(10)
If the forces P1, P2, ... be turned in the same sense through the same angle about the respective points A1, A2, ... so as to remain parallel, the value of θ is alone altered, and the resultant Σ(P) passes always through the point
| x = | Σ(xP) | , y = | Σ(yP) | , |
| Σ(P) | Σ(P) |
(11)
which is determined solely by the configuration of the points A1, A2, ... and by the ratios P1 : P2 : ... of the forces acting at them respectively. This point is called the centre of the given system of parallel forces; it is finite and determinate unless Σ(P) = 0. A geometrical proof of this theorem, which is not restricted to a two-dimensional system, is given later (§ 11). It contains the theory of the centre of gravity as ordinarily understood. For if we have an assemblage of particles whose mutual distances are small compared with the dimensions of the earth, the forces of gravity on them constitute a system of sensibly parallel forces, sensibly proportional to the respective masses. If now the assemblage be brought into any other position relative to the earth, without alteration of the mutual distances, this is equivalent to a rotation of the directions of the forces relatively to the assemblage, the ratios of the forces remaining unaltered. Hence there is a certain point, fixed relatively to the assemblage, through which the resultant of gravitational action always passes; this resultant is moreover equal to the sum of the forces on the several particles.
| Fig. 23. |
The theorem that any coplanar system of forces can be reduced to a force acting through any assigned point, together with a couple, has an important illustration in the theory of the distribution of shearing stress and bending moment in a horizontal beam, or other structure, subject to vertical extraneous forces. If we consider any vertical section P, the forces exerted across the section by the portion of the structure on one side on the portion on the other may be reduced to a vertical force F at P and a couple M. The force measures the shearing stress, and the couple the bending moment at P; we will reckon these quantities positive when the senses are as indicated in the figure.
If the remaining forces acting on the portion of the structure on either side of P are known, then resolving vertically we find F, and taking moments about P we find M. Again if PQ be any segment of the beam which is free from load, Q lying to the right of P, we find
FP = FQ, MP − MQ = −F·PQ;
(12)