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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

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    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

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    Staff: Mentor

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