Smooth plane at 30 degrees

Problem

A ball slides down a smooth inclined plane of $\theta = 30^\circ$. What forces act on it, and why does it accelerate less than $g$?

Smooth plane: only the weight $m\vec g$ and the normal reaction $\vec N$ act.

Solution

The forces at play. The plane is smooth, i.e. frictionless. Only two forces therefore act on the ball: the weight $m\vec g$ (vertical, downward) and the normal reaction $\vec N$ (perpendicular to the plane). There is no force along the plane besides the component of the weight.

Choice of axes and resolution. Take $x$ along the plane (positive downward) and $y$ perpendicular. The weight resolves into: $P_x = mg\sin\theta \qquad P_y = mg\cos\theta$

Perpendicular direction. The ball neither sinks into nor lifts off the plane, so along $y$ there is equilibrium: the normal reaction balances the pressing component of the weight: $N = mg\cos\theta$

Direction along the plane. The only driving force is $P_x$. Newton’s second law gives: $ma = mg\sin\theta \;\Longrightarrow\; a = g\sin\theta$

Why it accelerates less than $g$. Only the component of the weight parallel to the plane pulls the ball forward; the other component ($mg\cos\theta$) is cancelled by the constraint $\vec N$. Since $\sin\theta < 1$ for $\theta < 90^\circ$, the acceleration is reduced: $a = g\sin 30^\circ = g\cdot\tfrac{1}{2} = \frac{9{,}81}{2} \approx 4{,}9\ \mathrm{m/s}^2$

$\ev{a = g\sin\theta = \frac{g}{2} \approx 4{,}9\ \mathrm{m/s}^2,\qquad N = mg\cos\theta}$

Only with a vertical plane ($\theta = 90^\circ$) would one get $a = g$: free fall.

Links

Topics: Dinamica Concepts: Forza peso · Forza normale · Seconda legge di Newton Skills: Scomposizione sul piano inclinato Methods: Metodo del diagramma di corpo libero Objects: Piano inclinato