Consider a block kept on an inclined plane (inclined 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 :

Question

G1 Physics · Accepted Answer

For impending upward motion, F₁ = mg sin45° + μmg cos45° = (mg/√2)(1 + μ). For impending downward motion, F₂ = mg sin45° - μmg cos45° = (mg/√2)(1 - μ). Given F₁ = 2F₂, so 1 + μ = 2(1 - μ), giving μ = 1/3 ≈ 0.33.