# Friction Help finding the coefficent of static friction

1. Sep 15, 2008

### Jazzybelle4

1. The problem statement, all variables and given/known data
A force of 48.0 N is required to start a 5.0 kg box moving across a horizontal concrete floor. (a) what is the coefficient of static friction between the box and the floor?
(b) if the 48.0N force continues the box accelerates at 0.70m/sec2. what is the coefficient of kinetic friction

I have no idea how to set up up problem (a), all i know is the Ffr=us(mg)

2. Sep 15, 2008

### Kurdt

Staff Emeritus
You assume the force that moves the box is only just enough to overcome static friction and thus you can set the max force of static friction equal to the force applied to the box.

3. Sep 15, 2008

### Jazzybelle4

so then the equation set up would look like:
Ffr max= us(mg)
48N=us(mg)

4. Sep 15, 2008

### Kurdt

Staff Emeritus
Yes. You can then find the coefficient.

5. Sep 15, 2008

### Jazzybelle4

I found the coefficient it is .98 now i do not know what to do at all for the second part of the question (b)

6. Sep 15, 2008

### Kurdt

Staff Emeritus
What force would be required to accelerate the box at that rate without any friction?

7. Sep 15, 2008

### Jazzybelle4

any force about 48.0? there will always be friction. the coefficient needs to be smaller at .98

8. Sep 15, 2008

### Kurdt

Staff Emeritus
What about Newton's second law. You can work out the force required to make a mass the same as the box accelerate at the value you're given. How does that compare to the force you're having to use?

9. Sep 15, 2008

### Jazzybelle4

second law is
F=ma
48=m(.70)
I am so lost... my teacher doesn't explain anything

10. Sep 15, 2008

### Kurdt

Staff Emeritus
You're finding the force required to accelerate a 5 kg mass at 0.7 ms[su[]-2[/sup]. You don't need the 48 in there. Hang in there all will become clear.