Posted by Steph on .
A 3.0 m rod is pivoted about its left end.A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a ccw torque, and a force of 5.2 is applied at the end of the rod 3.0 m from the pivot. The 5.2 N is at an angle of 30 degrees to the rod and causes a cw torque. What is the net torque about the pivot?
I don't have a clue how to even start this question.
PLEASE HELP!!!

physics classHelp!!! 
drwls,
Add up the torques about the pivot, treating counterclockwise torques as positive and clockwise as negative.

physics classHelp!!! 
Steph,
torque is radius X Force sin theta
Torque 1 and 2 have angles of 90 degrees making it equal to 1.
Torque 1:0.6 (radius of 1.2) X 6.0 N
Torque 2:1.5 (radius of 3)X 5.2 N
Torque 3:1.5 (radius of 3)X 5.2 N sine 30 degrees
Torque 1 + Torque 2 + Torque 3 = Net Torque
3.6 + 7.8 + 3.9 = 15.3
Is this correct? 
physics classHelp!!! 
drwls,
No, it isn't, but thanks for showing your work. Actually there are only two torques applied. I don't see why you are applying a factor of 1/2 to each radius
Torque 1 is +1.2 x 6.0 = +7.2 Nm
Torque 2 is 3.0 x 5.2 sin 30 = 7.8 Nm
The net torque is 0.6 Nm