Forces on an Object on a Slope

Splits the forces on an object resting on a slope. Resolving gravity along and across the slope gives mg·sinθ down the slope and a normal force of mg·cosθ. The angle at which it starts to slide is also shown.

This resolves the weight of an object on a slope into a part along the slope and a part at right angles to it.

F_{\parallel} = mg\sin\theta, \quad N = mg\cos\theta

$m$ is the mass, $g$ the gravity and $\theta$ the angle of the slope. The part along the slope tries to slide the object down; the normal force is what generates friction.

The angle at which it slides

Sliding begins when the force along the slope beats the maximum static friction. Rearranging $mg\sin\theta > \mu mg\cos\theta$ gives $\tan\theta > \mu$ , and both $m$ and $g$ drop out. The angle at which an object slides depends only on the coefficient of friction, not on its weight. A heavier object is no more secure than a light one.

Example

A 10 kg object rests on a 30 degree slope. Its weight is 98 N, the force along the slope is $98 \times \sin 30° = 49$ N and the normal force is $98 \times \cos 30° = 84.87$ N. With a coefficient of 0.3 the maximum static friction is only 25.46 N, so the object slides. It starts to slide at $\arctan 0.3 = 16.7$ degrees, and 30 degrees is past that.

Notes

At 0 degrees the surface is level: nothing acts along it and the normal force is the full $mg$ . At 90 degrees the surface is vertical, the normal force is 0, and the object is in free fall.