IMAGE EVALUATION TEST TARGET (MT-3) k {/ Z I/. 1.0 !.l 12.5 ? lis IIIIIM 18 \M IIIIIJA 111 1.6 V <^ /; Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 L17 V iV \\ Lv 2d t ^ CIHM/ICMH Microfiche Series. CIHM/ICIVIH Collection de microfiches. Canadian Institute for Historical Microreproductions / Institut Canadian de microreproductions historiques Technical and Bibliographic Notes/Notes techniques et bibliographiques The Institute has attempted to obtain the beat original copy available for filming. Features of this copy which may be bibiiographically unique, which may alter any of the images in the reproduction, or which may significantly change the usual method of filming, are checked below. □ Coloured covers/ Couverture de couiaur □ Covers damaged/ Couverture endommagie D □ D D D n Covers restored and/or laminated/ Couverture restaur^ at/ou peiliculAe □ Cover title missing/ Le titre de couverture manque D Coloured maps/ Cartes g^ographiques en couieur Coloured ink (i.e. other than blue or black)/ Encre de couieur (i.e. autre que bleue ou noire) Coloured plates and/or illustrations/ Planches et/ou illustrationa en couieur Bound with other material/ Relit avec d'autres documents Tight binding may cause shadows or distortion along interior margin/ La re liure serrie peut causer de I'ombre ou de la distorsion le long da la marge inttrieure Blank leaves added during restoration may appear within the text. Whenever possible, these have been omitted from filming/ II se peut que certaines pages blanches ajouties lors d'une restauration apparaissent dans le texte, mais, lorsque cela ttait possible, ces pages n'ont pas 6ti filmies. Additional comments:/ Commentaires supplAmentaires; L'Institut a microfilm^ le meilleur exemplaire qu'il lui a iti possible de se procurer. Les details de cet exemplaire qui sont peut-^tre uniques du point de vue bibliographique, qui peuvent modifier une image reproduite, ou qui peuvent exiger una modification dans la mtthode normale de filmage sont indiquAs ci-dessous. □ Coloured pages/ Pagea de couieur Q Pages damaged/ Pages endommagies Pages restaurdes et/ou pelliculies Pages discoloured, stained or foxe< Pages dicolories, tachettes ou piquees Pages detached/ Pages ditach^es Showthrough/ Transparence Quality of prir Qualiti intgale de ('impression Includes supplementary materii Comprend du materiel suppltmentaire Only edition available/ Seule idition disponible T» to I I Pages restored and/or laminated/ r~T| Pages discoloured, stained or foxed/ l~~| Pages detached/ [~7] Showthrough/ r~l Quality of print varies/ r~1 Includes supplementary material/ r~| Only edition available/ D Pages wholly or partially obscured by errata slips, tissues, etc., have been refilmed to ensure the best possible image/ Les pages totalement ou partiellement obscurcies par un feuillet d'errata, une pelure, etc., ont it6 filmies A nouveau de facon a obtenir la meilleure image possible. Tl po of ffil Or be th 8l( oti fir 8l( or Th 8h Til wl Mi dif en be( rig rec mc This item is filmed at the reduction ratio checked below/ Ce document est film* au taux de rMuction indiqui ci-dessous. 10X 14X 18X 22X 26X 30X y 12X 16X 20X 24X 28X 32X The copy filmed here he* been reproduced thanks to the generosity of: Douglas Library Queen's University The Images appearing here are the best quality possible considering the condition and legibility of the original copy and In Iceeping with the filming contract specifications. Original copies in printed paper covers are filmed beginning with the front cover and ending on the last page with n printed or illustrated impres- sion, or the back cover when appropriate. Ail other original copies are filmed beginning on the first page with a printed or illustrated Impres- sion, and ending on the last page with a printed or illustrated impression. The last recorded frame on each microfiche shall contain the symbol — ^ (meaning "CON- TINUED"), or the symbol V (meaning "END"), whichever applies. Maps, plates, charts, etc., may be filmed at different reduction ratios. Those too large to be entirely included in one exposure are filmed beginning in the upper left hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method: L'exemplaire film* fut reproduit grAce A la gAn6roslt6 de: Douglas Library Queen's University Las images suivantes ont 4t6 reprodultes avec le plus grand soin, compte tenu de la condition at de la nettet* de l'exemplaire film*, et en conformity avec les conditions du contrat de f llmage. Les exemplalres originaux dont la couverture en papier est imprimte sont fllmfo en commenpant par le premier plat et en termlnant solt par la derniire page qui comporte une empreinte d'Impression ou d'lllustration, solt par le second plat, salon le cas. Tous les autres exemplalres originaux sont filmte en commengant par la premiere page qui comporte une empreinte d'Impression ou d'illustratlon et en termlnant par la derniire page qui comporte une telle empreinte. Un des symboles sulvants apparaltra sur la derniire image de cheque microfiche, seion le cas: le symbols — ► signifle "A SUIVRE", le symbols V signifle "FIN". Les cartes, planches, tableaux, etc.. peuvent dtre film6s d des taux de reduction diffirents. Lorsque le document est trop grand pour dtre reproduit en un seui cliche, il est film* d partir de Tangle supirieur gauche, de gauche d droite, et de haut en bas, en prenant le nombre d'images nicessaire. Les diagrammes suivants illustrent la m6thode. 1 2 3 1 2 3 4 6 6 1 S' PC IIA LOYELL'S SERIES OF SCHOOL BOOKS. NOTES AND EXERCISES nr MTDBAL PHILOSOPII, INCLUDING STATICS, HYDROSTATICS, PNEUMATICS, DYNAMICS, AND HYDRODYNAMICS, DESIONBD FOR THE USE OF NORMAL AND GRAMMAR SCHOOLS, AND THE HIGHER CLASSES IN COMMON SCHOOLS. BY JOHN HERBERT SANGSTER, ES(t., MATHEMATICAL MASTER AND LECTURER IN CHEMISTRT AND NATURAL PHILOSOPHY IN THE NORMAL SCHOOL FOR UPPER CANADA ; AND AUTHOR OF AN ARITHMETIC IN THEORY AND PRACTICE. indcrrL PRINTED AND PUBLISHED BY JOHN LOVELL ; AND SOLD BY R. & A. MILLER. tloronto : R. & A. MILLER, 87 YONGE STREET. 1860. Entered, according to the Act of the Provincial Parliament, in the year one thousand eight hundred and sixty, by John LoTBLL, in the Office of the Registrar of the Provlnoe of Canada. > « a. .n'ir PREFACE. The fpllowing Treatise was originally designed to serve as a hand-book or companion to the lectures on Natural Philosophy, delivered to the junior division in the Normal School. Although numerous text-books on the subject were already in existence, it was found that they were either too abstruse and technical for beginners, or too general and superficial to be of much practical use. The aim of the present little work is to occupy a position between these extremes — to present the leading facts of the science in a form so concise as to be readily remembered, and at the same time to give that thorough drilling upon the principles which is absolutely essential to their full comprehension. As a hand-book to lectures fully illustrated by apparatus, it was not necessary to introduce many (wood-cuts,^ and accordingly they have been given only where absolutely necessary. The chief peculiarity of the following Treatise is the introduction to a large extent of problems calculated to impart that intimate and practical knowledge of the 4 PREFACE. facts and principles of the subject, without which the student^s information on the science of Natural Philosophy is, comparatively speaking, useless. How frequently do we meet with a pupil who has read carefully through one of the common text-books on the subject without acquiring, to any very great extent, clear and definite ideas of the science t And what should we say of a work professing to teach the principles of arithmetic or algebra by mere rules and explanations, without an appropriate selection of examples and problems? The exercises are therefore deemed an important feature in the present little book, and it is thought that the science may be taught by their aid more thoroughly and in less time than otherwise. Toronto, January, 1860. ■ '■:■■, ,-<: ■■; ]■ ■A- 4 ,',,«' ,i".vv; :,,,!- r, ... i , '1 -_,. '. "■■--'. >■< , 1 ■■ :\ 1 tJ • j.j:' 1.. .'•■ •v-s [■; 1' ■:'. '-•? i,-: J.'V -A: ^■\ .':A'' ■*'.:. >r- ERRATA. ■ '^C''' - ' i ' ^ L Page 29, Quea. 44, for 1T60 lbs. read 880 lbs. Page at, Ques. 68, for lOX3«XlO = 720,read 10x36= lOX "729. Page 39, Ques. 75, for Ana. 9f , read Am. 181 lbs. ■.""".f ,' •■ :-:■■. '.{^'f i I''»" t^; .■■;■'■ .CONTENTS. CHAPTER I. Paob General Sub-divisions of Natural Science, 9 Sub-divisions of Natural Philosophy, 10 Properties of Matter, 10 Table of Tenacity, 13 Attractions, 13 Problems in Attraction of Gravity, 14 CHAPTER n. Sub-divisions of the Science of General Mechanics,. ... 15 Statics, 15 Parallelogram of Forces, 17 Parallel Forces 18 Centre of Gravity, 19 CHAPTER III. Mechanical Powers, 20 Virtual Velocities, 20 The Lever, 22 The Compound Lever, 25 The Wheel and Axle, 26 The Diflferential Wheel and Axle, 28 Wheel Work, 30 The Pulley, 33 The Inclined Plane, 38 The Wedge, m..^. 40 ''W^.-W- 2 f } G CONTENTS. V PAOl The Screw, 41 Tho Differential Screw, 44 The Endless Screw, 45 Friction, 47 Table of Friction, 48 CHAPTER IV. Unit of Work, 49 Work of different agents, 50 Work on a horizontal plane, 53 Table of traction of tho horse, 54 Work of atmospheric resistance, 57 Work on an inclined plane, 59 Modulus of machines, 64 Table of moduli, 65 The Steam Engine, 65 Work of the Steam Engine, 68 Source of work in the Steam Engine, 71 Pambour's Experimental . Table, 72 CHAPTER V. Hydrostatics, 76 Liguid pressure, 76 Weight of cubic inch, gallon, and cubic foot of water,.. 77 Pressure against a vertical or inclined surface, 78 Pressure against a vertical or inclined surface at a given depth 81 Bramah's Hydrostatic Press, 84 Hydrostatic paradox, 87 Hydrostatic Bellows, 87 Specific Gravity, 88 To find the specific gravity of a solid, 89 To find the specific gravity of a liquid, 91 ' ■i CONTENTS. 7 PAOl To find the specific gravity of a gas, 92 Table of specific gravities, 93 To find the weight of a given mass of any substance,... 93 To find the mass of a given weight of any substance,... 94 CHAPTER VI. Pneumatics, 95 Composition of Atmospheric Air, 95 Gaseous diffusion, 96 Aqueous vapour, 96 Physical properties of atmospheric air, 97 Weight of air, 97 Density of air, 98 Pressure of air, 99 Mariotte's Law, 100 The Air Pump, 101 Pressure and elasticity of air, ,.102 The Barometer, 103 Use of the barometer as a weather glass, 105 To ascertain the height of mountains, &c., by the bar- ometer, 105 The Common Pump, 107 The Forcing Pump, 108 The Syphon, 108 CHAPTER VII. Dynamics, 109 Momentum, ..110 Laws of Motion, /. 114 Reflected Motion, 114 Descent of bodies freely through space, 115 Analysis of the motion of a falling body, 116 Table of formulas for descent of bodies through space,. 117 8 ^ CONTENTS. ProMoms, 118 DoHCont on inoliiiud plunoH, 124 DcHcont in curvi^H, 125 TubluH of tbrinulafl for doKoont on inclined plnnoH, 12G ProbloiMH 127 PnyectiloH, 129 Parabolic theory, l.'JO Modern parabolic theory 131 Velocity of »hot and shell, 133 Circular motion, 1 34 Centrifugal force, 1 34 ProbleniH, 136 Accumulated work, 137 The Pendulum, 139 The centres of suspension and oscillation, 140 Laws of the oscillation of the pendulum, 1 40 Tables of lengths of second's pendulums, 141 Problems, 142 CHAPTER VIII. Hydrodynamics, 146 Torricelli's theorem, 146 Discharge of water through an orifice, 146 The vena contracta^ 146 Problems, 147 The random of spouting fluids, 149 Velocity of water flowing in a pipe or channel, 149 Upright water-whccls, 160 To find the horse powers of water wliocls, 151 Miscellaneous Problems, 152 Examination Papers, 160 Answers to Examination Papers, 169 Examination Questions, 171 NATURAL PHILOSOPHY. CHAPTEU I. SUnDIVISIONS— GENERAL PROPEREIES OP MATTER- ATTRACTION. 1. Natuml Sciunce, in its widest sense, embraces the study of all crented oljectB and beings, and t!te laws bj wliicli tliey are governed. 2. Natural objects are divided into two great classes, viz : organic and inorganic, the former being distinguished from the latter by the exhibition of vital power or life. 8. Organic existences are separated into animals and rrgctahleSf the former distinguished from the latter by the possession of HenHibility and volition. 4. The difTerent subdivisions of natural science and their objects are as follows : — Zoology describes and classifies animals. Botany teaches the classification, use, habits, structure, &c., oi plants. Mineralogy describes and classifies the various mineral constituents of the earth's crust. Astronomy investigates the laws, (kc, of celestial phe- nomena. Geology has for its object the description, ■ 24. Attraction is that power in virtue of which particles and masses of matter are drawn towards each other. 26. Attraction is of several kinds — viz: I, Attraction of Gravity. ' II. Attraction of Cohesion. . - ; ' III. Attraction of Adhesion. ^ IV. Capillary Attraction. V. Electrical Attraction. VI. Magnetic Attraction. VII. Chemical Attraction. ,. — 28. Attraction of Gravity (Lat. gravitas "weight") is that force by which masses of matter tend to approach each other. It is sometimes spoken of as gravitation^ or when applied to the force by which bodies are drawn towards the centre of the earth, terrestrial gravity, 27. The intensity of the force of gravity varies directly as the mass of the bodies, and inversely as the square of their distance apart. u ATTRACTION. [ABTB. 28, 29. , :• XoTs.— If we suppose two spheres of any kind of matter, lead, for example to be placed lu presence of each other, and undw such oondiMons that being themselves flree to move in any dveotlon they are entirely nninftu- enoed by any other bodies or circumstances they will approach each other and : — 1st. If their masses are equal their velocities will be equal. 2nd. If one contain twice as much matter as the other, its velocity will be only half as great as that of the other. 3rd. If one be iniinitely great in comparison with the other its motion will be inflnitely small in comparison with that of the other ; and 4th. The more nearly they approach each other the more rapid will their motion become. ,^ 28. By 8a3nng the intensity of the force of gravitation varies inversely as the square of the distance between the attracting bodies, we merely mean that if the attractive force exerted between two bodies at any given distance apart be represented by the unit 1, then, if the distance apart be doubled, the force of attraction will be reduced to \ of what it was before ; if the distance between the bodies be increased to three times what it was, the force of gravity will be decreased 9 tiraes, or will be only \ of what it was, &c. Example 1. — If a body weigh 981 lbs. at the surface of the earth, what will it weigh 8000 miles from the surface ? 80LX7TIOW. Here since the distance of the body in the first case is 4000 miles firom the centre of the earth and in the latter case 1200(Ki. e. 8000+4000) the distance apart has been trebled. a i > **^ ■< - . , ij =j Then weight = ^ = ?^ = 109 lbs. Ana. 8* 9 Example 2. — The moon is 240000 miles from the (centre of) earth, and is attracted to the earth by a certain force. How much greater would this force become if the moon were at the surface of the earth ? SOLVTIOir. Here ^ !^^, = ^^ = 60, and 602 = 8600 times. An^. Earths radius 4000 ^ EXERCISES. 3. If a mass of iron weigh 6700 at the surface of the earth how much would it weigh at the distance of 12000 miles from . the surface ? Ans. 418| lbs. 4. If a piece of copper weigh 9 lbs. at the distance of 36000 miles from the earth's surface, what would it weigh at the surface of the earth ? Am, 900 lbs. 29. Attraction of Cohesion is that force by which the constituent particles of the same body are held together. AeTS. 30-86.] STATICS. 15 NoTB.— The attraction of cohesion acts only at insensible distances ; i. e., at distances so minute as to be incapable of measurement. The attracts n of gravity, on the other hand, acts at sensible distances. 30. Attraction of Adhesion is that force by which the particles of dissimilar bodies adhere or stick together. 31. Capillar^/ Attraction (Lat, capillar " a hair") is the force by which fluids rise above their level in confined situations, such as small tubes, the interstices of porous substances, n 64. Tlie object of all Mechanical contrivanceB is Ist. To gain power at the expense of velocity ; or 2nd. To gain velocity at the sacrifice of force. 66. The relative gain and loss of power and velocity is regulated by that principle in philosophy known as the Law of Virtual Velocities, or the Equality of Moments. 66. The Law of Virtual Velocity may be thus enun- ciated : — ' If in an\f macJdne the power and loelght he in equilibrium and the wliolc he put in motion, then the power multiplied hy the units of distance through which it m^ovcs is equal to the weight multiplied hy the units of distance tJwough which it mx)ves. ,,. -- . u.; Or if F=:power, W=zweight, S=space moved through by P, and vzuspace through which W moves. ^ , ^ ,-!..« ThenV : W :: s : S. W X s Hence P = S W X 8 P X s ; S = — p — , W = — - — and s PXS. ' W ExAMPLK 5. — A weight of 700 lbs. is moved through 90 feet by a certain power moving through 5100 feet. Required the power SOLUTION. ! •<; Here IK=700, s=«0 and S = 6100. Hence P = —5— = Wxs 700 X 90 6100 = 12^ lbs. Ans, Example 6. — A weight of 500 lbs. is moved by a power of 20 lbs., through how many feet must the power mpve in order to f^ise tHe weight through 16 feet ? AuT9. rt7, tw. MKCIIANICAL I'OWEKS. 31 s cnun- 80LUTI0N. Horc »r= 600, P = 20 and » = 10. Wxa 500X16 HoncR • Primary Mechanical Elements, . The Pulley and Cord) The Wheel and Axle) The Wedge >• Secondary Mechanical Mementt. The Screw ) S2 THK LEVER. [Abti.6»-7II. 69. In reality however, there are but two simple me- chanical elements, viz. the Lever and the Inclined Plane, The Wheel and Axle and the Pulley are merely modifica- tions of the Icver^ while the Wedge and the Screw are both formed from the inclined plane. 70. In theoretical mechanics levers are assumed to be perfectly riff id and imponderable — cords, ropes and chains are regarded as having neither thickness, stiffness nor weight, they are assumed to bo mere mathematical lines^ infinitely flexible and infinitely strong. At first no allow- ance is made for friction, atmospheric resistance, Ta pair of scales, a crowbar when used forpry iug,&c. V i ■w 76. In levers of the second class the weight is between the fulcrum and the power. Nutcrakers, an oar in rowing, a cowbar when used in lifting, &c., are examples of levers of the second kind. ^ Fig. 2. T V AnTn.7fl,77.] THE LEVLH. isd I 76. In levers of tlie third class llie pnwrr is between the fulcrum and the weight. Fig. 3. P A pair of common tonKs, nhoop- Nh(!ani, tun truadio of a foot lathe, a door whon oiwncd or closed by placing thu hand near the hincro, afford examples of levers of the third cla88. Note.— In levers of the first class y ■ ■ ' ' the power may bo oithor (greater or IfiHs than tho weight ; in levers of the second nlasH. the power is at- ways leas than tho weight ; and in lovers of the third class, tho power is alwayt greater than tho weight. Hence fevers of tho third class are ' called losing levers, and are used mornly to secure extent of motion. Most of tho levers in tlie animal economy are lovers of the third kind. 77. That portion of the lever included between the ful- crum and tho weight is termed the arm of the weight; that portion between tho fulcrum and tho power is termed ihQ arm of the power. The power and the weight in the lever are in equilibrium when the power is to the weight as the arm of the weight is to the arm of the power. Or let P = power ^ W = the weighty A = the arm of the power, and a = the arm of the weight. Then P : W :: a : A. Hence P = Wxa W = PXA a a = PXW and A =: Wxa Example 13. — The power-arm of a lever is 11 feet long, the arm of the weight 3 ft. long, the weight is 93 Iba , Required the power ? HeroTr=93. A: SOLUTIOS'. :11 and a =3. „ TFXo__ 9?>X3_ „,,>,. J Then P= — -r— = -— — = 25^\ Ibs.^w*. 11 Example 14.— The power-arm of a lever ia 17 feet long, thf arm of the weight is 20 feet long, the power is 110 Ibi. What it the weight ? SOLUTION. Here P = 110 lbs., A= 17 and a = 20. Then Tr=^^: 110X17 80 =r93ilbi. An$, i 24 THE LEVEU. [Abts. 78, 79. ExAMPLK 16. — By means of a lever a power of 4 oz. is made to balance a weight of 7 lbs. Arour. ; the arm of the weight is 2i inches long. Required the arm of the power. SOLUTION. Here P=:4 oz., Tr= 7 lbs. = 112 oz„ and a == 2\. Ans. Then A =2 — i,- = — - — == 70 inches. r 4 17 18 — EXERCISES. 16. The power-arm of a lever is 16 feet long, the arm of the weight 2 feet long, and the waight is 260 lbs. Required the power? Ans. 31 i lbs. The power-arm of a lever is 20 feet long, the arm of the weight 70 feet ; what power will balance a weight of 5 cwt. ^ Ans. 17J cwt. The power-arm of a lever is 60 inches long, the arm of the weight 90 inches long, the power is 76 lbs. Required the weight? Ans. 50§ lbs. 19. The power-arm of a lever is 17 feet long, the arm of the weight 19 ft. ; what power will balance a weight of 950 lbs ? Ans. \QQ\\} \h%. 20. The power-arai of a lever is 12 ft. long, the power is 10 lbs. and the weight 75 lbs. Required the length of the arm of the weight. Ans. If feet. 21. By means of a lever a power of I25 lbs. is made to balance a weight of 93 lbs. ; the arm of the weight being 6J feet, what is the length of the arm of the power ? Ans. 47^^ ft. 78. When the power and the weight merely balance each otlier, i, e., when no motion is produced, there is no difference between the second and third classes of levers since neither force can be regarded as the mover or the moved. In order to produce motion, one of these forces must prevail, and the lever then belongs to the second or third class, according as the force nearer to or farther from the fulcrum prevails. 79. If the arms of the lever are curved or bent, their effective lengths must be ascertained by perpendiculars drawn from the fulcrum upon the lines of direction of the power and the weight ; the same rule must; be adopted when the lever is straight, if the power and weight do not act parallel with one another. ^BT3. 78, 70. z. is made eight is 2 i irm of the quired the IS. 31i lbs. xm of the t of 5 cwt. f. ITJ cwt. irm of the iqiiired the IS. 50§ lbs. irm of the of 950 lbs? 061\^\h». r is 10 lbs. the arm of . Ig feet. to balance g 6i feet, . m^ ft. balance ere is no of levers r or the se forces econd or her from nt, their idiculars )n of the adopted t do not ^ Akts.80,81.] . THE LEVER. THE COMPOUND LEVER. S6 80. Two or more simple levers acting upon dne another constitute what is called a Compound Lever or Corn- Fig. 4. ▼jt/ "ST T \ ■• '^ position of Levers. In such a combination the ratio of the power to the weight, is compounded of the ratios existing between the several arms of the compound lever. 81. In the compound lever if Wzziweight, V=power, a a' a" the anns of the ioeight, and A A' A" the arms of the power. Then ?: "W Is a X a' X a" : A X A' X A" .. . , ^ Wxaxa'xa" PxAxA'xA^ Hence V=:-^^^^ and W = —^^^^^^JT- Example 22. — In a combination of levers the arms of the power are 6, 7, and 11 feet, the arms of the weight 2, 3, and 3 J feet, the weight is 803 lbs. ; what is the power ? Hero Tr=803lb9.,a-i SOLUTION. : 2, a'= 3, a" = 3i. A=z6, ALz^ 7, A" = 11. mv. D WXaXa"Xa" 803X2X3X3^ „^, ,. ^ ^^^^^ ^ = ^X^'X^ ~ = -6^X11- = 3^ ^^'' ^'^' Example 23. — In a compound lever the power is 17 lbs., the arms ot the power 9, 7, 6, 5, and 4 ft., and the arms of the weight 2, 3, 1, 1, and i ft. Required the weight. SOLUTION. Here P = 17 lbs., A = 9.A'='7, A" = 6, A'" = 5, A"" = 4, a = 2, a' = 8. a" = 1. a'" = 1, and a"" = i. ._ PXAXA'XA"XA"'XA"" _ 17X9X7X6X5X4 _ 12862« 'Xa"> Alls. Then W = 642J0 lbs. aXa'Xa"Xa'"Xa"" 2X3X1X1X| "~ 2 EXERCISES. 24. In a compound lever the arms of the power are 9 and 17 ft. the arms of the weight 3 and 4 ft., the power is 19 lbs What is the weight? jins. 242^ lbs 2 \ 26 THE WHEEL AND AXLE. [ASTS. 82-M. 36. la a compound lever the arms of the power are 6, 8, 10, and 12 ft., the arms of the weight, 7, 6, 3, and 1 ft., the weight iB 700 lbs. Required the power ? Ans, 12J|. 26. In a compound lever the arms of the weight are 11, 13, and 9 ft., the arms of the power are 4, 7, and 2 ft., the weight is 560 lbs. What is the power? .in». 12870 lbs. - THE WHEEL AND AXLE. 82. The wheel and axle consists of a wheel with a cylindrical axle passing through its Fig. 5, centre, perpendicular to the plane ^ of the wneel. The power is applied to the circumference of the wheel, and the weight to the circumfer- ence of the axle. 83. The wheel and axle is merely a modification of the lever with un- equal arms ; the radius of the wheel corresponding to the arm of the power and the radius of the axle to the arm of the weight. 84. The wheel and axle is sometimes called the con- tinual or perpetwil levevy because the power acts continu- ally on the weight. 86. The power and weight in the wheel and axle are in equilihrium when the power is to the weight as the radius of ike axle is to the radius of the wheel, 86. For the wheel and axle — let F=the powers W = the weight, r := radius of the axle, B =: radius of the wheel. Then P : W R. Hence P = Wxr B W = -y- ; HxR Wxr r = -T^— ; and E= — p- . ExAMPLB 27.— In a wheel and axle the radius of the axle is 7 inches, the radius of the wheel is 36 inches, what power will balance a weight of 643 lbs. ? ,1 , [Arts. 82-8<. 6, 8, 10, and , the weight e 11, 13, and he weight is 8. 12870 lbs. eel with a 5. the con- continu- ;fe are in radius ie weight, Wxr P • |azle is 1 rer wilj Abt.86.3 THE WHEEL AND AXLE. 87 BOLUTIOir. Here fr= 643 lbs., £ = 30 inches, and r = 7 inches. Then P = Wxr^ 643X7 ,„„ ^ Example 28. — In a wheel and axle the radius of the axle is 6 inches, the radius of the wheel is 2*7 inches. What weight will be balanced by a power of 123 lbs. ? SOLUTION. Here P = 123 lbs., ^ = 27 in., and y = 6 in. Then Jr=?^ = l?2^ = 653i lbs. ^«*. ' Example 29. — By means of a wheel and axle a power of 11 lbs. is made to balance a weight of 117 lbs., the radius of the axle is 3 inches. Required the radius of the wheel ? solution. ,1 Here >r= 719 lbs., P = 1 1 lbs., and r = 3 In. Wxr 719X3 _,„«,. 1. . ~c^ ., — := IW^-v Inches. Ant. ThenJB= — k^=. 11 'TT EXERCISES. 30. In a wheel and axle the radius of the axle is 7 inches, the radius of the wheel is 70 inches. What power will balance a weight of 917 lbs.? ^n«. 91^V lbs. In the whsel and axle the radius of the axle is 5 inches, and the radius of the wheel 1 7 inches. What power will balance a weight of 5&'50 lbs. ? Jlns. 1750 lbs. 31. 32. 33. In a wheel and axle the radius of the axle is 9 inches and the radius of the wheel is 37 inches. What power will balance a weight of 925 lbs. ? jins. 225 lbs. In a wheel and axle the radius of the axle is 11 inches and the radius of the wheel is 45 inches. What weight will a power of 17 lbs. balance ? jins. 69^^B 6X32 160 „»„,.,, ^ Then d = ,-,■ ■ = . .^ ' = — — = lag of an inch. Ans. iW i of 729 i|a TT7 EXERCISES. 40. In a differential wheel and axle the radii of the axles are . 7^ and 7^ inches, and the radius of the wheel is 85 inches, what power will balance a weight of 6900 lbs. ? Ms. ff^ lbs. 41. In a differential wheel and axle the radii of the axles are 17 and 16 inches, and the radius of the wheel is 130 inches, what weight will a power of 17 lbs. balance ? Ms. 4420 lbs. 42. In a differential wheel and axle, the radii of the axles are 2h and 2} inches, and a power of 23} oz. balances a weight of 6400 oz. Required the radius of the wheel ? Jlns. 6^ inches. 43. In a differential wheel and axle, the radii of the axles are 4^ and 5 inches, the radius of the wheel being 120 inches, what power will balance a weight of 2430 oz. ? ^ns. 8^ff or. In a differential wheel and axle, the radii of the axles are If and If feet, the radius of the wheel is 12} feet, what weight will a power of-i46^1bs. balance. Ms. 146880 lbs. 89. Since the wheel and axle is merely a modification of the lever, a system of wheels and axles is simply a modification of the compound lever, and the conditions of 44 30 WHEEL WORK. TAkts. 90-04. equilibrium are the same, i. e., the ratio of the power to the weight is compounded of the ratios of the radii of the axles to the radii of the wheels. In toothed gear^ however, owing to the difficulty in determining the effective radii of wheel and axle, the ratio of the power to the weight is determined by the number of teeth and leaves upon the wheel and pinion. 90. Axles are made to act on wheels by various methods — as by the mere friction of their surfaces, by straps or endless bands, &c. ; but the most common method of transmitting motion through a train of wheelwork is by means of teeth or cogs raised upon the circumferences of the wheels and axles. 91. When cogged wheels and axles are employed, that part of the axle bearing the cogs is called sipin- ^^' ion. The cogs raised upon the pinion are called leaves^ those upon the wheel are termed teeth. 92. Wheel work may be used eiiher to con- centrate or diffuse power. The power is concen- trated when the pinions turn the wheels, as is the case in the crane^ which is used to gain power. The power is diffused when the wheels turn the pinions, as is the case in the fanning mill, threshing machine, &c., where extent of motion is sought. 98. In a system of toothed wheels and pinions, the conditions of equilibrium are that, — the power is to the weight as the continued product of all the leaves is to the continued product of all the teeth. 04. For a train of wheel work let P = the power ^ W = the weight , t *f' = the teeth of the wheel, and 1 1' T = the leaves of shepiniott^. <^'—J^><80X8<»<80_ 8704000 „„„„„,. . Then Tr=— y^-^^p: ^^^^^ = -nT-^"^^^^ ^^'- ^'^' Example 46. — In a train of wheel work there are four wheels and four axles, the iSrst wheel and the fourth axle plain, (t. «. . without cogs), and having radii respectively of 10 and 2 feet. The second wheel has 60, the third 90 and the fourth 70 teeth, the first axle has 7, the second 6 and the third 9 leaves. What power will hold in equilibrium a weight of 20000 lbs. ? BOLVTIOir. Here we have a combination of the simple wheel and axle and a system of cogged wheels and axles. ?r=20000 lbs. 26=10, r=2, fc=60, <'=90, <"=70, fc=7, Z'=5 and l"=9. Then cogged wheels and axles acting alone F= 20000X7X8X9 = 161 60X90X70 lbs., and so far as the action of the plain wheel and axles is concerned this 16| lbs., becomes the weight. Then P = TTXr 161X2 = -TT=8Hb8.ii>M. B 10 ExAMPLB 47. — ^In a train of wheel work there are three wheels and axles, the first wheel and the last axle plain, an . having a radius of 9 and 3 feet respectively — the cogged wheels have respectively 80 and 110 teeth, and the pinions 11 and 8 leaves. What weight will a power of 100 lbs. sustain ? SOLUTION. Here P=100 lbs., JB=9, r=3, fc=80, t'=110, 7=11 and V=», Pxtxtf 100X80X110 Then for cogged wheel work acting alone 7r= 880000 IXl 11X8 88 -= 10000 lbs. T. ii 1. 1 J 1 1 ,^_ PX 22^10000 X9_ 90000 ^ For plain wheel and axle alone 7^2= r= j— = .—— = 80000 lbs. Ans, 98 WHEEL WOllK. TAUTS. 'J5 V»s, EXERCISES. 48. In a system of wheel work there are five wheels and pinions ; the wheels have respectively 100, 90, 80, 70 and 60 teeth, and the pinions respectively 9, Y, 11, 9 and T leaves — with such an appliance, what weight would be sustained by a power of 11 lbs. ? Jlns. 5333333J lbs. In a train of four wheels and axles the wheels have respec- tively 10, 65, 60 and 50 teeth, and the axles respectively 9, 8, 1 and 6 leaves ; with such an instrument, what power could support a weight of 13000 lbs. ? Jlns. 2|f lbs. In a train of wheel work there arc three wheels and three axles, the first wheel and last axle plain, and having radii respectively 6 and 2 feet. The second and third wheels have respectively 80 and 50 teeth, and the first and second pinions respectively 5 and 8 leaves. With such a machine what weight will be balanced by a power of 11 lbs. ? Jlns. 33000. 49. 50. \.f 95. In ordinary wheel work it is usual, in any wheel and pinion that act on each other, to use numbers of teeth that are prime to each other so that each tooth of the pinion may encounter every too«h of the wheel in succes- sion that thus, if any irregularities exist, they may tend to diminish one another by constant wear. This odd tooth in the wheel is termed the hunting cog. Thus if a pinion contain 10 leaves and the wheel 101 teeth, it is evident that the wheel must turn round 101 times and the pinion 10 X 101 or 1010 times before the some leaves and the teeth will be again engaged. 96. Wheels are divided into crown, spw and bevelled gear, 97. The croion wheel has its teeth perpendicular to its plane ; the spur wheel has its teeth, which are continua- tions of its radii placed on its rim ; the levelled wj/teeZhas its teeth obliquely placed, i. e. raised on a surface inclined at any angle to the plane of the wheel. 98. To communicate motion round parallel axes spur- gear is employed, bevelled gear is used when the axes of motion are inclined to one another at any proposed angle. Where the axes are at right angles to one another a crown wheel working in a spur pinion or a crown pinion working in a spur wheel is usually employed. AKT8.90-10H.1 rilK VULLEY. 39 to its ;inua- leZhas ilined 99. Bevelled wheels are always frusta of cones channel- led from their apices to their bases. NoTB.— When bovolled wheels of different diameters are to work together the sections of the cones of which they are to bo ftiista are found in the following manner:— -p. „ Let ABbo the dia- * ig- o- meter of the largo wheel and B C that of the smaller. Place A B and B C so as to include the pro- posed angle. Bisect A B in D and B C in E. Draw perpen- diculars D F, E F meeting in F and join FA, FB and FO. Then FAB and FBO are sections of the required cones. Also drawing H G paraU . lei to A B and G P A parallel to B C, we obtain H A B G, and G B C P any required frusta. THE PULLEY. 100. The Pulley is a circular disc of wood or iron, grooved on the edge and made to turn on its axis by means of a cord or rope passing over it. 101. The pulley is merely a modification of the lever with equal arms, and hence no mechanical advantage is gained by using it — the theory of its use being just as perfect if the cord be passed through rings or over perfectly smooth surfaces. The real advantage of the pulley and cord as a mechanical power is due to the equal tension of every part of the cord, i. e., is founded upon the fact that the same flexible cord, free to run over pulleys or through smooth rings in every direction must always un- dergo the same amount of tension in every part of its length. 102. The pulley is called either fixed or movable ac- cording as its accis is fixed or moveable. _ . 103. Movable pulleys are used either singly, in which case they are called runners^ or in combination. Systems of pulleys are worked either by one cord or by several cords. Pulleys worked by more than one cord are called Spanish Bartons. m THE PULLEY. fARTS. lW-10^. 104. The pulley is often called a 8heaJ\ and the case in which it turns a block. A block may contain many sheaves. A combination of ropes, blocks and sheaves is called a tackle. 106. In the single fixed pulley the power must be equal to the weight, i. e., a fixed pulley does not concentrate force at all. And hence the only mechanical advantage derived from its use is, that it changes the direction of the power. Fiff. 9. 106. lib a system of pulley moved by one cord the conditions of equilibrium are that the power is to the weight as 1 is to twice the number of movuhle pulleys. This is evident from the fact that the weight is sustained equally by every part of the cord, and, neglecting the last fold or that to wliich the power is attached, there are tw6-folds of cord for every mov- able pulley. Thus in Fig. 9 the weight is sustained by A and B, each bearing i of it ; and since B passes over a fixed pulley, the power attached to C must be equal to the tension exerted on }{ = ^ the weight. 107. For a system of pulleys moved by one cord let P = the power, W = the weight and n = the number of movable pulleys. Then P : W :: 1 : 2n. W W Hence P = 2^, W = P x 2u, n = ^p- Example 51. — In a system of pulleys worked by a single cord there are 4 movable pullevs. What power will support a weight of 804 lbs. ? SOIUXION. Here Tr= 804 and » = 4. Example 52. — In a system of 1 movable pulleys worked by a single cord, what weight will be supported by a power of Itl bs.? f SOLUTION. Here P = 17 and w = 7. Hence jr=i* X 2 Xnri 17X2X7=^17 X U = 238lbs. Ant Am. 108.1 THE PULLEY. 35 tlxAMPLB 63. — 111 a system of movable pulleys worked bj a single cord a power of 7 lbs. balance a weight of 84 lbs. ; how many movable pulleys arc there in the combination ? SOLUTIOK. Here P=71b«.and ir=84 lbs. Honcen=~ = -?i=-=0. Ana. 2XP 2X7 H EXERCISES. 54. In a system of six movable pulleys worked by one cord the weiglit is 700 lbs. What is the power ? jins. 58^ lbs. 65. In a system of eleven movable pulleys worked by one cord the weight is 2563 lbs. Required the power ? Jns. 116i lbs. 56. In a system of 1 movable pulleys worked by one cord, the power is 37 lbs. Required the weight? Jns. 592 lbs. 57. In a system of eight movable pulleys worked by a single cord, the power is 13 lbs ; what is the weight ? ./?n». 182 lbs. 58. In a system of movable pul- leys worked by a single cord, pj_^ ^q a power of 35 lbs. supports a weight of 7000 lbs. How many movable pulleys arc there in tlie combination ? jlns. 100. 108. In system of pulleys such as represented in Fig. 10, where each movable pulley hangs by a separate cord, one extremity of each cord being attached to a movable pulley and the other to a hook in a beam or other fixed support, each pulley doubles the eflfect, and the conditions of equi- librium are that the power is to the weight as 1 is to 2 raised to thejyowcr indicated hy the mim- her of movable pulleys. Note.— This will become evident by attentively examiuing the diagram and following up the several cords. The figures at the top show the portion of weight borne by the several' parts of the beam, those attached to the cords show the portion of the weight sustained by each part of the cord. 16 lbs. 96 THE PULLEY. rARTS. 100, 110. ! 109. For a sy stem of pulleys tuch at exemplified in Fig. 10 let P =: the power, W = the weighl, and n = the number of movable pulleys, W Then P : W :: 1:2". Hence P = ^t. end W «a P x 2". Example 50. — In a system of pulleys of'tlio form indicated in Fig. 10, there arc 5 moveable pulley.'? and a weight of 128 lbs. What is the power ? SOLUTION. Hero n' = 128 lbs. and « = B. Thou P = ^ = ?# = ~ =4 U.H. Ans. 2" 2" 32 Example 60. — Tn such a systi-m of pulloy.q as is shewn Fig. 10 there are 7 movable pulleys. What weight will a power of 11 lbs. balance ? SOLUTIOW. ^ Here P=: 11 and ;» = 7. Ilenco »'= P X 2" = 11 X 2 ' =11 X 128 = 1408 11)8. ^w*. EXERCISES. 61. In the system of pulleys represented in Fig. 10, where there are movable pulleys ; what power will sustain a weight of 8000 lbs.? ./?;?s. 125 l))s, 02. In such a system when there arc 10 movable pulleys, what power will sustain a weight of 48000 lbs. ? ^ns. 40} lbs. G3. In such a system when there are 7 movable pulleys, what power will support a weight of 4564 lbs. ? Jlns. 355-i lbs. 64. In such a system when there are 3 movable pulleys, what weight will be sustained by a power of 17 lbs. ? Jns. 136 lbs. 65. In such a system what weight will a power of 70 lbs. support when there are 5 movable pulleys ? ,^ns. 2240 lbs. 66. In such a system what weight^will a power of 100 lbs. support when there are 1 1 movable pulleys ? Jns. 204800 lbs. 110. In a system of pulleys such as represented in Fig. 11, where the cord passes over a fixed pulley at- tached to the beam instead of being Fig. 11. 2 18 rs. 100. 110. Ann. Ill, 112.] THE PULLEY. 87 t^. 10 let ^ movable < 2". catcd in 128 11)3. Fijf. 10 r of 11 '0 there weight 25 lbs. fastened to a hook in the boHm, each movable pulley triphs the otlect, and tho conditions of equilibrium are that the power is to the weight » = 4. mi D ^^ 4600 __ 4500 ... Then P=-=^= --=355 lbs. Ana, ExAHPLB G8. — In such a system when there are 6 moyable pulleys, what weight will a power of 10 lbs. support ? Jlnt. 11 lbs. W SOLUTION. Horo P = 10, and m:=0. Then Wz=P X Sn = 10 X 36 SCIO)^ 729 = 7200 lbs. An». EXERCISES. 69. In the system of pulley represented in figure 11 there are 5 movable pulleys ;what weight may be supported by a power oflllbs. ? ^ns. 2430 lbs. 10. In such a system there are *I movable pulleys and the weight is 24057 lbs. Required the power ? 'l. In such a system are all 9 movable pulleys — through how many feet will the power descend in order to raise the weight 10 feet ? ^ns. 196830 feet. 112. If the lines of direc- tion of the power and weight make with one [another an angle greater than 120°, the power will require to be great- Fig. 12. 38 THE INCLIUED PLANE. [Aetb. 113-117. er than the weight; and as this angle approaches 180°, the difference between the power and weight will ap- proach oc . Hence it is impossible for any power -P, however great, applied at P, to pull the cord ABC mathematically straight, and that however small the weight W may be. - • •* ,? INCLINED PLANE. 113. The Inclined Plane is regarded in mechanical science as a perfectly hard, smooth, injlexihle plane, in- clined obliquely to the weight or resistance. 114. There are two ways of indicating the degree of inclination of the inclined plane : Ist. By saying it rises so many feet, inches, &c., in a certain distance. 2nd. By describing it as rising at some stated angle with the horizon, 116. In the inclined plane the power may be applied in any one of three directions : 1st. Parallel to the plane. 2nd. Parallel to the base. 3rd. Inclined at any angle to the base. 116. In the inclined plane the conditions of equilibrium are as follows : — Ist. 1/ the power act parallel to the plane: — the power is to the weight as the height of the plane is to its length. 2nd. If the power act parallel to the base: — the power is to the weight as the height of the plane is to its base. NoTB.— The third case does not come within the design of the present work. 117. For the inclined plane Id P = the power, W = the weight, L = length of the plane, H = height of the plane and B = base of theplane. 41 Aet. 117.] THE INCLINED PLANE. 39 ThenF : W :: H : L. Hence P = — = — PXL PXL WXH jihoP : W :: H : B. /fence P = — g— ; W = — jy— H = ,y and B =r — = — . Example '72.— On an inclined plane rising 7 feet in 200, what power acting parallel with the plane will sustain a weight of 4000 lbs.? SOLUTION. , Here W= 4000 lbs., L = 200, and i/= 7 IVxJI 4000X7 28000 ThenP = 200 200 140 lbs. Am. Example) 13. — On an inclined plane rising 9 feet in 170— what weight will support a power of 180 lbs. acting parallel to the plane? SOLUTION. Here P = 180 lbs., X = 170 and i/= 9. Then W= PxL 180X170 =3400 lbs. Jft«. H 9 Example 74.— On an inclined plane a power of II lbs. acting parallel to the piano supports a weight of 150 lbs.- how much does the plane rise in ^00 feet? SOLUTION. . . fr' Here P = 11 lbs., TV— 150 lbs., L = 200 feet. Then H= PX£_11X200 W ~ 160 =14 feet 8 inches. Ans, Example 75.— The base of an inclined plane is 40 feet and the height 3 feet, — what power acting parallel to the base will sup- port a weight of 250 lbs. ? SOLUTION. Here Tr= 250 lbs., //= 3, and jB = 40. TTxfl 250X3 Then P= B 40 ="*9iJb8. Anf. 1'^% Example 76. — On an inclined plane a power of 9 lbs. acting parallel to the base supports a weight of 700 lbs. — the height of the plane being 18 feet what is the length of the base ? ' I f 40 THE INCLINED PLANE. [Akts. 118, 119. SOLUTION. Hero P=0 lbs. W- 700 lb8. and /f =18 feot. Then B = WxH 700X18 9 = 1400 feot. Alls, EXERCICES. 77. On an inclined plane rising 1 foot in 35 feet what power acting parallel to the plane will support a weight of 17500 lbs. ? Jlns. 500 lbs. 78. On an inclined plane rising 9 feet in 100 feet what power acting parallel to the plane will sustain a weight of 4237 lbs.? ^ns. 381-i3(fJjylb3. 79. On an inclined plane whose height is 11 feet and base 900 feet what power acting parallel to the base will sustain a weight of 27900 lbs. ? ^ns. 341 lbs. 80. On an inclined plane rising 7 feet in 91 feet what weight will be supported by a power of 1300 lbs. acting parallel with the plane ? Jlns. 16900 lbs. 81. On an inclined plane a power of 2 lbs. acting parallel to the plane, sustains a weight of 10 lbs.— what is the inclination of the plane ? »^ns. Plane rises 1 foot in 5 feet. 82. On an inclined plane a power of 7 lbs. acting parallel to the base sustains a weight of 147 lbs. — if the base of the plane be ] 7 feet what will its height be ? Jlns. ^ j- feet. 83. On an inclined plane rising 2 feet in 109 feet what weight will be sustained by a power of 17 lbs. acting parallel to the plane ? Ms. 926 J lbs. 84. On an inclined plane a power of A} lbs. sustains a weight of 223-i*f lbs. ; the power acting parallel to the plane what is the degree of inclination ? Ms. Plane rises 341 feet in 17189 feet. 85. What weight will be supported by a power of 60 lbs. acting parallel to the base of an inclined plane whose height is 7 feet and base 15 feet. jins. 1320 lbs. THE WEDGE. 118. The wedge is merely a movable inclined plane or a double inclined plane, i. e. two inclined planes joined to- gether by their bases. 119. The wedge is worked either by pressure or by jpercussion. NoTH.— When the wedge is worked hy percussion, the relation between the power and weight cannot be ascertained since the force of percussion difTers so completMy from continued forces as to admit of no comparison with them. ^i Aaie. 120-122.] THE SCREW. il ?9 feet, acting htis 7 Olbs. 120* In the wedge the conditions of equilibrium are tliat the power is to the weight as half the width of the hack of the wedge is to its length. NoTB 1.— IJulike all the other mechanical powers, the practical use of the wedge depends on Mction, as, were it not prevented by Mction, the wedge would recoil at every stroke. NoTB 2.— BAzors, knives, soisson, chisels, awls, pins, needles, ftc, are exantples of the application of the wedge to practical purposes. 121. For the wedge, let P zz power or pressure, Wz=:the weighty L = the length of the wedge, and B = the width of the back. Then P:W::l B:L. Hence P = WxiB and Wz=, PXL Example 86. The length of a wedge is 24 inr'ic~, and its thickness at the back 3 inches, what weight would be raised by a pressure of 750 lbs. ? - , SOLVTIOK. Here P = 760 lbs., £ = 24 inches, and i JB =-: li inches. Then W - PxL 780X24 = 750 X 16 = 12000 lbs. Ans. \B '■ li Example 87. In a wedge, the length is 17 inches, thickness of hack 2 inches, and the weight to be raised is 11000 lbs. Re- quired the pressure to be applied 7 SOLVTIOK. HerelFssllOOO, L = 17 inches, audi J? = 1 inch. ,• / mu « Wx^B 11000X1 „^„ , ,. . Then P = ' = — rz — — 647-^^ lbs. Ans. , .. EXERCISES. 88. The length of a wedge is 30 inches and the thickness of its back 1 inch, what weight will be raised by a pressure of 97 lbs. wdPn«. 5820 lbs. 89. The length of a wedge is 19 inches and the thickness of its back 4 inches, what pressure will be required to raise a weight of 864 lbs. ? jlns. 90 jf lbs. 90. The length of a wedge is 23 inches and the thickness of its back 3 inches — ^with this instrument what pressure would be required to raise a weight of 1771 lbs. ? .Sns. 115J lbs. THE SCREW. 122. The screw is a modification of the inclined plane and may be regarded as being formed of an inclined plane wound round a cylinder. NoTB.— The lerew bean the same rehition to an ordinary inclined plant that a circular staircase does to a straight one. 42 THE SCREW, [AkTB. 12a-128. - Fig. i3. 128. The threflds of the screw are either triangular or square. The distance of a thread and a space when the tljread is square, or the,di8tance between two contiguous triangular threads, is called the pitch. 124. The screvf is commonly worked by pressure against the threads of an external screw, called the box or nut. The power is applied either to turn the screw while the ljut is fixed, or to turn the nut while the screw is kept immovable. 126. In practice, the screw is seldom used asasimpie mechani- cal power, being nearly always combined with some one of the others — usually the lever. 126. The conditions of equili- brium between the power and the weight in the. «crew are the same as for the inclined, plane, where the Dower acts parallel to the base, i.e. The power is to the weight as the pitch {i. e. height') is to the circumference of the base (i. c, length of the plane). When the screw is worked by means of a lever, the con- ditions of equilibrium are : — The power is to the weight as the pitch is to the circum- ference of the circle described by the power. \ 127. The eflSciency of the screw as a mechanical power may be increased by two methods : Ist. By diminishing the pitch. • 2nd. By increasing the lengtli of the lever. 128. For the screw, let P = the power, W = the weight, p = the pitch, and I = hngth of the lever. Then since the lever forms the radius of the circle described by the power, and the circumference of a circle is 3*1416 times the diameter, and the diameter is tunce the radius, P:lV::p:l'X.2x 3-1416. Wy.p Hence P = „, PX?X2X3-1410 , PXiX2X3*1416 W= andp=- ZX2X3*1416 " P """'' W NoTB.— The pitch and the length of the lever must be both ezpreesed in wnita of the same denominations, i. e. both feet, or both inches. 'f '■*>■ re. 123-128. Art. 128.] THE SCREW. 43 Example 91. What power will sustain a weight of 70000 lbs. by means of a screw having a pitch of -^(ih of an inch, and th« lever to which the power is attached 8 ft. 4 in. in length 7 SOLUTION. Hero 1V= 70000 lbs., p = -j^^ in., and i = 8 ft. 4 in. = 100 in. Here P— WX P 7000OX-14- 5000 800000 ,„.. „ =«i;;s::^=sss^=^7 9871bs. Atu. iX2X31416 100X2X3-M18 628*32 62832 Example 92. — What weight will be sustained by a power of 5 lbs. by means of a screw having a pitch of ^^th of an inch, the power lever being 50 inches in length ? SOLUTIOir. Hero P = 5 lbs.-, p = -jij inch, and 1 = 60 inches. mt. Mr PX?X2X3-1416 5X50X2X8'1416 1570*8 Then }V= = — -i^ff tV a5708lb8. Am. Example 93. By means of a screw having a power lefer 6 ft. 10 inches in length, a power of 6 lbs. sustains a weight of '80000 lbs. ; what is the pitch of the screw ? SOLUTION. Here P = 6 lbs., W=^ 80000 lbs., and I = 70 inches. PxiX2X31418 6X70X2X3*1416 2614*944 Then p=: 33 about 1000 W of an inch. Ans. 80000 80000 =•0326868 inches, or Example 94. What power will sustain a weight of 96493 lbs. by u^eans of a screw having a pitch of ]^th of an inch, the power lever being 25 inches in length? SOLUTION. Here W= 96493 lbs., p = -^^^th inch, and I — 25. ThenP = 108*403 lbs. WXP iX2X3*1416 Ans. 96493Xfr 25X2X3*1416 EXERCISES. '!4 saaiya 157*08 170281764 167*08 95. What power will support a weight of 87000 lbs. by means of a screw having a pitch of -,%th of an inch, the power lever being 6 ft. 3 inches long ? ^ins. 31*83 lbs. 96. What weight will be sustained by a power of 200 lbs. acting on a screw having a pitch of ^%th of an inch — the power lever being 15 inches long? jin$. 314160 lbs. 97. By means of a screw having a power lever 50 inches in length, a weight of 9000 lbs. is supported by a power of 12 lbs. Required the pitch of the screw 7 An*. '41888, or rather over f of an inch. 44 THE DIFFERENTIAL SCREW. [Ahtb. 129, 180. 'I ' ■' a. Wb^t power will lupport a weight of 11900 lbs. bj means of a screw having a pitch of ^th of an inch, the power lever being 10 ft. in length ? ^na. 3-713 lbs. 99. By means of a screw having a power lever 1 ft. 6 inches in length, a power of 10 lbs. supports a weight of 65400 ; what is the pitch of the screw ? Ana. -0864 of an inch. 100. What weight will be supported by a power of 60 lbs. act- ipg on a screw with a pitch of :^th of an inch— the power lever being 8 ft. 4 inches in length ? Ana. 418880 lbs. Pig. 14. THE DIFFERENTIAL SCREW. 129. The differential screw, (invented by Dr. John Hunter,) like the differential wheel and axle, acts by dimi- nishing the distance through which the weisfht is moved in comparison with that traversed by the power. It consists of two screws of dif- ferent pitch, wcjrldng one within the other (Fig. 14), so that at each revolution of the power lever the weight is raised through a space only equal to the d^erence be- tween the pitch of the exterior screw and the pitch of the inner screw. It follows that the mechan- ical effect of the differential screw is equal to that of a single screw having a pitch equal to the differ- ence of pitch of the two screws. For iiuiianoe, in Fig. 14, the part B works within the part A. Now, if B have a pitch of i^h of an inch and A a pitch of iV* then at each revolu- tion of the handle the weight will be raised through ^—i^d=^ST of an inch, and th'j whole instrument has the same mechanical effect as a single screw having a pitch of i itrth of an inch. 130. For the differential acrew, let P =: power, W" = weight, I ±: length of lever, and d = difference ofpitih df the two acrewa. Then P : TT:: (i : /X 2X31416. , ' ., .= . Wxd . .,^ PX/X2X31416 Hence P=^^2X31416^"^^= d • rs. 129, 130. AftT. 131.] THE ENDLESS SCREW. 46 by means wrer lever )-7l3 lbs. inches in •f 65400 ; ' an inch. I lbs. act- he power 8880 lbs. h, John by dimi- s moved Now. if B ch revolu* ^ijs of Ml B a single ibeightf screws. I I Example 101.— What power will exert a pressure of 20000 lbs. by means of a differeatial screw having a power lever 60 inches in length, the exterior screw a pitch of ^ of an inch, and the inner screw a pitch of j/^th of an inch ? * ' SOLUTIOX. Here W= 20000, 1 - 50 in., and d = A — A = ift/^ - A'o = --= Wxr liOOOX.1 42000 IXt ' 20X36 1120 EXERCISES ---;i7ilbs. Ans. 108. In an endless screw the length of the winch is 18 inches, tho radius of the axle is 2 inches, the wheel has 48 teeth, and the power is 120 lbs. Required the weight. ^ns. 51840 lbs. 109. What power will support a weight of a million of lbs. by means of an endless screw having a winch 25 inches long, an axle with a radius of 1 inch, and a wheel with 100 teeth ? jins. 400 lbs. 110. What weight will be raised by a power of 40 lbs. by means of an endless screw in which the winch is 20 inches long, the raditti of the axle 2 inches, and tho number of teeth in th« wheel dO ? Mi. 82000 lbs. Arts. Ml 139.1 FRICTION. 47 134. Tho tbeorotical results obtained by tbe foregoing rules are in practice very greatly inodifi«»'' cy fieveral retarding forces. Thus friction has to be taken into ac- count in each of the mechanical powers — the weight of the instrument itself in the lever and in the movable pulley — the rigidity of cordage in tho pulley and in the wheel and axle, portion. BrOLLIVO FBICTI05. VIII. Friction caused by one body rolling on another is directly proper* tioual to the pressure, and inversely to the diameter of the rolling body. That is, if a cylinder rolling along a plane have its pressure doubled, its friotion will also be doubled ; but if its diameter be doubled, the friction will be only half of what it was. The firiction of a wooden cylinder of 32 inches in diameter rolling upon rollers of wood is ts7 of the pressure. The friotion of an iron axle turning in a box of brass and well coated with oilis ^ of the pressure. - ••" " ' ' ' • v- - • - - ,.. >0Uj iH Adts. 140, Ul.j UNIT OP WORK. 49 OHAPTEH IV. UNIT OF WORK, WORK OP DIFFERENT AGENTS, HORSl POWER OF LOCOMOTIVES, STEAM ENGINES, AND WORK OF STEAM. UNIT OF WORK. 140. In comparing the work performed by different iigont9, or by the same agent under diflfcrent circumBtances, it becomes necessary to make use of some definite and dis- tinct unit of work. Tlic unit commonly adopted for this purpose in England and America is the labor requisite to raise- the weight of one pound through the space of one foot. Thus in raining 1 lb. through 1 foot. 1 unit of work is prrformed. If 2 lb. be raiaed 1 ft., or if l lb. be raised 8 ft., 2 units of work are IMjrformed. 'J 7 lba.be raised through 9 ft., or if 9 lbs. be raised through 7 ft., 6S units of work are performed, &o. 141. The units of work expended in raising a hody of n given weight are found hy multiplying the weight of the hody in lbs. by the vertical space in feet through which it is raised. \ Example 111. — How many units of work are expended in raia- inff a weight of 642 lbs. to a height of 70 ft. ? SOLUTION. .i/w. Units of work=642 X T0=44940. fixAMPLB 112. — How many units of work are expended in raising a weight of 423 lbs. to a height of 267 ft. ? , ? SOLUTION. ^ .4us. Units of work=423X 267=112941. Example 113. — How many units of work are expended in raising 11 tons of coal from a pit whose depth is 140 ft. ? SOLUTION. Here, 11 ton8=llX2000c=22000 lbs. Then 22000X 140 = 3080000 An$. Example 114.— How many units of work are expended in raising 7983 gallons of water to the height of 79 ft. ? SOLUTION. Here, since a gallon of water weighs 10 lbs., 7983 gal8.=379830 lbs. Then units of work =3 79830X79 =s 6306570. ilH«. ExAitPLi 115.— How many units of work are expended ia raialng'^O )cubie feet of water frdin d wiell irbotae fleptH W»0 feat 7 50 INIT OF WOKK. Art. 11!!. I mivtinv. Hitter A luliic foul of Wilier woIkIis Olii H*'*., •!(» riiliir ffol noiuli 02J <«M)-j: .1750 Um. Thou tiMits of woi'U ^37.10 -'im •TlT.'iOO. vi/i.t. KXKRCISEH. llrt. How nuirli work would bo required to pinup GOOOO giillonH ol' wiitor from a niiuo wlio.so Uoptli in HOO ft. 7 Jni. 51(5000000 uiiitH. llT. How mauy uulta of work would bo cx|)cud4Ml in pumping 8000 cubic foot of wutcr from u luino whoso drpth is G7y foot? .^Ins. ;i:{9600000 units. 118. JIow much work would bo expended in ralHinpf tho ram of a pile driving ongino — tho ram woighlng 'J touB, and tho height to which it ia rai.iod being 29 ft. ? Jlns. IIGOOO finilH. 119. How much work w"' 'i bo roquirod to raise 17 tons of coals from a mine whoso depth in 300 foot? Jlns. 10200000 units. 120. How much work would bo expended in raising GOO cubic foot of wnter'to tho height of 293 feet ? .Ins. 10987500 unit,?. ¥■■ 142. Tiie iiioi^t important soiuco^ of laboring; I'orco arc animals, waier^ windy and steam. Tlio laborincf force of ftnimalft is niodilied by various circumstances, tho most important of which are tlio duration of tho labor, and the mode by which it is applied. Tho following tablo shows the amount of ofFo(;tivo work that can uo performed under different cirmimiPtaiiccs by the more common living agents: TABLE SHRWLNO TUB WORK DONR PER MINUTE DT VARIOUS AQBNTS. Duration of labor eight hours per day. Horse 33000 units Mule 22000 " Ass 8250 ' Man, with wheel and axle 2600 drawing horizontally 3200 raising materials with a pulley 1600 throwing earth to the height of 6 ft. . . 660 (( (t (( (i I VI n Of i'.- (( U i( (( AitT. IIJ.I W(/KK or I.IVINU AOKNTS. 61 Mai), working with liin iirtnH atul h>gH nn in rowirtg 4000 unitM " raising water from a well with n pail and ropo 1 054 " " raising water from a well with an upright (haiii pinnp l7.')0 *' NoTK.- Till) work nNHiKiKMl by Wnlt to tlio liorso per niiitiito WM 33000 iitiitH, hut tlih \h known to Im> nixxit \ too gruat. A liurno of avoraK<* NtroDKth tii'i'loriiiN iihoiit 2'iOOO tiiiilx of work nor iniiiuto. Tlin niinilMT KJvoii ill tiio tublr is, liow( _44000X360_„ . Hence.H.P.=g555^— .=8. win,. BxAMPLi 127. — How many cubic feet of water will an engine of 15 horse powers pump each hour from a mine whose depth is ^oofeet? ■ ' ' ' BOLUTIOV. Units of work performed by engine perliour == S3000X60X1S. Units of jwork requh^ed to raise 1 cubic foot =: 62*6 X 900. M «- * 1.J * i. _53000X60X16 ■„„ V Hence, number of cubic feet ==———— — — = 528. An$. oa o,X800 ■' ' ^XAKjPLK 128[. — ^What must be the horse powers of an engine in brder that working 12 hours per day it may supply 2300 ^mi- lies with 50 gallons of water each per day — taking the mean height to which the water is raised as 80 feet, and assuming that I of the work of the ehgine is lost In transmission ? •"^^■'roiTJTioir. '"''^"^^''^^'^'" Weight of water pumped per day = 2SQQX SOX 10. Unite of work requU^ed di^^ Units of work in one horse power per day = SSOOOX 12X60. But since i of the work of the engine is lost in transmission, Useful work of one U,. P. per day=|x33000Xl2X60. -, _ ^ 2800X 50X10 X80 ■ «. . Hence. H. P. = --^^5^5-^ = 4.64. Ani. 129. How many cubic feet of earth, each weighing 100 lbs., will a man raise by means of a pulley from a depth of 30 feet in a 4ay of 8. hours? Ans. 256 cubic feet. 130. How many cubic feet of water per hour will an engine of 20 H. P. raise from a* mine Whose depth is 450 roet, MJWWi^g ^9^ i Pf tfee; vorf^ of tb^ engine if IPAtin IrapfpiiMjoij? • 4ps.um' oniric ««e^ CAST. 142. ▲mn. 1«, 144.] WORK ON A LEVEL PLANE. 53 ofllOft.= e to raise }Ofeet? a engine this ^*Pf ine ^enffuK e mean suming .will )0 feet C ftet. mgine feet, o«rin 131. "Vhrai &Qkt be the H. P. of ain engine in oirdet that it miiy raise 11 ions of material per hour from a deptii of tOQ ft. ? Jtu. 1-11 U.F. 132. A forge hammer weighing 890 lbs. makes 50 lifts of 4 feet eaqh per minute— -What must be the horse powers of the engine that works the hammer ? Jin*. H. P.=S-39. 133. An engine of 8 horse powers works a forge hammer, cans- ing it to make 50 lifts per minute, each to the height of 6 feet. What is the weight of the hammer? Jin»' 9,80 lbs. 134. An engine of 8 horse powers giv^s mptlon to a fpfge ham- mer, which weight 300 lbs., and makes 30 liifts per minute of 2 feet each ; and at the same tiI^e r^^ses 2 tons ,of coal per bpur from the bottom of a mine. Required, tbe. depth of the mine ? J^m 3990 feet. Note.— The work of tbe'englne=S3000x 8 units per minnte. From ihls nibtractthe units of wock reauired ny the hsimner ; thftremaln^.wiU be the wprk expended per minute in. raising the ooal, : Multiphrbig th|s bi ^ givJBs us the work required per hour for the ooial; and tkis last is the product of the weight in lbs. by the depth in feet, of which the former is given. WORK EXPENDED IN MOYINQ A OARRIAQE OR RAIL- WAY TRAIN ALONG A HORIZONTAL PLANE. ^ 143. In moving a carriage, &c., along a level plane, a q^rt^ii amount of power is expended in overcoming the friction of the road. This is rolling friction, and amounts, as belbre stated (Art. 138), to from j\ to ^j of the entire load on common roads, and from ^j^ to y}^ of the load on railway tracks. In the cafle of rfdly^ay trains, friction is usually tsken as 7 lbs. per ton of 2000 lbs. 144. In running carriages of any descri|>tion, irdik is employed to overcome the reisistances. These resistances itfe: — Ist. Friction — ^which on the same roacl >n4 w|ik the same load is the same for all yejiocitjes. Ascent of inclined plan^ — in which, since Umi Ibpd has to £e lifted vertically through the height of tho plane, the work is the same, whatever may be the velocity of the motion. , The Renitdnce of the il<»ton)Ae/'£— which depends ii^ii the «tteiit of BtitfMe, wA increases as the square of the velocity. 2(i. 3d. 54 WORK ON A LEVEL PLANE. rAnis.US.ue. i i! tfi 146. When a railway train is set in motion, the work of the locomotive engine at first far exceeds the work of re- sistances^and the motion is consequently rapidly accelerated. But as the velocity of the train increases, the atmospheric resistance also increases, and with such rapidity as very soon to equalize the work of resistances to the work of the loco- motive. When this occurs, i. e., when the work applied by the locomotive is exactly equal to the continued work of resistances (atmospheric resistance and friction), the velocity of the train will be uniform. In this case the train is said to have attained its greatest, or maximum speed. 146. The traction or force with which an animal pulls depends upon the rate of his motion. A horse, for example, moving only 2 miles an hour, can draw with a far greater force than when running at the rate of 6 miles an hour. The following table shows the relation between the speed and the traction of a horse : ;- "^'■' ' TABLE, or TRACTION OF A HORSE. .A,ir..i. spg^a. "■'-'''■• '' . ' j Traction. A horse moving 2 miles per hour, can draw with a force of 166 lbs. 125 " 104 " " 83 " 62J " 41§ " ExAMPLB 135. — What gross load will a horse draw travelling at the rate of four miles per hour on a road whose friction is i^ of the whole load ? ,' . . , SOLUTION. Here from the table the traction is 83 lbs., which by the conditious of the question is ^J^ of the gross load. Henoe load = 83 X 20 =1660 lbs. il»*. ' / -"- ,; ■^. EzAUPLB 136. — At what rate will a horse draw a gross load of 1800 lbs. on a road whose coefficient of friction is -jV? SOLUTION. Here, traction =: -^f^ = 100 lbs., whence by the table the rate mu8t be rather over 3i miles per hour. f ^ . ' ; . {■" ExAiiPLB 137. — If a horse draw a load of 2500 lbs. upon a road whose coefficient of friction is -3^, what traction will he exert and how many units of work will he perform per minute ? " 5 ' . 3 U 3J 'ir. ' 4 <( 4J « 5 « (( ft li % T8. U5,146. aht.146.] work on a level plane. 55 he work nk of re- elerated. ospheric ery soon ihe I060- applied ed work )n), the he train 'peed. ml pulls xauple, greater 1 hour, e speed TractuMt, 166 lbs. 125 " 104 " 83 " 62j " 41§ « ivelling is^of itious of IS load dust be pon a ill he inute ? SOLUTIOV. Here, traction == iJ^^a =: 83i lbs., and hence he moveii at a rate of four miles per hour. 4X6280* Then distance moved per minute = — ;.. — = 352 feet. 60 Hence units of work = 83i X 362 = 29333i. Aru, Example 138. — What must be the effective horse powers of a locomotive engine to carry a train weighing 70 tons upon a level rail at the steady rate of 40 miles per hour, neglecting at- mospheric resistance and taking -^his ^^ ^hc coefficient of friction ? BOLUTIOK. Here, weight of train = 70 tons =:: 140000 lbs. Space passed over per minute : ^% miles = ^!^^** = 3520 feet, 140000 Work of flriction to 1 foot = ^^ of 140000 = -^—- = 7«0 units. Work of friction per minute ■ Units of work in one H. P. = ' 200 3 700X3520 ~ 2464000 units. 33000. 700X3520 2464000 Therefore H. P. of locomotive = — „„„„r- = = 74*66. Ans. 33000 3:)000 Example 139. — A train weighing 120 tons is carried with a uniform velocity of 30 milen per hour along a level rail ; assum- ing the friction to be 11 T'S. \ 1- ton, and neglecting the resistance of the atmosphere, what ai horse powers of the locomotive ? ' wXION. 30X6280 Space passed over per minute = |§ miles = — — — = 2640 feet. Work of friction to each foot = 120X11 = 1320 units. Work of friction per minute — 1320X2640 = 3484800 units. Hence H. P. = -^~ = 106'6. Ans. ssooo Example 140. — At what rate per hour will a train weighing 90 tons be drawn by an engine of 80 horse powers, neglecting the resistance of the atmosphere and taking 7^17 as the coefficient of friction ? SOLUTION. Work done by the engine per hour = 33000X60X80. Weight of train in lbs. = 90X2000 = 180000. Units of work requured to move the train through 1 foot = iy|^ of 180000 = 720. Work expended in moving the train through 1 mile =720X6280. . Number of miles per hour ; 38000X60X80 = 41"66, Ans. 720X6280 Example 141. — A train moves on a level rail with the uniform speed of 35 miles per hour ; assuming the H. P. of the locomotive to be 50, the friction equal to 9 lbs. per ton, and neglecting atmospheric resistance, what is the gross weight of the train ? ♦ 52S0 is the number of feet in pne rail?.. I''!' m I f I 56 WORK ON A LEVEL PLANE. 80LUTI0K.- [AhT. 146. Work of engine per hour s= 33000X80X50. Feet moved over per hour = 36X6280. Work expended per hour in moving 1 ton = 36X5280X9. . w 1 vi. «A 1 J i. 83000X60X50 ^^ ,„„ . . • . Weight of train in toni = ^^^ = 59-828. Atis. ExAMPLB 142. — In what time will an engine of 100 H. P. move a train of 90 tons weight through a journey of 80 miles along a level rail, assuming friction to be equal to 10 lbs. per ton and neglecting atmospheric resistance ? SOLUTION. Work expended in moving the train through 1 foot = 90X10 = 900 unitB. Work expended on whole journey in moving the train ~ 900X5280X80. Work of engine per minute = 33000X100. .Number of minutes = -:::r:rrrrmr- = 116^ minutes = 1 hour 56^ minutes. Ans, 33000X100 EXERCISES. 143. What gross load will a horse draw travelling at the rate of 2 miles per hour on a road whose coefficient of friction is i^ ? Ms. 2988 lbs. 144. What must be the H. P. of a locomotive in order that it may draw a train whose gross weight is 130 tons, at the uniform speed of 25 miles per hour, allowing the friction to be 7 lbs. per ton and neglecting atmospheric resistance ? Jlns. H. P. 60-66. 145. A train weighs 75 tons and moves with the uniform speed of 30 miles per hour on a level rail ; taking ^^^j as the coeffi- cient of friction and neglecting the resistance of the atmos- phere, what are the horse powers of the engine ? Jlns. H. P. = 48. 146. In what time will an engine of 160 H. P. moving a train whose gross weight is 110 tons complete a journey of 150 miles, taking friction to be equal to 7 lbs. per ton, neglect- ing atmospheric resistance and assuming the rail to be on a level plane throughout? jins. 1 hour 55| minutes. 147. At what rate per hour will a horse draw a load whose gross weight is 220C lbs. on a road whose coefficient of friction is ^V? -Ans. Rather over 3 J miles per hour. 148. From the table given (Art. 145) ascertain at what rate per Itpur a hprse must travel, when drawing a load, in order to dp th^e greatest amount of work ? Ms. 3 miles per hour. 149. M what rat^ per hour will a locomotive of 50 H. P. draw a train whose gross weight is 70 tons, neglecting atmos- pheric resistance, taking j^^ as the coefficient of friction and ABBuming the rail to be level? Jnt. 26-78 miles. \BT. 14(6. Ants. U>, 148.1 WORlC ON A LEVEL PLANE. 57 H. P. 30 miles lbs. per OOOunitB. >280X80. hour 56^ iie rate of ion is -^ ? 2988 lbs. jr that it IS, at the riction to tance? P. 60-66. trm speed the coeffi- ie atmos> 147. When a body moves through the atmosphere or any other fluid, it encounters a resistance which increases : 1st. In proportion to the surface of the moving body ; 2nd. In proportion to the square of the velocity. Thus 1st. If a board presenting a surface of 1 sq. foot in moving through the air meet with a certain resistance, a board haviug a sur* face of 2 sq. feet will meet with double that resistance ; a board having a surface of 8 square feet will meet with three times that resistance, &c. 2nd. If a body moving 2 miles per hour meet with a certain rcsis* tance, a body of the same size moving 4 miles per hour will meet with (4)'i or 2', or 4 times that resistaure. If the velocity be increased 3 times; i. e., to 6 miles per hour, the resistance will be increased 9 times (i. e., n* times). If the velocity be increased 7 times, i e., to 14 miles per hour, the resistance will be increased 7' times, i. «., 49 tunes, &c. 148. In the case of railway trains, the atmospheric re- sistance is about 33 lbs. when the train is moving at the rate of 10 miles per hour. It has been found, however, by recent experiment, that the atmospheric resistance encoun- tered by a train in motion depends very much upon the length of the train. • r j Example 150. — When a train is moving at the rate of 10 miles per hour, it encounters an atmospheric resistance of 33 lbs.; what will be the resistance of the atmosphere when the train moves at the rate of 60 miles per hour ? SOLUXIOK. Here the velocity increases ^ times, i. e., S times. w-r . , Hence the resistance increases 6' times —25 times. . '. Resistance = 33 X 25 = 825 lbs., i. e., 826 units of work are expended every foot in overcoming the atmospheric resistance. ExAMPLB 151. — If a train moving 1 miles per hour meet with an atmospheric resistance equal to 5 lbs. ; what resistance will it encounter if its speed be increased to 49 miles per hour ? SOLUTION. , Here the velocity increases 7 times, (t. g.. *^). i - Hence the resistance increases 7' =49 times. . '. Resistance — 5X49 = 245 lbs. ; L e., 245 units of work are expended bVery foot in overcoming the atmospheric resistance. Example 152. If a railway train moving at the rate of 10 miles per hour encounters an atmospheric resistance of 33 lbs.; what must be the horse powers of the locomotive in order that the train may move 60 miles per hour, neglecting friction and asBumixg the rail to be level? . B i M i i ! t ■ I 68 WORK ON A LEVEL PLANE. CAST. 143. SOIVTIOW. Here the velocity is increued 6 times, since f ^ = 6. Then the reftistanee is increased 86 times (Art. 147). Hence atmospheric resistance =: 33X36 = 1188 lbs. ; 1. e., 1188 units of work are expended in moving the train through 1 ft. 60X6280 Number of feet train moves through in a minute =: Units of work required per minute "> _ , , aa v Mon to overcome atmospheric resistance } ~* **«»'» ^^ «>*«»«. 60 = 5280. H.P «i it 1188X6280 ,^„„ . of locomotive = - ,„^^^ — = 190*08. Ans. 33000 Example 153. — What must be the H. P. of a locomotive to more a train at the rate of 40 miles per hour on a level rail, taking atmospheric pressure as usual, (i. e., 33 lbs. ^hen train moves 10 miles per hour,) and neglecting friction? soLxrrioN. Here velocity increases 4 times, and hence resistance increases 16 times. Then resistance encountered = 33X16 = 628 = units of work required per foot. Feet moved over per hour = 5280X40; hence units of work per hour =: 6280X40X628. ™„ - « « 828X40X6280 „.„, . Therefore H. P. = 33000^^^ = 56-32. Ans. Example 154. — What must be the H. P. of a locomotive to draw a train whose gross weight is 80 tons, along a level rail, with the uniform velocity of 40 miles per hour, taking atmos- pheric resistance and friction as usual ? soLuizoir. . ^ 40X6280 Feet passed over per minute = — — — = 3620. Work of flriction per minute = 80X7X3620 = 1971200 units. Work of atmospheric resi8tanoe=33X16X3520=1858660 units. Therefore H. P. Work of friction -f work of atmospheric resistahce Work of one U. P. 33000 83000 Example 155. — What must be the H. P. of a locomotive to draw a train, whose gross weight is 125 tons, along a level rail with the uniform velocity of 42 miles per hour, taking friction as usual, and assuming that the atmospheric resistance encountered by the train is equal to 10 lbs. when moving at the rate of 7 miles per hour ? aoLVTioir. 42XS280 Feet moved over per minute = =lii^i£:? — 3690. 60 Work of friction per minute = 126X7X3696 = 3234000 units. Work of atmospherio;re8i8tanceperminute=10X36X3696=ri330560 units. -.„___ Work of ftricMon 4- work of a^ospheric resistance _ inen n.i'. — Work of one ff. 1*. ' ~ 8234000+1380660 4864860 ,.„., . — ^ 1S8.32, Aus. 33000 830OO iBT. 148. Abts. 14». ISO.l WORK ON AN INCLINED PLANE. 69 units of 5280. lOtive to 3vel rail, en train 16 times, t required er hour = oaotive to evel rail, ig atmos- resiatatice notiye to level rail riction as lountered rate of t OB60 units, si stattce _ 157. EXERCISES. 156. If a train •neonnters an atmospheric resistanet of 8 lbs. when moring at the rate of 6 miles per hour, what re- sistance will it encounter when its speed is increased to 45 miles per hour ? ytfn«. 648 lbs. What must be the H. P. of a locomotive to draw a train at the rate of 30 miles per hour on a level rail, assuming that the atmospheric resistance is equal to 9 lbs. when the train moves 6 miles per hour, and neg^'^ct*'* 'riction 7 fn«. H. P = 36. What must be the H. P. of a locomotive to draw a train weighing 140 tons along a level rail with the uniform velocity of 36 miles per hour, taking friction as 7 lbs. per ton, and the nresistance of the atmosphere 12 lbs. when the train moves 9 miles per hour? Jlns. H. P. =r 112'512. A train weighing 200 tons moves along a level rail with a uniform speed of 30 miles per hour ; what are the H. P. of the engine—friction and atmospheric resistance being as usnal? ^ns. H. P. = 135-76. 168. 159. 149. If a body be moved along a surface without fric- tion or atmospheric resistance, the units of work performed are found by multiplying the weight of the body in lbs. by the vertical distance in ^et through which it is raised. Thus, if a body weighing 12 lbs. be moved 200 feet along an inclined plane having a rise of 19 feM in 100, the units of work performed will be 12X19X2 = 456, because in moving up the plane 800 feet, the body is raised through 19X2 = 38 feet. 160. When a train is moving along an inclined plane, and- the inclination is not very great, the pressure on the plane is very nearly equal to the weight of the body. Hence we find the work due to friction by Arts. 143-146, the work due to atmospheric resistance by Art. 148, and the work due to gravity by Art. 149. ExAMPLS 160. — A train Weighing 90 tons is drawn up a gradient having a rise of 3 feet in every 1000 feet^ with the uniform speed of 40 miles per hour— neglecting friction and atmospheric resistance, what are the H. P. of .the engine 7 eoLusioir. Weight of trahi in lbs. =s 00X2000 = 180000. I'eet travelled per minute = — tz — « 3520. Vertical distance moved through per minute ss ^^ oi 8620 at lO'fie ft. tJnits of work due to gravity per minute =s 10'66 X laoooo. ...H.p^J225>^:i. 57-6.-1^. V . 33000 -i A . J 60 WORK ON AN INCLINED PLANE. rAm. 150. I i< Example 161. — ^A train weighing 140 ions moves up a gradient having a rise of 3 feet in 1100 feet, with the uniform velocity of 36 miles per hour— neglecting atmospheric resistance and taking friction as usual, what are the H. P. of the locomotive ? SOLUTION. Hereweightof train in lbs. =E 140X2000 =s 280000; audsneod pprminute 86X6280 .,„„, . = — — — = 8168 feet. 60 The units of work due per minute to friction = 140X7X3168 = 3104640. Height to which train is raised per minute == j~^ of 3168 = 8*64 ft. Then units of work due per minute to gravity = 8*04X280000 — - 2410200. work diio gravity+work due friction 3104640+2419200 .-. H.P.= 6528840 WorkofoneH. P. 33000 83000 = 167*889. Arts. ExAMPLB 162. — A train weighing 100 tons moves up a gradient with a uniform velocity of 30 miles per hour, the rise of the plane being 3 feet in 1000 feet, and taking friction and atmospheric resistance as usual, ivhat are the H. P. of the locomotive? SOLUTION. Here weight of train in lbs. ■■= 100 x 2000 = 200000 ; space passed per => 2640 ft., and elevation of train per minute minute nfW ThenH,P.= . . „ „ 1848000+784080+1684000 60 of 2640 = 7*92 ft. Work of friction per minute = 100X7X2640 = 1848000 units. Work of atmospheric resistance per minute =33X9X2640 =784080 units. Work of gravity per minute = 7*92X200000 = 1584000 units. _ Workdu« tofrio. permln.+wnrkdaftoiitinoi. re»Ut. parmln.+workd ueto gr»T. rermlll. Uoiu of work in one U. If, 4216080 ,„^.^„ . 83000 = IsoUr* ^27*76. ^«,. * Example 163. — A train weighing 130 tons descends a gradient having a rise of 1 ft. in 2000 ft. with the uniform velocity of 60 miles per hour — taking atmospheric resistance as usual, and the coeflBcient of friction 7^0", what are the horse powers of the loco- motive ? SOLUTION. Here weight of train in lbs. = 130X2000 = 260000 ; space passed over per minute = 60X6280 60 =6280 ft. ; increase in the velocity = ^§ = 6 ; and verti- cal fall of train per minute = j^ of 6280 ft. = 18*48 ft. Then work of friction per minute = j^'X.26(mQX5280 = 1300X5280 ^ 6864000 units. Work of atmospheric resistance per minute=33X 36X6280=6272640 units. Wbrk of gravity per minute=18*48X 260000=4804800 units. Then, since the train descends the gradient, ^rravi^j^ acts with the engine. n -a -a Work of flriction+work of atmos. resist. —work of gravity . Work of one H. P H. P. =; 68g*ooo+ogy^g^o-"^Q*so<> S3000 S8000 km. 150. radieut Dcity of 1 taking r minute tl0464(«. *64ft. 2410200. kl9200 gradient he plane ospheric ussed per ~ rooiT )80 units. {r«T. r«rtBlw. gradient ty of 60 and the le loco- over per ,nd verti- X5280 == 340unitft. e engine. gravity. \Jnt. AEI.150.J WORK ON AX INCLINED PLANE. 01 ExAifPLB 164.— A train weighing 80 tons moves along a gradient with the uniform speed of 40 mjles per hour— assuming the inclination of the gradient to be 3 ft. in 1000 ft., and taking friction and atmospheric resistance as usual, what will be the H. P. of the locomotive : 1st. If the train move up the gradient, and 2d. If the train move down the gradient ? BOLUTIOir. Here weight of train in lbs. := 80X2000=160000 ; space passed over per minute = — — — = 3520 ft.; velocity is increased f£ = 4 times, and ver. tical ascent or descent of train j-^^ of 3620 = 10*50 ft. Worlc of friction = 80X7X3520 — 1971200 units per minute. Worlc of atmospheric resistance = 33X16X8520 = 1858660 units per min. Work of gravity = 10'56X 160000 = 1689600 units per minute. Work of friction+work of atmos.rcsist.iwork of gpravity. Then H. P. = Train ascending, H. P. = Train descending, H. P. = Work of one H. P. 1071200+18585604-1689600 33000 19712004-1868560— 1689600 6519360 33000 2140160 = 167-253. = 64-853. 33000 33000 Example 165. — A train weighing 110 tons ascends a gradient having a rise of | in 100 — taking friction as usual, and neglecting atmospheric resistance, what is the maximum speed the train will attain if the li. P. of the locomotive be 120? 80LUTI05. Hereweightof train in lbs. = 110X2000 = 220000. - Work of friction in one mile = 110X7X5280 = 4065600 units. Work of gravity in one mile= Tolf of 5280 =6-6X220000 = 1452000 units. Total work of resistance in 1 mile = 4065600+1462000 = 6617600 units. Total work of engine per hour = 33000X60X120 = 237600000 units. 237600000 Number of miles per hour = • : 43*06 Ans. 6517600 Example 166. — If a horse exert a traction of 120 lbs., what gross load will he pull up a hill whose rise is 17 feet in 1000 ft., assuming the coefficient of friction to be -^^ ? SOLUTION. ; . . Workofhorse in moving the load over 1000 ft. = 120xi000 =126000 units. Work of friction in moving 1 lb. over 1000 ft. = IXiVxiOOO = 100 units. Work of gravity in moving 1 lb. over 1000 ft. = 1 x 17 — 17 units. Total work in moving 1 lb. over 1000 ft. = work of friction + work of gravity = 100+17 = 117 units. . • . Number of lbs. drawn by horse = i^j)^ — 1025*641. Ans. Example 167. — What backward pressure is exerted by a horse in going down a hill which has a rise of 7 feet in 100, with a load whose gross weight is 2000 lbs,, assuming -gV to be coeffi^ cient of friction? 1, ll 6d WORK ON AN INCLINED PLANE. [Aii.161. •oiVTioir. thtk oilftl«v«lpl»iieth«fHottonwouUlb«|>r of 806« )bf.s=Bri« Ibt.o: ttiiiUi of work for eaott foot. Work of RTftvity =: T?iv of 8000 = 140 uniti to eaeh fbot. Therefore, the baokwMrd preMure ii 144>— S7'14 =3 srM lb*. Am. EXERCISES. 168. What backward pressure will a horse exert in going down a hill which has a rise of 9 feet in 100, with a load whos« gross weight is 1200 lbs., assuming the coefBcient of frio* tion of the road to be ^? Jm. 68 lbs. 169. What gross load will a horse exerting a traction of 150 lbs. draw up a hill whose inclination is 3 in 100— assuming the coefficient of friction to be iV ? ^^n'- 1651*73 lbs. 1 70. What will be the maximum speed attained bj a train weighing 200 tons, drawn by a locomotive of 160 H. P. up a gradient having a rise of ^ in 100— taking ftriotion as usual and neglecting atmospheric resistance 7 Jnt. 29*032 miles per hour. 171. A train weighing 88 tons moves up a gradient having a rise of i in 100 witn the uniform velooitj of 20 miles per hour —taking friction and atmospheric resistance as usual, what are the H. P. of the lo.r:ite per rain.iu ft. w 'mm weight of train in tons, then 2000u; >■ weight of train in lbs. h mm rise of the plane in every 100 feet. /— friction per ton. i^aM given atmospheric resistance at given speed, «. Then units of work due per minute to friction =/u>X88r. <♦ '♦ to gravity =s 2000to X Ti^z- X 88 r =s 20 Aw X 88 r. Units of work due per min. to atmos. resist. = R i—\ x 88r. Units of work per min. in given H. P. = H. P. x 33000. henoeH.P.X33000=s/M>X88r-fil(Yj x 88r±20Awx 88r, and, factoring this, we get : H. P. X 33000 =:{fw ■\' R irjf^ 20Aw) 88r. „/r\« 88r Therefore H.P.= (> + ^^yj ± 20ftw) -^^^. Or H. P. = 0+il(y) ± 20 hvf)-^ (I). From this we obtain by transposition and reduction, and neglecting atmospheric resistance, H.P. X3y5 (/'±20A)r ("-^ H. P_21ill *'-(/±20A)ii, ("^>- Bince/ is commonly = 7, Ji = 8:i. and » ~ 10, these formulas become reBpcctively, H.P.= (7w+-38r«i20Atc)^ (IV.) H.P.X875 .'^'^ : (7taO*)r Ki) . . . H.y.xa76 (7_+20*)ic (H.) 64 THE MUDULUH OF MACHINES. ! ARi*. iM, loa ExAMPLB 174.— A traia weighing 140 tons moves along a gradient harlng a ride of | in 100 with tho uniform speed of 30 miles per hour ; taking friction and atmospheric rcsintanco us usual, wiiat are tho H. P. of tlio locomotive ; Ist, when tho truin moves up the gradient? 2d, when tho train moves down tho gradient ? SOLUTION. Uoro w -is 140, ;• ~ 30, h - |. U. P. - (7w+'33r2 iidhw)—;^ . . = (Txi-to+'asxao' i mklxiia) ^i^\ — (080-1-207 1 700)-^ 1977 X« or 25 577X2 25 20 =- 168'16 or tO'lO. Ans. ExAMPLB 175.— A train drawn by a locomotive of 80 II. P. moves along an inclined plane having a rise of j^ in 100 with a uniform velocity of 45 miles per hour; taking friction as usual and neglecting atmospheric resistance, what is the weight of the train ? SOLUXIOX. Here H. P. — 80, r - 45, and 7i = ^ Then by formula (V.) to = "~™ 80X375 .■iOOOO 30000 soono .10000 30000 (7:h20xJ)46 (7±3p45 10} X45 *"• -83^4^^ -4flr-°''"n«r"= «*■" *°"» *' ^^^ *"^*" *» 8°*"» up the gradient, or 181*81 tons if the train is KoiuR down the gradient. For practice in the application of these formulas, work any of the fore- going problems. THE MODULUS OF A MACHINE. 162. Tho Modulus of a raachine is the fraction which expresses the vahie of the work done compared with the work applied, the latter being expressed by unity. Thus if T of tho work applied to a machine be lost in transmitision, the modulus or useful work of that machine is ^ ; if § be lost in transmission, the modulus of the machine is f) &c. 153. The amount of work lost depends on fiiotion, rigidity of cordage, #. the stelMaa genemtied is Ui^ iii quantity and tfosseosed of obnipaatbehr Itttla elastic force ; if the pressure be high, the steam generated w (tf smaU quan* tity, but of high elastic force. 164. High pressure engines are commonly used where it is desirable to have the engine as simple, chciftp, compact, and light as possible, as the condensing apparatus renders the engine more costly and cumbrous. The high pressure engine is, however, far more liable to burst and get other-' wise out of repair, 166. The units of work performed per minute by a steam engine are found by nmUiplyiny together the pret* sure per square inch on the boiler , the area of the piston in inches, the length of thx stroke of the piston in feet^ and the numher of sitroka per minute. Thus let the pressure exerted on nch square inch of the piston be 80 lbs., and let the piston mue 40 strokes perminute of 8 ft. each, also let the area of the puiton be 100 sqiiare inches s Now if a weight of 80 lbs. be placed on each square inch of the surf^e of the piston, the elastic force of the steam will be Just sufficient to lift the loaded piston through the length ^f the 8tA)ke in opposition to gravity, then the work performed oH 1 sq. in. or the piston would be 80X8 for each stroke. Work performed oh whole piston wotud be 30X8X100 forleach stroke. Work " " " " " 80X3X100X40 per minute. 166. In the high pressure engine, the pressure of the atmosphere, about 15 lbs. to the square inch, acts in op- position to the pressure of the steam; and in the low-pres- sure or condensmg engine a pressure of about 4 lbs. to the square inch of the piston is exerted by thn vapour in the condensing chamber. Besides these, a resistance of 1 lb. per square inch is commonly allowed for the friction of the piston. Deducting thdiealbwsnces f)fl>m the total pres- sure, we obtain the emctive pressure ; and we must further make an allowance of | of this for the friction of thf whole engine. liMlli iiliili..tt 68 WOKK OF STEAM. [Aet. 167. Thus in the high pressure engine : - Load-^\ load+ + 16 — whole pressure, . lu the condensing engine : Load + I load + 1 + 4 ■— tohole pressure. For cxainplo,— if the whole pressure be B8 lbs. per square inch, Then for the high pressure engine 68—1—16=42 is the working pres- sure on the piston, and 42 is f (i. c, load + | load) of the useful pressure* and hence useful or effectlvo prossure=:42-f-*— 36^ . Por tl^e low pressure engine 88—1—4 = 63 = working pressure on the piston, and 63 is i^ of the useful pressure. Therefore useful or efrectlvc pressure Is 63 -f- f = 46^^ 167. For finding the H. P. of astoam engine, let p = useful pressure in lbs. on each square inch of the piston, a ma area of piston, / =» length of piston stroke in feet, and »«*■ number of strokes per minute. pain « Then H. 1^. p. a n I 33000 H. P. X 33000 a In H. P. X 33000 pin H. P. X 83000 pal H. P.X 33000 pan ■ (11.) • (I") • (IV.) .(V.) Example 1*78. — The piston of an engine bas an area of 260 inches,and makes 110 stroke;^, of 6 feet each, per minute — taking tbo useful pressure of the steam as 28 T s. per sq. inch, what are the H. P. of the engine? soiuTioir. Here p — 28,a = 260, n = 110, and 1 — 5, 28X250X110X8 Then (formula I.) H. P. » 83000 llOf. Ant. ExAMPLB 179. — The piston of a high pressure engine bas an airea of 1200 inches and makes in each minute 30 strokes of 7 feet each — taking the gross pressure of the steam as 48 lbs. per square inch, what are the H. P. of the engine ? [Aet. 107. WVftM Hl/'«"-«»TS- Aet. 167. 1 WORK OP STEAM. 60 king pres- 1 prosBuroi tre on the r effeotivo lot p = > piston, in feet, of 260 taking lat are has an 98 of 7 bs. per SOLUTION. Hero 48 — p+\p'\-U-\-l,or^^p =: 32, and hone© p ~ 32-T-f Then p — 2»,a — 1200, n — 30, and 1 — 7. 28X1200X30x7 2« lbs. By formula I., H.r. 33000 — 213-.S1. An9. Example 180. — The piston of a low pressnre enplnc has a diameter of 20 ins. and makes 60 strokes of 4 ft. each, per minute — the pressure of the steam on the boiler is 45 lbs. to the sq. inch, what are the H. P. of the engine? SOLUTIOjr. Hero 43 — p-!-|p+44-l, or fp — 40, and henco p — 40-r-f s= S.*}- n* -.- 10« x;j-1410 ~ 100X3'1416 = 314'16. Tlion p -= 35, a --= 314-16, n — 00, and I — 4. 35X314-10X60X4 H. P. =^ 33000 = 70*908. Ant. Example 181. — In a steam engine of 32 horse power, the area of the piston is 600 inches, the length of the stroke 4 feet, and the useful pressure of the steam 33 los. to the sq. inch, how manj strokes does the piston make per minute ? BOLXTTIOW. Here, H. P. — 32, a -- 500, 1 = 4, and p = 38 m.. /» .1 T^x H.P.X33000 32X33000 Then (formula IV.) n = 10 Ana, pal 600X4X33' Example 182.— In a low pressure steam engine of 190 H. P. the area of the piston is 1000 inches, the length of stroke 6 feet, and the number of strokes per minute 110, what is the useful pressure per square inch on the piston, and also, what is the gross pressure of the steam ? SOLUTION. Here, H. P. = 190, a — 1000, i = 6 and w = 100. Then (Formula II.) p = i^^g^^o =** ^^'' ~ "'®'"^ pressure. And pressure on boiler (Art. 166) = 04+ j of 9J+4+1 = 16lf lbs. Example 183.-^In a high pressure engine the piston has an area of 800 inches, and makes 40 strokes per minute, of 10 feet each, what must be the pressure of the steam on the boiler in order that the engine may pump 120 cubic feet of water per minute from a mine whose depth is 400 feet — making the usual allowance for friction and the modulus of the pump 7 * When the diameter of the piston is given, its area i« found by multiply* ing the square of half the diameter by 8*1416. to ^ Work op stEAHf. tABT.ier. I ■■ 80LVTI0V. Here, work done per minute = l<0X6t'8X400 ss 8000000 units. ~ Work applied. «. «.. work of engine = 8000000-f-l = 4BO0O0O units s: H. P. X88000. Then by Pormul. II. p ^ H. P. X 88000 ^ 4600000 j. , ibg.=u«,fui aln 800X10X40 ^» pressure. And Art. 166, gross pressure — 1*^+1 of 14^g-fl6+l ^ Sirh ^' ^^*' ExAMPLi 184.—- The piston of a high pressure engine has an area of 600 inches, and makes 20 strokes per minute, each 8 ft. in length, gross pressure of the steam 52 lbs. to the square inch. How many gallons of water per minute will this engine pump from a mine whose depth is 600 feet, making the usual allow- ance for friction and the modulus of the pump ? BOLVTIOV. Herea = 600, I = 8,n sr 20, and since 62 =!> + 13> 4* 18 -f 1 ; fP ~ S6, and p =. 81i. Work of ei)gine per niittute=:pa2fi = 311X600X8X20 = 8024000. TTsefUl work per minute s= 30$400pX| =» 20111000. Work of pumping 1 g^on of water to height of 600 feet = 10X500 ^ 6000 units. . •. No. of gallons pumped per minute = * Vlft»^^ = 408i. Ans. EXERCISES. 186. The piston of a low pressure steam engine is 40 inches i& diameter and makes 40 strokes of 6 feet each per minute ;— the gross pressure of the steam is 37 lbs. per square inch ; wbAt are tlie &. P. of the engine ? jin*. 2 13'248. 186. The piston of a high-pressure engine is 20 inches in diame- ter and makes 50 strokes of 4 feet per minute ; taking the gross pressure of the steam as 40 lbs. per square inch and making the usual allowance for friction, what are the H. P. of the engine ? Ant. 39 '984* 187. The piston of an engine has an area of 2400 inches and makes 16 strokes per minute, each 10 feet in length ; the useful pressure of the steam on the piston is 20 lbs. per jr. 'pqjBLMp inch, what are the H. P. of the engine 7 • ,r Jlru. 33;2-72. 188. In a high p.reasure engine of 140 H. P. the piston has an area of lOOO inches, and makes 20 strokes, of 5 feet each, per minute ; what is the useful pr^sure of the steam on the piston and also the g'rose preslrare per square inch? - Jtni» Useful pressure r: 46*2 lbs. per sq. in. Gross pressure =: 68'8 ll^s. per iq. in. I ii 2 ii I Art. 168 J WOUK OP STEAM. U 189. In a low preiisure engine of 100 H. P. the piston has an area of 200 inches and makes 40 strokes per minute ; the gross pressure of the steum is 45 lbs. per square inch. Required the length of the stroke made by the piston? An^. 11*786 feet. 190. In a high pressure engine of 80 H. P. the piston makes 44 strokes per minute, each 6 feet in length, and the gross pres- sure of the steam is 66 lbs. per square inch. What is the area of the piston? Aiu. 286* 714 Bq. in. 191. How many cubic feet of water may be pumped per minute from a mine whose depth is 600 feet by an engine in which the piston has an area of 2000 inches, and makes 30 strokes per minute, each 8 feet in length, the useful pressure of the steam being 40 lbs. per square inch, and the usual allow- ance being made for the modulus of the pump? Jlns. 409*6 cubic feet. 168. In all the modifications of the Bteam engine, the real source of work is the evaporating power of the boiler; the amount of work done hj the engine depending not only upon the rapidity with which the water is evaporated) but also upon the temperature^ and consequently tne pressure under which the steam is produced. The foUowmg is a specimen of an experimental table, given by Pambour, showing the relation between the pressure, temperature, and volume of the steam produced by one cubic fiot of water. By means of this table^ we are enabled to ascertain the volume of the steam produced by a given quantity of water, when we know the pressure or temperature under which it is formed. NovB 1.— ^e first column gives the inresiore In lbs. to the square inch Under which the steaiti is produced) the second column shows the correspond* ing temperature, an indicated by Fahrenheit's thermometer; and the third OoTumn, the volume of the steam commred with the volume of the water which produced it. It will be observed that the lower the temperature, or what amounts to the same thins, the less the pressure under which the steam is ftarmed. the greater itis volume. Thus under the usual atmospheric Eressure of 15 lbs. to the square inch (or at the common temperature of oiling water, 212^ or 213^ Fahr.), a cubic foot of water produces 1609cubib ISset at steam. If, however, the pressure be deoroasod to l lb. to the square inch, the steam is formed at the temperature of 103" Fahr, and occupies 20064 enble fSseli ; while if the pressure be increased to 30 lbs. to the square inch, the temperature required for the production of the steam rises to 251' Fahr. and the steam only occupies 882 cubic feet. Non 2.—^ has been shown by nuaierotts experiments that the quantity of Aiel requisite for the evaporation of a given quantiigr of vafeor is in> variably the same, no matter what may be the premie under which the steam u produced. Hence it Is obvious that it irmbst atdHmiaipBr to employ steam of a bJj|h prewnve. i; I . 1 1 I ; 't ■ I f 72 WORK OF STEAM. TABLE [Art, 160 SITOWIZrO TBB VOLUME OF 8TBAM FBODUCBD BT ONE CUBIC BOOT Of WATBB AT TUB COBRBSPOITDINO FBBBSUBB AND TBMFBRATUBE. if s 1 6 10 16 20 86 30 36 40 46 60 lOS* 161«» 192° 213" 228« 24l» 251« 260» 268» 276° 282«> 3^2 20954 4624 2427 1669 1280 1042 882 765 677 608 652 so £: MS Yolume of one cubic foot of water in the form of steam under the given pressure p. Then to find a, /, w, p^ c, or v, when the others are given, we proceed as follows : » :^ When p ig given, v is found by the table. Art. 160. Akt. 169. 1 WOBK OF STEAM. 73 FOOT or ruBE. to I inches. piston. minute. q. incli ted per f water der the given, Now the cubic feet of water evaporated per minute = ev, cubic feet of steam used at each stroke of the piston »- J— - ^ 144* nn.1 . ' . cubic feet of water used in n strokes >» — ao also 144 * the steam evaporated or used per minute. cVy and from this by reduction we obtain 144cv 144CV naJ nal Hence nal "144 144C2; W: a. naJ na ' al ^ nl When V is known p may be found by the table. Example 192. — The piston of a steam engine has an area of 200 square inches and makes a stroke 4 feet in length, the boiler evaporating -^s of a cubic foot of water per minute, under a pressure of 40 lbs. to the square inch. What number of strokes per minute does the piston make ? SOLUTION. Here a = 200, ? = 4, c = -^ = '3, and p — ifS', also from table v — 877. _, lUcv 144X-8X677 ,^.„„ ,„, ^ Then » = -^- = toOX4 = ^ "® ""^ = *«*• -*»'• Example 193 — The piston of a steam engine has an area of 1000 inches and makes 10 strokes per minute, each 3 feet in length, the boiler evaporates '4 of a cubic fcot of water per minute. What is the pressure under which tht) steam is gene- rated ? 80LUTI0W, Here a =1000, 2 = 3, n =10, and c = '4. _, nal 10X1000X3 ,„, a. ». *i. * i,, , -^ x Then v = rrr- = = 621, whence by the table, p is between 80 and 65, or about 63 lbs. Example 194. — The piston of a. steam engine has an area of 80 inches, and make 20 strokes per minute ; the boiler evaporates iV of a cubic foot of water per minute under the pressure of 50 lbs. to the square inch ; required the length of the stroke mad* by the piston. 80LUTI0W. Here a=80, n=20, a?rt 47.25 X 200 X 6 X 19.488 Then Art. 167, H. P. - ^^rxx = — 3:J'48. Ant. 33000 33000 Example 202. — What are the eflfective horse powers of a low pressure engine in which the piston has an area of 288 inches and makes every minute 16 strokes, the boiler converting \ of a cubic foot of water per minute into 304 cubic feet of steam? bOIiUTION. Since i of a cul)ic foot of water produces 304 cubic feet of steam, 1 cnl)ic foot of water would produce 608 cubic feet of steam, and hence (Table) tho gross pressure of the steam is 45 lbs. to the square inch. Then (Art. 166) 45 =7 jp + 4 + 1 or f P = 40 whence p = 35. Also (Art. 169) i = 144c« 144X'5X608 na :94 ft. 288X16 Then a — 288, i = Oi, n = 16, and p = 35. X. 1 T **,/.* TT Ti P«^» 35X288X94X16 ^„.,„^ . I lence Formula I Art. 167. H. P. = ^— = g^^ = 46*42». An$ EXERCISES. 203. What are the effective horse powers of a high pressure engine in which the piston has an area of 360 inches and makes 20 strokes per minute, — the boiler evaporating I of a cubic foot of water per minute under a pressure of 40 lbs. to the square inch ? w5n«. H. P. = 46-528. 204. The piston of a low pressure steam engine has an area of 432 inches and makes strokes 10 feet in length. Now, if the boiler evaporates -9 of a cubic feet of water per minute under a pressure of 25 lbs. to the square inch, what are the useful H. P. of the engine ? Ana. H. P.= '71-G13, 205. In a high pressure engine the area of the piston is 600 inches, the length of stroke is 6 feet, the effective evapo- ration of the boiler is i of a cubic foot per minute and the pressure of the steam in the cylinder 80 lbs. to the square inch ? Required the H. P. Ans. H. P. = 32-897. 76 HYDROSTATICS. [Abii. 171-170. Ill CHAPTER V. HYDROSTATICS. 171. Fluidity consists in tlio transmission of pressuie in all directions, or, a fluid raay be defined to be a body whose particles are so free to raove among one another til. it they yield to any pressure, however small, that may be applied to them. 172. The term lluid is ooinmonly applied to bodies in both the liquid and ga«cous slate. 173. Fluids are divided into two classes: — Ist. Elastic fluids, of which atinosphcric air is the type, 2d. Non-Elastic fluids, of which water is the represen- tative. Note.— Water wa.s formerly thought to be absolutely incompressible' but recent oxpoJ imeuts show that water is diminished in volume t2 olTO of its bulk for c: .ch atmosph -re of pressure upon it ; or in other words a pressure of 20i)C atmosphere or 30000 lbs. to the square inch would compress 11 cubic fyet iito 10 cu'jic feet. Alcohol is about twice as compressible as water. 174. Liquids, by which leini we mean non-ehistic fluids, differ from gases chiefly in luiving less elasticity and com- pressibility. 175. Liquids ditt'er from F^olids chiefly in the fact that their particles are less under the influence of the attraction of cohesion, and therefore have a freer motion among them- selves, in consequence of which each atom is drawn sepa- rately towards the earth by the force of gravity ; hence : — /. .4 liquid, confined in any vessel, presses equally in all direc- lions — upwards, downvmrds, and laterally. 11. The surface of a liquid in a state of rest is always level. 111. A liquid rises to the same height in all the tubes connected with a common reservoir, whatever may be their form or capacity Note.— The facjt that a liquid exerts a downward pressure is self-evident and reqires no ilhistration. The lateral pressure of liquids is shown Viy their spouting from holes pierced in thi side of the vessel in which they are contained. The upwar.. pressure is shown by taking a glass cylinder, open at both ends, and having one end ac(!urately ground. A plate of ground glass i" held to this end bv nieanH of a piece of string passing through the cylinder and the closed end of the instrument then emereed in water to a small AlTi. 17»-178.J HYDROSTATICS. 77 at both glass i«! cylinder a small depth. lT))on lettlriK fo tlio string tho plate Is still hold against the cylin- dur by the upward pressure of tho water, it wili even sustain any woiKht, which, together with the plate itself, is not greater than the weight of the water that would enter the cylinder if the plate were removed. 176. When two liquids of different densities are placed in the opposite branches of an inverted syphon or bent tube — their heights in the two legs above the point of con- tact will be inversely as their densities. Note.— This may easily be proved by placing mercury and water in a bent graduated glass tube, when it will be found that tho column of water will be 13i times as high as th« column of mercury since tho latter is about \^ times as heavy as the former. 177. The amount of downward pressnre exerted by a liquid in any vessel is equal to that of a column of the same liquid, whose base is equal to the an a of the bottom of the vessel, and whose height is equal to tho depth of tho liquid, whatever ntay be the form or capacity of tho vessel. Note 1.— To illustrate this fact we procure three vcssesl, having bottoms of the same area, and sides, in tho first pcrpetidicular, in the second conver- ging towards tho top, and the third diverging towards the top. The bottoms arc hinged and are iield in their places by a cord passing over a pulley and terminating in a scale pan in which are placed weights to a certain amount. Water is then carefully poured into the vessel having the perpendicular •ides until its downward pressure is just sufficient to force out the bottom when its depth is accurately measured. Upon using either of the other vessels it is found that the bottom remains fixed until the water reaches this depth and is then forced open. This arises from the fact that when the sides are perpendicular the bottom support tho whole weight of the water. When the vessel is wider at top than at bottom a portion of the downward pressure is sustained by the sides, while, when the vessel is wider at tho bottom than at top, the particles near the bottom are pressed upon by the whole column of lihole gate= 16X12X6X62-5=72000 lbs. Pressure on upper third = whole pressure X {i)*=72000X^ I. II. 8000 lbs. III. Pressure on loMi-er half = whole pressure x f 1— {i)« j=72000Xf= 64000 lbs. IV. Pressure on upper two-fifth8= whole pressure X (|)* =72000 X ■^^; = 11620 lbs. V. Pressure on lower two-elevenths - . whole pressure X f l~(-ft-) ' j == 2000 X -^jQj- = 23801'6528 lbs. In III we take the given fraction i from unity this leaves i which we square and a);ain subtract from unity and thus obtain } for the multiplier. In y we take the given fraction -^j from unity, this gives us -^^ whioh are square and again subtract from unity tlius obtaining -^^ for the multi- plier. ExAMPLK 229. If a flood gate be placed as in last example what pressure will be exerted on the upper ^ and what on the lower I of the gate if it be 10 feet wide and 12 feet deep ? ■ SOLUTIOir. We Itrst find the pressure on the whole gate by Art. 179. Then for the upper ^ we multiply the whole pressure by the square of ^. For the lower 2. we subtract ^ from 1, this gives us g which we square and thus obtain -^, then we subtract -^i,. from 1 and thus obtain ^.lastly we multiply the whole pressure by this ^^j "Whole pressure = 10X12X6X62'5 = 45000 lbs. Pressure on upper ^ = 45000 X'^ff = 8265^ lbs. Pressure on lower I = 45000 xM = 28800 lbs. EXERCISES. 230. The flood-gate of a canal is 30 feet wide and 10 feet deep, and is placed vertically in the canal, the water being on one lide only and level with the top, required the presBure — Ist. ill' AST. 181.] HYDROSTATICS. On the whole gate ; 2Dd. On the upper half of the gate ; 3rd. On the lower half of the gate ; 4th. On the lowest two- sevenths of the gate. Am. Pressure on whole gate = 93750 lbs. " upper half = 23437} " lower half = 70312} " lowest two-sevenths = 45918^1 " 231. A hollow globe has a surface of 7 square feet, and is sunk in water to the depth of 150 feet. Required the total pres- sure it then sustains. Ans. 65625 lbs. 232. What pressure is exerted against one square yard of an embankment if the upper edge of the square yard be 11 ft. and the lower edge 13 feet beneath the surface of ths water? Ans. 6750. 233. A hollow glass globe is sunk in water to the depth of 400 feet, at which point it breaks. Required the extreme pressure to the square inch which the vessel was capable of sustaining. Ans. 173'568 lbs. 234. Required the pressure sustained by the body of a man at a depth of 100 yards beneath the surface of water — assuming the man's body to have a surface of 15 square feet? Ans. 281250 lbs. 235. A flood gate 16 feet long is submerged to the depth of 9 feet in water of it? what pressure is exerted against each side Ans. 40500 lbs. 230. A mill dam is 120 feet long and 11 wide, the water being exactly level with the top of the dam and the lower edgt of the dam 7 feet beneath the surface. 1st. What will be thj pressure exerted against the whole dam. 2nd. What pres- sure will be exerted agamst the upper part of the drou. 3rd. What pressure will be exerted against the lower iaif of the dam? Ans. Against whole dam 288750 lbs. " upper half 72187} lbs. « lower half 216562} lbs. 237. A flood gate 26 feet wide is submerged perpendicularly to the depth of 12 feet ; find 1st. The pressure against one side of the whole part submerged. 2nd. The pressure against the lower half. 3rd. The pressure against the lowest third. 4th. The pressure against the lowest sixth ? Alls. 117000 lbs. whole gate. 87750 lbs. lower half. 65000 lbs. lowest third. 35750 lbs. lowest sixth f^i Htl>R08TAttCS. [Ants. l««, 18.1, m 162. If wnter bo conflnofl in n vessel siiul n proHMiro to jiny ftinoiMit bo oxerlinl npoii niiy ono sqimro inch of tba surface of tlmt w?iter, a pressure lo an e(pnil amount will bo transmitted to every s-jnani incli of the interior surface of tbo vessel in wliieli llie water is condniMl. rii?. If,. NoTB.— la tlio n^TotiiimiiyiiiK Okh''*' ciuppoHn tb(» ))istoti /* iiaH tin an-ji of 1 Hqimrn inoh, niut tho piNton >>' an area of n>0 nquaro iiichcH, thon \f I lb. pros- nnro h<> appHcMi to /' as wimkIiI or loi) lb. nuirtt bo appUcnl to p' in ordc^r to maintain oquilinriuni. It iH this pro- pcrty of ,'(inal and instant transniix- Bion of jircHNiiro wiiioli (MiablrH ns to innko UNO of li.vtlroNtalic prcsHnr<> as a nio('bani«'al ijowcr, and it is upon tliis nrlnolplo tliat Uraniaii's Jl.vdrostatic VroNN IS constnu'tcd. \ 183. Hram all's II, ydrostatic Tross consists of two stronjr motaH'o cylinders /I ajid (t, one many times as large as tlur other, conned ed «. ^^ together by a tube. The small cylinder jci c3a. is supplied with L„ ., a strong forcing pump *•', and the larg<>r one with a tightly fitting pis- ton ^S^ attached, a firm jil.rtforin or strong head /\ Hoth thecvlinders and the commu- nicating tube con- tain water, an< when downward ]>ressure is applied lo the water in the smaller cylinder, by means of the attached forcing i)U!np, the piston in the larger is forced upward by a ])ressure as much greater tlian the downward pressure in the smaller, as the sectional area of the larger cylinder is greater than that of the smaller. m\fo to V [ of tlie Hit Avill 8nrfa B stioiiix ''0 ns tlio ylindcr, I in tlic jnjrcater ectional of the AXTS. IH 185. ) HYDROSTATICS. 86 For (ixani|)I(N if Ihn Nmallor cylitidRr haro an aroa of half a aquare inch, and thtf largn ryliniior an ar(>a of ftOO Hquaro innlu'H than thn upward preNHuro In tho lattur will be lOUti tlniuH a* great as the downward pruMuro in tho former. 184. liratnali'rt llydroHtatic. Press is used f«)r prcBsinn; papor, cotton, cloth, gunpowder, and other tilings — also for tcHting tho strength of ropes, for uprooting trees and for other pnrpoHCs. 186. lo find tho relation between the force applied and tho pressure obtained in IJramah's Hydrostatic Press. RULE. /, 1/ the power be applied by means of a lever, find the amount oj downward pressure in the smaller cylinder by the rule in Jlrt. 11. If. Divide the sectional area of the lar^e cylinder by that of the the smaller cylinder and multiply the quotient by the power applied to the smaller cylinder. ExAMi'LB 238. — In a Uydroatatic Press the force pump has a sectional urea of one square inch ; the hirgc cylinder a sectioDal area of one square foot, tlie force pump is worked by means of n lever whose arms are to one another aa 21 : 2. If a power of 20 lbs be applied to tho extremity of the lever wliat will be the upward pressure exerted against the piston in the large cylin- der? BOIirTION. 20X21 I'ower a|)i)li(!(l to force i)Uinp ~ 2 210 lbs. Spctlonal area of sinallor cylinder =- 1 inch and of largw cylinder ~ l^i inches. Then UK 1^1111X210- 302410 Ibw Ana. KxAMi'LB 239. — In a Hydrostatic Press the sectional areas of the' cylinders are J of an inch and 150 inches, and the power lever is so divided that its arras are to one another as 7 to 43. What pressure will be exerted by a power of iOO lbs. applied at the extremity of the long arm of the lever? SOLUTION. ,, , , „ ,. , 100X13 Dowuwiml pressure in Huiall cylinder = — = — Upward i)rcM8urc in largo cylindor — -'- YMii 27«M28fll)s, Ans. >s nU'j lbs. -450X614^ Example 240. — Tho area of tho small piston of a Hydrostatic Press is ^ an inch and that of the larger one 300 inches, the lever is 30 inches long and tho piston rod is placed 5 inches from tho fulcrum (so as to form a lever of tho second order) what power must be applied to the end of the lever in order to produce au upward pressure in the cylinder of 1000000 lb»? ' I 86 HYDROSTATICS. [Abt. 1M. BOLUTIOK. 300 Downward proMurc in oinKlIor oylinderrrrlOOOOOO lbs — -i- -- 1000000 Ibii. f- 600 - 10061 Ibu. •HI Then power applied - lOflOl lbs. -r -""-IflOOl-: C-STTJ lbs. Ant. 241. 242. 243. 244. 245. EXERCISES. In a Hydrostatic Press tlio area of tho Biiinll cylinder is 1 inch, and that of tho largo ono 300 inches, tlio forco pump is worked by a lever of tho second order 30 inches long, having the piston rod 2 inchea from tho fulcrum, if a rn'ssiirc of 60 lbs. bo applied to tho lover what upwaru pressure will bo produced in the largo cylinder? .4ns. 225000 Ib.s. Ir. aTiij'.l;oatatic Press tho forco pump has a sectional area of half tin inch; the largo cylinder a sectional nrea .4" ?,00 iiches ; tho force pump is worked by meuna if ii lov r whose arms are to one another as 1 to 50. Now !■ i !io60 a forco of 50 lbs. be apj)liod to tho o-x- tremn; the Ivvcr what will bo the upward proMsuro exerted against the piston in tho largo cylinder? Jns. 1000000. In a Uydrostatic Press tho small cylinder has an area of one inch, and the largo ono an area of 500 inches, the pump lever is so divided that its arms are to one another as 1 to 25 what will be the upward pressure against tho pioton in tho largo cylinder produced by a forco of 100 lbs. acting at the extremity of tho lever I Jliis. 1250000. The area of the small piston of a Hydrostatic Press is J of an inch and that of tho large one 120 inches — the arms of tho lover by which the force pump is worked arc to one another as 40 to 3. Required the upward pres- sure exerted against the piston of the large cylinder by a power of 11 lbs. applied at the extrepvty of the lover. 'jns. 3G2GGr, ll'S- Tho area of tho small piston of a Hydrostati ' 'ress is IJ inches, and that of the lario one 200 inches — the arms of tho lever by which the force pump is worked are to ono another as 20 to li. What power applied at the extremity of the lever will produce a pressure of Y50000 lbs? ^«s. 421|lbs. hi [ I I- 1 186. Since the pressure of water upon a given base depends upon the height of tho liquid and not upon its quantity, it follows that : — I i ABT. 1M. 00000 lbs. (lor is 1 10 force inclios fulcrum, ?r what ^Minder? 5000 lbs. >nal area mil iirca Y mciina 1 to 50. tho ox- prcHSuro 1000000. nroa of hos, the another linst tho brco of llUfjOOOO. [rcss is 4 hes — the •kcd arc Jrd pros- Idcr by a lover. \6Cy?i lbs. 'rcss is lies — the rked are Id at the 750000 |21| lbs. in base )on its AmTt.l87,lS8.] HYDROSTATICS. 87 .1- 1 Any quantity of water how$ver amallj may be made to balance the pressure of any other quantity however greaty or to raise any weight however large* NoTR.— TIiiB is wliat is pommonly called tho Ifydrontatin Paradox. In ronlitv, howuvcr, thoro Ih iiothiiiK at alS paradoxical in it : Nince, althouich a nound of water may bn niadu to balance 10 !\)h., or 1000 lbs., or l(M)0O0 Ids,, it doits it upon procisoly tho same iirinciplo that the powor balances tho weight ill tho lover and other nicchanioal powern. Thus in order to raise 20 lbs. of water by tho deacf^iidinK force of t \\>., the latter must dcHcwnd 20 inches in ordor to raiio the former 1 inch. Honco what is called the hydrostatic paradox Ih in strict conformity to tho principle of virtual velo- cities. 187. This principle is illustrated by an instrument called tho Hydrostatic Bellows, which consists of a pair of boards united together by leather as in the common Fig. 18. bellows and made water-tight. From the upper board there rises a long tube, //, finished with a funnel-shaped termin- ation, c. NoTB. — When water is poured into tho tube an upward pressure is exerted against the u))per board as much greater than the weight of the water in the tube as the area of the board is greater than the sectional area of the tube. For oxaniplo, if tho sectional area of tho tube be \ of an inch, and the area of tho board bo 250 inches, then tho area of tho board will bo 1000 times aH Kreat as that of tho tube, rnd conscquontly 1 lb. of water in the tube will exert a pressure of 1000 lbs. against the upper board of the oellows. 188. To find the ui>ward pressure exerted against the board of a hydrostatic bellows by the water contained in the tube. RULE. Divide the sectional area of the board by that of the tube^ and multiply the result by the weight of the water in the tube. NoTB. — The weight of water in the tube is found by multiply- ing the sectional area of the tube by the height of the water in inches and the product J vhich is cubic inches of water ^ by 0.03616 Ibs.^ the weight of one cubic inch of water . ExAMPLB 246.— The upper board of a Hydrostatic Bellows hai an area of 1 foot, the tube has a sectional area of \ an inch and is filled with water to the height of *l feet. What upward preasure is eierted againit the top board of the bellows 7 ■i— ^, M ! ) i I • [ 11 }>■'> 8§ HYDROSTATICS. [AAXB. 169-199. 144 = 1.61872X288 SOLUTIOir. Cubic inches of water in the tube =i X 84 = 42. Weight of water in tube = 0.08616 X 42 =: 1.61872 Ibv Upward pressure against bellows board =: 1. 61872 X =437.39 lbs. ii«a, ExAHPLB 247. — In a Hydrostatic Bellows the board has an area of 200 inches and the tube a sectional area of | of an inch. What upward pressure is exerted on the board by 7 lbs. of water in the tube ? SOLUTION. Upward pressure = 7 X*— r* = 7 X 800 = 5000 lbs. An$. EXERCISES. 248. In a Hydrostatic Bellows the board has an area of 260 inches, the tube has a sectional area of li inches, and contains 1 1 Iba. of water. What is the amount of upward pressure exerted against the board of the bellows ? wfn«. 2200 lbs. 249. The board of a Hydrostatic Bellows has an area of 300 inches, the tube a sectional area of 1 inch and is filled with water to the height of 10 feet — what pressure will be exerted against the upper board of the bellows ? I . Ms. 1301.76 lbs. 250. The tube of a Hydrostatic Bellows has a sectional area of •72 of an inch and is filled with water to the height of ! ' 50 feet — what weight will be sustained on the bellow's board if the latter have an area of 3 feet ? jins. 9372.672. 189. A body immersed in any liquid will either float, sink, or rest in equilibrium, according as it is specifically lighter, heavier, or the same as the liquid. 190. A floating body displaces a quantity of liquid equal to its own weight. 191. A body immersed in any liquid loses a portion of its weight equal to the weight of the liquid displaced, and, hence, by weighing a body first in air and then in water, its relative weight or specific gravity may be determined. 192. The specific gravity of a body is its weight as compared with the weight of an equal bulk or volume of some other body assumed as a standard. 193. Pure distilled water at the temperature of 60^ Fahr. is taken as the &tandarvl with which to compare all ▲ftfl.lM»10S.] HTDR08TATICS. 89 Bolids and liquids, and pure dry atmospheric air al a tempe- rature of 32<> Fahr., and a barometric pressure of 30 inches is taken as the standard with which all gases are compared. 194. To find the specific gravity of a solid heavier than water : — RULE. Divide the weight of the body in air by iti lo$i of weight «u toater, the result will be its Specific Gravity. Example 251. — A piece of lead weighs 226 grains in air and only 205 grains in water; required its specific gravity. SOLUTION. Loss of weight = 225—205 = 20 grains. Htnce specifio gravity =: 225-f-20 = 11.250. Ant. ExAMPLB 252. — A piece of sulphur weighs 97 grains in air and but 50-5 grains in water; what is its specific gravity? SOLVTIOK. Loss of weight in water = 97— 60"6 = 4i6*5 grains. Then specific gravity = y? -4- 46'6 = 2*008 Ams. EXEROIOES. 263. A piece of silver weighs 200 grains in air and only 180 grains in water ; required its specific gravity. ^ns. 10.000. 254. A piece of platinum weighs 154J oz. in air and only 147J OB. in water; required its specifis gravity. jlns. 22.0'71. 255. A piece of glass weighs 193 oz. in air and but 130 oz. in water ; required its specific gravity. Ans. 3.063. 195. To find the specific gravity of a solid not i^ufficiently heavy to sink in water. RULE. To the body whose specific gravity is sought attach some other body sufficiently heavy to sink it, and of which the weight in air and loss of weight in water are known. Then weigh the united mass in vmter and in air^ from its loss of weight deduct the loss of weight of the heavier "body in water, arid divide the absolute weight of the lighter body by the remainder, the quotient will be the specific gravity of the lighter body. ExAMPLB 256. — A piece of wood which weighs 55 oz. in air has attached to it a piece of lead which weighs 45 oz. in air and 41 in water, the united mass weighs 30 oz. in water ; required^ the specific gravity of the piece of wood. .-M'' 90 UYDROSTATICS. AST. 19% ^, ^1%."^... .^JK^J^. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 gia 1^ II I.I f '^ i^ 11-25 11.4 11.6 Photographic Sciences Corporation 33 WEST MAIN STREET WEBSTER, N.Y. M580 (716) 872-4503 %^ ^V ■^ N> "% ^ # v «.^ J z %^ %' r .\ o^ w WW ]: 1 tj \ ii It ill d!3 fltDROSTATICS. SOLVTIOir. 47.8 and «'=^ 10.6. CASTI.m,19S Here to = 68.2, w': Then s = iilli^ X «'=^ifei?::5 x 10.6:-. ''\^^''' =rimAr,.. w 58.2 EXERCISES. 68.2 265. A piece of copper (Spec. Grav. 8.850) weighs 446.3 grains in liquid, and 490 grains out of it, required the Specific Grav. of the liquid. Jln$. .789. 266. The Thousand-grain Bottle filled with Olive oil weighs 915 grains — what is the Specific Gravity of Olive Oil? ^ns. .915. 207. The Thousand-grain Bottle filled with mercury weighs 13596 grains — what is the Specific Gravity of mercury ? Ans. 13.596. 268. A piece of cast-iron (Spec. Grav. 7.425) weighs 34.61 oz. in a liquid and 40 oz. out of it — what is the Specific Gravity of the liquid? Am. 1.000 nearly. 269. A piece of gold (Spec. Grav. 19.360) weighs 139.85 grains in a liquid and 159.7 grains in the air, required the Specific Gravity of the liquid ? Ans. 2.406. 270. A piece of marble (Spec. Grav. 2.850) weighs 30 lbs. in a certain liquid, and 35.9 lbs. in the air, required the Specific Gravity of the liquid ? ^ns. .468. 197. The Specifiic Gravity of gases is found by exhaust- ing a flask of atmospheric air and filling it with the gas in question previously well dried. This is accurately weighed and its weight compared with the weight of the same volume of dry atmospheric air at the temperature of 60** Fahr. and under a barometric pressure of 30 inches. 108. The following table gives the Specific Gravities of the most common substances : — ASTI.199.200.] HYDROSTATICS. 93 TABLE OP SPECIFIC GRAVITIES. OASES. Siiu Atmospheric Air, Hydrogen, Oxygen, Nitrogen, • Ammoniacal Gas, .... Carbonic Acid Gac, . . Sulphurous Acid Gas,. Chlorine, LIQUIDS. Distilled Water, Mercury, Sulphuric Acid, Nitric Acid, Milk Sea Water, Wine, Olive Oil, Spirits of Turpentine, . Pure Alcohol, Ether, Prussic Acid, SOLIDS. Platinum, Gold, . Silver, Lead, 1.000 .069 1.106 .972 .596 1.529 2.234 2.470 1.000 13.596 1.841 1.220 1.030 1.026 .993 .915 .869 .792 .715 .696 22.050 19.360 10.500 11.250 Copper, 8.850 Brass, 8.300 Iron, 7.788 Tin, 7.293 Zinc, 7.190 Diamond, 3.530 Flint Glass, S.3S0 Sulphur, 2.086 Slate 2.840 Brick, 2.000 Common Stone, 2.460 Marble, 2.850 Ivory, 1.825 Phosphorus, 1 . 770 Lignum Vitse, 1.350 Boxwood, 1.320 Potassium, 875 Sodium, 972 Pumice stone, 914 Dry Pine, 657 Dry Poplar 383 Ice, 865 Living Man, 891 Cork, 240 Graphite, 2.500 Bituminous Coal, 1 . 250 Anthracite Coal, 1 . 800 199. A cubic foot of pure distilled water at the tem- perature of 60° Fahr. weighs exactly 1000 ounces. Hence if the Specific Gravity of any substance be known the weight of a cubic foot, &c., may be easily found. For example.— the Spec. Grav. of Mercury is 13.596 water being 1.000, and a cubic foot of water weighing lOUO ounces it follows that a cubic foot of Mercury weighs 13596 ounces. 200. To find the solid contents of a body from its weight : — RULE. w Contents in feet ==^ "^i; where w z=z whole weight, andv/:=weight of a cubic foot as ascertained from its Spec, Grav. ExAMPLi 271. — How many cubic feet are there in 2240 lbs, of dry oak (Spec. Grftv, '926.) ? 94 HYDROSTATICS. [Abt. SOI. BOLVTIOir. Hero IV 2240 lbs. 35840 to' —im^cnhictect. 925 oz 025 Example 2Y2. — How many cubic feet are there in a mass of iron which weighs 17829 lbs. ? , SOLUTIOIT. Specific Gravity of iron — 7*788. Therefore 1 cu. ft. weighs 7788 o». Then cubic feet in mass = 17829 lbs. -?- 7788 oz. =8G'628 Ans. 201. To find the weight of a body frora its solid con- tents : — RULE. u)=icontents inft.xio'. Where w and v/ are same as in last rule. Example 273. — What is the weight of a block of dry oak 10 ft. long, 3 ft. thick, 2 J ft. wide. Here 10 X 3 X 2i = 75 cubic feet. Then «j=m)' x 75=925 oz. X 75=69375 oz.=4335 || lbs. Jnt. Example 274. — What is the weight of a block of marble 8 ft. long, 2 ft. wide, and IJ ft. thick. SOLUTION. i' CJubic feet of marble = 8X2X1^=24. Spec. Grav. of marble— 2*850. Therefore one cubic foot weighs 2850 ob Tlicn weight of block — ■ 2850 X 24 =68400 oz. = 4275 lbs. Ans. EXERCISES. 275. What is the weight of a mass of copper which contains 29 cubic feet ? Jlns. 16040 lbs. 10 oz. 276. How many cubic feet are there in a mass of lead which weighs seven million pounds? jlns. 9955.55 cub. ft. 277. How many cubic feet of sulphuric acid are there in 78124732 lbs? w«ns. 678976-48 cub. ft. 278. What is the weight of the mercury containe<1 in a rectangu- lar cistern 6 feet long, 4 feet wide and 1 et deep, the mercury filling it ? .. 203940 lbs. 279. If a block of zinc be 11 feet long by 3 feet wide and 2 feet thick, how much does it weigh? ^ns. 29658| lbs. 280. What is the weight of a squared log of dry pine 44 feet long and 18 inches square. Jlns, 4065 lbs. 3 oz. [ABT. Ml. ABTI. 202-20S.] PNEUMITICS. 95 masi of 88 ox. ' >lid con- ' dry oak lbs. jin$. ,rble 8 ft. i;hs 2850 OS s. ntains 29 bs. 10 oz. id which cub. ft. there in cub. ft. ectangu- eep, the 3940 lbs. nd 2 feet 658} lbs. 44 feet lbs. 3 oz. CHAPTER VI. PNEUMATICS. 202. Pneumatics treats of the mechanical properties of permanently elastic fluids^ of which atmospheric air may be taken as the type. 203. The atmosphere (Greek atmoi " gases") or sphere of gases is the name applied to the gaseous envelope which surrounds the earth. 204. It is supposed, from certain astronomical con- siderations that the atmosphere extends to the height of about 45 miles above the surface of the earth. NoTB.— The height of the atmosphere is only ^ of the radius of the earth, BO that upon an artificial globe 12 inches in diameter the atmosphere would be represented by a covering i^T of an inch in thickness. 206. Atmospheric air is a mechanical mixture chiefly of two gases, Oxygen and Nitrogen in the proportion of 1 gallon of the former to 4 gallons of the latter. Its exact composition, omitting the uqueous vapour, is as follows : — Composition BY Volume. Nitrogen, '79*12 per cent. Oxygen, 20-80 Carbonic acid, '04 Carburetted Hydrogen, -04 Ammonia, Trace. "S OTIS.— Oxygen is the sustaining principle of animal life and of ordinary combustion. When an animal is placed in a vessel of pure oxygen its heart beats with increased energy and rapidity and it very soon dies fh>m excess of vital action. Many svibstances, also, that are not at ail combus- u In its chemical nature it is distinguished chiefly by its negative properties. In the atmosphere it serves the imjrortant purpose of diluting the oxygen and thus fitting it for the function it is designed to perform in the animal economy. Carbonic acid is a highly poisonous gas, formed by the union of oxygen and carbon (charcoal) . It is produced in large quantities during the process of animal respiration, common combustion, fermentation, volcanic action and the decay of animal and ve^table substances. Although when inhaled, it rapidly destroys animal life it constitutes the chief source of food to the plant. Animals take into the lungs air loaded with oxygen and throw it off so charged with carbonic acid as to be incapable of again senriiig tov 96 PNEUMATICS. [ABtt.206,M7. M ' tho purposes of respiration. Tho green parts of plants on the contrary, absorb air, dcconiposo the carbonic acid it contains, retain the carbon and ffive oir air contaninK no carbonic acid but a largo amount of oxygen. This IS a most beautiful illustration of tho mutual dependence of tho different orders of created boinga upon one another. Wero it not for plants the air would rapidly become so vitiated as to cause the total extinction of animal life ; were it not for animals, plants would not thrive for want of the food now supplied in tho form of carbonic acid by the living animal. As it is, the one order of beings pronarcs the air for the sustenance and support of tho other, and so admiranly is the matter adjusted that the composition of the air is, within very narrow limits, invariably the same. The amount of carbonic acid varies fh)m 3'7 as a minimum to 6*2 as a maximum in 10000 volumes. Carburetted Hydrogen is produced during the decay of animal and vege* table substances. It is one of the chief ingredients of common illuminat- ing gas, and is poisonous to animals when present in the airs in large quan- titios. 206. One of the most remarkable characteristics o^ gases, is the property they possess of diffusing themselves among one another. Thus if a light gas and a heavy one are once mixed they exhibit no tendency to separate again and no matter how long they may be allowed to stand at rest, they arc found upon examination intimately mingled with each other. Moreover if two vessels be placed one upon the other, the upper being filled with any light gas (hydro- gen) and the lower with any heavy gas (carbonic acid) and if the two gases be allowed to communicate with one another by a narrow tube, or a porous membrane, a remarkable interchange rapidly takes place, i, e., in direct opposition to the attraction of gravity the heavy gas ascends and the light gas descends until they become perfectly mixed in both vessels. NoTE.-The property of gaseotis diffusion has a veiy intimate bearing upon the composition of the air. If either of the constituents of the air were to separate from the mass, the extinction of life would soon follow. Besides were it not for the existence of this property, various vapours would accumulate in certain lucalitles, as large cities, manufacturing districts, volcanic regions, &o., in such quantities as to render them totally unin* habitable. 207. In addition to the gases already mentioned, atmospheric air always contains more or less water in the form of invisible vapour. This is derived partly from combustion, respiration and decay, but chiefly from spon- taneous evaporation from the surface of the earth. The amount of invisible vapour thus held in solution depends upon the temperature of the air being as high as ^^ of the weight of the air in very hot weather, and as low aa j\-^ in cold. .:! tTI. 106, 107. AmT0. 806-tlO.] PNBUtfATICS. •» e contrary, carbon and xygen. This lio different Jlanti the inction of for want of niiK animal, beuanoe and ted that the r the same. I to 6*2 as a al and vego* 1 illuminat- large quan- ristics o' lemselves leavy one ate again nd at rest, ^led with )ne upon .8 (liydro- lic acid) with one brane, a in direct 9 ascends perfectly aringupon air were T. Besides urs would : districts. Ally unin> ntioned, in the y from spon- The lepends ,3V Of low aa n 1. 208. The blue color of the sky is due to light that has suffered prolarization, and which is, therefore reflected light, like the white light of the clouds. The air appears to absorb to a certain extent the red rays and yellow rays of solar li^lit and to reflect the blue rays. In the higher regions the blue becomes deeper in colour and is mixed with black. The golden tiuts of sunset depend upon the largo amount of aqueous vapour held in solution by the air. 209. Air like all other material bodies possesses the properties of impenetrability, extension, inertia, porosity, compressibility, elasticity, &c. (See Arts. 11-18. Note l.— The impenetrability of atmosphorio air is illustrated by various experiments, among which are the following : I. If an inverted tumbler be immersed in water the liquid does not rise in the interior of the tumbler, because the latter is full of air and the water cannot enter until the air has been displaced. II. If th« two boards of a bellows be drawn asunder and while in ttaa^ position the nozzle of the bellows be closed, the boards cannot be pressed together because the bellows is ftiU of air. III. If an india-rubber bag or a bladder be inflated with air. and prefl> sure applied, it is found that there is a material something within which keeps the sides asunder,— that material something is atmospheric air. Note 2.— The Inertia of atmospheric air is shown :— I. By the force of wind, which is nothing more than air in motion. II. By attempting to run on a calm day, carrying an open umbreUa. III. By the apparent current of wind experienced on a perfectly calm day by a person standing on the deck of a steamboat, or the platform of a rail- way car when in rapid motion, which current is caused by the body dis- placing the air. IV. By causing a feather and a ball of lead to fall in avacuum, when it is observed that they fall with the same velocity. In the atmosphere, however, the ball falls faster than the featRer because it contains a greater amount of matter with the same extent of surface as the feather, and meets hence with less resistance fh>m the inertia of the air. 210. Air,. in common with all other forms of matter, is acted on by the attraction of gravity, and hence possesses weight. Note 1.— This is the fundamental fact in the science of pneumatics. To prove it we take a glass globe capable of containing 100 cubic inches, and after weighing it accurately, withdraw fW)m it, by means of an air pump, all the air it contains. When we weigh it again we find that its weight is about 31 grains less than when filled with air. 1 00 cubic inches of Atmospheric Air weigh 31 grains. 100 " Oxygen " 34 " 100 •• Nitrogen " 80 " 100 " Carbonic Add " 47* " 100 ** Hydrogen " 2 " Note t.— Although a small quantity of air when examined appears to be almost imponderable, the aggregate weight of the entire atmoapn?r« is mofi mouiibeini; equal to : Ui ! 98 PNEUMATICS. CASTS. Ill-tlS. m I. Five thouMnd millions of millions of tons, or II. A glob« of lead 66 miles in diameter, or III. An ooean of water covering the whole surface of the earth to the depth of 32 feet, or IV. A stratum of mercury covering the entire surface of the globe to the depth of 80 inches. 211. Since the air is ponderable and also compressible, and since tbe lower stratum has to sustain the pressure of the superincumbent portion, it necessarily follows that the air is denser near the surface of the earth than in the higher regions of the atmosphere. 212. The density of the air decreases in geometrical progression, while the elevation increases in arithmetical progression. That is at the height of 2'1 miles, the atmos- pheric pressure is reduced to one-half, at twice that height to one- fourth, at three times that height to one-eighth, &c. NoTB.— The following table exhibits the density, elasticity and pressure of the aii> at the different elevations given. Halley fixed the height at which the pressure decreased to o ic-half at 8^ milci, but a more careful collection, by Biot and Arago, of tho observations made on the Andes and in balloons, respecting the upward decrease of pressure and temperature, has led to the adoption of 2*7 miles as the point at which we may say that one- half of the atmosphere is beneath us. HTHGHT IS MILES. DENSITY. HEIGHT IN INCHES OP COLUMN OB MERCUEY. PEE88URE IN Ibs. TO THE 8Q. INCH. 2.7 i 16 7.5 5.4 i 7.8 8.76 8.1 i 3.75 1.876 10.8 ^s 1.876 .987 18.6 i^ .937 .468 16.2 'ek .468 .284 18.9 tW .234 .117 21.6 sis . .- "^ .058 24.8 rk ' > .058 .029 j 27.0 To^rr .029 .014 29.7 fif-r? .014 • .007 218. The pressure of the air is a necessary consequence of its weight, and is equal, at the level of the sea, to about 15 lbs, to the square inch. 1X8.111-118. uiih to the riobe to the pressible, essure of that the 1 in the •metrical bmetical e atmos' it heigfht hth, (fee. 1 pressure height at >re careful des and in 'atui'e. has that one> tr lbs. TO [KCH. ▲111.214-118. J PNEUMATICS. 99 [uence aboiU NoTB.— By saying that the pressure of the atmosphere is equal to 15 lbs to the sq. inch, we mean that it is capable of baUncing a column of meronr/ SO inches in height, and a column of mercury 80 inches in height and having a sectional area of 1 sq. inch weighs 15 lbs. Or in other words, that a co- lumn of air having a sectional area of 1 sq. Hch, and extending from the level of the sea to the top A the atmosphere v/eighs 15 lbs. 214. Air at 60® F. is 810 times as light as water, and 10466 times as light as mercury. It follows that the pressure of the atmosphere is equal to that of a column of air of the same density as that at the surface of the earth 810 times 32 feet or 10466 times 30 inches in height. That is, if the air were throughout of the same density that it is at the level of the sea, it would extend to the height of about 5 miles. 216. The particles of elastic gases, unlike those of solids or liquids, possess no cohesive attraction, but on the contrary a powerful repulsion, by means of which they tend to separate from one another as far as possible. 216. Permanently elastic fluids such as {«tmospheric air, and certain gases, are chiefly distinguished from non- elastic fluids, such as water, by the possession of almost perfect elasticity and compressibility. Note.— Air and certain gases as Oxygen, Hydrogen, Nitrogen, ^c, are called permanently elastic to distinguish them fh>m a number of others as CarDonlo Acid, Nitrous Oxide, &c., which under great pressure and intense cold pass first into the liquid and finally into tne solia state. 217. If a liquid be placed in a cylinder under the piston it will remain at the Fame level, no matter to what height the piston may be raised above it, but if a portion of air or any other elastic gas bo thus placed in the cylinder and the piston be air tight, the confined air will expand upon raising the piston and will always fill the space beneath it, however great this may become. This expansibility or tendency to enlarge its volume so as entirely fill the space in which it is inclosed is termed elasticity. NoTB.— It is obvious that the elasticity of air is due to the repulsive power possessed by the particles. 218. The law determining the density and elasticity of gases under diflerent pressures was investigated by Boyle in 1060, and afterwards by Mariotte. 100 PNEUMATICS. LAlT.tl9. Fig. 20. ' ( NoTB.— To lUuitrat« thif Uw we t»k« a bent glaii tube Fif^. IuitIdk one limb A muoh loncer than the other. Trhe longer limb it open »ud the snorter fur- nished with % nop-coek. Both ends beinc open a quantity of mrircury is poured into the tune and of course rises to the same level in iKtth legs— the surfkco of the morriivy at A a, sustaininK the weight of a column of air extending to the top of the atmosphere. We now close the stop* cook and thus shut on the pressure of the atmosphere above that point, so that the surfsrce a, cannot bo affected by tne weight of the atmosphere—/, e., cannot be influenced by atmospheric pressure. We find, however, tlmt the mercury in both limbs remains at the same level, fk'om^which wo infer that the clastic force of the air conflned above a is equal to the weight of the whole column on a before the stop cock was closed. Hence the elasticity ef tho air is equal to its weight, which is equal to a column of mercury 80 inches high. If now wo pour mercury into the tube until ino air conflned above a is compressed into half its former volume, i. e., until the mercury rises to b in the shorter tube, we shall find that the column of mercury bB\H exactly SO inches in length, or in other words, we have doubled the pressure on the air con>lned in the shorter tube and have decreased Its volume to one«half its former dimensions, and at the same time!doubled its elastic force since it now reacts against the surface of the mercury with a force equal to 60 lbs. to the square inch. If we increase the height of the mercury in the longer leg to 60 inches above its height in the shorter leg, we shall compress the air into one>third its original volume and at the same time treble its elasticity, and ■o on. Hence the law of Mariotte. 219. Mariotte's law may be thus enun- ciated. I. The density and elasticity of a gas vary directly a$ the pressure to which it is subjected. II. The volume which a gas occupies under different pressures varies inversely as the force of compression. NoTB.— Becent researches tend to prove that Muriotte's law is true only within certain limits, and that all gases vary from the law when subjected to very greatpressures, their density increasing in a greater ratio than their elasticity. With atmospheric air tne law holds good to a far greater extent than with any other gas, the correspondence being found to be rigidly exact when the air in expanded to SOO volumes, and also when it is compressed into ^ ofits primary volume. Mariotte's law would require the air to be indefinitely expansible while we know that there is, beyond all doubt, an upward limit to the atmos- phere. Dr. Wollaston imagines that when the particles of air are driven a certain distance apart by their mutual repulsive power, the weight of the individual particles comes at last to balance this repulsive force l»pd thxa prevent their further divergence. If this be the case as is Am. no-ill.] l»WBTjlfATIC8. 101 probable from various oonsidorations, there li a limit to the rareflM* tion of a ni, arriving at which tlie gas ceases to expand further and oomes to have a true uppor surflMo like a liquid. As has been already remarked this pxaot limit and uppor surftMe of the atmosphere is supposed to be at an elevation certainly not greater than 46 mites— Blot Axes it at 30 miles. 220. The air pump, as its name implies, is an instrument used for pumping out or exhausting the air from any closed vessel. 221. The bell-shaped glass vessel usually attached 'to the air-pump is called a Receiver^ and when the air is exhausted as fiir as practicable from this a vacuum is said to have been produced. NoTB.— The air pump was invented by Otto Guericke, a celebrated JhirKoniastcr of Magdnburg, in the year ISOO. At the close of the Imperial Diet in 1564, ho exhibited hiH first public experiments with it before the omporor and assemblitd princes aiici nobles of Germany. On this occasion hoexhaustod the air from two 12-inoh hemispheres fitted together by ground edges and greatly astonished his noble audience by showing that the oom« l)incd strength of 12 horses was insufflciont to puu them asunder. The exhausting syringe of Otto Guericke was so imperfect in its action that while using it no was compelled to keep it immersed in water to pre> vent the inward leakage of the air. Since his time, however, the attention of many eminent men has been directed to the subject, and the form and construction of the air-pump have been very greatly improved. 222. The exhausting syringe which is the essential part of an air-pump, consists of a brass cylinder ahcdy supplied with an air-tight piston efy and an arrangement of valves hk^ by means of which the air is permitted to pass out from the receiver q and through the piston ef^ but not in the contrary direction. Note.— When the piston ef is raised the valve h closes, and ai the piston in its ascent produces a partial vacuum be- neath it, the air contained in the receiver o opens the valve k by its expansive power and thus refills the cylinder abed. Now when the {listen is forced down again, the air contained n the cylinder tends to rush back into the re- ceiver but in doing so closes the valve k, and has therefore no other mode of escape than through h, thus passing above the piston to bo lifted out at the next stroke. In this manner the air continues to bo exhausted until what remains in the receiver has not sufficient ex- pansive power to open the valve k, when the exhaustion is said to be complete. Pig. 21. •a f 5' 229. The principle upon which the air-pump acts is the elasticity or expansibility of the air, and since in order to enable the pump to act, the air contained in the receiver must possess sufficient elastic force to raise the ralve, it )lll t 109 PNEUMATICS. UBTt.ni,MO. followt lliat n perfect vAcuum cnnnot bo Hecured by the air-pump. Thus, puin|)8 of common cotiHtruction will not withdraw more than ^^^ of the containod air, but the improvod form is said to oxhauRt i^^niVr* NoTR.— ir we suppoie thn oylindcr of thfloxhauNtinfr nrriiifrn tn htvn ilin iianiP offootive (»p»oUy m thn roorlver, and that the uiHtoii jikmivi at «aottoni. onn-hair of tho air will have i)aHMfld fhim tho reoolvnr ; tlin muiaining half ooinpletnlv Ailing it, but having only half an much doniity and ('InHHicity ax before. Tlio Hooond «troku of thn piHton will nHluoo the quantity, douMity. and elasticity, to ono-fourth, the third to ono-eighth, and no on an eihlbitou by tho following table :— HTROKR. nOKBOUT. l8t, and, 3rd, 4th, Rth, 6th, 7th, 8tb, 9th, 1 I \ of of of of 0fT»» Of Vf Of if* =» ofiir = LBFT IR VKH8B1., BLASTIO rORCR Of THB KBMAtNPKK. = \ = k It \ of fig =rl» IR in. of mercury, or 7.35 lbs per sq. in. 7Jin.of 3) in. of 1.876 in. of 0.937 in. of 0.468 in. of 0.234 in. of 0.1l7in. of 0.068 in. of II It or 3.676 or 1.837 or .918 .469 .229 .114 .067 .028 or or or or or do. do. do. do. do. do. do. do. 224. The condensing tyringey which is used for forcinp; air into a receiver or condensing chamber, differs from an exhausting syringe only in tiio fact that its valves open inward towards tlic chamber instead of outward. 226. Tho Air-pump is cinefly employed to illustrate tho pressure and elasticity of the air. NoTR 1.— The preamre of the atmosphoro may bo shown by innunierablo cxperinionts among which are the following :— I. Whon the air is exhausted fTom the receiver of an air-pump the re« ceiver is lirnily fastened to the plate and cannot be reraovea until the air is ro-admitted. II. The hand placed on the open end of the receiver it pressed inward with a force sufflcieutly groat to cause pain. III. Thin square glass*tubes are crushed when the air is exhausted flroui them. IV. In the surgical operation ot cupping, the air is removed from a small cup which is then placed over an opened vein ; tho pressure of the air on the surrounding parts causes the blood to flow rapidly into the cup. V. When a cask of boor is tapped, the beer docs not run until a small hole called the vent-hole bos been made in the upper part of the cask. Through this the atmospheric air enters and pressing on the surface of the boer with a force of 16 lbs. to the square inch, forces it through the tap. VI. The useful small glass instnunents cMed pipettes act upon theprin- oiple of atmospheric pressure. Til. A hole is usually made in the lid of a tea-pot so as to bring into play the presaure of the atmosphere and thua cause tho beTerage to flow more rapioljr. lTt.fUillO. hI by the I will not V but the ' to liAVfl tlin iNOH Rt ua«^h raiitliiK tliH llio lM)ttoni, ittinliiK >ial^ 'laHMlcity ax ity, (liMiNlty. M eiliiliitoil INDKK. poraq. in. do. do. do. do. ♦ do. do. do. do. r forcinpf from All ves open llu»trato iiuniorablo ^ the r«> ktil th« air ;d inward sted flroui ma small )he air on ip. a small the cask. ,co of the le tap. the prin- into play low more ABTi.tM.tr.] PKEUMATICS. 103 VIII. Flioii walk on fflaas or on thn CftlllnR by prodndniv a facuum under eaoh foot whloh in thuit prcnnod affaluMt tho Hiirfkoo with a force ■ufllclnnt to sustain the weight of tlio inm><n- toenth century by TorriooUi, a pupil of the celebrated Galileo. 227. The essential parts of a barometer are : — let. A well formed glass tube 33 or 34 inches long, closed at one end and having a bore equal throughout, of two or three lines in diameter. The tube contains pure mercury only, and is so arranged that the mercury is sup- ported in the tube by the pressure of the atmosphere; and 1^ ^HVmUTiOS. [▲«n.ttH8l' I' > i \ i i 2nd. An Attached graduated scale and various appliances for protecting the tuDc and ascertaining the exact height of the column of mercury. NoTB.— The vacant space between the top of the column of mercury and the top of the tube is called the Torricellian vacuum, in honor ot the inventor of the barometer, and in a good instrument is the most perfect vacuum that can be produced by art. 228. The excellency of a barometer depends principally upoii the purity of the mercury in the tube and the perfect- ness of the Torricellian vacuum. The value of the instrument may be tested :— 1st. Br the brightness of the column of mercury, and the absence of any ■peck, flaw, or dullness on its surface. 2nd. Bs the barometric light ; i. e., flashes of electric light produced in the dark in the Torricellian vacuum by the friction of the mercury against ihe glass. Srd. By the clearness of the ring or clicking sound produced by making the mercury strike the top of the tube, and which is greatly raodifica when any particles of air are present above the column. 229. The cause of all the oscillations in the barometer is to be fouud in ' the unequal and constantly varying dis- tribalion of heat over the earth's surface. If the air is much heated at any spot it expands, rises above the mass of air, and rests upon the colder portions surrounding it. Tlie ascended air consequently flows off laterally from above, the pressure of the air is decreased in the warmer place and the barometer falls. In the colder surrounding places, however, the barometer rises, because the air that ascended in the warmer region is diffused over and presses upon the atmosphere of these cooler parts. NoTK.— It is found that the fluctuations in the height of the barometer var^ greatly in extent in different latitudes— being so small in tropical regions as almost to escape notice, and comparatively so fitful and extreme in the temperate and firigid zones as to defy all attempts at reducing them to any system. In our climate the column varies in height fVom a little over SO inches as a maximum to a little over 27 inches as a minimum. Within the torrid zone the column of mercunr scarcely ever exhibits any disturbance greater than what would occur in Canada before a slight thun- der storm— but such a disturbance is there the sure and rapid precursor of one of those mighty atmospheric convulsions which sometimes desolate vast regions and which are frequently as disastrous in their efiTects as the most violent earthquakes. 230. Besides the irregular fluctuations depending upon the weather, the barometer is subject to regular semi-diurnal oscillations depending upon atmospheric tides, caused by the heat of th« sun— the two maxima of pressure always occurring at about 9 a.m. and 9 p.m. and the two minima at about 3 a.m. and 3 p.m. Abts. 231. 232.] PNEUMATICS. SV 105 XoTB.— The semi-diurnal oscillation is greatest at the equator, where it averages one-tenth of an inch — diminishing to six-hundredtht of an inch in lat. 30**, beyond which it still decreases, and in our climate becomes completely masked by the irregular fluctuations peculiar to the temperate and frigid zones. 231. USE OF THE BAROMETER AS A WEATHER GLASS. I. The state of the weather to he expected depends not so much upon the absolute height of the column of mercury as upon the RAPIDITY AND EXTENT OF ITS MOTION whether rising or falling. Note.— If the mercury have« convex surface the column is rising; if the surfiMse is concave the column is falling, when the surface is flat the column is usually changing ft'om one of these states to the other. II. A fall in the barometer generally indicates approaching rain, • high winds, or a thunder storm. III. j1 rise in the mercury commonly indicates the approach of fine weather ; sometimes, however, it indicates the approach of a snow storm. TV. A rapid rise or fall in the mercury indicates a sudden change of weather. V. A steady rise in the column, continued for two or three daysf is generally followed by a long continuance of fine settled weather. VI. A steady fall in the column, continued for two or three days, is commonly followed by a long continuance of rainy weather. VII. A fluctuating state in the height of the mercury coincides with unsettled weather. Note.— The barometer is far more valuably as a means of ascertaining approaching changes in the state of the wind than in foretelling the ap- proach of wet or dry weather. 232. To ascertain the height of mountains, &c., by the barometer. • HALLEY'S RULE. I. Find the logarithm corresponding to the number which ex* presses the height in inches of the column of mercury in the baro- meter at the level of the sea. II. Find also the logarithm corresponding to the number which expressesHn inches the height of the column in the barometer at the top of the mountain or other given elevation, III. Subtract the latter of these logarithms from the former i, multiply the remainder by' the constant number ^ 62170, and the result will be the elevation in English feet > Note.— The nimiber 62170 in this rule, and 63046 in the following, were selected by Halley from certain mathematical reasons into whicn it is unnecessary to enter. Example 281. — On the top of a certain mountain the barometer stands at the height of 21*793 inches, while on the surface of 8 S i 1 ! 1 1 ill i I i ; I l;i I- \\ 106 ** PNEUMATICS. [ABT. 282. the oartb. it stands at 29* 780 inches ; required the height of the mountain? ^ - -^ SOIUTION. Logarithm of 29*780 = 1*473925 aud logarithm of 21*793 = 1*828317. Then from 1*473925 Subtract 1*328317 Remainder = '145608X62170 = 9062 feet. Ant. RULE WITH CORRECTION FOR TEMPERATURE. I. Obtain^ as before, the difference between the logarithms of the numbers expressing the heights at which the mercury stands at the surface of the earth and on the summit of the mountain. ' II. Multiply this difference by the constant number, 63946 — the result IS the elevation in feet, if the mean temperature of the surface of the earth and the elevation is 69*68° Fahr. III. If the mean temperature of the two elevations be not 69*68° Fahr., add j^t^ of the whole weight found for each degree above 69*68°, or subtract the same quantity if the mean temjterature be below. ExAMPLR 282.T-Humboldt found that at the level of the sea, near the foot of Chimborazo, the mercury stood at the height of 30 inches,'while at the summit of the mountain it was only 14*85 inches. At the same time the temperature at the base of the mountain was 87'=' Fahr., and at the top 50*40^ Fahr. What is the height of Chimborazo ? \ SOLUTION. Log. of 80 = 1*477121, log. of 14.85 = 1*171724 and mean temperature = 87°+50*4'' 137*4" _ „ 2 ^^ ~ ^'^^ • Then 1*477121 — 1*171724 = *306897. And •306397X63946= 19539 feet. Since the mean temperature of the two stations is 1^ less tlian 69*68**, we deduct j^ of the elevation found. fhs ot 19539 = 40*7 ft. and 19539 — 40'7 == 19498*3 ft. Ans. LESLIE'S RULE. rOR HKASmtlNa HBIOBTB BT THK barometer WITHOUT THE USE OF LOGARITHMS. I. Note the exact height of the column of mercury at the base and at the summit of the elevation. II. Then say, as the sum of the two pressures is to their difference ^ so is the constant number 62000 to the answer in feet. Example 283. — The barometer in a balloon is observed to stand at a height of 22 inches, while at the surface of the earth it stands at 29*8 inches ; what is the elevation of the balloon 7 « Abt. 233.] PNEUMATICS. 107 empcrature =: Ihan eO'CS", we THB rSB OF 22+29-8 ! Or, 61-8 : 7*8 29*8 52000 BOLnioir. - 22 : : 52000 : 52000X7*8 Ans. ' = 7837.8 ft. Ans. 51-8 EXERCISES. 284. At what height would the mercury stand in the barometer at an elevation of 29'T miles above the earth's surface ? jlns. 0-0146 inches. NoTB.— Divide 29*7 by 2*7 (See Art. 212), the quotient is 11, then divide 80 inches by 2", t. e. 2C48, and the result is tlio answer. 285. At what height will the barometer stand in a balloon which is at an elevation of 16} miles? ^ns. '46875 inches. 286. *It is observed that while the barometer at the base of a mountain stands at a height of 30 inches, at the top of the mountain it stands at a height of only 18 inches, required the height of the mountain. 287. *While the mercury at the base of a mountain stands at the height of 29-5 inches, at the summit of the mountain the barometer indi- cates a pressure of only 20-4 inches, what is the height of the moun- tain ? ^ns. 9482-9 feet. 288. fWhile in a balloon the barometer indicates a pressure of only 19 inches, at the surface of the earth the pressure is 29-94 inches — tak- ing the mean temperature of the two stations as 72-50">, what is the elevation of the balloon ? • ^ns. 12708 feet./ 233. The common pump consists of a barrel SB, a tube AS, which descends into the writer reservoir, a piston cdj moving air-tight in the barrel and two valves, v and .r, which act in the same manner as in the ex- jausting syringe of the air pump. NoTfi 1.— "When the machine begins to act the piston is raised and produces a vaoiuim below it in the barrel, and the atmospheric Jns. 13000 feet. Fig. 22. * Use Leslie's rule. t Use Halley'a rule with correction for temperature; i. e., the second <^f the rules given. l08 PNEUMATICS. [AbTS. 284. 235. I -v I 1 ■ pressure on the water in the reservoir forces it up the tube and throueh the valve x into the lower part of the barrel. As the piston descends tno valve .r closes and the water contained in the barrel passes through the valve V above the piston, to be lifted out at the next stroke. Hence the common pump is sometimes called a lifting pump. Note 2.— Since the specific gravity of mercury is 13"596 and the i)ressure of tlie atmosphere sustains a column of mercury 30 inches in height— it follows that atmospheric pressure will sustain a column of water SOX 13*590 inches, or 34 ft. in neight. Hence the vertical distance of the valve x above thesurfaceofthewaterinthereservoirmustbeless Fig. 23. than 34 feet, or taking the variations in atmospheric pressure into account, about 82 feet. 234. The forcing pump consists of a suction pump Aj in which the piston /* is a solid plug without a valve. When the piston P descends the valve V closes and the water is forced through the valve v' into the chamber MN". The upper part of this chamber is filled with compressed air, which, by the pressure! it exerts against the surface of the water, ww' drives it with considerable force through the pipe or tube HG. "PfF^ NoTK.— Sometimes the forcing pumn is used without the air chamber, MN. Fig. 23 exhibits the A arrangement of the valves, &c., in a common ftre engine with the exception that there is another f similar forcing pump on the other side of the air chamber. HO represents the tube leading to the hose. 235. Tbe Syphon is a bent tube oi glass or other material having one leg somewhat longer than the other and is used for transferring liquids from one vessel to another. Note.— The machine is set in operation by immers- ing the shorter leg in the liquid to be decanted, and sucking the air out of the tube, when the pressure of the atmosphere forces the liquid into the syphon over the bend and down through the longer leg instead of sucking the air out of the syphon, the instrument may be set m operation by first filling it with the liquid Fig. 24. and, while thus full, placing the finger over each end, and immersing the shorter leg in the liquid. Note 2.— In order to understand why one limb must be shorter than the other, it is only necessary to remember that the pressure of the atmosphere acts as much at one extremity as at the other. If we raise the column ofliquid as far as B, by sucking at the ex- tremity C, and then withdraw the mouth, the water falls back into tho "S I !i ..». ABT8. 23A-241.] DTNAMICS. 109 vessel F. The column will likewise run hook if wc get it no farther than L, which is the level of the water in the vessel F, because at that point the upward pressure of the atmosphere prevails over the downward pressure of the liciuid, but if wo got the column below L, the downward pressure of the liquid exceeds the upward pressure of the atmosphere and the liquid will flow. Thus the motion of the fluid in the syphon is similar to the motion of a chain haneing over a pulley,— if the two parts of the chain bo equal, the fluid remains at rest, but if one end be longer than the other, it moves in the direction of the longer, and fresh linics, so to speak, are added con- tinuously to the fluid chain by the atmospheric pressure exerted on the surface of the water. CHAPTER Vlf. ^ DYNAMICS. 236. When the forces which are the subject of inves- tigation are balanced, the consideration of them properly comes under the science of Statics, but when they cease to be bahinced, and the body acted upon is set in motion other principles become involved, and the investigation of these constitutes the more complex science of Dynamics. 237. Statics is a deductive science, ^mcQ all its facts arc deducible, like those of Arithmetic and Geometry from abstract truths ; dynamics is an inductive, experimental or physical science, many of its principles being capable of proof only by an appeal to the laws of nature. 238. Force may be defined to be the cause of the change of motion, i. e., force is required : — 1st. To change the state of a body from rest to motion or from motion to rest. 2nd. To change the velocity of motion. 3rd To change the direction of motion. 239. Forces are either m^tontoneows or continued, and continued forces are either accelerating, constant or retard- ing. 240. Motion may be defined to be the opposite of rest or a continuous changing of place. 241. Motion has two qualities, direction and velocity, and is of three kinds, m. ! f r'l ' 110 DYNAMICS. [AST8. 24^251. Ist. Directf 2nd. Rotatory or Circular ; and 3rd. Vibratory or Oscillatory, 242. An accelerating^ constant or retarding force pro- duces an accelerated^ uniform or retarded motion. 243^. Velocity is the degree of speed in the motion of a body and may be either uniform or varied. It is uniform when all equal spaces great or small, are passed over in equal times. 244. The principles of the composition and resolution oi force are equally applicable to motion, 245. Momentum or Motal Force or Quantity of Motion is the force exerted bv a mass of matter in motion. 246. The momenta of bodies are proportional to their weights, multiplied by their velocities. 247. When the velocities of two moving bodies are equal, their momenta are proportional to their masses. 248. When the masses of two moving bodies are equal, their momenta are proportional to their velocities. 249. When neither the masses nor velocities of two moving bodies are equal, their momenta are in proportion to the products of their weights by their velocities. Note.— When we speak of multiplying a velocity by a weight, we refer to multiplying^ the number of units of weight by the number of units of velocity, and it makes no difference what units of each kind are employed for the product, thus obtained, means nothing by itself, but only by com- parison with otlier products similarly obtained by the use of the same units. For example, when we say that a weight of 11 lbs. moving 6 feet per second, lias a momentum of 66, all we mean is, that in this case the weight strikes a body at rest with 66 times the force that a body weighing one lb. and moving only one foot per second would exert. 250. If a moving body M, having a velocity F, strike another m at rest, so that the two masses shall coalesce, and move on together with a velocity v, then Jf X F= (M-|-7?i)Xv; or whatever momentum may be acquired by the body m must be lost by M. 251. If a moving body M having a velocity F", strike another body m moving in the same direction with a velocity v, so that the two may coalesce, and move on tI8. 242-261. AHT8. 252, 258.] DTNAMICS. Ill together with a velocity vel^ — then Myc F+mXvrs {M^-{'m)Xvelj or \n other words, the two hodies united nave the same momentum that they separately had before impact. 252. If a moving body Jlf having a velocity F", strike another body m moving with a velocity v, in the opposite direction, so that the two masses shall coalesce and move on together with a velocity vel — then MXV<^mXv=z (Jl/4"''*)Xve^ or in other words the body moving with least force will destroy as much of the momentum of the other as is equal to its own momentum. 253. If a moving body Jf, having a velocity F, strike another body m moving obliquely towards it with a velocity V, so that the two masses shall coalesce and move on together, then by representing their momenta, just before impact by lines in the direction of their motion and completing the parallelogram, the diagonal will represent the quanti- ty and direction of the momentum of the combined mass. Example 289. — What is the momentum of a body weighing 78 lbs. and moving with a velocity of 20 feet per second ? SOLUTION. Momentum = 78 X 20 := 1560. Ans. That is, the momentum of such a body is 1560 times as grc9 tiO, m = ^0, V=* 100 and v = 20. rri, Tr » _ ^ X F + wt X t>_ BO X 100 + -to X 20 _ BOOO + 80 6800 Then vol ^^- _ ^^— _ - _ — -■ 64| ft. per see., and momentum = (60 + 40) X 64| = 6800. Ant, Example 292. — If u body weighing 120 lbs., and moving to the east with a velocity of 40 feet per second, come into contact with a second body weighing 90 lbs. and moving to the west, with a speed of 80 feet per second, so that the two bodies coal- esce and move onward together, in what direction will they move, with what velocity, and what will be their momentum ? SOLUTION. From Art, 252. if M and m be the bodies, and V and v their respective velocities, and vel. the velocity of the united mass after impact :— Then ( Jf + m) x vel. =Mx V r^m Xt» and hence , af X v^ky.v vel. = iTT In tUs example M = 120, w = 90, F = 40 and v = 80. _. ^ , ilfxr^wtX» (120 X 40) r^ (90 X80) Thenuei. = - — Jf+«» 120 + 90 4800^7200 210 2400 = —^ = llf feet per second = the velocity. 11 ^ X (120 + 90) = 11^ X 210 = 2400 = momentum. And since 90 X 80. the momentum of the body moving to the west is greater than 120 X 40, the momentum of the body moving lo the east, the united mass moves to the west. EXERCISES. 293. What is the momentum of a body weighing 79 lbs. moving with a velocity of 64 feet per second ? Ans. 5056. 294. Which would strike an object with greatest force, a bullet weighing one ounce and propelled with a velocity of 2000 feet per second, or a ball weighing 5 lbs. and thrown with a velocity of 28 feet per second ? Ans. momentum of bullet = 125. " of ball = 140. Therefore the ball would exert most force of impact. 295. Which has the greatest momentumi a train of cars weighing 1*70 tons and moving at the rate of 40 miles per hour, or a steamer weighing 790 tons and moving at the rate of 9 miles per hour ? Am. momentum of train = 6800, of [Abt. 268. Iieir separate m 1-800 moying to ito contact ) the west, odies coal- i will they nentum ? ir respective t:— 4800 ^ 7200 210 0) = 11^ X the west is le east, the moving ns. 5056. a bullet locity of i thrown t=125. 1 = 140. d exert eighing 10 ur, or ate of 9 800, of AST. 258.3 DYNAMICS. 113 steamer = 7110, and therefore the latter has most mo- mentum. 290. If a body weighing GO lbs. and moving at the rate of 86 feet per second, come in contact with another body weigh- ing 400 lbs., and moving in tlie same direction at the rate of 12 feet per second, so that the two bodies coalesce and move on together ; what will be the velocity and mo- mentum of the united mass? Arts, velocity = 21^^ feet per second ; momentum =: 9960. 297. If a body weighing 56 lbs. and moving with a velocity of 80 feet per second come in contact with a body at rest, weighing 70 lbs., so that the two bodies coalesce and move on together ; what will be the velocity of the united mass ? jlns. 35-/*u^j feet per second. 298. If a body weighing 77 lbs. and moving from south to north, with a velocity of 40 feet per second, come in contact with another body weighing 220 lbs. and moving from north to south, with a velocity of 14 feet per second, so that the two bodies coalesce ; in what direction and with what velocity does the united mass move? Ans. Their momenta exactly neutralize each other and the bodies come to a state of rest. 299. If a body weighing 70 lbs., moving to the south with a velocity of 70 feet per second, come in contact with another body which weighs 80 lbs. and is moving to the north with a velocity of 60 feet per second, so that the two bodies coalesce and move on together ; in what di- rection will they move and with what velocity and mo- mentum ? Jins. To the south with velocity of 8 inches per second. Momentum of united mass = 100. 300. If a body weighing 600 lbs. and moving to the west with a velocity of 40 per second, come in contact with a second body weighing 50 lbs. and moving to the east with a velocity of 20 feet per second, and after the two have coalesced they come in contact with a third body which weighs 100 lbs., and is moving in opposite direction with velocity of 150 feet per second, and the three then coa- lesce and move on together ; in what direction will their motion be and what will be the velocity and momentum of the united mass ? Ans. Direction, west. Velocity = 10§ feet. Momentum = 8000. 114 DYNAMICS. CABTI.U4-SM. 264. When force is communicated by impact to a body at rest, the body will remain at rest until the force is dUtrihuted throughout all the atoms of the mass, unless a fragment be broken off by the force of impact, in which case this fra^ nent alone moves. LAWS OF MOTION. 255. The first law of motion. — Every body must persevere in a state of rest or of uniform motion in a straight line, unless it be compelled to change that state by force impressed upon it. 256. The second law of motion. — Every change of motion must be in proportion to the impressed force, and must be in the direction of that straight line in which the impressed force acts, 267. Tnino law of motion. — ^11 action is attended by a corres- ponding re-action, which is equal to it in force and opposite in direction. These laws are commonly known as Sir I. Newton's laws of motion— in reality however the flritt ia duo to Kepler, the second to Newton and the third to Galileo. \ 258. When a moving elastic body strikes against the surface of another body, the direction of its motion is changed, and the motion thus resulting is said to be reflected. Here : — Ist. The angle at which the moving body strikes the surface of the other is called the Angle of Incidence ; 2nd. The angle at which the moving body rebounds is called the Angle of Reflection ; and 8rd. The Angle of Reflection is always equal to the Angle of Incidence. i 259. In a vacuum all bodies, whatever may be their form or density, fall towards the centre of the earth in vertical lines and with equal rapidity ; but in ordinary circumstances, i. e., falling through the air, only heavy bodies fall in vertical lines, and the density and form of a body materially affect its velocity. 260. The resistance which a body encounters in moving through the atmosphere or any other fluid, varies : — 1st. Directly as the surface of the moving body. 2nd. Inversely as the square of the velocity of the moving body (See Art. 147). kSTI. SS4-M0. ▲bti. sei-t64.] DTNAHICS. 115 NoTB.— In the oaso of heavy bodies fallinfr through the atr, therosiBtsnce of the atmosphero producos a considerable dincropancy lM)twoon the actuU fall of bodies and the distance through which they should theoretically fall. Thus, it hOM born found by experiment that a ball of lead dropped from the lantern of St. Paul's Cathedral required 4i seconds to roach the pave- ment, a distance of !272 foot. Hut in H soconds the ball ought to have lallen 324 feet by theory, the differonco of 02 feet being due to the retarding force of the atmosphere. 261. A heavy body falling from a lieigLt moves with a uniformly accelerated motion, since the attraction of gravity which causes the descent of the body never ceases to act, and the falling body gains at each moment of its descent a new impulse, and thus an increase of velocity, so that its final velocity is the sum of all the infinitely small but equal increments of velocity thus communicated. 262. Hence the velocity of a falling body at the end of the second moment of its descent is twice that which it had at the end of the first second ; at the end of the third second, three times that which it had at the end of the first ; at the end of the fourth, four times f ifc. » 203. Hence also a heavy body starting from a state of rest and falling during any time, acquires a velocity, which would in the same space of time carry it through twice the space it has passed over. 264. It has been ascertained by numerous and careful experiments, that a falling body acquires at the end of the first second of its descent, a velocity equal to that of 82| feet per second, and hence during the first second of its descent a body falls through one half of 32} feet, i. e., through 16^3 feet. Note 1.— The average speed of the falling body is the arithmetical mean between its initial anof terminal velocities, or in the case of the first second of its fall, between and 32^, and this is 16^. Note 2.— In the following exercises wo shall use 32 and 16 in place of 32^ and 16^^, since the fractions materially increase the labour of making the calculations without illustrating the principles any better than the whole numbers used alone. IIG DYNAMICS. [A£T8. M5, 860. 200. ANALYSIS OF TU£ MOTION OF A FALLING BODY. It Number OF Skconds. Space passed over each Second. Terminal Velocities. Total Space. 1 1 2 1 9 3 4 4 8 5 4 1 8 10 5 10 25 11 12 30 7 13 14 49 , 8 15 10 04 ' 17 18 81 10 19 20 100 Note.— Tho nutnbcra in the second, third and fourth columns moan so many times Itt feet. From this it is evident that : — I. The spaces through which the body descends in equally succes- sive portions of time increase as the odd numbers, 1, 3, 5, 7, 9, £fc., and hence the space through which the body falls during any second of its flight, is found by multiplying 16 feet by the odd number which corresponds to that second; i. e., one less than twice the number of the second. II. The final velocity acquired by a falling body at the end of successive equal portions of time, varies as the even numbers, 2, 4, 6, 8, ^c, and hence the final velocity acquired by a body at the end of any second of its fall, is found by multiplying 16 feet by twice the number of seconds. III. The whole space passed over by a body falling during equal successive portions of time, varies as the square of the num- bers 1, 2, 3, 4, SfC, and hence the whole space passed over during any given number of seconds, is found by multiplying IQfeet by the square of the number of seconds. 266. Let t = the time of descent in seconds, v = the terminal velocity, i, e., the velocity acquired at the end of tho last' second of its fall, a = whole space passed over, and^ = 32, i. e., the measure of the attraction of gravity. Then Art. 2G3, the time is equal to the space di\idcd by half the terminal velocity, oxt=- s 2s V AST8. Bes, 260. 80DY. KL Space. 1 4 10 26 30 40 04 81 100 kmns mean so tally succes' 5, 1, 9, 4rc., any second )dd number twice the at the end lumbers, 2, body at the 6 feet by ng during f the num- ber during feet by the il velocity, its fall, s attraction c terminal ABTI. 267, MS.] 0TKAMIC8. 117 Anln (Art. tflS, III) tho wholo spare paiwpd orcr f« ''nnal to 16, i. f,, half of the gravity, g, multiplied the square of the time or s -- J^^. AIho (Art. 2nn, I) the terminal velocity i« equal w will be as long ascending as de- scending, and hence the problem is reduced to finding the distance through which the arrow will fall in half of 20 seconds, i. e., in 10 seconds. Then formula I. s = \gt^= iX32 X 10* = 16X100 = 1600 feet. Ans. Example 313. — If a cannon ball be fired vertically with an in- itial velocity of 1600 feet per second to what height will it rise ? SOLUTION. First, the time it ascends is equal to the time it would require if descend- ing to acquire a terminal velocity of 1600 feet. By formula VIII. < = — = -r^r = 50 seconds = time of ascent g 32 Then formula XI. s= Vt-^gt^ = 1600X50 — iX32 X 50* = 80000 — 16 X 2500 = 80000 — 40000 = 40000 feet. Ans. Example 314. — If a body be shot upward with an initial Velocity of 1200 feet per second, at what height Will it be at the end of the 10th second, and also at the end of the 70th second of its flight ? SOLUTION. Formula XI. s=.Vt--\gt^ •=. 1200X 10 — iX32Xlo' = 12000—1600 = 10400 feet = elevation at end of 10th second. Also 1200X70— ix32X70* =84000 — 16X4900=84000 — '78400= 6600 feet ■s elevation at end of the 70th second. 120 ^ DYNAMICS. [AST. 268. IM i 'k ExAMPLB 316. — If a cannon ball be fired vertically with an initial velocity of 2400 feet per second — 1st. In liow many seconds will it again reach the ground ? 2nd. How far will it rise ? .3rd. Where will it be at the end of the 40th second ? * 4th. What will be its terminal velocity ? » ' 5th. In what other moment of its flight will it have the same velocity as at the end of the 19th second of its ascent ? SOLUTION. Since the initial velocity =r terminal velocity = 2400 ft, I. Formula VIII. time of ascent = — = — — = 75 seconds, and since g 32 it IS as long ascending as descending, it again reaches the ground in 160 sec* II. Formula I. sz=z^gt^ = iX32X75*^ = 16X5625 = 90000 ft. = height to which it rises. III. Formula XI. s = Fi{ — i gt^ = 2400X40 — iX 32X 40*=96000— 16X 1600 = 96000 — 25600 = 70400 ft. = elevation at end of 40th second. IV. Terminal velocity =: initial velocity = 2400 feet per second, V. Since the whole time of flight = 150 seconds, and, since at all equal spaces of time from the moment it ceases to ascend and begins to descend, the velocity is the ikame in rising as in falling, it follows that the moment in which the body has the same velocity as at the end of the lyth second of Its ascent is 19 full seconds before it again reaches the grouud, or in 150 ~ 19 =: 131st second, i. c., in the end of the 13lst second. Example 316. — If a body is thrown downwards from an ele- vation with an initial velocity of TO feet per second, how far will it descend in 27 seconds ? SOKJTIOIf. rormula X. «= Vt+igt^ = 70X27+iX32X27* = 1890+16X729 = 1890 +11664 = 13554 ft. Ans. Example 317.— If a body is thrown down from an elevation with an initial velocity of 140 ft. per second, what will be its velocity at the end of the 30th second? SOLUTION. V = V+tg = 140+30X32 = 140+960 = 1100 feet per Second. Ana. Example 318. — If a body be projected vertically with an ini* * tial velocity of 400 feet per second, what will be its velocity at ' the end of tlie 12th second? SOLUTION". . Formula Xlll.v=V—fgz= 100—12 x 32=400— 384=:16 feet per second. Ans . Example 319.-- If a cannon ball be fired vertically upward with an initial velocity of 1800 feet per second : — 1st. In how many seconds will it again reach the ground? 2nd. What will be its terminal velocity ? 3rd. How far will it rise? 4th. Where will it be at the end of the 90th second ? 6th. In what other moment of its flight will it have the same velocity as at the end of the 27th second of its ascent? (i [ABT. 268. \y with an le ground ? nd? re the same its ascent ? ds, and since id in 160 sec* ft. r= height =96000— 16X cond. !ond, 3 at all equal s to descend, the moment >th second of , or in 150 — om an ele- d, how far <729 = 1890 elevation rill be its Ans. ith an ini'> elocity at ;cond.^«s. y upward round ? Ithe same ;nt? AbT. 268.] DYNAMICS. SOLVTIOir. 121 \. V ISOO 1. t=z - = ——=661 =:time of ascent or descent, hence whole time of flight = 56J X 2 = 112i seconds. II. Terminal velocity = initial velocity =1800 feet per second. III. Formula 1, S= i flr <« = i X 32 X (56J)2 =16 X 8164.0626=60625 ft. IV. Formula XI. 8= rt—\g t^ =1800 X 90-i X 32 X 90^ =162000—16 X 8100=162000—129600=32100 ft, = elevation at end of the 90th second. V. 112 J— 27= 85i = middle of 86//* second of flight. ExAUPLB 320. — A stone is dropt into the shaft of a mine and is heard to strike the bottom in 9 seconds ; allowing sound to travel at the rate of 1142 ft. per second, and taking gr=: 32^; required the depth of the shaft. SOLUTIOX. Lot a* := time stone takes to fall. Then (9— x) = time sound takes to reach the top and x^y. 16 iV = depth of shaft = (9— a?) x 1142 feet. Therefore ^^ = 10278 - 1142x. 198«2 + 13704a? =123336. 148996*2 + 1057948837 + 187799616= 95215392 + 187799616 = 283015008. 386x + 13704 =16823 + 886aJ=3119. X =: 8.0803 = number of seconds body was falling. 9— a; =9—8.0803 = .9197 = time sound travelled. And 1142 X .9197 = 1050.2974 feet = depth of shaft. Example 321. — A body has fallen through m feet when an- other body begins to fall at a point n feet below it ; required the distance the latter body will fall before it is passed by the for- mer? FIEST SOltrilON. -T- t- At cud of m ft. < = 1 ^=J-;r' *"^ ^ —Qt—g !2w»-:Vaw^and since »-= distance to be traversed t=J==; hence S = ^gt^ = ig X I / — 1 ~ Igx - — = — -. Ans. "^ \ Nimgf 2mg 4j» i'lUS*.'! >-' . ■> I (1-..I i:; .( ! !•>. fi 11(1') :■ '■'■■>[ ■>; '■ : 122 ^ DYNAMICS. [AST. 268. iii III 111 ! i BECOND SOLUTION. , . ,, . v^ las— lair.and \i(m+n+x) Let a? = distance. Then (of 2nd body) < = ^y — ^— ^ :s entire time taken by the first body to pass through whole space. Then J2("'+"+-^)-J'^ = J^ »"*^ multiplying all by V^ V2(m+n+x)—V!J»« = ViJ*. V'iCw+n+oj) = ^/2x + Vam. and squaring. 2(wi+n+a;) = 2a? + 2j» + i^/imx. 2>» + 2n + 2x = 2x + 2w + 2 V4im;c, 2» = WwJ?. n = 2 mx. n^:=Z'imx, •EXERCISES. 322. Through how many feet will a body fall during the 3tth second of its descent? ^n«. 1168 ft. 323. Through what space will a body descend in 25 seconds? ^ns. 10000 ft. 324. With what velocity does a body move at the close of the 20th second of its fall ? Ans. 640 ft. per sec. 325. During how many seconds must a body fall in order to ac- quire a terminal velocity of 1100 ft. per sec? ^ns. 343 sec. 326. Through what space must a falling body pass before it ac- quires a terminal velocity of 1700 ft. per sec ? jins. 45156i ft. 327. What will be the terminal velocity of a body that has fallen through 25000 ft. ? Jns. 1264.8 ft. 328. If a body is projected upwards with an initial velocity of 6000 ft. per second, where will it be at the end of the 40th second? Jlns. At an elevation of 214400 ft. 329. If a body be thrown downward with an initial velocity of 120 feet per second, through how many feet will it fall in 32 seconds? ^n«. 20224 ft. 330. A cannon ball is fired vertically, with an initial velocity of 1936 feet per second : — * In all cases, when not otherwise directed, use ^ = 82 ft. [AST. 268. 2(»i+rt+x) ABT. 268.] DYNAMICS. 123 a taco. g the 3Yth \s. 1168 ft. eccnds? 10000 ft. ose of the t. per sec. der to ac- 342 sec. fore it ac- 45156i ft. has fallen 12G4.8 ft. [elocity of the 40th 114400 ft. jloclty of Ifall in 32 1 20224 ft. (elocity of ) 1st. How far will it rise? 2nd. Where will it bo at the end of the 6th second? 3rd. In how many seconds will it again reach the ground? 4tli. What will be its terminal velocity? 5th. In what other moment of its flight will it have the same velocity as at the end of the 13th second of its ascent? Ans. 1st. 58664 ft. 2nd. At an elevation of 11040 ft. 3rd. 121 seconds. 4th. 1936 ft. per second. 5th. At end of lOSth second of flight. 331. If a body be projected vertically with an initial velocity of 4000 feet per second, taking gravity to 32^ feet : — 1st. How high will the body rise ? 2nd. Where will it be at the end of the 50th second? 3rd. Where will it bo at tlio end of the 100th second? 4th. Where will it bo at the end of the 200th second? 5th. In what time will it again reach the ground ? Arts. Ist. 248704.66 ft. 2nd. At an elevation of 159790.83 ft. 3rd. " 239166.66 ft. 4th. " 156661.00 ft. 5th. 248.70 seconds. 332. If a cannon ball be lired vertically with an initial velocity of 1100 feet per second, what will be its velocity at the end of the 7th second, at the end of the 20th second, and at the end of the 33rd second ? Jim. End of 7th sec. vel. = 876 ft. 20th " = 460 ft. 33rd " =44 ft. 333. If a stone be dropped into a well and is seen to strike the water after the lapse of 5 seconds how deep is the well ? • Ans. 400 ff. 334. If a stone be thrown downward with an initial velocity of 250 ft. per second, what will be its velocity at the end of the 3rd, the 9th, the 30th, and the 90th seconds of its des- cent? w^ns. End of 3rd sec. vel. = 346 ft. per sec. " 9th " = 538 ft. " " 30th " =1210 ft. " 90th «« =3130 ft. li n 336. A stone is dropt into the shaft of a mine and is heard to strike the bottom in 12-76 seconds, assuming that sound travels at the rate of 1100 ft. per second, what is the depth of the mine? wtfn». 1936 ft. 124 DESCENT ON INCLINED PLANES. [Abtb. 2(»-271. 336. A body has fallen through 400 feet, when another body ; begins to fall at a point 2500 feet below it ; through what space will the latter body fall before the former overtakes it? ^ns. 3906i feet. 337. A body A has fallen during m seconds, when another body B begins to full, / feet below it ; in what time will A overtake B'i /».../ Alls. 32 m DESCENT ON INCLINED PLANES. i! If ii 269. AVhen a body is descending an inclined plane a portion of the gravity of the body is expended in pressure on the plane and the remainder in accelerating the motion of the descending body. 270. The following are thu laws of the descent of bodies on inclined planes : — /. The pressure on the inclined plane .s to the weight of the body as the base of the plane is to its length. II. The terminal velocity of the descendinr; body is that which it would have acquired in falling freely through a distance equal to the height of the plane. III, The space passed through by a body falling freely, is to that gone over an inc ined plane, in equal times, as the length of the plane is to its height. IV. If a body which has descended an inclined plane meets at the foot of it another inclined plane of equal altitude, it will ascend this plane with the velocity acquired in coming down the former, it will then descend the second and re-ascend the former plane, and will thus continue oscill'iting down one plane and up the other. Note.— The same takes place if the motion be made in a curve instead of on an indhied plane. In practice, however, the resistance of the atmos- phere and friction retard the motion very greatly at each oscillation and very soon bring the body to a state of rest. 271. The final velocity, neglecting friction, on arriving at the bottom of the plane is dependent solely on the height of the plane, and will be the same for all planes of equal h( i<','^t, however various may be their lengths and the tim' of descent are exactly proportional to the lengths of the planes. I H 8. 2d(h271. ler body igh what ivertakes I06i feet. her body e will A IS. -JL — 32 m plane a pressure motion jcent of U of the is that distance to that of the ts at the ascend former, ine, and ther. instead e atmos- tion and rnving height equal id the fths of Arts. 272-275.] DESCENT ON INCLINED PLANES. 125 272. If in a vertical semicircle any number of cords bo drawn from any points Fig- 25. whatever and all meeting in the lowest point of the semicircle, and a number of bodies be allowed to stait along these cords at the same instant they will all arrive at the bottom at the same instant, and at every instant of their descent they will all be in the cir- cumference of a smaller circle. Thus in the accompanying figure if ADP be a semicircle and liP,CP, DP, HP, FP, any cords, and balls be allowed to start simultaneously from A, B, C, D. E and F, they will all arrive at P at the same instant. At the end of one-fourth the entire time they take to fall to P, A will have arrived at g, and the other bodies will be in the circumference flfP; at the end of one-half the time of descent all will be in the circum- ference h, &c. 273. Bodies descending curves are subject to the same law as regards velocity as those on inclined planes, i. e., the terminal velocity is due only to the pei^>endicular fall. 274. T\\Q Bmcliystoc1hrone{QvQQk hrachistos,^^ shortest," And cJironos, "time,") or curve of quickest descent, is. a curve somewhat greater than a circular curve, being what mathematicians denominate a cycloid, or that which is de- scribed by a point in the circumference of a carriage-wheel rolling along a plane. 275. Since, Art.. 270, the effect of gravity as an accelerating force on a body descending an inclined plane is to the effect of gravity on a body freely falling through the air as the height of the plane is to its length ; we have accelerating force of gravity or inclined plane: gr:: A:;; and hence accelerating force of gravity on inclined planes = -r-, v where A=height of plane, I = length. g =. effect of gravity=32. Substituting this value of the effect of gravity in the formulas in Art. 266, wo get the following formulas for the descent of bodies on inclined planen. 126 DESCENT ON INCLINED PLANES. CABT.27e. \ I) I M til FORMULAS FOR DESCENT OP BODIES ON INCLINED PLANES. No. 2 3 8 OIVEW. g, h, I, t s, t S, h, h i «, V g, h, I, V S} ^> h «• TO FIND. V FORMULAS. ght 2 21 8 =• 2gh 8 = J CORRESPONDING FORMULA IN ART.2G6. I II III V =: 2s T ght I J 2gi 2ghs _2s V t = gh _ J 2ls — 'J'TT Sh VI IV V VII VIII IX 276. When the body is projected down an inclined nht^ plane with a given initial velocity V ; 8= Vt + -—r- (10.) and vz=z Vt-{- ^-(11.) When the body is projected up an inclined plane with a given initial velocity V ; szuVt ->ti_^ (12) andv = Fit— ^(13.) Note.— When a body is thrown up an inclined plane, the attraction of gravity acts as a uniformly retarding force as when a body is projected CAbt.276. LINED 'ONDINO INABT.20d.' I II inclined (10.) cted up Sz=Vt raction of projected Abt. 276.] DESCENT ON INCLINED PLANES. 127 vertically Into the air. In the case of the inclined plane the body will continue to rise with a constantly retarded motion until Vt =^;n~ ^*»®** it will remain ntationary for an instant and then commence to descend. It will occupy the same lime in cominx down as in eoing up: its terminal velocity will bo the same as its initial velocity, and it will have the same velocity at any given point of the plane both in ascending and descending. ExAMPLi 338. — Through how many feet will a body fall in 15 seconds on an inclined plane which rises 7 feet in 40 ? SOLUTIOIf. Here < = 16, ft, = 7, ? = 40, and flf = 32. Then a = -^i — = — alTko =030 feet. Antt ExAMPLB 339.— Through how many feet must a body have fallen on an inclined plane, having a rise of 3 feet in 32, in ordei* to acquire a terminal velocity of 1700 feet per second ? SOLUTION. Here gr = 32, t> = 1700, ft = 3, ? = 32. Then » = -—,= Iv^ 32 X 17002 = 4816663 feet. 3 Ans, Igh^ 2 X 32 X 3 Example 340. — ^What will be the velocity at the end of the 20th second, of a body falling down an inclined plane, having an inclination of 7 feet in 60 ft. ? SOLUTIOW. :7,and2 = 60. __ ght __ 32X7X20 _ , Ana. Herefl' = 32, < = 20.ft: Then formula 6. t> = •^y—=-^^^^^^^p^ = 74| feet per second. ExAB(PL<" 341. — On an inclined plane rising 3 ft. in 17, a body has fallen through one mile, what velocity has it then acquired ? SOLUTION. Here a—\ mile = B280 ft., ft = 3, ? = 17 and g — 2,2. mu e 1-TTT ighs _^ 2X32X3X5280 j Then formula VI. v =J^ — == J ^-= = V5963294 =s * V' « V 17 24417 feet per second. Ana. Example 342. — In what time will a body falling down an inclined plane, having a rise of 7 feet in 16, acquire a terminal velocity of 777 feet per second ? BOLUTlOir. fi. Here ^r = 32, ft= 7, ? = 16, and v = 111. Then formula 8. ._?«__ 16X777 _„. , ^ * ~ ft^~"32x7~ ^^* seconds. Ana. Example 343. — In what time will a body fall through 4780 on an inclined plane, having a rise of 3 feet in 4 ? Here g — 32, ft Then formula 9. SOLUTIOir. 3, i! = 4,ands = 4780. t. \2^__ ^/ gh~'' 2X4X4780 V 32X3 = V398*3 = 19*9 seconds * II if: 128 DESCENT ON INCLINED PLANES. TART. 270. ExAMPLS 344.— If a body bo projected down an inclined plane, having a rise of 8 feet in 15, with an initial velocity of 80 feet per second, tlirougk what space will it pass in 40 seconds ? BOLUTIOIT. Hero i> = 80, flf = 32, A = 8, ? = 16, and < = ^0. Then fonimlaio, s=n + -2^1 = 10 x 80 -}- ^^5^^^ = 3200+l:W53i n 2x10 = 18858i ft. Jn#. Example 345. — If a body be projected up an inclined plane having a rise 5 feet in 16, with an initial velocity of 2000 ft. per second : 1st. How far will it rise ? 2nd. When will it again reach the bottom of the plane ? 3rd. What will bo its terminal velocity ? 4th. Where will it be at the end of the 100th second ? 5th. In what other moment of its flight will it have tlic same velocity as at the end of the 11th second of its ascent? Hero h Then formula 8 SOLUTION. 5, 1 = 16, flr = 32 and v = 2000. ?» _ 16X2000 gh 32X5X200» 2X16 „, =: 200 seconds. 6X32 Ist. Formula 12. s=:Vt — ^^ = 200 X 2000 — '■ — 200000 = 200000 ft. Ans. 2nd. Ascent ^ 200 sec. + descent 200 sec. = 400 sec. Ana. 8rd. Terminal velocity := initial velocity = 2000 feet per sec. Ans. —Vt ^!Mi— • 32X5X1002 -; 400OOO 4th. Formula 12. s ; 100 X 20000 — =200000 %l " 2X16 — 60000 = 160000 = elevation at end of 100th sec. Ans. 5th. 400 — 11 = 389th second. Am. EXERCISES. 346. On an inclined plane rising 5 ft. in 19 through what space will a body descend in half a minute? Ans. 3789^^^ ft. 347. On an inclined plane rising 3 ft. in 13, what velocity will .iiT: . a descending body acquire in 39 seconds? Ans. 288 ft. second. 348. What time does a body require to descend through 3800 ft. on a plane rising 19 ft. in 32 ? Ans, 20 seconds. 349. If a body be projected down an inclined plane, having a fall of 7 in 11 with an initial velocity of 50 feet per . tf second, what will be its velocity at the end of the 44th second? Ans. 946 ft. per second. 350. If a body be thrown down an inclined plane having a rise of 13 feet in 32 with an initial velocity of 100 feet per second, through how many feet will it descend in 130 sec. ? Vh.i . .#• Ans. 122850 ft. TART. 27fl. ABT8.877,27f).| PROJECTILES. 129 i inclined city of 80 seconds ? 200+l.'W53j icd plane r 2000 ft. no? 1? tlio same 3 nscent? =•" 400O0O ins. 2 -=200000 Ii what rssfcrft. ity will 288 ft. ?h 3800 econds. iving a eet per le 44th second. a riso eet per sec. ? K850 ft. H51. If a body be projected up an inclined plane, having a fall of 6 feet in 8, with an initial velocity of 800 feet per second :^ 1st. How far will it rise ? 2nd. In how many seconds will it again reach bottom of the plane ? 3rd. What will be its terminal \elocity ? 4th. Where will it be at the end of the 68th second ? 0th. In what other moment of its flight will it have the same velocity as at the end of the 37th second of its ascent ? .4ms. 1st. Rise = IGOOO ft. ; 2nd. Time of flight = 80 seconds 3rd. Terminal velocity = 800 feet per second; 4th. Ele- vation at end of 68th sec. = 8160 ft. ; 0th. At the end of the 43rd second. 352. A body rolls down an inclined plane, being a riso of 7 ft. in 20^when it has descended through / feet, another body commences to descend at a point m feet beneath it. Through how many feet will the second body descend before the first body passes it ? 5,^8 PROJECTILES. 277. A projoctilo is a solid body to which a motion has been communicated near the surface of the earth, by any force, as muscular exertion, the action of a spring, the explosive eflfects of gunpowder, &c., which ceases to act the moment the impulse has been given. 278. A projectile is at once acted upon by two forces : — 1st. The projectile force which tends to make the body move over equal spaces in equal times ; and 2nd. The force of gravity, which tends to make the body move towards the centre of the earth over spaces which are proportional to the squares of the times. Under the joint influences of these two forces the pro- jectile describes a curve, which in theory is the parabola, but which in practice departs very materially from that figure. Note l.— The parabola ia that curve which is produced by cutting a cone parallel to its side. Note 2.— The parabolic theory is based upon three suppositions, all of which are more or less inaccurate. 1st. That the force of ifravity is the same in every part of the curve described by the projectile. 'k * w^ 130 PROJECTILES. [ABT8. 279, 880. I I i 2n(l. That tho forco of gravity acts In parallel lines. ^ Sril. Tliiit tlio projcc'tilt) inuvos lliroiiKh a noi)-ru,si!«ting modiuni. 1\w (Irst and hi-coiuI of those siippositionH dilFiir soinnonslbly from truth that thi'y Muiy hi) nsMnnu'd to ht> ahsoliitt'iy ctirroct, hut tho r«»iMtanoo of till) atrii>>spli'(>r<' so niuturially iill'ft^t.s tho niotiunn of all bodioN, *>Npocially wluMi llioir velocities aro con'^tidorahU), that it renders tho parabolic thoory prncticaliy usflesH. 279. Wlieii i\ body is projoctcil horizoiitjiUy forward, tho liorizonl.il motion dot's not intoifero with the action of i,'ravity, — tho projoctilo descending witli the same rapii velocity not exceeding 1280 per second. Note l.— This arises from tho fact that atmospheric air flown into a vacuum with a velocity of only 128i» foot per second, mu timt when a ball moves with a Kfcater velocity tlian this, it leaves a vacnum behind it into which tho strongly compressed air in front tends powerfully to forco it. Note 2.— From experiments made witli great care, it has been ascertained that when tho velocity of a ball or other projectile is 2(MI0 Pftet per second, tho ball meets with an atmosphorlu roiiistance equal to lUU times its own weight. NoTB 3. — Another groat irregularity In tho flring of balls arises from tho fact that tho ball deviates more or less to tho right or loft, sometimes crossing tho direct lino several times in a very short course. This deflet^tion sometimes amounts from J to ^ of the whole rango, or as much as SUO or 400 yards in a railo when there is considorablo wlndago ; 1. c, when tho ball is too small for tho calibre of tho gun. 282. The motion of projectiles has recently been inves- tigated with much care, with the view of deducing a new theory in which tho resistance of the air should be takefi into account. The following are the most important results: — : - WHEN THE BODY IS THROWN VERTICALLY UPWARDS INTO THE AIR. I. The time of ascent is less thaji the time of descent. II. The velocity of descent is liss than that of ascent. III. The terminal velocity is less than the initial velocity. IV. The velocity of descent is not infinitely accelerated, since when the velocity becomes very great, the resistance of the atmos- phere becomes so great as to counterbalance the accelerating force of gravity, and the velocity of the descending body is thenceforth uniform. WHEN THE PROJECTILE IS THROWN AT AN ANGLE OF ELEVATION. I. The ascending branch of the curve is longer than the descend- ing branch. II. The tvne of describing the ascending branch is less than that of describing the descending branch. III. The descending velocity is less than the ascending, • IV. The terminal velocity is less than the initial. V. The direction of the descending branch is constantly approxi- mating to a vertical linCf which it never reaches. 132 PROJECTILES. [Arts. 288-288. VI. The descending velocity is not infinitely accelerated, hut, as in case of a body falling vertically, becomes constant after reaching a certain limit. VII. The limit of the velocity of descent is different in different bodies, being greatest when they are dense, and increasing with the diameter of spherical bodies. 283. The explosive force of gunpowder, fired in a piece of ordnance, ia equal to 2000 atmospheres, or 30000 lbs. to the square inch, and it tends to expand itself with a velocity of 6000 feet per second. Note.— Gunpowder is an intimate mixture of 6 parts'saltpetre, 1 part charcoal, and 1 part sulphur. In firing good perfectlj' dry gunpowder, the ignition takes place in a space of time so short as to appear instantaneous. 1 cubic inch of powder produces 800 cubic inches of cold gas, and, as at the moment of explosion the gas is red hot, we may safely reckon the expansion as about 1 into 2000. 284. The greatest initial velocity that can be given to a cannon ball is little more than 2000 feet per second, and that only at the moment it leaves the gun. Note.— The velocity is greatest in the longest pieces ; thus Hutten found the velocity of a ball of given weight, fired with a given charge of powder, to be in proportion to the fifth root of the length of the piece. 285. The velocities communicated to balls of equal weights, from the same piece of ordnance, by unequal weights of poAvder, are as the square roots of the quantities of powder. 286. The velocities communicated to balls of different weights and of the same dimensions, by equal quantities of powder, are inverselv proportional to the square roots of the weights of the bans. 287. The depth to which a ball penetrates into an obstacle is in proportion to the density and diameter of the ball and the square root of the velocity with • which it enters. Note 1.— An 18-pound ball with a velocity of 1200 feet per second pene- trates 34 inches into dry oak, and a 24-pouna ball with a velocity of 1300 ft. per second penetrates 13 feet into dry earth. Note 2.— The length of guns has been much reduced in all possible cases. Field pieces are now seldom made of greater length than 12 or 14 calibre8(dia- meter of the ball). The maximum charge of powder has also been dimi- nished very greatly— now seldom exceeding one-third, and often being as low as one-twelfth of the weight of the ball. 288. The following rule, obtained from experiment, has been given, to find the velocity of any shot or shell, when ;AeT8. 283-288. 'ated, but, as fter reaching t in different sing with the 1 in a pieco '000 lbs. to 1 a velocitv Itpetre, 1 part n powder, the xstantaneous. and, as at the bhe expansion e given to JCond,an(l lutton found of powder, of equal unequal quantities ' different luantities are roots into an ;er of the which it !ond pene- of 1300 ft. ble cases. ibres(dia- )een dimi- being as ent, has 1, when ABT.288.J PROJECTILES. 133 the weight of the charge of powder and also that of the shot are known : — RULE. Divide three times the weight of powder by the weight of the shot, multiply the square root of the quotient by 1600, and the pro- duct will be the velocity per second in feet. Orif p:=i charge of powder in lbs., w = weight of ball in lbs., and V = velocity per second in feet ; then v = 1600 X V 3p w Example 353. — What is the velocity of a ball weighing 48 lbs., fired by a charge of 4 lbs. of powder? Here p = 4 and «<; == 48. BOLUTIOW. Then v = 1600 X V w X i = 800 feet per second. Ans. ^^ = 1600 X ^ ?^ = 1600 X 48 VT = 1600 Example 354. — With what velocity will a charge of 7 lbs. of powder throw a ball weighing 32 lbs. ? Hero p = 7 and w = 32. Then v = 1600 x . ) ^ = V w SOLUTION. I 3X7 ; 1600 X J — - V 82 X "81 = 1296 feet per second. Ans. = 1600 X V '65626 = 1600 Example 355. — If 4 lbs. of powder throw a ball 16 lbs. in weight with a velocity of 1200 ft. per second, vrhat amount of powder would throw the same ball with a velocity of 600 feet per second ? SOLUTION. Art. 285. vel. : vel. : : V weight of powder : V weight of powder ; or 1 200 : 600 : : V ~4 ' V"^ and hence a? = lib. Ans. Example 356. — If 3 lbs. of powder throw a ball 6 inches in diameter and weighing 32 lbs., with a velocity of 850 feet per second, with what velocity will the same charge throw another ball of the same dimensions but weighing only 9 lbs. ? SOLUTION. Art. 286. VoT VsF: : 850 : a?, or 3 : 5.65 : And hence x = 1600 feet. Ans. 860 : X. EXERCISES. 367. With what velocity will a charge of 11 lbs. of powder throw a cannon ball weighing 24 lbs. ? Ans. 1875 feet per second. : f H 134 358 CIRCULAR MOTION. [Abt. 280-202. With what velocity will a charge of 9 lbs. powder throw a ball weighing 3G lbs. ? Jns. 1387 feet per second. 359. If 1 lb3. of powder throw a ball with a velocity of 1000 feet per second, what charge will throw the same ball with a velocity of 1500 feet per second ? Jns. 151 lbs. 360. If a certain charge of powder throw a 10-inch ball weighing 20 lbs. with a velocity 973 feet per second, with what velocity will the same charge throw a ball. of the same dimensions weighing only 25 lbs.? J71S. 863 feet per second. »> CIRCULAR MOTION. 289. Centrifugal force (Lat. centrum, *' the center,' and/i/yto, " I flee") is that force by which a body mov- ing in a circle tends to fly oft' from the centre. Note.— Since a body moving in a circle would, if not restrained by other forces, lly oil' iji :a tangent to that circle, centrifugal force is sometimes called tangential force, 290. Centripetal force (Lat. centrum, *' the centre," and "^cto, I seek or rush to ") is that force by which a body moving in a circle is held or attracted to the centre. 291. Wlicn a body is at once acted upon by both cen- trifugal and centripetal force, it moves in a curve, and the form of this curve depends upon the relative intensities of the two forces: /. e., if the two be equal at all points, the curve will be a circle, and the velocity of the boily will be uniform ; but if the centrifugal force, at difttirent points of the body's orbit, be inversely as the square of the distance from the centre of gravity, the curve will be an ellipse, and the velocity of the body will be variable. 292. When a body rotates upon an axis, all its parts revolve in equal times ; hence the velocity of each particle increases with its perpendicular distance from the axis, and so also does its centrifugal force. Note l.— As long as the centrifugal force is less than the cohesive force by which the particles are hold together, the body can preserve itself ; but, as soon as the centrifugal force exceeds the cohesive, the parts of the rotating mass lly off in directions which are tangents to the circles in which they were moving. Note 2. — We have examples of the effects of centrifugal force in the destructive violence with which rapidly revolving grindstones burst and fly to pieces— the expulsion of water from a rotating mop, the projection ol a stone from a sling, the action of the conical pendulum or governor in regulating the supply of steam in an engine, SiC, olc. [Am. 289-292. vder throw a per second. city of 1000 le same ball Ans. 153 lbs. all weighing , with what of the same t per second. fie center," body mov- *ained by other ) ia sometimes entre," and ich a body entre. y both cen- curvc, and intensities b all points, f the body at ditfercnt uarc of the will be an iable. all its parts ich particle the axis, cohesive force vo itself; but, )f the rotating n which they 1 force ill the nea burst and he projection )r governor in Aets. 293-207.] CIRCULAR MOTION. 136 293. When the velocity and radius are constant, the centrifugal force is proportional to the weight. 294. When the radius is constant, the centrifugal force varies as tlie square of the velocity. Note. — At the equator the centrifugal force of a particle is uiy of its gravity or weight and from the equator it diminishfs as wo approach the poles where it becomes 0. It follows that if the earth were to revolve 17 times faster thau it does, the centrifugal force at tl" <' Example 361. — ^What is the centrifugal force exerted by a body weighing 10 lbs. revolving with a velocity of 20 feet per second in a circle 8 feet in diameter ? Here «j = 10,« = 20, r = wi>» 10X20» Then c -~ -—■-. gr SOLUTION. : 4, and g = 32. 10X400 = 3U lbs. Ans. 32X4 ~ 32X4 fi Example 362. — What centrifugal force is exerted by a body weighing 15 lbs. revolving in a circle 3 feet in diameter and making 100 revolutions per minute ? u I » rf 136 CIRCULAR MOTION. MOLUTION. TAbt. S97. Hero w ~ 13, r = 1'5, « = W = 1*. Then formula \.: c — wrn^ Xl-2345 ~ ISXl'oX (l!)« X l-23i5 = 77'15628 lbs. Ana. Example 363. — A body weighing 40 lbs. revolves in a circle 4 feet in diameter ; in order that its centrifugal force may be 1847 lbs., what must be its velocity-and number of revolutions per second ? SOLUTION. Here w = 40 lbs., »• = 2, and c = 1847. Then formula VIll. : » = 18'li7 — £ I — '.^ — = ViHor9 VwrXl-234i5~ V40X2X 1*2345 = 4'28 = number of revolutions per second, and hence revolutions per minute = 25t>*8. Also V = 4iX3-1416X4-32 = 5378 feet per second. ExAMPLB 364. — The diameter of a grindstone is 4 feet, its weight half a tbn, and the centrifugal force required to burst it is 45 tons : with what velocity must it revolve, and how many revolutions must it make per minute in order to burst ? Here w = 4, c = 45, and r : Then formula VIII. : SOLUTION. 2. _£ I 45 : 1-2345 ~ V ' EXERCISES. 365. If a ball weighing 4 lbs. be attached to a string 2J feet long and whirled round in a circle so as to make 120 revolutions per minute, — what must be the strength of • the string in order to just keep the ball from flying oflf? Ans. 49-38 lbs. 366. A ball weighing 2 lbs. is attached to a string 3J feet long and capable of resisting a strain of 200 lbs. ; if the ball be whirled in a circle with the whole length of the string as radius, how many revolutions per minute must it make in order to break the string? Ans. 288^ revolutions. 367. A ball is whirled in a circle, with a velocity of 64 feet per second, by means of a string 4 feet in length and capa- ble of resisting a strain of 840 lbs ; what must be the weight of the ball in order to break the string ? »/?n». 26i lbs **~ VtfrX 1-2345 ~ '\/iX2xr2345 " V**'*^'* V = 6*03 — revolutions per second, and hence 6*03X60 = 361'8 = the revolu- tiors per minute. Also velocity =- 4X3*1416X6-03 = 75-775 feet per second. 1( t( •fe. TAXT. S97. ,5 = 77'15626 in a circle rce may be revolutions = Vl8'70i9 olutions per 4 feet, its [ to burst it how many it? ^36-452 the revolu- ing 2i feet make 120 trength of ying off? 49-38 lbs. feet long if the ball the string ist it make volutions. 64 feet per and capa- st be the ? ». 26^ lbs Abt8.298.299.] ACCUMULATED WOBK. 137 368. What is the centrifugal force exerted by a body weighing 20 lbs. revolving in a circle 10 feet in diameter nnd mak- ing 2*8 revolutions per second? Jns. 967-848 lbs. 369. What is the centrifugal force exerted by a body weighing 8 lbs. and revolving in a circle 20 feet in diameter with a velocity of 100 feet per second ? ^m. 250 lbs. ACCUMULATED WORK. 298. Work is required to set a body in motion or to bring a moving body to a state of rest. For example, when a common engine is first set in action a considerable portion of the work of the engine goes to give motion to the fly-wheel and other parts of the machinery ; and before the engine can come to » state of rest, all of this accumu- lated work must be destroyed by friction, atmospheric resistance, &c. 299. To find the work accumulated in a moving body : — , . RULE. /. Find the height in feet from which the body must have fallen to have acquired the given velocity. II. Multiply the number thus found by the weight of the body in pounds. Or let IT = units of work accumulated^ v = velocity^ w = the weight in Ibs.^ and g = 32. V* Then Art. 266, since s:=zh=z-^ s tr=Aw=-X «> = v*w ExAifPLB 370.— A ball weighing 10 lbs. is projected on stnooth ice with a velocity of 100 feet per second : assuming the friction to be -^ of the weight of the ball, and neglecting atmospheric resistiince, over what space will it pass before coming to a state of rest? SOLUTION. Here t> =e 100, w — 10, and g = Z2. Then TJ a= "sir == v*to 100* XIO 100000 64 == 15621 = units of work accu- se ~ 2X32 mulated in the ball. Also "^ XlOXl = I = units of work destroyed by friction iu moving the ball through 1 foot. Therefore the number of feet s= 1562i -f f = 23431 Ans. 10 -^ 138 ACCUMULATED WORK. [AbT. 29^. ( Example 3 '71. —A train weighs 100 tons, and has a velocity of 40 miles per hour when the steam is turned oS: how far will it ascend a plane having an inclination of i in 100, taking fric- tion as 11 lbs. per ton, and neglecting the resistance of the atmosphere? SOLUTION. 40X5280 Hero V ^ 40 miles per hour = gyxoo" ^^ ^^^ ^^^^ P®^ second, w = 100 tons = 200UOO Iba., and ^ = 32 Then U = -27 = «««? (68|)«X200U00 Sm'rX 200000 2X32 64 = 3411^ X 3125 = 10703555^ = units of work accumulated in the train. Work of friction = 100 X 1 1 = 1100 units to each foot. - Work of gravity =: 7uuX2u0000 = 1000 usits to each foot. Work destroyed by resistances, i. e. friction and gravity, in moving the train uvur one fout = llou+iOOO = 21U0 units. 10755565^ Therefo"^ > number of feet = — ^;^^ — = 6121*69 fect=nearlyonemile 2100 Example 372. — If a car weighing 3 tons, and moving at the rate of 10 feet per second on a level rail, pass over 500 feet before it comes to a state of rest, what is the resistance of fric- tion per ton ? SOLUTION, 10«X6000 Work accumulated in car = gxsa Work of friction = friction X 600. _ 600000 __ 5375 ^„it3, ~" 04 "" Therefore friction X 600 = 9376, and hence friction = --— = i82 lbs. on whole car. Then friction per ton = 18i-j-3 — 6i lbs. ^JM. EXERCISES. -869960S6 = = 160835 adulam in = son iu. ds at the tin and is required flOLUTlON. Here » = 60, / ~ 30*11, and V = 1-28. Then formula X.: n'= r-r= =0'058 = thoimmb«rofvibra- a I ^ X Stf 1 1 tions gained ; hcnco tha number of vibrations made = 00*038. Ant, ExAMPLi 391.— What will bo the velocity acquired by a heavy body falling during one entire second in the latitude of Spitz- bergen ? SOLUTION. Here tzsl, and by the tablo Art. 810. 1 = SOZt, Thenflf = lirt— 39-2lX(3'l'llO)« — 30-21 XOMOCS EXERCISES. 1 380038 inches. Ant. i inches, 392. What must be the length of a pendulum in the latitude of Canada in order that it shall vibrate once in 3 seconds ? ^ns. 351-99 inches. 393. A pendulum that vibrates seconds at the surface of the earth is carried to the summit of a mountain 3 miles in height : how many seconds will it lose in 24 hours ? jint. 64*8. 394. In what time will a pendulum 10 inches in length vibrate ? Ans. '505 seconds. 395. What velocity will a heavy body falling in the latitude of New York acquire in one entire second? wfns. 385*903. 396. If a clock lose 10 minutes in 24 hours, how much must the pendulum be shortened in order that it shall keep correct time? Jins. -543 or over I of an inch. 397. If a seconds pendulum be shortened 5 inches, how many vibrations will it make in a minute? ^ns. 63*83. 398. A pendulum which vibrates seconds at the surface of the earth is carried to the summit of a mountain, where it is observed to lose 30 seconds in 24 hours : required the height of the mountain ? Ans. 7333*3 feet, 399. In what time will a pendulum 100 inches long vibrate ',' ^ns. 1*69 seconds. 400. Required the length of a pendulum which makes 120 vibra- tions per minute? ^ns. 9*77 inches. 401. Through how many feet will a body fall in one second, and what will be its terminal velocity at the end of that por- tion of time in the latitude of Paris ? ^ns. Terminal velocity = 186*1 in. ' Space passed over =: 16*0875 ft. 146 [Axn. 322-82ft. HYDRODTNAMICS. CHAPTER VIII. HYDRODYNAMICS. 322. Hydrodynamics treats of the motions of liquids and of tlie forces which they exert upon the bodies when their action is applied. 323. The particles of a fluid on escaping from an orifice possess the same veloiiiy as if they had fallen freely in vacuo from a height eqtial to that of the fluid surface above the centie of the oriOce. NoTB.— This is known as Torricelli's theorem. Since all bodies falling tn vacuo from the same height acquire the »ame velocity, density has no effect in increasinu: the velocity of a liquid escapinu from an orifice in the side or in the bottom of a veHScl. Thus water, alcohol, and mercury will all flow with the same rapidity ; for though the pressure of the mercury is ISj times greater thatf that of water, it has ISi times as much matter to move • 324. When a liquid flows from an orifice in a vessel which is not replenished but the level of which continually descends, the velocity of the escaping liquid is uniformly retarded, being as the decreasing series of odd numbers 9, 7, 6, 3, (fee, so that an unreplenished reservoir empties itself through a given aperture in twice the time the same quantity of water would have required to flow through the same aperture had the level been maintained constantly at the same point. 325. The quantity of fluid discharged from a given aperture in a given time is found by multiplying the area of the aperture by the velocity of the escaping liquid. Note.— Experiments do not agree with this theory as regards the quan- tity of liquid discharged. The whole subject has been carefully investigated by Bossut, and he has shown that ' ActtMl discharge : Theoretical discharge : : -62 : 1 or as 6 : 8. Hence the theoretical discharge must be multiplied by | to obtain the true quantity. This discrepancy arises ft*om the fact that the escaping jet diminishes in diameter just after leaving the vesseL forming what is known as the vena contracta or contracted . The minimum diameter of the vein is found at a Fig. 27. vein distance about equal to half the diameter of the aperture as at c e' Fig. 27. This effect arises from the fact that just above the orifice the lateral particles of fluid move ai well as the descending portions. iiiii BTI. 822-Stt. quids and hen their nn orifice freely in ace above Ddics falling 181 ty has no riflce in the cury will all sn-ury is 13| ter to move • I a vessel )ntinually miformly imbers 9, empties the same ough the instantly a given the area uid. the quan- ivestigated :8. obtain the ig. 27. Aaxs.8S6.] HTDBODTNAMICS. 147 826. Istt Q = the quantity discharged in 1 tecond^ a == area of aperture, k = height of fluid level above the centre of the or0c9t g = accelerating force {^gravity, and v ^ velocity. ^ — — Q Then Jirt. 266 v = V2gA, ' ' Note.— Since g — 32, 2g = 64, and V^=: 8. formulas I, II and III become respectively « 1= sVa Q=Sa^/h,oxida = ^~l^ ExAMPLs 402. — With what velocity does water issue from a small aperture at the bottom of a vessel filled to the height of 100 feet? 1 SOLUTION. Fonnula 1 v— sV* = sVlOO = 8X10 - 80 feet per second. Ans. ExAUPLB 403. — What quantity of water will be discharged ia one minute from an aperture of half an inch in area — the height of the water in the vessel being kept constant at 10 feet above the centre of the orifice ? SOLUTION. Hera a=si square inch =s ^^ J ^ of a square foot. The cubic feet discharged in 1 secoud = SaV^ Cubic feet discharged in 1 minute =60X8axVA = 60 X yItX VlO= 60X Vs X 3*162 = 6*27 cubic feet — the theoretical quantity, and 6*27x1 = 3*29 cubic feet = true quantity. Example 404. — What must be the area of an orifice in the side of a vessel in order that 40 cubic feet of water may issue per hour — the water in the reservoir being kept constantly at the level of 20 feet above the centre of the aperture ? HereQ: 40 60X60 SOLUTION. : ^ of a cubic foot, and since this is only f of the theoretical quantity, Q =: f of ^^ = j^j of a cubic foot. Also h = 2). Then formula III, a ■■ inch. An9. _Q_7^ •J5T V'^ 8V20" = 35=776 =nPj? °^ * '°°*= H *'*" EzAUPLB 405. — An upright vessel 16 feet deep is filled with water and just contains 15 cubic feet. Now if a small aperture i of an inch in area be made in the bottom, in what time will the vessel empty itself ? 148 HYDRODYNAMICS. CAbt.827. I li ' BOLUTIOW. - • V Iloro h — ld ft., a — i of an iuoh, and Q == 15 cubic foot. llonco the theoretical quantity =» 16X Jf = 21 cubic foot. Then velocity at commoncflmont = sV'* = sVw = 32 ft* Quantity dischargt^d in 1 socend = 32 X ^ i| ^ := ^f^^ = -,i^ of a cubic foot. Time required to disoharffo 21 cubic foot — 21-f-,'jf = 432 seconds. lint, Art. 321, ^vhen a vessel onintioa itself, the time required to dischargA given quantity of water is double tliat requisite for discliarging quantity when the level is maintained, llcuc* time — 432X2 = 881 seconds =: 16'-1 minutes. Ans. EXERCISES. rging the same 406. With what velocity does water issue from a small aperture in tho side of a vessel iillud to the height of 25 feet above the centre of tho orifice ? ^ns, 40 feet per second. 407. With what velocity does water flow from a small aperture in tho side of a vessel filled with water to the height of 17 fee t above tho centre of tho orifice ? I ^ns. 32 '984 feet per second. 408. In the last example, if tho water flows into a vacuum, what is its velocity ? Jins. 56 feet per second. Note.— Since tho pressure of tho atmosphere is equal to that of a column of water 32 feet high, the elTective height of the column of water is 17+32 — 40 feet. 409. How much water is discharged per minute from an aperture having an area of ^ of an inch— the surface of tho fluid being kept constant at 36 feet? >dns, 2} cubic feet. 410. What must bo the area of tho aperture in the bottom of a vessel in order that 90 cubic feet of water may issue per hour — the level of the water in the vessel being constantly kept at 20 foet above tho centre of the orifice ? jlns. *161 or about i^g of an inch. A vessel contains 20 cubic feet of water, which fills it to the depth of 30 feet — now if an aperture having an area of I of an inch bo made in tho bottom of the vessel, in what time will it empty itself? »dns. 2 minutes 30) seconds. 411. 327. When waterspouts from several apertures in the side of a vessel, it is thrown with the greatest random from the orifice nearest the centre, the jet issuing from the centre will reach a horizontal distance equal to the entire height of the liquid, and all jets equally distant from the centre will be thrown to an equal horizontal distance. CAbt.827. )f a cubio foot. )cond8. d to disoharRA ;iiig tlio saino all aperturo 5 foot abovo per second, ill aperturo e height of per second, cuum, what per second. 1 to that of a leight of the :in aperture the fluid cubic feet. ottom of a ' issue per constantly )f an inch. mis it to g an area vessel, in seconds. GS in tlio om from le centre e height centre AaiS. 328-330.] nYDRODTNAMICS. Fig. 28. 149 NoTW.~Lot VA bo a veiiHol flilod with water, having its Hide AH Dor|M)ndicu1ar to tho horizontal piano HM. On Ali desoribo tho scinioircle liDA. lUsoct Ali in C'and in All take any points 1) and // equally distant from B, also C aud c' equally dlHtant from E. Draw alHo CO, 1)1), HE, &c., porpon- dicular to A Hand produced to tho circumforcnco A ItC. Then if Hmal I oriflcoii t)e pierced i n Ihn >-. J« of tho voHHol at C',D',E',1)' . d a, the liquid from E wilL__ upout to twico EE'.-Alt-nM-?k= tho liquid from U or C" will BTiout to // — twico VC or C'C and that from Dor D will roach K = twice DD or D'lJf. 328. When water flows in any bed, as in the channel of a river or in a , ipe, the velocity becomes constant when tho lengtli of » d bears a large proportion to i(a sec- tional area. TI -. ' ■ pipes of more than 100 feet in length or in rivers whose course is unopposed by natural obstacles, the velocity of the body of the stream is the same throughout Wiicn this occurs tho liquid is said to be in train. 329. The velocity of the liquid flowing in a pipe or channel is not the same in every part of its section^ being greatest in tho centre of tlie section of tho pipe or in the middle of the surface of the stream. NoTB 1.— This arises from tho friction cxertod against the fluid by tho Interior surface of tho plpo or the banks of the Htroain. !• a str«ain, on account of the middle part having tho greatest velocity, tho surface Is always more or less convex. NoTB 2.- ways :— Ist. An open tube bent at right angles is placed in a stream with one of its logs opposed to tho current and tho other branch vertical— the velocity of tho stream is measured by tho height to which tho water rises in the vortical log. 2nd. A float Is thrown into the stream and tho time occupied by It in passiug over a known distance observed. Srd. Tho convexity of tho surface may bo measured by a levelling instru* mcnt, and its velocity thus determined. 330. To fmd the velocity of efflux f and hence the quantity of water discharged in a given time from a reservoir of given height through a pipe of given length and diameter: — -The velocity of a stream may bo determined in throe different 150 HYDRODYNAMICS. tABT8.8Sl-8S$. i (i ■i Let d = diameter of pipe, I = lengthf h = height, and v = velocity. Then, all the dimensions being in feet, v =: 48V ) iTTTi \ Not B.— This is tlio formula of M. Foncelot and la regarded as strictly accurate. 331. Water is frequently made to drive ma.;hinery by its weit^ht or rnoinentuin exerted on a vertical watei-wliecl. 332. There are three varieties of vertical water-wheels, viz : the undershot, the overahot, and the breast wheel. Fig. 29. o Breast wheel. Undershot wheel. Overshot wheel. NoTK.— The mode in which the water is made to act on these is repre- sented in Fit?. 29. It will be observed that the undershot wheel is moved by the moraentuui of the water — the breast wlieel and overshot wheel by its weight aided by its momentum. An overshot wheel will produce twice the effect of an undershot wheel, the dimensions, fall, and quantity of water beiiii< the same. The breast wheel is found to consume twice the quantity of water required by an overshot wheel to do the same work. 333. In all water-wheels the greatest mechanical effect is produced when the velocity of the water is 2j- times that of the wheel. d ITS. 891-3SS. AST. SS4.3 ^TDItODTKAMlCd. 151 -velocity. ^} " as strictly inery by 5i-wliecl. pwheels, eel. rcpre- lovecl by 1 by its 56 twice ntity of nee the ark. effect times 334. To find the horse powers of a vertical water- wheel — Let b = breadth of stream in feet, d = depth of stream, V = mean velocity in feet of stream per minute, h = height of fall, $ =r weight of one cubic foot of water, m ^ modulus of the wheel, and U = units of work. «... mbdvsh , Then horse powers = ^^^^^ . ExiuPLK 412.— A water-wheel is worked by a stream 6 feet wide and 3 feet deep, the velocity of the water is 2i feet per minute, and the height of the fall 30 feet, required the horse powers of the wheel, the modulus being -7. J5r.p.= SOLUTIOir. mhdvsh _ II.*' '^ :t' i-p and 4 nute, the ay cubic ;he depth s. 15120. ie and 3 feet per of water feet, the gallons. to a city liameter, and how : gallons. ,titude of . the arm ent what cylinders pump is another >d to the will be locity of (round. he same lent. set wide 27 feet et, how ammer stroke are 3^ weight MISCELLANEOUS PROBLEMS. 153 7. How far may an empty vessel capable of sustaining a pres- sure of 1 59 lbs. to the square inch be sunk in water before breaking? 8. In a screw the pitch is -j^f of an inch, the power lever 9 feet 2 inches long and the weight is 44000 lbs., what is the power? 9. How many units of work are ' "tended in raising 70 cubic feet of water to the L' ' .h^ '3 feet? 10. The piston of a low pressure Sic^m engine has an area of 360 inches and makes 13 strokes of 7 feet each per minute, the pressure of the steam on the boiler being 40 lbs. to the square inch, required the horse powers of the engine. 11. Through how ^nany feet will a power of 7 lbs., moving through 120 feet, carry a weight of 29 lbs. ? 12. A locomotive weighing 75 tons is drawn along an inclined plane with a uniform velocity of 40 miles per hour, as- suming the inclination of the plane to be | in 100, and taking friction and atmospheric pressure as usual, what is the horse powers of the engine? 1st. If the train is ascending the plane ? ■ ^ 2nd. If the train is descending the plane ? 13. If a body weighing 7 lbs. at the surface of the earth be carried to a distance of 30000 miles from the earth, what will be its weight? 14. With what velocity per second will water flow from a small aperture in the side of a vessel, the fluid level being kept constantly 12 feet above the centre of the orifice? 15. In a Hydrostatic Bellows the tube has a sectional area of 1} inches, the area of the board is 37 inches, and the tube is filled with water to the height of 28 feet, what upward pressure is exerted against the board of the bellows ? 16. In a differential wheel and axle the radii of the axles are If and 2\ inches, the radius of the wheel is 40 inches, what power will sustain a weight of 8700 lbs. ? 17. A elock is observed to lose 17 minutes in 24 hours, how much must 'the pendulum be lengthened in order that it may keep correct time ? 18. At what height will the mercury stand in a barometer at an elevation of 30'5 miles ? 19. An upright flood gate of a canal is 17 feet wide and 13 feet deep, the water being on one side only and level with the top ; required the pressure. Ist. On the whole gate. 2nd. On the lowest three-fifths of the gate. 11 I i I I6i MISCELLANEOUS PBOBLEMS. 3rd. On the middle threenal area a of 2^ a lever a power '^er what m in the tions of common 1. To what depth may an empty glass vessel capable of sus- taining a pressure of 197 lbs. to the square inch be sunk in water before it breaks ? 8. In a system of pulleys consisting of eight movable pulleys worked by eight cords, the upper end of each fastened to the beam, the power is 7i lbs., what is the weight ? 9. How many gallons of water per hour will an engine of 7 horse powers pump from a mine 67 feet in depth, making the usual allowance for the modulus of the pump? 10. The piston of a low pressure engi-^e has an area of 400 inches and makes 20 strokes, each 8 feet in length, per minute, the boiler evaporates -731 of a cubic foot of water per minute, what are the useful horse powers of the engine ? VIII. 1. Explain the difference between the simple and compound pendulum — also what is meant by the " centre of oscil- lation" and by the "centre of percussion." 2. What velocity will a heavy body falling freely in the latitude of London acquire in one entire second, the London second's pendulum being 39*13 inches long? 3. In a Hydrostatic Bellows the tube is filled with water to the height of 13 J feet; what uptv^ard pressure is exerted against the board of the bellows if the area of the latter be 3-,\ feet ? 4. In a differential screw the exterior screw hab a pitch of iV of an inch, the interior screw a pitch of -5,^ of an inch, the power lever is 50 inches long; what pressure will be exerted by a power of 130 lbs. ? 5. A train weighing 100 tons moves up a gradient having an inclination of f in 100 with a uniform speed of 20 miles per hour ; taking friction and atmospheric resistance as usual, what are the horse powers of the locomotive ? 6. When a body has fallen through 2500 feet what velocity has it acquired ? 7. Explain what is meant by gaseout diffusion and show the important influence it has in maintaining the composition of atmospheric air constant at all places. 8. In a common wheel and axle the radius of the axle is 1 1 inches and the radius of the wheel 47 in. : what power will, with this machine, sustain a weight of 793 lbs. ? 168 EXAMINATION PAPERS. n 9. A flood gate is 22 feet wide and 20 feet deep, the water being on one side only and level with the top ', required the pressure— 1st. Against the whole gate. 2nd. Against the lowest three-sevenths. 3rd. Against the upper four-ninths. 4th. Against the middle three-elevenths. 6ih. Against the lowest three-tifths. 10. Give the different rules for finding the specific gravity of liquids. IX. 1. In a differential wheel and axle the radii of the axles are 2f and 2-,3f inches, the radius of the wheel is 90 inches ; what weight will be sustained by a power of 7 lbs. ? 2. The tube of a Hydrostatic Bellows is filled with water to the height of. 50 feet; if the board of the bellows has an area of 6} feet, what upward pressure is exerted against it ? 3. How many vibrations per minute will a pendulum 9 yards long make ? 4. Give the principal laws of the descent of bodies on inclined planes. 5. A body has fallen through 3600 feet when another body begins to fall at a point 4000 feet beneath it ; through what space will the latter body fall before the first over- takes it ? 6. The piston of a steam engine has an area of 440 inches and makes 11 strokes per minute, each 9i^f feet in length, the boiler evaporates '9 of a cubic foot of water per minute ; what is the volume of the steam produced per minute and what is the pressure under which it is gene- rated ? 1. Give the most important consequences that result from the fact that each atom of a liquid is separately drawn towards the centre of the earth by the force of gravity. 8. What gross load will a horse exerting a traction of 74 lbs. draw on a road whose coefficient of friction is it^f ? 9. What are the conditions of equilibrium between the power and weight in the inclined plane ? 10. Through how many feet must a body fall in order to acquire a velocity of 250 feet per second ? ANSWERS TO EXAMINATION PAPERS. 169 ANSWERS AND REFERENCES PAPERS. I. TO EXAMINATION 1. H. P. = 29008 or 43-68. 2. Art. 66. 3. -1619 lbs. 4. Arts. 227,229. 5. 5522-34375 lbs. 6. 96000 lbs., 82S00 lbs., and 137l4t lbs. 7. Art. 205. 8. 151-2 tons. 9. Arts. 9, 19, and 10. 10. 125 tons or 750 tons. II. 1. H.P = 161-28 or 38-08. 6. Art. 44. 2. Arts. 25, 27. 7. Art. 205. 3. 1-921. 8. 6205 lbs. ■ ; 4. 929430 lbs. 9. Arts. 9, 19, and 10. 5. 184} lbs. 10. 8025 lbs. per ton. III. 1. Power arm 13-^ times as 6. 28flbs. .: great as the arm of the 7. Arts. 57, 58. . weight. 8. 1400. 2. Arts. 25, 26, and 27. 9. 194 lbs. 3. 14918-4 lbs. 10. Velocity = 6-366 feet per 4. 688 feet. second. 5. Arts. 192, 195, and -57584. Quantity = 7962071 cubic - feet per hour. IV. ': 1. Art. 212. 6. 1600 feet. 2. 259796J lbs. 7. Art. 88. 3. 12^V 8. 730700144 lbs. 4. Art. 62. 9. 9721-2 feet. 5. H.P. = 176-26 or 69-6. 10. 33^ to the square inch. V. 1. 15312Jlb3.,7500lbs.,6562J lbs., and3213H|lbs. 2. 64^7f lbs. 3. Art. 219, density 4 times as great, volume 1 qt. and elasticity 60 lbs. to the sq. inch. 4. 16|§lb3. 6. 3340} cubic feet. e. 7. 8. 9. 257. 10. 12 900 feet. Arts. 255, 256, 2111. 27500 feet. At elevation of 17600 feet. 100 seconds. 1100 feet per second. At the end of the 33rd sec. 1020000 lbs. 170 ANSWERS TO EXAMINATION PAPERS. ri Jl 11 VI. 1. 19724 feet. 2. Art. 282. 3. 24330 feet. 4. 45-36 H. P. 6. ^YA ^^^^ ''^®^' 6. 7. 8. e. 10. Arts. 183 and 182, Note. •618. 3990 Iba. Arts. U33 and 234. 2163-4 feet. VII. 1 1040GJ lbs. e. Arts. 129, 126. 2. 1-303 incheg. 7. 454 feet. 8 Art. 235. 8. 1920 lbs. 4 594 lbs. 0. 13 791 'J gallons 6. 626933i lbs. 10. II. P. = 67-87. VIII. 1. Arts. 301, 302, and 308. 8. 1851? lbs. 2. 386-n inches. 0. 275000 lbs. 3. 3022-68672. 185204A lbs. 4. 14660K5-6 lbs. 64320gV lbs. 6. 133-t>62. 75000 lbs. 6. 400 feet per second. 231000 lbs. 7. Art. 206. 10. Art. 196. - IX 1. 8085 lbs. 2. 21479-04 lbs. 3. 20-8 4. Art. 270. 6. llllHe®*' e. Volume = 339 Pressure = 85 7. Art. 175. 8. 1776 lbs. 9. Art. IIG. 10. 976^6 feet. '5 cub. feet! lbs. the sq. [inch. i 'I EXAMINATION QUESTIONS. 171 rote. lib feet; the sq. [inch. QUESTIONS TO BE ANSWERED ORALLY BY THE PUPIL. Note.— TA« mtmh&rs following the questiong refvr to the articles in the work where the annwcra nitty be found, 1. What is Natural ScUsncc? (1) 2. Into what classcH an; all iintiirni objncts divided and how arc these distiiiKuiNhod from each other V (2) How areaiiiiiialMdisliitKuiiihod from vcgotablos? (3) . ^ What is ZooloK'i' ? (4) Wl)atis Hot»iiy¥ (4) ; Wljati8MineraloKy?(4) » " (4) 8. 4. 6. 6. 7. Wliat is AHtronomy ? 8. Wliati8Gcoloiry?(4) 0. What is ChcnniHt try MO What is tho object of Natural Philosophy? (4) What am tho Huhdivisious of Natural Philosophy ? (5) In what separate forms dooH matter exist V (0) Dollno what is moaiit by tho essential properties of matter. (9) Enumerate tho ossuiitial properties of matter. (1(») What is extension ? ( 1 1 ) 10. 11. 12. 1.}. 14. 18. 16. What is impenetrability P Give some illustrations. (12) 17. What is divisibility? (i;{) 18. Docs tho property of divisibility belong to masses or to particles of matter or to both ? (13) 19. Give some illustrationsoftheoxtrcmc divisibility of matter? (13,Note\ SO. What is Indesfcruotibility ? (14) 21. What is Poro.sity ? {U) 22. What is Compressibility? (10) 23. What is Inertia? (17) 24. If l)0(lies cannot bring themsclvfjs to a state of rest, how is it that all bodies movin;< upon the earth soou come to rest? (17, Note) 28. What is elasticity ? (18) 26. Name tho dilferent kinds of elasticity as applied to solida. (18, Note) 27. What aro the accessory properties of matter? (19) 28. Enumerate some of the most important of tho accessory properties of matter. (20) "" WhatismalleaV)ility ? Which aro thcmost malleable What is ductility P Name the most ductile metals. || SO. 31. 82. 33. 84. 85. 38 87. 38. 89. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 60. 61. 62. Whatistenacit;^'? (23) is attraction? (24) •of the metals? (21) (22) (25) (31) What Enumerate the dilVercnt kinds of attraction. What is the attraction of gravity ? (20) what is the law of variation in the intensity of gravity? (27) Explain what is meant by saying tho force of gravity varies in/erwly as the square of the distance. (28) What is the attraction of cohesion? (29) What is the attraction of adhesion? (30) What is capillary attraction ? Give some examples. What is ehictrical attraction ? (32) What is magnetic attraction P (33) What is chemical attraction ? (34) What is the derivation of the word Statics 7 (36) what is the object of tho science of Statics? (36) what is the derivation of the word llydroHatics? (.JO) What is tlie oVyect of the science of Hydrostatics P (:J«) What is tho derivation of the word Dynamics? (36) What is the object of the science of Dynamics? (36) Wh„t is the derivation of the word Hydrodynamics ? (36) what is the object of the science of Hydrodynamics? (30) what is the derivation of the word Pneumatics 1 (36) What is the object of the science of i'neumatics r (36) ti 172 EXAMINATION QUESTIONS. S3. M. 65. 50. 57. 68. SO. 60. fll. 62. 63. 64. 65. 60. «S. •49. 70. 71. n. 73. 74. 75. .76. 77. 78. 79. 80. 81. 82. «3. 84. 65. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. Whon is a b^y Mid to be in nquillbriura ? (37) WImt aro Htatical forcoH or pruMSurcii V (:i8) What aro tli« clMmeiits of a fonio V (39) Whnt aro tiio dilffri'iit inodcH of roproscntlriR a foro P (40) Whoii Hcveral forcoH act upon tlio samu iioint of a body, how manj niotionH oaii thoy givo it V (41 ) DiNtiiiKuiHh hotwo«'n component and remtltant forces. (42) If N(>v<>ral forcoH act unon a point in tlio humu HtraiKlit lino and in ths Haiiio diroction, to wliat \» tlivir rcHultant equal V (43) Wlion Novcral forcott act upon a point in tlio Hanio straight lino but in opixiHito diroctionH, to what itt their rcHultant c<{ual? (43) Enunciato the principle of tho paralleloKrani of forccH (4-1) Wlicn Huvoral forcett act on a point in any direction whatever, state how the reeiultant may bo found. (45) What ia tho distinction botwocn the parallelogram of forces and the parallelopipcd of forcea? (4«J) Wliat in tiio rosulti iltant of two parallel forces which act on difTcrent points of a body but in tho same direction? (47) what is tho resultant of two parallel forces which act on different t>ointa of a body and in difi'orent directions P (48) How do wo find the resultant of any number of parallel forces? (49) What is a couple I (50) DistiuKiiish between tho composition of forces and tho resolution of foror are curved or bent, how uiuht their effective lengths be determined? (70) 103. Wliat iH a compound lever or eompoHitlon of levcrsl' (80) 10k. Deduce rules for tiutliug the power or the weight in a compound lever. (Hi) 105. Describe the wheel and axle. (h2) lOrt. Why is thu wheel and axle sometimes eallod a pcrpefual leverP (84) 107. AVhat are the conditions of equilibrium in the wheel and nxb*? (NR) 108. Deduce a set of rules for finding the power, the weight, the radius of the axleor the ra7. When is a pulley said to be fixed ? (102) 128. What is a single movable pulley called ? (103) 120. What are Spanish Bartons? (10;J) 130. Explain the meaning of the worHs sheaf, block, and iachle? (104) 131. What is the only mechanical aflvantage derived from the use of a fixed pulley? (105) 132. In a system of pulleys worked by a single cord, what are the conditions of equilibrium ? (106) 133. Deduce a set of rules for a system of pulleys worked by a single cord ? (107) 134. What are tho conditions of equilibrium in a Spanish Barton when the separate cords are attached directly to the beam ? (108) 135. What are the conditions of equilibrium when tho separate cords are attached to the movable pulleys? (109) 136. Deduce in each of these last two cases a set of rules for finding the ratio between the power and the weight ? (llO and ill) B';i.f ■t.;'- " ^ imjy, i ^jp i ^f^i.4j i ^^ ^P^ 174 EXAMINATION QUESTIONS. 137. l.'W. 139. W). 141. 142. 143. 141. 145. 145. 147. 14S. 149. 150. 151. 152. 153. 134. JOJ. 150. 157. 15S. IS'J. lf)0. 161. Hi2. 103. 164. 165. 166. 1(57. 168. 160. 170. 171. 172. 173. 174. 175. 17.'). 177. 178. 17U. ISO. IS]. 182. 183. 184. 185. If tho lines of direction of the power and weight malce with one unotlicr an angle (jroater than 120"^, what is the relation betweon the power and the weight? (112) In th(!oretical mechanics now is the Inclined plane regarded ? (1T3) Wiiat are the modes of indicating the inclination of tlio pl.'uieP (il4) In the inclined plane how may the power be applied ? (115) What are the conditions of cqnilibrium in the uiclined plane? (116) Dcdnco a set of rnl»-s for the inclined plane ? (117) What is the wedge? (118) How is the w(Hlge worked P (110) Wiiat are ih-.' conditions of equilibrium in the wedge when it is worked l)y proasnre? (120) In what imi)ortant particular does the wedge differ from all the other mechanical powers? (120, Note 1) Give some examples of the application of the wedge to practical pur- poses? (120, Note 2) Deduce a set of rules for the wedge P (121) Describe the screw ? (122) How is the screw related to an ordinary inclined plane ? (122, Note.) What is the pitch of the screw ? (123) How is tlio screw commonly worked ? (124 and 125) What are the conditions of equilibrium in the screw? (126) How mav the clficiency of the screw as a mechanical power be increas- ed ? (127) Deduce a set of rules for the common screw ? (128) By whom was the differential screw invented P (129) Upon what principle does the differential screw act ? (129) To what is the differential screw, in effect, equivalent ? (129) Deduce a sot of rules for tlie differential screw P (130) Describe the endless screw? (131) What are the conditions of equilibrium in the endless screw P (132) Deduce a set of rules for the eniUess screw ? (139) How does friction affect the relation between the power and the weight in the incchanical elements? (135) What are the different kinds of friction ? (136) Wliat is meant by the coefficient of friction ? (137) What is the (iooilicient of sliding friction ? (138) AVhat is the coefficient of friction on railways ? (138) AVhat is the coefficient of friction on good macadamized roads ? (138) What is meant by the force of traction ? (138) Enumerate the different expedients in common use for diminishing friction? (139) Give Coulomb's conclusions as reirards sliding friction ? (139) Give Coulomb's ^lonclusions as regards rolling friction? (139) What is Iho unit of work 1 (140) How are tli(^ units of work expended in raising a body found ? (141) What are the most important sources of laboring forces? (112) How many units of work are there in one horse i)owcr ? (142) Wliat is m(!ant by the Table in Art. 142 ? What is tiio true work of tlie horse per minute? (1 12. Note) In moving a carriage along a horizontal plane, for what purpose is work expended ? In the case of railway trains what is the amount of friction ? (143) In the case of railway trains when does the velocity become uniform? Upon what does the traction or force with which an animal pulls de- pend? (Ifti) At what rate per hour must a horse travel to do most wck ? (146) Upon what docs the amount of atmospheric resistance ca^ crienced by a moving body depend P (147) Explain wliat is meant by tiiis P (147) ith one /wn the (113) P (iU) (116) I worked he other ical pur- , Note.) iiicreas- EXAMINATION QUESTIONS. 17$ ';* (132) md the (138) inishing (141) ; is work U:\) iiiform? nils de- 14(5) need by 186. 187. 188. 189. 190. 191. 192. 19S. 191. 195. 196. 197. 198. 199. 200. 201. 202. 203. 204. 205. 206. 207. 208. 209. 210. 211. 212. 213. 214. 215. 216. 217. 218. 219. 220. 221. 222. 223. 224. What is the amount of atmospheric resistance experienced by a train of medium length moving at the rate of 10 miles per hour P (148) If a body be moved along a surface without friction or atmospheric resistance, how may the units of work performed be found P (149) When a train is moved along an inclined plane, how is the work per- formed by the locomotive found P (150) Deduce a set of formulas for finding the horse power, weight, maximum speed, &c., of trains ? (151 ) What is meant by the modulus of a machine P (162) Of machines for raising water, which has the greatest modulus P (163) How is the work performed by water falling from a height found ? (154) How is steam converted Into a source of laboring force? (155) What are the two principle varieties of the steam engine P (156) What are the essential parts of the high pressure engine P (157) How does the low pressure differ from the high pressure engine f (168) What are the varieties of the low pressure engine P ^159) How do these dillcr from each other P (160, 161) In the high pressure engine, at what part of the stroke does atmos- pheric pressure act against the piston ? (162) Give the leading ideas that enter into the construction of the steam engine? (163) In wliat respects is the low pressure engine preferable to the non-con- densing engine ? (164) How are the units of work performed by an engine found ? (165) Knowing the pressure of the steam on the boiler, how do we obtain the useful pressure on the piston P ( 166) Give the rules for finding the H. P., &c., of engines ? (167) What is the real source of work in the steam engine ? (168) Why is it most advantageous to employ steam of high pressure P (168) Give formulas for iinding tlie area of the piston, length of stroke, pres- sure, effective evaporation, &c., in the steam engine P (169) Define what is meant by ajluicl ? (171) How is the term fluid commonly applied P (172) Into what classes are fluids divided ? Name the type of each. (173) To what extent is water compressible P Alcohol ? (173, Note) How do liquids chiefly difter from gases ? (174) ; In what respects do liquids chiefly differ from solids? (175) Give the most important consequences that flow from this fact? (176) How would you illustrate tlie upward and lateral pressure of liquids ? (175, Note) What relation exists between the respective heights of two liquids of different densities placed in an inverted syphon? (176) What is the amount of downward pressure exerted by a liquid confined in any vessel? (177) How would you illustrate this fact? (177, Note.) Show that weight and pressure are not to be confounded with one another? (177, Note 2.) What are tho weights respectively of a cubic inch, a cubic foot, and a gallon of water, at the teinperature of 60° Fahr. ? (178) To what is the pressure exerted by water on a vertical or inclined surface equal? (179) Give a rule for finding tho lateral pressure exerted by water? (179) How do you find the pressure exerted by water against a vertical or inclined surface at a ^iveu depth beneath the water ? (180) How do you find the pressure exerted against any fraction of a vorti- cal surface when the upper edge is level with tho surface of the water! a81) 225. Explain what is meant by transmission of pressure by liquids? (182) 226. Describe Braraah's Hydrostatic Press, and lUustratv. by a figure ? (183) 227. Explain the pri ciple upon which Bramah's Press acts F (182, Note) 176 EXAMINATION QUESTIONS. 228. For %hzt purposes is Bramah's Press used ? (184) S29. How do we And the relation between the power applied and the preS' sure obtained by Bramah's I'tckh? (185) 2S0. Describe what is meant bjr the Hydrostatic Paradox? (186) 231. Show that it is not in reality a i)ai-adoxf (186, Note) 232. Describe the Hydrostatic Bellows ? (187) 233. Give the rule for linding the upward pressure against the board of a Hydrostatic Bellows? (188) 234. "When will a body float, sink, or rest in equilibrium in a fluid ? (189) 235. What weight of liquid docs a floating body displace? (190) 236. What portion of its weight is lost by a body immersed in a liquid f (191) 237. What is the specific gravity of a body? (192) 238. What is the standard of comparison for solids and liquids ? (193) 239. What is the standard of comparison for all gases? (193) 240. How do we find the specific gravity of asolid heavier than water? (1P4) 241. How do we find the specific gravity of a solid not sufficiently heavy to sink in water? (195) 242. What is the first method of finding the specific gravity of a liquid P (19«) 2i^3. What is the second method of finding the specific gravity of a liquid ? (196) 244. How is the specific gravity of a liquid determined by means of the Hydrometer? (196) 245. Describe the Hydrometer ? (196) 246. What difference is there between liydrometers designed for deter- mining the specific gravity of liquids specifically lighter than water, and those for ascertaining the specific gravity of liquids specifically heavier than water ? (I9t}) 247. How is the specific gravity of gases found P (197) 248. How may the weight of a cubic foot of any substance be found when its specific gravity is known ? (199) 249. How may the solid contents of a body be found from its weight ? (200) 250. How may the weight of a body be found from its solid contents ? (201) 251. 252. 263. 254. 255. 256. 257. 258. 259. 260. 261. 262. 263. 264. 265. 266. 267. 868. 260. 270. 271. 272. What is Pneumatics P (202) What is the derivation of the word atmosphere P (203) What is the atmospliere ? (203) To what height does the atmosphere extend ? (204) Give the exact composition of atmospheric air ? (205) What purpose is served by the oxygen in the air P (205, Note) What purjiose is served by the nitroeon P (205, Note) Describe the principal properties of carbonic acid ? (205, Note) What are the chief sources of carbonic acid ? (205, Note) What is the maximum and what the minimum amount of carbonic acid in the air? (205, Note) Describe the mode by which the air is kept sufficiently pure to sustain animal life. (205, Note) Describe the property of gaseous diffusion. (206) Explain how the property of gaseous ditt'usion affects the composition or the atmosphere. (206, Note) Upon whatdoestheauiouutof aqueous vaporpresentin the atmosphere depend? (207) What is its maximum amount P What its minimum amount ? (207) To wliat is the blue color of the sky due ? To what the golden tints of sunset? (208) Which of tlie essential properties of matter belong to air P (209) How would you illustrate the impenetrability of air ? (209, Note) How would you illustrate the inertia of the air ? (209, Note 2) Why does air possess weight ? (210) What may be taken as the fundamental fact of pneumatics ? (210, Note) What is tne weight of 100 cubic inches of each of the following gases, viz., oxygen, hydrogen, nitrogen, atmospheric air, carbonic air ? EXAMINATION QUESTIONS. 177 epres- rd of a (189) liquid? 3) 'P (1P4) Davy to liquid? iquid P of the deter- water, iiically when ' (200) ISJOl) 273. Give some illustrations of the aggregate weight of the atmosphere? (210, Note 2) 274. How in it that the lower strata of air are denser than the upper P (211) 275. By what law does the density of the atmosphere decrease as we aa^ cendP (212) •bonic istafii sition phere [)7) nts of ) Jfote) cases. 276. 277. 278. 279. 280. m\. 282. 28;j. 284. 285. 286. 287. 288. 289. 290. 291. 292. 293. 294. 295. 296. 297. 298. 299. 300. 301. 802. 303. 304. SOS. .%6. 307. 808. 309. 310. 311. 312- 313. 314. 315. 316. 317. 318. 319. 320. From what does the pressure of the air result ? (213) What do we mean by sayinR the pressure of the air is equal to 16 lbs. to the square inch ? (2l3,Note) If the air were of the same density throughout to what height would it extend? (214) How is this known ? (214) How are perraaiientiy elastic gases chiefly distinguished from non- elastic gases V (216) What is meant by permanently elastic gases ? (216, Note) Illustrate what is meant by tiie elasticity of a gas. (217) To what is the elasticity of gases due P (217, Note) Eiuuiciate Mariotte's law ? (219) Illustrate it by a bent tube as in Art. 218. To what extent is Mariotte's law true P (219, Note) What is the air-pump P (220) By whom and when was it invented ? (221, Note) Describe the exhausting syringe. (222) Draw a sketch of the air-pump and describe its mode of action. (222) Upon what principle does the air-pump act ? (223) How perfect a vaoium can be secured by the air-pump P (223, Note) Describe the condensing syringe, (224) For what purpose is the. air-pump chiefly used? (223) Give some illustrations of the pressure of the air ¥ (225, Note) Give some illustrations of the elasticity of the air. (225, Note) What is the barometer P (226) By whom and when was it invented P (226, Note) What are the essential parts of a barometer? (227) What is meant by the Torricellian vacuum? (227, Note) How may the excellency of a baronicter be tested? (228) What is the cause of th» oscillations of the barometer y ?29) In what regions of the earth are the oscillations of the biu ometer most fitful and ectensive? (229 Note) To what regular oscillations is the barometer subject ? (£,;vO"; At what hours are the two maxima of pressure V (2;Wy) At what hours are the two minima of pressure V (230V In what region are the serai-diurnal oscillations grciitiibtP (230, Note) Give some idea of their extent in tropical countries aad cxplaii? why they arc not observed in our climate. (230, Note> How may the weather to be expected be foretold u.y ;he oscillatioiis m the height of the barometric column? (251) What does a fall in the barometer denote P (231, II.) What does a rise in the barometer indicate? (231, III.) What does a sudden change in the height of the mercury in the barometer denote? (231, IV.) What does a steady rise in the column denote? (231, V.) What does a steady fall in the column denote? (231, VI.) What does a fluctuating state in the height of the column of mercery denote? (231, VII.) Give Halley's rule for ascertaining the height of mountains, Ac, bv the barometer. (232) Give Halley's rule with correction for temperature. (232) Give Leslie's rule. (232) Describe the essential parts of a common pump and illustrate by i diagram. (233) Explain why the common pump is sometimoa called a lifting pump. (233, Note) 178 EXAMINATION QUESTIONS. 32 1 . Explain tho principle upon which the common pump acts. (2.'53, Note 2) o22. Explain why tlie lower valve must be within 32 feet of tho water in tha reservoir in order that the pump may act at all times. (234, Note 2) 321. Descrribo the forcing pump. (231') 32 1. Describe tlio essential purts of a Are engine. (234, Note) 325, Describe the syplion. (235) 32(). How is tho syphon set in operation? (235, Note 1) 327. Explain upon what principle the syphon acts. (235, Note 2) 328. 329. 330. 331. 332. S33. 334. 333, 336, 337. 338, 339. 340, 341. 342. 313, 344, 3i5. 340. 347, 348. 349. 350, 351, 352. 353, 354. 355. 35(5, 357, 358, 359, 3(50, 361. 302, 803, S64. 305. 366. Wlien does the consideration of forces come under tho science of statics V (230) What kind of forces are considered in dynamics? (236) Why is statics called a deductive science!' (237) Why is dynamics called an inductive or experimental science? (237) What may force be doQned to be? (238) For wliat purposes is force required? (238) What are the diflerent kinds of forces as regards duration? (239) What are the dill'ereut kinds of contirmed forces ? (239^ What may motion be defined to be? (2l0) What are the qualities of motion ? (241) What arc the dilforent kinds of motion ? (241) What kind of a motion is produced by an accelerating, constant, or retarding force? (2'<'') What is velocity? (243) Of how many kinds is velocity ? (243) Wh«!n is velocity said to be uniform ? (243) What is momentum or motal force? (245) To what are the momenta of bodies proportional ? (240) When tho velocities of two moving bodies are equal, to what are their momenta proportional ? (247) When the masses of two moving bodies are equal, to wliat are their momenta proportional ? (248) Wlien we speak of multii) lying a velocity by a weight, what do we mean ? (249, note) When force is communicated by impact to a body at rest, how long will the body remain at rest ? (254) Give the first general law of motion. (255) Whose law is this? (257, Note) Give the second law of motion. (256) Whose law is this? (257, Note) Give the third law of motion. (257) Whoso law is this ? (257, Note) What is reflected motion? (258) What is the angle of incidence ? (258) Wliat is the angle of reflection ? (258) What proportion exists between the angle of incidence and the angle of reflection? (258) How would all bodies fall in a vacuum? (259) Upon what does the resistance encountered by a body moving through tho atmosphere depend? (20(>) What is the nature of the motion of a heavy body falling from a height? (2«n) WMiat velocity is acquired by a heavy body in falling thrcngh one second? (2154) Through how many feet docs a body fall during the first second of its descent? (2(;5j Deduce^ a set of formulas for tho descent of bodies freely through space. (206) When a body is projected upwards what is tho nature of its motion? (207) Give the formulas for the motion of a body projected upwards or downwards ? (208) I '^ EXAMINATION QUESTIONS. 179 Note 2) r in the rote 2) once of ' (237) 9) ant, or 3 their ! their do we V long angle ough loin a one if its Jugh ' its s or i 367. When ahody is descending an incline how is the gravity expended? (289) 808. What are the laws of descent on inc^litied planes? (270) 369. Upon what is the final velocity of a body falling down an iucline dependent? (271) 370. What are the laws of descent in curves? (273) 371. What is the braeliystoehrone? 372. What is a cycloid? (274) 373. Deduce a set of formulas for descent on inclines, (27r), 276) 374i. 375. 376. 877. 378. 379. 3S0. 381. 3S2. 383. 384. 385. 386. 387. 388. 389. 390. 391. 392. 393. 894. 395. 396. Whatisaproji'ctile? (277) What forces iiillucnco projectiles? (278) What is the theoretical path of a projectile? (278) What is a parabola? (278, Note 1) Upon what erroneous suppositions is the parabolic theory based? (278, Note 2) Show that when ahody is projected horizontally forward the horizontal motion does not interfere with thi; action of gravity. (279. Note) What are th(! thret; conclusions of the parabolic theory V (2S0) What is the greatest horizontal range of a projectile? (2S0, Note) To wliat is tlie velocity of projection speedily reduced, no matter what it may liave been originally ? (2si) , How do you explain tliis? (281, Note 1) What is th.' atmospheric resistance encountered by a ball or other projectile having a velocity of 2000 feet per second? (281, Note 2) When a ball has considerable windage, what is the amount of deflection in its course? (281, Note 3.) What are the most important laws regarding the motion of projoctilcB thrown vertically into the air? (282) What are the most important laws regarding the motion of projectiles thrown at an angle of elevatiim? (282) To what is the explosive force of gunpowder exploded in a cannon equal ? (283) With what velocity does exploded gunpowd'ir tend to expand ? (283) What is the c mposition of gunpowder ? (283, Note) What is the geatest initial velocity that can be given to a cannon ball ? (284) To what is the velocity of a ball of given weight fired with a given charge of powder projiori ional ? (284, Note) To what are the velocities of balls of equal weight fired by the same charge of i»owder proportional ? (2S5) To what a '. f he velocities of balls of different weights but of the same dimensiui.s tired by ecjiial quantities of powder proportional ? (28(5) To what is the depth which a ball penetrates into an obstacle propor- tional ? (287) Give the rule for finding the velocity of any shot or shell when its weight and also that of the charge of powder are known ? (288) 307. 398. 399. 400. 401. 402. 403. 491. 405. 406. 407. 408. 409. Wliat is centrifugal force ? (2S9) Wiiyt is sometimes called tang(;ntial force ? (289, Note) What isoeutiipetal force? (29i)) When does a body move in a circle? (291 ) When does a body move in an ellipse? (2'.)1) How long can a rotating mass jjreserve itself? (292, Note 1) Giv(! some examples of the elllv'ts of centrifugal force. (292, Note 2) If the velocity ami radius are constant, to what is the centrifugal force proportional? (-I'M) When the radius is constant how doe»^ the centrifngal force vary? (294) What is the amount of centrifugal force at the equator? (294, Note) How rapidly must the earth revolve in order that the centrifugal force at the equator may equal gravity? (29t, Note) ■W hen the velocity is constant how does the centrifugal force vary? (295) AVhon the number of revolutions is constant to what is the centrifugal force proportional? (296) 180 EXAMINATION QUESTIONS. 410. Give a set of formulas for calculating centrifugal force. (297) . 411. Give a rule for liudinK the work accumulated in a moving body. (200) 412. "Wh.it is a pendulum P (300) 413. "What is a simple pendulum ? (301) 414. What is a compound or material pendulum? (302) 415. "What is an oscillation or vibration P (303) 416. "What is the amplitude of the arc of vibration ? (304) 417. "What if t^^ duration of a vibration? (305) 418. "What l:i tht length of a pendulum? (306) 419. "What is the centre of suspension? (307) r 45SO. "What is the centre of oscillation ? (3(i8) 421. What is the centre of percvission ? (308, Note) 422. What is meant by saying the centres of oscillation and suspension are interchanjjeableP (369) 423. How is the duration of a vibration afTected by its amplitude P (310) 424. What is meant by saying the vibration of the pendulum is isochronous P (31 i>, Note) 425. vVhal relation exists between the lengths and times of vibrations of pendulums? (314) 4J16. n.ve the chief laws of the oscillations of the pendulum. (311-316) 427. ^» iiy docs the seconds pendulum vary in length in different latitudes ? •;n6, Note) 4i>S. "Wiiat is the length of a seconds pendulum in Canada? (316, Note a) 4ii9. '^<.- what purposes is tie pendulum applied? (317) 430, ^ is the pendUlum used as a measure of time? (317, Note) 481. ; ■■■■> V IS the pendulum used as a staudard of measure? (317, Note) 432. il.'v do we find the length of a pendulum to vibrate in a given time (319) 433. How do we find the number of vibrations lost by a pendulum of given length when the force of gravity is decreased ? (320) 434. How do we find the number of vibrations gained by a pendulum of given length when it is shortened? (321) 435. "What isthe science of Hydrodynamics? (322) 436. Enunciate Torricelli's theorem. (323) 437. In what time does a full vessel empty itself through an orifice in the bottom? (324) 438. How is the quantity of fluid discharged through an orifice of given size found? (325) 489. What is the vena contracta^ (325, Note) 440. What relation exists between the theoretical discharge and the actual discharge ? (325, Note) 441. Give the rule for finding the velocity and quantity of fluid discharged through an aperture of given size. (326) 442. When water spouts from an aperture in the side of a vessel how is the horizontal distance to which it is thro.vn found? (327) S43. When a liquid flows through a pipe or channel, which part has the greatest velocity ? (329) 441. How is the velocity of a strear. ;• iterii. ned ? (329, Note 2) 445. What are the principal varietie^^ of water wheels? ^",32) 446. In water whoels, wlieu is the grc i; *pst mechanical «;i' x ^t produced P (333) 447. Give the rule for finding the.ho:>.>! powers of upright water wheels. (334) \ TtiE END. LOVELL'8 SEBIE8 OF SCHOOL BOOKS. IN THE PRESS: To be published early in the Spring of 1860 : BT J. GEORGE HODGINS, M. A. BmbelUshed with about 40 superior Maps and 100 beautiful Engravings. mmS GEOGRAPHY, while it is designed to furnish a satis- factory resumSe of geographical knowledge of all parts of the world, will give equal prominence to the BRITISH COLO- NIES, concerning which such meagre information is generally found in works of this kind. It is designed and will be a suitable Text Book for children in CANADA, NOVA SCOTIA, NEW BRUNSWICK, NEWFOUNDLAND, PRINCE EDWARD'S ISLAND, the EAST AND WEST INDIES, AUSTRALIA, &c. The Maps and Wood-Cuts are now in course of preparation and every effort will be made to expedite publication. The work will be in the Bookstores in the Spring of 1860. PRICE $1. JOHN LOVELL, Publisher. Caitada Dikectort Office, Montreal, January, 1860. LOVELL'S 8EBIES 07 BOHOOL BOOKB. IN PEEPARATION: EA.8Y LESSOISrS \ IK GENERAL GEOGRAPHY, POB THK USE OP YOUNGER PUPILS; BEINQ INTRODUCTORY TO LOVELL'S GENERAL GEOGRAIPHY. To be Embellished with Maps and Illustrations, BY J. GEORGE HODGINS, M.A. Canada Directobt Office, Montreal, January', 1860 JOHN LOVELL, Publisher. IN PREPARATION: A SCH OOL HISTORY OF CANADA, AND OF THE OTIIEE BRITISH PROVINCES. With Illustrations. BY J. GEORGE HODGINS, M.A. JOHN LOVELL, Publisher, Canada "Directotit Opftcb, Montreal, Jauuax:y, 1860. LOVELL'8 SEBIES OF SCHOOL BOOKS. ARITHMETIC, IN DESIGNED FOR THE USE OF CANADIAN SCHOOLS. BY 3. TL. SANGSTER. Cawada Dieectort Officb, Montreal, January, 18G0. JOHN LOVELL, Publisher. IN PREPARATION : FIRST BOOK Of ARITHMETIC IN BECIMAIi CURRENCY, &c., &c. BY J. H. SANGSTER. JOHN LOVELL, Publisher Canada Dieectoet Oppicb, Montreal, January, I860. lOYELL'S SERIES OF SCHOOL BODXS. A NEW SPELLING BOOK V THE CLASSICAL ENGLISH SPELLING BOOK; IN WHICH THB HITHBKTO Dl PIOULT ART OF ORTHOaRAPHT 18 RENDERED BABY AND i LEASANT, AND SFSEOILY AOQUIUBD. COMPRi NO ALL THB IMPORTANT ROOT-WORDS PROM THE ANGLO- SAXON, TUB LATIN, AND THE GREEK ; AND Several hundred Exercises in Derivations and in Verbal Distinctions. BY <■. G. VASEY. W'- Canada Liebctoht Office, Moutreal, January, 1800. JOHN LOVELL, Publisher, IN THE PRESS, Jtnd shortly will be published : GRAMMAR MxVDE EASY ^ TO THB CAPACITY OF CHILDREN: IK WHICH, Br PLEASANT EXERCISES, A KNOWLEDGE OF THE PARTS OF SPEECH IS EASILY ACQUIRED. BY G. G. VASEY. Canada Directory Office, Montreal, January, 1860. JOHN LOVELL, Publisher. V. . )K. OK; NDBBBD EASY THE ANGLO- d in , Publisher, ASY THE FASTS ublisher.