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Lever Problem

  1. Mar 16, 2007 #1
    1. The problem statement, all variables and given/known data

    a uniform 12.0m long board has mass 500kg. it rest 5.5m over a cliff. How far can a 80kg man walk on the board before it tips?

    2. Relevant equations

    equilibrium and torque

    3. The attempt at a solution

    im not quite sure about any portion of my work, because i resorted to canceling units to produce the one that made sense.

    [tex]W_{board}=500kg\cdot9.8=4.9\cdot10^{3}N[/tex]
    [tex]W_{man}=80kg\cdot9.8=784N[/tex]
    the fulcrum is at at position: 12.0m-5.5m = 6.5m

    torques must be equal so,

    [tex]\tau_{clockwise}=\tau_{counterclockwise}[/tex]
    [tex]\left(L_{1}\right)\left(784N\right)=\left(4.9\cdot10^{3}N\right)\left(.5m\right)[/tex]
    [tex]\left(L_{1}\right)=\frac{\left(4.9\cdot10^{3}N\right)\left(.5m\right)}{784N}=3.1m[/tex]


    so, the man can stand 3.1m from the fulcrum on the side over the cliff?
     
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 16, 2007 #2
    actually i just realized that if any addition weight cases it to tip, that means the fulcrum is in the center, which it is not.
     
  4. Mar 16, 2007 #3

    Doc Al

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    Staff: Mentor

    (1) Recalculate the weight of the board--you are off by a decimal.
    (2) Where does the weight of the board act? How far is that from the fulcrum?

    In calculating torques, always measure distances from the fulcrum.
     
  5. Mar 16, 2007 #4
    okay, i recalculated the weight of the board, but i got the same answer, i did discover the weight of the man was wrong however. so the weight of the board acts at a point .5m from the fulcrum, so the ground has an equal force in the opposite direction, but its at that same point, so the first torque equation, i used should equal the weight of the board for F1. hows is it now?
     
  6. Mar 16, 2007 #5

    Doc Al

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    Staff: Mentor

    You have the weight of the board equal to 500*9.8 = 490; that's not right. (Exponent problem.) The weight of the man is fine.

    Yes, the weight of the board acts 0.5m from the fulcrum. Set the torque created by the weight of the board equal to the torque created by the weight of the man.
     
  7. Mar 16, 2007 #6
  8. Mar 16, 2007 #7

    Doc Al

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    Staff: Mentor

    Your torque equation doesn't make sense:

    You know both forces, so why are you treating F_1 as an unknown? The unknown is the distance from the fulcrum to the man. And why are you multiplying the weight of the board by 6.0 m?? You just told me that that force acts 0.5 m from the fulcrum.

    [tex]F_1 D_1 = F_2 D_2[/tex]
     
  9. Mar 16, 2007 #8
    okay, the number i got now is much more reasonable, but is it right?
     
  10. Mar 16, 2007 #9

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
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