Question

In the system shown in the fig. there is no friction and string is light. Mass of movable pulley $P_2$ is $M_2$. If pulley $P_1$ is massless, what should be value of applied force F to keep the system in equilibrium?

Answer 1

$(M_2 - m)g$

Answer 2

Explanation of the solution:

1. Identify the movable pulleys: P2 (mass $M_2$) and P3 (massless, as not specified).
2. Identify the mass 'm' attached to P3.
3. Identify the applied force F.
4. Assume a continuous string (tension $T'$) starts from the ceiling, passes under P2, then over P1, then under P3, and is finally attached to the ceiling.
5. Draw FBD for (P3 + m): The downward force is $mg$. The upward force is $2T'$ (two string segments). For equilibrium, $2T' = mg \implies T' = mg/2$.
6. Draw FBD for P2: The downward force is $M_2g$. The upward forces are $2T'$ (two string segments) and the applied force F (acting directly on P2's axle, transmitted via P1). For equilibrium, $2T' + F = M_2g$.
7. Substitute $T'$ from step 5 into the equation from step 6: $2(mg/2) + F = M_2g \implies mg + F = M_2g$.
8. Solve for F: $F = (M_2 - m)g$.