Question 18: Pulley System Equilibrium
The system balances forces on the pulleys. The tension \( T \) is constant throughout the rope.
The central mass \( m \) hangs from a movable pulley supported by two rope segments. $$ 2T = mg \implies T = \frac{mg}{2} $$
The side mass \( M \) hangs from a pulley supported by one vertical rope and one angled rope (angle \( \theta \)).
Upward Force: \( T + T \cos\theta \)
Downward Force: \( Mg \)
$$ T(1 + \cos\theta) = Mg $$
Substitute \( T = mg/2 \): $$ \frac{mg}{2} (1 + \cos\theta) = Mg $$ $$ \frac{m}{M} = \frac{2}{1 + \cos\theta} $$
To prevent the side pulleys from slipping against the wall, the friction torque must counteract the torque from the vertical tension.
Condition derived in note: \( \mu (T \sin\theta) \ge T \)
$$ \mu \sin\theta \ge 1 \implies \mu \ge \frac{1}{\sin\theta} $$
If the middle block is pulled down, \( \theta \) increases, \( \cos\theta \) decreases. The supporting force on the side blocks decreases relative to gravity, or net restoring forces pull the system back.
Equilibrium is Stable.