Mechanics/Equilibrium?

Question

I need to determine the location of a movable weight (180 lb(m)) hanging from a beam that will result in equal vertical reactions at either end of the beam.

The left side of the beam (A) is attached to a hinge (or pin if you like). 3ft. in the +x direction there is 420lb(f) in the -y direction. 6ft from that (again in the +x direction) there is 640 lb(f) in the -y direction. Then another 6ft from that is the end of the bar which rests on a roller (point B). So the total length of the beam is 15ft.

Neglecting the weight of the beam I figure the total force in the -y direction is -1240 lb(f). So the reactions at A and B should each be 620 lb(f) in the +y direction right? So then, the total moments are: 620(0)+620(15ft)-420(3ft)-640(9ft)-180(xft)=0 right? I determined that the weight should be 12.666667 ft from the hinge. Did I do this right?

Also, there are no forces in the x direction.

Thanks for any help.

Standard gravity (32.174 ft/s^2) exists so the 180 lb(m) weight would be 180 lb(f) right?

Okay, I feel better seeing that I'm going in the right direction, but I'm a bit confused with the result.

The 640 lbs is at 9ft, then the 180 lb weight is at 12.6ft. Just looking at a diagram I drew, it would seem there is much more force on the right part of the beam than the left.

Answer 1

Yes, your answer is correct.

The total downward force with the 180 lb weight included would be 1240 lb. If this is to be equally supported at A & B, then each must support 620 lb. If you set up the equation with the sum of moments around A set to 0 and the location of the 180 lb weight as the unknown, you'll get:

Sum M[A] = 420(3) + 640(9) + 180(x) - 620(15) = 0

Solving for x, you'll get 12.66667 ft.

Answer 2

Add all the wights together including the movable one. Divide by 2 to get the desired reaction at each end. Then sum moments around the hinge point and set equal to zero. For the movable weight use x as the moment arm. Then solve your equation for x.