Question

An athlete trains by dragging a heavy load across a rough horizontal surface. The athlete exerts a force of magnitude f on the load at an angle of 25° to the horizontal. (a) once the load is moving at a steady speed, the average horizontal frictional force acting on the load is 470 n. Calculate the average value of f that will enable the load to move at constant speed.

Answer 1

The force applied f by the athlete under steady speed is 518.76 N

Explanation:

The force f applied by the user has two components which are given by,

vertical component = f sinФ

horizontal component = f cosФ

In case of steady speed the net horizontal force applied will be equal to the friction force.

Given friction force = 470 n,

Ф = load angle = 25°

Hence,

f cos 25 = 470

=> f = 470/cos 25

=> f = 470/0.906

=> f = 518.76 n.

Hence the total force applied under steady speed is 518.76 N.