Maximum angle for pushing a box across floor with friction

[prettynerd]

while moving in, a new homeowner is pushing a box across the floor at a constant velocity. the coefficient of kinetic friction between the box and the floor is 0.41. the pushing force is directed downward at an angle beta below the horizontal. when beta is greater than a certain value, it is not possible to move the box, not matter how large the pushing foce is. find beta.


this is a bonus question i got for hwk. an equation or two might help. i drew a free body diagram. the question just confuses me.

thanks.

/prettynerd.

 

[Diane_]

The model used for friction says that the force is equal to the coefficient of friction times the normal force. The normal force, in this case, will be the weight of the box plus any perpendicular component of the applied force.

Now: if you think about it, the harder you push, the harder you're going to be pushing the box into the floor. Does that make sense? The steeper the angle, the more of the force you apply will simply go into pushing the box into the floor. The harder you push it into the floor, the greater the normal force. The greater the normal force, the greater the force of friction. At some point, every additional Newton of force you exert will do more to increase friction than it will to move the box across the floor. You're looking for the angle at which the frictional force is exactly equal to the parallel component of the applied force. Any angle greater than that will give you a friction force too large to be overcome, and the box doesn't move.

Does that make sense?