You have a 120-g meterstick. Where should you place the meterstick atop your finger so it balances? Where is the balance point if you place an additional 100-g mass at the 20-cm mark and a 200-g mass at the 80-cm mark?

Answer 1

(1) In the middle of the meterstick.

(2) Require that the net torque about the balance point be zero.

100(x - 20) = 200(80 - x) + 120(60 -x)
Solve for x.

Answer 2

does x = 60?