# Coefficient of Friction formula

1. Mar 29, 2007

### zmike

I am trying to do some coefficient of friction questions but I am very confused with this concept. I don't understand the purpose of finding the coefficient of friction (is there an anology that can help me understand it?)?

and

I am unsure of the units to use for the result of the formula Ff/N (shouldn't it be in newtons but doesn't newtons cancel out when you divide the two forces?)?

thanks

2. Mar 29, 2007

### BishopUser

Perhaps you could list a problem that you are having trouble with and explain what you don't understand.

The frictional force is calculated by F = (mu)*N where mu is the coefficient of friction (either kinetic or static depending on the situation) and N is the normal force. Coefficients of friction are unitless and N is already in newtons so the units work out. The higher the coefficient of friction is the higher the frictional force will be (if N remains constant). The coefficient of friction is usually unique between two different materials (rubber on cement, ice on ice, wood on rubber, etc). Why is coefficient of friction important? Well, if you were an engineer working for a tire company you would probably want to experiment with different materials that gave you the highest coefficient of friction against snow/pavement/mud so that you could find the safest tire. There are many applications for where coefficients of friction would be important.

3. Mar 31, 2007

### zmike

Theses were the 2 problems, I had problems with.

1. Determine the coefficient of friction. It takes 59 N to move a 22 kg leather case (static friction).

Since normal force is = gravity force, 22*9.81=215.82 N

Ff=215.82 N *u - I got stuck here

2. Determine the magnitude of Fk of the average acceleration and stopping distance during the skid of a car (values know below).

V1= 26.8m/s
V2=0
Distance= 39.3 m
mass=1580
u=1.07

I got the acceleration of 9.14 m/s^2 then I got the average stopping force with the friction formula and got 1.44 x 10^4 but I don't know what to do from here.

thanks again

Last edited: Mar 31, 2007
4. Mar 31, 2007

### zmike

I solved the 1st one but I still can't figure out 2.

2. Determine the magnitude of Fk of the average acceleration and stopping distance during the skid of a car (values know below).

V1= 26.8m/s
V2=0
Distance= 39.3 m
mass=1580
u=1.07

I got the acceleration of 9.14 m/s^2 then I got the average stopping force with the friction formula and got 1.44 x 10^4 but I don't know what to do from here.

5. Mar 31, 2007

### denverdoc

re part 2)

the 9.14 looks good. The frictional force is as you posted above:

1580*g*1.07

Last edited: Mar 31, 2007
6. Apr 1, 2007

### zmike

But how do I find out the stopping distance during the skid of a car. The distance I have is before the car skids.

Last edited: Apr 1, 2007
7. Apr 1, 2007

### Feldoh

It's not. Since it's a stationary (non-moving) object, according to Newton your Net Force is going to be 0.

Last edited: Apr 1, 2007
8. Apr 1, 2007

### denverdoc

the way these types of problems are worded is that the applied force, that just sufficient to make it to move, is equal to the frictional force. Other problems that specify an acceleration, displatement, etc are done by newtons second usu using kinetic coefficient.. In this case the static friction is given so you want to apply just enuf force to overcome the frictional resistance.
ma=Fapp-Ff=0.
59=9.8*22*mu
mu=?

Last edited: Apr 1, 2007