this line passes through the point T and, being parallel to OR, makes an angle with the axis Ox, its equation is

67. Any number of forces act in one plane at different points of a rigid body; to find the conditions that they may balance each other.

Let the system of forces be that of Art. 64, then we have to consider the two cases of Art. 65. In the second of these cases the resultant is the force R' (= R) acting at T; there cannot be equilibrium unless this force vanish, or R = 0. But if this be the case, the second case coincides with the first; and the resultant is a couple whose moment = Σ(Xy - Yx); there cannot be equilibrium unless this couple also vanish. Consequently the conditions of equilibrium are

=

ΣΧ = 0, ΣΥ = 0, Σ(Xy - Y) = 0.

These are both necessary and sufficient.

By referring to Art. 64 we perceive that Xy, - Yx, is equal to the moment of F, about the point 0, consequently Σ(Xy - Yx) is equal to the sum of the moments of all the forces about O. If then we remember that the point 0, and the directions of the axes Ox, Oy, were arbitrarily chosen in the plane of the forces, we may enunciate the conditions of equilibrium as follows:

The algebraic sums of the components of the forces parallel to any two lines at right angles to each other in the plane of the forces are each equal to zero; and the sum of the moments of all the forces about any point in the plane of the forces, or about any axis at right angles to the plane of the forces, is also equal to zero.

68.

Suppose that there is in the plane of the forces a fixed point, or perpendicular to the plane of the forces a fixed axis; to find the conditions of equilibrium.

Let the fixed point, or the point where the fixed axis cuts the plane of the forces, be taken for the point O in the investigation of Art. 64. Then the force R which acts at O can produce no effect since it acts on an immoveable point, it is not necessary then that R should be equal to zero. But the couple whose moment is Σ(Xy – Yx), if it exist, will turn the body about O, and therefore that there may be equilibrium it is necessary and sufficient that Σ(Xy - Yx) = 0.

There is therefore only one necessary condition of equilibrium, viz;

That the sum of the moments of all the forces about the fixed point or axis should be equal to zero.

REMARK. When there is equilibrium, the pressure on the fixed point is due entirely to the force R which acts directly upon it. Hence the pressure on the fixed point is the same as if all the forces which act on the body were transposed to the fixed point without altering their directions.

CHAPTER IV.

ON FORCES, NOT IN ONE PLANE, WHICH ACT UPON
DIFFERENT POINTS OF A RIGID BODY.

69. IF the directions of the forces all pass through a point we may transfer them to that point, and find their resultant by Chapter I. or II.

70. In the present Chapter we shall meet with couples of which the planes are not parallel. We can however always reduce them to other couples in the planes of rectangular co-ordinates. It is necessary therefore only to observe that when a couple acts in a co-ordinate plane, it will be considered a positive couple when its axis stands on the positive side of that plane. Thus a positive couple

in the plane ys has its axis coinciding with + Ox,

71. Parallel forces not in one plane act on different points of a rigid body; to find their resultant.

Take any point 0 (fig. 17) in the rigid body, from which draw Ox parallel to the direction of the proposed forces, which take for axis of z. Draw Ox, Oy in any directions at right angles to each other and to Ox, which take for the axes of x and y. Let Z1, Z2... Z be the forces; P the point where the line in which Z, acts cuts the plane xy. OM = x1, MP = y, the co-ordinates of P. Complete the parallelogram OMPN and join OP. At the point O apply two

19

19

n

pairs of opposite forces Z', Z" each equal and parallel to Z; these do not affect the system.

1

Now it has been shewn in Art. 55, that when two equal parallel forces act in the same direction at the extremities of one of the diagonals of a parallelogram, they may be transposed to the extremities of the other diagonal. Let us on this principle transpose Z, to M, and Z' to N. We have then, one force Z' acting at O in the direction Ox, and two couples in the plane xx, yx whose arms are OM, ON, the former couple being negative. By this means we have transposed the force Z1 to O, retaining its proper direction, and have introduced the couples - Z11, + Z1y, in the planes xx, yz respectively. Proceeding in the same manner with the remaining forces Z2, Z3....Z,, we shall have, instead of the original system, the forces Z1, Z2....Z, acting in the line Ox, which (by Art. 23) have a resultant

n

R = Z. Z.........(1) ;

and, in the plane xx, a set of couples, which (by Art. 52) are equivalent to a single couple in that plane whose moment

and, in the plane yx, a set of couples which are equivalent to one whose moment

= Σ(Zy).

If G be the moment of the resultant of these two couples, and the angle which its arm makes with Ox, we shall have from Art. 55,

G. cos 0 = (Za), and G. sin 0 = Σ (Zy) ;
.. G2 = (Σ. Zx)2 + (Σ. Zy)3.........(2),

and tan

=

Σ.Zy
Σ.Ζω

(3).

Equation (1) gives the resultant force acting at the origin of co-ordinates; and equations (2) (3) give the magnitude and position of the plane of the resultant couple.

72. We have determined the position of the arm of the resultant couple. That result supposes, as in fig. 17, a negative force acting at that extremity of the arm which

is at O, and a positive force at the other extremity.

It will

sometimes be more convenient to know the position of the

In which equations G is to be accounted positive.

73. The results obtained in Art. 71 are perfectly general, but they admit of reduction to a single force except when R or ΣΖ = 0.

(1) When ΣZ = 0, there is no force acting at 0, and the only resultant is the couple whose moment is G.

(2) When ΣZ is not equal to zero, change the couple whose moment is G into an equivalent couple which has each of its forces R' R" equal to R or EZ; its arm will

G

be equal to ; place this couple as in fig. 18, so that one

R

of its forces R" balances the resultant R. By this mode the whole are reduced to a single force R' (= ΣZ) acting at a point P whose co-ordinates a'y are known from the equation x = OP cos 0, and y' = OP sin

These equations are free from ambiguity.

74. To find the conditions that the system of forces in Art. 71 may balance each other.

We must consider the two cases mentioned in the last Article. In the second of these cases the resultant is the force ZZ acting at a point whose co-ordinates are