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Normal Force at the bottom of a Ferris Wheel

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data

    A student of weight 678 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force N on the student from the seat is 565 N. (a) What is the magnitude of N at the lowest point? If the wheel's speed is doubled, what is the magnitude FN at the (b) highest and (c) lowest point?

    2. Relevant equations



    3. The attempt at a solution

    So M = 678N, NTop = 565N. Fc = mg - Ntop = 6079.4
    So Nbottom = Nbottom - mg = 6079.4 which leads Nbottom to = 12723.8 but this is incorrect. Where am I going wrong?
     
  2. jcsd
  3. Feb 18, 2008 #2

    Doc Al

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    Staff: Mentor

    At the top, normal force, weight, and acceleration all point down:
    N + mg = mv^2/r; so N = mv^2/r - mg

    At the bottom, normal force and acceleration point up, but weight points down:
    N - mg = mv^2/r; so N = mv^2/r + mg
     
  4. Feb 18, 2008 #3
    I really don't need all that though do I? If I do how am I going to solved for v^2 or r? I only have one equation and two unknowns. At best I'd have Ntop+Nbottom = mv^2/r.
     
  5. Feb 18, 2008 #4

    Doc Al

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    Staff: Mentor

    Yep. It's the easy way!
    No need to solve for those.

    Examining those expressions for N, how does Nbottom compare to Ntop? (Hint: What's Nbottom - Ntop?)
     
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