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5.2MechanicsStretch

Inclined Planes

On a slope it is far easier to resolve forces parallel and perpendicular to the plane rather than horizontally and vertically. Weight then splits neatly into a component down the slope and one into it.

26 min Forces and Friction
Edexcel A Level Maths: 5.2 Inclined PlanesCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Resolve parallel and perpendicular to a slope
  • Split weight into mg sin θ and mg cos θ
  • Apply Newton’s second law along the plane
  • Find the normal reaction
1

Resolving on a slope

For a slope at angle θ\theta, the weight mgmg resolves into mgsin⁡θmg\sin\theta down the slope and mgcos⁡θmg\cos\theta into the slope. The normal reaction balances the perpendicular component.

W∥=mgsin⁡θ,W⊥=mgcos⁡θW_{\parallel} = mg\sin\theta, \quad W_{\perp} = mg\cos\theta
Weight components on a slope.
Example 1A 2 kg block on a smooth 30° slope. Find the acceleration down the slope (g = 9.8).
1Down-slope force =mgsin⁡θ=2(9.8)sin⁡30°=9.8= mg\sin\theta = 2(9.8)\sin30° = 9.8 N.
2a=F/m=9.8/2=4.9a = F/m = 9.8/2 = 4.9 m/s².
Answer
4.94.9 m/s² down the slope.

Tip — Down the slope uses sin θ; into the slope uses cos θ.

Formula recap

mgsin⁡θmg\sin\theta
Component down the slope.
mgcos⁡θmg\cos\theta
Component into the slope (= normal reaction if no other ⊥ forces).

Common mistakes to avoid

Resolving horizontally and vertically on a slope.
Resolve parallel and perpendicular to the plane.
Swapping sin and cos for the weight components.
Down the slope: sin θ; into the slope: cos θ.

Key takeaways

  • Resolve parallel and perpendicular to the slope.
  • Weight: mg sin θ down the slope, mg cos θ into it.
  • Normal reaction balances the perpendicular component.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. On a slope you resolve:

    • parallel and perpendicular to the plane
    • horizontally and vertically
    • only vertically
    • along the weight
    Show answer and explanation

    Answer: parallel and perpendicular to the plane

    Simplifies the algebra.

  2. The weight component down a slope at θ is:

    • mgsin⁡θmg\sin\theta
    • mgcos⁡θmg\cos\theta
    • mgtan⁡θmg\tan\theta
    • mgmg
    Show answer and explanation

    Answer: mgsin⁡θmg\sin\theta

    Along the slope.

  3. The component into the slope is:

    • mgcos⁡θmg\cos\theta
    • mgsin⁡θmg\sin\theta
    • mgmg
    • 00
    Show answer and explanation

    Answer: mgcos⁡θmg\cos\theta

    Perpendicular.

  4. A 2 kg block on a smooth 30° slope accelerates at (g=9.8):

    • 4.94.9 m/s²
    • 9.89.8 m/s²
    • 2.452.45 m/s²
    • 19.619.6 m/s²
    Show answer and explanation

    Answer: 4.94.9 m/s²

    g sin30°.

  5. The normal reaction balances the:

    • perpendicular component
    • down-slope component
    • whole weight
    • friction
    Show answer and explanation

    Answer: perpendicular component

    No motion into the slope.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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Related lessons

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