11_InstSolManual_PDF_Part2

11_InstSolManual_PDF_Part2 - 11-2 Chapter 11 Solve(a lower...

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Solve: (a) lower wire: upper wire: (b) lower wire: upper wire: 11.5. Set Up: Solve: relaxed: maximum tension: 11.6. Set Up: The force applied to the end of the rope is the weight of the climber: Solve: 11.7. Set Up: For steel, Solve: gives Reflect: The thinner the wire the more it stretches for the same applied force. The length of this wire changes by 0.12% 11.8. Set Up: A 5.0% elongation means For a spring, Solve: (a) (b) so The change in length is gives (c) and 11.9. Set Up: so and From Problem 11.8, for the natural Achilles tendon. Solve: (a) so so and the diameter is 2.2 mm. (b) The natural tendon has and diameter 10.0 mm. The artificial tendon’s diame- ter is much smaller. Reflect: The artificial tendon has a larger Y and therefore a smaller diameter. r 5 " 1 78.1 mm 2 2 / p 5 4.99 mm r 5 " A / p 5 1.1 mm A 5 p r 2 A 5 kl 0 Y 5 1 4.6 3 10 5 N / m 21 0.25 m 2 30 3 10 9 Pa 5 3.8 3 10 2 6 m 2 k 5 YA l 0 A 5 p r 2 k 5 4.6 3 10 5 N / m k 5 YA l 0 . F T 5 1 YA l 0 2 D l Y 5 F T / A D l / l 0 x 5 F T k 5 9555 N 4.6 3 10 5 N / m 5 2.08 cm F 5 13 mg 5 13 1 75 kg 21 9.80 m / s 2 2 5 9555 N k 5 F T x 5 5.76 3 10 3 N 1.25 3 10 2 2 m 5 4.6 3 10 5 N / m F T 5 kx x 5 D l 5 1 0.050 21 25 cm 2 5 1.25 cm. F T 5 stress 3 A 5 1 7.37 3 10 7 Pa 21 78.1 3 10 2 6 m 2 2 5 5.76 3 10 3 N stress 5 F T / A stress 5 Y 3 strain 5 1 1474 3 10 6 Pa 21 0.050 2 5 7.4 3 10 7 Pa F T 5 kx . D l / l 0 5 0.050. Y 5 stress strain . d 5 Å 4 A p 5 Å 4 1 1.6 3 10 2 6 m 2 2 p 5 1.43 mm A 5 l 0 F ' Y
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