Question

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

• 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. $\theta <{45}^{\circ }$, and a frictional force acts on the block towards $Q$

Answer 1

The correct option is C $\theta >{45}^{\circ }$, and a frictional force acts on the block towards $Q$

From the FBD of small block, while at rest
Along the plane,
$\sin (90-\theta )=mg\sin \theta -f_s$
$\Rightarrow 1\cos \theta =mg\sin \theta -\mu R$ ... (1)
Perpendicular to the plane,
$R=mg\cos \theta +\cos (90-\theta )$
$R=mg\cos \theta +1\sin \theta$ ... (2)
From (1) & (2) we get
$\cos \theta =mg\sin \theta -\mu \times mg\cos \theta -\mu \sin \theta$
$\cos \theta +\mu \sin \theta =mg(\sin \theta -\mu \cos \theta )$
$\frac{\cos \theta +\mu \sin \theta }{\sin \theta -\mu \cos \theta }=\frac{mg}{1}=1$
For a frictionless surface $\mu =0$
$\therefore \theta ={45}^{\circ }$

Since for $\theta ={45}^{\circ }$ no friction is required to keep the block stationary, therefore as the value of $\theta$ becomes more than ${45}^{\circ }$ , the component of gravitational force ($mg\sin \theta$) increases and friction acts towards Q to keep the block stationary.