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Diving board and torque

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data

    A uniform diving board, of length 5.0 m and mass 52 kg, is supported at two points; one support is located 3.4 m from the end of the board and the second is at 4.6 m from the end (see the figure below). What are the forces acting on the board due to the two supports when a diver of mass 63 kg stands at the end of the board over the water? Assume that these forces are vertical. [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]

    2. Relevant equations

    Torque = Force x Distance

    3. The attempt at a solution

    There are two pivot points involved. The one that is 3.4m from the diver is going to experience force upward while the support 4.6m away will experience force downward.

    I take the weight of the diver*weight of the board*distance from support (3.4m) and come up with a wrong answer. How should I go about solving this?
  2. jcsd
  3. Oct 30, 2007 #2

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    There's no picture. So, it's not clear which side of the board the diver is standing on. First, write the problem precisely.
  4. Oct 30, 2007 #3
  5. Oct 30, 2007 #4
    The sum of all torque must be equal to zero, correct?
  6. Oct 30, 2007 #5

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    About any point, yes, the sum is equal to zero.
  7. Oct 30, 2007 #6

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    Consider the extreme pivot p1, which is 0.4 m from the inside end. Suppose the vertical force acting downward on the board at that point is N1. Similarly N2 at p2. You know all the other vertical forces and where they act. Take the sum of the torques about any point.
  8. Oct 30, 2007 #7
    I was given answers to this problem, but I am still not sure how to solve this problem.

    For example, point p1 is solved by:
    ((mass of board)(9.81)(4.6-2.5)+(mass of diver)(9.81)(4.6))/1.2
  9. Oct 30, 2007 #8

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    Take the sum of all the torques about the p2, for example.

    Torque due to weight of diver about p2 = wt of diver*dist of pt. of application of force from p2. Proceeding like this,

    weight of diver*(3.4) + wt of board* (0.9) – N1*(1.2) = 0. (why is the last one negative?)

    Do the same for the other one.
  10. Oct 30, 2007 #9
    It is negative because the force goes in opposite directions between p1 and p2. Thanks for your help.
  11. Oct 30, 2007 #10

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    Hmm...I don't know exactly what you meant by that. Let me just say that the weight of the board and the weight of the diver tend to rotate the board about p2 in the opp dircn as does N1.

    It's not because N1 is negative. N1 points downward, as does the weights of the diver and board. It’s the distance here which is negative, because you are going to the opp side of p2.

    Even if you had not put in the negative sign, the torque due to N1 would have come out to be negative, signifying that the torque due to the force at p1 is opp in dircn to the other two torques.
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