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323 325.

OF THE FALL OF HEAVY BODIES,

139

feel be multiplied by so many of the ndd 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 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 2. From what hcight mugt a body fall to acquire the ve- a budy fall to acquire a velocilocity of a cannon ball, which ty of 1200 feet per second ? ja about 660 feet per second ?

Ans 22500 feet. 660:-882,5, and 82.5X 82.5=806.25ft128 niles, 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 allen 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

5X4320, and 20x20=400 | descending bodies are accelerfeet, Ans.

ated; therefore, if a ball, fired

upwards, return to the earth 2. A stone, 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.

and 32X32-10241t., Ans.

325. The velocity per second given to finil 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 20:45

Aris. 123 seconds. seconds, Ans.

326. The space given to And 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 2. In how many seconds will a body fall 400 feet?

a bullet fall through a space of .00=20, and 20:4–5. 11625 feet? seconds, Ans.

Ans. 261 seconds. 327. To find the velocity per second, with rohich 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 10 feet, so is the square of any other distance from the earth's cencre, 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 As 4000X 4000 : 16 : : 8000 X 8000 : 4 feet, Ans.

328. To find the velocity acquired by a falling body, per second, at the end of any given period of time.

Rulc.-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 5.1 seconds, with what velocity 225 feet?

will it strike ? 225X64=14400, and

Ans. 176 ✓14400=120, Ans.

329. The velocity with which a boily strikes given to find the pace 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 2. If a stream move with a with a velocity of 56 feet per velocity of 12.049 feet per recond, from what height did second, what is its perpendicu.

? 56X56:6449 feet, Ada

Ana, 23 feet

$30 $33.

or PENDULUMS.

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$30. To find the force with thich a fulling body will strike. RULE-Multiply its weight by its velocity, and the product will be the force.

1. If a rammer for driving 2. With what force will a piles, weighing 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?

Ans. 50401b. V10X61-25.3 -velocity, and 25.3X 4500=1138501b. Ans.

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331. The time of a vibration, in a cycloid, is to the time of a heavy body's descent through halt iis length as the circumference of a circle és its diameter; therefore to find the length of a peululum vibrating, seconds, siuce a falling body, descends 193.5 inches in the first second, say, as 3.1416X3,1416:1X1 :: 193.5,19.6 inchies= the length of the pendulum, and 19.6X2=39.2 inches, the length.

332. To find the length of a penduluin 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 & seconds, seconds, and two seconds ?

.5X.5X39.2--9.8 in. for seconds. IX1X39.2—3.1.2in. for seconds. Ane.

2x2x39.2=156.8in. for 2 seconds. 2. What is the length of a pendulum, which vibrates 4 times in a second ?

.25 8.25X39.2=2.42 inches, Ang 3. Required the lengths of 2 pendulums, which will respootively swing minntes and hours?

60X60X39.?-141120in.=2m. 1960 fect. 3600x3600 X:33.L.SEC320002301sm. 960 feet

.

Ans. 333. To ford the time which a pendulum of a given length will sing.

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 ?

9.6 39.25, 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 i was the height of the steeple? 2.45:39.2= 258. and .25% 102.59.; then 2.5x4=10, and 10x10=100 feet, Ans.

334. To find the depth of a well by 'dropping a stone into ito

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 (716X4), 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?

773088 X 4+1304161-1142121.53, and 121.53X121.53; 643230,77 feet, Ans. Then 230.77-1142=2 of a second, the sound's ascent, and 4-2—3.8 seconds, stone's descent.

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Of the Lever. 836. It is a principle in mechanics that the power is to the weigbe 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.

ROLE.-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 suppor337, 338.

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, bean more than his harness ?

Ans.

lb, in the second.

4. Of the wheel and Arle. 337, RUE. 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 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 Id.

5. Of the scrtw. 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 firmit. the circumference of the circle ; multiply twice the length of the lever by 3.1416; then cay, as the circumference is to the distance between the threads of the screw, 80 is the weight to be raised to the power which will raise it.

1. The threads of a screw are 1 inch azunder, 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 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 threado 1 inch sonder what is the length of the lever?

Ans. 30 inches, boasty.

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