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Tension in 3 cords

  1. Oct 21, 2009 #1
    This is my first post...is there a way to add a picture from my own computer to a post?

    1. The problem statement, all variables and given/known data

    similiar to this problem but with different numbers
    P5.26a.gif
    In my problem, C=T3, B=T2, and A=T1. The angle on the left is 30 and the one on the right is 45



    2. Relevant equations

    [tex]
    \vec{F}_{net} = \Sigma \vec{F} = m \vec{a}
    [/tex]





    3. The attempt at a solution

    What I know...
    1. T_c=w
    2. T_a sin 30 + T_b sin 45 = w
    3. T_a cos 30 - T_b cos 45 = 0
    4. sin 45 = cos 45

    T_a sin 30 + T_b sin 45 = w
    T_a sin 30 + T_b cos 45 = w
    becuse of # 3 above, I get
    T_a sin 30 + T_a cos 30 = w
    T_a(sin 30 + cos 30) = w
    T_a(1.366) = w


    ...I'm stuck here. Please help direct me to find T_a in terms of W. As shown above I came up with "T_a=w/1.366" but the book came up with "T_a=.732w" Thanks is advance for the help.
     
    Last edited: Oct 21, 2009
  2. jcsd
  3. Oct 21, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi dbakg00! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    erm :redface: … they're the same! :smile:
     
  4. Oct 21, 2009 #3
    w/1.366 is the same as 0.732w, so it is both correct
     
  5. Oct 21, 2009 #4
    I'm not seeing how those two are the same, would someone mind clarifying for me?
     
  6. Oct 21, 2009 #5

    tiny-tim

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    Science Advisor
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    w/1.366 = w (1/1.366) = w(.732) :wink:
     
  7. Oct 21, 2009 #6
    thank you...i've been staring at this problem for so long, even the obvious becomes oblivious to me!
     
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