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Free body diagram

  1. Oct 7, 2007 #1
    Hi everyone, I just need some help on setting up a free body diagram with the appropriate forces acting on it.

    The question is:
    An engineer must design a curved exit ramp for a highway in such a way that a car, exiting at the posted speed limit of 17.88 m/s (40 mi/hr), does not depend on friction to round the curve without skidding. The radius of the curve is 208.0 m. At what angle with respect to the horizontal must the curve be banked (in degrees)?

    I m confused as to whether to draw the angle theta with respect to the x-axis or y-axis? I already got the answer I am just reviewing for a test I am having.

    Thank you!
  2. jcsd
  3. Oct 7, 2007 #2
    I've seen questions like this, and when I drew a FBD I always drew it lengthwise to the car, so on a graph, looking from the front or back of the car, the x axis is the ground, and the car's direction is headed in or out of the page. Therefore, the angle theta would be with respect to the x axis. In any case, the other angle WRT the y axis is just 90 degrees - theta!!!
  4. Oct 7, 2007 #3
    Thats what I thought, the x-axis, but then the normal force becomes ncos(theta) and its supposed to be sin. So what am I doing wrong?
  5. Oct 7, 2007 #4
    The normal force is just directional , and how you calculate it you will not necessarily use sin always... you may have to use cos sometimes, depending on how you drew your component forces, and as long as you can prove in your FBD that it is cos, then that's what it is. It's not necessarily wrong, it just so happens that in most examples the instructor may have used sin.
  6. Oct 7, 2007 #5
    Then again you did say you have the answers...does it show a diagram of how the FBD was presented? if so then obviously it should show why sin was used.
  7. Oct 7, 2007 #6
    No it doesn't, thats why I am confused about it.
  8. Oct 7, 2007 #7
    Maybe the angle was of the one not to the x axis, but the angle up that forms with the line perpendicular to the x axis. If that's so then it'd be sin theta. But like I said, as long as you end up with the right answer - regardless of which angle you used and whether you used sin or cos, as long as it makes sense and you get the correct value of the normal force, then you know you did it correctly.
  9. Oct 8, 2008 #8
    How would I go about solving this same problem, only with a different radius?
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