Question: A meter stick balances horizontally on a knife-edge at the 50 cm mark. With two 5.00 gm coins stacked over the 12.00 cmmark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?

• The meter stick balances on the knife-edge before coins are stacked at the $x_3=50.0\text{cm}=0.50\text{m}\text{mark}.$
• The meter stick balances on the knife-edge when coins are stacked at the$x_2=45.5\text{cm}=0.455\text{m}\text{mark}.$
• Mass of both coins, m = 10 .00 gm
• The coins are stacked over the meter stick at the$x_1=12.0\text{cm}=0.12\text{m}\text{mark}$

Using the condition for static equilibrium, you can write that the net torque point acting at the pivot point is zero. From that, you can get an equation. After solving it, you will get the mass of the meter stick. The formula used is given below.

At static equilibrium:

${\tau }_{net}=0$

The free body diagram:

At static equilibrium, the torque about point becomes zero.

So,

$$\begin{gathered} Mg\left(x_3-x_2\right)-mg\left(x_2-x_1\right)=0 \\ Mg\left(x_3-x_2\right)=mg\left(x_2-x_1\right) \\ M=\frac{\left\{mg\left(x_2-x_1\right)\right\}}{\left\{g\left(x_3-x_2\right)\right\}} \\ =\frac{m\left(x_2-x_1\right)}{x_3-x_2} \\ =\left(10\right)\left(\frac{0.455-0.120}{0.5-0.455}\right) \\ =74.4\text{g} \end{gathered}$$

Hence, the mass of the meter stick is 74.4 g