... (2-c) kN 1. 5m kN kN 3m 3m 3m 2m 4m 4m (3- a) (3- b) 1. 2 m kN 4kN 5kN kN 1. 6 m (4-a) 0.7 m 0.9 m 0.9 m 0.7 m 0.9 m 0.9 m 0.9 m 2m (3- c) 1. 2 m 1. 2 m 3m 3m (4-b) (4-c) kN kN 2m 2m kN 1m 1m 1m 1m 2m 2m ... Fy = 0, F5 (3/ 5) – F4 (3/ 5) –F1= 0, F1= –6 kN 38 Truss Analysis: Force Method, Part I by S T Mau Joint Σ Fx = 0, 4 F6(4/5) + F3(4/5) =0, F3 = kN Σ Fy = 0, F6 F6 (3/ 5) – F3 (3/ 5) –F2= 0, F3 F2 F2= –6 ... F2= –6 kN Joint Σ Fx = 0, F1 Rx1+ F3(4/5) =0, Σ Fy = 0, Ry1+F3 (3/ 5) +F1= 0, F3 Rx1 Ry1 Ry1= kN Joint Σ Fx = 0, 5 Rx5 Rx5 – F4(4/5) =0, Σ Fy = 0, F2 F4 Rx1 = –4 kN Ry5+F4 (3/ 5) +F2= 0, Rx5 = kN Ry5...