The diagram below shows a cross member of a bridge. The mass of the member is M

Question

The diagram below shows a cross member of a bridge. The mass of the member is M = 141 kg and its center of mass is at its geometrical center. Two forces, (1) F ? (with both x and y components) acting at A and (2) R ? (also with both x and y components) acting at O, represent the contacts at the ends of the member. The angle ? = 41.7° and Ry = +2Mg.

What is Rx (N)?
What is Fx (N)?
What is Fy (N)?

I don't even know where to begin how to solve for Rx and Fx. I know that Rx=-Fx. As for Fy:

Fy+Ry-Fg=0

Fy=Fg-Ry=mg-2mg

Fy=-mg=-141kg(9.8m/s^2)=-1381.8N

Please help me with how to solve for Rx and/or Fx. Here is the link to the image:
http://i43.tinypic.com/16iw70y.jpg

Explanation / Answer

taking moment at R
therefore moment equation we get is
l/2cosFg+lcosFx=lsinFy

by cancelling l (length in both sides)

Fx=-1922.03 N

taking forces in X and Y direction

Rx+Fx=0 (there is no gravitational force in X direction)

in y direction Fy+Ry-Fg=0

Rx= -Fx= 1922.03 N

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