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Equilibrium Problem

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    The entire problem can be found in this image link.
    http://www.freeimagehosting.net/image.php?58acd0e92b.jpg

    2. Relevant equations
    Equilibrium so the usual:
    Total F = 0, Total moment = 0


    3. The attempt at a solution
    I'm not sure how we're supposed to find out the answer because there seems to be too many unknowns. I'm assuming the rope is an unknown tension force, and we also have 2 unknowns at point a and b.
     
  2. jcsd
  3. Nov 12, 2009 #2

    ehild

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    How many unknowns you have got exactly? How many relations do you find among them?

    ehild
     
  4. Nov 12, 2009 #3
    so there's the reaction force of point a, reaction force of point b, tension force of the rope, and the 400N force. so a total of 4 unknowns when one known.

    Fy = 0 = -400N + Ay + T + By
    Fx = 0 = Ax + Bx
    M(@ rope) = (-1.2)(Ay) + (1.5)(-400N) + (2.5)(By)

    which seems to be unsolvable...
     
  5. Nov 12, 2009 #4

    ideasrule

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    The reaction forces at A and B are assumed to be perpendicular to the respective surfaces, so you can express Ax, Bx, Ay, and By in terms of A and B and the angles (45 degrees and 30 degrees). That will result in only 3 unknowns. With 3 equations, you can solve for the unknowns.
     
  6. Nov 12, 2009 #5

    ehild

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    It seems unsolvable if you do not want to solve it.

    The 400 N force is known, don't count it as unknown.
    You can express the components of the reaction forces with their magnitude and angle. The angles are given, so you need to find only the magnitudes. That is two unknowns. The tension is the third.
    You have written the torque with respect to the point where the rope is attached. It is correct, but you know it as the beam is in equilibrium.

    M(@ rope) = (-1.2)(Ay) + (1.5)(-400N) + (2.5)(By)=??

    ehild
     
  7. Nov 12, 2009 #6
    Okay guys, I think I got it...

    Here's my work:
    Mc = 0 = (1.5m)(-400N) + (2.5)(Bcos30) + (-1.2)(Acos45)
    Fy = 0 = (-400) + (Bcos30) + (Acos45) + T
    Fx = 0 = (Bsin30) + (Asin45)

    A = -152.3355
    B = 215.4349
    T = 321.15

    I can't help but feel like I did wrong cause the answer is so specific but I don't really find anything wrong with the work.
     
  8. Nov 12, 2009 #7

    ehild

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    Ax and Ay are opposite: Fx = 0 = (Bsin30) - (Asin45)

    ehild
     
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