Question

A meter stick balances horizontally on a knife-edge at the $50.0$ cm mark. With two $5.00$ g coins stacked over the $12.0$ cm mark, the stick is found to balance at the $45.5$ cm mark. What is the mass of the meter stick?

Answer 1

The x-axis is along with the meter stick, with the origin at the zero position on the scale. The forces acting on it are shown in the diagram below. The nickels are at $x=x_1=0.120$ m, and m is their total mass. The knife-edge is at $x=x_2=0.455$ m and exerts force $\overrightarrow{F}$. The mass of the meter stick is M and the force of gravity acts at the center of the stick, $x=x_3=0.500$ m. Since the meter stick is in equilibrium, the sum of the torques about $x_2$ must vanish:
$Mg(x_3–x_2)–mg(x_2–x_1)=0$.
Thus,
$M=\frac{x_2-x_1}{x_2-x_2}m=(\frac{0.455m-0.120m}{0.500m-0.455m})\left(10.0g\right)=74.4$ g