A) \[F=mg\]
B) \[F=\mu Mgf\]
C) \[Mg\le F\le Mg\sqrt{1+{{\mu }^{2}}}\]
D) \[Mg\,\ge F\,\ge =\mu Mg\sqrt{1+{{\mu }^{2}}}\]
Correct Answer: C
Solution :
[c] Maximum force by surface when friction works \[\sqrt{{{f}^{2}}+{{R}^{2}}}=\sqrt{{{(\mu R)}^{2}}+{{R}^{2}}}=R\sqrt{{{\mu }^{2}}+1}\] Minimum force = R when there in 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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