# The safely curve

1. Mar 27, 2007

### quickclick330

A car is safely negotiating an unbanked circular turn at a speed of 23 m/s. The maximum static frictional force acts on the tires. Suddenly a wet patch in the road reduces the maximum static friction force by a factor of three. If the car is to continue safely around the curve, to what speed must the driver slow the car?

i have it so where mu_s = v^2/r but I don't know if thats right, help please! thanks :-)

2. Mar 27, 2007

### PhanthomJay

I think you mean mu_s(g) = v^2/r. But in any case, if mu_s is reduced by 3, then v^2 must be reduced by what factor to keep the car from slipping?

3. Mar 27, 2007

### quickclick330

3/r maybe?

4. Mar 27, 2007

### mbrmbrg

The radius of the turn is constant. If you divide $$\mu_s$$ by 3, what must you do to balance your equation?

5. Mar 27, 2007

### quickclick330

then you must divide the other side of the equation to balance it, so would it be ((v^2/r)/3)?

6. Mar 27, 2007

### mbrmbrg

Exactly. But the only thing you're allowed to change is v.
What should you do to v to get the whole expression to equal $${\frac{v^2}{3r}$$?

7. Mar 27, 2007

### quickclick330

I'm not sure...not having the radius measurement is throwing me off

8. Mar 27, 2007

### mbrmbrg

It doesn't matter what the radius is. It doesn't change.
You have $$\frac{v^2}{r}$$
You need to divide all of that by three, but you're NOT allowed to touch r. All you're allowed to change is v. What might you do to v so that

$$\frac{v^2}{3r}=\frac{(v')^2}{r}$$ ?

9. Mar 29, 2007

### quickclick330

I tried to divide by 3 again...what is v prime??

10. Mar 29, 2007

### hage567

The speed the driver must slow down to in order to continue along the curve safely. This is what you are trying to solve for.

11. Apr 1, 2007

### mbrmbrg

v prime is your new velocity. The one that makes the origianl exprssion (v^2/r) get divided by 3.