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Statics: Forces in Cartesian Vector form

  1. Jul 6, 2013 #1
    1. The problem statement, all variables and given/known data
    For cable AD it is known that the magnitude is 14 kips, x-component has a value of -6.216, the direction angle in the z-direction is 83.63°, and Fy is less than zero. Find forces in Cartesian vector form, coordinates of point D if it lies on the x-z plane and point A is (0, 20, 0).


    2. Relevant equations



    3. The attempt at a solution
    I know that the z component of TAD is 14000(cos83.63°)= +1553.28[itex]\hat{}k[/itex] but I don't know how to solve for the x and y components. Any help would be greatly appreciated.
     
  2. jcsd
  3. Jul 6, 2013 #2

    haruspex

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    But it tells you the x-component, no? And you know the z component and the overall magnitude, so the magnitude of the y-component should be easy. Is it the sign that has you stumped?
     
  4. Jul 7, 2013 #3
    im just not sure if this means that its -6.216i for the position vector of AD or -6.216i for the actual x component for the tensile force...I just don't see how it could be -6.216 for the tensile force when the magnitude is 14000 lbs.

    Having the x component of the position vector of AD doesn't really help me because I cant solve for the magnitude of the position vector. Maybe im just making it more complicated than what it really is.
     
  5. Jul 7, 2013 #4

    haruspex

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    In that case you will need to state the question exactly as given. I have the impression you have reworded it somewhat.
     
  6. Jul 7, 2013 #5
    I wrote it exactly as it was given.
     
  7. Jul 7, 2013 #6

    haruspex

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    Then it is poorly worded.
    This seems to be leaping into the middle of a question. There has been no mention of any tension in the cable, so what is this supposed to be the magnitude of? It could be windshear! If we assume it refers to tension in AD, then it is logical to suppose the 'x-component' is a component of that tension. I agree the mismatch in precision is a worry, but I see no other basis on which to proceed.
     
  8. Jul 7, 2013 #7
    ok thank you very much....
     
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