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 $m$.
$(a)$ At what angle of inclination $\theta $ of the plane to the horizontal will the box just start to slide down the plane ?
$(b)$ What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to $\alpha > \theta $ ?
$(c)$ 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 ?
$d)$ What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$ ?
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 $m$.
$(a)$ At what angle of inclination $\theta $ of the plane to the horizontal will the box just start to slide down the plane ?
$(b)$ What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to $\alpha > \theta $ ?
$(c)$ 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 ?
$d)$ What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$ ?
Consider diagram shown in figure,
$(a)$ For box just starts to sliding down slope,
$f=m g \sin \theta$
$\mathrm{N}=m g \cos \theta$
$\tan \theta=\frac{f}{\mathrm{~N}}=\frac{m g \sin \theta}{m g \cos \theta}$
$\frac{f}{\mathrm{~N}}=\tan \theta$
$\frac{f}{\mathrm{~N}}=\mu$
$\mu=\tan \theta$
$\theta=\tan ^{-1}(\mu)$
$(b)$ When angle of inclination increased to $\alpha>\theta$. Resultant force be $F_{1}$
$\mathrm{F}_{1} =m g \sin \alpha-f$
f $=\mu \mathrm{N}$
$=\mu m g \cos \alpha$
$=m g \sin \alpha-\mu m g \cos \theta$
$=m g(\sin \alpha-\mu \cos \alpha)$
$(c)$ To keep box stationary or moving with uniform speed upward force ir required. Here friction force would be in downward direction.
$\mathrm{F}_{2} =m g \sin \alpha+f$
$=m g \sin \alpha+\mu \cos \alpha$
$=m g(\sin \alpha+\mu \cos \alpha)$
$(d)$ When box is to be moved with acceleration a upward along the plane net force be $F_{3}$ Friction would be in downward direction.
$\mathrm{F}_{3}=m \mathrm{~g}(\sin \alpha+\mu \cos \alpha)+m a$
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