Question Number 19822 by virus last updated on 16/Aug/17
Answered by ajfour last updated on 16/Aug/17
![[xsin θ_1 −(l−x)sin θ_2 ]g=la a=[(x/l)sin θ_1 −(((l−x))/l)sin θ_2 ]g .](Q19824.png)
$$\left[\mathrm{xsin}\:\theta_{\mathrm{1}} −\left({l}−\mathrm{x}\right)\mathrm{sin}\:\theta_{\mathrm{2}} \right]\mathrm{g}={l}\mathrm{a} \\ $$$$\:\mathrm{a}=\left[\frac{\mathrm{x}}{{l}}\mathrm{sin}\:\theta_{\mathrm{1}} −\frac{\left({l}−\mathrm{x}\right)}{{l}}\mathrm{sin}\:\theta_{\mathrm{2}} \right]\mathrm{g}\:\:. \\ $$
Commented by virus last updated on 16/Aug/17
$${thanks}\:{sir} \\ $$