is at Paris, and near the equator it is five parts in 1000 less than it is near the pole. This is due in part to the earth's rotation and in part to the fact that the earth is not a perfect sphere and that in going from the equator toward the pole we are coming nearer and nearer to the center of the earth. We see, therefore, that the weight of one gram of mass is not an absolutely definite unit of force. One gram of force is, strictly speaking, the weight of one gram of mass in latitude 45° at sea level.

COMPOSITION AND RESOLUTION OF FORCES

A'

76. Graphic representation of force. A force is completely described when its magnitude, its direction, and the point at which it is applied are given. Since the three characteristics of a straight line are its length, its direction, and the point at which it starts, it is obviously possible to represent forces by means of straight lines. Thus, if we wish to represent the fact that FIG. 58. Graphic a force of 8 pounds, acting in an easterly direction, is applied at the point A (Fig. 58), we draw a line 8 units long, beginning at the point A and extending to the right. The length of this line then represents the magnitude of the force; the direction of the line, the direction of the force; and the starting point of the line, the point at which the force is applied.

representation of a single force

77. Resultant of two forces acting in the same line. The resultant of two forces is defined as that single force which will produce the same effect upon a body as is produced by the joint action of the two forces.

If two spring balances are attached to a small ring and pulled in the same direction until one registers 10 g. of force and the other 5, it will be found that a third spring balance attached to the same point and pulled in the opposite direction will register exactly 15 g. when there is equilibrium;

that is, the resultant of two parallel forces acting in the same direction is equal to the sum of the two forces.

Similarly, the resultant of two oppositely directed forces applied at the same point is equal to the difference between them, and its direction is that of the greater force.

78. Equilibrant. In the last experiment the pull in the spring balance which registered 15 g. was not the resultant of the 5 g. and 10 g. forces; it was rather a force equal and opposite to that resultant. Such a force is called an equilibrant. The equilibrant of a force or forces is that single force which will just prevent the motion which the given forces tend to produce. It is equal and opposite to the resultant and has the same point of application.

R

C

FIG. 59. Direction of resultant of two

equal forces at right angles

of 10 lb. (repre

79. The resultant of forces acting at an angle (concurrent forces). If a body at A is pulled toward the east with a force of 10 lb. (represented in Fig. 59 by the line AC) and toward the north with a force sented in the figure by the line AB), the effect upon the motion of the body must, of course, be the same as though some single force acted somewhere between AC and AB. If the body moves under the action of the two equal forces, it may be seen from symmetry that it must move along FIG. 60. The resultant lies

A

B

nearer the larger force

R

a line midway between AC and AB, that is, along the line AR. This line, therefore, indicates the direction as well as the point of application of the resultant of the forces AC and AB.

If the two forces are not equal, as in Fig. 60, then the resultant will lie nearer the larger force. The following experiment will show the relation between the two forces and their resultant.

Let the rings of two spring balances be hung over nails B and C in the rail at the top of the blackboard (Fig. 61), and let a weight W be tied near the middle of the string joining the hooks of the two balances. The weight W is not supported by the

pull of the balance E or by that of F; it is supported by their resultant, which evidently must act vertically upward, since the only single force capable of supporting the weight W is one that is equal and opposite to W. Let the lines OA and OD be drawn upon the blackboard behind the string, and upon these lines lay off the distances Oa and Ob, which contain as many units of length as there are units of force indicated by the balances E and F respectively. Similarly, on a vertical line from O lay off the exact distance OR required to represent the force that supports the weight. This, as noted above, represents the resultant. Now let a parallelogram be constructed upon Oa and Ob as sides. The line OR already drawn will be the diagonal.

FIG. 61. Experimental proof of parallelogram law

Hence, to find graphically the resultant of two concurrent forces, (1) represent the concurrent forces, (2) construct upon them as sides a parallelogram, and (3) draw a diagonal from the point of application. This diagonal represents the point of application, direction, and magnitude of the resultant.

ท

B

m

A

R

FIG. 62. Component of a force

80. Component of a force. Whenever a force acts upon a body in some direction other than that in which the body is free to move, it is clear that the full effect of the force cannot be spent in producing motion. For example, suppose that a force is applied in the direction OR (Fig. 62) to a car on an elevated track. Evidently OR produces two distinct effects upon the car: on the one hand, it moves the car along the track; and, on the other, it presses it down against the rails. These two effects

might be produced just as well by two separate forces acting in the directions OA and OB respectively. The value of the single force which, acting in the direction OA, will produce the same motion of the car on the track as is produced by OR, is called the component of OR in the direction OA. Similarly, the value of the single force which, acting in the direction OB, will produce the same pressure against the rails as is produced by the force OR, is called the component of OR in the direction OB. In a word, the component of a force in a given direction is the effective value of the force in that direction.

81. Magnitude of the component of a force in a given direction. Since, from the definition of component just given, the two forces, one to be applied in the direction OA and the other in the direction OB, are together to be exactly equivalent to OR in their effect on the car, their magnitudes must be represented

force is to have the

n

FIG. 63. Horizontal component of pull on a sled

same effect upon a body as two forces acting simultaneously, it must be represented by the diagonal of a parallelogram the sides of which represent the two forces. Hence, conversely, if two forces are to be equivalent in their joint effect to a single force, they must be sides of the parallelogram of which the single force is the diagonal. Hence the following rule: To find the component of a force in any given direction, represent the force by a line; then, using the line as a diagonal, construct upon it a rectangle the sides of which are respectively parallel and perpendicular to the direction of the required component. The length of the side which is parallel to the given direction represents the

magnitude of the component which is sought. Thus, in Fig. 62 the line Om completely represents the component of OR in the direction OA, and the line On represents the component of OR in the direction OB.

Again, when a boy pulls on a sled with a force of 10 lb. in the direction OR (Fig. 63), the force with which the sled is urged forward is represented by the length of Om, which is seen to be but 9.3 lb. instead of 10 lb. The component which tends to lift the sled is represented by· On.

To apply the test of experiment to the conclusions of the preceding paragraph, let a wagon be placed upon an inclined plane (Fig. 64), the height of which, bc, is equal to one half its length ab. In this case the force acting on the wagon is the weight of the wagon, and its direction is downward. Let this force be represented by the line OR. Then, by the construction of the preceding paragraph, the line Om will represent the value of the force which is pulling the carriage down the plane, and the line On the value of the force which is producing pressure against the plane. Now, since the triangle ROm is similar to the triangle abc (for ≤mOR = 4abc, RmO = 2acb, and ≤ ORm = Zbac), we have

m

FIG. 64. Component of weight parallel to an inclined plane

that is, in this case, since bc is equal to one half of ab, Om is one half of OR. Therefore the force which is necessary to prevent the wagon from sliding down the plane should be equal to one half its weight. To test this conclusion let the wagon be weighed on the spring balance and then placed on the plane in the manner shown in the figure. The pull indicated by the balance will, indeed, be found to be one half the weight of the wagon.

The equation Om/OR = bc/ab gives us the following rule for finding the force necessary to prevent a body from moving down an inclined plane, namely, the force which must be applied to a body to hold it in place upon an inclined plane bears the same ratio to the weight of the body as the height of the plane bears to its length.