Question

A force of $100N$ is applied on a block of mass $3kg$ 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. $25N\text{downloads}$
• B. $20N\text{downloads}$
• C. $15N\text{downloads}$
• D. $30N\text{upwords}$

Answer 1

The correct option is B $20N\text{downloads}$

Draw free body diagram of the given problem.

Find the frictional force acting on the block.
It is clear from the diagram that'
$N=100\text{cos}{30}^{\circ }=100.\frac{\sqrt{3}}{2}=50\sqrt{3}N$
Net driving force $=100\text{sin}{30}^{\circ }-mg$
$=(100\times \frac{1}{2}-30)$
= $20N\text{(upward)}$
$f_{lim}=\frac{1}{\sqrt{3}}\left(50\sqrt{3}\right)=50N$
$\therefore \text{Friction}=20N\text{downward}$
Final Answer : $\left(C\right)$