a uniform pole 20ft long ang weighing 50lbs is used to carry a load of 180lbs.Two men,one supporting twice as much as the other support the pole to keep it in a horizontal position.How far from the center of the pole should the load be hung.?

Answer 1

sum of forces:
50+180=f1+F2 where f2 and f1 are the men pushing upward.


let f1=2F2 (given) f3 is the lighter load man.

50+180=3F2
f2=230/3
f1=460/3

sum moments (about the center of pole).
10F1-10F2+x180=0
you know f1, f2, solve for x