Banking Angle, Normal/Centripetal Force?

1. Jan 9, 2008

physicsma1391

[SOLVED] Banking Angle, Normal/Centripetal Force?

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

Curves are often banked on highways so that the horizontal component of the normal force provides the required centripetal force. What is the proper banking angle of a 1200 kg car making a turn of a radius of 30 m at a speed of 20 km/hr?

2. Relevant equations

I'm not sure how to get to the angle part of the problem, or if i am even doing this correctly.

3. The attempt at a solution

Fc = Fny - Fg
[1200(20^2)] / 30 = Fny - 1200(9.81)
Fny = 27772

2. Jan 9, 2008

HallsofIvy

Staff Emeritus
The normal component of force is the force perpendicular to the roadway. If the centripetal force (horizontal) is F and the bank angle is $\theta$, then the component of force perpendicular to the roadway is F sin($\theta$).

3. Jan 9, 2008

physicsma1391

So how should my equation be set up?

Fc = Fnsin(theta)?

4. Jan 9, 2008

cryptoguy

Yep. Since Fn sin(theta) is the only force acting to keep the object in circular motion, that is right