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Mechanics · Forces & Newton's laws

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Forces on an inclined plane

Resolving the weight on a slope into components along (mg sinα) and perpendicular (mg cosα) to the surface, the normal reaction R = mg cosα, and applying F = ma along the slope (acceleration g sinα on a smooth slope).

Mechanics · Forces & Newton's laws

Forces on an inclined plane

Resolving the weight on a slope into components along (mg sinα) and perpendicular (mg cosα) to the surface, the normal reaction R = mg cosα, and applying F = ma along the slope (acceleration g sinα on a smooth slope).

Why it works

When a body sits or slides on a slope, gravity still pulls it straight down — but the motion is along the slope, so resolving horizontally and vertically is clumsy. The clean choice is to resolve along and perpendicular to the slope instead.

For a slope at angle α\alpha to the horizontal, the weight mgmg splits into: down the slope: mgsin⁡α,into the slope: mgcos⁡α.\text{down the slope: } mg\sin\alpha, \qquad \text{into the slope: } mg\cos\alpha. (A useful sanity check: a flat surface, α=0\alpha = 0, gives no component down the slope and full weight pressing in; a vertical wall, α=90∘\alpha = 90^\circ, gives full weight down the slope and none into it.)mmgRThe perpendicular direction has no acceleration (the body stays on the surface), so the forces perpendicular to the slope balance. With nothing else pressing on it, the normal reaction is R=mgcos⁡α.R = mg\cos\alpha.

Along the slope is where the motion happens. On a smooth slope the only force along it is mgsin⁡αmg\sin\alpha, so by F=maF = ma, mgsin⁡α=ma  ⇒  a=gsin⁡α down the slope.mg\sin\alpha = ma \;\Rightarrow\; a = g\sin\alpha \text{ down the slope}. Notice the mass cancels — every smooth slope of angle α\alpha gives the same acceleration gsin⁡αg\sin\alpha. With a driving force, a string, or friction, you simply add those forces along the slope and apply F=maF = ma to the total.