Consider a block kept on an inclined plane (incline at 45°) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane(μ) is equal to :

Solution

Physics sir · 4 steps

Friction flips direction between the two cases: it fights you going up and helps you holding it.

1. Push up. When the block is just about to move up, friction acts down the incline along with mg sin θ. So the force needed is F₁ = mg(sin θ + μ cos θ).
2. Prevent sliding. When the block is just about to slide down, friction acts up the incline and helps your force. So F₂ = mg(sin θ - μ cos θ).

   sin θ + μ cos θ = 2(sin θ - μ cos θ) ⇒ μ = (tan θ)/3

3. Put in numbers. Setting F₁ = 2F₂ and cancelling mg gives 3μ cos θ = sin θ. With θ = 45°, tan 45° = 1, so μ = 1/3.
4. So. μ = 1/3 ≈ 0.33, which is option (b).

The idea to keep. On a rough incline, the force to just push up is mg(sin θ + μ cos θ) and to just hold is mg(sin θ - μ cos θ). If their ratio is n, then μ = (n-1)/(n+1)tan θ.

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