# Circular Curve of a Highway problem

1. Sep 25, 2008

### mossfan563

1. The problem statement, all variables and given/known data
A circular curve of highway is designed for traffic moving at 95 km/h. Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is 110 m, what is the correct angle of banking of the road?

2. Relevant equations
F_y = N sin(theta) = (mv^2)/r
F_x = N cos(theta) = mg

3. The attempt at a solution

I assumed that if N = mg then I could cancel out m from either equation since I don't know it initially.
Then I would be left with:
sin(theta) = ((mv^2)/r)/N
cos(theta) = (mg)/N

I used my sin equation. With v = 26.38 m/s and r = 110 m and g = 9.8 m/s^2
I ended up with: sin(theta) = .64589 which is the coefficient of friction.
Doing the inverse sin of that equation, I got an angle of 40.23 degrees.
It was incorrect. What am I doing wrong?

2. Sep 25, 2008

### tiny-tim

Hi mossfan563!
sin(theta) ? friction??

and what happened to g?

Hint: tan(theta) = … ?

3. Sep 25, 2008

### mossfan563

There is no friction.

4. Sep 25, 2008

### tiny-tim

i know!!

you mentioned it!

try the tan(theta) thing

5. Sep 25, 2008

### mossfan563

Sorry I thought LowlyPion replied to my question. I got an email with his response asking if there was friction or not.

Thanks for the hint! It worked!

6. Sep 25, 2008

### LowlyPion

Sorry. I withdrew my post when I saw someone else was helping you already. Too many cooks and all that. You found the Cosθ term ... so carry on.