A rectangular box lies on a rough inclined surface. The coefficie… - Vidyadip

Question

A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $\mu$. Let the mass of the box be m.

At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a $\theta>$?
What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a?

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Answer

As the box just start to slide down the plane then $\mu=\tan\theta$ (by angle of repose) $\theta-\tan^{-1}(\mu)$
If angle$\alpha>\theta$ the angle of inclination of the plane with horizontal it will slide down (f upward) as $\theta$ is the angle of repose. So net force downward

$\text{F}_1=\text{mg}\sin\alpha-\text{f}=\text{mg}\sin\alpha-\mu\text{N}.$

$=\text{mg}\sin\alpha-\mu\ \text{mg}\cos\alpha$

$\text{F}_1=\text{mg}[\sin\alpha-\mu\cos\alpha]$

To keep the box either stationary or just move it up with uniform velocity (1 =0)upward (f downward)

$\text{F}_2-\text{mg}\sin\alpha-\text{f}=\text{ma}$

or $\text{F}_2-\text{mg}\sin\alpha-\mu\text{N}=0(\because\text{a}=0)$

$\text{F}_2=\text{mg}\sin\alpha-\mu\text{N}=0$

$\text{F}_2=\text{mg}(\sin\alpha-\mu\cos\alpha)$

The force applied $\text{F}_2$ to move the box upward with acceleration a

$\text{F}_3-\text{mg}\sin\alpha-\mu\cos\alpha=\text{ma}$

$\therefore\text{F}_3=\text{mg}(\sin\alpha+\mu\cos\alpha)+\text{ma}$

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