A string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by

Step 1: Identify the forces acting on the block.

The block of mass $M$ is in static equilibrium, meaning the tension $T$ in the string balances the weight of the block.

$$T = Mg$$

Step 2: Identify the forces exerted by the string on the pulley.

The string passes over the clamped pulley, pulling it horizontally to the left and vertically downwards with equal tension.

$$T_{\text{horizontal}} = Mg$$ $$T_{\text{vertical}} = Mg$$

Step 3: Account for the weight of the pulley.

The pulley has a mass $m$, which exerts a downward gravitational force.

$$W_{\text{pulley}} = mg$$

Step 4: Determine the net horizontal and vertical forces on the pulley (excluding the clamp).

Sum the forces acting horizontally (to the left) and vertically (downwards) on the pulley.

$$F_{x} = -Mg$$ $$F_{y} = -(M + m)g$$

Step 5: Calculate the components of the force exerted by the clamp.

Since the pulley is in static equilibrium, the clamp must exert an equal and opposite force to balance these components.

$$F_{\text{clamp}, x} = Mg$$ $$F_{\text{clamp}, y} = (M + m)g$$

Step 6: Find the magnitude of the clamp force.

Use the Pythagorean theorem to calculate the magnitude of the resultant force.

$$F_{\text{clamp}} = \sqrt{(F_{\text{clamp}, x})^2 + (F_{\text{clamp}, y})^2}$$ $$F_{\text{clamp}} = \sqrt{(Mg)^2 + ((M + m)g)^2}$$ $$F_{\text{clamp}} = \sqrt{(M + m)^2 + M^2}\,g$$

Answer: $\sqrt{(M+m)^2 + M^2}\,g$