Physics - NLM Question with Solution | TestHub

Let the tension in the cable at angle $\theta$ be $T_2$ and the one at $60^{\circ }$ be $T_1$. Horizontal: $T_1\cos 60^{\circ }=T_2\cos \theta$. Vertical: $T_2\sin \theta +T_1\sin 60^{\circ }=W$.

$T_2\sin \theta +T_2\cos \theta \tan 60^{\circ }=W\Rightarrow T_2(\sin \theta +\sqrt{3}\cos \theta )=W\Rightarrow 2T_2\cos (\theta -30^{\circ })=W$.

$T_2$ is minimum when $\cos (\theta -30^{\circ })$ is maximum, i.e. $\theta -30^{\circ }=0\Rightarrow \theta =30^{\circ }$.