A 3.0 m rod is pivoted about its left end.A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a ccw torque, and a force of 5.2 is applied at the end of the rod 3.0 m from the pivot. The 5.2 N is at an angle of 30 degrees to the rod and causes a cw torque. What is the net torque about the pivot? My assumption is: torque is radius X Force sin theta Torque 1 and 2 have angles of 90 degrees making it equal to 1. Torque 1:0.6 (radius of 1.2) X 6.0 N Torque 2:1.5 (radius of 3)X 5.2 N Torque 3:1.5 (radius of 3)X 5.2 N sine 30 degrees Torque 1 + Torque 2 + Torque 3 = Net Torque 3.6 + 7.8 + 3.9 = 15.3 Is this correct?