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Why does vertical component of normal force

  1. Oct 31, 2005 #1
    For car to be able to drive trough banked curve on the road with no friction, the vertical component of a normal force must be equal in size but opposite in direction to the force of gravity.

    Horizontal component F[h]

    F[h] = m * g * tan(alpha) = m * v^2 / r

    So there is only one speed at which F[h] will equal centripetal force at specific angle of the curve and and specific radius.
    But why does vertical component of normal force equal the force of gravity when F(h) = m * v^2 / r ?

  2. jcsd
  3. Oct 31, 2005 #2

    Chi Meson

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    The vertical componant of the normal force has to balance the weight, otherwise the car will move up or down the banked curve, thus changing the radius of the turn. There is only one speed at which F[h] will provide centripetal force while simultaneously F[v] will balance weight. That's why this problem is so unrealistic
  4. Nov 1, 2005 #3
    What do you mean by unrealistic?

    Forget that you ever heard of car making a turn on a banked curve.
    You only know that for a car to make that turn:

    A- vertical force must cancel out force of gravity

    B- only possible centripetal force at certain angle of a curve and certain radius must equal m * v^2 / r and thus speed must be v^2=r*g*tan(alpha).

    But your problem is that you don't know if A and B can happen at the same time ( obviously they do since cars do make the turns but suppose you never knew that )
    Based on what would you figure out if those two things can happen at the same time?
    Last edited: Nov 1, 2005
  5. Nov 1, 2005 #4


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    First of all the problem is unrealistic because you always have to stay at the same radius and cause there's no friction.
    >How do you tell if the two things can happen? - make two equations and compare them.
  6. Nov 1, 2005 #5

    Chi Meson

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    Perfect uniform circular motion is very difficult. For constant speed in a perfect circle to happen, the following things must be in perfect harmony:

    the relationship between force, mass, speed, and radius of curvature MUST be exactly thus: F = (mv^2)/r . If any one of these variables is a tiny bit higher or lower, then there is no circular motion. It's a nearly impossible balance to achieve for anything other than a very controlled experimental setup.

    Also, the force must contiunally change direction so as to always be exacly perpendicular to the direction of motion at one instant. This is also nearly impossible to attain.

    For the car to turn on a *Frictionless* banked turn at a constant radius, it must start the turn at the perfect speed, and remain at this perfect speed, that is predetermined by the given situation. The given situation (through balancing two necessary formulas, one for the vertical compnant, one for the horizontal componant of normal force) dictates that there will be only one speed for a given radius on a given incline where the exact centripetal force will be supplied AND the exact balancing force to gravity will be supplied by the normal force.

    It's possible, but there is no real situation that uses this, hence: "unrealistic." A real banked turn will use static friction to allow for other speeds.
  7. Nov 1, 2005 #6
    I forgot that normal force (when equal to gravity force) is already included in

    F[h]=m*g*tan(alpha) with N*sin(alpha)= ( m*g/cos(alpha) ) * sin(alpha)

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