Forces on inclined planes

The idea

Resolve a force $F$ at angle $\theta$ into $F\cos\theta$ along the direction and $F\sin\theta$ perpendicular to it. On a slope at angle $\alpha$, weight splits into $mg\sin\alpha$ down the slope and $mg\cos\alpha$ into it.

1. Draw the forces; choose directions along and perpendicular to the slope.
2. Resolve every force in each direction.
3. Perpendicular: forces balance ($R = mg\cos\alpha$ if nothing else acts). Along: $F = ma$.

Watch out The normal reaction on a slope is $mg\cos\alpha$, not $mg$.