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Normal force on inclined planes.

  1. Aug 25, 2006 #1
    If a car is going up or down an inclined plane, the normal force on the car is just mg cos (theta).

    However, if a car is turning on an inclined curve, the normal force is mg/cos(theta). Why are the two normal forces not the same?

    To clarify, here is an example problem:

    A highway curve of radius 70m is banked at a 15 degree angle. At what speed can a car take this curve without assistance from friction?

    I've attached a force diagram of the problem.

    Thanks for the help!
     

    Attached Files:

  2. jcsd
  3. Aug 25, 2006 #2

    Doc Al

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    Consider the car's acceleration. In one case, the acceleration has no component normal to the surface, but in the other case, it does.

    To find the normal force, one must apply Newton's 2nd law. Start by identifying all the forces acting on the car and the car's acceleration.
     
  4. Aug 25, 2006 #3

    HallsofIvy

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    First, you have the arrow for "normal force" vector pointing the wrong way. The normal force isn't going to lift the car off the road! As to why the normal force isn't weight*cos(theta), I can't answer that- because it is! cos(theta) is a decreasing function of theta. When theta=0, a horizontal road, the normal force is equal to the weight. Okay, that would work for either weight*cos(theta) or weight/cos(theta) because cos(0)= 1. But as theta increases from 0 to pi/2, weight*cos(theta) declines to 0 while weight/cos(theta) increases to infinity. Increasing the slope does not increase the normal force! As rock climbers know, on a vertical slope you are not forced against the slope by an infinite normal force!
     
  5. Aug 25, 2006 #4
    I'm sorry, but I'm still confused. How does that explain why the two cars have different normal forces (the only difference being that turning car has centripetal accleration)?
     
  6. Aug 25, 2006 #5

    Doc Al

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    First tell me how you know that the normal force on the car going up or down an inclined plane is mg cos (theta). (You have to figure it out using Newton's 2nd law!)
     
  7. Aug 25, 2006 #6
    I thought that the normal force, in this case, is the force of the ground against the car. This is why I have the normal force vector pointing toward the car.

    And yes, I also think the normal force is weight cos(theta). However, the book I have states that the normal force is weight/cos(theta). Why is there a discrepancy?

    Thanks.
     
  8. Aug 25, 2006 #7

    Doc Al

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    The arrow is drawn correctly. The road most certainly does push up on the car!

    That's only true when the acceleration is parallel to the road surface (or zero).

    When the car's going around a curve you have centripetal acceleration to contend with. The steeper the angle, the faster you have to go to keep from sliding down. The faster you go, the greater normal force needed to provide that centripetal acceleration.
     
  9. Aug 25, 2006 #8
    Assuming the only force acting on the car is gravity (I'm going to say that the car has constant velocity), then the only force from the road against the car is weight cos(theta).

    I've attached an image of my work.
     

    Attached Files:

  10. Aug 25, 2006 #9

    Hootenanny

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    It may be useful to note that if the car is travelling around a banked corner, it is undergoing circular motion. (I can't comment on your diagram because as yet I cannot see it)
     
  11. Aug 25, 2006 #10

    Doc Al

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    Here's what your reasoning should be: Since the car does not accelerate in a direction perpendicular to the plane, the net force in that direction must be zero. From that you can conclude that normal force = mg cos(theta).

    But that's not the case when the car is going around a corner. In that case there is a component of acceleration perpendicular to the plane. For that problem, you want to realize that the vertical component of acceleration must be zero, while the horizontal component must equal the centripetal acceleration.
     
  12. Aug 25, 2006 #11
    I'm sorry for the belated reply. I couldn't access the site for some reason.

    The first part of your logic makes perfect sense. However, I'm still not clear on the second part. What is the component of accleration perpendicular to the plane? If there is a net accleration perpendicular to the plane, shouldn't the car be being lifted into the air? Thanks.

    Ultimately, here is the discrepancy I'm confused about. I've attached the image.
     

    Attached Files:

  13. Aug 25, 2006 #12

    Doc Al

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    You realize that if the car is turning a corner it is undergoing circular motion. And that something moving in a circle is accelerating towards the center. Correct? (If not, review circular motion and centripetal acceleration pronto.)

    The car accelerates toward the center of the circle it is turning on. That acceleration is horizontal and has a component perpendicular (and parallel) to the plane.

    No. The vertical acceleration is zero. But realize the car is accelerating.

    Understand circular acceleration. For example: Twirl a ball tied to a string in a circle. The ball has an acceleration towards the center, yet it doesn't move towards the center.

    Note that the net acceleration is horizontal. But there is a component of acceleration perpendicular to the plane.

    Your diagram shows the forces on the car (weight and normal force) but doesn't show the acceleration of the car. You need the acceleration to figure out the normal force (if you want to know it). The acceleration--a centripetal acceleration--will be a horizontal vector pointing to the left in your diagram.
     
  14. Aug 25, 2006 #13
    Thank you for your patience! I now see the perpendicular component and can solve the problem. I'm teaching myself physics right now, and good help is hard to come by. Again, I really appreciate it.
     
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