Question

The pulleys and strings shown in the figure are smooth and massless. For the system to remain in equilibrium, the angle $\theta$ should be:

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${30}^{\circ }$

The tension ($T$) in both strings i.e connecting block $B$, $A$ and $B$,$C$ will be equal due to symmetry of arrangement.


For the system to be in equilibrium, all the three masses $A,B,C$ should be in equilibrium.

$\Rightarrow$ From equilibrium of mass $A$ and $C$:
$T=mg.......\left(i\right)$

Resolving the tension $T$ acting on $B$ in horizontal and vertical direction. The net force in horizontal direction is balanced on mass $B$.

$\therefore$ Applying equilibrium condition for mass $B$ in vertical direction :

$\sqrt{3}mg=2T\cos \theta ...........\left(ii\right)$

From Eq. $\left(i\right)$ and $\left(ii\right)$:
$\sqrt{3}mg=2\left(mg\right)\cos \theta$
$\Rightarrow \frac{\sqrt{3}}{2}=\cos \theta$
$\therefore \theta ={30}^{\circ }$