# hw 3 - 4-2* Determine the moments of the 225-N force shown...

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Unformatted text preview: 4-2* Determine the moments of the 225-N force shown in Fig. P4-2 about points A, B, and C. SOLUTION |F|dA 225(0.600) 135.0 N'm 5 Ans. F d 225(0.400) 90.0 N-m 5 Ans. B 180.0 N-m E 22 II II II lFIdc 225(0.3001 130.0 N-m 5 Ans. 4-8 Two forces are applied to a beam as shown in Fig. P4-8. Determine the moments of forces F1 and F2 about point A. SOLUTION M H 8(3) = 24 0 kN'm A1 ‘F1ldA1 3A1 = 24 0 kN-m Q Ans. 3(6 sin 60°) = 15.588 kN-m a 15.59 kN-m leldAz EAZ = 15.59 kN-m 5 Ans. 4-12 Determine the moment of the 750-N force shown in Fig. P4—12 about points A, B, and C. Fig. P4-12 SOLUTION 750(0.013 cos 25°) 12.235 N-m 2 12.24 n-m = |F|dA 750(0.013 cos 25° + 0.025 20.159 N-m 3 20.2 N-m (Fla. 20.2 N-m D 750(0.025 sin 25°) 7.924 N'm % 7.92 N‘m IFldC 7.92 N-m 3 4-24* Two forces #1 and F2 are applied to a beam as shown in Fig. P4-24. Determine (a) The moment of force F1 about point A. (b) The moment of force F2 about point B. Fig. P4-24 SOLUTION (a) + C MA = F1 cos 30° (175) + F1 sin 30° (600) = 2.5 cos 30° (175) + 2.5 sin 30° (600) = 1129 kN'mm = 1.129 kN'm HA = 1.129 kN-m j (b) +CMB —F2 cos 45° (50) — F2 sin 45° (425) -4 cos 45° (50) — 4 sin 45° (425) -1344 kN-mm -1.344 kN-m H B 1.344 kN‘m D 12.24 N-m D ‘ Ans. Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-27* Determine the moments of the 50—lb force shown in Fig. P4-27 about points A and B. SOLUTION + C MA = 50 cos 38°(20 cos 30°) - 50 sin 38°(20 sin 30°) = 374.6 in.-lb a 375 in.'lb HA 2 375 in.-lb 5 Ans. + C ME = 50 cos 38°(20 cos 30°) + 50 sin 38°(20 sic 30°) = 990.3 in.'lb E 990 in.'lb as g 990 in.'lb 5 4-28* Determine the moments of the 450-N force shown in Fig. P4-28 about points A and B. SOLUTION + q MA = -450(0.400 + 0.200 cos 32°) = -256.3 N'm a ~256 N'm EA 2 256 N'm Q + C ME = —450(0.400.— 0.200 cos 32°) = -103,68 N-m a -103.7 N-m ‘ EB a 103.7 N'm D )7] 4—55 A 750—lb force is applied to a pipe bracket as shown in Fig. P4-55. Determine (a) The moment of the force about point 0. (b) The direction angles associated with the moment vector. SOLUTION (a) F = = 310.49 1 - 338.72 3 - 592.76 2 lb 11 T + 14 J in. x F = (11 i + 14 3) x (310.49 1 - 338.72 3 - 592.76 R) i 3 E 11 14 0 310.49 —338.72 -592.76 = —8299 i + 6520 3 - 8073 E in.-1b g -8.30 i + 6.52 3 - 8.07 E in.°kip V (-8299)2 + (6520)2 + (-8073)2 = 13,300 in.'lb 128.61° z 128.6° O 60.640 a 60.6 - 127.37° 2 127.4° I79 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-59 A force with a magnitude of 580 lb acts at a point in a body as shown in Fig. P4-59. Determine (a) The moment of the force about point B. (b) The direction angles associated with the moment vector. SOLUTION (a) F = 580 (—7)2 + ( 15)2 + (10)2 = —209.94 E - 449.87 3 + 299.91 E lb = -14 T + 8 3 in. (—14 i + 3 j) x (-209.94 E — 449.87 3 + 299.91 E ) i j E -14 8 0 -209.94 —449.87 299.91 = 2399 i + 4199 3 + 7978 E in.-1b z 2.40 i + 4.20 ‘ + 7.98 E in.-kip = /(2399)2 + (4199)2 + (7978)2 = 9329 in.-1b a 9.33 in.-kip -1 x _ -1_2399 _ o _ cos E; — cos §§§- 75.10 - 75.1 0 63 24° 2 63.2° 31.22° a 31.2° ...
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## This note was uploaded on 07/16/2009 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.

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hw 3 - 4-2* Determine the moments of the 225-N force shown...

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