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Car on a Banked Curve (no friction)

  1. Jun 10, 2010 #1

    If we have a car traveling in a circle on a banked curve without friction, the forces acting on the car would only be the normal force and gravitational force.

    Using a x-y coordinate system, the horizontal component of normal (Nsintheta) provides the centripetal force Fc and it's vertical component (Ncostheta) cancels out the force of gravity. Everything works out here.

    However, if you were to the draw the forces in a coordinate system that had axes parallel and perpendicular to the plane, you would find the normal force being canceled out by the force of gravity perpendicular to the plane (Fgcostheta) and leaving us with just a net force of the parallel force of gravity down the plane. What's stopping the car from sliding down the curve now and what's providing the centripetal force on this diagram?

    Thank you.
  2. jcsd
  3. Jun 11, 2010 #2
    This is where you were wrong.
    This component of gravity force is canceled, but the normal force isn't, which leaves a component perpendicular to the plane. It, along with the parallel one, forms the net force, which is the centripetal force.
  4. Jun 11, 2010 #3
    No matter what coordinate system you select you'e still going to get the normal force having a horizontal component of N sin theta (which must be equal to centripetal force mv^2/r) and a vertical component of N cos theta (which must be equal to the weight mg).

    A handy way to get the banking angle that you need, for a given radius of curvature and speed, is to divide those two equations, which results in tan theta = v^2/rg.
  5. Jun 11, 2010 #4
    The important thing here is to analyse why the vertical component of normal force is higher than gravity.
    Place a stationary block on the same banked curve. It will slide down. In that case the vertical component of normal force would get defeated by gravity. Drawing force diagrams isn't the only part of physics in this situation. You just found that it had misled you.
  6. Jun 11, 2010 #5

    Doc Al

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    Staff: Mentor

    No, the normal force isn't 'canceled out'. You are tacitly assuming that the acceleration perpendicular to the plane is zero, but it's not. The acceleration is horizontal, which has components parallel and perpendicular to the plane.
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