Free body diagram friction homework

[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?

 

[quicknote]

I can provide a response to this question by using the principles of Newton's laws of motion and the concept of free body diagrams. In this scenario, the box is being pushed across the floor at a constant velocity, which means that the net force acting on the box is zero. This is because the box is not accelerating, therefore the sum of all the forces acting on it must be equal to zero.

Let's break down the forces acting on the box. We have the pushing force, which is directed downward at an angle beta below the horizontal. This force can be resolved into two components - one parallel to the floor and one perpendicular to the floor. The perpendicular component is balanced by the normal force exerted by the floor on the box, which is equal in magnitude but opposite in direction. This normal force is what prevents the box from falling through the floor.

The parallel component of the pushing force is what overcomes the force of kinetic friction between the box and the floor. The coefficient of kinetic friction, given as 0.41, tells us the ratio between the force of friction and the normal force. This means that the force of friction is equal to 0.41 times the normal force.

Now, if we consider the maximum value of the pushing force that can be applied without causing the box to move, we can set up an equation as follows:

Maximum pushing force = 0.41 x normal force

We know that the normal force is equal to the weight of the box, which is given by the mass of the box (m) multiplied by the acceleration due to gravity (g).

Therefore, the maximum pushing force can be expressed as:

Maximum pushing force = 0.41 x m x g

Now, let's consider the angle beta. When the angle beta is greater than a certain value, the maximum pushing force will not be enough to overcome the force of friction and the box will not move. This means that the maximum pushing force must be equal to the force of friction, which can be expressed as:

Maximum pushing force = force of friction = 0.41 x m x g

Setting these two equations equal to each other, we can solve for the angle beta:

0.41 x m x g = 0.41 x m x g x cos(beta)

Simplifying, we get:

cos(beta) = 1

This means that beta must be equal to 0 degrees, or in other words,