Question

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $\alpha$ should be

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

Answer 1

The correct option is C ${45}^{\circ }$

Let us first draw the FBD of the string-pulley system given in the question.

Let $T$ be the tension in the strings over the pulleys.

For equilibrium of both blocks of mass $m$ :
$T=mg$ ....(i)

For equilibrium of block of mass $\sqrt{2}m$
$\sqrt{2}mg=2T\cos \alpha$
$\Rightarrow T\cos \alpha =\frac{1}{\sqrt{2}}mg$ ..(ii)
(from the figure, by symmetry, the angle made by the other section of the string with the vertical should also be $\alpha$)

Using (i) and (ii), we get
$\cos \alpha =\frac{1}{\sqrt{2}}$
$\Rightarrow \alpha ={45}^{\circ }$