Density and pressure

Welcome to the study notes on Density and pressure (section 3.2.4). These foundational concepts are essential for understanding how forces act throughout materials, whether you are analyzing the structural integrity of a solid block, the depths of the ocean, or the lift acting on a hot-air balloon.

What you'll learn:

How to define and calculate density and pressure in different physical contexts.
How to derive and apply the equation for hydrostatic pressure in a column of liquid.
How to explain the origin of upthrust and use Archimedes' principle to solve fluid mechanics problems.

1. Density ( $ρ\rho$ )

Density is a fundamental characteristic property of a material, describing how closely its mass is packed into a given volume.

Definition

Density

Density ( $ρ\rho$ ) is defined as mass per unit volume of a substance.

\rho = \frac{m}{V}

Where:

$ρ\rho$ is the density in kilograms per cubic metre ( $m−3\text{kg}\,\text{m}^{-3}$ )
$m$ is the mass in kilograms ( $kg\text{kg}$ )
$V$ is the volume in cubic metres ( $m3\text{m}^3$ )

The standard SI unit for density is kilograms per cubic metre ( $m−3\text{kg}\,\text{m}^{-3}$ ), but you will often see it given in grams per cubic centimetre ( $cm−3\text{g}\,\text{cm}^{-3}$ ).

Common Mistake

Unit Conversions

Be careful when converting units of density! Volume conversions are scaled in three dimensions:

$m1\text{ cm} = 10^{-2}\text{ m}$
$m31\text{ cm}^3 = (10^{-2}\text{ m})^3 = 10^{-6}\text{ m}^3$

Therefore, to convert from $cm−3\text{g}\,\text{cm}^{-3}$ to $m−3\text{kg}\,\text{m}^{-3}$ , you must multiply by $1000$ . For example, water has a density of $cm−31.0\text{ g}\,\text{cm}^{-3}$ , which is equal to $m−31000\text{ kg}\,\text{m}^{-3}$ .

Determining Density Practicals

In practical work (linked to PAG 1), you are expected to determine the density of both regularly and irregularly shaped objects:

Regular solids: Measure mass using a digital balance. Measure dimensions using a micrometer screw gauge or vernier calipers to calculate volume.
Irregular solids: Measure mass using a digital balance. Find the volume by completely submerging the object in water inside a displacement (Eureka) can and measuring the volume of water displaced using a measuring cylinder.

Example

Calculating the density of a brass alloy sphere

A solid sphere of a brass alloy is measured to have a diameter of $cm5.40\text{ cm}$ and a mass of $g695\text{ g}$ . Calculate the density of the brass alloy in $m−3\text{kg}\,\text{m}^{-3}$ to three significant figures.

Convert the given quantities into SI base units: The radius $r$ is half of the diameter:

\frac{5.40\text{ cm}}{2} = 2.70\text{ cm} = 0.0270\text{ m}

The mass $m$ in kilograms is:

695\text{ g} = 0.695\text{ kg}

Calculate the volume of the sphere: Use the volume formula for a sphere, $\frac{4}{3}\pi r^3$ :

\frac{4}{3} \times \pi \times (0.0270\text{ m})^3 \approx 8.245 \times 10^{-5}\text{ m}^3

Calculate the density: Substitute the mass and calculated volume into the density formula:

\rho = \frac{m}{V} = \frac{0.695\text{ kg}}{8.245 \times 10^{-5}\text{ m}^3} \approx 8430\text{ kg}\,\text{m}^{-3}

2. Pressure ( $p$ )

Pressure is a measure of how concentrated a force is over a surface area.

Definition

Pressure

Pressure ( $p$ ) is defined as the normal force exerted per unit cross-sectional area.

\frac{F}{A}

Where:

$p$ is the pressure in Pascals ( $Pa\text{Pa}$ ) or Newtons per square metre ( $m−2\text{N}\,\text{m}^{-2}$ )
$F$ is the normal force (acting perpendicular to the surface) in Newtons ( $N\text{N}$ )
$A$ is the cross-sectional area in square metres ( $m2\text{m}^2$ )

The SI unit of pressure is the Pascal ( $Pa\text{Pa}$ ), where $m−21\text{ Pa} = 1\text{ N}\,\text{m}^{-2}$ . Pressure applies to solids, liquids, and gases:

Solids: A solid block resting on a table exerts downward pressure due to its weight distributed over its contact area.
Liquids and Gases (Fluids): Pressure acts in all directions. Fluid particles are in constant, random motion, colliding with the walls of their container and exerting a cumulative normal force over the contact area.

Common Mistake

Using the wrong force component

The force $F$ in the pressure equation must be perpendicular to the surface. If a force is applied at an angle $θ\theta$ to the normal of the surface, you must first resolve the force to find its perpendicular component, $F⊥=Fcos⁡θF_{\perp} = F\cos\theta$ , before calculating the pressure.

3. Pressure in Fluids ( $h\rho g$ )

The pressure at any point in a fluid depends on the depth below the surface. This is because the fluid at a deeper point has to support the weight of all the fluid sitting directly above it.

Deriving $h\rho g$

Consider a vertical column of fluid with a constant density $ρ\rho$ and a horizontal cross-sectional area $A$ . The column has a height $h$ .

Fluid Column Derivation

Let's derive the pressure exerted at the base of this column step-by-step:

Volume of the fluid column:

\times h

Mass of the fluid column: Using the mass-density relation ( $\rho V$ ):

\rho A h

Weight of the fluid column: The weight $W$ represents the downward force $F$ exerted by the fluid column on its base:

\rho A h g

Pressure at the base: Since pressure is force divided by area:

\frac{F}{A} = \frac{\rho A h g}{A}

The area $A$ cancels out, leaving:

h\rho g

Key Idea

Hydrostatic Pressure Formula

The pressure $p$ due to a fluid column of depth $h$ is given by:

h\rho g

This formula shows that fluid pressure depends only on depth, fluid density, and gravitational field strength. It is independent of the shape of the container or the surface area of the base.

Tip

Total vs. Hydrostatic Pressure

The equation $h\rho g$ calculates the pressure due to the fluid itself. If a container or pool is open to the atmosphere, the total pressure at depth $h$ is the sum of the hydrostatic pressure and atmospheric pressure ( $patmp_{\text{atm}}$ ):

p_{\text{total}} = p_{\text{atm}} + h\rho g

Atmospheric pressure at sea level is approximately $kPa101\text{ kPa}$ ( $1.01 \times 10^5\text{ Pa}$ ).

Example

Calculating pressure at the bottom of a diving pool

A diving pool has a depth of $m5.00\text{ m}$ . The density of the fresh water is $1.00 \times 10^3\text{ kg}\,\text{m}^{-3}$ , and atmospheric pressure is $1.01 \times 10^5\text{ Pa}$ . Calculate the total pressure experienced by a diver at the bottom of the pool. Use $9.81\text{ m s}^{-2}$ .

Calculate the hydrostatic pressure of the water:

p_{\text{water}} = h\rho g = 5.00\text{ m} \times (1.00 \times 10^3\text{ kg}\,\text{m}^{-3}) \times 9.81\text{ m s}^{-2}

p_{\text{water}} = 4.905 \times 10^4\text{ Pa}

Add the atmospheric pressure to find the total pressure:

p_{\text{total}} = p_{\text{atm}} + p_{\text{water}} = (1.01 \times 10^5\text{ Pa}) + (4.905 \times 10^4\text{ Pa})

p_{\text{total}} = 1.5005 \times 10^5\text{ Pa}

Round to a sensible number of significant figures: Since the input values are given to three significant figures, we quote the final answer as:

p_{\text{total}} \approx 1.50 \times 10^5\text{ Pa} \quad (\text{or } 150\text{ kPa})

4. Upthrust and Archimedes' Principle

Have you ever wondered why a heavy steel ship can float, while a small steel coin sinks? The answer lies in upthrust and pressure differences.

The Origin of Upthrust

When an object is submerged in a fluid, its upper surface is at a shallower depth than its lower surface.

Because pressure increases with depth ( $h\rho g$ ), the fluid pressure acting upwards on the bottom of the object is greater than the pressure acting downwards on the top of the object. This pressure difference creates a net upward force, which we call upthrust.

Upthrust on Submerged Block

Let's derive the expression for this upward force:

Downward force on the top surface:

F_{\text{top}} = p_{\text{top}} A = h_{\text{top}} \rho g A

Upward force on the bottom surface:

F_{\text{bottom}} = p_{\text{bottom}} A = h_{\text{bottom}} \rho g A

Net upward force (Upthrust, $U$ ):

F_{\text{bottom}} - F_{\text{top}} = (h_{\text{bottom}} - h_{\text{top}}) \rho g A

Relating to volume: The difference in depth $(hbottom−htop)(h_{\text{bottom}} - h_{\text{top}})$ is simply the height $x$ of the block. The volume of the block is $V = A x$ .

\rho g A = V \rho g

Since $VρV\rho$ is the mass $mfluidm_{\text{fluid}}$ of the fluid that has been pushed aside (displaced) by the block, $VρgV\rho g$ is equal to the weight of that displaced fluid. This brings us directly to Archimedes' principle.

Definition

Archimedes' Principle

Archimedes' Principle states that:

An object submerged or floating in a fluid experiences an upward upthrust force equal to the weight of the fluid it displaces.

V_{\text{sub}} \rho_{\text{fluid}} g

Where:

$U$ is the upthrust force in Newtons ( $N\text{N}$ )
$VsubV_{\text{sub}}$ is the volume of the object submerged in the fluid in cubic metres ( $m3\text{m}^3$ )
$ρfluid\rho_{\text{fluid}}$ is the density of the fluid in kilograms per cubic metre ( $m−3\text{kg}\,\text{m}^{-3}$ )
$g$ is the acceleration of free fall ( $s−29.81\text{ m s}^{-2}$ )

Sinking vs. Floating

An object sinks if its weight is greater than the maximum upthrust (which occurs when the object is fully submerged). This happens if the average density of the object is greater than the density of the fluid ( $ρobject>ρfluid\rho_{\text{object}} > \rho_{\text{fluid}}$ ).
An object floats in equilibrium if its weight is equal to the upthrust. In this state, it only needs to displace a volume of fluid whose weight matches the object's total weight.

Example

Calculating the submerged volume of a floating ice block

An ice block of mass $kg450\text{ kg}$ floats in freshwater of density $1.00 \times 10^3\text{ kg}\,\text{m}^{-3}$ . Calculate the volume of the ice block that is submerged below the water level.

State the condition for floating: For a floating object in equilibrium, the upward upthrust $U$ must balance the downward weight of the object $W$ :

U = W

Calculate the weight of the ice block:

450\text{ kg} \times 9.81\text{ m s}^{-2} = 4414.5\text{ N}

Set up Archimedes' principle equation for upthrust:

V_{\text{sub}} \rho_{\text{water}} g

Since $U = W$ :

V_{\text{sub}} \rho_{\text{water}} g = mg

Notice that $g$ cancels out from both sides:

V_{\text{sub}} \rho_{\text{water}} = m

Rearrange and solve for the submerged volume ( $VsubV_{\text{sub}}$ ):

V_{\text{sub}} = \frac{m}{\rho_{\text{water}}} = \frac{450\text{ kg}}{1.00 \times 10^3\text{ kg}\,\text{m}^{-3}} = 0.450\text{ m}^3

Exam technique

In the exam

Always check your units: Ensure volume is in $m3\text{m}^3$ (not $cm3\text{cm}^3$ or litres) and density is in $m−3\text{kg}\,\text{m}^{-3}$ before carrying out substitutions.
Memorize the $h\rho g$ derivation: This is a classic 3-to-4 mark exam question. Practice writing down the algebraic steps clearly, defining what each letter represents.
Use free-body diagrams for upthrust questions: Draw the forces acting on the submerged object (Weight downwards, Upthrust upwards, and sometimes Tension in a string or Drag if the object is moving). Set up your equilibrium equations ( $Fnet=0F_{\text{net}} = 0$ ) from the diagram.
Be precise with vocabulary: Use the term "fluid" rather than just "liquid" when discussing pressure and upthrust, as these principles apply equally to gases (like balloons floating in air).

Self review

Check yourself

A metal cylinder has a radius of $cm1.5\text{ cm}$ and a height of $cm12\text{ cm}$ . If its mass is $kg0.72\text{ kg}$ , what is its density in $m−3\text{kg}\,\text{m}^{-3}$ ?
Explain why the pressure equation $h\rho g$ proves that a thin, tall cylinder of water exerts the same pressure on its base as a wide, tall tank filled to the same height.
A fully submerged concrete block experiences an upthrust of $N250\text{ N}$ in seawater. If the density of the seawater is $m−31025\text{ kg}\,\text{m}^{-3}$ , what is the volume of the block?

What you'll learn:

How to define and calculate density and pressure in different physical contexts.
How to derive and apply the equation for hydrostatic pressure in a column of liquid.
How to explain the origin of upthrust and use Archimedes' principle to solve fluid mechanics problems.

1. Density ( $ρ\rho$ )

Density is a fundamental characteristic property of a material, describing how closely its mass is packed into a given volume.

Definition

Density

Density ( $ρ\rho$ ) is defined as mass per unit volume of a substance.

\rho = \frac{m}{V}

Where:

$ρ\rho$ is the density in kilograms per cubic metre ( $m−3\text{kg}\,\text{m}^{-3}$ )
$m$ is the mass in kilograms ( $kg\text{kg}$ )
$V$ is the volume in cubic metres ( $m3\text{m}^3$ )

The standard SI unit for density is kilograms per cubic metre ( $m−3\text{kg}\,\text{m}^{-3}$ ), but you will often see it given in grams per cubic centimetre ( $cm−3\text{g}\,\text{cm}^{-3}$ ).

Common Mistake

Unit Conversions

Be careful when converting units of density! Volume conversions are scaled in three dimensions:

$m1\text{ cm} = 10^{-2}\text{ m}$
$m31\text{ cm}^3 = (10^{-2}\text{ m})^3 = 10^{-6}\text{ m}^3$

Determining Density Practicals

In practical work (linked to PAG 1), you are expected to determine the density of both regularly and irregularly shaped objects:

Regular solids: Measure mass using a digital balance. Measure dimensions using a micrometer screw gauge or vernier calipers to calculate volume.
Irregular solids: Measure mass using a digital balance. Find the volume by completely submerging the object in water inside a displacement (Eureka) can and measuring the volume of water displaced using a measuring cylinder.

Example

Calculating the density of a brass alloy sphere

Convert the given quantities into SI base units: The radius $r$ is half of the diameter:

\frac{5.40\text{ cm}}{2} = 2.70\text{ cm} = 0.0270\text{ m}

The mass $m$ in kilograms is:

695\text{ g} = 0.695\text{ kg}

Calculate the volume of the sphere: Use the volume formula for a sphere, $\frac{4}{3}\pi r^3$ :

\frac{4}{3} \times \pi \times (0.0270\text{ m})^3 \approx 8.245 \times 10^{-5}\text{ m}^3

Calculate the density: Substitute the mass and calculated volume into the density formula:

\rho = \frac{m}{V} = \frac{0.695\text{ kg}}{8.245 \times 10^{-5}\text{ m}^3} \approx 8430\text{ kg}\,\text{m}^{-3}

2. Pressure ( $p$ )

Pressure is a measure of how concentrated a force is over a surface area.

Definition

Pressure

Pressure ( $p$ ) is defined as the normal force exerted per unit cross-sectional area.

\frac{F}{A}

Where:

$p$ is the pressure in Pascals ( $Pa\text{Pa}$ ) or Newtons per square metre ( $m−2\text{N}\,\text{m}^{-2}$ )
$F$ is the normal force (acting perpendicular to the surface) in Newtons ( $N\text{N}$ )
$A$ is the cross-sectional area in square metres ( $m2\text{m}^2$ )

The SI unit of pressure is the Pascal ( $Pa\text{Pa}$ ), where $m−21\text{ Pa} = 1\text{ N}\,\text{m}^{-2}$ . Pressure applies to solids, liquids, and gases:

Solids: A solid block resting on a table exerts downward pressure due to its weight distributed over its contact area.
Liquids and Gases (Fluids): Pressure acts in all directions. Fluid particles are in constant, random motion, colliding with the walls of their container and exerting a cumulative normal force over the contact area.

Common Mistake

Using the wrong force component

3. Pressure in Fluids ( $h\rho g$ )

The pressure at any point in a fluid depends on the depth below the surface. This is because the fluid at a deeper point has to support the weight of all the fluid sitting directly above it.

Deriving $h\rho g$

Consider a vertical column of fluid with a constant density $ρ\rho$ and a horizontal cross-sectional area $A$ . The column has a height $h$ .

Fluid Column Derivation

Let's derive the pressure exerted at the base of this column step-by-step:

Volume of the fluid column:

\times h

Mass of the fluid column: Using the mass-density relation ( $\rho V$ ):

\rho A h

Weight of the fluid column: The weight $W$ represents the downward force $F$ exerted by the fluid column on its base:

\rho A h g

Pressure at the base: Since pressure is force divided by area:

\frac{F}{A} = \frac{\rho A h g}{A}

The area $A$ cancels out, leaving:

h\rho g

Key Idea

Hydrostatic Pressure Formula

The pressure $p$ due to a fluid column of depth $h$ is given by:

h\rho g

This formula shows that fluid pressure depends only on depth, fluid density, and gravitational field strength. It is independent of the shape of the container or the surface area of the base.

Tip

Total vs. Hydrostatic Pressure

p_{\text{total}} = p_{\text{atm}} + h\rho g

Atmospheric pressure at sea level is approximately $kPa101\text{ kPa}$ ( $1.01 \times 10^5\text{ Pa}$ ).

Example

Calculating pressure at the bottom of a diving pool

Calculate the hydrostatic pressure of the water:

p_{\text{water}} = h\rho g = 5.00\text{ m} \times (1.00 \times 10^3\text{ kg}\,\text{m}^{-3}) \times 9.81\text{ m s}^{-2}

p_{\text{water}} = 4.905 \times 10^4\text{ Pa}

Add the atmospheric pressure to find the total pressure:

p_{\text{total}} = p_{\text{atm}} + p_{\text{water}} = (1.01 \times 10^5\text{ Pa}) + (4.905 \times 10^4\text{ Pa})

p_{\text{total}} = 1.5005 \times 10^5\text{ Pa}

Round to a sensible number of significant figures: Since the input values are given to three significant figures, we quote the final answer as:

p_{\text{total}} \approx 1.50 \times 10^5\text{ Pa} \quad (\text{or } 150\text{ kPa})

4. Upthrust and Archimedes' Principle

Have you ever wondered why a heavy steel ship can float, while a small steel coin sinks? The answer lies in upthrust and pressure differences.

The Origin of Upthrust

When an object is submerged in a fluid, its upper surface is at a shallower depth than its lower surface.

Upthrust on Submerged Block

Let's derive the expression for this upward force:

Downward force on the top surface:

F_{\text{top}} = p_{\text{top}} A = h_{\text{top}} \rho g A

Upward force on the bottom surface:

F_{\text{bottom}} = p_{\text{bottom}} A = h_{\text{bottom}} \rho g A

Net upward force (Upthrust, $U$ ):

F_{\text{bottom}} - F_{\text{top}} = (h_{\text{bottom}} - h_{\text{top}}) \rho g A

Relating to volume: The difference in depth $(hbottom−htop)(h_{\text{bottom}} - h_{\text{top}})$ is simply the height $x$ of the block. The volume of the block is $V = A x$ .

\rho g A = V \rho g

Definition

Archimedes' Principle

Archimedes' Principle states that:

An object submerged or floating in a fluid experiences an upward upthrust force equal to the weight of the fluid it displaces.

V_{\text{sub}} \rho_{\text{fluid}} g

Where:

$U$ is the upthrust force in Newtons ( $N\text{N}$ )
$VsubV_{\text{sub}}$ is the volume of the object submerged in the fluid in cubic metres ( $m3\text{m}^3$ )
$ρfluid\rho_{\text{fluid}}$ is the density of the fluid in kilograms per cubic metre ( $m−3\text{kg}\,\text{m}^{-3}$ )
$g$ is the acceleration of free fall ( $s−29.81\text{ m s}^{-2}$ )

Sinking vs. Floating

An object sinks if its weight is greater than the maximum upthrust (which occurs when the object is fully submerged). This happens if the average density of the object is greater than the density of the fluid ( $ρobject>ρfluid\rho_{\text{object}} > \rho_{\text{fluid}}$ ).
An object floats in equilibrium if its weight is equal to the upthrust. In this state, it only needs to displace a volume of fluid whose weight matches the object's total weight.

Example

Calculating the submerged volume of a floating ice block

An ice block of mass $kg450\text{ kg}$ floats in freshwater of density $1.00 \times 10^3\text{ kg}\,\text{m}^{-3}$ . Calculate the volume of the ice block that is submerged below the water level.

State the condition for floating: For a floating object in equilibrium, the upward upthrust $U$ must balance the downward weight of the object $W$ :

U = W

Calculate the weight of the ice block:

450\text{ kg} \times 9.81\text{ m s}^{-2} = 4414.5\text{ N}

Set up Archimedes' principle equation for upthrust:

V_{\text{sub}} \rho_{\text{water}} g

Since $U = W$ :

V_{\text{sub}} \rho_{\text{water}} g = mg

Notice that $g$ cancels out from both sides:

V_{\text{sub}} \rho_{\text{water}} = m

Rearrange and solve for the submerged volume ( $VsubV_{\text{sub}}$ ):

V_{\text{sub}} = \frac{m}{\rho_{\text{water}}} = \frac{450\text{ kg}}{1.00 \times 10^3\text{ kg}\,\text{m}^{-3}} = 0.450\text{ m}^3

Exam technique

In the exam

Always check your units: Ensure volume is in $m3\text{m}^3$ (not $cm3\text{cm}^3$ or litres) and density is in $m−3\text{kg}\,\text{m}^{-3}$ before carrying out substitutions.
Memorize the $h\rho g$ derivation: This is a classic 3-to-4 mark exam question. Practice writing down the algebraic steps clearly, defining what each letter represents.
Use free-body diagrams for upthrust questions: Draw the forces acting on the submerged object (Weight downwards, Upthrust upwards, and sometimes Tension in a string or Drag if the object is moving). Set up your equilibrium equations ( $Fnet=0F_{\text{net}} = 0$ ) from the diagram.
Be precise with vocabulary: Use the term "fluid" rather than just "liquid" when discussing pressure and upthrust, as these principles apply equally to gases (like balloons floating in air).

Self review

Check yourself

A metal cylinder has a radius of $cm1.5\text{ cm}$ and a height of $cm12\text{ cm}$ . If its mass is $kg0.72\text{ kg}$ , what is its density in $m−3\text{kg}\,\text{m}^{-3}$ ?
Explain why the pressure equation $h\rho g$ proves that a thin, tall cylinder of water exerts the same pressure on its base as a wide, tall tank filled to the same height.
A fully submerged concrete block experiences an upthrust of $N250\text{ N}$ in seawater. If the density of the seawater is $m−31025\text{ kg}\,\text{m}^{-3}$ , what is the volume of the block?

What you'll learn:

1. Density ( $ρ\rho$ )

Determining Density Practicals

2. Pressure ( $p$ )

3. Pressure in Fluids ( $h\rho g$ )

Deriving $h\rho g$

4. Upthrust and Archimedes' Principle

The Origin of Upthrust

Sinking vs. Floating

Review Density and pressure by teaching Genie

Remember key concepts with flashcards

Density and pressure

What you'll learn:

1. Density ( $ρ\rho$ )

Determining Density Practicals

2. Pressure ( $p$ )

3. Pressure in Fluids ( $h\rho g$ )

Deriving $h\rho g$

4. Upthrust and Archimedes' Principle

The Origin of Upthrust

Sinking vs. Floating

Review Density and pressure by teaching Genie

Remember key concepts with flashcards

Density and pressure

What you'll learn:

1. Density (ρ\rhoρ)

Determining Density Practicals

2. Pressure (ppp)

3. Pressure in Fluids (p=hρgp = h\rho gp=hρg)

Deriving p=hρgp = h\rho gp=hρg

4. Upthrust and Archimedes' Principle

The Origin of Upthrust

Sinking vs. Floating

Review Density and pressure by teaching Genie

Remember key concepts with flashcards

Density and pressure Revision Guide

Density and pressure

What you'll learn:

1. Density (ρ\rhoρ)

Determining Density Practicals

2. Pressure (ppp)

3. Pressure in Fluids (p=hρgp = h\rho gp=hρg)

Deriving p=hρgp = h\rho gp=hρg

4. Upthrust and Archimedes' Principle

The Origin of Upthrust

Sinking vs. Floating

Review Density and pressure by teaching Genie

Remember key concepts with flashcards

Density and pressure Revision Guide

1. Density ( $ρ\rho$ )

2. Pressure ( $p$ )

3. Pressure in Fluids ( $h\rho g$ )

Deriving $h\rho g$

1. Density ( $ρ\rho$ )

2. Pressure ( $p$ )

3. Pressure in Fluids ( $h\rho g$ )

Deriving $h\rho g$