Question 291: Laws of Motion

Question

A small block of mass 0.1 kg lies on a fixed inclined plane PQ which makes an angle $$	heta$$ 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/s
2
)

Accepted Answer

The forces acting on the block are F = 1 N towards the left, weight mg = 0.1 $$	imes$$ 10 = 1 N downwards, normal force N, and the frictional force f. Resolve F and mg along and perpendicular to the inclined plane.
When $$	heta$$ = 45$$^\circ$$, the net force that brings the block down is
$${F_d} = mg\sin 	heta  - F\cos 	heta  = {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 0$$.
Thus, the block is stationary if $$	heta  = 45^\circ $$. When $$	heta  > 45^\circ $$, 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.

Answer

$$	heta$$ = 45$$^\circ$$

Answer

$$	heta$$ > 45$$^\circ$$ and a frictional force ats on the block towards P.

Answer

$$	heta$$ > 45$$^\circ$$ and a frictional force ats on the block towards Q.

Answer

$$	heta$$ < 45$$^\circ$$ and a frictional force ats on the block towards Q.

Laws of Motion

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/s 2 )

Choose the correct answer: