Solution
Step 1: Analyzing the final state (After ice melts)
Initially, we have block B and ice C. In the second state, the ice C has melted. When the ice melts, it becomes water and mixes with the tank water. Block B remains, and we are told it is now completely submerged in water and the system is still in equilibrium.
Let $m_A$, $m_B$, $m_C$ be the masses of load A, load B, and ice C respectively. Let $\rho_w$, $\rho_s$, $\rho_i$ be the densities of water, steel, and ice.
In the final state (Case 2):
- Block A is in air: Tension $T = m_A g$.
- Block B is fully immersed in water.
The equation of equilibrium for Block B is:
$$T + F_{B,buoyancy} = m_B g$$ $$m_A g + V_B \rho_w g = m_B g$$Since $V_B = \frac{m_B}{\rho_s}$, we can rewrite this as:
$$m_A = m_B – \frac{m_B}{\rho_s}\rho_w$$ $$m_A = m_B \left( 1 – \frac{\rho_w}{\rho_s} \right)$$Substituting the given values: $m_A = 0.68 \text{ kg}$, $\rho_w = 1000 \text{ kg/m}^3$, $\rho_s = 7800 \text{ kg/m}^3$.
$$0.68 = m_B \left( 1 – \frac{1000}{7800} \right) = m_B \left( \frac{6800}{7800} \right)$$ $$m_B = 0.68 \times \frac{78}{68} = 0.78 \text{ kg}$$Step 2: Analyzing the initial state
In the initial state (Case 1):
- Block B is half submerged ($V_{sub, B} = V_B/2$).
- Ice C is affixed to the bottom of B. Since the bottom half of B is submerged, C is also fully submerged.
The equilibrium equation is:
$$T + F_{buoyancy} = (m_B + m_C)g$$ $$m_A g + (\text{Buoyancy on B} + \text{Buoyancy on C}) = (m_B + m_C)g$$ $$m_A + \rho_w \frac{V_B}{2} + \rho_w V_C = m_B + m_C$$Substitute volumes $V_B = \frac{m_B}{\rho_s}$ and $V_C = \frac{m_C}{\rho_i}$:
$$m_A + \rho_w \left( \frac{m_B}{2\rho_s} \right) + \rho_w \left( \frac{m_C}{\rho_i} \right) = m_B + m_C$$Rearranging to solve for $m_C$:
$$m_C \left( 1 – \frac{\rho_w}{\rho_i} \right) = m_A – m_B + \frac{\rho_w m_B}{2\rho_s}$$Substitute values ($m_A=0.68, m_B=0.78, \rho_w=1000, \rho_i=900, \rho_s=7800$):
$$m_C \left( 1 – \frac{1000}{900} \right) = 0.68 – 0.78 + \frac{1000 \times 0.78}{2 \times 7800}$$ $$m_C \left( \frac{-100}{900} \right) = -0.10 + \frac{1000}{20000} \times 1$$ $$-\frac{m_C}{9} = -0.10 + 0.05 = -0.05$$ $$m_C = 0.05 \times 9 = 0.45 \text{ kg}$$