Banking Angle, Normal/Centripetal Force?

[physicsma1391]

[SOLVED] Banking Angle, Normal/Centripetal Force?

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

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

 

[HallsofIvy]

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($\theta$).

 

[physicsma1391]

So how should my equation be set up?

Fc = Fnsin(theta)?

 

[cryptoguy]

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