting along 5's line of action. Ejc. 258. — Let a horizontal line be draSyn from a point A to the right, and let forces of 5 lbs., 12 lbs., and 19 lbs. act vertically upwards on it, and of 10 lbs. and 20 lbs. act vertically downwards on it, the former at distances of 2 ft., 5 ft., and 14 ft,, and the latter at distances of 8 ft. and 20 ft. from A. Determine completely their resultant. Am. 6 lbs. acting upwards through a point 21 ft. to the lefc of A, Ex. 259. — If in addition to the forces in the last example, one of 6 lbs. acts at a distance of 10 ft. to the right from A, determine the resultant (1) when the force acts vertically upwards ; (2) wheij it acts vertically down- wards. Ans. (1) 12 lbs, acting vertically upwards 7 ft. to the left of A. (2) A couple whose moment is —204. i59. The reaultcmt of any number of forces acting in one plane at a point can be found by a very simple con- struction called the ' polygon of forces.' — Let the forces P, Q, B, s act at a point. in the directions op, OQ, o E, o s, and let it be required to find their resultant, Draw any line a b proportional to p in the direction p ; from B draw B c proportional to Q and in the direction Q^from H 2 100 PEACTICAL MECHANICS. . Pia. 69. C draw c D proportional to R and in tke direction o B, and Fio. 69a, finally D E propor- tional to s and in the -direction s. Then, if A E be joiiaed, the result- ant (t) of p, Q, B, and s will be a force acting at o in the direction a e, and proportional to ae. This is evident,. since by the tri&ngle of forces (Art. 36) the force at o represented by A e is equi- valent to two forces at o represented by A D and D E ; these to three forces at o represented by A 0, c D, D e ; and th^se in turn by four forces at o represented by a B, B c, c d, D E, i.e. p, Q, R, s. It is immaterial in' what order we take the forces ; for instance (fig. 59a), we may draw a b to repre- sent p, then B c to represent E, then c d to represent Q, and finally D E to represent s ; the resultant will be, as before, a force at o represented by A E, When the polygon is drawn, if it is found that E co- incides with A, the magnitude of the resultant is zero, and the forces acting at are in equilibrimn. To render the calculation of the magnitude and direc- tion of the resultant intelligible it is necessary in the first place to explain what are the rectangular components of a force. Let ox,oy he two rectangular axes, and let p be a iTo- ««• force acting at o along the line o P ; let o A be the line which represents the force p, and let the angle it makes with the axis of x, viz. a!0 A, ^ equal 6 ; now, if the parallelogram o B A c be completed, p will be equi- valent to two forces respectively, represented by B and c, and since these forces are at right angles to one another, they are called the rectangular components of p with re- spect to the axes ox and oy; again, since oc = OA sin 5 RESULTANT OF n FORCES. 101 and OB=OA cos^, it is plain that the rectangular components of P are p cos 6 along the axis o x and P sin 6 along the axis o y. If we always measure 8 in the same direction, viz. upwards from 33, so that it increases in a direction opposite to that in which the hands of a watch move, p cos 6 and p sin will give not only the mag- nitudes of the components, but also the directions in which they act : — thus if we suppose p to act towards o, the line which represents the force is o A, so that 6 is not xoF, but a; A, indicated by the dotted arc; and. then, since 6 lies between 1 80° and 270°, both p sin 6 and p cos will be negative, as they ought to be. Proposition 13. To determine the resultant of any system of forces acti/nq in one pla/ne at a point : and to vnfer the condi- tions of equuilibrium of such a system of forces. (a) Let Pj, Pj, P3, .... be the forces acting at any given point ; through dra^ two rectangular axes o x and oy, and let ^,, 62, 6^, . . . be the angles that the lines representing the forces make with the axis of x. Then these forces can be replaced by their rectangular compo- nents along the axes of x and y, i.e. by Pj cos ^1, Pj cos ^2, P3C0S ^3, • . . along the axis of x, and Pj sin ^1, Pj sin 6^, P3 sin 6^, . . . along the axis of y. Now, the former set is equivalent to a single force x acting along the axis of x, and the latter to a single force y acting along the axis of y, provided x = P, cos ^i+Pa cos ^j + Pg cos ^3 + . . . . Y=Pj sin ^i + Pj sin B^ + V^ sin 6^ + . . . . lOS ■ PRACTICAL kECHANICS, ■ Now, if R be the resultant of x and Y, and ^ the angle ^hich the line representing it makes with ox, we must have E cos ^=x (1) and R sin ^=X (2) which equations determine R and =15-86-60 =— 71'60 y hence b = 76-17 lbs. and <(> = 289° ,57', i.e. e acts as indicated in the diagram; this result may he p^ r,^ Terified by construction. Ete. 26l.^Let Pj, p^i *3 te three forces each of 100 lbs., let the angle a; o p, be 135° ; find their resultant by the above method. ,/l»s. E=41-41bs. ^ = 315°. 60. Tromsfer of a force vn a 'parallel dvrection. — Let A B and c D be two parallel lines, and p the length of thei^ perpendicular o N drawn from o Fig. es. in A B to CD; then if a force p acts ^ v« n p d from A to B along A B, it will be equi- valent to an equal parallel force acting along c D in the same direction, and a couple whose moment is p^, the sign of the couple being positive if o N is to the left of the direction of the force (as in the diagram), and negative if to the right. For if two opposite forces p', p", each, equal to p, act along CD, they will be in equilibrium, and the three will be equal to p ; but P and p" constitute a couple with a positive moment pjp, hence p is equivalent to p' and that couple. Hence also we can determine the resultant of a force P, acting along a line A b, and a couple pia. 64. whose moment is M ; for let M equal pj9, a p» o v t ' from in A B draw a perpendicular o N equal to p, and to the right of p's di- rection, if the moment of the couple is positive ; make the arm of the couple coincide with N then the couple will consist of the forces p' and p", each equal to p, acting as sbown in the figure ; hence the force and the couple are ' equivalent to the three forces p, p', and p", but P tod p" are in equUibfium, therefore the force p and the couple are equivalent to p'. 10,4 PRACTICAL MECHANICS. Em. 262, — If A, B, c, d are the corners of a square taken in order, and if forces act along three of the sides, viz. p from a to b, p from a to d, and p from c to D, shpw that the three are equivalent to a single force p acting from B to c. Proposition 14. To determine the resultant of any system of forces acting i/n a plane. Take ox, oy, any two rectangular axes, and let p„ Pj, Pj, .... be the forces, acting along given lines ; from o let fall perpendiculars j5j,j32,j93, .... on these lines ; then P, is equivalent to an equal parallel force acting in the same direction through o, and a couple whose moment is P J9„ the Hke is true of Pj, Pj, . . . . ; let 6^, 6^, 0^, , . . . be the angles made with the axis of x by the lines representing the transferred forces. Now, let R be the resultant of the transferred forces, and let ^ be the angle which the line representing it makes with the axis of x. Therefore, R cos <^=P, cos ^i+Pj cos (^sj + Pg cos 0^ + . . . . (1) R sin <^=Pi sin ^i+Pj sin 0^ + 7^ sin 6^+ .... (2) also let R r be the moment of the resultant of the couples, therefore, B.r=Tjp^ + T^^ + V^P3 + . ... (3) The equations (1) and (2) completely determine R. Hence the given system of forces is reduced to a known force and a couple of known moment; by compounding these we obtained the required resultant. Oor. When equations (1) (2) and (3) are formed, if we obtain p, cos ^i + Pj cos 02 + ^3 cos 03 + . . . =0 p, sin ^j+Pj sin'^j + Pg sin ^3+. . . =0 P J9i + P2P2 + PaPa + . • • = a finite quantity the system manifestly reduces to a couple. RESULTANT OF n FORCES. 105 Ex. 263. — A B o is a triangle right-angled at A, its sides a b and a c are each 10 ft. long. The forces p„ p^, Pj, each of 100 lbs., Fig. 65. act as shown in the figure : find their resultant by the method of Prop. 14. The force p, is equivalent to an equal parallel force whose direction passes through A, and a couple whose moment is 500 ^2. Hence the three given forces are equivalent to thethreeforcesof &. 261,andtotheabove couple. Now the latter three forces are equivalent to e, acting through A parallel to c B, where e equals 1 00( ^2 — 1), and the couple is equivalent to the two forces e' and b" each equal to e acting as shovm in the figure where the line A N is drawn at right angles to A e, and equals S00./2-Hl00( a/2-1) or 6(2 + V'2) ft. in length. The required resultant is therefore the force e". FiQ. 6e. Ex. 264. — ^In the last case if p, equals 200 lbs., show, by the method of Prop. 14, that the resultant equals 100(2— V2) lbs. and acts parallel to e' (fig. 66) along a line which cuts n Aproduced at a distance of 10( a/2 + 1) ft. from A. Ex. 265. — If A B c is a triangle, each of whose sides is 10 ft. long, and if a force p acts from a to b, an equal force from B to c, and another equal force from c to A, show that the three are equivalent to a couple, whose moment is Spa/S, Ex. 266. — If ABCD is a square, and if a force equal to 2p acts from A to B, an equal force from B to c, 3p from c to D, and an equal force from D to A, show by the method- of Prop. 14 that the resultant equals Pa/2, and acts in a direction parallel to the diagonal c A, along a line which cuts the diagonal b d produced in a point whose distance from d equals 2 b i>, . Ex, 267. — Let A b c be an equilateral triangle, draw A d at right angles to B c, in B produced take n e equal to d a, let equal forces (p) act from a to n, from B to c, from c to A, and from D to A respectively ; show that their resultant equals p, and acts through b in direction parallel to d a. Ex. 268, — In the last case determine the resultant if the foirrth force had acted from a to n. Ex. 269. — If three parallel forces are in equilibrium, they consist of two couples having equal moments of opposite signs. Ex, 270. — If A B c is any triangle, and if a force p acts from A to b, a from B to c, and e from c to A ; and if p : q : b : ; ab : bc : CA, show that the resultant of the three forces is a couple whose moment is represented by twice the area of the triangle. 106 PEACTIOAL MECHANICS. Proposition 15. To determme the coniMtions of equiUbrimn of a/w^ system of forces acting m one pUme on a rigid body. Adopting the notation of Prop, 14, let K be the result-^ ant of Pj, Pg, Now, the necessary and sufficient condition of equilibrium is that p, shall be equal and op- posite to R. But if we transfer p, to the point o, and then resolve it along o x and y, we obtain a force P, cos 0^' acting along o a;, a force P, sin 6^ acting along y, and a couple whose moment is p J3, : and in like manner by transferring R we shall obtain r cos , p, sin 6^ to R sin ^, and Fjjo^ to R r, i.e. it is necessary and sufficient for the equilibrium of the system that p, cos ^1 +R cos B,+P3a!3 (4) and _ 2/(r, + P3) = B,^/, + P3^3 or 2/(P,+P2 + P3)=(i'i+i'2)2/i + F32/3 (5) Hence, adding together (2) and (4), and also (3) and (5), we obtain «(p,+Pj+P3)=p,a;i+P2a;2+P3a;3 (6) 2/(p, + Fj + P3)=p,2/.+Pj2/,+P32/3 (7) As before, x and y undergo no change if thes lines of action of the forces are turned through equal angles and continue parallel. They are therefore the co-ordinates of the centre of the parallel forces. The same proof can evidently be extended to four, five, or any number of forces. Q. E. D, Cor. 1. If the points of , application of the forces had been situated in space of three dimensions, and referred to three co-ordinate planes, a precisely similar proof would have given us »(Pi + ^a + ^3 + - • • )=Pia'i+i'2a'2 + P3a'3 + - • • y{.^i + ^2 + ^3 + - ■ • )=Pi2/i+Pa2/2 + 1*32/3 + • • • «:(P,+ Pj + P3-|-. . . ) = PiZi.fV^Z^ + FsZs+. . . It will be remembered that the same values of x, y, z would be obtained in whatever order the forces had been taken, consequently a system of parallel forces can- not have more than one centre. It of course follows from this that a body or system of bodies cannot have more than one centre of gravity. CENT'EE OF PAKALLEL EORCES. Ill Cor. 2, If the case should arise in which but ?^x^ + 7JXf^ + F3X^ + . . . = A and Pi2/i+P23/2 + i'32/3 + - • •=» where one at least of A and B has some determinate finite .value, the system reduces to a couple ; and in this case there is %o centre of parallel forces in finite space, If the forces are the weights of parts of a body they act in the same direction, and therefore their sum can never be zero, so that every body and system of bodies must have one, ■and only one centre of gravity, which can be determined by the above equations. N.B,— For examples on this Proposition see Art. 6&, 112 PRACTICAL MECHANICS. CHAPTER V. OF THE CENTRE OF QEAVITY. 63. Dejlrdtion of the centre of gravity^— It has been already remarked that the weight of a body is an instance of a distributed force, and that it can be treated as a single force by supposing it to be collected at a certain point, called its centre of gravity. The centre of gravity of any system of particles is the centre of the system of parallel forces composed of the weights of those particles. If the particles form a solid body, it is plain that, if the centre of gravity be supported, the body will rest in any position under the action of gravity only, since the re- sultant of the applied forces will in all cases pass through the fixed point. It is also plain that no point but the centre of gravity has this property. That, as a matter of fact, every body has a centre of gravity, is shown in the corollary to Proposition 16. In determining the centre of gravity of any figure, it is assumed that a heavy line is made up of particles, a heavy plane of heavy parallel lines, and a solid of heavy parallel planes. It is also assumed that every figure is of uniform density, unless the contrary is specified. Ex. 271 .—Determine the centre of gravity of a uniform straight line a b. The line ab may he conceived to be made up of a number of equal particles distributed uniformly along it (like heads on a vidre) ; now if we take the two extreme particles, the resultant of their weights will pass through the middle point of a b, and in like manner that of each successive pair ; consequently the weight of the whole will act through the middle pdint of a b, which is therefore the centre of gravity of the whole, or of the heavy line A b. 64. Method of determining the centre of gravity of a plane area.-^Let a b c D be the plane area ; we may CENTRE OF GRAVITY. 113 conceive it to be made up of a set of parallel heavy lines, such as B D, E F drawn in any FlQ. 69. This method Pig. 70. direction. If we can find a set of paral- lel lines all bisected by a single line a c, the centre of gravity of each line must be in ac, and therefore that of the whole figure must be in A c. If, more- over, we can determine a second line bisecting another set of parallel lines, we know that the centre of gravity must also be in this second line, and must there- fore be at its point of intersection with a c. enables us to determine the centre of gra^- vity of many simple figures : it also sug- gests a practical means of determining the centre of gravity of any plane area whatever. Suppose the figure to be cut out carefully to the required shape in cardboard or tin ; suppose it to be suspended by a fine thread • from any point b ; now the forces in equi- librium are the tension of the string and the weight of the body; they must therefore act along the same Hne, so that the required centre of gravity must be in the prolongation B c of A B ; this prolongation can easily be marked by suspending a plumb-line from a. Again, suspend the body by a fine thread de fastened to any other point E, and draw the prolongation of this line, viz. E f ; the centre of gravity must be in E f, and therefore at G, the point of intersection of e f and b c. Ex. 272. — Show that the centre of gravity of the area of a circle is at its centre. Since any diameter bisects all lines in the circle drawn perpendicularly to it, the centre of gravity must 1)6 in any diameter, and thereforo %t the centre of the circle. I 114 PKAOTICAL MECHANICS. Ex. 273. — Show that the centre of gravity of an ellipse must be at its centre. Ex. 274. — Determine the centre of gravity of a triangle. j.jg 7j_ Let A B c be any triangle, bisect b c in d and join ^ ad; draw any line K l parallel to b c cutting A d in H ; then by similar triangles we have EH : ha::bd : ba HA : hi.::b A : dc .*. (ex sequali) kh:hi.::bd:dc But B D is equal to D c, therefore k h is equal to H L, or K L is bisected by a d ; and the same being true of any line drawn parallel to b c, the centre of gravity of the triangle must be in ad. Again, if AC is bisected in e and b b is drawn, the centre of gravity will be in b e, and therefore must be at g, the point of intersection of a d and b e. It can be easily proved that g d = J a d. For j oin e d, then because A e = e 0, and B D = D c we have AE : ec::bd : do and therefore e d is parallel to A 5 ; hence the triangle d e o is similar to ABO and EDCtoABc: therefore do : de::ga : ab and de;dc::ab:bc therejore (ex sequali) do : dc::qa : bc But dc = ^bc.*.dg = ^ga = 1da. Ex. 275. — Show that the centre of gravity of a parallelogram is at the intersection of the diagonals. 65. Centre of gravity of solids. — The . above method can easily be extended to the case of solids ; we may sup- pose them to be made up of heavy parallel planes : if we can show that the centres of gravity of these aU lie along a line, we know that the centre of gravity of the solid must be in that line, and if two such lines can be found, the centre of gravity of the solid must be at their point of intersection. Ex. 276. — Show that the centre of gravity of a sphere is at its centre. Ex. 277.— Show that the centre of gravity of a cylinder is at the middle point of its axis. [It may be regarded as evident that the same rule will hold good of any prism.] Ex. 278.— Show that the centre of gravity of a parallelepiped is at the point of intersection of its diagonals. CENTEE OF GRAVITY. 115 66. Centre of gravity of a figure consisting of two or nnore simple figures. — Let Wj and Wj be the weights of the simple fio. ra. figures and Gj, G^ their centres of gravity, join GiG^, such a manner that Then is G the required centre of gravity. If there were a third body weighing M'3 whose centre of gravity is G3, we can find the centre of gravity of the three bodies by joining G G3 and dividing it into parts inversely proportional to Wj+Wjand W3; and of course we could continue the same construction to a fourth or a fifth weight, &c. Ex. 279. — Two spheres whose radii are respectively 4 and 5 in. touch one another ; determine the distance of the centre of gravity from the centre of the siiialler sphere when the former is of copper and the latter of cast iron. Ans. 5'54 in. Ex. 280. — A solid sphere 4 in. in radius touches a hollow sphere 5 in. in radius and 1 in. thick; they are of the same material; show that their centre of gravity js 4'392 in. from the centre of the solid sphere. Ex. 281. — Determine by construction the centre of gravity of the todies shown in fig. «, where A B is a beam 20 ft. long, and its section 1 ft. square ; c and D the centres of two cylinders 1 ft. thick, the radii of whose bases "are respectively 6 ft. and 4 ft. ; they are of the same material as the beam, and rest with their centres of gravity vertically over the axis of the beam, at distances of 6 in. from a and b respectively. Construct the figure to scale ; this is done in fig. e, to the scale of 1 in. for 5 ft. — ^join c D, then the weights of the cylinders being in the proportion of 9 to 4, divide cD into parts do, and q,c respectively proportional to 9 and 4 ; this will give the centre of gravity of the two cylinders. The con- struction may be made as follows, by Eucl., Bk. VI. — Take d h any line contaiijing 13 equal parts (in the figure each part is |th of an inch) and measure off D k containing 9 of them, join h c and draw k o, parallel to h c ; then CO, : o,D : :hk : KD i.e. : : 4 : 9. JFind B the centre of gravity of the beam, join eg, ; now the united weight of the cylinders is to the weight of the beam very nearly in the ratio 163 : 20, hence, divide bg, in o so that E G : G G, : : 163 : 20, and the point g is the centre of gravity required. 1 2 n" H« PRACTICAL MECHANICS. Ex.. 282. — At points 120° apart on the edge of a round table weights of 84 lbs. and 112 lbs. are respeotivdy hung. Find where aweight of 22i lbs. should be placed so as to bring the centre of gravity of the three weights to the middle of the table. Ex. 283. — A disc of cast iron 12 in. in radius and 2 in. thick rests on a disc of lead 24 in. in radius and 3 in. thick ; the circumference of the upper disc passes through the centre of the lower ; determine by construction the centre of gravity of the whole. Ex. 284. — Show that the centre of gravity of any quadrilateral a b o D is given by the following construction r — take o the middle point of the diagonal b d ; in o a take o p a third of o a, and in o c take o a a third of o c ; jp^ p a cutting d b in h ; in p q take p a equal to a h ; the centre of gravity is at o. 67. The centi'e of gravity of points lyi/ng vn a straight lit^. — The method above explained of finding the centre P(j*. 73. of gravity of a collection of two or more 9, s S'. bodies can be applied to all cases ; how- ever, if there are only two bodies, or if the centres of gravity of three or more >yi " 1.2 ijodies lie in a line, itiscojmmonlymore c^avisoient to determine its distance frooa some fixed point ip,that line. Let g,, g^ be the centres of gravity of the 1^4 "^llJodies whose weights are w, and w^ respectively ; thgfl. the distafloe G of the centre of gravity of Wj and w^ fr(Sm is determined by the equation OG (Wj + W2) = G, X Wj + Gjj X Wj The method of teeating three or more weights is exactly t^e same. It is also plain that if we know o g and Gj, the same equation will give us o Gj. Ex. 285. — How far from the one end of the handle is the centre of gra»*f of the hammer described in Ex. 9 situated, if we suppose the other end to "fit scpiare with iJie face of the hammer ? *''*• '■*• [If the annexed figure represent the ham- mer, we have oa = 42 in. ab = 2 in., so that if G, is the centre of gravity of the handle and Oj that of the head, we have 001 = 21 in. oSj = 41 in. Also the weight of the handle is 4-46 lbs. and of the head 8'37 lbs. Hence 00 x 12-83 = 21 x 4-46 + 41 x 8-37 ,■.00 = 34 inches] ^' CENTRE OF GRAVITY. 117 Ex.. 286. — How far from the ead of the handle is the centre of gravity of the hammer described in Ex. 12 ? Ans. ,72^ in. Ex. 287. — Let A B be the diameter of a cirAlar disc of cast iron 12 in. in radius ; out of the disc is cut a circular hole (whose Ka. 76. centre is in a b) 4 in. in radius ; the shortest distance between the circumferences is one inch ; find the dis- tance of 6, the centre of gravity of the remainder, from A. Ans. Il| in. Ex. 288. — If in the last example the hole were filled up with lead, determine the distance of the centre of gravity of the body from A. Ams. 12-42 in. Ex. 289. — The gnomon A b o is cut out of a parallelogram A c ; determine the distance of its centre of gravity from b ; having given that D e and d b are respectively 20 and 15 ft. ' ' in length. Ans. 6-786 ft. Ex. 290. — If A B is the axis of a cross made up of six squares, the side of each being 3 in. long ; find the distance of the centre of gravity from a. Ans. 6J in. 'Ex. 291. — A rod capable of turning round a fixed point is kept in equili- brium by two weights suspended by strings of given length from the respective ends. Show that the centre of gravity of the weights is fixed whatever angle the rod makes with the horizon. Ex. 292.— Weights of 7, 7, and 6 lbs. respectively are placed at the angular points of a triangle ; find their centre of gravity relatively to that of the triangle. Ex. 293. — Out of an isosceles triangle cut a square having twO angles on the base and one on each of the equal sides. Find the centre of gravity of the remainder. Ex. 294, — A piece of wire of uniform thickness is bent so as to form three sides of a triangle ; show that the centre of gravity is the' centre of the circle inscribed in the triangle formed by joining the middle points of the original triangle. 68. Remark. — The following examples of the determi- nation of centres of gravity are similar to those contained in the former article, hut involve somewhat greater geo- metrical difficulties ; in many- cases it will be well if the reader bear in mind, that when bodies are of the same substance their weights are proportional to their volumes, so that it frequently happens we may reason upon their volumes instead of their weights. 118 1>EA0TICAL MECHANICS. Fig. 77. Ex. 295. — To find the centre of gravity of a triangular pyramid. Let ABCD bathe pyramid; bisect bd in H,join a h and h c ; take r h = iA-H and H E= J H c ; draw p o and A e, then these lines being in the same plane, viz. A c H, will intersect, let them do so in g ; this point will be the required centre of gravity, and b g will equal jth part of A E. For draw any plane bed parallel to B c D cutting the plane A c H in A c, the line A e in e, and a H in h ; then k is the middle point of bd; and it is evident by similar triangles that he : a^::hb : ah and A A : Ac:: ah : Hc .■.(exseq.)Ae::A(!::HB : HC but HE=iHC .'.he=^hc, and e is the centre of gravity of the triangle bed; and the same being true of every other parallel section, the centre of gravity of the pyramid must be in a e ; in the same manner it can be proved that the centre of gravity of the pyramid must be in of; therefore it must be at G the point of intersec- tion of A B and c F. Next, to show that e g = jA E. Join p b ; then since HB = iEcand HP=JrA, we have he : ec::hp : fa, and therefore PEis parallel to A c ; hence the triangles g e p and o a c are similar, and we have GE : ga::ef : ao::eh : ch but eh = |ch.'.ge=|ba = ^ab. Hence the centre of gravity of a triangular pyramid is found by the rule : Draw the line joining the centre of gravity of the base and the vertex of the pyramid, divide it into four equal parts ; the first point of section above the base is the centre of gravity. Ex. 296. — If the middle points of any two edges of a triangular pyramid which do not intersect are joined by a straight line, the middle point of that line is the centre of gravity of the pyramid. Ex. 297. — Show that the centre of gravity of any pyramid or cone is found by the same rule as the centre of gravity of a triangiUar pyramid. Ex. 298.— If out of any cone a similar cone is cut, so that their axes are in the same line and their bases in the same plane ; show that tho height of h*—h'* the centre of gravity of the remainder above the base equals f . .3_,„ where h is the height of the original cone, and h' the height of that which is cut away. Ex. 299.— If out of any right cylinder is cut a cone of the same base and height ; show that the centre of gravity of the remainder is |ths of the height above the base CENTKE OF GRAVITY. 119 Ex. 300. — ^Find the centre of gravity of a trapezoid in terms of the lengths of the two parallel sides, and of the line joining their middle points. Let A B c D be the trapezoid, of which a b and d c are the parallel sides ; produce A d and b c to meet in e ; bisect A B in p, join E F cutting d c in ' '*' H, which is its middle point. Take fGi = |fe, hQ2 = 5Hb; then g, is the centre of gravity of the whole tri- angle ABE and G2 of the part c d e ; therefore G, the centre of gravity of the remainder, will lie in p b. Now, we have given a b = a, d c = 4, and p H = A, and are to find ^a = x. Since the weights are in the same proportion as the areas of the tri-j angles A b e and c d b, we have FG, XABEsFGXABCD + FQjXODE Now, fg, = |fe andpG2 = A + lHE = A + |(pE-A) = |* +iFE therefore x x abcd = ^pex abe— (IA + Jfe) x cde But by similar triangles (Euc. 19 — ^VI.) ABE : oDE::a= ; V therefore a b c d : c d e : : a''— 6^ ; 6^ therefore x {a'^-b') = |fb x a2_ (|a + 1 j, e) i^ Again, by similar triangles, FE ; he:;ae ; dex^^ ; h therefore fb ." pe— HElIa : a—b ha and PE~ ^ a—b therefore x{a^-b') = ika(a + b)-^hb' = |(a + 2i)(<5-J) " h,a + 2b therefore x=-r r- 3 a + b Ex. 301.— Show that the centre of gravity of the frustum of a pyramid is situated in the line joining the centres of gravity of the ends and at a distance from the lower end, given by the formula x = —.- %Ji — , 130 PEACTICAL MECHANICS. where a and 6 are any pair of homologous sides of the ends, and h is the length of thfl line joining the centres of gravity of the ends. Piu. 79. ^'''- 302. — If a segment of a sphere is described by the revolution of A b c round b o, show that the centre of gravity of the surface of the segment is in the middle point of B c. [It can be easily proved that if B c is divided into any number of equal parts, and planes are drawn perpendicularly through the points of section, they will divide the surface of the segment into equal Mnes — the weight of each can be collected in b c ; and as these equal weights will be uniformly distributed along bc, the required centre of gravity will be in its midille iPoint.] JSc. 303. — Show that the centre of gravity of the spherical sector formed by the revolution of the sector A b o (fig. 79) round b o is at a distance from o = f oB— Sbc or |(ob + oc). [It must be remembered that the spherical sector may be conceived to be made up of an indefinitely great number of equal pyramids having a common vertex o, whose bases form the spherical surface ; the weights of each of these can be collected at its centre of gravity, distanced f ob from o, and the question is reduced to a case of the last example.] Ex. 304. — Determine the position of the centre of gravity of the volume of the spherical segment formed by the revolution of A b c round b o. And when A B c is a quadrant, show that the centre of gravity of the hemisphere generated by its revolution, is at a distance of |ths of the radius from the centre of the sphere. 69. Applications of the forTnuloe of Prop. 16. — When a body consists of parts, and we know the weights of the several parts, and the co-ordinates of their centres of gravity ; the co-ordinates of the centre of gravity of the body will be found by means of the formulse of Prop. 16. Ex. 305. — A, B, c, D are the angular points taken in order of a square (one of whose sides is a) and b the inters^ etion of its diagonals ; weights of 3, 8, 7, 6, and 10 lbs. are placed at these points respectively. Find their centre of gravity. Ans. If A B and A d are the axes of x and i/, S4x = 20a, Sit/ = 18a. Ex. 306. — ^Weights of 1, 2, 3, 4, 5, and 6 lbs. are placed respectively at the angular points of a regular hexagon (one of whose sides is a) taken in order. Find their centre of gravity. Ane. If the lines .ioining the points at which 1 and 2 and 1 and 3 are placed be the axes of a: and y, lix = 5a, liv = 9a^/3. Ex. 307, — A B C is an isosceles triangle right-angled at c ; parallel forces APPLICATIONS OF PROP. XVI. 121 of i, 6, and 8 lbs. act at A, b, and c respectively. Find their centre when the two former act in the same direction and the latter in the opposite direction. Likewise when the third force is 10 lbs. Ans. (1) If c Aand CBare the axes of a; andy,a = 2a,y = 3a. (2) Centre at an infinite distance, forces redncing to a couple. Ex. 308. — A, B, c, D are the angular points taken in order of a square, one of whose sides is a; parallel forces of 5, 9, 7, and 3 lbs. act at the angular points respectively. Find their centre— (1) supposing 5 and 9 to act in the same direction, and 7 and 3 in the opposite direction ; (2) supposing 5 and 7 to act in the same direction, and 9 and 3 to act in the opposite direction. Alls. (1) If AB and ad are the axes of x and p, then 2x = o, 1y= ~ 5a. (2) Centre at an infinite distance, forces re- ducing to a couple. Ex. 309. — Parallel forces of 5 lbs. apiece act in the same direction through the angular points of a square, and a parallel force of 20 lbs. acts through the intersection of the diagonals in the opposite direction. Find the centre. Ans. Centre indeterminate, forces being in equilibrium. Ex. 310. — ^Find the co-ordinates of the centre of gravity of the trapezoid ABCD, having given 0B= 7 ft., oc= 19 ft., AB=12ft., D c = 18 ft. ; the angles at b and c being right angles. ^"'- ^''• [If A N is drawn parallel to B c dividing the figure into a triangle and a square, the co-ordinates of the centre_of gravity of each can be easily found, and if X and t/ are the required co-ordinates, it will appear that they are determined by the equations 180'x = 13x 144 + 15x36 180y= 6x144 + 14x36] ^ras. J=13f, y = 7f. Ex. 311. — Let ABC D represent the section of Pig.81. a ditch : the breadth a d is 20 ft. and the depth 8 ft. ; the slope of A b is 1 in 1 and of D c is 2 in 1 ; determine the horizontal distance from A of the centre of gravity of the section. Ans. lOf ft. Ex. 312. — If in the last example the breadth ad is a feet, the depth of the ditch h feet, and if a b has a slope of m in 1 nnd d c oi « in 1, show that if¥be the horizontal distance of the centre of gravity of the section from A ; then x will be found by the formula 122 PKAOTICAL MECHANICS. Ex. 313. — If ABCD represents the section of a wall of -which bc is vertical and equal to A, a b = a and d c = J ; then if w is the weight of a cubic foot of the material, the moment of 1 foot of the length of the wall round a and b respectively are given hy the forraulse wh{2a'' + idb - ¥) M = and M = "'^("' + a* + !>') 6 Ex. 314. — The engine-room of a steam-vessel is 30 ft. long, 20 ft. wide, and 15 ft. high ; at 10 ft. from one side, 6 ft. from one end, and 5 ft. from the floor, is situated the centre of gravity of the hoiler, the weight of which is 2 tons ; at 4 ft. from the same side, 11 ft. from the same end, and 7 ft. from the floor, is the centre of gravity of the beam of the engine, which weighs I a ton ; at 9 ft. from the side, 7 ft. from the end, and 3 ft. from the floor, is the centre of gravity of the furnace, which weighs 1^ ton ; at 6 ft. from the side, 11 ft. from the end, and 10 ft. from the floor, is the centre of gravity of the cylinder, which weighs 1 ton ; where is the centre of gravity of the whole ? Jns. 8'1 ft. from the side, 7'8 ft. from the end, 5'6 ft. from the floor. 70. On stable and unstable equilibrium. — Bearing in mind that when forces are in equilibrium any one of them is equal and opposite to the resultant of all the rest, it is plain that when a heavy body is supported by any forces their resultant must act vertically upward through the centre of gravity. Suppose, then, that a body is supported at one point, the reaction of the fixed point and the weight of the body are in equilibrium, therefore the direc- tion of the reaction must pass vertically through the centre of gravity, consequently the conditions of equilibrium are fulfilled when the line joining the centre of, gravity and the fixed point is vertical, or, which comes to the same thing, when the centre of gravity is vertically under or vertically over the fixed point. Practically speaking, there is the greatest possible dif- ference between these two cases, for a body could scarcely be made to rest in the latter position, and , could be dis- placed from it by the smallest possible force and caused to take up the former position. In fact, the former case — STABLE AND UNSTABLE EQUILIBEIUM, 123 centre of gravity under the point of support — is said to be a position of stable equilibrium, while the latter — centre of gravity above the point of support — is said to be one of unstable equilibrium. The distinction between stable and unstable equilibrium is thus stated : Suppose a body to be in equilibrium under the action of given forces, and suppose it to be ■slightly displaced, if the forces tend to bring the body back again to the original position, that position was one of stable equilibrium, but if they tend to make it move farther from its original position, that position was one of unstable equilibrium. If the student will draw a figure of a body suspended from a point he will see at once that the two positions of equilibrium are stable and unstable according to the terms of the definition. It is obvious that there may be an intermediate case in which, after the body has been displaced, the forces have no tendency to move it either backward or forward. In this case the body is said to have been in a pos-ition of neutral equilibrium. If, in the example already given, the point supported had been the centre of gravity the equilibrium would have been neutral. A sphere of uniform density on a horizontal plane is in a position of neutral equilibrium ; if it be loaded at the top of a vertical dia- meter its position becomes one of unstable equilibrium, if loaded at the lower end of a vertical diameter it is in a position of stable equilibrium. Ex. 315. — A liemisphere (whose radius is r) and a cone (the radius of whose bar is r and height h) of equal and uniform density are fastened together so that their bases coincide. They are placed on a horizontal plane, and are in equilibrium resting on the lowest point of the hemisphere ; show that the equilibrium is stable, neutral, or unstable,, according as rVZ> =or , also since the body is on the point of sliding from b to a, the direction of e must make with d w an angle h d w equal to (p. Then we have w d e = 1 80** — , BDP = 90— o + ^, and pDW = 90 + o, therefore (Prop. 7) p:w:e ::sin : cos (0—0) : cos o. Ex. 338. — In the last example determine p and e if the mass m, weighing .750 lbs., is of wrought iron, on oak, and the direction of p inclined to the horizon at an angle of 16°. -dns. p = 413-3 lbs. e= 756-9 lbs. Ex. 339. — What would be the required force p in the last case if its direction were horizontal ? Ans. p = 465 lbs. Ex. 340. — Show that when a body rests on a horizontal plane the smallest force that will bring it into the state bordering on motion will act in a direction inclined upwards from the horizon at an angle equal to the limiting angle of resistance. 74. Conditions under which a body acted on by cer- tain forces 'will neither he overthrown nor slide. — Let a mass A B rest on a horizontal plane c D, and let the forces concerned be its weight acting vertically along the line E w EQUILIBRIUM OF A BODY ON A PLANE. 135 and a force p acting along the line E p : find R the result- ant of these forces ; in order that the body may be at rest it is necessary that R be balanced fig. 89. by a reaction equal and opposite to it ; this cannot happen if the direction of R cuts c D outside the base ; hence the condition that the body be not over- thrown is that the direction of the resultant fall within the base; if this condition be ful- filled, the body will slide or not, according as the direction of R makes with the normal to the point where it cuts the surface, an angle greater or less than the limiting angle of resistance. The question may be asked, if A B be pulled along the line E p by a continually increasing force, will it slide before it topples, or vice versa ? This is readily answered by joining A E ; then if A e w be less than the limiting angle of resistance, the body will topple before it slides, since r's direction will fall without the base before its direction makes with the perpendicular an angle greater than the limiting angle of resistance ; if, however, A E w be greater than the limiting angle of resistance, the body will slide before it topples. In the intermediate case, when A E w equals the limiting angle of resistance, the body will be on the point of toppling and sHding for the same value of P. It obviously follows from the above reasoning that when a body stands on a horizontal plane a vertical line drawn through its centre of gravity must cut the plane within its base. If a body rest upon points its base is the polygon formed by joining the points in succession. It is to be observed, however, that if any points would fall inside the polygon formed by joining the rest, they are not to be reckoned. 136 PEACTICAL MECHANICS. Ex. 341. — ^A rectangular mass of oak the base of -which is 2 ft. square and height 7 ft. rests endwise on a floor of oak, a rope is fastened to it at a certain height above the floor and is pulled by a force in a direction inclined upward at an angle of 20° to the horizon ; it is found to be on the point both of toppling and sliding ; find the height of. the point of attach- ment from the floor, and the magnitude, of the force. Am. (1) 2-68 ft. (2) 648-7 lbs. [It is manifest, referring to fig. 89, that e will be found by making the angle e a w equal to the complement of the limiting angle of resistance when the circumstances are those mentioned in the question.] Ex. 342. — A cylinder of copper the radius of whose base is 2 in. and height 3^ in. rests on a horizontal oak table, it is pulled by a horizontal force whose direction coincides with a radius of the upper end ; find the force that will just make the body move, and determine whether the motion will be one of sliding or toppling. Ans. (1) 8 lbs. (2) The body wiU topple. Ex. 343. — Work the last eicample supposing the cylinder to be of oak, the fibres being parallel to the axis of the cylinder. Ans. (I) 10-2 oz. (2) The body will slide. Ex. 344.— A round table stands on four legs, one at each angle of the inscribed square. It weighs 120 lbs. ; find the least weight which hung from its edge would overthrow it. Am, 290 lbs. Ex. 345. — ^A rectangular box is overthrown by turning round a horizontal edge ; given the lengths of the edges ; determine the height through which its centre of gravity must be raised. 75. Friction and the la/ws of friction. — Let A B be a ta,ble ; M a mass which, in consequence of the action of Pig. 90. certain forces, is on the point of sliding in the direction B A ; then the reaction b' will be equal to their resultant, and its direction will be inclined to the perpen- dicular to A B at an angle equal to the limiting angle of resistance ; let c r' be the direction of this re- action ; draw c D at right angles to A B, then the angle dcr' is equal to ^ ; take c E to represent e', and complete the LAWS OF FRICTION, , 137 rectangle H k ; we may replace e' by two components R and F, of which E acts along c D and f along a b ; these compo- nents are represented by H and c k respectively ; now it is evident that tan <6=— i.e. tan The tangential reaction f is commonly called the Frio- tion, and tan (which is generally denoted by the letter jti) is called the coefficient of friction ; so that when a body resting on a plane is in the state bordering on motion, the friction equals the normal reaction multiplied by the co- efficient of friction ; it will be remarked that unless the body is in the state bordering on motion the whole of the ■friction is not called into play, but only so much of it as is sufficient to produce equilibrium. If in any particular case we are required to determine the relation between the forces which keep a body in the state bordering on motion, and amongst these forces is the reaction of a rough surface, we may treat this reaction in either of two ways : — First, we may consider the re- action (e') to be a single force making an angle • • PfiijiBBdicnIar 29°40' 0-67 0-60 24°15' 0-45 0-41 Wnm^ inm on oak . Parallel 31°50' 0-62 0-53 31°50' 0-62 0-53 Cast iron on oak . Parallel 33°0' 0-65 0-65 Copper on oak . ParaUel 31°50' 0-62 0-53 31°5Q' 0-62 0-53 Wrought iron on cast — 10°45' 0-l9 0-19 10°10' 0-18 0-18 Cast iron on cast 9°5' 0-16 016 8°30' 015 016 Oak on calcareous oolite * . ■ Endwise 32°10' 0-63 0-53 20°50' 0-38 0-36 Wrought iron do. — 26°10' 0-49 0-44 34°40' 0-69 0-57 Brick do. . — 33°50' 0-67 0-56 Calcareous oolite on do. . — ' 36°30' 074 0-59 32°40' 0-64 0-54 each other. The experimental results from which this value was deduced are as follows : — Surface of Contact Normal Pressure Pressure on point of causing Motion Coef. Friction ;»,. 0-947 ft. 121 lbs. 283 „ 495 „ 1995 „ .2525 „ 67 lbs. 151 „ 252 „ 1171 „ 1287 „ 0-55 0-53 0-51 0-68 0-61 0043 ft. 389 „ 403 „ 1461 „ 204 „ 213 „ 855 „ 0-52 0-53 0-52 * The stone employed in M. Morin's experiments seems to have been a soft oolitic stone from the (jaarries at Jaumont near Metz. i 140: PRACTICAL MECHANICS. It is to be observed that in the above Table the numerical values of 11 were ascertained by experiment ; the values of

= 0-07. 76. The vnclined plane. — 'The principles which regu- late the equilibrium of a body resting on a plane inclined to the horizon are the same as those which regulate the equilibrium of a body resting on a horizontal plane — a case which has been already considered;^ — the applica- tions of the former case are, however, very numerous and very important, it will therefore be discussed at some length. It is scarcely necessary to observe that the in- clined plane is commonly reckoned amongst the ' Mecha- nical Powers.' Ex, 346. — A mass whose weight is w rests on aplane a b (fig../), inclined to an angle o to the horizon AC; it is acted on by a force p in a direction (n p) making an angle ;8 with A b : determine the relation between the forces p and w when p is on the point of making the body slide up the plane. TaJce G the centre of gravity of the body, and through it draw the vertical line G w, cutting p N in D, both lines being produced if necessary. Now, th& only forces acting on the body are its weight w along d w, the force V along D p, and the reaction (e) of the plane a b ; e's direction must pass through D, and must be inclined to a perpendicular to A b at an angle equal to ip, the limiting angle of resistance ; draw d M at right angles to a b, and make M D E equal to ; then b will act along the line b b. (The line e n i s drawn as in the figure, since the reaction e tends to oppose the sliding of the body.) Hence we have p : w:: sin wde : sin edp:: sin wdb : sin edp But yiT>s = a + , and bdp = 90 + j3 — ^ Therefore p : w:: sin (o + ((>) : cos (/3— ^) In the same manner it can be shown that w : B : : cos (3 — ) : cos (o + p) If the question is solved by the general equations of equilibrium (Prop. 15) Fig. /; page 140'. THE INCLINED PLANE. 141 ■we may call b' the normal reaction, acting at a point -whose distance from M is a; ; the friction will be /i r', acting from b to a ; also we may represent DM by^. Then, if -we resolve the forces along and at right angles to ab, and measure momehts round d, we shall obtain W sin + /4B,'— pcos ;8 = (1) — WC0SO+ K' + psin^ = (2) a!ii,'—fipji.' = (3) Equations (1) and (2), when solved, give relations between p and w, and between w and b', equivalent to those already obtained ; equation (3) shows that b' will act through the point e. With numerical data, a solution can be obtained by construction, as indicated in the diagram, by the parallelogram h e, in which, if d h repre- sents the given weiglit, D K will represent the required force, and e d the reaction. Ex, 347. — If o be greater than p, show that when the body is on the point of sliding down the plane p : w:: sin (o— ^) : cos (;3 + ^) w : B : : cos (|8 + ^) : cos (o + ;8) Ex. 348. — Show that if a< the body wUl remain at rest without support. Ex. 349. — A mass of wrought iron weighing 500 lbs. rests on a plane of oak inclined at an angle of 20° to the horizon, a force p acts upon it so as just not to pull it up the plane in a direction inclined to the plane at an angle ol 12°; find p. Ans. 417-9 lbs. [In fig. / the construction is shown by which this example was solved, the scale being 1 in. to 200 lbs. ; the result obtained by the construction was 416 lbs., the correct answer being 417'9 lbs.] Ex. 360. — In the last example suppose p to act along p d as a pushing force ; find its magnitude that it may just not push the body down the plane. Ans. 142-1 lbs. Ex. 351. — Referring to Ex. 349 and 360 : first, if p had been a force of 200 lbs. acting up the plane; next, if p had been a force of 100 lbs. acting down the plane ; and, lastly, if there were no force p ; find the magnitude and direction of the reaction of the plane. Am. (l)428-71bs. PDB = 81° 18'. (2) 6594 lbs. p n e = 1 30° 42 '. (3) 500 lbs. acting vertically upward. Ex. 352. — Show that the direction of the smallest force which will make a body slide either up or down an inclined plane makes an angle (p with the plane. Ex. 353. — What is the least force that will draw a cubic foot of cast iron down a plane of oak inclined to the horizon at an angle of 14° ? Ans. 146-7 lbs. 14? PRACTICAL MECHANICS. Kt. 354. — In the last example what would have been the least force necessary to support the mass had the plane been of cast iron ? Ans. 38-6 lbs. Ex. 356. — What would be the horizontal force that would just push the body up the inclined plane in the last case ? Ans. 192 lbs. Ex. 356. — If the body represented in fig. / is a cylinder the radius of whose base is r and height 2h, and if p acts at a point n so chosen that for the same value of p the body is on the point of turning round x when it is also on the point of sliding up the plane, show that (rcos a + h sin o) cos (/8— *) cos/B sin (a + (the limiting angle of resistance between the sides of the wedge and the obstacle): then, since the wedge is on the point of moving forward, the mutual action between the surfaces of contact at e and f will act along these lines, and the wedge is kept at rest by w antf re- actions b' and b', equal and opposite to r and B ; the directions of these three forces must pass through a common point G ; therefore e' : w : : sin c G E : sin e g e Now, if A c G equals a, we have COB equal to 90 — ((/> + o), and E G E equal to 180 — 2 (_

—a) w, = w , — ii 1 sm (i^ + o) Ex. 369. — An iron wedge whose vertical angle is 13° is driven into a. mass of oak by a force of 1 cwt. : — what force will be necessary to extract it? Ans. 77-27 lbs. Ex. 370. — Sliow that the wedge will start if the force be withdrawn, provided the angle of the wedge be greater than 2^. Ex. 371. — An iron wedge whose angle is 7° is driven into a mass of oak ; find what fraction of the driving force is consumed by friction. Jjis, If w' is the force on the smooth edge which exercises the same normal pressure on the block as that produced by w on the rough edge, then w' = 009 w. Ex. 372. — In Ex. 366, if a b c is not isosceles, and if the limiting angles of resistance at e and p are

) Ex. 375. — ^Ih the last example, if all the surfaces are rough, show that p cos (o + (/>) cos (i/>, + (f)^) = « cos (p^ sin (a + ((> + ^,) Ex. 376. — In the last example, if Q, is the force that will just drive p out, show that a sin (0 + ^ + * cos («—<(>) cos (.i—i) = ,) cos (0 + ^) cos (, + ip^) What is the smallest slope of A e at which it will be possible for this to happen ? 79. The form- of the helix or the thread of the screw. FIG. 95. — Let ABC be a right- angled trianglcj and defg a cylinder, tte circum- ference of whose base is equal to the base (a c) of the triangle ; if we sup- • pose this triangle to be wrapped round the cylin- der so that A and c come together, as indicated by the small letters acb, the hypothenuse A b will take the form of a curve called the helix, i.e.''the curve to which the thread of a screw would be reduced if it became merely a line. Ex. 377. — If the distance measured parallel to the axis between two turns of a thread of a screw (or its pitch) is h and the radius of the cylinder is r, show that the length of n turns of the thread is n.^Wr' + h?. Ex. 378.— Show that if h is the pitch and r the radius of the cylinder, then if* 9 is the angle of inclination of the thread of the screw we shall have tan 9=--. 2w Ex. 379. — The length of a screw is 1 J ft., in which space the screw makes .36 turns, the radius of the cylinder is \\ in. ; determine the angle of incli- nation of the thread and its length. Ans. (1) 3° 2' 12". (2) 3397 in. 80. The form of screw with a square thread. — In THE SCREW. 147 the last Article we considered the form of the geometrical curve called the helix. If we suppose that instead of the triangle ACBwehave a solid, suchthat, when j,,^ gg it surrounds the cylinder, its upper face projects at right angles to the cylinder at every point, as shown in the annexed figure ; this upper surface will have the form of the upper surface of the square-threaded screw ; if now the lower part of this pro- jection be cut away, so as to leave a project- ing piece of uniform thickness, we shall obtain a screw with a square thread, as shown in fig. 97,* a section of which made by a plane passing through the axis of the cylinder is shown in fig. 98. The student will remark that the thread of a screw, though a very common object, has a very remarkable form ; for instance, the curve ri3. 97. Fig. 98. 'p a a' (fig. 97), which when prolonged passes through the points a, a^, a^ (fig. 98), is a helix, as also is the curve b b' (fig. 97), which when prolonged will pass through the points b, 6,, b^ (fig. 98). Now imagine a cylinder to be described whose axis coincides with that of the screw, and * When there is a considerable distance between two consecutive turns of the thread, as is the case with the screw represented in the figure, it is usual to have a second intermediate thread running round the cyljader. This is done for the purpose of distributing the pressure exerted bo.,ween the thread and its companion over a larger area, and thereby decreasing the risk of breaking the thread. L 2 us PRACTICAL MECHANICS. whose surface cuts the thread between a and b (fig. 97), the curve of section will be a helix, as indicated by the dotted line ; the triangles whose hypothenuses form these helices will all have the same height, viz. a a^ or 6 b^ (fig. 98), but their bases will be the circumferences of the bases of their respective cylinders. ^. 380. — If h is the height between two turns of the thread of a screw (or its pitch), r and r, the radii of the external and internal cylinders, and 8 and fi, the angles of inclination of the external and internal helices, show that tan (9.-9)= ^"f ('•-'•■> and show that the formula gives a correct result when r, = 0. Ex. 381. — If the thread of the screw in Ex. 379 were cut half an inch deep, determine the difference between the lengths of the interior and exterior helices, and the inclination of the mean helix. Aiis, (1) 112-8 in. (2) 3° 38'. Ex. 382. — The external and internal radii of the thread of a square- threaded screw are r and r, ; its thickness (measured parallel to the axis) is a ■ show that the volume of one turn of the thread is ir (y'^—r^) a. Ex. 383. — A wrought-iron screw is 1 ft. loJig, and 1^ in. in radius, the thread makes 3 turns in 2 in., its thickness is | in., its depth ^ in. ; find its weight, and the weight of the part cut away when the screw was made. Ana, (1) 276-1 oz. (2) 106-2 oz. ^«>- »9- 81. The screw- press. — The most familiar application of the screw occurs in the screw-press, and as it is very desirable that the student should get a clear conception of the mode of action of the forces in the case of the screw, he will do well to examine a screw -press; its most usual form is i a r^ n K A a 11 M \ .. f ■ SCREW-PEESS. 149 Fig. 100. represented in the annexed figure, and can be sufficiently described as follows : F F F f is a strong frame ; at a in the middle of the cross -piece is a hoUow nut, on whose interior surface is cut a groove, called the companion screw, which the thread of the screw b c exactly fits ; the end c of the screw is fixed to the piece de in such a manner that the screw is free to turn, while the piece d e- can only move in a vertical direction in consequence of the guides FF and ff; it moves downward when the screw is turned by the handle G H in one direction, and upward when the screw is turned in the opposite direction ; in the. former case a pressure is exerted on the mass m which it is the purpose of the machine to compress. The action of the forces in this case will be understood by considering the annexed figure, in which AAA A represents a section of the nut, B c of the screw, F F the guides, D E the movable piece, Y Y the thread of the screw, X x the groove of the companion ; the force p is equivalent to the pressure at the end of the arm which tends to turn the screw ; Q is the reaction against de which balances p; the frictions called into play in this case are the following : (1) be- tween the thread and the groove, (2) between the end of the screw and the piece D E, (3) between the guides F F and the sides of the piece D E, (4) between the cylindrical surfaces of B and A. It is not easy to obtain the relation between p and Q in tha 150 PRACTICAL MECHANICS. state bordering on motion when all the frictions are taken "into account ; * the frictions marked (3) and (4) are, how- ever, small, and in the following pages will be neglected. Ex. 384. — Show that in the case of the screw-press the relation between V and a is given by the formula pa=Q?' tan (a + ). Now^i, p^ p„ . , . have the same tendency as p to turn the screw round its axis, and therefore the principle of moments gives us pa = p,r+^2r+/)jr+ .... * If 25 equals d e, p the radius of the end of the screw, /j, /, jj." the co- efficients of friction between screw and nut, screw and d e, and guides and DE respectively, and the remaining notation the same as that employed in Fx. 384, the following, it is believed, will be found to be the correct formula for the relation between p and q : — (urcos(2a + (()) \ bn I ^ , ,„,\ Vl + ^''cos=ocos(a + *)/ i + i/;tt"pV ^ ^> ^v ) evidently differs but little from the formula of Ex. 393. SCREW-PEESS. 151 also sin<:e the pressures g,, q^, q^, , , , are all parallel to q's direction, we Jiave therefore pa = ar tan (o + . 386. — There is a screw with a square thread the radius of which is 1 in., the pitch is | in., the nut is of cast iron and the screw of wrought iron, their surfaces are well greased ; determine the pressure that would be pro- duced on the substance in the press if we neglect all the frictions but that between the thread and the groove, when the screw is turned by a force of 150 lbs. acting at a distance of 3 ft. from the axis of the screw. Am. 35,275 lbs. ' Ex. 387. — In the last example determine Q if the screw is not greased. Am. 22,007 lbs. Ex. 388. — Find the number of turns per foot which the thread of a per- fectly smooth screw will make whose power is the same as that of the screw described in Ex. 386. Ans. 12^ nearly. Ex. 389. — If in any screw reckoned perfectly smooth a force p were re- quired to compress a substance with' a force a, and if p' were the additional force required in consequence of the friction between the thread and the groove, show that p* = tJ^, very nearly, where a is the angle of inclination of the thread of the screw, and ju the coefficient of friction — neither being large. Ex. 390. — If the screw described in Ex. 386 has to exert a pressure q, find both from first principles and from the formula in the last example the value of^. Ans. (1) 1-885. (2) 1-880. p Ex. 391. — The diameter of the screw of a vice is 1 in. and the thread -makes 4 turns to the inch, the whole is of cast iron and the screw is well greased ; the handle by which it is turned is 6 in. long and is urged by a force of 100 lbs. ; the jaws of the vice hold an ungreased piece of wrought iron by friction only ; find the force requisite to extract it. Ans. 2530 lbs. 82. Friction on the end of the screw.—Let A B c be a cylinder or pivot, the end of which is urged against a rough plane by a force Q acting along its axis c ; the cylinder 152 PKACTICAL MECHANICS. Fig. 101, is supposed to be on the point of turning round the axis, and is opposed by the friction ; it is required to determine the moment of the frictions with respect to the axis o c. It may be assumed that the inequalities of the surfaces ■will wear away, and that th^ pressure will be equally dis- tributed ; consequently if p is the radius of the pivot (say in inches), -^ will be the pressure per square inch, and consequently ^^ will be the friction per square inch ; •jrp hence if we consider a small ring enclosed between two circles, whose radii op and Of are respectively r and r + Sr-, its area will ultimately equal 2'n-rSr, and the &ic- tion on it will equal -^ rBr. Now the friction at every point of this ring acts in a direction perpendicular to the radius at that point, and hence the sum of the moments of the frictions on this ring with respect to the axis will ultimately equal -^ r^Br ; the same will be true of any other ring, and therefore we shall obtain the required moment if we divide the area into a great number of rings, and ascertain the limit of the sum of the moments of the frictions on each ring; this can be done as follows: Take de = /3 and at right angles to it draw ef=jo, perpendicularly to both draw eh=-^ , complete the P rectangle e f G h, and complete the pyramid d e f g h ; take DP=r and pp — Br, and through p and p draw planes parallel to the base enclosing the lamina pes; then it is FEICTION ON END OF SCEEW. 153 con- plain by similar triangles that p s=r and p Es=-i^ r, P sequently the volume of the lamina is ultimately equal to Fig. 102. p p 2^.Q r^Sr, i.e. the moment of the friction on the ring is correctly represented by the volume of the lamina, and the same being true of any other lamina, we shall have the moment of the whole correctly represented by the volume of the pyramid,* i.e. the moment equals 2uQ ^pxpx -^— or moment of friction = |joQ/t. Ex. 392.^If the screw rests on a hollow pivot whose internal and ex- ternal radii are respectively p, and p, show that the moment of the friction round the axis of the screw is given by the formula and show from this formula that when p, is very nearly equal to p tli6 friction is very nearly equal to pej/u. Ex. 393. — In the screw when the friction on the end as well as the fric- tion on the thread is taken into account we have v=— tau(o + <(>) + §.£-(j;U a a where p is the radius of the end on which the screw rests. * The student who understands the Integral Calculus will perceive that the above construction is equivalent to integrating the expression -^r'dr between the limits of r = and rsp. 15t PEACTICAL MECHANICS. [Referring to Ex. 384 the equation deduced from the prineiple of moments will become Fa = rpt+rp, + rp,+ .... +|pcy*] Ex, 394. — It is required to compress a substance ^ith a force of 10,000 lbs. ; thescrewwith which this is done has a diameter of 3 in., and its thread makes 1 turn to the inch ; the arm of the lever is 2 ft. long j determine the force p that would be required — (1) if all frictions were neglected ; (2) if the friction between the thread and groove were taken into account ; (3) if the friction on the end of the screw, which is 1 in. in radius, were also taken into account ; the surfaces being iron on iron well greased, ^»s. (1) 66-3 lbs. (2) 129-6 lbs. (3) 157-5 lbs. Ex. 395. — An iron screw 4 in. in diameter communicates motion to a nut ; the force is applied at the extremity of a lever 1 ft. long ; the inclination of the threiid of the screw is 6° ; determine the relation between the force applied and the weight raised by the nut, taking into, account the frictions between the thread and groove, and the end of the screw whose diameter is 3 in. — ^the surfaces are cast iron — (1) when well greased, (2) when ungreased. ■ ^M«. (1) p=0-0427a. (2)p = 0-0583«. Ex. 396. — If the angle of the screw were 12°, the diameter of the screw and of its end 4 in., and the lever by which it is turned 2 ft. long, the sur- faces being of cast iron and ungreased, what weight will a force of 1 cwt. overcome ? Ans. 2730 lbs. Ex. 397. — Determine the force required in Ex. 394 if the surfaces are ofungreased oak. Ans. 488 lbs. {The fibres may be reckoned to rest endwise between the thread and the groove as well as between the end and the movable piece.] Ex. 398. — Given q the pressure to be produced by the screw, r the radius of the mean thread, n the length of the arm, h the pitch, /x the coefficient of friction between the thread and the groove, if the friction between the thread and the groove is the only one taken into account, show that the force to be applied at the end of the arm is given by the formula * T h + 27riir -i- . — Q 83. Tlie endless screw. — It is not very unusual to make a screw work with a toothed wheel ; the arrangement of the pieces when this is done will be sufficiently understood by an inspection of the annexed diagram ; the screw A B may * This is the formula given in General Morin's Aide-Mehwire, p. 309. , ENDLESS SCREW. 155 be mounted in a frame, and turned by a winch; the teeth of the wheel (c) work with the worm of the . screw, on turning which the wheel is j,,^ j^g caused to revolve ; as the screw has no forward motion, it will never go out of action with the whed, and is, on that account, termed an endless screw. The reader will find in Mr. Willis's ' Principles of Mechanism ' * a discussion of the form that must be given to the teeth in order to secure equable working. When the machine is employed, it commonly happens that the screw drives the wheel ; sometimes, however, the screw is driven by the wheel, as in the case of the fly of a musical box. In the former case, if p is the force at the end of the arm which turns the screw, and Q the force exerted by the screw on the wheel in a direction parallel to the axis, it is easily shown that the relation between p and Q is the same as that determined in Ex. 384. Ex. 399. — If a force p acting on the thread of a' screw in a direction parallel to its axis is on the point of driving a force Q acting along a tangent to its base, show that « = p tan (a — 0) where a is the inclination of the thread of the screw at the working point, and (p the limiting angle of resistance between the driving and driven surfaces. Ex. 400. — If the action of an endless screw is reciprocal, i.e. if it will act whether wheel or worm is driver, show that the inclination of the thread ot the screw must be greater than ij) and less than its complement. Ex. 401. — An endless screw consists of a cylinder of cast iron the radius jf whose base is 3 in. ; the thread makes one turn in 4 in. ; what is the greatest extent to which the thread can project if the tooth by which it is driven is of cast iron and is ungreased ? Ans. 0'98 in. Ex. 402.- — In the last example, if the depthof the thread be 1 in. what is the least pitch with which the machine can work if the surfaces are greased ? Ans. 2-513 in. * P. IfiO. 156 PRACTICAL MECHANICS. 84. Friction of guides. — One or two instances of the Fig. 104. friction of guides have been given already {Ux. 373-6) ; the following case will still further illustrate the' subject : — E F is a beam constrained to move in a vertical direction by the four guides A, B, c, D ; a projection G H at right angles to E F works with a tooth or cam, K, revolving on a wheel : by the action of the cam the beam is lifted and then allowed to fall by its own weight, thereby serving as a ^ hammer. In the fundamental case the forces act as in th.e figure t and we treat the beam as a straight line, the guides as points, and represent A c by a, G H by 6, H c by x.* Ex. 403. — In the above case show that p < a— 2Jju— (a— 2x)jn/ K =ow Ex. 404. — In the above case if n'x > b show that the foMes. will not act exactly as shown in the figure, and that * For a fuller discussion of this case, see Traitt de Mecamque appliq. aux Machines; par J. V. Poncelet, vol. i. pp. 234-238. 157 CHAPTEE VII. OF THE EQUILIBRIUM OF BODIES RESTING ON AN AXLE, AND OF THE RIGIDITY OF ROPES; WHEEL AND AXLE, PULLEY. Section I. 85., Fundamental condition of equilibriv/m in the state bordering on motion, of a body capable of revolving rov/nd an axle. — All the forces acting on the body can be reduced to a single resultant, to which, rie. 105. ■when the body is at rest, the reaction of the bearing must be equal and opposite 5 let the annexed figure represent the axle resting on its bearing ; let B be the re- sultant of the forces acting on the body,/ and let its direction cut the circumference \ of the bearing at the point p ; take the centre of the bearing and join o p ; this line is the normal at the point of contact ; the body will therefore be in the state bordering on motion when the angle P R equals the limiting angle of resist- ance, the motion being about to ensue in the direction indicated by the arrow-head. This consideration enables us to give a very simple construction, which will apply to all cases in which the forces act on the body along parallel lines. Take the centre of the bearing (fig. 106), draw a line A parallel to the directions of the forces ; if the body is about to move in the direction indicated by the arrow-head, make the angle aop equal. to the limiting 158 PEACTICAL MECHANICS. Fig. lOG. angle of resistance; then the resultant force must act along the line E p parallel to A, since this is the only line drawn parallel to o A which will cut the circumference in a point p such that the angle ope equals the limiting angle of resistance ; hence if we measure mo- ments round p, we shall obtain the re- quired relation between the forces, the sum of those moments being equal to zero by Art. 58. Of course if the motion is about to ensue in a contrary direction, the angle A o p must fall on the other side of A. It will be remarked that the radii of the axle and its bearing are sensibly equal, so that though in the diagram they are repre- sented as different, that difference never enters the question. 86. Friction of axles. — ^When the body is in the state bordering on motion, the values of the coeflScient of fric- tion are the same as those given in the last chapter ; the same is also true in cases of motion where no unguent is interposed ; in nearly all cases of motion, however, an axle is kept well greased, both to prevent wear and to diminish the resistance ; the unguent may be supplied at intervals, as in the case of a common cart-wheel, or continuously, as in the case of the wheel of a railway carriage ; as might be expected, a continuous supply of unguent is found to be the most effective means of diminishing the resistance. The following table gives the values of the coefficients of fric- tion, and the limiting angle of resistance for the axles and bearings most commonly uSed ; the coefficients of friction^ are taken from the experimental determinations of Gr^eral' Morin,* from which the limiting angle of resistance has * Notions Fondammtales, p. 309. To avoid ambiguity, themeans of some of Gen. Morin's results h«ve been taken ; thus, instead of 007 to 0-08, the following table gives 0'075. FRICnON OF AXI-ES. 159 been calculated — those cases have been selected in ■which the unguent is most effective in diminishing friction. Table XII. FRICTION OF AXLES MOVING ON THEIE BEARINGS. Axle Eenewed at intervals Benewed continuously Unguents Bearings « u, tan tan <|> or sin or sine/) "/> Oast iron on Oilof olives, tallow, 0-075 (mean) 4° 20' 0-054 3° 6' cast iron or hog's lard Wrought iron Do. 0075 (mean) 4" 20' 0-054 3° 6' on cast Wrought iron Do. 0'075 (mean) 4° 20' 0054 3° 6' on brass Wrought iron Oil, or hog's lard Oil 6° 20' on liguum- ■vitse Brass on brass Do. 0095 (mean) 5° 30' Brass on oast Oil or tallow • « ■ • - , k 0-0485 2° 47' iron (mean) Ex. 405. — Let A B (fig. g) be a beam movable about a wrought-iron axle which rests on a cast-iron bearing, and whose axis passes at right angles through the axis of the beam ;* the centre c of the axle is 12 in. from a, and 30 in. from the centre of gravity of the beam and axle, the radius of the axle being 3 in. ; the weight of the whole (i.e. of the beam and axle) is 400 lbs. : find the weight which, when hung at a, will just cause the end a to descend. Draw the figure to scale ; draw through c the vertical line c d, and make the angle- d c Q equal to the limiting angle of resistance (10° 45') ; draw the * Of course there are in reality two bearings situated symmetrically with reference to the length of the beam, each of which supports half the united Pig. lor. S^ pressures p and w ; theptanof the machine being shown in the aecompanjing figure. 160 PRACTICAL MECHANICS. vertical line q r cutting A b in » ; then this being the direction of the "reaction the principle of moments gives us px A»j=wx»a but since no is very small, it is desirable to construct the axle on a larger scale ; this is done in fig. h, from which we obtain en equal to 0'd7 in. ; hence we find p equal to 10698 lbs. ; a result precisely the same as that obtained by calculation. If A c is represented by p, c g by 5, c q by p, and if <(> is the limiting angle of resistance between the axle and its bearing, we shall have c» = p sin , and therefore A3i=p—p sin and tia = q + p sin ^, whence generally V (p—p sin') = w(q + p sijKf)') In future p WJU be used to denote the radius of any axle that may be under consideration. Ex. 406. — In the last example determine the value of p which will just prevent the beam from falling when no unguent is used. Ans. 936-5 lbs. Ex. 407. — Determine the magnitudeand position of theresultant pressure in Ex. 405 if we suppose p = 1020 lbs.; and determine the magnitude of the angle its direction makes with the normal to the point of its application. ^»«. (I) 1420 lbs. (2)(»s = lfin. (3) CQ» = 3° 13' 47". Ex. 408. — ^There is a beam of oak a b whose length is 30 ft., depth 2 ft., and thickness 1 ft. ; at right angles to its face passes an axle of wrought iron the part of which within the beam is 8 in. square, the projecting part on each side is 6 in. in diameter and 6 in. long (so that its total length is 2 ft.), itn axis is situated 10 ft. from the end a, at which end is exerted a force of 6000 lbs. ; find the force at B which will just keep the beam from turning and the amount to which that force must be increased if it is on the point of overcoming the force at A ; the axle rests on an oaken bearing ungreased. Am. (1) 1550 lbs. (2) 1700 lbs. Ex. 409. — If a string were wrapped round the grindstone described in Ex. 16, determine the greatest weight that could be tied to the end of the string without causing motion, supposing the bearing to be of cast iron well greased. Ans. 4'8 lbs. jElr. 410. — If p and Q are two parallel forces acting in contrary directions ^nd keeping a body in equilibrium, and if p, the one more remote from the axle, is on the point of causing motion, show that v(p + p sin (p) = (i{q + p aintj)) [If we gradually increase p while o, continues constant, it is plain that tlieir resultant will be made to act at a continually increasing distance from a. Consequently, in the case supposed in the question, the resultant apts along a line as remote from q as is consistent with equilibrium.] 87. Wheel and axle, pulleys. — The wheel and axle and Fig. h, page 160. WHEEL AND AKLE. 161 the pulley are familiar examples of bodies capable of mov- ing round a fixed axle ; they may be sufficiently described as follows : — (1) The wheel amd axle. — Let A b represent a cylinder of wood or some other material called the axle, to the end of which is firmly fixed a fig. los. cylinder of a large diameter E c called the wheel ; they rest on a pair of bearings by means of a small cylindrical axis, one end of which is D, the geometrical axes of all these cylinders being coin- cident ; ropes are wrapped in op- posite directions round the wheel and axle respectively, to the ends of which weights p and Q are at- tached ; if p is so large as to descend, it will do so by turn- ing the machine ; this will wind up q's rope, and thereby cause that, weight to ascend. It is usual to describe the wheel and axle in the above form, in order to give defini^- ness to the calculation ; in practice, however, a winch com- monly supplies the place of the wheel. (2) The pulley is simply a thin cylinder with a groove cut in its circumference, on which a rope can rest : the cylinder is capable of turning round an axis, which is supported by a piece called a block ; this well- known machine is represented in the accompanying diagram. When several pulleys are combined into a single machine, they constitute what is called a system of puUfeys ; the system most commonly used is called the block and tackle; it consists of two blocks containing pulleys M Fig. Ifl9. \y 163 PEACTIGAL MECHANICS. are Fig. 110. Fig. 111. (under these circumstances called sheaves) which either equal in number, or else the upper block contains one more sheave than the lower ; the upper block is fixed, while the lower car- ries the weight; one. end of the rope by which the weight is raised is fastened to one of the blocks, and passes in succession round each of the sheaves, as represented in fig. 110; but it must be added that the sheaves in each block are commonly made equal, and ar- ranged one behind the other on a com- mon axis. Another ■ fil system of pulleys, Lm'^ called the Barton, is sometimes employed ; it consists of one fixed and any number of movable pulleys ; to the block containing each movable pulley is fastened a rope^ which after passing under the next pulley (thereby sup- porting if) is fastened to a fixed beam. The last of these pulleys carries the weight to be raised ; the rope which carries the first movable pulley passes over the, fixed pulley ; on shortening this rope the pulleys, and with them the weight, are raised ; the arrangement is shown in fig. Ill ; it rarely happens that more than one movable pulley is employed. It is to be observed that the rigidity of the cords, i.e. their want of perfect flexibility, plays an importaiit part RIGIDITY OF ROPES. 163 ■ in calculations concerning the mechanical power of the wheel and axle, and of the pulley ; we will therefore proceed to explain the method of taking that resistance into account. 88. Rigidity of ropes. — Let abc represent a drum or pulley, movable about an axis c, and let a rope A b d -pass fc>ver it, to whose ends are fig. 112. applied forces p and Q re- spectively, the friction of the rope being sufficient to pre- vent sliding; if one of the forces p overcome the other Q, it must do so by causing the drum to revolve, thereby winding on the rope A B d. ' Now the portion A B being circular, and B D being straight, the rope must be bent at the point B, and the rope not being perfectly flexible will' offer a resistance to being thus bent, and a certain portion of the force p will be expended in overcoming the re- sistance. It is found that this ' rigidity ' of the rope can be taken account of by supposing Q to act along the axis of the rope, i.e. at a distance from c equal to ^ of the sum of the diameters of the rope and drum, and then increas- ing Q by a certain force ; it is found by experiment that this additional force consists of a part depending only op the rope, and another part proportional to Q ; it is also found that, when other circumstances are the same, this additional force is greater as the curvature of the axis of the rope is greater, and therefore it can be correctly, represented by the formula A + BQ E where A and B are constants to be determined by experi- ment, and R is the effective radius of the drum, i.e. half the sum of the diameters of rope and drum. H 2 Iff4 PRACTICAL MECHANICS. The principal experiments on the rigidity of ropes are due to M. Coulomb,* whose results have been discussed by various -writers. M. Morin considers that M. Coulomb's experiments are sufficient for the construction of empirical formulae only in the cases of new dry ropes and of tarred ropes : from a discussion of the experiments f he obtains values of A and B which, after reduction, give the follow- ing values of the above formula : — (1) For new dry ropes, the resistance due to rigidity in lbs. equals - 1 0-062994 + 0-253868cH0-034910q I (2) For tarred ropes, the resistance due to rigidity in lbs. equals - I 0-222380 + 0-185525cHO-028917q I where Q is estimated in lbs., c is the circumference of the rope in inches, and R the effective radius of the drum or pulley in inches. From these formulae the following table has been calculated : — Table XUI. RIGIDITY OF ROPES. —^ — \ Eadiusot Eope Ciroumt of Eope New Dry Hopes Tarred Eopes A B A B 016 in. 1 in. 0-32 0-031910 0-41 0-028917 0-24 1-5 1-43 0-078543 1-44 0-066068 ,0-32 2 4-31 0-139640 3-86 0-115668 0'40 2-5 10-31 0-218183 8-64 0-180731 0-48 3 2M3 0-314190 17-03 0-260253 0-66 3-5 38-87 0-427643 .SO-66 0-354233 0-64 4 66-00 0-558560 61 -05 0-462672 72 4-5 105-38 0-706723 80-08 0-685569 080 6 160-23 0-872760 121-60 0-722925 * An abstract of Coulomb's Memoirs is given in Young's Kat. Phil. Vol. ii. p. 171. t Notions Fondamentales, pp. 316-332, RIGIDITY OF ROPES. 1C5 Rule. — Multiply b by Q in lbs., add the product to a,' divide this sum by the effective radius of the drum or pulley in inches, the quotient is the resistance in lbs. If the resistance added to Q give q', the relation between p and Q will be the same as that which obtains between p and q', acting by means of a perfectly flexible thread on a drum or pulley whose radius equals the effective radius. It is to be remarked, that the resistance due to rigidity is only called into play when the rope is wound on to a drum ; there is no resistance when the rope is wound off. For example : If the diameter of a pulley is 11 in. and a new dry rope 3 in. in circumference is used to lift a weight of 500 lbs., we have the effective radius of pulley 5*98 or 6 in,, and hence A + BQ _ 21-13 + 0-31419 X 500 _gQ ^^^ E 6 so that we may consider that a weight of 530 lbs.. has to be raised by means of a perfectly flexible string over a pulley 6 in. in radius. Ex. 411. — To determine tlie relation between p and q in the ease of the wheel and axle. ■ . ' ; In the annexed, figure, let c A, the radius of the wheel, be repre- sented by p; CB, the radius of the axle, by g ; c d, the radius of the axis, by p ; the power p ai^d the weight a act vertically at A and B, and the weight of the machine w acts vertically through c. If p is on the point of preponderating over a, make w o D equal to

sin ; also if we take into account the rigidity of the rope, the effective value of a, is A+Ba a + 2 Hence the required relation between p and q is p(jj— p sin ^)= ( a+ ) (ff + P sin ) + wp sin If no account he taken of the rigidity of the rope, the relation between p and Q will be p (p — p sin ) = Q (y + p sin <^) + wp sin (^ Ex. 412. — Awheel and axle weigh 1 cwt., the radius of the wheel is 2 ft., of the axle 6 in., the radius of the axis is 1 in., it is of wrought iron, and rests in a hearing of cast iron well greased ; if a equals 1000 lbs. find the magnitude of p (1) when it will just support, (2) when it is on the point of raising q — the rope being considered perfectly flexible. Ans. (1) 244-3 lbs. (2) 255-7 lbs. Ex. 413. —In the last example, if a is supported by a new dry rope 3 iri> in circumference, determine the value of p when on the point of raising a. Am. 290 lbs. [The increase of the radius of the axle due to the thickness of the rope must not be overlooked.] Ex. 414. — If p and a are two parallel forces, and p is on the point of drawing up a over a pulley whose effective radius is r, and weight w, show that p(r— p sin ) = Q(r + psin^) +wpsin0 where the positive sign is used if p and a act downward, and the negative sign if they act upward ; and that when the rigidity of the rope is taken into account the formula becomes p(r-psin (f) = Q |1 + ?. j (r + psin ip) + - (»• + p sin 0) ± wp sin In the following questions a and b will have these values, and it will be understood in every question re- lating to combinations of pulleys that the effect of the weight of the pulley on the friction of the axle is neg- lected ; it must also be remembered that this is not the same thing as neglecting the weight entirely. Ex. 415. — A pulley 6 in. in radius has an axle of 1 in. in radius of wroiigl.t iron, turning on an ungreased bearing of cast iron; aweightof albs, attached to a rope 3 in. in circumference is on the point of being raised over the pulley by a -sreight of p lbs. attached to the other end of the rope : show that p=l-1117a+3-4 Ex. 416. — If p is on the point of lifting a by means of a Barton consist- ing of one fised and one movable pulley, as shown piQ_ ii4_ in the annexed figure, determine the relation be- * tween p and q. *-'- [Let T, and T^ represent the tensions of the portions of the rope against which they are written; then since the rope is the same and the pulleys like one another, we shall have : — since p is on the point of overcoming t,, and t, on the point of overcoming Tj, and both t, and Tjj together lift Q, p = OT, + 5 T,= flTj + fi ^ a = Ti + Tj J Therefore (1 + a) p = a'a + (1 + 2a) J.] Ex. 417. — If the pulleys and ropes are of the kind specified inEx, 41S, and if the whole weight , lifted is 1000 Ibl., determine p ; also determine p supposing that all passive resistances are neg- kcted, Ans. (1) 590 lbs. (2) 500 lbs. [The weight of 1000 lbs. of course includes the weight of the lower block.] / \ 168 PKACTICAL MECHANICS, Ex. 418. — If Q is raised by means of a block and tackle each containing » single sheave, show that the same relation exists between p and Q as that given in Ex. 416. Ex. 419. — If pis on the point of raising Q by means of abloek and tackle containing in all n equal sheaves — the parts of the rope being all parallel — find the relation between p and a. [See fig. 110. If <„ t^ tg, . . . ■ a-rrangcil aa jr. ^g annexed diagram Fig, 116. (fig. 115) show that the relation be- a I V I J tweeu p and a is given by the follow- [ ing formula : p(l + a + flKi,) = a2a, + 6(1 + 11 + 2(io,) + aJ,(l+a) ■where a, b refer to the smaller pul- leys and a„ 6, to the large pulley. Ex. 423.— If a pair of similar pulleys is arranged as shown in the annexed diagram (fig. 116), where A and B represent immovable beams, show that { o'q , , aw P = — -■ +i a+1 0+1 \phere w is the weight of the movable pulley. CAPSTAN. 169 Ex. 424.— In the last exeunple suppose each pulley to be similir to that described in Ex. 421, and the movable pulley -with its block to weigh 50 lbs. ; the rope being dry and 4 in. in circumference, find the force re- quired to raise a weight q of 1000 lbs. and determine the corresponding value of p when the passive resistances are neglected. Ans.il) 6b8lhB. (2) 475 lbs, Ex. 425. — If two equal pulleys are employed to raise a ^i»- H'- weight Q, in the manner indicated in fig. H7, show that tl (2a + 1) p = a'a + i (2a + !)—««) and determine p when q weighs 1000 lbs., the pulleys and ropes being the same as in Ex. 424 ; and when passive re- sistances are neglected, ^ns. (1) 432 lbs. (2) 317 lbs. Ex. 426. — In the case of a tackle with three equal sheaves show that the force p which will just support a weight Q is given by the formula J (B-l)a 36 c .^ T-- D L o(a»-l) a(a»-l) a-1 and show that when the passive resistances are neglected^ the equation reduces to 3p = a. 90. The copsiflwi.— This machine in one of its common- est forms consists of a cylindrical mass of wood, c D, along the axis of which is fig. iis. cut a cylindrical aper- ture, which receives an axis A b (commonly of metal) on the top of which it rests; in the upper part of the capstan holes are cut, into which are in- serted arms, such as e f, by means of which the capstan is turned, thereby winding up the rope G H which carries the weight. Ex. 427. — A capstan is turned by two equal parallel forces p acting in opposite directions at equal distances a from the geometrical axis- of the figure, which are on the point of overcoming a force a ; let i be the radius of the cylinder round which the rope is wrapped, r the radius of the metal axle, |U, the coefBcient of friction between the top of the axle and the cap- E 170 PEACTIOAL MECHANICS. Stan, and n or tan ^ that between the side of the axle anJ tte capstan; show that when the friction on the top of the axle is neglected 2pa = ( J + r sin <(>) I a + — g — I and when the friction on the top of the axle is taken into account where w is the weight of the capstan. [For friction on top of axle, see Art. 82.] 91. EquilibriuTn of two forces acting in given direct tions on a body capable of turning round an axle. — ^Let p and Q be the forces whose directions intersect in A, and FiQ. 119. let p be on the point of pre- ponderance ; let o be the centre and p the radius of the axle, and

sin (, LP sin <1>\ PI J TWO- WHEELED CARRIAGE. 171 [Observing that a ho k is a quadHlateral, about which a cixcle can be described, it is plain that ohk = ;3 and o kh = o, consequently l=/> cos 3 + q cos o.] Ex. 430. — A weight ei hangs from one end of a rope, which after passing over a pulley (whose weight is neglected) takes a horizontal direction ; it is now sxipportect by. » equal pulleys, placed at equal distances apart ; show- that the force p applied to the other end of the rope, which is on the point of lifting Q, is given approximately by the formula \ »• / rV2~psm^ r + p' where r, p, belong to the first pulley, /, p', tp' to the remaining ra pulleys, w is the weight of one of the m pulleys, and w the weight of the rope which rests upon them. 92. The two-wheeled carriage. — In this case we may consider that the weight of the carriage is equally dis- tributed upon each wheel. i'iq.im. Now it will be observed that J at each instant the wheel is lifted over a small obstacle a; then if o is the centre of the axle, and b the point of contact with the road, the angle a o B must have a cer- tain magnitude, which we will denote by the letter 7. We will also denote the in- clination of the road by a, and the angle between the direction of the traction and the road by /3. Then the forces concerned are, the trac- tion T, the weight w, and the reaction E, of the point A, which, when t is on the point of moving w, must cut the circumference of the axle in a point D, such that D R=<^ ; then if we denote the arigle o A R by ^, the relation between T and w will be easily obtained by the triangle of forces. Ex. 431. — When the wheel, as above explained, is on the point of moving, show that T-w Sin(a + y+9 ) eos(/8-7-fl) 172 .PRACTICAL MECHANICS. Ex. 432. — If A is the length of the are A b, rand p the radii of the wheel and axle respectively, and if the road and the direction of traction are hori- zontal, show that ri = w (a + p ^) very nearly. Semark. — It appears from the experiments of General Morin that the traction is sensibly proportional to the weight directly and the radius of the wheel inversely, when the roads are paved or hard macadamised, and both the road and direction of traction are horizontal ; * consequently it appears that for such loads, under the circumstances assigned in Ex. 432, the trac- tion, as found by experiment, equals — , where A is a constant quantity ; r but from the example it appears that ^ = a + p ^, and hence the length of the arc a must be very nearly the same for the same road whatever be the radius of the wheel. * Morin, Notions Fondamentalee, p. 353. The account of the carriage wheel given in the text is taken from Mr. Moseley's Mechanical Principles of Engineerinff, pp. 395, 6, 7. The general results of M. Morin's experi- ments will be found in the Appendix to Mr. Moseley's work. The reader will find a great deal of condensed information on the subject of carriage wheels in Dr. Young's Natural Philosophy, Lecture 18. 173 CHAPTEK VIII. THE STABILITY OF WALLS. The general principles which regulate the relations that exist between the dimensions of a wall and the pressure it can sustain on its summit have been already discussed CArts. 42, 43) ; in the present chapter we shall extend the application of the same principles to a few other cases. Several questions intimately connected with the snbject of the present chapter are not discussed, as being too difficult for a purely elementary work — such are the conditions of the equilibrium of arches, vaults, domes, the more compli- cated forms of roofs, &c. 93. The Ime of resistance. — Let a b L m represent any structure divided into horizontal courses by the lines c n, ET, G H . . . . and let it be subjected fis. 121. to the action of any pressure P along the line Pa ; produce Pa to meet c D in a' ; if the mass abcd were without weight the pressure on G D would act on the point a' ; but the total pressiu-e on CD is the resultant (r,) of p and the weight of a b c ; the direction of this resultant mustcut CDatsome determinate point between a' and D, say at h, and let the direction of Ej be hh' ', now the total pres- sure on E F will be the resultant (^Rj) of El, and the weight of c d f E, which will cut E F at a determinate point c, between V and f; in the same 174 PRACTICAL MECHANICS. manner, the pressure on the joint o H will act through a determinate point d, and on l M through a point e. Now if we join the points a,h,c,d . . . we shall obtain a poly- gonal line which cuts each joint in the point through which the direction of the resultant pressure on that joint passes ; and if, further, we suppose the number of joints to be indefinitely great, the polygonal line will become a curved line, which is then called the line of resistance. It will be remarked that the directions of the resultants do not coincide with the sides of the polygon ah, he, . . ^ .. . and therefore the line of resistance determines only the point at which the pressure on each joint acts, not the direction of the pressure at that point. The line of resistance can be determined without much difficulty in a large number of cases : when this has been done, the condition of equilibrium — so far as the tendency of the structure to turn round any of its joints is concerned — is that this line cut each joint at a point within the structure ; and, of course, the stability of a structure about any joint will be greater or less according as the intersec- tion of the line of resistance with the joint is at a greater or less distance within the surface to which it is nearest. It is plain that since the resultant of the pressures that act on a wall passes through the point of intersection of the line of resistance within its base, the algebraical sum of the moments of the pressures, acting on the wall taken with respect to that point must equal zero. It may also be re- marked that, in the case of most walls of ordinary shapes, the line of resistance continually approaches the extrados or outward surface ; and hence, if the wall possess a cer- tain degree of stability with reference to its lowest joint, it will possess a greater degree of stability with reference to any higher joint. Most of the following questions can, accordingly, be solved without the actual determination of the line of resistance. EQUILIBRIUM OF WALLS. ITS' ^ Ec. 433.^A wall of Portland stone 30 ft. high and sustain on each foot of its length a pressure equal to the weight of 3 cubic ft. of the stone actina; in a direction inclined to the Tertical at an angle of 45°. Find the point of a ■ bracket to which this force must be applied that the line of resistance may cut the base 6 in. within the extrados. [Let the annexed figure represent a sec- tion of the wall ; let the force act along the line X N, and let a x equal x ; take b q, equal to&-in.; then the condition of equilibrium is that the moments of the force and of the weight of the wall round o, be equal. Draw a N perpendiculay to x n ; it can be easily- shown that QN = AC cos AXN — QC sin AXN — AX sin AXN 285-a; 2 ft. thick has ria. 122. .e. QS — - VI Whence we obtaiu 28-5-a V2 X 3 = 60 X J .•.a;= 14-36 ft. It maybe remarked tha' the determination of a perpendicular resembling Q N occurs in many of the following questions. It may also be added that it is sometimes convenient to resolve the pressure into its horizontal and vertical components at x and obtain the moment of each.] .&. 434. — Determine the point of application of the pressure in the last article if the line of resistance cut the base 3 in. within the extrados. Am. 7-04 ft. fir. 435. — A roof, whose average weight is 20 lbs. per square foot, is 40. ft. in span and has a pitch of 30°, i.e. the rafters make an angle of 30° with the horizon ; the walls of the building are of brickwork, and are 50 ft. "high and 2 ft. thick ; they are supported by triangular buttresses reaching to the top of the wall ; the buttresses are 2 ft. wide, and 20 ft. apart from centre to centre. Determine their thickness at the bottom that the line of resistance may fall 6 in. within their extrados : determine also the answer that, results from neglecting the weight of the buttress. Am, (1) 1;1675 ft. (2) M754 ft. JSt. 436. — A roof weighing 20 lbs. per square foot has a pitch of 60° ; the distance between the walls that support it is 30 ft. ; they are of Portland Stone and are 1\ ft. thick ; the pressure of the roof being received on the 176 PRACTICAL MECHANICS. inner edge of the summit, what is the extreme height to which the walls can be built ? Ans. The -walls can be can-ied to any height whatever, Ex. 437. — If the weight of each square foot of a roof is 15 lbs., its pitch 22J°, and the length of the rafters 30 ft., determine — (1) the thrust along the rafters, supposing them to be 4 ft. apart ; (2) the tension of the tie- beam if one is introduced; (3) the magnitude and direction of the pressure on each foot of the length of the wall-plate,* if there is no tie-beam ; (4) the thickness of the wall, which is of brickwork and 20 ft. high, when the line of resistance cuts the base 2 in. within the extrados, the pressure of the roof being received on the inner edge of the summit ; (5) the distance from the axis of the wall at which the pressure of the roof must act if the line of resistance cuts the base of the wall 3 in. within the extrados. ^Jis. (1) 2352 lbs. (2) 2173 lbs. (3) 705 lbs. at an angle ef . 60° 21' 40" to the vertical. (4) 3 ft. (6) 27 ft. Ex. 438. — If w is the weight supported by each rafter of an isosceles roof whose pitch is o, show that the thrust on each rafter is and the 2sma tension of the tie , 2 tan a 94. The pressure produced against a wall by water, — The following construction can be easily proved from Pig. 123. the principles of hydrosta- tics. Let A B represent a — II_ section of the wall made by a vertical plane, CD the surface of the water ; \^ draw the vertical line B e; "'''-,^ draw B F, at right angles ^,,!l>i' to AB and equal to BE; \ I ^,^''^' join cf; then the pres- \ L,--'" ' ' sure on any length of the " wall will equal the weight of a prism of water whose base is c b f and height the length of the wall ; or, in other words, the pressure on each foot cf the length of the wall will be the weight of as many cubic feet of water as the triangle B c F contains square * The wall-plate is the beam on which the feet of the rafters rest : its office is to distribute the pressure along the wall. RIVER WALLS. 177 feet ; this pressure will act perpendicularly to the face of the wall through a point p, where B p = Jbc. Ex, 439. — There is a wall supporting the pressure of water against its vertical face '; determine the pressure produced by the water on each foot of its length when 20 ft. of its height are covered. Ans. 12,500 lbs. Ex. 440. — In the last case determine the pressure on the lower 10 ft. of the wall. Ans. 9375 lbs. Ex. 441. — An embankment of brickwork has a section whose form is a right-angled triangle A B o ; the base bc is 6 ft. long; the height ab is 14 ft. ; will the embankment be overthrown when the water reaches to the top, if A b is the face which receives the pressure ? Ans. Yes; the excess of the moment of pressure of water is 9767. Ex. 442. — In the last case wiU. the embankment be overthrown if a c is the face which receives the pressure ? Ans. Yes ; excess of moment of pressure of water 8675. Ex. 443. — In Ex. 441 what horizontal pressure applied at a would keep the embankment steady ? Ans. 698 lbs. Ex. 444. — If the section of a river wall of brickwork have the form shown in the accompanying diagram, in which ab = 5 ft., DC = 15 ft., and bc equals 50 ft.; bc being vertical, and the angles b and c right angles, find the height to which the water must rise against b c to overturn it. Ans. 37'2 ft. Ex, 445. — If in the last example the dimensions were B o equal to 30 ft., A b equal to 3 ft., and d c equal to 10 ft., would the wall be overthrown if the water rose to the summit ? Am. Yes. Ex. 446. — There is the cofferdam sustaining the pressure of 26 ft. of water, supported by props 20 ft. long, 20 ft. apart, one end of each is placed |rds below the surface of the water and the other end on the ground ; determine the thrust on each prop. Ans. 468,800 lbs. Ex. 447. — If the section of an embankment of brickwork were of the form shown in fig. 1 24, and the dimensions were A B equal to 4 ft., D c equal to 12 ft., and b c equal to 24 ft., would it support the water when it rises to the top and presses on the face as? Ans. Yes ; excess of moment of weight of wall 5184. Ex. 448. — If the coefficient of friction between the courses of brickwork in the last example be 0'76, will the wall slide on its lowest section? Ans. No ; defect of horizontal pressure 2628 lbs. Ex. 449. — In Ex. 446 what vertical pressure must by some means be supplied that equilibrium may be possible? Ans. 203,100 lbs. N 178 PRACTICAL MECHANICS. Ex. 450. — There is a river wall of Aberdeen granite 1 5 ft. high and having a rectangular section ; the water comes to the distance of one foot from the top. of the wall ; find its thickness when the line of resistance cuts the base 6 in. within the extrados. Ans. 5 '34 ft. Ex. 451 . — In the last example if the wall had a section of the form shown in fig. 1 24, where A B is 1 ft. long, the vertical face of the wall being towards the water ; determine the width at the bottom when the line of resistance cuts the base 6 in. within the extrados. If the walls in this example and the last are 200 ft. long, determine the solid contents of each. Ans. (1) 5-86 ft. (2) 10,290 and 16,020 cub. ft. Ex. 452. — In each of the last examples determine the distance from the extrados of the point at which the line of resistance cuts a horizontal joint 8 ft. below the surface of the water. Ans. (1) 1-98 ft. (2) l'75ft. [The point will, of course, be that round which the moment of the weight of the incumbent portion of the wall equals the moment of the pressure of the water on the eight feet.] Ex. 453. — A river wall whose section is a right-angled triangle just supports the pressure of water when its surface is on a level with the top of the waU ; show that the thickness of the base = height X /_ V n if the hypothenuse of the triangle is turned towards the water ; but when the perpendicular is turned towards the water the thickness of the base = height X /_ w where w is the weight of a cubic foot of water, and w, that of a cubic foot of the material of the wall. And show from hence that in the former case the thickness of the base is greater or less than in the latter according as the specific gravity of the wall is greater or less than 2. Ex. 464. — ^A wall of brickwork is to be built round a reservoir 20 ft. deep ; its slope is inward ; it is 1 ft. thick at top ; what must be its thick- ness at the bottom, that when the reservoir is full, the line of resistance may cut the base 6 in. within the extrados ? Ans. 10-74 ft. Ex. 455. — The wall of a reservoir fall to the brim is of brickwork and is 20 ft. high and 2 ft. thick ; it is supported by props at intervals of 6 ft. ; the length of each is 20 ft., and its inclination to the horizon 30°: determine the thrust on each prop, its weight being neglected. Am. 54,632 lbs. Ex. 456. — In the last example determine the thickness of the wall that would just support the pressure of the water if the props were removed. If the wall stand on its lowest section without the aid of cement, what must be the coefficient of friction between the surfaces ? Ans. (1) 8'6 ft. (2) 0-65. PRESSURE OF EARTH. 179 Ex. 457. — A reservoir is divided by a brickwork wall 12 ft. high and 2 ft. thick ; the water on one side of the wall is 10 ft. deep ; what must be the depth on the other side if the wall is just overthrown ? Arts. 10'4 ft. Ex. 458. — A cofferdam sustains the pressure of 26 ft. of water, and is Buppotted at intervals of 10 ft. by props D E and c p ; given that b c and b d are respectively 4 ft. and 18 ft. and that d e ^ and c F are respectively 30 ft. and 18 ft. ; find the thrust on each prop. And what must be the weight of the struts, and of 10 ft. of the length of the cofferdam, that the whole be not overthrown ? The thickness of the cofferdam and the adhesion at b are to be neglected. Ans. (1) Thrust on D e = 88,020 lbs. ; on E c = 144,400 lbs. (2) 84,900 lbs. / ./"c / / 95. The pressure of earth. — ^Let a b represent a section of a wall supporting earth, whose surface is A c, it is re- quired to determine the pressure produced fig. lae. on A B by the earth. Now, it must be remembered that two extreme cases may come under consideration : the first arises when the earth is thoroughly penetrated with water, in which case the pressure is the same as would result from hydrostatic pressure ; the second arises when the co- hesion of the earth is so considerable that it would stand with its face vertical even if the wall were removed. Dismissing these two extreme cases, let us sup- pose the wall A B removed, the following result will then ensue : the earth being friable wiU weather and break away until its surface has taken a slope b c, inclined to the horizon at an angle equal to the limiting angle of resist- ance ; when reduced to this state it will have no further tendency to break away, and, unless washed down by rain or removed by some other extrinsic cause, will remain permanently at rest at that slope, which is therefore called n2 180 PRA.CTICAL MECHANICS. its natural slope. Hence, in the case we are considering, the wall is required to give a certain degree of support to the wedge of earth a b c ; this wedge is generally supported in some degree by the cohesion of its parts with each other and with the earth below B c, so that the wall will be sufficiently strong if it will support the earth, on the sup- position that the cohesion is quite destroyed, unless (which is not contemplated) the earth should be saturated with water. The angle of the natural slope of fine dry sand is about 35°; of dry loose shingle about 40°; of common earth, pulverised and dry, about 45°.* Proposition 20. If w is the weight of a cubic foot of earth, and )=^wx^ cotan 6 tan (0—^) Now, according as 6 has different values E will have differ- ent values, and if we determine the value of for which E is greatest, the wall cannot be called on to supply a greater reaction, and this must therefore equal the pres- sure which A X actually sustains. But cot tan (-g-d^- "°^ ^ "^" (g-'^ )_ sm (2g-<^)-sin cot tan {U ^) - ^-^-—^^-^^ _ ^.^ ^20 _ ^^ ^ ^.^ ^ _ , _ 2 sin (f> sin (20—(f)) + sin(j) •which is manifestly greatest when the fractional part of the expression is least, i.e. when 20— (f> equals -, so that the required value of is — + 5, and, therefore, the required value of the pressure is ^x^cotan(j + |) tan (j-|)=i^«'Han^ (j-|) acting through a point N which is below A by a distance equal to fx; but this is the same as the pressure that would be produced by a fluid each cubic foot of which weighs It) tan^ T--^). Therefore, &c. Q. E. D. Ex. i59. — A mass of earth the specific gravity of which is 17, whose surface is horizontal, presses against a revetement wall whose top is on the level of the ground and height 20 ft., the natural slope of the earth being 45° ; determine the pressure of the earth on each foot of the length of the ^all. Ans. 3646 lbs. Ex. 460. — If the wall in the last example is of brickwork and has a rect- 182 PRACTICA.L MECHANICS. angular section, determine its thickness to enable it to sustain the pressure of the earth. . ^»s. 4'65 ft. Ex. 461. — ^The vertical face of a rev^tement wall of brickwork sustains the pressure of 20 ft. of earth, the surface of which is horizontal and 2 ft. below the summit of the wall ; the thickness of the wall at top is 1 ft. : what must be its thickness at bottom if it just sustains the earth, the spe- cific gravity of the earth being 2 and its natural slope 45° ? Also deter- mine the thickness that would enable the wall to sustain the pressure if the earth were thoroughly permeated with water.* Jm. (1) 5-47 ft. (2) 9-6 ft. Ex. 462. — If a pressure p is applied against a wall supported on the op- posite side by earth with its surface horizontal ; show that when p is on the point of causing the earth to yield, the resistance of the earth is the same as that of a fluid the weight of a cubic foot of which equals (weiglit«f«abie foot of earth) x tan? ( — + — ). [The reasoning in this case is step by step the same as that given in Prop. 20, except that now the wedge of earth is on the point of being forced wp, so that the direction of e, will be on the other side of the perpendicular to X T.] Ex. 463. — A reservoir wall of brickwork is 4 ft. thick and 15 ft. above the surface of the ground; the foundations are 15 ft. deep; the natural slope of the earth is 45° and it weighs 100 lbs. per cubic foot ; when the reservoir is full (so that the water presses against the whole 30 ft. of wall) will tlie wall stand, supposing the adhesion of the cement perfect ? Arts. Yes ; excess of the moment of the greatest pressure that could support the wall over that of the pressure of the water 73,480. *Ex. 464. — If A B is a section of a rectangula.r wall, p the pressure ap- Ra. 128. p « A B * It is common for re vetement walls to sustain a surcharge of earth, as shown in the accompanying diagram ; an investiga- tion of the pressure in this case will be found in Mr. Moseley's Mechanical Prin- ciples of Engineering, p. 453. The follow- ing practical formul.a (Morin, Aide-Me- moire, p. 417) gives the thickness (x) of a rectangular wall for a given height (h) of the revfitement (q m) and a surcharge (p q) whose height is h, viz. ■' (M) VI being the weight of a cubic foot of earth and to, that of a cubic foot of masonry. « = 0-865 (h + V w. tan LINE OF EESISTANCE. 183 t)lied to every foot of its length at a, the inner edge of its summit ; deter- mine the equation to the line of resistance. [Take any horizontal section of the wall m n ; let A n = a;, B c = a, then the weight w of A-SM = a!cw, where w is the Fis. 129. weight of a cubic foot of the wall ; now, if the direction of the resultant cuts mn in e, this will he a point in the line of resistance, and if rk =y we are to determine a relation between a; and y. The relation in question can easily be shown to be awx(j/—-\=v (x sin a—y cos o) y where a is the inclination of p's direction to the Vertical.] *Ex, 465. — In the last example show that the curve is a hyperbola and determine its asymptotes ; and show that if the thickness of the wall 1 y R a equals^' 2 p sin a w it may be carried to any height whatever with safety. *Ex. 466. — If the wall in Ex. 464 has to support the pressure of earth or water reaching to the top of the wall, show that the line of resistance is a parabola with its axis horizontal, and show that in the latter case its focus is in the summit of the wall at a distance from the iutrados equal to -(l H — J, where w is the weight of a cubic foot of masonry and «/, of water. *Ex. 467. — A B c D is the section of a reservoir wall the vertical fac^ of which (b c) is towards the water ; the width of the top of the wall (a b) is a ; the inclination of a d to the vertical is 0, and s is the specific gravity of the wall ; show that when the water reaches to the top of the wall the equation to the line of resistance is — x and y being measured as in Ex. 464— x'(- + tan^'fl) — Sxy tan fl + 3aa; tan 9 - 6m/ + 3a'=0 *Ex. 468. — Show that if the wall in the last example stand, whatever be the depth of the water whose pressure it sustains, then tan 9 must be>-J=. *Ex. 469. — ^Determine the equation to the line of resistance in a river wall of Aberdeen granite, the thickness of which is 4 ft., and which sus- tains the pressure of water whose surface is on the level of the top of the wall. ^ns. 3-^= 63 (y-2.) *Ex. 470. — Determine from the equation in the last example the height of the wall when the line of resistance intersects the base at a distance of 4in. within the extrados. .4«». 10'2 ft. 184 PRACTICAL MECHANICS. CHAPTER IX, ON THE DEFLECTION AND ETJPTURE OF BEAMS BY 96, 'Neutral surface a/nd neutral line of a beam.— If we consider a long beam of wood A d supported at its j,jg J3Q two ends, the effect of its weight will be to beud it into such a shape as that shown in the figure; it is evident that the under surface c d will suffer extension, and the upper surface A B com- pression : so that there will be a section p Q interme- diate to the compressed and extended parts, which will undergo neither compression nor extension ; this surface is called the neutral surface. Forces may act on the beam in such a manner .that the whole of it is either compressed or extended ; in such a case the neutral surface will not have a real existence, but there will exist without the body an imaginary surface bearing the same relation to the compressions or extensions as that borne by the actual neutral surface in other cases. In what follows we shall assume that the forces act in a plane containing the geometrical axis of the beam con- sidered as a prism and at right angles to the axis in its undeflected state ; that the cross section of the beam is * This chapter cannot he read with advantage by any student who has not some acquaintance with the Integral Calculus. D — +- BENDIKG MOINIENT. 185 symmetrical with respect to the plane contaiiiiiig the forces ; and that the deflection is so small as not to change sensibly the moments of the forces. The plane in which the forces act will cut the neutral surface along a line called the neutral lime. It will appear in sequel that under these circumstances the neutral line coincides with the axis of the beam. Before going further the student should make himself acquainted with the simpler cases of moments of inertia. They are given in Part II., Chap. V. 97. The bending moment. — Let a b be a uniform rod held at rest by three forces p, Q, E acting at right angles to its length ; suppose the rod pj^ i3j to be so strong as to sustain the action of these forces with- ' ^ out being much bent, so that T c~" the distances A o, B c are not V sensibly altered. We have to * '. do with two sets of forces — (a) the external forces P, Q, K ; (6) the internal forces due to the elasticity of the rod, which are called into play by its extension or compres- sion. The first question to be considered is : — ^What is the tendency of the forces to break the rod at any assigned point D ? Now (Art. 60) P may be replaced by an equal force acting in the same jjg i33_ direction at D, and a negative couple i A D . p ; the Kke is true of the other ^ ^'^t ^ forces. Consequently, if we suppose an J j ? imaginary plane drawn across the rod ^ |j * at D, as shown in fig. .132, where for |q convenience the thickness is magni- fied, we have on the left of D e a force E— P, and a couple whose moment is — ad.p + cd.e; on the right of D E we have the force Q and a couple whose moment is B D . Q. It thus becomes plain: — First, that the forces tend to 186 ' PEAOTICAL MECHANICS. cause A E D and B E D to slide in opposite directions along D E, i.e. they produce a shearing stress measured by either of the equal forces E— P or Q. Secondly, the forces tend to make the parts a e d and bed turn in opposite direc- tions, the measure of this tendency being the moment BD.Qor — AD.p + CD.K, which, however, have contrary signs. Either of these moments is called the bending moment at the point D. In the cases to be considered in the present chapter the shearing stress can be put out of the question ; the bending moment alone comes under consideration* We may say, therefore, that the tendency of the forces to break the rod at any point (d) is measured by the sum o? the moments taken with respect to d of all the forces on one side of that point, whatever be the number of forces acting on the beam. Er. 471. — In fig. 131 suppose the rod to rest on two points nnder A and B, and let the force e be produced by a weight w hung at c ; then if A B is denoted by a, and a c by b, and e d by x, the bending moment at d is — . n, if the weight of the rod is neglected. If, however, d is between a A and c, the bending moment is ~ ' — (a—ic), and at c it is ^ ~ ' .' a a Ex. 472. — If the weight of the rod only is considered the bending moment is ^wax—^wx', where w denotes the weight of a unit of the length of the rod. Ex. 473. — In Ex. 471 if the weight of the rod is taken into account the bending moment is ( + Iwa J x—^wx^ or /(a-i)w^^^,^\ (a-x)-iw(a-xy according as d is between b and c or c and A. Ex. 474. — In the last example either formula gives for the bending moment at c the value Ex. 475. — In Ex. 472 the value of the bending moment is greatest at the middle point, and there equals Iwa'. BENDING MOMENT. 187 Ex. 476.— In Ex. 473 if - >b+ ''—, the bending moment will be 2 aw greatest at a point between o and tbe middle of the rod, and the greatest . value will be _ | — + — ) ; but otherwise its greatest value is at c, and 2 \ 2 awj is that given in Ex. 474. Ex. 477. — A rod is supported on many points all on the same horizontal Ime, so that the only forces acting are weights and the vertical reactions of the fixed points ; let A and b be any two consecutive fixed points, B to the right of A ; let the distance a b be denoted by I, the load on A B (including the weight of the rod) per unit of length by w ; and let p be any point in A B at a distance x from A. Show that the bending moment at p is M + Q {l—x) + \w (l—x)' where m is the bending moment at b, and o, the sum of the forces acting at and to the right of b. [Let there be any number of parallel forces, viz. E acting at b, and E„ B^, B3 . . . . to the right of e at distances r^, r^, r, the sum of their moments with respect to b will be M=r,E, + r2K2 + r3B3 + . . . . With respect to a point at a distance b to the left of b the sum of their moments will bo 6E + (6 + ?-,)B,+(6+r2)E2 + (J+r3)E5 + = 6(e+E, + E2 + E3 + . . . .) + M.] 98. To obtain a clear view of the reactions by whicli the rod resists the tendency of the forces to break it, let us suppose it to be divided along ^^_ jj3 D E, and the connection to be ^ re-estabhshed as shown in the | ^ ~ figure by means of two small pieces of an elastic material D f and EG whose unstretched lengths are equal. The couple acting on "" A D tends to turn it (as we have already seen) in the same direction as that of the motion of the hands of a watch, and that acting on B f in the opposite direction, so that D F will be compressed and E G stretched. Consequently D F will react with equal forces (r and Ej) against d and F respectively, as shown in the figure ; while E G reacts with B. d 188 PRACTICAL MECHANICS. equal forces (b' and E'j) against E and G respectively. The two forces E and e' form a couple with a positive moment, which balances the couple that tends to turn ad e. In like manner the forces E, and E'l form a couple which balances the couple that tends to turn B F G. Consequently, whether we consider the forces acting on A d e or those acting on B F G, we obtain the relations e=e' and EXDE = the bending moment. Of course D f would be compressed and E G stretched to just the extent needed for calling into play the forces R and e'. If now we suppose d E G F to be a portion of the beam included between two planes at right angles to its axis in its undeflected state, the only difference will be that the part D H K F above a certain line H K will be compressed, while the other part H E G K will be stretched. The com- pression wiU increase gradually from H K to d f and the extension from H K to eg. Thus there will be a distri- buted force which gradually increases from H to D and acts in e's direction, and in like manner from H to e a dis- tributed force acting in the direction of e'. It is plain that the forces on D e must reduce to a couple with a positive moment, and as it balances the external forces on A D e we must have the moment of this couple equal to the bending moment. The same conclusion would follow if we reasoned on the forces distributed along f g which must balance the couple acting on B f. Proposition 21. If a cyli/ndrioal or prismatiG beam has a cross section symmetrical with respect to a plane containvng the axis, a/nd is acted on by forces in that plane at right angles to the axis in its undeflected state, the neutral line will coincide with the axis of the beam (i.e. will pass through FLEXUEE OF BEAMS. 189 the centre of gravity of each cross section), and the re- ciprocal of its radius of curvature at any poi/nt will equal (the bending 7noment)-T-E a k^, where E denotes the modulus of elasticity and a k% the moment of inertia of the cross section taken with reference to an axis passing through the centre of gravity at right angles to the plane of symmetry in which the forces act. Fig. 134 corresponds exactly to fig. 133, p. 187 ; and fig. 135 represents the cross section tlirougli d h e at right angles to the plane of sym- metry, -which is the plane of the paper. Let E d and G F be produced to meet in o. Then as hk is unstretched it is part of the neutral line, ' ' "^' ^^^' and if HK becomes in the limit indefinitely small, ho becomes the radius of curva- ture (p) of the neutral line at the point H. Let h L be denoted by z, and let us con- sider a lamina hi/ of the beam the area of whose cross section IT is denoted by a — the lamina is, of course, supposed to be between two planes very near together, parallel to H K and at right angles to the plane of sym- metry. Now, as the uncompressed length of l l' equals H K, it follows from Art. 6 that the reaction of L l' due to its compression equals HK — ll' Eax HK But p ; />— 0::hk : ll' or p : z::nK : hk— ll' therefore the reaction is . This force is distributed P unifoimly over 1 1', and therefore may be supposed to act 190 PEACTICAL MECHANICS. at L. If a point is taken below H the lamina correspond- ing to it is stretched, and therefore the reaction at that point is in the opposite direction to that exerted at L ; this, however, is provided for by z being negative in that case, so that the formula just given applies to all points of D E. Hence if a^,a^.... be the areas of any other portions of D A. E A' taken in the same way as a at distances z^,z^.... from H, the resultant of all these reactions will be ■pi -{az + aiZ^ + a^^ + . . . .) P Now (Prop. 16) this equals -xohEh'xz, where z is the distance of the centre of gravity of the cross section from H. As the resultant is a couple, this expression must be zero ; but neither e nor the area d A E A' is zero, nor can p be infinite (except at particular points), for then there would be no flexure. Consequently z must be zero, i.e. H is the centre of gravity of the cross section. Therefore the neutral axis passes through the centre of gravity of evbry cross section, i.e. it coincides with the axis of the beam. The sum of the moments of the reactions taken with respect to h h' is lx(az^+ayZ^^ + a^z^^ + ) and this must equal the bending moment. Now the quantity within the brackets ia the moment of inertia (ak'') of the cross section taken with reference to hh'. Hence bending moment = - P 1 _ bending moment p eTk^ In the ordinary case of a rectangular beam it appears from FLEXUEE OF BEAMS. 191 Part II., Chap. V., that A k' equals ^ b% i.e. a twelfth part of the width multiplied by the cube of the depth, Ex. 478. — Let the beam be ield firmly at pne end, and a force p applied at the other at right angles to its length ; it is required to determine the equation to the neutral line, neglecting the weight of the beam. [Let I. L, be the neutral line, l g the position of the beam's axis when unbent, f any point in the neutral line, Fig. 13G. p the radius of curvature at f, x and y the co-ordinates of F, viz. li s and b f ; then, since the bending moment at f is p (a—x), we have l^^ 12p (a-a) Now, since the curvature is small, -£- is small, and there- dx fore/^j can be omitted ; consequently dx^ therefore whence d^y 12p , . 12p fa^ a^\-, ^ eJ'o V 2 ~ 6 j-' Ex. 479. — Show that the deflection of the beam in the last example 4p a? Ex, 480. — If in the last example a force is applied to the end of the beam and gradually increased up to p, show that the number of units of work expended in producing deflection equals 2p^ o^ [Compare Ex. 149.] Ex. 481. — The end of a beam of oak is firmly embedded in masonry; the length of the projecting part is 15 ft., its breadth is 3 in., and its depth 6 in. ; a force of 2 cwt. is applied perpendicularly at its end ; determine the deflection, and the work expended in producing that deflection — the weight of the beam being neglected. Ans. (1) 5'5 in. (2) 51 ft.-pds. of work. Ex. 482. — If a beam is held firmly by one end in a horizontal position and is bent simply by its own weight, show that -= i-^ — '-, where p ECO' 192 PRACTIOAIi MECHANICS. w is the weight of a unit length of the beam ; and that the deflection is 3 via a' [These results are true -when the beam is loaded uniformly at the rate of w per unit of length.] Ex. 483.— If the beam in Ex. 482 were of elm, were 5 ft. long, 1 ft. broad, and 1 ft. deep, and had to support the pressure of brickwork 14 in. thick and 10 ft. high, determine the depression. Ans. 0'15 in. Ex. 484. — If a horizontal beam A b is supported at its ends and is loaded by a weight w at its middle point, and if p is the radius of curvature at a point in the neutral line whose distance from the middle point of the beam is X ; show that 1 _ 3w (a- 2a;) p Eci' and that the deflection at the middle of the beam is —j- . — . [If the centre of the beam is taken as the origin of co-ordinates, x being measured horizontally andy yertically, the bending moment is fw (|o— x), and the value of r/ at either end equals the required deflection.] Ex. 486. — If the beam in the last example were bent by its own weight, which is w per unit of length, show that — = "* q 7. ^ ) ^'^d that the depression at the middle point is— - . _=- . ^ . Ex. 486. — A fir batten 3 in. deep, IJ in. broad, is placed horizontally between two props 6 ft. apart and loaded with a weight of 135 lbs. in the middle ; its own weight being neglected, determine the depression ; determine also the depression if it were fixed at one end and loaded with the same 18 288 weight at the other end. Am. (1) -— in. (2) -— in. loo loo Ex. 487. — A spar of oak 3"2 in. square is placed horizontally between two props 12"8 ft. apart and loaded with 268 lbs. in the middle ; determine the deflection, neglecting the weight of the beam. Ans. 1-697 in. Ex, 488. — A piece of elm 2 in. square is placed horizontally between two supports 7 ft. apart, it is loaded in the middle with a weight of 125 lbs. ; determine the deflection when its own weight is neglected. Ans. 1-66 in. Ex. 489. — There is a beam of larch 6 in. deep, 4 in. wide, and 12 ft. long, it is supported on a fulcium whose distance from one end is 4 ft. ; the shorter end carries a weight of 2 cwt. ; determine the deflection of each arm of the beam, its own weight being neglected. .4»s. (1) 0'109 in.. (2) 0-437 in. Ex. 490. — A rod, whose weight can be neglected, has a weight tied to each end, and is then placed on a fulcrum so that the weights balance each FLEXURE OF BEAMS. 193 other ; show that the droops at the end§ of the rod are inversely propor- tional to the squares of the weights. Ex. 491. — Pind the force which being applied vertically to the end of the team in Ex. 482 exactly neutralises the droop. Why should not this force equal half the weight of the beam 1 Ans. 3 wa-i-8, Ex. 492.^-The ends of a beam rest on horizontal supports, it is deflected by its own weight and a vertical force w acting through its middle point ; determine the total deflection, and show that it equals the sum of the separate deflections produced by its own weight and by w, if w act vertically downward, and their difference if w act vertically upward. Ex. 493. — If A B, A c are the principal rafters of a roof the feet of which are fastened together by a tie-beam B o, the middle point of which is d ; if A and D are joined by a ' king-post ' which exaptly neutralises the bending in the middle of the tie-beam caused by its weight, show that the tension of the king-post equals | of the weight of the tie-beam. Ex. 494. — In Ex. 487 determine the deflection when the weight of the spar is taken into account. Aas. 1'8 in. . Ex. 495. — A beam of larch supported at each endmeasures 20 ft. between the points of support, it is 6 in., wide and 10 in. deep, it sustains a wall of brickwork 30 ft. high and 1 ft, thick throughout its whole length ; And the deflection. Ans. 23'13 in. Ex. 496.^If the beam in the last example is supported by a column which exactly neutralises the deflection of the middle point, find the pres- sure on the column, Ans. 42,170 lbs. Ex. 497, — If in the last example the under surface of the beam in its un- deflected state is 12 ft. from the ground, the middle point is supported by a column of cast iron 3 inches in diameter, which in its uncompressed state is exactly 12 ft. long ; determine the deflection of the beam at its middle point and the pressure on the column. Ans. (1) 0°0d in. (2) 42,077 lbs. [The column being compressibfe- will allow the middle of the beam to descend, whereby the thrust on the column will be diminished : the question to be answered is — At what degree of compression will the tendency of the column to recover its form upward exactly balance the tendency of the beam to deflect downward ?] Ex. 498. — In the last example suppose the measurements to be made at . 50° Fahrenheit, at what temperature would there be' no deflection at the middle point of the beam ? Ans. 107° F. Ex. 499. — If a hollow cylinder (whose weight is neglected) the radii of whose section are j-j and r be supported horizontally at two points whose distance is a; show that, when it sustains a weight w at its middle point, the radius of curvature of the neutral line at a point distant x from the middle is given by the formula 1 _ w(a— 2j;) "p irB(r,*— r*) ;194 PBAOTIOAL MECHANICS. -and tHe deflection at the middle point by the formula -_ w a? ~12irB(r,*-?-*) [The moment of inertia of the space between two concentric circles with respect to a diameter is J ir (r,*— r*) ; see Ex. 760.] Ex. 500. — If in the last example the cylinder sustains throughout its length a uniform load of w lbs. per unit of length, then p 2irB(ri*— r') , , 5 ro a* *nd = -—-, — - 96irE {ri* — r*) Ex. 501. — If an iron girder* has a section of the form shown in the annexed . diagram, of the following dimensions, ae = c,, ae = 6„ C¥=c, CTi = b, the lower end o h being of the same dimensions as the upper, show that when this girder sustains a uniform pressure throughout the whole of its length the deflection at the middle point is given by the formula 5 wa* 5 = - FlG. 137. A 32{6 {b + bi)''byCt + 2bt'Ci + b'c}E- Ex. 502.— If there are two beams containing the. same amount of materials, of the same length and the same depth, and sustain- ing the same weight, the one has a rectangular section, tbe other a section of the form shown in the last example; given that J = 4 in., c=l in., b, = 1 in., c, = 4in., show that the deflection of the rectangular beam will be ^f of the deflection of the other beam. 99. Equation df three Tnoments, — A very interesting application of Prop. 21 is to the determination of the pressures exerted by a beam or a rod on its points of support when there are more than two of them. For this purpose it is convenient to investigate a relation between the bending moments at any three * In practice the lower flange is commonly made much larger than the upper, since cast iron offers more resistance to pressure than to tension, and of course the greatest economy of materials is effected when the load that would tear the lower flange would also crush the upper. To discuss this question would lead us beyond our present limits. — See Mr. Moseley's Mechanical Principles, p. 566 ; Mr. Eankine's Applied Mechamcs, p. 319 ; see also Mr. Fairbaim's Useful Information, Append. I. EQUitTION OF THREE MOMENTS. 195 consecutive points of support, which may be done as follows : — Let A be any one of the points, Aj and A^ the points next to it on either side, and Pi<3 138 let the bending moments at them respectively be M, Mj, M^ ; — tc + j^ — + — ji — let Aj A be denoted by J„ a a^ by l^, the corresponding weights per unit of length by Wj and Wj, and the sum of the forces to the right of A, (in- cluding the reaction of Aj) by Qj. Then it follows from Ex. 477 that M = M^ + QiZi + iWjZ,' (1) If we reckon the moments of the weights positive and if AP be denoted by x, the bending moment about p {Ex. 477) gives the equation In integrating this equation we must remember that the neutral axis at A need not be horizontal ; aU that we know is that the curve has some determinate but unknown in- clination at that point, which we wiU denote by a ; hence, when a; = Owe have ■M=Oand-^=tan a; also when x=l, ax we have y=0. On integrating twice, therefore, we obtain EAK''y=E AK^ic tan a + ^Mja;^ + Qj {Wx^—^d^) + ^w,{^l,^x'-^l,x^+ ^x*) and therefore 0=24EAKnana + 12M,Zi + 8Qj?iH3'M;,?i' (2) If now Q2 is the sum of the forces to the left of Aj we have from Ex. '^Il M = M2 + Q2i2 + ilDj?j,^ (3) o2 196 PEACTICAL MECHANICS, and if we denote a F^ by x^ we shall obtain as before, by considering the bending moment about Fj — On integrating this twice we shall obtain, as before — = 24 E A KUan /3 + 12 u^l^ + 8 Q^l^^ + 3 w^l^^ (4) Now, as there is no sudden change of direction of the curve at A, we havQ a + 0=18O°, and consequently tan a + tan y8= 0. Therefore, by adding (2) and (4), we obtain 12 (Mi^i + Mj^ + S (Q,?i2 + Q,^i^2)4.3 {wj,,^ + wj,^^)=0 From (1) and (3) we obtain M (?i + y = Ml?, + MjZjj + Qi?,2 + Q Zj* + i (wi^i^ + w^y) and therefore 8 M {\ + l2) + 4:{u^l^ + ^^l^)=Wjli^ + w^l. an equation which gives the required relation and is called the equation of three momenta. It should be observed that the reasoning goes upon the assumptions that the points of support are accurately in a horizontal line, and that the moment of inertia of the cross section of the beam is the same at all points of its length, but the load need not be at the same rate per unit of length on the parts of the beam between different points of support. By means of this equation the reaction at each point of support can be determined without ambiguity. Suppose there are n points of support 1, 2, 3, 4, ... . we can ex^ press the bending moment at each point in terms of the weights and the reactions, and we can apply the equation of three moments first to 1, 2, 3, then to 2, 3, 4, then to 3, 4, 5, and so on, thereby obtaining n — 2 relations be- tween the reactions and known quantities. The two equations of equilibrium between the parallel forces acting EQUATION OF THEEE MOMENTS, l97 on the beam (Prop. 12) give two more equations, and thus we haye n equations between the n unknown reac- tions, which are thereby completely determined. In most particular cases the process can be abridged. When the reactions have been determined, the circum- stances of the flexure of any portion of the neutral line can be determined by means of the equation (a). Ex. 503. — K team is supported on five equidistant props (a, b, c, d, e), one being under each end ; find the pressures (p, a, e, s, t) on the points of support. Here we need five equations, and they could be easily formed as above explained, but we may assume as evident that p is equal to T, and Q to s ; so that 2p + 2Q + E=4aK) where ia is the length of the beam and w its weight per unit of length. How if M, M„ Mj, Mj, M, are the bending moments at A, b, c, and D we shall have M = = Mj, m, = \aNi — a p = m„ and m^ = 1a?vi —an— 2ap. The equation of three moments must now be applied to M, m„ and M^, and then to m„ M,, and Mj. In doing this the student must observe the unavoidable change of notation ; in Art. 99 M is the moment at the intermediate point, and M, And Mj at the extreme points, so that, as ?, = 2^= a, the first application gives 16 M,a + 4 (m+Mj) a = 2«;a' 6r 8m, +2M2=Kia' the second application gives 16 MjO + 4 (m, + M,) a = 2iva' or 8M2 + 4M, = «;a' Hence 12p + 2(j = 7 aw •and 20p + 8Q = 17aw Therefore 28p=11ow, 7q=8o«i, 14E=13a«). Ex, 504. — A beam is supported on three points, one under each end and one in the middle ; find the pressure on each point of support [If the pressures are p, e, p, and the moments are denoted as in Art. 99, we have 2p + (j = 2ara M, = M2 = and M:=^a%o— ftp ; and then, on applying the equation of three moments, we obtain 8p = Saw, and in = Saw.'] Ex. 506. — In the last example required the equation to the neutral line, the point at which the deflection is greatest, and the amount of the same. 108 PRACTICAL MECHANICS. [If the middle point is taken as the origin of co-ordinates, it can be easily shown that the bending moment at a point distant x from the, origin is it« (a'— 6ffla: + 4x'), whence the equation to the neutral line is easily determined, observing that when x = we have y=0 and —2 = 0, and that dx at the point of greatest deflection we shall also have ^ = ; whence we dx shall obtain 16x = (15— ^33) a for the position of the required point, and for the amount of greatest deflection (d) we have (nearly) ISSeak^d = V) a*.] Ex. 606. — A beam rests on four points, one under each end and the other two equidistant from the middle point ; find the pressures on the points of support. [If the distances between the props are b, 2a, and b, and the pressures on them p, Q, (j, p, it is evident that p + a=(a + J)«) and 'it follows from the equation of three moments that 8 6(3o + 6)p = «)(3 6'+12oJ2_8aS).] Ex. 507. — In the last example explain the result arrived at by making 3b = 2a. 100. Strength of hearths. — On this subject we may ask either of two questions — (a) What is the greatest load applied in a given way that the beam will support with safety ? (6) What is the load applied in a given way that will break the beam ? In either case we must ascerr- tain the point of the beam at which the bending moment is greatest, for it is evident that if the beam is strong enough at this point it is strong enough at all points; and if the load is gradually increased it wiU break at this point. With regard to the first question, suppose it to be ascertained that the material can be safely stretched 1 — 'n-th part of its natural length. Suppose that fig. 134 shows the section at which the bending moment is greatest, E G is the fibre that is most stretched, and if the load is such that E G is longer than H K by 1 — Tith part of H K the load is the greatest consistent with safety. Now eg:hk::eo:oh therefore eg— hk:hk::eh:oh STRENGTH OF BEAMS. 19^ or denoting e h by 6, and o h by p, we have and therefore (by Prop. 21) n 6j (greatest bending moment) =e a k* Kow if p is the greatest tension per unit of area of cross section that can be applied to the material with safety we have (Art. 6) wp = e; therefore 6i (greatest bending moment) = p A K* We may reason in the same manner on the greatest pres- sure (q) per unit of area of cross section that the material can sustain safely, and we shall obtain b^ (greatest bending moment) = QAK* where b^ denotes hd. Whichever of these equations gives the smallest value of the bending moment will give' the greatest value of the load that can be borne safely. If we suppose the cross section of the beam to be rectangular, and that the greatest pressure per unit of area equals the greatest tension per unit of area that the' material can support safely, we shall have, denoting either byQ greatest bending moment =^QA.b With regard to the second question, let there be two beams of the same material acted on by any transverse" forces, and at the sections where the bending moment is greatest suppose the fibres which are most elongated to be equally stretched, i.e. suppose n to have the same value in both cases. Then if p' and b\ denote in the case of the second beam the quantities correspondiiig~to p and b^ in ; the first we must have p V, p' 200 PRACTICAL MECHANICS. and therefore by Prop. 21 6, (bending moment) _ b\ (bending moment) AK" a'k'2 If either of these beams is on the point of breaking, the other will be on the point of breaking also, for at the ■weakest part of the beam both are equally stretched. Suppose the second beam to be a foot long, and to have for a cross section a square of one inch ; let it be supported at two points, one under each end, and let the force (p) that wiU just break it when applied to its middle point be found by actual experiment; then, taking all the measurements in inches, the right-hand side of the above equation reduces to 18 p. If we denote this by S we have what is called the modulus of rupture for the material, and we see that when the beam is about to yield at any point by cross breaking the bending moment=s A y =- It will be observed that the reasoning by which this formula is arrived at assumes that the immediate cause of the rupture is the yielding of the under Side of the beam to tension. If the beam gave way by the yielding of the upper side to pressure, precisely similar reasoning could be applied, but by would denote the depth of the neutral axis below the upper surface. Table XIV .* MODULUS OF EUPTUEE. Substance Lbs. per Square lucb Substance Lbs. per Square Inch "Oak (English) Larch 10,032 4,992 Fir (Riga) Elm 7,110 6,078 * Prom Mr. Mossley's Mechanical Principles of Engineering; p. 622. STEENGTH OF SEAMS. 201 JEx. 508. — A rectangular beam 6 in. deep and 3 in. wide rests horizon- tally on two points 10 ft. apart; it can be safely subjected to either pressure or tension at the rate of 1,000 lbs. per square inch. What is the greatest weight that can be safely hung from its middle point, its own weight being neglected ? Let p denote the required weight ; the greatest bending moment, being at the middle point, will be 30f, the units being pounds and inches ; hence 3x30p=1000xl8xix6« therefore p= 600 lbs. It wiU be found that the depression caused by p at the middle point is ftiis of an inch, if the modulus of elasticity is 1,000,000 lbs. per square inch. Ex. S09. — In the case of a rectangular beam, whose weight is neglected, ehow that the breaking load is as follows : — (o) ?— £ . _, when it is held 6 a firmly at one end and loaded at the other ; (i) _?_ ? . _ ., when supported 3 a under the two ends and the load is applied at the middle. Ex. 510.— In the last example, if the loading is distributed uniformly over its length, show that the breaking load is (a) ? — - . — ; (6) — — - . 3 a 3 a Ex. 511. — Given a cylindtical log of wood, show that the strongest rectangular beam that can be cut out of it is one whose sides are in the ratio of 1 : ^/% Ex. 5l2. — A beam of oak is supported in a horizontal position on points 20 ft, apart, it is 3 in. deep and 4 in. wide; determine the weight that can be suspended at a distance of 6| ft. from one point of support withput breaking it. What would be the magnitude of the weight if the depth ■were 4 in. and breadth 3 in. ? Ans. (1) 1128-6 lbs. (2) 1504-8 lbs. Ex. 513. — What must be the depth of a beam of Riga fir 4 in. wide and 30 ft. long, that will just sustain a weight of i a ton at its middlOj taking into account its own weight? Ans. 4-6 in. 202; PRACTICAL MECHANICS. A CHAPTER X. VIETUAL VELOCITIES — MACHINES IN A STATE OF UNIFORM MOTION — TOOTHED WHEELS. 101. The principle of virtual velocities. — Let p be a Fia. 139. force acting at the point A along tlie line A p, and let it be repre- sented by A c (Art. 25). Suppose I "e p's point of application to be shifted through an indefinitely, small distance to B, draw Bw at right angles to AC or c A produced, and let A n be denoted hyp, which is commonly reckoned positive when n falls between a and c, and negative when it falls on c A produced, then p is called the virtual velocity of p, and Fp its virtual moment or virtual work. The principle of virtual velocities is as follows : — If a system of forces in equilibrium act on any machine which i:eceives any small displacement — consistent with the connection' of the parts of the machine — the algebraical' sum of the virtual works of the forces will equal zero. If Pj, Pj) P3 are the separate forces, and ^j, p^, p^ their virtual velocities, the principle is expressed algebraically by the following equation, which is commonly called the equation of virtual velocities : It must be remarked that in the above definition the line A B is considered a small quantity of the first order ■ VIETUAL VELOCITIES. 203 (App. Art. 3), and consequently the virtnal works Pi Pi, Pa Pii ^aPs • • • are in general of the first order ; if, however, the virtual velocity of the point of applica- tion of any one of the forces be of the second order, the virtual work of that force will vanish in comparison with the virtual works of the other forces and will dis- appear from the above equation ; this will happen in the following cases : — (a) "When A B is ultimately at right angles to A c — e.g. when A c is the normal to a curve of which AB is a chord — hence the virtual work of the reaction of a smooth surface equals zero when the body slides along the surface; (b) when the points A and b coincide, e.g. when A c is a portion of a rigid body in the act of turning round the point A, i.e. the virtual work of the reaction of a fixed axis is zero provided the axis can be treated as a line ; hence also when an incompres- sible body rolls without sliding on any surface, rough or smooth, the virtual work of the reaction equals zero. The principle now enunciated will be seen from the following pages to be one of very great importance in the theory of machines ; as the general proof is not by any means easy it will be iiseful for the student to prove from first ■principles that it holds good in a few elementary cases. Ex. 514. — If X and T are the rectangular components of a force p, show that the virtual work of p equals the sum of the virtual -works of x and T.' Let i. be the point of application of p, and let A be transferred to b ; Complete the rectangle m n, and draw Bp and mg a,b right angles to A p ; then AJ9, Am, An, are the virtual velocities of p, x, and T, and we have to prove that p . Ap = X . Am + T . A» Let X A p be denoted by 6, then it is evident that Ap = Aq + gp = Am.coB 8 + An Bm tlierefore p . A^) = A m . p cos 9 + A « . p sin 9 or p. A^ = X.ATO + T. A» {V) If p had acted in the contrary direction, x, T, and p would have been in 204 PRACTICAL MECHANICS, eiluilibriuta ; the Tirtual work v), up, vr, are the virtual velocities of the forces, and of them D ^ is positive and the others negative ; the equation of virtual velocities therefore be- Fio. 1«. I)^ = W.D«) + E.Dr and this the student is required to prove.] Ex, 517. — Verify the principle of virtual velocities in the last case, as- suming that the plane (and with it the body) is so moved that b describes a straight line at right angles to d e. Ex. 618. — ^Verify the principle of virtual velocities in the case of two forces in equilibrium on a straight bar capable of turning round a fixed point. [Let p and o be the forces which balance on the rod a b round the fixed point c ; Suppose the rod to turn through a small angle and to come into the position a' b' ; draw Fra. 142. 1 t, ilm tA right angles to A/ ..^— ■""'"xi Ap and B'n at right angles to Ba, then Am is the virtual velocity of p and B » of 0, the latter being negative; also the virtual work of the reaction of c is zero (Art. 101) ; the equation to be proved is therefore p. A«i = a. B» The student must remember that a a' «i and bjb' » are Ultimately right- Bngled triangles.] • VIRTUAL VELOCITIES. 205 Eie, 619. — Verify the principle in the case of two parallel forces P and a which keep a heam at rest round a rough axle of finite dimensions (as in Ex. 405), the motion heing given to the beam round the axle. [Using the notation of Ex. 405 and calling fl the small angle through which the beam is turned, the virtual works are severally p^ 9, w j 6, and B p 8 sin ip.l Ex. 520. — ^In the last example how would it be possible to move the system so that the reaction b should disappear from the equation of virtual velocities? [Round the point «, fig. ^.] Ex. 521. — In Ex. 619 show that when the axle is smooth the reaction wiU disappear from the equation of virtual velocities. 102. Proof of the principle of virtual velocities. — ^The following proof applies to the case of any system of forces acting on a single rigid body and in one plane, in whicli the displacement is supposed to be made : it gan be easily extended so as to include every case of forces that act on any machine. Lemma. — Let a avd x he any two points in a given 1/me, let the line he transferred fig, la, to any consecutive position o t, so that A coTnes to b and xtoY; then if BY equals A x, and ifBn and T Tn, are drawn at right angles to a x, the line A n wiU ultimately equal x m. For nrn, equals B T cos 0, i.e. it ultimately differs from BY, and therefore from AX, by a small quantity of the second order ; take away the common part A m, then a n and xm ultimately differ by a small quantity of the second order, but they are themselves of the first order, and therefore are ultimately equal. (See App. I., Art. 3.) N.B, — If A X be transferred to b y in such a manner that either A«. or x w. is of an order higher than the first, then will the other also be of an order higher than the first; e.g. if A is a small quantity of the first order, and B A Q a finite angle, A b and A n are bo€h of the second order ; likewise A x Y is ultimately a right angle, and consequently ^ m is also of the second order. 206 PKACTICAIi MECHANICS. Cor. — ^Hence if a force act along a certain line, and if two points in the line be rigidly connected, its virtual velocity ■will be the same at whichever point we suppose it to act ; also if there be two equal and opposite forces, their virtual works will be equal and have contrary signs, whether we suppose them to act at the same point or each at one of two rigidly connected points, e.gy Suppose p to act along ax (fig. 143); if it act at A its virtual work is p . A n, if it act at x its virtual work is P . X m ; consequently in either case its virtual work is the same. If a to is of the second or some higher order, X m. is not of the first order, and in either case the virtual work is zero. We can now proceed with the general demonstration required, and this is given in the three following steps : — (a) If a system of parallel forces acting in a given plane have a resultant, and if the points on which the forces and their resultant are supposed to act be rigidly connected, then the sum of the virtual works of the forces wiU equal the virtual work of the resultant. Let Xj, X2 be the forces, x their resultantj draw a line (o y) at right angles to their directions, and cutting them in Nj, N2, . . • N, and suppose these points to be rigidly connected with those at which the forces are supposed to be applied, then the virtual works of the forces in the required case are severally equal to their virtual works if supposed to act at Nj, Nj . . . . n. Now, suppose these points to receive any small displacement consistent with their rigid connection, and suppose them to be transferred to M,, M^ . . . M, these points will be in a straight line (0 y') and their mutual distances will b6 FlS. 114. "VIRTUAL VELOCITIES. 207 the same as before ; the two lines will (generally) inter- sect in some point 0. Draw Mj m,, M^ m,^, .... m m, at right angles to the directions of the forces, then their virtual velocities are respectively N,mj, 'S^w,^, . . . HTn, Let the angle yoy'he denoted by 6, and o N,, o Nj . . . . N, by y^, 2/2, .. . y, then it is plain that ultimately * Ti(^m^=y^e,s^m^=y^e, fim=y0 But by Prop. 12 we have Xi2/i+Xj2/2+ =xy and therefore x,J'i^ + X2 2/2 0+ =xyd i.e. the sum of the virtual works of the forces equals the virtual work of their resultant in the case specified. (6) Next, let us consider the case of any system of forces P,, P2, P3 . . . . acting in one plane on points rigidly copnected. Kesolve the forces in directions respectively parallel to two rectangular axes, then P, will be equivalent to its two components Xj, y,, and similarly p^ to x^, Yj, P3 to Xj, Y3, &c., and the original system is divided into two systems of parallel forces, viz. Xj, x^, Xj . . . and Yj, y^, Y3 . . .; let X be the resultant of the former system and Y of the latter, and let their directions intersect at a certain point A, then the direction of their resultant (e) will pass through A, and B will be the resultant of Pj, Pjj P3 Suppose A to be rigidly connected with the other points, and sup- pose X, Y, and R to axjt at A. Now, if the points of applica- tions of the forces receive any displacement whatsoever, the virtual work of B equals the sum of the virtual works of X and Y {Ex. 514), i.e. (by a) equals the sum of * For let oy cut NX in *, we shall have N»i=y tan fl— mot tan 9, but M m and tan 6 are small quantities of the first order, so that their product is of the second order, and can therefore be neglected, i.e. H m ultimately equals y tan 9 ov y 6. ■30S PRACTICAL MECHANICS. the virtual works of x,, x^, Xj . . . and of Y,, y^, Y3 . . . ; but (Ex. 5,1 4) the virtual work of p, equals the sum of the virtual works of x, and Y„ and similarly of p^, P3, . . , ; hence the virtual work of R equals the sum of the virtual works of P„ Pjj P3 . « » ; or Br = P,j9i+P2J32 + P3j93, + (c) If p, Pj, Pj, Pg, . . . . are forces in equilibrium acting in one plane at points of a rigid body, and if that body receive any small displacement, the sum of the virtual works of the forces will equal zero. For let R be the resultant of P,, P^, P3, . . . . and let it act on the body at atiy one point in its direction, then (by 6) P,J9, + P2i3j+P3i53 + . . . =Br But R is equal and opposite to p, since the given forces are in equilibrium, and hence, since R and p act on rigidly connected points, we have by the corollary to the lemma P^-f-Rr=0 and therefore, by addition, Q. E. D. 103. The work done by a force.-^-li the student turn to Art. 11 he will see that the definition therein given might be stated more generally as follows : — a unit of work is the work done by a force of one unit when its point of application moves through a unit of distance in the direction of the force ; it follows that, when the point of application of a force of p units moves through s units of distance in the direction of the force, Ps units of work are done. When the units are pounds and feet it is convenient to call the unit of work a foot-pound. We have now to consider the extension of the defi- nition which must be made to meet the case of a force DEFINITION OF WORK. 209 whose point of application moves in any manner what- soever. The required extension will be readily made by observing that if the point of application of a force receives any sTnall displacement, the vi/rtual work of the force is the work done by the force during the displace- = Q, sin (a + ) or p = Q (sin a + ju cos a) where a denotes the angle b a c, and /i or tan ip the coefficient of friction between M and A B. Now, if M is in motion along A B, under the action of p and ft it will move uniformly, and the work done by p will equal the work expended on Q ; but the work done by p is p x a b, therefore the work expended on q Q X A B (sin o + ju cos a) or Q X (b + /i X Ac) But At a X A c is the work required to drag m along A o, if /i is the coefficient of friction between jiandAC, andaxBcis the work that must be expended in lifting Q from o to b, therefore the number of units of work is as stated. By an exactly similar process it may be shown that the number of units of work required to drag a body dovm a rough inclined .plane equals the number required to drag it along the base supposed equally rough diminished by the number required to lift the body through the height of the plane. Ex. 523. — If a train weighs 80 tons and the friction is 7 lbs. per ton, determine the number of foot-pounds of work that must be, expended in. 314 PEACTICAL MECHANICS. drawing it for 4 miles up an incline of 1 in 200 ; and determine the horse-' power of the engine that will do this in 10 minutes with a uniform velocity. Jns. (1) 30,760,720 ft.-pds. (2) 93^ H.-P. Ex. 524.— In the last example over what distance on a horizontal plane would the same engine hare drawn the train in the same time ? Ans. lOf miles. Ex. 525.— How long would it take the engine in Ex. 623 to draw the same train with a uniform velocity over a space of 4 miles up an incline of 1 in 100? " Ans. lO^min. Ex. 626. — A train is drawn with a uniform velocity up an incline 3 miles long of 1 in 250, on which the resistances are 7 lbs. per ton ; deter- mine the distance on a horizontal plane over which the same train could be drawn with a uniform velocity by the same expenditure of work. Ans. 6|f miles. Ex. 527. — In Ex. 346 if the body is in the state of uniform motion up the J)lane, show that the relation between u, the work done by P, and u^ the work expended on w, is given by the equation n, sin a cos {$ — )=w sin (a + i(>) Now, if s, is the distance through which p's point of application moves measured in the direction of that force s, = 2 cos |8 and if Sj is the distance through which Vs point of application moves when similarly measured iJ2 = 2 sin a where I is the length of the plane, hence s, sin a=:S2 cos whence the relation between u, and Uj is at once found.] Ex. 528. — If a pivot sustaining a pressure a is made to revolve once, show that the number of units of work expended on the friction of the end equals Itt ;« p Q. [See Art. 82.] Ex. 529. — In the case of a single fixed pulley the number of units of work expended in raising a weight a through a height q is given by the formula v = aciq + bq where a and b have the values assigned in Art. 89. Ex. 630. — In the ease of a tackle of n sheaves show that the number of units of work expended in raising a weight a through a height q is given by the formula * a°— 1 \a' — l a-1' ^ [iSee Ex. 419.] MODULUS OF A MACHINE. 215 Ex. 331. — In Ex. 421 determine the number of foot-pminds of work ex- pended on the passive and on the useful resistances when the weight of 1000 lbs. is raised through 50 ft. Ans. (1) 67,000. (2) 50,000. Ex. 332. — ' It is said that in a pair of blocks with five pulleys in each two-thirds of the force are lost by the friction and rigidity of the ropes.' * Determine the degree of truth in this statement when each sheave is 4 in. in radius, and turns on an axle a of an inch in radius, the axle being of wrought iron and the bearing of east iron, and the rope 4 in. iu circum- ference ; the weight to be raised being 1000 lbs. J " Work expended on passive resistance s _ 19 , Work done ~29 Ex. 533. — In the capstan Ex. 427 show that the work that must be done by the forces in order to move the weight a through a height g is given by the formula .= (1.^4-*) (l.S) a.+ V^ (i .^_^^).^^.l Ex. 534. — A rope passes over a single fixed pulley in such a manner that its two parts are at right angles to each other ; the one end carries a weight O; the radius of the pulley is r and of the axle p, the angle such that sin j3= "- — — i ; then, the weight of the 'pulley being neglected, show that if p is the force that will just raise % we have P = ^ Q^- ^Lt£«\ tan (45° + ;8) Ex. 533. — In the last example show that the relation between p and u may be very nearly represented by the formula P = a('l+S+P^sin ^)-^f(l-^P^ sin A Ex. 536. — A weight of 500 lbs. has to be raised from a depth of .^0 iathoms ; it is fastened to a rope which passes over a fixed pulley in such a manner that the parts of the rope are at right angles to each other ; the rope is wound up by means of a capstan which is turned by two equal parallel forces acting at the end of equal arms ; the rope is 3 in. in circumference, the pulley 6 in. iu effective radius, its axle half an inch in radius, and of wrought iron turning upon cast ; the capstan weighs 4 cwt., its axle is 4 in. in radius, oak moving on wrought iron, the effective radius of the capstan 15 in. ; determine the number of foot-pounds of work that must be done in order to raise the weight (not weight and rope), and the number expended on passive resistances. Aiis. (I) 201,356. (2)54,356. Ex. 637. — There is a fixed pulley 20 inches in radius (r) moving on an axle 1 in. (p) in radius (sin = 0'16) '; a weight of 500 lbs. is raised from a depth of 300 feet (Z) by means of a rope 3 in. in circumference which passes * Dr. Young's Lectures, vol. i. p. 206. 216 PRACTICAL MECHANICS, over it ; the end of the rope falls as the weight rises ; determine the error that results from neglecting the weight of the rope in calculating the foot- poun4s of work required to raise the weight — the united length of the two hanging parts of the rope being reckoned at 300 ft. Am. Error = O-^^^Ppsm^ ^ 2^^_ r [Compare Ex. HI and 158.] Ex. 538 . — In the last example determine the error that would result from neglecting the weight of the rope if the end were not allowed to fall. Ans. Error 19,000. Ex. 539. — If a weight o is raised through a height q by means of a screw, show that if the same notation is employed as in Ex. 393 the number o units of work expended is given by the formula u = ag' I tan (o + <^) +| ■ — M > cotan a where all frictions are neglected except those between thread and groove and on the end of the screw. Ex. 340. — An iron screw 4 in. in diameter communicates motion to an iron nut, the screw thread is inclined to its base at an angle of 18°, the diameter of the end of the screw is 2 in. ; all the surfaces are of cast iron ; determine the number of foot-pounds of work that must be expended in raising a weight of 3 tons through a height of 2 ft. by means of this screw. Ans. 23,358. Ex. 541. — Determine through whatheight a man working with this screw could raise a weight of 1 ton in a day ; and what would be the best length of the arm of the screw on which he works — pushing horizontally ; deter- mine also the part of his work which is expended in overcoming friction. Asis. (1) 384 ft. (2) 71 ft. (3) f. 106. The end to be attavned by cutting teeth on pjheels. — The problem to be solved is this : — Griven an axle A, moving with a uniform angular motion roimd its geometrical axis, it is required to connect it in such a manner with a parallel axis b, as to communicate to it a uniform angular motion which shall have a given ratio to the former. Suppose the axle A to revolve w, times in one minute, and it is required to make the axle B revolve n times in one minute ; join the centres A and B, divide A B into m + n equal parts, and take a g equal to n of these parts, and therefore B c will contain m of them, so that AC . CB;:n : m TOOTHED WHEELS. 217 with centres A and B, the radii AC, BC respectively, describe circles touching at c; if these circles are fixed each to its own axle, and revolve with them, and if their circumferences are rough, so that they roll on each other, the problem is solved ; for take on the circumferences G 1 O / respectively points c' and c" which were in contact at c, then must the arc c c' equal the arc c c", since the several points of the arcs have been successively in contact each with each, and this is true whatever be the lengths of those arcs. Now, in one minute the point c' describes an arc whose length is 27rA c . m, and therefore c" describes an arc whose length is 27r A c . -m., i.e. an arc whose length is 27r B c . TC, since A c . 7n=BG . n ; but 27r BG .nisn times the circutoference of the circle whose radius is BC, and therefore the axle B makes n turns while A makes m turns, i.e. B moves in the required manner. It is evident that the angular motions will have the same ratio whatever be the time, and therefore when the time is very short ; hence if the angular motion of the axle A varies from instant to instant, that of the axle B will also vary, but the ratio of the angular motions will remain constant. 218 PRACTICAL MECHANICS, It is also plain that the directions of the angular motions will be contrary, as indicated by the arrow heads. It may be remarked that the wheel A c is called the driver, and B c the follower. Ex. 542 If in the last article a single wheel moving on a parallel axle with its centre in the line a b -were interposed between a c and b c, it woiild cause the follower to revolve in the same direction as. the driver, and would not produce any change in the ratio of their angular motions, the radii A c and B c being unchanged. 107. Practical objection to the above solution. — It is evident that the above solution faUs if the surfaces of the wheels rub smooth, so that the motion becomes par+ly one of sliding and partly one of rolling contact ; and also that it wiU fail if the centres A and B are slightly displaced, since then the contact ceases : one method, in common use, of obviating this objection is to pass a strong band of leather tightly over the wheels ; this method is com- monly used when the centres a and B are so considerable a distance apart that the wheels would be inconveniently large if in immediate contact ; the most eifectual means, and the only one with which we are here concerned, is to cut teeth on the circumferences of the wheels ; when this is properly done the uniform revolution of the wheel A can be made to communicate a uniform revolution to the wheel B. The problem we have to solve is therefore twofold : — (1) To determine the form that must be given to the teeth of wheels, in order that any uniform motion of the driver round its axis shall communicate to the follower a uniform motion round its axis. (2) As this cannot be done withoxit causing the teeth of the one wheel to slide over those of the other, it is required to determine what amount of work is lost by the friction of the teeth when work is transmitted from one axle to the other. The limits of the present work will not allow us to do THE EPICYCLOID. 219 more than give one solution of the former question, and an approximate solution of the second. Readers who desire further information on this very important subject will be able to obtain it by reference to Mr. Willis's Principles of Mechanism, and to Mr. Moseley's Jfe- chanical Principles of Engineervng : * the former work treats only of the question of form ; the latter also con- tains a very full discussion of the question oi force. 108. Definition and properties of the epicycloid. — If a circle carrying on its circumference a pencil-point be made to roll on the pig. 149. outside of the circum- ference of a fixed circle, the point will trace out a cmrve called an epicycloid: the fixed circle is called the base ; the moving circle is called the generating circle. Thus if Q is a point on the generating circle a d q, and A p c is the base or fixed circle, then if Q were in contact with A p c at p, the point Q will trace out the epicycloid p Q. (a) It is evident that the length of the arc A Q equals that of the arc A p. (6) It is evident that the point Q is at the instant moving in a circle of which the centre is A, and radius A Q, so that the Hue A Q is the normal to the epicycloid at the point Q, and if D Q be joined that line is a tangent to the- curve at Q. (c) It is evident that the form and dimensions of the curve are independent of the particular point Q occupies * A very clear elementary discussion of the forms of the teeth of wheels ■will be found in Mr. Goodeve's Elements of Mechanism. 220 PBACTICAL MECHANICS, on the generating circle, so that if we take a succession of points Q, Qj, Qj . . . on the generating circle, and describe with them a succession of epicycloids Q p, Qj Pj, q^ ^2 • • • they will all be exactly like one another, and if p' Q' be any epicycloid described on the same base with the same generating circle as the others, it too will be exactly like the rest : if we now suppose aU the former to remain fixed, and the circle p' A c to revolve round its centre, carrying p' q' with it, then when p' comes to Pj, the curve p' q' will exactly cover P2 Qj, and in like manner it will successively cover Pj Qj and p Q. Proposition 22. An epioycloidal tooth can be Tuade to work correctly with a straight tooth. Let PQ be the tooth described on the base ap, the centre Fig. 150. of which is Oj, by a circle whose diameter is A o ; suppose the base to revolve round Oj and let the tooth assume successively the positions p, q\, Tilv IP3I3 emit- ting the circle ad o in points g'l, q^, q^, then since the straight linesoqy,oq^,oq^ , .. . touch the epicycloid in the points ^j, q^, q^ .... it is plain that a straight line whose length is OA, and, which is movable round 0, will, if driven by the tooth, come successively EULE FOE FOEM OF TEETH. 221 into the positions a„ o a^, oa^, . . . passing Ihrough the points g,, 5'2» 33 . . . . respectively. Now, if we suppose the angles A o^jp^^-p^ Oj ^2 .... to be equal, the arcs Ap^ VilPiilPilPs ... axe equal, and therefore (Art. 108 (a)) the arcs A.q^, q^q^, g'aS's ■ • • a^re equal, and the angles they sub- tend at the centre c will be equal, and their halves will be also equal, i.e. the angles A a,, a, o a^, a^ o a^, are equal ; so that if the circle p A O, move with a uniform angular motion, it will communicate a uniform angular motion to a straight line A o movable about the point 0, i.e. the straight Hne works truly with the epicycloidal tooth. Ex. 543.^If with centre o and radius o A a circle te described, show that 'if this circle work with a p by friction, any one of its radii will have the same angular velocity as if it had been driven by the tooth p ft. 109. Practical rule for the form of teeth* — Let 0, o, be the centres of the two toothed wheels ; draw the line of centres o, ; when the point ^^- '^'^^ of contact of any two teeth is on the Hne of centres let it be at A; with centres o and Oj .and radii oa and OjA respec- tively describe circles, a A a', hA.h' ; these are called ihepitch circles of the respective wheels, i;e. the two circles which rolling by friction would move with the sam.e angular motions as the wheels. Now, if there are to be m teeth in the wheel o, there must be m-j in the ■OTj is given by the proportion o A OjA wheel Oj, where : : m. : m,. Divide the circumference of a A a! into m equal parts, * This rule, though not the best, is — or, at all events, used to be — very generally employed in practice. (See 'Willis, p. 106). 222 PRACTICAL MECHANICS. of which paxts let a Aj be one ; the chord of this axe is called the pitch of the wheel ; divide it into two (nearly) equal parts, of these A E (the smaller) is the breadth of a tooth, and E A; the space between two teeth ; then the flanks B A, D E of a tooth (i.e. the parts of its outline within the] pitch circle) are straight lines converging to the centre o ; and the faces of the tooth a c, e f (i.e. the parts of its outHne on the outside of the pitched circle) are por- tions of epicycloids described on the pitch circle as a base by a generating circle whose diameter equals the radius of the pitch circle of the wheel with which it is to work, viz. Oj A. The teeth of the wheel Oj are cut upon the same principle ; the circumference of the pitch circle 6 A 6' is divided into m, equal parts, and each is divided into a tooth and a space ; the flanks of the teeth converge to o„ the faces are epicycloids described on the pitch circle as a base by a generating circle whose diameter equals the radius OA. That the two wheels thus constructed will work truly, follows immediately from Prop. 22 ; thus, if the wheel o revolve uniformly, the tooth b a c driving the tooth b' a c', the epicycloid a c will cause the straight line A b', and therefore the wheel 0,, to revolve uniformly : on the other hand, if the wheel 0; moving with a uniform motion drive 0, the epicycloid ^ c' will cause the straight RULE FOR LENGTH OF TEETH. 223 line A B, and therefore the wheel 0, to revolve uniformly. This is of course true whether the wheels move in the direc- tions indicated by the arrow heads in fig. 151, or in direc- tions opposite to them. In order to prevent the locking of the teeth, it is usual to make A E less than A, E by ^ of the pitch A Aj ; and to cut the space a b' deeper than the perpendicular length of the tooth a c in such a manner that the distance from c to the centre is less than the distance from b' to the same centre by ^ of the pitch A A, ; if, however, the workmanship is very good, the differences can in both cases be made smaller. The rule for determining the length of the teeth com- monly adopted by millwrights is to make the length of the tooth beyond the pitch circle (i.e. AC or AC') equal to •^ of the pitch.* This rule is, however, a very bad one ; the following, though not perhaps the best, is very much better : — Suppose o to be the driver, and suppose a pair of teeth to be in contact on the line of centres, the face of the next tooth shoidd be so long that its extreme point c^ should just be on the circumference of the generating circle a Xj, as shown in the figure ; the length of the tooth of the follower is determined by a similar rule ; the extreme point of the following tooth Cj should (under the same circumstances) be on the circumference of the gene- rating circle A x o. The reason of this rule is as follows : — It may be considered that when the wheels are in motion the pair will bear the whole or nearly the whole stress which at any instant will be the next to go out of contact : so that, the above construction being employed, the one pair of teeth is just going out of contact when the next pair comes to the line of centres, and consequently the working stress is not thrown upon any pair of teeth until » Willis's Principles of Meclianism, p. 98. The rule which follow^ is given both by Mr. Moseley, Mechanical Principles, p. 267, and by Gen. itorin, Aide-Mimoire, p. 280. 224 . PRACTICAL MECHANICS. it comes to the line of centres ; hut it appears that prac- tically the friction between a pair of teeth is very much more destructive when they are in contact before the line of centres than when in contact behind the line of centres; by following, therefore, the rule above given, the fric- tion between any pair of teeth is diminished. (Compare Ex. 566.) In practice the teeth of a wheel are all cut from a pattern ; in constructing a pattern the epicycloidal curve may be drawn from the actual rolling of a circle of the proper size ; or an approximation may be obtained by means of circular arcs. Kules proper for this purpose will be found in Mr. Willis's Treatise above referred to. Ex. 544. — To determine the radius of the pitch circle of a wheel ■vrhich shall contain » teeth of given pitch a. a — "■ „ ■ 180° 2 sin — 1- Ex. 645.— If a wheel of m teeth drive another of n teeth ; then if the driver make ^ revolutions per minute, the follower will make ^ revolutions » per minute. Ex. 546. — There are three parallel axes, a, b, c ; a makes p revolutions per minute, it carries a wheel of m, teeth which works with a wheel of «, teeth on B ; B also carries another wheel of m^ teeth which works with a wheel of n, teeth on c : show that c makes — ! . — 5 . » revolutions per minute.* Ex. 647. — A winding engine is worked in the'foUowing manner : — A steam engine causes a crank to make 30 revolutions per minute ; the axle of the crank has on it a wheel containing 36 teeth, which works with a wheel containing 108 teeth ; the latter wheel is on the same axle as the drum, which is 5 ft. in radius ; determine the number of feet per minute described by the load. Ans. 314 ft. * The above arrangement is to be found in most cranes ; if the student is not acquainted with the arrangement of a train of wheels he will do well to examine a good crane, such as is to be seen at most railway stations : the train of wheels in a clock is also a good example, but cannot commonly be studied without taking the clock to pieces. TOOTHED WHEELS. 225 110. The hunting cog. — If wheels have to do heavy workj and the precise ratio between the velocities is not of great importance, an additional tooth — called a hunting cog — is introduced into one of the wheels, so that the same pair of teeth may seldom work together; by this means they are kept from wearing une(^ually. Foi* instance, if in the last example we denote the teeth of the driver by the successive numbers 1,2,3, . . , 36, and the teeth of the follower by the successive numbers 1, 2, 3, . , . 108 ; then in every revolution 1 will work with 1, 37, and 73; 2 will work with 2, 38, and 74; and 36 will work with 36, 72, and 108. If now we introduce a hunting cog into the driving-wheel, so that it contains 37 teeth, then on'the first revolution 1 will work with 1, 38, and 75 ; in the next revolution with 4, 41, and 78 ; in the third with 7, 44, and 81, and not until the 38th re- volution will it work with 1 again. Ex. 548. — If in the last example a 'hunting cog' were introduced into the driver so that it contains 37 teeth, determine the number of feet pei: minute the load will now travel. Ans. 323 ft. Ex. 549. — If in Ex. 546 there are i + 1 axles and the drivers contain m teeth, and the followers contain n teeth a-pieee, show that the number of revolutions made by the last axle will be p / — J Ex. 560. — If in the last example it is required to multiply the number of revolutions 200 times, how many axles must we use— (1) if we take m = 2n; (2) if we take ot = 4» ; (3) if we take m=6n, and determine the number of teeth employed, in each case using the nearest whole numbers ? Ans. Axes (1) 8. (2) 4. (3) 3. Teeth (1) 24». (2) 20m. (3) 21». Ex. 551. — If each driver has m teeth, and each follower n teeth, and if K is the total number of teeth in the train, and if the last axle makes j revolutions while the first axle makes one revolution, show that ■(;) 226 PRACTICAL MECHANICS. * Ex. 552. — In the last example show that for given values of M and n we shall obtain the greatest value of q by making »i = 3'69 .n nearly.* [It is easily shown that log ( — ) = 1 h > whence the result stated.] Ex. 553. — In the case of a pair of wheels with epicyeloidal teeth show that the distance through which the surfaces of each pair of teeth slide one upon the other while in contact and after passing the line of centres is ap- proximately represented by the formula — (— + Z-\ or JUl (!L + !!L | n \n %/ n, \n n^J where r and r, are the radii of the driver and follower respectively, and n and », the number of teeth in those wheels respectively. [The motion of one tooth on the other is partly a sliding and partly a rolling motion. Now, if we refer to fig. 152, it is evident that the pair of teeth just going out of contact touch at c^; it is also evident that the two points Aj and a'j were in contact at a, so that the space through which the surfaces have slidden over each other is Aj a'^, which is very nearly equal to the sum of the versed sines of the arcs a k„ and A a' 2ir I.e. to r vers — 29r '+ »•, vers — ; whence the value assigned in the question.] Ex. 554. — A weight p balances a weight Q under the following circum- stances : p is tied to a rope which is wrapped round an axle whose radius is ^ ; Q is tied to a rope which is wrapped round an axle whose radius is g ; to the former is attached a concentric rough wheel, whose radius is ♦•, to the latter in like manner a concentric rough wheel, whose radius is Fie. 163. these two wheels are in contact on the line of centres so that r + r^ equals o o, ; show that if we neg- lect the magnitude of the axes and the rigidity of the cords, we shall have p = } r.s f a sin A ' p, sin *, 1 * + _!_.' w c i + w, "li ^ V Ex. 558i — In the last example if we suppose the rough wheels to be re- placed by a pair of toothed wheels whose pitch circles have the same radii as the wheels ; then if the wheel o contains n teeth, and the wheel o, con- tains «i teeth, show that when ft is raised through a distance s the work lost by the friction of the teeth is approximately represented by the f6rmula fiQS (— + — 1 , where /i is the coefficient of friction between the teeth. 2— [If the wheel o,A revolves through an angle — the distance through; n^ which the surfaces of the driving and driven teeth slide is -^ f * + ?LY », \n nj and therefore, supposing e, the mutual pressure, to ccmtinue constant during the contact of the teeth, the number of units of work expended on friction equals /ie -^ ("""'" ~ r ■^°^' '''PP™^™a'tely, a.r,=Qfl, and therefore the work expended on one pair of teeth equals ltd ^^ ( JT + !L ^ ; but -^ is the distance through which q is raised during J^, \n nJ m, * If p instead of being a weight were a force acting vertically upvrard it is easily shown that the third term of this equation is \2 n/ Os( — + — jp, sini(>i ft2 228 PRACTICAL MECHANICS. the action of one pair of teeth, and the same being true of every pair of teeth, we obtain the result stated in the question. Of course, the addition of the expression contained in the present question to that obtained in the last is the correct approximate formula for the work expended in raising a weight through the intervention of a pair of toothed wheels.] Ex. 669. — ^A force p acting at the end of an arm o a, two feet long, causes the toothed wheel OB to make 10 turns per minute ; this wheel working with the wheel o, b turns the drum o, c and raises the weight a ; given that p does at the point a 330,000 foot-pounds of work per minute, determine approximately the weight q that will be raised by the drum, having given the radius of o b to be 1 ft., o, b to be 3 ft., the number of teeth in o b to be 40, and the radius of the drum 5 ft. ; the teeth, axles, and bearing are all of cast iron without unguents ; the radii of the axles are 3 in., the weight of the axles and appendages of u is 3600 lbs., and that of o, is 5400 lbs. Ans. 2752 IhB. [See Note to Ex. 557.] Ex. 660. — Show that in a train of p pairs'of wheeb and pinions* the work lost by friction between the teeth is given by the formula liQgvi — + — + —+ . . . + I L n^ 7I2 Jtg n^p J where »„ n^, n, . . .n^ are the number of teeth in the sufcessive wheels and pinions. Ex. 561. — There is a train of p equal pairs of wheels and pinions ; the numbers of teeth are such that the last axle revolves m times faster than the first ; show that if r is the number of units of useful work yielded, the- work lost by the friction between the teeth is represented by the formula n where n. is the number of teeth in each wheel. *Ex. 562. — ^It is required to make the last axle move m times faster than the first, show that the loss of work is least when p, the number of pairs of wheels and pinions, is given by the formula i_ \_ m~''+log «!~''+i=o *£r. 563. — If in the last example it is required to multiply the velocity 100 times, show that the proper number of pairs of wheels and pinions is 3 or 4, i.e. show that the equation in the last exeimple gives a value of p between 3 and 4 ; and determine the number of teeth employed in each case if the first pinion have 20 teeth, using the nearest whole numbers. Ant. (1) 339. (2) 333. * When a small wheel drives a large one the former is frequently called a pinion and the latter a wheel. FRICTION BETWEEN TEETH OF WHEELS. 229 Ex. 564.— If in the pair of wheels already described (Art. 109) all but a single pair of teeth be cut away, so that the remaining teeth act on each other while the nrheel o moves through an angle ?^ before coming to the line of centres, and also while it moves through an equal angle after having passed the line of centres, and if we suppose p and « to act on the pitch circles of their respective wheels, show that when the point of contact is in Such a position that the wheel o has to revolve through an angle fl before the point of contact comes to the line of centres we have P {?•, — (}• + »•,) tan 9tan<()} =qj', and that when the point of contact is so situated that the driver has re- volved through an angle 9 from the line of centres we have pr = Q { »• + (?• + r,) tan — tan } [If in the accompanying figure x is the point ol contact of the teeth be- fore they come to the line of centres, that point x will be on the circum- FlQ. 164. ference of a circle whose diameter is o A ; if then we draw a line B b' such that the angle b x A equals ^, this will be the line of the mutual action of the teeth ; remembering that the angle A » x equals it is easily shown that the perpendiculars on e b' from o and o, are respectively equal to T cos fl cos ^ and (»■ + ''i) cos (fl + <())— r cos fl cos ^ whence the first equation is obtained ; the second is obtained in a similat manner, by determining the relation between p and q when the follower has revolved through an angle fl' which will be found to be pr = a {r + ()• + »•,) tan fl* tani^} whence we obtain the answer. 1 230 PRACTICAL MECHANICS. Ex, 565. — If A B be any diameter of a circle A p B ; if c be any point taken in the prolongation df ab (so that b is between a and c), and if a p, b p, c p be joined, show that B c = A c tan p A B tan b p o and hence explain the action of the forces which produces the result that follows from the first equation in Ex. 564, viz. that when r^ = {r + r^ tan tan ip the force p must be infinitely large to bring o, into the state bordering on motion. Ex. 566. — If the driver be not greater than the follower, show from the equations of Ex. 564, that for a given value of a the value of p is greater when the driving tooth is in a given position before it comes to the line of centres than when it is in a corresponding position after having passed the line of centres. [If m be written for the ratio of r to r, (so- that m cannot be greaiter than unity) the equations in Ex. 564 can be written thus : — p { 1— (l+a8)/ttanfl } =q and p'=a{l+n+ i-J/ttan wifl} consequently P— p'=Q { (1 + J»)/ttan 9— I 1 + — J/t tan »i9 + positive terms } and this, on expanding in powers of 6, is found to equal /:*Q[(l+m)e{|(I-m2)fl2 + ^(l_m*)fl' + . . .} + positive terms] 231 PAET n. DYNAMICS. CHAPTEE I. INTRODUCTORY.* Ill, Vehoity. — Before considering /orce as the cause of change of velocity, it will be necessary to define accurately the means of estimating velocities numeri- cally. Def. — A body moves uniformly or with a uniform velo- city when it passes over equal distances in equal times. The units of distance and time commonly employed are feet and seconds :f and whenever, a body is said to be moving with any particular velocity, e.g. 5 or 6, this will always mean with a velocity of 5 or 6 ft. per second. Def. — When a body moves with a variable velocityj that velocity is measured at any instant by the number of units of distance it would pass over in a unit of time if it continued to move uniformly from that instant. It will be seen from the definition that variable velocity is measured in a manner that exactly falls in with the * The student is particularly recommended to make himself thoroughly master of this chapter before proceeding further. t To prevent mistake, it may be stated that the time referred to is meaii solar time. 232 PRACTICAL MECHANICS. ordinary way of speaking : thus, when we say that a train is moving at the rate of 40 miles an hour, we mean that if it were to keep on moving uniformly for an hour, it would pass over 40 miles. Again, if we were to drop a small heavy body, we should find that at the end of a second it is moving at the rate of about 32 feet per second, or, as it is commonly stated, it acquires in a second a velocity 32, meaning that if it were to move uniformly from the end of that second it would pass over 32 feet in each successive second. 112. Relation between uniform, velocity, time, and distance. — In the case of a body moving with a imiform velocity, it is evident that the number of feet (s) passed over in t seconds must be t times the number of feet passed over in one second (v), .•.s = vt. The distance s can, of course, be represented geometri- cally by. the area of a rectangle whose sides severally re- present on the same scale the velocity and the time. Ex. S67. — ^A body moves uniformly over 21 miles in half an hour ; determine its velocity. Ans. 7k Ex. 568. — A body moves at the rate of 12 miles an hour; determine its velocity. Am. 17f. Ex. 569.— The equatorial diameter of the earth is 41,847,000 ft., and the earth makes one revolution in 86,164 seconds; determine the velocity of a point on the earth's equator. Ans. 1526. Ex. 570.— A body moves with a velocity of 12 ; how many miles will it pass over in one hour ? "What would be its velocity if we used yards and minutes as units instead of feet and seconds ? .4»js. (1) 8A. (2)240. 113. The velocity acquired by falling bodies.— It appears as the result of the most careful experiments that at any given point on the earth's surface, a body falling ^freely in vacuo acquires at the end of every second a FALLING BODIES. 233 certain constant additional velocity : * this velocity is sUghtly different at different places, but is always the same at the same place, and never differs greatly from 32 ; so that if at any instant the falling body has a velocity v, it will have at the end of the next second a velocity v + 32. This additional velocity is the accelerative effect, or, as it is sometimes called, the accelerati/ng force, or simply the acceleration, of gravity, and is denoted by the letter g ; — in all the following examples it will be assumed that g equals 32, unless the contrary is specified. From what has been said it is plain that if a body is let fall, it acquires a velocity g at the end of the first second, 2g at the end of the second second, 3g at the end of the third second, and so on : consequently, if v is the velocity acquired at the end of t seconds, we shall have v=gt. By the same reasoning it appears that if the body is thrown downward with a velocity v, and if v is its velo- city after falling t seconds, then v=Y+gt Moreover, when a body is thrown upward so as to move in a direction opposite to that in which gravity acts, it appears that it loses in every second a velocity g ; conse- quently in that case v=v—gt Ex. 571. — A body falls for 7 seconds ; ■with what velocity is it moving at the end of that time ? Ans. 224. Ex. S72. — If a body is let fall, how longvriU it take to .acquire a velocity of 200 ft. per second ? Ahs. 6j sec. Ex. 673. — A body is projected downward with a velocity of 80 ft. per second ; determine the velocity it will have at the end of 5 seconds, and the number of seconds that must elapse before its velocity equals twice its initial velocity. Am. (1) 240. (2) 21 see. * It may be remarked, that the difference between the velocities with which a feather and a bullet descend is entirely due to the resistance of the air. 234 PRACTICAL MECHANICS. Ex. 674. — A body is thrown down-ward with a velocity of 160 ft. per second; determine its velocity at the end of 4 seconds, and the number of seconds in which a body that is merely dropped would acquire that velocity. Am. (I) 288. (2) 9 sec. Ex. 675. — A body A is projected downward with a velocity of 160 ft. per second; at the same instant another body B is projected upward with an equal velocity ; determine how much faster A will be moving than B at the end of 4 seconds. Ans. 9 times. Ex. 576. — A body is thrown upwards with a velocity of 96 ft. per second ; with what velocity will it be moving at the end of 4 seconds ? [The formula gives —32, i.e. it will be moving downiwmd -wMh. avelocity of 32 ft. per second.] Ex. 677. — In the last case how long will it take the body to reach the highest point ? [It will be at the highest point when v = 0, i.e. after 3 seconds.] Ex. 578. — A body is at any instant moving upvjard with a given velocity o— V ; show that it will be moving downwards with an equal velocity after — I- 9 seconds ; and that it will reach its highest point after— seconds. 9 Ex. 579. — A body is thrown up with a velocity mg ; after how loqg will it be descending with a velocity ngt Arts, m + n sec. 114. The distarice described in a given time by a fall- ing body. — It admits of proof that if a body is allowed to fall freely from rest for t seconds the number of feet (s) which it will pass over is given by the formula s = igt^ If, however, it is thrown downward with a velocity V, we shall have and if upward with a velocity v, it will, at the end of t seconds, be s feet above the point of projection, where s=vf-ig'i2 Ex. 580. — How many feet will be described in 4 seconds by a body that moves freely from rest under the action of gravity ? Ans. 266 ft. Ex. 681. — Through how many miles would a body falling freely from rest descend in one minute? Ans. lOls mi. Ex. 682. — A body is projected downward with a velocity of 20 ft. per second ; how far will it fall in 1 J second ? Ans. 66 ft. FALLING BODIES. 235 Ex. 583. — A body is projected upward with a velocity of 100 ft. per second ; liow high will it hare ascended in 3 seconds? A7is. 156 ft. Ex. 584. — Show that the greatest value of rt—igt' is found by making ^ = — ■ [Compare this result with Ex. 578.] Ex. 586. — If a body be projected upward with a velocity of 96 ft. per second, where will it be at the end of 7 seconds, and what will be the whole distance it will have described ? Am. (1) 112 ft. below the point of projection. (2) 400 ft. Ex. 586. — A body is projected upward with a velocity of 100 ft. per second ; determine where the body will be, with what velocity, and in what direction, the body will be moving at the end of 4 seconds. Ans. (1) 144 ft. above the point of projection. (2) 28 ft. per sec. downward. Ex. 587. — A body is projected upward with a velocity v; show that it will return to the point of projection after — sees. 9 [Compare this result with Ex. 578.] Ex. 588. — A body falls for a time t, and has a velocity v at the begin- ning, and V at the end of that time : show that it describes the same dis^ tance as another body describes in the same time with a uniform velocity |(v + u). 115. Relation between velocity acquired and distance passed over by a falling body. — The above relations between the velocity (v) which the body has at the end of a time (f) and between the distance (s) which it describes in the same time (i) enable us to determine the relation between v and s ; thus, if the body is simply let fall we have v=gt and s=^gt^ whence v'^=2 g s an equation which gives the velocity acquired in falling from rest through s feet. In like manner if we take the equations v=y+gt and s=vt + ^gt^ we see that 2gs=2v gti-g^t^ 236 PRACTICAL MECHANICS, and y^+2ga=v'' + 2ygt+g^t^ ={v+gtf therefore v^ = v^-{-2g8 This equation gives the velocity (v) which the body has after falling through s feet from the point at which it was moving downward with a velocity v. Similarly we can show that v^=Y'^—2g8 in which v is the velocity which it has when it is 8 feet ahove the point at which it was moving upwards with a velocity V; whether the direction of the velocity v is upward or downward does not appear from the equation, and must be determined by other considerations. When a body is moving with a given velocity (v), a certain height (s) can always be found such that if a body fell down it freely from rest it would acquire the given velo- city ; under these circumstances v is said to be the velo- city due to the height H. These quantities are, of course, connected by the equation v^=2gB. Ex. 589.— If a body is thrown upward with a velocity t, show that it will ascend through — feet. ' Ex. 590. — If a, body is thrown upward with a velocity of 200 ft. per second, find its greatest height. Ans. 625 ft. Ex. 591.— If a body falls freely through 150 ft., find the velocity it acquires. Ans. 98. Ex. 592. — A body is projected vertically upward with a velocity of 200 ft. per second ; how long will it take to reach the top of a tower 200 ft, high, and with what velocity will it reach that point? Ans. (1) 1-1 sec. (2) 164-9. Ex. 593. — Let A be the highest point of a vertical line a b : at the same instant one body is dropped from a and another thrown up from B, they meet at the middle point of A B ; find the initial velocity of the second body. Alls. Vff x a B. Ex. 694. — A stone (a) is let fall from a certain point ; one second after another stone (n) is let fall from a point 100 ft.- lower down ; in how many UNIFORM ACCELERATION. 237 peconda from the beginning of its motion will A overtake b, and what distance will it have described ? Am. (1) 3| sec. (2) 210J ft. Ex. 695.— A stone (a) is let fall from the top of a tower 350 ft. high ; at the same instant a second stone (b) is let fell from a window 50 ft. below the top ; how long before A will B strike the ground ? Ans. 0"35 sec. Ex. 696. — A stone (a) is projected vertically upwards with a velocity of 96 ft. per second ; after 4 seconds another stone (b) is let fall from the same point ; how long will b move before it is overtaken by A, and at what point will this happen ? Ans. (1) i sec. (2) 256 ft. below the point of projection. Ex. 597. — In the last example if only 3 seconds had elapsed when B was let fall, would a ever have overtaken it ? Ans. No. Ex. 898. — The point a is 128 ft. above b; a body is thrown upward from A with a velocity of 64 ft. per second, and at the same instant another is thrown npward from B with a velocity of 96 ft. per second ; show that after i seconds they will both be at A ; moving downward with velocities 64 and 32 respectively. Ex. 899. — Determine the heights to which velocities of 20, 69, and 760 &. per second are respectively due. Ans. (1) 6^ ft. (2) 84|| ft. (3) 9028 ft, 116. Other cases of uniformly accelerated motion. — The velocity of a body is said to be uniformly accelerated when it is increased by equal amounts in equal intervals of time. Thus, taking feet and seconds as the units of distance and time, if the velocity of a body is, during any second of its motion, changed from V ft. a second at the beginning of the second to V + 20 ft. a second at the end of the second, the acceleration is said to be 20 in feet and seconds. In like manner if the acceleration is / in feet and seconds, this means that if at any instant the body is moving at the rate of V feet a second, its velocity will be-^ come at the end of a second v+/ feet a second. The velocity is said to be uniformly retarded when it is diminished by equal amounts in equal intervals of time ; thus, if at any instant the velocity is V feet a second, and at the end of a second it becomes V— / feet a second, the velocity is said to undergo a uniform retardation /. It is usual to reckon retardation as a negative acceleration. If we write / for g in the formulae of the preceding article? ass PRACTICAL MECHANICS. they will apply to the rectilineal motion of bodies whose velocities undergo any uniform acceleration or retai;dation. Thus, if V is the initial velocity, / the acceleration, t the time at the end of which the body has a velocity v, and is at a distance s from the starting point, we shall have v=v-{-ft Ex. 600. — At the distance of the moon the aocelerative effect of grarity is reduced to about jjj ; if a body fell freely from this distance for one hour, with what Telocity per minute would it then be falling ? and in how many seconds would a body falling in the neighbourhood of the earth's surface acquire the same velocity ? Am. (1) 1928f. (2) 1 sec. very nearly. Ex. 601. — If a body were to begin to fall to the earth from the distance of the moon, how many yards would it fall through in half an hour ? Ans. 4821 yards. Ex. 602. — In the last example if a body were thrown upward with a Telocity of 4 miles an hour, how long would it take to return to the point of projection ? Ans. ISli Be6. 117. The acceleration of the motion of a given body produced by a gi/uen force. — In most cases the moving body is acted on by several forces, which to a certain extent neutralise each other, and its motion is caused by their resultant. Suppose that a body is placed on a smooth horizontal plane and moved by a force (p) acting horizon- tally ; the forces acting are the weight of the body, the re- action of the plane, and the force P; of these the two former neutralise each other, and the latter produces the motion. So long as the force producing motion remains unchanged, it will uniformly accelerate (or retard) the motion. The amount of the acceleration is determined by the fundamental principle * — If any given body is acted on " The eTidence for this principle, as for all the other fundamental principles of dynamics, is experiment, though it is Tery difficult to deTise experiments which shall exhibit them in a state of isolation: Galileo, who discoTered most of them, possessed a rare sagacity in detecting the ^orfa of RELATION BETWEEN FOECE AND MOTION. 239 successively by two forces, the accelerations due to the action of the forces are i/n the same ratio as the forces. Now, if the weight of a body be w lbs., we know that the sensible attraction the earth exerts on it at London is a force of w lbs. — the term pound being used to denote the unit of force, as in Art. 23. Also, if this body fall freely in vacuo in London it has been ascertained (see Table XV., p. 250) that its velocity is increased in each second by a velocity of 32*1912 ft. per second ; this is, therefore, the acceleration of that body's velocity when acted on by a force of w lbs. Suppose the same body to be acted on by a force of p lbs. and the corresponding acceleration to be/; — suppose, as in the above instance, that the body is placed on a smooth horizontal plane, and urged along a straight line on the plane by the force P; — then in each successive second of its motion its velocity will be increased by a velocity of/ feet per second, where/ is given by the proportion w : p:: 32-1912 :/ In the following examples 32 will be used as an approxi- mate value of 32-1912. It foUows from the remark already made (Art. 116) that the formulae previously given for falling bodies will be true in the present case when / has been substituted for g. Thus we shall have V=ft s = \ft^ v^=2fs &c. Ex. 603.' — A body weigliing 30 lbs. slides along a smooth horizontal plane under a constant force of 15 lbs.; determine — (1) the additional velocity it acquires in every second ; (2) the velocity it Tpill have at the end of 5 seconds ; (3) the distance it will pass over in 5 seconds. ^»s. (1) 16. (2)80. (3) 200 ft. a phenomenon which were due to disturbing causes, and thus was enabled to get at the fundamental principles. The experimental verification of these principles is nearly always indirect, and consists in comparing actual cases of motion (e.g. that of planets, of pendulums, &c.) with the secondary principles which have been derived from them. 240 PRACTICAL MECHANICS. Ex. 604. — A mass -weighing w lbs. is urged along a rough horizontal plane by a force of p lbs. acting in a direction parallel to the plane : the coefficient of friction is /i ; if the body's velocity is increased in every second by/, show that ■w Trhere ^ denotes 321912 or (approximately) 32. Ex. 605. — A weight of 100 lbs. is moved along a horizontal plane by a constant force of 20 lbs. ; the coefficient of friction is 017 ; determine — (1). the distance it will describe in 10 seconds ; (2) the time in which it will describe 200 ft. ^ms. (1)48 ft. (2) 20-4 see. Ex. 606. — A train weighing fiO tons is impelled along a horizontal road by a constant force of 560 lbs. ; the friction is 8 lbs. per ton ; what velocity will it have after moving from rest for ten minutes, and what distance will it describe in that time ? * Am. (1) 17^ miles per hour. (2) 7714 ft. Ex. 607. — If in the last example the steam were cut off at the end of the 10 minutes, how many seconds will elapse before the train stops, and how far will it go ? ^«3. (1) 225 see. (2) 2893 ft. Ex. 608. — A train is observed to move at the rate of 30 miles per hour, the steam is cut off, and it then runs on a horizontal plane for 10,000 ft. ; find how many lbs. per ton the resistances amount to supposing them inde., pendent of the velocity. [It is easily shown that/=0'0968 ; then the resistance (p) in lbs. per ton (w) is found to equal 6'776 lbs.] Ex. 609. — A sphere lies on the deck of a steamer and is observed to roll back 20 inches ; if the resistance to rolling is the ^th part of its weight, determine the change in the velocity of the steamer. Am. 2'309 ft. per see. 118. The motion of connected bodies. — The meaning of the term reaction has been already explained in Art. 28, where the law is stated that when a body (a) acts on another body (b) the action is mutual ; whatever force a exerts on B, B exerts an equal opposite force on A. It f If the resistances which oppose the motion of the train were constant, it would be possible to attain any velocity, however great ; in reality the resistance of the air always imposes a limit to the velocity that can be' attained by a train moved by a force that exceeds the frictions by any given amount ; thus Mr. Scott Eussell's formula for the resistance con- tains a term involving the square of the velocity of the train (Eankine, p. 620). CONNECTED BODIES, 241 must be understood that this law is perfectly general, and is true whether the bodies are at rest or in motion. Sup- pose that two bodies, whose weights are p and Q, are connected by a very fine weightless thread, supported by a smooth point on which it hangs. If p is the heavier of the two, it will descend, and in doing so it will di"aw Q up. If the question is to determine the velocity acquired and the distance described by the bodies in a given time, we may proceed thus : — The system whose weight is P + Q is moved by the excess of the weight of P over that of Q ; hence if/ is the accelerative effect of that force we have -p + Q : 7-Q::g : f and as this proportion gives / the question can be an- swered easily. But if the question is to find the force which causes p's motion or q's motion, we must proceed as follows : — If p acts on Q with a force t, q will react on p with an equal opposite force, and, as both these forces are transmitted along the thread, T is the tension of the thread. Now the velocities of the bodies are always equal, therefore the acceleration of p's velocity must equal that of q's velocity. But p moves downwa,rd under a force of P — T pounds, and Q upward under a force of T — Q pounds. Therefore if / is the required acceleration we have /=I:^ and Z^^-HS g p 9 Q whence we obtain , v—Q J 2pq /= .g and t = — -^ P + Q "^ p+Q The value of / is the same as that previously deter- mined, and T is the value in lbs. of the force with which p acts on Q, and of that with which Q reacts on P. Ex. 610. — If in the case explained in the last article p and q weigh 12J lbs. and 11^ lbs. respectirely ; find (1) the acceleration of p's and a's R 242 PEACTICAL MECHANICS. velocity, (2) the distance they would describe in 5 sec. from a state of rest, (3) the tension of the thread. Jm. (1) 11. (2) 16| ft. (3) llg lbs. Ex. 611. — ^A weight q is tied to a string, and rests on a rough horizontal table ; to the other end of the string is tied a weight p which hangs ver- tically over the edge of the table ; if the weight of the string and its fric- tion against the edge of the table are neglected, show that when r falls it accelerates q's velocity in every second by/, where ■' p + nd 353-1 ft, Ex. 664. — A body during the 2nd, 6th, and' 7th second of its motion describes respectively 16, 24, and 46 feet. Is this consistent with uniform acceleration? Ex. 666. — A body moves from rest under the action of a certain force ; at the end of 6 sec. the force ceases to act, in the next 4 sec. the body describes 1 80 ft. Pind the accelerative effect of the force. Atis. 9. 1 --. COMPOSITION OF VELOCITIES. 257 126. Composition of velocities. — Suppose a body to be at any instant at the point A, and suppose it to be moving with such a velocity as would in a ria.i69. certain time carry it to B along the line AB; suppose that at that instant another velocity were communicated to it such as would in the same time carry the body along the line AC to c, if it had moved with that velocity only ; complete the paraUelogram A B c D and join a d, then at the end of the given time the body will arrive at d, having moved along the line ad. That this is so appears at once from the well-known fact, that when a ship is in a state of steady motion, a man can walk across her deck with precisely the same facihty as if she were at rest ; thus if he were to walk across the deck when the ship is at rest he would go from a to c ; but if we suppose the ship to have such a velocity as will in the same time carry the point A to b, he wiU come to the point D ; and if the velocities have been uniform he will have moved along the line a d. Now, let v and u be the two velocities, then AB : Ac::u : v and if V is the velocity compounded of them AD : ab::v : u So that if AB and AC represent the given velocities in magnitude and direction, a d will represent the velocity compounded of them in magnitude and direction. Hence the rule for the composition of velocities is the same, mutatis mutandis, as that for the composition of forces. If the velocities vary from instant to instant, the rule will give the magnitude and direction of the component velocity at any instant ; this is the case which commonly happens, for example, when a body is thrown in any s 258 PEACTICAL MECHANICS. direction transverse to the action of gravity. The method of determining the motion of the body may be described in general terms as follows : — Conceive the time to be divided into a great number of intervals, and suppose the velocity that is actually communicated by gravity during each interval to be communicated at once,* then, by the com- position of velocities, we can determine the motion during each interval, and therefore during the whole time ; the actual motion is the limit to which the motion, thus determined, approaches when the number of intervals is increased. Proposition 25. A body is thrown in vacuo with a given velocity (v) in a direction making any angle with the horizon, to determine its position at the end of any given time (t). Let the body be projected along the line an ; take an equal to v t, and divide t into n parts, each equal r ; then if A N is divided into the same number of equal parts in pio. 160. Nj, Nj, Nj . . . each part will equal v t. Now, the effect of gravity is to increase the velocity of a falling body by 5^ T in a time t ; we may therefore conceive the body to move through each interval with the velocity it has at the beginning of that interval, and at the end the velocity to be compounded with gr. During the first interval the body will move over the dis- tance AN, ; draw N, m, vertical and equal to grr x t ; complete • It is immaterial whether ■we conceive it to be communicated at the beginning or at the end of the interval. MOTION OF A PROJECTILE. 259 the parallelogram n, n^ ; then since the sides Ttf^n j and N, Nj are proportional to the velocities g r and V, the body will, during the next interval, move along the line N, Q, and at the end of the interval will arrive at a point Q vertically under Nj; the actual velocity with which the body has moved being, of course, equal to Nj q-t-t. At the point Q we have to compound this velocity with gr; to do this we must produce n, q to r, making Qr equal to N, Q ; take QWj equal to grxr, and complete the parallelogram, then the sides of this figure are proportional to the component velocities, a,nd therefore the diagonal is in the same pro- portion to the velocity compounded of them ; at the end of the third interval therefore the body will be at E verti- cally under Nj ; the same construction will apply to any number of intervals, and the required point p will be vertically under N. To determine N p ; produce n,q torn,, Q R to Tn^, R s to mg, &c., then will N p equal the limit of the sum of Nm,, m^in^, m^m^, &c. ; but by similar triangles N TTi/j is the same multiple of N^ Q that Nj N is of N, Nj, therefore Nmj equals (w — 1) gr t^, similarly m, m^ equals (»— 2) gT^, in^'m^ equals (n — 3) gr^, &c., and therefore their sum equals {n-l)g7^-\-{n-2)gT^ + {n-3)gT^ + . . . + 2gT^ + gT' =Sr t''{(w-1) + (w-2) + . . . + 2-M} _ t^ nin-l) _, ,f 1\ ~^n' 2 '^ V nJ Now, however great the number of intervals, Q, R, s, &c., will remain vertically imder N^, N3, N^, &c., so that in the limit P will remain vertically under N.. Also the limit of igt^[l ) is igt^; therefore the true position of the s 2 260 PBACTICAL MECHANICS. body will be found by measuring downward from N a dis- :tance equal to ^g f* Ex. 666. — A. body moves along a smooth horizontal plane with a velocity of 3 ft. per second ; at the end of 2 seconds a velocity of 8 ft. per second is impressed on it in a direction at right angles to its motion ; after how long will its distance from the starting-point he 20 ft. ? Ans. i sec. Ex. 667. — ^A body is projected in vacuo with a given velocity in a given direction ; determine its range on the horizontal plane passing through the point of projection and the time of flight. Let A be the point of projection, an the direction of projection, ab the horizontal plane through A, let the velocity of projection be denoted by v, and the angle nab by a. Let the body strike the plane at b after T seconds ; we have to determine a b and t ; draw B N at right angles to A b, then (Prop. 25) Fig. 161. and NB = ipound equals 32-1912 absolute units of work. Let a particle whose mass is m pounds be moving at any instant with a velocity of vfeet a second, and suppose it to be brought to rest after describing a distance of h feet from that instant by a force of p absolute units acting in a direction opposite to that of its motion. If / be the accelerative effect of p on M we have the following rela- tions : — p=m/ and 2fh=v^ therefore Fh = ^uv^ Now P A. is the number of absolute units of work done against p, and consequently -^ M v^ is the kinetic energy of the body,- or its capacity of doing work estimated in absolute units of work, when M is in poimds and v in feet per second. If the kinetic energy is required in foot- pounds its value in absolute units of work must be divided by 32-1912 (Art. 119). It will of course follow from this that if the velocity of a particle is increased from v to v, a force or forces must have acted which have done on it ^ M (v^—v^) abso- lute units of work ; and if its velocity is diminished from V to v it must have done against a force or forces ^ M {y^—v'^) absolute units of work. In the following examples, which illustrate the subject of Arts. 128, 129, the numerical value of g in London is taken strictly as 32'1912 ; in other parts of the book the appproximate value 32 is used except where the contrary is stated.* * The pound has been used as the unit of mass throughout the remain- ing pages of the hook. This accounts for the answers to several of the questions differing from those given in some of the previous editions. THE ABSOLUTE UNIT OF FOECE. 271 Ex. 687. — A force estimated in British absolute units is 90 ; what would be the numerical value of the force if the unit of mass were 1 cwt. and the units of distance and time a yard and a minute ? Ans. 964-3. Ex. 688. — How many British absolute units would there be in an abso- lute unit of force if the kilogramme, m&tre, and second were employed as the units of mass, distance, and time ? A kilogramme equals 2-2066 lbs. and a mitre equals 39-37043 inches. Ana. 7-236. Ex. 689. — Show that a British absolute unit or poundal equals 1 3828 dynes, and that a foot-poundal equals 421394 ergs; having given that a metre equals 3937043 inches and a gramme 16-43235 grains. Ex. 690. — A mass of 500 lbs. is acted on by a force of 126 absolute units ; what distance will it describe from rest in 8 seconds ? Ans. 8 ft. Ex. 691, — A cubic foot of cast iron is observed to increase its velocity by 5 ft. per second in every second of its motion ; determine the force which produces this acceleration, assuming that a cubic foot of water weighs 1000 oz. Ans. (1) 2252-2 absolute units. (2) 6996 grav. units. Ex. 692. — A body whose mass is 10 lbs. moves from rest along a straight line imder the action of a constant force ; during the first second of its motion it describes a distance of 60 ft. What is the force which pro- duces the motion ? What is the mass of the body which that force would just support in London ? What in Trinidad ? (Table XV.) ^»s. (1) 1000 absolute units. (2) 31-06 lbs. (3) 31-16 lbs. Ex. 693. — The aceelerative effect of the moon's attraction on a particle situated on its surface is about 5-4.* A man can jump to a height of 6 ft. on the earth's surface ; how high could he jump on the moon's surface ? Ans. 29-6 ft. [In both cases the mass of the man's body is the same, and the force exerted by the muscles is the same, quite irrespectively of the force of gravity ; consequently, in the act of jumping, the velocity with which he leaves the ground is the same in both cases.] Ex. 694. — Suppose that near the surface of the moon, a mass of 5 lbs. were placed on a plane, and it were observed that the plane descended with a uniform acceleration of 3-6 ft. per second ; what pressure would the mass produce oq the plane in absolute units ? How much matter would a force equal to this pressure support in London? ^»s. (1) 9. (2) 4-47 oz. ^jp_ 695. — If near the moon's surface two bodies are connected by a fine thread passing over a smooth fixed cylinder, the former, containing 30 lbs. of matter, falls and pulls up the latter, which contains 20 lbs. of matter ; find the tension of the thread (1) in absolute units, (2) in gravitation units, Ans. (1) 129-6. (2) 4-C2. * Herschel's Outlines of Astronomy, Art. 608. .272 PRACTICAL MECHANICS. Ex. 696. — A mass of 7 lbs. falls through 100 ft. near the surface of the moon ; what kinetic energy does it acquire in the fall ? Ana. 3780 abs. units, or 117'4 ft. -pounds. Ex. 697. — If equal forces (p) act on two unequal bodies for the same time, show that the bodies will acquire equal momenta. Ex. 698. — Show that momenta of the bodies in the last example will be equal when p varies from instant to instant, provided the forces are the game at the same instant throughout their time of action. [The results of the last two examples are of considerable importance ; they are almost self-evident and therefore liable to be forgotten— for this reason the student's attention is particularly directed to them.] • Ex. 699. — When the powder in the bore of a cannon is ignited, the pressure on the ends of the bore and on the shot are at each instant equal : a shot weighing 6 lbs. is fired from a gun quite free to more and weighing 6 cwt. ; the velocity with which the shot leaves the gun is 1000 ft. per second ; what is the velocity of the gun's recoil ? Ans. 8-93 ft. per sec. Ex. 700. — Show that the kinetic energy of the gun is always small compared with that of the shot ; and ascertain their values in foot-pounds in the case suggested in the last example. Ans. 93,760 ft.-pds. in the shot and 837 in the gun. Ex. 701. — If the trunnions of the gim in Ex. 699 are supported on two parallel smooth planes inclined at an angle of 30°, determine how far it will move along these planes. Ans. 2-5 ft. Ex. 702. — The mercurial barometer (when placed on the sea level) stands at any given height ; suppose the force of gravity to undergo any change either of increase or decrease ; explain why this, unaccompanied by any other change, would not alter the height of the barometer. Ex. 703. — There are two mercurial barometers placed on the sea-level ; the height of the mercury in the one is 30 in., and there is a perfect vacuum above it ; the height of the mercury in the other is 28 in., and there is an imperfect vacuum of 6 in. above it ; suppose the force of gravity to change from 32 to 24, at what height will the mercury now stand in the second barometer ? Ans. 32 — 2 a/5 in. 130. Motion on an inclined plane. — Let a particle whose mass is M lbs. slide on a smooth inclined plane. The. force of gravity on the body, which is ag absolute units, may be resolved into two forces M gr sin a along, and M g cos a at right angles to, the plane, where a denotes the inclination of the plane to the horizon. The latter of MOTION ON AN INCLINED PLANE. 273 these forces is neutralised by the reaction of the plane, and consequently the motion is caused by the former force; If, therefore, / denotes the acceleration of the velocity of the body when moving down the plane, m/ is the force causing motion,, and therefore f=g cos a If the plane is rough there will be a friction (lUg cos a, where fi stands for the coefficient of friction between the body and the plane ; consequently in this case the force causing motion \sM.g sm a— fiug cos a; and as this must equal m/ we must have f=g sin a— fig cos a In like manner if the body moves up the plane the retar- dation (/) is given by the equation f=g sin a + fig cos a If the angle of friction is denoted by ^, the last two equa- tions may be written /cos ^=g sin (a—=g sin (a + ) The reader will not have failed to observe the distinc- tion between the case of a body moving upward against gravity and that of a body sliding on a rough plane against friction. In the former case, when the upward velocity of the body is exhausted, it immediately begins to move downward. In the latter case the body simply comes to rest. Friction has no tendency to reverse the direction of the motion ; it tends merely to stop the motion. Ex. 704.— Pind the velocity acquired by a body in descending a smooth inclined plane 50 ft. long and having an inclination of 23° ; determine also the velocity that would be acquired if the limiting angle of resistance -were 16°. Ans. (1) 35-4 ft. per sec. (2) 21-5 ft. per sec. T 274 . PEACT'ICAL MECHANICS. Ex. 705. — There is a plane 50 ft. long and inclined to the horizon at an angle of 30° ; the limiting angle of resistance bet-ween it and a given hody is 15° ; determine the velocity the hody must have at the foot of the plane so as just to reach the top, and the time it ■will take to get there. Ans. (1) 48-4 ft. per see. (2) 2-07 sec. Ex. 706. — A body just rests on a plane which is inclined at an angle of 30° to the horizon. Find the velocity acquired and the distance described from rest in 2 sec. by the body ■when the plane is inclined at an angle of 60° to the horizon. Am. (1) 36-9 ft^per sec. (2) 36-9 ft. Ex. 707. — If a body slides do^wn a gentle incline of 1 foot vertical to m horizontal, show that the acceleration very nearly equals ( — — /*) 5'. And if OT = 100 sho^w that the error equals about ^mss V^^^ °^ 'he whole. ' Ex. 708. — A train moving at the rate of 24 miles an hour comes to the top of an incline of 1 ft. in 350 ; the resistances are 8 lbs. per ton ; the steam is Cit off at the top of the incline, and the train comes to rest at its foot ; determine — (1) the retardation of the train's velocity ; (2) the length of the incline ; (3) tjie time of motion. Am. (1) iIj. (2) 27,104 ft. (3) 1640 sec. Ex. 709. — At the slide at Alpnach the first declivity has an inclination of 20° 30' and is SOO ft. long ; being kept continually wet the limiting angle of resistance is 14° ; in how many seconds would a tree descend this first declivity were it not for the resistance of the air ? Ans. 14-3. sec. Ex. 710. — A body slides from rest down a plane whose inclination is i and length L ; it passes ■with the velocity acquired during the descent of the first plane to a second whose inclination i is less than the limiting angle of resistance ip; if t is the distance through which it slides before Com- ing to rest show that sin ((p—i) Ex. 711. — A train beginning to run down an incline with a given velo- city, will go before stopping » times as far as it would do if it had begun to run up the incline with the same velocity ; show that the tangent of the angle of inclination of the plane is /t (» -^ 1 ) 4- (» + 1). Ex. 712. — In the last example, if the friction is 8 lbs. a ton and the in- cline 1 in 420, show that it will run down the plane before stopping five times farther than up the plane ; that the time which the train takes in coming to rest is also five times as great in the latter case as in the former; Ex. 713. — Two planes equally inclined to the horizon ascend from the same point on opposite sides of the vertical line ; a particle slides down the former and passes without change of velocity on to the seconds, up ■which it slides till its velocity is exhausted ; it then elides down the secondand MOTION ON AN INCLINED PLANE. 275 up the first ; if l and I are the lengths of the first plane described in the descent and ascent respectively, show that L sin" (o— (()) = Z sin" (a + 0) and that the time of the motion is 2f s in acos (^ f ^ /sva{a—if)\ sin(o + ^) "L V sin (o + <())J where t is the time of descent of the first plane. Ex. 714. — ^A body whose mass is M lbs. is pulled up an inclined plane by a force of vg absolute units that acts parallel to the plane ; show that the acceleratire effect equals ( -- °'° (° + '^A n where o is the angle of in- Vm cos ^ / ° clination, and i^ the limiting angle of resistance. Ex. 715. — Let AC, CB be two planes sloping downward in contrary di- rections from the point c, and inclined to the horizon at angles A and b respectively ; a mass p slides down c a and draws a mass a up c b by means of a fine cord which passes over c and is tied to each body : if the limiting angle of resistance between the' bodies and the planes is ) (p + a) cos <^ Ex. 716. — In the last case if the inclines are equal and small, being 1 in m, show that Ex. 717. — If the resistances are 8 lbs. per ton and the incline 1 in 140, and a set of full trucks is required in their descent to pull up the incline an equal number of similar empty trucks, show that the contents of each truck should on the average be more than double the weight of the truck. Ex. 718. — If a circle be placed with its plane vertical, and through its highest point any chord be drawn, a body will descend along that chord (supposed to be smooth) in the same time as down the vertical diameter. Ex. 719. — If through any point there is drawn a vertical line and any number of inclined planes on the same side of the line, and having a common limiting angle of resistance ip ; then if bodies begin to slide from the point down these planes at the same instant, show that after any interval they will be found in the arc of the segment of a vertical circle cut off by the vertical line which subtends at the centre an angle equal to v-2tp. 131. If we are only reqtiired to find a relation between the velocity of the body and, the distance it describes on T 2 276 PRACTICAL MECHANICS. an inclined plane, we may reason on its kinetic energy ; thus : — Suppose it is at any instant moving with a velocity V, and after it has moved up the plane a distance h let its velocity have changed to v ; the change in its kinetic energy is -^MV^— -^ M-y^, and this by Ex. 522 must equal the number of absolute units of work done against gravity (^gh sin a) and the number done against friction {fiug h cos a) during the motion. Therefore ^ M (y^—v^) =M.gh (sin a + /i cos a) ; and we may reason similarly in other cases. Ex. 720 A train moving at the rate of 15 miles an hour comes to the foot of an incline of 1 in 300, resistances 8 lbs, per ton ; if the steam is cut off, how far wiU it go before stopping ? Am. 1095 ft. Ex. 721. — A body slides do'wn an inclined plane the height of which is 12 feet and length of base 20 feet; find how far will it slide along a hori- zontal plane at the bottom, supposing the coefficient of friction on both planes to be |, and that it passes from one plane to the other without loss of velocity? ^»«. 52 ft. Ex. 722. — A train weighing 90 tons comes to the foot of an incline of 1 in 160 with a velocity of 30 miles an hour; the resistances are 7 lbs. per ton, the length of the incline 2 miles ; the train has at the top of the incline a velocity of 20 miles an hour ; how many foot-pounds of work have been done by the steam in getting the train up the incline ? and through how great a distance would the same expenditure of work have taken the train with a uniform velocity along a horizontal line ? Am. (1) 16,570,400 ft.-pds. (2) 26,302 ft. Ex. 7i3. — If a train begins to descend the incline in the last example with a velocity of 20 miles an hour, how far will it descend by its own ■weight before acquiring a velocity of 30 miles an hour? Ans. ,5378 ft. Ex. 724.— There are two points a and b on a railroad four miles apart on the same horizontal line ; the railroad is in two equal inclines, one up and the other down, of 1 in 160 ; the train which weighs 50 tons and ex- periences resistances equal to 7 lbs. per ton, has a velocity of 30 miles an hour at A and b, and a velocity of 20 miles an hour at the top of the incline ; the velocity being supposed to change uniformly from 30 to 20, and again from 20 to 30, and when the latter velocity is attained further acceleration is checked by putting on the break; determine— (1) theloss of work infoot- pounds ; (2) the loss of time due to the inclines. Ans. (1) 1,810,000 foot-pounds. (2)72J sec. Ex. 72.5. — A chest 6 ft. long and 2 ft, square stands on its end on the NEWTON'S LAWS OF MOTION. 277 deck of a ship, one face being perpendicular to the direction of the motion ; the ship is suddenly brought to rest ; what must have been its Telocity if the chest is just overthrown, it being supposed that all sliding is prevented 1 Ans. 2-2 miles per hour. [If V is the required velocity and m the mass of the chest in pounds, its kinetic energy in absolute units is Jiiv^ ; and to overtliruw the chest re- quires Mjr ( a/10 - 3) absolute units of work.] Ex. 726. — Show from the principles of the present article that the velo- city acquired by the bodies in Ex. 715 while movingfrom rest over a length I of the planes is given by the formula ^2 = 2flZ . Psin(A-y)-asin(B + ( 1— cos -^) = cd sin=— V T / T but 2 CN.CH = cp2 and 2cD .CH = c a' therefore « = «, sin ] T Ex. 742. — If V is the velocity at c, and v that at a time f, show that 2irt D = veos — - T Ex. 743. — In Ex. 741 and 742 obtain the value of s and v, when t is reckoned from the instant the body is at a. [For t write t + Jt.] Ex. 744. — A body vibrates in a small circular arc, the velocity at the lowest point being v ; when at its lowest point it has communicated to it a velocity v at right angles to its plane of vibration ; show that its plane of vibration is turned through an angle of 45°, its arc of vibration is increased from s, to s, Vi, and its time of oscillation is sensibly unchanged. VIBRATIONS OF A ROD. 287 Ex. 745. — In the last case, if the velocity is communicated to the body when at A, show that it will now desorihe a horizontal circle round c, whose radius is s, and time of description 2ir • / Ex. 746. — If the angle A. s c (fig. t68) is denoted by 6, the mass of the body by m, and the tension of the thread when the body is at c by T, show that T = Mg (3 — 2 cos 8). Ex. 747. — In Ex. 741 show that the acceleration along the curve is^- Hence when a particle, whose mass is m, vibrates under the action of a force H s, where a is the distance of m from the middle point of the arc or line of vibration, show that the time of vibration equals tt . /— . 133. Longitudinal vibrations of a rod. — A rod sus- pended by one end whose length is L, area of section K, and modulus of elasticity E, has attached to the other end a particle whose weight is Q (which we will suppose so large that the weight of the rod can be neglected) ; if the rod is allowed to lengthen slowly, Q will descend through a small distance I equal to L q-t-k e and will continue at rest (see Art. 6 and Ex. 149) ; if, however, Q is allowed to descend at once, a certain number of units of work will be accumu- lated in it when it has descended through the distance I, so that it will continue to descend till the resistance to further elongation shall have neutralised them; a contrac- tion will then ensue, and thus Q will vibrate in a vertical line about the point (a), at which in the former case it would have come to rest. Ex. 748. — Show that when the particle, which contains a lbs. of matter, is at a distance a from a it is moving under a force that varies as a, and that the time in which it proceeds from the highest to the lowest point is 9^ Ex. 749. — ^In the last example suppose Q to be at a distance a below A ; determine the number of units of work accumulated in it at that instant, and show that its velocity (1/) is given by the equation 288 PRACTICAL MECHANICS. [See Ex. 149. The student must bear in mind that the quantity gx.s is the Modulus of Elasticity reckoned with reference to the whole cross- section of the rod and in absolute units. The same is true of the answer to Ex. 748.] Ex. 750. — If a cylinder whose height is h and specific gravity s floats with its axis vertical in a fluid whose specific gravity is «„ show that if it is depressed through any distance the time in which it will rise from its point of greatest depression to its greatest height is constant, and will be given by the formula /T 9 h t=. Pi 280 CHAPTER V THE MOMENT OF INERTIA. Bef. — If we conceive a body to consist of a large number of particles, and multiply the mass of each by the square of its perpendicular distance from a given line or axis, the sum of all these products is the moment of inertia of the body with respect to that axis. 134. Properties of the imoTnent of inertia. — It will appear hereafter that the moment of inertia is a quantity that enters into nearly every question in which the rotatory motion of a body is concerned ; the present chapter will be devoted to proving some of its properties, and ascertain- ing its magnitude in certain particular cases. The first property we shall notice is one that follows immediately from the definition. Since the mass of a particle and the square of its perpendicular distance from a given axis are essentially positive, their product must be so too ; consequently if we conceive any group of" particles to consist of two or more subordinate groups, the sum of the moments of inertia of these separate groups with respect to a given axis will equal that of the whole group with respect to the same axis : hence if a body can be divided into a certain number of parts, and their respective mo- ments of inertia are knOwn with respect to a certain axis, that of the whole body, with respect to that axis, is found by adding them together. u 290 PEACTICAL JIECHANICS. Proposition 30. If I is the tnommi of vnertia of cmy hod/y whode mass is a, about am, axis passing through its centre of gravity, Fig. 169. cmd ij the moment of i/nertia of the same hod/y about a pa/rallel axis situated at a perpemUeula/r dAetta/nce h from, the former, then Suppose the axes tx> be perpendicular to the plane of the paper, let the axis which passes through the centre of gravity meet that plane in 0, and let the other axis meet it in Oi ; let p be one of the particles of which the body is conceived to be made up, and let its mass be m, j join p o, p Oj, and o Oj, and draw p N at right angles to o o, ; then (Eucl. II. 13) 0,P^=OP' + 0,^ — 2 0, . ON. Let OPssr-,, 0, P=r/, o N=a!„ and o o,=A, then m,^r^"' =m,^r^^ + m^h^~2mjwci^ (1) and the same algebraical formula will be true whatever be the position of p ; hence if m^, r^, r^, x^, m,^, r^, r,, x^, &c., . . . are the magnitudes corresponding to other parti- cles we shall have m^r/^ = m^r^ + mji^ — 2mjix^ m^r^'^ = m,^^ + m,^^ — ^m,^^ (2) (3) and so on for every particle. Now by the definition m,-;r-l^ + m^^"^ + m,^^"^ + . m^r^->rm^^-\-m{r^ +. , also mj + m2+ma+ . . . and by the properties of the centre of gravity (Prop. 16) m, a;, H-mjXj \-m,^x^ . . .=0 since the weight of each particle is proportional to its mass. = 1 = M MOMENT OF INERTIA. 291 Therefore by adding the equations (1), (2), (3), &c., we obtain Ij=i + mA2 Q. E. D, Fig. 170. y e X^ N « Proposition 31. If a/ny number of particles Ue in a plcme, cmd if ij and ij are respeativeVy thei/r moments of inertia about two rectangular axes in that plane, and if i is their moment ofi/nertia about an axis perpendicular to the two others, and passing through their point of i/ntersection, then 1=1, + 1^ For let oXfOyhe the two axes, the third being perpen- dicular to the plane of the paper, and passing through o ; let p be one of the particles whose mass is m ; draw P M and p N at right angles to oy and o x, join o P, and let p M = x, vs=y, 0'P=r, then .•,7nr^=mx^ + my\ (1) Similarly, if other particles are taken, and the corre- sponding magnitudes are m,, r„ x-^, y^, m^, r^, x^, y^ , we shall have m^r^^=m^x^^ + m,^j^ and so on, whatever be the number of particles. mr^ + m^r^^ + m2r^'^+ . . . . =i 'mx^ + m,^x^^ + m^x^^ + . ... =i, my^ + m^y^^ + m^y^^ + . . . . =1^ Therefore, adding together (1), (2), (3), &c., we obtain 1=1, + I2 Q. E. D. (2) (3) Now 292 PRACTICAL MECHANICS. rio. 171. Ex. 751.— If i: is the area of the section of a tbin rod, p the density of the material, and I its length, show that its moment of inertia about an axis pass- ing through one end and perpendicular to it equals g p ^ ^'. [If A B is the line, and a pyramid is constructed whose base bd is a square the side ofwhieh equals ab and its plane perpendicular to A b (compare the end of Art. 82) ; then if we consider a lamina contained by planes drawn parallel to the base through the extremities of any small portion vp of a b, its volume will ultimately equal ■pp xap^; now, the moment of inertia of vp equals mass of rp x ap^ i.e. it equals p& x vol. of lamina ; and hence the moment of inertia of the rod equals p hx volume of the pyramid.] Ex. 752. — The moment of inertia of the rod in the last example about an axis perpendicular to its length and passing through its middle point equals r- . p A ?. [See Prop. 30.] Ex. 753. — There is a rectangular lamina whose thickness is k, and sides a and b ; show that with reference to an axis parallel to a and passing through the middle point of b the moment of inertia equals — .pkab'. [The lamina can be divided into a number of lines whose length is b ; the moments of inertia of these can be added together by Art. 134.] Ex. 754.— If in the last example the axis is perpendicular to the plane and passes through the centre of gravity, show that the moment of inertia of the lamina equals r^.pkab (a^ + b'). [See Prop. 31.] Ex. 755. — There is a rectangular parallelopiped whose edges are a,b,c; an axis is drawn through the centre of gravity, and parallel to the edge c; show that the moment of inertia about that axis equals r^-pabo (a' + b'). [See Alt. 134.] Ex. 756. — There is a right prism whose base is a right-angled triangle, the sides containing the right angle of which are a and b, the height of the prism is c. Show that if an axis be drawn through the centres of gravity of the ends the moment of inertia about that axis equals -r^.paJc xCd' + i"). MOMENT OF INERTIA. 293 [By Art. 134 and Ex. 755 the amount of inertia about a parallel axis joining the middle points of the hypothenuses of the ends is J- . p a J c X (a^ + ¥) ; the result is then obtained by Prop. 30.] _ -Er. 757. — There is a right prism -whose height is c and base an isosceles triangle, the base of which is a and height 4 ; if an axis be drawrn passing through the centres of gravity of the ends its moment of inertia about that axis equals —.pabc(— + ^. 12 "^ \i z) [This prism can jje divided into two resembling that in the last example.] Ex. 768. — There is a right prism whose mass is m and base a regular polygon, the radius of whose inscribed circle is r, and length of one side a ; show that its moment of inertia about its goometrical axis is -5 • »r (S-)- [This prism can be divided into prisms like that in the last example.] Ex. 759. — If there is a cylinder whose height is h and radius of base r, show that its moment of inertia about its geometrical axis equals xp h r*. [If the cylinder reduces to a circular lamina whose thickness is k, the same formula is of course true.] Ex. 760. — There is a thin dicular lamina whose radius is r and thick- ness k ; show that the moment of inertia about a diameter equals -jpjirr'. [See Prop. 31.] Ex. 761. — There is a drum tie length of which is a, the mean radius of the end r, and the thickness t ; show that its moment of inertia about its axis very nearly equals 2 irpoi 9^; and that if? equals — .the error in the above determination of the moment of inertia is the — th part of that quuntity. Ex. 762. — ^There is a cylinder the length of which is h and the radius of whose base is r ; show that its moment of inertia about a diameter of one end equals irpAHf — "'"'t)- [If we consider a lamina contained between two planes parallel to the end and at distances s and x + Sx, it appears from Ex. 760 and Prop. 30 that the moment of inertia of the lamina equals i.Tpr'Sx + irpHjrSSx; whence the required moment of inertia equals the mass of a line the mass of each unit of whose length is j^ . t p r', together with the moment of inertia about one end of a line the mass of each unit of whose length is x p H.] 294 PKACTICAL MECHANICS. Ex. 763. — Determine the moment of inertia of a cylinder about a gene- rating line. Ex. 764. — Tliere is a cone the height of which is h and radius of hase r; show — (1) that its moment of inertia about its axis equals iirpAr'; (2) that its moment of inertia about an axis drawn through the vertex and perpendicular to the axis of the cone equals \ir phrH ^^ + -3- )• Ex. 765. — Show that the moment of inertia of a sphere about any dia- meter equals ^ t p r'. [The results in the last two examples cannot be easily obtained without the aid of the integral calculus.] Ex. 766. — In the mass of iron described in Ex. 1 2 let an axis be drawn passing through the end of the longer rectangular piece and bisecting those sides of the end which are 6 in. long ; determine the moment of inertia of the mass with respect to that axis.* Ana. 73,900. Ex. 767. — There is a cast-iron cone 16 in. high, radius of base 8 in.; -determine its moment of inertia — (1) about an axis through its centre of gravity and parallel to its base ; (2) about a parallel axis distant 4 ft. from the former. ^tss. (1) 37-27. (2)4509. Ex. 768. — ^Determine the moment of inertia about a vertical edge of the oak door described in Ex. 17. Ant. 460. Ex. 769.— There is a cube of oak whose edge is 8 in. long ; through the middle of it at right angles to one of its faces passes a cylinder of oak 4 ft. long and 3 in. in diameter ; the centres of gravity of the two figures coin- cide ; determine the moment of inertia of the whole about an axis passing through the common centre of gravity and perpendicular to the axis of the cylinder and also to a face of the cube. Ans. 16'54. Ex. TIO. — ^Determine the moment of inertia of the hollow leaden cylinder described in Ex. 15 about a diameter of its mean section. Ans. -6462. Ex. 771. — If a cylinder like that in the last example is fitted to each arm of the figure described in Ex. 769, determine the moment of inertia of the whole about the specified axi^ — (1) when the ends of the leaden cylinders coincide with those of the arms ; (2) when the other ends of the cylinders are in contact with the cube. Ans. (1) 200-8. (2) 28-16. Ex. 772. — Determine the moment of inertia, about the axis, of a grind- stone 4 ft. in diameter and 8 in. thick. Ans. 2244. Ex. 773. — ^There is a easb-iron fly-wheel consisting of a rim, four spokes at right angles to each other, and an axle ; the external and internal radii of the rim are 4 and 3^ ft, respectively, and its thickness 8 in. ; the sections * In this, as in all examples of moments of inertia, mass is reckoned in pounds and distance in feet. KADIUS OF GYRATION. 5395 of the spokes are each i square in., the axle 12 in. in diameter and 18 in. long ; determine the moment of inertia of the ■whole about the geometrical axis of the axle ; and also determine the error if the spokes and axle were neglected and the moment of inertia of the rim calculated by Ex. 761. Ans. (1) 50,750. (2) 1000. Ex. 774. — If the moment of inertia of any body with reference to an axis coinciding with a given line of particles be represented by i, and if the body be uniformly expanded by heat, so that the linear dimensions be- fore expansion are to those after in the ratio of 1 : 1 + n, show that the moment of inertia with reference to the axis coinciding with the same line of particles becomes (1 + a)^i. 135. The radius of gyration, — It is e\-ident from the definition of the moment of inertia of a body with respect to a given axis, that there will be, with respect to that axis, a line of a certain determinate length k, such that where i is the moment of inertia, and M the mass of the body ; the line k is called the radius of gyration with respect to that axis, and may be defined to be that distance from the axis at which the whole mass of the body may be supposed to be collected without producing any change in the moment of inertia. Thus, it is evident that in Ex. 751, 754, and 759 the values of the radius of gyration are respectively ^73' A/ ^^^3^ ' ^^^ "72 ' ^°^^'°'^^^' i^ ^ ^^ the radius of gyration of a body with reference to an axis passing through the centre of gravity, and fc, its radius of gyration with reference to an axis parallel to the former, and at a perpendicular distance from it equal to h, then it is evident from Prop. 30 that k,''=k'' + h^ It is to be observed that the moment of inertia is essen- tially a mechanical magnitude, while the radius of gyra- tion is simply a line ; now, suppose fc to be the radius of 200 PRACTICAL MECHANICS. gyration of any lamina of uniform density, the area of one face of which is A, it is not unnsual to speak of that area as having a moment of inertia ; when this is done it means that For instance, in the case of a rectangular area whose sides are 6 and c, the moment of inertia, with respect to an axis bisecting the two parallel edges c, is as is plain from Ex. 753. It is in this sense that the term Moment of Inertia is used in Prop. 21. Strictly speaking, an area has a moment of inertia no more than it has weight. 297 CHAPTER VI. d'alembert's peinciple. 136. Account of problem, to be solved. — The manner in which the dimensions of a body influence its motion may be illustrated as follows : — If we suppose a rigid bar to be suspended by one end and to oscillate, the velocities with which the different points are, at any instant, moving stand to one another in a fixed relation ; thus the free end moves twice as fast as the middle point ; moreover, with one exception, each point has a different velocity from what it would have if it were detached from the rest, and swung freely at the same distance from the centre of suspension ; this difference must depend upon the cohe- sive forces which bind the parts of the bar together. The consideration of this simple case points out the two chief additional conceptions required for the investigation of the motion of a body whose form has to be taken into account. (1) A means must be obtained for comparing the velo- cities of different points of a rigid body revolving round an axis, which is done by introducing the conception of Angular Velocity. (2) A principle is required by means of which we can avoid the consideration of the cohesive forces which hold together the parts of the body: this is generally called D'Alembert's Principle. 137. Angular velocity. — If a rigid body revolves round an axis, it is plain that the perpendiculars let fall from each 298 PRACTICAL ' MECHANICS. point of the body on the axis will, in any given time, de- scribe equal angles ; hence arises the following Def. — If a body revolves uniformly round an axis, the angle (estimated in circular vneasure) described in one second by the perpendicular let fall from any point on the axis of rotation is called the angular velocity of the body. If the velocity is variable, it is measured at any instant by the angle that would be so described if, from that in- stant, the velocity continued uniform for one second. In the following pages a and 11 are used to denote angular velocity. Kx. 775. — A body makes 30 uniform reTolutions in one minute ; what is its angular velocity ? Am, ir. Ex. 776. — A point moves at the rate of 12 ft. per second in a circle whose radius is 15 ft. ; what is its angular velocity? Am. |. Ex. 777. — Determine the angular velocity of the earth round its axis. Ane . 43,082 [See Ex. 569.] Ex, 778. — If a body has an angular velocity 2'6, how many revolutions will it make per hour ? Ans. I432'4. F,x. 779. — If a body has a uniform angular velocity a, show that the accelerative effect of the centrifugal force of a point in it, situated at a dis- tance r from the axis, is r ^ {mz + m■^z^+m^3^ + . . . } =0)21112 parallel toNjS; where y, z are the co-ordinates of Gj, and M, is the mass of the lamina d E ; if we compound these two forces, we shall obtain w^ . M, 03, as their resultant acting along Gj Np (2) Let the masses of the several laminse into which B c is divided be respectively M,, M^, M3, and let the respective distances of their centres of gravity from A o RESULTANT OF EFFECTIVE FORCES. 303 be aij, x^, Xj, . . . . then their efifective forces are severally «D^M,a;„ Qj^MjOJi,, w^Mj^g, Now, it follows from the symmetry of the figure, that all these centres of gravity are in the same plane, viz. the plane of the paper ; the effective forces are therefore parallel, and their resultant will equal their sum, viz. ; a rope (whose weight is to be neglected) is wrapped round th? pulley, and carries at its end a body whose mass is p ; given m the mass of the pulley and m k^ its moment of inertia; determine the angular velocity acquired by the pulley when p has fallen through h feet. [If the force of gravity on a mass p' at the end of the rope could just overcome the friction of the axle we should have p'^ (r—p sin i()) = M9p sin ^. SME AXON'S MACHINE, 311 This is, of course, the equation estaWished in Ex. 41 1, when the rigidity of the rope is put out of account, and o is equal to zero ; the forces, however, are expressed in absolute units. The force therefore which produces motion equals the force of gravity on a mass p— p' i.e. it equals (p-p')5', or / _ Mpsin/ in absolute units ; and this must equal the kinetic energy of the system viz. ^ (m k'' + Tr') a\ Therefore ^,_ 2ffh(pr-(F + u) pBin 6, with an error not exceeding ^th part of the true value, is due to M. Poncelet ; it may be proved as follows :— Let a = r sin €, b = r cos 8 .'. r= Va' + b'', and 6 must have some value between 45° and 90°. Now, if r' = 0-96a + 0-46 we have / = 9'(0-96 sin 9 + 0-4 cos 0); but 0-4 = 0-96 tan 22° 30', therefore r' = 0.9e Sin(9 + 22°30') ^j^^^^ ^ g increases from 45° up to 67° 30', r' cos 22° 30' will increase from 0-96 r to l'04r, and as fl increases from 67° 30' up to 90°, / decreases from l'04r to 0'96r, and consequently /never differs from r by more than __. 316 PRACTICAL MECHANICS. Ex. 8H.— A bodywhose weight is w is caused to slide on a rough hori- zontal plane by a force t ; after moving through s ft. it acquires a velocity V : show that the coefficient of friction Qi.) is given by the equation 148. Oompov/nd peTidulwma. — The terms centre of suspension and centre of oscillation have already been explained (Art. 123) ; their properties are proved in the following propositions. Proposition 34. If k[ is the radius of gyration of a body with reference to its axis of suspension, and h the distance of the centre k ^ of gravity below the centre of suspension, then -j- is the lb distance of the centre of oscillation from the latter poi/nt. Let A B be the body (whose mass is m) oscillating about an axis passing through s at right angles to the plane of the paper, which also contains the centre of gravity G ; join s G, draw the vertical line s 0, let Gj be the position of G at the commencement of the motion, draw G, Mj and G m at right angles to s c, and denote Gj s and G s c by ^, and re- spectively. Now, when SGj falls to SG the centre of gravity de- scends through a ver- tical height M, M or h (cos ^— cos 0,). If then g denotes the accelerative effect of gravity at the station, and w the angular velocity acquired, we have (Prop. 33, Cor. 2) Fig. 178. Fio. 179. COMPOUND PEKDULUMS. 317 a)''=T2L(cos ^— cos 6,), Let D p be a simple pendulum oscillating about D, draw the vertical line d e, and let p, be the position from wbich P begins to move ; draw pn and PjN, at right angles to de, and let d p be denoted by I, and let Pi DE equal 6^, and pde equal ; then if v is the velocity acquired by the point in falling from f^ to P, we have -y^ = 25r X N N, = 2gl {cos 6— cos 0^) and therefore, if w' is the angular velocity of P, we have a,'2=^ (cos ^-cos 0^). V k ^ Now, if I equals -^, a' will equal w for all values of 0, and since A B and D p are moving at each instant with the same angular velocity, their oscillations will be performed in h ^ the same time, and therefore -J- is the length of the simple pendulum oscillating in the same time as a b ; hence, if in s Gr produced a point o be taken, such that s equals J-, that point will be the centre of oscillation. Cor. — The time of a small oscillation of B a will equal ^/gh - ^ by Prop. 29. Proposition 35. The centres of oscillation cmd suspension are reci- procal. Let a B be the body, G its centre of gravity, s a centre of suspension through which the axis of rotation passes at right angles to the plane of the paper, and o the corre- sponding centre of oscillation, it is to be proved that Fia. 180. 318 PRACTICAL MECHANICS. these points are reciprocal, i.e. if o is made the centre of suspension, s will be the corresponding centre of oscillation. Let k be the radius of gyration round a parallel axis through the centre of gravity, let s G, G be respectively denoted by h and x, , I. J^^ + h^ h or hx=k^ Next, let be" the centre of suspension, and y the distance from G to the corresponding centre of oscillation, then k^ + x^ X or yx=k^ and therefore y=h,ox sis the centre of oscillation. Q. E. D. Ex. 812. — A thin rod of steel 10 ft. long oscillates about an axis passing through one end of it : determine the time of a small oscillation; the num- her of oscillations it makes in a day ; and the number it will lose in a day if the temperature is increased by 15° F. Ant. (1) 1-434 sec. (2) 60,254. (3) 3. Ex. 813. — A pendulum oscillates about an axis passing through its end; it consists of a steel rod 60 in. long, ■with a rectangular section J by J of an inch ; on this rod is a steel cylinder 2 in. in diameter and 4 in. long ; when the ends of the rod and cylinder are set square, determine the time of a small oscillation. Ans, 1-174. Ex. 814. — Determine the radius of gyration with reference to the axis of suspension of a body that makes 73 oscillations in 2 minutes, the distance of the centre of gravity from the axis being 3 ft. 2 in. Ans. 5-267 ft. Ex. 815. — Determine the distance between the centres of suspension and oscillation of a body that oscillates in 2 j sec. Ans. 20-264 ft. Ex. 816. — If -J- is the length of a simple pendulum corresponding to an h oscillating rod, show that if it expands uniformly in the proportion of 1 + a : 1 that the length of the simple pendulum becomes (1 + o) -J-. h CENTRE OF OSCILLATION. 319 Ex. 817. — Miaran determined the length of the seconds pendulum at Paris to be 39-128 inches ; he employed a ball of lead 0-533 inch in dia- meter, suspended by an exceedingly fine fibre whose -weight cotild be neg- lected ; " supposing the measurements made with perfect accuracy, upon the supposition that the distance from the point of suspension to the centre of the ball is the length of the pendulum, show that the error is less than the 0-001 of an inch. Ex. 818. — A pendulum consists of a brass sphere 4 in. in diameter sus- pended by a steel wire jg of an inch in diameter ; the centre of the sphere is 40 in. below the point of support : f determine the number of oscillations it will make in a day, and what number would be obtained on the supposi- tion that the centre of oscillation coincides -with the centre of the sphere (5' = 32). ^»s. (1) 8S,76e. (2)85,212. Ex. 819. — If a sphere whose radius is r is suspended successively from two points by a very fine thread, and if the distances of the centre of the sphere from the points of suspension are respectively h and h', and if I and I' are the distances of the corresponding centres of oscillation from the points of suspension, show that Ex. 820. — If t and f are the times of a small oscillation of the pendulum in the last example corresponding respectively to I and I', show that the acoelerative effect of gravity is given by the equation -rXh-h') / 2r^\ " P-t'- y-~6hh') 149. M. Bessel's determiTiation of the accelerative effect of gravity. — The last two examples contain the principle of the method by which M. Bessel determined the accele- rative effect of gravity at Konigsberg.f The pendulum was first allowed to swing from a point of support at a distance h above the centre of the sphere, and the num- ber of oscillations made in a given time was noted, by which t was determined with great accuracy; the wire was then grasped firmly at a point lower down, so that the oscillations were now performed about a point dis- tant h' from the centre of the sphere, and f noted as before; now h—h' being the distance between two fixed » Airy, Fiffwe of the Earth, p. 224. j; Jbd.-p. 225. { Ibd. p. 223. 320 PRACTICAL MECHANICS. points can be very accurately determined; the lengths h and hf cannot be determined without some liability to error, but as they only appear in the small term -yy/, that Oftft error will hardly affect the determination of g, which can by this method be ascertained with extreme accuracy. Ex. 821 . — In the last example let r, h, and W be respectively reckoned 1 , 50, and 40 inches, so that h—h' is exactly 10 in., but it is doubtful whether the separate values of h and h' are not as much as ^th of an inch longer than the values assigned : determine the possible error in the value of ^. Am. S. . 1,115,000 150. Captavn Kater's tnethod of determvnvng the ae- celerative effect of gravity. — This method depends on the reciprocity of the centres of oscillation and suspension ; the pendulum has two axes (or ' knife edges,' as they are called, though they are really wedges of very hard steel), by either of which it can be suspended ; now, if the time of oscillation about either axis be the same, the distance between the edges (T) will be the length of the simple pendulum; the distance, being that between two fixed points, admits of very accurate measurement, and then g is obtained by the formula The diflficulty of giving the edges their exact position is overcome as follows : — On the pendulum rod is placed a weight that can be moved up or down by screws; the edges are fixed as nearly as possible in the right position ; and then by moving the weight up or down, the values of ky and h can be changed until h^-r-h equals the distance between the edges, i.e. until the number of oscillations made in a given time about either edge is the same. 821 CHAPTEE VIII. ON IMPACT. 151, Impulsive action. — Suppose a sphere A to overtake a sphere B, their centres moving 'in the same line; it is a matter of common observation that they will strike, and then separate, A moving after impact with a le^s, and b with a greater, velocity than before ; the problem we are to solve is this: — Given the masses of the bodies and their velocities at the instant before impact, to determine the velocities they will have at the instant after impact. Now, it will be observed that though the bodies are in contact during a very short time, yet that time is really finite, and the pressure which the one exerts on the othei' must increase from zero at the instant of contact, till it attains a very considerable magnitude, and must then decrease down to zero at the instant of separation.- More- over, it appears from Ex. 698, that if a exerts at each instant against b a pressure equal to that which B exerts against A — in other words, if the action and reaction are equal and opposite forces — then the momentum lost by A must equal that gained by b, and the total amount of momentum, in A and B before impact must equal the total amount after impact. Now, that this is a fact was ascer- tained by numerous experiments made by Newton,* and this we shall take as our fundamental principle, viz, tJiab the moinentuTn lost during the impact by one body equals that gained by the other. To prevent misunderstanding, * Introduction to the Frincvpia, Y 822 PRACTICAL MECHAKICS. it may be added that the sum of the momenta of the two bodies means their algebraical sum. 152. The mean pressure exerted duri/ng impact. — ^The following example is intended to illustrate the fact that during impact there is really called into play a very large force which is exerted during a Very short time. Ex. 822. — A hard mass weighing SO lbs. falls from a height of 6 ft. on to a plane surface which at the instant of greatest compression has yielded to the extent of Jgth of an i nch — the mass itself being supposed to be entirely uncompressed ; determine the mean mutual pressure, and the duration of compression supposing it produced by the Ttiean pressure.* Ans. (1) 72,000 ff abs. un. (2) 0-000425 sec. . [The pressure by acting through ^th of an inch brings the mass to rest.] 153. Impact of vnelastic bodies. — When a overtaikes b, it is plain that so long as A moves faster than b, the two surfaces of contact will be compressed^ and the compression will continue to increase until A and B are moving with the same velocity ; if the mutual action then ceases, the bodies are said to be inelastic. Now, let the masses be denoted by a and b respectively, let R be the momentum lost by the one and gained by the other during impact, and let their velocities before impact be V and u, and their common velocity after impact be v ; then we obtain from the fundamental principle (Art. 151) ai;=^av — R {^l) Bli = BU + R (2) whence R=f2ilZ-Jl) (3) A + B ^ ' J AV + BU ,.^ and 'v= — - — (4) A + B ^ ^ In working examples the student is recommended to proceed from the general principle, or, in other words, to * Poncelet, Introd. a la Mec. ind. p. 166. IMPACT OF INELASTIC BODIES. 323 form and then solve the equations (1) and (2), and not to substitute particular values in (3) and (4). If a meet B, one of the velocities must be reckoned negative, and the bodies will move after impact in the direction of that velocity if v be negative. Esp. 823.-^If A weighing 2 lbs. and moving -with a velocity of 20 ft. per second overtakes b weighing S lbs. and moving with a velocity of 5 ft. per second, determine the common velocity after impact. Atts. 9| ft. per see. JEx. 824. — In the last example if the bodies had met, determine the common velocity after impact. Jsis, 2y ft. per sec. in a's direction, Ex. 825.^-In Art, 153 show that the number of absolute units of work lost during impact equals .~ . '^ ^ ;2(a + b; Ex. 826. — If a shot weighing p lbs. is fired with a velocity v into a mass of wood weighing q lbs. which is quite free to move, show that the velocity with which the wood begins to move is ; and state why this case must be one of inelastic impact. Ex. 827. — If in the last example q=«p, show that, in consequence of the impact, » units of work are lost in every 54+1. 154. Impact of elastic bodies. — It commonly happens that the mutual action does not entirely cease with the compression, but when that ends the bodies begin to re- cover their shapes, and thereby continue to press on each other till the impact terminates. Now^ let K be the mo- mentum lost by the one body and gained by the other during compression, and r' that lost and gained during expansion ; then the whole momentum lost by the one body and gained by the other will equal R + e'. But it is found by experiment that for the same substances R bears to e' a fixed ratio 1 : X;* therefore r' = \r, and r + r'= (l-|-\) r; where \ is a constant quantity depending on the materials of the impinging bodies called the coefficient of restitution ; it is often called the coefiacient of elasti- city, but must on no account be mistaken for the modulus * This follows from Newton's experiments already referred to. T 2 324 PRACTICAL MECHANICS. of elasticity (Art. 6). In the two extreme cases of inelas- ticity and perfect elasticity, \ equals and 1 respectively ; in other cases \ is a proper fraction. We have already seen that if a body whose mass is A, moving with a velocity v, overtakes another whose mass is B, moving with a velocity u, then the momentum lost by the one and gained by the other at the end of compression equals - — ^ -'. Hence the total momentum gained and lost. will equal (l+X) x ■^ ~ ' . And therefore if v and « are their respective A + B velocities after impact, we shall have A?;=AV — (1h-X)r bu = bu + (1+X)r , (I-I-X)b(v-u) or v=\— N- — -J—^ '' a + b and w = u+ 'i^ '- — ^^ '- a + b It may be added that the remarks made in Art. 153, rela- tive to the working of examples, are applicable to the case of elastic bodies. Ex. 828. — Show that v and u are given by the following formulae — AT + BTI \b(v— U) V = — ^ c A+B A+B AVH BTJ ^ Kk (V — tt) A+B A+B Ex. 829. — Determine the velocities after impact of a hall (a) weijjhing 20 lbs. which, moving with a velocity of 100 ft. per second, overtakes a ball (b) weighing 50 lbs. and moving with a velocity of 40 ft. per second, their coefficient of restitution being \. Ans. a's velocity Z^. b's veloci'y e.if. Ex. 830. — In the last case suppose the heavier body (b) to be at rest : determine the velocities after impact. Ans. A rebounds with a velocity 7y ; B moves forward with a velocity 42f . IMPACT OF ELASTIC BODIES. 325 Ex. 831. — Obtain the velocities after impact in Ex. 829, upon the sup- position that the bodies meet. Ana. A rebounds with a velocity 60, and b with a velocity 20. Ex. 832. — If there are two perfectly elastic balls A and e of equal masses, and a moving with a velocity v impinges on b at rest, show that A is brought to rest and p takes the velocity v. If there is a number of equal and perfectly elastic balls, e, c, d, e, placed in a line, what would be the result of A striking b, the direction of the impact coinciding with the line? Ex. 833. — If a ball whose weight is a moving with a velocity v meets a ball whose weight is b moving with a velocity u, show that in the case of perfect elasticity the velocities of rebound are given by the following con- struction: — Draw any line A b, divide it in G in the inverse ratio of the weights of A and b, and in c in the ratio of their velocities ; on the other side of G measure off a D equal to a c, then a's veloeitiy of rebouad : b's velocity of rebound : : A D : b d.* Ex. 834. — Two balls weighing respectively 12 and 8 lbs. are suspended by threads in such a manner that their centres are 4 ft. below the points of support ; when at rest the line joining their centres is horizontal ; if the smaller one is raised so as to fall through a quadrant, detePnine the anglp described by the other after impact, if the coefficient of restitution equals |. Ans. 66° 14'. Ex. 835. — If A and b are the weights of two perfectly elastic balls, if v and V are their velocities before impact and v and a their velocities after impact, show that AV'' + BU'' = A«'' + E»' Ex. 836. — If a ball impinges perpendicularly on a fixed plane with a velocity v, show that the velocity of rebound equals A. v. [It must be remembered that at the end of compression the velocity is entirely destroyed, consequently = a v -- r ; hence, if v is the velocity at the end of the impact a« = av— (I+\) k, whence « = — \ v.] Ex. 837. — If bodies are dropped from equal' heights on to a fixed hori- zontal plane, show that their coefficients of restitution are in the same ratio as the square roots of the heights to which they rebound. Ex. 838. — A ball is dropped from a height h : show that the whole- distance it describes before coming to rest equals 1-A.2 » It was in this form that the problem of impact was originally solved by Sir C. Wren (vide\MbMi!Mcfe, voL'ii. p. 411). 326 PRACTICAL MECHANICS. Ex. 839. — A ball (a) is thfown up-#ard with a velocity of 160 ft. per second ; when it has reached a height of 300- ft. it is met by an equal ball (p) which has fallen from a height of 100 ft. ; determine the time after the instant of impact in which each will reach the ground, assuming that \ equals unity. Aiia. A. after 2J sec. b after 7J sec. 155. Oblique impact of smooth bodies. — Suppose a smooth ball A, moving with a velocity V, to impinge obliquely on a smooth ball b, moving with a velocity V ; draw the line of centres, and resolve V into com- ponent velocities v, and v^, the former along, the latter at right angles to, the line of centres ; in like manner resolve u into Ui and v^ ; now v^ and u^ will remain un- changed by the impact, but v, and Vy will be changed into v^ and u^ exactly as if the bodies had impinged directly with the velocities Vj and Ui : hence, by com- pounding v^ and Vg and al^O «, and U2, we obtain the required velocities after impact. The general formulae commonly given for these velocities are of very little value, as any particular example is much more easily worked by proceeding ftom first principlies: the following example will sufficiently exhibit the method of treating these cases. , Ex. 840. — Lot A and b be two perfectly elastic balls which at the instant of impact are moving along the lines pa and (m,y,H-m23/2 + m3i/3+ . . .) —'Y = m^x^w + 10(1^^(0 + 711^x^0)+ . . . = (o{mipa^ + m^x^ + m^x^+ . . .) a E = m^r■^^(o + m^T^a + m^r^w + = m{m^r^ + m^T^ + m^r^+ . Or by Prop. 16 and Art. 134— E+X=M^ft) — Y = Ma;6a .) 830 PRACTICAL MECHANICS. •"=s-v ^'^ — ^~ (2) -=-^-0 (3) The first of these equations gives the angular velocity communicated to the body; the second and third equations give the components of the reaction of the axis, which is of course equal and opposite to the blow sustained by the axis. N.B. — It will be an instructive exercise for the student to ascertain for what positions of the centre of gravity the reactions of the axis will be as indicated in the figure : it will commonly happen, as he will find, that the reactions will in reality act in the contrary directions to those in- dicated. Ex. 848. — A uniform rod 12 ft. long and weighing 10 lbs. is suspended at one end ; it receires at the other, in a, direction perpendicular to its length, a blowwhose momentum is 32 : determine — (1) the angular velocity .with which it begins to move ; (2) the impulsive pressure on the axis ; and (3) find how manj times this impulse exceeds the blow given by a weight of 1 of a .pound which has fallen through a height of 4 ft. Ans. (1) 0-8, (2) 16. (3) 4 times. Ex. 849, — A beam of oak 10 ft. long and 1 ft. square is suspended by an axis perpendiciilar to one face and passing through the axis of the beam, at a distance of 1 ft. from the end ; it is struck at a point 8 ft. below the axis by a bullet weighing 1 lb. and moving with a velocity of 1000 ft. per second : determine— (1) the impulse on the axis ; (2) the angular velocity communicated to the beam ; (3) the angle through which the beam will re- volve. Ane. (1) 310. (2) 0-56. (3) 14° 5'. Ex. 850. — A hammer's head weighs 10 lbs. and makes 60 strokes per minute on an anvil : if the time of ascending equals that of descending, and the blow is entirely due to the velocity it acquires in falling, compare that blow with the impulse on the axis in the last example. Ans. One half. THE CENTRE OF PERCUSSION. Sal E.T. 851. — Determine the impulse on the axis if the mass of cast iron ih Ex. 791 strikes an anvil after falling through the 30°, the blow on the anvil being supposed to be given by the extreme edge of the cube. Ans. 31 SO. [It will be otserved that in this Case the impulse on the axis is greater than thut which would be produced by a shot weighing 3 lbs. and moving at the rate of 1000 ft. per second ; it is obvious that a succession of such impulses would tear to pieces the masonry on which the axis of such a hammer is supported ; and accordingly it becomes a point of great practical' importance to suspend a tilt hammer in such a manner that there shall be no impulse on the axis. The following article explains the principle on which this is done.] , 157. The centre of percussion. — Keferring to the equations (2). and (3) of Prop. 35, we see that if the blow is delivered in such a manner that x equals „ ,„ _ ^ Fig. 183. zero, and P equals a y, then x and Y equal zero separately, and there is no impulsive pressure on the axis of suspension ; hence if O be the centre of suspension, G the centre of gravity of the body, and a point o^ be taken in o G pro- duced so that 00,=^ OG then if the body be struck by a blow whose direction passes through Oj at right angles to Oj, there will be no impulsive pressure on the axis, and the point o, is there- fore called the centre of percussion ; it evidently coincides with the centre of oscillation with respect to the centre of suspension o. It must be remembered that the body is supposed to be symmetrical with regard to the plane of the paper, as specified in the enunciation of Prop. 35. 158. Axis of spontaneous rotation. — Since the body in the last article when struck begins to rotate round the axis through o without any constraint, it follows that if the body were entirely free, it would begin to move round that axis, which is therefore called the axis of spontaneous S33 PRACTICAL MECHANICS. rotation. If it is given that a body is struck by a blow R along a given line, the axis of spon tan sous rotation is de- termined as follows : — Consider the plane passing through ,G the centre of gravity and the direction of the blow ; through G draw a line at right angles to this plane, and let h be the radius of gyration of the body with respect to it: through the centre of gravity draw a line at right angles to the direction of the blow and cutting it in Oj, and on the other side of the centre of gravity take in the line a point o such that OG . GOi = P then an axis through o perpendicular to the given plane is the axis of spontaneous rotation, provided the body is symmetrical with reference to that plane. It will be observed that if the axis of spontaneous rotation is to pass through the centre of gravity, we must have in equations (2) and (3) of Prop. 35, both x = and y=0, and therefore R=0 ; but from equation (1) to having a finite value a E must also have a finite value ; or in other words the body must be struck by an impulsive couple whose moment is a R, and whose plane passes through the centre of gravity of the body ; it will then begin to revolve with an angular velocity — j- round an axis at right angles to the plane of the couple, and passing through the centre of gravity. Ex. 852. — A hammer turns round a given axis, the weight of the head is w, and its radius of gyration is k with respect to an axis parallel to the given axis and passing through its centre of gravity ; the weight of the handle is w„ its radius of gyration with respect to the axis is k^, and the distance of its centre of gravity from the axis a. If the head of the hammer is so placed that its centre of gravity is at the same distance (x) from the axis as the centre of percussion of whole hammer, then w, A,'+wF BALLISTIC PENDULUM. 333 Ex. 863. — If the head of the hammer in Ex. 851 is shifted so as to fulfil the conditions of the last example, determine the distance of its centre of gravity from the axis of rotation. Ans. 5"35 ft. Ex. 854. — A sledge hammer ab is movable round an axis through ^-^ it is 6 ft. long and weighs 4 cwt., it is held in a horizontal position by a weight of 3 cwt. attached to the end of a string which after passing over a small pulley is fastened to b (the parts of ^he string being vertical) ; the hammer when allowed to fall into a vertical position makes 50 oscillations per minute round a : determine — (1) the centre of percussion, and (2) the radius of gyration about an axis parallel to the axis of suspension and pass- ing through its centre of gravity. Ans. (1) 4-67 ft. (2) 0-87 ft. Ex. 855. — A cylindrical bolt of cast iron 4 in. in diameter and 8 in, long is struck simultaneously by two equal blows in contrary directions, each at right angles to an extremity of a diameter of its mean section ; in consequence the bolt rotates 250 times in a second : determine the magni- tude of each blow, and compare it with tha't which the bolt itself would give if moving with a velocity of 1000 ft. per second: Ans. (1) 1715. (2) ^. 159. Robin's ballistic penduluTn. — This machine is employed to ascertain the velocity with which a shot leaves the mouth of a cannon. The principle on which it is con- structed will be most easily understood by describing it in its priginal form ; at present the gun itself is suspended and the recoil observed ; but at first it was constructed as follows : — A large mass of wood is carefully suspended so as to turn freely round a knife edge (Art. 150); the shot is tired into this mass, which is backed with iron plates to prevent the ball passing through or shivering it, so that the shot stays in it, and by the blow causes it to revolve through a certain angle {d\ the magnitude of which can be ascertained by a riband attached to a point of the pendulum which is pulled through a spring suflficiently strong to keep the riband straight while the mass moves up, and also to prevent any of it returning when the mass moves back; it is evident that the length of the riband gives the chord of the arc described by the point to which it is fastened, and thus is observed ; the' weight w of the 334 PRACTICAL MECHANICS, pendulum includes that of the shot w ; the distance h of the centre of gravity of w from the knife edge is deter- mined in the manner suggested by Ex. 854. The radius of gyration is inferred from n, the number of small oscilla- tions made in a minute ; the distance (a) below the point of support of the point in which the shot strikes the pen- dulum is measured ; and it is (of course) endeavoured that this point should as nearly as possible coincide with the centre of percussion. From these data the velocity v of the shot can be found, Ex. 836, — In the ballistic pendulum shcm that 120oAw . 9 v=s— — " em ~ APPENDIX. ON LIMITS. Throughout the present work particular geometrical limits have been used instead of the formulse of the differential and integral calculus — at least, this has been done as far as possible : if the reader has not been accustomed to reason on limits, he may perhaps find a difficulty in understanding the propositions in which they occur : should this be so, the following remarks may prove useful. 1. Definition of a limit. — Let there be any variable magnitude X, and let there be a fixed magnitude a ; also suppose that x in the course of its successive changes continually approaches A, but never becomes equal to it, though the difference between the two magnitudes can be made less than any assigned mag- nitude, however small ; a is then said to be the limit of x. Thus, suppose that x denotes the area of a polygon of n sides inscribed in a circle whose area is A 4 if we continually increase the number of ddes, x will continuaily approach a ; also if we assign any magnitude, say one square inch, a polygon with a certain nuinber of sides can be found, whose area will differ from A by less than one square inch ; in like manner if -jJ^, y^-g^, &c., of a square inch had been assigned ; therefore the area of a circle is the limit of the area of the inscribed polygon. The simplest form which the reasoning on limits can_assvime is the following : — Suppose it can be proved that two variable quantities x and Y remaiu equal throughout their variation^ and suppose that x continually approaches a limit a, while y apy 336 PRACTICAL MECIIAMCS, preaches B, then it follows that A must equal b. Thus it can be proved that the area of the regular polygon inscribed in a circle equals the rectangle between the semi-perimeter and the per- pendicular let fall from the centre on one side ; now the limit of the former is the area of the circlCj and of the latter the rect- angle between the semi-circumference and semi-diameter, and therefore the area of the circle equals that rectangle ; not, the reader will observe, nearly equals it, but actually equals it. Prop. 1 supplies a good example of the same form of reasoning. 2. On ultimate ratios. — -Suppose there are two variable mag- nitudes X and y whose separate limits are zero ; what, it may be asked, is the limit of their ratio - ? The value of this limit y depends upon circuinstances, and in different cases may have 'values differing to any extent whatever. Suppose x denotes the sine of an arc, and y the length of that arc, when x continually diminishes y continually diminishes, and their separate limits are zero ; it is capable of proof that ia this case the limit of - is 2/ unity ; but if x denotes the base and y the hypotenuse of a right-angled triangle, whose dimensions continually diminish in such a manner that the angle (a) between x and y continues unchanged, then although the separate limits of x and y are zero, the limit of - is cos a ; in the former case x is frequently said to be ultimately equal to y; in the latter, x ultimately equals y cos A. As this point is of gi-eat importance, we will illustrate it by the following case : — Let a p a be a semicircle j take P any point in Fig. 184. Jts circumference, join p with the centre o, and draw p n at right angles to a o ; take q a point between a and p, draw q m q and APPENDIX, 337 p^ parallel to a a ; let p t be a tangent to the circle at p and produce M Q to meet p t in t. Now, suppose q to move along the circumference up to p, then it is plain that the limiting values of p M, p Q, p T, M Q, M T, and Q T are separately zero, while p j» is the Umiting value of g m, g q, and g-T. Under these circum- stances it is commonly stated that p m q is ultimately a triangle similar «o o P N ; this means that the limitof ^ equals ^ from PM ^ on' whence it will of course follow that the limit of — equals — p Q op' and that of ^ equals!^. Now it will be remarked that ^ P Q O P p jj P N equals — under all circumstances, and therefore in the limit ■ O N ^ SO that what we have to prove will be done if we can show that the limit of ^equals that of—, i.e. equals that of^Q + Q^ PM ^ PM ^ PM PM or, in other words, we have to show that the limit of ^ is PM zero. But