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Dynamics exam prep -- forces explanation in FBD please

  1. Apr 30, 2017 #1
    • Thread moved from the technical forums, so no Homework Help Template is shown
    I am doing a review for a Dynamics exam, but I am having trouble with this one question (below).

    Please note the top of the picture contains the answers.

    What I have done is make a free body diagram that includes the tension of the cable and the weight of the rider/swing. So according to my calculations for part i:

    T= mg cos (30) = 680N

    The tangential acceleration is 0

    Which according to the provided solutions is incorrect. Am I missing a force?

    I figured the tension is integral to finding the velocity of the rider. So I want to get my tension right before I proceed with finding the velocity, though I am also having trouble figuring out how to relate tension and velocity of the acceleration of the rider is 0

    I would appreciate an explanation so I can understand.

    nSd3YBe.jpg
     
  2. jcsd
  3. Apr 30, 2017 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. :smile:

    if the rotational speed is zero, then theta is zero, and T=mg.

    As the rotational speed increases, theta increases. So if T=mg cos(theta), the tension would be decreasing with increasing theta, which clearly is not the case. What other force acts on the rider that pulls them out? How does that force add to the tension over and above mg?
     
  4. May 1, 2017 #3

    haruspex

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    Science Advisor
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    With that equation you are saying forces balance along the line of the cable, so there is no acceleration component in that direction.
    What acceleration is occurring? In what direction does that have no component?
     
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