How Do You Calculate Net Torque with Different Forces and Angles?

[mmark]

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?

My assumption is:
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?

 

[LowlyPion]

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?

My assumption is:
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?

You might want to mind your CW and CCW's. They don't add positively. Moreover there are only 2 forces (which = only 2 torques) by the statement of the problem. (Draw a careful diagram for yourself.)

Your answer should be of the form magnitude and CW or CCW.

 

[Doc Al]

Is this correct?

No.

(1) There are only two forces acting. You counted the 5.2 N force twice.
(2) Clockwise and counterclockwise torques have opposite signs.