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# hw 4 - game M56611 MumÂ@4069 4-113 Determine the magnitude...

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Unformatted text preview: game! M56611. MumÂ» @4069 4-113* Determine the magnitude and direction of the resultant of the two forces shown in Fig. P4-113 and the perpendicular distance dR from point A to the line of action of the resultant. 27s 15 Fig. P4-113 SOLUTION 750 cos 35Â° 614.36 15 750 sin 35Â° - 275 = 155.18 15 i + R: = â€œ\$14.36)2 + (155.18)2 = 633.66 lb z 634 lb R _ â€”1 614.36 -1 â€”x R ' Â°Â°S 633.66 cos = 14.177Â° z 14.18Â° 634 15 a 14.18Â° 750 cos 35Â° (8) + 750 sin 35Â° (8) 8356.6 in.-lb = 8356.6 in.-lb j RdR = 633.66dR = 8356.6 in.'lb 13.188 in. a 13.19 in. 245 ENGINEERING MECHANICS - STATICS W. F. RILEY AND L. D. STURGES 4-122 Replace the three forces shown in Fig. P4-122 by an equivalent force- couple system at point C. Fig. P4-122 SOLUTION = 350 cos 450 + 475 cos 450 + 625 = 1208.4 N = -350 sin 450 - 475 sin 450 = -583.4 N /(1208.4)2 + (533.4)2 1341.9 N z 1342 N, ï¬‚ = 1342 N s 25.80 -475 cos 45Â° (0.100) - 625(0.100) -96.09 N-m a -96.1 N-m CC a 96.1 N-m D 225KB ENGINEERING MECHANICS - STATICS W. F. RILEY AND L. D. STURGES Four forces and a couple are applied to a rectangular plate as shown in Fig. P4â€”125. Determine the magnitude and direction of the resultant of the force-couple system and the distance xR from point 0 to the intercept of the line of action of the resultant with the x-axis. SOLUTION 60 30 N 50 = 25 N {(30)2 + (25)2 = 39.05 a 39.1 N -1_30 O 0 cos 39.05 â€” 39.80 â€” 39.8 39.1 N a 39.8Â° 10 - 75(0.300) - 90(0.250) - 50(0.300) - 60(0.250) - -65.0 N-m = 65.0 N-m D ENGINEERING MECHANICS - STATICS W. F. RILEY AND L. D. STURGES 4-153 A 2500-lb jet engine is suspended from the wing of an airplane as shown in Fig. P4-153. Determine the moment produced by the engine at point A in the wing when the plane is (a) On the ground with the engine not operating. (b) In flight with the engine developing a thrust T of 15,000 lb. F%,P4453 SOLUTION (a) H = wau E = 2500(8 sin 35Â°) E = 11,472 E ft'lb s 11.47 ft'kip ' Ans. (b) H (Wdu - TdT) E = [2500(8 sin 35Â°) - 15,000(3 cos 35Â°)1 E = -86,827 E ft-lb a -86.8 ft-kip Ans. 4-154 Determine the moment of the 1650â€”N force shown in Fig. P4-154 about point 0. F=1650N Fig. P4-1 54 SOLUTION â€˜130 1 + 130 + 30 E â€”1113.1 1 + 1113.1 7 + 494.7 E N (-130)2 + (130)2 + (302 - F x F , = (0.350 1 + 0.240 3 + 0.400 E) x (-1113.1 1 + 1113.1 3 + 494.7 E) -327 i - 523 j + 668 E N-m Ans. 2253! ENGINEERING MECHANICS â€” STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-159* A Z-section is loaded with a system of forces as shown in Fig. P4â€”159. Express the resultant of this force system in Cartesian vector form. Fig. P4-1 59 SOLUTION I? 1+ZFj+2FE y z (80 - 30) i + (150 â€” 150)_3 + (100 â€” 100) E 2 U C=2M â€™1â€˜+EM j+2M E . x y z A = -(100)(15) 1 - (100)(12) j - (150)(6) E - (80)(8) E -1600 i â€” 1200 j - 1540 E in.-lb fâ€”â€”â€”l /(-1600)2 + (â€”1200)2 + (-1540)2 = 2524.2 in.-lb -1 -1600 _ 129.30 = 118.4Â° 127.6Â° 125343 ...
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hw 4 - game M56611 MumÂ@4069 4-113 Determine the magnitude...

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