Friction on an inclined plane — find minimum angle for block to start sliding

Question

A block of mass $m$ rests on a rough inclined plane. The coefficient of static friction between the block and the surface is $\mu_s = 0.5$ . Find the minimum angle $\theta$ at which the block just begins to slide down.

(JEE Main 2023, similar pattern)

Solution — Step by Step

Three forces act on the block: weight $mg$ downward, normal reaction $N$ perpendicular to the incline, and static friction $f_s$ along the incline (opposing the tendency to slide, so directed up the incline).

Resolve $mg$ into components: $mg\sin\theta$ along the incline (pulling down) and $mg\cos\theta$ perpendicular to the incline (pressing into the surface).

Perpendicular to the incline:

N = mg\cos\theta

Along the incline, the block is on the verge of sliding, so friction is at its maximum value:

f_s = \mu_s N = \mu_s mg\cos\theta

The block just starts sliding when the component pulling it down equals the maximum static friction:

mg\sin\theta = \mu_s mg\cos\theta

Mass $m$ and $g$ cancel from both sides:

\tan\theta = \mu_s

\theta = \tan^{-1}(\mu_s) = \tan^{-1}(0.5)

\boxed{\theta \approx 26.57°}

This angle is called the angle of repose — a term that appears frequently in PYQs.

Why This Works

The beauty of the angle of repose is that it depends only on the coefficient of friction, not on the mass of the block. When we set the net force along the incline to zero (at the verge of sliding), $m$ and $g$ cancel out completely.

Physically, a steeper incline means a larger $mg\sin\theta$ component pulling the block down. The friction force $\mu_s mg\cos\theta$ actually decreases because the normal force drops. At the critical angle, these two competing effects balance exactly.

This is why heavy and light blocks on the same rough incline begin to slide at the same angle — a fact that often surprises students but is a direct consequence of the cancellation.

Alternative Method

You can also think in terms of the friction angle. The resultant of $N$ and $f_s$ makes an angle $\phi$ with the normal where $\tan\phi = \mu_s$ . The block slides when the weight vector falls outside the friction cone, which happens exactly when $\theta > \phi = \tan^{-1}(\mu_s)$ .

Remember: angle of repose = $\tan^{-1}(\mu_s)$ . This one-liner solves at least one question per JEE Main session. If the question gives $\mu_s$ and asks for $\theta$ (or vice versa), this is all you need.

Common Mistake

Students often write $mg\sin\theta = \mu_s mg$ instead of $mg\sin\theta = \mu_s mg\cos\theta$ . They forget that friction equals $\mu_s N$ , not $\mu_s mg$ . On an incline, the normal force is $mg\cos\theta$ , NOT $mg$ . The normal force equals $mg$ only on a flat surface. Always resolve forces properly before applying the friction formula.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next