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HW9 - 5555515555555 IVTwo forces are applied t-o a beam as...

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Unformatted text preview: 5555515555555 _ IVTwo forces are. applied t-o a . beam as Show gDetermlne the moment5 of - - " '3 153517.515 ' 5.253. 1347.711 2 .. .. \$055510N‘ 'iai'M;;i" ' 5 25013 sin 50° 1 549. 5 55 15 5 550 55 15 ~-ﬁA1.5555o 55-15-5 gahsr l75l3)= 525 ft 1b % 525 ft°lb 'f 315‘ 525 55-15 Q 3’555{ I”Two-fordes 5ré 5551555 .to- a- beam as shown in Fig. P4- 8 Determine - the moménts of forces .F1 an'd F_2 abou_t- pgint A. E'SOLUTIQN-FT' '35235~}Esid51 = 5(3) .24 5 kN n 1|5215A2'_ 315 515'5o°1;=15.593 kN 5 5 15 59 55-. = g15 5-9 RN 5 5 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-15 Two forces are applied to a bracket as shown in Fig. P4w15. Determine the moments cf forces F1 and ﬁg about poinis A and B. Fig. 94-15 SOLUTION 625 :9 cos 23° « 3 sin 28°) IF‘I ldA‘l 2619 in.'lb E 2.62 in.-kip H = 2.62 in.'kip 5 74519 cos 22° + a sin 22°} 8449 in.'lb % 8.45 in.'kip 625(9 cos 28°) 4967 in.'1b a 4.97 in.-kip 745(9 cos 22°) 6317 in.°1b g 6.22 in.'kip M3 _ ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-25 Three forces ﬁh, ﬁg, and E, are applied to a beam FA 5 200 15 as shown in Fig. P4»25. ' Determine (a) The moments of forces FA and Fe about point 0. (b) The moment of force ?B about point D. Ft: 250 lb Fig. [14-25 SOLUTION O . O (a) + C MOA 5 —FA cos 50 (9} + FA Sln 60 (30) = “200 cos 60°‘(9) + 200 sin 60° (30} 4296 in.°lb E 4.30 in.'kip ﬂ 0A FC cos 30° (9} - FC sin 30° (50) 250 cos 30° (9) , 250 sin 30° (60) -5551 in.'lb = "5.55 in.'kip 30C 0. 1n. = —FB cos 45° (9 + 4).+ FE sin 45° (50) = ”300 cos 45° (13) + 300 sin 45° (60) 9970 in.'lb = 9.97 in.-kip RD 2 9.97 in.'kip D Ans. I67 ...
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HW9 - 5555515555555 IVTwo forces are applied t-o a beam as...

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