Problem 16: Composite Rod Equilibrium
Setup: A rod of length $L=100\,\text{cm}$ is composed of two halves:
Lower half (Heavy): Density $\rho_2 = 2.50\,\text{g/cm}^3$.
Upper half (Light): Density $\rho_1 = 0.50\,\text{g/cm}^3$.
Water density $\rho = 1.00\,\text{g/cm}^3$.
Let’s check the condition for torque balance about the bottom contact point when the rod is fully submerged.
Torque due to Gravity (Clockwise):
$$ \tau_g = (m_2 g \frac{L}{4} + m_1 g \frac{3L}{4}) \cos \theta $$
Masses are proportional to density. Let Volume of half rod be $V_{half}$.
$$ \tau_g \propto V_{half} g [ \rho_2 \frac{L}{4} + \rho_1 \frac{3L}{4} ] \cos \theta $$
Torque due to Buoyancy (Counter-Clockwise):
Buoyancy acts at the geometric center ($L/2$).
$$ \tau_B = (V_{total} \rho g) \frac{L}{2} \cos \theta = (2 V_{half} \rho g) \frac{L}{2} \cos \theta $$
Equating torques:
$$ \rho_2 \frac{1}{4} + \rho_1 \frac{3}{4} = \rho \cdot 1 $$
$$ \frac{2.5}{4} + \frac{1.5}{4} = 1 $$
$$ \frac{4.0}{4} = 1 $$
The equation holds true! This means that if the rod is fully submerged, the net torque about the pivot is zero regardless of the angle. It is in neutral equilibrium. For the rod to exist at an equilibrium of $\theta = 53^\circ$, it must be fully submerged. If it were only partially submerged, the buoyant torque would be less, and the rod would sink to the bottom ($\theta=0$).
Thus, the water depth $h$ must correspond to the height of the rod tip when fully submerged at $53^\circ$.
Answer: The depth of the water is 80 cm.