Question

Second Condition of Quilibrium

Two men are carrying a 15.00m telephone pole that weighs 100.00kg. if the center gravity of the pole is 6.00m from the right end, and the men lift the pole at the ends, how much weight must each man support?

Answer 1

W_1 is weight at the left end, W_2 is the weight at the right end.
Total weight is
W=W_1+W_2=mg
Second Condition of Equilibrium:
W_1 (l-d)=W_2 d
W_1 (l-d)=(mg-W_1)d
(mg-W_1 )=W_1 (l-d)/d
W_1=mg 1/(1+(l-d)/d)=mg d/l=(100)(9.81) 6/15=392 N.
W_2=(mg-W_1 )=(mg-mg d/l)=mg(1-d/l)=(100)(9.81)(1-6/15)=589 N.