Fluid Pressure

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Pressure is defined as the force per unit area:

\[P = \frac{F}{A}.\]

If an object is immersed in a liquid at a depth \(h\), the fluid pressure is given by the constant depth formula

\[P = \rho gh,\]

where \(\rho\) is the fluid density and \(g\) is the acceleration due to gravity.

Fluid pressure is a scalar quantity. It has no direction, so a fluid exerts pressure equally in all directions. This statement is known as Pascal's law discovered by the French scientist Blaise Pascal (\(1623 - 1662\)).

Consider the case when a vertical plate bounded by the lines

\[x = a,\;\; x = b,\;\; y = f\left( x \right),\;\; y = g\left( x \right)\]

is immersed in a liquid.

Since the different points of the lamina are at different depths, the total hydrostatic force \(F\) acting on the lamina is determined through integration:

\[F = \rho g\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]xdx} .\]

This formula is often referred as the variable depth formula for fluid force.

Solved Problems

Click or tap a problem to see the solution.

Example 1

A cylindrical tank with height of \(3\,\text{m}\) and base radius \(1\,\text{m}\) is filled with gasoline. Calculate the hydrostatic force exerted on the wall of the tank if the density of gasoline is \(800\,\frac{{\text{kg}}}{{{\text{m}^3}}}.\)

Example 2

A rectangular swimming pool is \(H\) meters deep, \(a\) meters wide, and \(b\) meters long. Calculate

The fluid force \(F_{ab}\) acting on the bottom of the pool;
The fluid force \(F_{aH}\) acting on each \(\left({a \times H}\right)\text{m}\) side;
The fluid force \(F_{bH}\) acting on each \(\left({b \times H}\right)\text{m}\) side;

Example 3

A triangular plate with base \(a\) and height \(H\) is submerged vertically in water so that its base lies at the surface of the water. Find the hydrostatic force acting on each side of the plate.

Example 4

A cube with side \(a\) is submerged in water so that its top face is parallel to the water surface and \(H\) meters below it. Find the total hydrostatic force acting on the cube.

Example 1.

Solution.

We choose the \(x-\)axis directed vertically downward with origin at the top base of the tank.

Consider a thin layer at a depth of \(x.\) If its thickness is \(dx,\) the lateral surface area of the layer is given by

\[dA = 2\pi Rdx.\]

The fluid pressure at the depth \(x\) is \(P = \rho gx,\) so the force exerted by the fluid on the lateral surface is

\[dF = PdA = 2\pi \rho gRxdx.\]

To find the total hydrostatic force \(F,\) we integrate from \(x = 0\) to \(x = H:\)

\[F = \int\limits_0^H {dF} = 2\pi \rho gR\int\limits_0^H {xdx} = \left. {\frac{{\cancel{2}\pi \rho gR{x^2}}}{\cancel{2}}} \right|_0^H = \left. {\pi \rho gR{x^2}} \right|_0^H = \pi \rho gR{H^2}.\]

Substituting the given values into the formula, we have

\[F = \pi \times 800 \times 9.8 \times 1 \times {3^2} \approx 221671\,\text{N} \approx 222\,\text{kN}.\]

Example 2.

A rectangular swimming pool is \(H\) meters deep, \(a\) meters wide, and \(b\) meters long. Calculate

The fluid force \(F_{ab}\) acting on the bottom of the pool;
The fluid force \(F_{aH}\) acting on each \(\left({a \times H}\right)\text{m}\) side;
The fluid force \(F_{bH}\) acting on each \(\left({b \times H}\right)\text{m}\) side;

Solution.

The pressure at the bottom of the swimming pool is \(P = \rho gH,\) so the hydrostatic force acting on the bottom is given by
\[{F_{ab}} = PA = \rho gHA = \rho gabH.\]
To determine the force on the \(\left({a \times H}\right)\text{m}\) side of the pool, we take a thin strip of thickness \(dx\) at a depth \(x.\)
Figure 4.
The area of the strip is \(dA = adx.\) Since the water pressure at depth \(x\) is \(P = \rho gx,\) the force acting on the elementary strip is
\[dF = PdA = \rho gaxdx.\]
The total force on the \(\left({a \times H}\right)\text{m}\) side is obtained by integration:
\[F_{aH} = \int\limits_0^H {dF} = \int\limits_0^H {dF} = \rho ga\int\limits_0^H {xdx} = \left. {\frac{{\rho ga{x^2}}}{2}} \right|_0^H = \frac{{\rho ga{H^2}}}{2}.\]
Similarly we can find the force acting on \(\left({b \times H}\right)\text{m}\) side of the pool:
\[{F_{bH}} = \frac{{\rho gb{H^2}}}{2}.\]

Example 3.

A triangular plate with base \(a\) and height \(H\) is submerged vertically in water so that its base lies at the surface of the water. Find the hydrostatic force acting on each side of the plate.

Solution.

From similar triangles we have

\[\frac{W}{a} = \frac{{H - x}}{H},\;\; \Rightarrow W = a - \frac{a}{H}x.\]

The area of the elementary horizontal strip at depth \(x\) is

\[dA = Wdx = \left( {a - \frac{a}{H}x} \right)dx.\]

The water pressure at depth \(x\) is \(P = \rho gx,\) so the force acting on the strip is written as

\[dF = PdA = \rho gx\left( {a - \frac{a}{H}x} \right)dx = \rho gax\left( {1 - \frac{x}{H}} \right)dx.\]

The total force is determined through integration:

\[F = \int\limits_0^H {dF} = \rho ga\int\limits_0^H {x\left( {1 - \frac{x}{H}} \right)dx} = \rho ga\int\limits_0^H {\left( {x - \frac{{{x^2}}}{H}} \right)dx} = \rho ga\left. {\left[ {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{{3H}}} \right]} \right|_0^H = \rho ga\left( {\frac{{{H^2}}}{2} - \frac{{{H^3}}}{{3H}}} \right) = \frac{{\rho ga{H^2}}}{6}.\]

Example 4.

A cube with side \(a\) is submerged in water so that its top face is parallel to the water surface and \(H\) meters below it. Find the total hydrostatic force acting on the cube.

Solution.

Using the the constant depth formula, it is easy to find the force acting on the top face:

\[{F_{top}} = {P_{top}}A = \rho g{a^2}H.\]

Similarly, the force on the bottom face is written as

\[{F_{bottom}} = {P_{bottom}}A = \rho g{a^2}\left( {H + a} \right) = \rho g{a^2}H + \rho g{a^3}.\]

To determine the side force, we consider a thin horizontal strip of thickness \(dx\) at depth \(x.\) Its area is \(dA = adx.\) The water pressure at this depth is \(P = \rho gx,\) so the hydrostatic force \(dF\) acting on the strip is given by the expression

\[dF = PdA = \rho gaxdx.\]

Then the force acting on the entire face of the cube is obtained by integration:

\[{F_{side}} = \int\limits_H^{H + a} {dF} = \rho ga\int\limits_H^{H + a} {xdx} = \left. {\frac{{\rho ga{x^2}}}{2}} \right|_H^{H + a} = \frac{{\rho ga}}{2}\left[ {{{\left( {H + a} \right)}^2} - {H^2}} \right] = \frac{{\rho ga}}{2}\left( {\cancel{{H^2}} + 2aH + {a^2} -\cancel{{H^2}}} \right) = \rho g{a^2}H + \frac{{\rho g{a^3}}}{2}.\]

The total hydrostatic force acting on the cube is given by

\[F = {F_{top}} + {F_{bottom}} + 4{F_{side}} = \rho g{a^2}H + \rho g{a^2}H + \rho g{a^3} + 4\left( {\rho g{a^2}H + \frac{{\rho g{a^3}}}{2}} \right) = 6\rho g{a^2}H + 3\rho g{a^3} = 3\rho g{a^2}\left( {2H + a} \right).\]