Page 47 - eEISBN FINALIZE

Tan θ = 5 . 0
. 0 525
1 - 
θ = tan  5 . 0 

 . 0 525 
= 43.60 0

0
0
F BC kos 43.60 + F AC kos1 43 . 13+ 38.07 kos 270 = 0
0 =
) ( 0
. 0
F BC ( 72 + − 8 . )F AC + 38 . 07 ( ) 0
0.72F BC − 8 . 0 F CD = 0
− . 0 72 F
F CD = − BC
8 . 0
F CD = 9 . 0 F BC .......... .......... .... ( ) 1

substitute F CD = 9 . 0 F EC
0
0
F BC sin 43.60 + F AC sin1 43 . 13+ 3001 . 86 sin 270 = 0

0.68F BC + 0.6 F CD − 3001 . 86 = 0
0.68F + 0.6 F = 3001 . 86 .......... .......... ... ( ) 2
BC CD
=
0.68F + 0.6 ( 9 . 0 F ) 3001 . 86
BC BC
0.68F BC + . 0 54 F BC = 3001.86
1.22F BC = 3001 . 86

F BC = 3001 . 86
1.22
= 2460 . 54 N
so , F CD = 9 . 0 F EC

= 0.9 (2460 . 54 )
= 2214 . 49 N

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