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Vertically inserted into the water I have a rectangle 6 feet wide and 4 feet high that is submerged under the water with 2 feet of water above it.

Using a riemann sum how do I find the pressure? I am not able to use the triangle ratio trick so what do I do now?

I tried to do the ratio and I got

$$\frac{a}{4-x}$$

$$a = 6 - \frac{3x}{2}$$

$$\int_2^6 19620 (6x - \frac{3x^2}{2})dx$$

This is wrong but I do not know why.

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In the title and in the question instead of "pressure" you should write "force", because the hydrostatic pressure $P$ depends on the depth $d$ of the water below the surface, while the total hydrostatic force $F$ depends not only on $d$ but also on the shape. In this case the shape is a rectangle 6 feet wide and 4 feet high and $d$ varies from 2 to 2+4 feet. –  Américo Tavares Jun 17 '12 at 22:18
    
I have no idea what all that means. –  user138246 Jun 17 '12 at 22:21
    
The pressure $P$ is a force per unit area. When $P$ is constant, the force $F$ is just $F=PA$, where $A$ is the area of the shape. However in the case at hand $P$ is not constant, rather a function of the depth $2\le d\le 6$. –  Américo Tavares Jun 17 '12 at 22:26
    
If the rectangle was inserted horizontally into the water, the pressure $P$ would be constant, but the rectangle is vertical. –  Américo Tavares Jun 17 '12 at 22:38

4 Answers 4

up vote 6 down vote accepted
+100

The hydrostatic pressure $P$ "is the pressure exerted by a fluid at equilibrium due to the force of gravity". "For water and other liquids $P$ can be calculated according to the following formula" $$ P=\rho gh,\tag{1} $$

where

  • $P$ is the hydrostatic pressure at a generic point $H(\text{measured in }\textrm{Pa } \equiv $ Pascal above the sea-level atmospheric pressure$^1$)
  • $\rho $ is the liquid density ($\textrm{kg/m}^{3}$),
  • $g$ is the gravitational acceleration ($\mathrm{m/s}^{2}$),
  • $h$ is the height of the fluid column above $H\; (\textrm{m}).$

[Remark: Had we considered $P$ as an absolute pressure, we would have to add to $(1)$ the term $P_{\text{atm}}$, i.e. the mean sea level atmospheric pressure$^1$

$$ P=\rho gh+P_{\text{atm}},\tag{1a} $$

and adapt the numeric computation below accordingly. For the sake of simplicity we consider in this problem equation $(1)$ only.]

Since the water density $\rho \approx 1000 \textrm{kg/m}^{3}$ and the gravitational acceleration $g=9.81 \textrm{m/s}^{2}$, we have $$ P(h)=9810h.\tag{2} $$

So the hydrostatic pressure is a function of the single variable $h$, the height of the water column (in meters), i.e. the depth below the surface of $H$. Since $$ 1\text{ }\mathrm{ft\ }=0.3048\text{ }\mathrm{m}, $$ the hydrostatic pressure at $a=2 \textrm{ft }=2\times 0.3048 =0.6096\; \textrm{m}$ below the surface is $$ P(a)=9810\times 0.609\,6=5980.176\text{ }\mathrm{Pa}\; $$ and at $b=2+4=6 \textrm{ft }=6\times 0.3048=1.828\,8 \textrm{m},$ $$ P(b)=9810\times 1.828\,8=17940.528\text{ }\mathrm{Pa}. $$

If the rectangle was placed horizontally at e.g. $b$ the hydrostatic force $F$ (in Newtons) would be the product of $P(b)$ by the rectangle area $A=6\times 4=24\;\textrm{ft}^{2}=24\times 0.3048^{2}$ $=2.22967296\;\mathrm{m}^{2}$ $$ F_{\text{horizontally at }b}=P(b)\times A=17940.528\times 2.22967296=40001.51\text{ }\mathrm{N} $$ If it was placed horizontally at $a$, a similar computation would give $$ F_{\text{horizontally at }a}=P(a)\times A=13333.84\text{ }\mathrm{N} $$

enter image description here

Since the rectangle is vertical the hydrostatic pressure exerted on it is not constant, but depends on the variable $h$ (see sketch). Hence, to find the total hydrostatic force $F$ we need to add all the contributions of the hydrostatic force element $$dF=P(h)\;dA$$ exerted on the area element $$dA=w\;dh,$$ where $w=6\; \textrm{ft }=\textrm{ }1.8288 \textrm{m}$ is the rectangle width and $dh$ is the height element $$ dF=P(h)\;dA=P(h)\,w\;dh=9810h\,w\;dh,\quad h\in \left[ a,b\right]\tag{3} $$ So, the total hydrostatic force (in Newtons) exerted by the water on the rectangle is the defined integral$^3$ $$\begin{eqnarray*} F &=&\int_{a}^{b}dP=\int_{a}^{b}P(h)w\;dh=\int_{a}^{b}9810h\,w\;dh=9810\,w \int_{a}^{b}h\;dh \\ &=&\left. 9810\,w\times \frac{h^{2}}{2}\right\vert _{a}^{b}=9810\times 1.828\,8\left( \frac{b^{2}}{2}-\frac{a^{2}}{2}\right) \\ &=&\frac{9810\times 1.828\,8}{2}\left( 1.828\,8^{2}-0.609\,6^{2}\right) \\ &=&26667.67\text{ }\mathrm{N}=5995.1\text{ }\mathrm{lbf\quad }\text{(conversion of force units as per foot-note 2)}.\tag{4} \end{eqnarray*}$$

This value is of course less than $40001.51$ and greater than $13333.84$ computed above. Due to the linearity of the formula $(2)$, we have the following relation:

$$F=\frac{F_{\text{horizontally at }a}+F_{\text{horizontally at }b}}{2},$$

which means that we could have found $F$ without evaluating the integral $(4)$.

--

$^{1}$ Mean sea level pressure: $P_{\text{atm}}=101325\; \textrm{Pa} =14.7\; \textrm{psi}$

$^{2}$ Conversion of units: $1$ pound-force $=1\; \textrm{lbf }\equiv g\times 1\; \textrm{lb} \approx 4.448221615\; \textrm{N}$

$^3$ Note: this integral is the limit of a Riemann sum.

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The geometry here is simpler than in your previous triangle question. We give a highly informal description that should lead you to the answer.

Divide the rectangle into a large number of horizontal strips of width $\Delta x$. Consider such a strip, which is more or less at depth $x$ feet beneath the water. The area of the strip is $6\,\Delta x$. We want to compute the force against it, and add up over all the strips. Then if we wish we can divide by the area to find the pressure (force per unit area) against the rectangular plate.

The force against the strip is approximately $6\,\Delta x$ multiplied by a certain constant $K$ times $x$. The constant is $g$ times the mass of a cubic foot of water. I recall that the acceleration due to gravity is (in United States units) about $32$ feet per second$^2$ but cannot recall the mass of a cubic foot of water. (Life is easier in metric!)

We will "add up" the pressures on the strips, from $x=2$ (the depth of the top of the rectangular plate) to $x=2+4=6$ (the depth of the bottom of the rectangle). More precisely, we let $\Delta x \to 0$, and the limit of the sums is an integral. So the force is $$\int_2^6 6Kx\,dx.$$

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@mixedmath: Thanks. I confused the dimensions with the $4$ wide $5$ high dimensions of an earlier question by the OP. –  André Nicolas Jun 13 '12 at 0:56
    
I don't understand how the force is 6, that seems wrong since the size of a strip needs a variable. –  user138246 Jun 17 '12 at 19:27
    
@Jordan: You are thinking of the similar question about a triangle, where the width of the triangle at depth $x$ depended on $x$, and you needed similar triangles to detrmine that width. In our current case, the width of the object at depth $x$ is $6$ feet, independent of $x$. –  André Nicolas Jun 17 '12 at 20:03

The pressure at a certain depth is given by $P = \rho g d$, where $\rho$ is the denstity of the fluid, $g$ is gravitational acceleration, and $d$ is depth.

Hydrostatic force $F$ can be calculated from the equation $F = PA$, where $P$ is the pressure and $A$ is the area.

The idea here is that water pressure changes based on depth, and so the force changes at different depths. So you want to sum across infinitesimally small changes in depth. The idea is that for an infinitesimally small band, you can assume the whole band is at the same depth. The area will be this infinitesimally small change in height times the width of the plate. And you sum across the depths.

This is, in effect, a word problem describing your situation. For more, including two worked examples, I refer you to Paul's Online Math Notes again. I also note that I have taught from Stewart, and it's covered in there as well.

As a final note, I see that your constant $19620 = 2 \cdot 9810$ which is the constant in the metric system. But the problem is presented in feet, not meters.

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The pressure difference from sea level at depth $x$ is given by $P(x) = \rho g x$, where $\rho$ is the density, and $g$ is the acceleration due to gravity. The force on a element '$dx$' tall at depth $x$ is $P(x) w \, dx$, where $w$ is the width.

Hence the total force in the plate is:

$$ F = w\int_2^6 P(x) \, dx = \rho g w \int_2^6 x \, dx = 96 \rho g.$$

Substitute your value of $\rho g$ (salt or fresh water?) for a final answer; I have substituted linear dimensions in feet already.

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Where does this 96 come from? I have no clue. I get 62.5 * 32.174 for my constants. –  user138246 Jun 17 '12 at 19:39
    
You should be getting 16, copper.hat. –  Pedro Tamaroff Jun 17 '12 at 20:17
    
$\int_2^6 x \, dx = \frac{1}{2}(36-4) = 16$. $w = 6$. So I get $w \int_2^6 x \, dx = 6 \cdot 16 = 96$. Am I missing something here? –  copper.hat Jun 18 '12 at 6:03

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