# 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

## Answers and Replies

Homework Helper
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