A) \[F=Mg\]
B) \[F=\mu Mg\]
C) \[Mg\le F\le Mg\sqrt{1+{{\mu }^{2}}}\]
D) \[Mg\ge F\ge Mg\sqrt{1+{{\mu }^{2}}}\]
Correct Answer: C
Solution :
Maximum force applied \[F=\sqrt{{{f}^{2}}+{{R}^{2}}},\] where f = force of friction Since, \[f=\mu R\] Then, \[f=\sqrt{{{\mu }^{2}}{{R}^{2}}+{{R}^{2}}}=R\sqrt{{{\mu }^{2}}+1}\] Now, if \[\mu =0,\] \[F=R\]s which is minimum force \[\because \] \[R=mg\] \[\therefore \] \[Mg\le F\le Mg\sqrt{{{\mu }^{2}}+1}\]
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