Question

Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are0.50L, 2.6 g/cm³; 0.25 L, 1.0 g/cm³; and0.40 L, 0.80 g/cm³ . What is the force on the bottom of the container due to these liquids? One liter , 1L=1000 cm³. (Ignore the contribution due to the atmosphere.)

Short answer

The force on the bottom of the container due to three liquids is 18.3 N.

Step-by-step solution

The given data

i) Volume and density of liquid 1 is $V_1=0.50\mathrm{L},{\mathrm{\rho }}_1=2.6\mathrm{g}/{\mathrm{cm}}^3$

ii) Volume and density of liquid 2 is $V_2=0.50\mathrm{L},{\mathrm{\rho }}_2=1.0\mathrm{g}/{\mathrm{cm}}^3$

iii) Volume and density of liquid 3 is$V_3=0.40\mathrm{L},{\mathrm{\rho }}_3=0.80\mathrm{g}/{\mathrm{cm}}^3$

Understanding the concept of density and weight

We can use the formula for force (weight) in terms of density and volume to find the forces on three liquids separately. Then adding them, we can get the force on the bottom of the container due to three liquids.

Formula:

Weight of a body in terms of force,

$$\begin{gathered} W=mg \\ =\mathrm{\rho }Vg, \end{gathered}$$

where, (i)

$$\begin{gathered} \mathrm{\rho }=\mathrm{density}\mathrm{of}\mathrm{material}, \\ V=\mathrm{volume}\mathrm{of}\mathrm{materialm} \\ g=\mathrm{acceleration}\mathrm{of}\mathrm{gravity}\left(9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{or}980\mathrm{cm}/{\mathrm{s}}^2\right) \end{gathered}$$

Calculation of force (weight) on the bottom of the container

Force on the bottom of the container due to liquids:

We know that$1\mathrm{L}=1000{\mathrm{cm}}^3$

For first liquid, using equation (i)

role="math" localid="1657193879159" $$\begin{gathered} W_1=2.6\times \left(0.50\times 1000\right)\times 980 \\ =1.27\times {10}^6\mathrm{g}·\mathrm{cm}/{\mathrm{s}}^2 \\ =1.27\times {10}^6\mathrm{dyne} \\ =12.7\mathrm{N}\left(\because 1\mathrm{N}={10}^5\mathrm{dyne}\right) \end{gathered}$$

Similarly, we can find the weight of second liquid using equation (i)

role="math" localid="1657194157414" $$\begin{gathered} W_2=0.25\times 1.0\times 1000\times 980 \\ =2.45\times {10}^5\mathrm{g}·\mathrm{cm}/{\mathrm{s}}^2 \\ =2.45\times {10}^5\mathrm{dyne} \\ =2.45\mathrm{N}\left(\because 1\mathrm{N}={10}^5\mathrm{dyne}\right) \end{gathered}$$

For third liquid, using equation (i) we have

$$\begin{gathered} W_3=0.40\times 0.80\times 1000\times 980 \\ =3.14\times {10}^5\mathrm{g}·\mathrm{cm}/{\mathrm{s}}^2 \\ =3.14\times {10}^5\mathrm{dyne} \\ =3.14\mathrm{N}\left(\because 1\mathrm{N}={10}^5\mathrm{dyne}\right) \end{gathered}$$

Total force on the bottom of the cylinder is

$$\begin{gathered} F=W_1+W_2+W_3 \\ =12.7+2.45+3.14 \\ =18.29\mathrm{N} \\ \approx 18.3\mathrm{N} \end{gathered}$$

Therefore, the force on the bottom of the container due to three liquids is$18.3\mathrm{N}$