stretching of the wires allows us to find a relationship between FAand FB. If wire Aoriginally had a length LAand stretches by ΔLA, then ΔLFLAEAAA=/, where Ais the cross–sectional area of the wire and Eis Young’s modulus for steel (200 × 109N/m2).Similarly, ΔLAEBBB=/. IfAis the amount by which Bwas originally longer than Athen, since they have the same length after the log is attached, ABLLΔ=Δ+A. This means FLAEAE=+A.We solve for FB:FLAELB=−A.We substitute into FA+ FB– mg= 0 and obtain FmgLAEAB=++A.The cross–sectional area of a wire is Ar==×=×−−ππ2326120 104 52 10..mm2ch.BothLAand LBmay be taken to be 2.50 m without loss of significance. Thus 262923(103kg)(9.8m/s )(2.50m)+(4.52 10 m )(200 10 N/m )(2.0 10 m)2.50m+2.50m866 N.
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