Solution: Rotational Equilibrium in Fluids
Let the distance of the hinge $O$ from the end $A$ be $x$. Consequently, the distance of the hinge from end $B$ is $l – x$. For the rod to remain horizontal just as the bodies leave the bottom, the torques produced by the tension in the strings about the hinge must balance.
Figure 1: Free body diagrams of the sphere (left) and hemisphere (right). Note the fluid forces.
1. Analysis of the Sphere (Left Side)
The sphere has radius $r$ and the water level is also $r$. This means the sphere is submerged exactly up to its center (half submerged).
- Weight ($W_s$): Acting downwards. $W_s = V \rho g = \frac{4}{3}\pi r^3 \rho g$.
- Buoyant Force ($F_{B1}$): Since half the volume is submerged, the buoyant force is equal to the weight of the displaced water. $$ F_{B1} = \frac{1}{2} \left( \frac{4}{3}\pi r^3 \right) \rho_0 g = \frac{2}{3}\pi r^3 \rho_0 g $$
For the sphere to just leave the bottom, the tension $T_1$ must support the net weight: $$ T_1 = W_s – F_{B1} $$ $$ T_1 = \frac{4}{3}\pi r^3 \rho g – \frac{2}{3}\pi r^3 \rho_0 g $$ $$ T_1 = \frac{2}{3}\pi r^3 g (2\rho – \rho_0) \quad \dots \text{(i)} $$
2. Analysis of the Hemisphere (Right Side)
The hemisphere is fully submerged, but the problem states there is no water under the hemisphere. This changes the fluid force calculation significantly. The fluid exerts pressure only on the curved surface, pushing it downwards. There is no upward buoyant force acting on the flat bottom face.
We calculate the downward force due to water ($F_P$) using the concept of missing upthrust:
Here, $\text{Force}_{up} = (\text{Pressure at bottom}) \times (\text{Base Area}) = (\rho_0 g r)(\pi r^2) = \pi r^3 \rho_0 g$.
The standard Archimedes’ upthrust for a hemisphere is $\frac{2}{3}\pi r^3 \rho_0 g$.
Therefore, $\frac{2}{3}\pi r^3 \rho_0 g = \pi r^3 \rho_0 g – F_{down}$.
$$ \implies F_{down} = \frac{1}{3}\pi r^3 \rho_0 g $$
So, the forces acting on the hemisphere are:
- Weight ($W_h$): $W_h = \frac{2}{3}\pi r^3 \rho g$ (acting down).
- Fluid Pressure Force ($F_P$): $F_P = \frac{1}{3}\pi r^3 \rho_0 g$ (acting down).
For the hemisphere to lift off, Tension $T_2$ must overcome both: $$ T_2 = W_h + F_P $$ $$ T_2 = \frac{2}{3}\pi r^3 \rho g + \frac{1}{3}\pi r^3 \rho_0 g $$ $$ T_2 = \frac{1}{3}\pi r^3 g (2\rho + \rho_0) \quad \dots \text{(ii)} $$
3. Torque Equilibrium
Taking moments about the hinge $O$: $$ T_1 \cdot x = T_2 \cdot (l – x) $$
Substituting values from (i) and (ii): $$ \left[ \frac{2}{3}\pi r^3 g (2\rho – \rho_0) \right] x = \left[ \frac{1}{3}\pi r^3 g (2\rho + \rho_0) \right] (l – x) $$
Canceling common terms $\frac{1}{3}\pi r^3 g$: $$ 2(2\rho – \rho_0) x = (2\rho + \rho_0) (l – x) $$ $$ (4\rho – 2\rho_0) x = l(2\rho + \rho_0) – x(2\rho + \rho_0) $$ $$ x(4\rho – 2\rho_0 + 2\rho + \rho_0) = l(2\rho + \rho_0) $$ $$ x(6\rho – \rho_0) = l(2\rho + \rho_0) $$
Solving for $x$: $$ x = l \left( \frac{2\rho + \rho_0}{6\rho – \rho_0} \right) $$
4. Numerical Substitution
Given: $l = 116 \text{ cm}$, $\rho = 5.0 \text{ g/cm}^3$, $\rho_0 = 1.0 \text{ g/cm}^3$.
$$ x = 116 \left( \frac{2(5) + 1}{6(5) – 1} \right) $$ $$ x = 116 \left( \frac{10 + 1}{30 – 1} \right) $$ $$ x = 116 \left( \frac{11}{29} \right) $$
Since $29 \times 4 = 116$: $$ x = 4 \times 11 = 44 \text{ cm} $$