Posted by Emily on Thursday, December 6, 2007 at 7:54pm.
This is how the problem starts out:
A meter stick is placed with the fulcrum at the 50cm mark.
If a 120g mass is placed at the 25cm mark and a 25g mass at the 10cm mark, where should a 500g mass be placed to balance the system?
I thought that since the two masses (the 120 and the 25 ones) were on the same side of the fulcrum, you could add the two masses together and just use the longest distance from the fulcrum. Apparently (what my physics teacher told me) this is wrong.
Thanks!

Physics  Damon, Thursday, December 6, 2007 at 9:00pm
No, you may not add the two masses
the 25 grams is 40 cm from the fulcrum
the 120 grams is 25 cm from the fulcrum
so on that side you have
25* 40 + 120 * 25 = 4000
That is NOT the same as 145 * 40 !!!
on the other side you have 500 grams a distance "x" from the fulcrum
so
500 x = 4000
solve for x
then remember that x is measured from the fulcrum at the 50 cm mark so the 500 grams must be located at the 50 + x cm mark 
Physics  Emily, Thursday, December 6, 2007 at 9:31pm
Okay. So in equation form it would look like this:
d(3)=((120*25)+(25*40))/500
which equals 8.
So the fulcrum would be placed at the 58 cm mark, right?
This is exactly what my friend did, I just couldn't remember what he wrote down or how he explained that concept. Thanks for all of your help, Damon! 
Physics  Emily, Thursday, December 6, 2007 at 9:37pm
Oops, I meant the 500g mass is placed at 58 cm. I was thinking fulcrum because of my next problem, and I got the two terms mixed in my head. Thanks again!