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4 see-saws in a balancing act

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A team of Mexican acrobats are developing a new act. Four see-saws are arranged in a square so that the ends of the see-saws overlap as shown in the plan view below. Each member of the team has a mass of exactly 50kg.

    (a) Calculate the contact forces between the see-saws when the acrobats, under the sombreros are positioned as shown in the digram - {linked}

    (b) What will happen if one member of the team over-eats and becomes heavier than the three others?
    Can the other three acrobats position themselves such that the system is balanced?
    2. Relevant equations

    [tex]\sum{V_{forces}}=0[/tex]

    [tex]\sum{M_{pivotPoint}}=0[/tex]

    3. The attempt at a solution


    [tex]R_A=\text{ Force exerted on see-saw in question on the left}[/tex]
    [tex] R_B=\text{ Force exerted on see-saw in question on the right}[/tex]
    [tex] N=\text{ The reaction force due to the fulcrum in the middle of the see-saw}[/tex]
    [tex] L =\text{ The length of the see-saw}[/tex]

    [tex]\sum{M_{Fulcrum}} =+R_A*L/2 - 50*g*L/3 +R_B*L/2[/tex]

    This is just for one of the four see-saws. For part (a) I suppose each see-saw should be the same. Is this right? Or do I have to include the forces the see-saw in question exerts on the other see-saws?
     

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    Last edited: Oct 11, 2014
  2. jcsd
  3. Oct 11, 2014 #2

    haruspex

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    Yes, by symmetry you can deduce the reactions between see-saws are all the same.
     
  4. Oct 11, 2014 #3
    Can the system be seen as a set of two x two see-saw balancing acts, where the reaction force = 0 where they meet? N must be equal to 50g newtons right? Can the reaction forces = 0?
     
    Last edited: Oct 11, 2014
  5. Oct 11, 2014 #4

    haruspex

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    Sure, but by the symmetry you can reduce it to a single see-saw with equal and opposite forces applied at the ends.
     
  6. Oct 12, 2014 #5
    That's great thanks!
    I got the reaction force = [tex]50g/3[/tex]
    for part (a)
     
    Last edited: Oct 12, 2014
  7. Oct 12, 2014 #6

    haruspex

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    Looks right.
     
  8. Oct 12, 2014 #7
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