An aquarium of length $L$, width (front to back) $W$, and depth $D$is filled to the top with liquid of density

a. Find an expression for the force of the liquid on the bottom of the aquarium.

b. Find an expression for the force of the liquid on the front window of the aquarium.

c. Evaluate the forces for a$100cm$-long, $35\text{-cm-wide,}40-\mathrm{cm}$deep aquarium filled with water.

Expression for hydrostatic pressure at depth $D$is given by

Here $p_0$is the surface pressure where $D=0andp$is the density of the liquid.

Let force acted upon by the liquid on bottom of the aquarium, $F_a$Force acted upon by the liquid on the bottom of the aquarium equals to the gravity of the liquid because the bottom of the aquarium is the only support in vertical direction:

$F_a=mg$

The volume $V$equals to length times width times depth,

$V=DLW$

Density of the liquid is given by,

$\rho =\frac{m}{V}$

Re arrange equation for $m$

$m=\rho V$

Substitute $V=DLW$in the equation $m=\rho V$

Substitute $m=\rho DLW$in the equation $F_a=mg$

$F_a=mg$

Therefore force acted upon by the liquid on bottom of the aquarium is

$F_a=\rho gDWL$

The hydrostatic pressure at depth $D$is given by

$p=p_0+\rho gD$

So in the front window, at spot depth $D$, there is the pressure indicated above pushing the windows outward. There is also a pressure due the atmosphere that pushes the window inward, which cancels the first term in the above equation. So the net pressure is the one only related to the liquid:

$p=\rho gD$

Here $\rho$is the density of the liquid; $d$is the distance from the point of interest to the surface of the liquid. And $g$ is the gravitational acceleration. The total force on the window is to integrate the pressure in the $x-y$ plane,

$F_b={\int }_0^D {\int }_0^L \rho gddxdy$

At depth $d$in our coordinates is actually $(D-Y)$so we have,

$F_b=\rho g{\int }_0^D {\int }_0^L (D-y)dxdy$

The function does not depend on $x$.So we can integrate the above equation $x$

$F_b=L\rho g{\int }_0^D (D-y)dy$

We can continue to do the integration over $y$

$F_b=L\rho g{\left[Dy-\frac{1}{2}{y}^2\right]}_0^D$

The result of inetgration of

$F_z=\frac{1}{2}\rho g{D}^{2}L$

Conversion of units of length of the aquarium from cm to $m$

$L=100cm$

$\mathbf{=}\mathbf{\left(}\mathbf{100}\mathbf{cm}\mathbf{\right)}\left(\frac{0.01m}{\mathrm{cm}}\right)$

$=1m$

Conversion of units of breadth of aquarium from $cm$to $m$

$W=35cm$

$=\left(35\mathrm{cm}\right)\left(\frac{0.01\mathrm{m}}{\mathrm{cm}}\right)$

$0.35m$

Conversion of units of depth of the aquarium from $cmtom$

$D=40cm$

$=\left(40\mathrm{cm}\right)\left(\frac{0.01\mathrm{m}}{\mathrm{cm}}\right)$

$=40m$

Substitute $0.40\mathrm{m}\text{for}{D}_2,0.35\mathrm{m}\text{for}{W}_{\text{, and}}0.10,\text{and}0.1\mathrm{m}\text{for}L\text{, and}1000\mathrm{kg}/{\mathrm{m}}^3\text{for}\rho$

$9.8\mathrm{m}/{\mathrm{s}}^2$for in equation $F_a=\rho gDWL$

$F_a=\rho gDWL$

$\begin{array}{r}=\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.4\mathrm{m}\right)\left(0.35\mathrm{m}\right)\left(1\mathrm{m}\right)\\ \end{array}$

$\begin{array}{r}=\left(1.4\times {10}^3\mathrm{N}\right)\left(\frac{\mathrm{kN}}{{10}^3\mathrm{N}}\right)\\ =1.4\mathrm{kN}\end{array}$

Therefore the force acting on aquarium is $1.47KN$

Using part (b) expression for force acting on aquarium in $xy$plane

$F_y=\frac{1}{2}\rho gL{D}^2$

Substitute $1000\mathrm{kg}/{\mathrm{m}}^3\text{for}\rho ,9.8\mathrm{m}/{\mathrm{s}}^2\text{for}g\text{, and}1\mathrm{m}\text{for}L\text{, and}0.4\mathrm{m}\text{for}D\text{in the}$equation

$F_b=\frac{1}{2}\rho gL{D}^2$

$=\frac{1}{2}\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1\mathrm{m}\right)(0.4\mathrm{m}{)}^2\mathrm{N}$

$=784N$

$=\left(0.784\times {10}^3\mathrm{N}\right)\left(\frac{\mathrm{kN}}{{10}^3\mathrm{N}}\right)$

$=0.78\mathrm{kN}$

$xy$plane is $0.78\mathrm{kN}$