Physics
posted by Lila .
If a 100g mass was placed at the 25cm mark, and a 20g mass at the 10cm mark, where should a 500g mass be placed to balance the system?

100*(25 distance from fulcrum) + 20*(40 distance from fulcrum)= 500(x distance from fulcrum)
3300=500x Solve for x. then add to 50 because it's in the opposite side of the fulcrum. 
100*(25 distance from fulcrum) + 20*(40 distance from fulcrum)= 500(x distance from fulcrum)
3300=500x Solve for x. then add to 50 because it's in the opposite side of the fulcrum.
At the 50.66 cm mark the 500g mass should be placed 
the answer should be 56.6 cm because 3300/500 is 6.6.