Question 13: Rotational Equilibrium
1. Torque Condition:
Taking Torques about the left pulley. If the rod has to stay in rotational equilibrium, the weight of the rod must counter the lifting torque of the tension at the far end. $$ mg \frac{L}{2} \ge T L $$ $$ mg \ge 2T $$
Taking Torques about the left pulley. If the rod has to stay in rotational equilibrium, the weight of the rod must counter the lifting torque of the tension at the far end. $$ mg \frac{L}{2} \ge T L $$ $$ mg \ge 2T $$
2. Tension Calculation:
Calculate acceleration of the hanging masses: $$ a = \frac{(m_2 – m_1)g}{(m_2 + m_1)} $$ Calculate Tension \( T \) (using free body diagram of \( m_1 \)): $$ T = m_1 (g + a) $$ $$ T = m_1 g \left[ 1 + \frac{m_2 – m_1}{m_2 + m_1} \right] $$ $$ T = \frac{2 m_1 m_2 g}{m_1 + m_2} $$
Calculate acceleration of the hanging masses: $$ a = \frac{(m_2 – m_1)g}{(m_2 + m_1)} $$ Calculate Tension \( T \) (using free body diagram of \( m_1 \)): $$ T = m_1 (g + a) $$ $$ T = m_1 g \left[ 1 + \frac{m_2 – m_1}{m_2 + m_1} \right] $$ $$ T = \frac{2 m_1 m_2 g}{m_1 + m_2} $$
3. Final Result:
Substitute \( T \) into the equilibrium inequality: $$ m \ge 2 \left( \frac{2 m_1 m_2}{m_1 + m_2} \right) $$ $$ \therefore m \ge \frac{4 m_1 m_2}{m_1 + m_2} $$
Substitute \( T \) into the equilibrium inequality: $$ m \ge 2 \left( \frac{2 m_1 m_2}{m_1 + m_2} \right) $$ $$ \therefore m \ge \frac{4 m_1 m_2}{m_1 + m_2} $$