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Trying to solve 3x3 determinant with two zeros

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    3 wires and unknown tensions: Tab, Tbc, Tbd
    Flowerpot being suspended in equilibrium

    Code (Text):

    Fx =  .05Tab - 0.728Tbc - 0.728Tbd = 0 lbs
    Fy = -.05Tab + 0.485Tbc + 0.485Tbd = 20 lbs (weight of flower pot)
    Fz =    0Tab + 0.485Tbc - 0.485Tbd = 0 lbs
     

    2. Relevant equations

    Is it possible to find tension in one of these wires?

    3. The attempt at a solution

    Using Cramers rule, came up with

    Code (Text):
    -14.1232/0.117855
    which was incorrect.
     
  2. jcsd
  3. Sep 10, 2011 #2

    vela

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    You seem to have made an arithmetic error somewhere. Your answer for Tab is off by a factor of 10.
     
  4. Sep 11, 2011 #3
    I went over it again today and got the same answer :(
    back of the book lists 17 lbs for Tab. Here is the determinant I used,

    2n0q0i9.png

    resulting in

    [tex]\frac{-7.0616-7.0616}{(-0.1176125)+(0.17654)-(-0.017654)-(0.1176125)}[/tex]

    = 119.83 lbs, Tab
     
  5. Sep 11, 2011 #4

    vela

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    Are the coefficients of Tab equal to 0.5 (third post) or 0.05 (original post)? That's the factor of 10 difference between our answers. So the good news is you're basically calculating Tab correctly using Cramer's rule. The bad news is, if Tab is supposed to be 17 lb, your equations are wrong.
     
  6. Sep 11, 2011 #5
    0.5, sorry. Not sure where the equations may have gone wrong.
     
  7. Sep 11, 2011 #6

    vela

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    Just a guess, but I'd expect the coefficients of Tab should be either [itex]\pm 1/\sqrt{2}[/itex] or one of them should be [itex]\pm \sqrt{3}/{2}[/itex].

    More generally, if you form a vector from the coefficients of Tab, the norm of the vector should be 1.
     
    Last edited: Sep 11, 2011
  8. Sep 14, 2011 #7
    I went over the numbers with someone else in my class who also verifies that the coordinates used are correct. It sounds like this problem can be solved using a graphing calculator, but its strange that I was unable to solve it using cramers rule.
     
  9. Sep 14, 2011 #8

    vela

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    It's not strange that Cramer's rule doesn't work. What's strange is you concluding Cramer's rule only works some of the time rather than concluding your equations are wrong or that you are making some other mistake. If a proved result, like Cramer's rule, doesn't seem to be working, that's an indication that there's something else is wrong.

    I can tell you that if you and your classmate got exactly the same equations, you and your classmate are wrong. If your classmate claims to have gotten the right answer from those equations, he or she is making yet another mistake.

    I can read off the components of Tab from your equations. According to them, the force Tab is equal to[tex]\vec{T}_{ab} = \frac{T_{ab}}{2} \mathbf{\hat{i}} + \frac{T_{ab}}{2} \mathbf{\hat{j}}[/tex]where Tab=|Tab|. But this implies that[tex]
    \lvert \vec{T}_{ab} \rvert = T_{ab} \sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2} = T_{ab}\frac{\sqrt{2}}{2} \ne T_{ab}[/tex]which is a contradiction, which in turn indicates your equations are wrong.
     
    Last edited: Sep 14, 2011
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