hw7_sol

# hw7_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 4-114* Determine the magnitude and direction of the resultant of the two forces shown in Fig. P4~114 and the perpendicular distance dR from point A to the line of action of the resultant. F¢.P4414 SOLUTION R = = 300 cos 45° + 480 cos 30° = 981.38 N X = 800 sin 45° - 480 sin 30° = 325.69 N I 2 2 / , 2 2 x Ry = (981.38) + (325.69) = 1034.01 N a 1034 N R -1 981.38 cos" 51 = cos ——————— = 18.3590 2 18.360 1034.01 = 1034 N a 18.360 300 sin 45° (0.320) — 800 cos 45° (0.120) 400 cos 30° (0.240) - 400 sin 30°(0.640) —140.23 N-m = 140.23 N-n D RdR = 1034.0dR = 140.23 N'm 0.13562 m § 135.6 mm 23‘343 ENGINEERING MECHANICS - STATICS, 2nd. F. RILEY AND L. D. STURGES 4-115 Replace the three forces Fh=lﬂow Shown in Fig. P4—115 by an equivalent force-couple E :3 P! system at point B. FD=3me Fig. 134-115 SOLUTION -100 lb 200 - 300 = -100 lb {-100}2 + {—10012 = 141.42 1b 3 141.4 lb R x -1 -100 _ _ O _ - E- w 003 141_42 — 133.00 — 130.0 —1 COS 0 = 141.4 lb y 45.0° C = 2MB = 100(10) - 300(14} + 200(20} = 800 in.-lb a 800 1n.-1b j 22‘9'7 ENGINEERING MECHANICS - STATICS, 2nd. Replaca the three forces shown in Fig. P4—116 by an equivalent force—couple system at point 0. SOLUTION = BF X — -400 + %(750} = 200 N 2F _ §(?50) + 100 = Y D 550 N Ed. W. F. RILEY AND L. D. STURGES (200}2 + (550}2 = 585.2 N e 535 N -‘1 -1 200 COS .1 _ _ R ‘ COS 585.2 ' = 585 N a 7o.0° EM = 400(0.160) + 100(U.360} EZ‘IiB ?0.02° e 70.0° 100.0 N'm Fig. P4416 ENGINEERING MECHANICS - STATICS W. F. RILEY AND L. D. STURGES 4—123* Replace the three forces shown in Fig. P4-123 by an equivalent force-couple system at point 0. Fig. P4-1 23 SOLUTION R = 20 = 200 cos 60° + 300 cos 45° + 250 cos 30° 523.54 lb R = 2F = 200 sin 50° + 300 sin 45° - 250 sin 30° 260.34 lb 3" ‘1’ ﬁ = /R: + a: = {(528.64}2 + (260.34}2 = 589.2? lb s 539 lb R _ -1 528.64 _ o o n 003 589.27 — 25.22 a 26.2 -1 x G : -— x cos R ﬂ = 589 lb a 26.2° —200 cos 60° {9) + 200 sin 60° (30; + 250 cos 30° (9) 250 sin 30° (60) - 300 cos 45° (9) + 300 sin 45° (90) 15,928 in.'lb 3 15.93 in.°kip CO e 15.93 in.-kip 3 25¢; ENGINEERING MECHANICS - STATICS W. F. RILEY AND L. D. STURGES Four forces and a couple are applied to a frame as shown in Fig. P4—126. Determine the magnitude and direction of the resultant and the perpendicular distance dR from point A to the line of action of the resultant. Fig. P4-126 SOLUTION 16_ o 30 — 28.07 cos 28.070 — 50 - 75 = —54.41 lb sin 28.o7° — 90 = —52.36 lb = y’(-54.41)2 + (-52.36)2 = 75.51 lb 2 75.5 lb ‘1 '54041 S 0 O 75.51 — -136.10 — -136.1 CO 75.5 lb y 43.9° 75(12) + 50(30) — 100 - 80 cos 28.070 (30 — 10 cos 28.090) + 80 sin 28.07° (16 + 10 sin 28.070) - 90(16 + 18 tan 28.090 + 14) —1979.3 in.°lb = 1979.3 in.-lb D — 26.21 in. 3 26.2 in. 257 ...
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## This note was uploaded on 03/26/2008 for the course CVEN 221 taught by Professor - during the Spring '08 term at Texas A&M.

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hw7_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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