A square Lamina OABC of length 10 cm is pivoted at ' O′. Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of F is :

Solution

Physics sir · 3 steps

Take torques about the pivot O: its own force drops out, and several forces pass right through O.

1. Why torque about O. The pivot pushes on the lamina with an unknown force, but that force acts at O and gives zero torque about O. For the lamina to stay at rest, the net torque about O must be zero. Take anticlockwise as positive and the side as 0.1 m.

   Force	Lever arm about O	Torque about O
   10 N up at C	0 (line passes through O)	0
   F to the left at C	0.1 m	+0.1F (anticlockwise)
   10 N up at B	0.1 m	+1 N m
   10 N right at B	0.1 m	-1 N m
   10 N right at A	0 (line passes through O)	0
   10 N down at A	0.1 m	-1 N m (clockwise)

2. Add the torques. The two forces at B cancel each other's torque. Only F at C and the downward 10 N at A are left.

   0.1F - 10 × 0.1 = 0

3. So. F = 10 N, which is option (a).

The idea to keep. With a pivot, take torques about the pivot itself. Any force whose line of action passes through it contributes nothing, so cross those out first and balance what remains.

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