Friction on an Inclined Plane with an Oblique Push
A 2.0 kg block sits on a rough plane inclined at 30° to the horizontal. The coefficients of friction between block and plane are μs = 0.60 and μk = 0.50. A horizontal push of magnitude F is applied to the block, directed into the incline, so that its line of action makes 30° with the slope: part of the push presses the block onto the surface and part acts up the slope. Take g = 10 m/s² and treat the block as a particle. Two students disagree. Anil argues that any push produces motion, so he simply writes friction = μk N and computes an acceleration. Bina insists the block may not move at all, and wants to compare the net force along the slope (gravity minus the up-slope component of the push) with the limiting static friction computed from the actual normal force. No friction value can be assumed before the normal force is found, because the push itself changes the normal force.
1. With F = 10 N, decide whether the block moves, and give the friction force (magnitude and direction).
Read the worked answer
Resolve perpendicular and parallel to the slope.
- Perpendicular: N = mg cos30° + F sin30° = 2(10)(0.866) + 10(0.5) = 17.32 + 5.00 = 22.32 N
- Limiting static friction: f_max = μsN = 0.60 × 22.32 = 13.39 N
- Along the slope, before friction: gravity down-slope = mg sin30° = 10.0 N; push up-slope = F cos30° = 8.66 N; net = 1.34 N down-slope.
Since 1.34 N < 13.39 N, the block stays at rest and static friction adjusts to the exact demand: 1.34 N directed up the slope, not μsN.
This is the trap: friction is not automatically μsN. The limiting value only fixes the maximum; the actual value is whatever is needed for equilibrium.
2. Find the minimum horizontal push needed to just start the block moving up the slope.
Read the worked answer
At impending motion up the slope, the up-slope drive just equals limiting friction:
F cos30° − mg sin30° = μs(mg cos30° + F sin30°) 0.866F − 10 = 0.60(17.32 + 0.5F) = 10.39 + 0.30F 0.566F = 20.39 → F ≈ 36 N
Check at F = 36 N: N = 17.32 + 18.0 = 35.3 N, f_max = 21.2 N; up-slope drive = 31.2 − 10.0 = 21.2 N - exactly balanced.
Below 36 N the block stays at rest no matter how long the push is applied. The answer is not simply μs mg cos30°, because the push also raises N.
3. Repeat for F = 40 N: does the block move, and if so with what acceleration?
Read the worked answer
First test whether motion starts: N = 17.32 + 40(0.5) = 37.32 N; f_max = 0.60(37.32) = 22.4 N; up-slope drive = 40(0.866) − 10.0 = 24.6 N > 22.4 N, so it does move up the slope.
Once sliding, kinetic friction acts down the slope: f_k = 0.50 × 37.32 = 18.7 N.
Net force = 24.6 − 18.7 = 5.9 N; a = 5.9/2.0 ≈ 3.0 m/s² up the slope.
Common errors: using μs after motion has begun, or forgetting that N includes F sin30°.