How Is Torque Calculated on a Triangular Plate with Pivoted Forces?

[Soaring Crane]

A light triangular plate OAB is in a horizontal plane. The 3 forces, F1 = 7 N, F2 = 3 N, F3 = 8 B, act on the plate, which is pivoted about a vertical axes through point O. Consider the counterclockwise sense as positive. The sum of the torques about the vertical axis through point O, acting on the plate due to forces F1, F2, and F3, is closest to:

a. 0.30 N*m-----------b. 1.2 N*m---------c. –0.30 N*m------d. –1.2 N*m-------e. zero

In the attachment, please note that the angle between F1 and the dotted line is 30 degrees. The angle between the dotted line and F3 is 45 degrees.
Angle 1 = 30 degrees
Angle 3 = 45 degrees

Hypotenuse = 1.0 m

If you cannot see the attachment, please tell me.

Since the plate is rotated around point O,

The torque where F3 acts is 0 m*N since r = 0 m??

Therefore, only forces 1 and 2 are left.

Is Force 2 counterclockwise, which makes it positive? Torque 2 = F*r*sin theta = 3N*0.8 m*sin 90 = +2.4 m*N??

Is Force 1 clockwise, which makes it negative? Torque 1 = F*r*sin theta = 7N*sin 90*0.6 m = -2.1 m*N?

Sum of torques = 2.4 m*N – 2.1 m*N = 0.30 m*N??

Thanks.

 

[radou]

Unfortunately, I cannot see the attachment, but it is possibly due to my computer [:grumpy:]. But yes, if the force vector intersects with the axis, the torque produced by that force must equal zero.

 

[Soaring Crane]

I tried saving it as a Word document. Hopefully, it is accessible now.

 

[Soaring Crane]

Are my signs for the torque correct?

Thanks again.

 

[Doc Al]

Yes, your signs are correct.