question_answer

A body of mass m rests on a horizontal floor, with which it has a coefficient of static friction u. It is desired to make the body move by applying the minimum possible force F. the magnitude of F is:

A) \[\mu \operatorname{mg}\]                                 

B) \[\sqrt{\frac{1+{{\mu }^{2}}}{\mu }}\operatorname{mg}\]

C) \[\mu \sqrt{1+{{\mu }^{2}}}\operatorname{mg}\]         

D) \[\frac{\mu \operatorname{mg}}{\sqrt{1+{{\mu }^{2}}}}\]

Correct Answer: D

Solution :

\[\operatorname{F} + F sin\theta = mg\] \[\operatorname{R} = mg -F sin\theta \] \[\operatorname{Force} of friction \left( {{f}_{r}} \right)= \mu R =\mu \left[ mg - Fsin\theta  \right]\]Block will move \[\operatorname{Fcos}\theta \ge fr\] \[\operatorname{F} cos\theta  \ge  \mu  \left( mg -F sin\theta  \right)\] \[\operatorname{F}\ge \frac{\mu mg}{\cos \theta +\mu \sin \theta }\]                               ...(1) \[\operatorname{F} will be minimum when cos\theta + psin\theta  = max\] \[\frac{d}{d\theta } \left[ cos\theta  +\mu sin\theta  \right] =0\,\,\,\Rightarrow  \mu = tan \theta \] \[\cos \theta  =\frac{1}{1+{{\mu }^{2}}}, sin\theta =\frac{\mu }{\sqrt{1+{{\mu }^{2}}}}\] \[\operatorname{this} value put in \left( 1 \right) we get {{F}_{min}} =\frac{\mu \operatorname{mg}}{\sqrt{1+{{\mu }^{2}}}}\]