Ответ: FA≈ 2 кН.
\[ V_T = 15 \, \text{дм}^3 = 0.015 \, \text{м}^3 \]
\[ F_A = \rho_ж V_T g \]
\[ F_A = 13600 \, \frac{\text{кг}}{\text{м}^3} \cdot 0.015 \, \text{м}^3 \cdot 10 \, \frac{\text{м}}{\text{с}^2} \]
\[ F_A = 2040 \, \text{Н} \approx 2 \, \text{кН} \]
Ответ: FA≈ 2 кН.