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Ferris wheel normal force

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A chemistry student with a mass of 75 kg is riding a steadily moving Ferris wheel. When she is at the top of the Ferris wheel, the normal force from the seat on to her body has a magnitude of 567 N.

    What is the magnitude of the normal force on her body when she is at the bottom of the Ferris wheel's arc?

    3. The attempt at a solution

    This is what I have so far:

    Top: Ntop + mg = m(v^2/r)
    Bottom: Nbot - mg = m(v^2/r)

    This is what I got from the free body diagram. It doenst look right at all to me and I have struggled with this problem for a while. Help!

    NEW!!! Just added!
    Part 2:

    What would the normal force be on the student at the top of the wheel if the wheel's velocity were doubled?

    I multiplied the acceleration (a = -2.24) by 4 and plugged it into the equations I figured out for the first problem:

    Top: Ntop - mg = -m(v^2/r)

    I got 63N. Which is wrong. Any ideas?
    Last edited: Oct 1, 2009
  2. jcsd
  3. Sep 30, 2009 #2
    Then your free body diagram is wrong. Rethink about what forces are acting up and down when the student is at the top of the wheel, and then do the same for the forces at the bottom of the ferris wheel.

    Remember, the normal force is the force that the seat exerts on the student, such that the net force will equal 0 at that point.
  4. Oct 1, 2009 #3


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    Most interesting! The force will be larger at the bottom than at the top (from experience). The mg reverses direction. I'd say those two equations are correct! But I'm open to enlightenment.
  5. Oct 1, 2009 #4
    Will it equal 0 even at the top?
  6. Oct 1, 2009 #5


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    No, the question says it is 567 N at the top.
    Incidentally, I like your equations better when solved for N:
    Top: Ntop = m(v^2/r) - mg
    Bottom: Nbot = m(v^2/r) + mg
    Maybe both terms in the bottom equation should be negative. But certainly the magnitude of the normal force is gravity and centrifugal force combined in the same direction.
  7. Oct 1, 2009 #6
    It's centripetal force. He's in an inertial frame of reference.

    The normal force at the top is the SUM of the centripetal force and gravity. Centripetal force is pulling you downwards along with gravity. Thus, the normal force has to be equal the sum of those two.

    That said, think about what happens to the normal force at the bottom. Again, the normal force exists to make the sum of the forces acting on the student equal to 0, otherwise the student would break the ferris wheel from where he or she is sitting.

    Also, the NET force at the top will be equal to zero, not the normal force.
  8. Oct 1, 2009 #7
    The force of gravity never reverses direction. It will pull downwards. The centripetal force reverses direction because it will always point towards the center.
  9. Oct 1, 2009 #8


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    No. Since the student in the ferris wheel is always accelerating, the net force is always mv2/r, though it does change direction.

    This is actually a good start to solving the problem. The problem is with incorrect + and - signs that you'll need to fix.

    Think about the following:

    mg always acts ____ (upward or downward?), and therefore should have a __(+ or -?) sign in each equation.

    The net force mv2/r acts ____(upward or downward?) at the top, and ____(upward or downward?) at the bottom? What should the signs (+ or -) be on the mv2/r terms in the equations?
  10. Oct 1, 2009 #9
    Thanks for the help so far everyone!

    Ah ok, I was thinking earlier that I needed a negative sign somewhere in there.

    Top: Ntop - mg = -m(v^2/r)
    Bottom: Nbot - mg = m(v^2/r)

    Now, I think it is a matter of solving for the unknowns.
    Last edited: Oct 1, 2009
  11. Oct 1, 2009 #10
    Phew! Thanks for the help! I finally got it. In fact I actually like that problem.
  12. Oct 1, 2009 #11
    I need some guidance on part 2 please. See my first post.
  13. Oct 1, 2009 #12


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    You have Fc - mg = 567. You know mg, so you can find Fc.
    That was a smart observation to see that the acceleration and Fc quadruple when the speed is doubled. But I don't think a = 2.24.
  14. Oct 1, 2009 #13
    By the way for acceleration I did this:

    Ntop -mg = -m(v^2/r)
    567 - (75)(9.8) = -75(a)

    a calculated to be -2.24

    ..and for part 2:

    Ntop -mg = -m(v^2/r)

    Ntop -(75)(9.8) = (75)(-8.96)

    Ntop came out to 63N. Which is not correct.
  15. Oct 1, 2009 #14


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    I was working from the rider's point of view where he feels centrifugal force, reducing his weight on the seat. F = Fg - Fc = 567 at the top.
    From the inertial (outside) point of view, gravity provides the centripetal force plus a bit more so the person is pressed down on the seat with 567 N. Same equation.
    I don't know why I said acceleration was not 2.24; I get 2.25 now.
    Used g = 9.81.

    Oops, was thinking "toward the center at the top" and "away from the center at the bottom". But better to use up to be positive.

    For the part (b), according to my theory
    Fc = 169 -> 169x4 = 675 N. (g = 9.81 and 4 digits in calcs)
    Force on the seat = Fg - Fc = 735.75 - 675 = 60.75 N.
  16. Oct 1, 2009 #15
    Ah ok, i was on the right track. Thanks!
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