Question

Q2 (33.33%) Find ye, TAC, TBC For the cable shown. Hoft cable Length. c 10pt 15 st

Answer 1

Answer:

1 02 Yc FCAM TFCB 02 = ) 3okips => AC= 103 tuj BC = 15;²+4)² cable length Whos y J² + (15) 7 (GJ² y 2 +$225+ 4² y = 15.49ft = Act BC HO WO 11oot =) 0i = tant لا Yc 15 =tan(15.49 15 O H5 ss' 02 = - tani Yc To tan -1/15.49 10 @z = 57'qi as At jointc, E Fx 20 FCA COSO, FCB COSO, - FEB COSMEss' - FCA costa FCB - 0.779 FCA

2 & Fy Fog sino, + FCA sino = 30 0.779 FCA sin 45.55't FCA sin 57°91 = 30 0.559 FCA +0.84 FCA = 30 1.399 FCA 30 IFCA = 31. Hoe kip FCB 0.7798 21.44 FCB = 16.701) kip