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Line tension

  1. May 17, 2010 #1
    I need help trying to figure this problem out.

    Attached Files:

  2. jcsd
  3. May 17, 2010 #2


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    Hi drkidd22! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
  4. May 17, 2010 #3
    So I have this so far trying to find Tbd.
    It will be the same for Tbe as they are the same distance away from A.

    Attached Files:

  5. May 17, 2010 #4
    distance on horizontal=√(18^2+18^2)=18*√2 ft
    distance on vertical=21 ft
    ∴θ=angle in between pole & cable=tan-1(18*√2/21)=50.479 degree
    Σall horizontal force=0
    1680=2T sinθ
    ∴T=1088.944 lb
    Σall vertical force =0
    R=2T cosθ [R= reaction force at A ]
    ∴R=1385.924 lb

    Not sure if it's correct
  6. May 17, 2010 #5
    How many components are there to the reaction at A (at most)? You ask if it is correct. But you could determine this yourself by recognising that there are at most 6 equilibrium equations for the whole thing.....
  7. May 17, 2010 #6
    There are three forces at A
  8. May 17, 2010 #7
    So have you calculated any of them so far?

    (By the way your physicsp148.doc document has just one line and it is sloppy, or incorrect or inconsistent. If T is a force, why do the units have ft at the end?)
  9. May 17, 2010 #8
    when you say "1680=2T sinθ ", exactly what free body diagram are you considering the equilibrium of?
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