= Gravity components on an inclined plane
{title2=$W_{\parallel}=mg\sin\theta,\quad N=mg\cos\theta$}
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 give
$$
\mathbf W=mg\sin\theta\,\mathbf e_d-mg\cos\theta\,\mathbf e_n.
$$
The <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.
\Image[/inclined-plane-weight-and-contact-geometry.png]
{title=Actual contact forces and a separate decomposition of gravitational weight on an inclined plane}
{description=Original geometry for a block sliding downhill at $30^\circ$. Left: its <free-body diagram> shows <gravitational weight>, <normal force>, and uphill <kinetic friction>. Right: the <gravitational weight> is resolved into downhill and inward-normal components; the dashed arrow is their sum. Arrow lengths are illustrative.}
Continued contact with a stationary straight <inclined plane> gives zero perpendicular <acceleration>. The perpendicular component of <Newton's second law> therefore gives
$$
N-mg\cos\theta=0,\qquad N=mg\cos\theta.
$$
This 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.<University Physics Volume 1 by William Moebs and others/6.2 Friction>[University Physics Volume 1 by William Moebs and others; “Friction and the Inclined Plane”, Example 6.11 strategy and Figure 6.14; perpendicular balance and $N=mg\cos\theta$]{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.
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