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F = MA 2010 # 11 (Static equilibrium, 3 masses on a string)

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    See:
    http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf
    Question 11


    2. Relevant equations
    T_net = 0
    F_net = 0

    3. The attempt at a solution
    Start with F_net = 0:
    Let T1 be the force of tension on the left, T2 on the right
    T1cosθ=T2cosθ
    Y components:
    T1sinθ+T2sinθ = mg
    Since T1 = T2, as established by the x-component force equation:
    2T1sinθ= mg
    T1sinθ = mg /2

    Now let's look at T_net = 0
    Take the leftmost point to be the PoR
    This is where I got stuck.
    T1cosθb = T2sinθa
    T2sinθ=T1sinθ = mg/2
    T1cosθb = mg/2a
    a/b = mg/2T1cosθ

    I'm really confused about how to find the torque:
    I get that it is rxF or rFsintheta but I'm struggling to make the equation.
     
    Last edited: Jan 28, 2013
  2. jcsd
  3. Jan 28, 2013 #2

    tms

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    There is no tension in question 12 at that link.
     
  4. Jan 28, 2013 #3
    Sorry I meant question 11.
     
  5. Jan 28, 2013 #4

    tms

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    What is [itex]\vartheta[/itex]? If you are assuming that the angle on each side is the same, you have to justify that.
    You don't need torque in this problem.
     
  6. Jan 28, 2013 #5
    Huh? How would you do the problem then?
    Ok, instead of justifying that, I'll just split it into components.
    T1x = T2x
    I'm not at all sure how to proceed. Could you give me a hint?
     
  7. Jan 28, 2013 #6

    tms

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    Start with a free-body diagram. Look at the leftmost mass. What are the forces on it? What are the forces on the other masses? What can you say about the tensions in the two strings?

    The next step is, as you say, breaking the forces on the center mass into components.
     
  8. Jan 28, 2013 #7
    Left-most mass:
    Down: mg
    Up: T0

    Middle Mass:
    Left: T1x
    Right: T2x
    Up: T1y, T2y
    Down: mg

    Right-most mass:
    Down: mg
    Up: T3

    Thus we know T1x = T2x and that T1y + T2y = mg
    Also T0 = T3 = mg
    so T1y + T2y = T0 = T3

    How could we relate this to a,b?
     
  9. Jan 28, 2013 #8

    tms

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    Up to here everything is okay, and [itex]T_0 = T_3[/itex] is okay, too, [strike]but the first term is a problem[/strike]. If the tension at one end of a string is [itex]T[/itex], what is the tension at the other end?
    Trigonometry.
     
    Last edited: Jan 28, 2013
  10. Jan 28, 2013 #9
    I don't see why T1y + T2y = mg is wrong.
    The two y-components of the two tensions need to add up to equal the weight of the middle sphere.

    Using the trig:
    T2sinθ=b
    T1sinθ2 =a

    ??
     
  11. Jan 28, 2013 #10

    tms

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    It isn't. I don't know what I was thinking.
    By now you know that all the tensions are equal, so you can drop the subscripts.

    Next you should look at the horizontal components of the forces on the central mass.
     
  12. Jan 28, 2013 #11
    We know that 2Tcosθ = mg
    This implies that Tcosθ = .5mg
    So θ is logically 60 degrees.
    I'm still stuck. Could you just show me?
     
  13. Jan 28, 2013 #12

    tms

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    You are again assuming that the angles are equal. They are, but you have to demonstrate it first; you can't just assume it.

    Which angle is θ? Between the string and the horizontal, or between the string and the vertical? You should look at the horizontal forces first; that will be a big help.
    That's against the rules of PF.
     
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