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 the minimum possible force $F$. The magnitude of $F$ is

• A. $\mu mg$
• B. $\frac{\sqrt{1+\mu^{2}}}{\mu} m g$
• C. $\mu \sqrt{1+\mu^{2}} m g$
• D. $\frac{\mu m g}{\sqrt{1+\mu^{2}}}$

Answer 1

Answer: (D)

Suppose the force $F$ is applied at an angle $\theta$ with horizontal, as shown in figure. Then

$R+F \sin \theta=m g$ or $R=m g-F \sin \theta \,\,\,\, ...(i)$
Force of friction, $f=\mu R=\mu(m g-F \sin \theta)$
The block will move, when
$F \cos \theta \geq f \text { i.e., } F \cos \theta \geq \mu m g-\mu F \sin \theta\,\,\, ...(ii) $
$\text { or } F(\cos \theta+\mu \sin \theta) \geq \mu m g \text { or } F \geq \frac{\mu m g}{\cos \theta+\mu \sin \theta}$
Now, $F$ will be minimum, when $\cos \theta+\mu \sin \theta=$ maximum.
For which
$\frac{d}{d \theta}(\cos \theta+\mu \sin \theta)=0$
or $-\sin \theta+\mu \cos \theta=0$ or $\tan \theta=\mu$
$\therefore \sin \theta=\frac{\mu}{\sqrt{1+\mu^{2}}} ; \cos \theta=\frac{1}{\sqrt{1+\mu^{2}}}$
From $(ii) $,
$F \geq \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}}} $
$\therefore F_{\min } =\frac{\mu m g}{\sqrt{1+\mu^{2}}}$