HW11 - ENGINEERING MECHANLCS STATICS 2nd Ed W F RILEY AND L...

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Unformatted text preview: ENGINEERING MECHANLCS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-88 A plate is loaded with a system of forces as shown in Fig. P4h88. Exprsss the resultant of the force system in Cartesian vector form. 150 N 150 N Fig. P4-88 ‘ SOLUTION An examination of Fig. P4fi88 indicates that the force system consists of a system of three couples in a plane. A scalar analysis yields: With counterclockwise moments positive: For forces A and B: M1 = FAd1 = 250(0.150) = 37.50 N'm For forces C and D: M2 = ”chz : ~350(0.150) = -52.50 N°m For forces E and F: FEdB = 150(0.320) % 40.00 N'm +M2+M3 37.50 » 52.50 + 48.00 = 33.00 N-m = 33.0 N'm E 2 33.0 N-m 5 = 33.0 E N-m 226 ENGINEERING MECHANICS _ STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4w91 Three couples are applied to a bent bar as shown in Fig. P4-91. Determine the magnitude of the resultant .‘u couple C and the direction angles associated with the resultant couple vector. fig. P4-91 SOLUTION The scalar components of the resultant couple C are: Cx = ~80(3) = —240 ft‘lb CY = ~?5(4) = "300 ft'lb C = ~100(8) = "800 ft-lb 2 The resultant couple C eXpressed in Cartesian vector form is: C = —240 i - 300 j u 800 E c a (~240)2 + (—300)2 + (—800)2 = 887.5 ft’lb E 887 ft'lb Ans. -1 Ex ~1 —z40 o 9X = cos 6“ = cos §§?T§ = 105.7 Ans. —1 Cy —1 —300 0 9y 2 cos 6— =-cos 887.5 = 109.8 Ans. -1 CZ -1 “300 o 92 = cos 6- 2 cos §§?T§ - 154.3 Ans. 12'2‘? D. STURGES ENGINEERING MECHANKCS - STATICS, 2nd. Ed. W. F. RILEY AND L. 4-107 'Replace the SOD—lb force shown in Fig. P4~107 by a force at point A and a couple. Express your ahswer in Cartesian vector form. Fig. P4-1 07 SOLUTION F A F : 300(cos 30° 3 + sin 30° 3) = 259.8 f + 150.0 3 1b e 200 i + 150 0 3 1b F = (20 — 10 tan 30°) 3 —10 j in. = 14.223 3 , 10 3 in. A F x F = (14.220 1 — 10 3) x (250.3 3 + 150.0 3) = 4732 E in.°lb e 4.73 E in.'kip 4-108 Replace the 450*N force shown in Fig. P4—108 by a force at point A and a couple.’ Express your answer in Cartesian veetor n " flMmm“ form. ' Fig. 24-300 SOLUTION FA = fi = 450(cos 58° I * sin 58 = 233.5 E - 331.0 3 N a 239 i — 382 3 F = (0.200 + 0.400 cos 329) i + 0.400 sin 32° 3 = 0.3392 3 + 0.2120 3 m EA = F x F a (0.5392 3 + 0.2120 3) x (230.5 3 - 331.0 3) = —250.3 E N-m a —250 fi_N-m ...
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