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Answers: PROBLEM. A. 254.3322mm In the assembly shown, sags of s = 0.9944 m and 1.5 B. m.=6.0776kq, my=20.8777kg s resulted when the two cylinders were suspended C. Two=170.6868N, Tec=164.2107N from the rings at A and B, respectively. If s = 0 when D. TM=123.4631N the cylinders are removed, determine the following: E. K. 235.5319N/m A. The horizontal distance x (mm) moved by point B. B. The mass (kg) of the two cylinders. Given: $= 0. 9944m C. The forces (N) in the two springs. D. The tension (N) in cable AB. E. The spring constant Kac. Horizontal Distance of BD Note: Use g = 9.81 m/s2 and Kbd = 134 N/m. 15+ 15 (0 944) = 2.9916 2 m -1 m- 2 m 24x Hx This is not your 2. 9916 My property to do LX 1.5 m 2.9916-tan37 = 2+x X= 2.9916tan37-2 KAC = 0.154 m or 254. 3312mm ) 1.54 Solving force at spring BD 2m Indined length BD - a 1.50- 35 Cay 84 = 2.9416 q= 2.9416 -= 3445889m O F = - kox = 134N/m(3745629-2.5) Sokehim of angle of Tenum AB ToD= 170. 6468 NG 7 1.55 - 5 =0.5S FBD al junchim B 2170-6865 time=X -10.134 41 091 0.55 0.5(04944) by component methed me(9.81) FAD at junction A EFX=0 14 36-7714 LAB fin 56.4052 3 140. 6868 gin37 363051 TAB= 14.4631 NOD *423.4631 . N MA (9.31) TABCOS 27- 091. +770-686808137 - m.(980)=0 129 4631 /(01 56.3052 1 170-68686\\37 By component Method 9.81 my= 20817 kq) TAC Sin 98. 74 = 123,463] sin 56.3052 TACE 123 4631 sin 5C.3052 1234631 68 56.3052 )=W Sim 38.7014 W = 59 6213= mg TAC= 164. 2107 N 59. $ 213 9:51 KAC= E = 164.2107 AX 6.697 In = 60716kg KAC- 295 . 5319 W/m HOW TO GET THE FBD AT JUNCTION B AND the other red circle