Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass ‘m’ and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is :

Solution

Physics sir · 3 steps

Take torques about B, so the tension at B drops out and only the tension at A is left to find.

1. List the forces. Three forces act downward: the rod's weight mg at its centre (50 cm) and the hanging weight 2mg at 75 cm. Two act upward: tensions T_A at 0 cm and T_B at 100 cm.
2. Choose the pivot. Take torques about B. T_B passes through B, so it has no torque about B. The distances from B are 1 m for T_A, 0.5 m for mg and 0.25 m for 2mg.

   T_A × 1 = mg × 0.5 + 2mg × 0.25 = 0.5 mg + 0.5 mg

3. So. T_A = 1 mg, which is option (d). As a check, T_B = 3mg - mg = 2mg, larger because the extra weight hangs closer to B.

The idea to keep. For a rod held by two strings, take torques about one support. That string's tension drops out, and you get the other tension in one line.

Answer from the NTA final answer key. Solution by PrepWiser.