Frictionless pulley:

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

Sol.: Here system is jib-crane. Thus Member CA is in compression and AB is in tension. As shown in the given figure.

Cable DG goes over frictionless pulley, so Tension in AD = Tension in AG = 20KN

Free body diagram of the system is as shown in the given figure 

∑H = 0

P sin30° - T sin60° - 20 sin60° = 0

0.5 P - 0.866T = 17.32KN 

Pcos30° + Tcos60° - 20 - 20cos60° = 0

0.866P + 0.5T = 30KN                                                                                       ...(ii)

Multiply by the equation (i) by 0.5 , we obtain

0.25P - 0.433T = 8.66                                                                                        ...(iii)

Multiply by the equation (ii) by 0.866 , we obtain

0.749P + 0.433T = 25.98                                                                                       ...(iv)

Add the equation (i) and (ii), we obtain

P = 34.64KN                                          .......ANS

Putting the value of P in equation (i), we obtain

17.32 - 0.866T = 17.32

T = 0                                                         .......ANS