1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum velocity on a banking curve formula

  1. Nov 19, 2008 #1

    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 ?

    Thank you for your help.
  2. jcsd
  3. Nov 19, 2008 #2

    Doc Al

    User Avatar

    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.
  4. Nov 19, 2008 #3
    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.
  5. Nov 20, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Right, that's the component of the net force perpendicular to the incline.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Maximum velocity on a banking curve formula