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Projection of F onto a Line

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data

    The turnbuckle is tightened until the tension in the cable AB equals 2.9 kN. Determine the vector expression for the tension T as a force acting on member AD. Also find the magnitude of the projection of T along the line AC.


    2. Relevant equations



    3. The attempt at a solution

    I was able to find the vector expression for T without any real trouble:

    AB = <2.6, 1.7, -6.1>

    |AB| = sqrt(2.62+ 1.72 + (-6.1)2)

    T*AB/|AB| = 0.42364<2.6, 1.7, -6.1>

    T = 1.101i + 0.7201j - 2.584k

    I know this part is correct but I'm having difficulty doing the projection.

    I tried taking taking the dot product of the T vector with the unit vector AC:

    AC = ,2.6, -2.8, -6.1>

    |AC| = sqrt(2.62 + (-2.8)2 + (-6.1)2)
    |AC| = 7.1979

    AC/|AC| = 0.3612, -0.389, 0.8475

    T°AC/|AC| = 1.101*0.3612 +(-0.389*0.7201)+(-0.8475*-2.584)
    = 2.2746

    But it says my answer is wrong and I'm not sure where I'm going wrong.

    Advice would be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Sep 22, 2012 #2

    Simon Bridge

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    Isn't the z component of the unit vector AC/|AC| supposed to be negative?
     
  4. Sep 22, 2012 #3
    Yes it is; it's a typo, but I did include the negative in the calculation below that.
     
  5. Sep 22, 2012 #4

    Simon Bridge

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    Code (Text):
    octave:20> T=[1.101;0.7201;-2.584] [color=green]< the tension vector[/color]
    T =

       1.10100
       0.72010
      -2.58400

    octave:21> C=[2.6;-2.8;-6.1] [color=green]< this is AC vector[/color]
    C =

       2.6000
      -2.8000
      -6.1000

    octave:22> sqrt(C'*C) [color=green]< |AC| = √AC.AC[/color]
    ans =  7.1979 [color=green]<---<<< agrees with your value[/color]

    octave:23> C=C./sqrt(C'*C) [color=green]< set up unit vector[/color]
    C =

       0.36122
      -0.38900
      -0.84747

    octave:24> T'*C
    ans =  2.3074 [color=green]<---<<< does not agree with yours[/color]
    ... it follows, unless I messed something up above, that you've made a mistake in your arithmetic someplace - go over it again, slowly.

    I think your method is sound ... the projection ##p## of ##\vec{u}## on to ##\vec{v}## is $$p=\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|}$$
     
    Last edited: Sep 22, 2012
  6. Sep 22, 2012 #5
    I ran my calculations again. You answer is correct. I must have made a mistake when I typed in the numbers into my calculator. Thank you.
     
  7. Sep 22, 2012 #6

    Simon Bridge

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    Those are the hardest mistakes to spot - especially if you make the same number-punch error repeatedly.
     
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