Inclined planes

Objective. Resolve forces parallel and perpendicular to a slope.

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Learn: the key idea

On inclined planes, forces must be resolved into components parallel and perpendicular to the slope, since weight mg acts vertically but motion occurs along the incline. The component of weight parallel to the slope is mg sinθ, and perpendicular to the slope is mg cosθ, where θ is the angle of incline. For example, a block on a 30° slope has a parallel weight component of mg sin30°=0.5mg, driving it down the slope, while mg cos30° balances the normal reaction if there is no friction. Always draw perpendicular axes along and across the slope before resolving.

Worked example

A 2kg block rests on a smooth inclined plane at 30° to the horizontal. Find the component of its weight acting parallel to the plane, taking g=9.8m/s².

1 $Parallel component of weight is mg sinθ$
2 $Here m=2, g=9.8, θ=30°$
3 $Component=2\times9.8\timessin30°=2\times9.8\times0.5$
4 $Component=9.8N$

Answer: 9.8N

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Common mistake. Students swap sine and cosine, using mg cosθ for the component along the slope instead of mg sinθ. Always check that the parallel component uses sine of the angle measured from the slope's horizontal base.

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