Question

A block of mass $15\text{kg}$ is acted upon by a force $25\text{N}$ at an angle ${37}^{\circ }$ with the horizontal in a direction as shown in figure. The coefficient of static friction($\mu$) between the block and horizontal surface is $0.2$ . Find the frictional force acting on the block, so that it doesn't move. Take $(g=10{\text{m/s}}^2)$.

• A. $33\text{N}$
• B. $13\text{N}$
• C. $10\text{N}$
• D. $20\text{N}$

Answer 1

The correct option is D $20\text{N}$

Mass of block is $m=15\text{kg}$
$F=25\text{N}$
$\mu =0.2$
For the block to be in equilibrium $\sum \overrightarrow{F_{\text{net}}}=0$, hence balancing the forces in
y and x - directions:
$N=mg+F\sin {37}^{\circ }...\left(1\right)$
$F\cos {37}^{\circ }=f...\left(2\right)$


Maximum available value of frictional force :
$f_{\text{max}}=\mu N$
$\therefore f_{\text{max}}=\mu N=\mu (mg+F\sin {37}^{\circ })$
$f_{\text{max}}=0.2\left(\right(15\times 10)+(25\times \frac{3}{5}\left)\right)$
$f_{\text{max}}=33\text{N}.......\left(3\right)$
From equation $\left(2\right)$:
$F\cos {37}^{\circ }=f$
$\therefore f=25\times \frac{4}{5}=20\text{N}$
$\because f\le f_{\text{max}}$, frictional force acting on block is sufficient to prevent slipping, so block will not move.