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Banked curve

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A curve with a 135 m radius on a level road is banked at the correct angle for a speed of 20m/s.
    If an automobile rounds this curve at 30m/s, what is the minimum coefficient of static friction between tires and road needed to prevent skidding?


    2. Relevant equations

    N=mg(cos\theta)
    mgsin(\theta)=centripetal force=mv^2/r

    Quite confused with the workings for this question, anyone can help?
     
  2. jcsd
  3. Feb 13, 2009 #2

    Kurdt

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    Since the curve is banked for an optimum velocity of 20m/s that means that the centrifugal force is cancelled by the component of the normal force which acts along the horizontal. Can you work out what angle the curve must be banked at for that to occur? Draw a free body diagram with the forces to help you.
     
  4. Feb 13, 2009 #3
    The angle that i got is 16.8degrees
    After getting the angle, do i take the static force/normal force?
    Do i take the value of the centrifugal force as my static force?
    Also when should i use 30m/s?
     
  5. Feb 13, 2009 #4

    Kurdt

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    OK so you know the banking angle of the curve. Now what forces will be present when the car takes the curve at the wrong velocity? What are the forces parallel to the bank's surface? How much force will stop the car sliding off the track at that velocity?
     
  6. Feb 13, 2009 #5
    The forces that will be present are normal force, friction, weight and centrifugal force.
    Forces that are parallel to bank's surface would be centrifugal and friction.
    To calculate the force that prevented the car from sliding, do i use the centrifugal force-static force?
     
  7. Feb 13, 2009 #6

    Kurdt

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    Can you draw a free body diagram and analyse the horizontal and vertical components with the friction present?
     
  8. Feb 13, 2009 #7

    tiny-tim

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    Hey guys! :smile:

    why are you talking about centrifugal force? :confused:

    it's centripetal acceleration, times mass, and it equals the other horizontal forces! :wink:
     
  9. Feb 13, 2009 #8

    Kurdt

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    Indeed. I had a 50/50 and got it wrong. :redface:
     
  10. Feb 13, 2009 #9
    Sorry i have made mistakes in my previous reply. I hope i'm not wrong this time wrong. The horizontal forces are centripetal force, friction and centrifugal force. Whereas the vertical force would be the weight and the normal and the vertical friction.

    To prevent the car from sliding up, there must be a force acting in opposite direction and that would be friction. Since the car is sliding up, thus the friction must be acting towards the left? Therefore centripetal force + friction = centrifugal force?

    Is this correct?

    Horizontal: Nsin(a) + Fscos(a) = mv^2/r
    Vertical: mg + Fssin(a) = Ncos(a)
     
    Last edited: Feb 13, 2009
  11. Feb 13, 2009 #10

    Kurdt

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    Its entirely my fault actually, no need to apologise.

    Your horizontal and vertical equations are fine. Substitute for the friction force in terms of the coefficient of friction into the equations and solve them by eliminating a variable. You can them manipulate the equations to find the coefficient of friction.

    Just a note on centrfugal and centripetal as tiny-tim pointed out. Centripetal force is a real kinematic force directed toward the centre of circular motion, whereas centrifugal force is a pseudo-force to do with rotating reference frames.
     
  12. Feb 13, 2009 #11

    tiny-tim

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    centripetal acceleration

    Hi zerogoal! :smile:

    The left hand side of your equation are forces …

    the right hand side is (mass times) centripetal acceleration …

    instead of the usual Newton's second law equation, in which the acceleration is zero, and so the forces have to add to zero,

    in this case there is an acceleration, and the forces have to add to that acceleration (times mass) …

    Fnet = ma

    there is no centrifugal force …

    there is no centripetal force …

    there is only centripetal acceleration. :smile:
     
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