A) \[F=Mg\]
B) \[F=\mu Mgf\]
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 by surface when friction works \[F=\sqrt{{{f}^{2}}+{{R}^{2}}}=\sqrt{{{(\mu R)}^{2}}+{{R}^{2}}}=R\sqrt{{{\mu }^{2}}+1}\] Minimum force \[=R\] when there is no friction Hence ranging from R to \[R\sqrt{{{\mu }^{2}}+1}\] We get, \[Mg\le F\le Mg\sqrt{{{\mu }^{2}}+1}\]
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