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Circular Motion-Quick Question

  1. Nov 10, 2007 #1
    Circular Motion--Quick Question

    A ferris wheel has a radius of 15 m and a period of 60 s is shown below. Find the Fnet at points B and D if the rider has a mass of 60 kg.

    Ok, so this is a solved problem from a physics text book.

    This is the solution given in the text book.
    At B
    Fnet = Fc - Fg (since they act in the same direction)
    At D
    Fnet = Fc + Fg (since they act in the different directions)

    Solution
    At B:
    Fnet = Fc - Fg = 686 N - 9.9 N = 676 N
    At D:
    Fnet = Fc + Fg = 686 N + 9.9 N = 696 N

    They found Fc by using 4pi^2rm/T^2
    Heres where the problem arises.
    I think that this solution is wrong.
    How I solved it:
    I took the direction in wich tension acts as positive.

    So at B(at top of the wheel)

    Net force= Fc+Fg

    And at D(at the bottom of the wheel)

    Net force= Fc-Fg

    Am I right?
     
  2. jcsd
  3. Nov 10, 2007 #2

    G01

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    Gold Member

    The equations you have for the net force at the top and bottom of the wheel are correct. The book is incorrect.
     
  4. Nov 10, 2007 #3

    Doc Al

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    Staff: Mentor

    I'm confused. Does the question ask for net force, or the normal force? I understand that Fc = centripetal force and Fg = weight. (At the top and bottom positions, the centripetal force is the net force.)
     
  5. Nov 10, 2007 #4
    uh..the question asks for net force.
     
  6. Nov 10, 2007 #5

    G01

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    Hmm good point Doc Al. I was taking the net force to mean the force caused by the beams of the ferris wheel pulling the seat toward the center of the wheel. Now I'm confused...
     
  7. Nov 10, 2007 #6

    Doc Al

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    The net force on the rider, which is the sum of the normal force and the weight, is the centripetal force. The only difference at the top versus the bottom (assuming a uniform speed) is the direction of the net force.

    What the book (and you) are presumably solving for seems like the normal force (the force exerted on the rider by the seat). That normal force is greater at the bottom of the wheel, since it acts up while gravity acts down. (Except for calling that the "net force", the book is correct.)
     
  8. Nov 10, 2007 #7
    uh..but, if I take the direction the tension acts as positive, the book makes no sense to me. I mean the At point B(top) both Fg and the tension act in the same direction. So they should have the same sign.
     
  9. Nov 11, 2007 #8

    Doc Al

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    At the top they do have the same sign, regardless of the chosen sign convention. So does the centripetal force. All three act downward. So?

    Assuming that you really mean to calculate the tension (or normal force), not the net force:
    At the top: Fc = Fn + Fg
    Thus: Fn = Fc - Fg
     
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