Page:Somerville Mechanism of the heavens.djvu/87

with it the angles a, b, c, will be given by (1). If one of the component forces as Z be zero, then

c=90°, F=$\scriptstyle {\sqrt {{\text{X}}^{2}+{\text{Y}}^{2}}}$, X=F cos a, Y=F cos b.

38. Velocity and force being each represented by the same space, whatever has been explained with regard to the resolution and composition of the one applies equally to the other.

 

The general Principles of Equilibrium.

 

39. The general principles of equilibrium may be expressed analytically, by supposing o to be the origin of a force F, acting on a particle of matter at m, fig. 11, in the direction om. If o' be the origin of the co-ordinates; a, b, c, the co-ordinates of o, and x, y, z those of m; the diagonal om which may be represented by r, will be

$\scriptstyle r={\sqrt {{({x-a})}^{2}+{({y-b})}^{2}+{({z-c})}^{2}}}$

But F, the whole force in om, is to its component force in

oA::r:a-x,

hence the component force parallel to the axis ox is

F(x-a)/r.

In the same manner it may be shown, that

F(y-b)/r; F(z-c)/r

are the component forces parallel to oy and oz. Now the equation of the diagonal gives

δr/δx=(x-a)/r δr/δy=(y-b)/r; δr/δz=(z-c)/r;

hence the component forces of F are

F$\scriptstyle \left({\frac {\delta {r}}{\delta {x}}}\right);$ F$\scriptstyle \left({\frac {\delta {r}}{\delta {y}}}\right);$ F$\scriptstyle \left({\frac {\delta {r}}{\delta {z}}}\right).$

Again, if F' be another force acting on the particle at m in another direction r', its component forces parallel to the co-ordinates will be,

F'$\scriptstyle \left({\frac {\delta {r'}}{\delta {x}}}\right);$ F'$\scriptstyle \left({\frac {\delta {r'}}{\delta {y}}}\right);$ F'$\scriptstyle \left({\frac {\delta {r'}}{\delta {z}}}\right).$