$A$ 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 $a$?

(N/A) Consider the diagram shown in the figure.
$(a)$ For the box to just start sliding down the slope:
$f = mg \sin \theta$
$N = mg \cos \theta$
Since $\mu = \frac{f}{N}$,we have $\mu = \frac{mg \sin \theta}{mg \cos \theta} = \tan \theta$.
Therefore,$\theta = \tan^{-1}(\mu)$.
$(b)$ When the angle of inclination is increased to $\alpha > \theta$,the net force $F_1$ acting down the plane is the component of gravity minus the kinetic friction:
$F_1 = mg \sin \alpha - f_k = mg \sin \alpha - \mu N = mg \sin \alpha - \mu mg \cos \alpha = mg(\sin \alpha - \mu \cos \alpha)$.
$(c)$ To keep the box stationary or moving with uniform speed upward,the applied force $F_2$ must overcome both the component of gravity and the friction force (which now acts downwards):
$F_2 = mg \sin \alpha + f_k = mg \sin \alpha + \mu mg \cos \alpha = mg(\sin \alpha + \mu \cos \alpha)$.
$(d)$ When the box is to be moved with acceleration $a$ upward along the plane,the net force $F_3$ must account for the acceleration:
$F_3 = mg(\sin \alpha + \mu \cos \alpha) + ma$.