When an extended object -- think a meterstick, a bridge, or a plank -- is in equilibrium, a problem will often require setting counterclockwise torques equal to clockwise torques. The question that often bugs first-year students is how to deal with the weight of the meterstick, bridge, or plank itself.

When the fulcrum is at the extended object's center of mass, no problem -- just ignore the object's weight.

But imagine a meterstick that's pivoted about the 70-cm mark. Now 70 cm hang to the left, and 30 cm hang to the right of the fulcrum. We need to include a torque produced by the weight of the meterstick itself when calculating, say, where to apply a known force to balance the stick. Torque is force times distance from a fulcrum. What force do we use, and what distance?

The correct answer is that we use the ENTIRE weight of the meterstick, and we use the distance from the meterstick's center of mass (i.e. the 50 cm mark) to the fulcrum. If the pivot is at the 70-cm mark, then that distance would be 20 cm. That torque would be acting in the direction pivoting toward the center of mass.

Students don't initially like that answer. "But 30 cm is hanging on the other side! Shouldn't I take that portion of the meterstick's weight into account?" Well, you did. Since you used the entire weight of the meterstick, and you used the position of the entire stick's center of mass, no further work is neccessary.

(Sure, you

*could*treat the left and right sides separately, apportioning the correct percentage of the stick's weight on each side and then indicating that weight acts at the center of mass of each side. Whew, I don't think that even I understand that last sentence. Better stick [ha!] with the simple method.)Today Jim Brice, a member of my group at the University of Georgia AP Summer Institute, shared a clever method of assuaging the students' worries about treating the whole meterstick as a unit. He puts a meterstick on an electronic balance. Of course, part of the stick hangs off of each side of the balance's platform. Nevertheless, he asks his class: Does the scale read the ENTIRE weight of the meterstick, or just the part actually touching the platform?

I was looking all over the internet for such elegant answer, and I found it here.

ReplyDeleteThanks a lot.

Ravi Adik