Unit 2 · Lesson 4 of 5

Pressure builds force one submerged strip at a time.

A deeper gate experiences more pressure and a force that acts below its center. Change gate height, width, and top depth to see all three effects.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its total gauge-pressure force.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase gate height h? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

Force density

Gate height H2 mTop depth d1 mGate width b1 m

total gauge-pressure force

F=9.81(1)\left[1(2)+\frac{2^2}{2}\right]=39.24\,\mathrm{kN},\quad y_{CP}=1.16667\,\mathrm m

Pressure grows with depth. The force acts below the gate center; the dot on the gate marks the center of pressure relative to its top.

Assumptions, units, and limits

Water ρ=1000 kg/m³, $g=9.81 \mathrm{m}/\mathrm{s}$ ², vertical rectangular gate. Atmospheric pressure cancels because the other side is exposed to the same atmosphere.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

At depth $d+y$ , gauge pressure is ρg( $d+y$ ). A horizontal strip has area b dy, so its force is pressure times strip area. Integrating gives total force; integrating the moment and dividing by force locates its resultant.

p(y)=\rho g(d+y),\quad dF=p(y)b\,dy,\quad F=\rho gb\left(dH+\frac{H^2}{2}\right)

Keep the reasoning visible

y_{CP}=\frac{\int_0^H y(d+y)\,dy}{\int_0^H(d+y)\,dy}

Pressure grows with depth. The force acts below the gate center; the dot on the gate marks the center of pressure relative to its top.

What makes this model useful?

Water ρ=1000 kg/m³, $g=9.81 \mathrm{m}/\mathrm{s}$ ², vertical rectangular gate. Atmospheric pressure cancels because the other side is exposed to the same atmosphere.

Uniform pressure at the gate midpoint gives the correct total force for this rectangle because the pressure profile is linear, but it gives the wrong force location.

A worked example

Calculate the starting scenario

Use gate height h = 2 m; top depth d = 1 m; gate width b = 1 m.
$F=9.81(1)\left[1(2)+\frac{2^2}{2}\right]=39.24\,\mathrm{kN},\quad y_{CP}=1.16667\,\mathrm m$
Pressure grows with depth. The force acts below the gate center; the dot on the gate marks the center of pressure relative to its top. The checked quantity is total gauge-pressure force.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the total gauge-pressure force using gate height h 2 m, top depth d 1 m, gate width b 1 m. State the meaning of its sign and units.

Checked quantity: total gauge-pressure force · kN. Follow the definition for finite-cutoff and numerical-model questions.

Hint

At depth d+y, gauge pressure is ρg(d+y). A horizontal strip has area b dy, so its force is pressure times strip area. Integrating gives total force; integrating the moment and dividing by force locates its resultant.

Explore the visual solution

Force density

Gate height H2 mTop depth d1 mGate width b1 m

total gauge-pressure force

F=9.81(1)\left[1(2)+\frac{2^2}{2}\right]=39.24\,\mathrm{kN},\quad y_{CP}=1.16667\,\mathrm m

Pressure grows with depth. The force acts below the gate center; the dot on the gate marks the center of pressure relative to its top.

Assumptions, units, and limits

Water ρ=1000 kg/m³, $g=9.81 \mathrm{m}/\mathrm{s}$ ², vertical rectangular gate. Atmospheric pressure cancels because the other side is exposed to the same atmosphere.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

p(y)=\rho g(d+y),\quad dF=p(y)b\,dy,\quad F=\rho gb\left(dH+\frac{H^2}{2}\right)

Calculate and explain

Apply the idea to a new scenario

For a gate $H=1.5$ m, top depth $d=0.5$ m, width $b=2$ m, find water force in kN.

Checked quantity: total gauge-pressure force · kN. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Force density

Gate height H1.5 mTop depth d0.5 mGate width b2 m

total gauge-pressure force

F=9.81(2)\left[0.5(1.5)+\frac{1.5^2}{2}\right]=36.7875\,\mathrm{kN},\quad y_{CP}=0.9\,\mathrm m

Pressure grows with depth. The force acts below the gate center; the dot on the gate marks the center of pressure relative to its top.

Assumptions, units, and limits

Water ρ=1000 kg/m³, $g=9.81 \mathrm{m}/\mathrm{s}$ ², vertical rectangular gate. Atmospheric pressure cancels because the other side is exposed to the same atmosphere.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

p(y)=\rho g(d+y),\quad dF=p(y)b\,dy,\quad F=\rho gb\left(dH+\frac{H^2}{2}\right)

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

Force density

Gate height H2 mTop depth d1 mGate width b1 m

total gauge-pressure force

F=9.81(1)\left[1(2)+\frac{2^2}{2}\right]=39.24\,\mathrm{kN},\quad y_{CP}=1.16667\,\mathrm m

Pressure grows with depth. The force acts below the gate center; the dot on the gate marks the center of pressure relative to its top.

Assumptions, units, and limits

Water ρ=1000 kg/m³, $g=9.81 \mathrm{m}/\mathrm{s}$ ², vertical rectangular gate. Atmospheric pressure cancels because the other side is exposed to the same atmosphere.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. At depth $d+y$ , gauge pressure is ρg( $d+y$ ). A horizontal strip has area b dy, so its force is pressure times strip area. Integrating gives total force; integrating the moment and dividing by force locates its resultant.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.