 Gravity components on an inclined plane

Choose a unit vector ${\mathbf{e}}_d$ pointing downhill and a perpendicular unit vector ${\mathbf{e}}_n$ pointing outward from the inclined plane. The gravitational weight points vertically downward. Its scalar projections onto these directions giveThe angle between the downward vertical and the inward normal is $\theta$: the inward component uses cosine, and the downhill component uses sine. These are two components of one force. They replace its vertical arrow in a component calculation; adding them to that same arrow would count gravitational weight twice.

Continued contact with a stationary straight inclined plane gives zero perpendicular acceleration. The perpendicular component of Newton's second law therefore givesThis is why the normal force generally differs from $mg$: it balances the inward component, rather than the whole gravitational weight. An extra lifting force, a curved surface, or motion of the support would require a new perpendicular balance.^[ref]

The same geometry explains the ramp's usefulness. A frictionless block pulled uphill at constant velocity needs a parallel force $P=mg\sin \theta$. Travelling a distance $L$ raises it through height $h=L\sin \theta$, so the pulling work is $PL=mgh$, its gain in gravitational potential energy. The smaller force costs a larger displacement; the work-energy theorem keeps the accounting consistent.