Question

Three liquids that will not mix are poured into a cylindrical container. The amount and densities of the liquids are (0.50 L, ${2.6g/cm}^3$and 0.40L) Find the total force on the bottom of the container. (Ignore the contribution due to atmosphere.) Does it matter whether the fluids mix?

Answer 1

We note that the container is cylindrical, the important aspect of this being that it has a uniform cross-section (as viewed from above); this allows us to relate the pressure at the bottom simply to the total weight of the liquids. Using the fact that $1L=1000c{m}^3$, we

find the weight of the first liquid to be

$W_1=m_{1}g={\rho }_1{V}_{1}g=\left(2.6g/c{m}^3\right)\left(0.50L\right)\left(1000c{m}^3/L\right)\left(980cm/s^2\right)=1.27\times {10}^{6}g.cm/s^2=12.7N$

In the last step, we have converted grams to kilograms and centimeters to meters. Similarly, for the second and the third liquids, we have $W_2=m_{2}g={\rho }_2{V}_{2}g=\left(1.0g/c{m}^3\right)\left(0.25L\right)\left(1000c{m}^3/L\right)\left(980cm/s^2\right)=2.5N$

and

$W_3=m_{3}g={\rho }_3{V}_{3}g=\left(0.8g/c{m}^3\right)\left(0.40L\right)\left(1000c{m}^3/L\right)\left(980cm/s^2\right)=3.1N$