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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern