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Find the forces in CF and CD by method of section and the remaining by Method of Joints.

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$∑f_x=0 ∴ H_A+R_B=0 -----(1) \\ ∑f_y=0 ∴ V_A=5+6+7=18N \\ M_A=0 ∴ -5×3-6×6-7×9+R_B×4=0 \\ R_B=28.5KN $

From eq (1)

$H_A= -28.5KN \\ θ_1=\tan^{-1} (\dfrac 49)=23.96^0 \\ θ_2=\tan^{-1} (\dfrac {3\tan(23.96)}3) =18.43^0 \\ θ_3 = \tan^{-1} (\dfrac {6\tan(23.96)}3) = 41.63^0 $

By method of section

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$M_F=0 ∴F_{CD} ×1.333-7×3=0 \\ F_{CD} =15.75KN \\ ∑f_x=0 F_{CF} \cos 18.43^0+F_{FG} \cos 23.96= -15.75 ------- (1) \\ ∑f_y=0 F_{CF} \sin 18.43^0 + F_{FG} \sin 23.96^0= 13 ------- (2) \\ F_{CF}=8.134 KN \space and\space\space F_{FG} = -25.68 KN $

By method of Joint

Joint E:

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$∑f_y=0 ∴ -F_{EF} \sin 23.96-7=0 \\ F_{EF}= -17.23 KN \\ ∑f_x=0 ∴ -F_{DE} - F_{EF} \cos 23.96^0 \\ F_{DE}=15.75 KN $

Joint D

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$∑f_y=0 \\ F_{DF}= -6KN $

Joint C

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$∑f_x=0 \\ F_{AC} = F_{CD} + F_{CF} \cos18.43^0 \\ F_{AC} =23.47 KN \\ ∑f_y=0 \\ F_{CG}=-5-8.134 \sin 18.43 \\ F_{CG}=-7.57KN $

Joint G

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$∑f_x=0 \\ F_{AG} \cos 41.63+F_{BG} \cos 23.96= -25.68 \cos 23.96 ----(1) \\ ∑f_y=0\\ F_{AG} \sin 41.63-F_{BG} \sin 23.96=7.57+25.68 \sin 23.96 \\ F_{AG} \sin 41.63-F_{BG} \sin 23.96=18 ----(2) \\ $

From (1) and (2)

$F_{AG} =7.6 KN \\ F_{BG} =-31.897 KN \\ $

Joint A

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$∑f_y=0 \\ F_{AB} = -7.6 \sin 41.63+18 \\ F_{AB} =12.95 KN$

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