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MAE 130A HW1 prob 2.35 soln

# MAE 130A HW1 prob 2.35 soln - PROBLEMESL 1.65 Knowing that...

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Unformatted text preview: PROBLEMESL 1.65 Knowing that the tension in cable BC is 145 1b, determine the resultant of the three forces exerted at point B of beam AB. SOLUTION Cabie BC Force: F: = 4(145 lb)% = —105 lb 2‘: 4'2] H, Fy = (1451b)% = 1001b 0( 100-11; Force: F; = —(1001b)% == —60 1b F), = —(ioo 1b)? = —80 lb K= ‘ 4°H° 1564:; Force: F, = (156 “9% = 1441b 5 _ =—156lb—=—601b szzFx=—2llb, Ry=2Fy=—40lb R = (—21113)2 +(~401b)2 = 45.17711) tam—£1 21 a = tan—'19 2 623° 21 R = 45.2 lb 7623‘“ < ”1.0.9 05 Fm) 04 +5 max; «Marian-«ﬁnk P¢m\\an§i 9) mﬁﬁbni‘fqbz cat-ﬁne {‘Cw\\t«n\-, sch a") if), =6) v-——> QOK‘o-‘r 7MB 51nd ' 130% aback =0 “"” binoL’ JV" 0.0504“; \ 44.5de \1l++“‘ ) . MAW" Ct0\\:> + 70% + _ mm; \ :0 _ (WE (-‘Il 3. *1 70“} (+3 ~ v30 kb = 40 \b A" HT: (7000 09 — @0qu a @MQILC\+—\-‘\ LHOO 16051} ~ moon? C+\ 4— Moo“? = 6100M" * 8Loo\b"(+‘\ 67.00“.“ (+7“) + new“? (*3 ”QCCDCBWI “O -|- - .LH ‘Bm (CON "F" Jrqﬁ OK. OK ‘" *am-‘G-3 , M-‘LMI-‘uﬂ (32 9 7093 c950" +130“: SUN '< Kw mom. b3 5-6)! "" 'Fa . ’ 1"?“ CL\ Te.n\$_[0r~ i“ he 1? mama‘- (3:11th «\mi #13, 5% Mgnl’ruba a? mm\\m\-. 'TM. = Room TED =- \100\b ‘ka‘o @529 + “(200% abs 56"” “The coszs": 0 Kg = \Q‘quHB 53 F2 “ 5-6;: 400K; 5;.‘15 .— tzocﬂb cage, — GEM-“‘19 Siq’LS" = F9. hm = mouuow } ...
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MAE 130A HW1 prob 2.35 soln - PROBLEMESL 1.65 Knowing that...

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