hw6_sol

# hw6_sol - 4-32 Determine the moment of the 675—N force...

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Unformatted text preview: 4-32* Determine the moment of the 675—N force shown in Fig. P4-32 about point 0. SOLUTION A A 575(cos 22° 3 — sin 22° 3) = 625.8 1 — 252.9 J N A 0.230 T + 0.250 J m A F x F = (0.230 T + 0.250 3) x (625.8 1 — 252.9 3) = —58.17 E — 156.45 R - —214.62 R N-m a -215 E Nim a 215 N~5 D I73 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 4-38* Two forces F1 and F2 are applied to a bracket as shown in Fig. P4-38. Determine (a) The moment of force F1 about point 0. (b) The moment of force F2 about point A. SOLUTION A (a) F1 = 5(cos 45° 1 + sin 45° 3) = 3.536 E + 3.536 3 kN 300 T + 500 3 mm = rC/o x F1 = (300 1 + 500 J) x (3.536 1 + 3.536 J) = -707.2 E kN'mm a -0.707 R kN-m a 0.707 kN-m D A = 3(cos 45° 1 — sin 45° 3) = 2.121 f — 2.121 3 kN rC/A = (300 - 35) i + (500 — 125) 3 = 265 i + 375 mm A - I‘C/A x F2 = (265 1 + 375 j) x (2.121 1 - 2.121 J) = -1357.4 K kN'mm a -1.357 E kN-m a 1.357 kN'm D [77 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 4-45* A force with a magnitude of 970 1b acts at a point in a body as shown in Fig. P4-45. Determine the moment of the force about point 0. 301m. Fig. 94—45 SOLUTION = -14.904 2 — 12,420 3 - 18,630 E in.-lb A g —14.90 i — 12.42 3 - 18.63 E in.-kip ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4—57* Determine the moment of the 1000-1b force shown in Fig. P4-57 about point 0. SOLUTION -28 1 - 26 ' 3 E ] = —73o.55 i — 678.36 3 — 78.27 E lb (-23)2 + (—26)2 + (-3)2 - 12 E in. A — 12 3) x (-730.55 1 ~ 678.36 3 — 78.27 E) -12 0 -12 -730.55 -678.36 —78.27 = -8140 i + 7827 3 + 8140 E in.-lb A g -8.14 1 + 7.83 /(—8140)2 + (7327)2 + (8140)2 = 13,921 in.-lb 3 13.92 in.'lb + 3.14 E in.'kip L4) 125.780 a 125.80 55.79° a 55.8° 54.22° a 54.2° )‘76 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 00. Express the results in Cartesian vector form. SOLUTION A = 114.42 i + 133.48 J — 95.35 E 15 (18 i + 15 3) x (114.42 T + 133.48 3 — 95.35 E) i j E 18 15 0 114.42 133.48 —95.35 -1430.25 3 + 1716.30 3 + 686.34 E in.'lb = cos 30° 3 + sin 30° E = 0.8660 3 + 0.5000 E m> R '6 0 OC (~1430.25 i + 1715.30 3 + 686.34 E)-(0.3660 j + 0.5000 E) 1829.5 in.'lb E0C : MOCeOC 1829.5(0.8660 j + 0.5000 R) = 1584 j + 915 E in.-lb 222. ...
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hw6_sol - 4-32 Determine the moment of the 675—N force...

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