Please answer this Physics question.

6. A massless bar of length L = 6.0 m is placed on a pivot under its centre. A block of mass m1 = 2.4 kg is placed at a distance of T1 = 0.50 m to the right of the pivot. A second block of nlass m2 = 1.8 kg is placed at a distance of T2 = 1.5 In to the right of the pivot. (a) What is the net torque due to the masses ml and mg? (13) Where must a third block of mass m3 = 1.5 kg be placed for the bar to be in static equilibrium? Note: static equilibrium means that the net torque is zero Take 9 = 9.81 III/s2 and show all your work. J. H