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Согласно правилу моментов: М₁ = М₂ + М₃, где M₁ = m₃gl, M₂ = m₂gl, M₃ = P × 2l ; P = mg - Fa • Плотность спирта - 800 кг/м³ Ответ: m₃ = 11.6 кг
Большое спасибо!
![m_{3}gl = m_{2}gl + (m_{1}g - \rho_c g V_{1}) \\ \\ m_{3} = \dfrac{m_{2}gl + (m_{1}g - \rho_c g V_{1}) * 2l}{gl} \\ \\ m_{3} = \dfrac{8*10*0.2 + (5*10 - 800*10*4*10^{-3})*0.4}{10*0.2} = 11.6 \ [kg]](https://tex.z-dn.net/?f=m_%7B3%7Dgl+%3D+m_%7B2%7Dgl+%2B+%28m_%7B1%7Dg+-+%5Crho_c+g+V_%7B1%7D%29+%5C%5C+%5C%5C+m_%7B3%7D+%3D+%5Cdfrac%7Bm_%7B2%7Dgl+%2B+%28m_%7B1%7Dg+-+%5Crho_c+g+V_%7B1%7D%29+%2A+2l%7D%7Bgl%7D+%5C%5C+%5C%5C+m_%7B3%7D+%3D+%5Cdfrac%7B8%2A10%2A0.2+%2B+%285%2A10+-+800%2A10%2A4%2A10%5E%7B-3%7D%29%2A0.4%7D%7B10%2A0.2%7D+%3D+11.6+%5C+%5Bkg%5D "m_{3}gl = m_{2}gl + (m_{1}g - \rho_c g V_{1}) \\ \\ m_{3} = \dfrac{m_{2}gl + (m_{1}g - \rho_c g V_{1}) * 2l}{gl} \\ \\ m_{3} = \dfrac{8*10*0.2 + (5*10 - 800*10*4*10^{-3})*0.4}{10*0.2} = 11.6 \ [kg]")