A body of mass $m$ is suspended by two strings making angles $ heta_{1}$ and $ heta_{2}$ with the

Question

A body of mass $m$ is suspended by two strings making angles $	heta_{1}$ and $	heta_{2}$ with the horizontal ceiling with tensions $T_{1}$ and $T_{2}$ simultaneously. $T_{1}$ and $T_{2}$ are related by $T_{1}=\sqrt{3} T_{2}$, the angles $	heta_{1}$ and $	heta_{2}$ are\

Professor PC Thomas Classes · Answer

(c) : Given, $T_{1}=\sqrt{3} T_{2}$\ By Lami's theorem, we have, 
$ \begin{aligned} \frac{T_{1}}{\sin \left(90+	heta_{2}ight)} & =\frac{T_{2}}{\sin \left(90+	heta_{1}ight)}  & =\\frac{m g}{\sin \left(180-\left(	heta_{1}+	heta_{2}ight)ight)} \end{aligned} $
So, $\frac{T_{1}}{\cos 	heta_{2}}=\frac{T_{2}}{\cos 	heta_{1}}=\frac{m g}{\sin \left(	heta_{1}+	heta_{2}ight)}$\ $\frac{\sqrt{3} T_{2}}{\cos 	heta_{2}}=\frac{T_{2}}{\cos 	heta_{1}}$\ (Using eqn. (i))\ $\sqrt{3} \cos 	heta_{1}-\cos 	heta_{2}=0$\ $\sqrt{3} \cos 	heta_{1}=\cos 	heta_{2}$\ It is possible if $	heta_{1}=60^{\circ}$ and $	heta_{2}=30^{\circ}$\ So, $\frac{\sqrt{3} 	imes T_{2}}{\cos 30^{\circ}}=\frac{m g}{\sin 90^{\circ}} ; T_{2}=\frac{m g}{2}$\