Question

The frictionless pulley A is supported by two bars AB and AC which are hinged at B and C to a vertical wall. The flexible cable DG hinged at D goes over the pulley and supports a load of 20KN at G. The angle between the various members shown in fig(a) Determine the forces in AB and AC. Neglect the size of pulley.

Answer 1

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.