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.
If a 100-g mass was placed at the 25-cm mark, and a 20-g mass at the 10-cm mark, where should a 500-g mass be placed to balance the system?
4 answers
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
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.
why ? 40- distance from fulcrum
2 answers