• # question_answer A block of weight crests on a horizontal floor with coefficient of static friction $\mu .$It is desired to make the block move by applying minimum amount offered. The angle $\theta$ from the horizontal at which the force should be applied and magnitude of the force F are respectively.               [JEE ONLINE 19-05-2012] A) $\theta ={{\tan }^{-1}}\left( \mu \right),F=\frac{\mu W}{\sqrt{1+{{\mu }^{2}}}}$ B) $\theta ={{\tan }^{-1}}\left( \frac{1}{\mu } \right),F=\frac{\mu W}{\sqrt{1+{{\mu }^{2}}}}$ C) $\theta =0,F=\mu W$ D) $\theta ={{\tan }^{-1}}\left( \frac{\mu }{1+\mu } \right),F=\frac{\mu W}{1+\mu }$

 [a] Let the force F is applied at an angle $\theta$ with the horizontal. For horizontal equilibrium,$F\cos \theta =\mu R$(i) For vertical equilibrium, $R+F\sin \theta =mg$or$R=mg-F\sin \theta$       (ii) Substituting this value of R in eq. (i), we get $F\cos \theta =\mu (mg-F\sin \theta )$ $=\mu \,mg-\mu \,F\sin \theta$ or,$=F(\cos \theta +\mu \sin \theta )=\mu mg$ or,$F=\frac{\mu mg}{\cos \theta +\mu \sin \theta }$                                (iii) For F to be minimum, the denominator $(\cos \theta +\mu \sin \theta )$ should be maximum. $\therefore$$\frac{d}{d\theta }(\cos \theta +\mu \sin \theta )=0$ or$-\sin \theta +\mu \cos \theta =0$or$\tan \theta =\mu$ or$\theta ={{\tan }^{-1}}(\mu )$ Then, $\sin \theta =\frac{\mu }{\sqrt{1+{{\mu }^{2}}}}$ and$\cos \theta =\frac{1}{\sqrt{1+{{\mu }^{2}}}}$ Hence,${{F}_{\min }}$ $=\frac{\mu }{\frac{1}{\sqrt{1+{{\mu }^{2}}}}+\frac{1}{\sqrt{1+{{\mu }^{2}}}}}=\frac{\mu w}{\frac{1}{\sqrt{1+{{\mu }^{2}}}}}$