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Weight distribution

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data
    In the illustration there is a hawser attached to the ceiling at points A and B. Attached to point C is a mass of 400kg.

    Find the tension in the hawsers AC and BC.


    2. Relevant equations



    3. The attempt at a solution
    I have no idea how to go about this one. I can't find any materials in English about this that are understandable and in my own language things are vague to begin with :/
    What is tension? This much I have figured out in my head that it's a force and its unit is 1N or that would make sense.
    But how do I find the tension.

    I know that the weight of the mass is 3920N and it should be distributed between those 2 holding hawsers in some proportion. Does it mean that the sum of the tensions in the hawsers will have to be mg?
     

    Attached Files:

  2. jcsd
  3. Oct 17, 2012 #2
    Hello,

    I hope my attached photo helps.
     

    Attached Files:

  4. Oct 17, 2012 #3
    So what I understand is that point B is pulling the weight attached to point C by F1 and point A is pulling it by F2. Considering the mass is still then according to Newton the sums of these forces will have to be 0.
    so F1 + F2 + mg = 0 (sum of vectors)

    If that's true I understand the idea, but how do I put it on paper with this assignment.

    Can I use this?
    the sine law or theorem or how it s called in English:
    A/sina = B/sinb so in this assignment F2/sin30 = F1/sin50 and using the a/b = c/d => ad = cb rule I get that
    F1/F2 = sin50/sin30
    sin50/sin30 = 1.532 so F1 = 1.532F2

    But something doesn't feel right, isn t F2 supposed to be > F1, I mean, it's closer to being in a vertical position compared to F1 so isn t it supposed to have more tension not less :s
     
    Last edited: Oct 17, 2012
  5. Oct 17, 2012 #4
    hell,

    i would use sum on x and sum on y.

    In this example F1x=F2x; F1*cos30=F2*cos50....
     

    Attached Files:

  6. Oct 17, 2012 #5
    Oh okay, so F1 = F2*cos50/cos30 - so my assumption is correct, the closer the hawser gets to a vertical position, the bigger the tension gets?
    Thank you, mishek, I understand how you approached this, cheers.
     
    Last edited: Oct 17, 2012
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