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# HW15 - 6-75 A 500~lb homogeneous circular plate is...

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Unformatted text preview: 6-75 A 500~lb homogeneous circular plate is supported by three cables as shown in Fig. P6-75. Determine the tensions in the three cables. SOLUTION From a free-body diagram for the piate: For moment equilibrium: SEA : (rBM x TB) + {rcxn x TC} + (rGM x W) = {(12 i + 30 j} x (TB 211 + {1n24 i + 36 31 x (Tr E11 4 11~9 i + 21 j) x (”500 E1} 2 13028 + 36TC - 10.500) i + (-12TB + 242C — 4500) j = 6 Solving yields: if TB = 78.13 lb e 78.1 lb T 78.1 E 1b = 73.1 lb T' Ans. 227 2 lb 2 227 1b 1 Ans. Ii TC = 226.56 lb 2 227 lb T For force equilibrium: ﬁzTA+TB+TC+W = (TA + TB + TC — 5001 E = (TA + 73.13 + 226.56 ~ 500) E = 6 TA = 196.31 lb 3 195.3 lb TA = 195.3 E 1b 195.3 lb T Ans. 44\$ W. F. RILEY AND L. D. STURGES E ENGINEERING MECHANICS - STATICS, 2nd. 6‘34 The block W shown in Fig. P6-84 has a mass of 250 kg. . 2 Bar AB rests against a ***. -& smooth vertical wall at end“ B and is supported at end A ‘ with a ball and socket _ .¢“¢ - joint. The two cables are 2w 3 - n._ attached to a point on the "“ “tn. momm bar midway between the ends. 30.91““ D t ' 'h t' t l~«7?r>*wl;- A e ermine t e reac lens a /« -“Hgﬁ“*2mnmn A, . supports A and B and the " ~~—~—~ /'/ “M -‘ r . _ ' SUI] mm ‘3' tens1ons In the two cables. . : J) 200 mm SOLUTION T —0.200 t + 0.200 3 + 0.500 E c0 " 00 0/0 to 2 2 "mm? {{"0.200) + {0.200) + (0.500) : TCD{~0.3482 i + 0.3402 3 + 0.8704 ﬁ) 1-13 l h-EI a» n W = mg : 250(9.007)(HE) = —2452 E N From a free-body diagram for the bar: a For moment equilibrium: SE 3 (r010 x Ten) + (rB/A X E} + {rD/A x w) = [{—0.300 3 + 0.800 E) x 00.34820CD i + 0.3482TCD j + 0.0704Tfp E): + [(0.400 t — 1.000 3 + 0.000 ﬂ» x {B 3)] + [40.200 3 a 0.500 3 + 0.300 Q) x ("2452 E)] = 00.53971“CD — 0.0003 + 1226.0) i + (—0.2786TCD + 490.4) 3 + (-0.10446TCD + 0.4000) E = ﬁ 475’ §ET=E+B§+T +§§ = Ax E + av 3 + AZ E + 4590? 3 _ 612.9 E + 61209 3 + 1532,: E ~ 2452 E = (AV , 61209) i + gay + 1072u6} 3 + (AZ & 919q9) E z (’3 9:3 K = 61259 E « 13?2°6 3 + 919:9 g i g 613 E - 1073 3 + 920 Q N Ansa g A = 1(61301} + {-10?3°@}2 + (92093)2 = 1540.8 N 2 1 41 N 0.31 ? ENGINEERING MECHANICS ~ STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6~90 The plate shown in Fig. P6-90 has a mass of 75 kg. The brackets at supports A and B exert only force reactions on the plate. Each of the brackets can resist a force along the axis of pins in one direction only. Determine the reactions at supports A and B and the tension in the cable. Fig. £15-99 SOLUTION 3 T = T g z T —1.00 i - 1.40 cos 30° 3 + {1.20 + 1.40 sin 30°) E C C D/C C 2 2 2 /{—1;00) + (\$1.40 cos 30°) + (1.20 + 1.40 sin 30°) A = "0.4056TC 1 — 0.4917T, 3 + 0.7706TC ﬂ W = m§ = 75(9.81)(—E) : —735.?5 ﬁ N a —735.8 E N 4 From a free-body diagram ’ for the plate: For moment equilibrium: 235 z (re/B x TC) + (PA/B x A} + {re/B x W) = [(1 i + 1.2 E) x (-0.4056TC i — 0.4917TC 3 + 0.7706TC 3)] + {(2 i) x (AV 3 + A2 §)] + {(1 i + 0.70 cos 30° 3 u 0.70 sin 30° E) x (“735.8 E)} (0.590011C u 446.0) E + (-1.2573TC ~ 2A2 + 735.8) 3 + (~0.4917TC + 2A } E = 6 Y 484 I ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES i ' 6-90 (Continued) I Solving yields: TC 3 755.9 N g 756 N ' Ans. TC 3 755.9(~0.4056 i — 0.4917 3 + 0.7?06 ﬁ) ~305.5 E a 371.7 3 + 582.5 E N A = 185.84 N . Y A2 = ~107.30 N K = 135.34 3 - 107.30 g N a 135.3 3 - 102.3 Q N Ans. A = «(135.3412 + {—107.30)2 = 215.59 5 2 215 5 For force equilibrium: 2? z E + E + TC + w = 135.34 3 1 107.30 E + 3x 1 + By 3 + 32 ﬁ — 305.5 2 m 371.7 3 + 532.5 E — 735.3 E H (Bx - 306.6) E + (By + 185.84 - 371.7) 3 + (32 - 107.30 + 532.5 ~ 735.8) E = 6 X 3 = B i + 3y 3 + 32 E a 305.5 E + 135.55 3 + 250.5 g N El? 307 1 + 135.3 3 + 251 E N Ans. B : f(306.6)2 + (185.86)2 + (260.6)2 = 443.2 N % 443 N 457 ...
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HW15 - 6-75 A 500~lb homogeneous circular plate is...

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