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hw6_sol

# hw6_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-31* Determine the oment of the 375-lb force shown in Fig. P4-31 F=3751b about point 0. SOLUTION A = 375(cos 49° 1 + sin 49° 3) = 246.0 3 + 233.0 3 1b = 7 i + 5 3 in. F x F = (7 i + 5 j) x (246.0 E + 283.0 3) = 1981 E - 1230 E = 551 E in.-lb = 751 in.-lb 5 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-35 A 250—lb force is applied to a beam as shown in Fig. P4-35. Determine the moment of the force about point A. SOLUTION 250(cos 60°_i + sin 60° 3) = 125.0 E + 216.5 lb 31‘ ft F x F = (3 i) x (125.0 E + 216.5 3) = 649.5 E ft-lb a 650 E ft-lb a 650 ft-lb 5 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-39 Two forces F1 and F2 are applied to a bracket as I F2=6mnh Shown in Fig. P4—39. Tm Determine L I u ; i (a) The moment of force F1 about point B. {b} The moment of force ?2 about point A. SOLUTION O (a) F1 800{—cos 35° 3 + sin 35 3) r1/B -14 1 + 8 3 1n. H3 F x F1 = {—14 i + 3 31 x (—555.3 1 = —1132.2 2 in.-lb e —1182 E in.-lb e 1182 in.-1b D A 600(cos 20° i + sin 70° 31 = 205.2 1 + 563.8 3 lb 14 i + 8 3 in. F x F2 = (14 i + a 3} x {205.2 E + 563.8 31 = 5252 E in.'lb a 5250 § in.-lb a 6250 in.-lb b J78 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-40 A 450—N force is applied to a bracket as shown in Fig. P4—40.Determine the moment of the force (a) About point B. (b) About point C. Fig. P4-40 SOLUTION F 450(-cos 60° 3 - sin 60° 3) = -225.0 E - 339.7 3 N 300 T + 100 3 mm x F = (300 i + 100 j) x (-225.0 3 - 339.7 3) = —94,410 E N'mm a -94.4 E N'm a 94.4 N-m D 300 i - 300 3 mm x F = (300 i - 300 j) x (—225.0 i — 389.7 3) = -184,410 E N'mm a —134.4 E N'm a 184.4 N'm D Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-57* Determine the moment of the 1000-lb force shown in Fig. P4-57 about point 0. SOLUTION A - -730.55 E — 678.36 J - 78.27 i lb (-23)2 + (—26)2 + (-3)2 A. -12 i - 12 E in. \L A FB/o x F = (-12 1 - 12 3 x (—730.55 E - 678.36 3 — 78.27 E) i j E -12 0 -12 —730.55 —678.36 -78.27 = -8140 i + 7827 3 + 3140 E in.-lb A a -3.14 1 + 7.83 + 8.14 E in.'kip = {(-8140)2 + (7827)2 + (8140)2 = 13,921 in.- lb 3 13.92 in. l V“? - 125.780 a 125.80 M -1 J M0 = COS 55.79° = 55.3° 54.22° z 54.2° ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-73* A ZOO-lb force is applied to a lever-shaft assembly as shown in Fig. P4—73. Determine the moment of the force about line 0C. Express the results in Cartesian vector form. SOLUTION - 114.42 E + 133.48 3 — 95.35 E lb (18 i + 15 j) x (114.42 T + 133.48 3 - 95.35 E) i 3 E 18 15 0 114.42 133.48 -95.35 -1430.25 T + 1716.30 3 + 686.34 E in.°lb a = cos 30° 3 + sin 30° E = 0.8660 3 + 0.5000 E eOC (-1430.25 3 + 1716.30 3 + 686.34 E)-(0.8680 j + 0.5000 E) 1829.5 in.'lb 0C OCeOC 1829.5(0.8660 j + 0.5000 2) = 1584 3 + 915 E 1n.-1b ...
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hw6_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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