A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from the pivot causing a ccw torque, and a force of 2.60 N is applied at the end of the rod 3.00 m from the pivot. The 2.60-N force is at an angle of 30.0o to the rod and causes a cw torque. What is the net torque about the pivot? (Take ccw as positive.)
a. 16.4 N·m b. –8.58 N·m
c. 8.58 N·m d. –16.4 N·m
e. 4.68 N·m
c
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