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 Video by Zeeshan Zamurred Forces and Friction

Edexcel A Level Maths: 5.2 Inclined PlanesWatch the full walkthrough 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

Resolving on a slope

For a slope at angle $θ$ , the weight $m g$ resolves into $m g sin θ$ down the slope and $m g cos θ$ into the slope. The normal reaction balances the perpendicular component.

W_{∥} = m g sin θ, W_{⊥} = m g cos θ

Weight components on a slope.

1Down-slope force

= m g sin θ = 2 (9.8) sin 30° = 9.8

a = F / m = 9.8/2 = 4.9

m/s².

Answer

4.9

m/s² down the slope.

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

Formula recap

m g sin θ

Component down the slope.

m g cos θ

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

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