5. (15 points) Find the ratio $m_{1} / m_{2}$ of two hanging spheres, given that the tensions in the upper and lower threads differ by a factor of two.
Answer: $\frac{m_{1}}{m_{2}}=1$

Solution. For the lower ball: $m_{2} g=T_{H}$. For the upper ball: $m_{1} g+T_{H}=m_{1} g+m_{2} g=T_{B}$. At the same time, $T_{B}=2 T_{H}$. We obtain that

![](https://cdn.mathpix.com/cropped/2024_05_06_cf00a53ce06ae27d2795g-2.jpg?height=491&width=140&top_left_y=1057&top_left_x=1780)
$m_{1} g+m_{2} g=2 m_{2} g$, which means $\frac{m_{1}}{m_{2}}=1$.