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Unit Vector Magnitudes and Forces

  1. Jan 14, 2013 #1
    1. The problem statement, all variables and given/known data

    http://prntscr.com/p7dxt

    Here's a screenshot of the revision question.

    2. Relevant equations



    3. The attempt at a solution

    Co-ordinates of each point
    A, (0, 60, 0)
    B, (40, 0, 0)
    C, (-40, 0, 40)
    D, (-60, 0, -60)

    Position Vectors
    rAB, 40i + -60j + 0k
    rAC, -40i + -60j + 40k
    rAD, -60i + -60j + -60k

    Unit Vectors corresponding these position vectors
    eAB, 0.5547i + -0.8321j + 0k
    eAC, -0.4851i + -0.7276j + 0.4851k
    eAD, -0.5774i + -0.5774j + -0.5774k

    This is all I know, I'm not sure how to complete the other questions.

    For question would I have to put the two position vectors and do a cross product?

    Something like...

    i j k
    -0.4851 -0.7276 0.4851
    -0.5774 -0.5774 -0.5774

    [(-0.7276)(-0.5774) - (0.4851)(-0.5774)]i + [(-0.4851)(-0.5774) - (0.4851)(-0.5774)]j + [(-0.4851)(-0.5774) - (-0.7276)(-0.5774)]k

    If not, I'm totally lost, and need help!
     
    Last edited: Jan 14, 2013
  2. jcsd
  3. Jan 14, 2013 #2

    haruspex

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    I see no reason to consider cross products here. It's a simple statics question.
    What components of the forces in the cables are of interest? What equations can you write down which say they balance out?
     
  4. Jan 16, 2013 #3
    So essentially, balance out the forces...

    Basically something like...

    ForceTotal = ForceAB + ForceAC + ForceAD

    2kN(eAB) = ForceAB = 1.1094i - 1.6642j + 0k
     
    Last edited: Jan 16, 2013
  5. Jan 16, 2013 #4

    haruspex

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    Forces in the y direction are not interesting. Whatever they add up to in tensions will be balanced by compression in the tower.
    Your resolution of the 2kN into i and j looks right, but I'd rather you stuck with the algebraic symbols, like 'cos(θ)', not plugging in actual numbers until the end. It makes it much easier to follow what you're doing and spot any errors.
    Create unknowns for the other tensions, write out their resolutions into i, j, k and hence the balance of forces equation.
     
  6. Jan 16, 2013 #5
    So do you mean like...
    ForceTotal = (eABi |FAB| + eACi |FAC| + eADi |FAD|)i + (eABj |FAB| + eACj |FAC| + eADj |FAD|)j + (eABi |FAB| + eACk |FAC| + eADk |FAD|)k

    Which then becomes something like...

    eACi |FAC| + eADi |FAD|)i = eABi |FAB|
    eACj |FAC| + eADj |FAD|)j = eABj |FAB|
    eACk |FAC| + eADk |FAD|)k = eABk |FAB|
     
  7. Jan 16, 2013 #6

    haruspex

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    Yes, except that as I mentioned you cannot write a useful equation for the vertical forces. That would involve the compression in the tower, which you don't care about.
     
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