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Moment of a pole of unknown height

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Utility Pole BC is guyed by cable AB as shown. The tension in cable AB (TAB) acting at point B creates a moment about point O (MO) equal to -900i+Myj-315k lb-ft, where My is unknown.
    a) What is the length of the pole (LBC)?
    b) What is the magnitude of the tension, TAB?

    Diagram.png

    2. Relevant equations

    MOABxF
    TAB=uAB|TAB|

    3. The attempt at a solution
    Vector BO <0,-(h+1),6>
    Vector BA <4,-h,-6>
    |BA|= [itex]\sqrt{h^{2}+52}[/itex]
    TBA= TBA[itex]\frac{<4,-h,-6>}{|BA|}[/itex]

    I cross multiplied and got TBA<[itex]\frac{12h+6}{|BA|}[/itex],[itex]\frac{24}{|BA|}[/itex],[itex]\frac{4h+4}{|BA|}[/itex]>

    In order to solve for the height of the pole, I need TBA. Where i get stuck is here, and a lack of confidence that my vectors BA and BO are correct. In order to solve for the height of the pole, I need the tension, but that is the second part of the problem.

    EDIT: tips on how to fix the itex tags would be appreciated,too.
     
    Last edited: Sep 22, 2013
  2. jcsd
  3. Sep 22, 2013 #2

    TSny

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    Do you need vector BO or vector OB?

    Check the diagram carefully to see if the y-displacement from B to A is -h.

    You will have to solve 2 equations simultaneously for h and TAB.

    When writing latex, don't use the superscript and subscript buttons on the tool bar. Instead, use standard latex notation: h^2, for example.
     
  4. Sep 22, 2013 #3
    Vector OB should be the one I need, which is <0,h+1,-6> Vector BA should be <4,0.5-h,-6>.

    This gives a new cross product of T[itex]_{BA}[/itex]<[itex]\frac{-12h-3}{|BA|}[/itex],[itex]\frac{-24}{|BA|}[/itex],-[itex]\frac{4h-4}{|BA|}[/itex]>

    Am I right in thinking I can then set the i and k components of M[itex]_{O}[/itex] equal to the i and k components of the cross product? If so then I get:

    From the i component:
    T[itex]^{2}_{BA}[/itex]:[itex]\frac{90000(h^{2}-h+52.25)}{16h^{2}+h+52.25}[/itex]

    From the k component:
    T[itex]^{2}_{BA}[/itex]:[itex]\frac{6202(h^{2}-h+52.25)}{h^{2}+2h+1}[/itex]

    Do I set those equal to each other? If so, I get h=14.74 feet. Plugging that value for h into
    -900=T[itex]_{BA}[/itex][itex]\frac{-12h-3}{|BA|}[/itex] or
    -315=T[itex]_{BA}[/itex][itex]\frac{-4h-4}{|BA|}[/itex]
    gives me T[itex]_{BA}[/itex]=84.6 lb
     
    Last edited: Sep 22, 2013
  5. Sep 22, 2013 #4

    TSny

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    Are you sure about the negative sign in the numerator of the last term? Shouldn't it be 4h + 4 instead of 4h - 4?

    I don't agree with your denominator here.

    I agree with your value of h ( I get 14.75 ft ).

    I get somewhat less than 84.6 lb. But your approach seems correct.
    [EDIT: Note that if you take the ratio of your last two equations, you get a simple equation for finding h.]
     
  6. Sep 22, 2013 #5
    4h+4 is correct, I dropped a negative.

    The denominator was a typo. I have it on my paper as [itex]\frac{90000(h^{2}-h+52.25)}{16h^{2}+8h+1}[/itex].

    I redid the calculation for the tension and got 79.5 lb.
    It would seem I got sloppy, this was the 8th time working through the problem.Thanks for the help!
     
  7. Sep 22, 2013 #6

    TSny

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    OK. I got 79.9 lb. Good work.

    Pensacola?
     
  8. Sep 22, 2013 #7
    I get that when I spend the extra 15 seconds to calculate |BA| further than 15.9.

    Yep, home to beaches and Naval Aviation.:biggrin:
     
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