Torque on a see-saw type object

1. Jan 8, 2006

Captain Zappo

Torque on a "see-saw" type object

I had this question on my Physics test last week. I missed it, and it's now time to do test corrections. I have read through the section once again, but I keep on getting the same answer.

A uniform plank XY is supported by two equal 120-N forces at X and Y, as shown. the support at X is then moved to Z (half-way to the plank center). The supporting force at Y then becomes:
A. 40
B. 160
C. 60
D. 80
E. 240
I keep getting 60. I know that the net torque must remain 0, so as to keep the object from rotating. That being said, I cannot seem to come up with the correct answer (D).
Here's what my logic was:
If you move the point X, halfway to the center, it will produce 1/2 the amount of torque. So, the force at Y must decrease by 1/2 so as to keep the object in rotational equilibrium.

2. Jan 8, 2006

Fermat

If the force at Z is half the distance (x) that the force at Y is (2x) from the centre,
and the torques must be equal for balance,
then the reaction at Z must be twice that of the reaction at Y.

And, of course, the sum of these two reactions is 240 N.