A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is __________ N. (Round off to the Nearest Integer) [Take g = 10 ms⁻² ]

Solution

Physics sir · 5 steps

Take the boy and the wood as one system: the same tension lifts them and also drags them sideways.

1. One rope, one tension. With ideal pulleys, the tension is the same all along the rope. So if the boy pulls with force F, the tension everywhere is T = F.
2. Forces on the system. Treat the boy and the wood together, with total weight (4 + 5) × 10 = 90 N. The rope at the boy's hands pulls the system up with T, taken as vertical as the problem intends. The rope tied to the wood pulls it sideways with T.
3. Normal reaction. The upward pull lightens the load on the floor, so R = 90 - T. A smaller R means the floor can offer less friction.

   T ≤ μ R = 0.5 (90 - T)

4. Solve. At the limit, T = 45 - 0.5T, so 1.5T = 45 and T = 30 N.
5. So. The boy can pull with at most 30 N before the wood starts to slide.

The common slip: 45. Taking the normal reaction as the full weight, 90 N, and ignoring the upward pull of the rope gives 0.5 × 90 = 45 N, not 30.

The idea to keep. Whenever a rope pulls a body partly upward, it reduces the normal reaction. Write R first, then set the sideways pull equal to μ R for the limiting case.

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