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Solving force systems

  1. Jan 29, 2012 #1
    I have tried everything on this one that I can think of.

    I understand that the forces in the horizontal are 0, but I don't know where to start.

    says: determine the angle for which the resultant of the three forces is vertical. See Picture.
     
  2. jcsd
  3. Jan 29, 2012 #2
    Please see this picture for above question.
     

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  4. Jan 29, 2012 #3

    PeterO

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    What is the sum of the horizontal components of the two, fully defined forces?
     
  5. Jan 29, 2012 #4
    The sum of the horizontal components, in the x direction, is:

    70(cos140) + 80(cos165) = -131lb

    to take it further;

    70(sin140) + 80(sin165) = 66lb

    r= (-131)2 + (66)2 = 147 lbs

    inverse tan = 66/-131 = -27 degrees


    (I think I get it now) the resultant is vertical up.. so the sum of the left side is the same as right side...
     
    Last edited: Jan 29, 2012
  6. Jan 29, 2012 #5

    PeterO

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    Exactly, and with your creative use of 140o and 165o angles, you could simply have said

    70(cos140o) + 80(cos165o) + 150(cosθ) = 0
     
  7. Jan 29, 2012 #6
    Thanks lots Peter,

    At first I was trying to rearrange " 70(cos140o) + 80(cos165o) + 150(cosθ) = 0" for the unknown angle and was frustrated as to why that wouldn't work.

    tricky trick question.

    Thanks, now I can sleep, lol.
     
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