All subjects · Mathematics · WME01 (M1)

Motion on a rough inclined plane

On a rough slope, friction acts along the plane, opposing the motion. The direction of friction depends on which way the particle is moving, so a particle sliding up and then back down has two different accelerations.

Two directions

In every case R = mg cos α if no other force has a perpendicular part, so F = μmg cos α.

Does it slide back?

At the top the particle stops. It slides back only if mg sin α > μ mg cos α, that is, if tan α > μ. Otherwise friction holds it at rest.

Constant velocity

If a particle slides down at constant velocity, the forces balance: mg sin α = μ mg cos α, so μ = tan α.

Worked example

A particle is projected up a rough slope (tan α = 3/4, μ = 0.5) at 10 m/s. Find how far it travels up the slope.

  1. sin α = 0.6, cos α = 0.8.
  2. Up the slope: -(mg × 0.6 + 0.5 × mg × 0.8) = ma, so a = -9.8 m/s2.
  3. 0 = 100 - 2(9.8)s.

Answer: s = 5.10 m

Key idea: Friction μmg cos α opposes motion: up the slope it adds to mg sin α, down the slope it subtracts. Slides back only if tan α > μ.

Check yourself

  1. A particle slides down a rough plane inclined at 30° with μ = 0.2. Find its acceleration, to 3 significant figures.
    Show answer

    3.20 m/s2. a = g(sin 30° - 0.2 cos 30°) = 3.20 m/s2.

  2. The particle in the worked example (tan α = 3/4, μ = 0.5) comes to rest and then slides back. Find its acceleration down the slope, to 3 significant figures.
    Show answer

    1.96 m/s2. a = g(0.6 - 0.4) = 0.2g = 1.96 m/s2.

  3. A particle slides down a rough plane inclined at 25° at constant velocity. Find μ, to 3 significant figures.
    Show answer

    0.466. μ = tan 25° = 0.466.

  4. A 4 kg crate is pulled up a rough plane inclined at 30° (μ = 0.3) by a force of 40 N parallel to the plane. Find its acceleration, to 3 significant figures.
    Show answer

    2.55 m/s2. 40 - 4g sin 30° - 0.3(4g cos 30°) = 4a, so a = 2.55 m/s2.

  5. A particle starts from rest and slides 5 m down a rough plane with sin α = 0.6, cos α = 0.8 and μ = 0.25. Find its speed, to 3 significant figures.
    Show answer

    6.26 m/s. a = g(0.6 - 0.2) = 3.92 m/s2. v2 = 2(3.92)(5) = 39.2, v = 6.26 m/s.

  6. A particle projected up a rough slope stops. It stays at rest if:
    • μ < tan α
    • μ ≥ tan α
    • μ = sin α
    • μ = 0
    Show answer

    μ ≥ tan α. Friction can hold it if the maximum friction μ mg cos α is at least mg sin α, that is μ ≥ tan α.

Practise with instant feedback

Sources

Last updated 7 November 2026. Spotted a mistake? Tell us.