lab question?

Question

We have a meter stick balanced with the support at the 52.6 cm mark, a .2 kg at the 10 cm mark, a .1 kg at the 50 cm mark, and a .2 kg at the 98 cm mark.

We found the Tcc to be 88.49 N*cm and the Tc to be 88.98 N*cm by finding the distances each weight (including the weight of .943 N for the meterstick) was from the 52.6 cm support and multiplying that distance by each weight. These came out approximately equal as predicted.

Now, we are supposed to make the pivot point at 0 cm and show that the pivot point location is arbitrary. So, I added each (force*distance) (10*1.96)+(50*.98)+ (50*.941)+(98 *1.96) and got 307.73 cm . Supposedly the forces add up to the same?? What am I missing here with the 0 pivot point stuff??? 88.49+88.98 do not equal 307.73, so what am I doing wrong at this step?

Answer 1

I'm not following exactly, but I think your problem is that you have not considered things to the left of zero to be negative. The 10 cm mark is now the -42.6 cm mark. The 50 cm mark is now the -2.6 cm mark, and the 98 cm mark is now the 46.4 cm mark.