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: 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 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. 5040 lb. 710 64325.3=velocity, and I 25:3 X4500=113850 lb. Ans.!

2. Of Pendulunis. 331. The time of a vibration, in a cycloid, is to the time of a heavy body's descent through hall 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 iaches in the first second, say, as 3.1416 X 3.1416:1X1::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 gioen 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 ?

..5X.5 X 39.2=9.8in. for 1 seconds. )

1x1x 39.2=39.2in. for seconds. Ans.

2x2x39.2=156.8in. for 2 seconds.) 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 respec. tively swing minutes and hours ?

60 60 >39.2=141120in.=2.n. 1200 feet.) - 3600X 3600%39.2=508032000=8018m. 960 feet.

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.

1. In what time will a pendulum 9.8 inches in length, vi. brate ?

19.8.-39.25,5, or second, Ans.

334, 335, 336.



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 ? V2.45:39.2= 25s. and 25% 10=2.5s. then 2.5 >4=10, and 10% 10=100 feet, Ans.

334. To find the depth of u 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 tho 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 stope, dropped into a well, is heard to strike the bottom in 4 seconds, what is the depth of the well ?.

773088 X4+1304164–1142=121.53, and 121.53 4121.535 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.

3: Of the Lever. 335. It is a principle in mechanics that the power is 10 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 weigbing 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 al 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, Ang. 3. In giving directions for inaking 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 suppos

sing two passengers to weigh 3 cwt. and the body of the chaise i cwt. more, what will the horse, in both these cases, bear, Lore than his harness?

5116 lb. in the first. 105. 777 lb. in the second.

1. Of the vvheel and arle.

837. 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 bala ance 1268 lb. on the axle ? 48:6::1268: 158 H. 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 balapce ?

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 Ib.

5. Of the crew. 3:38. 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 leyer by 3.1416; then say, as the cirference 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 top=2240. Ib. what power must be applied to turn the screw ? 30*2=60, and 60»<3.1416=188.496 inches, the circ.

Then 188.496 : 1 :: 2240 : 11.88 lb. Aos. 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 ?

f: 188.496 : :11.88 : 2240 lb. nearly, Ane, 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 tbreads 1 inch asunder, what is the length of the lever ?

Ans. 30 inches Dearlya



339. 1. What number taken from the square of 48 will leave 16 times 54 ?

Aas. 1440. 2. Wbat number added to the 31st part of 3813, will make the sum 200 ?

Ans. 77. 3. What will 14cwt. of beef cost, at 5 cents per pound?

Ans. $78.40. 4. How much in length that is 8 inches wide, will make a square foot ?

Ans. 1713 inches. 5. What number is that to which if 4 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 4 of the time till midnight ; what was the time?

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. 5_5_m.

10. A young hare starts 40 yards before a grey-hound," 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 oo view makes after it at the rate of 18. In what time and distance will the dog overtake the bare ?

Ans. 60,5 s. time, 530 yds. distance. 11. What part of 3d. is part of 2d. Ans. g.

12. A hare is 50 leaps before a grey hound, and takes 4 leaps to the grey hound's 3; but 2 of the grey hound'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::1:1 the hound's gain. Then 1-1=1, and 5: 1 :: 7 : 329=300, Ans. .; 13. A post is in the sand, in the water, and 10 feet above the water; what is its length ?

Aos. 24 feet.

14. A man being asked how many sheep he had, said, if he had as many more, half as many more, and 7 sheep, he should have 20 ; how many had he? Ans. 5.

15. In an orchard f the trees bear apples, 1 pears, 1 plums, and 50 of them cherries ; how many trees are there in all ?

Ans. 600. 16. A can do a piece of work alone in 10 days, B can do it in 13, in what time will both together do it? Ans. 515 days.

17. What is the difference between the interest of £350 at 4 per cent. for 8 years, and the discount of the same sum at the same rate, and for the same time? Ans. +27 353.

18. Sound moves at the rate of 1142 feet in a second; if the time between the lightning and thunder be 20 seconds, what is the distance of the explosion ?

Ans. 4.32-miles. 19. If the earth's diameter be 7911 miles, and that of the moon be 2180, how many moons will be required to make one earth ?

Ans. 47.788+ 20. If a cubic foot of iron were drawn into a bar 4 of an inch square, what would be its length, supposing no waste of metal ?


- - =27648in. =2304ft. Ans. 21. A lent B a solid stack of hay, measuring 20 feet every way; sometime after, B'returned a quantity measuring every way 10 feet; what proportion of the lay is yet due ? Ans. 7.

22. A general disposing his army into a square, finds he has 284 soldiers over and above, but increasing each side by one soldier, he wants 25 to fill up the square ; how many soldiers had he ?

Ans. 24000. · 340. 23. Supposing a pole 75 feet high to stand on a horizontal plane, at what height must it be cut off, so as that the top of it may fall on a point 55 feet from the bottom, and the end where it was cut off, rest on the stump or upright part ?

RULE.- From the square of the levgth of the pole, (i. e. the sum of the hypothe- 75x75-55*<55=71 Bft. Ans. nuse and perpendicular) take the square of the base; the divide the remainder

75x2 by twice the length of the pole, and the quotient will be the height at which it raust be cut off.

24. Suppose a ship sail from lat. 43° N. between N. and E. till her departure from the meridian be 45 leagues, and the sum of her distance and difference of latitude be 135 leagues; what is the distance sailed, and the latitude come to ? 135% 135—45% 45. lea. m.

135-60=751. dis. s'd. / -=60=180=3° of lat. 43°+3=46° come to. 135X2


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