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) \[\operatorname{F} will be minimum when cos\theta + psin\theta = max\] \[\frac{d}{d\theta } \left[ cos\theta +\mu sin\theta \right] =0\,\,\,\Rightarrow \mu = tan \theta \] \[\cos \theta =\frac{1}{1+{{\mu }^{2}}}, sin\theta =\frac{\mu }{\sqrt{1+{{\mu }^{2}}}}\] \[\operatorname{this} value put in \left( 1 \right) we get {{F}_{min}} =\frac{\mu \operatorname{mg}}{\sqrt{1+{{\mu }^{2}}}}\]You need to login to perform this action.
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