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

\[{x = a,\;\;}\kern0pt{x = b,\;\;}\kern0pt{y = f\left( x \right),\;\;}\kern0pt{y = g\left( x \right)}\]

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.

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 \(x = H:\)

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

Substituting the given values into the formula, we have

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

Solution.

10. The pressure at the bottom of the swimming pool is \(P = \rho gH,\) so the hydrostatic force acting on the bottom is given by

    \[{F_{ab}} = PA = \rho gHA = \rho gabH.\]

11. 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 \(x.\)

    

    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:

    \[{F_{aH} = \int\limits_0^H {dF} }={ \int\limits_0^H {dF} }={ \rho ga\int\limits_0^H {xdx} }={ \left. {\frac{{\rho ga{x^2}}}{2}} \right|_0^H }={ \frac{{\rho ga{H^2}}}{2}.}\]

12. Similarly we can find the force acting on \(\left({b \times H}\right)\text{m}\) side of the pool:

    \[{F_{bH}} = \frac{{\rho gb{H^2}}}{2}.\]

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:

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

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

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

Solution.

By Pascal’s law, the fluid pressure at a depth \(x\) is \(P = \rho gx\) in any direction. So if we take a small strip on the plate at depth \(x\) corresponding to the increment \(dx,\) the force acting on the strip is given by

\[{dF = PdA }={ \rho gx \times \frac{{adx}}{{\sin \alpha }} }={ \frac{{\rho gaxdx}}{{\sin \alpha }}.}\]

To total hydrostatic force is obtained by integration:

\[{F = \int\limits_H^{H + b\sin \alpha } {dF} }={ \frac{{\rho ga}}{{\sin \alpha }}\int\limits_H^{H + b\sin \alpha } {xdx} }={ \frac{{\rho ga}}{{\sin \alpha }}\left. {\frac{{{x^2}}}{2}} \right|_H^{H + b\sin \alpha } }={ \frac{{\rho ga}}{{2\sin \alpha }}\left[ {{{\left( {H + b\sin \alpha } \right)}^2} – {H^2}} \right] }={ \frac{{\rho ga}}{{2\sin \alpha }}\left( {2bH\sin \alpha + {b^2}{{\sin }^2}\alpha } \right) }={ \rho gab\left( {H + \frac{b}{2}\sin \alpha } \right).}\]

Solution.

If we choose the vertical \(x−\)axis directed downward, the fluid pressure at a depth \(x\) is written as

\[P = \rho gx.\]

A thin horizontal strip of width \(dx\) at depth \(x\) can be approximated by a rectangle with the area equal to

\[dA = Wdx,\]

where the width \(W\) of the trapezoid at depth \(x\) is determined from similar triangles and is given by

\[W = a – \left( {a – b} \right)\frac{x}{H}.\]

Hence, the hydrostatic force acting on the strip is expressed by the formula

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

The total force exerted on the dam due to hydrostatic pressure is given by

\[{F = \int\limits_0^H {dF} }={ \rho g\int\limits_0^H {x\left[ {a – \left( {a – b} \right)\frac{x}{H}} \right]dx} }={ \rho g\int\limits_0^H {\left( {ax – \frac{{a – b}}{H}{x^2}} \right)dx} }={ \rho g\left. {\left[ {\frac{{a{x^2}}}{2} – \frac{{\left( {a – b} \right){x^3}}}{{3H}}} \right]} \right|_0^H }={ \rho g\left[ {\frac{{a{H^2}}}{2} – \frac{{\left( {a – b} \right){H^2}}}{3}} \right] }={ \rho g{H^2}\left( {\frac{a}{6} + \frac{b}{3}} \right).}\]

Now we can easily calculate the value of the force:

\[{F = 1000 \times 9.8 \times {3^2} \times \left( {\frac{{6.4}}{6} + \frac{{4.2}}{3}} \right) }={ 217560\,\text{N} }\approx{ 218\,\text{kN}.}\]

Solution.

We have the following proportion from similar triangles:

\[{\frac{W}{{H – x}} = \frac{R}{H},}\;\; \Rightarrow {W = \frac{{R\left( {H – x} \right)}}{H} = R\left( {1 – \frac{x}{H}} \right).}\]

The surface area of the elementary cone strip at the point \(x\) is given by

\[{dA = 2\pi Wdx }={ 2\pi R\left( {1 – \frac{x}{H}} \right)dx.}\]

The pressure in any direction at depth \(x\) is \(P = \rho gx,\) so the force on the strip is equal to

\[{dF = PdA }={ 2\pi \rho gRx\left( {1 – \frac{x}{H}} \right)dx.}\]

The total force is obtained by integrating from \(x = 0\) to \(x = H:\)

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

Solution.

The vertices of the parallelogram \(ABCD\) are

\[{A\left( {0,0} \right),\;\;}\kern0pt{B\left( {0,b} \right),\;\;}\kern0pt{C\left( {a\sin \alpha ,b + a\cos \alpha } \right),\;\;}\kern0pt{D\left( {a\sin \alpha ,a\cos \alpha } \right).}\]

Determine the equation of the side \(AD.\) Using the two-point form of straight line equation, we have:

\[{\frac{{x – {x_A}}}{{{x_D} – {x_A}}} = \frac{{y – {y_A}}}{{{y_D} – {y_A}}},}\;\; \Rightarrow {\frac{{x – 0}}{{a\sin \alpha – 0}} = \frac{{y – 0}}{{a\cos \alpha – 0}},}\;\; \Rightarrow {\frac{x}{{a\sin \alpha }} = \frac{y}{{a\cos \alpha }},}\;\; \Rightarrow {{y_1} = x\cot \alpha .}\]

The other side \(BC\) is shifted \(b\) units upwards along the \(y-\)axis, so its equation is given by

\[{y_2} = b + x\cot \alpha .\]

Now we use we apply the variable depth formula:

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

This gives the total force acting on the plate:

\[{F = \rho g\int\limits_0^{a\sin \alpha } {\left( {{y_2} – {y_1}} \right)xdx} }={ \rho g\int\limits_0^{a\sin \alpha } {\left( {b + \cancel{x\cot \alpha} – \cancel{x\cot \alpha} } \right)xdx} }={ \rho gb\int\limits_0^{a\sin \alpha } {xdx} }={ \left. {\frac{{\rho gb{x^2}}}{2}} \right|_0^{a\sin \alpha } }={ \frac{{\rho gb{a^2}{{\sin }^2}\alpha }}{2}.}\]

Solution.

Consider a thin horizontal strip of thickness \(dx\) at a depth \(x.\) The width of the strip is

\[W = AB = 2\sqrt {{R^2} – {x^2}},\]

so its area is

\[{dA = Wdx }={ 2\sqrt {{R^2} – {x^2}} dx.}\]

The force on the strip is approximately

\[{dF = PdA }={ \rho gxdA }={ 2\rho gx\sqrt {{R^2} – {x^2}} dx.}\]

The total hydrostatic force is given by the integral

\[{F = \int\limits_0^R {dF} }={ 2\rho g\int\limits_0^R {x\sqrt {{R^2} – {x^2}} dx} .}\]

We evaluate this integral using the change of variable:

\[{I = \int {x\sqrt {{R^2} – {x^2}} dx} }={ \left[ {\begin{array}{*{20}{l}} {z = {R^2} – {x^2}}\\ {dz = – 2xdx} \end{array}} \right] }={ \int {\sqrt z \left( { – \frac{{dz}}{2}} \right)} }={ – \frac{1}{2}\int {\sqrt z dz} }={ – \frac{{{z^{\frac{3}{2}}}}}{3} }={ – \frac{{\sqrt {{z^3}} }}{3} }={ – \frac{{\sqrt {{{\left( {{R^2} – {x^2}} \right)}^3}} }}{3}.}\]

Hence, the force \(F\) is given by

\[{F = – \frac{{2\rho g}}{3}\left. {\sqrt {{{\left( {{R^2} – {x^2}} \right)}^3}} } \right|_0^R }={ – \frac{{2\rho g}}{3}\left( {0 – {R^3}} \right) }={ \frac{{2\rho g{R^3}}}{3}.}\]

Solution.

First we determine the equation of the parabola given its base \(2a\) and height \(H.\) The initial equation is \(x = H – k{y^2}.\) Since \(y = a\) at the point \(x = 0, \) the coefficient \(k\) is equal to

\[{0 = H – k{a^2},}\;\; \Rightarrow {k = \frac{H}{{{a^2}}}.}\]

This yields:

\[{x = H – \frac{H}{{{a^2}}}{y^2} }={ H\left( {1 – \frac{{{y^2}}}{{{a^2}}}} \right).}\]

By solving this equation for \(y,\) we get

\[{\frac{x}{H} = 1 – \frac{{{y^2}}}{{{a^2}}},}\;\; \Rightarrow {{a^2} – {y^2} = {a^2}\frac{x}{H},}\;\; \Rightarrow {{y^2} = {a^2}\left( {1 – \frac{x}{H}} \right).}\]

So, the parabola segment is bounded by the curves

\[{y = g\left( x \right) = – a\sqrt {1 – \frac{x}{H}} ,\;\;}\kern0pt{y = f\left( x \right) = a\sqrt {1 – \frac{x}{H}} .}\]

To calculate the hydrostatic force, we apply the variable depth formula:

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

In our case,

\[F = 2\rho ga\int\limits_0^H {\sqrt {1 – \frac{x}{H}} xdx} .\]

Make the substitution

\[{1 – \frac{x}{H} = {z^2},}\;\; \Rightarrow {x = H(1 – {z^2}),\;\;}\kern0pt{dx = – 2Hzdz.}\]

When \(x = 0,\) \(z = 1,\) and when \(x = H,\) \(z = 0.\) Hence

\[{F = – 4\rho ga{H^2}\int\limits_1^0 {z\left( {1 – {z^2}} \right)zdz} }={ 4\rho ga{H^2}\int\limits_0^1 {\left( {{z^2} – {z^4}} \right)dz} }={ 4\rho ga{H^2}\left. {\left( {\frac{{{z^3}}}{3} – \frac{{{z^5}}}{5}} \right)} \right|_0^1 }={ 4\rho ga{H^2}\left( {\frac{1}{3} – \frac{1}{5}} \right) }={ \frac{{8\rho ga{H^2}}}{{15}}.}\]