A box of mass \(2\ \text{kg}\) is placed on an inclined plane that makes \(30^\circ\) with the horizontal. The coefficient of friction between the box and inclined plane is \(0.2\). A force \(F\) is applied on the box perpendicular to the incline to prevent the box from sliding down. The minimum value of \(F\) is
\[
(\text{acceleration due to gravity }=10\ \text{m s}^{-2})
\]
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When an external force is applied perpendicular to an inclined plane, it changes the normal reaction. Since friction is
\[
f=\mu N,
\]
increasing \(N\) increases the maximum friction available to prevent sliding.
Step 1: Identify forces on box. Weight component along incline: \( mg\sin 30^\circ = 2\times10\times0.5 = 10\text{ N} \). Normal force \( N = mg\cos 30^\circ + F \) (F pushes into slope). Friction \( f = \mu N \) acts up the slope.
Step 2: Equilibrium along incline. \( \mu(mg\cos 30^\circ + F) = mg\sin 30^\circ \) won't hold since \( \mu\cos30^\circ < \sin30^\circ \); box would slide down, so friction acts up: \( \mu(mg\cos30^\circ + F) \geq mg\sin30^\circ \) \( 0.2(2\times10\times\frac{\sqrt{3}}{2}+F) = 10 \) \( 0.2(10\sqrt{3}+F) = 10 \Rightarrow F = 50 - 10\sqrt{3} \approx 50 - 17.3 = 32.7\text{ N} \)
