# Fluid Pressure

## Solved Problems

Click or tap a problem to see the solution.

### Example 5

A rectangular plate with sides $$a$$ and $$b$$ $$\left({a \gt b}\right)$$ is submerged in water at an angle $$\alpha$$ to the water surface. The longer side is parallel to the surface and lies at a depth $$H$$. Find the force acting on each side of the plate.

### Example 6

A dam has the shape of an isosceles trapezoid with upper base $$a = 64\,\text{m},$$ lower base $$b = 42\,\text{m},$$ and height $$H = 3\,\text{m}.$$ Find the force on the dam due to hydrostatic pressure.

### Example 7

A right circular cone with base radius $$R$$ and altitude $$H$$ is submerged, vertex downwards, in water so that its base is on the surface of the water. Find the force due to hydrostatic pressure acting on the lateral cone surface.

### Example 8

A plate in the shape of a parallelogram with sides $$a, b$$ and angle $$\alpha$$ is submerged vertically in water, so that the side $$b$$ is at the water surface. Calculate the hydrostatic force acting on each side of the plate.

### Example 9

A disk of radius $$R$$ is half submerged vertically in liquid of density $$\rho.$$ Find the hydrostatic force acting on one side of the disk.

### Example 10

A plate in the shape of a parabolic segment is submerged vertically in water as shown in Figure $$12.$$ The base of the segment is $$2a,$$ the height is $$H.$$ Find the force due to hydrostatic pressure acting on each side of the plate.

### Example 5.

A rectangular plate with sides $$a$$ and $$b$$ $$\left({a \gt b}\right)$$ is submerged in water at an angle $$\alpha$$ to the water surface. The longer side is parallel to the surface and lies at a depth $$H$$. Find the force acting on each side of the plate.

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

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

The total hydrostatic force is obtained by integration:

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

### Example 6.

A dam has the shape of an isosceles trapezoid with upper base $$a = 64\,\text{m},$$ lower base $$b = 42\,\text{m},$$ and height $$H = 3\,\text{m}.$$ Find the force on the dam due to hydrostatic pressure.

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:

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

### Example 7.

A right circular cone with base radius $$R$$ and altitude $$H$$ is submerged, vertex downwards, in water so that its base is on the surface of the water. Find the force due to hydrostatic pressure acting on the lateral cone surface.

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 $$x = H:$$

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

### Example 8.

A plate in the shape of a parallelogram with sides $$a, b$$ and angle $$\alpha$$ is submerged vertically in water, so that the side $$b$$ is at the water surface. Calculate the hydrostatic force acting on each side of the plate.

Solution.

The vertices of the parallelogram $$ABCD$$ are

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

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:

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

### Example 9.

A disk of radius $$R$$ is half submerged vertically in liquid of density $$\rho.$$ Find the hydrostatic force acting on one side of the disk.

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

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

### Example 10.

A plate in the shape of a parabolic segment is submerged vertically in water as shown in Figure $$12.$$ The base of the segment is $$2a,$$ the height is $$H.$$ Find the force due to hydrostatic pressure acting on each side of the plate.

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

$y = g\left( x \right) = - a\sqrt {1 - \frac{x}{H}} ,\;\; y = f\left( x \right) = a\sqrt {1 - \frac{x}{H}} .$

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

$1 - \frac{x}{H} = {z^2},\;\; \Rightarrow x = H(1 - {z^2}),\;\; dx = - 2Hzdz.$

When $$x = 0,$$ $$z = 1,$$ and when $$x = H,$$ $$z = 0.$$ Hence

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