Question

The center of gravity of a 45gram meter stick is at the 49 cm. mark. The stick is part of a lever which has the fulcrum at the 73 cm. mark. Where must an 86gram object be hung in order to have equilibrium? Show solutions.

Answer 1

$M=Fl,$

$0.045\cdot(0.73-0.49)=0.086l,$

$l=12.5~cm,$

$x=73+12.5=85.5~cm.$