Problem No. 5 (15 points)

The system shown in the figure is in equilibrium. It is known that the uniform rod $AB$ and the load lying on it have the same mass $m=10$ kg. The load is located exactly in the middle of the rod. The thread, passing over the pulleys, is attached to one end of the rod and at a distance of one quarter of the rod's length from its left end. Determine the mass $m_2$ of the second load, suspended from the center of one of the pulleys. The threads and pulleys are weightless, and there is no friction in the pulley axes.

![](https://cdn.mathpix.com/cropped/2024_05_06_3c4b08bfdadd8858901bg-4.jpg?height=625&width=523&top_left_y=1252&top_left_x=801)

Answer: 80 kg

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# Solution and Evaluation Criteria:

Diagram with forces correctly placed.

Tension force in the string: $T=\frac{M g}{2}$

The lever rule, written relative to point v:

$m g \cdot \frac{1}{2} A B+m g \cdot \frac{1}{2} A B+T \cdot \frac{3}{4} A B=T \cdot A B$

As a result, we get: $M=8 m=80$ kg.