ch12-p045 - 45. (a) Let FA and FB be the forces exerted by...

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stretching of the wires allows us to find a relationship between F A and F B . If wire A originally had a length L A and stretches by Δ L A , then Δ LF L A E AA A = / , where A is the cross–sectional area of the wire and E is Young’s modulus for steel (200 × 10 9 N/m 2 ). Similarly, Δ L A E BB B = / . If A is the amount by which B was originally longer than A then, since they have the same length after the log is attached, AB LL Δ= Δ+ A . This means FL AE AE =+ A . We solve for F B : F L AE L B =− A . We substitute into F A + F B – mg = 0 and obtain F mgL AE A B = + + A . The cross–sectional area of a wire is Ar == × = × −− ππ 23 2 6 120 10 4 52 10 .. mm 2 ch . Both L A and L B may be taken to be 2.50 m without loss of significance. Thus 26 2 9 2 3 (103kg)(9.8m/s )(2.50m)+(4.52 10 m )(200 10 N/m )(2.0 10 m) 2.50m+2.50m 866 N.
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