140 OF THE FALL OF HEAVY BODIES. 326-329, 326. The space given to find the time the body has been falling. RULE. Divide the square root of the space fallen through by 4, and the quotient will be the time. 1. In how many seconds will a body fall 400 feet? 400-20, and 2045 seconds, Ans. 2. In how many seconds will a bullet fall through a space of 11025 feet? Ans. 26 seconds. 327. To find the velocity per second, with which a body will begin to descend at any distance from the earth's surface. RULE.-As the square of the earth's semi-diameter is to 16 feet, so is the square of any other distance from the earth's centre, inversely, to the velocity with which it begins to descend per second. 1. Admitting the semi-diameter of the earth to be 4000 miles, with what velocity per second will a body begin to descend, if raised 4000 miles above the earth's surface? As 4000x4000: 16: : 8000 X80004 feet, Ans. | 2. How high above the earth's surface must a ball be raised, to begin to descend with a velocity of 4 feet per second? Ans. 4000 miles. 328. To find the velocity acquired by a falling body, per second, at the end of any given period of time. RULE.-Multiply the perpendicular space fallen through by 64, and the square root of the product is the velocity required. 1. What velocity per second does a ball acquire by falling 225 feet? 225X64 14400, and 14400 120, Ans. 2. If a ball fall 484 feet in seconds, with what velocity will it strike? 5 Ans. 176. 329. The velocity with which a body strikes given to find the space fallen through. RULE. Divide the square of the velocity by 64, and the quotient will be the space required. 1. If a ball strike the ground with a velocity of 56 feet per second, from what height did it fall? 56×56÷64—49 feet, Ans. 2. If a stream move with a velocity of 12.649 feet per second, what is its perpendicular fall? Ans. 24 feet. 330. To find the force with which a falling body will strike. RULE. Multiply its weight by its velocity, and the product will be the force. 1. If a rammer for driving piles, weighing 4500 pounds, fall through the space of 10 feet, with what force will it strike? 10X64 25.3=velocity, and 25.3X4500 113850lb. Ans. 2. With what force will a 421b. cannon ball strike, dropped from a height of 225 feet? Ans. 5040lb. 2. Of Pendulums. 331. The time of a vibration, in a cycloid, is to the time of a heavy body's descent through half its length as the circumference of a circie to its diameter; therefore to find the length of a pendulum vibrating seconds, since a falling body descends 193.5 inches in the first second, say, as 3.1416×3.1416:1×1:: 193.5,19.6 inches the length of the pendulum, and 19.6X2=-39.2 inches, the length. 332. 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 the product will be the length required in inches. 1. What are the lengths of three pendulums, which will swing respectively seconds, seconds, and two seconds? .5X.5X39.29.8 in. for seconds. Ans. 2. What is the length of a pendulum, which vibrates 4 times in a second? 25.25X39.2-2.42 inches, Ans. 3. Required the lengths of 2 pendulums, which will respectively swing minutes and hours?" 60X60X39.2=141120in.-2m. 1200 feet. 3600X3600X39.2-508032000-8018m. 960 feet. Ans. 333. To find the time which a pendulum of a given length will swing. RULE.-Divide the given length by 39.2, and the square root of the quotient will be the time in seconds. 0 1. In what time will a pendulum 9.8 inches in length vibrate ? √9.8÷39.2—.5, or second, Ans. 2. I observed that while a ball was falling from the top of a steeple, a pendulum 2.45 inches long, inade 10 vibrations; what was the height of the steeple ? 2.4539.225s. and .25X 10—2.5s.; then 2.5×4=10, and 10×10—100 feet, Ans. 334. To find the depth of a well by dropping a stone into it. RULE. Find the time in seconds to the hearing of the stone strike, by a pendulum; multiply 73088 (=16X4X1142; 1142 feet being the distance sound moves in a second), by the time in seconds; to this product add 1304164 (the square of 1142), and from the square root of the sum take 1142; divide the square of the remainder by 64 (=16×4), and the quotient will be the depth of the well in feet; and if the depth be divided by 1142, the quotient will be the time of the sound's ascent, which, 'taken from the whole time, will leave the time of the stone's descent. 1. Suppose a stone, dropped into a well, is heard to strike the bottom in 4 seconds, what is the depth of the well? 73088X4+1304164-1142-121.53, and 121.53×121.53÷ 64-230.77 feet, Ans. Then 230.77-1142-2 of a second, the sound's ascent, and 4-.2-3.8 seconds, stone's descent. B. Of the Lever. 335. It is a principle in mechanics that the power is to the weight as the velocity of the weight is to the velocity of the power. 336. To find what weight may be balanced by a given power. RULE. As the distance between the body to be raised or balanced, and the fulcrum, or prop, is to the distance between the prop and the point where the power is applied, so is the power to the weight which it will balance. 1. If a man weighing 160 lb. rest on a lever 12 feet long, what weight will he balance on the other end, supposing the prop to be 1 foot from the weight? 1:11:: 160: 1760 lb. Ans. 2. At what distance from a weight of 1440 lb. must a prop be placed, so that a power of 160 lb. applied 9 feet from the prop may balance it? 1440 160: 9:1 foot, Ans. 3. In giving directions for making a chaise, the length of the shafts between the axletree and back band being settled at 9 feet, a dispute arose whereabout on the shafts the centre of the body should be fixed; the chaise maker advised to place it 30 inches before the axletree; others supposed that 20 inches would be a sufficient incumbrance for the horse. Now sunpos ing two passengers to weigh 3 cwt. and the body of the chaise cwt. more, what will the horse, in both these cases, bear, more than his harness? S 116 lb. in the first. Ans. 777 lb. in the second. 4. Of the Wheel and Axle. 337. RULE. 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 on the axle. 1. If the diameter of the axle be 6 inches, and that of the wheel be 48 inches, what weight applied to the wheel will balance 1268 lb. on the axle ? 48:6: 1268: 158 lb. Ans. §. 2. If the diameter of the wheel be 50 inches, and that of the axle 5 inches, what weight on the axle will 2 lb. on the wheel balance? 5:50::2: 20 lb. Ans. 3. If the diameter of the wheel be 60 inches, and that of the axle 6 inches, what weight at the axle will balance 1 lb. on the wheel? Ans. 10 lb. 5. Of the Screw. 338. The power is to the weight which is to be raised, as the distance between two threads of the screw, is to the circumference of a circle described by the power applied at the end of the lever. To find the circumference of the circle; multiply twice the length of the lever by 3.1416; then say, as the circumference is to the distance between the threads of the screw, so is the weight to be raised to the power which will raise it. 1. The threads of a screw are 1 inch asunder, the lever by which it is turned, 30 inches long, and the weight to be raised, 1 ton=2240 lb.; what power must be applied to turn the screw? 30×260, and 60×3.1416-188.496 inches, the circ. Then 188.496 : 1 :: 2240:11.88 lb. Ans. 2. If the lever be 30 inches (the circumference of which is 188.496), the threads 1 inch asunder, and the power 11.88 lb., what weight will it raise? 1: 188.496: : 11.88: 2240 lb. nearly, Ans. 3. Let the weight be 2240 lb., the power 11.88 lb., and the lever 30 inches; what is the distance between the threads? 4. If the power be 11.88 lb., the threads 1 inch asender what is the Ans. 1 inch, nearly. weight 2240 lb., and the length of the lever? Ans. 30 inches, nearly SECTION IV. MISCELLANEOUS QUESTIONS. 339. 1. What number taken from the square of 48 will leave 16 times 54? Ans. 1440. 2. What number added to the 31st part of 3813, will make the sum 200? Ans. 77. 3. What will 14 cwt. of becf cost, at 5 cents per pound? Ans. $78.40. 4. How much in length that is 88 inches wide, will make a square foot? Ans. 171 inches. 5. What number is that to which if of be added, the sum will be 1? Ans. 6. A father dividing his fortune among his sons, gave A 4 as often as B 3, and C 5 as often as B 6; what was the whole legacy, supposing A's share $5000? Ans. $11875. 7. A tradesman increased his estate annually by £100 more than part of it, and at the end of 4 years found that his estate amounted to £10342 3s. 9d.; what had he at first? Ans. £4000. 8. A person being asked the time of day, said the time past noon is equal to of the time till midnight; what was the Lime? Ans. 20 minutes past 5. 9. The hour and minute hand of a clock are together at 12 o'clock; when are they next together? Ans. 1h 55m. 10. A young hare starts 40 yards before a greyhound, and is not perceived by him till she has been up 40 seconds; she scuds away at the rate of 10 miles an hour, and the dog on view makes after it at the rate of 18. In what time and distance will the dog overtake the hare? Ans. . Ans. 60s. time, 530 yds. distance. 11. What part of 3d. is part of 2d.? 12. A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound's 3; but 2 of the greyhound's leaps are as much as 3 of the hare's; how many leaps must the hound take to catch the hare? If 3:1:1: the hare's gain. 2:1:1: the hound's gain. Then, and ::: 50: 300-300, Ans. 13. A post is in the sand, in the water, and 10 feet above the water; what is its length? Ans. 24 feet. |