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

Answer 1

Answer: (D)

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

$R+F\sin \theta =mg$ or $R=mg-F\sin \theta ...(i)$
Force of friction, $f=\mu R=\mu (mg-F\sin \theta )$
The block will move, when
$F\cos \theta \ge f\text{ i.e., }F\cos \theta \ge \mu mg-\mu F\sin \theta ...(ii)$
$\text{ or }F(\cos \theta +\mu \sin \theta )\ge \mu mg\text{ or }F\ge \frac{\mu mg}{\cos \theta +\mu \sin \theta }$
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 $(ii)$,
$F\ge \frac{\mu mg}{\frac{1}{\sqrt{1+{\mu }^2}}+\frac{{\mu }^2}{\sqrt{1+{\mu }^2}}}=\frac{\mu mg}{\sqrt{1+{\mu }^2}}$
$\therefore F_{\min }=\frac{\mu mg}{\sqrt{1+{\mu }^2}}$