An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is (Take g = 10 m/s²)

Solution

Physics sir · 4 steps

Take torques about the pivot C: the wall's force drops out and only the cable's tension is unknown.

1. Choose the pivot. The wall pushes on the rod at C with an unknown force. Its torque about C is zero, so balancing torques about C leaves only the tension T to find.
2. Clockwise torques. The 8 kg block hangs at D, 1 m from C: 80 × 1 = 80 N m. The rod's own weight acts at its centre, 0.5 m from C: 20 × 0.5 = 10 N m. Together that is 90 N m.
3. Anticlockwise torque. The cable pulls at B, 0.6 m from C, at 30° to the rod. Only the perpendicular part, T sin 30°, turns the rod, so its torque is T sin 30° × 0.6 = 0.3 T.

   T sin 30° × 0.6 = 80 × 1 + 20 × 0.5 = 90 N m

4. So. 0.3 T = 90, which gives T = 300 N, option (d).

The idea to keep. For a hinged rod in equilibrium, take torques about the hinge so its unknown force disappears. Use only the force component perpendicular to the rod, and include the rod's own weight at its centre.

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