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Max speed of a car at a curve banked road.

  1. Mar 21, 2012 #1
    If given
    radius=r meters
    weight=mg
    Friction=μ
    Banking angle=θ

    For the y direction,
    NCosθ-mg-µNSinθ=0

    My question.
    What is the x-direction equation?
    I know its equal to mv^2/r.
    Always mixed up between "real" force like mg and acquired force like N and friction.
    If the car at rest on the banked road, N=MgCosθ.

    Thank you.
     
    Last edited: Mar 21, 2012
  2. jcsd
  3. Mar 22, 2012 #2

    tiny-tim

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    hi azizlwl! :smile:

    the important thing in these problems is to find the directions in which you know the acceleration

    one is horizontal radial …

    what is the other? :wink:
     
  4. Mar 22, 2012 #3
    thank you.
    yes i know the object is accelerating towards the center.
    What bother me the directions of frictional force and the weight of the object.
    In a equilibrium state on an inclined plane, the gravity(MgSin[x]) is pulling the object down the plane and the friction force(UMgCos[x] pull it up.
    But above situation the friction is towards the center means opposite direction when it in equilibrium state.
     
    Last edited: Mar 22, 2012
  5. Mar 22, 2012 #4

    tiny-tim

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    the vertical acceleration is zero, isn't it? :wink:

    so the extra equation you need comes from the vertical F = ma equation :smile:
     
  6. Mar 22, 2012 #5
    Then can I use the pseudo force, centrifugal force in this equation.
    Normal + friction= centrifugal force.
    Newton 3rd law. Action= Centripetal and reaction=centrifugal.

    Then i see the friction going out(not towards center) as in equilibrium state of object on inclined plane.
     
  7. Mar 22, 2012 #6

    tiny-tim

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    i'm not sure what you're doing :confused:

    and it's a very bad idea to use centrifugal force (unless you're in a rotating frame) …

    it'll confuse you, and it's difficult to know where to put the plus and minus signs :redface:

    always use centripetal acceleration and F = ma (in an inertial frame)

    show us your two F = ma equations, one vertical, and the other radial horizontal :smile:
     
  8. Mar 22, 2012 #7
    sorry for all the wandering. Try to fully understand.
    Another conflicting facts. I see the frictional force only cause by mg not N or its the same.

    A curve on a highway has a radius of curvature r . The curved road is banked
    at θ with the horizontal. If the coefficient of static friction is μ,
    (a) Obtain an expression for the maximum speed v with which a car can go
    over the curve without skidding.
    (b) Find v if r = 100 m, θ = 30◦, g = 9.8m/s2, μ = 0.25
    http://img29.imageshack.us/img29/998/bankedz.jpg [Broken]
    http://img341.imageshack.us/img341/5640/bank2k.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  9. Mar 22, 2012 #8

    PeterO

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    The statement in red is true, but unfortunately in this case the car is not at rest!!
     
  10. Mar 22, 2012 #9

    tiny-tim

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    hi azizlwl! :smile:

    yes, those equations (1) and (2) in your image are the two F =ma equations you need

    what is it about those two equations that's worrying you?​
     
  11. Mar 22, 2012 #10
    At rest friction =μmgCosθ
    But as Mr. Peter0 reply it is only applicable at rest not for above case.

    As you see from the equations from the diagram, the friction is constant=μMgCosθ. But intuitively i feel seated more deeply which shows exert more force to road which increases the friction.

    I've seen different equation too.
    NCosθ-mg-µNSinθ=0 ..here friction depends on N which also depends of (mv^2)/r.
     
    Last edited: Mar 22, 2012
  12. Mar 22, 2012 #11

    PeterO

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    The diagram does not show the friction force - which will be acting parallel to the slope, and down the slope.

    The three forces acting are
    weight - vertically down - that one is on your diagram.
    Normal Force - perpendicular to the slope - that is also on your diagram.
    Friction - parallel to the slope, down the slope.

    Your diagram best covers the case when no friction was needed at all, as W and N would add vectorially [join the arrows head to tail] to give the F you show.

    If you translate the N vector down until it is on top of the dotted green line, you will see what I mean.

    You need to have F larger than in your diagram.
    N is also larger, and will cross the F vector.
    When you add one more vector [friction] parallel to the slope, pointing down and left, you will have the three acting vectors adding to produce the required resultant - the centripetal force.
    Clearly then N will be longer than it is when the car is stationary.

    The size of the friction force is of course μN.
     
  13. Mar 22, 2012 #12

    tiny-tim

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    hmm … yes, that is different from the equation in your diagram :rolleyes:

    the equation in your diagram is wrong :redface:

    Ncosθ - mg - µsNsinθ = 0 is correct

    that's the vertical components of N, W, and µsN (at the maximum speed before slipping)
    i see … you feel that if we increase the speed, that increases the centripetal acceleration, and so N must also increase, to compensate?

    yes, you're correct! :smile:

    remember, the friction is not µsN

    it's ≤ µsN ! :wink:

    as we increase the speed, both the friction force and N increase to match the increasing centripetal acceleration, and the ratio friction/N increases …

    when that ratio reaches µs, that's the maximum speed :smile:

    moral: with questions like this, don't worry about what happens at lower speeds, just answer the question as asked … that'll be much simpler! :wink:
     
  14. Mar 22, 2012 #13
    And finally from Wikipedia which different from my first image. Which one is correct?
    http://img829.imageshack.us/img829/6870/bank3y.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  15. Mar 22, 2012 #14

    tiny-tim

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    wikipedia ! :smile:
     
  16. Mar 22, 2012 #15

    PeterO

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    I suppose it depends who put the posting on wikipedia !
     
    Last edited by a moderator: May 5, 2017
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