5.5.1a Pressure in a fluid

What counts as a fluid

Definition

Fluid

A fluid is a substance that can flow because its particles move past one another; it can be a liquid or a gas.

In an answer, say liquid or gas, not just “a liquid”.
A solid cannot flow because its particles are held in fixed positions and can only vibrate.
In a liquid the particles touch but slide over each other; in a gas they are far apart and move quickly in random directions.
Both take the shape of their container and both press on every surface they touch.

Pressure in a fluid acts normal to a surface

The pressure in a fluid causes a force normal (at right angles) to any surface it touches.
This follows from the particle model: the fluid’s particles constantly move and collide with any surface, each collision exerting a tiny force.
The collisions come from every direction, so the sideways pushes cancel and the resultant force acts at right angles to the surface.
So fluid pressure acts in all directions, not just downwards: a gas presses outwards on the top, sides and bottom of its container, and water presses inwards on every face of a submerged object.
The direction of the force depends only on how the surface is orientated: tilt the surface and the force stays at right angles to it.

Key Idea

The pressure in a fluid causes a force normal (at right angles) to any surface.

Calculating the pressure at a surface

Definition

Pressure

Pressure is the force normal to a surface divided by the area of that surface.

The equation is $p=FAp = \dfrac{F}{A}$ $p = \frac{F}{A}$ .
1. $p$ is the pressure in pascals ( $Pa\text{Pa}$ ).
2. $F$ is the force normal to the surface in newtons ( $N\text{N}$ ).
3. $A$ is the area of that surface in metres squared ( $m2\text{m}^2$ ).
$N/m21\ \text{Pa} = 1\ \text{N/m}^2$ , so a pressure of $Pa250\ \text{Pa}$ means each square metre feels a force of $N250\ \text{N}$ at right angles.
Rearranged, the equation gives $\times A$ and $\dfrac{F}{p}$ .
Areas are often given in $cm2\text{cm}^2$ , but the equation needs $m2\text{m}^2$ : divide an area in $cm2\text{cm}^2$ by $00010\,000$ , because $cm21\ \text{m}^2 = 10\,000\ \text{cm}^2$ .

Example

Compressed gas pushes on a piston with a force of $N4500\ \text{N}$ normal to the piston. The piston area is $m20.030\ \text{m}^2$ . Calculate the pressure of the gas at the piston.

Write the equation: $\dfrac{F}{A}$ .
Check the units: force is in $N\text{N}$ and area is in $m2\text{m}^2$ , so no conversion is needed.
Substitute the values: $\dfrac{4500\ \text{N}}{0.030\ \text{m}^2}$ .
Calculate the pressure: $150\,000\ \text{Pa} = 1.5 \times 10^5\ \text{Pa}$ .

Finding a force from a pressure

When the pressure and area are given, rearrange to $\times A$ to find the force normal to the surface.
Convert the area to $m2\text{m}^2$ before substituting.

Example

Water in a tank exerts a pressure of $Pa12\,000\ \text{Pa}$ on an inspection hatch in the side of the tank. The hatch area is $cm2250\ \text{cm}^2$ . Calculate the force the water exerts on the hatch.

Rearrange the equation: $\times A$ .
Convert the area: $\dfrac{250}{10\,000} = 0.025\ \text{m}^2$ .
Substitute the values: $12\,000\ \text{Pa} \times 0.025\ \text{m}^2$ .
Calculate the force: $300\ \text{N}$ , acting normal to the hatch (horizontally outwards, at right angles to the side of the tank).

Exam technique

Do not treat pressure and force as the same: a large force over a large area gives a small pressure, and a small force over a tiny area gives a huge pressure, so always ask “over what area?”.
Do not assume fluid pressure only pushes downwards; it acts in every direction and, on any surface, at right angles to it.
Convert areas in $cm2\text{cm}^2$ to $m2\text{m}^2$ before substituting, or the answer will be $00010\,000$ times too big.
Only the force normal to the surface goes into $\dfrac{F}{A}$ , and $A$ is the area of that same surface, not the whole object.
To describe the direction, use “the pressure in a fluid causes a force normal (at right angles) to the surface”; “downwards” or “outwards” alone will not score.

Self review

What is a fluid?
In which direction does the force from fluid pressure act on a surface, and why?
State the equation for pressure and the units of each quantity.
What is $Pa1\ \text{Pa}$ in $N/m2\text{N/m}^2$ , and how many $cm2\text{cm}^2$ are in $m21\ \text{m}^2$ ?
How do you find the force on a surface from the pressure and area?

5.5.1b Pressure due to a column of liquid and upthrust

Pressure due to a column of liquid

Definition

Pressure due to a liquid column

The pressure caused by the weight of the liquid above a point.

The pressure due to a column of liquid is $p=hρgp = h\rho g$ $p = h ρ g$ .
1. $p$ is the pressure in pascals ( $Pa\text{Pa}$ ).
2. $h$ is the vertical height (depth) of the liquid column in metres ( $m\text{m}$ ).
3. $ρ\rho$ is the density of the liquid in kilograms per metre cubed ( $kg/m3\text{kg/m}^3$ ).
4. $g$ is the gravitational field strength in newtons per kilogram ( $N/kg\text{N/kg}$ ), and its value is given in the question.
The height $h$ is the vertical depth of the point below the surface; at the bottom of a swimming pool, $h$ is the distance from the water’s surface to the bottom.

Why pressure increases with depth and density

A deeper point has a taller column of liquid above it, which contains more liquid and so has a greater weight, giving a greater pressure.
$h\rho g$ shows pressure is directly proportional to depth: double the depth, with $ρ\rho$ and $g$ constant, and the pressure doubles.
A denser liquid has more mass in the same volume, so its column weighs more and produces a greater pressure at the same depth.

A diagram of a container filled with water with three spouts at different depths. The water jet from the bottom spout travels the furthest (strongest jet), while the jet from the top spout travels the shortest distance (weakest jet), demonstrating that liquid pressure increases with depth.

Differences in pressure

For the pressure difference between two points in the same liquid, first find the difference in depth: $Δh=h2−h1\Delta h = h_2 - h_1$ .
Then use $ρg\Delta p = \Delta h\,\rho g$ .

Example

Two points in water are at depths of $m0.40\ \text{m}$ and $m1.70\ \text{m}$ . The density of water is $kg/m31000\ \text{kg/m}^3$ and $9.8\ \text{N/kg}$ . Calculate the difference in pressure between the two points.

Write the equation: $ρg\Delta p = \Delta h\,\rho g$ .
Find the difference in depth: $m\Delta h = 1.70 - 0.40 = 1.30\ \text{m}$ .
Substitute the values: $N/kg\Delta p = 1.30\ \text{m} \times 1000\ \text{kg/m}^3 \times 9.8\ \text{N/kg}$ .
Calculate the pressure difference: $Pa\Delta p = 12\,740\ \text{Pa} = 1.27 \times 10^4\ \text{Pa}$ , with the deeper point at the greater pressure.

Upthrust

Definition

Upthrust

Upthrust is the resultant upward force exerted by a fluid on an object that is partially or totally submerged.

Liquid pressure acts on every submerged surface; the bottom is deeper than the top, so the pressure on the bottom is greater.
The upward force on the bottom is therefore greater than the downward force on the top, and this pressure difference produces a resultant upward force called upthrust.

Key Idea

Upthrust is caused by the difference in liquid pressure between the lower and upper surfaces of a submerged object.

Floating and sinking

Whether an object floats or sinks depends on its weight compared with the upthrust acting on it.
1. If weight is greater than upthrust, there is a resultant downward force and the object sinks.
2. If upthrust is greater than weight, there is a resultant upward force and the object rises.
3. If upthrust equals weight, the forces are balanced and the object floats at a constant level or stays suspended.
Density matters too: an object with a lower average density than the liquid can float, while one with a greater average density sinks.
Increasing the liquid’s density increases the upthrust and makes floating more likely.
A floating object settles where the upthrust equals its weight; a heavier object displaces more liquid, so it floats with more of its volume submerged.

Exam technique

A floating object is not force-free: it has an upward upthrust and a downward weight, which are equal when it floats at rest.
Use the vertical depth below the surface for $h$ , and convert centimetres to metres before using $h\rho g$ .
To explain upthrust, give the full chain: the bottom is deeper, so the pressure there is greater, so the upward force on the bottom is larger than the downward force on the top, leaving a resultant upward force.
For floating and sinking, compare upthrust with weight and state the direction of the resultant force.

Self review

State the equation linking pressure, depth, density and gravitational field strength.
Why does liquid pressure increase with depth?
Why does a denser liquid give greater pressure at the same depth?
How do you calculate a pressure difference between two depths?
How does a pressure difference create upthrust, and how do you predict floating or sinking?

5.5.1a Pressure in a fluid

What counts as a fluid

Definition

Fluid

A fluid is a substance that can flow because its particles move past one another; it can be a liquid or a gas.

In an answer, say liquid or gas, not just “a liquid”.
A solid cannot flow because its particles are held in fixed positions and can only vibrate.
In a liquid the particles touch but slide over each other; in a gas they are far apart and move quickly in random directions.
Both take the shape of their container and both press on every surface they touch.

Pressure in a fluid acts normal to a surface

The pressure in a fluid causes a force normal (at right angles) to any surface it touches.
This follows from the particle model: the fluid’s particles constantly move and collide with any surface, each collision exerting a tiny force.
The collisions come from every direction, so the sideways pushes cancel and the resultant force acts at right angles to the surface.
So fluid pressure acts in all directions, not just downwards: a gas presses outwards on the top, sides and bottom of its container, and water presses inwards on every face of a submerged object.
The direction of the force depends only on how the surface is orientated: tilt the surface and the force stays at right angles to it.

Key Idea

The pressure in a fluid causes a force normal (at right angles) to any surface.

Calculating the pressure at a surface

Definition

Pressure

Pressure is the force normal to a surface divided by the area of that surface.

The equation is $p=FAp = \dfrac{F}{A}$ $p = \frac{F}{A}$ .
1. $p$ is the pressure in pascals ( $Pa\text{Pa}$ ).
2. $F$ is the force normal to the surface in newtons ( $N\text{N}$ ).
3. $A$ is the area of that surface in metres squared ( $m2\text{m}^2$ ).
$N/m21\ \text{Pa} = 1\ \text{N/m}^2$ , so a pressure of $Pa250\ \text{Pa}$ means each square metre feels a force of $N250\ \text{N}$ at right angles.
Rearranged, the equation gives $\times A$ and $\dfrac{F}{p}$ .
Areas are often given in $cm2\text{cm}^2$ , but the equation needs $m2\text{m}^2$ : divide an area in $cm2\text{cm}^2$ by $00010\,000$ , because $cm21\ \text{m}^2 = 10\,000\ \text{cm}^2$ .

Example

Compressed gas pushes on a piston with a force of $N4500\ \text{N}$ normal to the piston. The piston area is $m20.030\ \text{m}^2$ . Calculate the pressure of the gas at the piston.

Write the equation: $\dfrac{F}{A}$ .
Check the units: force is in $N\text{N}$ and area is in $m2\text{m}^2$ , so no conversion is needed.
Substitute the values: $\dfrac{4500\ \text{N}}{0.030\ \text{m}^2}$ .
Calculate the pressure: $150\,000\ \text{Pa} = 1.5 \times 10^5\ \text{Pa}$ .

Finding a force from a pressure

When the pressure and area are given, rearrange to $\times A$ to find the force normal to the surface.
Convert the area to $m2\text{m}^2$ before substituting.

Example

Rearrange the equation: $\times A$ .
Convert the area: $\dfrac{250}{10\,000} = 0.025\ \text{m}^2$ .
Substitute the values: $12\,000\ \text{Pa} \times 0.025\ \text{m}^2$ .
Calculate the force: $300\ \text{N}$ , acting normal to the hatch (horizontally outwards, at right angles to the side of the tank).

Exam technique

Do not treat pressure and force as the same: a large force over a large area gives a small pressure, and a small force over a tiny area gives a huge pressure, so always ask “over what area?”.
Do not assume fluid pressure only pushes downwards; it acts in every direction and, on any surface, at right angles to it.
Convert areas in $cm2\text{cm}^2$ to $m2\text{m}^2$ before substituting, or the answer will be $00010\,000$ times too big.
Only the force normal to the surface goes into $\dfrac{F}{A}$ , and $A$ is the area of that same surface, not the whole object.
To describe the direction, use “the pressure in a fluid causes a force normal (at right angles) to the surface”; “downwards” or “outwards” alone will not score.

Self review

What is a fluid?
In which direction does the force from fluid pressure act on a surface, and why?
State the equation for pressure and the units of each quantity.
What is $Pa1\ \text{Pa}$ in $N/m2\text{N/m}^2$ , and how many $cm2\text{cm}^2$ are in $m21\ \text{m}^2$ ?
How do you find the force on a surface from the pressure and area?

5.5.1b Pressure due to a column of liquid and upthrust

Pressure due to a column of liquid

Definition

Pressure due to a liquid column

The pressure caused by the weight of the liquid above a point.

The pressure due to a column of liquid is $p=hρgp = h\rho g$ $p = h ρ g$ .
1. $p$ is the pressure in pascals ( $Pa\text{Pa}$ ).
2. $h$ is the vertical height (depth) of the liquid column in metres ( $m\text{m}$ ).
3. $ρ\rho$ is the density of the liquid in kilograms per metre cubed ( $kg/m3\text{kg/m}^3$ ).
4. $g$ is the gravitational field strength in newtons per kilogram ( $N/kg\text{N/kg}$ ), and its value is given in the question.
The height $h$ is the vertical depth of the point below the surface; at the bottom of a swimming pool, $h$ is the distance from the water’s surface to the bottom.

Why pressure increases with depth and density

A deeper point has a taller column of liquid above it, which contains more liquid and so has a greater weight, giving a greater pressure.
$h\rho g$ shows pressure is directly proportional to depth: double the depth, with $ρ\rho$ and $g$ constant, and the pressure doubles.
A denser liquid has more mass in the same volume, so its column weighs more and produces a greater pressure at the same depth.

Differences in pressure

For the pressure difference between two points in the same liquid, first find the difference in depth: $Δh=h2−h1\Delta h = h_2 - h_1$ .
Then use $ρg\Delta p = \Delta h\,\rho g$ .

Example

Write the equation: $ρg\Delta p = \Delta h\,\rho g$ .
Find the difference in depth: $m\Delta h = 1.70 - 0.40 = 1.30\ \text{m}$ .
Substitute the values: $N/kg\Delta p = 1.30\ \text{m} \times 1000\ \text{kg/m}^3 \times 9.8\ \text{N/kg}$ .
Calculate the pressure difference: $Pa\Delta p = 12\,740\ \text{Pa} = 1.27 \times 10^4\ \text{Pa}$ , with the deeper point at the greater pressure.

Upthrust

Definition

Upthrust

Upthrust is the resultant upward force exerted by a fluid on an object that is partially or totally submerged.

Liquid pressure acts on every submerged surface; the bottom is deeper than the top, so the pressure on the bottom is greater.
The upward force on the bottom is therefore greater than the downward force on the top, and this pressure difference produces a resultant upward force called upthrust.

Key Idea

Upthrust is caused by the difference in liquid pressure between the lower and upper surfaces of a submerged object.

Floating and sinking

Whether an object floats or sinks depends on its weight compared with the upthrust acting on it.
1. If weight is greater than upthrust, there is a resultant downward force and the object sinks.
2. If upthrust is greater than weight, there is a resultant upward force and the object rises.
3. If upthrust equals weight, the forces are balanced and the object floats at a constant level or stays suspended.
Density matters too: an object with a lower average density than the liquid can float, while one with a greater average density sinks.
Increasing the liquid’s density increases the upthrust and makes floating more likely.
A floating object settles where the upthrust equals its weight; a heavier object displaces more liquid, so it floats with more of its volume submerged.

Exam technique

A floating object is not force-free: it has an upward upthrust and a downward weight, which are equal when it floats at rest.
Use the vertical depth below the surface for $h$ , and convert centimetres to metres before using $h\rho g$ .
To explain upthrust, give the full chain: the bottom is deeper, so the pressure there is greater, so the upward force on the bottom is larger than the downward force on the top, leaving a resultant upward force.
For floating and sinking, compare upthrust with weight and state the direction of the resultant force.

Self review

State the equation linking pressure, depth, density and gravitational field strength.
Why does liquid pressure increase with depth?
Why does a denser liquid give greater pressure at the same depth?
How do you calculate a pressure difference between two depths?
How does a pressure difference create upthrust, and how do you predict floating or sinking?

5.5.1 Pressure in a fluid

5.5.1a Pressure in a fluid

What counts as a fluid

Pressure in a fluid acts normal to a surface

Calculating the pressure at a surface

Finding a force from a pressure

5.5.1b Pressure due to a column of liquid and upthrust

Pressure due to a column of liquid

Why pressure increases with depth and density

Differences in pressure

Upthrust

Floating and sinking

Review 5.5.1 Pressure in a fluid by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.5.1 Pressure in a fluid

5.5.1a Pressure in a fluid

What counts as a fluid

Pressure in a fluid acts normal to a surface

Calculating the pressure at a surface

Finding a force from a pressure

5.5.1b Pressure due to a column of liquid and upthrust

Pressure due to a column of liquid

Why pressure increases with depth and density

Differences in pressure

Upthrust

Floating and sinking

Review 5.5.1 Pressure in a fluid by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.5.1 Pressure in a fluid

5.5.1a Pressure in a fluid

What counts as a fluid

Pressure in a fluid acts normal to a surface

Calculating the pressure at a surface

Finding a force from a pressure

5.5.1b Pressure due to a column of liquid and upthrust

Pressure due to a column of liquid

Why pressure increases with depth and density

Differences in pressure

Upthrust

Floating and sinking

Review 5.5.1 Pressure in a fluid by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.5.1 Pressure in a fluid Revision Guide

5.5.1 Pressure in a fluid

5.5.1a Pressure in a fluid

What counts as a fluid

Pressure in a fluid acts normal to a surface

Calculating the pressure at a surface

Finding a force from a pressure

5.5.1b Pressure due to a column of liquid and upthrust

Pressure due to a column of liquid

Why pressure increases with depth and density

Differences in pressure

Upthrust

Floating and sinking

Review 5.5.1 Pressure in a fluid by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.5.1 Pressure in a fluid Revision Guide