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if a body be projected vertically upwards, it will ascend to a certain point, and then return upon its path, in such a manner, that the velocities in ascending and descending will be equal at the same points.

EXAMPLES.

1. Through what distance will a body fall from a state of rest in vacuum, in 10 seconds, and through what space will it fall during the last second? Ans. 1608 ft., and 305 ft.

2. In what time will a body fall from a state of rest through a distance of 1200 feet? Ans. 8.63 sec.

3. A body was observed to fall through a height of 100 feet in the last second. How long was the body falling, and through what distance did it descend?

SOLUTION.

If we denote the distance by h, and the time by t, we shall have,

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4. A body falls through a height of 300 feet. Through what distance does it fall in the last two seconds?

The entire time occupied, is 4.32 sec. The distance fallen through in 2.32 sec., is 86.57 ft. Hence, the distance required is 300 ft. 86.57 ft. =213.43 ft. Ans.

5. A body is projected vertically upwards, with a velocity of 60 feet. To what height will it rise? Ans. 55.9 ft. 6. A body is projected vertically upwards, with a velocity of 483 ft. In what time will it rise to a height of

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The smaller value of t gives the time required; the larger

value of t gives the time occupied in rising to its greatest height, and returning to the point which is 1610 feet from the starting point.

7. A body is projected vertically upwards, with a velocity of 161 feet, from a point 2143 feet above the earth. In what time will it teach the surface of the earth, and with what velocity will it strike?

SOLUTION.

The body will rise from the starting point 402.9 ft. The time of rising will be 5 sec.; the time of falling from the highest point to the earth will be 6.2 sec. Hence, the required time is 11.2 sec. The required velocity is 199 ft.

8. Suppose a body to have fallen through 50 feet, when a second begins to fall just 100 feet below it. How far will the latter body fall before it is overtaken by the former?

117.

Restrained Vertical Motion.

Ans. 50 feet.

We have seen that the entire force exerted in moving a body is equal to the acceleration, multiplied by the mass (Art. 24). Hence, the acceleration is equal to the moving force, divided by the mass. In the case of a falling body, the moving force varies directly as the mass moved; and, consequently, the acceleration is independent of the mass. If, by any combination, the moving force can be diminished whilst the mass remains unchanged, there will be a corresponding diminution in the acceleration. This object may be obtained by the combination represented in the figure. A represents a fixed pulley, mounted

on a horizontal axis, in such a manner that the friction shall be as small as possible; W and Ware unequal weights, attached to a flexible cord passing over the pulley. If we suppose the weight W greater than W', the former will descend and draw the latter up. If the difference is very small, the motion will be very slow, and if the instrument is nicely constructed,

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Fig. 99.

W

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А

we may neglect all hurtful resistances as inappreciable. Denote the masses of the weights W and W', by m and m', and the force of gravity, by g. The weight W is urged downwards by the moving force mg, and this motion is resisted by the moving force m'g. Hence, the entire moving force is equal to mg - m'g, or, (m — m')g, and the entire mass moved, is m +m', since the cord joining the weights is supposed inextensible. If we denote the accel eration by g', we shall have, from what was said at the beginning of this article,

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Fig. 99.

(83.)

By diminishing the difference between m and m', we may make the acceleration as small as we please. It is plain that g' is constant; hence, the motion of Wis uniformly varied.

m m'

If we replace g by m + m'I, in Equations (73) and (74),

they will make known the circumstances of motion of the body W. This principle is employed to illustrate the laws of falling bodies by means of ATWOOD's machine.

Had the two weights under consideration been attached. to the extremities of cords passing around a wheel and its axle, and in different directions, it might have been shown that the motion would be uniformly varied, when the moment of either weight exceeded that of the other. The same principle holds good in the more complex combinations of pulleys, wheels and axles, &c. In practice, however, the hurtful resistances increase so rapidly, that even when the moving force remains constant, the velocity soon attains a maximum limit, after which the motion will be sensibly uniform.

EXAMPLES.

1. Two weights of 5 lbs. and 4 lbs., respectively, are suspended from the extremities of a cord passing over a

fixed pulley. What distance will each weight describe in the first second of time, what velocity will be generated in one second, and what will be the tension of the connecting cord?

SOLUTION.

Since the masses are proportional to the weights, we shall have,

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Hence, the velocity generated is 3.574 ft., and the space passed over is 1.787 ft. To find the tension of the string, denote it by x. The moving force acting upon the heavier body, is (5x)g, and the acceleration due to this force,

5

5

g; the moving force acting upon the lighter body,

is (x-4)g, and the corresponding acceleration, (4) g.

But since the two bodies move together, these accelerations must be equal. Hence,

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.. x= 44 lbs., the required tension.

2. A weight of 1 lb., hanging on a pulley, descends and drags a second weight of 5 lbs. along a horizontal plane. Neglecting hurtful resistances, to what will the accelerating force be equal, and through what space will the descending body move in the first second?

SOLUTION.

The moving force is equal to 1 × g, and the mass moved g is equal to 6. Hence, the acceleration is equal to = 5.1944 6

ft., and the space described will be equal to 2.5972 ft.

3. Two bodies, each weighing 5 lbs., are attached to a string passing over a fixed pulley. What distance will each

body move in 10 seconds, when a pound weight is added to one of them, and what velocity will have been generated at the end of that time?

SOLUTION.

The acceleration will be equal to g 2.924 ft. = g'. But, s = g't', v = g't. Hence, the space described in 10 seconds is 146.2 ft., and the velocity generated is 29.24 ft.

4. Two weights, of 16 oz. each, are attached to the ends of a string passing over a fixed pulley. What weight must be added to one of them, that it may descend through a foot in two seconds?

SOLUTION.

Denote the required weight by x; the acceleration will

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118.

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gg'. But s=g't': making s=1

32 + x
2, we have,

2x

32+x

× 321; .. x 0.487 oz. Ans.

Atwood's Machine.

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ATWOOD's machine is a contrivance to illustrate the laws of falling bodies. It consists of a vertical post AB, about 12 feet in height, supporting, at its upper extremity, a fixed pulley A. To obviate, as much as possible, the resistance of friction, the axle is made to turn upon friction rollers. A fine silk string passes over the pulley, and at its two extremities are fastened two equal weights C and D. In order to impart motion to the weights, a small weight G, in the form of a bar, is laid upon the weight C, and by diminishing its mass, the acceleration may be rendered as small as desirable. The vertical rod AB, graduated to feet and decimals, is provided with two sliding stages E and F; the upper one is in the form of a ring, which will permit the

Fig. 100.

D

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