Load Distribution/Rigid Body Equilibirum Problems

[jhchex]

Homework Statement

The problem statement is attached to the image.

Homework Equations

ƩFx = 0
ƩFy = 0
ƩMa = 0.

The Attempt at a Solution

So I summed the forces in both x&y and found the resultant force to be = -48i -36j [N].
I summed to moments and got -308N*m, but I am not sure if that makes sense. I don't know how I am supposed to show an equivalent loading at A.

 

[kjohnson]

I think you may need to check your moment summation again, but i could be wrong (i just glanced at the problem quick)...

Anyway it seems like you are more stumped on the concept. What you are doing is taking a system (bar in this case) that is being loaded in two ways:
1. applied forces- that also create moments
2. applied moment
and converting it to an equivalent system. To make it an equivalent system the same net force must act on it and the same net moment moment must act on it.

If we want to take an equivalent system about 'a' then we need to "shift" the applied forces to 'a'. This will take care of the net force being equal. BUT we also need to account for the moment the forces caused about 'a'. To do this we simply sum moments about 'a' due to the forces. Lastly we still need to account for the applied moment in the middle of the beam which simply adds (or subtracts) to the moment from the forces.