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 net torque is calculated by subtracting the clockwise torque from the counterclockwise torque. The counterclockwise torque is 10 N * 1.2 m = 12 N·m. The clockwise torque is 5.2 N * 3.0 m * sin(30°) = 7.8 N·m. 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