Acceleration on Smooth Inclined Plane

The situation

A body slides down a smooth (frictionless) inclined plane of angle $\theta$. What is its acceleration?

Derivation

Forces on the body:

• Weight $mg$ vertically downward
• Normal reaction $N$ perpendicular to the surface
• No friction (smooth surface)

Resolve weight along and perpendicular to the incline:

• Component along incline (down the slope): $mg\sin\theta$
• Component perpendicular to incline: $mg\cos\theta$

Perpendicular to incline (no acceleration in this direction):

$N = mg\cos\theta$

Along the incline (Newton's Second Law):

$mg\sin\theta = ma$

$\boxed{a = g\sin\theta}$

Understanding the result

The acceleration depends only on $\theta$ and $g$ — not on the mass. A heavy block and a light block slide down the same frictionless incline with exactly the same acceleration. This is the same result as free fall — mass cancels because both the gravitational force and inertia are proportional to mass.

How $\theta$ affects acceleration

At $\theta = 90°$, the incline becomes a vertical wall and the body is in free fall — consistent.

Normal force on the incline

$N = mg\cos\theta$

At $\theta = 0°$: $N = mg$ (flat surface, full weight supported) At $\theta = 90°$: $N = 0$ (vertical wall, no support)

The normal force decreases as the slope steepens.