RBD BYU 9 - CHATUP707

Solution Q9

Question 9: Balancing a Rod

Step 1: Calculate Mass of the Rod (Using Case 2)

Load A is removed. The rod balances at pivot distance $ l/3 $ from the center.
Torque Balance about Pivot:

$$ M \cdot g \cdot \left(\frac{l}{3}\right) = m_B \cdot g \cdot \left(\frac{l}{2} – \frac{l}{3}\right) $$ $$ M \left(\frac{l}{3}\right) = 30 \left(\frac{l}{6}\right) $$ $$ \frac{M}{3} = 5 \implies M = 15 \text{ kg} $$

Step 2: Calculate Mass of Load A (Using Case 1)

Load A is added. The pivot moves to distance $ l/4 $ from the center.
Torque Balance about Pivot (Note: Load A is at one end, so distance from center is $ l/2 $):

$$ \text{Torque}_{CCW} = \text{Torque}_{CW} $$ $$ m_A \cdot \left(\frac{l}{2} + \frac{l}{4}\right) + M \cdot \left(\frac{l}{4}\right) = m_B \cdot \left(\frac{l}{2} – \frac{l}{4}\right) $$ $$ m_A \left(\frac{3l}{4}\right) + M \left(\frac{l}{4}\right) = m_B \left(\frac{l}{4}\right) $$

Canceling common terms ($ l/4 $):

$$ 3m_A + M = m_B $$

Step 3: Final Answer

Substitute $ M = 15 $ and $ m_B = 30 $:

$$ 3m_A + 15 = 30 $$ $$ 3m_A = 15 $$ $$ m_A = 5 \text{ kg} $$

Answer: 5 kg (Option b)