Page 45 - eEISBN FINALIZE

0 0
F EG kos 60 + F EC kos (180+ 45 ) + 20 kos 270 = O
F EG kos 60 0 + F EC kos 225 + 20 kos 270 = O
0
0 =
0.5 F EG + ( 0.71 )F + 20 ( ) 0
-
EC
0.5 F EG - 0.71F = 0 .......... .......... .......... .. ( ) 1
EC
0
F EG sin 60 0 + F EC sin (180+ 45 ) + 20 sin 270 = O
F sin 60 0 + F sin 225 + 20 sin 270 = O
0
EG EC
−
−
) 0
0.87 F + ( 0 . 71 )F + ( 20 =
EG EC
0.87 F EG − . 0 71 F − 20 = 0
EC
(2)-(1) 0.87 F EG − . 0 71 F = 20 .......... .......... .... ( ) 2
EC

0.37 F EG + 0 = 20
20
F =
EG . 0 37
Substitute in (1) = 54.05N

0.5 ) 0.71F EC = . 0
-
(54.05
27.03 - 0.71F EC = 0

- 0.71F = − 27.03
EC −
03
F EC = − 27 . 0 . 71

= 38.07N

FEG =38.07N

FEG

Ɵ=36.87
0
45 0

Tan θ = 3
4
0
0
1 -  3  F kos 45 + F kos1 43 . 13+ B kos 270 = 0
θ = tan   EC CD
 4  38 . 07 ( 71.0 ) ( 0−+ 8 . )F CD + B ( ) 00 =
= 36.87 0 B 27 . 03− 8 . 0 F = 0
CD
F = 26 . 92
CD 8 . 0
= 33 . 65 N
0
0
F sin 45 + F sin1 43 . 13+ B sin 270 = 0
EC CD
38.07 sin 45 + 0.6 F CD = B
0

26.92+ 0.6 (33.65 ) = B
B = 47.11N
36