X = Σ(Xr), Y = Σ(Yr), Z = Σ(Zr),
(6)
and a couple G whose components are
L = Σ(Lr), M = Σ(Mr), N = Σ(Nr),
(7)
where r = 1, 2, 3 ... Since R2 = X2 + Y2 + Z2, G2 = L2 + M2 + N2, it is necessary and sufficient for equilibrium that the six quantities X, Y, Z, L, M, N, should all vanish. In words: the sum of the projections of the forces on each of the co-ordinate axes must vanish; and, the sum of the moments of the forces about each of these axes must vanish.
If any other point O′, whose co-ordinates are x, y, z, be chosen in place of O, as the point to which the forces are transferred, we have to write x1 − x, y1 − y, z1 − z for x1, y1, z1, and so on, in the preceding process. The components of the resultant force R are unaltered, but the new components of couple are found to be
| L′ = L − yZ + zY, M′ = M − zX + xZ, N′ = N − xY + yX. |
(8)
By properly choosing O′ we can make the plane of the couple perpendicular to the resultant force. The conditions for this are L′ : M′ : N′ = X : Y : Z, or