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Deriving Formula Help

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Derive the formula to calculate the angle at which the road should be banked to prevent the car from skidding(Ignore force of friction).

    2. Relevant equations
    tan x = v^2/gr ----- answer
    mv^2/r = fc

    3. The attempt at a solution
    I get how they derived the formula due to this website http://batesvilleinschools.com/physics/phynet/mechanics/circular%20motion/banked_no_friction.htm [Broken]

    My question is how can I do it, by putting mg in terms of horizontal and vertical components, instead of Fn.(I know the website explains why it makes more sense to do it that way, but I still want to know).

    My attempt:

    Forces in Y = 0:
    Fn - mg cos x = 0
    Fn = mg cos x

    Forces in X:
    mg sin x = mv^2/r
    g sin x = v^2/r
    sin x = v^2 / gr --stuck not the same answer

    If someone can please explain what I am doing wrong much help appreciated.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 24, 2013 #2


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    Hi, Brute. Welcome to PF!
    If the Y axis is perpendicular to the plane (road), then the sum of the forces in the Y direction will not be zero. The car is going in a circle at constant speed. Think about the direction of the acceleration of the car and then think about whether or not that acceleration has a non-zero Y-component.
    Is v2/r the X-component of acceleration?
    Last edited: Mar 24, 2013
  4. Mar 24, 2013 #3
    Im not sure if I understand you correctly. But what i got is centripetal acceleration goes towards the circle, making it not the X-component of acc. But(assuming I understood right) it goes directly opposite the Fn force.

    So after I drew a new diagram I got:

    Fn = mv^2/r + mg cos x

    Fx = 0

    mg sin x = 0

    But I am totally stuck after that.(Thank you for the quick response)
  5. Mar 24, 2013 #4


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    Yes, the acceleration is toward the center of the circle. How is the direction of "towards the center" related to the direction of your X and Y axes? (Towards the center is not directly opposite Fn.)
  6. Mar 24, 2013 #5
    Im so confused now, been working on it for over 2 hours almost in total. I'm assuming that there is a X and Y component of centripetal force that I don't understand, but I'm unsure of how I would draw it. I also realize that Fnet is equal to mv^2/r but I don't know what to do now. If you could assist me a bit more it would I would be very thankful.
  7. Mar 24, 2013 #6


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    In the attached figure, imagine drawing the acceleration vector of the car. Which way would it point? What angle does the acceleration make to the X axis? How would you find the X and Y components of the acceleration?

    Attached Files:

  8. Mar 24, 2013 #7
    Hopefully I go this right.

    For the forces in Y:

    Fn = mv^2/r sin X + mg cos X

    For forces in X:

    Fn Tan x = mv^2/ r cos x + mg sin x
  9. Mar 24, 2013 #8


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    That looks correct.

    For forces in X:
    How did you get the term on the left? Does Fn have an X component?
  10. Mar 24, 2013 #9
    Ohh yeah I see it now so
    For forces in the x:

    mg sin x = mv^2/r cos x
  11. Mar 24, 2013 #10


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    Yes, that's right. If you use this equation to find the angle in terms of the speed, does it give the correct answer?
  12. Mar 24, 2013 #11
    OMG your a genius, it works perfectly. Thanks for the all the help man, this made me understand a lot of other problems too.
  13. Mar 24, 2013 #12


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    Good. As a general rule of thumb, if you're applying Newton's law to an object and if you know the direction of the acceleration, it's best to choose one of the coordinate axes along the direction of the acceleration. Then you only have one nonzero component of acceleration and that usually makes working with the component equations easier. But as you've shown, you'll still be able to get the correct answer even if you orient your axes some other way.
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