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 50-cm mark.

If a 120-g mass is placed at the 25-cm mark and a 25-g mass at the 10-cm mark, where should a 500-g 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!

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