Finding moment, relative to a different axis

[Dell]

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

and knowing that
|T_CB|=380N
|T_CA|=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=M_TCA+M_TCB
and i found that
M_TCA=A×T_CA
M_TCB=B×T_CB


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 T_x (which is the same force T_x i used before)
2) the shortest distance d_z between T_x and my new z axis
3) force T_z
4) the shortest distance d_x between T_z 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.

 

[Dell]

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 don't 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??

 

[nlightner]

I would like to commend you for your approach in solving this problem. Your use of vector notation and breaking down the moment into components is correct. To find the moment relative to the axis OB, you are correct in creating a new axis system where OB is the y-axis. However, there is a simpler way to find the moment using this new axis system.

First, let's define the new axis system as follows:
- x-axis: parallel to the original x-axis
- y-axis: along OB
- z-axis: perpendicular to both x and y-axis

Now, let's consider the force TCA acting at point C. This force has a component along the new y-axis (Ty) and a component along the new z-axis (Tz). The moment of TCA about the new y-axis is simply Ty multiplied by the distance from the point of application (C) to the new y-axis (OB). Similarly, the moment of TCA about the new z-axis is Tz multiplied by the distance from C to the new z-axis (which is the same as the distance from C to the original z-axis). So, the total moment of TCA about the new axis OB is:

MTCO = Ty * OB + Tz * CB

Similarly, the moment of TCB about the new axis OB is:

MTCO = Tx * OB + Tz * CA

Therefore, the total moment relative to the axis OB is the sum of these two moments:

M = MTCO + MTCB = (Ty * OB + Tz * CB) + (Tx * OB + Tz * CA)

I hope this helps you to find the moment relative to the axis OB. Keep up the good work!