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

F = \frac{g}{4}(m+2M_2)

Answer 2

To determine the value of the applied force $F$ to keep the system in equilibrium, we need to analyze the forces acting on each movable part of the system. We are given that there is no friction, the string is light (massless), pulley $P_1$ is massless, and movable pulley $P_2$ has mass $M_2$. The mass $m$ is attached to pulley $P_3$. We assume pulley $P_3$ is massless, as its mass is not specified.

Let's trace the string and identify the tensions. Due to the string being light and frictionless, the tension is uniform throughout the continuous string. Let $T$ be the tension in the continuous string.

Since $F$ is applied to one end of this continuous string, the applied force $F$ is equal to the tension $T$ in the string. So, $F = T$.

Now, let's draw Free Body Diagrams (FBDs) for the movable parts:

1. FBD for Pulley $P_3$ and Mass $m$:

• Pulley $P_3$ is movable. The mass $m$ is attached to its axle.
• Downward forces:
  • Weight of mass $m$: $mg$.
• Upward forces:
  • Two segments of the string pull pulley $P_3$ upwards. Each segment has tension $T$.
• For equilibrium, the sum of upward forces equals the sum of downward forces: $2T = mg$ $T = \frac{mg}{2}$

2. FBD for Pulley $P_2$:

• Pulley $P_2$ is movable and has mass $M_2$.
• Downward forces:
  • Weight of pulley $P_2$: $M_2g$.
• Upward forces:
  • The string segment where force $F$ is applied pulls pulley $P_2$ upwards. This segment has tension $T$.
• For equilibrium, the sum of upward forces equals the sum of downward forces: $T = T + M_2g$

Revised Interpretation of the Diagram:

There are two independent strings in the system, or the way the string is wrapped around $P_1$ and $P_2$ is different.

• String 1: The string that goes around $P_3$ and supports mass $m$.

  • One end is fixed to the ceiling (left side).
  • It goes down and around pulley $P_3$.
  • The other end is attached to the axle of pulley $P_2$. Let the tension in this string be $T_A$.
  • FBD for $P_3$ and $m$:
    • Downward: $mg$.
    • Upward: Two segments of string $T_A$ pull $P_3$ up.
    • Equilibrium: $2T_A = mg \implies T_A = \frac{mg}{2}$.

• String 2: The string to which force $F$ is applied.

  • One end is attached to the ceiling (right side).
  • It goes down and around pulley $P_1$ (fixed, massless).
  • Then it goes down and around pulley $P_2$ (movable, mass $M_2$).
  • Then it goes up, and force $F$ is applied to this end. Let the tension in this string be $T_B$.
  • Since $F$ is applied to this string, $F = T_B$.

Now, let's consider the FBD for pulley $P_2$.

• Downward forces on $P_2$:
  • Weight of pulley $P_2$: $M_2g$.
  • The tension from string 1, $T_A$, pulling down on the axle of $P_2$ (since string 1 is attached to the axle of $P_2$).
• Upward forces on $P_2$:
  • Two segments of string 2 pull pulley $P_2$ upwards. Each segment has tension $T_B$.
• For equilibrium of $P_2$: $2T_B = M_2g + T_A$ Substitute $T_A = \frac{mg}{2}$: $2T_B = M_2g + \frac{mg}{2}$ $T_B = \frac{M_2g}{2} + \frac{mg}{4}$

Since $F = T_B$, we have: $F = \frac{M_2g}{2} + \frac{mg}{4}$ $F = g \left( \frac{M_2}{2} + \frac{m}{4} \right)$ $F = \frac{g}{4} (2M_2 + m)$

This interpretation resolves the contradiction and is consistent with common pulley system designs where movable pulleys support other parts of the system.

The final answer is $\frac{g}{4}(m+2M_2)$.

Explanation of the solution:

1. Identify that the system involves two separate strings.
2. String 1 connects the ceiling, goes around pulley $P_3$, and its other end is attached to the axle of pulley $P_2$.
3. String 2 connects the ceiling, goes around fixed pulley $P_1$, then around movable pulley $P_2$, and force $F$ is applied to its free end.
4. Apply equilibrium conditions:
   • For pulley $P_3$ and mass $m$: The two segments of string 1 support $mg$. So, $2T_1 = mg \implies T_1 = mg/2$.
   • For pulley $P_2$: Its weight $M_2g$ and the tension $T_1$ from string 1 (pulling down from its axle) are balanced by the two segments of string 2 pulling it up. So, $2T_2 = M_2g + T_1$.
5. Substitute $T_1 = mg/2$ into the equation for $P_2$: $2T_2 = M_2g + mg/2$.
6. The applied force $F$ is equal to the tension $T_2$ in string 2.
7. Solve for $F$: $F = T_2 = \frac{M_2g}{2} + \frac{mg}{4} = \frac{g}{4}(2M_2+m)$.