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Finding moment, relative to a different axis

  • Thread starter Dell
  • Start date
  • #1
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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.
 
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Answers and Replies

  • #2
590
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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:

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