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On a slope, weight splits into mg sin theta down the incline and mg cos theta into it. Friction is proportional to that second component, so it fades as the slope steepens while the driving component grows. A 50 kg mass on 25 degrees with mu = 0.2 has 207 N pulling it down and 89 N of friction, so 118 N must be held.
Components of weight
along slope = mg sin(theta); into slope = mg cos(theta)
Holding force
F = mg sin(theta) - mu.mg cos(theta)
When tan theta exceeds mu. For mu = 0.2 that is about 11.3 degrees, so 25 degrees is well past it.
Because the surface is pressed together by mg cos theta, which shrinks as the slope steepens. Vertically there is no normal force and no friction at all.