Two masses are hanging on a meter stick suspended at its center. On the right side an object with a mass of?

Two masses are hanging on a meter stick suspended at its center. On the right side an object with a mass of?


mass of 200 square nuts is 10cm from the fulcrum. And on the left side an object that balances 100 square nuts is 20 cm from the fulcrum.





QUESTION: Now the mass of the 100 square nut objet is decreased by an amount m. How much farther must the diminished object be moved from the fulcrum to maintain the balance?Two masses are hanging on a meter stick suspended at its center. On the right side an object with a mass of?
The mass times the distance gives you the force applied to the meter stick. Therefore d X (100 - m) = 10 cm X 200 sn. Where m is the amount the mass is reduced and d is the distance from the fulcrum to place the mass to maintain equilibrium.





sn = square nuts





If m = 50 then





d X (100 sn - 50 sn) = 10 cm X 200 sn


d X 50 sn = 2000 cm sn


divide by 50 sn


d = 2000 cm sn / 50 sn


d = 40 cm