A 3.0m rod is pivoted about its left end. A force of 6.0N is applied perpendicularto the rod at a distance of 1.2m from the pivot causing a ccw torque, and a force of 5.2N is applied at the end of the rod 3.0m fromt eh pivot. The 5.2N is at an angle if 30 degrees to the rod and causes a cw torque. what is the net torque about the pivot?

Question

A 3.0m rod is pivoted about its left end. A force of 6.0N is applied perpendicularto the rod at a distance of 1.2m from the pivot causing a ccw torque, and a force of 5.2N is applied at the end of the rod 3.0m fromt eh pivot. The 5.2N is at an angle if 30 degrees to the rod and causes a cw torque. what is the net torque about the pivot?  
can someone help me out im totally lost on this and don't even know where to start.

bobpursley · Answer

Is there any reason why you would not sum torques about the piviot point? Remember torque=force*distance*sinTheta

natali · Answer

i got 15 is that correct?