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 0 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 `a 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 diagram shown in figure

(a ) For bx just starts to sliding down slope
`f = mg sin theta`
`N= m g cos theta`
`(f )/(N ) = tan theta`
`mu = tan theta`
`theta = tan ^(-1) (mu)`
( b) When angle of inclination increased to
`alpha gt theta`
Resultant force be `F_(1)`
`F_(1) = mg sin alpha - f`
(c ) To keep box stationary or moving with uniform speed upward force ir required Here friction force would bed in downward direction
(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
`F_(3) = mg (sin alpha + mu cos alpha ) + ma`