Page:Somerville Mechanism of the heavens.djvu/88

12 And any number of forces acting on the particle m may be resolved in the same manner, whatever their directions may be. If &Sigma; be employed to denote the sum of any number of finite quantities, represented by the same general symbol

is the sum of the partial forces urging the particle parallel to the axis ox. Likewise $$\scriptstyle \Sigma. \text{F}. \left( \frac{\delta r}{\delta y} \right); \,\Sigma. \text{F} \left( \frac{\delta r}{\delta z} \right);$$ are the sums of the partial forces that urge the particle parallel to the axis oy and oz. Now if F$i$ be the resulting force of all the forces F, F', F'', &amp;c. that act on the particle m, and if u be the straight line drawn from the origin of the resulting force to m, by what precedes

are the expressions of the resulting force F$i$ resolved in directions parallel to the three co-ordinates; hence

or if the sums of the component forces parallel to the axis x, y, z, be represented by X, Y, Z, we shall have

If the first of these be multiplied by &delta;x, the second by &delta;y, and the third by &delta;z, their sum will be

40. If the intensity of the force can be expressed in terms of the distance of its point of application from its origin, X, Y, and Z may be eliminated from this equation, and the resulting force will then be given in functions of the distance only. All the forces in nature are functions of the distance, gravity for example, which varies inversely as the square of the distance of its origin from the point of its application. Were that not the case, the preceding equation could be of no use.

41. When the particle is in equilibrio, the resulting force is zero; consequently

which is the general equation of the equilibrium of a free particle.