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Minimum Radius of Curvature

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data
    A civil engineer is asked to design a curved
    section of roadway that meets the following
    With ice on the road, when the coefficient of
    static friction between the road and rubber is
    0.1, a car at rest must not slide into the ditch
    and a car traveling less than 80 km/h must
    not skid to the outside of the curve.
    The acceleration of gravity is 9.81 m/s2 .
    At what angle should the road be banked?

    What is the minimum radius of curvature of
    the curve?
    Answer in units of m
    2. Relevant equations

    3. The attempt at a solution
    I already got the angle which is 5.7105931. What I couldn't get is the radius.
    I drew the diagram and I switched the diagram, so that Ff is going to left, Fn is going straight up, and gravity is going south-west at the angle of 5.

    I set up the equation as:
    Fc = (m*v^2)/r = Ff + mgsin (5.7105931)
    Ff = u(s) * Fn
    Fn = mgcos (5.7105931)

    So m cancels out:

    (v^2)/r = u(s) * gcos (5.7105931) + gsin (5.7105931)

    Last edited: Sep 27, 2011
  2. jcsd
  3. Sep 27, 2011 #2
    :D, better be someone you already hate, right :)?
    Did you remember to convert your velocity to m/s?
    The angle you found is correct, and I see no fault in your reasoning. Make sure you get 22.22 m/s for v.
    Let us know if that yields the right answer,
  4. Sep 27, 2011 #3
    Daniel I already converted that, but I still have no idea what is wrong T.T
    What is the answer you got??
    Last edited: Sep 27, 2011
  5. Sep 27, 2011 #4
    Roughly 253 m. What was yours?
  6. Sep 27, 2011 #5
    I got 253.2155189 so im guessing we got Same answers. And its still wrong. Sigh thank you for the help
  7. Sep 27, 2011 #6
    Are you sure you have everything properly stated?
    Perhaps you're using the wrong answer key.
    What do they claim is the right answer, by-the-way? Was the angle indeed correct?
  8. Sep 27, 2011 #7


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    When you drive round a banked track, the Normal reaction force is greater than when you park in a banked track.
    [In a similar way to the reaction force when you drive through a dip, which is stronger than when you park in the dip].
    Thus it is not sufficient to merely turn the friction around and retain the size. It will be higher!!
    Let me know if you need more help [this may have been told by some of the later posts - I only read the first few.]

    EDIT: Oh and the Normal Reaction force is of course perpendicular to the road, not vertical. That's why it is called the Normal Reaction Force. Normal as in perpendicular, not Normal as in "the usual thing".
  9. Sep 27, 2011 #8
    Yes it said angle is right but for radius it is wrong. I believe I stated and everything ... I hate online hw
  10. Sep 27, 2011 #9
    Wait... Normal force is bigger?? I do not understand this
  11. Sep 27, 2011 #10
    Hi Peter,
    It's best observed I think, that the normal force would be perpendicular to the Centrifugle force, and thuswise to the friction force as well, so none of their components, but for gravity could ever be factored in there...
    Or am I getting the geometry wrong?
  12. Sep 27, 2011 #11


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    I got 249 m [249.1788 but rounded] using the higher Normal Reaction force.


    If this worked you should see a diagram of what you need.

    The vertical components of the Normal Reaction Force and the Friction force have to balance the weight force, and the horizontal components of them equal the centripetal force - enagbling you to calculate R - using the speed of 22.222222 m/s of course.
    Last edited: Sep 27, 2011
  13. Sep 27, 2011 #12


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    I see no mention of a free body diagram in this thread.

    A free body diagram will be a big help. Try it.
  14. Sep 27, 2011 #13


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    Don't call it the Normal force!!!!!!!!!!!!! Call it by its correct name. The Normal Reaction force, and being a Reaction force it is as big or as small as it needs to be to maintain the situation you are in.
  15. Sep 27, 2011 #14
    Then how do i calculate that reaction normal force? Is it still perpendicular to Ff? Frankly i never heard of normal reaction force
  16. Sep 27, 2011 #15


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    I think the word you were trying for is centrifugal, and there is no such thing. And I don't understand your Geometry.
  17. Sep 27, 2011 #16
    Thank u for the diagram peter. I have that drawn exactly. Wait... Shouldnt it be Fn = Fgy + Ffy?? So Fncos (theta) =mg + Ffsin (theta) no?
    Why is it Fn + Ffy = Fg??
    Last edited: Sep 27, 2011
  18. Sep 27, 2011 #17


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    Note my edit to post #11
  19. Sep 27, 2011 #18
    Thank u for the diagram peter. I have that drawn exactly. Wait... Shouldnt it be Fn = Fgy + Ffy?? So Fncos (theta) =mg + Ffsin (theta) no?
    Why is it Fn + Ffy = Fg for u??
  20. Sep 27, 2011 #19
    Oh yes, do excuse my typo(keyboard issues, you see).
    And you could've effectively equated that with the Centripetal force, which should act on the same plane as the friction force, and thus perpendicular to the normal force.
    But then again, I may've misunderstood the whole thing, but with relation to the drawing you posted, it could still work.
    By-the-by, though the force be fictitious, that doesn't mean it simply doesn't exist. It has all sorts of useful application in rotating reference frames, so it can very well be incorporated.
  21. Sep 27, 2011 #20


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    The bit I highlited in red above is true.

    Not sure what you mean by the bit I made blue.

    That maths behind this - when using all these components is horrendous.

    It leads to the right answer but I was able to do it more simply without resolving the forces at all. Unfortunately I just can't explain it. I might try to prepare a power point. COuld take a day. It is 1:00 am here now. I am off to bed.
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