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JEE Advance - Physics (2012 - Paper 1 Offline - No. 14)

A small block of mass 0.1 kg lies on a fixed inclined plane PQ which makes an angle $$\theta$$ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block remains stationary if (take g = 10 m/s2)

IIT-JEE 2012 Paper 1 Offline Physics - Laws of Motion Question 6 English

$$\theta$$ = 45$$^\circ$$
$$\theta$$ > 45$$^\circ$$ and a frictional force ats on the block towards P.
$$\theta$$ > 45$$^\circ$$ and a frictional force ats on the block towards Q.
$$\theta$$ < 45$$^\circ$$ and a frictional force ats on the block towards Q.

Explanation

The forces acting on the block are F = 1 N towards the left, weight mg = 0.1 $$\times$$ 10 = 1 N downwards, normal force N, and the frictional force f. Resolve F and mg along and perpendicular to the inclined plane.

IIT-JEE 2012 Paper 1 Offline Physics - Laws of Motion Question 6 English Explanation

When $$\theta$$ = 45$$^\circ$$, the net force that brings the block down is

$${F_d} = mg\sin \theta - F\cos \theta = {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 0$$.

Thus, the block is stationary if $$\theta = 45^\circ $$. When $$\theta > 45^\circ $$, the force $${F_d} > 0$$, and hence the block has a tendency to move down. Thus, the frictional force f acts on the block upwards i.e., towards Q.

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