A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2.

#### Solution

(a) To make the block move up the incline, the applied force should be equal and opposite to the net force acting down the incline.

Applied force = μR + 2g sin 30° (1)

(where μ is the coefficient of static friction)

R = mg cos 30°

Substituting the respective values in Equation (1), we get

`=0.2xx(9.8)sqrt3+2xx9.8xx(1/2)`

3.39 + 9.8 ≈ 13 N

With this minimum force, the body moves up the incline with a constant velocity as the net force on it is zero.

(b) Net force acting down the incline is given by

F = 2g sin 30° − μR

`=2xx9.8xx1/2-3.99`

= 6.41 N

Because F = 6.41 N, the body will move down the incline with acceleration, hence the force required is zero.