3. (physics) Two weightless glasses with different amounts of water are balanced on a lever of unequal-arm scales with a length of $L$. A mass of water $m$ is transferred from one glass to the other, and the scales lose their balance. To restore balance, the fulcrum of the lever needs to be moved by a distance $l$ relative to its old position. Find the total mass of water in the glasses. Neglect the mass of the lever of the scales.
Solution. Let the mass of water in the left glass be $m_{1}$, in the right glass be $m_{2}$, and the lever arms of the glasses be $l_{1}$ and $L-l_{1}$, respectively. Then the condition

![](https://cdn.mathpix.com/cropped/2024_05_06_f912147981c7262840e0g-2.jpg?height=137&width=557&top_left_y=865&top_left_x=1338)
for equilibrium of the balance in the first case gives

$$
m_{1} l_{1}=m_{2}\left(L-l_{1}\right)
$$

Suppose we pour a mass of water $m$ from the left glass to the right glass. Then the condition for equilibrium of the balance in the second case gives

$$
\left(m_{1}-m\right)\left(l_{1}+l\right)=\left(m_{2}+m\right)\left(L-l_{1}-l\right)
$$

Subtracting the first equation from the second, we find the total mass of water in the glasses

$$
m_{1}+m_{2}=\frac{m L}{l}
$$