Track II · Forces

Lesson 3 · 16 min

Friction and ramps

The inclined plane is not a puzzle. It is a fixture, a hopper, a road grade, a wedge.

Resolve the weight

On a ramp of angle θ, weight mg splits into mg sinθ down the plane and mg cosθ into the plane. The normal is usually N = mg cosθ (no other perpendicular forces). Kinetic friction is μk N, opposite velocity. Static friction is ≤ μs N, opposite impending slip.

Sliding down a ramp
a = g (sin θ − μk cos θ)

If sinθ > μs cosθ, static friction cannot hold it. That is the angle of repose: θc = arctan μs.

Worked example

Pallet on a loading dock

  • Given: θ = 12°
  • Given: μs = 0.30
  • Given: μk = 0.22
  • Given: m = 80 kg

Find: Does it sit? If not, a?

  1. Compare tanθ to μs. tan 12° ≈ 0.213, μs = 0.30. 0.213 < 0.30, so it sits.
  2. If someone greases it so μs drops to 0.18: tanθ > μs, it goes.
  3. Then a = g(sin12° − 0.18 cos12°) ≈ 9.81(0.208 − 0.176) ≈ 0.31 m/s²

Sits at μs = 0.30. With μs = 0.18, slides at 0.31 m/s².

Check

The critical angle at which a block starts to slide depends on

Drive the numbers. Then go back to the algebra.

Bench

Incline

θc = arctan μs

N
43.3 N
mg sinθ
23.0 N
fs max
17.3 N
a
2.01 m/s²

Angle of repose 21.8°. The block is sliding.