point upon the direction of the force. The moment, as we shall see in the next chapter, measures the tendency of the force to produce rotatory motion about the fixed point.

11. PROP. The moment of the resultant about any point in the plane of the forces equals the sum of the moments of the forces.

Let the forces P and Q, acting at A, be represented by the lines AB, AC, and their resultant R by AD, the diagonal of the parallelogram drawn upon AB, AC. Let O be the point about which the moments are taken; join OA, and draw AEFG perpendicular to OA; draw Ol,

A

E

F

G

m

n

B

H

R

D

P

Om, On, respectively perpendicular to AB, AC, AD; and BE, CF, DG, perpendiculars to AEFG; and CH parallel to that line.

Now the triangles OlA, OmA, OnA, are respectively similar to the triangles AEB, AFC, AGD.

but AECH = FG .. AF+AE = AG

or, AB. Ol+AC. Om = AD. On

or, P. Ol+Q. Om = R. On

Or the sum of the moments of the components equals the moment of the resultant about any point in their plane, or about an axis perpendicular to their plane.

If the point O fell within the angle formed by the forces, we should have the moment of one of the forces tending to cause ro

of the resultant. In the annexed figure we have the proof the same as above, except that AG = AF – AE,

By compounding R with another force acting at A, we should obtain a like result: or the moment of the resultant of these forces acting at A equals the algebraic sum of the moments of the forces. The proposition may, in the same manner, be extended to any number of forces acting at a point.

EXAMPLES.

1. Shew that if O be the angle between two forces of given magnitudes, their resultant is the greatest when = 0, least when, and intermediate for intermediate values of 0. If the component forces be P and Q, what is the magnitude of the resultant when 0 = 0, and also when = π? Ans. (P+Q) and (P-Q).

2. If two equal forces (P) meet at an angle of 60°, shew that their resultant = P √3.

3. If two equal forces meet at an angle of 135°, shew that their resultant = P (2-2).

4. If three forces, whose magnitudes are 3m, 4m, and 5m, act at one point and are in equilibrium, shew that the forces 3m and 4m are at right angles to each other.

5. If two equal forces are inclined to each other at an angle of 120°, shew that their resultant is equal to either of them.

6. If the magnitudes of two forces are 6 and 11, and the angle between their directions 30°, shew that the magnitude of their resultant is 16-47 nearly.

7. Shew, that in the last question the resultant makes with the force 6 the angle whose sine is 3339, and with the force 11 the angle whose sine is 1821, which are the sines of 19° 30′ and 10° 30′ nearly.

8. Apply the proof of the polygon of forces to the case of five equal forces represented by the sides of a regular pentagon taken in order.

9. Enunciate all the propositions requisite to prove that the resultant is in every respect mechanically equivalent to the component forces.

10. A cord PAQ is tied round a pin at the fixed point A, and its two ends are drawn in different directions by the forces P and Q shew that the angle between these directions is found 3 (P2 + Q2) - 2 PQ from the expression cos. when the

=

8 PQ

11. A cord whose length is 27 is tied at the points A and B in the same horizontal line, whose distance is 2 a: a smooth ring upon the cord sustains a weight w: shew that the force of tension in the cord =

พ

a2

CHAPTER II.

ON FORCES WHOSE DIRECTIONS ARE PARALLEL.

THOUGH the propositions in the last chapter will not apply at once to forces acting at different points, of which the directions are parallel, yet we can reduce the proof of the method of finding their resultant to that of two forces acting at one point in different directions.

12. PROP. If two parallel forces P and Q act at points A and B respectively, then their resultant equals their algebraic sum in magnitude, and acts at a point C in the same straight line with A and B, such that P× AC = Q × BC.

forces act in opposite directions, we suppose Q greater than P, so that the resultant of Q and S' will lie nearer Q than the re

sultant of P and S does to P; and therefore the directions of the resultants will meet at some point, as D, in both figures. We may suppose the whole of the forces to act at the point D. Resolving the forces parallel to the original directions, we shall have forces S and S' parallel to AB, which will destroy each other; and forces P and Q acting parallel to their original directions, giving a resultant R = P+Q in the upper figure, and R=Q-P in the lower figure. To find the point C in the line AB, or AB produced, where R acts, we have,

from triangle ACD, P:S:: CD: AC

from triangle BCD, S': Q:: BC: CD

Compounding these ratios, we have

P: Q:: BC: AC

or, Px ACQxBC

13. If AB be perpendicular to the direction of the forces, AC and BC are called the arms of the forces, and the products P.AC, Q.BC, are the moments of the forces, about the point C. If the forces are inclined at an angle a to AB, then the perpendiculars from C on the lines of action of the forces are AC.sin. a and BC.sin.a, and the moments are P.AC.sin.a and Q.BC.sin. a.

14. If C be a fixed point, its resistance will destroy the effect of the resultant force; so that P and Q will be in equilibrium about such a point, when their moments, tending to cause rotation opposite ways about it, are equal to each other.

15. To find AC in terms of AB, we have

PX AC = Q(AB-AC), in the upper figure;

-AB and similarly, BC =

AB

and P× AC=Q(AC—AB), in the lower figure; Q P or, AC= Q-P Q-P The point C is determined in both cases; unless in the latter P=Q, when AC=infinity; but then the resultant R=Q-P=0. This is a peculiar case; the effect of two equal and parallel forces which do not act at the same point being to produce rotatory motion only. Such forces constitute what is called a

C