given forces P and Q + R acts at A or at F; A and F are therefore two points in its direction; and we have therefore proved that if the proposition of the parallelogram of forces be true for two forces P and Q, and also for two forces P and R, it must necessarily be true for P and Q + R.

But the proposition has been seen to be true for any pair of equal forces as p and p: and also for a second pair p and p, and therefore by the preceding for p and 2p: again, since it is true for Ρ and P, and also Ρ and 2p, it is therefore true for p and 3p; similarly it may be shewn by successive deduction to be true for p and mp where m is any whole number. Since it is true for mp and p, and also for mp and p, it is therefore true for mp and 2p; and by proceeding as before, it may be shewn to be true for mp and np when m and n are any whole numbers; i.e. since p is indeterminate, it is true for any commensurable forces whatever.

17. To extend the proposition to incommensurable forces, let AB, AC represent two such forces acting simultaneously at A. Complete the parallelogram ABDC.

If AD be not the direction of the resultant of the forces represented by AC, AB, let some other line as AE be it, cutting CD in E. Suppose AC to be divided into a number of equal

B

parts each less than ED, and suppose distances of the same magnitude to be marked off along CD; then one of these divisions must necessarily fall between E and D, let it be F; complete the parallelogram ACFG. Then if we conceived two commensurable forces to act at A, which should be represented by the lines AC, AG, the direction of their resultant would be AF, and it would lie farther away from AC than AE, the direction of the resultant of the forces represented by AC and AB, does, although AG is less than AB; which is evidently absurd. Hence AE is not the direction of the resultant of the given forces. In like manner it can be shewn that no other

line except AD is in that direction, and therefore the proposition is true for incommensurable as well as for commensurable forces.

The second step in our proof is to shew that the proposition is true as regards the magnitude of the resultant.

As before, let AB, AC represent the two forces P and Q acting at the point A; and let AE represent a third force R, which also acting at A will keep P and Q in equilibrium; then, by Art. (12), R must be equal and opposite to the resultant of P and Q. For the same reason P must be equal and opposite to the resultant of Q and R; therefore, completing the parallelograms BACD, CAEF, we must have DA in the same straight line with AE, and BA with AF: hence

AFCD is a parallelogram, and AD equals FC, also FC is by construction equal to AE; therefore AD = AE, in other words AD, the diagonal of the parallelogram described upon AC, AB which represent the forces P and Q, represents a force which is equal in magnitude to the resultant of P and Q. Q. E. D.

18. From the proposition of the Parallelogram of Forces follows immediately the proposition of the Triangle of Forces :

That if three forces acting together upon a point be in equilibrium, they are respectively proportional to the three sides taken in order of the triangle formed by drawing lines parallel to their directions. And conversely, that if three forces acting at a point

be proportional to and in the direction of the respective sides of a triangle taken in order, they must be in equilibrium.

This is in fact only another method of wording the parellelogram of forces. For that asserts that three forces acting upon a point, keeping each other in equilibrium, are proportional to the two sides and diagonal of a parallelogram whose directions are parallel to those of the forces, and conversely. But the sides and diagonal form a triangle, as may be seen by reference to triangle ACD of the last figure; and it is therefore the same thing whether we say of three straight lines that they are the sides and diagonal of a parallelogram, or that they will form a triangle. The parallelogram and the triangle of forces are therefore identical propositions.

19. Since in any triangle the sides are proportional to the sines of the opposite angles, our proposition shews us, that of three forces keeping any point in equilibrium, each is proportional to the sine of the angle contained between the directions of the other two.

Thus, in the annexed figure, which represents three forces P, Q, and R keeping the point A in equilibrium, if ABC be

Р

a triangle having its sides parallel to the forces, we have by

triangle of forces,

5

P: QR::AB: BC: AC,

:: sin ACB sin BAC: sin ABC,

:

:: sin QAC: sin BAC: sin PBC,

:: sin QAR : sin RAP: sin PAQ.

20. In a similar manner, if any number of forces acting upon a point are parallel and proportional to the sides taken in order of a polygon, they must keep each other in equilibrium. Let AB‚Â ̧Â ̧Â, be the polygon to whose sides the forces P1P,..P, acting at A, are respectively parallel and proportional. Join AB,, then the triangle ABB, has its sides AB1, BB2 respectively parallel and proportional to the forces P,P2; therefore the third side AB, is in the direction of, and proportional to, their resultant. Similarly AB, is in the direction of, and proportional to, the resultant of the forces represented by AB, and B,B,; but BB, is parallel and proportional to P,, and AB2 is parallel and proportional to resultant of P, and P2, therefore AB, is parallel and proportional to the resultant of P1, P2, and P. By proceeding in the same way it could be shewn that AB, is

P4

P5

P3

B4

Pe

BL

B2

parallel and proportional to the resultant of P1, P2, P ̧, and P ̧, but the same line is by hypothesis parallel and proportional to the last force P, in a direction opposite to AB. Therefore the resultant of the four first forces is equal and opposite to the fifth and last force, hence the five forces P,P,...P, keep each other in equilibrium.

5

In the same way, this proposition, which is called the polygon of forces, could be shewn to be true of any number of forces acting at a point. It is evidently not necessary that the forces should lie in one plane.

The converse of the polygon of forces cannot, as the triangle of forces, be affirmed; for parallel-sided polygons have not necessarily their sides in the same proportion, whereas triangles have.

21. The Parallelogram of Forces enables us to substitute for a given force acting upon a particle its two resolved parts in two given directions.

The same may be done with each of any number of given forces acting in the same plane upon a point. Hence any such system of forces may be resolved into known components acting in two given directions.

22. A further application of the parallelogram of forces leads us to the conclusion, that if the given forces be in equilibrium, then must their components in any two given directions separately vanish.

From this proposition is deduced the analytical method of forming the "Equations of Equilibrium" of a material particle under the action of given forces.

Examples to Section II.

(1). A knot in a cord is rendered immoveable by a tack driven through it, the two extremities of the cord are then pulled with forces P and Q respectively, in such a manner as to include an angle a between their directions; required the magnitude and direction of the resultant force upon the tack.

Let A be the tack, AP, AQ the directions in which the two strings are pulled by the forces P and Q respectively;

B

R

= α.

then PAQ In AP, AQ take AB, AC respectively proportional to P and Q; complete the parallelogram, and draw