59. The Resultant of two parallel forces. By Art. 56, a force in the same plane, whose resolved part in each of two directions in that plane equals the algebraical sum of the resolved parts of the two forces in the same direction, and whose moment about some point in the plane equals the algebraical sum of the moments of the two about that point, is the resultant.

Let A, B be two points where two parallel forces, P, Qrespectively act. The first two conditions are satisfied

1 1/1

Fig.33

Q-P

by a force which acts in the same direction as P and Q, and is equal to their algebraical sum. The required force must then be parallel to the other two, and at such distances from them that their moments about any point in it are equal in magnitude but opposite in sign. It will therefore be between them if the signs of P and Q are the same, but not otherwise: its distances from them must be inversely proportional to their magnitudes. Hence if C be the point where its line of action meets AB, P. AC Q.BC. When P and Q act in opposite directions, the greater force will clearly lie between the less and the resultant.

=

Cor. The position of C is independent of the direction of the forces, so long as they remain parallel.

If the forces P and Q are equal in magnitude and ·

opposite in sign, the preceding solution fails, and we can find no single force, whose effect is equal to that of the two together. Two such forces constitute a couple.

Ex. 1. Four forces, P, 2P, 3P, and 4P act along the sides taken in order of a square: find their resultant.

Ans. 2P√2, acting parallel to the diagonal through the corner where 2P, and 3P meet, and at a distance from it√2 times a side of the

square.

Ex. 2. A uniform beam 4 ft. long is supported in a horizontal position by two props which are 3 feet apart, so that the beam projects one foot beyond one of the props: shew that the pressure on one prop is double the pressure on the other.

Ex. 3. If a bicycle and its rider weigh 60 lbs. and 10 stone respectively, find how the pressure on the ground is divided between the two wheels, whose points of contact with the ground are 3 ft. 6 inches apart, while the points through which the weights of the bicycle and rider act, are distant horizontally 7 in. and 6 in. respectively from the centre of the driving wheel. Ans. 170 lbs. and 30 lbs.

60. Since a rigid body under the action of any system of coplanar forces, can be fixed by two forces of constraint acting in that plane at two arbitrarily chosen points in it, the system must be equivalent to the forces of constraint reversed: but two forces in one plane can be replaced by a single one, unless they form a couple. Hence

Any system of forces in one plane is equivalent to a single force or a couple.

61. Prop. If three forces maintain equilibrium, their lines of action must be in one plane, and either all meet in one point or be all parallel.

Let P, Q, R be the three forces, Aa, Bb, Cc, their respective lines of action.

Since the algebraical sum of the moments of a system of forces in equilibrium about any line is zero, that of the moments of P, Q, R about AB vanishes: but as P and Q both intersect this line, each of their moments about it is zero, hence that of R about it must also be zero, i.e. Cc meets AB or is parallel to it.

Similarly we can shew that Cc meets, or is parallel to Ab. Therefore Cc either lies in the plane ABb or passes through A. In the first case R and Q are in the same plane, in the second R and P. But if two of the forces are in one plane, the third must also be in it, as its moment about every line in the plane must be zero. Hence all three forces are in one plane.

If the forces are not all parallel, two of them meet and can be replaced by a single force, which is counterbalanced by the third force, and is therefore in the same straight line with it, i.e. the third force passes through the point of intersection of the other two.

Cor. Two forces, whose lines of action are not in one plane, cannot be equivalent to a single force.

62. Def. The moment of a couple is the algebraical sum of the moments of the two forces which form it, about any point in their plane.

This moment can easily be shewn to be independent of the position of the point and to be equal to the product of either force into the arm, i.e. the perpendicular distance between the lines of action of the forces.

For let P acting at A, and P acting in the opposite direction at B, form the couple. Then the algebraical sum of the moments of the two forces about is

where Oa, Ob are the perpendiculars from O on the lines of action of the forces.

If the body on which the couple acted were only free to turn round O, the tendency of the couple in all the above figures is to turn the body in the direction in which the hands of a watch move; the couples are said therefore to have moments of the same sign, or to be like; were the tendency of one of them to turn the body in the opposite direction, its moment would be of the opposite sign, and it would be unlike the other two.

63.

Prop. Two like couples of equal moment, in the same or parallel planes, are equivalent to one another. (i) When the couples are in the same plane.

In this case the two couples form two systems of forces in one plane, such that the algebraical sums of their moments about any point whatsoever in the plane are the same; therefore the systems are equivalent to one another (Art. 56).

(ii) When the couples are in parallel planes.

2

Let P,, P, be the two equal forces forming one of the couples, acting at the points A, B respectively.

Fig.38

2P

2

In the plane in which the other couple acts, draw CD equal and parallel to AB. Join AD, BC, intersecting in 0; then O bisects both AD and BC,