***PLEASE SAY IF THEY ARE COMPRESSION OR TENSION AS WELL.Previous forces found are Fcb = 700lbs(C), Fcd = 606lbs(T) Fdb = 630 lbs (T) Fde = 606 lbs (C). Consider the truss shown in the diagram. The applied630 lb and P₂ = 350 lb and theforces are P₁=distance is d = 8.50 ft.(Figure 1)FigureE30°dBDP₁30°1 of 1C Having found FCD in the analysis of joint C and FDB in the analysis of joint D, there are now only two unknownforces acting on joint B. Determine the two unknown forces on joint B.Express the magnitudes of the forces to three significant figures. Separate your answers with a comma.

Consider the truss shown in the diagram. The appliedforces are P1 =630 lb and P2 = 350 lb and the distance is d = 8.5 ft.Having found FCD in the analysis of joint C and FDB in the analysis of joint D, there are now only two unknown forces acting on joint B. Determine the two unknown forces on joint B. Draw FBD of joint C,Take sum of all the vertical forces is equal to zero.Sign conventions- (upward direction= +, Downward direction= -)We get, Fbc sin30° = P1Fbc sin30° = 350Fbc = 700 lb. (Tension)Take sum of all the horizontal forces is equal to zero.Sign conventions- (right side direction= +, left side direction= -)We get, Fcd = Fbc cos30°Fcd = 1260 cos30°Fcd = 606.2177 lb (Compression)Draw FBD of joint D,Take sum of all the vertical forces is equal to zero.Sign conventions- (upward direction= +, Downward direction= -)We get,Fdb = P1Fdb = 630 lb (Tension)Take sum of all the horizontal forces is equal to zero.Sign conventions- (right side direction= +, left side direction= -)We get,Fed = FcdFed = 606.2177 lb (Compression)Draw FBD of joint B,Take sum of all the vertical forces is equal to zero.Sign conventions- (upward direction= +, Downward direction= -)We get,Fab sin30° + Feb sin30° = Fdb + Fcb sin30°Fab + Feb = 2 Fdb + Fcb Fab + Feb = 2 × 630 + 700Fab + Feb = 1960 _________(1)Take sum of all the horizontal forces is equal to zero.Sign conventions- (right side direction= +, left side direction= -)We get,Feb cos30° + Fcb cos30° = Fab cos30°Fab - Fed = 700 __________(2)Adding equation (1) and (2), we get,2 Fab = 1960 + 700Fab = 1330 lb (Tension)Put, Fab = 1330 lb in equation (1)1330 + Feb = 1960Feb = 630 lb (Compression)All member forces are,Fbc = 700 lb. (Tension)Fcd = 606.2177 lb (Compression)Fdb = 630 lb (Tension)Fed = 606.2177 lb (Compression)Fab = 1330 lb (Tension)Feb = 630 lb (Compression)