A 3.0-m rod is pivoted about its left end.A force of 10 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 N is applied at the end of the rod 3.0 m from the pivot.The 5.2-N force is at an angle of 30^o to the rod and causes a cw torque.What is the net torque about the pivot? (Take ccw as positive. ) A) 18 N·m B) −18 N·m C) 4.2 N·m D) −4.2 N·m

Question

Quizplus · Accepted Answer

The torque from the 10 N force is 10 N * 1.2 m = 12 N·m (ccw, positive). The torque from the 5.2 N force is 5.2 N * 3.0 m * sin(30°) = 7.8 N·m (cw, negative). The net torque is 12 N·m - 7.8 N·m = 4.2 N·m.

A 30-M Rod Is Pivoted About Its Left End