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() is equal to :
- A.1. 0.50
- B.2. 0.33
- C.3. 0.60
- D.4. 0.25
Show answer and solution
Correct option: B
Solution
For a block on a 45° incline, let N = mg cos45°.
Force to push up: Fup = mg sin45° + μmg cos45° = mg cos45°(tan45° + μ) = mg cos45°(1 + μ)
Force to prevent sliding down: F_down = mg sin45° − μmg cos45° = mg cos45°(1 − μ)
Given Fup = 2 × F_down:
(1 + μ) = 2(1 − μ)
1 + μ = 2 − 2μ
3μ = 1
μ = 1/3 ≈ 0.33
Force to push up: Fup = mg sin45° + μmg cos45° = mg cos45°(tan45° + μ) = mg cos45°(1 + μ)
Force to prevent sliding down: F_down = mg sin45° − μmg cos45° = mg cos45°(1 − μ)
Given Fup = 2 × F_down:
(1 + μ) = 2(1 − μ)
1 + μ = 2 − 2μ
3μ = 1
μ = 1/3 ≈ 0.33
[JEE Main 2023 · 25 Jan · Shift 2]