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) \[\theta ={{\tan }^{-1}}\left( \frac{\mu }{1+\mu } \right),F=\frac{\mu W}{1+\mu }\]
Correct Answer: A
Solution :
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, \[=\frac{\mu }{\frac{1}{\sqrt{1+{{\mu }^{2}}}}+\frac{1}{\sqrt{1+{{\mu }^{2}}}}}=\frac{\mu w}{\frac{1}{\sqrt{1+{{\mu }^{2}}}}}\]You need to login to perform this action.
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