What is the Coefficient of Kinetic Friction for a Box Pushed at an Angle?

[physicsphail]

A box of books weighing 325N moves with a constant velocity across the floor when it is pushed with a force with 425N exerted downward at an angle of 35.2^o below the horizontal. Find u_k between the box and the floor.

Please help me I have a test soon and I'm not exactly understanding how to solve that. I skipped a few classes because I traveled for a volleyball tournament.

 

[LowlyPion]

A box of books weighing 325N moves with a constant velocity across the floor when it is pushed with a force with 425N exerted downward at an angle of 35.2^o below the horizontal. Find u_k between the box and the floor.

Please help me I have a test soon and I'm not exactly understanding how to solve that. I skipped a few classes because I traveled for a volleyball tournament.

Welcome to PF.

Draw a force diagram.

Your box exerts a downward force on the floor.

The force at 35.2° has 2 components - 1 is horizontal and the other is vertical down.

It moves at constant velocity so there is no net force horizontally. The horizontal component of the force is equal then to the weight down and the component down both multiplied by the coefficient of friction.

 

[thomate1]

I would approach this problem by first understanding the concept of kinetic friction. Kinetic friction is the force that opposes the motion of an object when it is in contact with a surface. It is dependent on the coefficient of kinetic friction (uk), which is a measure of the roughness of the surfaces in contact.

To solve this problem, we need to use the formula for calculating kinetic friction:

Fk = uk * N

Where Fk is the force of kinetic friction, uk is the coefficient of kinetic friction, and N is the normal force (perpendicular force) exerted by the surface on the object.

In this scenario, the normal force is equal to the weight of the box, which is 325N. The force pushing the box is 425N at an angle of 35.2o below the horizontal. To find the horizontal component of this force, we can use trigonometry:

Fh = 425N * cos(35.2o) = 347.6N

Now, we can plug in the values into our formula:

Fk = uk * N
347.6N = uk * 325N

Solving for uk, we get:

uk = 347.6N / 325N = 1.07

Therefore, the coefficient of kinetic friction between the box and the floor is 1.07. This means that the surfaces in contact are quite rough, causing a high amount of friction that opposes the motion of the box. To reduce this friction, we can try using a lubricant or smoothing out the surfaces. I hope this helps with your understanding and preparation for your test. Good luck!