A body of mass m is kept on a rough horizonta | vedclass

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(C) The normal reaction $N$ acting on the body is $N = mg$.The frictional force $f$ acting on the body is a static frictional force,which balances the

Vedclass · Accepted Answer

$|\overrightarrow{F}| \leq mg \sqrt{1 + \mu^{2}}$

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(C) The normal reaction $N$ acting on the body is $N = mg$.The frictional force $f$ acting on the body is a static frictional force,which balances the applied horizontal force. Since the body does not move,$f \leq \mu N = \mu mg$.The resultant force $F$ of the normal reaction $N$ and the frictional force $f$ is given by the vector sum $\overrightarrow{F} = \overrightarrow{N} + \overrightarrow{f}$.Since $N$ and $f$ are perpendicular to each other,the magnitude of the resultant is $|\overrightarrow{F}| = \sqrt{N^{2} + f^{2}}$.Substituting $N = mg$ and $f \leq \mu mg$,we get $|\overrightarrow{F}| = \sqrt{(mg)^{2} + f^{2}}$.Since $f^{2} \leq (\mu mg)^{2}$,it follows that $|\overrightarrow{F}| \leq \sqrt{(mg)^{2} + (\mu mg)^{2}} = mg \sqrt{1 + \mu^{2}}$.

$A$ body of mass $m$ is kept on a rough horizontal surface (coefficient of friction $= \mu$). $A$ horizontal force is applied on the body,but it does not move. The resultant of the normal reaction and the frictional force acting on the object is given by $F$,where $F$ is:

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