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326. The space given to find the tiine 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 V400=20, and 20:45 11025 feet? seconds, Ans.

Ans. 264 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-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 descend second will a body begin to with a velocity of 4 feet per descend, if raised 4000 miles second ? above the earth's sursace ?

Ans. 4000 miles. As 4000X4000 : 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.

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 | 54 seconds, with what velocity 225 feet?

will it strike ? 225X64=14400, and

Ans. 176. V14400—120, Ans.

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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 2. If a stream move with a with a velocity of 56 feet per velocity of 12.649 feet , per second, from what height did second, what is its perpendicuit fall?

lar fall ? 56X56-;-64–49 feet, Ans.

Ans. 23 feet.

330. To find the force with which a falling body will strikc. RULE.- Multiply its weight by its velocity, and the product will be the force.

1. If a ramnier 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, 5040lb. 10X61325.3=velocity, and 25.3X4500=1138501b. Ans.

2. Of Pendultms.

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331. The time of a vibration, in a cycloid, is to the time of a heavy body's descent thrvugh half iis length as the circumference of a circie in its diameter; thercfore to find the leng!h of a peu lulum vibrating, seconds, since a falling body, descends 193.5 inches in the first second, say, as 3.1416 X3.1416:1X1 :: 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.

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 s seconds, seconds, and two seconds ?

.57.57339.2—9.8 in. for s seconds. 1x1x39.239.2in. for seconds. Ans.

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

.25X.25X39.2=2.42 inches, An3. 3. Required the lengths of 2 pendulums, which will respectively swing minutes and hours? 3600X3000X33.32508032000-2.501sm. 960 feet

; } Ans. 333. To find 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 ?

9.8; 39.2.5, or second. Ans.

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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? 2.45_-39.9.-= 25$. and .25% 10=2.5s. ; then 2.5x4-10, and 10x10=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 equare 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?

173088X4+1304164–1142=121.53, and 121.53X121.535 84-230.77 feet, Ans. Then 230.77-1142—2 of a second, the sound's ascent, and 4-2-3.8 seconds, stone's descent.

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8. 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 sett 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 incuimbrance 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 ?

777 lb. in the second.

Ans. { 1162 lb. in the first.

4. Of the Tuheel 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 bal. ancc 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 tile 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 ton2240 lb.; what power must be applied to turn the screw? 30x2=60, and 60X3.1416–188.196 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 ?

Ans. 1 inch, nearly. 4. If the power be 11.88 lb., the weight 2240 lb., and the threads 1 inch asinoer what is the length of the lever ?

Ans. 30 inches, nearly

SECTION IV.

MISCELLANEOUS QUESTIONS.

Ans.

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 8f inches wide, will make a square foot ?

Ans. 1714 inches. 5. What number is that to which if . of f be added, the sụm will be 1 ?

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 4 part of it, and at the end of 4 years found that his estate amounted to £10342 33. 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 ime ?

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. lh 53m.

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. 607s. 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::1:1 the hare’s gain.

2:1 1 3 the hound's gain. Then =, and : 1 :: 7 : 390=300, Ans. 13. A post is 1 in the sand, in the water, and 10 feet above the water; what is its length ?

Ans. 24 feet.

Ans. g.

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