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

(a) As the box tends to slide down force of friction will act up the plane
As is clear from box will just start slidding down , when
` mg sin theta = f = mu R = mu mg cos theta`
or ` tan theta = mu or theta = tan^(-1) (mu) `
(b) When angle of inclination is increased to `alpha gt theta` then net force acting on the box down the plane is
` F_(1) = mg sin alpha - f = mg sin alpha + f = mg (sin alpha+ mu cos alpha)`
(In this case friction would act down the plane ) (d) If the box is to be moved with an upward acceleration a then upward force needed
` F_(3) = mg (sin alpha+ mu cos alpha) + ma `
.