A body of mass 5 kg is placed on a rough inclined plane of angle 30° and coefficient of friction √3/2. Find the force required to push the body downward at constant velocity.

Solution:

Weight component along incline:

mg sin30°

= 5 × 10 × 1/2

= 25 N

Normal reaction:

N = mg cos30°

= 50 × √3/2

= 25√3

Friction:

f = μN

= (√3/2)(25√3)

= 37.5 N

For constant velocity:

F + 25 = 37.5

F = 12.5 N

Answer: 12.5 N