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Normal force on a banked curve?

  1. Feb 23, 2007 #1
    Normal force on a banked curve????

    In every textbook I've looked at, I can't find an answer to this question. When determining the centripetal force on an object on a banked curve, it is stated that the banking angle for a given speed and radius is found by tan θ = v^2/rg

    It is found as follows (there is an attachment as well):

    The normal force on the object is resolved into components. The x-component (the one providing the centripetal force) is:

    Fn * sin θ = mv^2/r (1)

    Then, the y component is set equal to mg and it is found that the normal force:

    Fn = mg / cos θ (2)

    Substitute Fn from (2) into (1) and get:

    tan θ = v^2/rg

    I'm fine with that.


    Here is where I'm confused...When resolving the forces of an object resting on an inclined plane:

    The component down and parallel to the plane due to gravity, Fp, is as follows:

    Fp = mg * sin θ

    The component representing the force of gravity into (perpendicular) to the plane is:

    mg * cos θ

    The normal force is equal to this component into the plane by Newton's 3rd Law, so Fn = mg * cos θ


    Why in the first scenario (banked curve) is Fn = mg / cos θ, HOWEVER, in the second (block on plane) Fn = mg * cos θ ????

    How can this be? There are two different values for Fn

    I'm stumped. I have no solution.

    Thank you for any help.

    Attached Files:

  2. jcsd
  3. Feb 23, 2007 #2
    I think I understand your question. In the second scenario, you assumed that the component of the force of gravity perpendicular to the plane is equal to the the normal force. Is this valid?
  4. Feb 23, 2007 #3
    Yes, isn't it?
  5. Feb 23, 2007 #4
    Is the object accelerating?
  6. Feb 23, 2007 #5
    In the first scenario, yes.

    In the second scenario, no.
  7. Feb 23, 2007 #6
    Sorry, I just caught that. But anyway, in the first case the normal force is larger than in the second case right? Why would that be?
  8. Feb 23, 2007 #7
    I don't know. I thought that they should be equal.

    Oh, is it because there is the component to the normal force that is actually responsible for the centripetal acceleration? Thus, if the object increases its speed, that component will increase even more, and then normal force will increase as well.

    If the above is correct, you've helped me.
  9. Feb 23, 2007 #8
    yes! good.

    of course we might have to bring friction into the picture if we want to generalize more.
  10. Feb 23, 2007 #9
    sheesh... I think the point was attained but the second situation of course needs friction if that object is at rest. sorry- time to sign off!
  11. Jan 18, 2010 #10
    Re: Normal force on a banked curve????

    I've been struggling with this question for well over an hour by this point!

    Thank you so much for having posted the same question.

    I was trying to find an answer in the Fundamentals of Physics text (Resnick), but was becoming quite frustrated.

    I think that the idea is summed up as follows: "the normal force will be greater than just the component of the gravitational force that is perpendicular to the surface (mgcosθ)."

    However, I am uncomfortable with this. Why exactly does this happen? Let's consider the example of a car turning on a banked curve.

    Is the normal force greater than mgcosθ because:

    a) the normal force also is responsible for the centripetal acceleration, so it needs to be greater?


    b) the car is not sliding down the curve, so the normal force is greater because of the translational equilibrium requirement?

    The textbook that I have (Resnick, Fundamentals of Physics) seems to suggest (b), since they use the equation for equilibrium in the Y direction.

    But if that is the case, then what is physically creating this "extra" normal force (as compared to a sliding car)? The problem setup states explicitly that friction is ignored. But then where else is the force coming from?

    Shouldn't there be a clear and precise way of calculating the normal force? This leaves me unsettled.
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