Solving Simple Statics Problem - Find Force Fb & Fa

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The discussion centers on solving a statics problem involving a bar with two reaction forces, Fb at point B and Fa at point A, and a torque applied at point C. The user struggles with the sum of forces, mistakenly concluding that Fb equals Fa, despite their different distances from point C. The correct approach involves calculating moments around points A and B to find the magnitudes of Fb and Fa. By applying the sum of moments at these points, the user can derive the correct relationships between the forces. The conversation emphasizes the importance of understanding torque and distance ratios in static equilibrium problems.
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Guys
I need some help with a simple statics problem.

I have a bar with a length of Z from point A to point B. A torque is input at point c which is located a length of Y from point A. There are two reaction forces: Fb located at Point B and Fa located at point A. no other forces are input. I am having problems doing my sum of forces and moments.

I know for zero rotation the sum of moments is at point c
T=Fb(Z-Y) + Fa(Y)

However, I can't seem to find the correct sum of forces in one direction let's just say x.

With the sum of forces I get FB = FA I know this isn't true because they are two different distances from Point C and there will have different magnatudes. any help would be appreciated.

I am thinking I need to some how convert T to a force with some sort of ratios of the lengths. any assistance would be appreciated.

I can't believe this simple of a problem is kicking my butt.

Thanks
 
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Try doing the sum of the moments about each end (A and B). About point A, there is no moment due to Fa and then you only have the moment around C and then the moment caused by Fb a distance of Z units away. You solve for Fb. Repeat the process for Fa, but take the moment about point B.

Hope this helps.
 
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