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Sum of the Forces at a point

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data

    If the weight of the boom is negligible compared with the applied 45-kN load, determine the cable tensions T1 and T2 and the force acting at the ball joint at A.


    2. Relevant equations



    3. The attempt at a solution

    I have successfully calculated the tensions in T1 and T2 to be 71.1512 kN and 60.749kN respectively.

    But I've been having a lot of difficulty calculating the forces at point A. My attempt is as follows:

    ƩFx = 0

    -T2 + T1cos(θx) + Ax

    θx = arccos(9/15.81) where 15.81 is the length of the cable T1

    Ax = 60.749 - 71.1512cos(55.304)
    Ax = 20.2481

    ƩFy = 0

    θy = arctan (15/9)
    θy = 59.036

    0 = t1cos(59.036) - 45 + Ay
    Ay = 8.393

    ƩFz = 0
    θz = arctan(10.2956/12) where 10.2956 is the length from A to C.
    θz = 40.62857

    0 = Az - 71.1512cos(40.62857)
    Az = 53.999

    A = [itex]\sqrt{}20.2481^2 + 8.393^2 + 53.999^2[/itex]
    A = 58.278

    It says my answer is wrong and I don't see where I'm going wrong. Help would be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Oct 12, 2012 #2

    tiny-tim

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    Homework Helper

    Hi Northbysouth! :smile:

    (try using the X2 button just above the Reply box :wink:)
    yes :smile:, but there was no need to find the actual angle

    all you needed to know in -T2 + T1cos(θx) + Ax was cosx) …

    which you know is 9/15.81 ! :wink:

    i don't understand why you then used tan for the y and z directions, instead of the cos you used for the x direction :confused:

    try it the same way :smile:

    (if it helps, you can visualise the tension T1 as replaced by forces of relative magnitudes 9:5:-12 along the x y and z axes)
     
  4. Oct 12, 2012 #3
    I'm not sure I completely understand what you're saying.

    When I was calculating the angles I imagined that the origin was at point B. From there I then used trigonometry to determine what angle the force T1 made with each axis.

    So, when I was determining θy I visualized a right angle triangle that went from B to C and then directly back to the y axis at B.

    Are you saying that I should use arccos instead of arctan when I'm calculating the angles at θy and θz?
     
  5. Oct 12, 2012 #4

    tiny-tim

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    Science Advisor
    Homework Helper

    not following you :confused:

    let's see what you did …
    where does the 15 come from?

    (i can see a 15, but it's a hyptoneuse, and it's the wrong angle)

    the unit vector along T1 is (9,5,-12)/15.81 …

    the cosines are the individual coordinates :smile:
     
  6. Oct 12, 2012 #5
    I understand what you mean when you said that I didn't need to calculate the angles.

    Using the unit vector of T1 = <9, 5, -12>/[itex]\sqrt{}250[/itex]

    I multiplied the components i, j and k by the force to get the magnitude of the force in the x, y and z direction. From there it was simple to calculate Ax, Ay and Az

    The answer I got for A is 61.9053 kN, which is correct.

    Thank you for your help.
     
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