Question

A small block of mass of $0.1kg$ 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 center of mass as shown in `the figure. The block remains stationary if (take $g=10m/s^2$)

• A. $\theta ={45}^{\circ }$.
• B. $\theta$ > ${45}^{\circ }$ and a frictional force acts on the block towards P.
• C. $\theta$ > ${45}^{\circ }$ and a frictional force acts on the block towards Q.
• D. ${45}^{\circ }$ and a frictional force acts on the block towards Q.

Answer 1

Answer: (A), (C)

The correct options are
A $\theta ={45}^{\circ }$.
C $\theta$ > ${45}^{\circ }$ and a frictional force acts on the block towards Q.

The net force along the plane PQ acting on the mass is given by:
$F=mg\sin \theta -f-1\times \cos \theta wheref=frictionalforcetowardsQm=0.1kgg=10m/s^2$
For the mass to remain stationary, the net force along the plane should be zero.
Thus,
$\sin \theta -f-\cos \theta =0$
In the absence of friction,
$\sin \theta =\cos \theta i.e.\theta ={45}^{\circ }$
In the presence of friction,
$\sin \theta =f+\cos \theta$
When $\theta >{45}^{\circ },\sin \theta >\cos \theta$ and both are positive. Hence, $f>0$.Thus, a positive frictional force act on the mass towards Q.