CHAPTER V.

FORCES IN DIFFERENT PLANES.

66. PROP. To find the magnitude and direction of the resultant of any number of parallel forces acting upon a rigid body, and to determine the centre of parallel forces.

Let the points of application of the forces be referred to a system of rectangular coordinate axes.

Let m1, m,...

be the points of application; x111, x,y,..... their coordinates; P, P...... the forces acting at these points, those being reckoned positive which act in the direction of P1, and those negative which act in the opposite direction.

Join mm; and take the point m on mm, such that

2

2

1

then the resultant of P, and P, is P1 + P2, and it acts through m parallel to P. (Arts. 36, 37).

Draw ma, mb, mc perpendicular to the plane of (x, y), meeting that plane in a, b, c; draw m.de parallel to abc meeting mb in d and m ̧c in e.

Then, by similar triangles,

mb

=

P1 + P2

This gives the ordinate parallel to the axis of z of the point of application of the resultant of P, and P.

2

1

2

Then, supposing P, and P, to be replaced by P1+ P2 acting at m, the ordinate of the point of application of the resultant of P+ P2 and P ̧

and this process may be extended to any number of parallel forces. Let R denote the resultant force and z the ordinate

of its point of application; then

Similarly, if x, y be the other coordinates of the point of application of the resultant,

The values of x, y, z are independent of the angles which the directions of the forces make with the axes.

Hence

if these directions be turned about the points of application of the forces, their parallelism being preserved, the point of application of the resultant will not move. For this reason this point is called the centre of the parallel forces.

67. DEF. The moment of a force with respect to a plane is the product of the force into the perpendicular distance of its point of application from the plane.

In consequence of this definition, the equations for determining the position of the centre of parallel forces shew that the sum of the moments of any number of parallel forces with respect to any plane is equal to the moment of their resultant.

68. If the parallel forces all act in the same direction, the expression ΣP cannot vanish; hence the values of the coordinates of the centre of parallel forces found in Art. (66) cannot become infinite or indeterminate, and we are certain that the centre exists. But if some of the forces are positive and some negative, ΣP may vanish, and the results of Art. (66) become nugatory. In this case, since the sum of the positive forces is equal to the sum of the negative forces, the resultant of the former will be equal to the resultant of the latter. Hence the resultant of the whole system of forces is a couple, unless the resultant of the positive forces should happen to lie in the same straight line as the resultant of the negative forces.

We shall give another method of reducing a system of parallel forces.

69. PROP. To find the resultant of a system of parallel forces acting upon a rigid body.

Let P1, P.............. denote the forces. Take the axis of z parallel to the forces. Let the plane of (x, y) meet the

direction of P, in M1, and suppose x, y the coordinates of this point.

Draw MN, perpendicular to the axis of a meeting it in N1. At the origin 0, and also at N1, apply two forces each equal and parallel to P, and in opposite directions. Hence the force P at M, is equivalent to

(1) P1 at 0.

(2) a couple formed of P, at M, and P, at N.

1

(3) a couple formed of P, at N, and P1 at 0.

1

1

1

The moment of the first couple is Py,, and without altering its effect it may be transferred to the plane of (y, z), which is parallel to its original plane. The moment of the second couple is P,x,, and it is in the plane (x, z).

If we effect a similar transformation of all the forces, we have, as the resultant of the system,

(1) a force EP acting at 0,

(2) a couple ΣPy in the plane of (y, z),

(3) a couple ΣPx in the plane of (x, z).

The first couple tends to turn the body from the axis of y to that of z, and the second from the axis of x to that of z. We may therefore take Ox as the axis of the first couple according to the definition in Art. (41). For the second couple, however, we must either take Oy' as the axis, or consider it as a couple turning from z to x, of which the moment is Px and the axis Oy. Adopting the latter method, we may replace the two couples by a single couple of which the moment is G, where

G = (ΣPx)2 + (ΣPy)*,

and the axis is inclined to the axis of x at an angle a given by the equations

cos α =

ΣΤΥ
G

sin a=

-ΣPa
G

70. PROP. To find the conditions of equilibrium of a system of parallel forces acting upon a rigid body.

A system of parallel forces can always be reduced to a single force and a couple. Since these cannot balance, and neither of them singly can maintain equilibrium, they must both vanish. That is,

ΣP=0, and G = 0;

the latter requires that

ΣΡ = 0 and ΣΡΙ = 0.