# Problem №8 (15 points)

A load was decided to be weighed on unequal-arm scales. When the load was placed on one of the pans of these scales, a weight of mass $m_{1}=1$ kg had to be placed on the other side for balance. In the situation where the load was placed on the other pan of the scales, it had to be balanced by a weight of mass $m_{2}=2$ kg. Determine the mass of the load, given that the ratio of the arms of the scales is $1: 2$.

Answer: 1.333 kg
# Solution and evaluation criteria:

![](https://cdn.mathpix.com/cropped/2024_05_06_3c4b08bfdadd8858901bg-6.jpg?height=123&width=800&top_left_y=1692&top_left_x=662)

From the condition, we can conclude that the scales have their own mass, which must be taken into account.

The lever rule for the first weighing:

$m_{z} g \cdot l_{a}=m_{1} g \cdot l_{n}+m_{\text {ececoco }} g \cdot\left(\frac{l_{n}+l_{n}}{2}-l_{n}\right)$.

The lever rule for the second weighing:

$m_{2} g \cdot l_{n}=m_{z} g \cdot l_{n}+m_{\text {acecos }} g \cdot\left(\frac{l_{n}+l_{n}}{2}-l_{n}\right)$.

As a result, we get: $m_{e}=\frac{m_{1} l_{n}+m_{2} l_{n}}{l_{n}+l_{n}}=\frac{2 m_{1}+m_{2}}{3}=1.333$ kg