476. The third law states that action and reaction are equal and opposite. A man can not lift himself by his bootstraps, for the reason that he presses downwards with the same force that he pulls upwards; the downward reaction equals the upward action, and is opposite to it.

In springing from a boat we must exercise caution or the reaction will drive the boat from the shore. When we jump from the ground, we tend to push the earth from us, while the earth reacts and pushes us from it.

EXAMPLE. Two men pull on a rope in opposite directions, each exerting a force of 100 pounds; what is the force which the rope resists ? SOLUTION.-Imagine the rope to be fastened to a tree, and one man to pull with a force of 100 pounds. The rope evidently resists 100 pounds. According to Newton's third law, the reaction of the tree is also 100 pounds. Now, suppose the rope to be slackened, but that one end is still fastened to the tree, and the second man to take hold of the rope near the tree, and pull with a force of 100 pounds, the first man pulling as before. The resistance of the rope is 100 pounds, as before, since the second man merely takes the place of the tree. He is obliged to exert a force of 100 pounds to keep the rope from slipping through his fingers. If the rope be passed around the tree, and each man pulls an end with a force of 100 pounds in the same and parallel directions, the stress in the rope is 100 pounds, as before, but the tree must resist the pull of both men, or 200 pounds.

477. A force may be represented by a line; thus, in

Fig. 63, let A be the point of application of A

B

the force; let the length of the line AB represent its magnitude, and let the arrow- FIG. 63. head indicate the direction in which the force acts, then the line A B fulfils the three conditions (see Art. 469), and the force is fully represented.

CENTER OF GRAVITY.

478. The center of gravity of a body is that point at which the body may be balanced, or it is the point at which the whole weight of a body may be considered as concentrated.

This point is not always in the body; in the case of a horseshoe or a ring it lies outside of the substance of, but within the space enclosed by, the body.

In a moving body, the line described by its center of

FIG. 64.

W

center of gravity of a body may be applied to a system of bodies if they are considered as being connected at their centers of gravity.

If w and W, Fig. 64, be two bodies of known weight, their center of gravity will be at C. The point C may be readily determined as follows:

Rule 75.—The distance of the common center of gravity, from the center of gravity of the large weight, is equal to the weight of the smaller body multiplied by the distance between the centers of gravity of the two bodies, and this product divided by the sum of the weights of the two bodies.

EXAMPLE.-In Fig. 64, w= 10 pounds, W = 30 pounds, and the distance between their centers of gravity is 36 inches where is the center of gravity of both bodies sit

ters of gravity are known, by the following rule:

W2

W

Rule 76.-Find the center of gravity of two of the bodies as W, and W, in Fig. 65. Assume that the weight of both

bodies is concentrated at C,, and find the center of gravity of this combined weight at C, and the weight of W; let it be at C; then find the center of gravity of the combined weights of W, W, W, (concentrated at C), and W,; let it be at C; then C will be the center of gravity of the four bodies.

480. To find the center of gravity of any parallel

FIG. 66.

ogram:

Rule 77.-Draw the two diagonals, Fig. 66, and their point of intersection C will be the center of gravity.

481. To find the center of gravity of a triangle, as A B C, Fig. 67:

F

Rule 78. From any vertex, as A, draw a line to the middle point D of the opposite side B C. From one of the other vertexes, as C, draw a line to F, the middle point of the opposite side A B ; the point of intersection O of these two lines is the center of gravity.

B

D

FIG. 67.

It is also true that the distance DO=D A, and that F OF C, and the center of gravity could have been found by drawing from any vertex a line to the middle point of the opposite side, and measuring back from that side of the length of the line.

The center of gravity of any regular plane figure is the same as the center of the inscribed or circumscribed circle.

482. To find the center of gravity of any irregular plane figure, but of uniform thickness throughout, divide one of the parallel surfaces into triangles, parallelograms, circles, ellipses, etc., according to the shape of the figure; find the area and center of gravity of each part separately, and combine the centers of gravity thus found in the same manner as in rule 76; in this case, however,

dealing with the area of each part instead of its weight. See Fig. 68.

01

FIG. 68.

EXAMPLE.-Suppose that the two balls shown in Fig. 64 were 5′′ and 10" in diameter, and weighed 10 lb. and 80 lb., respectively. If the distance between their centers were 40", and they were connected by a steel rod 1" in diameter, where would the center of gravity be, taking the weight of a cubic inch of steel as .283 lb.?

SOLUTION. The length of the rod = 40

5

10

-

2 2

=

324". Its volume

is 12.7854 × 324 = 25.53 cu. in. 25.53 X .283 = 7.22 lb. The rod being straight, its center of gravity is in the middle at a distance of 32.5 5 32.5 10 + = 18" from the center of the smaller weight, and + = 2 2 2 2 211" from the center of the larger weight. Now, considering the weight of the rod to be concentrated at its center of gravity, we have three weights of 10, 7.22, and 80 lb., all in a straight line, and the distances between them given to find the center of gravity, or balancing point, of the combination, by rule 76. We will first find the center of gravity of the two smaller weights by rule 75, as follows: 7.22 × 18 = 135.38. 107.22 17.22. 135.38 ÷ 17.22 = 7.86" = distance from the center of the 10-lb. weight. Considering both of the smaller weights to be concentrated at this point, we find the center of gravity of this combined weight and the large weight as follows. 407.86 32.14" distance

=

between the center of gravity of the two small weights and the center of gravity of the 80-lb. weight. Applying rule 75, 17.22 × 32.14 =

553.45.

17.228097.22.

553.45 ÷ 97.225.693" distance from the center

of the 80-lb. weight. Ans.

=

483. Center of Gravity of a Solid.—In a body free to move, the center of gravity will lie in a vertical plumb

FIG. 69.

line drawn through the point of support. Therefore, to find the position of the center of gravity of an irregular solid, as the crank, Fig. 69, suspend it at some point, as B, so that it will move freely. Drop a plumb line from the point of suspension, and mark its direction. Suspend the body at another point, as A, and repeat the process. The intersection C of the two lines will be directly over the center of gravity.

Since the center of gravity depends wholly upon the shape and weight of a body, it may be without the body, as in the case of a circular ring whose center of gravity is the same as the center of the circumference of the ring.

EXAMPLES FOR PRACTICE.

1. A spherical shell has a wrought-iron handle attached to it. The shell is 10" in diameter, and weighs 20 lb. The handle is 14" in diameter, and the distance from the center of the shell to the end of the handle is 4 ft. Where is the center of gravity? Take the weight of a cubic inch of wrought iron as .278 lb. Ans. 13.612" from center of shell.