A body of mass m = 2 kg lies on an incline of angle α = 30^°…

A body of mass $m = 2\ \mathrm{kg}$ lies on an incline of angle $\alpha = 30^\circ$ with friction coefficient $\mu = 0.2$. What is the smallest horizontal force that keeps it from sliding down?

Explanation of Method

Smallest horizontal force holding a body on an incline.

Step 1: Components

The horizontal force $F$ (pushing toward the slope) has a component $F\cos\alpha$ up the slope and $F\sin\alpha$ into it. At the minimum force the body is on the point of sliding down, so friction $\mu N$ points up the slope:

\[ N = mg\cos\alpha + F\sin\alpha, \]

\[ F\cos\alpha + \mu N = mg\sin\alpha. \]

Step 2: Solve

\[ F = mg\,\frac{\sin\alpha - \mu\cos\alpha}{\cos\alpha + \mu\sin\alpha} = 19.62 \cdot \frac{0.3268}{0.966}. \]

\[F_{\min} \approx 6.64\ \mathrm{N}\]

$30^\circ$ $F$ $mg$ $N$ $\mu N$

The horizontal push has a part up the slope, $F\cos\alpha$, and a part into the slope, $F\sin\alpha$, which increases the normal force and so the friction. At the minimum force the block is just about to slide down, so friction points up the slope at its maximum.

Result

\[\boxed{\,F_{\min} \approx 6.64\ \mathrm{N}\,}\]

Remark

Check. The original result is correct. Without friction it would take $mg\tan\alpha = 11.3\ \mathrm{N}$; friction does part of the work.