Question

A body of mass $m$ rests on a horizontal floor with which it has a coefficient of static friction $\mu$. It is desired to make the body move by applying a minimum possible force $\vec{F}$ as shown in the diagram. The values of $\theta$ and $F_{\min }$ shall be respectively equal to

• A. $\tan ^{-1} \mu \frac{\mu m g}{\sqrt{\left(1+\mu^{2}\right)}}$
• B. $\tan ^{-1} \mu, \frac{m g}{\sqrt{\left(1+\mu^{2}\right)}}$
• C. $\tan ^{-1} \mu, \frac{\mu m g}{\sqrt{\left(1-\mu^{2}\right)}}$
• D. $\tan ^{-1} \mu, \frac{m g}{\sqrt{\left(1-\mu^{2}\right)}}$

Answer 1

Answer: (A)

$\tan ^{-1} \mu \frac{\mu m g}{\sqrt{\left(1+\mu^{2}\right)}}$

Answer 2

$\mu=\tan \theta$
$\therefore \theta=\tan ^{-1} \mu$
$\therefore \sin \theta=\frac{\mu}{\sqrt{1+\mu^{2}}}$
and $\cos \theta=\frac{1}{\sqrt{1+\mu^{2}}}$
$F_{\min }=\frac{\mu m g}{\cos \theta+\sin \theta}$
$=\frac{\mu m g}{\frac{1}{\sqrt{1+\mu^{2}}}+\frac{\mu^{2}}{\sqrt{1+\mu^{2}}}}$
$=\frac{\mu m g}{\sqrt{1+\mu^{2}}}$