Forces on Inclined Planes

Consider a slope inclined at $\theta$ to the horizontal. The weight acts vertically downwards, and the perpendicular to the slope is tilted from the vertical by exactly $\theta$ — the same angle, because both are rotations from the horizontal and vertical respectively.

So the angle between the weight and the inward perpendicular is $\theta$. The component $\text{along}$ that perpendicular is therefore the adjacent one, $mg\cos \theta$, and the component down the slope is the opposite one, $mg\sin \theta$.

A quick check confirms the assignment. At $\theta =0$ the surface is flat, and the formulas give no force down the slope and the full weight pressing in — correct. At $\theta =90^{\circ }$ the surface is vertical, giving the full weight down the slope and nothing pressing in — also correct.

On a smooth slope the only force along it is the weight component $mg\sin \theta$, so applying $F=ma$ gives $mg\sin \theta =ma$ and the mass cancels.

The acceleration is therefore $g\sin \theta$, independent of mass — the same reason all objects fall at $g$, since the driving force and the inertia both scale with $m$.

Perpendicular to the slope there is no acceleration, so the normal reaction balances the perpendicular weight component: $R=mg\cos \theta$.

With friction present, the maximum friction is $\mu R=\mu mg\cos \theta$ — note that it depends on the angle, because the normal reaction does.

For a body sliding down, friction acts up the slope and the equation along the slope is $mg\sin \theta -\mu mg\cos \theta =ma$. The mass cancels again, giving $a=g\left(\sin \theta -\mu \cos \theta \right)$.

For a body being pushed or pulled up, friction acts down the slope and adds to the weight component instead — the direction of friction always opposes the motion, so it must be decided before writing the equation.

A body on a rough slope stays put if the weight component down the slope does not exceed the maximum friction: $mg\sin \theta \le \mu mg\cos \theta$.

The mass cancels, leaving $\tan \theta \le \mu$. So whether an object slides depends only on the angle and the coefficient — not on how heavy it is, which is a genuinely surprising and examinable result.

The critical angle at which sliding just begins satisfies $\tan \theta =\mu$, and is called the $\text{angle of friction}$. Tilting a plank until an object starts to slide is a standard way of measuring $\mu$ experimentally.