What are the coordinate direction angles for the resultant couple moment?

[Bismuth]

Homework Statement

[PLAIN]http://img25.imageshack.us/img25/5139/physicsproblem.jpg

If F1 = 110 lb and F2 = 300 lb, determine the magnitude and coordinate direction angles of the resultant couple moment.

Homework Equations

M_R = \Sigmar x F, M = Fd

The Attempt at a Solution

Unfortunately, the couple on the angle is too much for my tiny brain to tolerate. I have found the moments to be:
Couple moment of the forces on image M₂₅₀ = 500 lb*ft
M₁ = 220 lb*ft (i)
M₂ = 600 lb*ft (j)

I tried to further break up the angled moment M₂₅₀ into its x and y components based on the hypotenuse it lies perpendicular to of the 3-4-5 triangle within the box, which gave me M = -400(i) - 300(j) + 0(k). Summing all the moments together gives me M_R = -180(i) + 300(j) + 0(k), from which I get \alpha = 121^o and \beta = 31.0^o

I'm pretty sure I'm over complicating things and making a stupid mistake. Any guidance would be greatly appreciated. Thank you.

 

[PhanthomJay]

Your image does not appear.

 

[Bismuth]

Your image does not appear.

How very computer illiterate of me. Thank you for letting me know!

 

[PhanthomJay]

Homework Statement

[PLAIN]http://img25.imageshack.us/img25/5139/physicsproblem.jpg

If F1 = 110 lb and F2 = 300 lb, determine the magnitude and coordinate direction angles of the resultant couple moment.

Homework Equations

M_R = \Sigmar x F, M = Fd

The Attempt at a Solution

Unfortunately, the couple on the angle is too much for my tiny brain to tolerate. I have found the moments to be:
Couple moment of the forces on image M₂₅₀ = 500 lb*ft
M₁ = 220 lb*ft (i)
M₂ = 600 lb*ft (j)

I tried to further break up the angled moment M₂₅₀ into its x and y components based on the hypotenuse it lies perpendicular to of the 3-4-5 triangle within the box, which gave me M = -400(i) - 300(j) + 0(k). Summing all the moments together gives me M_R = -180(i) + 300(j) + 0(k), from which I get \alpha = 121^o and \beta = 31.0^o

I'm pretty sure I'm over complicating things and making a stupid mistake. Any guidance would be greatly appreciated. Thank you.

You didn't overcomplicate it at all, you did it perfectly! Note that the resultant couple makes an angle of 121 degrees ccw with the positive x axis, which I assume is your beta angle. Nice work!

 

[Bismuth]

You didn't overcomplicate it at all, you did it perfectly! Note that the resultant couple makes an angle of 121 degrees ccw with the positive x axis, which I assume is your beta angle. Nice work!

It wants alpha as the angle between the moment axis and the x axis, and beta as the angle between the moment axis and the y axis. I have tried inputting the following:

\alpha = 121, \beta = 31
\alpha = 149, \beta = 31
\alpha = 31, \beta = 59
\alpha = 59, \beta = 31

With no success. Given the resultant moment I found, I believe it should be \alpha = 121, \beta = 31. Thanks for your help!