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 normal reaction and the frictional force acting on the object is given by F, where $F$ is, (1) $|\vec{F}|=\mathrm{mg}$ (2) $\overrightarrow{\mid} \mid=m g+\mu m g$ (3) $|\mathrm{F}|=\mu \mathrm{mg}$ (4) $|\vec{F}| \leq m g \sqrt{1+\mu^{2}}$ - Noon Academy

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 normal reaction and the frictional force acting on the object is given by F, where $F$ is, (1) $|\vec{F}|=\mathrm{mg}$ (2) $\overrightarrow{\mid} \mid=m g+\mu m g$ (3) $|\mathrm{F}|=\mu \mathrm{mg}$ (4) $|\vec{F}| \leq m g \sqrt{1+\mu^{2}}$

Physics | Physics

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 normal reaction and the frictional force acting on the object is given by F, where $F$ is, (1) $|\vec{F}|=\mathrm{mg}$ (2) $\overrightarrow{\mid} \mid=m g+\mu m g$ (3) $|\mathrm{F}|=\mu \mathrm{mg}$ (4) $|\vec{F}| \leq m g \sqrt{1+\mu^{2}}$

Answer (4)
Sol. Since body does not move hence it is in equilibrium.
$\mathrm{f}_{\mathrm{r}}=$ frictional force which is less than or equal to limiting friction.
Now $\mathrm{N}=\mathrm{mg}$
Hence $\overrightarrow{\mathbf{F}}=\overline{\mathbf{N}}+\overline{\mathbf{f}}_{r}$ $|\vec{F}| \leq(\mathrm{mg})^{2}+(\mu m g)^{2}$ $|\vec{F}| \leq m g \sqrt{1+\mu^{2}}$

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