2.4 Pressure and pressure differences in fluids — Learn

Prerequisites: Section 2.1, especially forces and the unit newton; Section 2.3, work done; the area of a rectangle; changing between square centimetres and square metres.

By the end of this section you can: explain what pressure means in a fluid; calculate pressure, force or area; describe atmospheric pressure and why it changes with height; and, at Higher tier, calculate the pressure at a depth in a liquid and explain upthrust.

1. Pressure in a fluid

A fluid is a liquid or a gas. The particles of a fluid move around and collide with anything in contact with the fluid. The pressure in a fluid causes a force normal to any surface, which means at right angles to the surface.

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

p = \frac{F}{A}

where $p$ is the pressure in pascals, $F$ is the force normal to the surface in newtons, and $A$ is the area of the surface in square metres.

One pascal is one newton per square metre.

1\,\text{Pa} = 1\,\text{N/m}^2

A large pressure means a large force on a small area. This is why a sharp knife cuts well: the same push acts on a very small area.

Worked example 1. A force of 300 N acts on an area of 0.015 m². Calculate the pressure.

\begin{aligned} p &= \frac{F}{A} \\ &= \frac{300}{0.015} \\ &= 20000\,\text{Pa} \\ &= 2.0 \times 10^4\,\text{Pa} \end{aligned}

Worked example 2. A pressure of 8000 Pa acts on an area of 0.25 m². Calculate the force normal to the surface.

\begin{aligned} F &= p A \\ &= 8000 \times 0.25 \\ &= 2000\,\text{N} \\ &= 2.0 \times 10^3\,\text{N} \end{aligned}

Worked example 3. A force of 900 N produces a pressure of 30000 Pa. Calculate the area.

\begin{aligned} A &= \frac{F}{p} \\ &= \frac{900}{30000} \\ &= 0.030\,\text{m}^2 \end{aligned}

Watch out. The area must be in square metres before you use $p = F/A$ . One square metre is $10000\,\text{cm}^2$ , so a contact area of $12\,\text{cm}^2$ is $12 \div 10000 = 0.0012\,\text{m}^2$ . Dividing by 100 instead of 10000 gives a pressure 100 times too large. Check that your area is much smaller than a square metre before you divide.

Try it: pressure and area

The figure shows a force pushing on a surface. The reader sets the force and the area of the surface.

Predict first. Do not move the sliders yet. Write down your answers.

If the force stays the same and the area is halved, what happens to the pressure?
A wide tyre and a narrow tyre carry the same force. Which one has the larger pressure on the road?

Test. Set the force to 400 N and the area to 0.20 m². Read the pressure. Now keep the force the same and halve the area. Read the pressure again.

Explain. Why does halving the area double the pressure for the same force? Why does a sharp point cut better than a blunt one? Write your answer in two or three sentences.

2. Atmospheric pressure

The atmosphere is a thin layer of air around the Earth. The air is made of molecules that move around and collide with surfaces. Each collision pushes, and the total push is the atmospheric pressure.

The atmosphere gets less dense as you go higher, and there is less air above a high surface. So the number of molecules above a surface, and the weight of the air above it, both decrease as the height increases. Fewer collisions give a smaller pressure, so atmospheric pressure decreases as you go higher.

3. Pressure at a depth in a liquid

The pressure due to a column of liquid depends on how deep you are and how dense the liquid is.

p = h \rho g

where $h$ is the height of the column in metres, $\rho$ (rho) is the density of the liquid in kilograms per cubic metre, and $g$ is the gravitational field strength in newtons per kilogram. Density is the mass of one cubic metre of the substance.

Worked example 4. Calculate the pressure due to a column of water 5.0 m high. The density of water is 1000 kg/m³ and $g = 9.8\,\text{N/kg}$ .

\begin{aligned} p &= h \rho g \\ &= 5.0 \times 1000 \times 9.8 \\ &= 49000\,\text{Pa} \\ &= 4.9 \times 10^4\,\text{Pa} \end{aligned}

The pressure at a point in a liquid increases with the height of liquid above that point, and with the density of the liquid. To find the difference in pressure between two depths, use the difference in depth as $h$ .

Worked example 5. Calculate the pressure difference between depths of 1.0 m and 3.0 m in water of density 1000 kg/m³. Take $g = 9.8\,\text{N/kg}$ .

\begin{aligned} h &= 3.0 - 1.0 = 2.0\,\text{m} \\ p &= h \rho g = 2.0 \times 1000 \times 9.8 \\ &= 19600\,\text{Pa} \\ &= 2.0 \times 10^4\,\text{Pa} \end{aligned}

Upthrust and floating

A submerged object has a greater pressure on its bottom surface than on its top surface, because the bottom is deeper. The upward force on the bottom is therefore larger than the downward force on the top. The resultant upward force is called upthrust.

An object floats if its average density is less than the density of the liquid, and sinks if its average density is greater. A block of wood averages less than water, so it floats. A block of steel averages more than water, so it sinks.

4. Progress check

Return to the chapter problem. The front tyre carries a force of 540 N and the contact patch has an area of 12 cm². Work out the pressure in the tyre in pascals. Then decide what happens to the pressure if the tyre is pumped harder.

You should now be able to

explain why the pressure in a fluid causes a force normal to a surface;
calculate pressure, force or area, using $p = \dfrac{F}{A}$ ;
change an area from square centimetres to square metres;
describe a simple model of the atmosphere and explain why atmospheric pressure decreases with height;
calculate the pressure at a depth in a liquid, and explain upthrust and floating (Higher tier).