Physics · Mechanics

Pressure: normal force per area

Pressure describes normal force per area; with nonuniform loading $\frac{F}{A}$ gives only average normal compressive loading.

BasicExam-relevantLower secondaryUpper secondary

Create flashcards or a quiz I already have an account

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Formula

Calculate

p = \frac{F}{A}

p in pascals [Pa] = [N/m²] · F in newtons [N] · A in m² (1 bar = 100,000 Pa)

Variables & units

Variables & units
Symbol	Meaning	Unit
$p$	Pressure	Pa
$F$	Force acting perpendicularly	N
$A$	Area on which the force acts	m²

LaTeX

p = \frac{F}{A}

QUANTA STUDYquanta-study.de

A 150 N force acts on a horizontal surface at 60° from the dashed normal. Components show 75 N normal and about 129.9 N tangential. The correct quotient uses 75 N and 3 cm², not total force. — 150 N at 60° from the normal gives 75 N normal force. Over 3 cm² this produces 250 kPa average normal contact loading. The tangential component does not enter that pressure quotient.

Pressure calculator

Find pressure, force or area when a force pushes perpendicular to a surface; the result is the mean pressure on that surface.

Find

Perpendicular force F⊥

Force pushing perpendicular to the surface, e.g. 600 N, the weight of a person of just over 60 kg.

Area A

Area the force pushes on, e.g. 300 cm² for two shoe soles together.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for p, substitute, evaluate

1. Rearrange for p
$p=\frac{F_\perp}{A}$
2. Substitute values (in SI units)
- A = 300 cm² = 0.03 m²
$p=\frac{600}{0.03}=20000\,\mathrm{Pa}$
3. Result
p = 20 kPa

Result: Pressure p

p = 20 kPa

With F⊥ = 600 N and A = 300 cm², p = 20 kPa.

Quick tips

Formula triangle: F on top, p and A below. Cover the quantity you need and the rest is the calculation, e.g. A = F ÷ p.
Unit check: 1 Pa is 1 N per m². Convert cm² first, 1 cm² = 0.0001 m², otherwise the pressure comes out 10,000 times too small.

Assumptions and limits

Nonnegative mean normal compressive pressure on a positive area. Force is the normal magnitude of this selected loading, not arbitrary total or tangential force. Specify reference pressure and other surface forces separately.

Derivation & background

In a static fluid, local pressure is a scalar and force on an area element acts normal to it. F⊥/A gives an area average; uniform pressure yields F⊥ = pA. Net pressure force on an ideal equal-area piston is (p₁ − p₂)A. Absolute pressure refers to vacuum; gauge pressure to a named reference. Pascal’s principle concerns pressure changes; hydrostatic elevation differences remain.

Exam blueprint

Validity range

$\overline p=\frac{F_{\perp}}{A}$ gives average normal loading on a planar area; uniform local pressure requires uniform distribution. Net piston pressure force uses the pressure difference across it.

Derivation steps

Normal pressure loading is defined per unit area; total force follows by summing area elements.

1Specify a planar area A and normal force F⊥.
2Do not count tangential force as pressure.
3A small element obeys dF⊥ = p dA.
4Integrate over the surface: F⊥ = ∫p dA.
5Uniform pressure gives F⊥ = pA and thus $p = F$ ⊥/A.
6Varying pressure makes the ratio an area average; a piston uses pressure difference between its sides.

Rearrangements

Force from pressure and area

F = p \cdot A

The principle of hydraulics: a large piston gives a large force at the same pressure.

Area from force and pressure

A = \frac{F}{p}

This is how bearing surfaces are sized against sinking in.

Hydrostatic pressure

p = \rho \cdot g \cdot h

About 1 bar extra per 10 m of water depth.

Task variant

An ideal hydraulic cylinder of effective area $A = 0.02 m^{2}$ has a pressure difference of $5\times{}10^{5} Pa$ across its sides. Find net pressure force.

Approach: $F_{\mathrm{net}}=\Delta pA$ .
Substitution: $F_{\mathrm{net}}=5\cdot10^5\cdot0.02\,\mathrm N=10000\,\mathrm N$ , toward the lower-pressure side.

Check: $\frac{F_{\mathrm{net}}}{A}=500000\,\mathrm{Pa}$ . Actual lifting capacity additionally depends on weight, friction and the system.

At what water depth is the hydrostatic pressure $1 bar (\rho{} =\frac{ 1\,000 kg}{m^{3}})$ ? Use $g =\frac{ 9.81 m}{s^{2}}$ .

$h=\frac{p}{\rho g}=\frac{100000\,\mathrm{Pa}}{1000\,\mathrm{kg\,m^{-3}}\cdot9.81\,\mathrm{m\,s^{-2}}}\approx10.2\,\mathrm m$ .

Common mistakes

Equating pressure and force.

Pressure is force per area; the same force can be harmless or cutting.

Substituting areas in $cm^{2}$ and reading the result as pascals.

$1 m^{2} = 10\,000 cm^{2}$ , convert the area to $m^{2}$ first.

Confusing bar and pascal.

$1 bar = 100\,000 Pa = 1\,000$ hPa.

Exam context

Classics: the hydraulic press (force multiplication), hydrostatic pressure in diving, comparing snowshoes with heels.

Formula cluster

Pressure and fluids

Pressure, density and hydrostatic pressure form the basis of fluid mechanics.

ρ = m/V

Density

pV = nRT

Ideal Gas Law

W = F·s·cos(α)

Mechanical Work

p + ½ρv² + ρgh = const.

Bernoulli Equation

Worked example

A normal load of 600 N over a total $0.003 m^{2}$ gives average contact loading $200000 Pa = 2 bar$ . This does not determine local distribution or the surface material’s response.

Applications

Hydraulic presses and brakes, tyre pressure, diving physics, blood pressure measurement, weather maps

How do you calculate with Pressure?

Here is how to work through a typical Pressure (p = F/A) task step by step:

1
Task
An ideal hydraulic cylinder of effective area $A = 0.02 m^{2}$ has a pressure difference of $5\times{}10^{5} Pa$ across its sides. Find net pressure force.
Solution path
1. Approach: $F_{\mathrm{net}}=\Delta pA$ .
2. Substitution: $F_{\mathrm{net}}=5\cdot10^5\cdot0.02\,\mathrm N=10000\,\mathrm N$ , toward the lower-pressure side.
Check: $\frac{F_{\mathrm{net}}}{A}=500000\,\mathrm{Pa}$ . Actual lifting capacity additionally depends on weight, friction and the system.
2
Task
At what water depth is the hydrostatic pressure $1 bar (\rho{} =\frac{ 1\,000 kg}{m^{3}})$ ? Use $g =\frac{ 9.81 m}{s^{2}}$ .
Solution path
$h=\frac{p}{\rho g}=\frac{100000\,\mathrm{Pa}}{1000\,\mathrm{kg\,m^{-3}}\cdot9.81\,\mathrm{m\,s^{-2}}}\approx10.2\,\mathrm m$ .

Understand and apply the equation

The same force can spread over a large area or concentrate on a small patch. Pressure describes normal force per area. Starting from this idea, follow the steps through area units, absolute and gauge pressure, and force multiplication in a hydraulic system.

How do you calculate pressure from normal force and area?

Supporting a load on a broad face distributes its normal force differently from using a narrow tip. A smaller contact area increases average normal loading at the same total force. Whether the material yields is a separate question depending on its strength and local loading distribution. In a stationary fluid, pressure is a local state quantity; the resulting forces act normal to each surface.[1]

The ratio $p=\frac{F_\perp}{A}$ uses only force perpendicular to the area. Forces parallel to a surface tend to slide material layers relative to one another; this shear loading is not part of the pressure ratio. In a stationary fluid, pressure is a number with units rather than a force with its own direction. The resulting force acts normal to the surface element being considered. Differently oriented walls therefore experience forces in different directions.[1]

\overline p=\frac{F_\perp}{A},\qquad F_\perp=\int_Ap\,\mathrm dA

For varying pressure, F⊥/A is an area average. This scalar integral assumes a planar surface.

For nonuniform solid contact, $\frac{F_\perp}{A}$ gives only average normal compressive loading, not a uniform local distribution. Sinking also depends on material strength, shape and actual contact area.

Smaller area, greater average loading

QUANTA STUDYquanta-study.de

At a fixed normal force of 120 N, average loading is 400 kPa over 3 cm² and only 100 kPa over 12 cm². The curve covers areas from 3 to 12 cm²; it specifies neither local distribution nor a material failure limit.[1]

Free with a Quanta account. Your download stays ready after registration.

How do you rearrange for force or area?

Divide a planar surface into small elements. Each has $\mathrm dF_\perp=p\,\mathrm dA$ . Uniform p can be taken outside the integral; the area elements sum to A, giving $F_\perp=pA$ . This is a definition and its summation, not an empirical proof of universal material load capacity.

F_\perp=p\int_A\mathrm dA=pA\quad\Longrightarrow\quad p=\frac{F_\perp}{A}

The compact expression gives a uniform pressure only when the distribution is uniform.

For force F at angle θ to the surface normal, $F_\perp=F\cos\theta$ . If the angle is measured from the surface itself, the normal component is $F\sin\theta$ . Draw the angle first; do not interchange these conventions.

Converting pascals, bar and square centimetres correctly

1\,\mathrm{Pa}=1\,\mathrm{N/m^2},\quad1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2},\quad1\,\mathrm{bar}=10^5\,\mathrm{Pa}

Area conversion squares the length conversion factor.

A square with 2 cm edges has area $4\,\mathrm{cm^2}=4\cdot10^{-4}\,\mathrm{m^2}$ , not 0.04 m². $1\,\mathrm{kPa}=1000\,\mathrm{Pa}$ and $1\,\mathrm{hPa}=100\,\mathrm{Pa}$ . A value in bar specifies the unit but not whether it is absolute or gauge pressure.

F_\perp=pA,\qquad A=\frac{F_\perp}{p}

Solving for area requires p > 0 for a positive pressure magnitude.

How do absolute pressure, gauge pressure and net force differ?

Absolute pressure is referenced to vacuum. Gauge pressure is the difference from a stated reference, often ambient pressure. A negative gauge pressure means an absolute pressure below ambient, not necessarily a negative absolute pressure. The ideal gas law requires absolute pressure; a gauge reading relative to ambient must first be combined with its reference pressure.[2][4]

p_{\mathrm{gauge}}=p_{\mathrm{abs}}-p_{\mathrm{ref}},\qquad F_{\mathrm{net}}=(p_1-p_2)A

For an ideal flat piston with equal effective area on both sides, use the pressure difference.

Original check: 250 kPa absolute inside and 100 kPa outside produce net pressure force $150000\cdot0.0008\,\mathrm N=120\,\mathrm N$ on 8 cm². Multiplying only inside pressure gives 200 N and omits the opposing external 80 N. Piston weight and friction would add other forces.

Angled force and hydraulics: calculating normal force and stroke

Average loading of a flat contact area

Given: Force magnitude 150 N at 60° to the surface normal; loaded area 3 cm².
Find: Find average normal compressive loading.

Approach: Only the force component normal to the loaded area contributes to average normal compressive loading.
$\begin{aligned}F_{\perp}&=F\cos\theta\\\overline p&=\frac{F_{\perp}}{A}\end{aligned}$
Find the normal component.
$F_\perp=150\cos60^\circ=75\,\mathrm N$
Convert area to square metres.
$A=3\cdot10^{-4}\,\mathrm{m^2}$
Form the ratio.
$\overline p=\frac{75}{3\cdot10^{-4}}=250000\,\mathrm{Pa}$

Result: Average normal loading: 250 kPa = 2.5 bar.

Check: The normal component is 75 N. The tangential component is not counted as pressure; the average does not determine local distribution.

Ideal hydraulics with two pistons

Given: Input force 120 N; areas A₁ = 3 cm² and A₂ = 45 cm²; equal piston elevation and external pressure. Neglect friction, piston weight and storage changes; incompressible liquid. Input stroke 12 cm.
Find: Find output force and output stroke.

Approach: With equal piston elevation and external pressure, combine the transmitted pressure difference with incompressible volume conservation.
$\begin{aligned}\Delta p&=\frac{F_1}{A_1}=\frac{F_2}{A_2}\\A_1s_1&=A_2s_2\end{aligned}$
Find the generated pressure difference.
$\Delta p=\frac{120}{3\cdot10^{-4}}=400000\,\mathrm{Pa}$
Multiply this pressure by output area.
$F_2=400000\cdot45\cdot10^{-4}=1800\,\mathrm N$
Apply displaced-volume conservation.
$\begin{aligned}A_1s_1&=A_2s_2\\{} s_2&=\frac{12\cdot3}{45}=0.8\,\mathrm{cm}\end{aligned}$

Result: Output force is 1800 N; output stroke is 0.8 cm.

Check: Input work 120·0.12 = 14.4 J equals output work 1800·0.008 = 14.4 J. Force multiplication does not multiply energy.

How do hydraulics multiply force without gaining energy?

An enclosed static fluid transmits an added pressure change. This does not make absolute pressure identical at every elevation: hydrostatic differences remain. The simple force ratio $\frac{F_2}{F_1}=\frac{A_2}{A_1}$ therefore needs matching conditions, particularly piston elevations and the relevant pressure differences. Mechanical work explains the stroke exchange in the existing ideal example: greater output force accompanies a shorter output displacement when volume is conserved and losses are neglected.[3]

Fluid compressibility or a trapped gas bubble may initially store energy and delay piston motion. Seal friction and pressure losses change useful force. The ideal calculation is a checkable approximation; a real actuator needs its operating data.

Area alone does not determine load capacity

Larger support area reduces average normal loading at fixed force. Whether a surface supports the load also depends on material behaviour; $\overline p=\frac{F_\perp}{A}$ supplies no universal threshold for snow, soil or tissue. Area may change with load on a deformable surface.

For curved surfaces, integrate pressure forces as vectors. Equal pressure on opposite sides can give small net force despite large individual forces. Treating pressure as one directed force loses both area dependence and this force balance. Hydrostatic pressure instead describes depth dependence in a stationary fluid under gravity; the simple linear-height expression assumes constant density and approximately constant g.[1]

Five problems with solutions

Calculate · Foundation

A normal force of 360 N acts over 12 cm². Find average loading.

Show hint

12 cm² = 0.0012 m².

Learning goal: Find average normal loading with area conversion.

Show solution and calculation

Approach: Divide normal force by total contact area, converting area to square metres using the squared conversion factor first.
Result: 300 kPa average normal loading.

Check: This pressure times 0.0012 m² gives 360 N again. The average does not determine a local maximum.

\begin{aligned}p&=\frac{F_\perp}{A}\\A&=12\cdot10^{-4}\,\mathrm{m^2}\\p&=\frac{360}{0.0012}\,\mathrm{Pa}=300000\,\mathrm{Pa}=300\,\mathrm{kPa}\end{aligned}

[1]

Calculate · Application

An ideal piston of effective area 20 cm² has 180 kPa absolute inside and 100 kPa outside. Find net pressure force.

Show hint

Use the pressure difference.

Learning goal: Find piston net force from pressure difference.

Show solution and calculation

Approach: The absolute pressures act on opposite piston faces.
Subtract them first, then multiply by effective area.
Result: 160 N toward the lower-pressure side.

Check: The inner force is 360 N and the outer force 200 N; their difference confirms the net value. Additional friction or weight forces are outside the stated ideal model.

\begin{aligned}F_{\mathrm{net}}&=(p_{\mathrm{in}}-p_{\mathrm{out}})A\\&=(180000-100000)\cdot0.002\,\mathrm N\\&=160\,\mathrm N\end{aligned}

[1]

Rearrange · Foundation

What area distributes a normal 600 N force at an average 15 kPa?

Show hint

Convert kilopascals to pascals.

Learning goal: Infer contact area from force and mean pressure.

Show solution and calculation

Approach: Rearrange the definition for area and convert kilopascals to pascals.
Result: 0.04 m², or 400 cm².

Check: Multiplying by 15 000 Pa recovers the given 600 N. This is average loading; an uneven contact distribution may produce higher local values.

\begin{aligned}A&=\frac{F_\perp}{p}\\&=\frac{600\,\mathrm N}{15000\,\mathrm{Pa}}=0.04\,\mathrm{m^2}\\&=400\,\mathrm{cm^2}\end{aligned}

[1]

Interpret · Application

A flat contact consists of two equal 1 cm² regions with uniform normal loadings of 100 and 300 kPa. Find total normal force and average loading. Can the average rule out a local maximum of 300 kPa?

Show hint

Sum forces over regions before dividing by total area.

Learning goal: Distinguish average contact loading from local maxima.

Show solution and calculation

Approach: $F_\perp=\sum_i p_iA_i$ gives 10 N+30 N=40 N.
Average loading over 2 cm² is 200 kPa.
The average is below the actual maximum and does not determine the distribution.

Check: 200 kPa times 2 cm² also gives 40 N. Many other distributions could yield that total force; a local material limit therefore requires checking local loading.

\begin{aligned}F_\perp&=100000\cdot10^{-4}+300000\cdot10^{-4}=40\,\mathrm N\\\overline p&=\frac{40\,\mathrm N}{2\cdot10^{-4}\,\mathrm{m^2}}=200\,\mathrm{kPa}\end{aligned}

[1]

Error analysis · Challenge

A closed sphere of radius R is surrounded by uniform static pressure p. A solution claims a resultant pressure force p×4πR² in one direction. No other forces or pressure gradients are present. Why is net force zero despite nonzero pressure?

Show hint

Local forces point inward along different normals.

Learning goal: Sum pressure forces vectorially over a curved closed surface.

Show solution and calculation

Approach: each element experiences $-p\hat{\vec n}\,\mathrm dA$ .
Diametrically opposite elements have equal opposite forces, so the vector sum vanishes. p×4πR² sums magnitudes rather than a force component.

Check: the integral of outward normals over a closed surface is zero. Pressure can load a shell without driving centre-of-mass translation. A hydrostatic gradient would be a different assumption.

\mathbf F=-p\oint_{\partial V}\hat{\mathbf n}\,dA=\mathbf0

[1]

Key points

Mean pressure uses the force component normal to the selected area. At the same force, a smaller area gives a higher mean. Local pressure distribution and total contact force are different quantities; also check pressure reference and convert area before substituting.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made pressure calculation, rearrangements, square-area units, pressure differences and hydraulic stroke accessible through learning questions; retained contact/hydraulic examples and added contextual work, absolute-pressure gas-law and hydrostatic-pressure links.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the learning introduction and solution paths; independently checked model assumptions and calculations against primary sources.

Cite this reference page

Quanta. Pressure. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/druck

When citing a specific scientific statement, also cite the specialist source linked to that statement.

Suggest a correction with a reference

Frequently asked questions about Pressure

How do you calculate pressure with

p =\frac{ F}{A}

Divide the perpendicular force in newtons by the area in square metres; the result is the pressure in pascals. Example: a person with a weight of $F = 600 N$ stands on heels with a combined $A = 0.003 m^{2}$ : $p =\frac{ 600}{0.003} = 200\,000 Pa = 2 bar$ . The most common mistake is the area unit: $1 m^{2} = 10\,000 cm^{2}$ , so one square centimetre is $10^{-4} m^{2}$ . For everyday values remember $1 bar = 100\,000 Pa = 1\,000$ hPa; air pressure is about 1,013 hPa. Crucially, only the perpendicular force component counts; slanted forces must be decomposed first.

Why do you not sink in with snowshoes but do with heels?+

At the same normal load, larger contact area reduces average compressive loading. An assumed 700 N over $0.25 m^{2}$ gives $\frac{F}{A} = 2800 Pa$ . The same load over $1 cm^{2}$ gives seven million pascals, 2500 times as much. This calculation explains area dependence but does not prove that a particular snow layer supports the load. Snow density, strength, temperature and load distribution also matter; contact area may change during sinking. Allocate the load to the shoes or heels actually bearing it. The ratio is an average, not necessarily uniform local loading and not a universal threshold for damage.

How does a hydraulic press work?+

An ideal hydraulic system transmits an added pressure change through an enclosed static fluid. With equal piston elevations and external pressures, negligible piston weights and negligible friction, $\frac{F_{1}}{A_{1}} =\frac{ F_{2}}{A_{2}}$ . A larger output area therefore gives more force. For 120 N on $3 cm^{2}$ and an output area of $45 cm^{2}$ , ideal output is 1800 N. Displaced volume stays equal, so output travel is only one fifteenth of input travel. Pascal’s principle does not imply identical absolute pressure at all elevations: hydrostatic differences remain. Real systems also need pressure losses, seal friction and possible energy storage from compressibility considered.

What is hydrostatic pressure and how do you calculate it?+

Hydrostatic pressure arises from the weight of the liquid column above you and grows linearly with depth: $p = \rho{}\cdot{}g\cdot{}h$ . In water $(\rho{} =\frac{ 1\,000 kg}{m^{3}})$ about 1 bar is added per 10 m of depth: $p = 1\,000 \times{} 9.81 \times{} 10 \approx{} 98\,100 Pa$ . A diver at 20 m thus experiences about 2 bar of hydrostatic pressure plus 1 bar of air pressure from above, roughly 3 bar in total. Remarkable is what does not appear in the formula: the shape of the vessel and the amount of water. In a thin pipe the pressure at 10 m depth is the same as in the ocean (the hydrostatic paradox). Water towers rest on the same principle: the height of the reservoir sets the mains pressure in the houses.

Why do you not feel the air pressure of about 1 bar?+

Atmospheric pressure imposes normal loading on surfaces; it is not a single directed force on the entire body. Uniform external pressure on a closed body gives cancelling force vectors. That alone does not prevent deformation: internal pressure and mechanical strength also matter. A gas-filled space can respond sensitively to a pressure difference, while liquids and tissue transmit forces differently. Changes in ambient pressure become noticeable at the middle ear when equalisation is delayed. In an ideal piston model, net force equals pressure difference times area. Identify which internal and external pressures are involved rather than assuming identical pressure everywhere inside a body. The body’s response also depends on its structure.

Quanta exam set

Curated exam set for "Pressure":

Question (front)

Which formula describes Pressure?

Answer in your set

Question (front)

How do you rearrange Pressure for Force from pressure and area?

Answer in your set

Question (front)

Which common mistake happens with Pressure?

Answer in your set

+ 7 more cards: units, variables, derivation, example, exam task

The formula set contains 10 cards. After importing, review them and start practicing; your ratings determine later review intervals.

Create flashcards or a quiz I already have an account

Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §14.1 Fluids, Density, and Pressure

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

Spaced-repetition apps

Compare scheduling workflows

Study evidence

Read methods, results and limits

Compare flashcard apps

FSRS support and study features

Create a formula exam set

From Essential in Quanta

Retain Pressure for exams

Create a curated FSRS exam set for p = F/A: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Create flashcards or a quiz Already registered? Continue directly

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Quanta Study editorial team

Pressure: normal force per area

Pressure describes normal force per area; with nonuniform loading

\frac{F}{A}

gives only average normal compressive loading.

BasicExam-relevantLower secondaryUpper secondary

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

p

Pressure

F

Force acting perpendicularly

A

Area on which the force acts

m²

Derivation & background

Exam blueprint

Validity range

Derivation steps

Normal pressure loading is defined per unit area; total force follows by summing area elements.

1Specify a planar area A and normal force F⊥.
2Do not count tangential force as pressure.
3A small element obeys dF⊥ = p dA.
4Integrate over the surface: F⊥ = ∫p dA.
5Uniform pressure gives F⊥ = pA and thus $p = F$ ⊥/A.
6Varying pressure makes the ratio an area average; a piston uses pressure difference between its sides.

How do you calculate with Pressure?

Here is how to work through a typical Pressure (p = F/A) task step by step:

Task

An ideal hydraulic cylinder of effective area $A = 0.02 m^{2}$ has a pressure difference of $5\times{}10^{5} Pa$ across its sides. Find net pressure force.

Solution path

Approach: $F_{\mathrm{net}}=\Delta pA$ .
Substitution: $F_{\mathrm{net}}=5\cdot10^5\cdot0.02\,\mathrm N=10000\,\mathrm N$ , toward the lower-pressure side.

Check: $\frac{F_{\mathrm{net}}}{A}=500000\,\mathrm{Pa}$ . Actual lifting capacity additionally depends on weight, friction and the system.

Task

At what water depth is the hydrostatic pressure $1 bar (\rho{} =\frac{ 1\,000 kg}{m^{3}})$ ? Use $g =\frac{ 9.81 m}{s^{2}}$ .

Solution path

$h=\frac{p}{\rho g}=\frac{100000\,\mathrm{Pa}}{1000\,\mathrm{kg\,m^{-3}}\cdot9.81\,\mathrm{m\,s^{-2}}}\approx10.2\,\mathrm m$ .

How do you calculate pressure from normal force and area?

\overline p=\frac{F_\perp}{A},\qquad F_\perp=\int_Ap\,\mathrm dA

For varying pressure, F⊥/A is an area average. This scalar integral assumes a planar surface.

Smaller area, greater average loading

QUANTA STUDYquanta-study.de

Free with a Quanta account. Your download stays ready after registration.

How do you rearrange for force or area?

F_\perp=p\int_A\mathrm dA=pA\quad\Longrightarrow\quad p=\frac{F_\perp}{A}

The compact expression gives a uniform pressure only when the distribution is uniform.

Converting pascals, bar and square centimetres correctly

1\,\mathrm{Pa}=1\,\mathrm{N/m^2},\quad1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2},\quad1\,\mathrm{bar}=10^5\,\mathrm{Pa}

Area conversion squares the length conversion factor.

F_\perp=pA,\qquad A=\frac{F_\perp}{p}

Solving for area requires p > 0 for a positive pressure magnitude.

How do absolute pressure, gauge pressure and net force differ?

p_{\mathrm{gauge}}=p_{\mathrm{abs}}-p_{\mathrm{ref}},\qquad F_{\mathrm{net}}=(p_1-p_2)A

For an ideal flat piston with equal effective area on both sides, use the pressure difference.

Angled force and hydraulics: calculating normal force and stroke

Average loading of a flat contact area

Given: Force magnitude 150 N at 60° to the surface normal; loaded area 3 cm².
Find: Find average normal compressive loading.

Approach: Only the force component normal to the loaded area contributes to average normal compressive loading.
$\begin{aligned}F_{\perp}&=F\cos\theta\\\overline p&=\frac{F_{\perp}}{A}\end{aligned}$
Find the normal component.
$F_\perp=150\cos60^\circ=75\,\mathrm N$
Convert area to square metres.
$A=3\cdot10^{-4}\,\mathrm{m^2}$
Form the ratio.
$\overline p=\frac{75}{3\cdot10^{-4}}=250000\,\mathrm{Pa}$

Result: Average normal loading: 250 kPa = 2.5 bar.

Check: The normal component is 75 N. The tangential component is not counted as pressure; the average does not determine local distribution.

Ideal hydraulics with two pistons

Given: Input force 120 N; areas A₁ = 3 cm² and A₂ = 45 cm²; equal piston elevation and external pressure. Neglect friction, piston weight and storage changes; incompressible liquid. Input stroke 12 cm.
Find: Find output force and output stroke.

Approach: With equal piston elevation and external pressure, combine the transmitted pressure difference with incompressible volume conservation.
$\begin{aligned}\Delta p&=\frac{F_1}{A_1}=\frac{F_2}{A_2}\\A_1s_1&=A_2s_2\end{aligned}$
Find the generated pressure difference.
$\Delta p=\frac{120}{3\cdot10^{-4}}=400000\,\mathrm{Pa}$
Multiply this pressure by output area.
$F_2=400000\cdot45\cdot10^{-4}=1800\,\mathrm N$
Apply displaced-volume conservation.
$\begin{aligned}A_1s_1&=A_2s_2\\{} s_2&=\frac{12\cdot3}{45}=0.8\,\mathrm{cm}\end{aligned}$

Result: Output force is 1800 N; output stroke is 0.8 cm.

Check: Input work 120·0.12 = 14.4 J equals output work 1800·0.008 = 14.4 J. Force multiplication does not multiply energy.

How do hydraulics multiply force without gaining energy?

Area alone does not determine load capacity

Five problems with solutions

Calculate · Foundation

A normal force of 360 N acts over 12 cm². Find average loading.

Show hint

12 cm² = 0.0012 m².

Learning goal: Find average normal loading with area conversion.

Show solution and calculation

Approach: Divide normal force by total contact area, converting area to square metres using the squared conversion factor first.
Result: 300 kPa average normal loading.

Check: This pressure times 0.0012 m² gives 360 N again. The average does not determine a local maximum.

\begin{aligned}p&=\frac{F_\perp}{A}\\A&=12\cdot10^{-4}\,\mathrm{m^2}\\p&=\frac{360}{0.0012}\,\mathrm{Pa}=300000\,\mathrm{Pa}=300\,\mathrm{kPa}\end{aligned}

[1]

Calculate · Application

An ideal piston of effective area 20 cm² has 180 kPa absolute inside and 100 kPa outside. Find net pressure force.

Show hint

Use the pressure difference.

Learning goal: Find piston net force from pressure difference.

Show solution and calculation

Approach: The absolute pressures act on opposite piston faces.
Subtract them first, then multiply by effective area.
Result: 160 N toward the lower-pressure side.

Check: The inner force is 360 N and the outer force 200 N; their difference confirms the net value. Additional friction or weight forces are outside the stated ideal model.

\begin{aligned}F_{\mathrm{net}}&=(p_{\mathrm{in}}-p_{\mathrm{out}})A\\&=(180000-100000)\cdot0.002\,\mathrm N\\&=160\,\mathrm N\end{aligned}

[1]

Rearrange · Foundation

What area distributes a normal 600 N force at an average 15 kPa?

Show hint

Convert kilopascals to pascals.

Learning goal: Infer contact area from force and mean pressure.

Show solution and calculation

Approach: Rearrange the definition for area and convert kilopascals to pascals.
Result: 0.04 m², or 400 cm².

Check: Multiplying by 15 000 Pa recovers the given 600 N. This is average loading; an uneven contact distribution may produce higher local values.

\begin{aligned}A&=\frac{F_\perp}{p}\\&=\frac{600\,\mathrm N}{15000\,\mathrm{Pa}}=0.04\,\mathrm{m^2}\\&=400\,\mathrm{cm^2}\end{aligned}

[1]

Interpret · Application

A flat contact consists of two equal 1 cm² regions with uniform normal loadings of 100 and 300 kPa. Find total normal force and average loading. Can the average rule out a local maximum of 300 kPa?

Show hint

Sum forces over regions before dividing by total area.

Learning goal: Distinguish average contact loading from local maxima.

Show solution and calculation

Approach: $F_\perp=\sum_i p_iA_i$ gives 10 N+30 N=40 N.
Average loading over 2 cm² is 200 kPa.
The average is below the actual maximum and does not determine the distribution.

Check: 200 kPa times 2 cm² also gives 40 N. Many other distributions could yield that total force; a local material limit therefore requires checking local loading.

\begin{aligned}F_\perp&=100000\cdot10^{-4}+300000\cdot10^{-4}=40\,\mathrm N\\\overline p&=\frac{40\,\mathrm N}{2\cdot10^{-4}\,\mathrm{m^2}}=200\,\mathrm{kPa}\end{aligned}

[1]

Error analysis · Challenge

A closed sphere of radius R is surrounded by uniform static pressure p. A solution claims a resultant pressure force p×4πR² in one direction. No other forces or pressure gradients are present. Why is net force zero despite nonzero pressure?

Show hint

Local forces point inward along different normals.

Learning goal: Sum pressure forces vectorially over a curved closed surface.

Show solution and calculation

Approach: each element experiences $-p\hat{\vec n}\,\mathrm dA$ .
Diametrically opposite elements have equal opposite forces, so the vector sum vanishes. p×4πR² sums magnitudes rather than a force component.

Check: the integral of outward normals over a closed surface is zero. Pressure can load a shell without driving centre-of-mass translation. A hydrostatic gradient would be a different assumption.

\mathbf F=-p\oint_{\partial V}\hat{\mathbf n}\,dA=\mathbf0

[1]

Key points

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made pressure calculation, rearrangements, square-area units, pressure differences and hydraulic stroke accessible through learning questions; retained contact/hydraulic examples and added contextual work, absolute-pressure gas-law and hydrostatic-pressure links.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the learning introduction and solution paths; independently checked model assumptions and calculations against primary sources.

Cite this reference page

Quanta. Pressure. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/druck

When citing a specific scientific statement, also cite the specialist source linked to that statement.

Suggest a correction with a reference

Frequently asked questions about Pressure

How do you calculate pressure with

p =\frac{ F}{A}

Why do you not sink in with snowshoes but do with heels?+

How does a hydraulic press work?+

What is hydrostatic pressure and how do you calculate it?+

Why do you not feel the air pressure of about 1 bar?+

Pressure: normal force per area

Variables & units

Pressure calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Pressure and fluids

Worked example

Applications

How do you calculate with Pressure?

How do you calculate pressure from normal force and area?

Smaller area, greater average loading

How do you rearrange for force or area?

Converting pascals, bar and square centimetres correctly

How do absolute pressure, gauge pressure and net force differ?

Angled force and hydraulics: calculating normal force and stroke

Average loading of a flat contact area

Ideal hydraulics with two pistons

How do hydraulics multiply force without gaining energy?

Area alone does not determine load capacity

Five problems with solutions

A normal force of 360 N acts over 12 cm2. Find average loading.

An ideal piston of effective area 20 cm2 has 180 kPa absolute inside and 100 kPa outside. Find net pressure force.

What area distributes a normal 600 N force at an average 15 kPa?

A flat contact consists of two equal 1 cm2 regions with uniform normal loadings of 100 and 300 kPa. Find total normal force and average loading. Can the average rule out a local maximum of 300 kPa?

A closed sphere of radius R is surrounded by uniform static pressure p. A solution claims a resultant pressure force p×4πR2 in one direction. No other forces or pressure gradients are present. Why is net force zero despite nonzero pressure?

Key points

Sources for this explanation

Frequently asked questions about Pressure

Subject references and further reading

More formulas: Mechanics

Retain Pressure for exams

Pressure: normal force per area

Variables & units

Pressure calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Pressure and fluids

Worked example

Applications

How do you calculate with Pressure?

How do you calculate pressure from normal force and area?

Smaller area, greater average loading

How do you rearrange for force or area?

Converting pascals, bar and square centimetres correctly

How do absolute pressure, gauge pressure and net force differ?

Angled force and hydraulics: calculating normal force and stroke

Average loading of a flat contact area

Ideal hydraulics with two pistons

How do hydraulics multiply force without gaining energy?

Area alone does not determine load capacity

Five problems with solutions

A normal force of 360 N acts over 12 cm2. Find average loading.

An ideal piston of effective area 20 cm2 has 180 kPa absolute inside and 100 kPa outside. Find net pressure force.

What area distributes a normal 600 N force at an average 15 kPa?

A flat contact consists of two equal 1 cm2 regions with uniform normal loadings of 100 and 300 kPa. Find total normal force and average loading. Can the average rule out a local maximum of 300 kPa?

A closed sphere of radius R is surrounded by uniform static pressure p. A solution claims a resultant pressure force p×4πR2 in one direction. No other forces or pressure gradients are present. Why is net force zero despite nonzero pressure?

Key points

Sources for this explanation

Frequently asked questions about Pressure

Subject references and further reading

More formulas: Mechanics

Retain Pressure for exams

A normal force of 360 N acts over 12 cm². Find average loading.

An ideal piston of effective area 20 cm² has 180 kPa absolute inside and 100 kPa outside. Find net pressure force.

A flat contact consists of two equal 1 cm² regions with uniform normal loadings of 100 and 300 kPa. Find total normal force and average loading. Can the average rule out a local maximum of 300 kPa?

A closed sphere of radius R is surrounded by uniform static pressure p. A solution claims a resultant pressure force p×4πR² in one direction. No other forces or pressure gradients are present. Why is net force zero despite nonzero pressure?

A normal force of 360 N acts over 12 cm². Find average loading.

An ideal piston of effective area 20 cm² has 180 kPa absolute inside and 100 kPa outside. Find net pressure force.

A flat contact consists of two equal 1 cm² regions with uniform normal loadings of 100 and 300 kPa. Find total normal force and average loading. Can the average rule out a local maximum of 300 kPa?

A closed sphere of radius R is surrounded by uniform static pressure p. A solution claims a resultant pressure force p×4πR² in one direction. No other forces or pressure gradients are present. Why is net force zero despite nonzero pressure?