Question

In the pulley arrangement shown in figure, the pulley $P_2$ is movable. Assuming the coefficient of friction between $m$ and surface to be $\mu$, the maximum value of $M$ for which $m$ is at rest is :

• A. $M = \frac{\mu m}{2}$
• B. $m = \frac{\mu M}{2}$
• C. $M = \frac{m}{2\mu}$

Answer 1

M = 2μm

Answer 2

To determine the maximum value of $M$ for which mass $m$ remains at rest, we need to analyze the forces acting on both masses and the pulley system.

1. Free Body Diagram for mass $m$:

Mass $m$ is placed on a horizontal surface.

• Vertical forces:

  • Weight of $m$: $mg$ (downwards)
  • Normal force: $N$ (upwards)

  Since there is no vertical acceleration, $N = mg$.

• Horizontal forces:

  • Tension in the string: $T$ (to the right)
  • Frictional force: $f_s$ (to the left, opposing the tendency of motion)

For mass $m$ to be at rest, the tension must be balanced by the static friction. For the maximum value of $M$, mass $m$ will be on the verge of slipping. In this case, the static friction reaches its maximum value, $f_{s,max}$.

$f_{s,max} = \mu N = \mu mg$

Therefore, for equilibrium on the verge of motion:

$T = f_{s,max}$

$T = \mu mg$ (Equation 1)

2. Free Body Diagram for pulley $P_2$ and mass $M$:

Pulley $P_2$ is a movable pulley, and mass $M$ is attached to it. The string from pulley $P_1$ goes around pulley $P_2$ and is then fixed to the vertical support. The tension in this continuous string is $T$.

• Forces on the pulley $P_2$ and mass $M$ system:
  • Upward forces: Two segments of the string pull the pulley upwards, each with tension $T$. So, the total upward force is $T + T = 2T$.
  • Downward force: Weight of mass $M$: $Mg$. (Assuming the pulley $P_2$ is massless).

For the system to be at rest (in equilibrium):

Sum of upward forces = Sum of downward forces

$2T = Mg$ (Equation 2)

3. Combining the equations:

Substitute the expression for $T$ from Equation 1 into Equation 2:

$2(\mu mg) = Mg$

$2\mu mg = Mg$

Since $g$ is the acceleration due to gravity and is non-zero, we can cancel it from both sides:

$2\mu m = M$

So, the maximum value of $M$ for which $m$ is at rest is $M = 2\mu m$.