Page:Mécanique céleste Vol 1.djvu/33

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I. i. §1.] COMPOSITION OF FORCES. 3forms the angle θ 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};\qquad \qquad [1]

and we may substitute these two forces instead of the force x. We may like-

wise 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};\qquad \qquad [2]

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};\qquad \qquad [3]

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.

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.

* (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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