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 gt 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 ?

Explanation

Consider the adjacent diagram , force of friction on the box will act up the plane .
For the box to just starts sliding down mg
sin `theta = f = muN = mu "mg cos " theta`
or `" " tan theta = mu implies theta = tan^(-1) (mu)`
(b) When angle of inclination is increased to `alpha lt theta` , then
net force acting on the box , down the plane is
`F_(1) = " mg sin" alpha - f = "mg sin " alpha - mu N`
= mg (sin `alpha - mu cos alpha)`
(c) To keep the box either stationary or just move it up with uniform speed , upward force needed , `F_(2) = "mg sin" alpha + f = "mg" (sin alpha + mu cos alpha)` (In this case , friction would act down the plane .)
(d) In the box is to be moved with an upward acceleration a , then upward force needed , `F_(3) = mg ( sin alpha + mu cos alpha) `+ ma.