144 SECTION IV. PHILOSOPHICAL MATTERS. 1. Of the fall of heavy Bodies. Heavy bodies, near the surface of the earth, fall one foot the first quarter of a second, three feet the second quarter, five feet the third quarter, and seven feet the fourth quarter, equal to 16 feet in the first second. The velocities acquired by falling bodies, are in proportion to the squares of the times in which they fall; that is,if 3 bullets be dropped at the same time, and the first be stopped at the end of the first second, the second at the end of the second, and the third at the end of the third, the first will have fallen 16 feet, the second, (2×2=4) four times 16, equal to 64; and the third, (3×39) nine times 16, equal to 144 feet, and so on.. Or if 16 feet be multiplied by so many of the odd numbers beginning at 1, as there are seconds in the given time, these several products will be the spaces passed through in each of the several seconds, and their sum will be the whole distance fallen. 1. The velocity given, to find the space fallen through. RULE.-Divide the velocity in feet by 8, and the square of the quotient will be the space fallen through to acquire that velocity. 1. From what height must a body fall to acquire the velocity of a cannon ball, which is about 660 feet per second? 6608-82.5 and 82.5 82.5 -6806.25ft. 1,37 miles, Ans. 2. The time given to find the space fallen through. RULE.-Multiply the time in seconds by 4, and the square of the product will be the space fallen through in the given time. 1. How many feet will a body fall in 5 seconds? 54-20, and 20×20=400 feet, Ans. 2. A stone, dropped into a well, reached the bottom in 3 seconds; what was its depth? 3X4= 12, and 12X12-144 feet, Ans. re 3. Ascending bodies are tarded in the same ratio that descending bodies are accelerated; therefore, if a ball, fired upwards return to the earth in 16 seconds, how high did it ascend? The ball being half the time, or 8 seconds, its ascent; therefore, 84-32, and 32× 32=1024 ft. Ans. 3. The velocity per second given to find the time. RULE.-Divide the given velocity by 8, and one fourth part of the quotient will be the answer. 1. How long must a body be falling to acquire a velocity of 160 feet per second? 160-8-20, and 20-4-5 seconds, Ans. 2. How long must a body be falling to acquire a velocity of 400 feet per second ? Ans. 12 seconds. 4. 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 20-4-5 seconds, Ans. 2. In how many seconds will a bullet fall through a space of 11025 feet? Ans. 264 seconds. 5. 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 4000 × 4000 : 16:: 8000 8000: 4 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. 6. 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. 7. 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 sec 2. If a stream move with a velocity of 12.649 feet per second, ond, from what height did it fall? | what is its perpendicular fall ? 56 56-64-49 feet, Ans. Ans. 24 feet. 8. 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? 10×64 25.3-velocity, and 25.3 x 4500=113850 lb. Ans. 2. With what force will a 42lb. cannon ball strike, dropped from a height of 225 feet? Ans. 5040 lb. 2 Of Pendulums. 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 circle 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 3.1416x3.1416: 1x1 :: 193.5 19.6 inches the length of the pendulum, and 19.6×2-39.2 inches, the length. 1. 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 2 seconds ? .5.5×39.29.8in. for 2 seconds. 1159.239.2in. for seconds. 2 <2 <39.2=156.8in. for 2 seconds. } Ans. 2. What is the length of a pendulum, which vibrates 4 times in a second ? .25.25×39.2-2.42 inches, Ans. 3. Required the lengths of 2 pendulums, which will respectively swing minutes and hours? 60><60><39.2=141120in.-2m. 1200 feet. ? Ans. 3600 3600 39.2-508032000=8018m. 960 feet. S 2. 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. 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, made 10 vibrations; what was the height of the steeple ?. 2.45 39.2.258. and .25 × 10=2.5s. then 2.5 x 4-10 and 10× 10-100 feet, Ans. 3. 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 (=16×4×1142; 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? √73088×4+1304164—1142=121.53, and 121.53 × 121.53÷64 230.77 feet, Ans. Then 250.771142.2 of a second, the sound'sascent, and 4-.2-3.3 seconds, stone's descent. 3. Of the Lever. 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. 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 supposing 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? Ans. {11631b. in the first. 7771b. in the second.. 4 Of the Wheel and Arle. 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: 1584lb. 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::2:20lb. 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. 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 3.146; 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 x 250, and 60 × 3.1416-188,496 inches, the circumference. The 188.496:1::2.40: 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 ? Ans. 1 inch, nearly. 4. If the power be 11.88 lb. the weight 2240 lb and the threads 1 inch asunder, what is the length of the lever? Ans. 30 inches nearly. |