323 325. OF THE FALL OF HEAVY BODIES. 139 feet be multiplied by so many of the odd numbers, beginuing 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. 323. The velocity given to find the space fullen 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 | 2. From what height must a body fall to acquire the ye. a budy fall to acquire a velocilocity of a cannon ball, which ty of 1200 feet per second ? is about 660 feet per second ? Ans 22500 feet. 6608_82.5, and $2.5X 82.5-806.25f1=19 miles, Ans. 324. 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 3. Ascending bodies are rebody fall in five seconds ? tarded in the same ratio that 5X4–20, and 20x20=400 descending bodies are accelerfeet, Ans. ated ; therefore, if a ball, fired upwards, return to the earth 2. A etonc, dropped into a in 16 seconds, low high did it well, reached the bottom in 3 | ascend? The ball being half seconds; what was its depth? | the time, or 8 seconds, in its 3X4-12, and 12X12-144 | ascent: therefore 8X4332, feet, Ans. 1 and 32X321024t, Ans. 325. 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! 2. How long must a body be be falling to acquire a velocity falling to acquire a velocity of of 160 feet per second ? | 400 feet per second ? 160;8–20, and 20:4-51 Ars. 124 seconds poconds, Ans. . 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 bc the time. 1. In how many seconds will l 2. In how many seconds will a body fall 400 feet? a bullet fall through a space of V400—20, and 20: 4-5. 11025 feet? seconds, Ans. Ans. 261 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 18 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-diame-| 2. How high above the ter of the earth to be 4000 | earth's surface must a ball be miles, with what velocity per raised, to begin to dcscend second will a body begin to with a velocity of 4 feet per descend, if raised 4000 miles | second? above the earth's surface Ans. 4000 miles 328. To find the velocity acquired by a falling body, per seconda 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 2. If a ball fall 484 feet in does a ball acquire by falling 5.seconds, with what velocity 225 feet? will it strike ? 225X64–14400, and Ans. 176 ✓14400=120, Ans. *329. The velocity with which a body strikes given to find the pace fallen through Rule-Divide the square of the velocity by 64, and the quouent will be the space required. 1. If a ball strike the ground 2. If a stream move wiih a with a velocity of 56 feet per velocity of 12.049 feet pot second, from what height did second, what is its perpendicu. i fall? lar fa!? 56X56; 64A9 kiet, ADA 1 Ana, 23 feet $30, $13. Or PENDULUMS. 141 $30. To find the force with ethich a fulling body will sirik.. RULE.-Multiply its weight by its velocity, and the product will be the force. 1. If a rammer for driving 1 2. With what force will a piles, weigling 4500 pounds, 421b. cannon ball strike, dropfall through the space of 10 feet, ped from a height of 225 feet? with what force will it strike? Aus, 5040lb. V10x61325.3-velocity, and 25.3X 450=113850lb. Ans. 2. Of Pendultms. 331. The time of a vibration, in a cycloid, is to the time of a heavy body's descent through halt iis length as ihe circumference of a circle é its diameter; therefore to find the length of a pcurlulum vibrating seconde, siuce 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 19.6X2=39.2 inches, the length. 332. To find the length of a pendulum that will swing any given time. ROLE.--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 s seconds, seconds, and two seconds ? .5X.5X39.2--9.8 in. for seconds. ) 2X2X39.2-156.8in. for 2 seconds. ) 254.25X39.2=2.42 inches, Ang 3. Required the lengths of 2 pendulums, which will respootively swing minntes and liours? 60X60X39.?--141120in.-2m. 1960 fect. ? 3600X3600X39.24508032000=8018m. 960 feet." 333. To ford 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 } 79.6; 39. 25, or 4 second. And 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 i was the height of the steeple ? 2.45 +39.%= 258. and .25X 10-2.59.; then 2.5x4310, and 10x10=100 feet, Ans. 334. To find the depth of a well by dropping a stone into its ROLE. 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 =16X4), 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 ? 773088X4+1304164–1142_121.53, and 121.53X121.53; 64_230.77 feet, Ans. Then 230.77--11422 of a second, the sound's ascent, and 4-2=3.8 seconds, stone's descent 8. Of the Lever. 336. It is a principle in mechanics that the power is to the weight me the velocity of the weight is to the velocity of the power. 336. To find what weight may be balanced by a given proer. 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 leyer 12 feet long, what weight will he balance on the other end, supposing the prop to be I foot from the weight? 1:11:: 360 : 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, wow suppor337, 388. OF THE SCREW. 143 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 ? S1169 lb. in the first. 4. Of the una heel and Arle. 337, RUJE.-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. d. 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 diaspeter 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 Ib. 5. Of the Screw. 338. The power is to the weiglit 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 tie screw, go 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 ton2240 lb.; what power niust be applied to turn the screw? 30X2_60, and 60X3.14165188.496 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 incli 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 224Q Ib., the power 11.88 lb., and the Jever 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 :.' threado 1 inch Agndow what is the length of the lever? Ans. 30 inches, Deasly. |