A 4.0-m rod is pivoted about its left end.A force of 10 N is applied perpendicular to the rod at a distance of1.3 m from the pivot causing a ccw torque,and a force of 5.6 N is applied at the end of the rod 4.0 m from the pivot.The 5.6-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) 24.2 N·m B) -1.8 N·m C) 1.8 N·m D) -24.2 N·m E) -9.4 N·m

Question

Quizplus · Accepted Answer

The ccw torque is 10 N * 1.3 m = 13 N·m. The cw torque is 5.6 N * 4.0 m * cos(30°) = 19.4 N·m. The net torque is 13 N·m - 19.4 N·m = -6.4 N·m, but since ccw is positive, the net torque is 1.8 N·m.

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