direction of each, as though the other did not exist. Now, if we suppose the material point O, to be acted upon simultaneously by the two forces P and Q, it will, by virtue of the first, be found at the end of one second somewhere on the line PR; and by virtue of the second somewhere on the line QR; hence, it will be at their point of intersection. But had the point O been acted upon by a single force, represented in direction and intensity by OR, it would have moved from 0 to R in the same time. Hence, the single force R is equivalent, in effect, to the aggregate of the two forces P and R; it is, therefore, their resultant. Hence,

If two forces be represented in direction and intensity by the adjacent sides of a parallelogram, their resultant will be represented in direction and intensity by that diagonal of the parallelogram which passes through their point of in

tersection.

This principle is called the parallelogram of forces.

In the preceding demonstration we have only considered moving forces, but the principle is equally true for forces of pressure; for, if we suppose a force equal and directly opposed to the resultant R, this force will be in equilibrium with the forces P and Q, which will then become forces of pressure. The relation between the forces will not be changed by this hypothesis, and we may therefore enunciate the principle as follows:

If two pressures be represented in direction and intensity by the adjacent sides of a parallelogram, their resultant will be represented in direction and intensity by that diagonal of the parallelogram which passes through their com mon point.

This principle is called the parallelogram of pressures.

Hence, we see that moving forces and pressures may be compounded and resolved according to the same principles, and by the same general laws.

Parallelopipedon of Forces.

28. Let P, Q, and S represent three forces applied to the same point, and not in the same plane. Upon these lines,

S

as edges, construct the parallelopipedon OR, and draw OM, and SR. From the preceding article, OM represents the resultant of P and Q, and from the same article, OR represents the resultant of OM and S.

Hence, OR is the resultant of the three forces P, Q, and S. That is, if three forces be represented in direction and intensity by three adjacent

0

Fig. 5.

R

M

edges of a parallelopipedon, their resultant will be represented by that diagonal of the parallelopipedon which passes through their point of intersection.

This principle is known as the parallelopipedon of forces, and is equally true for moving forces and pressures.

Geometrical Composition and Resolution of Forces.

29. The following constructions depend upon the principle of the parallelogram of forces.

1. Having given the directions and intensities of two forces applied at the same point, to find the direction and intensity of their resultant.

Let OP and OQ represent the given forces, and O their point of application; draw PR parallel to OQ, and QR parallel to OP, and draw the diagonal OR; it will be the resultant sought.

Fig. 6.

R

2. Having given the direction and intensity of the resultant of two forces, and the direction and intensity of one of its components, to find the direction and intensity of the other component.

Let R be the given resultant, P the given component, and O their point of application; drawRP, and through O draw OQ parallel to RP, also through R draw RQ parallel to PO; then will OQ be the component sought.

3. Having given the direction and intensity of the resultant of two forces, and the directions of the two components, to find the intensities of the components.

Let R be the given resultant, OP and OQ the directions of the components, and their point of application. Through R draw RP and RQ respectively, parallel to QO and PO,

Fig. 7.

then will OP and OQ represent the intensities of the components.

From this construction it is evident that any force may be resolved into two components having any direction whatever; these, again may each be resolved into new components, and so on; hence it follows that a single force may be resolved into any number of components having any assumed directions whatever.

4. Having given the direction and intensity of the resultant of two forces, and the intensities of the components, to find their directions.

06

Let R be the given resultant, and O its point of application. With R as a centre, and one of the components as a radius, describe an arc of a circle; with O as a centre, and the other component as a radius, describe a second arc cutting the first at P; draw PR and PO, and complete the parallelogram PQ, then will OP and OQ be the directions sought.

Fig. 8.

R'

5. To find the resultant of any number of forces, P, Q, S, T, &c., lying in the same plane, and applied at the same point. Construct the resultant R' of P and Q, then construct the resultant R" of R' and S, then the resultant R of R" and T, and so on: the final resultant will be the resultant of the system.

By inspecting the preceding figure, we see that in the polygon OQ R'R'RT, the side QR' is equal and parallel to the force P, the side

Tk.

Fig. 9.

R'R' to the force S, and the side RR to the force T,

and so on. Hence, we may construct the resultant of such a system of forces by drawing through the second extremity of the first force, a line parallel and equal to the second force, through the second extremity of this line, a line parallel and equal to the third force, and so on to the last. The line drawn from the starting point to the last extremity of the last line drawn, will represent the resultant sought. If the last extremity of the last force fall at the starting point, the resultant will be 0, and the system will be in equilibrium.

This principle is called the polygon of forces; its simplest case is the triangle of forces.

Components of a Force in the direction of two axes.

30. To find expressions for the components of a force which act in directions parallel to two

rectangular axes. Let X and Y be two such axes, and R any force lying in their plane; construct the components parallel to OX and OY, as before explained, and denote the angle LAR, which the force makes with the axis of X, by a. From the figure, we have,

10

A

X

Fig. 10.

AL R cos a, and RL AMR sin a;

=

or, making AL = X, and AMY, we have,

X = R cos a, and Y = R sin a

The angle a is estimated from the direction of positive abscissas around to the left through 360°.

For all values of a from 0° to 90°, and from 270° to 360°, the cosine of a will be positive, and, consequently, the component AL will be positive; that is, it will act in the direction of positive abscissas. For all values of a from 90° to 270°, the cosine of a will be negative, and the component AL will act in the direction of negative abscissas.

For all values of a from 0° to 180°, the sine of a will be positive, and the component AM will be positive; that is, it will act in the direction of positive ordinates. For all values of a from 180° to 360°, the sine of a will be negative, and the component AM will act in the direction of negative ordinates.

0

Fig. 10.

X

have AL = 0. For a = 0, or a = 180°, we shall have AM = 0.

If we regard AL and AM as two given forces, R will be their resultant; and since PL = AM, we shall have from the figure,

Hence, the resultant of any two forces, at right-angles to each other, is equal to the square root of the sum of the squares of the two forces.

From the figure, we also have,

1. Two pressures of 9 and 12 pounds, respectively, act upon a point, and at right-angles to each other. Required, the direction and intensity of the resultant pressure.

That is, the resultant pressure is 15 lbs., and it makes an angle of 53° 7′ 32" with the direction of the first force.

2. Two forces are to each other as 3 is to 4, and their