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Resultant Force for circular motion on a banked track

  1. Feb 25, 2008 #1
    [SOLVED] Resultant Force for circular motion on a banked track

    1. The problem statement, all variables and given/known data
    A car travels 77 m/s around a circular track of radius 71.9
    mass of car= 2800 kg
    coefficient of friction = .1
    angle of track with horizontal = 22 degrees
    acceleration of gravity = 9.8 m/s^2
    What is the magnitue pf the resultant force on the 2800 kg driver and his car?

    2. Relevant equations
    normal force= mgcos(angle)
    force of friction = (normal force)(coefficient of friction)

    3. The attempt at a solution
    i know that the force of friction and normal force do not create the centripetal force necessary to cause uniform circular motion as
    Fn= 2800(9.8)cos(22)=25441.9N
    Ff= 25441.9(.1)=2544.19N
    and the necessary centripetal force= mv^2/r= 2800(77)^2/71.9= 230892.9068N

    so i tried making the resultant= the horizontal component of Ff+horizontal component of Fn= cos(22)(2544.19)+sin(22)(25441.9)= 11889.6N which was not the right answer
    Last edited: Feb 25, 2008
  2. jcsd
  3. Feb 25, 2008 #2


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    Gold Member

    You may ignore friction. At any moment the force will be the resultant of the centripetal force and gravity. The question isn't about whether the car will stick.
  4. Feb 25, 2008 #3
    never mind, the answer was just centripetal force
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