EXAMPLES.

1. A certain inclined plane is 16 feet in length, and 7 feet in perpendicular height; what weight might be drawn up this plane by a power, which, if exerted on a cord over a single fixed pulley,

would raise 25 lb. ?

As the height, 7: 16, length:: 25, power: 574 lb., weight.
Also, as 16: 57::7: 25 lb., power.

NOTE.-When any three of the preceding parts are given, in questions relating to this mechanical power, the other may be found by attending to Principle 1.

2. If a set of pulleys, 3 of which are movable, be applied to a weight upon an inclined plane of 50 feet in length, and 14 feet perpendicular height; what weight upon the plane would be sustained by 40 lb. at the power cord of the pulleys? Thus, as 14: 50:: 40X3X2: 857 lb., Ans.

THE LEVER.

1. Given, the power, the fulcrum's place, and the length of the lever, to find the weight that may be raised.

RULE. This is contained in Principle 2; hence, when any three parts be given, by a little consideration, the other may be readily found; also, see the following:

EXAMPLES.

1. If a man weighing 150 lb. hangs atthe end of a lever 10 ft. long, what weight will he balance on the other end, if the prop or fulcrum be 2 ft. from the weight?

As 2 ft.: 8 ft. (10 ft.-2 ft.):: 150 lb. : 600 lb., Ans.

2. In giving directions for making a chaise, the length of the shafts, between the axletree and back-band, being settled at 9ft, a dispute arose whereabout on the shafts the centre of the body should be fixed. A advised to place it 30 inches, B 20 inches before the axletree. If the body of the chaise and its burthen are 420 lb., what will the beast in both cases bear more than its harness?

9 ft 108 inches; then, as 108: 420::

3. A cheese in one scale weighed 76 lb., in the other scale 56 lb. Required its true weight.

The square root of the product of the different weights is the true weight.

Thus, /76X56-65,31 lb., Ans. The denominator, 130, is double the root, 65. (See page 143.)

4. A cheese in one scale weighed 90 lb., in the other scale 40 lb. Required its true weight, and the proportional lengths of the two arms of the balance beam on each side of the fulcrum,

Thus, 90X40-60 lb., its true weight; 90-60-30, and 60-40 20. Then, as 30 lb.: 3 ft.:: 20 lb.: 2 ft.; hence the arms of the scales are to each other as 2 ft. to 3 ft.

5. Two men carrying a burthen of 200 lb., hung on a pole, the ends of which rest on their shoulders; how much of this load is borne by each man, the weight hanging 6 in. from the middle, and the whole length of the pole being 6 feet?

4 ft. 48 in.; 6X2-12 in.; then 48-12-36 in.; 36÷2—18 in.; then, 18+12-30.

Then, as {48: 200 :: 30: 125 lb.
200 :: 18: 75 lb.

THE PULLEY.

Ans.

RULE. AS unity is to twice the number of movable pulleys, so is the power to the weight. Reverse the statement to find the

power.

When any three parts of this mechanical power are given, to find a required part, see Principle 3.

EXAMPLES.

1. If a cord which runs over 3 movable pulleys, be attached to an axle 4 inches in diameter, the wheel of the axle being 38 inches in diameter, and a power of 20 lb. be exerted at the circumference of the wheel; what weight would be raised under the pulleys?

Thus, as 4 in.: 38 in.:: 20 lb.X3×2: 1140 lb., Ans.

2. There is a power equal to 30 lb. and a weight of 200 lb.; if one end of a rope is fastened to a block of fixed pulleys, how many movable pulleys must there be so that the power shall raise the weight?

As 30 lb. of 30 lb., allowed for friction, is to unity, so is 200 lb. to twice the number of movable pulleys, Ans. 3.

THE WEDGE.

The wedge is of the same nature as the inclined plane, calling the breadth of the head of the wedge, the perpendicular height of the plane, and the length of its side, the length of the plane, and the force acting against the head of the wedge, the power.

Example. If a wedge be 12 inches long, and its head 14 inch broad; and if a screw whose threads are of an inch asunder, be applied to the head of this wedge; and if a power of 200 lb. were applied to the end of the lever, 16 feet long; what would be the force exerted on the sides of the wedge?

As .75 in. :3.141592×16×2×12 :: 200: 3216990.208.

As 15 in.: 12 in.:: 3216990.208 lb.: 2573592.116 lb., Ans.
See Principles 4 and 6.

THE WHEEL AND AXLE.

1. Given three parts of the wheel and axle, to find the fourth or required part.

RULE.-See Principle 5, and the following:

EXAMPLES.

1. I would make a windlass in such a manner that a pound of power should equal 10 lb. of weight suspended from the axle, it being 6 inches in diameter. Required the diameter of the wheel, Inversely, as 10: 6:: 1: 60 inches, Ans.

By a little consideration, when any three of the preceding parts are given, the other may be readily found.

2. There are two wheels; one of which is 6 ft. in diameter, with an axle of 9 inches diameter; and the other is 4 feet in diameter, with an axle of 7 inches diameter.

& Suppose the power cord of the smaller wheel to be coiled upon the axle of the larger; what weight on the axle of the smaller wheel would be balanced by 100 lb. at the power cord of the larger?

First, as 9 in.: 6 ft.:: 100 lb.: 800 lb., the power applied to the smaller wheel; then, as 7 in. :4 ft.:: 800 lb.: 5485§ lb., Ans.

THE SCREW.

When any three properties of the screw are given, to find required part, proceed by Principle 6.

FALLING BODIES.

PRINCIPLES.

1. Heavy solid bodies, near the surface of the earth, fäll about 16 feet in the first second, that is, one foot in the first quarter, 3 feet in the second quarter, &c., increasing each last quarter, for every next quarter, 2 feet in common air; but, in vacuo, falling bodies acquire to 16.1 feet in the second.

2. The space in feet through which a body has fallen, is equal to the square of the time in which it was falling, in fourths of a second.

3. The velocities acquired by falling bodies, in a different number of seconds, are in proportion to the squares of their times; hence, the number of feet a body falls in any second, beginning at one, may be found by multiplying 16 feet by the odd numbers, 1,3,5, &c.

Thus, in 5 seconds; 16×1=16; 16×3—48.

And, 16X580; 16×7=112; 16x9 144, &c.

4. Ascending bodies are retarded in the same ratio that descending bodies are accelerated.

1. The velocity of a body moving in any direction, being given, to find how far a body must fall to acquire the same velocity.

RULE.-Divide the velocity by 8, the square of the quotient will be the space fallen through.

Reason. Because the square root of the space fallen through, is always equal to one eighth of the velocity acquired at the end of the fall.

Example. If the velocity of a cannon ball is 660 ft. per second, from what height must a body fall to acquire the same velocity? 660-8-82.25, then 82.2582.25-68064 feet, Ans.

2. The time in which a body has been falling, being given, to find the distance or space fallen through.

RULE.-Multiply the time in seconds by 4, and the square of the product will be the distance required.

Reason. Because the square root of the feet fallen through. is always equal to four times the number of seconds the body has been falling.

EXAMPLES.

1. What is the difference between the depth of two wells into

each of which, should a stone be dropped at the same instant, one would reach the bottom in 5 seconds, and the other in 3 seconds? 5X4-20, and 20×20=400; 3×4=12; 12×12—144. Then, 400-144-256 feet, Ans.

2. A ball was seen to fall half the way from the top of a tower in the last second of time. How long was it in descending, and what was its height before its descent?

The 1=1,2 1.4142; and 1.4142-1-4142. Then, as .4142: 1.4142: : 1:3.414&c. seconds, time of its descent.

3.414X42-186.424&c. feet, the space fallen through. 3. Given, the velocity per second, to find the time a body has been falling.

RULE.-Divide the velocity by 8, and a fourth of the quotient is the time.

Reason. Because four times the number of seconds in which a body has been falling, is equal to one eighth of the velocity, in feet, per second, acquired at the end of the fall.

Example.-How long must a bullet be falling to acquire the velocity of 160 feet per second?

1608-20, and 20-4-5 seconds, Ans.

4. Given, the number of feet fallen through, to find the time of a body's falling.

RULE. Divide the square root of the distance fallen through by 4, and the quotient will be the answer in seconds.

Example. In what time will a ball, dropped from the top of a steeple 144 feet high, reach the ground?

Thus, 144-12, and 12÷÷4-3 seconds, Ans.

5. To find the velocity per second, with which a heavy body will begin to fall from any distance above the earth's surface.

RULE. As the square of the earth's semi-diameter, in miles, is to 16 feet, so is the square of any other distance, in miles from the earth's centre, inversely, to the velocity required.

Reasons-See Principle 1, Gravity or Weight, and Principle 1, Falling Bodies.

Example. With what velocity will a heavy ball begin to descend per second, if raised 1000 miles above the earth's surface? As 4000X4000: 16:: 5000×5000: 10.25 ft., Ans.

6. To find the velocity per second acquired by a falling body, or by a stream of water, at the end of any given time, in seconds, the perpendicular descent or fall being given.