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

![](https://cdn.mathpix.com/cropped/2024_05_06_cfead746905039171f35g-09.jpg?height=608&width=640&top_left_y=1843&top_left_x=731)
Answer: 10 kg

Solution. From the equilibrium condition for the large block, it follows that the tension force in the left thread: $T_{n}=\frac{m g}{2}$ (5 points). The equilibrium condition for the rod relative to the attachment point 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: $m_{1}=\frac{m_{2} l}{S}=\frac{2 \cdot 0.5}{0.1}=10$ kg (5 points).