Question

In the figure shown, the strings and pulleys are massless. The blocks are in equilibrium. If the force by the clamp on the pulley is $\sqrt{\frac{8}{3}}mg$, then $\frac{M}{m}$ is equal to ______.

Answer 1

3

Answer 2

To find the ratio $\frac{M}{m}$, we analyze the forces acting on the pulleys and blocks.

1. Right Pulley: Let $T_1$ and $T_2$ be the tensions in the strings supporting the two masses $m$. These strings make an angle $\theta$ with the vertical. The vertical components of the tensions balance the weights of the masses:

   $T_1 \cos\theta + T_2 \cos\theta = 2mg$

   The horizontal components of the tensions are balanced by the clamp force $F$:

   $(T_2 - T_1) \sin\theta = F = \sqrt{\frac{8}{3}}mg$

2. Left Pulley: The tension $T_0$ in the vertical string supporting the left pulley is equal to the weight of mass $M$ plus an extra downward force $F$ (from the connecting rope):

   $T_0 = Mg + F$

3. Relating Tensions: The tension $T_0$ is also the resultant force from the two support strings on the right pulley. Eliminating $T_0$ and $F$, we find:

   $Mg = 2mg$

   Therefore,

   $\frac{M}{m} = 3$

Thus, the correct answer is 3.