JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • 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) \[\theta ={{\tan }^{-1}}\left( \frac{\mu }{1+\mu } \right),F=\frac{\mu W}{1+\mu }\]

    Correct Answer: A

    Solution :

    [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,\[{{F}_{\min }}\]
                \[=\frac{\mu }{\frac{1}{\sqrt{1+{{\mu }^{2}}}}+\frac{1}{\sqrt{1+{{\mu }^{2}}}}}=\frac{\mu w}{\frac{1}{\sqrt{1+{{\mu }^{2}}}}}\]

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