weight C, to pass, but not the bar G; the lower one is in the form of a plate, which is intended to intercept the weight C. There is also connected with the instrument a seconds pendulum for measuring time.

Let us suppose that the weights of C and D, are each equal to 181 grains, and that the weight of the bar G, is 24 grains. Then will the acceleration be

= 1 sec., we shall

2 sec., we shall,

If, in these equations, we make t have h = 1, and v 2. If we make t = in like manner, have h = 4, and v 4. If we make t = 3 sec., we shall have h

=

9, and v =

6, and so on. To verify these results experimentally, commencing with the first. The weight C is drawn up till it comes opposite the 0 of the graduated scale, and the bar G is placed upon it. The weight thus set is held in its place by a spring. The ring E is set at 1 foot from the 0, and the stage F, is set at 3 feet from the 0. When the pendulum reaches one of its extreme limits, the spring is pressed back, the weight C, G descends, and as the pendulum completes its vibration, the bar G strikes the ring, and is retained. The acceleration then becomes 0, and the weight C moves on uniformly, with the velocity that it had acquired, in the first second; and it will be observed that the weight C strikes the second stage just as the pendulum completes its second vibration. Had the stage F been set at 5 feet from the 0, the weight C would have reached it at the end of the third vibration of the pendulum. Had it been 7 feet from the 0, it would have reached it at the end of the fourth vibration, and so on. To verify the next result, we set the ring E at four feet

from the 0, and the stage F at 8 feet from the 0, and proceed as before. The ring will intercept the bar at the end of the first vibration, and the weight will strike the stage at the end of the second vibration, and so on.

By making the weight of the bar less than 24 grains, the acceleration is diminished, and, consequently, the spaces and velocities correspondingly diminished. The results may be verified as before.

Motion of Bodies on Inclined Planes.

119. If a body be placed on an inclined plane, and abandoned to the action of its own weight, it will either slide or roll down the plane, provided there be no friction between it and the plane. If the body is spherical, it will roll, and in this case the friction may be disregarded. Let the weight of the body be resolved into two components; one perpendicular to the plane, and the other parallel to it. The plane of these components will be vertical, and it will also be perpendicular to the given plane. The effect of the first component will be counteracted by the resistance of the plane, whilst the second component will act as a constant force, continually urging the body down the plane. The force being constant, the body will have a uniformly varied motion, and Equations (67) and (68) will be immediately applicable. The acceleration will be found by projecting the acceleration due to gravity upon the inclined plane. Let AB represent a section of the inclined plane made by a vertical plane taken perpendicular to the given plane, and let P be the centre of gravity of a body resting on the given plane. Let PQ represent the acceleration due to gravity, denoted by g, and let PR be the component of g, which is parallel to AB, denoted by g', PS being the normal component. Denote the angle that AB makes with the horizontal plane by a. Then, since PQ is perpendicular to BC, and QR to

B

R

Fig. 101.

A

AB, the angle RQP is equal to ABC, or to a. have, from the right-angled triangle PQR,

g' = gsina.

Hence we

But the triangle ABC is right-angled, and, if we denote its height AC by h, and its length AB by 1, we shall have

sina =

h

This value of g' is the value of the acceleration due to the Substituting it for f in Equations (67) and

moving force.

If the body starts from rest at A, taken as the origin of spaces, then will v' = 0 and s′ =

0, giving,

(85.)

(86.)

To find the time required for a body to move from the

top to the bottom of the plane, make s

=

7, in (86); there

Hence the time varies directly as the length, and inversely

as the square root of the height.

For two planes having the same height, but different lengths, the radical factor of the value of t will remain con

stant. Hence, the times required for a body to move down any two planes having the same height, are to each other as their lengths.

To determine the velocity with which a body reaches the bottom of the plane, substitute for t, in Equation (85) its value taken from Equation (86). We shall have, after reduction,

v = √2gh.

But this is the velocity due to the height h (Art. 115). Hence, the velocity generated in a body whilst moving down any inclined plane, is equal to that generated in falling freely through the height of the plane.

EXAMPLES.

1. An inclined plane is 10 feet long and 1 foot high. How long will it take for a body to move from the top to the bottom, and what velocity will it acquire in the descent ?

substituting for its value 10, and for h its value 1, we have,

t = 2 seconds nearly.

From the formula v = √2gh, we have, by making h = 1,

v = √64.33 = 8.02 ft.

2. How far will a body descend from rest in 4 seconds, on an inclined plane whose length is 400 feet, and whose height is 300 feet?

3. How long will it take for a body to on a plane whose length is 150 feet, and feet?

Ans. 193 ft. descend 100 feet whose height is 60 Ans. 3.9 sec.

4. There is an inclined railroad track, 24 miles in length, whose inclination is 1 in 35. What velocity will a car attain, in running the whole length of the road, by its own weight, hurtful resistances being neglected?

Ans. 155.75 ft., or, 106.2 m. per hour.

5. A railway train, having a velocity of 45 miles per hour, is detached from the locomotive on an ascending grade of 1 in 200. How far, and for what time, will the train continue to ascend the inclined plane?

66 =

SOLUTION.

We find the velocity to be 66 ft. per second. Hence, √2gh; or, h 67.7 ft. for the vertical height. Hence, 67.7 × 200 = 13,540 ft., or, 2.5644 m., the distance which the train will proceed. We have,

2

t = 1

gh

410.3 sec., .or, 6 min. 50.3 sec.,

for the time required to come to rest.

6. A body weighing 5 lbs. descends vertically, and draws a weight of 6 lbs. up an inclined plane of 45°. How far will the first body descend in 10 seconds?

SOLUTION.

The moving force is equal to 5 - 6 sin 45°; and, conse. sequently, the acceleration,

Motion of a Body down a succession of Inclined Planes. 120. If a body start from the top of an inclined plane, with an initial velocity v', it will reach the bottom with a velocity equal to the initial velocity, increased by that due to the height of the plane. This velocity, called the terminal velocity, will, therefore, be equal to that which the body