✓15625= 125, and 1254 31.25 seconds, Ans. 10. In what time will a musket ball, dropped from the top of a steeple, 484 feet high, come to the ground? Ans. 5 seconds. 322. To find the velocity, per second, with which a heavy body will begin to descend, at any given distance above the earth's surface, RULE. As the square of any given distance from the earth's centre is to the square of the earth's semidiameter; so is 16 feet to the velocity required. 11. With what velocity, per second, will an iron ball begin to descend if raised 3000 miles above the earth's surface? 4000+3000 × 4000+3000: 4000 x 4000 :: 16: 5.22449 feet, Ans. 323. It is evident we may find the MEAN VELOCITY of a falling body, by dividing the SPACE FALLEN THROUGH by the number of seconds it has been falling. 12. A musket ball dropped from the top of a steeple 484 feet high in 5 seconds; required its mean velocity. Ans. 88 feet per second. 324. To find the velocity acquired by a falling body per second, (or by a stream of water, having the perpendicular distance given,) at the end of any given time, RULE. The VELOCITY acquired at the END OF ANY PERIOD is equal to twice the mean velocity during that period. Or, if we multiply the perpendicular space fallen through by 64, the square root of the product will be the velocity required. 13. If a ball fall through a space of 484 feet in 54 seconds, with what velocity will it strike? 4845.581, and 88 x 2 = 176, Ans. Or 484 X 6430976, and √30976 = 176 as before. 14. One end of a sluice or flume is 23 feet lower than the other; required the velocity of the stream per second? Ans. 12.649ft. 325. The velocity, with which a falling body strikes, given to find the space fallen through. This is evidently the reverse of the above, wherefore we have this RULE. Divide the square of the velocity by 64, and the quotient will be the height required. 15. If a ball strike the ground with a velocity of 56 feet per second, from what height did it fall? 56 X 566449 feet, Ans. 16. The mean velocity of a fluid or stream, is 12.649 feet per second; required the perpendicular fall of the stream. Ans. 2 feet. 326. THE MOMENTUM OF FORCE, with which a falling body strikes, is equal to its WEIGHT multiplied by its VELOCITY. 17. If the rammer, used for driving the piles of Charlestown bridge, weighed 24 tous, or 4500 lbs., and fell through a space of 10 feet, with what force did it strike the pile? By Art. 324, 10 × 64 = 25.3 = the velocity, and 25.3 × 4500= 113850 lbs. momentum, Ans. 327. Hence we may find the space fallen through, by dividing the momentum by the weight, for the velocity; and then the square of the velocity by 64. 18. If the aforementioned rammer weighed 4500lbs., and struck with a force of 113850lbs.; from what height did it fall? Ans. 10 feet. 328. When it is required to know with what quantity of motion, momentum or force, a fluid, moving with a given velocity, will strike against a fixed obstacle, the result may be obtained by the following RULE. Multiply the height of the fall by 62.5 lbs. Avoir. for clear water, by 63 for dirty water, and by 64 for sea water. 19. Suppose a stream of clear water to move at the rate of 5 feet per second, and to meet with a fixed obstacle (or bulk head) 15 feet wide, and 4 feet high; required the momentary pressure of the stream. 5 x5 ÷ 64.39 of a foot, the perpendicular fall of the water (Art. 325.) Then 52.5 X .3924.375 lbs. pressure upon each square foot, which multiplied by 60 (15 x 4) gives 1462.5 lbs.; and this by the given velocity, produces 7312.5 lbs., Ans.* 329. The velocity of water, spouting through a sluice, or aperture in a reservoir, or bulkhead, is the same that a body would acquire by falling through a perpendicular space equal to that between the top of the water and the aperture. 20. What is the velocity of water, issuing from a head of 5 feet deep? 330. The quantity of water discharged from a hole in a vessel, is as the SQUARE ROOT of the height of water above the APERTURE. 22. A miller has a head of water 4 feet above the sluice; how high must the water be raised above the opening, so that half as much again water may be discharged from the sluice in the same time? ✔2, then 2+ of 2 = 3, and 329 feet, Ans. 331. If a large vessel be kept constantly full, it is required to find how far water will spout from it through a hole D, on the plane BC?... RULE. BC will ever be equal to twice the SQUARE ROOT of A D x DB. NOTE. The fluid will spout farthest, if the hole be made half way between A and B. 23. How far will water spout on the plane B C, if the distances between the top and the hole, and bottom and the hole be 2 and 6? 5 and 3? 3 and 5? Ans. to last, 7.7459666. 4 and 4? 332. If the vessel be raised above the plane, the foregoing rule and note will apply equally well, only we must regard the distances given, as between the TOP and the orifice, or hole, and between the PLANE and the hole. 24. A vessel 10 feet deep, is raised 2 feet above a surface or plane; it is required to find how far water will spout through a hole 3 feet from the bottom? Ans. 11.8321596 feet. 333. I. If the WEIGHTS of any two bodies, put in motion, be equal, the FORCES by which they are moved, will BE AS THEIR VELOCITIES. II. If the VELOCITIES be equal, their FORCES will be as THEIR WEIGHTS. *Water, being a yielding substance, is said to lose of its power in producing effects. 1 III. If both their WEIGHTS and VELOCITIES be unequal, THE FORCES will be AS THEIR PRODUCTS. 25. If a body weighing 30lbs. be impelled by such a force as to send it 20 rods in a second, with what velocity would a body weighing 12lbs. move, if impelled by the same force? Ans. 50 rods per sec. 26. Suppose the battering ram of Vespasian weighed 60000lbs.; that it was moved at the rate of 24 feet per second; and that this was sufficient to demolish the walls of Jerusalem: with what velocity must a cannon ball of 42lbs., be moved to do the same execution? Ans. 342855 ft. per scc. MECHANICAL POWERS. 334. These are usually accounted six in number, namely, the LEVER, the WHEEL and AXLE, the PULLY, the INCLINED PLANE, the WEDGE, and the SCREW. 335. WEIGHT and POWER, when opposed to each other, signify the body TO BE MOVED, and the body that MOVES IT. 336. It is a PRINCIPLE in mechanics, that the POWER is to the WEIGHT, as the VELOCITY OF THE WEIGHT to the VELOCITY OF THE POWER. THE LEVER OR STEELYARD. 337. The distance between the body to be raised or balanc ed, and the fulcrum or prop, is to the distance between the prop and the point where the power is applied, as the POWER is to the WEIGHT it will balance. 1. If a man of 160lbs. weight, rest on the end of a lever 10 feet long, what weight will he balance on the other end, supposing the prop one foot from the weight? Ans. 1440lbs. 2. In giving directions for making a chaise, the length of the shafts between the axletree and backbond being settled at 9 feet, a dispute arose whereabout on the shaft the centre of the body should be fixed. The chaise maker advised to place it 30 inches before the axletree; others supposed 20 inches would be sufficient: now, supposing two passengers to weigh 3cwt., and the body of the chase cwt. more; what will the beast in each of these cases bear more than his harness? Ans. 116 lbs. in the former, and 777 in the latter. THE WHEEL AND AXLE.* 338. The proportion for the wheel and axle, (in which the power is applied to the circumference of the wheel, and the weight is raised by a rope, which coils about the axle as the wheel turns round,) is, as the diameter of the AXLE is to the diameter of the WHEEL, so is the POWER APPLIED TO THE WHEEL, to the WEIGHT SUSPENDED by the axle. * In this kind of machinery it is usual to abate of the effect for friction. 3. A mechanic would make a windlass in such a manner, that 1lb. applied to the wheel, shall be equal to 10lbs. suspended from the axle; now, supposing the axle to be 6 inches, required the diameter of the wheel? Ans. 60 inches. THE PULLY. 339. If a power sustain a weight by means of a FIXED PULLY, the power and weight are EQUAL; but if the pully be moveable, the power is but HALF the weight; hence for EVERY cord passing over a moveable pully, the power is double. 4. What weight will a person of 160lbs. raise by means of a rope passing over three fixed and three moveable pullies alternately? THE INCLINED PLANE. Ans. 960lbs. 340. The power gained by the inclined plane is, as the LENGTH of the plane to its PERPENDICULAR HEIGHT. NOTE. The double of the inclined plane forms the WEDGE. To obtain the greatest effect by this power, the body must be moved in a line parallel to the inclination. 5. What weight will a power of 150lbs. draw up an inclined plane, whose height is 10 feet, and length 30 feet? Ans. 450lbs. THE WEDGE. 341. When a wedge is in equilibrio, the power acting against the BACK, is to the force acting PERPENDICULARLY against EITHER SIDE, as the BREADTH OF THE BACK to the LENGTH OF THE SIDE. * NOTE. The above respects only one side; to calculate for both sides, we must double the result. 6. How large a mallet will it take to raise a ship of ten tons, by means of a half wedge, whose back is 5 inches, and inclined side 20, supposing it to be impelled at the rate of 200 feet per second? Ans. allowing for friction, it will take a mallet of 56lbs, weight. THE SCREW. 342. The POWER is to the WEIGHT to be raised as the distance between TWO THREADS of the screw, to the CIRCUMFERENCE OF A CIRCLE described by the POWER applied at the END OF THE LEVER. 7. What power applied at the end of a lever 5 feet long, will raise 9 tons, by means of a screw whose threads are an inch asunder? Abating for friction, Ans. 71.3+ lbs. PENDULUMS. 343. The time of a vibration, is to the time of a heavy body's descent through half the length of the pendulum, as the circumference of a cir * Here the friction is very great; and is, at least, equal to the force to be over come. cle to its diameter. Therefore, (as a body descends freely through the space of about 193.5 inches in one second of time,) To find the length of a pendulum vibrating seconds, to 19.6, the HALF LENGTH = 39.2 344. To find the length of a pendulum that will swing any given time, RULE. Multiply the square of the time in seconds by 39.2, and it will give the length in inches. 1. What are the lengths of pendulums, vibrating 4 seconds, seconds, seconds, minutes, and hours? Ans. to last, 508032000 inches. 2. What is the difference between two pendulums, the one vibrating half seconds, and the other 3 seconds? Ans. 343 feet. 345. To find the time a pendulum of any given length will swing, RULE. The SQUARE ROOT of the given length divided by 39.2, will give the TIME IN SECONDS. 3. How often will a pendulum of 2.45 inches vibrate in a second? Ans. 4 times, that is, it will vibrate seconds. 346. To find the true depth of a well by dropping a stone into it. RULE I. With a PENDULUM find the time from the dropping of the stone, till you hear it strike the bottom. II. Multiply 73088 (= 16 × 4 × 1142) by the NUMBER OF SECONDS thus found. III. From the SQUARE ROOT of this product plus 11422 take 1142. IV. The SQUARE of the remainder divided by 64, will give the depth of the well IN FEET. NOTE. The depth of the well divided by 1142, will give the time of the sound's ascent, which taken from the whole time will leave the time of the stone's descent. 4. When I dropped a stone into a well, a line with a bullet which measured to the middle of the ball 2.45 inches, made 8 vibrations before I heard the stone strike the bottom; what was the depth of the well? time of the stone's descent? and of the sound's ascent? Ans., the depth of the well was 60.645 feet; stone's descent 1.947, and sound's ascent .053 of a second. MENSURATION OF SUPERFICIES AND SOLIDS. (See pages 35, 49, 50, 116, 119, & 124.) 347. To measure a RHOMBOID. DEF. A RHOMBOID is a figure, whose opposite sides and op posite angles are EQUAL. |