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Suppose in the previous problem m3 balances the see-saw by pulling now

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose in the previous problem m3
    balances the see-saw by pulling now on the end 1.5 m from the ful-crum at an angle of 30° from the horizontal as shown.
    What force is necessary for balance?

    2. Relevant equations

    See picture for both questions
    The first question i got 31.4 kg

    second question it added the angle and i dont know what to do with that.

    Also I don't understand why does the mass of the seesaw doesn't matter?



    3. The attempt at a solution
     

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  3. Apr 16, 2013 #2

    PhanthomJay

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    Yes, looks good
    forget about the mass of the person m3 and just consider a pull force as shown. Find its torque about the pivot by breaking it into its components, then proceed as before.
    it is assumed that the see saw is uniform and its mass acts at the center (at the pivot), so if you use the pivot as the point about which to sum moments equal 0, does it make a difference?
     
  4. Apr 16, 2013 #3
    I actually dont understand how does the breaking component work...
    Fx cos 30 + Fy sin 30?
     
  5. Apr 16, 2013 #4

    PhanthomJay

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    If you denote the pulling force as F, then the horizontal x component is Fcos30 and the vertical y component is Fsin30. The x component does not produce any torque about the pivot, since there is no perpendicular moment arm. Only the y component does produce torque.
     
  6. Apr 16, 2013 #5

    oh make sense. but u said ignore m3? why? torque = rF , so the force does not include m3g?
     
  7. Apr 16, 2013 #6

    PhanthomJay

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    The problem isn't worded that good, but presumably, the person called 'm3' is standing on the ground as (s)he pulls, so his or her weight does not act on the seesaw (draw a free body diagram of the seesaw), only the pulling force F acts on it.
     
  8. Apr 16, 2013 #7

    oh ok thanks!
     
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