11_InstSolManual_PDF_Part2

11_InstSolManual_PDF_Part2 - 11-2 Chapter 11 Solve: (a)...

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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 5p 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 5p 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
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