What is the coefficient of Friction?

[1irishman]

Homework Statement

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

Homework Equations

Fn=Fg=mg
Ff=u*Fn

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?

 

[Mirole]

Did you draw a FBD?

 

[1irishman]

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

 

[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)
mu_k = .15, which is your answer.

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.

 

[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.

 

[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.

 

[1irishman]

and fk=uFpush right?

 

[Mirole]

and fk=uFpush right?

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)