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Incline Plane versus Banked Curve

  1. Jan 17, 2007 #1
    Okay, so I've been working on incline plane and then banked curve problems. From the incline plane problems and the definitions in my book I believed that normal force was just the y component of the weight, ie the weight times the cos of the angle of incline and no x component, ie parallel to the surface. This assumptions worked for the inclined plane problems, but when I got to the banked curve problems, a car going in a circular banked path with now friction, the centripetal force was now the normal force times sin of the angle of incline, the "x-component" of the normal force. Can any one explain?
     
  2. jcsd
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