# Homework Help: Static Friction Coefficient

1. Feb 23, 2010

### Bearbull24.5

1. The problem statement, all variables and given/known data To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 412 m. The car increases its speed at uniform rate of, at= 4.14m/s^2, until the tires start to skid. If the tires start to skid when the car reaches a speed of 37.7 m/s, what is the coefficient of static friction between the tires and the road? The acceleration of gravity is 9.8 m/s2.

2. Relevant equations
u(s)=(V^2)/(r*g)

3. The attempt at a solution

I believe the above equation is the one I am supposed to be using. I simply rearranged the equation to solve for Vmax to solve for the static coefficient. I have no idea why I need the tangential acceleration and I believe that 37.7 is the maximum velocity.

2. Feb 23, 2010

### kuruman

What must be true for the car to start skidding? Where does the formula that you plan to use come from?

3. Feb 23, 2010

### Bearbull24.5

For the car to start skidding doesnt that mean it overcomes the force of static friction? I got the formula from my notes and it was originally Vmax=sqrt((radius)(u(s))(g))

4. Feb 24, 2010

### kuruman

That is correct, but in what direction is the force of static friction that needs to be overcome in the present case?
Careful here! The formula from your notes was derived for a car that goes around a circle at constant speed, i.e. there is no tangential acceleration. When there is tangential acceleration, the net force of static friction that needs to be overcome is "mass times resultant magnitude of acceleration."

5. Feb 24, 2010

### Bearbull24.5

But there is no mass given

6. Feb 24, 2010

### kuruman

If no mass is given, call it m and proceed with the algebra. If you do things correctly, maybe you will not need to know the mass in the end.

7. Feb 24, 2010

### Bearbull24.5

Okay, I think I may be onto something here. f(s)=u(s)N. If I substitute ma(t) in for f(s) and solve for u(s) I end up getting that u(s) is equal to tangential acceleration times gravity since the mass in the forces cancel out.

8. Feb 24, 2010

### Bearbull24.5

Edit** It would be tangential acceleration divided by gravity

9. Feb 24, 2010

### Bearbull24.5

And that was not correct. Extremely lost, confused and only 2 more attempts left before I get no points at all for this problem

10. Feb 24, 2010