See Our New JEE Book on Amazon

**Question:**
A person tries to lock the door by putting a brick in Case-1 and a small wooden piece in Case-2. Which of the following forces gives a torque about the rotation axis of the door that opposes the torque of the force by the person trying to open the door in Case-1?

- the normal force by the brick on the door
- the frictional force by the brick on the door
- the weight of the door
- the weight of the brick

- the normal force by the wooden piece on the door
- the frictional force by the wooden piece on the door
- the weight of the door
- the weight of the wooden piece

**Solution 1:** The torque of the force by the person about the rotation axis of the door is vertically downward. The normal force by the brick on the door is opposite to the force by the person. The torque of this normal force about the rotation axis is vertically upward. This torque opposes the torque of the force by the person. Thus, option (A) is correct.

The frictional force by the brick on the door almost passes through the axis of rotation. The torque due to frictional force about the axis of rotation is zero.

The torque due to the weight of the door is perpendicular to the door. The torque due to the weight of the brick is also perpendicular to the door. This torque acts on brick.

**Solution 2:**
The normal force by the wooden piece on the door is vertically upward. The torque of normal force about the axis of rotation is perpendicular to the door.

The frictional force by the wooden piece on the door is opposite to the direction of its motion. This force is opposite to the force applied by the person. The torque of frictional force about the axis of rotation is vertically upward. Thus, option (B) is correct.

The torque due to the weight of the door is perpendicular to the door. The torque due to the weight of the wooden piece is also perpendicular to the door.

See Our Books