Question

A rod of length 6 m has a mass of 12 Kg. It is hinged at one end of the rod at a distance of 3 m below the water surface. What must be the mass of the block that is attached to the other end of the rod so that 5 m of rod's length is under water? ($\rho_{rod}$ = 1/6).

Answer 1

19 Kg

Answer 2

The problem requires balancing torques about the hinge. Let O be the hinge. The rod's weight ($W_{rod} = 12g$) acts at its center (3m from O). The block's weight ($W_b = m_b g$) acts at the end (6m from O). The buoyant force ($F_B$) acts at the center of buoyancy (2.5m from O, midpoint of submerged length).

Given $\rho_{rod} = \frac{1}{6} \rho_{water}$, the buoyant force on the submerged part (5m) is $F_B = \rho_{water} V_{sub} g = \rho_{water} (\frac{5}{6} V_{rod}) g$. Since $W_{rod} = \rho_{rod} V_{rod} g = \frac{1}{6} \rho_{water} V_{rod} g$, we have $\rho_{water} V_{rod} g = 6 W_{rod}$. Thus, $F_B = \frac{5}{6} (6 W_{rod}) = 5 W_{rod} = 5 \times 12g = 60g$.

For rotational equilibrium, the sum of clockwise torques equals the sum of counter-clockwise torques. Assuming the rod is inclined, the lever arms are multiplied by $\cos \alpha$. Torque due to $W_{rod}$: $W_{rod} \times (3 \cos \alpha) = 12g \times 3 \cos \alpha$. Torque due to $W_b$: $W_b \times (6 \cos \alpha) = m_b g \times 6 \cos \alpha$. Torque due to $F_B$: $F_B \times (2.5 \cos \alpha) = 60g \times 2.5 \cos \alpha$.

Equating torques: $12g \times 3 \cos \alpha + m_b g \times 6 \cos \alpha = 60g \times 2.5 \cos \alpha$ Dividing by $g \cos \alpha$: $36 + 6 m_b = 150$ $6 m_b = 114$ $m_b = 19$ Kg.