Resultant Force Exerted by Surface on Block
Category : JEE Main & Advanced
In the above figure resultant force \[S=\sqrt{{{F}^{2}}+{{R}^{2}}}\]
\[S=\sqrt{{{(\mu mg)}^{2}}+{{(mg)}^{2}}}\] \[S=mg\sqrt{{{\mu }^{2}}+1}\]
when there is no friction \[(\mu =0)\] S will be minimum
i.e. S = mg
Hence the range of S can be given by, \[mg\le S\le mg\sqrt{{{\mu }^{2}}+1}\]
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