Question

A block of mass $m$ is in equilibrium on an inclined plane as shown.


Find out the normal force acting on the block

• A. $\frac{2mg}{\sqrt{3}}$
• B. $\frac{4mg}{\sqrt{3}}$
• C. $\frac{\sqrt{3}mg}{4}$
• D. $\frac{mg}{2\sqrt{3}}$

Answer 1

The correct option is A $\frac{2mg}{\sqrt{3}}$

FBD of block :


Because the block is equilibrium, sum of all forces along the plane must be zero.
$mg\sin {30}^{\circ }-F\cos {30}^{\circ }=0$
$\Rightarrow mg\times \frac{1}{2}=F\frac{\sqrt{3}}{2}$
$\Rightarrow F=\left(\frac{1}{\sqrt{3}}\right)mg\dots \left(i\right)$
Sum of forces in perpendicular direction to plane :
$N-F\sin {30}^{\circ }-mg\cos {30}^{\circ }=0$
$\Rightarrow N=F.\frac{1}{2}+mg\frac{\sqrt{3}}{2}$
$\Rightarrow N=\frac{mg}{\sqrt{3}.2}+\frac{mg\sqrt{3}}{2}=\frac{4mg}{2\sqrt{3}}=\frac{2mg}{\sqrt{3}}$ (substituting from (i))