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# hw 10 - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY AND...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-83* Determine all forces acting on member BCD of the linkage shown in Fig. P7-83. SOLUTION Member AC is a two-force member;. therefore, the line of action of force C is known as shown on the free—body diagram for member BCD: + C 2MB = c cos 45° (2.0) — 40 cos 30° (3.5) = o c = 85.73 1b a 85.7 lb C a 85.7 lb 7 45° Ans. = Bx + 40 cos 30° — 85.73 cos 45° = 0 25.98 1b a 26.0 lb —+ = B - 85.73 sin 45° + 40 sin 30° = 0 V 40.62 lb 3 40.6 lb T AB”)2 + (By)2 = /(25.98)2 + (40.82)2 = 48.22 lb ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-88* A block with a mass of 150 kg is supported by a cable which passes over a 150-mm diameter pulley that is attached to a frame as shown in Fig. P7—88. Determine all forces acting on member BCD _of the frame. SOLUTION .r From a free-body diagram for the block: 147).} W = mg = 150(9.807) = 1471.1 N T 2 : — , = + Fy T 1471 1 0 ”7).,” T = 1471.1 N 5 From a free-body diagram i_____)( for the pulley at C: + —+ 223 c - 1471.1 cos 20° = 0 X C = 1382.4 N = 1382.4 N —# ' X + T 2Fy = cy - 1471.1 — 1471.1 sin 20° = 0 Cy = 1974.2 N = 1974.2 N T From a free-body diagram for the bar BCD: + g 2M5 = B(0.450) + 1332.4(0.300) - 1974.2(0.400) = 0 V/ B = 833.2 N a 333 N y/// + —+ 2F = D - 1382.4 - 833.2 = 0 X X D - 1974.2 = 0 D = 1974.2 N T L/ y y /(D )2 + (D )2 = /(2215.6)2 + (1974.2)2 = 2967.5 N z 2970 N x v -1 1974.2 _ o - _ 0 tan '2215.6 — 41.71 D - 2970 N E 41.7 Ans. /(c )2 + (c )2: /(—1332.4)? + (-1974.2)2 = 2410.1 N z 2410 N x y -1 -1974.2 _ _ o _ o - tan 31555:: - 125.01 C _ 2410 N r 55.0 Ans. JJZEB ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Forces of 50 lb are applied to the handles of the bolt cutter of Fig. P7-101. Determine the force exerted on the bolt at E and all forces acting on the handle ABC. Fm-P7401 SOLUTION From a free-body diagram for member CDE: +—>EFX=C =0 X C = 0 X + C ZMD = Cy13) - E(2) = o - 2 cy - 3 E From a free—body diagram for handle ABC: + C 2MB = cy(1) — 50(20) cy = 1000 1b = 1000 C = 1000 lb T By + 50 + 1000 = 0 —1050 1b 3 = 1050 lb ¢ 1.50y = 1.5(1000) = 1500 lb Force on the bolt: E = 1500 lb T (‘18 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-114 Figure P7-114 is a simplified sketch of the mechanism used to raise the bucket of a bulldozer. The bucket and its contents weigh 10 kN and have a center of gravity at H. Arm ABCD has a weight of 2 kN and a center of gravity at B; arm DEFG has a weight of 1 kN and a center of gravity at E. The weight of the hydraulic cylinders can‘ be ignored. Calculate the force in the horizontal cylinders CJ and EI and all forces acting on arm DEFG for the position shown. SOLUTION From a free-body diagram for the complete mechanism: + C 2MA = P(1.8 cos 30°) — 2(0.9 sin 30°) - 1(3.3 sin 30°) — 10(0.3 + 4.5 sin 30°) = 0 F = 18.00 kN <— Ans. "2"- 30° From a free—body diagram for 0.3!". the bucket and its contents: G: )OKN + C ZMG = E(1.2 cos 30°) - 10(0.3) = 0 E = 2.887 kN = 2.887 kN —+ (on DEFG) E e 2.89 kN —+ Ans. + -9 2Fx Gx - 2.887 = 0 G = 2.887 kN = 2.887 kN +- (on DEFG) X + 1‘ my = G — 10 = 0 G = 10.00 kN = 10.00 kN ¢ (on DEFG) ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-114 (Continued) 0 = /(Gx)2 + (0y)2 = /(-2.887)2 + (—10.00)'2 = 10.408 kN _ -1 —10.00 _ _ o o 66 _ tan _2.887 _ 106.10 G a 10.41 kN P 73.9 Ans. From a free-body diagram for arm DEFG: . o 0 = tan'1 9L§—§lE—§9— = 19.107° 3.0 cos 60o + C EMD = F[1.2 cos (30° — 19.1o7°)] - 2.887(1.2 cos 30°) - 1(0.6 sin 30° — 10(1.8 sin 30° = 0 F = 10.438 kN F a 10.44 RN 3 19.11° Ans. + —+ 23x =‘Dx + 2.887 + 10.438 cos 19.1o7° - 2.887 = o 0 = -9.863 kN X + T EFy = 0y + 10.438 sin 19.107° - 1 - 10 = 0 D = 7.583 kN Y D = (0x)2 + (my)2 = /(-9.863)2 + (7.583)2 = 12.441 kN _ -1 7.583 _ o — _ . ~ 0 . eD _ tan _9.863 - 142.44 D _ 12.44 1N a 31.6 Ans. ¢4££L ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-121 A block weighing 1000 lb is supported by a five-bar truss as shown in Fig. P7-121. Determine the force in each member of the truss. SOLUTION From a free—body diagram for the pulley at C: +—>2Fx=c —1000=0 X 1000 lb = 1000 lb -+ C - 1000 = 0 y = 1000 lb = 1000 lb T From a free—body diagram for joint C: + ‘\ ZFy, = 1000 sin 30° - TCD sin 30° - 1000 sin 60° = o + /” 2F , -1000 cos 30° - T — T cos 30° — 1000 cos 60° x BC CD -1000 cos 30° - TBC — (-732.05) cos 30° — 1000 cos 60° = o TBC = -732.05 N E 732 N (C) Ans. From a free—body diagram for joint B: + /” EFX, = —TAB - 732.05 = 0 TAB = -732.05 N E 732 N (C) + ‘\ EFy, = -TED = 0 Tan = 0 From a free-body diagram for joint D: + —» 2px - 732.05 cos 60° = o TAD = -366.0 N E 366 N (C) ...
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