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Angles of tension

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Three equal weights, two of them connected to the third by individual strings. If the mechanism is in equilibrium, what are the angles of the strings.

    2. Relevant equations
    EFi=ma


    3. The attempt at a solution
    T1=W
    so 2T1=t2 = t1
    T1=1/2W
    So what now?
     
  2. jcsd
  3. Oct 6, 2009 #2

    kuruman

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    Your problem statement is not very informative. Connected how by individual strings? Is there a picture that goes with this?
     
  4. Oct 6, 2009 #3
    Yes, but I don't know how to post it. In essence it is the letter M (with the left, right, and middle being the blocks, and the corners being the massless pulley system. However the only difference is that the center block is attached to a string and that string is what connects it to the other two strings. Think of a M with a Y for the inside (I know that's poorly worded, but I can't think of any other way to explain it.)
     
  5. Oct 7, 2009 #4

    kuruman

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    You have described it well enough. You need to draw a free body diagram of the "crotch" of the Y and say that the net force on it is zero. There are three forces but remember to add them as vectors. You can say that T1 = W for the bottom force, but it's the vertical components of T2 and T3 that add up to balance T1.
     
  6. Oct 7, 2009 #5
    And since the weights of the side blocks are the same as the weight on the center block, we could assume that the angles are 45 degrees each (representing one half of the total weight?
     
  7. Oct 7, 2009 #6

    kuruman

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    No, we cannot assume that. As I said before, you need to draw a free body diagram and balance the vertical components. Cutting corners is dangerous. See what comes out of the equations.
     
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