Question

A force of $100N$ is applied on a mass of $3kg$ as shown in the figure. The coefficient of friction between the surface and the block is $\mu =\frac{1}{\sqrt{3}}$. The frictional force acting on the block is

• A. $15N$ downwards
• B. $25N$ upwards
• C. $20N$ downwards
• D. $30N$ upwards

Answer 1

The correct option is C $20N$ downwards

FBD of the block is given below.


Along the horizontal direction,
$Fcos30=N=100\frac{100\sqrt{3}}{2}$

Along vertical direction,
$Fsin30-mg=50-30=20N$

So, friction must act in the downward direction to balance $20N$ and prevent relative motion between the block and wall.

Limiting friction $f_{max}=\mu N=\frac{1}{\sqrt{3}}\times 100\frac{\sqrt{3}}{2}=50N>20N$

Therefore, friction force will only be $20N$ downwards.