How Do You Convert Forces into a Wrench in Vector Mechanics?

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    Reduction Wrench
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The discussion focuses on converting forces into a wrench in vector mechanics, detailing the process of moving forces to a point and calculating their resultant. The user successfully decomposed the forces into Cartesian-vector form and calculated the sum of forces and moments about the x, y, and z axes. Initially confused about the next steps, they realized their moment about the y-axis was incorrect and corrected it. The final moments were confirmed as M_x=5.86 and M_y=-14.1, leading to the identification of M_perp and M_parallel. The user expressed satisfaction upon resolving their confusion and confirming their calculations.
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[SOLVED] Reduction to a Wrench

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So far I have moved the forces to point O where \sum F=F_r=-10.0\hat{j} I have also re-written each force in Cartesian-vector form where

F_1=10[\frac{6i-6j}{\sqrt{72}}=7.071i-7.071j F_2=-10j and F_3=-F_1=-7.071i+7.071j.

I am told to decompose each force into its components and then use scalar method to find the sum (which I did above) and to find the sum of the moments about each of the x,y,z axes. Then move the forces to get a wrench.

I have found each of the moments as follows:
\sum M_x=2(10)-2(7.071)=5.858
\sum M_y=-6(7.071)=-42.426
\sum M_z=6(7.071)-6(7.071)= 0

Now I am a little confused. I am just not sure where to go from here. I know I need to find M-perp and M-parellel which looks to be what I have just found.

Are my moments correct? I think they are. And where do I go from here?
 
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I GOT IT! Amidst all my typing I noticed that my moment about y-axis is incorrect.

M_x=5.86=M_perp--->distance from z =5.86/10=.586 ft.
M_y=-14.1=M_parellel

YAY!
 
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