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Tension/torque problem

  1. Nov 11, 2008 #1
    Hello, I've been unable to figure this question about tension and torque out.

    1. The question is: calculate the torque about the front support on the right of a diving board exerted by a 70 kg person 3.0 m from that support.

    This is a crude diagram of a person on a diving board, where
    D = a diver, and || means a diving board support:

    ||- - - - - - - - - ||
    ||- - - - - - - - - ||

    2. Relevant equations:
    τ = Fd
    F = ma
    Fup = Fdown / Fleft = Fright

    3. I get these equations:

    Fup = Fdown
    F1 + F2 = Mg + mg (where mg = downward force of diver, Mg = downward force of diving board at center of board)
    F1 + F2 = M(9.8) + 686

    τcw = τccw
    Mg(1.5) + mg(3) = F2(1) (assuming F2 is the upward force for the middle support?)
    Mg(1.5) + 70(9.8)(3) = F2(1)
    Mg(1.5) + 2058 = F2

    I assume the pivot point to be on the far left of the board. The main problem problem with both of these are that I end up with equations that have two variables. Also, I don't have M, which is the mass of the diving board - how would I figure this out?

    What's wrong in my process to get the answer?
  2. jcsd
  3. Nov 11, 2008 #2


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    Welcome to PF!

    Hi fp252! Welcome to PF! :smile:
    I don't think they're asking you for to look at the whole problem …

    I think they literally only want the torque of the diver at the front support … you can ignore everything else. :wink:
  4. Nov 13, 2008 #3
    Thank you for replying.

    Unfortunately, I'm still stuck. Here's what I have now:
    τcw = τccw
    mg(4) = F2(1)
    686(4) = F2(1)
    2744 N = F2

    That would seem like an answer, but the force is designated upwards, so it can't be right.

    Every other diving board problem I've seen has a setup where I either cannot base it off of it to get the answer of the question above. So far I haven't seen any questions that have a downward force on the front support of the diving board either.
  5. Nov 13, 2008 #4


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    Hi fp252! :smile:
    Yes, that's the right formula and the right answer, but for the wrong question …

    you've calculated the normal force at the front support, which is 2774 N CCW, or upwards.

    But the question is only asking you for the torque of the diver about the front support, which is (70g)3 J clockwise. :smile:

    I'm not following you … your own answer gives a positive CCW value (I assume that stands for "counter-clockwise"). :confused:

    The correct way of writing the answer is:
    Torque = force times perpendicular distance, clockwise = (70g)3 J clockwise. :smile:
  6. Nov 13, 2008 #5
    Sorry, I'm only half awake while trying to do this, so excuse me if I don't get something obvious.

    If Torque = force times perpendicular distance, clockwise, then why is it 70g(3), when it should be in newtons, not grams?
  7. Nov 13, 2008 #6


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    Get some sleep! :zzz:
    My g was gravity!

    And torque is force x distance, so it's in joules. :wink:
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