# Homework Help: What is the coefficient of Friction?

1. Oct 26, 2009

### 1irishman

1. The problem statement, all variables and given/known data
It says: A 20kg object is placed on a horizontal surface. A force of 3N is required to keep the object moving at a constant speed. What is the coefficient of friction between the two surfaces?
I figured that we know that Ff=uFn and Fn=Fg=mg therefore Fn=20*9.8=196N That's as far as i can go...i am not sure what to do with the 3N force. The answer in the books says that u is .15

2. Relevant equations
Fn=Fg=mg
Ff=u*Fn

3. The attempt at a solution
Fn=20*9.8
=196N
I'm lost from here...not sure what to do to solve for coefficient of friction given horizontal 3N force etc. Help please?

2. Oct 26, 2009

### Mirole

Did you draw a FBD?

3. Oct 26, 2009

### 1irishman

Yes I drew a free body diagram, but it does not seem to be helping me.

4. Oct 26, 2009

### Mirole

Are you sure it's 20kg and not 2.0kg?

Because:
Fpush - fk = 0
3 = mu_k*mg
3 = mu_k*(2.0)(9.8)

Now, using 20:
Fpush - fk = 0
3 = mu_k*mg
3 = mu_k*(20)(9.8)
mu_k = .015

Also, 20kg(44.1 pounds) would need a bit more than 3 newtons of force to push it constantly, or at least I would think.

Last edited: Oct 26, 2009
5. Oct 26, 2009

### 1irishman

Yes, it is 20kg in the text, but this text has numerous typing errors so it looks like your mass is correct at 2.0kg based on the answer of .15 for the u value being the correct answer listed in the text. Thanks for your help. Just a few more questions if I may?
Is the Fpush - fk = ma on the left hand side of the equation the Fnet value? What does mu_k represent? Thank you.

6. Oct 26, 2009

### Mirole

Yes, Fpush - fk is Fnet, and since the box is moving at a constant speed, ma = 0. So, Fpush - fk = 0, you know the rest.

mu_k is the co-efficient of friction, μk. μ is mu.

7. Oct 26, 2009

### 1irishman

and fk=uFpush right?

8. Oct 26, 2009

### Mirole

No, because fk = μkn.
Where n is the normal force.
Using the two vertical forces from your FBD:
(Fnet)y = n - Fg = 0(since your object is not moving up or down.)
(Fnet)y = n = Fg => n = mg
So, fk = (μk)(mg)