Weight of a beam; please help?

Question

A uniform beam of length L is hanging from a point a distance x to the right of its center. The beam weighs W and makes an angle of theta1 with the vertical. At the right-hand end of the beam a concrete block weighing w1 is hung; an unknown weight w hangs at the other end.

1.] If the system is in equilibrium, what is w? You can ignore the thickness of the beam.


2.] If the beam makes, instead, an angle of theta2 with the vertical, what is w?


THANKS in advance

Answer 1

w = [w1 (L/2 - x) - W x] / (L/2 + x).

The answer is INDEPENDENT of the angle theta; it DOESN'T matter whether it's theta1 or theta2 !

That's because the angle completely FACTORS OUT of the equation of equilibrium. This is OBVIOUS, but if you insist on seeing it done, just take moments of the forces about the point of suspension.

The moment of the weight w1 on the right side is w1(L/2 - x) sin theta1.

The weight of the uniform beam can be considered as acting through its mid-point. So :

The moments of the weights on the left side are W x sin theta1 + w (L/2 + x) sin theta1.

In equilibrium, these must balance one another, hence :

w1(L/2 - x) sin theta1 = W x sin theta1 + w (L/2 + x) sin theta1.

As I already noted, since sin theta1 is common to all three terms, it completely cancels out, leaving ABSOLUTELY NO DEPENDENCE on the angle theta1.

The value of w is therefore 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 relation holds, the beam is in NEUTRAL EQUILIBRIUM, no matter what angle it is placed at. Thus, if the problem was indeed posed as you stated it, I'm afraid that it was set as a test to see whether you would appreciate that fact!

Live long and prosper.