Question

A block of mass $20$ $kg$ is acted upon by a force $F=30$ $N$ at an angle ${53}^0$ with the horizontal in the downward direction, as shown in the figure. The coefficient of friction between the block and the horizontal surface is $0.2$. If the block remains stationary, the friction force acting on the block by the ground is $(g=10m/s^2)$

• A. 40.0 N
• B. 30.0 N
• C. 18.0 N
• D. 44.8 N

Answer 1

The correct option is C 18.0 N

Maximum frictional force:

$f_{max}=\mu N$
$=\mu (mg+Fsin{53}^0)$
$=0.2(20\times 10+30\times \frac{4}{5})$
$=$ $44.8N$
As applied horizontal force is: $F\cos {53}^0=30\times \frac{3}{5}=18$ and $f_{applied}<f_{max}$
Hence, friction force will also be $18N$.