1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook