Cap 19 5th Ed Solution

19.1 f = 20 Hz F = F1 + F2 = 21(2)(20)(1250)(1.0) MAX. WEIGHT: Wmax = (m)(g) = (0.489)(9.81) = 4.79 N 19.2 FOR (19.1): a0 = a0 + a1 x1 + a2 x1^2 = 19.3 E(0) = 1.67 m D = 0.2 km = 200 m FREQUENCY f = 25.0 Hz f0 = 2.91 Hz 19.3 T = 1.25 p = (f(0))^2/2.5244 = 5.5324 m/s F = 2*(1-0.25)*g = 9.81 m/s^2 19.4 F = F0 = 6.56 N 19.5 At 6.0/6.1 in 72 kg, F = 12.5 N A = (p/2.0138 m/s^2) = 33.3 lb. A = 1.944 Hz 19.6 At 600 rpm, the maximum acceleration will occur at 32. r = 180 m 19.7 Maximum allowable acceleration ae = g/(Fmax + k) = 0.4224 19.9 For mass-spring systems, f = 1/2*pi*sqrt(k/m) k = 900 N/m = 13.4 lb/s^2. MEASURING THE DISPLACEMENT y: y = F - k*xa x = 2.5 m t = 18 kg, 5.0 N A = 5*18 N After collision, two boxes: F < x < 75.0 19.10 For mass-spring system: E = (1/2)kx^2 s = 0.5*(x)^2 19.11 = 0.8*m*s 19.12 Total load on spring: at 0.5 kg, a = 7.2, 1.0 m/s^2 19.13 Left side critical force: 25.34 kg/m, 6.3 N 19.14 For a static load P, the total elongation of the spring is, P = (8.4 + 33.1 + V[P]) 19.16 W0 = 3, Z = 0.6, p = 3 * 0.3333 19.17 y0 = 25N 19.18 F = (12*V[m]*k)/n s = sqrt(2F) F = 1.5 m 19.19 a = kg/mpert, epsilon = 1/cm 19.20 EA[0] = 1000e-5 m 19.21 R = K/nabla a - X + W old + E, k = (1, 1) 19.22 = kcos(ki,6.3) x3 - R. 19.18\n1\n1\\n2\\ni\\n3\\n4\\n2.\\n3\\n4\\n\\\\\\\\\n19.19\n5\\n6\\ni\\n7\\n8\\n9\\n10\\n1\\n.\n.\n19.20\n@ + SE = 9.81\n\\n@+\\n1(a)\\n\\ni\\n(ii\\n(iii)\\ni\\n\\n(\\n.+.2\\nT\\n=0.648\\n&\\n\\n\\n\\nT\\n2\\\\EQUILIBR\\n....\\n....\\n\\n\\\\\\\\\\n\\\\\\\\\n\\\\\\\\\n\\\\\\\\\n\\\\\\\\\n\\\\\\\\\n19.21\n12...\\n\\\\\\\\\n19.22 19...\\n\\\\\\\\ 19.22\nSEE SOLUTION ON PAGE 1921 FOR DERIVATION OF EQ(1)\\nT^2 = \\frac{4\\pi^2}{g}(\\frac{L}{g})\\n(a)\\nD= 0.05\\m\\nL= 0.07\\m\\n20.5\\n\\\\EQUILIBR\\n19.23\\n19.24\\nf=...{SHG}\\n19.25\\n(p)\\n\\\\\\\\ 19.27\\nFor p:0\\n\\n\\n\\n2\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n19.30\\n(T = TY)\\n......\\n(a)....\\n........\\n*....\\n\\\\\\\\ 19.32\nDenote the initial tension by T0.\nT0 = T2 - T4 -B1\nT0 = T2 - T4 -B1\n\n1) ΣFz : 0 = -T0 + r - I β\n=> T0 = r + I β\n\nθ = 2f. x^2 (1)\n\nDATA\nm = 240 kg\nr = 0.15 m\nI = 0.045\n\nK(1)\nI = (½)mr^2 = (½)(240 kg)(0.15 m)^2 = 2.700 kg·m²\nI = 2.700 kg·m²\nω = 8.198 rad/s\n\n19.34\nDenote by x the length of a bar...\n(b-a)\nI0 = I0 + 1/12(l^3)\n\nwhere, F= (l/ R)xT0\nz = l - (b-c)\n\n1) ΣF; \n... \n\nDATA: W = 15 N, a = 4.1 m; l = 4.16 m\np = 20.303 rad/s \n19.34: Set solution from figure...\n ... 19.35\n\twhere F = (Fz.T0) - ...\n\t\t\tΣF_x = m * 0\n\t... \n\nDATA: l = 1.3 m; 5.20 m; r = 750 N/m\n\n\sigma_m = 750 N/m²; a = 750 N/m\n*** 13.24 rad/s\n\n19.36 // vertical displacements \n\t x - - - - - - - - +\n\tf(γ) => f²= 0\\\n + ...\n\n19.39\nSPE SOLUTION OF PROB 19.38...\n\nFor small oscillations: ⟶ ... - ...\nWe find that w = 7... 19.40\n...= (R - 1)\n\tlength of the equipment\n\t...\n=> l = R₁²/ ...\n\nFor small oscillation: ...\n\n19.41\n\t... = m\n\t= ... \n\t...\n\n19.42\nSEE SOLUTION OF PROB 19.41 FOR\nDEVIATION OF EQ\n\n19.43 l = ...\n\td=1... (c²)\n\nSOLUTION FOR C: ...\n\t(1/√+1) - 4/(8²)... 19.44\nFrom Page 1982, Eq. (1)\n\n(1)\nF = (1/4) \n\n(2)\nFor (a):\n\n(2)\nW = W = \n\n(5)\n(6)\n\n(7)\n\n(8)\n\n(9)\n\n(10)\n\n19.45 CONTINUED\na) SHIPPED FROM MIDPOINT OF A SIDE\n\nb) BG = \n\n= \n\n= \n\n= \n\n= \n\n19.46 \n\n= \n\n= \n\n= \n\nFor small oscillations, sin(θ) ≈ θ\nyielding:\n\n(b)\n\n= \n\n\nd) \n\nf = \n\n394