Asked by magic 8 ball

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.
Answered by bobpursley
Is there any reason why you would not sum torques about the piviot point?

Remember torque=force*distance*sinTheta

Answered by natali
i got 15 is that correct?
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