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Eng Mechanics: Resultants of force systems

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data
    In the figure , the moment of a certain force F is 180ft - lb clockwise about O and 90ft-lb counterclockwise about B. If its moment about A is zero, determine the force.
    [​IMG]


    2. Relevant equations
    Moment of a force
    Mo = Fy(x)
    Mo = Fx(y)


    3. The attempt at a solution
    Since its moment about A is zero then the force passes through A down to the right because the problem mentioned its directions:

    Moment of F w/ respect to O = 180ft - lb = Fx(y)
    180ft - lb = Fx (3)
    Fx = 30lb

    Moment of F w/ respect to B = 90ft - lb = Fy(x)
    90ft - lb = Fy(6)
    Fy = 15lb
    using Summation of components:
    F^2 = (30)^2 + (15)^2
    F = 61.8lb

    the answer at the book is 75lb down to the right at theta = 36.9 degrees
    is the book a typo?
     
  2. jcsd
  3. Nov 22, 2007 #2

    PhanthomJay

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    your math is off.....Fx = 60

    What about the conrtribution of Fx(y)??
    No
     
  4. Nov 24, 2007 #3
    thanks for replying Phantom Jay

    Moment of F w/ respect to O = 180ft - lb = Fx(3)
    Fx = 60lb

    I dont understand this part:
    Since F passes through A... and you compute its moment w/ respect to O
    Fx on this part is clockwise so it is a positive moment
    Fy on this part is counterclockwise so it has a negative moment

    Moment of F w/ respect to B = 90ft-lb = Fx(3) - Fy(6)
    90 = (60)(3) - Fy(6)
    Fy(6) = 90 - 180
    Fy = 30lb

    but when i put the counter clockwise as positive moment the answer would be 45 and the answer would be the same as of the back of the book....
    is it because of the signs?
     
    Last edited: Nov 24, 2007
  5. Nov 26, 2007 #4

    PhanthomJay

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    You have chosen clockwise as positive therfore counterclockwise is negative. Therefore, your equation should read, asuming Fy acts down,
    -90 = 60(3) -Fy(6)
    -270 = -Fy(6)
    Fy = 45 downward
     
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