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Test on monday (banked road problem)

  1. Sep 30, 2007 #1
    URGENT!!! test on monday!! (banked road problem)

    Okay, so I am doing a problem involving a car driving on a banked, frictionless, circular track (theta=31degrees) and i am supposed to find the maximum velocity that the car can drive. I know that to find the velocity, i have to find the centripetal acceleration by saying that (mv^2)/r = nsin(theta). Then, I have to solve for n by saying that ncos(theta)=mg. However, I am confused... why can't n=mgcos(theta)? My understanding is that two forces are equal in magnitude if the object doesn't move in either direction. The car doesn't move into the road or out of the road... or does it?? please help! I have a test on monday.
  2. jcsd
  3. Sep 30, 2007 #2

    Okay, so maybe I should make my question more clear... here goes...
    The problem:

    The Daytona International Speedway in Daytona Beach, FL, is famous for its races(blahblahblah)... Both of its courses feature four-story 31.0degree banked curves, with maximum radius of 316m. If the car negotiates the curve too slowly, it tends to slip down the incline of the turn, whereas if it's going too fast, it may begin to slide up the incline.
    (a) Find the necessary centripetal acceleration on this banked curve so that the car won't slip up or slide down the incline.
    (b) Calculate the speed of the race car.

    My problem with the solution:

    It says in the book to set Ncos(theta) = mg. However, I was wondering, why can't N be = to mgcos(theta)?? My understanding is that two forces on an object are equal when the object does not accelerate in either direction. The car isn't moving into the road or out of the road, so why doesn't N = mgcos(theta)?? Why does it HAVE to be Ncos(theta) = mg???

    I understand that centripetal force is equal to Nsin(theta). And I know that I have to solve for N. My only problem is understanding why it has to be Ncos(theta)=mg and why it can't be N=mgcos(theta).

    I have a test on Monday!! Help would be much appreciated.
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