Question

A rod of length L & uniform cross-section area A having volume mass density $\rho'$ = 6$\rho$x/L is immersed in a liquid of density $\rho$ by means of two massless strings 1 and 2 as shown. String 1 is connected to the rod at x = 0. Find the tensions in string 1.

Answer 1

$\frac{1}{2}\rho A L g$

Answer 2

1. Weight and Center of Mass: The mass of an infinitesimal element of length $dx$ is $dm = \rho'(x) A dx = \frac{6\rho A x}{L} dx$. The total weight of the rod is $W = \int_0^L dm \cdot g = \int_0^L \frac{6\rho A g}{L} x dx = 3\rho A L g$. The center of mass $x_{cm}$ is $x_{cm} = \frac{\int_0^L x dm}{\int_0^L dm} = \frac{2L}{3}$.

2. Buoyant Force: The volume of the rod is $V = AL$. The buoyant force is $F_B = \rho V g = \rho A L g$. It acts at the geometric center ($x=L/2$).

3. Force Equilibrium: $T_1 + T_2 + F_B = W$ $T_1 + T_2 + \rho A L g = 3\rho A L g \implies T_1 + T_2 = 2\rho A L g$

4. Torque Equilibrium (about x=0): $T_2 \cdot L - W \cdot x_{cm} + F_B \cdot (L/2) = 0$ $T_2 \cdot L - (3\rho A L g) \cdot (2L/3) + (\rho A L g) \cdot (L/2) = 0$ $T_2 \cdot L - 2\rho A L^2 g + \frac{1}{2}\rho A L^2 g = 0$ $T_2 \cdot L = \frac{3}{2}\rho A L^2 g \implies T_2 = \frac{3}{2}\rho A L g$

5. Solve for $T_1$: $T_1 + \frac{3}{2}\rho A L g = 2\rho A L g \implies T_1 = \frac{1}{2}\rho A L g$.