11_InstSolManual_PDF_Part14

11_InstSolManual_PDF_Part14 - 5 8.4 3 10 4 N 2 1 70 kg 21...

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11.63. Set Up: For steel, The elongation is a small fraction of the length, so use to calculate Solve: Apply to the mass when it is at the bottom of the circle. The acceleration is and is upward, so take upward. is the tension in the wire. so Reflect: When the mass hangs at rest at the end of the wire, and the wire is stretched 0.37 mm. The radial acceleration is large and is much greater than mg . 11.64. Set Up: so the slope of the graph in part (a) depends on Young’s modulus. is the total load, 20 N plus the added load. Solve: (a) The graph is given in Figure 11.64. Figure 11.64 (b) The slope is (c) The stress is The total load at the proportional limit is 11.65. Set Up: The height from which he jumps determines his speed at the ground. The acceleration as he stops depends on the force exerted on his legs by the ground. In considering his motion take downward. Solve: (a) (b) As he is stopped by the ground, the net force on him is where is the force exerted on him by the ground. From part (a), and gives since the acceleration is upward. a y 52 1.19 3 10 3 m / s 2 a 5 1.19 3 10 3 m / s 2 . F net 5 ma 8.33 3 10 4 N. F
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Unformatted text preview: 5 8.4 3 10 4 N 2 1 70 kg 21 9.80 m / s 2 2 5 F ' 5 2 1 4.2 3 10 4 N 2 5 8.4 3 10 4 N F ' F net 5 F ' 2 mg , F ' 5 YA 1 D l l 2 5 1 3.0 3 10 2 4 m 2 21 14 3 10 9 Pa 21 0.010 2 5 4.2 3 10 4 N 1 y F ' A 5 Y 1 D l l 2 . stress 5 80 N p 1 0.35 3 10 2 3 m 2 2 5 2.1 3 10 8 Pa 60 N 1 20 N 5 80 N. F ' / A . Y 5 1 l p r 2 2 1 2.0 3 10 4 N / m 2 5 1 3.50 m p 3 0.35 3 10 2 3 m 4 2 2 1 2.0 3 10 4 N / m 2 5 1.8 3 10 11 Pa 60 N 1 3.32 2 3.02 2 3 10 2 2 m 5 2.0 3 10 4 N / m. 3 3.2 3.4 3.6 3.8 4 4.2 4.4 l 1 cm 2 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 F 1 N 2 F ' F ' 5 1 YA l 2 D l F ' F ' 5 mg 5 147 N D l 5 l F ' AY 5 1 0.50 m 21 1.34 3 10 3 N 2 1 0.010 3 10 2 4 m 2 21 2.0 3 10 11 Pa 2 5 3.35 3 10 2 3 m 5 3.35 mm. Y 5 l F ' A D l F ' 5 m 1 g 1 a rad 2 5 1 15.0 kg 21 9.80 m / s 2 1 79.4 m / s 2 2 5 1.34 3 10 3 N. F ' 2 mg 5 ma rad . F ' 1 y a rad 5 r v 2 5 1 0.50 m 21 12.6 rad / s 2 2 5 79.4 m / s 2 g F y 5 ma y a rad . l 5 0.50 m v 5 2.00 rev / s 5 12.6 rad / s. Y 5 2.0 3 10 11 Pa. 11-14 Chapter 11...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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