Forces on Immersed Surfaces

Imagine a boat hull pushing against water, a dam holding back a lake, or your hand feeling pressure when you hold it underwater 🌊. In all of these situations, a fluid is pressing on a surface. In this lesson, students, you will learn how fluids create forces on immersed surfaces and how to calculate those forces using hydrostatic pressure.

What you will learn

By the end of this lesson, you should be able to:

explain what happens when a surface is immersed in a fluid
describe how pressure changes with depth in a stationary fluid
calculate the resultant force on a submerged plane surface
find the position of the center of pressure
connect these ideas to real engineering examples such as dams, tanks, gates, and ship structures

This topic is an important part of Fluid Statics because pressure in a fluid does not act only at a point. It acts over an area, and that creates a force. Understanding that force helps engineers design safe structures and devices.

Pressure in a fluid acts everywhere

In a fluid at rest, pressure acts equally in all directions at a given point. That means if a tiny flat surface is placed in the fluid, the fluid pushes on it normally, or perpendicular to the surface.

Pressure is defined as force per unit area:

$$p = \frac{F}{A}$$

So if pressure increases, the force on a given area increases too.

For a static fluid, pressure also increases with depth according to:

$$p = p_0 + \rho g h$$

where:

$p_0$ is the pressure at the fluid surface
$\rho$ is the fluid density
$g$ is the gravitational acceleration
$h$ is the depth below the surface

This relation is the key idea behind forces on immersed surfaces. The deeper the surface is, the larger the pressure on it. That means the bottom of a submerged surface usually experiences more force than the top.

Real-world example

Think about a swimming pool wall 🏊. Near the water surface, pressure is smaller. Near the bottom, pressure is bigger. That is why pool walls must be stronger lower down.

Resultant force on a plane immersed surface

A common problem in Thermofluids 1 is finding the total force on a flat surface submerged in a liquid. Examples include a rectangular gate, a hatch, or a tank wall.

If pressure varies across the surface, the total force is found by adding up the pressure over the whole area. In calculus form, this is:

$$F = \int_A p \, dA$$

For a plane surface in a fluid, this can be simplified using the average pressure at the centroid of the area. If the surface is fully submerged, the magnitude of the resultant hydrostatic force is:

$$F = p_c A$$

where:

$p_c$ is the pressure at the centroid of the surface
$A$ is the area of the surface

If the free surface is open to the atmosphere, gauge pressure is often used, so:

$$F = \rho g h_c A$$

where $h_c$ is the vertical depth of the centroid below the fluid surface.

Why does this work?

The pressure distribution on a flat surface changes linearly with depth because of:

$$p = p_0 + \rho g h$$

That means the resultant force is the same as the force produced by the pressure acting at the centroidal depth, even though the actual pressure is different at different points. This is a very useful engineering shortcut.

Example 1: Vertical rectangular gate

Suppose a vertical gate is $2\,\text{m}$ wide and $3\,\text{m}$ high, with its top edge at the water surface. The water density is $1000\,\text{kg/m}^3$.

First, find the area:

$$A = 2 \times 3 = 6\,\text{m}^2$$

The centroid is halfway down the gate, so:

$$h_c = 1.5\,\text{m}$$

Now calculate the force:

$$F = \rho g h_c A$$

$$F = 1000 \times 9.81 \times 1.5 \times 6$$

$$F = 88290\,\text{N}$$

So the resultant hydrostatic force is about $8.83 \times 10^4\,\text{N}$.

This force is large because water is heavy and the area is big. That is why even small dams or tank walls can experience major loading.

Center of pressure: where the force acts

The resultant force does not act at the centroid in general. It acts at a point called the center of pressure. This is the point where the total pressure force can be considered to act.

For a vertical or inclined plane surface, the center of pressure is always below the centroid, because pressure is larger at greater depth.

For a plane surface submerged in a liquid, the depth of the center of pressure is given by:

$$h_{cp} = h_c + \frac{I_G}{h_c A}$$

where:

$h_{cp}$ is the depth of the center of pressure below the free surface
$h_c$ is the depth of the centroid
$I_G$ is the second moment of area about the centroidal axis parallel to the free surface
$A$ is the area

This formula shows that the center of pressure depends on the shape and size of the surface.

Meaning of $I_G$

The quantity $I_G$ measures how the area is distributed around its centroid. A surface with area spread farther from the centroid has a larger $I_G$. That affects where the pressure force acts.

Example 2: Why the center of pressure is lower

For a rectangular plate submerged vertically, the top part is under less pressure than the bottom part. So the stronger push from the lower part pulls the resultant force downward. This is why the center of pressure is below the centroid.

This is important in gate design. If engineers place the support in the wrong location, the gate may rotate or fail because the pressure force creates a moment.

Inclined surfaces

Not all immersed surfaces are vertical. Some are tilted, like a dam face, a sluice gate, or a ship ramp. The same ideas still apply.

For an inclined plane surface:

pressure still depends on the vertical depth $h$
the resultant force is still found using the centroid depth
the center of pressure still lies below the centroid in vertical depth

What changes is the geometry. The area is measured along the sloping surface, but the pressure depends on how deep each point is vertically.

Helpful idea

If a surface is tilted, the deeper end experiences more pressure. You can imagine the fluid as pressing more strongly on the lower part of the slope, making the total force act lower down.

Example 3: A slanted hatch

Suppose a rectangular hatch is tilted under water. If its centroid is at depth $h_c$, then the resultant force still uses:

$$F = \rho g h_c A$$

To locate the center of pressure, you use the shape’s moment of area and the formula:

$$h_{cp} = h_c + \frac{I_G}{h_c A}$$

This is why geometry matters so much in fluid statics. The fluid properties set the pressure, but the surface shape decides how the force is distributed.

Forces on curved surfaces

Curved surfaces are more challenging because the pressure force at each point is normal to the surface, and the normal direction changes from point to point. Examples include pipes, tanks, domes, and curved gates.

For curved surfaces, the resultant force is usually found by splitting it into components:

horizontal force component
vertical force component

The horizontal component equals the hydrostatic force on the vertical projection of the curved surface.

The vertical component is equal to the weight of the imaginary fluid above the curved surface up to the free surface, for cases where the fluid is on one side and the region above is filled with fluid.

This method is very useful because it turns a difficult curved-surface problem into simpler parts.

Real-world example

A large water tank may have a curved wall. Engineers do not guess the force. They calculate the horizontal and vertical components separately to make sure the wall can resist the load safely.

Why this topic matters in engineering

Forces on immersed surfaces are used in many practical situations:

designing dams to hold back water
calculating forces on tank walls
sizing flood barriers and gates
analyzing submarine and ship surfaces
designing manometer and pressure-measurement devices that use hydrostatic balance

If a force is underestimated, a structure may bend, crack, or collapse. If it is overestimated, a design may become too expensive. Accurate calculations are therefore essential.

This topic also connects directly to pressure measurement. Manometers use fluid columns and the relation between pressure and depth. The same hydrostatic principle is what creates force on immersed surfaces.

Conclusion

Forces on immersed surfaces are caused by hydrostatic pressure acting over an area. Because pressure increases with depth according to $p = p_0 + \rho g h$, the force on a submerged surface is not uniform. For a plane surface, the resultant force is found using the centroid depth:

$$F = \rho g h_c A$$

The force acts at the center of pressure, which lies below the centroid because the lower parts of the surface experience more pressure. For curved surfaces, the force is usually split into horizontal and vertical components.

These ideas are central to Fluid Statics and are used in the safe design of dams, tanks, gates, and many other fluid systems. Understanding them helps you see how fluids push on real objects in everyday life and engineering practice.

Study Notes

Pressure in a static fluid increases with depth according to $p = p_0 + \rho g h$.
The force from a fluid on a surface acts perpendicular to that surface.
The resultant force on a plane submerged surface is found by integrating pressure over area:

$$F = \int_A p \, dA$$

For a fully submerged plane surface, the resultant force can be written as:

$$F = p_c A$$

or, using gauge pressure,

$$F = \rho g h_c A$$

The centroid depth $h_c$ is used to find the magnitude of the force.
The center of pressure is the point where the resultant force acts.
The center of pressure lies below the centroid for a submerged plane surface.
The depth of the center of pressure is:

$$h_{cp} = h_c + \frac{I_G}{h_c A}$$

For curved surfaces, split the force into horizontal and vertical components.
The horizontal force on a curved surface equals the force on its vertical projection.
The vertical force on a curved surface can equal the weight of the fluid above the surface in many standard cases.
This topic is important for dams, gates, tanks, ship structures, and pressure-measuring devices 🌊