Chapter 6.6, Problem 32E

To determine

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

Answer to Problem 32E

The total hydrostatic force on one side of the plate $98000N$

Explanation of Solution

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

• The width of the plate is $4m$ .
• The height of the plate is $5m$ .

Concept Used:

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

While the depth increases by $5m$ , the width increases by $4m$ . So, at any depth $h$ from the surface of water, the width of the plate is given by

  $w=\frac{4h}{5}$

So, the area $A_h$ of a thin horizontal strip of the plate of thickness $dh$ at this depth $h$ is

  $d{A}_h=width\times height=\frac{4h}{5}\cdot dh$

The hydrostatic pressure at the depth $h$ is

  $P_h=\rho gh$

Therefore, the force on this thin strip is

  $\begin{array}{l}dF=P\times A=P_{h}d{A}_h=\rho gh\cdot \frac{4h}{5}dh\\ =7840hdh\end{array}$

Now the total force on this lateral surface is the integral of $dF$ from $h=0$ (surface of the water) to $h=5$ (depth of the lower edge of the triangle). So

  $F={{\displaystyle \int }}_0^{5}dF={{\displaystyle \int }}_0^{5}7840hdh$

  $={7840\times (\frac{h^2}{2})]}_0^5$

  $=7840\left[\frac{25}{2}-0\right]$

  $=358200\times 8.64$

  $=98000$

Therefore, the total hydrostatic force on one side of the plate is $98000N$ .