Page:Lagrange - Œuvres (1867) vol. 1.djvu/474

on ajoutera donc cette quantité au premier membre de l’équation générale de l’Article précédent, et l’on aura celle-ci :

(P)

$\left\{{\begin{aligned}&-\int \left[\,\left(\mathrm {M} 'd{\frac {dx'}{dt}}+\mathrm {M} '\Pi 'dt-kdt\right)\delta x'\right.\\&\qquad +\left(\mathrm {M} 'd{\frac {dy'}{dt}}+\mathrm {M} '\varpi 'dt-{\frac {\operatorname {d} y'}{\operatorname {d} x'}}kdt\right)\delta y'\\&\qquad +\left(\mathrm {M} 'd{\frac {dz'}{dt}}+\mathrm {M} '\Psi 'dt-{\frac {\operatorname {d} z'}{\operatorname {d} x'}}kdt\right)\delta z'\\&\left.\qquad \qquad +\left(\operatorname {d} {\grave {\,}}\!x\delta {\grave {\,}}\!x+\operatorname {d} {\grave {\,}}\!y\delta {\grave {\,}}\!y+\operatorname {d} {\grave {\,}}\!z\delta {\grave {\,}}\!z\right){\frac {\mathrm {T} +k}{\operatorname {d} {\grave {\,}}\!x}}dt\right]\\&-\int \mathrm {S} \left[\left(\operatorname {d} {\frac {\mathrm {U} dt\operatorname {d} y}{\operatorname {d} x}}+\operatorname {d} {\frac {\operatorname {d} y}{\operatorname {d} x}}kdt+dm\,d{\frac {dy}{dt}}+dm\varpi dt\right)\delta y\right.\\&\qquad \,\ \ \left.-\left(\operatorname {d} {\frac {\mathrm {U} dt\operatorname {d} z}{\operatorname {d} x}}+\operatorname {d} {\frac {\operatorname {d} z}{\operatorname {d} x}}kdt+dm\,d{\frac {dz}{dt}}+dm\Psi dt\right)\delta z\right]=0.\end{aligned}}\right.$

Les termes affectés du double signe $\int \mathrm {S}$ donneront pour le mouvement du fil en général les mêmes équations de l’Article XXX qu’il est inutile de répéter. Les autres termes fourniront l’équation

${\begin{aligned}&\left(\mathrm {M} 'd{\frac {dx'}{dt}}+\mathrm {M} '\Pi 'dt-kdt\right)\delta x'\\+&\left(\mathrm {M} 'd{\frac {dy'}{dt}}+\mathrm {M} '\varpi 'dt-{\frac {\operatorname {d} y'}{\operatorname {d} x'}}kdt\right)\delta y'\\+&\left(\mathrm {M} 'd{\frac {dz'}{dt}}+\mathrm {M} '\Psi 'dt-{\frac {\operatorname {d} z'}{\operatorname {d} x'}}kdt\right)\delta z'\\+&\left(\operatorname {d} {\grave {\,}}\!x\delta {\grave {\,}}\!x+\operatorname {d} {\grave {\,}}\!y\delta {\grave {\,}}\!y+\operatorname {d} {\grave {\,}}\!z\delta {\grave {\,}}\!z\right){\frac {\mathrm {T} +k}{\operatorname {d} {\grave {\,}}\!x}}dt=0\end{aligned}}$

Or, si le corps $\mathrm {M} '$ est libre, en sorte que les différentiations $\delta x',\delta y',\delta z'$ demeurent indéterminées, on fera

${\begin{aligned}&\mathrm {M} '\left(d{\frac {dx'}{dt}}+\Pi 'dt\right)-kdt=0,\\&\mathrm {M} '\left(d{\frac {dy'}{dt}}+\varpi 'dt\right)-{\frac {\operatorname {d} y'}{\operatorname {d} x'}}kdt=0,\\&\mathrm {M} '\left(d{\frac {dz'}{dt}}+\Psi 'dt\right)-{\frac {\operatorname {d} z'}{\operatorname {d} x'}}kdt=0.\end{aligned}}$