Chapter 6.6, Problem 36E

To determine

To Calculate: The hydrostatic force on the lateral face of a square plate dipped in water

Answer to Problem 36E

The total hydrostatic force on one side of the plate is given by the expression $\frac{\rho g{a}^3}{\sqrt{2}}$ .

Explanation of Solution

Given: The tank contains water with density $1000kg/m^3$ .

The side of the square is of length a .

Concept Used:

• The hydrostatic pressure of a water of density $\rho$ at a depth $d$ from the surface of the water is $P=\rho gd$ .
• The force on a surface due to pressure $P$ exerted is $F=PA$ , where $A$ is the area of the surface being considered.

The length of the half of the diagonal of the square is âˆ�(a²+a² )/2=a/âˆ�2, which is also the depth of the center of the square form the surface of water.

So, the depth of a point on the vertical diagonal which is at a distance x from the center of the square is a/âˆ�2-x and the width of a horizontal line on the plate passing through this point is 2 × (a/âˆ�2-|x| )=aâˆ�2-2|x|.

So, the area $A_x$ of a thin horizontal strip of the plate of thickness $dx$ at this distance $x$ is

  $d{A}_x=width\times thickness=(a\sqrt{2}-2|x|)dx$

and the hydrostatic pressure at the depth $h$ is

  $P_x=\rho \times g\times depth=\rho g\left(\frac{a}{\sqrt{2}}-x\right)$

Therefore, the force on this thin strip is

  $dF=P\times A=P_{x}d{A}_x=\rho g\left(\frac{a}{\sqrt{2}}-x\right)\cdot \left(a\sqrt{2}-2\left|x\right|\right)dx$

Now the total force on this lateral surface is the integral of $dF$ from $x=-\frac{a}{\sqrt{2}}$ (lower tip of the plate) to $x=\frac{a}{\sqrt{2}}$ (upper tip of the plate). So

  $\begin{array}{l}F=\underset{-\frac{a}{\sqrt{2}}}{\overset{\frac{a}{\sqrt{2}}}{{\displaystyle \int }}}dF=\rho g\underset{-\frac{a}{\sqrt{2}}}{\overset{\frac{a}{\sqrt{2}}}{{\displaystyle \int }}}\left(\frac{a}{\sqrt{2}}-x\right)(a\sqrt{2}-2|x|)dx\\ =\rho g\left(\underset{-\frac{a}{\sqrt{2}}}{\overset{0}{{\displaystyle \int }}}\left(\frac{a}{\sqrt{2}}-x\right)(a\sqrt{2}+2x)dx+\underset{-0}{\overset{\frac{a}{\sqrt{2}}}{{\displaystyle \int }}}\left(\frac{a}{\sqrt{2}}-x\right)(a\sqrt{2}-2x)dx\right)\end{array}$

  $\begin{array}{l}=\rho g\left(\underset{-\frac{a}{\sqrt{2}}}{\overset{0}{{\displaystyle \int }}}(a^2-2x^2)dx+\underset{0}{\overset{\frac{a}{\sqrt{2}}}{{\displaystyle \int }}}2{\left(\frac{a}{\sqrt{2}}-x\right)}^{2}dx\right)\\ =\rho g\left({\left[a^{2}x-\frac{2x^3}{3}\right]}^0{}_{-\frac{a}{\sqrt{2}}}+{\left[\frac{2}{3}{\left(x-\frac{a}{\sqrt{2}}\right)}^3\right]}_0^{\frac{a}{\sqrt{2}}}\right)\\ =\rho g\left(\left(\frac{a^3}{\sqrt{2}}-\frac{a^3}{3\sqrt{2}}\right)+\left(\frac{a^3}{3\sqrt{2}}\right)\right)\\ =\frac{\rho g{a}^3}{\sqrt{2}}\end{array}$

Therefore, the total hydrostatic force on one side of the plate is given by the expression $\frac{\rho g{a}^3}{\sqrt{2}}$ .