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.
can someone help me out im totally lost on this and don't even know where to start.
Answers
bobpursley
Is there any reason why you would not sum torques about the piviot point?
Remember torque=force*distance*sinTheta
Remember torque=force*distance*sinTheta
natali
i got 15 is that correct?