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Using Normal Force on inclined plane vs banked curve

  1. Oct 6, 2008 #1
    On an inclined plane, "[tex]\Sigma[/tex]Fy = N - mg cos ø = ma = 0".
    And "[tex]\Sigma[/tex]Fx = mg sin ø = ma" (frictionless).
    So, in the y direction, "N = mg cos ø".


    But on a banked curve, "[tex]\Sigma[/tex]Fy = N cos ø - mg = ma= 0".
    And "[tex]\Sigma[/tex]Fx = N sin ø = ma = mv²/r" (frictionless).
    So, in the y direction, "N cos ø = mg".
    I know how to get "tan ø = v²/rg".
    The two free body diagrams look identical to me so why doesn't "N = mg cos ø" (from inclined plane) work to get "tan ø = v²/rg"?
     
  2. jcsd
  3. Oct 7, 2008 #2

    Kurdt

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    On a banked curve, circular motion is normally involved and so there is an extra centrifugal force. I don't see the problem except you've neglected to take that into account on the free body diagram.
     
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