Question

A force of $100\text{N}$ is applied on a block of mass $3\text{kg}$ on shown in figure. The co-efficient of friction between the wall and the block is $1/4$. The friction force acting on the block is
(Take $g=10{\text{m/s}}^2$)

• A. $15\text{N}$ downwards
• B. $15\text{N}$ upwards
• C. $20\text{N}$ downwards
• D. $20\text{N}$ upwards

Answer 1

The correct option is C $20\text{N}$ downwards


From diagram,

$N=F\cos {30}^{\circ }=100\times \frac{\sqrt{3}}{2}=50\sqrt{3}\text{N}$

And maximum static friction force will be

$f_{max}=\mu N=\frac{1}{4}\times 50\sqrt{3}=21.6\text{N}$

Here, net force in vertical direction excluding friction acting on the block is

$F_y=F\sin {30}^{\circ }-3g$

$F_y=100\times \frac{1}{2}-3\times 10=20\text{N}$

So, $20\text{N}$ force is acting in upward direction which is less than $f_{max}$. Therefore, friction of $20\text{N}$ downwards acts on block.

Hence, option $\left(c\right)$ is correct.