As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30°, with horizontal. For μₛ=0.25, the block will just start to move for the value of F : [Given g=10 ms⁻²]

Solution

Physics sir · 4 steps

The upward pull lifts part of the weight off the floor, so the normal force and the friction both drop.

1. Normal force. The vertical part of the pull, F sin 30° = F/2, acts upward and helps hold up the block. So the floor pushes up with less than mg: N = mg - F sin 30° = 100 - F/2 newtons.
2. Condition to move. The block just starts to move when the horizontal part of the pull equals the maximum static friction μₛ N.

   F cos 30° = μₛ(mg - F sin 30°)

3. Put in numbers. 0.866 F = 0.25(100 - 0.5 F) = 25 - 0.125 F. So 0.991 F = 25, which gives F = 25.2 N.
4. So. The block just starts to move at F ≈ 25.2 N, which is option (a).

The idea to keep. Whenever a force acts at an angle, find the normal force first. A pull with an upward part reduces N, and a push with a downward part increases it. Friction is μ N, never just μ mg.

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