# Maximum velocity on a banking curve formula

1. Nov 19, 2008

### oranboron

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/imgmech/carbank.gif

I had a problem in my textbook asking me to find the maximum velocity for a car turning on a banked curve.

After i drew my freebody diagram i had the same formula for centripetal force as hyperphysics above (Fx).

However when i tried getting my result my answer was wrong.... I didn't know what i was doing wrong so i searched hyperphyics and found their vmax formula turned incredibly different from mine. Then i saw that they had a second equation. I was able to finally figure out that they isolated for the normal force, n = mg/(costheta - usintheta) in Fy and substituted it into the Fx formula.

BUT the value for the normal force is different from what i had before. I had n = mgcostheta

Why couldn't i use that value for N in my version of Fx = Fcentripetal ?

2. Nov 19, 2008

### Staff: Mentor

The normal force would only equal mg cosθ if there were no component of acceleration perpendicular to the incline, which is not the case here.

3. Nov 19, 2008

### oranboron

Ok, so if i understand it correctly. Since there is now a net force being applied to the centre and it has a component perpendicular to the incline. This perpendicular force itself is the value of N - mgcosθ ?

Thank you for your patience and help.

4. Nov 20, 2008

### Staff: Mentor

Right, that's the component of the net force perpendicular to the incline.