Problem%20Set%2C%20Chapter%206%20Solutions

Problem%20Set%2C%20Chapter%206%20Solutions - Problem Set,...

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Problem Set, Chapter 6 1. Given : W = 12 lb, Gm = 80, h = 36” To find : k2, dm, τ dm = 2h/Gm = 2*36/80 = 0.9” Gm = √(2k2h/W2) or k2 = (Gm 2 *W2/2h) = [(80 2 *12)/(2*36)] = 1067 lb/in τ = π√(W2/k2g) = π√(12/1067*386.4) = 0.0169 s or 16.9 ms 2. Given : W = 500 lb, Gm = 5, h = 24” To find : dm, τ dm = 2h/Gm = 2*24/5 = 9.6” Gm = √(2k2h/W2) or k2 = (Gm 2 *W2/2h) = [(5 2 *500)/(2*24)] = 260.42 lb/in τ = π√(W2/k2g) = π√(500/260.42*386.4) = 0.2213 s or 221.3 ms 3. Continuation of problem #2 Given : product dimensions = 12”*12”*12” WL or dm = 0.5TL or thickness of cushion, T T = 2dm = 2*9.6” = 19.2” To find : interior dimensions of package Interior L, W, H = 12 + (19.2*2) = 50.4” Interior dimensions of box = 50.4” * 50.4” * 50.4” 4. Given : Gm = 40, dm = 5/8 T, h = 42”, product dimensions = 11”*14”*17” To find : dimensions of box dm = 2h/Gm = 2*42/40 = 2.1” T = 8/5 dm = 8/5*2.1 = 3.36” Interior L = 11 + (3.36*2) = 17.72” Interior W = 14 + (3.36*2) = 20.72” Interior H = 17 + (3.36*2) = 23.72” Interior dimensions of box = 17.72” * 20.72” * 23.72” 5. Given : Gm = 25, dm = 7/8 T, h = 18”, product dimensions = 3”*6”*2.5” To find : dimensions of box dm = 2h/Gm = 2*18/25 = 1.44” T = 8/7 dm = 8/7*1.44 = 1.65” Interior L = 3 + (1.65*2) = 6.3” Interior W = 6 + (1.65*2) = 9.3” Interior H = 2.5 + (1.65*2) = 5.8” Interior dimensions of box = 6.3” * 9.3” * 5.8” 12” 12” 12” T T T
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6. Given : W = 18 lb, k = 882 lb/in, h = 30” To find : Gm, τ Gm = √(2k2h/W2) = √(2*882*30/18) = 54.2 g’s τ = π√(W2/k2g) = π√(18/882*386.4) = 0.0228 s = 22.8 ms 7. Given : W = 19 lb, k = 931 lb/in, h = 24” To find : Gm, τ Gm = √(2k2h/W2) = √(2*931*24/19) = 48.5 g’s τ = π√(W2/k2g) = π√(19/931*386.4) = 0.0228 s = 22.8 ms 8. Given : See figure representing the product, h = 38”, dm = 0.5 T To find : dimensions of box dm1 = 2h/Gm1 = 2*38/25 = 3.04” dm2 = 2h/Gm2 = 2*38/42 = 1.81” dm3 = 2h/Gm3 = 2*38/18 = 4.22” T1 = 2 dm1 = 2*3.04 = 6.08” T2 = 2 dm2 = 2*1.81 = 3.62” T3 = 2 dm3 = 2*4.22 = 8.44” Interior L = 4 + (8.44*2) = 20.88” Interior W = 6 + (3.62*2) = 13.24” Interior H = 8 + (6.08*2) = 20.16” Interior dimensions of box = 20.88” * 13.24” * 20.16” 4” 6” 8” 25 g’s 42 g’s 18 g’s 1 2 3
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9. To better understand this problem see example solved in the textbook on page 91 Given : See figure representing any one of the 3 spring-mass systems To find : true calibration f1/f2 = 25/5 = 5 Looking at the table (figure 6.11 in textbook), Am = 1.083 True calibration, Ge = AmGm = 1.083*10 = 10.83 g’s/division A. Given : Gm = 100, τ = 25 ms To find : recorder reading f2 = 1/2 τ = 1/(2*0.025) = 20 Hz f1/f2 = 25/20 = 1.25 Looking at the table, Am = 1.71 Ge = AmGm = 1.71*100 = 171 g’s/division 170.6 g’s at the stylus causes the it to move, 171/10.83 = 15.79 divisions B. Given : Gm = 100, τ = 10 ms To find : recorder reading f2 = 1/2 τ = 1/(2*0.01) = 50 Hz
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This note was uploaded on 04/09/2008 for the course IT 403 taught by Professor Olsen during the Spring '08 term at Cal Poly.

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Problem%20Set%2C%20Chapter%206%20Solutions - Problem Set,...

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