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P13 039

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L└o Ribeiro
L└o Ribeiro

1) The forces exerted by wires A (FA) and B (FB) on a log are related by the stretching of the wires based on their original lengths (LA and LB), cross-sectional areas, and Young's modulus. 2) Equating the total force on the log (FA + FB - mg) to 0 allows solving for FA and FB in terms of the known parameters. 3) FA is calculated to be 866 N and FB is then calculated from the total force equation to be 143 N.

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39. (a) Let FA and FB be the forces exerted by the wires on the log and let m be the mass of the log. Since the log is in equilibrium FA + FB ? mg = 0. Information given about the stretching of the wires allows us to ?nd a relationship between FA and FB . If wire A originally had a length LA and stretches by ?LA , then ?LA = FA LA /AE, where A is the cross-sectional area of the wire and E 2 is Young¨s modulus for steel (200 〜 109 N/m ). Similarly, ?LB = FB LB /AE. If is the amount by which B was originally longer than A then, since they have the same length after the log is attached, ?LA = ?LB + . This means FA LA FB LB = + . AE AE We solve for FB : FA LA AE FB = ? . LB LB We substitute into FA + FB ? mg = 0 and obtain mgLB + AE FA = . LA + LB The cross-sectional area of a wire is A = πr2 = π(1.20 〜 10?3 m)2 = 4.52 〜 10?6 m2 . Both LA and LB may be taken to be 2.50 m without loss of signi?cance. Thus (103 kg)(9.8 m/s2 )(2.50 m) + (4.52 〜 10?6 m2 )(200 〜 109 N/m )(2.0 〜 10?3 m) 2 FA = 2.50 m + 2.50 m = 866 N . (b) From the condition FA + FB ? mg = 0, we obtain 2 FB = mg ? FA = (103 kg)(9.8 m/s ) ? 866 N = 143 N . (c) The net torque must also vanish. We place the origin on the surface of the log at a point directly above the center of mass. The force of gravity does not exert a torque about this point. Then, the torque equation becomes FA dA ? FB dB = 0, which leads to dA FB 143 N = = = 0.165 . dB FA 866 N

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P13 039

  • 1. 39. (a) Let FA and FB be the forces exerted by the wires on the log and let m be the mass of the log. Since the log is in equilibrium FA + FB ? mg = 0. Information given about the stretching of the wires allows us to ?nd a relationship between FA and FB . If wire A originally had a length LA and stretches by ?LA , then ?LA = FA LA /AE, where A is the cross-sectional area of the wire and E 2 is Young¨s modulus for steel (200 〜 109 N/m ). Similarly, ?LB = FB LB /AE. If is the amount by which B was originally longer than A then, since they have the same length after the log is attached, ?LA = ?LB + . This means FA LA FB LB = + . AE AE We solve for FB : FA LA AE FB = ? . LB LB We substitute into FA + FB ? mg = 0 and obtain mgLB + AE FA = . LA + LB The cross-sectional area of a wire is A = πr2 = π(1.20 〜 10?3 m)2 = 4.52 〜 10?6 m2 . Both LA and LB may be taken to be 2.50 m without loss of signi?cance. Thus (103 kg)(9.8 m/s2 )(2.50 m) + (4.52 〜 10?6 m2 )(200 〜 109 N/m )(2.0 〜 10?3 m) 2 FA = 2.50 m + 2.50 m = 866 N . (b) From the condition FA + FB ? mg = 0, we obtain 2 FB = mg ? FA = (103 kg)(9.8 m/s ) ? 866 N = 143 N . (c) The net torque must also vanish. We place the origin on the surface of the log at a point directly above the center of mass. The force of gravity does not exert a torque about this point. Then, the torque equation becomes FA dA ? FB dB = 0, which leads to dA FB 143 N = = = 0.165 . dB FA 866 N