6. (2025·南京鼓楼校级模拟)如图所示,置于水平桌面上的一个上大下小、底面积为$0.02\ \mathrm{m}^{2}$的薄壁容器内装有质量为$4\ \mathrm{kg}$的液体,一个质量为$6\ \mathrm{kg}$、体积为$8× 10^{-3}\ \mathrm{m}^{3}$的物体放入容器内,物体漂浮在液面时有一半的体积浸在液体中,容器质量为$2\ \mathrm{kg}$,$g$取$10\ \mathrm{N/kg}$。求:
(1) 物体受到的浮力。
(2) 液体的密度。
(3) 容器对桌面的压强。
答案：6．(1)物体处于漂浮状态，所以物体受到的浮力$F_{浮}=G=mg=6 \mathrm{kg} × 10 \mathrm{N/kg} = 60 \mathrm{N}$ (2)物体排开液体的体积$V_{排}=\frac{1}{2}V=\frac{1}{2} × 8 × 10^{-3} \mathrm{m}^3 = 4 × 10^{-3} \mathrm{m}^3$，根据阿基米德原理可得，液体的密度$\rho = \frac{F_{浮}}{V_{排} g} = \frac{60 \mathrm{N}}{4 × 10^{-3} \mathrm{m}^3 × 10 \mathrm{N/kg}} = 1.5 × 10^3 \mathrm{kg/m}^3$
(3)容器对桌面的压力$F = G_{总} = m_{总} g = (4 \mathrm{kg} + 6 \mathrm{kg} + 2 \mathrm{kg}) × 10 \mathrm{N/kg} = 120 \mathrm{N}$，容器对桌面的压强$p = \frac{F}{S} = \frac{120 \mathrm{N}}{0.02 \mathrm{m}^2} = 6000 \mathrm{Pa}$
7. 一轻质细杆与不吸水的正方体物块$A$连接,另一端固定在质量为$1\ \mathrm{kg}$、底面积为$200\ \mathrm{cm}^{2}$的容器底部,如图甲所示。容器放在水平桌面上,现向容器内缓慢地加水,物块$A$对杆的力$F$与容器内水的深度$h$的关系如图乙所示,杆的体积不计。求:(已知水的密度是$1.0× 10^{3}\ \mathrm{kg/m}^{3}$,$g$取$10\ \mathrm{N/kg}$)
(1) 物块$A$浸没时受到的浮力。
(2) 物块$A$的密度。
(3) 物块$A$对杆的力$F$为$2\ \mathrm{N}$时,容器对桌面的压强。
答案：7．(1)由图乙可知，正方体物块A棱长是$14 \mathrm{cm} - 4 \mathrm{cm} = 10 \mathrm{cm} = 0.1 \mathrm{m}$，物块A浸没时受到的浮力$F_{浮} = \rho_{水} g V_{排} = 1.0 × 10^3 \mathrm{kg/m}^3 × 10 \mathrm{N/kg} × (0.1 \mathrm{m})^3 = 10 \mathrm{N}$ (2)由图乙可知，未加水时，物块A对杆的力为$7 \mathrm{N}$，也就是物块A重$7 \mathrm{N}$，物块A的质量$m_{A} = \frac{G_{A}}{g} = \frac{7 \mathrm{N}}{10 \mathrm{N/kg}} = 0.7 \mathrm{kg}$，物块A的密度$\rho_{A} = \frac{m_{A}}{V} = \frac{0.7 \mathrm{kg}}{0.001 \mathrm{m}^3} = 700 \mathrm{kg/m}^3$ (3)容器底面积为$200 \mathrm{cm}^2$，即$0.02 \mathrm{m}^2$，从图乙知细杆高度为$0.04 \mathrm{m}$，物块A下方水的重力$G_{0} = m g = \rho_{水} S_{容} h g = 1.0 × 10^3 \mathrm{kg/m}^3 × 0.02 \mathrm{m}^2 × 0.04 \mathrm{m} × 10 \mathrm{N/kg} = 8 \mathrm{N}$。第一种情况：当重力大于浮力时，浮力$F_{1} = G_{A} - F = 7 \mathrm{N} - 2 \mathrm{N} = 5 \mathrm{N}$，物块A浸入水中体积$V_{1} = \frac{F_{1}}{\rho_{水} g} = \frac{5 \mathrm{N}}{1.0 × 10^3 \mathrm{kg/m}^3 × 10 \mathrm{N/kg}} = 5 × 10^{-4} \mathrm{m}^3$，物块A浸入水中深度$h_{1} = \frac{V_{1}}{S_{A}} = \frac{5 × 10^{-4} \mathrm{m}^3}{(0.1 \mathrm{m})^2} = 0.05 \mathrm{m}$，此时A底面以上水的重力$G_{1} = m_{1} g = \rho_{水} V_{上} g = \rho_{水} (S_{容} - S_{A}) h_{1} g = 1.0 × 10^3 \mathrm{kg/m}^3 × (0.02 \mathrm{m}^2 - 0.01 \mathrm{m}^2) × 0.05 \mathrm{m} × 10 \mathrm{N/kg} = 5 \mathrm{N}$，容器重$G_{容} = m_{容} g = 1 \mathrm{kg} × 10 \mathrm{N/kg} = 10 \mathrm{N}$，容器对桌面的压强$p_{1} = \frac{F_{总1}}{S} = \frac{G_{A} + G_{容} + G_{0} + G_{1}}{S} = \frac{7 \mathrm{N} + 10 \mathrm{N} + 8 \mathrm{N} + 5 \mathrm{N}}{0.02 \mathrm{m}^2} = 1500 \mathrm{Pa}$。第二种情况：当重力小于浮力时，浮力$F_{2} = G_{A} + F = 7 \mathrm{N} + 2 \mathrm{N} = 9 \mathrm{N}$，物块A浸入水中体积$V_{2} = \frac{F_{2}}{\rho_{水} g} = \frac{9 \mathrm{N}}{1.0 × 10^3 \mathrm{kg/m}^3 × 10 \mathrm{N/kg}} = 9 × 10^{-4} \mathrm{m}^3$，物块A浸入水中深度$h_{2} = \frac{V_{2}}{S_{A}} = \frac{9 × 10^{-4} \mathrm{m}^3}{(0.1 \mathrm{m})^2} = 0.09 \mathrm{m}$，A底面以上水的重力$G_{2} = m_{2} g = \rho_{水} V_{上} ' g = \rho_{水} (S_{容} - S_{A}) h_{2} g = 1.0 × 10^3 \mathrm{kg/m}^3 × (0.02 \mathrm{m}^2 - 0.01 \mathrm{m}^2) × 0.09 \mathrm{m} × 10 \mathrm{N/kg} = 9 \mathrm{N}$，容器对桌面的压强$p_{2} = \frac{F_{总2}}{S} = \frac{G_{A} + G_{容} + G_{0} + G_{2}}{S} = \frac{7 \mathrm{N} + 10 \mathrm{N} + 8 \mathrm{N} + 9 \mathrm{N}}{0.02 \mathrm{m}^2} = 1700 \mathrm{Pa}$
