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Finding moment, relative to a different axis

  1. Apr 27, 2009 #1
    given the following system:
    2 cords are attached to point C,
    http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE&pli=1&gsessionid=oBksCECRwcHSRoX-oRPaRA#5329455433140212770 [Broken]

    and knowing that
    |TCB|=380N
    |TCA|=450N

    find:
    1) the moment relative to x,y,z
    2) the moment relative to the axis 'OB'

    for the 1st part, what i did was
    M=MTCA+MTCB
    and i found that
    MTCA=A×TCA
    MTCB=B×TCB


    now comes the problem, how do i find the momentum relative to OB, i know that i need to make a new axis system, where OB is my "y" axis, and an axis 90degrees to it(63.43° from z axis) will be my new "z" axis, while x will stay the same,
    i found the angle of my new axis relative to the original system

    α=angle between 'y' and OB
    α=26.565°

    to find the moment, i need

    1) force Tx (which is the same force Tx i used before)
    2) the shortest distance dz between Tx and my new z axis
    3) force Tz
    4) the shortest distance dx between Tz and my new x axis

    i am not at all sure that this is correct, if you can see a better way let me know.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 28, 2009 #2
    am i right in saying:

    since the axis OB passes through the origin O, the moment relative to axis OB will be the same as the moment relative to O ????
    BUT THAT DOESNT LOOK RIGHT TO ME,
    i dont think that Mob should have any magnitude on the X axis since the line OB is on the y-z plane:

    i thought that maybe i could do this:

    since i can find the angle of OB, (comes to 26.565 degrees) i can say that since OB is on the YZ plane, the moment can only be My and Mz, so the moment around OB is:

    (My/cos(26.565))j + (Mz/sin(26.565))k

    is this correct??
     
    Last edited: Apr 28, 2009
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