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Normal reaction force on banked track

  1. Jan 3, 2016 #1
    If a car us on a frictionless banked track what causes the normal reaction force to be greater than the weight force?

    My understanding from what I've read is that the vertical component of the reaction force cancels the weight force which makes sense as the object doesn't leave the ground. And I understand that the horizontal component of the normal force (parallel to the plane) acts down the plane and provides the centripetal acceleration for the car to continue around the track. The question I have is, what is it that causes the normal to have this extra magnitude? I have read many different answers for this and by and large it is ignored in most explanations but I would love to get my head around this to properly understand the situation. Many thanks in advance.
  2. jcsd
  3. Jan 3, 2016 #2
    Have you ever been on a carnival ride called the Round Up? You and others are on a flat floor surrounded by a cylindrical wall. The ride rotates about the axis, and you feel forced against the wall. At some point, the floor drops out. What causes the normal force that the wall exerts on you, and allows friction in the vertical to keep you from falling through the opening?

  4. Jan 3, 2016 #3
    "And I understand that the horizontal component of the normal force (parallel to the plane) acts down the plane "

    I think you have made a mistake here..the horizontal component of the normal force does NOT ACT DOWN THE PLANE. It acts horizontally towards the centre of the circle. this is the centripetal force and its value is determined by the velocity, the angle of the track and the radius of the circle.
  5. Jan 3, 2016 #4


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    The normal force is electromagnetic repulsion at the contact patch. It adjusts to whatever value is needed to prevent penetration of the bodies in contact.
  6. Jan 3, 2016 #5
    No, the horizontal component acts horizontally.
  7. Jan 3, 2016 #6
    Ok, you are both correct the horizontal component of the normal force acts towards the centre of the circle for centripetal acceleration, but this is not my question. My question is, If the normal force on a flat surface is equal in magnitude to the weight force, to stop the object falling through the surface, then where does this "extra" component of the normal force come from on an inclined plane? Mathematically, if Ncos(θ) = mg then what causes the Nsin(θ) component of the normal that acts to the centre of the circle?
  8. Jan 3, 2016 #7
    Have you drawn a free body diagram, or do you think you have advanced beyond the point where you need to draw free body diagrams? If you have a free body diagram, let's see it.
  9. Jan 3, 2016 #8
    I'm sure there is something fundamental I'm not grasping here but from all the explanations I've read (many differing) I can't get my head around it, FBD attached.

    Attached Files:

    • FBD2.png
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  10. Jan 3, 2016 #9


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    Is this a circular banked track or just a banked track?
  11. Jan 3, 2016 #10
  12. Jan 3, 2016 #11
    Is it possible to redraw this diagram with the blue arrow as the hypotenuse of the triangle and the upper red arrow horizontal. This is what is really happening. Why? You need a horizontal component of force to accelerate the car toward the center of the circular track (centripetal acceleration). So the normal reaction force from the track has to have a vertical component to balance the weight, and it has to have a horizontal component large enough to match the centripetal acceleration.

    For the normal force to be perpendicular to the frictionless track, its components must be at the exact same angle as the bank angle. This means that the centripetal acceleration has to be just right, meaning that the velocity of the car around the track must be just right. If the velocity is too slow (on a frictionless track), the car will slide down the slope. If the velocity is too high, the car will slide up the slope and over the side.
  13. Jan 3, 2016 #12
    I think this is where the confusion lies, you say the normal force has to have a horizontal component to match/provide the centripetal acceleration but I think I'm having trouble with the cause/effect here? I'm thinking from the point of view that the horizontal component of the normal is caused by something and that in turn provides the centripetal force? Or am I confusing the issue? Sorry if i'm completely lost here I just can't seem to find an explanation for this anywhere, it is always addressed as, the horizontal component of the normal provides the centripetal acceleration, I'm always left questioning how and why?
  14. Jan 3, 2016 #13
    The car always wants to go in a tangential straight line. But, if it did, it would go right through the side of the track. So it is the car pressing laterally against the side of the track that causes a 3rd law horizontal reaction force from the track on the car. This redirects the car in a circular path, rather than allowing the car to continue going in a tangent straight line.
  15. Jan 3, 2016 #14
    Ok, I think this is getting me close to an understanding, It's just hard to get my head around where that extra reaction force comes from? If there is no friction, it can't be friction from the road, and from what i can understand, it can't be due to a reaction from the weight force because the Ncosθ component cancels that out? Can you explain how the extra component comes about? Thanks heaps for this, I have been so frustrated trauling the internet/books for an answer.
  16. Jan 3, 2016 #15
    This is why I was asking you about the Round Up. Your body is trying to go in a straight line, right through the wall. But the wall is solid and circular, so it prevents you from going in a straight line. Instead, your body has to press against the wall to change direction, and the wall pushes back at you. It's like you are pushing off the wall. This accelerates you inward toward the center of rotation.
  17. Jan 3, 2016 #16
    Ok cool that sounds reasonable, so am I correct in saying that the component of the normal caused by gravity is exactly cancelled by the weight force, and the reason that there is a horizontal component of the normal providing centripetal acceleration is due to the normal reaction of the banked wall resisting the force applied on it by the car?

  18. Jan 3, 2016 #17
  19. Jan 3, 2016 #18
    Thanks heaps Chet!
  20. Jan 4, 2016 #19


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    Both components of the normal force on the car have an equal but opposite force on the track. And there is no cause-effect relationship between those equal but opposite forces, because they act simultaneously.
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