A 2 kg block is pushed against a vertical wall by applying a horizontal force of 50 N. The coefficient of static friction between the block and the wall is 0.5. A force F is also applied on the block vertically upward (as shown in figure). The maximum value of F applied, so that the block does not move upward, will be :
(Given : g = 10 ms⁻²)

Solution

Physics sir · 3 steps

When the block is about to slide up, friction turns around and pulls down along with gravity.

1. Normal reaction. Horizontally the block does not move, so the wall pushes back with N = 50 N. The largest friction the wall can supply is fₘₐₓ = μₛ N = 0.5 × 50 = 25 N.
2. Direction of friction. Friction always opposes the motion the block is about to make. If F is large enough to push the block upward, friction acts downward, in the same direction as the weight mg = 2 × 10 = 20 N.

   Fₘₐₓ = mg + μₛ N = 20 + 25

3. So. Fₘₐₓ = 45 N. Any larger and the block starts moving up, so the answer is (d).

The idea to keep. Friction has no fixed direction. First decide which way the body is about to move, then put friction against that, at its full value μₛ N for the limiting case.

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