# Sliding friction

1. Oct 21, 2007

### lim

1. The problem statement, all variables and given/known data

You are driving a 2090 kg car at a constant speed of 22.2 m/s along an icy, but straight and level road. While approaching a traffic light, it turns red. You slam on the breaks. Your wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 64.9 m. What is the coefficient of sliding friction (μ) between your tires and the icy roadbed?

2. Relevant equations

vf^2=vi^2=2ax
F=ma

3. The attempt at a solution

vf^2=vi^2=2ax
0= 22.2^2+2a64.9
a=-3.796 m/s^2

F= sum of forces/ W
= -3.796(2090)/2090(9.8)
= -.387
?

help appreciated

2. Oct 21, 2007

### Shooting Star

Frictional force = (mu)*N, so (mu)= F/(mg)

Why have you written F= sum of forces/W?

What is actually the magnitude of that thing which you are calling F?

3. Oct 21, 2007

### lim

It meant sigmaF=ma/mg. I put w, becuase w=mg. The magnitude of F, force was 3.796(2090)=7933.64 N. Is my method incorrect? I was trying to find the acceleration and use F=ma, and then I use (mu)= F/(mg).

4. Oct 21, 2007

### Shooting Star

Sum of forces = ma. That's all. Here, F=ma, where F is the force of friction.

ma= F= (mu)N = (mu)W = (mu)mg.

So, (mu) = (ma)/(mg) = a/g.

What you've found is the value of the co-eff of friction, with a -ve sign.

5. Oct 21, 2007

### lim

Just for understanding, Why isn't the value negative or why would it be the magnitude? Is it because friction is slowing down so it would always be a negative number?

6. Oct 21, 2007

### Shooting Star

The co-eff of friction is defined to be the ratio between the magnitudes of the frictional force and the normal reaction. So, we’ll consider only the positive value of the frictional force, and so the co-eff will always be a +ve number.

Upto finding ‘a’, you were correct. After that, you have written F= sum of forces/w, which is wrong.

What you should write is (mu)=F/N=ma/(mg)=a/g=0.378.

(Unknowingly, what you have calculated is nothing but Force/Normal Reaction, so you have actually calculated the value of the co-eff itself. But you have taken the force to be negative. That’s why I said it’s that value with a –ve sign.)

The answer is that the co-eff of friction is 0.378.

Is it clear now?