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Seesaw length from pivot

  1. Oct 3, 2013 #1
    1. The problem statement, all variables and given/known data
    It's a balanced seesaw but the pivot isn't in the exact middle and two weights are on it. One at the very end of each side.
    The weight on the left is 4.8 kg.[m1]
    The weight on the right is 5.5 kg[m2]
    The length of the entire seesaw is 5m
    L1 is the distance from the pivot on the left side to the end of the board where m1 is.
    Calculate the distance L1 from the pivot when the board is balanced.


    2. Relevant equations
    Distance = Torque/force
    Sum of all moments = 0

    3. The attempt at a solution
    m1*g=47.088 N This force is acting counterclockwise
    m2*g=53.955 N This force is acting clockwise
    The normal force from the pivot is m1g+m2g = 101,043 N

    I'll use L2 to identify the length from the pivot to the right end since L1 is the length from pivot to the left end.
    53.955*L2 = 47.088*L1

    This is where I get stuck when trying to calculate the lengths of L1 and L2.
    I could calculate the length of L1 and L2 if I knew the torque by dividing it by the force but it's unknown isn't it?
     
  2. jcsd
  3. Oct 3, 2013 #2

    gneill

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    How are L1 and L2 related via the total length of the seesaw arm?
     
  4. Oct 3, 2013 #3
    you don't need to know the torque because the torque of the one end of the seesaw should be equal and opposite to the torque of the other just like you have in your one equation. You know that your total length of the seesaw is 5m so set up a second equation and then you have two equations and two unknowns.
     
  5. Oct 3, 2013 #4
    L1 < L2
    The entire length of the seesaw is 5m. In the picture the pivot isn't at 2.5m, it's closer to L1.



    I don't understand the last part.
    Also sorry if I am confusing, English is not my first language.
     
  6. Oct 3, 2013 #5

    gneill

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    Write an equation that relates L1, L2, and L, where L is the total length of the seesaw arm.

    Hint: If you happened to know what L1 is, how would you determine what L2 is?
     
  7. Oct 3, 2013 #6
    L = L1+L2
    5 = L1 +L2

    If I knew L1 then I would know what L2 is but how do I get L1?
    If L1 was 2m then L2 would obviously be 3m.
     
  8. Oct 3, 2013 #7

    haruspex

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    You have two equations relating L1 and L2, the one in the OP and the one for L1+L2 above. So it's a straightforward problem in simultaneous equations: use one equation, in the form L2= something, to substitute for L2 in the other.
    Btw, you've made the arithmetic unnecessarily complicated. Given m1 g L1 = m2 g L2 you can cancel out g.
     
  9. Oct 3, 2013 #8
    L1+L2 = 5m
    4.8L1 = 5.5L2

    L2 = 4.8L1
    L1+4.8L1 = 5
    Am i on the right path? I am not sure what to do next.

    Or is this another way to do it?
    L1*4.8 = L2*5.5
    L1 = 0.872L2
    0.872L2+0.872L2 = mg
    1.744L2 = mg
    L2 = 5.5/1.744 = 3.15
    L1 = 0.872*3.15 = 2.74
    Actually then the total becomes too high(5.8m)
     
    Last edited: Oct 3, 2013
  10. Oct 3, 2013 #9

    gneill

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    Rather than plugging in numbers so soon, why not manipulate the variables. You've got two equations:

    L1 + L2 = L
    m1*L1 = m2*L2

    Solve for L1. Then plug the given numbers into the resulting expression.
     
  11. Oct 3, 2013 #10
    I am pretty sure I understand what you mean but I get stuck after a while again.

    L1 = L-L2
    m1*L-L2 = m2*L2
    4.4*5-L2 = 5.5*L2
    22-L2 = 5.5*L2
     
  12. Oct 3, 2013 #11

    SteamKing

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    And how do we solve an equation in one variable? Collecting like terms on one side of the equal sign?

    Hint: algebra
     
  13. Oct 3, 2013 #12
    Is this correct?

    22-L2 = 5.5*L2
    22-L2-5.5L2 = 5.5L2-5.5L2
    22-6.5L2 = 0
    -6.5L2 = 22
    6.5L2/6.5 = 22/6.5
    L2 = 3.384 m

    which means 5-3.384 = 1.616
    L1 = 1.6 m
    L2 = 3.3 m
     
  14. Oct 3, 2013 #13

    haruspex

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    Yes
    How did you get that ?!
    i think you have that backwards
    Where did that come from?
     
  15. Oct 3, 2013 #14
    I tried using the method to calculate a hanging weight held up by two strings but I realize that's completely wrong here.

    I tried solving it with algebra here as suggested:
    L1 = L-L2
    m1*L-L2 = m2*L2
    4.4*5-L2 = 5.5*L2
    22-L2 = 5.5*L2
     
  16. Oct 3, 2013 #15

    SteamKing

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    22-6.5L2 = 0

    should simplify to

    6.5L2 = 22
     
  17. Oct 3, 2013 #16
    But the answer is correct?

    Also Thanks a lot everyone for your help!
     
  18. Oct 3, 2013 #17

    gneill

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    Watch your algebra! If you replace variable "L1" with "L - L2", then if L1 was multiplied by something in the equation then that multiplication distributes over both L and L2. Use parentheses around things that you substitute in order to avoid inadvertently "dropping" operations. So: m1*L1 = m2*L2 ----> m1*(L - L2) = m2*L2.

    But as I indicated above, since you're looking for L1, don't substitute it away! Replace L2 instead, leaving you with an equation in just L1.

    Also, your 4.8kg seems to have become 4.4kg...
     
  19. Oct 3, 2013 #18
    Simplifying
    4.8L1 = 5.5(5-L2)
    4.8L1 = (5 * 5.5-L2 * 5.5)
    4.8L1 = (27.5-5.5L2)

    Solving
    4.8L1 = 27.5 -5.5L2

    Divide each side by 4.8.
    L = 5.729166667 -1.145833333L2

    L1+L2 = 5
    5.72-1.14L2 + L2 = 5
     
    Last edited: Oct 3, 2013
  20. Oct 3, 2013 #19

    gneill

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    Nooooo. You've replaced L2 with (5 - L2). That can't be right.

    It really would be better to work with just the symbols until the very end. Use the symbols.

    So, what's L2 in terms of L and L2?
     
  21. Oct 3, 2013 #20
    L2 = L-L1

    4.8L1=5.5(L-L1)
    and then I'm not sure.
     
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