Hydrostatic Pressure and Force

Suppose that a thin horizontal plate with area A{A} square meters is submerged in a fluid of density ρ\rho kilograms per cubic meter at a depth d{d} meters below the surface of the fluid.

The volume of fluid above plate is V=Ad{V}={A}{d}, so its mass is m=ρV=ρAd{m}=\rho{V}=\rho{A}{d}.

The force exerted by the fluid on the plate is F=mg=ρgAd{F}={m}{g{=}}\rho{g{{A}}}{d} where g=9.8ms2{g{=}}{9.8}\frac{{m}}{{{s}}^{{2}}}.

The pressure on the plate is defined to be the force per unit area: P=FA=ρgd{P}=\frac{{F}}{{A}}=\rho{g{{d}}}.

The SI unit for measuring pressure is newtons per square meter, which is called a pascal (1Nm2=1Pa)\left({1}\frac{{N}}{{{m}}^{{2}}}={1}{P}{a}\right). Since this is a small unit, the kilopascal (kPa) is often used.

For example, since density of water is ρ=1000kgm3\rho={1000}\frac{{{k}{g}}}{{{m}}^{{3}}}, the pressure at the bottom of a swimming pool 3 meters deep is P=ρgd=1000kgm3×9.8ms2×3m=29400Pa=29.4 kPa{P}=\rho{g{{d}}}={1000}\frac{{{k}{g}}}{{{m}}^{{3}}}\times{9.8}\frac{{m}}{{{s}}^{{2}}}\times{3}{m}={29400}{P}{a}={29.4}\ {k}{P}{a}.

So we can determine the hydrostatic force against a vertical plate or wall or dam in a fluid. We again must use integrals, because the pressure is not constant but increases as the depth increases.

Example 1. Suppose we have a trapezoidal dam whose height is 15m. Lower base is 20 m and upper base is 30 m. Find the force on the dam due to hydrostatic pressure if the water level is 5 m from the top of the dam.

hydrostatic pressure and force

We need convention for x{x}. Let x=0{x}={0} corresponds to the surface of water and x=10{x}={10} be bottom of the dam. Then water is in interval [0,10]{\left[{0},{10}\right]}.

Now divide interval [0,10]{\left[{0},{10}\right]} into n{n} subintervals of equal length with endpoints xi{x}_{{i}} and we choose xi[xi1,xi]{{x}_{{i}}^{{\star}}}\in{\left[{x}_{{{i}-{1}}},{x}_{{i}}\right]}. The i-th horizontal strip of the dam is approximated by a rectangle with height Δx\Delta{x} and width wi{w}_{{i}},

To find wi{w}_{{i}} we need to find a{a} first.

From similar triangle we have that a10xi=515\frac{{a}}{{{10}-{{x}_{{i}}^{{\star}}}}}=\frac{{5}}{{15}} or a=10313xi{a}=\frac{{10}}{{3}}-\frac{{1}}{{3}}{{x}_{{i}}^{{\star}}}.

Therefore, wi=2(10+a)=2(10+10313xi)=80323xi{w}_{{i}}={2}{\left({10}+{a}\right)}={2}{\left({10}+\frac{{10}}{{3}}-\frac{{1}}{{3}}{{x}_{{i}}^{{\star}}}\right)}=\frac{{80}}{{3}}-\frac{{2}}{{3}}{{x}_{{i}}^{{\star}}}.

If Ai{A}_{{i}} is the area of the i-th strip then AiwiΔx=(80323xi)Δx{A}_{{i}}\approx{w}_{{i}}\Delta{x}={\left(\frac{{80}}{{3}}-\frac{{2}}{{3}}{{x}_{{i}}^{{\star}}}\right)}\Delta{x}.

If Δx\Delta{x} is small then pressure Pi{P}_{{i}} is almost constant, so Piρgxi=10009.8xi=9800xi{P}_{{i}}\approx\rho{{g{{x}}}_{{i}}^{{\star}}}={1000}\cdot{9.8}{{x}_{{i}}^{{\star}}}={9800}{{x}_{{i}}^{{\star}}}.

The hydrostatic force acting on the i-th strip is the product of the pressure and the area: FiPiAi9800xi(80323xi)Δx{F}_{{i}}\approx{P}_{{i}}{A}_{{i}}\approx{9800}{{x}_{{i}}^{{\star}}}{\left(\frac{{80}}{{3}}-\frac{{2}}{{3}}{{x}_{{i}}^{{\star}}}\right)}\Delta{x}.

Adding these forces and taking the limit as n{n}\to\infty, we obtain the total hydrostatic force on the dam: F=limni=1n9800xi(80323xi)Δx=9800010x(80323x)dx=9800010(803x23x2)dx={F}=\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{9800}{{x}_{{i}}^{{\star}}}{\left(\frac{{80}}{{3}}-\frac{{2}}{{3}}{{x}_{{i}}^{{\star}}}\right)}\Delta{x}={9800}{\int_{{0}}^{{{10}}}}{x}{\left(\frac{{80}}{{3}}-\frac{{2}}{{3}}{x}\right)}{d}{x}={9800}{\int_{{0}}^{{{10}}}}{\left(\frac{{80}}{{3}}{x}-\frac{{2}}{{3}}{{x}}^{{2}}\right)}{d}{x}=

=9800(403x229x3)0101.09×107 N={9800}{\left(\frac{{40}}{{3}}{{x}}^{{2}}-\frac{{2}}{{9}}{{x}}^{{3}}\right)}{{\mid}_{{0}}^{{{10}}}}\approx{1.09}\times{{10}}^{{7}}\ {N}.

Example 2. Find force due to hydrostatic pressure on circular plate of radius 4 that is 3 meters under water.force due to hydrostatic pressure

Again we need convention about x{x}. This time we will take origin to be at the center of circle and x{x} pointing to the right (standard rectangular system).

Now we split plate by n{n} horizontal strips of heigth Δy\Delta{y} and choose sample point yi{{y}_{{i}}^{\star}} from i-th strip.

We can approximate area of i-th strip by rectangle with height Δy\Delta{y} and width 216(yi)2{2}\sqrt{{{16}-{{\left({{y}_{{i}}^{\star}}\right)}}^{{2}}}}: Ai216(yi)2dy{A}_{{i}}\approx{2}\sqrt{{{16}-{{\left({{y}_{{i}}^{\star}}\right)}}^{{2}}}}{d}{y}.

i-strip is 7yi{7}-{{y}_{{i}}^{\star}} meters under water the pressure on each strip is Piρg(7yi)=9800(7yi){P}_{{i}}\approx\rho\cdot{g{\cdot}}{\left({7}-{{y}_{{i}}^{\star}}\right)}={9800}{\left({7}-{{y}_{{i}}^{\star}}\right)}.

Now force that acts on each strip is FiPiAi=19600(7yi)16(yi)2dy{F}_{{i}}\approx{P}_{{i}}{A}_{{i}}={19600}{\left({7}-{{y}_{{i}}^{\star}}\right)}\sqrt{{{16}-{{\left({{y}_{{i}}^{\star}}\right)}}^{{2}}}}{d}{y}.

If we now add force that acts on each strip and take limit n{n}\to\infty then we will obtain that F=1960044(7y)16y2dy=19600(44716y2dy44y16y2)dy{F}={19600}{\int_{{-{4}}}^{{4}}}{\left({7}-{y}\right)}\sqrt{{{16}-{{y}}^{{2}}}}{d}{y}={19600}{\left({\int_{{-{4}}}^{{4}}}{7}\sqrt{{{16}-{{y}}^{{2}}}}{d}{y}-{\int_{{-{4}}}^{{4}}}{y}\sqrt{{{16}-{{y}}^{{2}}}}\right)}{d}{y}.

First interval requires trigonometric substitution y=4sin(u){y}={4}{\sin{{\left({u}\right)}}}, second requires substitution v=16y2{v}={16}-{{y}}^{{2}}.

So, F=1097600π{F}={1097600}\pi.