Question

A small block of mass of $0.1\text{kg}$ lies on a fixed inclined plane $\text{PQ}$ which makes an angle $\theta$ with the horizontal. A horizontal force of $1\text{N}$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if
$(\text{Take}g=10{\text{m/s}}^2)$

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

Answer 1

Answer: (A), (C)

$\theta >{45}^{\circ }$and a frictional force acts on the block towards $\text{Q}$

$W=mg=0.1\times 10=1\text{N}$


$F_1$= Component of weight $=1\sin \theta =\sin \theta$

$F_2$= Component of applied force $=1\cos \theta =\cos \theta$

Now, at $\theta ={45}^{\circ }:F_1=F_2$ and block remains stationary without the help of friction.

For $\theta >{45}^{\circ }$

$F_1>F_2\to$ So friction will act towards $\text{Q}$.

For $\theta <{45}^{\circ }$

$F_2>F_1$ and friction will act towards $\text{P}$.

Hence, options $\left(A\right)$ and $\left(C\right)$ are correct.