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HW9-phy2048-Spring-2014-solution

# 165 db fa 866 n 13p8 using f gmmr2 we find that the

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Unformatted text preview: um, FA + FB – mg = 0. Information given about the stretching of the wires allows us to find a relationship between FA and FB. If wire A originally had a length LA and stretches by ΔL A , then ΔLA = FA LA / AE , where A is the cross- sectional area of the wire and E is Young’s modulus for steel (200 × 109 N/m2). Similarly, ΔLB = FB LB / AE . If ℓ is the amount by which B was originally longer than A then, since they have the same length after the log is attached, ΔLA = ΔLB + l . This means 3 PHY2048 Spring 2014 We solve for FB: HW9 FA LA FB LB = + ℓ. AE AE FA LA AEℓ − . LB LB We substitute into FA + FB – mg = 0 and obtain mgLB + AEℓ FA = . LA + LB The cross- sectional area of a wire is FB = A = π r 2 = π (1.20 × 10 −3 m ) = 4.52 × 10 −6 m 2 . 2 Both LA and LB may be taken to be 2.50 m without loss of significance. Thus (103kg) (9.8 m/s2 ) (2.50 m) + (4.52 ×10−6 m2 ) (200 ×109 N/m2 ) (2.0 ×10−3 m) FA = 2.50 m + 2.50 m = 866 N. (b) From the condition FA + FB – mg = 0, we obtain FB = mg − FA = (103kg) (9.8 m/s2 ) − 866 N = 143 N. (c) The net torque must also vanish. We place the origin on the surface of the...
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