A horizontal 1.20 m bar can rotate about a point that is 0.30 m from its right end. Three forces act on the bar:
1. 12 N directed 45° below the bar at the left end,
2. 20 N acting upward at a point 0.40 m to the left of the pivot, perpendicular to the bar,
3. 15 N directed horizontally toward the pivot at the right end.
Using the sign convention that counter-clockwise torques are positive, determine the net torque about the pivot.