Answer: 5.2 kg

Solution. The rod consists of two pieces with masses $m_{n}=\lambda \cdot A B=3 \cdot 7=21$ kg (2 points) and $m_{n}=\lambda \cdot B C=3 \cdot 5=15$ kg (2 points). The condition for equilibrium in this situation is: $m_{1} \cdot A B+m_{n} \cdot \frac{1}{2} A B=m_{2} \cdot B O+m_{n} \cdot \frac{1}{2} B O$ (3 points). As a result, we get: $m_{1}=\frac{m_{2} \cdot B O+m_{n} \cdot \frac{1}{2} B O-m_{n} \cdot \frac{1}{2} A B}{A B}=\frac{20 \cdot 4+15 \cdot \frac{1}{2} \cdot 4-21 \cdot \frac{1}{2} \cdot 7}{7} \approx 5.2$ kg (3 points).
6. A bent thin homogeneous rod $ABC$, with small loads $m_{1}$ and $m_{2}=20$ kg located at its ends, is in equilibrium relative to a support placed at point B. The mass per unit length of the rod is $\lambda=3$ kg. It is known that $AB=7$ m, $BC=5$ m, $BO=4$ m, $OC=3$ m. Find $m_{1}$.

$(10$ points)

![](https://cdn.mathpix.com/cropped/2024_05_06_f940deb12dfd641283d0g-7.jpg?height=277&width=848&top_left_y=2357&top_left_x=610)