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Banking Angle, Normal/Centripetal Force?

  1. Jan 9, 2008 #1
    [SOLVED] Banking Angle, Normal/Centripetal Force?

    1. The problem statement, all variables and given/known data

    Curves are often banked on highways so that the horizontal component of the normal force provides the required centripetal force. What is the proper banking angle of a 1200 kg car making a turn of a radius of 30 m at a speed of 20 km/hr?

    2. Relevant equations

    I'm not sure how to get to the angle part of the problem, or if i am even doing this correctly.

    3. The attempt at a solution

    Fc = Fny - Fg
    [1200(20^2)] / 30 = Fny - 1200(9.81)
    Fny = 27772
  2. jcsd
  3. Jan 9, 2008 #2


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    Staff Emeritus
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    The normal component of force is the force perpendicular to the roadway. If the centripetal force (horizontal) is F and the bank angle is [itex]\theta[/itex], then the component of force perpendicular to the roadway is F sin([itex]\theta[/itex]).
  4. Jan 9, 2008 #3
    So how should my equation be set up?

    Fc = Fnsin(theta)?
  5. Jan 9, 2008 #4
    Yep. Since Fn sin(theta) is the only force acting to keep the object in circular motion, that is right
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