Page:Mécanique céleste Vol 1.djvu/33

forms the angle $\theta$ with the force $x'$, and the angle ${\pi \over 2}-\theta$ with the force $x''$; we shall therefore have

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

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:

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

the two last, acting in contrary directions, destroy each other;^[1] the two first, acting in the same direction, are to be added together, and produce the resultant $z$; we shall therefore have^[2]

[3] $x^{2}+y^{2}=z^{2};$

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 $x^{2}+y^{2}=z^{2}$.

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directions $AZ,AE,$ respectively, so that the angle $ZAX=\theta$, and $XAE={\pi \over 2}-\theta$. Then, in the same manner in which the above values of $x$, $y$, are obtained from $z$, we may get $x'=x.\phi (\theta )$; $x''=x.\phi {\Big (}{\pi \over 2}-\theta {\Big )}$. If in these we substitute the values $\phi (\theta )={x \over z};\ \phi {\Big (}{\phi \over 2}-\theta {\Big )}={y \over z}$, deduced from the above equations, we obtain $x'={x^{2} \over z};x''={xy \over z}$. In like manner, if the force $y$, in the direction $AY$, be resolved into the two forces $y'$, $y''$, in the directions $AZ$, $AF$, making the angle $YAZ={\phi \over 2}-\theta ,\ YAF=\theta$, we shall have $y'=y.\phi {\Big (}{\pi \over 2}-\theta {\Big )};\ y''=y.\phi (\theta );$ which, by substituting the above values of $\phi {\Big (}{\pi \over 2}-\theta {\Big )},\ \phi (\theta )$, become $y'={y^{2} \over z},\ y''={xy \over z}$, as above.

1. *
2. †