Page:Mécanique céleste Vol 1.djvu/33

I. i. §1.] forms the angle $$\theta$$ with the force $$x'$$, and the angle $${\pi\over 2} - \theta$$ with the force $$x''$$; we shall therefore have

$$ x' = x. \phi(\theta) = {x^2\over z};$$ $$x'' = x.\phi\Big({\pi\over 2} - \theta\Big) = {xy \over z};$$

and we may substitute these two forces instead of the force $$x$$. We may likewise substitute for the force $$y$$ two new forces, $$y'$$ and $$y''$$, of which the first is equal to $${y^2\over 2}$$ in the direction $$z$$, and the second equal to $${xy\over z}$$ perpendicular to $$z$$; we shall thus have, instead of the two forces $$x$$ and $$y$$, the four following:

$${x^2\over z}, \ {y^2\over z}, \ {xy \over z}, \ {xy \over z};$$

the two last, acting in contrary directions, destroy each other; the two first, acting in the same direction, are to be added together, and produce the resultant $$z$$; we shall therefore have

$$ x^2 + y^2 = z^2;$$

whence it follows, that the resultant of the two forces $$x$$ and $$y$$ is represented in magnitude, by the diagonal of the rectangle whose sides represent those forces.


 * (3) For, by the preceding note, the force $$x = {xy \over z}$$, is in the direction $$AE$$, and the force $$y = {xy \over z}$$, is in the opposite direction $$AF$$, and as they are equal they must destroy each other.

† (4) The sum of the two forces $$x' = {x^2\over z}, \ y' = {y^2\over z}$$, in the direction $$AZ$$, being put equal to the resultant $$z$$, gives $${x^2\over z} + {y^2\over z} = z$$, which multiplied by $$z$$ becomes $$x^2 + y^2 = z^2$$.