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Unformatted text preview: to about 2.9 × 108 N/m2, this is approximately the yield strength for the material. 12P48. Since the force is (stress × area) and the displacement is (strain × length), we can write the work integral (Eq. 7 32) as W = ∫ Fdx = ∫ (stress) A (differential strain)L = AL ∫ (stress) (differential strain) which means the work is (wire area) × (wire length) × (graph area under curve). Since the area of a triangle (see the graph in the problem statement) is 1 2
(base)(height) then we determine the work done to be W = (2.00 × 10−6 m2)(0.800 m)⎛1⎞(1.0 × 10−3)(7.0 × 107 N/m2) = 0.0560 J. ⎝2⎠ 12P49. (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 equilibri...
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 Spring '14
 Physics, Force, Work

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