6. The construction shown in the figure is in equilibrium. It is known that the mass of the load $m_{1}=1$ kg, the length of the uniform rod $l=50 \mathrm{~cm}$. The distance between the attachment points of the left thread to the rod $S=10$ cm. Determine the mass $m_{2}$ of the rod. All threads are weightless and inextensible. The pulleys are weightless. (15 points)

![](https://cdn.mathpix.com/cropped/2024_05_06_cfead746905039171f35g-04.jpg?height=617&width=642&top_left_y=1936&top_left_x=730)
Answer: 0.2 kg

Solution. From the equilibrium condition for the large block, it follows that the tension force in the left thread $T_{n}=\frac{m_{1} g}{2}$ (5 points). The equilibrium condition for the rod relative to the point of attachment of the right thread: $T_{n} \cdot l=T_{n} \cdot(l-S)+m_{2} g \cdot \frac{1}{2} l$ (5 points).

As a result, we get: $m_{2}=\frac{m_{1} S}{l}=\frac{1 \cdot 0.1}{0.5}=0.2$ kg (5 points).