A uniform beam of length L hangs from a point at a distance x to the right of its center. The beam weighs W and makes an angle theta1 with the vertical. Hanging from the right end of the beam is a concrete block of weight w1; an unknown weight w hangs from the other end.
Answer 1
w = [w1 (L/2 – x) – W x] / (L/2 + x).
The answer is INDEPENDENT of theta angle; It doesn’t matter if it’s theta1 or theta2!
This is because the angle IS completely OUTSIDE of the equilibrium equation. It’s OBVIOUS, but if you insist on doing it, just take a few moments from the forces above the ellipsis.
The moment of weight w1 on the right side is w1(L/2 – x) sin theta1.
The weight of the uniform beam can be considered to act through its midpoint. So :
The moments of the weights on the left are W x sin theta1 + w (L/2 + x) sin theta1.
At equilibrium, they must balance each other, therefore:
w1(L/2 – x) sin theta1 = W x sin theta1 + w (L/2 + x) sin theta1.
As I have already noted, since sin theta1 is common to all three terms, it completely cancels out, leaving ABSOLUTELY NO DEPENDENCE on the theta1 angle.
The value of w is given by:
w = [w1 (L/2 – x) – W x] / (L/2 + x).
Note that since this equilibrium solution does not depend on ANY ANGLE, no angle can be determined by it. If this relationship holds, the beam is in NEUTRAL BALANCE no matter what angle it is placed. So if the problem was really posed as you said, I’m afraid it was posed as a test to see if you would appreciate that fact!
Live long and prosper.