Question

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!!!

Answer 1

Add up the torques about the pivot, treating counterclockwise torques as positive and clockwise as negative.

Answer 2

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?

Answer 3

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 N-m
Torque 2 is -3.0 x 5.2 sin 30 = -7.8 N-m

The net torque is -0.6 N-m

Answer 4

(6.0*1.2)+(3.0*5.2*sin30)
=15

Answer 5

-0.6