Laws of Motion - Result Question 37
37. A body of mass $m$ is kept on a rough horizontal surface ( coefficient of friction $=\propto$ ). A horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by $F$, where $F$ is,
[NEET Odisha 2019]
(a) $|\vec{F}|=m g$
(b) $|\vec{F}|=m g+\propto m g$
(c) $|\vec{F}|=\propto m g$
(d) $|\vec{F}| \leq m g \sqrt{1+\mu^{2}}$
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Answer:
Correct Answer: 37. (d)
Solution:
- (d) Since body does not move hence it is in equilibrium.
$f_r=$ frictional force which is less than or equal to limiting friction.
Now $N=m g$
Hence $\vec{F}=\vec{N}+ \vec{f} _r$
$ \begin{aligned} & |\vec{F}| \leq \sqrt{(m g)^{2}+(\mu m g)^{2}} \\ & |\vec{F}| \leq m g \sqrt{1+\mu^{2}} \end{aligned} $
