109. To deduce the Parallelogram of Forces from the Principle of the Lever.

If a force equal to the resultant of P and Q act along ro, it will with the forces P and Q keep a particle at O in equilibrium. Suppose these forces applied by means of rods, connected at 0; these rods will then be in equilibrium. Imagine the point C on the rod Or to become fixed; this will not disturb the equilibrium. The system can still turn round C, and it will do so unless the moments round Care equal and contrary. Thus if there is equilibrium we must have, by the Principle of the Lever,

Thus the forces P and Q are proportional to the sides OA and OB of the parallelogram OACB, and the diagonal OC represents the direction of their resultant.

This demonstrates the Parallelogram of Forces so far as relates to the direction of the resultant; then as in Art. 49 we can demonstrate it also for the magnitude of the resultant.

EXAMPLES. VII.

1. ABCD is a square; a force of 3 lbs. acts from A to B, a force of 4 lbs. from B to C, and a force of 5 lbs. from C to D: if the centre of the square be fixed, find the force which acting along DA will keep the square in equilibrium.

2. The length of a horizontal lever is 12 feet, and the balancing weights at its ends are 3 lbs. and 6 lbs. respectively: if each weight be moved 2 feet from the end of the lever, find how far the fulcrum must be moved for equilibrium.

3. If the forces at the ends of the arms of a horizontal lever be 8 lbs. and 7 lbs., and the arms 8 inches and 9 inches respectively, find at what point a force of 1 lb. must be applied perpendicularly to the lever to keep it at rest.

4. The arms of a lever are inclined to each other: shew that the lever will be in equilibrium with equal weights suspended from its extremities, if the point midway between the extremities be vertically above the fulcrum.

5. A weight P suspended from one end of a lever without weight is balanced by a weight of 1 lb. at the other end of the lever; when the fulcrum is removed through half the length of the lever it requires 10 lbs. to balance P: determine the weight of P.

6. A rod capable of turning round one end, which is fixed, is kept at rest by two forces acting at right angles to the rod; the greater force is 6 lbs. and the distance

between the points of application of the forces is half the distance of the greater force from the fixed end: find the smaller force. Shew that if any force be added to the smaller force, a force half as large again must be added to the greater force in order to preserve equilibrium.

7. ABC is a triangle without weight, having a right angle at C, and CA is to CB as 4 is to 3; the triangle is suspended from C, and two forces P and Q acting at A and B in directions at right angles to CA and CB keep it at rest: find the ratio of P to Q.

8. In Example 7 if the force P act at A at right angles to AC, and the force Q act at B at right angles to BA, find the ratio of P to Q in order that the triangle may be at rest.

9. The lower end B of a rigid rod 10 feet long is hinged to an upright post, and its other end A is fastened by a string 8 feet long to a point C vertically above B, so that ACB is a right angle. If a weight of one ton be suspended from A find the tension of the string.

10. If three weights P, Q, S hang from the points A, B, C of a straight lever which balances about a fulcrum D, shew that

Q× AB+S× AC=(P+Q+S) × AD.

11. A triangle can turn in its own plane round a point which coincides with the centre of the inscribed circle; forces acting along the sides keep the triangle in equilibrium; shew that one of the forces is equal to the sum of the other two.

12. A string passes through a small heavy ring, and the ends of the string are attached to the ends of a lever without weight shew that when the system is in equilibrium the ring is vertically under the fulcrum.

VIII. Centre of Parallel Forces.

110. Suppose we have two parallel forces acting respectively at two points; we know that their resultant is equal to the algebraical sum of the forces, and is parallel to them, and that it may be supposed to act at a definite point on the straight line which passes through the two points. See Arts. 60 and 61. Moreover this definite point remains the same however the direction of the two forces be changed, so long as they remain parallel. This point is called the centre of the two parallel forces. Hence we adopt the following definition:

The centre of a system of parallel forces is the point at which the resultant of the system may be supposed to act, whatever may be the direction of the parallel forces.

111. To find the resultant and the centre of any system of parallel forces.

Join AB, and divide it at L, so that AL may be to LB as Q is to P; then the resultant of P at A and Q at B is P+Q parallel to them, at L.

Join LC, and divide it at M, so that LM may be to MC as R is to P+Q; then the resultant of P+Q at L and R at C is P+Q+R parallel to them, at M.

Join MD, and divide it at N, so that MN may be to ND as S is to P+Q+R; then the resultant of P+Q+R at M and S at D is P+Q+R+S parallel to them, at N.

Thus we have found the resultant and the centre of four parallel forces; and in the same way we may proceed whatever be the number of the forces.

112. In the figure and language of the preceding Article we have implied that the forces are all like: it is easy to make the slight modifications which are required when this is not the case. We may, if we please, form two groups, each consisting of like parallel forces, and obtain the resultant and centre of each group; and then by Art. 61 deduce the resultant and centre of the whole system of parallel forces.

We shall always obtain finally a single resultant and a definite centre, except in the case where the algebraical sum of all the forces is zero; and then either the forces are in equilibrium or they form a couple: see Art. 95.

113. We have thus shewn how to determine geometrically the position of the centre of a system of parallel forces: we shall now shew how we may attain the same end by the aid of algebraical formulæ.

114. The distances of the points of application of two parallel forces from a straight line being given, to determine the distance of the centre of the parallel forces from that straight line; the straight line and the points being all in one plane.

First let A and B be the points of application of two like parallel forces, P and Q; their resultant is P+Q, parallel to them, and it may be supposed to act at the point Č, which is such that

a

A

B

Let AD, BE, CF be perpendiculars from A, B, C on any straight line which is in a plane containing A and B. Let AD=p, BE=q, CF=r: then we have to find the value of r, supposing the values of p and q to be known.