A box of mass $2 \mathrm{~kg}$ is placed on a inclined plane that makes $30^{\circ}$ with the horizontal. The coefficient of friction between the box and inclined plane is 0.2 . A force $F$ is applied on the box perpendicular to incline to prevent the box from sliding down. The minimum value of $F$ is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

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Aicharya · Accepted Answer

As, block has a tendency to slide downwards, friction acts in upward direction. Box will not slide if friction balances component of weight acting in downward direction. $\Rightarrow \quad f \geq m g \sin \theta$ Here, $f=$ friction $=\mu N$ or $\quad f=\mu[(m g \cos \theta)+F]$ As, applied force is perpendicular to surface of incline. Normal reaction, $N=$ net perpendicular to surface force $ =m g \cos \theta+F $ Hence, by Eqs. (i) and (ii)we have $ \begin{aligned} & \mu(m g \cos \theta)+\mu F \geq m g \sin \theta \\ & \text { or } F \geq \frac{m g \sin \theta-\mu m g \cos \theta}{\mu} \end{aligned} $ here, $ \begin{aligned} & m=2 \mathrm{~kg} \\ & g=10 \mathrm{~m} / \mathrm{s}^2 \end{aligned} $ $ \begin{gathered} \sin \theta=\sin 30^{\circ}=\frac{1}{2} \\ \mu=0.2 \end{gathered} $ So, $F \geq\left(\frac{2 \times 10 \times \frac{1}{2}-0.2 \times 2 \times 10 \times \frac{\sqrt{3}}{2}}{0.2}\right)$ $ \begin{aligned} & \Rightarrow \quad F \geq 50-17.32 \\ & \Rightarrow \quad F \geq 3267 \end{aligned} $ or $F \geq 327$ Hence, minimum value of $F$ is $32.7 \mathrm{~N}$

A box of mass $2 \mathrm{~kg}$ is placed on a inclined plane that makes $30^{\circ}$ with the horizontal.…

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