question_answer

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) \[{{\tan }^{-1}}\mu ,\frac{\mu mg}{\sqrt{(1+{{\mu }^{2}})}}\]

B) \[{{\tan }^{-1}}\mu ,\frac{mg}{\sqrt{(1+{{\mu }^{2}})}}\]

C) \[{{\tan }^{-1}}\mu ,\frac{\mu mg}{\sqrt{(1-{{\mu }^{2}})}}\]

D) \[{{\tan }^{-1}}\mu ,\frac{mg}{\sqrt{(1-{{\mu }^{2}})}}\]

Correct Answer: A

Solution :

[a]:\[N+F\sin \theta =mg\] \[N=mg-F\sin \theta \] Force of friction, \[f=\mu N=\mu (mg-Fsin\theta )\] The block will move, when              \[Fcos\theta \ge f\] \[Fcos\theta \ge \mu (mg-Fsin\theta )\] \[F\ge \frac{\mu mg}{\cos \theta +\mu \sin \theta }\]                                  (ii) F will be minimum, when \[\cos \theta +\mu \sin 9=\] maximum, for which \[\frac{d}{d\theta }(\cos \theta +\mu \sin \theta )=0\] or\[-\sin \theta +\mu \cos \theta =0\] or\[\mu \cos \theta =\sin \theta \] or\[\tan \theta =\mu \]or\[\theta ={{\tan }^{-1}}(\mu )\] \[\therefore \]\[\sin \theta =\frac{\mu }{\sqrt{1+{{\mu }^{2}}}}\]and\[\cos \theta =\frac{1}{\sqrt{1+{{\mu }^{2}}}}\] From (ii),\[F\ge \frac{\mu mg}{\frac{1}{\sqrt{1+{{\mu }^{2}}}}+\frac{{{\mu }^{2}}}{\sqrt{1+{{\mu }^{2}}}}}\] \[F\ge \frac{\mu mg}{\sqrt{1+{{\mu }^{2}}}}\therefore {{F}_{\min }}=\frac{\mu mg}{\sqrt{1+{{\mu }^{2}}}}\]