A body of mass 2 kg is situated on a rough inclined plane of inclination 30^ . Determine the magnitude of the force parallel to the incline needed to make the block move down the incline. The coefficient of static friction is 0.2 .

Step-by-step solution

The component of the weight of the block acting down the inclined plane is: W_ = mg 30^ W_ = 2 9.8 1 2 = 9.8 N The maximum static friction force acting up the incline is: f_ max = _s N = _s mg 30^ f_ max = 0.2 2 9.8 3 2 3.39 N Since the downward driving force ( 9.8 N ) is strictly greater than the maximum static friction ( 3.39 N ), the block will slide down the incline under its own weight. Therefore, no additional external force is required to make the block move down the incline. The required force is zero.