ï~~a r;. z k x a i d x rt ^ r *Zr, a y;, s S. r ' o.,.rci r, Y p; i1+ ' t a a}.,Y, "r1 x _ - M k. a ' r} p " a + T-T i_<... 77,,,.c ï~~.a' _,. ï~~<'e VWBOK ROeMi ï~~ ï~~P It INC IP LE S TN N\ rTU RA L PHILOSOPHY: MAT11IMTICL Y LLIs~f TST 1 RTD, A N 1) PRACTICALLY A t)PPJtE.1D ff o tle 115E of teroo C Is (11b 2Tcabcell f.. BY ALEXANDER C. ITUESTIS, A. M., PROIFSSOR OF '1A Fill;NMATICS AND N ATURAL PHIILOSOPHY IN FORT WAYNiE FEMALE COLLEGE CINCINN.ATI: WILLIAM H. MOORE AND CO., 110 MAIN ST.. BETFWEEN TIRD AND FOURTIT. MARK H. NEWMAN & CO., NEW YORK. 18S49. ï~~Entered, According to Act of Conigress, in the year 1848, BY ALEXANDElR C. IIUESTIS, Ini the Clerk's Office of the Di. trict Court of the United States for the District of Indiana. JAMES & CO., s TER.OTYPE.RS, Walnut street, between Fourth and Fifth. E. SHEPARD, Pa. ï~~PREFACE. IT is a fact pretty generally admitted, that, although we have many very excellent textbooks, on the subject of NATURAL PILosorPHY, yet, most of them are wanting in a sufficient number and variety of practical Examples. To be thorough, one must be practical. To memorize and to repeat abstract rules and principles, is one thing-to understand and reduce them to practice- is quite a different thing. A student may, with but little, or no difficulty, commit to memory the principles - that, bodies attract each other inversely as the squares of the distances; that the spaces passed over by falling bodies vary as the squares of the times; that the lengths of different pendulums vary as the squares of the times of vibration, 4c., 4-c.- but, unless he can apply these principles, and reduce them to practice, can he be said to understand them? The following pages have been written, not with the design to supersede any of the Philosophies now in use; but to supply their defects -to make the study of Natural Philosophy more practical, and, consequently, more useful ( iii ) ï~~iv PREFACE. than it has heretofore been. Those subjects only, which are susceptible of practical application or mathematical illustration, have been treated of- but so treated and arranged, that the book may be used either with, or without an accompanying work. Commencing with Gravity, each subject has been taken up in its proper order, Formulas and Rules- deduced, and practical Examples added; sufficient, in variety and number it is believed, to impress the various principles upon the mind of the Student, and enable him to reduce them to practice. THE AUTHOR. Fort Wayne Female College, 1848. ï~~CONTENTS. CHAPTER I. PAGE. Attraction - - - - - - 7 CHAPTER II. Intensity of Light - -.- -. - -10 CHAPTER III. Gravity below the Earth's Surface - - - - 11 CHAPTER IV. Mutual Attraction -.- - - - - 18 CHAPTER V. Falling Bodies - - - - - -..14 CHAPTER VI. Uniform Motion- - -. --- - 86 CHAPTER VII. Momentum - - - - - - - 38 CHAPTER VIII. The Pendulum- - - - - - 42 CHAPTER IX. The Lever -.. -. 53 CHAPTER X. False Balance - - - 0 (x) ï~~vi CONTENTS. CHAPTER XI. rAGE. The Wheel and Axle - - - - - - 72 CHAPTER XII. Compound Wheel and Axle - - - - - 81 CHAPTER XIII. The Pulley --- -- --- -87 CHAPTER XIV. Composition of Forces - - - - - - 103 CHAPTER XV. Inclined Plane - - - - - - - 106 CHAPTER XVI. The Screw - - - - - - - - 127 CHAPTER XVII. Hydrostatic Press - ----- - 142 CHAPTER XVIII. Specific Gravity - - - - - - - 149 CHAPTER XIX. Hydrostatics... 158 CHAPTER XX. Sound -- - - - --------- 163 CHAPTER XXI. Table of Formulas- - - -.-- - 164 CHAPTER XXII. Promiscuous Examples ------ 169 APPENDIX -- - - - - - - 181 ï~~PRACTICAL EXERCISES IN NATURAL PHILOSOPHY. CHAPTER I. ATTR ACTION. SARTICLE 1. Attraction is that force or property of matter which causes bodies, or their particles, to approach, or tend to approach each other. The attraction of gravity may be defined, That property of matter which causes bodies in masses to approach, or tend to approach each other. ARr. 2. The attraction of gravity varies directly as the quantity of matter, and inversely as the square of the distance. The earth being a sphere, the eflfective force of its attraction is the same as if it wer'e concentrated at the center. Hence the following Rc Ln.-Gravity at dferent distances from the earth,aries inversely, as the square of the distance from the center; the r ulius of the earth considered as the unit. 1,USTR.ATIQ.-At the distance of three times that of the surface of the earth from the center, a body would weigh not one third as much, but (the square Q UESTIONS.-1. Define attraction. What is meant by the attraction of gravity '! 2. How does the attraction of gravity vary? Repeat the rule for the attraction of gravity at different distances from the earth. Give the illustration. (7) ï~~8 EXERCISES IN NATURAL PHILOSOPHY. of 3 being 9) one ninth as much as at the surface; at four times the distance of the surface of the earth from the center (the square of 4 being 16), one sixteenth as much. So that, at the distance of the moon (240,000 miles), sixty times the distance of the surface of the earth from the center, a body would weigh (the square of 60 being 3,600) o part as much as at the surface of the earth. Ex. 1. If a body at the distance of 4 feet, be attracted with a force equal to 16 pounds, what will be its attraction at the distance of 16 feet? Ans. 1 pound. SoLTrO.-By squaring the distances and inverting, we have 256 ft.: 16 ft.:: 16 lbs.: 1 lb. That is, the square of the greater distance is to the square of the less, as the attraction at the less, is to the attraction at the greater distance. Or, since the ratio of the distances 4: 16= 1: 4, and the square of the ratio = 1: 16, the square of the greater distance is sixteen times that of the less; hence, the attractive force will be one sixteenth as much, and we shall have 16 lbs. X '= 1 lb. Or, squaring the ratio of the distances, and stating by proportion, we have 16 ft.: 1 ft.::16 lbs.: 1 lb. 2. If a body at the surface of the earth weigh 600 pounds, what will be its weight 4,000 miles above the earth? Ans. 150 lbs. NOTE---It should be remembered that the distances are to be reckoned from the center instead of from the surface of the earth. When the diameter of the earth is not given, it will be understood to be 8,000 miles, and consequently, the distance to the surface from the center, 4,000 miles. SOLUTIO or EXAMPLE 2.-The respective distances of the body from the center, when at the surface of the earth and 4,000 miles above it, are 4,000 and ï~~EXAMPLES IN ATTRACTION. 9 8,000 miles, which distances are as 1: 2, and their squares as 1: 4. Since the square of the greater distance is four times that of the less, the attraction 4,000 miles above the surface, will be one fourth as much as at the surface of the earth. Hence, 600 lbs. Xi= -150 lbs. 3. If a body at the surface of the earth weigh 14,400 tons, what will be its weight at the distance of the moon? Ans. 4 tons. 4. If a body at the level of the sea weigh 4,000 pounds, what will be its weight on the top of a mountain 5 miles high? Ans. 3,990.018--+lbs. 5. What part of its weight would a body lose at the distance of 1,000 miles above the earth? At the distance of 2,000 miles? 3,000 miles? 4,000 miles? 8,000 miles? Ans. At the distance of 1,000 miles,, of its weight. 2,000 miles, of its weight. 3,000 miles, 3 of its weight. 4,000 miles, & of its weight. 8,000 miles, 1, of its weight. 6. Suppose a body at the surface of the earth to weigh 2,500 pounds, how much would it lose when carried to a distance of 1,000 miles above the earth? Ans. 900 lbs. 7. Suppose a body at the surface of the earth to weigh 36 pounds; required its weight 8,000 miles above the earth. Ans. 4 lbs. 8. Required the comparative weights of a body at the distance of 20,000 and 40,000 miles above the earth. Ans. 121: 36. 9. An aeironaut finds his balloon and contents to weigh, on the earth, 1 ton, 3 quarters; required their weight at the hight of 6 miles, reckoning 25 pounds to the quarter. ï~~10 EXERCISES IN NATURAL PHILOSOPHY. CHAPTER II. INTENSITY OF LIGHT. ARTICLE 3. The intensity of light varies inversely, as the square of the distance from the radiant. ILLUSTRATION.-If, at the distance of 2 feet from a radiant, the intensity of light be represented by 4, then at the distance of 4 feet (since the distances 2 and 4 are as 1: 2, and their squares as 1: 4, the square of the greater distance being four times that of the less) the intensity will be one fourth as great as at the distance of 2 feet. But the intensity of light at the distance of 2 feet is represented by 4; hence, at the distance of 4 feet, it will be represented by 1. Ex. 1. Suppose the intensity of a radiant at the distance of 40 feet be represented by 36, what will be its intensity at the distance of 240 feet. Ans. 1. 2. Two ships are at sea, one at the distance of 25, and the other 45 miles from a lighthouse; required the comparative intensity of light at each ship. Ans. 81: 25, or 3: 1. SOLUTION.-The distances 25 and 45 are as 5: 9; by squaring and inverting, we have 81: 25. 3. Required the comparative intensity of the sun's light, at the planets Mercury and Venus; the former being 37,000,000 miles, and the latter 68,000,000 miles from the sun. Ans. 4,624: 1,369, or 3 1': 1 - 4. Required the comparative intensity of the sun's light at the earth and the planet Saturn; the QusTiroxs.-3. How does the intensity of light vary? Give the illustration. ï~~INTENSITY OF LIGHT. 11 former being 95,000,000 miles, and the latter 900,000,000 miles from the sun. Ans. 32,400: 361, or 89.7-+-: 1. 5. What will be the intensity of light at the distance of 2,000 miles and 4,000 miles from the sun? Ans. 4: 1. 6. Suppose a globe lamp to be placed in the center of a room 40 feet long and 30 feet wide; required the comparative intensity of light at the corners of the room, and at the middle of each of the longer sides. Ans. 9: 25, or 1: 21. That is, the intensity of light at the middle of each of the longer sides, will be 27 greater than at the corners of the room. CHAPTER III. GRAVITY BELOWV TIIE SURFACE OF THE EARTH. ARTICLE 4. The force of gravity below the suface of the earth, varies DIRECTLY as the distance from the center. ILLUSTRATION.-The attractive force 2,000 miles from the center is twice as much as at 1,000 miles fr'om the center; 3,000 miles three times as much; 4,000 miles, or at the surface, four times as much. Again; if a body at the surface of the earth weigh 8,000 pounds, 2,000 miles below the surface, it will weigh 4,000 pounds; 3,000 miles below the surface, or 1,000 miles from the center, 2,000 pounds; 500 miles from the center, 1,000 pounds. QuYSTIno.-4. What law governs the attraction of bodies below the surface? Give the illustrations. ï~~12 EXERCISES IN NATURAL PHILOSOPHY. Ex. 1. If a body, at the distance of 600 miles from the center of the earth, weigh 2,400 pounds, what will it weigh at the distance of 300 miles from the center? Ans. 1,200 lbs. SOLUTION.-If a body, at the distance of 600 miles from the center of the earth, weigh 2,400 pounds, then, since 300 is half of 600, at the distance of 300 miles, it will weigh half of 2,400 pounds, or 1,200 pounds. Or, if at the distance of 600 miles from the center of the earth, a body weigh 2,400 pounds, at the distance of one mile it will weigh 2,400 pounds X j; and, at the distance of 300 miles, 300 times as much as at one mile. Or 2 400 pounds X4, X -Â~=1,200 pounds. 2. If a body, 3,000 miles from the center of the earth, weigh 6,000 tons; required its weight 200 miles from the center. Ans. 400 tons. 3. If, at the distance of 800 miles from the center of the earth a body weigh 480 pounds, required its weight 2,000 miles from the center. Ans. 1,200 lbs. 4. A body, at the surface of the earth, weighs 6,740 tons; required its weight at the distance of half a mile from the center of the earth? Ans. 1 tons. 5. Required the comparative weights of a body, 2,600 and 3,400 miles from the center of the earth. Ans. 13: 17= 1: 1 4. 6. A body, situated 2,720 miles from the center of the earth, weighs 24,000 tons; required its weight at the center of the earth. 7. A body, at the surface of the earth, weighs 4,000 pounds; required the difference between its weight 600 miles and 1,000 miles from the center. Ans. 400 lbs. ï~~MUTUAL ATTRACTION. 13 CHAPTER IV. MUTUAL ATTRACTION. ARTICLE 5. Bodies when left to their own mutual attraction, in approaching each other, pass over spaces INVERSELY AS THEIR QUANTITIES OF MATTER. ILLUSTRATION.-If two bodies, one containing twice as much matter as the other, be left to their own mutual attraction, then the distance passed through by the larger body, will be one half that passed through by the smaller; if the larger contain three times the quantity of matter, it will pass through one third the distance. Ex. 1. Two bodies, one weighing 20, and the other 60 pounds, and situated 800 feet apart, are left to their own mutual attraction; required the distance passed through by each. Ans. Larger body 200 ft., smaller, 600 ft. 2. Required the space passed through by two bodies, which are left to their own mutual attraction, one weighing 4,000 tons, and the other 12,000 tons-the two bodies being situated 32,000 feet from each other. Ans. Larger body 8,000 ft., smaller, 24,000 ft. 3. Suppose two bodies, 40,000 miles apart, and whose weights are to each other as 3: 7, to be left to their own mutual attraction; through what space will each body pass? Ans. Larger 12,000 miles, smaller 28,000 miles. 4. Suppose two planets, whose weights are to each other as 3: 5, and situated from each other at a distance of -8,000,000 miles, to be left to their QUESTIo~s.-5. Through what spaces do bodies, left to their own mutual attraction, pass? Give the illustration. ï~~14 EXERCISES IN NATURAL PHILOSOPIHY. own mutual attraction; over what space will each planet move? Ans. Larger 3,000,000, smaller 5,000,000 miles. 5. Suppose two bodies, whose weights are to each other as 5: 8, and situated at the distance of 59,000 miles from each other, be left to their own mutual attraction; required the distance passed over by each body. Ans. Larger 24,000, smaller 35,000 miles. 6. Required the distance passed over by two planets, left to their own mutual attraction, whose weights are to each other as 3: 5; the two planets being situated at the distance of 248,000,000 miles from each other. Ans. Larger 108,000,000, smaller 140,000,000 miles. CHAPTER V. VELOCITY, TIME, AND SPACE OF FALLING BODIES. ARTICLE 6. In order to solve questions under this head, it is necessary to know the relation existing between the time, velocity acquired, and space of falling bodies. Now, by geometrical demonstration, it is easily proved that the time, velocity acquired, and space of a falling body, acted upon by gravity, or any other uniformly accelerating force, sustain the same relation to each other as the perpendicular, base, and area of similar right-angled triangles-the area answering to the space described by the falling body, the perpendicular to the time of falling, and the base, to the acquired velocity, but, The areas of similar right-angled triangles are to each other as the squares of the perpendiculars or bases, and the perpendiculars are as the bases. ï~~FALLING BODIES. 15 Hence, it follows, That the spaces described by bodies falling from a state of rest, acted upon by the force of gravity, are as the squares of the times, or velocities, and the times are as the velocities. ART. 7. DEMONSTRATION. - In uniform velocity, where bodies pass over equal spaces in equal times, the whole space described in any given time evidently equals the time multiplied by the velocity.. Thus a body moving at the rate of 20 feet per second, would, in 10 seconds, move over a space of 10 times 20 feet, or 200 feet. In a given right- Fig. 1. angled parallelogram, as A B A B D C, the area equals the base C D multiplied by the hight A C. If, therefore, the time of a body moving with uniform velocity, be represented by C D A C, and the velocity by C D, the parallelogram A B D C will represent the space passed through by the body. For the area has the same relation to the base and hight, of which it is the product, as the space has to the time and velocity, of which it is the product. Suppose a body in equal successive por- Fig. 2. tions of time (Fig. 2), B A represented by A C, E o C H, H K, to move C with the uniform velocities A B, C E, and C F, H G, then, from what has been shown L K above, the several parallelograms, A BD C, CE F H, and HGL K ï~~16 EXERCISES IN NATURAL PHILOSOPHY. will represent the space described in the time, AK. Now, suppose the body to receive an equal increase of velocity, at the end of each equal successive portion of time, so that, during the second interval, C E (Fig. 3), it will move with twice, and during the third interval, Fig. 3. E G, with three times the H" A velocity as in the first, A C. Then the entire disF c tance passed over in the time, A G, will be repreL N E sented by the right-angled triangle, A G, plus the MG three equal exterior triangles, AH F, FIN, and NLAL If the intervals of time and the corresponding increase of velocity be diminished one half, as seen in Fig. 4, then the whole space described in the time, A G, will be Fig. 4. represented by the 1 - A right-angled trianS gle, A G, plus Ic the sum of the exe ] terior triangles, A S k E Ha, a b c, c Ik, k e f, __ __ _ _ ffLg, and g h M, M_ _ _ G which is half the sum of the exterior triangles in Fig. 3. By continuing to diminish by one half, the intervals of time, a i, i C, &c., and the corresponding increments of velocity, A H, a b, &c., the exterior triangles will diminish, and the figure representing the space will approach, at each successive division, by one half the sum of the exterior triangles, nearer to the triangle A G M Continuing the division ad infi itum, A i (which ï~~FALLING BODIES. 17 represents the equal intervals of time, at the end of which the body receives equal increments of velocity), and A H (which represents the velocity with which the body begins to move), will become 0; the exterior triangles, also, will be reduced to 0, and the whole space actually described will be accurately represented by the right-angled triangle A G M. That is; a body falling from a state of rest, and constantly receiving equal increments of velocity, will pass through a space which may be represented by the surface of any right-angled triangle, whose perpenidicular represents the time of falling, and' whose base, the acquired velocity. Now the force of gravity is just such a force as above described; and hence, the space, time, and velocity of a body falling from a state of rest, and acted upon by gravity, will sustain the same relation to each other as do the area, perpendicular, and base of a rightangled triangle. But, we have already stated, that, the areas of similar right-angled triangles are to each other as the SQUARES Of the perpendicular and base, and the perpendiculars are as the bases. Consequently, The spaces described by bodies falling by the force of gravity, or any other uniformly accelerating force, are as the SQUARES of the times and velocities; and the times, AS the velocities. ART. 8. The space described by a falling body, moving uniformly with the last velocity, acquired by falling from a state of rest, is twice that described in the same time, by gravity. DEMONSTRATION.-Let A C represent the time, and C B the acquired velocity of a falling body; then (Art. 7) A C B will represent the space described in the time A C. QUESTIONS.-7. What relation do the velocity, time, and space of falling bodies sustain to each other? 2 ï~~18 EXERCISES IN NATURAL PHILOSOPHY. Make C D=A C, and complete the rectangle CB ED,which will reFig. A. present the space de-.scribed in the time A C sor C D, by a body moving uniformly with the Sacquired velocity C B. __ __ _ GBut the rectangle CB ED is twice the triangle A CB; Hence, The space deSscribed by a body moving uniformly wzth the last velocity, acquired by falling from a state of rest, is twice that described in the same time by gravity. It has been ascertained by actual experiment, that a body in a vacuum, and near the earth, falling from a state of rest, falls, the first second of time, 16Y feet; hence, if it should continue to move uniformly, for another second, with the velocity acquired at the end of the first, it would move over a space of 16., ft. X 2= 321 ft. The velocity, therefore, acquired by a body falling one second, from a state of rest, and near the earth, is 32! feet. And, generally, If a body, at any distance from the earth, fall from a state of rest, it will, during the first equal portion of time, acquire a velocity equal to twice the space fallen during that time. ART. 9(a). To find the distance fallen the first equal portion of time, at any given distance above the earth. It is only near the earth that a body, meeting with no resistance, falls, the first second, 16, feet, or 193 inches-as it is removed from the earth the attraction of gravity continually diminishes; and consequently, the space described, the first second ï~~FALLING BODIES. 19 or period of its fall, becomes less and less as the distance from the earth increases. But the attraction of gravity varies inversely, as the squares of the distances: hence, The spaces described, during the first period of time, at dafrrent distances above the earth, will vary inversely as the squares of those distances from the earth's center. ILLTsrATION.- At the distance of 4,000 miles above the earth, a distance twice as far from the center as the surface, a body will fall (the square of 2 being four), the first period of time, one fiurth as far as at the surface; at the distance of 8,000 miles, one ninth as far; at the distance of 12,000 miles, one sixteenth as far, &c., &c. Let s' represent the space fallen, the first equal portion of time, near the earth, and s, the distance fallen the first and like equal portion of time at the distance D, above the earth. D+4,000 miles, will represent the distance D from the center of the earth. We shall then have the following proportions: s: s':: 4000: (D-f-4000)1; Multiplying extremes and means, sX(D+4 000)2 = s'X4 0002; 4 000' Dividing by (D+4 000)2, S=(D 4000)2Xs'. (D' 4 000)2 Hence, To fjnd the space fallen, the first equal portion of time, at any given distance ABOVE the earth, we have the following formula and rule: 4 000 FORMULA,.. D+40002Xs'. (D+4 000) 's RuLE.-Divide the square of the earth's semidiameter by the square of the given distance from the center of the earth, and multiply the quotient by the distance fallen the first equal portion of time near the earth. ï~~20 EXERCISES IN NATURAL PHILOSOPHY. Ex. 1. How far will a body fall, the first second of time, 4,000 miles above the earth? 4 0002 X,=4 000X4000 SotUT.--For. s= -X s' - X OLUT. * (D+4 000)' 8 000 X8000 193 193 -- X- =48 in. 1 1 2. How far will a body fall, the first second, 8,000 miles above the earth? Ans. 214 in. 3. How far will a body fall, the first second, at the distance of 12,000 miles from the earth? Ans. 1211 in. 4. How far will a body fall, the first second, at the distance of 32,000 miles from the earth? at the distance of 96,000 miles? Ans. At the distance of 32,000 miles from the earth, a body will fall, the first second, 24i in.; at the distance of 96,000 miles, 1 " in. 5. How far will a body fall, the first second, at the distance of the moon, or 240,000 miles from the earth's center? Ans. 1 in. 6. How far will a body fall, the first second of time, at the distance of the sun, or 95,000,000 miles from the earth's center? ART. 9(b). To fnd the distance above the earth, from which a body will fall a given space, the first equal portion of time. From Art. 9(a). we have the following proportion: s: s': 4000': (D+4 000)2; Multiplying extremes and means, sX(D+4 000)2=s'X4 0002; QUESTIONS 9(a).-How far does a body near the earth fall the first second? How do the distances through which a body will fall, the first equal portion of time, at different distances above the earth, vary? Give the illustration. Deduce and repeat the formula and rule. ï~~FALLING BODIES. 21 Dividing by s, (D+4 000)'= - X4 0002; S f Extracting the sq. root, D-+-4 000= - X4 0002; s By transposition D= -X 4 0002-4 000. 5 Hence, To find the distance above the earth, from which a body will fall a given space, the first equal portion of time, we have the following formula and rule: FORMULA.-D = \X4 0002-4 000. RULE.-Divide the space fallen the first equal portion of time near the earth, by the space fcdallen the first, and like equal portion of time at the required distance, multiply the result by the square of the earth's semidiameter, and from the square root of the product subtract the semidiameter of the earth. Ex. 1. At what distance above the earth will a body fall, the first second, 481 inches? SOLUTIo.-Formula,. D= X4 000-4 000= 193X 13 X 4 000-4 000= 2 X4 000-4000 = 4,000 miles. 2. At what distance above the earth will a body fall, the first second, 214 inches? Ans. 8,000 miles. 3. At what distance above the earth will a body fall, the first second, 12;- in.? Ans.12,000 miles. 4. At what distance above the earth will a body fall, the first second, 24 in.? Ans. 32,000 miles. QUESTIONS 9(b).-Deduce and repeat the formula and rule for finding the distance above the earth, from which a body will ft a given distanm the first equal portion of time. ï~~22 EXERCISES IN NATURAL PHILOSOPHY. 5. At what distance above the earth will a body fall, the first second, e1 in.? Ans. 96,000 miles. ART. 10.-Let T represent any time, S the space fallen through during that time, V the acquired velocity, and s the space fallen through the first equal portion of time, as one second, one minute, one hour, &c. Since the spaces described by falling bodies (Art. 7) are as the square of the times, we shall have S (the space described during the whole time) to s (the space described the first equal portion of time) as T2 (the squaresof the whole time) is to 12 (the square of the first equal portion of time). That is, S: s:: T2: 12, Multiplying extremes and means, S= T2xs. Hence, To find the space described by a falling body, the time and the distance fallen the first equal portion of time being given, we have the following formula and rule: FORMULA,.. S=T2Xs. RULE.--3fultiply the square of the time into the distance fallen the first equal portion of time, and the product will be the required answer. Ex. 1. How far will a body, falling from a state of rest, descend in 4 seconds? Ans. 257k ft. SOLUTION. Formula, S=T2 Xs=42X16 =16X 16-=257 feet. NOTE.--In the solution of questions for which formulas are given, first, write the formula, substitute, and then reduce. When not otherwise stated, the falling body is supposed to fall from a state of rest, and near the earth. 2. Through what space will a body fall in 8 seconds? Ans. 1,0291 ft. SoLuTMoN.-Formula, S= T2 X s= 82X 161. Ans. 1,0293 ft. ï~~FALLING BODIES. 23 3. An individual wishing to ascertain the hight of a bridge above a stream of water, drops a stone, which reaches the surface of the water in 2 seconds. Required the hight of the bridge. Ans. 64 ft. 4 in. 4. An individual wishing to ascertain the difference between the depths of two wells, finds that a stone will fall to the bottom of one, in 4 seconds, and to the bottom of the other, in 6 seconds. What is the required difference? Ans. 321 ft. 8 in. 5. From a black cloud a flash of lightning was observed, and 24 seconds afterwards it began to rain. On the supposition that the rain commenced falling at the instant of the flash; required the distance of the cloud. Ans. 9,264 ft. 6. An aieronaut, falling from a balloon, reaches the earth in 30 seconds. Required the hight from which he fell. Ans. 2 miles 3,915 ft. 7. A stone, let fall from a church steeple, occupies 3 seconds in falling. Required the hight of the steeple above the ground. Ans. 197wr ft. 8. A body, the first second, falls through a space of 5 ft.; how far will it fall in 20 seconds? Ans. 2,000 ft. 9. Through what space will a body fall in 28 seconds, the distance fallen through the first second being 2 ft. Ans. 1,960 ft. 10. How far will a body fall in 4 seconds, at the distance of 4,000 miles above the earth? Ans. 64 ft. 4 in. NoTE.-By rule under Art. 9(a), we find the distance fallen, the first second, 4,000 miles above the earth, to be 48j, or t in.; then, by substitution in general formula (Art. 10), we obtain the answer to the above question. 11. How far will a body fall, in 20 seconds, at the distance of the moon, or 240,000 miles from the earth's center. Ans. 21T4 in. QUESTIONS.-10. Repeat the formula and rule for finding the space described by falling bodies. ï~~24 EXERCISES IN NATURAL PHILOSOPHY. ART. 11. Bodies projected perpendicularly upwards, occupy the same time in ascending as in descending. Ex. 1. An archer wishing to ascertain the hight of a tower, found that an arrow, sent to the top, occupied 10 seconds in going and returning. Required the hight of the tower. Ans. 402 ft. 1 in. SoLUTIoN.--Since a body, projected perpendicularly upward, occupies the same time in ascending as in descending, the arrow must have been 5 seconds in falling from the top of the tower to the ground. Hence, to solve the question, all that becomes necessary, is, to find the distance a body will fall in 5 seconds. FORMULA, S= T2Xs=- 52X16~=--402 ft. 1 in. 2. A rifle ball, fired perpendicularly upward, was gone 18 seconds, when it returned to the same place. To what hight did it ascend? Ans. 1,302 ft. 9 in. 3: To what hight will a body ascend, which, projected perpendicularly upward, descends in 30 seconds? Ans. 3,618 ft. 9 in. ART. 12. The space described by a body during the last, or any equal portion of time. Since the spaces described by falling bodies are as the squares of the times, the distance fallen through during the times represented by the numbers 1, 2, 3, 4, 5, 6, &c., are as the squares of these numbers; namely, 1, 4, 9, 16, 25, &c. By subtracting 1 (the comparative distance fallen the first equal portion of time) from 4 (the comparative distance fallen the first two equal portions of time), we obQUESTIONS.--11. How much longer does a body occupy in ascending than in descending? 12. How do the distances described in equal successive portions of time, vary I How may you find the odd number corresponding to any time? ï~~FALLING BODIES. 25 tain 3, as the comparative distance fallen the second equal portion of time. In a similar manner: 9- 4=: 5, comparative distance fallen the 3d equal portion of time. 1 6 - 9 =" 7, ",, 4 th a, - 25-16-== 9, " " 5th " " " 36-25-=1 1, " " " 6th " " * That is, a body falls, during the second equal portion of time, three times as far is during the first; during the third, five times as far; fourth, seven times as far, &c. Therefore, the distances described during equal successive portions of time, are as the odd numbers, 1, 3, 5, 7, 9, 11, &c. ILLUSTRTIO.---If a body, the first second, fall 16! feet, then, during the third second, it will fall through a space represented by 161 feet X 5; fifth second, 16j ft. X 9, &c. Or, if a body, the first hour of its descent, fall 4,600 miles, the third hour of its descent it will fall through a space represented by 4,600 miles X 5 (third odd number), &c. TiHE ODD NUMBER, CORRESPONDING TO ANY TIME, MAY RE FOUND, by doubling the number of equal successive portions of time, and subtracting 1 from the product. ILLtuSTRATmON. - The odd number, corresponding to the 8th second, is equal to (8X2)-1 =15; corresponding to the 10th second (10 X2)-1 =19. Also, the odd number corresponding to the 6th hour is equal to (6X2)-1=11, &c. ART. 13.-Let s represent the space fallen during the first equal portion of time; then, from what has already been shown, To find the space described during any equal portion QuVsS0s.-13. Deduce the formula, and. repeat the rule for fldiig the space described during any equal portion of times 3 ï~~26 EXERCISES IN NATURAL PHIIILOSOPHY. of time, we shall have the following formula and rule: FORMULA,.. S=sX(2 T-1): RULE.---Multiply the distance fallen through during the.jirst equal portion of time, by the odd number correspondiing to the given time. Ex. 1. Suppose a body has been falling for 4 seconds: required the distance fallen through during the last, or 4th second. Ans. 112 ft. 7 in. SoLUTIO.-Formula,. S=sX (2 T-1)= 16 1 X (8-1) =162X7=112g2 ft. 2. A body has been falling for 20 seconds; how far did it fall during the last second? Ans. 627 ft. 3 in. 3. Required the space described by a body during the 37th second of its fall. Ans. 1,174 ft. 1 in. 4. A body has been falling 30 minutes; required the distance described during the last minute, on the supposition that the body fell 12 feet the first second. Ans. 2,548,800 ft. SoLUTION.-Find (by Art. 10) the distance fallen the first minute. Thus, Formula, 8=T2Xs=60X60X12=43,200 ft. Hence, S=sX(2 T- 1)= 43 200X(2X30-1) -= 43 200 X59=2,548,800 ft. 5. A body has been falling one half hour; what was the distance fallen through during the last second? 6. A body has been falling 37 seconds; required the difference between the spaces fallen through during the last and twentieth second. Ans. 546 ft. 10 in. 7. How far would a body fall during the last of 24 hours, on the supposition that it fell during the first second 3 feet. ï~~FALLING BODIES. 27 8. A body has been falling for 3 days; how far did it fall the last hour of the third day, on the supposition that it fell 3 inches during the first half hour of its fall? ART. 14. ThIe velocity being given, to find the space. Let V represent the velocity acquired by a body falling through a space represented by S, s the space fallen through the first equal portion of time, and then 2s (Art. 8) will represent the velocity acquired during the first equal portion of time. Since, The spaces described by falling bodies (Art. 7) vary as the squares of the velocities, we have s:: V2 " 4s; Multiplying extremes and means, SX4s2= - V2X s; V2Xs _ V2Xs _ Dividing by 4s2, s2 4s s 4s 4s2 4XsXs 4s Hence, To find the space described by a falling body, having given the acquired velocity and the space fallcn through during the first equal portion of time, we have the following formula and rule: V2 FORMULA, V Fonsus.,... 8=-4s. 4s Itur,.-Divide the square of the velocity by four times the space fallen through the first equal portion of time. Ex. 1. Through what space must a body all, to acquire a velocity of 193 feet per second? Ans. 579 ft. SoLUTION.-Formula, V2 193X193 193 X 193 4s= 416 - =579 ft. 4s 4X 16z' 64} Qc Es rroxs.-14. Deduce the formula for finding the space, having given the acquired velocity and distance fallen the first equal portion of time. Repeat the rule. ï~~28 EXERCISES IN NATURAL PHILOSOPHY. 2. Through what space must a body fall, to acquire a velocity of 579 feet per second? Ans. 5,211 ft. 3. Required the hight of a tower, from the top of which a body falling acquires a velocity of 128j feet per second? Ans. 257 ft. 4 in. 4. A stone, in falling from a certain hight, acquires a velocity of 257 feet 4 inches per second; required the hight from which it fell. Ans. 1,029 ft. 4 in. 5. Through what space must a body fall to acquire a velocity of 400 feet per second, the distance fallen the first second being 10 feet? Ans. 4,000 ft. SoLUTION.-Formula, V 400X400 V s 4x 4,000 ft. 4s 4X10 6. Through what space must a body fall, to acquire a velocity of 6,000 feet per minute, on the supposition that it falls the first minute 400 feet? Ans. 22,500 ft. 7. What number will represent the comparative space through which a body must fall to acquire a velocity of 3,450, the distance fallen the first equal portion of time being 20? Ans. 148,781*. 8. To what hight will an arrow shot perpendicularly upwar'd, with a velocity of 514 feet 8 inches per second, ascend? Ans. 4,1171 ft. NOTE.---A body projected perpendicularly upward with a given velocity, will ascend to that hight from which it must fall to acquire the same velocity. 9. An individual wishing to ascertain the hight of a church steeple, finds that an arrow shot with the velocity of 130 feet per second, just reaches the top of the steeple; required its hight. 10. How far must a body fall to acquire a velocity of 640 feet per second, the distance fallen the first second being 8 feet? Ans. 12,800 ft. ï~~FALLING BODIES. 2 ART. 15.-The space described by a falling body, and the distance fallen the first equal portion of time being given, to find the time. By Art. 10, we have S: s:: T2: 12; Multiplying extremes and means, T Xs= S; S Dividing by s, T..... 2-S.. s Extracting the square root,... T = f. S Hence, The space described by a falling body, and the distance fallen the first equal portion of time being given, to find the time, we have the following formula and rule: FORMULA,. T=. 4S RuLE.-Divide the space by the distance fallen the first equal portion of time, and extract the square root of the quotient. Ex. 1.-In what time would a body fall from the top of a tower 402w feet high? Ans. 5 sec. SOLUTION.-Formula, T = s--- -- = -25=-5 sec. s.16 T 2. An aiironaut fell from a balloon, 3 miles 5,004 feet high; how long was he falling? Ans. 36 sec. 3. What time would it require a body to fall 1,302f ) feet? Ans. 9 see. SoLUTVON.-Formula, T-4S 1 302-' 15633 12 T =1 '- 1 = 1563 1X =81=9 see. s 161 12 193 4. Required the time it would take a stone to fall to the bottom of a well 257 feet 4 inches deep, Ans. 4 sec. QUzSTIOars.-15. Deduce the formula for fmiding the time. Repeat the rule. ï~~30 EXERCISES IN NATURAL PHILOSOPHY. 5. How long would a body be in falling through a space of 10 miles, falling 3 feet the first second? 6. A body the first second falls 8 feet; how long will it be falling 4,608 feet? Ans. 24 sec. SoLUTIo.-Formula, 18 44608 T =4 - =576= 24 sec. s 8 7. How long will a body be in falling 450 miles, on the supposition that it falls the first minute 2 miles? Ans. 15 min. 8. In how long a time will a body fall 256,000 miles, on the supposition that it falls 1,000 miles the first hour? Ans. 16 hrs. 9. In what time will a body, 4,000 miles above the earth, fall through a space of 579 feet? Ans. 12 sec. ART. 16. The velocity acquired by a falling body, and the distance fallen the first equal portion of time being given, to find the time. Let V represent the velocity acquired by a body falling through a space represented by S; s the space fallen the first equal portion of time, and then (Art. 8) 2s will represent the velocity acquired in falling the first equal portion of time. Then, since (Art. 7) The times of falling bodies are as the acquired velocities, we shall have T: 1:V:: 2s. Multiplying extremes and means, TX 2s= V; V Dividing by 2s,....... T-. 2s Hence, The velocity acquired by a falling body, and the distance fallen the first equal portion of time, being given, to fintd the time, we have the following formula and rule. Fo,.. T=. 2s ï~~FALLING BODIES. 31 RULE.-Divide the acquired velocity by twice the distance fallen the first equal portion of time. Ex. 1. The velocity acquired by a falling body is represented by 424, the distance fallen the first equal portion of time by 4; what number will represent the time of falling? Ans. 53. V 424 SoLUTIo.-Formula, T=- - X4 53. 2s 2X4 2. A body in falling has acquired a velocity of 193 feet per second; required the time. Ans. 6 sec. 3. How long must a body fall to acquire a velocity of 772 feet per second? Ans. 24 sec. 4. A body the first second falls 12 feet, and at the end of a certain time acquires a velocity of 72,000 feet; required the time occupied in falling. Ans. 50 min. 5. A ball was shot perpendicularly upward, with a velocity of 3,088 feet per second; required the time of ascent. Ans. 1 min. 36 sec. 6. An aieronaut falling from 4 balloon, acquired a velocity of 24,704 feet per second; required the time of falling. Ans. 12 min. 48 see. 7. If a body be projected perpendicularly upward with a velocity of 4,632 feet per second, what time will elapse before its return? Ans. 4 min. 48 sec. 8. A rifle ball is shot perpendicularly upward with a velocity of 2,000 feet per second; required the time of ascent. Ans. I min. 2 34 sec. AnRT. 17. The space described by a falling body, and the distance fallen the first equal portion of time being given, to find the.acquired velocity. Since the spaces described by falling bodies vary as the squares of the acquired velocities, we have (Art. 14),: s: V2: 4s2. QUESTIONS.-16. Deduce the formula for finding time, velocity, &c. being given. Repeat the rule. ï~~32 EXERCISES IN NATURAL PHILOSOPHY. Multiplying extremes and means, V2Xs= SX4s2; Dividing by s, V2 SX4s' SX4XsXs.= --X_.... SX4s=4XSXs; S S Extracting the square root, V=2,/SXs. Hence, The space described by a falling body, and the distance fallen the first equal portion of time being given, to find the acquired velocity, we have the following formula and rule: FO IULA,.. V=2,SXs. RULE.-Mtdtliply the space by the distance fallen the first equal portion of time, and take twice the square root of the product. Ex. 1. What velocity will a body acquire in falling 2,316 feet? Ans. 386 ft. per sec. SoLuTIN.-Formula, V=2JSXs=22316X16!v=2X193=386 ft. pr s. 2. A body has fallen through a space of 394 miles 4,080 feet; what velocity did it acquire, on the supposition that it fell 2 feet the first second? 3. With what velocity must a body be projected upward, to ascend 4,117 feet 4 inches? Ans. 514 ft. 8 in. per sec. 4. Suppose a body, the first second, to fall 1 foot, what velocity will it acquire in falling 32,400 feet? Ans. 360 ft. per sec. 5. The distance fallen by a body, the first equal portion of time, is represented by 4, and the entire space fallen by 1,024; what number will represent the velocity acquired? Ans. 128. 6. With what velocity must a body be projected upward, to just reach the top of a tower 402 feet 1 inch high? Ans. 160 ft. 10 in. per sec. Q(sTIoKS.-17. Deduce the formula for finding acquired velocity, sp &c. being given. Repeat the rule. ï~~FALLING BODIES. 33 7. What velocity would a stone acquire in falling to the bottom of a well 579 feet deep? Ans. 193 ft. per sec. 8. What velocity would a body acquire in falling 5,211 feet? Ans. 579 ft. per sec. ART. 18. To find the velocity acquired by a falling body, the time and distance fallen, the first equal portion of time being given. Since the times occupied by falling bodies vary as the acquired velocities, we have (Art. 16), T: 1: V: 2s; Multiplying extremes and means, V= TX2s. Hence, To find the acquired velocity, the time and distance fallen the first cqual portion of time being given, we have the following formula and rule: FORMULA,.. = TX 2s. RULE.--Multiply the time by twice the distance fallen the first equal portion of time. Ex. 1. What velocity will a body acquire in falling 6 seconds? Ans. 193 ft. per sec. 2. An arrow, shot perpendicularly upward, returned in 24 seconds; required the velocity of projection. Ans. 386 ft. per sec. 3. A body, the first second, falls 6 feet; what velocity will it acquire in falling 12 seconds? Ans. 144 ft. per sec. 4. A body is 8 seconds in falling from the top of a tower; what velocity does it acquire? Ans. 257 ft. 4 in. per sec. 5. Let 10 represent the time a body has been falling, and 2 the distance fallen the first equal portion of time; what number will represent the velocity? Ans. 40. QUESTIONS.---18. Deduce the fornula for finding velocity, time, c being given. Repeat the rule. ï~~34 EXERCISES IN NATURAL PHILOSOPIIHY. ART. 19. SPACE DESCRIBED BY BODIES PROJECTED DOWNWARD. The space fallen through in any given time, by a body projected downward with a given velocity, equals that which it would describe moving uniformly for the given time, plus that described by gravity in the same time. Fig. 6. DEMONSTRAQ A TtON.-Let A B represent the velocity of projection, BD the time of falling, and DE the velocity acquired by gravity in the same time. Then Sthe figure A GEB will represent the entire space passed through; but the rectangle A CDB, represents the distance passed through by the body moving uniformly for the given time, and the right-angled triangle BDE (Art. 7), represents the space described by the influence of gravity in the same time. Let V represent the velocity of projection, T the time of falling, and s the distance fallen the first equal portion of time. Then, The velocity with which a body is projected downward, and the time of falling being given, to find the space described, we have the following formula and rule: FORMULA,.. S=TXV+ T T2X s: QVusTroNs.-19. Repeat the formula for finding the space described ya ebody in a given time, projected downward with a given velocity. eathe ue. Deduce the formula. ï~~FALLING BODIES. 35 RULr.-i Multiply the time by the velocity, to which product add the square of the time multiplied by the distance fallen the first equal portion of time: OR, 7o the space which the body would describe in the given time with the given velocity of projection, add that described by gravity in the same time. Ex. 1. Suppose a body to be projected downward with a velocity of 100 feet per second: how far will it fall in 6 seconds? Ans. 1,179 ft. SOLuTON.--Formula,.. S= Tx V+ T2X s=6 X 100Â~62X 161= 1,179 ft. 2. How far will a body, projected downward with a velocity of 800 feet per second, fall in 12 seconds? Ans. 2 miles 1,356 ft. 3. A body falling by the force of gravity 12 feet the first second, is projected downward with a velocity of 400 feet per second: how far will it fall in 10 seconds? Ans. 5,200 ft. 4. Suppose a body to be projected downward with a velocity of 2,000 feet per second. Required the distance through which it will fall in one half minute. Ans. 14 miles 555 ft. 5. The time of a falling body is represented by 18, the velocity of projection by 140, and the distance fallen by force of gravity, the first equal portion of time, by 6. What number will represent the entire space described? Ans. 4,464. 6. A body falling by the force of gravity 6 feet the first second, is projected downward with a velocity of 12 feet per second: how far will it fall in 5 seconds? Ans. 210 ft. ï~~36 EXERCISES IN NATURAL PHILOSOPHY. CHAPTER VI. UNIFORM MOTION. ART. 20. A body is said to move with uniform velocity when it passes over equal spaces in equal times. A body moving with the uniform velocity of 10 feet per second, in three seconds would move over a space of 10 ft. X 3 =- 30 ft.; in five seconds, 10 ft. X 5 = 50 ft. Hence, if T represent the time, V the uniform velocity, and S the space; then, to find the space described by a body moving with uniform velocity, we have the following formula and rule: FORMULA,... S= Vx T: RULE.-Multiply the velocity by the time. Ex. 1. A body moves with the uniform velocity of 20 feet per second; how far will it move in 8 seconds? Ans. 160 ft. SoLUTioN.-Formula, S= VX T= 20 ft. X 8 =160 ft. 2. Over what space would a bird fly in 12 hours, flying with the uniform velocity of 20 feet per second? Ans. 163 miles 3,360 ft. 3. How far would a man walk in 120 days, walking uniformly at the rate of 2j miles per hour, 6 hours each day? Ans. 1,800 miles. ART. 21. The space and time of bodies moving with uniform velocity being given, to jind the velocity. If a body, moving uniformly, pass over a space of 100 feet in 10 seconds, in one second it will pass QuzasTi ows.--20. What is uniform motion? Deduce the formula and repeat the rule for finding the space. ï~~UNIFORM MOTION. 37 over a space of 100 ft.x;'; = 10 ft. Hence, To find the velocity, we have the following formula and rule: F,,RULA,... -S+ T: RlULE.-Divide the space by the time, and the quotient will be the required velocity. Ex. 1. A body in 8 seconds moves over a space of 9,600 feet. Required the velocity. Ans. 1,200 ft. per sec. SOLUTION.-Form., V =- S- T= 9 600-+-8 - 1,200 ft. 2. Required the velocity of a body which, in 20 seconds, moves over a space of 5,240 feet. Ans. 262 ft. per sec. 3. A body in 2 hours moves over a space of 11 miles. Required the velocity per second. Ans. 62 ft. per sec. ART. 22. The space and velocity of bodies moving uniform/y, being given, to find the time. A body moving at the rate of 10 feet per second, would in ' second pass over one foot; and 400 ft. in Y sec.X400= -oÂ~=400+10==40 sees. Hence, To find the time, we have the following formula and rule: FoRNIULA,.... T= S V: RuLE.-Divide the space by the velocity, and the quotient will be the required time. Ex. 1. In what time would a body, moving at the rate of 40 feet per second, pass over a space of 840 feet? Ans. 21 sees. SOLUTrIoN.-Form., T= S-- V- 840 Â~40= 21 sees. 2. In what time would a man, traveling 12 hours each day, at the rate of 6 miles per hour, go 3,780 miles? Ans. 52 d. 6 h. Q VST TNs.--21, 22. Deduce the formulas, and repeat the rules for finding velocity and time. ï~~38 EXERCISES IN NATURAL PHILOSOPHY. 3. What time would it require a body, moving with the velocity of 60 feet per second, to pass over a space of 4 miles? Ans. T=-5 m. 52 secs. It will be seen, by observing the foregoing rules of uniform motion, that two numbers are always given to find the third, and from which we have the following GENERAL RULE. To fnd the SPACE, take the product of the given numbers. To find either the TIME or VELOCITY, divide the space by the remaining given number, and in each case the result will be the answer sought. CHAPTER VII. MOMENTUM. ART. 23. By the momentum of a body is meant its force or quantity of motion. Since in any moving body each particle has the same velocity, the quantity of motion, or comparative momentum, will evidently depend upon the product of the quantity of matter or weight and velocity. If two balls, one weighing 5, and the other 10 pounds, be projected with the same velocity, the momentum of the larger ball will be twice that of the smaller; if their weights and velocities be equal, their momenta will be equal. If a body weighing 80 ounces be projected with a velocity of 10 feet per second, its comparative momentum will be 80; that is, its quantity of motion will be 80 times that of a body weighQr STIoNs.-23. What is meant by momentum?-Relative momentum?--Comparative momentum? Repeat the formula and rule for finding the comparative momentum. ï~~MOMENTUM. 39 ing 1 ounce, and moving at the rate of 1 foot per second; or, in other words, it will overcome 80 times the resistance. The force or resistance which a body is capable of overcoming, expresses its absohlute momentum. If we let M= the momentum of a body, Q the quantity of matter, and V the velocity, then, To find the momentum of a body, we shall have the following formula and rule: FORMULA,.... = Q X V: RULE. -TlMtiply the quantity of matter by the velocity, and the product will be the comparative momentum. Ex. 1. Suppose the weight of a body to be 18 ounces, and its velocity 6 feet per second. Required its comparative momentum. Ans. M=108. SoLUTION.-Formula, M= Q X V= 18 X 6 = 108. 2. Required the comparative momentum of a body whose weight is 124 pounds, and velocity 18 feet per second. Ans. 21=2,232. 3. Required the difference between the comparative momenta of two bodies, one weighing 24 ounces, and having a velocity of 18 feet per second; the other weighing 18 ounces, and having a velocity of 10 feet per second. Ans. Diff. of momenta =252. ART. 24. Momentum and quantity of matter being given to find the velocity. Since (Art. 23) M= Q x V, the momentum is composed of the product of the factors, quantity of matter and velocity, and by dividing M, the momentum, by Q, the quantity of matter, one of the factors, we shall obtain V, the velocity, the other factor. QUESTIO s.-24. Repeat the formula and rule for finding th&e velocity. ï~~40 EXERCISES IN NATURAL PHILOSOPHY. Hence, The momentum and quantity of matter being given, tofind the velocity, we have the following formula and rule: FORMULA,... V=_- -: RULE.-.Divide the momentum by the quantity of matter, and the quotient will be the velocity required. Ex. 1. With what velocity must a body weighing 2 oz., be thrown, to acquire the same momentum as a body weighing 14 oz., and moving with a velocity of 40 feet per second? Ans. V=280 ft. per sec. SOLUTION. -The comparative momentum (Art. 23) of a body weighing 14 oz., and moving at the rate of 40 feet per second=560. Hence, V= M+ Q= 560.2--=280 ft. 2. With what velocity must a body weighing 8 pounds and 4 ounces, be thrown, to acquire the same momentum as a body weighing 100 pounds, and moving at the rate of 24 feet per second? Ans. V=-2901 ft. per sec. 3. Suppose the batteringram of Vespasian weighed 10,000 pounds, and was propelled with a velocity of 20 feet per second, and that this force was found sufficient to demolish the walls of Jerusalem. With what velocity must a 32 pound ball move, to do the same execution? Ans. V=6,250 ft. per sec. ART. 25. The momentum and velocity being given to find the quantity of matter. Since the momentum is composed of the product of the velocity and quantity of matter, by dividing M, the momentum, by V, the velocity, one of its factors, we shall obtain Q, the quantity of matter, as the other factor. Qu-sTIoss.--25. Repeat the formula and rule for finding the quantity of matter, the momentum and velocity being given. ï~~MOMENTUM. 41 Hence, To find the quantity of matter, the momentum and velocity beinf given, we have the following formula and rule: FORMULA,... Q fV RULE.-Divide the momentum by the velocity, and thew quotient will be the quantity of matter. Ex. 1. Suppose a ball, whose weight is 30 lbs., to move with a velocity of 1,200 feet per second. Required the weight of a ball, moving with a velocity of 2,000 feet per second, that shall produce the same momentum. Ans. weight = 18 lbs. SoL UTO.-Since their momenta are equal, 30X 1,200 (36,000)expresses the comparative momentum of the ball, moving at the rate of 2,000 feet per second. Ience, we shall have by substituting in the formula, Q= MV = 36 000Â~2 000 = 18 lbs. 2. A ship of 150 tons, sailing at the rate of 10 knots per hour, is met by a steamboat moving at the rate of 5 knots per hour. On the supposition that they mutually destroy each other's motion, required the tonnage of the steamboat. Ans. 300 tons. 3. Two individuals, walking in opposite directions, meet each other with equal momenta. One weighs 180 lbs., and walks at the rate of 4 miles per hour; the other walks at the rate of 5 miles per hour. Required the weight of the latter. Ans. W=-- 144 lbs. 4. Suppose a ball, moving with the velocity of 1,200 feet per second, to enter a suspended log of wood, weighing 5 cwt., 2 qrs., and to give it a motion of 18 feet per second. Required the weight of the ball. Anm. W= 9- lbs. It will be seen, by observing the nature of momentum, and the rules which have been given for the same, that two numbers are always given to find 4 ï~~42 EXERCISES IN NATURAL PHILOSOPHY. a third, and that we have, for the solution of examples under the head of momentum, the following GENERAL RULE. To FIND THE COMPARATIVE MOMENTUM, take the product of the given numbers. To FIND EITHER THE WEIGHT or VELOCITY, divide the comparative momentum by the remaining given number, and, in each case, the result will be the answer sought. CHAPTER VIII. THE PENDULUM. ART. 26. A pendulum is a body suspended by a rod or cord, so as to swing backward and forward. The swinging of a pendulum is called its vibration. The vibrations of the pendulum depend upon its length and the attraction of gravity. If a pendulum be shortened, or its accelerating force (attraction of gravity) be increased, the vibrations will be more rapid; if lengthened, or the accelerating force diminished, they will be less rapid. By shortening, therefore, the pendulum of a clock, it will be made to run faster; by lengthening it, slower. A clock, also, at different distances above the earth's surface, wi4l run either faster or slower, as its distance is less or farther removed from the center of the earth. ART. 27. The above relations are expressed in the following propositions: QUESTION s.-25. Repeat the general rule for momentum, weight, and velocity. 26. What is a pendulum. Upon what does the vibration of a pendulum depend '! What effect is produced by shortening or lengthenming a pendulum? ï~~THE PENDULUM. 43 PROPOSITION 1.-The times of the vibrations of pendulums of d/fercnt lengths vary as the square root of the lengths: Or, The lengths of dfferent pendulums vary as the squares of the times of vibration. ILLUSTRATIO.-If the lengths of two pendulu be represented by the numbers 4 and 9, then the times of their vibrations will be as 2: 3. If the length of one pendulum be four times that of another, then (the square root of 4 being 2) it will require twice the time for it to perform a single vibration; if the length of one pendulum be nine times that of another, it will require three times as long for it to perform a single vibration, and so on. Again; Since the lengths of different pendulums vary as the squares of the times of vibrations, a pendulum, in order to vibrate once in two seconds, must be (the square of two equaling 4) four times as long as one that vibrates seconds; to vibrate once in 3 seconds, nine times as long; to vibrate in I second (the square of I being 4), one fourth as long as one that vibrates seconds; to vibrate in of a second, one ninth as long; in second, one sixteenth as long, and so on. Now, it has been ascertained, that a pendulum near the earth, beating seconds, is about 39 inches long; hence, a pendulum, to vibrate once in two seconds, must be 4 times 39 inches, or 156 inches long; to vibrate once in 3 seconds, 9 times 39 in. long; to vibrate in second, ith of 39 in. long; in 4 second,. th of 39 in. ART. 28. Let T represent the time it takes any given pendulum to perform a single vibration, and L the length of the pendulum. Then, since The times of the vibrations of penduQUESTIroNs.---27. Repeat prop. 1. Give the illustration. ï~~44 EXERCISES IN NATURAL PHILOSOPIIY. lums of different lengths vary as the square root of the lengths; or, The lengths of different pendulums vary as the squares of the times of vibration, we shall have the following proportion; L: 39:: T': 1"; ultiplying the extremes and means, L=T2x39. Hence, To find the length of a pendulum vibrating in any given time, we have the following formula and rule: FORMULA,.. L= T2X39. RULE.- Multiply the square of the time reduced to seconds, by 39, and the product will be the length of the required pendulum in inches. Ex. 1. Required the length of a pendulum that will vibrate once in 3 seconds. Ans. L=351 in. SoLUTiON.-Formula,. L= T2x39 ==3 X3 X39= 351 in. 2. Required the length of a pendulum that will vibrate once in -th of a second. Ans. L= i5 in. SoLuTION.-Formula,. L= T2X39 - X 5 X39= 3 9-11 4 in 3. Required the length of a pendulum that will vibrate once in 30 seconds. Ans. L=2,925 ft. 4. What must be the length of a pendulum vibrating once a minute. Ans. L=-2 m. 1,140 ft., or 21 m. SoLuT o.-Formula,..... L= T X39= 5 1 1 195 0 X 0 X 39 X - 88 --21 miles, or 2 miles and 1,140 ft. 88 QuasTiows.-27. What is the length of a pendulum beating seconds? 28. What does T represent?--L? Repeat the rule and formula for fnding the length of a pendulum vibrating in any given time. ï~~THE PENDULUM. 45 It will be observed, to reduce the length, which the formula expresses in inches, to miles, we have multiplwd by;, and then by. (which is equivalent to dividing by 12 and then by 5280). After which, we have canceled and reduced. Generally, it will be found much more convenient, and shorter, first, to express all multiplications and divisions as above, then, cancel and reduce. 5. Required the length of a pendulum that will vibrate once in 2 minutes. SOLUTION.-Since, The lengths of different pendulhams vary as the squares of the times of vibrations, the length of a p)endulum vibrating once in 2 minutes (the square of 2 being 4) will be four times as long as one vibrating once in one minute. But, by the last examl)le, a pendulum vibrating once a minute is found to be 'yy miles long. Hence, the formula 195 195 becomes L 195 X4.195 89 m. = 8 m. 4,560 ft. becomes, L- -- X4-- 22.8m45 ' 88 22 22 - Ot 6. Required the length of a pendulum that will vibrate once an hour. SOLUTION.-Formula, 15 30 195 0 0 87750 L--- X 1- mn.=79777r miles. 4 11 7. Required the length of a pendulum that will vibrate once in 24 hours, or one day. SoLUTION.-Formula, 87750 24 24 50544000 L= 11 X -= X-11 m. = 4,594,9091 m. 11 11 11 8. What must be the length of a pendulum to vibrate once a century? 9. What must be the length of a pendulum to ï~~46 EXERCISES IN NATURAL PHILOSOPHY. make 4 vibrations in a century, or to vibrate once in 25 years? 10. A pendulum, whose length is represented by 24, vibrates once a day; what number will represent the length of a pendulum vibrating once in 6 days? Ans. 864. 11. A pendulum, whose length is represented by 100, vibrates once a century; what number will represent the length of a pendulum vibrating once in 10 centuries? Ans. 10,000. ART. 29. The length of a pendulum being given, to find the time of vibration. Since (Art 27), The lengths of different pendulums vary as the squares of the times of vibrations, we have the following proportion: L: 39:: T: 1"; Multiplying extremes and means, T' X 39==L; Dividing both members by 39,. T2= L-39; Extracting the square roots,.: T =,,L-39. Hence, To find the time required for a pendulum of a given length to make a single vibration, we have the following foibrmula and rule: FORMULA,.. T= /L-39. RILE.-Divide the length reduced to inches by 39, and extract the square root of the quotient, the result will express the required time in seconds. Ex. 1. In what time will a pendulum 13 feet long make a single vibration?. Ans. T=2 sec. SOLUTION.-Formula, T= 4L-;-39 = 13X12X3 %=4=2 sec. 2. Required the time that it will take a penduQ4UsTIo1.-29. Repeat the formula and rule for finding the time required for a pendulum of a given length to make a single vibration. Deduce the formula and rule. ï~~THE PENDULUM. 47 lum 29 feet 3 inches long to make a single vibration? Ans. T=3 sec. 3. In what time will a pendulum 52 feet long make a single vibration? Ans. T=4 sec. SoLUTiON.-Formula, T= VL+39= 52X 12X I=16=4 sec. 4. Required the time that it will take a pendulum 263 feet 3 inches long to make a single vibration? Ans. T=9 sec. 5. In what time will a pendulum 5,200 feet long make a single vibration? Ans. T=40 sec. 6. In what time would a pendulum reaching from the earth to the moon, make a single vibration? 7. In what time would a pendulum reaching from the earth to the sun, make a single vibration? AaRT. 30. PROPrOSITION 2.- The times of vibration of the same pendulum, at diferent parts of the earth's surface, or at different distances above the earth's surface, vary as those distances from the center of the earth. Hence, The number of vibrations in a given time, are inversely as the distances from the center of the earth. ILLUSTRATION.-A pendulum 8,000 miles from the center of the earth, would be twice as long in making a single vibration as at the surface; that is, a pendulum at the surface of the earth beating seconds, will, at the distance of 4,000 miles from the surface, beat once in two seconds. At the distance of 12,000 miles from the center, a pendulum would be three times as long in making a single vibration as at the surface of the earth; at the distance of 16,000 miles, four times as long, &c. Consequently, the number of vibrations in a given QUESTIO.s.-30. Repeat proposition 2. Give the illustration. ï~~48 EXERCISES IN NATURAL PHILOSOPIIY. time at the earth's surface'would be twice as many as at 8,000 miles friom the center; three times as many as at 12,000 miles from the center, &c. If, therefore, the number of vibrations made int a given time, at any distance above the surface of the earth, by a pendulum making a given number of vibrations in the same time at the surface of the earth, be known, the distance of that point from the earth's center, and consequently above the surface of the earth, may be readily determined. Also, we may ascertain the number of vibrations during any given time, at any given distance above the earth, the number of vibrations in the same time at the surface being known; also, the number of vibrations at the surface in a given time, may be found, the number df vibrations made by the same pendulum in the same time, at a given number of miles above the earth, being known. ART. 31. Let D represent the distance of any point above the earth's surface, D+4000 miles the distance of that point from the center. Let n represent the riumber of vibrations made in any given time by a pendulum, at the surface of the earth; N the number of vibrations made by the same pendulum, in the same time, at the distance of D miles from the surface. Then, since The number of vibrations made in a given time by the same pendulum at different parts of the earth's surface, or at diferent distances above the earth, vary INVERSELY as those distances from the center, we have the following proportion: N: n:: 4000 m.: D-+-4000 miles. Multiplying extremes and means, NX(D-+4000)- 4000Xn; 4000Xn Dividing by D-+-4000,. N= 4000Xn D+4000" ï~~THE PENDULUM. 4g Hence, To find the number of vibrations made by a pendalum in a given time, at a given distance from the earth, the number of vibrations made in the same time at the surface being given, we have the following formula and rule: FORMULA,... D 4000 RuLE.-Multiply the semnidiameter of the earth by the number qf vibrations made in the given time at the surface of the earth, and divide the product by the given distance from the center, the quotient will be the required number of vibrations. Ex. 1. Suppose a pendulum at the surface of the earth to vibrate seconds, or to make 3,600 vibrations in an hour; required the number of vibrations it would make per hour, at the distance of 400 miles above the earth. Ans. AN=3,272, or 327- vibrations per hour less than at the earth's surface. SOLUTION.-Formula, N==4000 X n 4000 X 3600 3272 D--4000 4400 2. Suppose a pendulum at the surface of the earth to beat seconds, or in other words, to make 120 vibrations per minute; required the number of vibrations it would make per minute, at the hight of 6,400 miles above the earth. Ans. N =46. 3. Required the number of vibrations that would be made per hour at the distance of 3,240 miles from the earth, by a pendulum that vibrates seconds at the earth's surface. Ans. N= 7,955. 4 QUESTIONs.-31. What does D represent 1-N?--n?-D-4000 Repeat the formula and rule for finding the number of vibrations made by a pendulum in a given time, &c., at any given ditance above the earth? 5 ï~~50 EXERCISES IN NATURAL PHILOSOPHY. ART. 82. Since (Art. 31), The number of vibrations made in a given time by the same pendulum at diferent parts of the earth's surface, or at different distances above the earth, vaiy inversely as the distances from the center, we have N: n:: 4000 m.: D+4000 miles. Multiplying extremes and means, n X 4000= (D+ 4000) X N; (D+4000) X N Dividing by 4000, n= 4000 ' Hence, To find the number of vibrations made by a penddulum in any given time at the sutrface of the earth, the number of vibrations made in the same time at any given distance above the earth being given, we have the following formula and rule: FMULA, Â~ n(D+ 4000) xN FORMULA,. n 00 4000 RULE. -Multiply the given distance from the center of the earth, by the number of vibrations made in the given time at that distance, and divide the product by the semidiameter of the earth; the quotient will be the required number of vibrations. Ex. 1. Suppose a pendulum to be of such a length as to beat seconds at the distance of 420 miles above the earth; required the number of vibrations that it would make per hour at the surface of the earth. Ans. n=3,978. SoLUTIoN.-Formula, (D+4000) X N 4420 X3600 n- =3,978. 4000 4000 2. Required the number of vibrations per hour, that a pendulum at the surface of the earth would make, on the supposition that the same pendulum vibrates once in 6 seconds, at the distance of 12,400 miles above the earth? Ans. n=2,460. QUESTION-.-832. Repeat the rule and formula for finding the nunm. her of vibrations made in any given time at the surface of the earth. ï~~THE PENDULUM. 51 3. How many vibrations would a pendulum make per hour, at the surface of the earth, on the supposition that the same pendulum vibrates 3,000 times per hour, at the distance of 20 miles above the earth? Ans. n=3,015. ART. 33. By Art. 31, N: n:: 4000 m.: D+4000 miles. Multiplying extremes and means, (D+4000) X N=4000 Xn; 4000Xn Dividing by N, D+4000= 400Xn Subtracting 4000 from both members, 4000 X n4000 D = - 4000; IV Reducing to a common denominator, D_ ( X4000. Hence, To find the distance of any point above the carth, the number of vibrations made by the same pendulum in a given time at that distance, and at the surface of the earth being given, we have the following formula and rule: FORMULA,. D=(n )>4000. RULE.-fultiply the difference between the number of vibrations made at the surface of the earth, and at the required distance above the earth in the same time, by the semidiameter of the earth, and divide the product by the number of vibrations made at the required distance above the earth. Ex. 1. Required the bight of that point above the earth, at which a pendulum vibrating seconds at the surface, would make 3,200 vibrations per hour. Ans. D=- 500 miles. QUE TLON.-33. Repeat the rule and formula for finding the distance of any point above the earth. Deduce the formula. ï~~52 EXERCISES IN N4TURAL PHILOSOPHY. SOLUTION.-Formula, (n-N) X4000-3600-3200 D= X4000=- 320 X4000= 500 ms. 2. Required the hight of a mountain, on the top of which a pendulum, which vibrates seconds at the level of the sea, makes but 3,596 vibrations per hour, the mean diameter of the earth being reckoned 7,912 miles. Ans. HIight =4.40+ miles. 3. Required the hight of that point above the earth, at which a pendulum, vibrating seconds at the surface of the earth, makes 400 vibrations per hour. 4. Required the hight of that point above the earth at which a pendulum vibrating seconds at the surface of the earth, makes 250 vibrations per hour. REMARK.-The pendulum affords the means of ascertaining the shape of the earth. This is done by noting the number of vibrations made by the same pendulum or pendulums of the same length, at different points on the surface of the earth; from which the comparative distance of each observation from the center, is easily determined. By actual observation it has been found, that the number of vibrations made in a given time, increases from the equator to the poles. Hence, the polar diameter of the earth, is less than the equatorial diameter. 5. Required the difference between the equatorial and polar diameters of the earth, on the supposition that a pendulum vibrating seconds at the equator (4,000 miles from the center), makes 3,613 vibrations per hour at the poles. Ans. The difference between the semidiameters equals 14.39+ miles; consequently, the difference between the diameters is 28.78--+ miles. 6. Required the hight of a mountain, on the top of which a pendulum, which vibrates seconds at the level of the sea, makes but 3,598 vibrations per hour, the diameter of the earth being reckoned 7,912 miles. Ans. 2.19+ miles. ï~~THE LEVER. 53 CHAPTER IX. THE LEVER. / ART. 34. THE Lever is an inflexible bar or rod, moving about a fixed point called the fulcrum. It consists of three kinds, depending upon the position of the fulcrum with regard to the power and weight. The power is that force which gives, or tends to give motion; the weight that which receives it. In the first order of the lever, the fulcrum is situated between the power anid weight; in the second, the weight is between the fulcrum and power; in the third, the power is between the fulcrum and weight. ART. 35. LEVER OF THE FIRST AND SECOND ORDER. Fig. 7 repre- Fig. 7. sents a lever of s L the first order.Fig. S of the second order. In the lever of the P first and second order, the power is applied to the Fig. 8. long arm, the weight to the short arm. P If the long be twice that of the short arm, then a L power of 1 pound will balance a weight of 2 lbs.; if the long arm be W three times that of NoTE.--For a rigid demonAstration of the principle of the lever, see Appendix. ï~~54 EXERCISES IN NATURAL PHILOSOPHY. the short arm, then a power of 1 pound will balance a weight of 3 pounds, and so on.. Hence, The power and weight are to each other inversely, as the length of the arms to which they are attached. ART. 36. Let L represent the long arm of a lever, S the short arm, W the weight, and P the power, then we shall have P: W:: S: L; Multiplying the extremes and means, Px L= Wx S. fTherefore; The power multiplied into the length of the long arm, equals the weight multiplied into the length of the short arm. ) OR, Since under certain circumstances there is, in every machine, an equilibrium existing between the power and weight; if those numbers, whose product express this equilibrium, be considered as connected or associated with each other, we shall have, in reference to the mechanical powers, the following GENERAL RULE, OR, EQUATION OF EQUILIBRIUM. The POWER multiplied into that with which it is CONNECTED, equals the WEIGHT multiplied into that with which it is CONNECTED. The above general rule we shall denominate the Equation of Equilibrium, between the power and weight. It should be remembered, that, in accordance with the definition given of connected or assoQUESTIONS.-34. What is a lever? Of how many kinds is the lever? Give the position of the power, weight, and fulcrum to each other, in the three orders of the lever. 35. What relation do the power and weight sustain to each other? 36. What does P represent?- W?-L?-S? How does the product of the power into the long arm compare with that of the weight into the short arm? What is meant by connected or associated.numbers? Repeat the general rule expressing the equation of equilibrium? What is the power connected with? With what is the weight connected? Repeat the equation of equilibrium for the first and second orders of the lever. ï~~THE LEVER. 55 ciated numbers, in the first and second order of the lever, the power will be connected with the long arm, and the weight with the short artn. ART. 37. By Art. 36,.. PXL=--WXS; WxS Dividing both members by L,. P -= L " Hence, To find the power, the long arm, short arm, and weight being given, we have the following formula and rule: WxS FORMULA,... P W S RULE.-Multiply the weight by the length of the short arm, and divide the product by the length of the long arm; the quotient will be the required power. Or, since in the equation of equilibrium the products of the connected numbers are equal, it is evident, by dividing the product of any two connected numbers by one of the two remaining numbers, the quotient will be the fourth. Hence, we have, by means of which the above, together with all the subsequent formulas connected with the mechanical powers, may be readily called to mind, the following GENERAL RULE. SDivide the product of the two given numbers which are connected with each other, by the remaining given number, and the quotient will be the required answer. / ILLUSTRATION. -Suppose the power is required; here will be given the ong arm, short arm, and weight, of which the short arm and weight are connected; divide their product by the remaining given number, namely, the length of the long arm, and the quotient will be the required power. / QUESTIOxNS.-87. Repeat the formula and rule for finding the power. Repeat the general rule. Give illustrations. ï~~56 EXERCISES IN NATURAL PHILOSOPHY. If the long arm is required, there will be given the power, the weight, and short arm, of which the weight and short arm are connected; divide their product by the remaining given number, namely, the power, and the quotient will be the required length of the long arm. Ex. 1. Suppose, in a lever of the first or second order, the long arm to be 12 ft., short arm 9 ft., and weight 240 lbs. Required the power. Ans. P= 180 lbs. S WXS 240 X9 SoLUTIoN.-Formula, P= 12 180. L -- 12 =10 REMARK.-It will be seen, by observing the formula for the power, that if the short arm by increased, or the long arm diminished, the power, in order to balance a given weight, must be increased; if the short arm be lessened, or long arm increased, the power to balance a given weight must be diminished; and if the weight be increased or diminished, the long and short arm remaining the same, the power must also be increased or diminished. 2. Required the power necessary to be applied, at the end of a lever of the second order, to balance a weight of 4,800 lbs., placed at the distance of 6 ft. from the fulcrum, the length of the lever being 24 ft. Ans. P= 1,200 lbs. 3. Required the power necessary to be applied at the end of a lever 12 ft. 6 in. long, to raise a weight of 420 lbs., the fulcrum being placed at the distance of 3 ft. 6 in. from one end of the lever. Ans. P= 117 lbs. ART. 38. Since (Art. 36). WXS=PxL; By dividing both members by 8, W= Px. Hence, To find the weight, the power, long arm, and short arm b ing given, we have the following formula and rule: ï~~THE LEVER. 57 P xL FORMULA,... -- RULE.-Multiply the power by the length of the long arm, and divide the product by the length of the short arm; the quotient will be the required weight. Ex. 1. Suppose, in a lever of the first or second order, the power to be 28 lbs., long arm 16 ft., and short arm 4 ft. Required the weight. Ans. W= 112 lbs. PXL 28X16 SoLUTrioN.-Formula, W=P L 28X16 =112. 4 2. There is a lever, 60 ft. long, with a fulcrum 3 ft. from one end. What weight, at one end of the lever, will balance a power of 480 tons, at the other end? Ans. W= 9,120 tons. 3. Suppose a lever, of the second order, to be 20 ft. long, the short arm 3 ft. 4 in. Required the weight that a power of 18 lbs. 4 oz. would balance. Ans. W= 91 lbs. 4 oz. ART. 39. Since (Art. 36). P X L= WX S, Wx S Dividing by P,........ L=. Hence, To find the length of the long arm of a lever, the swhort arm, weight, and power being given, we have the following formula and rule: WxAS FORMULA,.. X. P QuIrsTIONs.-38. Repeat the equation of equilibrium. Deduce the formula for weight. Repeat the formula and rule for finding weight, the power, long arm, and short arm being known. When the weight is required, what three numbers will be given? Which two of the given numbers are connected? What is the remaining given number! Repeat the formula for weight, by calling to mind the principle expressed in general rule (Art. 37). 39. Deduce the formula for finding the length of the long arm; Repeat the formula and rule. ï~~58 EXERCISES IN NATURAL PHIILOSOPIHY. RuLE.-Multiply the weight by the length of the short arm, and divide the product by the power; the quotient will be the required length of the long arm. Ex. 1. Suppose, in a lever of the first or second order, the weight to be 3,620 lbs., the power 905 lbs., and the length of the short arm 5 ft. Required the length of the long arm. Ans. L=20 ft. WXS 3620X5 SoLUTION.-Formula, L = 95 20 ft. P 905 2. Suppose, in a lever of the second order, the weight to be 420 lbs., the power 20 lbs., and the length of the short arm 3 ft. Required the length of the long arm. Ans. L= 63 ft. 3. Suppose the length of the short arm of a lever to be 8 ft. Required the length of the long arm, in order that a power of 4 tons may just balance a weight of 241 tons. Ans. L=49 ft. ART. 40. Since (Art. 36) WXS= PXL, Dividing by W,..- PxL Hence, To find the length of the short arm, the length of the long arm, the power, and weight being given, we have the following formula and rule: Px L FORMULA,.. -PXL RULE.-Multiply the power into the length of the long arm, and divide the product by the weight, the quotient will be the required length of the short arm. Ex. 1. Suppose the long arm of a lever to be 18 ft. Required the length of the short arm, in order QimEsTIons.--40. Deduce the formula for fiding the length of the short arm. Repeat the formula and rule. When the long arm is required, what three numbers will be given? Of the given numbers what two are connected? ï~~THE LEVER. 59 that a power of 144 lbs. shall balance a weight of 1,296 lbs. Ans. S= 2 ft. PxL 144X18 SoLUTrIoN.-Formula, S= -- =-- 12-96-2 ft. W 1296 2. Suppose the long arm of a lever to be 12 ft. 6 in. Required the length of the short arm, in order that a power of 3 tons shall balance a weight of 9 tons. Ans. S--= 4 ft. 2 in. 3. Required the length of the short arm of a lever, the long arm being 13 ft. 4 in., the power 24 lbs. 12 oz., and the weight 99 lbs. Ans. S=3 ft. 4 in. SOLUTION. - 13 ft. 4 in= 131 ft; 24 lbs. 12 oz. = 244{ lbs. PXL 24 X 13: 99 40 1 FORMvLA,. S= ------ -- X- = W 99 4 3 99 31 ft. = 3 ft. 4 in. ART. 41. LEVER OF THE THIRD ORDER. Fig. 9. repre- Fig. 9 sents a lever of the third order. P In the lever of the third order, the weight is at- L tached to the long arm, and the power is applied to the w short arm. If the long arm be twice as long as the short arm, then the power will be twice the weight; three times as long, three QUESTIoNS.---41. Describe the lever of the third order. Repeat the Equation of equilibrium. With what is the power connected?Weight ï~~60 EXERCISES IN NATURAL PHILOSOPHY. times the weight, &c. Hence, it will be seen, that the power and weight are to each other inversely, as the lengths of the arms to which they are attached. The power, being applied to the short arm, acts under a mechanical disadvantage, it always being greater than the weight or resistance. Let L represent the long arm of a lever of the third order, S the short arm, W the weight, and P the power. Then, Since the power and weight are to each other, INVERSELY, as the lengths of the arms to which they are attached, we have the following proportion: P:W:: L: S; Multiplying extremes and means, P x S= =Wx L, equation of equilibrium; That is, The power, multiplied into the length of the short arm, equals the weight multiplied into the length of the long arm. It will be observed, from the equation of equilibrium, that, in the third order of the lever, the power is connected with the short arm, the weight, with the long arm. AaRT. 42. Since (Art. 41). PXS= WxL, WxL Dividing by 8,....... P -- Therefore, To find the power, the long arm, short arm, and weight being given, we have the following formula and rule: FORMULA,.. X. P L. S R.!E.-Multiply the weight into the length of the long arm, and divide the product by the length of the short arm, the quotient will be the required power. QusTIozs.--42. Deduce the formula for power. Repeat the formula on principle of general rule (Art. 37). ï~~THE LEVER. 61 Ex. 1. Suppose, in a lever of the third order, the long arm to be 8 ft., short arm 2 ft., and the weight 40 lbs. Required the power. Ans. P= 160 lbs. Wx L 40 X8 10 SoLuTrON.-Formula, P= WXL 408 160. AS 2 2. In a lever of the third order, the short arm is 12 ft., the long arm 18 ft., and the weight 620 lbs. Required the power. Ans. P=930 lbs. 3. What power, in a lever of the third order, will balance a weight of 41 tons, the long arm being 6 ft., and the short arm 2 ft. 6 in.? Ans. P= 10, tons. NOTE.---The answer, obtained to any example connected with the lever, is known to be correct, when the product of the connected numbers are equal. ART. 43.-Since (Art. 41) WX L=PXS8, Dividing by L,...... W PS L. Hence, To find the weight, the power, and the length of the long and short arms being given, we have the following formula and rule: PxS FORMULA,... =. RULE.-M/ltiply the power into the length of the short arm, and divide the product by the length of the long arm. Ex. 1. What weight, in a lever of the third order, will a power of 64 lbs. balance, the long arm of the lever being 12 ft., the short arm 3 ft.? Ans. W= 16 lbs. PX S 64 X3 SOLUTION.-Formula, W= ---= - 16 lbs. L 12 QUErSTIow.-43. Repeat the rule and formula for finding the weight. When the weight is required, what three numbers will be given? What two are connected? Repeat general rule (Art. 37). ï~~62 EXERCISES IN NATURAL PIHILOSOPHY. 2. In a lever of the third order, suppose the power to be 181 lbs., the long arm 81 ft., the short arm 21 ft. Required the weight. Ans. W= 63 lbs. 3. In a lever of the third order, the power is 18 tons, the long arm 36 ft., the short arm 12 ft. 6 in. Required the weight. Ans. W=6' tons. ART. 44. Since (Art. 41) PXS= WXL, Dividing by P,..... S L Therefore, To find the length of the short arm, the power, weight, and the length of the long arm being given, we have the following formula and rule: WX L FORMULA.... ' P RULE. -Aultiply the W ei ht into the length of the long arm, and divide the product by the power, the quotient will be the required length of the short arm. Ex. 1. Required the short arm of a lever of the third order, the long arm being 15 ft., the weight 45 lbs., and the power 135 lbs. Ans. S= 5 ft. ~x L 45 X15 SoLUTION.-Formula, S= XL 45X 5 -5. P 135 2. Suppose, in a lever of the third order, the weight to be 320 lbs., the power 2,560 lbs., and the long arm 24 ft. Required the length of the short arm. Ans. S= 3 ft. 3. Required the short arm of a lever of the third order, the long arm being 36 ft., the weight 400 lbs., and the power 1,800 lbs. Ans. S=8 ft. QUESTIO* s.-44. Deduce the forma for finding the short arm. Repeat the formula and rule. ï~~THE LEVER.. 63 Art. 45. Since (Art. 41) WxL= PX S, PXS8 Dividing by W,.... L P- X. Hence, To find the length of the long arm, the short arm, power, and weight being given, we have the following formula and rule: PxS FORMULA,... RuLE.- YIultIiply the power into the length of the short arm, and divide the product by the weight; the quotient will be the required length of the long arm. Ex. 1. Required the length of the long arm of a lever of the third order, the short arm being 4 feet, the weight 18 pounds, and the power 144 pounds. Ans. L=32 ft. PXS 144X4 SOLUTION.-Formula, L = = 18i = 32. TW 18 2. What must be the long arm of a lever of the third order, that a power of 420 pounds may be balanced by a weight of 60 pounds, the length of the short arm being 7 feet? Ans. L=49 ft. 3. Required the length of the long arm of a lever of the third order, the length of the short arm being 5 feet, the weight 181 pounds, and the power 934 pounds. Ans. L= 25 ft. ANALYSIS. ART. 416. The method of solving questions by analysis, cannot be valued too highly, inasmuch as it, above all other methods, tends to invigorate and strengthen the reasoning powers, and lead the mind on, step by step, in a regular and connected course QUESTos.-45. Deduce the formula for finding the long arm. Repeat the formula and rule. ï~~64 EXERCISES IN NATURAL PHILOSOPHY. of logical reasoning, until it arrives at required results. Great care should be taken to express properly the analytical solution of questions; for, to analyze is one thing, and to express properly the analysis, is quite a different thing. We will now give a few examples of the solution of questions by analysis, after performing which, the student will turn back and solve in a similar manner the preceding questions connected with the lever. Ex. 1. What power, in a lever of the first order, will balance a weight of 24 pounds, the length of the short arm being 3 feet, and that of the long arm 12 feet? Ans. P=-6 lbs. SoLUTION.-It is evident that the P and W will balance each other when they are equal, the length of the arms to which they are respectively attached, at the same time being equal. Hence, in the above example, a P of 24 pounds with a leverage of 3 feet, will just balance the W. If therefore, a leverage of 3 feet require a P of 24 pounds to balance the W, a leverage of 1 foot will require a P of 3 times 24 pounds =-72 pounds, and a leverage of 12 feet will require. as great a P as the leverage of 1 foot, or -'j of 72 pounds =6 pounds. Stated thus: 24 lbs. X3X j_=6 lbs. 2. What W in a lever of the first order will balance a P of 6 pounds, the length of the short arm being 3 feet, and that of the long arm 12 feet? Ans. W= 24 lbs. SOLUTrIoN.-6 lbs. X 12 X -=24 lbs. That is, if a leverage of 12 ft. require a TW of 6 lbs. to balance the P, a leverage of 1 foot will require 12 times as great a W as the leverage of 12 feet, and a leverage of 3 feet, one third as great a weight as the leverage of 1 foot. ï~~THE LEVER. 65 3. Suppose, in a lever of the first order, the W to be 24 lbs., the P 6 lbs., and the length of the long arm 12 ft.; required the length of the short arm. Ans. 3 ft. SOLUTION.. 12 ft. X6 X j = 3 ft. That is, if a weight of 6 lbs., to balance the power, require a leverage of 12 ft., 1 lb. will require a leverage 6 times as long, and 24 lbs. 2 as long as that required by 1 lb. 4. Suppose, in a lever of the first order, the weight to be 24 lbs., the power 6 lbs., and the length of the short arm 3 ft.; required the length of the long arm. Ans. 12 ft. SOLUTION.. 3 ft. X24X I= 12 ft. 5. What power, in a lever of the third order, will balance a weight of 400 lbs., the length of the long arm being 12 ft., and that of the short arm 8 ft.? Ans. 600 lbs. SOLUTION. 400 lbs. X 12 X k= 600 lbs. 6. In a lever of the third order, the power is 80 lbs., weight 60 lbs., and length of the short arm 18 ft.; required the length of the long arm. Ans. 24 ft. SOLUTION.. 18 ft. X80 X Ea=24 ft. 7. In a lever of the third order, the long arm is 8 ft., short arm 6 ft., and weight 180 lbs.; required the power. Ans. 240 lbs. SOLUTION. 180 lbs. X 8 X 240 lbs. 8. In a lever of the third order the power is 800 lbs., and the long arm to the short arm as 3: 8; required the weight. Ans. 300 lbs. SOLUTION. 800 lbs. X3 X =300 lbs. 6 ï~~66 EXERCISES IN NATURAL PHILOSOPHY. ART. 47. THE COMPOUND LEVER. iThc compound lever consists of the union of two or more simple levers, as represented in the figure. The simple levers composing a compound lever, may be of the same or of different orders. The following figure exhibits a compound lever composed of three simple levers of the first order. Fig. 10. o 0 It will be seen, by observing the figure, that the weight or pressure exerted by the first lever, acts as a power upon the second, and the weight or pressure exerted by the second lever, acts as a power upon the third, &c. ART. 48. Let P represent the power attached to the first lever, and W the weight attached to the third lever which this power will balance. Let p, p; p;' represent the arms of the levers on the side of the fulcrum next to the power, and w, w', w' the arms on the side next to the weight. Now we shall find (Art. 38) the weight or pressure exerted by the first lever to be X Ã~; and since this pressur; n i c ti r s sure acts as the power upon the second lever by PXp, multiplying by p, the long arm of the second w QuEsTionrs.-47. What is a compound lever? Of what order are the simple levers composing a compound lever? What acts as the P upon the second lever?---Third lever?-Fourth lever, &c.? 48. What does pp',p" and w, w, wu;' represent? Deduce the formulas for P and W. Repeat the rules. ï~~THE LEVER. 67 lever, and dividing the product by w', the short arm, we shall have P Xp xp' we shall have,, as the force exerted by the w xw second lever, or power acting upon the third lever. PXp X p' p Again, if w X7 be multiplied by p" the long w Xw arm of the third lever, and the product be divided by w, the short arm of the third lever, we shall have PxpXp'x >Kp V IX Xw X, as the weight attached to the third w"xw Xw lever which the power P, attached to the first lever, will balance. In a similar manner shall we find the power required to balance the weight, equal to W X uX w' x w" WXwXwXW" pXp' Xp" " Hence, In the compound lever, tojfind the power and weight, we have the following formulas and rules: F WXwXwXw" PXpXp' X p" FORMULAS. TL =_ pXpXP wXw'Xw" RULE 1.-Multiply the weight into the product of all the arms on the side next to it, and divide the result by the product of all the arms on the side next to the power; the quotient will be the required power. RULE 2.-Multiply the power into the product of all the arms on the side next to it, and divide the result by the product of all the arms on the side next to the weight, the quotient will be the required weight. ART. 49. If both members of the equation, WXw X w' X w" P XpX,,, expressing the formula for the power, be multiplied by pXpXp, we shall have power, be multiplied by p XTp,' Xp", we shall have ï~~68 EXERCISES IN NATURAL PHILOSOPHY. PXp Xp' Xp" = WVXw Xw' X w", as the equation of equilibrium. Therefore, The power and weight will balance each other, when the power multiplied into the product of all the arms on the side next to it, is equal to the weight into the product of all the arms on the side next to the weight. Hence (Art. 36) the powers will be connected, or associated with the product of all the arms on the side next to it, and the weight with the product of the arms on the side next to the weight. Ex. 1. In a compound lever of the first order, suppose the three long arms, or the arms on the side next to the P, to be 8, 12, and 16 ft., respectively; the short arms, or the arms on the side next to the weight, 2, 3, and 4 ft., and the P 30 lbs. Required the W. SoLUTION.--Formula, W _PXXp'Xp"30X8X12X161,920 lbs. wXw'Xw" 2X3X4 2. Suppose the three long arms of a compound lever of the first order, each to be 20 ft., the short arms each 4 feet long, and the P 12 tons. Required the W. Ans. TVW= 1,500 tons. 3. What W, in a compound lever of the first order, will a P of 16 lbs. balance, the three long arms each being 16 ft., and the short arms each 2 ft. long? Ans. W= 8,192 lbs. 4. What P, in a compound lever of the first order, will a W of 1,500 tons balance, the three long arms being each 20 ft., and the three short arms each 4 ft. long? Ans. P= 12 tons. QUEsTIos.-49. Deduce the equation of equilibrium. Repeat the equation of equilibrium. With what is the power in the compound lever connected? With what is the weight connected? When the power is required, what will be given?--When the weight? Repeat the formu. las for power and weight on principle of general rule (Art. 37). ï~~THE LEVER. 69 SOLUTION.-Formula, WXwXw'X w" 1500X4X4X4 1 - -....... = 12 tons. pXp'Xp " 20>420X20 5. What P, in a compound lever, composed of levers of different orders, will a W of 60 lbs. balance, the length of the arms, on the side of the P, being 12, 8, 4, aind 14 ft., respectively, and those on the side of the Wbeing 8, 16, 12, and 7 ft.? Ans. P= 120 lbs. SOLUTION.-Formula, SWXwXw'Xw"Xw"' 60X8X16X12X7120. P =,_- = -- - - -=120. p Xp'Xp"XP" 12X8X4X14 NOTE.-Since, in the preceding example, the P is greater than the W, or resistance, it follows (Art. 41) that one or more of the simple levers, composing the compound lever, must be of the third order. 6. What power, in a compound lever of the first order, will balance a weight of 24 tons, the length of the long arms being 4, 6, and 8 ft., respectively, and the length of the short arms 2, 3, and 4 ft.? Ans. P= 3 tons. 7. Required the weight which a power of 100 lbs. will balance, in a compound lever of the first order, the short arms being 3, 6, 12, and 18 ft., and the long arms 6, 12, 18, and 24 ft. long. Ans. W= 800 lbs. 8... What weight, in a compound lever consisting of 5 simple levers, will a power of 400 lbs. balance; the length of the arms, on the side next to the power, being 4, 6, 8, 12, and 24 ft., respectively, and the length of those on the side next to the weight, 1, 2, 4, 6, and 12 ft.? Ans. W= 38,400 lbs. 9. What power, in a compound lever, consisting of 6 simple levers of the first order, will balance a weight of 64,000 lbs.; the length of the long arms being as the numbers 1, 1, 2, 3, 4., and 5, the length ï~~70 EXERCISES IN NATURAL PHILOSOPHY. of the short arms as the numbers -, 1, 1, li, 2, and 21? Ans. P=1,000 lbs. 10. What power must an individual exert, at the end of a compound lever of the first order, to raise a weight of 2,400 lbs.; the long arms being 4, 6, 8, and 10 ft., respectively, and the short arms 2, 3, 4, and 5 ft. Ans. P= 150 lbs. CHAPTER X. THE FALSE BALANCE. ART. 50. If the fulcrum be placed at the center of a lever, it is evident that the two weights, in order to balance, must be equal. If, on the contrary, the fulcrum be situated nearer one extremity of a lever than the other, thereby making the arms of the lever unequal, the weights attached to the arms, must, to balance, be unequal; the greater weight being attached to the short arm, and the less weight to the long arm. Balances, for fraudulent purposes, are frequently constructed on this principle, having the arm of the lever from which the body to be weighed is suspended, longer than that from which the weight or counterpoise is suspended, thereby causing the body to balance a weight heavier than itself, or, in other words, to weigh more than its true weight. Such a balance is called afalse balance. To ascertain whether a balance is true or false, having previously balanced a body, cause it and the weights to exchange places. If, after the exchange, there is an equilibrium, the balance is true; if otherwise, false. QUSTiO.--50. Upon what principle is a false balance constructed? ï~~THE FALSE BALANCE. 71 ART. 51. Let us now proceed to deduce a formula and rule for finding, by means of a false balance, the true weight of a body. The annexed diagram exhibits a false balance, the dish, A, being suspended from the long arm, and the dish, B, from the short arm. Fig. 11. B A Let L represent the long arm of the balance, S the short arm, W the true weight of a body; 1 the weight placed in the dish A, to balance the body in the dish B, and s the weight placed in the dish B, to balance the body in the dish A. Then, when the body is placed in the dish B, we shall have (Art. 36) Lxl its weight, or.... W=; When placed in the dish A, SXs we shall find its weight, or W= L; Multiplying together the LXIXSXs two equations,.. SxLW -lXs; --SX L Extracting the square root, W-= dlxs. Hence, To find the TRUE weight of a body, by a false balance, we have the following formula and rule: QUESTIONs.--51. Deduce the formula for finding the true weight of a body by a false balance. Repeat the rule. ï~~72 EXERCISES IN NATURAL PHILOSOPHY. FORMULA,... = //>Xs. RuLE.-Find the weight of the body in each scale, multiply together the two weights found, and extract the square root of the product. Ex. 1. A body weighs 7 lbs., at one end of a false balance, and 91 lbs. at the other. Required the true weight of the body. Ans.Weight= 8 lbs. SoLUTIo.-Form., 'V=,flXs=,/7X9= -64= 8. 2. Required the true weight of a body which weighs 12 lbs. at one end of a false balance, and 10- lbs. at the other. Ans.Weight = 11 lbs. 3. Suppose a body weighs, at one end of a false balance, 16 lbs., and at the other 14 lbs. I oz. Required its true weight. Ans.Weight = 15 lbs. 4. An individual going to a store, orders 49 lbs. of sugar; mistrusting the honesty of the merchant, he orders that the sugar, after it has been weighed, be placed in the opposite scale; it there weighs only 36 lbs. Required the true weight of the sugar. Ans.Weight.= 42 lbs. CHAPTER XI., THE WHEEL AND AXLE. A.T. 52. The wheel and axle consists of a cylinder or axle passing through and attached to a larger circle of wood, termed the wheel. We have, in Fig. 12, an end view of the wheel and axle, A C representing the semidiameter of the axle, B C that of the wheel. The weight is ï~~THE WHEEL AND AXLE. 73 attached to the axle by Fig. 12. means of a rope passing around the axle; and in a similar manner, is the power applied to the wheel. D c d A It will be seen, by examining the figure, that the wheel and axle acts on the principle of the simple lever of the first order-C acting as the fulcrum, A C, or the semidiameter of the axle, acting as the short arm, and B C, or the semidiSv ameter of the wheel, as the long arm. ART. 53. If the semidiameter, or (since the ratio of the diameters, and also of the circumferences, is the same as that of the semidiameters), if the diameter or the circumference, of the axle be one half that of the wheel, then the power to produce an equilibrium, will be one half the weight; if the diameter or circumference of the axle be one eighth that of the wheel, the power will be one eighth the weight, &c. That is, in the first case, a power of one pound will balance a weight of two pounds; and in the second, a power of one pound will balance a weight of eight pounds. ART. 54. Let P represent the power, W the weight, D the diameter, and C the circumference of the wheel, and d the diameter, and c the circumference of the axle. QrUESTIOs.-62. Describe the wheel and axle. Upon what pinciple does it act? 7 ï~~74 EXERCISES IN NATURAL PHILOSOPHY. Now, since the wheel and axle acts on the prineciple of the lever of the first order, the equation of equilibrium will be the same as that of the lever, and we shall have (Art. 30)-remembering that the ratio of the diameters or circumferences is the same as that of semidiameters, the semidiameters acting as the arms of the leverPX D= WXd Equations of equilibrium of the and P X 0= WX c wheel and axle. That is, In the wheel and axle, there is an equilibrium when the power multiplied into the diameter or circumference of the wheel, equals the weight multiplied into the diameter or circumference of the axle. Hence, in the wheel and axle, the power is connected with the diameter or circumference of the wheel, and the weight with the diameter or circimference of the axle. ART. 55. Since (Art. 54) PxD= Wxd. (1); And....... PXC=Wxc. (2); Dividing (1) by D,... WX d.... I -., D. (3); Dividing (2) by C,.... P=. W c (4) Therefore, To find the POWER, the weight and the diameters or circumferences of the wheel and axle being given, we have the following formulas and rules: (=Wx d P= D ' FORMULAS,.. P- W Xc.-. Quzsrzows.---54. Repeat the equation of equilibrium. Deduce the formulas for finding the power. Repeat the rules for finding the power. With what is the power connected?-Weight? When the power is required, what numbers will be given? Repeat the formulas for power, weight, diameter of the wheel, and that of the axle, on principles of general rule, Art. 37. ï~~THE WHEEL AND AXLE. 75 RU LE.-1. Multiply the weight into the diameter of the axle, and divide the product by the diameter of the wheel. RULE.-2. Multiply the weight into the circumference of the axle, and divide the product by the circumference of the wheel. Ex. 1. In a wheel and axle, the diameter of the wheel is 8 ft., and that of the axle 2 ft.; what power will balance a weight of 400 lbs.? Ans. Power = 100 lbs. SoLUTIoN.-Formula, Wxd 400X2 100 lbs. P = -= S -100 lbs. D 8 2. What power, in a wheel and axle, will balance a weight of 1,200 pounds; the diameter of the wheel being 9 ft., and that of the axle 14 ft.? Ans. Power =200 lbs. 3. What power, in a wheel and axle, will balance a weight of 16,200 lbs.; the diameter of the axle being 7 in. and that of the wheel 5 ft. 10 in.? Ans. Power = 160 lbs. 4. Suppose, in a wheel and axle, the weight to be 840 lbs., the diameter of the wheel 6 ft., and that of the axle 8 in.; required the power. Ans. Power = 93) lbs. 5. In a wheel and axle, the circumference of the axle is to that of the wheel as 2 to 96, the weight 4,800 lbs.; required the power. Ans.. Power = 100 lbs. 6. What power, in a wheel and axle, the circumference of the axle being 24 in., and that of the wheel 192 in., will balance a weight of 2,560 lbs. Ans. Power =320 lbs. ART. 56. Since (Art. 54) WXd= P xD. (1); And............ Wxc=PX C. (2); ï~~76 EXERCISES IN NATURAL PHILOSOPHY. PxD Dividing (1) by d,..... W= d. (3); PXC Dividing (2) by c,.... W-=. (4). c Hence, To find the WEIGHT, the power and the diameters or circumferences of the wheel and axle being given, we have the following formulas and rules: PxD d FORMULAS,..: f PX C Wc RULE.-1. Multiply the power into the diameter of the wheel, and divide the product by the diameter of the axle. RULE.-2. Multiply the power into the circumference of the wheel, and divide the product by the circumference of the axle. REMARK.-It will be seen, by observing the formulas PXD PX C T4r - and W=, that the power remaining the D same, the weight will vary as -, that is, as the ratio of the C diameter of the wheel to that of the axle; or as -, the ratio of the circumference of the wheel to that of the axle. Ex. 1. What weight, in a wheel and axle, will a power of 40 lbs. balance; the diameter of the wheel being 6 ft., and that of axle 2 ft.? Ans. Weight =120 lbs. -SOLUTION.-Formula, PXD 40X6 W = D 2 =--120 lbs. d 2 QuESTroN.-56. Deduce the formulas for weight. Repeat the rule, 1.-Rule 2. The power remaining the same, how will the weight vary? ï~~THE WHEEL AND AXLE. 77 2. What weight, in a wheel and axle, will a power of 320 lbs. balance; the diameter of the wheel being 7. ft., and that of the axle 11 ft.? Ans. Weight =1,600 lbs.. What weight, in a wheel and axle, will a power of 6 lbs. balance; the diameter of the wheel being to that of the axle as 18: 2? Ans. Weight =56J lbs. 4. In a wheel and axle, the diameter of the wheel is to that of the axle, as 13: 1j, the power is 200 lbs.; required the weight. Ans. Weight =1,800 lbs. 5. In a wheel and axle, the circumference of the axle is to that of the wheel, as 3: 270, the power 30 lbs.; required the weight. Ans. Weight =2,700 lbs. 6. What weight, in a wheel and axle, the circumference of whose axle is 18 inches; and that of the wheel 360 inches, will a power of 10 lbs. balance? Ans. Weight =200 lbs. 7. In a wheel and axle the power is 400 lbs., the circumference of the wheel to that of the axle as 5: 40; required the weight. Ans. Weight =3,200. ART. 57. Since (Art. 54) PXD= TVWXd;. (1) And...... PXC=Wxc;. (2) Wx d Dividing (1) by P,.... D=;. (3) Dividing (2) by P,... 0=.. (4)c...P (4) Hence, To find the DIAMETER or CIRCUMFERENCE of the wheel, the weight, power, and the diameter or cirQUESTIoxs.-57. Deduce and repeat the formulas for finding the diameter and circumference of the wheel. Repeat the rules. The diameter or circumference of the axle remaining the same, how will the diameter or circumference of the wheel vary 1 ï~~78 EXERCISES IN NATURAL PHILOSOPHY. cumference of the axle being given, we have the following formulas and rules: Wx d JP FORMULAS,.. ---WX c C =.4x P RULE 1. To FIND THE DIAMETER OF THE WHEEL. - Multiply the weight into the diameter of the axle, and divide the product by the power. RULE 2. To FIND THE CIRCUMFERENCE OF THE WHEEL. Multiply the weight into the circumference of the axle, and divide the product by the power. WXd REMARK. -By observing the formulas D=---p and WXc - p, it will be seen, that d the diameter, or c the circumference of the axle remaining the same, D the diameter, or C the circumference of the wheel, will increase or dimiW nish as -p, the ratio of the weight to the power, increases or diminishes. Ex. 1. Suppose, in a wheel and axle, the diameter of the axle to be 2 ft., and the weight to the power as 12: 4; required the diameter of the wheel. Ans. Diameter =6 ft. WXd 12x2 SoLUTIoN.--Formula, D=- = 12X2 6 ft. P = 4 = 2. Suppose, in a wheel and axle, the power to be 20 lbs., the weight 100 lbs., and the diameter of the axle g in.; required the diameter of the wheel. Ans. Diameter =3 ft. 4 in. 3. Required, in a wheel and axle, the diameter of the wheel, the diameter of the axle being 10 in., and the weight to the power as 12: 1. Ans. Diameter =-6 ft. 8 in. ï~~THE WHEEL AND AXLE. 79 4. Required, in a wheel and axle, the diameter of the wheel; that of the axle being 6j in., and the weight being to the power as 3j: 1 1. Ans. Diameter =-1 ft. 7i in. 5. In a wheel and axle, the power is to the weight as 2: 36, the circumference of the axle 5 inches; required the circumference of the wheel. Ans. Circumference =90 in. 6. In a wheel and axle, the power is 40 lbs., the weight 360 lbs., circumference of the axle 8 in.; required the circumference of the wheel. Ans. Circumference =6 ft. 7. A person wishing to construct a wheel and axle, for the purpose of raising a large stone, has a stick of timber, from which he can procure an axle, 9 inches in diameter. He wants to know of what diameter the wheel should be made, in order that a power of 320 lbs., applied to it, shall balance the stone, weighing 2,560 lbs. Ans. Diameter = 6 ft. ART. 58. Since (Art. 54) WXc=-PX C;.. (1) And......... Wxd=PxD;.. (2) Px C Dividing (1) by W,.. c=; (3) tt t L4P XD (2) ".. d= PW (4) Hence, To find the cIRCUMFERENCE, or DIAMETER of the axle, the circumference or diameter of the wheel, and the power and weight being given, we have the following formulas and rules: PXC FORMULAS,. Px D W RuLE 1.-To FID THE CIOUMFERENCE OF THE AER* ï~~80 EXERCISES IN NATURAL PHILOSOPHY. Multiply the power into the circumference of the wheel, and divide the product by the weight. RULE 2.- To FIND THE DIAMETER OF THE AXLE. Multiply the power into the diameter of the wheel, and divide the product by the weight. REMARK.-By observing the formulas d=Px D, and c = W PX it will be seen that D, the diameter, or C, the circumW ference of the wheel, remaining the same, d, the diameter, or c, the circumference of the axle, will increase or diminish as P the ratio of the power to the weight, increases or diminishes. Ex. 1. Required, in a wheel and axle, the diameter of the axle; the power being to the weight as 3:9, and the diameter of the wheel being 6 ft. Ans. Diameter= 2 ft. PxD 3X6 SoLuTrio.-Formula, d = -- 2 ft. 2. In a wheel and axle, the power is 460 lbs., the weight 4,140 lbs., and the diameter of the wheel 5 ft. 3 in. Required the diameter of the axle. Ans. Diameter = 7 in. 3. In a wheel and axle, the power is to the weight as 24: 171; the diameter of the wheel is 2 ft. 11 in. Required the diameter of the axle. Ans. Diameter= 5 in. 4. In a wheel and axle, the power is to the weight as 1: 32; the circumference of the wheel 256 in. Required the circumference of the axle. Ans. Circumf. =8 in. QUESTION.--58. Deduce, and repeat the formulas for the diameter and circumference of the axle. Repeat the rule. The diameter or circnuf brence of the wheel remaining the same, how will the diaheter of the axle vary? ï~~THE WHEEL AND AXLE. 81 5. In a wheel and axle, the power is 400 lbs., weight 3,600 lbs., the circumference of the wheel 180 in. Required the circumference of the axle. Ans. Circumf.= 20 in. CHAPTER XII. COMPOUND WHEEL AND AXLE. ART. 59. The compound wheel and axle consists of a system of wheels, connected together by means of cogs or bands. The annexed figure exhibits a system of wheels connected by means of cogs. Fig. 13. The power is attached to the wheel A. This ï~~82 EXERCISES IN NATURAL PHILOSOPHY. wheel, in revolving, turns, by means of the cogs on its axle d, the wheel B, which, in its turn, and in a similar manner, turns the wheel C, around whose axle a rope is coiled, to which the weight is attached. Since every single wheel and axle (Art. 52) operates on the principle of the simple lever, it will readily be seen, that a system of wheels, connected as above, or by means of bands, acts as a compound lever, whose long arms are represented by the semidiameters of the wheels, and short arms by the semidiameters of the axles. ART. 60. LetP represent the power, Wthe weight, D, D', D", &c. (we use the diameters instead of the semidiameters, since their ratios are equal) the diameters, and C, C', C", &c. the circumferences of the wheels; or, in other words, the long arms of the lever; also, let d, d', d", &c. represent the diameters, and c, c', c", &c. the circumferences of the axles, or short arms of the levers. Now, since a system of wheels acts on the principle of the compound lever, we shall have (Art. 49), PxDxD'xD", &c.= WXdXd'Xd", &c.quatiois and, of P X C X C' X C", &c. = WX c X c'X c", &c.f Equilibrium. Hence, IN A SYSTEM OF WHEELS THERE IS AN EQUILIBRIUM: 1. When the power into the product of all the diameters of the wheels equals the weight into the product of all the diameters of the axles. 2. When the power into the product of all the circumferences of the wheels equals the weight into the product of all the circumferences of the axles. QuzsTroxs.--59. On what principle does the compound wheel and axle act? Explain the figure. 60. Deduce and repeat the equation of equilibriunm. With what is the power connected?!-Weight? ï~~COMPOUND WHEEL AND AXLE. 83 It will be seen, therefore, that in a system of wheels, the power is connected with the product of the diameters or circumferences of the wheels, and the weight with the product of the diameters or circumferences of the axles. ART. 61. Since (Art. 60) PxDxD'xD"= Wxdxd'Xd"; (1) And... P CXC'XC" =--WXcxc'xc";(2) Wxdxd'xd' Dividing(1) by DXD'XD", P= DXDXD " (2) "C X C'XC", P= WXcXc'Xc" " (4) (2)~O C 'XC, C" ) Hence, To find the POWER, the weight and the diameters, or circumferences of the wheels and axles being given, we have the following formulas and rules: SWxdxd'Xd" FoatLs, DXD'xD" ' FORMULAS,cxc'xc WX c x c'x c" 1=.. X C' X C" t RULE 1.---Multiply the weight into the product of the diameters of the axles, and divide the result by the product of the diameters of the wheels. RULE 2.-Mdultiply the weight into the product of the circumferences of the axles, and divide the result by the product of the circumferences of the wheels. REMARK.--By observing the above formulas, it will be seen that the power required to balance a given weight, will vary as the ratio of the product of the diameters of the axles to the product of the diameters of the wheels; or, as the ratio of the QUEsTIoN.--61. Deduce, and repeat the formulas for finding the power. Repeat the rules. How does the power, required to belanc a given weight, vary? ï~~84 EXERCISES IN NATURAL PHILOSOPHY. product of the circumferences of the axles to the product of the circumferences of the wheels. Ex. 1. Suppose, in a compound wheel and axle, the weight to be 7,200 lbs., the diameters of the axles, 1, 11, and 2 ft., respectively, and the diameters of the wheels, 2, 3, and 6 ft. Required the power necessary to produce an equilibrium. Ans. Power --=600 lbs. SoLUTIoN.-Formula, Wxdxd'xd" 7200X1x1IX2 P- 2600 lbs. D XD'XD" 2X3X6 2. What power, in a compound wheel and axle, will balance a weight of 2,400 lbs., the diameters of the axles being 1, 2, and 21 ft., and those of the wheels, 3, 4, and 5 feet? Ans. Power = 300 lbs. 3. Required the power, in a compound wheel and axle, necessary to sustain a weight of 100 tons, the diameters of the axles being as the numbers 2, 3, and 4, and the diameters of the wheels as the numbers 4, 5, and 9. Ans. Power = 131 tons. 4. What power, in a system of wheels connected by cogs or bands, will balance a weight of 480 lbs., the diameters of the axles being as the numbers 1, 2, 2, and 3, and those of the wheels as the numbers 2, 5, 6, and 12? Ans. Power = 5 lbs. 5. What power, in a compound wheel and axle, will balance a weight of 3,600 lbs., the,circumferences of the axles being as the numbers 3, 5, and 10, and those of the wheels as the numbers 12, 15, and 25? Ans. Power = 120 lbs. 6. What power, in a compound wheel and axle, would be required to raise a weight of 4,800 lbs., the circumferences. of the axles being 15, 20, and 45 inches, respectively, and those of the wheels 45, 80, and 225 inches? Ans. Power - 80 lbs. ï~~COMPOUND WHEEL AND AXLE. 85 ART. 62. Since (Art. 60) IVWxdx'd'Xd"=PXDXD'XD"; (1) And... WXcXc'Xc" =PX C XC'X C"; (2) PXDxD'xD" Dividing (1) by dxd'Xd", W= dxd'X d" (3) (2"PX C C'XC Ct (2) " c x c 'x c", wc=,. (4) cXc' Xc Hence, To find the WEIGHT, the power, and the diameters, or circum ferences of the wheels and axles being given, we have the following formulas and rules: ( _PXDXD'XD"FORMULAS,., Px C x C'xC" RULE 1.- ultiply the power into the product of the diameters of the wheels, and divide the result by the product of the diameters of the axcles. Rvxr 2.-iIdltiply the power into the product of the circumferences of the wheels, and divide the result by the product of the circumferences of the axles. REMARK.--By observing the above formulas, it wil be seen that the weight which a given power will balance, will vary as the ratio of the product of the diameters of the wheels to the product of the diameters of the axles; or, as the ratio of the product of the circumferences of the wheels to the product of the circumferences of the axles. Ex. 1. What weight, in a compound wheel and axle, will a power of 5 lbs. balance, the diameters of the wheels being 2, 5, 6, and 12 ft., respectively, and those of the axles., 2, 24, and 3 ft.? Ans. Weight = 480 lbs. QuasTiows.-62. Deduce, and repeat the formulas for finding the weight. Repeat rule 1.-Rule 2. The power remaining unchanged, how will the weight vary I ï~~86 EXERCISES IN NATURAL PHILOSOPHY. SoLUTION.-Formula, PxDxD'xD" 5X2X5X6X12 d= -- = 480 lbs. dXd'Xd" X2X2 X3 2. What weight, in a compound wheel and axle, will a power of 40 lbs. sustain, the diameters of the wheels being as the numbers 6, 8, and 12, and those of the axles as 2, 3, and 4? Ans. Weight = 960 lbs. 3. What weight, in a compound wheel and axle, will a power of 2 lbs. equipoise, the diameters of the wheels being 1, 2, 3, and 4 ft., respectively, and those of the axles 4, 6, 9, an3 16 inches? Ans. Weight = 288 lbs. 4. What weight, in a compound wheel and axle, will a power of 10 lbs. equipoise, the diameters of the wheels being as the numbers 4, 6, and 8, and those of the axles 1, 1, and 2? Ans. Weight = 1,920 lbs. 5. What weight, in a compound wheel and axle, will a power of 4,000 lbs. balance, the circumferences of the axles being 9, 12, and 16 inches, respectively, and those of the wheels 36, 60, and 96 inches? Ans. Weight = 480,000 lbs. 6. In a compound wheel and axle, the circumferences of the axles are as the numbers 1, 3, and 6, and those of the wheels as 6, 12, and 30, the power 140 lbs. Required the weight. Ans. Weight = 16,800 lbs. ï~~THE PULLEY. 87 CHAPTER XIII. THE PULLEY. ART. 63. A pulley is a flexible cord, attached to a weight, and passing around Fig. 14. a grooved wheel turning upon a center. The mechanical efficacy of the pulley, is owing to the fact, that the tension of the same flexible cord, is the same throughout its entire length. Pulleys are of two kinds, A fixed and movable. ART. 64. Fig. 14 represents a fixed pulley, in which there is no mechanical advantage; for since the parts of the string, A and B, have the same ten- P sion, the power, P, must be equal to the weight, W. Fig. 15 repre- Fig. 15. sents a single movable pulley, sometimes called a runner; in this, the power is half the weight. For, the whole A weight is sustained by the two parts of the string, A and B; and since the tension of the string is the same throughout its entire length, ï~~88 EXERCISES IN NATURAL PHILOSOPHY. the tensions of the parts A, B, and C, will be equal, and B will sustain one half the weight. Hence the power attached to the part C, whose tension is equal to B, must be one half the weight. ARt. 65. Fig. 16 represents a system of pulleys, consisting of two blocks, each of which has two simple pulleys. It will be seen, from the figure, that the weight is equally sustained by each of the four parts of rope A, B, C, and D, passing between the movable and fixed blocks. Hence, the power, P, attached to the part of the rope E, which has an equal tension with each of the parts A, B, C, and D, must be one fourth of the Fig. 16. weight, W. And, generally, In a system of pulleys having but one rope, and one movable block, the weight is as many times the power, as there are different parts E of rope engaged in supporting the movable block. P Let W represent the weight, P the power, and n the number A B c of parts of rope engaged in sustaining the movable block. Then, Tojind the weight, in a system of pulleys having but one rope, we have the following formula and rule: FORMULA, W P X n. RULE. l Multiply the power into the number of parts of rope employed in sustaining the movable block. The product will be the required weight. QvUESTx~o.-63. What is a pulley? To what is the mechanical efficacy of the pulley owing? Pulleys are of how many kinds? What are they? 64. Describe a fixed pulley. Why is there no mechanical advantage in a fixed pulley? Explain figures 15 and 16. ï~~THE PULLEY. 89 Ex. 1. What weight, in a system of pulleys containing a single rope, will a power of 40 lbs. balance, the number of ropes employed in sustaining the movable block being 6? Ans. Weight = 240 lbs. SoLUTIoN.-Form.,. W=P X n =40 X 6 =240 lbs. 2. What weight, in a system of pulleys containg a single rope, will a power of 64 lbs. balance, number of parts of rope employed in sustaining the movable block being 8? Ans. Weight = 512 lbs. NOTE.-The weight, W, includes both the suspended weight, and that of the movable block. 3. What weight, in a system of pulleys of which four are movable, and in which the same string goes around all the pulleys, will a power of 160 lbs. balance? Ans. Weight = 1,280 lbs. NOTE.-It will be seen, by referring to Fig. 16, that the number of parts of rope engaged in sustaining the movable block, is twice the number of movable pulleys; hence, in the above example, the number of parts which support the movable.block is eight. 4. What weight, in a system of pulleys of which six are movable, and in which the same string passes around all the pulleys, will a power of 8 lbs. counterpoise? Ans. Weight = 102 lbs. AnRT. 66. Since (Art. 65) PXn= W, Dividing by n,... p=W n Hence, To find the power in a system of pulleys containing but one rope, we have the following formula and rule: QuzsTIoxs.--65. Deduce the formula for finding the weight. Repeat the formula and rule. 8 ï~~90 EXERCISES IN NATURAL PHILOSOPHY. W FORMULA,.. W n RuLE.-Divide the weight by the number expressing the parts of rope engaged in supporting the movable block, and the quotient will be the required power. Ex. 1. What power, in a system of pulleys containing a single rope, will be required to balance a weight of 720 lbs., the number of parts of rope engaged in supporting the movable block, being 8 Ans. Power = 90 lbs. 2. What power, in a system of pulleys of which three are movable, and in which the same string passes around all the pulleys, will balance a weight of 2,400 lbs.? Ans. Power = 400 lbs. 3. Required the power necessary to sustain a weight of 3,780 lbs., in a system of pulleys containing a single rope and six movable pulleys. Ans. Power =315 lbs. 4. What power, in a system of pulleys containing a single rope, will counterpoise a weight of 4 tons, the number of movable pulleys being 8? Ans. Power = ton, or 560 lbs. ART. 67. Since (Art. 65) PXn= W; Dividing by P,..... n= That is, the number of parts of rope which act in sustaining the movable block, is equal to the weight divided by the power, or, in other words, is equal to the ratio of W: P. Now, since, in a system of pulleys, similar to that represented by Fig. 16, the number of movable pulleys is half the number of parts of rope used in sustaining the movable block-if we represent by I QuzsTWrs.-66. Deduce, and repeat the formula for finding the power. Repeat the rule. ï~~THE PULLEY. 91 m, the number of movable pulleys-then, To find the number of movable pulleys, having given the power and weight, we shall have the following formula and rule: W FORMULA,... m=W. RuLE.-Divide the weight by the power, and take half the quotient. Or, divide the weight by twice the power; the result in either case will be the required number of movable pulleys. Ex. 1. In a system of pulleys, in which the same string passes around all the pulleys, a weight of 198 lbs. is balanced by a power of 161 lbs.; required the number of movable pulleys. Ans. m=-6. W 198 198 SoLuTion.-Formula, m=- = --6. 2P 2X 16 33 2. Required the number of movable pulleys in a system containing a single rope, and in which the power is to the weight as 2: 12. Ans. m=3. 3. Required the number of movable pulleys in a system containing a single rope, and in which a power of 221 lbs. will counterpoise a weight of 270 lbs. Ans. m= 6. 4. In a system of pulleys, in which the same string passes around all the pulleys, a power of 480 lbs. balances a weight of 9,600 lbs.; required the number of movable pulleys. r Ans. m=10. ART. 68. In Fig. 17, we have represented a system of pulleys in which each movable pulley is supported by a separate string, one end of each string being attached to a fixed beam. QUZsTlows.-67. Deduce the formnula for finding the number of movable pulleys. Repeat the rule. ï~~o92 EXERCISES IN NATURAL PHILOSOPHY. Fig. 17. By this system, a greater mechanical advantage is obtained than by that in which the same string passes around each pulley. 8 D The part of the weight sustained by the fixed pulley, a, is equal to the power, F P. The movable o pulley, b, since it is ZX 2 P supported by the e strings B and D, the P tension of each of C which is equal to P, sustains a weig h t zxXP equal to twice P, or 2P. The tension of d the string C, is equal to the weight sustained by the pulley Sb. Hence, the movable pulley, c, supported by the strings C and F, the tension of each of which is equal to 2P, sustains a weight equal to twice 2P, or 2X 2P. The tension of the string E, is equal to the weight sustained by the pulley c. Therefore, the movable pulley d, supported by the strings E and G, the tension of each of which is 2 X2P, sustains a weight equal to twice 2 X 2P, or 2X2x2P. But, d, the last movable pulley, sustains the entire weight, W. Hence,.. W=--2X2X2P. Each additional movable pulley enables the QuzsIToWs.-68. Whi system of pulleys is Tepresented by Fig. 17? Explain the figure. ï~~THE PULLEY. 93 same power to sustain twice the previous weight. Hence, by referring to the expression for the weight, it will be seen, that the weight, W, equals the power, P, multiplied into 2 repeated as a factor as many times as there are movable pulleys. But repeating a number as a factor, any given number of times, is raising that number to a power denoted by the number of factors. ART. 69. Let n represent the number of movable pulleys in a system, in which each movable pulley is supported by a separate string fastened to a fixed beam. Then, To find the weight, the power, and number of movable pulleys being given, we have the following formula and rule: FORMULA,.. W=Px2". RULE.--Mtltiply the power into 2 raised to a power denoted by the number of movable pulleys, and the product will be the required weight. Ex. 1. What weight, in a system of pulleys represented by Fig. 17, will a power of 6 lbs. balance, the number of movable pulleys being 4? Ans. Weight =96 lbs. SoLUTION.-Formula, W=PX2n= 6X24=6X 16=96 lbs. 2. What weight, in a system of pulleys represented by Fig. 17, will a power of 244 lbs. balance, the number of movable pulleys being 5? Ans. Weight =7,808 lbs. 3. In a system of pulleys, represented by Fig. 17, the power is 24- lbs., the number of movable pulleys 6; required the weight. Ans. Weight =1,5461 lbs. 4. What weight, in a system of pulleys repreQuisTrows.-69. Deduce and repeat the formula for finding the weight. Repeat the rule. ï~~94 EXERCISES IN NATURAL PHILOSOPHY. sented by Fig. 17, will a power of 31 lbs. counterpoise, the number of movable pulleys being 7? Ans. 448 lbs. ART. 70. Since (Art. 69) PX2*= W; W Dividing by 2Q,.... 2P=R. Hence, To find the power, the weight and number of movable pulleys being given, we have the following formula and rule: W FORMULA,.. P=. RULE.-Divide the weight by 2 raised to a power denoted by the number of movable pulleys, and the quotient will be the required power. Ex. 1. What power, in a system of pulleys represented by Fig. 17, will counterpoise a weight of 640 lbs., the number of movable pulleys being 4? Ans. Power =40 lbs. SoLuTIo.-Formula, W 640 040 P= 2_4 =1o=40 lbs. 2" 24 1 6 2. What power, in a system of pulleys represented by Fig. 17, will balance a weight of 1,920 lbs., the number of movable pulleys being 5? Ans. Power =60 lbs. 3. What power, in a system of pulleys represented by Fig. 17, will counterpoise a weight of I ton, the number of movable pulleys being 7? Ans. Power =17 lbs. 8 oz. 4. What power, in a system of pulleys represented by Fig. 17, will balance a weight of 12 tons, the number of movable pulleys being 10? Ans. Power =26 lbs. 4 oz. Qu zsTros.-70. Deduce the formula for power. Repeat the rule. ï~~THE PULLEY. 95 ART. 71. The system of pulleys represented by Fig. 17 may be so modified, without increasing the number of ropes or movable pulleys, as to cause a given power to sustain a much greater weight. This modification is exhibited in Fig. Fig. 18. 18. Each rope, in- H stead of being fastened to the beam, H, passes around a fixed pulley, and is attached to the respective b block of each mov- a able pulley. The movable pulley B, B since it is supported by the three strings, d e I a, b, and c, the tension of each of which is equal to P, sustains P C a weight equal to 3P. The tension of the h string d, is equal to a the weight sustained by the pulley B. Hence, the mova- D ble pulley C, supported by the strings d, e, and f, the tension of w each of which is equal to 3P, sustains a weight equal to 3 X 3P. The tension of the string g, is equal to the weight sustained by the pulley C. Therefore the pulley D, supported by the strings g, h, and m, the tension of eqch of which is equal QUESTIONS.--71. How may a system of pulleys, represented by Fig. 17, be so modified, without increasing the number of ropes, as to increase its mechanical efficacy? Explain Fig. 18. ï~~96 EXERCISES IN NATURAL PHILOSOPHY. to 3X3P, sustains a weight equal to 3X3X3P. But the weight sustained by the pulley D, is equal to W. Hence,. W=3x3x3xP. That is, The weight equals the power multiplied by 3 repeated as a factor as many times as there are movable pulleys. ART. 72. Let n represent the number of movable pulleys in a system in which each movable pulley is supported by a separate string passing around a fixed pulley, and attached to the respective block of each movable pulley. Then, To find the weight, the power, and number of movable pulleys being given, we shall have the following formula and rule: FORMULA,.. W=Px3". RULE.-Multiply the power bry 3 raised to a power denoted by the number of movable pulleys, and the product will be the required weight. Ex. 1. What weight, in a system of pulleys represented by Fig. 18, will a power of 12 lbs. balance, the number of movable pulleys being 4? Ans. Weight =972 lbs. SoLUTION.-Formula, S W=PX3"- 12x34=-12X81 =972 lbs. 2. What weight, in a system of pulleys represented by Fig. 18, will a power of 300 lbs. balance, the number of movable pulleys being 5? Ans. Weight =32 -, tons. 3. What weight, in a system of pulleys represented by Fig. 18, will a power of 2 tons balance, the number of movable pulleys being 7? Ans. Weight =4,374 tons. 4. What weight, in a system of pulleys represented by Fig. 18, will a power of 42 lbs. balance, fe number of movable pulleys being 8? Ans. Weight =29,5241 lbs. ï~~THE PULLEY. 07 5. What weight could an individual capable of exerting a force of 300 lbs. raise, by means of a system of pulleys represented by Fig. 18, the number of movable pulleys being 3? Ans. Weight =8,100 lbs. ART. 73. Since (Art. 72) Px3Â~= W; W Dividing by 3.... P=--. Hence, To find the power, the weight and number of movable pulleys being given, we have the following formula and rule: W FORMULA,... --- 3n RuLE.-Divide the weight by 3 raised to a power denoted by the number of movable pulleys, and the quotient will be the required power. Ex. 1. What power, in a system of pulleys represented by Fig. 18, will balance a weight of 2,700 lbs., the number of movable pulleys being 3? Ans. Power = 100 lbs. SOLUTION.-Formula, W 2700 2700 P=-- 1-- --100 lbs. 3" 33 27 2. What power, in a system of pulleys represented by Fig. 18, will balance a weight of 243 tons, the number of movable pulleys being 4? Ans. Power =3 tons. 3. How much greater weight will a power of 400 lbs. sustain in a system of pulleys represented by Fig. 18, than in a system represented by Fig. 17, the number of movable pulleys being 8? Ans. Weight =1,125 tons. 4. How much greater weight will a power of 24 QVESTIONS.-72, 73. Deduce and repeat the formulas for finding the power and weight. Repeat the rules. 9 ï~~98 EXERCISES IN NATURAL PHILOSOPHY. lbs. balance$ in a system of pulleys represented by Fig. 18, and containing 4 movable pulleys, than in a system represented by Fig. 17, and containing 5 movable pulleys? Ans. Weight =1,176 lbs. ART. 74. We have represented, by Fig. 19, a powerful system of pulleys, in which each rope, after passing around its respective pulley, is attached to the weight. It will be seen, by obFig. 19. serving the figure, that...... the first rope sustains a \ - part of the weight equal to the power P. The tension of the second rope is twice that of the first, and hence, it susSP tains a part of the weight equal to 2P. The third rope, whose tension is twice that of AP the second, sustains a part of the weight equal to 4P. And, for a similar reason, the fourth rope sustains a part of the weight equal to 8P. Now, the entire weight, W, is sustained by the united tension of the P 2P. 4P P four ropes. Hence, W=P+2P +4P 8P;, WP being a factor of each term of the right hand member, W=PX (l+2+4+ 8). The numbers within the parenthesis form a geo ï~~THE PULLEY. 9 - metrical series, in which 1 is the first term, 2 the ratio, and the number of terms 4t equal to the number of ropes. The sum of the series 1+2+4+8, &c., is one less than 2 raised to a power denoted by the number of terms. If, therefore, n represent the number of ropes (number of terms), To find the weight, havirng the power and number of ropes given, we shall have the following formula and rule: FORMULA,.. W=P(2*-1). RULE.-From 2 raised to a power denoted by tie number of ropes, subtract 1, and multiply the remainder by the power-the quotient will be the required weight. Ex. 1. What weight in a system of pulleys represented by Fig. 19, will a power of 6 lbs. balance, the number of ropes being 4? Ans. Weight =90 lbs. SOLTION.-Formula, W= PX(2-1)= 6(24-1)=90 lbs. 2. What weight, in a system of pulleys represented by Fig. 19, will a power of 350 lbs. balance, the number of ropes being 7? Ans. Weight =-44,450 lbs. 3. Suppose a person to exert a force of 480 lbs. at the end of a rope passing around the first pulley, in a system represented by Fig. 19; what weight will he be able to sustain, the number of ropes being 8? Ans. Weight =-54A tons. 4. What weight, in a system of pulleys represented by Fig. 19, will a power of * ton balance, the number of ropes being 10? Ans. Weight =--2551 tons. QT.ATIOs.-74. Describe the system of pulleys represented by Fig. 19. ï~~100 EXERCISES IN NATURAL PHILOSOPHY. ART. 75. Since (Art. 74) P X (2-1I)= W; w Dividing by 2n-1,....... P= Hence, To find the power, the weight and number of ropes being given, we have the following formula and rule: w FORMULA,... W 2Â~-1 RULE.-From 2 raised to a power denoted by the number of ropes, subtract 1, and divide the weight by the remainder-the quotient will be the required power. Ex. 1. What power, in a system of pulleys represented by Fig. 19, will sustain a weight of 1 ton, 860 lbs., the number of ropes being 5? Ans. Power = 100 lbs. SoLUTIo.-Formula, TV 3100 3100 P= -. 100 lbs. 2"-1 2'-1 31 2. What power, in a system of pulleys represented by Fig. 19, will sustain a weight of 126 tons, the number of ropes being 6? Ans. Power= 2 tons. 3. What power must an individual exert, at the end of a rope passing around the first pulley, in a system represented by Fig. 19, to sustain a weight of 13 tons 16 cwt. 2 qrs. 20 lbs. the number of ropes being 7? Ans. Power = 2 cwt. 0 qr. 20 lbs. 4. What power, in a system of pulleys represented by Fig. 19, must an individual exert, to sustain a weight of ton, the number of pulleys being 8? Ans. Power = 2 lbs. QUESTIOws.-74, 75. Reduce the formulas for finding the weight and power. Repeat the rules for the same. ï~~THiE PULLEY. 101 ART. 76. The mechanical efficacy of the system of pulleys last described, may be greatly increased by passing each rope round a fixed pulley, connected to the weight, and finally attaching it to the block of the Fig. 20. pulley around which it first passed. The pulley, thus modified, is represented by Fig. 20. It will be seen, by an examination of the figure, that the first rope sustains a part of the weight equal to 2 P; the second rope sustains a part of the weight equal to three times the first, or 6 P; the third rope sustains a part of the weight equal to three times the second, or 18 P. The whole weight is sus- 3P tained by the three ropes; and 3 P hence, W= 2 P + 6P+ 18P; From which, W= P x(2+ 6 + 18). w The numbers within the parenthesis form a geometrical series, in which 2 is the first term, 3 the ratio, and the number of terms 3, equal to the number of ropes. The sum of the series 2+6+18, &c., is 1 less than 3 raised to a power denoted by the number of terms. If, therefore, we let n represent the number of ropes (number of terms), To find the weight, having the power and number of ropes given, we shall have the following formula and rule: FORMULA,.. W=PX(3Â~-1). QRQusrIows.---76. By what means may the mechanical efficacy of the system of pulleys represented by Fig. 19, be increased? Explain Fig. 20. Deduce the formula for finding the weight. Repeat the rule. ï~~102 EXERCISES IN NATURAL PHILOSOPHY. RULE.-l'om 3 raised to a power denoted by the number of ropes, subtract 1, and multiply the remainder by the power; the quotient will be the required weight. Ex. 1. What weight, in a system of pulleys represented by Fig. 20, will a power of 36 lbs. balance, the number of ropes being 4? Ans. Weight --2,880 lbs. SoLUTION.-Formula, W= PX (3"-1)= 36 X (34-1) =36 X 80= 2,880 lbs. 2. In a system of pulleys represented by Fig. 20, the number of ropes is 6, the power 500 lbs. Required the weight. Ans. Weight=162 tons 10 cwt. 3. What weight, in a system of pulleys represented by Fig. 20, will a power of I' ton balance, the number of ropes being 8? Ans. Weight=410 tons. A.RT. 77. Since (Art. 76) PX(3Â~-1)= W; Dividing by 3"-1,.... p. W 3 -1 Hence, To find the power, the weight and number of ropes being given, we shall have the following formula and rule: W FORMULA,... -. 3o-1 RULE.-From 3 raised to a power denoted by the number of ropes, subtract 1, and divide the weight by the result; the quotient will be the required power. Ex. 1. What power, in a system of pulleys represented by Fig. 20, will balance a weight of 208 lbs., the number of ropes being 3? Ans. Power = 8bs. QusrTxoN.--77. Deduce the formula for finding thepower. Repeat the rule. ï~~THE PULLEY. 108 SOLUToN.---Formula, W 208 208 P. = =8 lbs. 3"-1 3*-1 26 2. What power, in a system of pulleys represented by Fig. 20, will balance a weight of 720 lbs., the number of ropes being 4? Ans. Power= 9 lbs. 3. What power, in a system of pulleys represented by Fig. 20, will balance a weight of 2,420. lbs., the number of ropes being 5? Ans. Power=10 lbs. 4. What power, in a system of pulleys represented by Fig. 20, will balance a weight of 8,008 lbs., the number of ropes being 6? Ans. Power= 11 lbs. CHAPTER XIV. COMPOSITION OF POROES. ART. 78. A body, acted upon by a single force, moves in a straight line, and in the direction of the force. But, if a body be acted upon by two or more forces at the same time, and in different directions, it will not take the direction of either; but a direction called the resultant, resulting from the united action of all the forces. ART. 79. Suppose a body at A (Fig. 21) be acted upon by two forces, whose quantity and direction are.respectively represented by A B and A C; one QuicsTioIs.-78. In what direction does a body, acted upon by a single force, move?-In what direction, when acted upon by tw6 or more forces? What is a resultant? ï~~104 EXERCISES IN NATURAL PHILOSOPHY. Fig. 21. force, when S.A acting alone, causing the body, in a given time, to move over the line A B, the other, when D -............................... c acting alone, causing the body, in the same time, to move over the line A C. Now the body will not take the direction of either of these lines, but that of A D. and at the end of the given time will be found at D. This will appear evident from the following. Draw C D, equal and parallel to A B; join B and D. The force acting in the direction A B, has no tendency to prevent the body moving toward the line C D; also, the force acting in the direction A C, has no tendency to prevent the body moving toward the line B D. If, therefore, the two forces act at the same time, the body, at the end of the given time, will be found in the two lines CD and D B; and hence, at their point of intersection, that is, at D. The line C D is equal and parallel to A B; it may, therefore, be taken for the quantity and direction of the force acting in the line A B: and since A C represents the quantity and direction of the other force, and A D the resultant of the two forces, it follows, If one side of a triangle represent the resultant of two forces, the remaining sides will represent the quantity and direction of those forces. ArT. 80. If two forces act in lines perpendicular to each other, the resultant, and quantity, and direcQuvsTros.---79. Explain Fig. 21, and prove--If one side of a triangle represent the resultant of two forces, the remaining aides will represent, in quantity and direction, those forces. ï~~COMPOSITION OF FORCES. 105 tion of the forces will be represented, respectively, by the sides of a right-angled triangle: the hypotenuse answering to the resultant, and the base and perpend(licular to the acting forces. If, therefore, the numerical values of two of the numbers he given, the remaining numbers may easily be determined, from the following property of the right-angled triangle, namely, The square described on the hypotenuse, is equal to the sum of the squares described on the base and perpendicular. Hence, To FID THE RESULTANT (HYPOTENUSE), square the riven forces (sides), and extract the square root of the/ir sum. To FIND EITHER FORCE (BASE OR PERPENDICULAR), From the square of the resultant (hypotenuse), subtract the square of the given force (base or perpendicular), and extract the square root of the remainder. Ex. 1. A body is acted upon by two forces, in lines at right angles to each other; the numerical values of the forces are represented by the numbers 3 and 4: what number will represent the resultant? Ans. 5. 2. Two forces, represented by the numbers 6 and 8, act at right angles upon a body: required the number representing the resultant. Ans. 10. 3. The resultant of two forces, acting at right angles, is 15; one of the forces is 12: required the other. Ans. 9. 4. Two forces act at right angles; the resultant is 20, and one of the forces 16: required the other. Ans. 12. 5. Two forces, acting at right angles to each other, are represented by the numbers 9 and 12: required the resultant. Ans. 15. QUESTIONS.-80. Repeat the rules for finding the resultant, or either force, when the forces act at right angles. ï~~106 EXERCISES IN NATURAL PHILOSOPHY. 6. The resultant of two forces, acting at right angles, is 30; one of the forces is 18: required the other. Ans. 24. CHAPTER XV. THE INCLINED PLANE. POWER ACTING PARALLEL TO THE LENGTH OF THE PLANE. ART. 81. An inclined plane is any hard and smooth surface inclined to the horizon. This machine is used for raising large and heavy bodies. It is evident that the power (acting parallel to the face of the plane) required to sustain a given weight upon an inclined plane, will be less than the weight, inasmuch as the plane supports a part of the weight; and hence its mechanical advantage. The part of the weight supported by the plane, will evidently depend upon its elevation; the greater the elevation of the plane, the greater the power required to sustain a given weight; and the less the elevation, the less the power. In every inclined plane, there are three parts; the base, the hight, and the length. We will now investigate the relation which these sustain to the power, the weight, and the pressure of the weight upon the plane. ART. 82. Let the weight, W, be balanced by the power, P, acting parallel to A B, the length of the plane. From c, the center of gravity of the weight, let fall cd perpendicular to the base; draw dm Q*zsTrroxs.--81. What is an inclined plane? For what purpose is the inclined plane used? ï~~THE INCLINED PLANE. 107 Fig. 22. A o w P P parallel to B A, and let fall cm perpendicular to d m, it will also be perpendicular to A B. Now the body, W, is kept at rest by three forces; 1st. Its own weight or attraction of gravity, acting in the direction c d, perpendicular to the base of the plane; 2d. The power, P, acting in the direction A B or d m, parallel to the plane; 3d. The resistance of the plane, equal to the pressure of the weight upon the plane, acting in the direction m c, perpendicular to the plane. Since the weight, W, acting in the direction c d, resists the two forces, power and pressure upon the plane, c d may be taken as the resultant of these forces, and as representative of the weight, and then will dm represent the power, and m c the pressure of the weight upon the plane. But it can be shown that c d, d m, and c m, are, respectively, proportional to A B, A H, and B H. Hence, A B, or the length of the plane, will represent the comparative weight of the body upon the plane; A H, the hight of the plane, will represent the power required to sustain the weight, and B H, the base, the pressure of the weight upon the plane. ART. 83. We shall therefore have, if we let H represent the hight, L, the length, and B, the base Qu STwaxs.-82. Explain Fig. 22. ï~~108 EXERCISES IN NATURAL PHILOSOPHY. of the plane, and p, the pressure of the weight upon the plane, the following proportions: (1)... P: V::H:L; (2)... P: p:: H: B; (3)... W: p: L:B; That is; IN THE INCLINED PLANE, WHEN THE POWER ACTS PARALLEL TO THE PLANE; 1. The power is to the weight, as the hight of the plane to its length; 2. The power is to the pressure on the plane, as the hight of the plane to its base; 3. The weight is to the pressure on -he plane, as the length of the plane to its base. ART. 84. By multiplying together the extremes and means of each of the above proportions, we have: (1)... P xL= WxH; (2)... P XB=pXIH; (3)... WXB = p XL; Hence, IN THE INCLINED PLANE, WHEN THE POWER ACTS PARALLEL TO THE PLANE, THERE IS AN EQUILIBRIUM; 1. When the power, into the length of the plane, equals the weight into the hight of the plane; 2. When the power, into the base of the plane, equals the pressure into the hight of the plane; 3. When the weight, into the base of the plane, equals the pressure into the length of the plane. ART. 85. Since (Art. 84) PXL= WXH; WxH Dividing by L,....... P= L... Hence, In the inclined plane, to find the power, the weight, hight, and length of the plane being given, we have the following formula and rule: QuasTioxs.--83. Deduce the proportions (1), (2), and (3). 84. Repeat the three equations of equilibrium. ï~~THE INCLINED PLANE. 109 FoRMUa,.. P. RIE.- iltipl the iweight by the hight, and divide the product by the length of the plane; the quotient will be the required power. Ex. 1. Suppose the length of an inclined plane to be 9 ft., and the hight 3 ft.: Required the power necessary to sustain a weight of 12 lbs.? Ans. Power=4 lbs. SoLUTION.-Formula, VXH 12X3 P = 4 lbs. L 9 2. What power will sustain a weight of 40 lbs., on an inclined plane 8 ft. long and 2 ft. high? Ans. Power= 10 lbs. 3. What power will sustain a weight of 440 lbs., on an inclined plane 16 ft. long and 4 ft. high? Ans. Power= 110 lbs. 4. What force must an individual exert, to sustain a weight of 1,200 lbs., on an inclined plane 20 ft. long and 6 ft. 8 in. high? Aus. Power = 400 lbs. 5. What force must an individual exert, to sustain a weight of 2,800 lbs., on an inclined plane whose length is 45,ft., and hight 3 ft.? Ans. Power =21553 lbs. 6. A weight of 4,600 lbs. rests on an inclined plane whose hight is, to its length, as 2: 92. Required the power necessary to sustain the weight. Ans. Power= 100 lbs. 7. What power will sustain a weight of 100 lbs., on an inclined plane 12 ft. long and 8 ft. high? Ans. Power = 25 lbs. QUvST1oNs.---85. Deduce, and repeat the rule and formula for finding the power, the weight, hight, and length of the plane being given. When the power acts parallel to the length of the plane, with what is the weight connected?-Power?-Pressure?-Repeat the formulas for power, weight, pressure, length, hight, and base, on principle of general rule (Art. 37). ï~~110 EXERCISES IN NATURAL PHILOSOPHY. ART. 86. Since (Art. 84) WXH= P x L; PxL Dividing by H,...... W=-H -. Hence, To find the weight which a given power will sustain on an inclined plane, the length and hight of the plane being known, we have the following formula and rule: PXL FORMULA,.. W= H H" RULE. Multiply the power into the length of the plane, and divide the product by the hight; the quotient will be the required weight. Ex. 1. What weight, on an inclined plane 20 ft. long and 5 ft. high, will a power of 44 lbs. sustain? Ans. Weight= 176 lbs. SoLUTION.-Formula, PxL 44X20 W= 5-- 176 lbs. H 5 2. What weight, on an inclined plane 30 ft. long and 6 ft. high, will a power of 83 lbs. sustain? Ans. Weight=415 lbs. 3. What weight, on an inclined plane 40 ft. long and 8 ft. high, will a power of 100 lbs balance? Ans. Weight= 500 lbs. 4. An individual is able to draw a weight of 250 lbs. up a perpendicular wall 15 ft. high. Required the weight he would be able to draw up a plank 60 ft. long, sloping from the top of the wall to the ground. Ans. Weight= 1,000 lbs. 5. What weight can an individual, capable of exerting a force of340 lbs., sustain on an inclined plane 100 ft. long and 10 ft. high? Ans. Weight = 3,400 lbs. QUESTION.--8. Deduce the formula for finding the weight. Repeat the rule. ï~~TIH1 INCLINED PLANE. 111 ART. 87, Since (Art. 84) PxL=WxH; WX H Dividing by P,...... L=W -. Hence, To find the length of an inclined plane, the hight, weight, and power being given, we have the following formula and rule: WxH FORMULA,... L=. P RULE.-Multiply the weight into the hight of the plane, and divide the product by the power; the quotient will be the required length. Ex. 1. Required the length of an inclined plane, upon which a weight of 30 lbs. is just balanced by a power of 5 lbs., the hight of the plane being 3 ft. Ans. Length = 18 ft. WXH 30X3 SoLUTiO.-Formula, L-- -- -----=18 ft. P 5 2. Required the length of an inclined plane, upon which a weight of 6 tons is just balanced by a power of' I ton, the hight of the plane being 20 ft. Ans. Length =240 ft. 3. An individual, who is just able to lift a weight of 400 lbs., sustains, on an inclined plane 10 ft. high, a weight of 2 tons; required the length of the plane. Ans. Length =112 ft. 4. What must be the length of an inclined plane 24 ft. high, in order that an individual, by exerting a force of 280 lbs., shall sustain upon the plane a weight of 4 tons? Ans. Length =768 ft. 5. The power necessary to sustain a given weight upon an inclined plane, is to the weight as 5: 60, the hight of the plane is 8 ft.; required the length. - Ans. Length =96 ft. ï~~112 EXERCISES IN NATURAL PHILOSOPHY. ART. 88. Since (Art. 84) WxH= Px L; PX L Dividing by WV,..... H L Hence, To find the hight of an inclined plane, having given the length, power, and weight, we have the following formula and rule-: FORMULA, H H-PXL RULE.-Multiply the power into the length of the plane, and divide the product by the weight; the quotient will be the required hight. Ex. 1. Required the hight of an inclined plane, upon which a weight of 400 lbs. is just balanced by a power of 20 lbs., the length of the plane being 80 ft. Ans. Might =4 ft. PxL 2OÃ~8O SOLUTION. Formula, H XL --20 4 ft. W 400 2. What must be the hight of an inclined plane, upon which a weight of 144 tons may be sustained by a power of 6 tons, the length of the plane being 600 ft.? Ans. Hight =25 ft. 3.. The ratio of the power to the weight, is as 3: 4, and the length of the plane 24 ft.; required the hight. Ans. Hight = 18 ft. 4. An individual, by exerting a force of 400 lbs., is enabled to sustain a weight of 6 tons on an inclined plane 800 ft. long; required the hight of the plane. Ans. Hight =231 ft. ART. 89. Since (Art. 84) PxB=pxH; Dividing by B,.... P- p, H QuESTro1.-87, 88. Deduce the formulas for finding the length and hight of the plane. Repeat the rules for the same. ï~~TIHE INCLINED PLANE. 113 Hence, To find the power required to sustain any weight upon an inclined plane, the pressure of the weight upon the plane, the hight and base of the plane being given, we have the following formula and rule: FORMULA, P. PX H* Â~ " -- " RULE.-Multiply the pressure of the weight into the hight of the plane, and divide the product by the basethe quotient will be the required power. Ex. 1. A weight presses upon an inclined plane with a force of 20 lbs., the hight of the plane is 6 ft., and its base 18 ft.; required the power necessary to sustain the weight. Ans. Power = 6 lbs. pX pH 20%6 SoLUTIoN.-Formula, P= p6 = l6bs. B =18 ~ ls 2. The hight of an inclined plane is to the base as 4: 18; what power will sustain a weight whose pressure upon the plane is 72 lbs.? Ans. Power = 16 lbs. 3. The hight of an inclined plane is to its base as 5: 22; what power will sustain a weight whose pressure upon the plane is 264 lbs.? Ans. Power =- 60 lbs. 4. The hight of an inclined plane is 24 ft., and its base 60 ft.; what force must an individual exert to sustain a weight whose pressure upon the plane is 240 lbs.? Anis. Force =96 lbs. ART. 90. Since (Art. 84) pxH=PXB; QUESTIONS.-89. Deduce the formula for finding the power, the pressure, hight, and base of the plane being given. Repeat the rule. In the inclined plane, when the power acts parallel to the length of the plane, the power, pressure, hight, and base, being given; with what is the power connected?-Pressure? Repeat the formulas for power, pressure, hight, and base, on principle of general rule Art. 37. 10 ï~~114 EXERCISES IN NATURAL PHILOSOPHY. PxB Dividing by H,..... p= H " Hence, To find the pressure upon an inclined plane, the power required to sustain the weight, the base and hight of the plane being given, we have the following formula and rule: PxB FORMULA,.. p H. RuLE.-Miultiply the power into the base, and divide the product by the hight of the plane-the quotient will be the required pressure. Ex. 1. The power, necessary to sustain a weight upon an inclined plane is 16 lbs., the hight of the plane is 8 ft., and its base 24 ft.; required the pressure of the weight on the plane. Ans. Pressure =48 lbs. PxB 16X24 SOLUTION.-Formula, p = = 8 -48 lbs. 2. The power necessary to sustain a weight upon an inclined plane is 480 lbs., the hight of the plane is to its base as 3: 12; required the pressure of the weight on the plane. Ans. Pressure = 1,920 lbs. 3. A weight, resting upon an inclined plane, is balanced by a power of 2' tons, the hight of the plane is to its base as 2*: 9; required the pressure of the weight on the plane. Ans. Pressure = 10 tons. 4. A weight, resting upon an inclined plane, is sustained by a power of 4 ton, the hight of the plane is 15 ft., and its base 60 ft.; required the pressure of the weight on the plane. Ans. Pressure =3 tons. QuzsTros ---90. Deduce the formula for finding the pressure on the plane, the power, base, and hight being given. Repeat the rule. ï~~THE INCLINED PLANE. 115 ART. 91. Since (Art. 84) PXB=pXIH; Dividinig by P,... B= pXH Hience, To find the base of an inclined plane, having given the hglht of the plane, the power, and the pressure of the we zht upon the plane, we have the following ibrmula and rule: FonruL,.. B. P RuL.- Multiply the pressure into the hight of the plane, and divide the product by the power-the quotient will be the required base. Ex. 1. Upon an inclined plane 10 ft. high, a weight pressing with a force of 20 lbs. is sustained by a power of 10 lbs.; required the length of the base. Ans. Base =20 ft. pXH 20X10 SoLuTIo.--Formula, ------ = =20 ft. P 10 2. Upon an inclined plane 18 ft. high, a weight pressing with a force of 480 tons is sustained by a power of 60 tons; required the length of the base? Ans. Base = 144 ft. 3. The hight of an inclined plane is 24 ft., the pressure of the weight upon the plane is to the power as 3: 4; what is the length of the base? Ans. Base =18 ft. 4. An individual, by exerting a force of 400 lbs., sustains upon an inclined plane 30 ft. high, a weight whose pressure on the plane is equal to 1 ton; required the length of the base. Ans. Base =168 ft. QUESTIONgS.--91. Deduce the formula for finding the base of the inclined plane, the pressure, power, and hight being given. ï~~110 EXERCISES IN NATURAL PHILOSOPHY. ART. 92 Since (Art. 84) pxH=PxB; Px B Dividing by p,...... H-. ~P p Hence, To find the hight of an inclined plane, the base, power, and pressure of the weight upon the plane being given, we have the following formula and rule: PxB FORMULA,... H=. P RULE.--Multiply the power into the base, and divide the product by the pressure of the weight on the plane; the quotient will be the required hight. Ex. 1. The base of an inclined plane is 100 ft. long; a power of 300 pounds sustains a weight whose pressure on the plane is equal to 1,500 lbs.; required the hight of the plane. Ans. Hight =20 ft. 2. The length of the base of an inclined plane is 140 ft., the power is to the pressure of the weight on the plane as 24: 14; required the hight of the plane. Ans. Hight =28 ft. SoLUTION.-Formula, PXB 25X14028 H= 14 =28 ft. p 14 3. Upon an inclined plane whose base is 24 ft., a power of 36 lbs. sustains a weight pressing on the plane with a force of 48 lbs.: required the hight of the plane. Ans. Hight= 18 ft. 4. An individual, by exerting a force of 380 lbs., sustains on an inclined plane a weight whose pressure upon the plane equals 800 lbs.; the length of QuzsTos.-92. Deduce the formula for finding the hight of an inclined plane, the base, power, and pressure of the weight being given. Repeat the rule. ï~~THE INCLINED PLANE. 117 the base is 400 ft.: required the hight of the plane. Ans. Hight= 190 ft. ART. 93. Since (Art. 84) WXB=p X L; pxL Dividing by B,..... W. B. Hence, To find the weight, the pressure of the weight on the inclined plane, the length and base of the plane being given, we have the following formula and rule: FORMULA,.. W.. RULE.---Multiply the pressure of the weight into the length of the plane, and divide the product by the base; the quotient will be the required weight. Ex. 1. Upon an inclined plane whose base is to its length as 5: 26, a weight presses with a force of 600 lbs.: required the weight. Ans. Weight=3,120 lbs. SoLUTION.-Formula, Wp pL 600 X26 w=-B -- 3,120 lbs. B 5 2. Upon an inclined plane whose base is to its hight as 8: 46, a weight presses with a force of 64 lbs.: required the weight. Ans. Weight=368 lbs. 3. What weight will press upon an inclined plane with. a force of 2 tons, the length of the plane being 56 ft., and its base 8 ft.? Ans. Weight-14 tons. 4. An inclined plane, whose length is to its base as 6: 11, sustains a pressure of 31 tons: required the weight. Ans. Weight= 14 tons. QZsTrows.-93. Deduce the formula for finding the weight, having given the pressure, base, and length of the plane. Repeat the rule. ï~~118 EXERCISES IN NATURAL PHILOSOPHIY. ART. 94. Since (Art. 84) WxB=p x L; WxB Dividing by L,........p L Hence, To find the pressure of the weight upon an inclined plane, having given the weight, the length, and base of the plane, we have the following formula and rule: Wx B FORMULA,. p= BL RuLE.-Multiply the weight into the base, and divide the product by the length of the plane; the quotient will be the required pressure. Ex. 1. Upon an inclined plane, whose base is to its length as 3: 7, a weight of 720 lbs. rests: required the pressure of the weight on the plane. Ans. Pressure = 3087 lbs. SOLUTION.--Formula, WxB 720X3 Pr- L 7 = 3087 lbs. 2. With what force will 15 tons press on an inclined plane whose base is 18 ft. and length 72 ft.? Ans. Pressure=3& tons. 3. There is an inclined plane whose base is 100 ft., and length 350 ft. With what force will a weight of 16 tons press on the plane? Ans. Pressure -4 tons. 4. There is an inclined plane whose base is to its length, as 1": 71. With what force will 25 tons press on the plane? Ans. Pressure= 5 tons. ART. 95. Since (Art. 84) WX B= p X L; Dividing by W,..... B p XL QuEsTIONs.-94. Deduce the formula for finding the pressure, having given the weight, base, and length of the plane. Repeat the rule., ï~~THE INCLINED PLANE. 119 Ilence, To find the base of an inclined plane, having given the length, the weight, and its pressure on the plane, we have the following formula and rule: FORMULA,.. B xL RULE.- Mltd iply the pressure into the length of the plane, and divide the product by the weight; the quotient will be the required base. Ex. 1. Required the base of an inclined plane whose length is 400 ft., and the weight, to its pressure on the plane, as 16: 4. Ans. Base=100 ft. SoLUTION.-Formula, B pXL 4X400 W 16 -=100 ft. 2. The length of an inclined plane is 300 ft.; the ratio of the pressure to the weight as 1: 9. Required the length of the base. Ans. Base --50 ft. 3. The length of an inclined plane is 84 ft.; the ratio of the pressure to the weight as *: 9. Required the base of the plane. Ans. Base-=7 ft. 4. Required the base of an inclined plane whose length is 144 ft.; the weight being 36 lbs. and pressure 6 lbs. Ans. Base=24 ft. AaRT. 96. Since (Art. 84) p XL= x B; Dividing by p,...... L Wx B P Hence, To find the length of an inclined plane, having given the base, the weight, and its pressure, we have the following formula and rule: WXB FORMULA,... L WXB P p QuersTros.-95. Deduce the formula for finding the base,having given the length, the weight, and its pressure. Repeat the rule. ï~~120 EXERCISES IN NATURAL PHILOSOPHY. RULE.-Multiply the weight into the length of the base, and divide the product by the pressure of the weight; the quotient will be the required length of the plane. Ex. 1. Required the length of an inclined plane whose base is 75 ft., and the ratio of the pressure to the weight as 21:121. Ans. Length= 375 ft. SoLUTIO.-Formula, WX B 12 X75 L=W --- - 1 375 ft. p 2k 2. The base of an inclined plane is 28 ft. long; the weight 161 lbs., and its pressure on the plane 24 lbs.; required the length of the plane. Ans. Length =168 ft. 3. Required the length of an inclined plane whose base is 500 ft., and the ratio of the pressure to the weight as: 1g. Ans. Length -3,500 ft. 4. A body, weighing 5 tons, presses on an inclined plane with a force of 1I tons; the base of the plane is 181 ft. long; required the length of the plane. Ans. Length= 55 ft. ART. 97. The length, base, and perpendicular hight of an inclined plane represent, respectively, the sides of a right-angled triangle. The length of the inclined plane, or hypotenuse, answers (Art. 82) to the weight, the base to the pressure of the weight, and the hight or perpendicular to the power. Hence, from the known property of a right-angled triangle (Art. 80), two of the three numbers (weight, pressure, and power) being given, the remaining number may easily be determined. QuZsTIo1s.-96. Deduce the formula for finding the length of an inclined plane, the weight, pressure, and base being given. Repeat the rile. ï~~THE INCLINED PLANE. 121 ART. 98. From the property of the right-angled triangle, W= P'+p;2 Extracting the square root, W= VJ-P2+p2. Hence, Tofind the weight, the pressure of the weight on the plane, and power required to sustain the weight, being given, we have the following formula and rude: FORMULA,...W= JP /P2+p'. RULE.-Add the squares of the power and pressure together, and extract the square root of the sum. From the equation P'+p2 = W, we obtain, (1)... P= J -p'; (2)... p = J W-P. Hence, To find either the power or pressure, we have the following rule: RuL.-From the square of the weight subtract the square of the remaining given number, and extract the square root of the remainder; the root, thus obtained, will be the required answer. Ex. 1. The pressure of a certain weight, on an inclined plane, is 3 tons, the power required to sustain the weight, 4 tons; what is the weight? S Ans. Weight= 5 tone. 2. Required the pressure, on an inclined plane which sustains a weight of 10 lbs., balanced by a power of 6 lbs. Ans. Pressure= 8 lbs. 3. Required the power necessary to sustain a weight of 10 lbs. on an inclined plane, the pressure being 8 lbs. Ans. Power=6 lbs. QUESTIONS.-98. Having two of the numbers,-weight, power, and pressure-given, how do you find the remaining number? Deduce the formulas and repeat the rules. 11 ï~~122 EXERCISES IN NATURAL PHILOSOPHY. ART. 99. POWER ACTING PARALLEL TO THE BASE OF THE INCLINED PLANE. In considering the inclined plane, thus far, we have supposed the power to act parallel to the length of the plane. We will now investigate the. relation which the power, weight, and pressure sustain to the hight, length, and base of the plane, when the power acts parallel to the base of the plane. ART. 100. It will be seen, by examining the annexed figure, that the body, W, is kept at rest by the Fig. 23. Â~ B A b c action of three forces: first, its own weight acting in the perpendicular direction a b; second, the pressure or resistance of the plane in the direction c a, perpendicular to the face of the plane; third, the power P, in the direction a d, or b c, parallel to the base of the plane. The weight, W, acting in the direction a b, resists the two forces, power, and resistance of the plane; and' hence, a b may be taken as the resultant of these forces, and as representative of the weight; a c will then (Art. 80) answer to the pressure on the plane, and b c, to the power required QuESTIoN.-100. Explain Fig. 23. When the power acts parallel to the base of the plane, what answers to the weight?-To the power? -Pressure ï~~THE INCLINED PLANE. 123 to sustain the weight. Now, it can be shown, that a c, a b, and b c, are respectively proportional to A B, A C, and B C. Hence, A C, the base of the plane, will answer to the weight; B C, the hight of the plane, to the power; and A B, the length, to the pressure of the weight on the plane. ART. 101.-If, as in Art. 83, we let H represent the hight, L the length, B the base of the plane, and p the pressure on the plane, we shall then have the following proportions (1)... P: W:: H: B. (2)... P: p:: H: L. (3)... W:p:: B L. That is, In the inclined plane, when the power acts parallel to the base: 1. The power is to the weight, as the hik ht of the plane to its base. 2. The power is to the pressure on the plane, as the hight of the plane to its length. 3. The weight is to the pressure on the plane, as the base of the plane to its length. ART. 102. By multiplying together the extrenies and means of the above proportions we have (1).... PxB= WxH. (2).... PXL= p >H. (3).... WxL= p XB. Hence, In the inclined plane, when the power acts parallel to the base, there is an equilibrium: 1. When the power into the base equals the weight into the hight of the plane. 2. When the power into the length equals the QUESTTONS.-101. Repeat proportions (1), (2), and (3). 102. Re-. peat the three equations of equilibrium, when the power acts parallel to the base. With what is the power connected in the first equation of equilibrium?--Weight? With what is the power connected in the second equation of equilibrium?-Pressure? With what is the weight connected in the third equation of equilibrium?-Pressure? ï~~124 EXERCISES IN NATURAL PHILOSOPHY. pressure on the plane into the hight of the plane. 3. Whcn the weight into the length equals the pressure on the plane into the base. ART. 103. From the above equations of equilibrium, the student will easily deduce, algebraically, or from the principle of general rule Art. 37, the following formulas and rules, applicable to the inclined plane, when the power acts parallel to the base of the plane: FORMULAS. WxH (1). P-- B pxH (2). P = H P xL (3). P= H. WxL (4).. p= (5) H=P x B PxL (6).. H=P-- p (7).. W -- H WpB (8)..W=(9).. B=W-pH. WxÃ~L (10).. B=WxL (11).. L ----p. p (12).. L- P RULE 1.- The power equals the weight into the hight divided by the base. Or, The power equals the pressure into the hight divided by the length. RULE 2.- The weight equals the power into the base divided by the hight. Or, The weight equals the pressure into the base divided by the length. RULE 3.-The pressure equals the power into the length divided by the hight. Or, The pressure equals the weight into the length divided by the base. RULE 4.-The hight equals the power into the base divided by the weight. Or, The hight equals the power into the length divided by the pressure. ï~~THE INCLINED PLANE. 125 Rur n 5.-The base equals the weight into the length dividtd b1y the pressure. Or, The base equals the wcight into the hight divided by the power. Rtu. 6.- The length equals the pressure into the hight divided by the power. Or, The length equals the pressure into the base divided by the weight. Ex. 1. Required the power necessary to sustain a weight of 240 lbs., on an inclined plane whose hight is 10 ft. and base 20 ft. Ans. Power --=120 lbs. 2. The hight of an inclined plane is 30 ft., base 40; what power will sustain a weight whose pressure is 400 lbs.? Ans. Power =300 lbs. 3. The power necessary to sustain a weight on an inclined plane is 4 tons, the ratio of the hight to the length is as 3: 4; required the pressure on the plane. Ans. Pressure =5k tons. 4. An inclined plane whose length is 80 ft. and base 35 ft., sustains a weight of 860 lbs.; required the pressure on the plane. Ans. Pressure =1,965k lbs. NOTE.-It will be seen, from the answer to Example 4, that the pressure on the plain is greater than the weight of the body supported by the plane. 5. Required the hight of an inclined plane whose base is 18 ft.; the ratio of the power to the weight being as 2: 6. Ans. Hight =6 ft. 6. The base of an inclined plane is 84 ft., the ratio of the power to the weight as 12: 7; required the hight of the plane. Ans. Hight =144 ft. 7. The length of an inclined plane is 260 ft., the ratio of the power to the pressure sustained by the plane as 11: 191; what is the hight? Ans. Hight =20 ft. 8. The power necessary to sustain a weight on QuisTioss.-103 Repeat rules 1, 2, 3, 4, 5, and 6. ï~~126 EXERCISES IN NATURAL PHILOSOPHY. an inclined plane is 3 tons, the base 30 ft., and hight 18 ft.; required the weight. Ans. Weight =5 tons. 9. The pressure upon an inclined plane is 96 tons, the ratio of the base to the length as 14 101; required the weight. Ans. Weight =16 tons. 10. Upon an inclined plane, whose hight is 30 ft., a power of 40 lbs. sustains a weight of 60 lbs.; required the length of the base. Ans. Base =45 ft. 11. Upon an inclined plane, whose length is 300 ft., a weight of 28 tons produces a pressure of 35 tons; what is the length of the base? Ans. Base =240 ft. 12. The hight of an inclined plane is 80 ft.; the ratio of the pressure to the power as 121: 2-; required the length of the plane. Ans. Length =400 ft. 13. The base of an inclined plane is 600 ft., the pressure, 40 tons, and weight 30 tons; what is the length of the plane? Ans. Length =800 ft. REMARK.-The mechanical efficacy of the inclined plane is the greatest when the power, as in Fig. 22, acts parallel to the length of the plane; for the reason, that the power is then wholly employed in drawing the weight up the plane. When the power acts parallel to the base, as in Fig. 23, a part is spent in pressing the weight against, and a part in drawing it up the inclined plane. ï~~THE SCREW. 127 CHAPTER XVI. TIlE SCREW. ART. 104. The screw is a modification of the inclined plane, and consists of a spiral thread coiled around a cylindrical piece of wood or metal. When used, it is made to pass through a nut having a hollow spiral exactly fitted to the thread of the screw. Sometimes the screw is turned, and sometimes the nut, as circumstances may require; and at each revolution of the screw or nut, an advance is made, equal to the distance between the contiguous threads of the screw. ART. 105. That the screw is a modification of the inclined plane, may be made sufficiently evident by cutting a piece of paper in the form of an inclined plane, and winding it around a small cylinder of wood, when the edge of the paper will form the thread of the screw. The inclined plane, wound around the cylinder, may be considered as composed of the union of several inclined planes, each of which has for its base the circumference of the cylinder, and for its hight, the distance between the contiguous threads. ART. 106. Since, therefore, the screw acts on the principle of the inclined plane, the distance between the contiguous threads answering to the hight of the plane, and the circumference of the screw to the base, the power being supposed to act horizontally, and to be applied at the circumference of the QuesTios.-104. Define the screw. 105. How may it be shown that the screw is a modification of the inclined plane 106. What part of the screw corresponds to the hight of the inclined plane?-.What to the base? Deduce, and repeat the equation of equilibrium, the power being applied at the circumference of the screw. ï~~128 EXERCISES IN NATURAL PHILOSOPHY. screw; the power and weight will vary as in the inclined plane, when the power acts parallel to the base of the plane. Hence, by letting p denote the power applied at the circumference of the screw, W.the weight or pressure produced, D the distance between the contiguous threads, and c the circumference of the screw, we shall have (Art. 101), p: W::D: c; Multiplying the extremes and means, SpXc= WXD. Fig. 24. Hence, In the.Â~.-.......... ".. screw, the power.,"" '"" being applied at the S\ circumference, an So equilibrium is prof. Â~ duced, when the power, multiplied into I the circumference of.- the screw, equals the." / pressure or weight "-/, into the distance between the contiguous threads. ART. 107. We have, thus far, considered the power as being applied at the circumference of the screw; but, in practice, it is generally applied -- at the end of a W lever attached to the head of the screw, as seen in the figure. ï~~THE SCREW. 129 We will now investigate the relation which the power sustains to the weight, when applied at the end of a lever. Let P denote the power at the end of the lever, W the pressure produced, p the power at the circumference of the screw required to produce the same pressure, c the circumference of the screw, and C that described by the lever. From the following principle of the lever, that the power required to sustain a given weight will vary inversely as the length of the arm to which it is attached, we have (see the figure), P:p::ab: am; But, The diameters or semidiameters of circles vary as their circumferences; hence, IP:p "c" C;"./, Multiplying extremes and means, pXc=PX C.... By Art. 106, pXc= WxD; Hence,... PX 0= WXD. That is; In the screw, the power acting at the end of a lever, there is an equilibrium when the power, multiplied into the circumference of the circle described by the lever, equals the pressure into the distance between the contiguous threads: Or, in other words; When the power is to the weight, as the distance between the threads is to the circumference described by the lever. QUESTIONs.-107. In practice, how is the power applied? What relation does the power sustain to the weight, when the power is applied at the end of a lever? Deduce, and repeat the equation of equilibrium, when the power is thus applied. With what is the power connected?The weight? Repeat the formulas for power, weight, distance between the threads, and circumference described by the lever, by calling to mind the principle of general rule (Art. 37). ï~~130 EXERCISES IN NATURAL PHILOSOPHY. ART. 108. Since (Art. 107) P X C= WX D; Wx D Dividing by C,....... P=-- XD Hence, To find the power, having given the pressure, the distance between the contiguous threads, and the length of the lever or circumference described by the lever, we have the following formula and rule: Wx D FORMULA,... P --. RULE.-AMultiply the pressure into the distance between the contiguous threads of the screw, and divide the product by the circumference described by the lever; the quotient will be the required power. Ex. 1. What power applied at the end of a lever, by means of which a screw is turned, will produce a pressure of 240 lbs., the distance between the contiguous threads being 2 inches, and the circumference described by the lever 2 ft., or 24 in.? Ans. Power = 20 lbs. SoLUTION.-Formula, WxD 240X2 P_ C-=20 lbs. C 24 2. Suppose the lever, by means of which a screw is turned, to be 7 feet long; the distance between the contiguous threads of the screw, 1 inch; required the power necessary to produce a pressure of 10,560 lbs. Ans. Power-= 20 lbs. NOTE.-The length of the lever is half the diameter of the circle which it describes; also, the circumference of a circle is 3) times the diameter. Hence, To find the circumference of a circle described by a given lever, multiply twice the length of the lever by 31. The distance between the contiguous threads, and the circumference described by the lever, should both be reduced to the same denomination. QussTItos.:--108. Deduce the formula and repeat the rule for finding the power. ï~~THE SCREW. 131 3. The length of the lever by means of which a screw is turned, is 14 ft., the distance between the contiguous threads of the screw 3 inches; what power will produce a pressure of 16,840 lbs.? Ans. Power= 474 lbs. SoLUTION.-Formula, N Wx D 16840X3 N P 8 = 4737 lbs. C 88X12 4 4. An individual wishes, by means of a screw, the distance between whose contiguous threads is 4 inch, to produce a pressure of 7,920 lbs.; what power must he apply at the end of a lever 10- feet long? Ans.Power= 5 lbs. 5. An individual wishes, by means of 8 screws, to raise a building weighing 720 tons; the distance between the contiguous threads of each screw is 3 inches, and the length of each lever 17J ft.; what power must be applied, at the end of each lever, to raise the building? Ans. Power = 458; lbs. ART. 109. Since (Art. 107) Wx D=P X C; PÃ~ C Dividing by D,........ = D. Hence, Toj find the pressure, the length of the lever, the power, and the distance between the contiguous threads being given, we have the following formula and rule: PXC FORMULA,... Px. D RuLE.-Multiply the power into the circumference described by the lever, and divide the product by the distance between the contiguous threads of the screw the quotient will be the required pressure. QuxsTos.-109. Deduce the formula and repeat the rule.for finding the pressure. ï~~132 EXERCISES IN NATURAL PHILOSOPHY. Ex. 1. What pressure will a power of 20 lbs., applied at the end of a lever 7 ft. long, produce, the distance between the contiguous threads of the screw being z inch? Ans.Pressure= 21,120 lbs. SoLUTION.-Formula, WP C 20X7X2X3 X1221,120 lbs. D 2. What pressure will a power of 400 lbs., applied at the end of a lever 101 ft. long, produce, the distance between the contiguous threads of the screw being 4 inch? Ans. Pressure = 1884 tons. 3. Required the pressure produced by a power of 18 lbs., acting at the end of a lever 171 ft. long, attached to a screw, the distance between whose contiguous threads is 24 in. Ans. Pressure =8,640 lbs. 4. Suppose the distance between the contiguous threads of a screw to be 3 inches; required the pressure produced by a power of 50 lbs., acting at the end of a lever 101 ft. long. Ans. Pressure =12,672 lbs. ART. 110. Since (Art. 107) WxD=PX C; PX C Dividing by W, D=PX. Hence, To find the distance between the contiguous threads of a screw, having given the length of the lever or the circumference which it describes, the power, and the pressure, we have the following formula and ru le:... C PxCo FORMULA,.. Â~ D= QuESTIoNs.-110. Deduce the formula, and repeat the rule for finding the distance between the contiguous threads of the screw. ï~~THE SCREW. 133 RULE.-Multiply the power into the circumference - described by the lever, and divide the product by the pressure; the quotient will be the required distance between the threads of the screw. Ex. 1. By means of a screw, a power of 20 lbs., acting at the end of a lever 7 ft. long, produces a pressure of 21,120 lbs.; required the distance between the contiguous threads of the screw. Ans. Distance--= - inch. SOLUTION.-Formula, D PXC 20OX7X2X37 X12 inch. D== - 1 0 inch. W 21120 2. Suppose the lever, by means of which a screw is turned, to be 7 ft. long; the power 50 lbs., and the pressure 7,200 lbs.; what is the distance between the contiguous threads of the screw? Ans. Distance =3- inches. 3. An individual wishes, by means of 4 screws, to raise a building whose estimated weight is 792 tons; the length of each lever is 21 ft., and the power applied at the end of each lever, 560 lbs.; required the distance between the contiguous threads of each screw. Ans. Distance= 2 inches. 4. An individual, by exerting a force of 1 lb., at the end of a lever 101 ft. long, attached to the head of a screw, is enabled to produce a pressure of 1,584 lbs.; required the distance between the contiguous threads of the screw. Ans. Distance = I inch. ART. 111. Since (Art. 107) P x C= Wx D; WxD Dividing by P,.......... P QUESTIzo.S.- 111. Deduce the formula, and repeat the rule for finding the circumference described by the lever. ï~~134 EXERCISES IN NATURAL PHILOSOPHY. Hence, To find the circumference described by the lever, the power, pressure, and distance between the contiguous threads of the screw being given, we have the following formula and rule: Wx D FORMULA,.. C P RuLE.-Multiply the pressure into the distance between the contiguous threads, and divide the product by the power; the quotient will be the required circumference. Ex. 1. By means of a screw, the distance between whose contiguous threads is 2 inches, a power of 40 lbs. produces a pressure of 3,200 lbs.; required the circumference of the circle described by the lever. Ans. Circumf.-= 13 ft. 4 inches. SoLUTion.-Formula, WxD 3200X2 C- --= F ---40 160 in.= 13 ft. 4 in. P 40 2. In a screw, the distance between the contiguous threads is 21 inches, and the ratio of the power to the weight or pressure, as 3: 2,500; required the circumference of the circle described by the lever. Ans. Circumf.= 160 g ft. 3. Required the circumference described by a lever attached to the head of a screw, the distance between whose contiguous threads is 4 in., and the ratio of the power to the weight or pressure, as 1: 3,400. Ans. Circumf.=- 212 ft. 6 inches. NoTE.-If the length of the lever be required, divide the circumference by 37. 4. What must be the length of the lever by means of which a screw is turned, in order that a power of 4 lbs. shall produce a pressure of 1,760 lbs.; the ï~~HUNTER'S SCREW. 135 distance between the contiguous threads of the screw being 1 inch. Ans. Length of lever= 11 ft. 8 inches. HiUNTER'S SCREW. ART. 112. The mechanical efficacy of the screw depends, as has already been shown, on the ratio of the circumference described by the lever, to the distance between the threads of the screw. That is; the power remaining the same, the weight or pressure may be increased, by diminishing the distance between the threads, or by increasing the length of the lever, and thereby the circumference which it describes. There is, h.wever, in practice, a limit to both of these methods of augmenting the effect of the screw; for, if the lever, which turns the screw, be increased to any considerable length, the power is made to pass over a very great space; which is frequently accompanied with great practical inconvenience. On the other hand, when a great weight is to be raised, or pressure produced, if the threads of the screw be very near together, the thread will be too slender to sustain the resistance, and will be torn off in passing through the nut. ART. 113. These inconveniences are remedied by a contrivance known as Hunter's Screw, which, while it preserves all the requisite strength and compactness in the machine, gives it an almost unlimited degree of mechanical efficacy. Hunter's Screw, represented by the annexed figure, consists of two screws, one moving within the other. The screw B, turning within A, is fastened to the board H, which, in descending, QUEsTro.-112. Upon what does the mechanical efficacy of the screw depend? 113. Explain the principle of Hunter's Screw. ï~~136 EXERCISES IN NATURAL PHILOSOPHY. Fig. 25. exerts a pressure - "....', c on any substance " placed between it ' and the bottom. P When the external S/ screw, A, is turned S once around, it ad_.o "vances through the nut, a space equal to the distance beA twcen its threads; and, if B, at the same time, does not act, the board, H, will descend through the same B space; but if B is H supposed to act, then, at each revolution A will descend, and B ascend w a space equal to the distance between Stheir threads, and consequently, the board, H, will, at each revolution, descend through a space equal to the difference of the distances between the threads of the two screws. It is evident, therefore, that the same effect will be produced, as would have been produced by a single screw, with a space between its threads equal to the difference of the distances between the threads of the two screws, A and B. Hence, In the contrivance known as Hunter's Screw, there is an equilibrium, when the power is to the weight as the difference of the distances between the threads of QusToxs.--113. When is there an equilibrium in Hunter's screw? ï~~THE SCREW. 187 the two screws is to the circumference described by the lever. ART. 114. If, therefore, we let P denote the power, W the weight or pressure, C the circumference described by the lever, and d the difference of the distances between the threads of the two screws, we shall have the following proportion: P: W:: d: C; Multiplying extremes and means WXd= PX C; PxC Dividing by d,.......... W= d... Hence, To find the weight or pressure, having given the power, the lever, or circumference described by the lever, and the diffcrence of the distances between the threads of the two screws, we have the following formula and rule: Px C FoRLA,.... W. RuLE..-4ltiply the power into the circumference described by the lever, and divide the product by the difference of the distances between the threads of the two SCreWS. Ex. 1. What pressure will a power of 1 lb., applied at the end of a lever which describes a circumference of 40 inches, exert; one screw having 20 threads to the inch, and the other 21? Ans. Pressure = 16,800 lbs. SoLuTION.-The distance between the threads of the screw having 20 threads to the inch is z in.; the distance between the threads of the screw having 21 threads to the inch, is! in. Hence, - r= in. expresses the difference QVZSTiros.-114. Repeat the rule and formula for finding the pressure. 12 ï~~138 EXERCISES IN NATURAL PHILOSOPHY. of the distances between the threads of the two screws. By substituting, we have PX 1X40 " d =- =40X420=16,800 lbs. 2. Required the pressure produced by a power of 20 lbs., acting at the end of a lever 34 ft. long; the distance between the threads of one screw being 3I inch, and that of the other 3i inch. Ans. Pressure =4,910,400 lbs. 3. Required the pressure produced by a power of j ton; the difference of the distances between the threads of the two screws being to the circumference described by the lever, as 1: 720. - Ans. Pressure =180 tons. 4. In Hunter's screw the circumference described by the lever to which the power is applied, is to the difference of the distances between the threads of the two screws, as 4500 2; what pressure will a power of I ton exert? Ans. Pressure =450 tons. ART. 115. Since (Art. 114) PX C= WXd; Wxd Dividing by C,...... P= dc Hence, To find the power, having given the weight or pressure, the circumference described by the lever, and the difference of the distances between the threads of the two screws, we have.the following formula and rule: WXd FORMULA,.. P=-. RULE.-Multiply the weight or pressure into the difference of the distances between the threads of the two QusrTIoxs.-115. Repeat the rule for finding the power. Deduce the formula. ï~~THE SCREW. 139 screws, and divide the product by the circumference described by the lever. Ex. 1. Required the power necessary to produce a pressure of 600 tons; the circumference described by the lever being to the difference of the distances between the threads of the two screws, as 500: 1. Ans. Power =1 tons. 2. What power must be applied at the end of a lever 7 ft. long, to produce a pressure of 400 tons; one screw having 24 threads to the inch, and the other 30? Ans. Power =14 4 lbs. 3. In Hunter's screw, the distance between the threads of one of the screws is 2k in., and that of the other 2 '! in.; required the power acting at the end of a lever 101 ft. long, necessary to produce a pressure of 114,048 lbs. Ans. Power =10 lbs. 4. In Hunter's screw, the circumference described by the lever, at the end of which the power acts, is 20 ft., the distance between the threads of one screw j- in., and that of the other x in.; required the power necessary to exert a pressure of 9,600 lbs. Ans. Power =2 lbs. ART. 116. Since (Art. 114) PX C= Wxd; Wx d Dividing by P,...... -- d Hence, To find the length of the lever, or the circumference described by the lever, we have the following formula and rule: Wxd FOULA,... 0=-----. RuLE.--Multiply the pressure or weight into the difference of the distances between the threads of the two screws, and divide the product by the power. QuEsTxos.--118. Deduce the formula for finding the circumference. Repeat the rule. ï~~140 EXERCISES IN NATURAL eAHILOSOPHY. Ex. 1. Required the circumference described by the lever attached to a Hunter's screw; the power being to the weight as 2: 2400, and the difference of the distances between the threads of the two screws. inch. Ans. Circum. =60 in. SoLUTIo.-Formula, Wxd 2400 X '- 2400 1 C -- - 2 =2 X-=60 in. P 2 2 20 2. Required the circumference described by the lever; the power being to the pressure as 1: 8000, and the difference of the distances between the threads of the two screws '4 inch. Ans. Circum. =7 ft. 41 in. 3. In Hunter's screw a power of 12 lbs. is made to exert a pressure of 84,000 lbs.; the difference of the distances between threads of the two screws is 7~ in.; required the circumference described by the lever to which the power is aljplied. Ans. Circum. ---S ft. 4 in. 4. By means of Hunter's screw, a power of lb. is made to exert a pressure of 1,200 lbs.; the difference of the distances between the threads of the two screws being A in.; required the circumference described by the lever at the end of which the power is applied. Ans. Circum. =25 ft. ART. 117. Since (Art. 114) Wx d= P X C; PX C Dividing by W,... d=. Hence, To find the difference of the distances between the threads of the two screws, we have the following formula and rule: PXC FORMULA,... d=. RuLE.-Multiply the power into the circumference ï~~rHE SCREW. 141 described by the lever, and divide the product by the weight or pressure. Ex. 1. Required the difference of the distances between the threads of the two screws; the power being to the weight as 1: 24000, and the circumference described by the lever 80 inches. Ans. Differ. = -Q in. SoLUTIN.-Formula, Px C 1 X80 d =- W -= 40- in. d i 24000 u on 2. In Hunter's screw, the power is to the weight as 2: 7200, the distance between the threads of the larger screw is I inch, and the circumference described by the lever 25 ft.; required the distance between the threads of the smaller screw. Ans. Distance = in. 3. In Hunter's screw the power is to the weight as 1: 3600, the distance between the threads of the smaller screw is inch, and the circumference described by the lever 25 ft.; required the distance between the threads of the larger screw. Ans. Distance =4 in. 4. By means of Hunter's screw, a power of 20 lbs. is made to exert a force of 480,000 lbs., the circumference described by the lever is 5 ft.; required the difference of the distances between the threads of the two screws. Ans. Differ. = -- in. QUESTIOIs.--117. Deduce the formula and repeat the rule for finding the difference of the distances between the threads of the two screws. ï~~142 EXERCISES IN NATURAL IHILOSOPHY. CHAPTER XVII. THE HYDROSTATIC PRESS. Arr. 118. The Hydrostatic Press is one of the most powerful and effective machines for overcoming great resistances, that has ever been invented. Its original construction and application is due to a Mr. Bramah. It is used principally in pressing, and its mechanical efficacy is owing to the principle that, In consequence of the mobility of the particles of a fluid, any force communicated to a portion of a confined fluid, is distributed equally throughout the entire mass. Thus, if a pressure of 10 lbs. be made upon a square inch of water, confined in a vessel, a like pressure of 10 lbs. will be exerted over every square inch of the inner surface of the vessel; and if, at the same time, a body be immersed in the water, there will also be a pressure of 10 lbs. upon every square inch of its surface. If a cubic vessel, each side of which is a foot square, be filled with water, and a pressure of 50 lbs. be made upon one square inch, then the pressure upon each side (since the area of each side is equal to 144 square inches) will be 144 times 50 lbs., or 7,200 lbs. ART. 119. The hydrostatic press is represented by the figure on the opposite page. B is a small, and A a large cylinder. In the small cylinder works a piston, C, of a forcing pump, by means of which the water in the reservoir, D, is drawn up and forced through the valve, F, into the cylinder, A, where it presses up against the piston, L, moving through the collar, N, fitting so closely as to be ï~~THE I DROSTATIC PRESS. Fig. 26. 143 watertight. H is a crosspiece, upon which, whatever is to be pressed is placed. S is a valve, which being opened by means of a screw, the water in the cylinder, A, is drawn off into the reservoir, D, thus allowing the piston, L, to descend. P is the handle or lever to which the power is applied, by means of which the piston of the forcing pump is worked. QuEsTioIs.-118. Describe the hydrostatic press, and t#l1 upon what property of a fluid its effiacy mainly depends. ï~~144 EXERCISES IN NATURAL'HILOSOPHY. ART. 120. Now, whatever force is communicated to any portion of the fluid in the small cylinder, is transmitted to every like portion in the larger. Hence, if the area of the end of the piston, L, be one hundred times that of C, the pressure upon the former will be one hundred times greater than the pressure upon the latter. And, generally, The force exerted upon the two pistons, will be as the areas of their ends, or (since the areas of circles are as the squares of their diameters) as the squares of their diameters. ART. 121. If, therefore, we let P denote the force exerted upon the smaller piston, W, the pressure upon the larger, d, the diameter of the smaller, and D that of the larger, we shall have the following proportion: P: W:: d2: D; Multiplying extremes and means, P X D2= WX d2. Hence, In the Hydrostatic Press, there is an equilibrium, when the power is to the weight, as the square of the diameter of the smaller, is to the square of the diameter of the larger piston: Or, When the power into the square of the diameter of the larger piston, equals the weight into the square o the diameter of the smaller piston. AaRT. 122. Since (Art. 121) PxD2= Wxd2; WXdz Dividing by D',....... P-- D2 Hence, To jind the power acting upon the smaller piston, having given the pressure upon the larger, and the diameters of the two pistons, we have the following formula and rule: QUXSTIONS.-121. In the hydrostatic press, when do the power and weight balance? ï~~THE HYDROSTATIC PRESS. 145 FORMULA, Â~ -Wxd RuL.-M.1tiply the weight into the square of the diameter of the smaller piston, and divide the product by the square of the diameter of the larger piston. Ex. 1. What power must be applied to the smaller piston of a hydrostatic press, to produce a pressure of 1,600 lbs.; the diameters of the two pistons being, to each other, as 1: 4? Ans. Power=100 lbs. SoLUTto.-Formula, Wxd2 1600X1 100bs 1)2- = 100 lbs. 2. Required the power necessary to be applied to the piston of the forcing pump, in a hydrostatic press, to produce a pressure of 2,304 lbs., the diameter of the forcing piston being J inch, and that of the larger piston 1 foot. Ans. Power=4 lbs. 3. The diameter of the smaller piston of a hydrostatic press is 3 inches, and that of the larger 12 inches; what power must be applied to the smaller piston to produce a pressure of 1,440 lbs.? Ans. Power=90 lbs. 4. An individual wishes, by means of a hydrostatic press, to produce a pressure of 100 tons; the diameter of the smaller piston is 2 inches, and that of the larger, 1 ft. 8 inches; required the power. Ans. Power= 1 ton. ART. 123. Since (Art. 121) WXd2-- P x D; Dividing by d2,..I.... W= D2 QUsUTroS.---122. Deduce the formula for finding the power Repeat the rule. With what is the power connected I-Weight 13 ï~~146 EXERCISES IN NATURAL PHILOSOPHY. Hence, To find the pressure, having given the power, and the diameter of the two pistons, we have the following formula and rule: P xD' FORMULA,... W-- d 2 Â~ d2Â~ RULE.-Multiply the power into the square of the diameter of the larger piston, and divide the product by the square of the diameter of the smaller piston. Ex. 1. The diameters of the two pistons of a hydrdstatic press are as 1: 12; what pressure will a power of 400 lbs., applied to the smaller piston, produce? Ans. Pressure= 57,600 lbs. SoLUTION.-Formula, W= PX D2 _ 400X144 - 57,600 lbs. d2 1 - 2. What pressure will a power of 140 lbs., applied to the piston of the forcing pump of a hydrostatic press, exert, the diameter of the piston of the forcing pump being J inch, and that of the larger piston, 2 ft. 2 inches? Ans. Pressure =1,514,240 lbs. 3. Suppose, in a hydrostatic press, the power applied to the smaller piston to be -1 ton; the diameter of the smaller piston being 4 inches, and that of the larger 1 ft. 4 in.: required the pressure. Ans. Pressure=1 ton. 4. What pressure will an individual, by applying a force of -- ton to the smaller piston of a hydrostatic press, be able to exert, the diameter of the smaller piston being I inch, and that of the larger 1 ft. 6 in.? Ans. Pressure= 644 tons. ART. 124. Since (Art. 121) PXD2= Wxd2; QUESTIONS.-123. Deduce the formula for finding the pressure. Repeat the rule. ï~~THE HYDROSTATIC PRESS. 147 W X d2 Dividing by P,..... D== W; Extracting the square root,. D= P Hence, To find the diameter of the larger piston, having the pressure, power, and the diameter of the smaller piston given, we have the following formula and rule: FORMULA...D=4W d RuLE.-Multiply the pressure into the square of the diameter of the smaller piston; divide the product by the power, and extract the square root of the quotient. Ex. 1. Required the diameter of the larger piston of a hydrostatic press; the ratio of the power applied to the smaller piston, to the pressure upon the larger, being as 1: 6,400, and the diameter of the smaller piston I inch. Ans. Diameter=3 ft. 4 in. SoLuTLo.-Formula, D=J -- -- K=1600=40 in.= 3 ft. 4 in. 2. The diameter of the smaller piston of a hydrostatic press is 24 inches; the pressure 25,600 lbs., and the power applied to the smaller piston 16 lbs.: required the diameter of the larger piston. Ans. Diameter=- 8 ft. 4 in. 3. In a hydrostatic press, a power of 40 lbs., applied to the smaller piston, exerts a pressure of 324,000 lbs.; the diameter of the smaller piston is 3- inches; required the diameter of the larger piston. Ans. Diameter= 26 ft. 3 in. QUTESTrONs.-124. Deduce the formula for finding the diameter of the larger piston. Repeat the rule. ï~~148 EXERCISES IN NATURAL PHILOSOPHY. 4. In a hydrostatic press, the ratio of the power to the pressure is as I: 800; the diameter of the smaller piston i inch; required the diameter of the larger piston. Ans. Diameter=1 ft. 8 in. ART. 125. Since (Art. 121) Wx d'=- P xD 2; Dividing by W,......d2= P, Extracting the square root,.. d= W. Hence, To find the diameter of the smaller piston, having given the power, pressure, and the diameter of the larger piston, we have the following formula and rule:,/PXD2 FORMULA,... d =; RULE. Muldtiply the power into the square of the diameter of the larger piston; divide the product by the weight, and extract the square root of the quotient. Ex. 1. In a hydrostatic press, the power, applied to the smaller piston, is 18 lbs.; the pressure upon the larger piston 1,800 lbs.; the diameter of the larger piston 7 inches; what is the diameter of the smaller piston? Ans. Diameter= 1 in. SoLuTrIoN.-Formula, d=IPXD2 /18X7X74 in......W - 1800 TV-' -1 5 in 2. In a hydrostatic press, the pressure upon the smaller piston is, to that upon the larger, as 4: 18,000; the diameter of the larger piston is 1 ft. 8 inches; what is the diameter of the smaller piston? Ans. Diameter= 1 in. QUESTIONS.--125. Repeat the formula and rule for finding the diameter of the smaller piston. ï~~THE HYDROSTATIC PRESS. 149 3. An individual wishes, by applying a force of 50 lbs. to the smaller piston of a hydrostatic press, to produce a pressure of 80,000 lbs.; the diameter of the larger piston is 8 inches; required the diameter of the smaller piston. Ans. Diameter-, in. 4. In a hydrostatic press, the power applied to the smaller piston is;' ton; the pressure exerted 64 tons; the diameter of the larger piston 8 in.,; required the diameter of the smaller piston. Ans. Diameter= 1 in. CHAPTER XVIII. SPECIFIC GRAVITY. ART. 126. By the specific gravity of a body is meant, its weight compared with the weight of another body of equal bulk, taken as a standard. The standard of comparison for solids and liquids is distilled water, at a temperature of 620 Fahrenheit, the barometer standing at 30 inches. The specific gravity of a solid or liquid is said to be 4, when its weight is 4 times that of an equal bulk of water; 5, when 5 times that of an equal bulk of water, &c. Now, it has been ascertained, by experiment, that a body weighs as much less in water than in air, as the weight of water which it displaces. Hence, the difference between the weight of a body in air and water, expresses the true weight of a volume of water equal in bulk to that of the body. ART. 127. If a body in air weigh 12 grains, and in water 10 grains, the difference, 2 grains, will exQuarrTms.--126. What is meant by the specific gravity of a body? What part of its weight does a body lose when weighed in water? ï~~150 EXERCISES IN NATURAL PHILOSOPHY. press the weight of water equal in bulk to that of the body. Now, by dividing 12, the weight of the body in air, by 2, its loss in water, we obtain 6 as its specific gravity. Hence, To find the specifc gravity of any body, heavier than water, we have the following RULE.-Divide the weight of the body in air by its apparent loss of weight in water. Ex. 1. Required the specific gravity of a body whose weight in air is 20 grs., and in water 16 grs. Ans. 5. SOLUTION. 20 grs. -16 grs. =4 grs., the loss of the weight in water. Hence, 20.4=5, the specific gravity of the body. 2. Required the specific gravity of a body whose weight in air is 30 grs., and in water 20 grs. Ans. 3. 3. What is the specific gravity of a body, which in air weighs 8.25 grs., and in water 4.5 grs.? Ans. 2.2. 4. Required the specific gravity of a mineral, which in air weighs 17.75 grs., and in water 14 grs. Ans. 4.73+. ART. 128. SPECIFIC GRAVITY OF BODIES LIGHTER THAN WATER. Suppose the weight, in water, of a heavy body be ascertained, then, if a body lighter than water be attached to it, and both be weighed in water, the -lighter body will not only lose all of its own weight, but will, by its buoyancy, cause the heavy body to weigh less than when previously balanced in water. Now, it is found, that the weight, in air, of the QusTxozs.-127. Repeat and explain the rule for finding the specific gravityof a body. ï~~SPECIFIC GRAVITY. 151 light body, added to the loss thus occasioned in the heavy one, equals the weight of a volume of water equal in bulk to that of the light body. Hence, To find the specific gravity of a body lighter than water, we have the following RuLE.-Divide the weight of the body by its weight added to the loss of weight which it occasions in a heavy body, previously balanced in water. Ex. 1. Required the specific gravity of a body weighing 20 grs., which, being weighed with a heavy body in water, causes it to weigh 5 grs. less than when previously balanced in water. Ans. 0.8. SOLUTION. 20 grs. +5 grs. -25 grs., the weight of a volume of water equal in bulk to that of the light body. Hence, 20--25= 0.8, the specific gravity of the body. 2. A piece of lead weighs, in water, 122 grs.; a piece of cork whose weight, in air, is 12 grs., being attached to the lead, causes it to weigh, in water, but 84 grs. What is the specific gravity of the cork? Ans. 0.24. SOLUTION. 122 grs. 4-84 grs. =38 grs., the loss of weight in the lead occasioned by the cork. Hence, 12"grs. +38 grs. =50 grs., the weight of a volume of water equal in bulk to that of the cork; therefore, 12-50=0.24, the specific gravity of the cork. 3. An individual, wishing to ascertain the specific gravity of camphor, finds that a piece weighing 12 grs., being weighed with a heavy body in water, causes it to weigh 0.115 grs. less than when previously balanced in water; required the specific gravity of the camphor. Ans. 0.99-. Qursnows.-128. Repeat and explain the rule for finding tle specific gravity of bodies lighter than water. ï~~152 EXERCISES IN NATURAL PHILOSOPHY. 4. A piece of lead weighs, in water, 24 grs.; a piece of beech wood, whose weight, in air, is 6 grs., being attached to the lead, causes it to weigh, in water, but 22.96 grs.; required the specific gravity of the beech wood. Ans. 0.852-+-. - ART. 129. SPECIFIC GRAVITY OF LIQUIDS. Since (Art. 126) the specific gravity of any body, is its weight compared with that of an equal bulk of the standard body; and, since the standard of comparison for liquids as well as solids, is distilled water, To fnd the specific gravity of liquids, we shall evidently have the following RULE 1.- Weigh equal volumes of water and the liquid whose specfic gravity is sought, and divide the weight of the latter by that of the former. Or, since a body weighs as much less in any fluid, as the weight of the fluid which it displaces, it is evident, if the same body be weighed in two different fluids, the loss of weight in each will express the weight of equal bulks of the two fluids. Hence, To find the specific gravity of liquids, we may also have the following RULE 2.-Divide the weight which a'solid body loses, when weighed in the liquid whose specific gravity is required, by the loss of weight which the same body sustains when weighed in water. Ex. 1. Required the specific gravity of acetic acid, a given volume of water weighing 125 grs., and an equal volume of the acid 132.75 grs. Ans. 1.062. SOLUTION.. 132.75-125 = 1.062. 2. A given bulk of distilled water weighs 62.5 QuZsTxows.-129. Explain and repeat the two rules for finding the specific gravity of liquids. ï~~SPECIFIC GRAVITY. 158 grs., and an equal bulk of muriatic acid 75 grs.; required the specific gravity of the acid. Ans. 1.2. 3. Required the specific gravity of alcohol, a given volume of which is found to weigh 213.75 grs., and an equal volume of distilled water 250 grs. Ans. 0.855. 4. Required the specific gravity of the water of the Dead Sea, a given bulk of which weighs 155 grs., and an equal bulk of distilled water 125 grs. Ans. 1.24. 5. What is the specific gravity of Port wine, a given quantity of which is found to weigh 497.5 grs., and an equal quantity of distilled water 500 grs.? Ans. 0.995. 6. A merchant, wishing to ascertain the specific gravity of Bordeaux wine, weighed a solid body in it, and found it weighed 15.3915 grs. less than in air, and also, that the same body, weighed in distilled water, sustained a loss of 15.5 grs.; required the specific gravity of the wine. Ans. 0.993. 7. What is the specific gravity of vinegar, in which a solid body being weighed, sustains a loss of 16.875 grs., the same body weighed in distilled water losing 15.625 grs.? Ans. 1.08. 8. A body, being weighed in sea water, sustained a loss of 32.125 grs., the same body weighed in distilled water, weighed 31.25 grs. less than in air; required the specific gravity of the sea water. Ans. 1.028. ART. 130. SPECIFIC GRAVITY OF GASES. In obtaining the specific gravity of gases, atmospheric air is taken as the standard of comparison. We say, therefore, that the specific gravity of a gas is 2.5, or 1.5, &c., when a given volume of the gas weighs 2.5, or 1.5 times as much as an equal volume of atmospheric air. ï~~154 EXERCISES IN NATURAL PHILOSOPHY. Hence, To find the speefic gravity of any gas, we shall, evidently, have the following RULE.- Weigh equal volumes of atmospheric air and the gas whose specfIc gravity is sought, and divide the weight of the latter by that of the former. Ex. 1. A given volume of atmospheric air weighs 20 grs. and an equal volume of sulphureous acid 44.44 grs.; what is the specific gravity of sulphureous acid? Ans. 2.222. SOLUTION. 44.44-20 =2.222, the specific gravity of sulphureous acid. 2. Required the specific gravity of oxygen gas, a given volume of which weighs 17.776 grs., and an equal volume of atmospheric air 16 grs. Ans. 1.111. 3. Required the specific gravity of ammonia, a given volume of which weighs 4.720 grs., and an equal volume of atmospheric air 8 grs. Ans. 0.590. 4. A given volume of hydrogen weighs 2.070 grs., and an equal volume of atmospheric air, 30 grs.; required the specific gravity of hydrogen. Ans. 0.069. 5. A given volume of sulphureted hydrogen weighs 47.200 grs., and an equal volume of atmospheric air 40 grs.; required the specific gravity of sulphureted hydrogen. Ans 1.180. 6. Required the specific gravity of chlorine gas, a given volume of which weighs 62.500 grs., and an equal volume of atmospheric air 25 grs. Ans. 2.500. QuxsTros.-130. What is the standard of comparison for ascertaining the specific gravity of gases? Explain and repeat the rule for finding the specific gravity of gases. ï~~SPECIFIC GRAVITY. 155 Aar. 131. Weights of given volumes of water and air, the barometer standing at 30 in., and thermometer at 620. Weight of a cubic ft. (1,728 cub. in) of water: 62.32 lbs. avoirdup.,i,, i iK -=997.12 oz. ", - in. " =-252.45 gr. troy i 100 cub. in. of air..... =. 30.49 " " AaRT. 132. The principle of specific gravity affords a very easy means of ascertaining the weight of any body; for, since the specific gravity of a body, is its weight compared with that of an equal bulk of water, if we multiply the weight of a cubic inch, or cubic foot of water, by the specific gravity of any body, the product will denote the weight of a cubic inch or cubic foot of that body. Hence, To find the weight of any body, its specific. gravity being known, we have the following RULE.-Find the weight of a volume of water, equal in bulk to that of the body, and multiply this by the specific gravity of the body. Ex. 1. Required the weight of 2 cubic inches of gold, its specific gravity being 19.25. Ans. 1 lb. 8 oz. 4 pwt. 23.325 grs. SOLUTION.-252.45 grs. X2 =504.90 grs.,the weight of 2 cubic inches of water. Hence, 504.90 grs.X 19.25=9719.325 grs.=-1 lb. 8 oz. 4 pwt. 23.325 grs. 2. Required the weight of 10 cubic in. of silver, its specific gravity being 10.5. 3. Required the weight of a cubic ft. of iron, its specific gravity being 7. 4. Required the weight of 2 cubic ft. of mercury, its specific gravity being 13.5. 4rsmrrows.-132. Explain, and repeat the rule for finding the weight of bodies, their specific gravities being known. ï~~156 EXERCISES IN NATURAL PHILOSOPHY. 5. Required the weight of 20 cubic ft. of tin, its specific gravity being 7.2. 6. Required the weight of 100 cubic in. of chlorine, its specific gravity being 2.5. 7. Required the weight of a cubic ft. of carbonic acid, its specific gravity being 1.5. 8. Required the weight of 4 cubic ft. of rock crystal, its specific gravity being 2.6. 9. Required the weight of 8 solid ft. of ivory, its specific gravity being 1.8. 10. Required the weight of a cord of ash wood, its specific gravity being 0.845. 11. Required the weight of a cord of oak wood, its specific gravity being 0.925. ART. 133. Heat expands all bodies; liquids more than solids, and gases more than solids or liquids. The expansion of gases and liquids is very apparent, 1,000 parts of water expanding, between the freezing point, 32Â~, and the boiling point, 2120, to 1,046 parts. It is evident, therefore, that the weight of equal bulks, of the same body, will increase as its temperature decreases, and decrease, as its temperature increases. Hence the necessity of employing some uniform standard, for ascertaining the true weight of bodies. Water, at the temperature of 400, is at its greatest density; and the weight of equal bulks of water, at this temperature, is always the same. Therefore, the avoirdupois pound, of such magnitude that a cubic foot of pure water, at its maximum density, shall weigh 1,000 ounces, or 621 lbs., is taken as the.unit of measures of weight. We give below a Table of magnitudes, corresponding to given weights of water, at its maximum density, or at a temperature of 400. QuESTIoSs.-183. Repeat the table of cubic contents, corresponding to given weights of water. ï~~SPECIFIC GRAVITY. 150 s. ft. s. illn. Cub. cont. of vol. of wat. weigh. 1 oz. av. =1a or 1;.. " " " 1 lb. " =_ 1 or 27-; " " " " " 1,000 oz. " = 1 or 1728; 432 1 gr. troy =....T-ays"f ART. 134. MAGNITUDE OR CUBIC CONTENTS OF IRREGULAR BODIES. The magnitude of any irregular body may be easily determined, by noting the loss of weight which it sustains when weighed in water; for, its loss of weight is the weight of a volume of water equal in bulk to that of the body. If, for example, a fragment of iron is found, when weighed in water, to lose 10 oz. of its weight, then, since the cubic contents of a volume of water, weighing 1 oz., is equal to 1, s. in., that of the body, or a volume of water weighing 10 oz., will be 10 times 1y'- s. in.=17 27 s. in. Hence, To find the magnitude of any irregular body, we have the following IRULE.-J ultiply thse cubic contents of a volume of water, weighing 1 oz., 1 lb., 4-c., by the number of ounces or pounds, 4c., representing the loss of weight, which the body when weighed in water, sustains. Ex. 1. Required the number of solid in. in an irregular piece of gold, which, being weighed in water, sustains a loss of 437.5 grs. Ans. 1 -"T s. in. SOLUTION.-- T W"ffi s. in. X 437.5= 1 z s. in. 2. Required the number of solid ft., in an irregular rock of granite, which, when weighed in water, was found to lose a part of its weight equal to 1,000 lbs. Ans. 16 s. ft. QussTrxoxs.-134. Repeat, and explain the rule for finding the. maguitude of irregular bodies. ï~~158 EXERCISES IN NATURAL PHILOSOPHY. 3. Required the number of solid ft., in an irregular body, which, when weighed in water, sustained a loss of weight equal to 2,500 lbs. Ans. 40 s. ft. 4. A silversmith, wishing to ascertain the number of solid inches in a gold chain, finds that it loses, when weighed in water, a part of its weight equal to 2,02555 grs.; required the number of solid inches in the chain. Ans. 8 s. in. CHAPTER XIX. HYDROSTATICS. EQUAL PRESSURE OF FLUIDS. ART. 135. All the particles of a fluid at rest, at the same depth, exert and sustain an equal pressure in every direction, upward, downward, and laterally. Were this not the case, then, since the particles of a fluid are free to move, they would move in that direction in which the pressure was least. But the fluid, by supposition, is at rest; and hence the truth of the proposition follows. ART. 136. DOWNWARD PRESSURE OF FLUIDS. The downward pressure of any fluid, of uniform density, is proportioned to its depth. For example, at the depth of 2 feet, the pressure upon a square foot is equal to the weight of two cubic feet of the fluid; at the depth of 3 feet, three cubic feet; at the depth of 4 feet, four cubic feet, &c. The presQUESTIONS.--136. What is the downward pressure of a fluid proportioned to? Repeat and explain the rule for finding the pressure of any fluid upon a horizontal surface. ï~~HYDROSTATICS. 159 sure of any fluid, therefore, upon a horizontal surface, is equal to the weight of a column of the fluid, found by multiplying the area of the base into its depth below the surface. Hence, To find the pressure of a fluid upon a horizontal surface, we have the following RULE.-Midtiply the weight of a solid inch or solid foot of the fluid, by the number of cubic inches, or cubic feet found by multiplying the area of the base into its depth below the surface. The following table exhibits the pressure of water (its density being supposed to be uniform) upon a square foot, at different depths. Depth in ft. Press. on sq. ft. Depth in ft. Press. on sq. ft. 1 - - 621 lbs. - - 24 - - 1,500 lbs. 4 - - 250 " - - 28 - - 1,750 " 8 - - 500" - - 32 - - 2,000 10 - - 625" - - 40 - - 2,500 " 12 - - 750".- - 50 - - 3,125" 14 - - 875 " - - 60 - - 3,750 " 16 - - 1,000 " - - 70 - - 4,375 " 20 - -1,250 " 1 m. or 5,280 ft. 330,000 " Ex. 1. Required the pressure on the bottom of a vessel filled with water, the depth of the vessel being 8 ft., and the area of the base 40 square ft. Ans. 20,000 lbs. SOLUTION. 40X8= 320 solid ft.; and 62} lbs. X320=20,000 lbs. 2. Required the pressure upon the ttom of a cylindrical vessel filled with water; the dineter of the vessel being 7 ft., and its hight 10 ft. Ans. 24,062.5 lbs. NoTE.-The circumference of a circle is 3, times the diameter; and the area of a circle is found by multiplying the circumference into one fourth of the diameter. ï~~160 EXERCISES IN NATURAL PHILOSOPHiY. 3. The surface of a Greenland whale is 3,400 square ft.; required the pressure which it will sustain at the depth of 4,200 ft. Ans. 398,4371 tons. 4. A cylindrical vessel, 31 ft. in diameter, and 6 ft, deep, is filled with mercury whose specific gravity is 13.5; required the pressure on the bottom of the vessel. 5. A cylindrical vessel, 2 ft. in diameter, and 10 ft. high, is filled with Burgundy wine whose specific gravity is 0.99; required the pressure upon the bottom of the vessel. ART. 137. PRESSURE OF FLUIDS UPON OBLIQUE OR PERPENDICULAR SURFACES. If a fluid press upon an oblique or perpendicular surface, all parts of this surface will not be at equal depths below the surface of the fluid; hence, the average depth, or in other words, the depth of the center of gravity must be taken. Hence, To find the pressure of a fluid upon an oblique or perpendicular surface, we have the following RULE.-M ltiply the weight of a solid inch, or solid foot of the fluid by the number of cubic inches or cubic feet found by multiplying the area of the surface into the depth of its center of gravity below the surface of the fluid. Ex. 1. Required the pressure of water upon a sloping dam; the length of the dam being 300 ft., the slant hight 20 ft., and depth of the water 14 ft. Ans. 2,625,000 lbs. SOLUTION. 300 X 20= 6,000, number of square feet in the area of the dam; 6000X7 =42,000, number of cubic feet in a volume of water whose weight is the pressure upon the dam. QtrBsTIons.-137. Repeat and explain the rule for finding the pressure of fluids on oblique or perpendicular surfaces. ï~~HYDROSTATICS. 161 Hence, 621 X42000= 2,625,000 pounds, pressure upon the dam. 2. A cylindrical vessel, whose diameter is 7 ft., and hight 24 ft., is filled with water; required the pressure against the upright sides. Ans. 16,500 lbs. 3. Required the pressure against the sides of a cubic vessel filled with water, each side of the vessel containing 16 square ft. Ans. 12,000 lbs. 4. A milldam, whose slant hight is 30 ft., runs horizontally across a river; the average depth of the river is 14 ft., and its breadth 400 yds.; required the amount of pressure on the dam. Ans. 70,312.5 tons. ART. 138. FLUIDS IN MOTION. The following exhibit two of the most important laws that govern fluids in motion. LAW 1.-- Thie velocity with which a fluid issues from an orifice in the side or bottom of a vessel, is equal to that acquired by a body falling freely through a vertical hliht, equal to the depth of the orifice below the surface of twhe fluid. Ex. 1. In a vessel containing water, an orifice is made 9 ft. below the surface; supposing a body to fall freely under the influence of gravity 16 ft. the first second, with what velocity will the water issue per second? Ans. 24 ft. SOLUTION.-By Art. 17, V=24fSX16; substituting, we have V=2/9X16=241:i_=2X12=24 ft. velocity per second. 2. In a containing vessel an orifice is made, 12 ft. below the surface; required the velocity of the spouting liquid, on the supposition that a body falls QuzsTros.-138. Repeat law 1. Repeat the formula and rule for finding the velocity. 14 ï~~162 EXERCISES IN NATURAL PHILOSOPHY. freely under the influence of gravity 16a ft. the first second. Ans. 27.6+ ft. per sec. 3. A cylindrical vessel, 20 ft. high, is filled with water; required the velocity with which the water will issue from an aperture in the bottom of the vessel, supposing a body to fall freely under the influence of gravity 16 ft. the first second. Ans. 35.8+ ft. per sec. ART. 139. LAW 2.- The quantity of a fluid issuing from an orifice in the side or bottom of a vessel, varies as the square root of the depth of the orifice below the surface of the fluid. Ex. 1. In a vessel, filled with water, two equal orifices are made, one 9 ft., and the other 16 ft. below the surface. The first orifice, in a given time, discharges 15 galls.; required the number of gallons discharged by the second orifice in the same time. Ans. 20 galls. SOLUTION.. /9:,16 15: Ans. 3 4:15: 20, Ans. 2. In a vessel filled with water, three equal orifices are made; one 4 ft., one 9 ft., and the other 25 ft. below the surface. The quantity of water discharged by the first orifice, in a given time, is 100 galls.; required the quantity of water discharged by the other orifices in the same time. Ans. 150 and 250 galls. 3. In a vessel kept constantly full, an orifice, situated 16 ft. below the surface, discharges, in a given time, 800 galls.; at what depth must another equal orifice be made, to discharge 1,200 galls. in the same time? Ans. 36 ft. QUESTIONS.-139. Repeat law 2. ï~~SOUND. 163 CHAPTER XX. SOUND. ART. 140. Sound, or the vibratory motions of the atmosphere, by which sound is produced, moves at the rate of 1,142 ft. per second. By considering the velocity of light instantaneous, that of sound being known, the distance of the sonorous body, in many instances, may be easily determined. If, for example, a flash of lightning is seen 4 seconds before the thunder is heard, then the distance of the cloud will be 4 times 1,142 ft. =4,568 ft. Ex. 1. How far will sound travel in 8 sec.?-in 9 sec.?-in 10 sec.? Ans. In 8 sec. 9,136 ft., in 9 sec. 10,278 ft., in 10 sec. 11,420 ft. 2. Observing a cloud during a thunder storm, 15 sec. after I saw the flash of lightning I heard it thunder; required the distance of the cloud. Ans. 17,130 ft. 3. Required the distance of a cloud from which the thunder is 24 sec. in reaching the ear. Ans. 27,408 ft. 4. Standing upon a bank of a river I fired a pistol, and in 6 sec. afterward heard the echo, the sound being reflected back by a projecting rock on the opposite side of the river; what was the width of the river? Ans. 3,426 ft. 5. A stone let fall from a high eminence was heard to strike the ground 7 sec. after it was seen to reach it; required the hight of the eminence. Ans. 7,984 ft. 6. A scout, standing upon a high hill, saw the QuEsTows.-140. With what velocity does sound move? How may the distance of a sonorous body be found? ï~~164 EXERCISES IN NATURAL PHILOSOPHY. flash from one of the enemy's cannon, and 20 sec. afterward heard the report; required the distance of the enemy. Ans. 4 m. 1,720 ft. CHAPTER XXI. TABLE OF FORMULAS. ART. 141. FALLING BODIES. Space described by falling bodies. s=T2Xs (Art. 10). S= V +4s (Art. 14). Space fallen the first equal portion of time at any given distance above the earth. 40002 s-(D+4000) 2Xs' (Art. 9a). Distance above the earth from which a body will fall a given space the first equal portion of time. D= X40002-4000 (Art. 9b). Space described during any equal portion of time. S-=s(2T-1) (Art. 13). Time of falling bodies. T= 4 (Art. 15). V T=- (Art. 16). Velocity of falling bodies. V=2 VSXs (Art. 17). V=TX2s (Art. 18). Space described by bodies projected downward. S= TX V+"T 2Xs (Art. 19). ART. 142. UNIFORM MOTION. = VX T; V=t 20+T; T =,-. V (Arts. 20, 21, 22). ï~~TABLE OF FORMULAS. 165 ART. 143. MOMENTUM. M= QX V; V=M--Q; Q=--aV (Arts. 23, 24, and 25). ART. 144. PENDULUM. Length of pendulum. L= T2X39 (Art. 28). Time of vibration. -,4 L-.39 (Art. 29). Number of vibrations made in a given time at a given distance above the earth. 4000Xn N-- (Art. 31). -D- 4000 Number of vibrations made in a given time at the level of the sea. D--4000 n= 4000 XN (Art. 32). Distance above the earth at which a given pendulum will, in a given time, make a given number of vibrations. (n-N D= N )X4000 (Art. 33). ART. 145. LEVER OF THE FIRST AND SECOND ORDER. WX S PX L Wx X L (Arts. 37, 38, 39, and 40). ART. 146. LEVER OF THE THIRD ORDER. WxL PXS - (Arts. 42, 43, 44, and 45). PXS _. W4 ART. 147. COMPOUND LEVER. WXwXw'Xw". P- pXp'Xp" ' PXPX'XP" (Art. 48) x p w'x w" ï~~166 EXERCISES IN NATURAL PHILOSOPHY. ART. 148. To obtain the true weight by a false balance. W=V -lxs (Art. 51). ART. 149. THE WHEEL AND AXLE. Wxd PXD P = D-; W ---; WXd P XD (Arts. 55, 56, 57, and 58). Wxd PXD1 D- p; d- W ART. 150. COMPOUND WHEEL AND AXLE. WxdXd'xd" P-- DXD'XD" (Art. 61). P PxDxD' X D" - dxd' X d" (Art. 62). ART. 151. THE PULLEY. To find the power, weight, &c., in a system of pulleys, in which the same rope passes around all the pulleys. W--Xn; P= W-n; Â~(Arts. 65, 66, and 67). m= W--2P To find the power and weight in a system of pulleys in which each movable pulley is supported by a separate string, one end of each string being fastened to a fixed beam. W=PX2" (Art. 69). P= W-2n (Art. 70). To find the power and weight in a system of pulleys in which each rope, instead of being fastened to a fixed beam, passes around a fixed pulley, and is attached to the respective blocks of each movable pulley. W=PX 3";,.(Art. 72). P= W-+3" (Art. 73). To find the power and weight in a system of pulleys in which each rope, after passing around its respective pulley, is attached to the weight. W=P X (2n-1) (Art. 74). P= W-2"-1 (Art. 75). To find the power and weight in a system of pulleys in which each rope, after passing around its respective pulley, is ï~~TABLE OF FORMULAS. 167 made to pass around a fixed pulley connected to the weight, and finally attached to the block of the pulley around which it first passed. W=PX (3n-1) (Art. 76). P= W--3U--1 (Art. 77). ART. 152. THE INCLINED PLANE. To find the power, weight, pressure, &c., when the power acts parallel to the face of the plane. WXH ppxH P L B PX L pX L H-; B= -, Wx H WX B L-- p, L---- P (Arts. 85, 86, 87, &c.). PXL PX B W pi px H pxL B=p, B -, PXB WXB p= H ' F-- L W= JP2+P' =p W2- P' (Art. 98). p=4 W 2-P2 ART. 153. To find the power, weight, pressure, &c., when the power acts parallel to the base of the plane. WX H pXH... B L PXB pxB H WL PXL WXL P= H ' B; WX WX L (Art. 103). B=--; B=, PXB PXL - H p ) L pXH L=pXB P, W ï~~1 68 EXERCISES IN NATURAL PHILOSOPHY. ART. 154. THE SCREW. FWX D WPX C ___ W~D (Arts. 108, 100, 110,&c.). A.RT. 155. HUNTERS SCREW. Px C Wyd W d-;P= C__ WXd (Arts. 114, 115, &c.). ART. 156. HYDROSTATIC PRESS. WXd2 PX D,? ____ fD2' (Arts. 122, 123, &c.). ï~~PROMISCUOUS QUESTIONS. 169 CHAPTER XXII. PROMISCUOUS QUESTIONS. ART. 157. Ex. 1. A body was observed to fall for 10 seconds, and then to move on a horizontal plane for 6 seconds, with its last acquired velocity. What was the whole distance, perpendicular and horizontal, through which the body moved? 2. A cannon ball, fired perpendicularly upward, was gone 10 seconds, when it returned to the same place. To what hight did it ascend, and what was the velocity of projection? 3. A body is projected downward with a velocity of 45 feet per second. How far will it fall in 8 seconds, and through what distance will it fall the last second? 4. A body, at the surface of the earth, weighs 4,800 lbs.; what will be its weight 2,000 miles above the earth?-2,400 miles below the surface of the earth? 5. Will a body weigh less, or more, 3,000 miles above, than 3,000 miles below the surface of the earth; and how much? 6. What part of its weight will a body lose 7,480 miles above the earth?---2,450 miles below the surface of the earth? 7. Compare the intensity of the sun's light at the planets Jupiter and Saturn; the former being 490,000,000 miles, and the latter 900,000,000 miles from the sun. 8. How far must a body fall to acquire a velocity of 800 feet per second, on the supposition that it fell, the first second of time, 4 feet? 9. Required the length of a pendulum that will vibrate once in 3 seconds. 15 ï~~170 EXERCISES IN NATURAL PHILOSOPHY. 10. In what time will a pendulum, 29 feet long, make a single vibration? 11. In a system of pulleys, in which the same string passes around all the pulleys, the weight is, to the power, as 24: 3; required the number of movable pulleys. 12. What weight, in a system of pulleys represented by Fig. 17 (see Art. 68), will a power of 450 lbs. counterpoise; the number of movable pulleys being 10? 13. What power, in a system of pulleys repre-. sented by Fig. 18 (Art. 71), will balance a weight of 8,100 lbs.; the number of movable pulleys being 4? 14. Required the difference between the weights counterpoised by a power of 10 lbs., in the two systems of pulleys represented by Figs. 17, and 18 (Arts. 68 and 71); the number of movable pulleys, in each, being 10. 15. What weight will an individual balance, by exerting a force of 320 lbs., at the end of the rope passing around the first pulley, in a system represented by Fig. 19 (Art. 74); the number of ropes being 10? 16. Required the difference between the weights counterpoised by a power of 100 lbs., in the two systems of pulleys represented by Figs. 19 and 20 (Arts. 74 and 76); the number of ropes, in each, being 12? 17. A body is acted on by two forces, represented by the numbers 6 and 8, at right angles to each other; what number will represent the resultant? 18. Two forces act at right angles; the resultant is 40, and one of the forces 32; required the other force. 19. What power, acting parallel to the face of an inclined plane, will sustain a weight of 3,800 ï~~PROMISCUOUS QUESTIONS. 171 lbs., the hight of the plane being 12 feet, and the length 30 feet? 20. Suppose a body, at the level of the sea, to weigh 60 tons; required its loss of weight on the top of a mountain 5 miles high. 21. Suppose the stone, which David threw at Goliath of Gath, to have weighed 2 ounces; that Goliath weighed 350 lbs.; and, when hit, was walking toward David at the rate of 60 feet per minute; with what velocity must the stone have been thrown to have prostrated the giant? 22. Compare the momenta of two ships, one of 200 tons, moving at the rate of 6 miles per hour; the other of 400 tons, moving at the rate of 9 miles per hour. 23. Suppose a globe of the same density of the earth, and 1,200 miles in diameter, to be situated at the distance of 8,027 miles from the earth; now, if both were left to their- own mutual attraction, through what space would each move? NOTE.-The solid contents of spherical bodies are as the cubes of their diameters. 24. A man fell into a pit 360 feet deep; how long was he in falling? What velocity did he acquire? And through what space did he fall the last second? 25. Suppose a ball, fired with a velocity of 1,200 feet per second, to enter a suspended stick of timber weighing 48 lbs., and to give it a motion of 3 feet per second; required the weight of the ball. 26. A meteoric stone fell upon a projecting stick of timber with a momentum, which, from the motion given to the stick, was estimated at 18,400 lbs. It occupied, in falling, 10 seconds. Required the hight from which it fell, and the weight of the stone. 27. An individual wishing to ascertain the velo ï~~172 EXERCISES IN NATURAL PHILOSOPHY. city of a cannon ball, suspends, in the manner of a pendulum, a block of wood, and finds that the ball, on entering the block, gives it a motion of 60 feet per second; required the velocity of the cannon ball, its weight being 18 lbs., and that of the block 600 lbs. 28. Required the difference between the momenta of two ships, one of 360 tons, sailing 8 knots per hour, and the other 180 tons, sailing 16 knots per hour. 29. Suppose 2 grains of light, emanating from the sun with the velocity of 20,000 miles per second, to impinge against a mass of ice, floating in a river, at the rate of 4 feet per second; what weight of ice would the light stop? 30. A rifle ball, weighing 3 ounces, was shot perpendicularly upward, and, as it issued from the gun, entered a suspended log, weighing 60 lbs. 5 oz., and gave it a motion of 6 feet per second; how far would the ball have ascended, and what time would have elapsed before its return, on the supposition that it had not entered the log? 31. A boy, standing on the top of a tower, threw a ball perpendicularly upward, which returned in 8 seconds, when he gave it an additional force downward,- of 15 feet per second; it then reached the ground in 3 seconds; required the hight of the tower. 32. In what time would a man fall from a balloon 4 miles high?-And what velocity would he acquire? 33. A body has fallen 2,316 feet; required the space fallen the last second? 34. A body is projected, perpendicularly upward, 9,264 feet; required the velocity of projection, and the distance the body ascended the 12th second. 35. A body, projected perpendicularly upward, ascends 5,211 feet; required the time of ascent; the ï~~PROMISCUOUS QUESTIONS. 173 velocity of projection, and the difference between the distances passed through the 10th second of its ascent, and the 10th second of its descent. 36. Suppose a body, during the last second of its fall, to pass through a space of 498 ft. 7 in.; required the time of falling, and the whole distance described. NOTE.-It is evident, that the time of falling will be indicated by the reverse of rule under Art. 13. 37. The distance fallen by a hailstone, the last second of its fall, was 1,013 ft. 3 in.; required the time of falling;-the hight of the cloud from which it fell;-and the velocity acquired during its fall. 38. In what time will a pendulum, 31 feet long, make a single vibration? 39. How long would a pendulum, I mile in length, be in making a single vibration? 40. Suppose a pendulum, at the surface of the earth, to beat J seconds; what number of vibrations would it make per minute, at the distance of 10,000 miles from the earth? 41. A pendulum, 8,400 miles from the earth, makes 2,400 vibrations per hour; required the number of vibrations that the same pendulum would make per hour, at the surface of the earth. 42. At what distance above the earth would a pendulum, vibrating J seconds at the surface, make 840 vibrations per hour? 43. With what velocity must a body be projected perpendicularly upward, to ascend to that hight at which a pendulum, vibrating seconds, at the surface of the earth, would make 3,560 vibrations per hour? 44. Required the time it would take a body to fall to the earth, from that point at which a pendulum 39 in. long, makes 3,591 vibrations per hour. 45. What power, in a compound lever consisting ï~~174 EXERCISES IN NATURAL PHILOSOPHY. of four simple levers, will balance a weight of 4,600 lbs., the length of the arms, on the side next to the power, being as the numbers 6, 8, 12, and 14, and the length of the arms on the side next the weight, as 3, 5, 7, 9? 46. Required the true weight of a body, which weighs 8 lbs., in one scale of a false balance, and 121 lbs., in the other. 47. In a wheel and axle, the power is, to the weight, as 3: 8; the diameter of the wheel 9 feet; required the diameter of the axle. 48. In a wheel and axle, the diameter of the axle is, to that of the wheel, as 5: 19; the power 340 lbs.; required the weight. 49. What power, in a wheel and axle, will balance a weight of 8,640 lbs., the diameter of the axle being, to that of the wheel, as 31: 17? 50. Which, with the same power, will overcome the greater resistance, a compound wheel and axle the diameters of whose axles are represented by the numbers 2, 3, 5, and 7, respectively, and those of the wheels by 4, 7, 10, and 24; or a screw, the distance between whose threads is 4 in., and the lever by means of which-the screw is turned, 7 feet long? 51. In a compound wheel and axle, the ratio of the diameters of the first wheel and axle is 3; that of the diameters of the second 41; of the third 8; required the weight which a power of 400 lbs. will balance. 52. What weight, in a system of pulleys (see Art. 65) of which 6 are movable, and in which the same string goes around all the pulleys, will a power of 3 lbs. balance? 53. Which will overcome the greater resistance, and how much, a power of 5 lbs. in a system of pulleys of which five are movable, the same string passing around all the pulleys; or, a power of 3 ï~~PROMISCUOUS QUESTIONS. 175 lbs., applied at the end of a lever 15 feet long, which turns a screw, the distance between whose threads is 1- in.? 54. What power, in a system of pulleys of which 8 are movable, the same string passing around all the pulleys, will balance a weight of 6,300 lbs.? 55. The hight of an inclined plane is, to its length, as 5: 9; what power, acting parallel to the face of the plane, will balance 12 tons? 56. The hight of an inclined plane is, to its base, as 5: 18; what power, acting parallel to the face of the plane, will sustain a weight whose pressure upon the plane is 2,800 lbs.? 57. A weight, resting on an inclined plane, is sustained by a power of 4 tons acting parallel to the face of the plane; the hight of the plane is, to the base, as 2: 5; required the pressure of the weight upon the plane. 58. The base of an inclined plane is 150 feet long, the power acting parallel to the plane is, to the pressure of the weight on the plane, as 8: 19; required the hight of the plane. 59. Compare the force exerted by a power of 10 lbs., acting parallel to a plane whose hight is, to its length, as 1: 281, with that exerted by a power of 3 lbs., applied at the end of a lever 15 feet long, turning a screw the distance between whose threads is $ in. 60. Which will exert the greatest force, a power of 15 lbs., acting parallel to an inclined plane whose base is, to its hight, as 2: 29; or a power of 5 lbs., acting upon a compound lever, the length of the arms on the side next to the power being as the numbers 6, 7, 8, and 9, respectively, and the lengths of those on the side next to the weight 12, 15, 40, and 60; or, a power of 3 lbs., acting at the end of a lever which turns a screw, the distance between whose threads is 2 inches? ï~~176 EXERCISES IN NATURAL PHILOSOPHY. 61. What power, acting parallel to the base of an inclined plane, will sustain a weight of 1,800 lbs., the hight being 6 feet, and the base 24 feet? 62. What power, acting parallel to the base of an inclined plane, will sustain a weight whose pressure upon the plane is 3,800 lbs., the hight of the plane being, to its length, as 3: 27? 63. Required the difference between the weight sustained by a power of 10 lbs., acting parallel to the face of an inclined plane whose hight is, to its length, as 2: 9, and that sustained by a power of 18 lbs., acting parallel to the base of an inclined plane whose base is, to its hight, as 24: 8. 64. What force must an individual exert, parallel to the base of an inclined plane whose hight is 10 feet, and length 50 feet, to sustain a weight whose pressure upon the plane is 240 lbs.? 65. Required the length of an inclined plane, upon which a weight of 380 lbs. exerts a pressure of 180 lbs., the base of the plane being 10 feet. 66. Upon an inclined plane whose hight is 10 feet, a power of 24 lbs., acting parallel to the base, sustains a weight of 72 lbs.; required the length of the base. 67. The base of an inclined plane is, to its hight, as 12: 4; what force will a power of 80 lbs., acting parallel to the base of the plane, sustain? 68. What must be the diameter of the wheel, that a power of 200 lbs., attached to it, shall balance a weight of 900 lbs., suspended from an axle 8 in. in diameter? 69. Suppose a lever, by means of which a screw is turned, to be 15 feet long, the distance between the threads of the screw 2 in.; what force will a power of 100 lbs., applied at the end of the lever, exert? 70. By means of a screw, whose threads are 3 in. ï~~PROMISCUOUS QUESTIONS. 177 apart, a man, by exerting, at the end of the lever, a force of 300 lbs., is enabled to produce a pressure of 24 tons; required the length of the lever. 71. Required the power necessary to produce a pressure of 2,500 lbs.; the distance between the threads of the screw being, to the circumference described by the lever, as 2: 400. 72. What pressure, with a hydrostatic press, the diameters of whose pistons are as 2: 40, can be produced by an individual capable of exerting a force of 100 lbs.; on the supposition that this force is applied at the end of the smaller piston. 73. The length of a lever which turns a screw, is 15 feet, the ratio of the power to the weight as 2: 90; required the distance between the threads of the screw. 74. In Hunter's screw, what pressure will a power of 20 lbs., acting at the end of a lever 7 feet long, exert, one screw having 30 threads to the inch, and the other 31? 75. Required the power necessary, by means of Hunter's screw, to sustain a pressure of 12,000 lbs., the distance between the threads of one screw being h in., and that of the other 1 in. 76. In Hunter's screw, the circumference described by the lever at the end of which the power acts, is 30 feet; the distance between the threads of one screw is - in., and that of the other _ in.; what pressure will a power of 100 lbs. exert? 77. By means of Hunter's screw, a power of 20 lbs. is made to exert a pressure of 8,400 lbs.; the circumference described by the lever is 21 feet; required the difference of the distances between the threads of the two screws. - 78. Which will exert the greater force, and how much; a power of 20 lbs., acting at the end of a lever 21 feet long, and turning a screw, the distance ï~~i'8 EXERCISES IN NATURAL PHILOSOPHY. between whose threads is in.; or, a power of 2 lbs., acting at the end of a lever turning a Hunter's screw, the length of the lever being 14 feet, the distance between the threads of one screw being I in., and that of the other I in.? 79. The diameters of the two pistons of a hydrostatic press are as 2: 18; what force will a power of 240 lbs. exert? 80. The diameter of the smaller piston of a hydrostatic press is 6 inches, that of the larger 2 feet; required the power necessary to produce a pressure of 9,600 lbs. 81. In a hydrostatic press, the power is to the pressure as 1: 64; the diameter of the smaller piston 2 inches; required the diameter of the larger piston. 82. In a hydrostatic press, the power is to the pressure as 1: 400; the diameter of the larger piston 8 in.; required the diameter of the smaller piston. 83. Required the specific gravity of a body whose weight, in air, is 12 grs., and, in water, 10 grs. 84. Required the specific gravity of a body which, in air, weighs 18.5 grs., and, in water, 16.5 grs. 85. Required the specific gravity of a body weighing 24 grs., which, being weighed in water with a heavy body, causes it to weigh 6 grs. less than when previously balanced in water. 86. Required the weight of 15.75 cubic inches of gold, its specific gravity being 19.25. 87. Required the weight of 148 cubic feet of iron, whose specific gravity is 7. 88. Required the weight of 400 cubic inches of chlorine, its specific gravity being 25. 89. Required the weight of 10 cords of wood, whose specific gravity is 0.88. 90. Required the weight of 40 cubic feet of oak wood, its specific gravity being 0.925. ï~~PROMISCUOUS QUESTIONS. 179 91. Required the cubic contents of an irregular piece of gold, which, being weighed in water, sustains a loss of 594.5 grs. 92. Required the pressure on the bottom of a vessel filled with water, the depth of the vessel being 7 feet, and the area of the bottom 25 square feet. 93. If a given bulk of distilled water weigh 125 grs., and the same bulk of muriatic acid 150 grs., what will be the specific gravity of the acid? 94. Required the specific gravity of alcohol, a given quantity of which is found to weigh 885 grs., and the same bulk of'distilled water, 1,000 grs. 95. If the flash of a cannon be seen at the moment it is fired, and the report be heard 40 seconds after; what will be the distance of the observer from the cannon? 96. The muscle connecting the back arm to the fore arm, is inserted one tenth part as far below the elbow as the hand: not taking into consideration the oblique action of the muscle, what force must it exert, to sustain in the hand, held in a horizontal position, a weight of 30 lbs.? 97. In a vessel containing water, an orifice is made 16 feet below the surface. Suppose a body to fall freely, under the influence of gravity, 16 feet the first second; with what velocity will the water issue? 98. How far will a body fall, the first second, 14,000'miles above the earth? 99. A body, 40,000 miles from the center of the earth, falls from a state of rest; how far will it fall in 30 seconds? 100. At what distance above the earth will a body fall, the first second, - in.? ï~~" r ï~~APPEND I X. WHEN TWO PARALLEL FORCES ACT IN THE SAME DIRECTION: 1st. The resultant is parallel to the two components; 2d. It is equal to their sum;, 3d. It divides a line, drawn at any inclination to the parallel components, into segments, which are to each other inversely as the adjacent components. K L M DEMONSTRATION.-Let A C, and B D represent, in intensity and direction, two parallel forces, acting at any inclination, upon the inflexible line A B. ï~~182 APPENDIX. At the points A and B, apply the two equal forces, A E and B F, acting in the same line A B, but in opposite directions. Since they exactly counteract one another, the resultant of the four forces, A C, B D, A E, and BF, is the same as that of A C and B D, the original parallel forces. But, completing the parallelograms E AC andD B F, the two forces A E, A C, are equivalent to A G; and B F and B D are equivalent to B H. Hence, A G and B H are equivalent to A C and B D, the original parallel forces. Produce the lines GA and H B till they meet at K; and let K become the point of application of the two oblique forces A G and B H, instead of the points A and B; the lines of the diagram representing an inflexible frame-work, the effect will be the same as before. Make KL equal to A G, and KN equal to B H. Draw L 0 parallel to KB, KR parallel to A C or B D, and L M parallel to A B. Join 0 N. Since A C and KR are, by construction, parallel, the exterior angle G A C is equal to the interior, and opposite angle, L KM. In like manner, G C and L M being parallel, the angles KL M and A G C are equal. K L is also, by construction, equal to A G. But when two triangles have two angles, and the included side of the one equal to two angles and the included side ofthe other, the two triangles are equal; and those parts similarly situated are equal, each to each. Hence KM is equal to A C, one of the component parallel forces; and L M is equal to G C, which is equal to E A or B F. Let us next compare the two triangles, L MO, B FH. Since the two sides, L M, MO, are parallel to B F, FH, and open in the same direction, the angle L MO is equal to the angle B F H. For a similar reason, the angles ML 0 and F B H are equal; and these angles include the sides L M and B F; ï~~APPENDIX. 183 which are, by demonstration, equal. Hence the triangles, L MO, B HF, are equal; and MO is equal to F II, or B D, the second of the original parallel forces. L 0 is also equal to B H, which, by construction, is equal to KN. Hence L O and KN, being equal, as well as parallel, the two remaining sides of the quadrilateral, KL and O N, are equal, and parallel; and the figure L KNO is a parallelogram. Hence, KO is the resultant of KL and KN, and also of their equivalents, A C and B D, the original parallel forces. But KO is, by construction, parallel to A C and B D, and, by demonstration, equal to them. It yet remains to prove that the force AC the force BD BR: AR. By comparing triangles A G C, and KA R; also B D H and KR B, we obtain the two proportions, AC: GC:: KR: AR; DH=BF=AE=GC: BD:: BR: KR; Multiply these two prdportions together, and suppress common factors, and we obtain,... AC: BD:: BR: AR. Q,E,D. To apply this proposition to the lever, and to the center of gravity, let us suppose the point, R, supported by a fulcrum, or otherwise. In this case, the resultant would be entirely counteracted; and, of course, the two parallel forces, A C and B D, would also be entirely counteracted; in other words, the forces, A C and B D, equilibrate about the point R; which divides the line A B into ï~~184 APPENDIX. segments inversely proportional to the adjacent forces. The weights of two bodies, the one at A, the other at B, may likewise be represented by the two parallel forces A C, B D. R would manifestly be the center of gravity of the two bodies. ï~~ ï~~ ï~~ ï~~ ï~~ ï~~S 3 il. ^ Tti - 4 t,e s rc { '; say s _ L 4 A 1c iR ' i"A L i f. aciq ewr r"c. sba-,.Y...e! aM y.^nh,.,-+5..P nrz., r# -.r c 4 {' m z aÂ~ 3",.;. s:; v _h ten, - -Â~, -v,: s 'rag, x Fs? 4:w I:" IN t f at" tl vet, R 8 x 3 j7 t t a z a 5fi wA k41y yt L I ' z; 1.. - * y, - -YV'(cn r....:: _ F. A"3-. + +YD'^ mow, a. _ "aa }'.. r.. '",.. 'F. 'ti.....:'.s iii _ i W 1A ^ -