$∑ f_x=0 \\ H_A- T_{BC} \cos 60=0 →(1) \\ ∑f_y=0 \\ V_A+ T_{BC} \sin⁡ 60-3000=0 \\ V_A+ T_{BC} \sin⁡ 60=3000 →(2) \\ M_A=0 \\ -3000 ×3 \cos30+T_{BC} \sin 60 ×6 \cos 30-T_{BC} \cos 60 × 6 \sin 30=0 \\ T_{BC} =2.598 KN \\ \text {From eq } (1) , H_A=1.300 KN \\ \text { From eq }(2), V_A=0.75 KN$

$CB=R_C+ R_B=5+6=11 \\ NB=18-R_B- R_C=18-6-5=7 \\ θ = \cos^{-1}⁡ (\dfrac 7{11})= 50.47^\circ \\ BN=R_A+ R_B=4+6=10 \\ MB=18-R_A- R_B=18-4-6=8 \\ θ = \cos^{-1} ⁡(\dfrac 8{10} )=36.87^\circ $

FBD:

From FBD of cylinder C

$\dfrac {R_1}{\sin ⁡140.47} = \dfrac {R_2}{\sin ⁡90} = \dfrac {20 ×9.81}{\sin⁡ 129.53} (\text { using Lami ' s Theorem }) \\ R_1=161.90 N \space\&\space R_2=254.37N $

From FBD of cylinder B,

$∑f_x=0 \\ -R_3+ R_4 \cos 36.87 + R_2 \cos ⁡50.47 \\ -R_3+ R_4 \cos 36.87= -254.37 \times\cos 50.47 →(1) \\ ∑f_y=0 \\ R_4 \sin⁡ 36.87-40 ×9.81-R_2 \sin⁡ 50.47\\ R_4 \sin 36.87=40 ×9.81 + 254.37 × \sin ⁡50.47 \\ R_4 \sin 36.87 =588.59 \\ R_4=980.98 N $

From eq (1)

$R_3=946.68 N $

From FBD of Cylinder A,

$∑ f_x=0 \\ R_6- R_4 \cos⁡ 36.87=0 \\ R_6=784.78 N \\ ∑ f_y=0 \\ R_5- 15×9.81-R_4 \sin ⁡36.87\\ R_5=735.73 N $