The equilibrium conditiions for the pole is
F1+F2=PF_1+F_2=PF1+F2=P and Pl/2=F2(l−x)Pl/2=F_2(l-x)Pl/2=F2(l−x) then x=l(P/2−F1)P−F1x=\frac{l(P/2-F_1)}{P-F_1}x=P−F1l(P/2−F1) hence x≈3.8mx\approx3.8mx≈3.8m
where l=10m,F1=100N,P=mg=490Nl=10m,F_1=100N,P=mg=490Nl=10m,F1=100N,P=mg=490N
(b) 3.8
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