11_InstSolManual_PDF_Part4

# 11_InstSolManual_PDF_Part4 - 11-4 Chapter 11 Dp DV 52 1 atm...

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11.16. Set Up: Solve: (a) (b) The depth for a pressure increase of is 1.5 km. Unprotected dives do not approach this depth so bone compression is not a concern. 11.17. Set Up: For steel the shear modulus is Solve: A is the area of each edge of the plate: 11.18. Set Up: is the component of the force tangent to the surface, so 1360 N. must be in radians, Solve: 11.19. Set Up: is in radians. with Solve: Reflect: The shear modulus of cartilage is much less than the values for metals given in Table 11.1. 11.20. Set Up: The tensile force equals the weight w of the object suspended from the wire. The maximum weight that can be supported in one that makes the stress equal to the breaking stress. Solve: (a) so (b) so (c) so 11.21. Set Up: The free-body diagram for the elevator is given in Figure 11.21. is the tension in the cable. Figure 11.21 mg a x F ' y F ' F ' A 5 1 3 1 2.40 3 10 8 Pa 2 5 0.80 3 10 8 Pa. w 5 1 11.0 3 10 8 Pa 21 0.040 3 10 2 4 m 2 2 5 4.4 3 10 3 N F ' / A 5 11.0 3 10 8 Pa D l 5 l 0 F ' YA 5 1 5.00 m 21 1.44 3 10 3 N 2 1 2.0 3 10
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Unformatted text preview: 1 2.0 3 10 11 Pa 21 0.040 3 10 2 4 m 2 2 5 9.0 3 10 2 3 m 5 9.0 mm. Y 5 l F ' A D l w 5 1 3.60 3 10 8 Pa 21 0.040 3 10 2 4 m 2 2 5 1.4 3 10 3 N. F ' A 5 3.60 3 10 8 Pa F ' f 5 F i AS 5 8 mg sin 12° 1 10 3 10 2 4 m 2 21 12 3 10 6 Pa 2 5 0.1494 rad 5 8.6° p rad 5 180° m 5 110 kg. F 5 8 mg , f F i 5 F sin 12°. S 5 F i A f . S 5 1360 N 1 0.0925 m 2 2 1 0.0216 rad 2 5 7.36 3 10 6 Pa f 5 1.24° 5 0.0216 rad. f F i 5 1 1375 N 2 cos 8.50° 5 F i S 5 F i A f . F i 5 SA 1 shear strain 2 5 1 0.84 3 10 11 Pa 21 1.00 3 10 2 3 m 2 21 0.0400 2 5 3.4 3 10 6 N A 5 1 0.100 m 21 0.0100 m 2 5 1.00 3 10 2 3 m 2 . S 5 shear stress shear strain 5 F i / A shear strain . S 5 0.84 3 10 11 Pa. 1.5 3 10 7 Pa D p 5 2 B D V V 5 2 1 15 3 10 9 Pa 21 2 0.0010 2 5 1.5 3 10 7 Pa 5 150 atm 1 atm 5 1.01 3 10 5 Pa. D V V 5 2 D p B . 11-4 Chapter 11...
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