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Angle needed on banked corner of road.

  1. Jul 16, 2010 #1
    1. The problem statement, all variables and given/known data

    this example is in the txtbook so i know the answer, but i dont really get the second part.

    The road needs to be at a banked angle so the car travelling at 13.4m/s can go around with no friction needed, the radius of the corner is 35m.

    2. Relevant equations

    the given contructed equations are
    Fr=nsinө = (mv^2)/r i get this part, just the horizontal component of the normal force, which is the centripedal force.

    then they use
    Fy=ncosө – mg = 0
    which i can see from the diagram is the vertical component of the normal force, less the weight force of the car. Why do they use this though?

    they then divide the two equations.

    3. The attempt at a solution

    i got the first equation, and initially tried to write n, the normal force as mgcosө, and then substitute it into the first equation. Why is this not the normal force? like for an inclined plane example? and why do they subtract the car weight force from the verticle component of the normal force?

  2. jcsd
  3. Jul 16, 2010 #2
    Try N = mg/cos θ... :wink:
  4. Jul 16, 2010 #3


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    Is the car accelerating in the vertical (y) direction? If it is not accelerating in the y-direction, then the y-components of all the forces involved must add up to zero.
    The car is accelerating in the horizontal direction (which can be expressed as the resultant force). This resultant force has to come from somewhere. So the normal force not only involves the force of gravity (in the vertical direction) but the resultant force too (in the horizontal direction).

    I suggest starting with a free body diagram (FBD). By looking at the forces in the FBD, it should be more clear. :smile:
    In most inclined plane examples, the objects are not accelerating with respect to the surface of the plane (they may or may not be accelerating parallel to it, but there usually is no acceleration component perpendicular to the plane). This problem is different.
    Is the car accelerating up or down? :wink:
  5. Jul 16, 2010 #4
    thanks for the replies. No, the car is not accelerating in the verticle direction, so i can see why the forces in the verticle must equal zero. I have been looking at a link at the bottom of the page to a question posted on here over a year ago which was asking the same question i am.
  6. Jul 16, 2010 #5
    The first equation comes from centripetal force which you already figured out.
    The second equation is the equation to keep The Car from moving in the air.
    So, consider a box on floor => for that you would do N = mg
    In this case. N = mgcostheta. because you don't want car to fly from the banked surface.

    Just look at diagram(red thing is car(best i could draw in paint))
    first eqn. is eqn for x-comp.
    second eqn is eqn for y-comp

    Also note: Normal is always perpendicular to surface and Force due to Gravity is always downwards

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  7. Jul 16, 2010 #6


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    Sorry utkarsh1, but you haven't got your FBD quite right. The normal force is definately not mgcosθ. That would only be true if there was no other acceleration happening that has a component parallel to the normal force. In this problem, the car is accelerating such that the acceleration is not perpendicular to the normal force. So mgcosθ, by itself, is not the normal force for this problem.

    Lachlan1, Try this FBD (if your text book doesn't already have its own FBD).

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