ME 242 Selected Answers Chapter 5

# ME 242 Selected Answers Chapter 5 - 5.1 5.1.2 5.1.4 5.1.6...

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5.1 Force Balance and Linear Momentum 5.1.2 * r G = (2 * ı + 2 3 * )ft 5.1.4 * r G = (0 . 138 * ı - 0 . 108 * - 0 . 0538 * k ) m 5.1.6 * L = m * v G = - 20 * kg · m/s 5.1.8 ¯ v = 15 . 4 m / s 5.1.10 v 3 = 1 . 13 m/s 5.1.12 Explosion at t = 2 s, velocity in * ı direction. 5.1.14 The prosecution is correct. 5.1.16 * a A = 0 . 746 * ı m/s 2 5.1.18 Masses do not move together as a single unit. 5.1.20 * v 1 ( t 2 ) = ( - 5 * + 7 . 071 * ı ) m/s, * v 2 ( t 2 ) = ( - 5 * - 7 . 071 * ) m/s 5.1.22 * F = - 72 * ı N 5.1.24 m p 2 m 5.1.26 m P = 2 m = 2(60 kg) = 120 kg 5.1.28 Δ t = 9 . 74 × 10 - 3 s 5.1.30 T 2 = 242 lb, T 3 = 363 lb 5.2 Angular Momentum 5.2.2 (0 . 2 * ı + 11 . 6 * - 4 . 0 * k )kg · cm 2 /s 5.2.4 * H P = ( - 1 . 1 × 10 - 3 * ı + 1 . 0 × 10 - 3 * +8 . 6 × 10 - 3 ) kg · m 2 /s 5.2.6 * H O = ( - 12 * j - 16 . 5 * k ) kg · cm 2 /s 5.2.8 * H O = - 15 * i kg · cm 2 /s 5.2.10 * H P = (1 . 72 × 10 - 3 * ı - 1 . 24 × 10 - 3 * - 1 . 6 × 10 - 4 * k )kg · cm 2 /s 5.2.12 * H P = 11 . 5 * k kg · cm 2 / s 5.3 Work and Energy 5.3.2 75% reduction in rotational kinetic energy 5.4 Systems with Mass Infow and Outfow 5.5 Non-constant Mass Sys- tems Systems with Mass Infow and Outfow 5.5.2 F = 4 . 64 × 10 3 N 5.5.4 c = 0 .
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