A factory worker pushes a 31.5-kg crate a distance of 4.4 m along a level floor at constant velocity by pushing horizontally on it

kinetic-friction
magnitude-of-force

#1

A factory worker pushes a 31.5-kg crate a distance of 4.4 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.23. Suppose the worker pushes downward at and angle of 31° below horizontal.
What magnitude of force must the worker apply?

Answer:

the horizontal component of the pushing force (F) must overcome the frictional force (f)
… the frictional force is increased by the downward pushing force

f = [(31.5 * g) + (F sin(31º))] * 0.23

F cos(31º) = f

F cos(31º) = [(31.5 * g) + (F sin(31º))] * 0.23

F cos(31º) = (31.5 * g * 0.23) + (F * sin(31º) * 0.23)

F cos(31º) - (F * sin(31º) * 0.23) = (31.5 * g * 0.23)

F [cos(31º) - (sin(31º) * 0.23)] = (31.5 * g * 0.23)

F = (31.5 * g * 0.23) / [cos(31º) - (sin(31º) * 0.23)]