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Tension of sign

  1. Feb 16, 2008 #1
    [SOLVED] tension of sign

    1. The problem statement, all variables and given/known data
    find force of tension in the chain and determine the force of compression in the support bar.
    [​IMG]
    box weighs 20.4 kg or approx. 200N


    ok so it doesn't give a chain length bar length nothing, so would the tension on the chain have to be equal to the weight of the box at 200N? and for the force of compression that just means the force the wall exerts on the bar? im confused about a few problems but ill start with this one.
     
  2. jcsd
  3. Feb 16, 2008 #2

    PhanthomJay

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    but it does give an angle,
    the system is in equilbrium , so that it is the vertical component of the tension that must equal the weight
    yes, or the force that the horizontal component of the tension exerts inwardly on the bar.
    Hint: Draw a free body diagram of the joint where the chain and bar meet, and apply Newton 1. Note that this is a truss, and the (massless) bar can only support an axial (horizontal) force.
     
  4. Feb 16, 2008 #3
    am I able to use Pythagorus' theorem to find the tension on the chain if i use 200 n as the y tension and then use tan theta =o/a to find the horizontal than go a(2)+b(2)=c(2) c being the tension of the chain?
     
  5. Feb 16, 2008 #4

    PhanthomJay

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    Yes that will work. Personally, I like Tsin theta = 200, then solve for T, and then T cos theta is the compressive force. Either way works, though.
     
  6. Feb 16, 2008 #5
    so tension is 533.89N and the compressive force is 495N alright so i have a couple more problems just give me a moment
     
  7. Feb 16, 2008 #6

    PhanthomJay

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    Ok. Be sure to post different problems under a new topic.
     
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