6. A bent thin homogeneous rod $ABC$, with small loads $m_{1}=2$ kg and $m_{2}$ located at its ends, is in equilibrium relative to a support placed at point $B$. The mass per unit length of the rod $\lambda=2$ kg. It is known that $AB=7$ m, $BC=5$ m, $BO=4$ m, $OC=3$ m. Find $m_{2}$.

(10 points)

![](https://cdn.mathpix.com/cropped/2024_05_06_f940deb12dfd641283d0g-3.jpg?height=300&width=899&top_left_y=2437&top_left_x=607)
Answer: 10.75 kg

Solution. The rod consists of two pieces with masses $m_{l}=\lambda \cdot A B=2 \cdot 7=14$ kg (ball) and $m_{n}=\lambda \cdot B C=2 \cdot 5=10$ 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: $m_{2}=\frac{m_{1} \cdot A B+m_{n} \cdot \frac{1}{2} A B-m_{n} \cdot \frac{1}{2} B O}{B O}=\frac{2 \cdot 7+14 \cdot \frac{1}{2} \cdot 7-10 \cdot \frac{1}{2} \cdot 4}{4}=10.75$ kg (3 points).