Question

A body of mass $m$ is suspended by two strings making angles $\theta_1$ and $\theta_2$ 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 $\theta_1$ and $\theta_2$ are

• A. $\theta_1 = 30^\circ, \theta_2 = 60^\circ$
• B. $\theta_1 = 60^\circ, \theta_2 = 30^\circ$
• C. $\theta_1 = 45^\circ, \theta_2 = 45^\circ$
• D. $\theta_1 = 30^\circ, \theta_2 = 30^\circ$

Answer 1

Answer: (B)

$60^\circ$ and $30^\circ$

Answer 2

For static equilibrium, the horizontal components of tensions must balance: $T_1 \cos \theta_1 = T_2 \cos \theta_2$. The vertical components must balance the weight: $T_1 \sin \theta_1 + T_2 \sin \theta_2 = mg$. Given $T_1 = \sqrt{3}T_2$, substituting this into the horizontal equilibrium equation gives $\sqrt{3}T_2 \cos \theta_1 = T_2 \cos \theta_2$, which simplifies to $\sqrt{3} \cos \theta_1 = \cos \theta_2$. If we test $\theta_1 = 60^\circ$, then $\cos \theta_1 = 1/2$, so $\cos \theta_2 = \sqrt{3}/2$, implying $\theta_2 = 30^\circ$. This pair of angles satisfies the conditions.