Reduction of resultant Moment/Force

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Saladsamurai
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[SOLVED] Reduction of resultant Moment/Force

So I have calculated the resultant force correctly but my solution for distance is incorrect. I keep getting 6.08 ft but the correct solution is 6.57 feet from B.
Isometric line drawing of a hollow cylindrical shell with a rolled-back flap, showing inner radius R1, outer radius R2, and labels Q1 and Q2 on the wall.


I found [itex]F_r=798.499[/itex] which is correct. Then I used,

[tex]\sum M_B=\frac{3}{5}(500)(9)+6(1200)+\frac{12}{13}(260)(4)=4860 lb\cdot ft[/tex]

Then [itex]d=\frac{M_r}{F_r}=6.09 ft[/itex] Where is my error?
 
F_r is the total magnitude of the resulting force, yes. But in the formula d=M/F (where M is the torque, and d is distance from axis), you want the F to be component of the total force that exerts torque. That would be the vertical component only of the total force.