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 $1N$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if: (take $g=10\frac{m}{s^2})$

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

FBD of the block
Weight of the block
$W=0.1\left(10\right)=1N$
The net pulling force on the block along the incline is,
$F=sin\theta -\cos \theta$, friction acts opposite to net pulling force.

If $\theta ={45}^{\circ }\Rightarrow F=0\Rightarrow$ no frictional force is required.
If $\theta >{45}^{\circ }\Rightarrow \sin \theta >\cos \theta \Rightarrow F$ is down the incline and friction acts towards $Q$.
If $\theta <{45}^{\circ }\Rightarrow \sin \theta <\cos \theta \Rightarrow F$ is up the incline and friction acts towards $P$.