Here the system is jib–crane. Hence Member CA is in compression and AB is in tension. As shown in fig (b)
Cable DG goes over the frictionless pulley,
so Tension in AD = Tension in AG
= 20KN
FBD of the system is as shown in fig(b)
∑H = 0
P sin 30° – T sin 60° – 20 sin 60° = 0
0.5P – 0.866T = 17.32KN .........(i)
∑V = 0
P cos 30° + T cos 60° – 20 – 20 cos 60° = 0
0.866P + 0.5T = 30KN ...(ii)
Multiply by equation (i) by 0.5 , we get
0.25P – 0.433T = 8.66 ......(iii)
Multiply by equation (ii) by 0.866 , we get
0.749P + 0.433T = 25.98 ...(iv)
Add equation (i) and (ii), we get
P = 34.64KN
Putting the value of P in equation (i), we get
17.32 – 0.866T = 17.32
T = 0.