COLLEGE PHYSICS • FLUIDS & HYDROSTATICS

Fluids and Newton's Laws

How Newton's classical laws govern pressure, buoyancy, and fluid behavior at rest and in motion.

Historical Context & Motivation

The study of fluids represents one of the oldest and most consequential threads in the history of physics, stretching from the legendary bath of Archimedes to the rigorous mathematical formulations of Euler and Navier. At the heart of fluid mechanics lies a deceptively simple idea: a fluid is simply a continuous distribution of matter, and every parcel of that matter must obey Newton's laws of motion. When we apply Newton's second law (F = ma) to infinitesimal fluid elements, we recover the foundational equations of hydrostatics, buoyancy, and eventually the full Navier–Stokes equations that describe virtually all fluid flow on Earth.

For centuries, engineers and natural philosophers struggled to reconcile the behavior of water, air, and other fluids with the particle-based mechanics Newton had so elegantly formalized for rigid bodies. The breakthrough came from recognizing that fluids differ from solids not in their obedience to Newton's laws, but in how they transmit and redistribute forces internally. A solid resists shear stress at rest; a fluid does not—it flows. This single distinction, combined with the principle that every fluid element must be in mechanical equilibrium (or accelerating according to F = ma), generates the entire theoretical edifice of fluid statics and dynamics.

~250 BCE
Archimedes' Principle
Archimedes of Syracuse discovered the law of buoyancy: a body immersed in a fluid experiences an upward force equal to the weight of the displaced fluid. This was the first quantitative application of force balance to fluids.
1586
Stevin's Hydrostatic Paradox
Simon Stevin demonstrated that the pressure at the bottom of a container depends only on the height and density of the fluid, not on the shape of the container—an insight that foreshadowed Pascal's law.
1653
Pascal's Law
Blaise Pascal formalized the principle that pressure applied to a confined fluid is transmitted undiminished throughout the fluid, laying the groundwork for hydraulic systems.
1687
Newton's Principia
Isaac Newton published his three laws of motion and introduced the concept of a Newtonian fluid—one in which shear stress is proportional to the velocity gradient—connecting fluid behavior to his broader mechanical framework.
1757
Euler's Equations of Fluid Motion
Leonhard Euler applied Newton's second law to inviscid fluid elements, deriving the Euler equations—the first complete set of differential equations governing fluid flow and the precursor to the Navier–Stokes equations.

This lesson addresses a central question: How do Newton's laws—originally stated for point particles—extend to continuous fluids? We will develop the key results of hydrostatics (pressure variation with depth, Pascal's law, Archimedes' principle) by applying force and momentum balance to fluid parcels, and then see how these ideas generalize to moving fluids. By the end, you will understand that every equation in introductory fluid mechanics is simply Newton's second law, rewritten for a medium that flows.

Core Principles & Definitions

Before we can apply Newton's laws to fluids, we must establish the key quantities and assumptions that distinguish fluid mechanics from particle mechanics. A fluid is any substance that deforms continuously under an applied shear stress, no matter how small. Both liquids and gases qualify. In the continuum approximation, we treat the fluid as a smooth, continuously distributed medium rather than tracking individual molecules, which allows us to define macroscopic fields like pressure and density at every point in space.

1

Pressure (P)

The normal force per unit area that a fluid exerts on any surface, including internal surfaces within the fluid. Pressure is a scalar quantity; at any point in a fluid at rest, it acts equally in all directions (Pascal's isotropy). SI unit: pascal (Pa = N/m²).
2

Density (ρ)

Mass per unit volume of the fluid, ρ = m/V. For incompressible fluids (most liquids under normal conditions), density is treated as constant. Density connects the mass of a fluid element to its volume, enabling application of F = ma to specific parcels.
3

The Fluid Element

An imaginary infinitesimal parcel of fluid large enough to contain many molecules (validating the continuum assumption) but small enough that pressure and density are uniform across it. Newton's second law is applied to this element as a free body.
4

Hydrostatic Equilibrium

A fluid is in hydrostatic equilibrium when every fluid element has zero acceleration—the net force on each element vanishes. This condition, which is simply Newton's first law applied to a fluid parcel, leads directly to the equation relating pressure to depth.
5

Buoyant Force

The net upward force exerted by a fluid on an immersed or floating object. Archimedes' principle states that this force equals the weight of the displaced fluid, a result derivable from integrating pressure forces over the object's surface—again, a direct consequence of Newton's laws.
KEY TAKEAWAY
Think of a fluid as a crowd of people in a packed stadium. No individual can resist being pushed sideways—they simply move. A solid is like that same crowd locked arm-in-arm: it resists deformation. When you apply Newton's second law to each person (fluid element) individually—accounting for the pushes from their neighbors (pressure) and gravity—you recover all of hydrostatics. The equations aren't new physics; they are Newton's laws wearing a different outfit.

Visualizing Pressure & Force Balance

The most illuminating way to understand how Newton's laws generate hydrostatic pressure is to examine a free-body diagram of a thin, horizontal fluid slab within a container. The following diagram isolates a rectangular fluid element of cross-sectional area A and infinitesimal thickness dy, located at depth y below the surface. Three forces act on this element: the pressure from the fluid above pushing down on its top face, the pressure from the fluid below pushing up on its bottom face, and the element's own weight pulling it downward.

A thin fluid slab of area A and thickness dy at depth y. The green arrow represents the upward pressure force on the bottom face, the red arrow shows the downward pressure force on the top face, and the amber arrow is the element's weight. Setting the net force to zero (Newton's first law for equilibrium) yields the fundamental hydrostatic relation dP/dy = −ρg.

In the diagram above, the key physical insight is that the pressure on the bottom face exceeds the pressure on the top face by exactly the weight of the fluid slab. This is Newton's first law (ΣF = 0 for a fluid element at rest) written in differential form. If the fluid were accelerating—say, in a rocket or a centrifuge—we would instead write ΣF = ma, and the pressure distribution would change accordingly. The equation dP/dy = −ρg is not an independent postulate; it is a direct consequence of applying force balance to a continuous medium under gravity.

Mathematical Framework

We now formalize the force-balance argument introduced visually in Section 3. Consider a fluid element of cross-sectional area A and infinitesimal height dy, situated at a depth y measured downward from the free surface. Three forces act on the element in the vertical direction: the downward pressure force on the top face, the upward pressure force on the bottom face, and the downward gravitational force. By choosing the positive y-axis pointing downward (into the fluid), Newton's second law for the element in static equilibrium becomes:

HYDROSTATIC DIFFERENTIAL EQUATION
dP/dy = ρg
Here P is the absolute pressure, y is depth measured downward from the surface, ρ is the fluid density (kg/m³), and g is the gravitational acceleration (9.81 m/s²). Note: with y measured downward, dP/dy is positive because pressure increases with depth.

For an incompressible fluid (constant ρ), we integrate from the surface (y = 0, where P = P₀) to an arbitrary depth h to obtain the most widely used result in hydrostatics:

PRESSURE AT DEPTH (GAUGE + ATMOSPHERIC)
P = P₀ + ρgh
P₀ is the atmospheric (or surface) pressure, and h is the depth below the free surface. The term ρgh is called the gauge pressure—the pressure in excess of atmospheric.

Deriving Archimedes' Principle from Pressure Integration

Consider a body of arbitrary shape submerged in a fluid. The fluid exerts a pressure force on every infinitesimal patch dA of the body's surface. The horizontal components of these pressure forces cancel by symmetry for any closed surface. The vertical (buoyant) component is found by integrating the pressure difference between the bottom and top surfaces of the body. Because the pressure difference across any vertical slice of height Δh is ρgΔh, the net upward force is ρg times the total volume V of fluid displaced by the body. This is precisely Archimedes' principle:

ARCHIMEDES' PRINCIPLE (BUOYANT FORCE)
F_b = ρ_fluid × g × V_displaced
Fb is the buoyant force (upward), ρfluid is the density of the surrounding fluid, and Vdisplaced is the volume of fluid displaced by the object. This result follows entirely from Newton's laws applied to the fluid around the object.
PASCAL'S LAW
ΔP₁ = ΔP₂ → F₁/A₁ = F₂/A₂
A change in pressure applied to any point in an enclosed, incompressible fluid is transmitted undiminished to every other point. This principle underlies hydraulic presses and braking systems: a small force on a small piston generates a large force on a large piston, with F₂ = F₁ × (A₂/A₁).

Applications & Classification of Fluid Forces

Newton's laws applied to fluids produce a rich family of phenomena that can be classified by whether the fluid is at rest (hydrostatics) or in motion (hydrodynamics), and by whether the fluid is compressible or incompressible. The diagram below maps the major branches of introductory fluid mechanics, showing how each topic traces back to Newton's second law applied to fluid elements.

A concept map showing how Newton's second law branches into hydrostatics (a = 0) and hydrodynamics (a ≠ 0). Each sub-topic—pressure at depth, Pascal's law, Archimedes' principle, continuity, Bernoulli's equation, and the Euler/Navier–Stokes equations—derives from applying ΣF = ma to fluid elements under different conditions.

The left branch of the diagram—hydrostatics—covers the cases in this lesson where every fluid element has zero acceleration. The right branch—hydrodynamics—previews topics you will encounter in subsequent chapters: the continuity equation (conservation of mass for flowing fluid), Bernoulli's equation (energy conservation along a streamline, derivable from work-energy considerations on a fluid parcel), and ultimately the Euler and Navier–Stokes equations (the full vector form of F = ma for a fluid continuum, including viscous stresses). The key message is that none of these results require new physical postulates beyond Newton's laws and the constitutive properties of the fluid.

Mapping fluid-mechanics results to their Newtonian origins
PrincipleNewton's Law AppliedKey ConditionResult
Pressure at depth1st Law (ΣF = 0)Static, incompressible fluidP = P₀ + ρgh
Pascal's law1st Law (ΣF = 0)Enclosed, incompressible fluidF₁/A₁ = F₂/A₂
Archimedes' principle1st Law (pressure integration)Body immersed in fluidF_b = ρ_fluid g V_disp
Bernoulli's equation2nd Law (ΣF = ma)Steady, inviscid flow along streamlineP + ½ρv² + ρgy = const
Euler equations2nd Law (full vector form)Inviscid fluid, any flowρ(Dv/Dt) = −∇P + ρg

Worked Example: Buoyant Force on a Submerged Sphere

A solid aluminum sphere of radius r = 0.10 m and density ρAl = 2700 kg/m³ is fully submerged in freshwater (ρw = 1000 kg/m³). Determine (a) the buoyant force acting on the sphere, (b) the apparent weight of the sphere while submerged, and (c) the acceleration of the sphere if released from rest.

Buoyancy and Apparent Weight of an Aluminum Sphere
1
Step 1 — Compute the Volume of the SphereThe volume of a sphere is V = (4/3)πr³. Substituting r = 0.10 m: V = (4/3) × π × (0.10)³ = (4/3) × π × 1.0 × 10⁻³ m³
V ≈ 4.19 × 10⁻³ m³
2
Step 2 — Calculate the Buoyant Force (Archimedes' Principle)By Archimedes' principle, Fb = ρw × g × Vdisplaced. Since the sphere is fully submerged, Vdisplaced = V. Fb = 1000 × 9.81 × 4.19 × 10⁻³
Fb ≈ 41.1 N (upward)
3
Step 3 — Compute the True Weight of the SphereW = ρAl × g × V = 2700 × 9.81 × 4.19 × 10⁻³
W ≈ 110.9 N (downward)
4
Step 4 — Determine the Apparent Weight While SubmergedThe apparent weight is the net downward force: Wapp = W − Fb = 110.9 − 41.1
Wapp ≈ 69.8 N
5
Step 5 — Find the Acceleration if ReleasedApplying Newton's second law: ΣF = ma, where the net force is W − Fb = 69.8 N downward, and the mass m = ρAl × V = 2700 × 4.19 × 10⁻³ = 11.31 kg. a = (W − Fb) / m = 69.8 / 11.31
a ≈ 6.17 m/s² (downward) — the sphere sinks, but more slowly than free fall because buoyancy partially supports it.
💡 Physical Insight
Notice that the downward acceleration (6.17 m/s²) is less than g (9.81 m/s²). The ratio a/g = 1 − ρfluidobject = 1 − 1000/2700 ≈ 0.63. An object sinks when ρobject > ρfluid, floats when ρobject < ρfluid, and is neutrally buoyant when the densities are equal.

Strengths & Limitations of the Hydrostatic Model

The hydrostatic framework—Newton's laws applied to fluids at rest—is remarkably powerful for a wide range of engineering and scientific applications, but it rests on assumptions that break down under certain conditions. Understanding these boundaries is essential for knowing when to apply the simple P = P₀ + ρgh model and when more sophisticated treatments are required.

Evaluating the hydrostatic model derived from Newton's laws
StrengthsLimitations
Exact for any static, incompressible fluid under uniform gravity—no approximations involvedFails when the fluid is in motion (requires Bernoulli or Navier–Stokes extensions)
Pressure depends only on depth, not container shape—simplifies design of dams, tanks, and submerged structuresAssumes constant density; breaks down for gases at large height differences or compressible fluids under extreme pressures
Pascal's law enables enormous force multiplication in hydraulic systems with minimal moving partsIgnores surface tension effects, which become dominant at small scales (capillary tubes, droplets)
Archimedes' principle provides a universal criterion for floating/sinking based solely on density ratiosDoes not account for dynamic lift (e.g., airplane wings) or viscous drag—these require fluid dynamics
Directly derivable from Newton's first law with no empirical constants—fully predictive from first principlesNon-inertial reference frames (rotating or accelerating containers) require modified body-force terms
KEY TAKEAWAY
The hydrostatic equations are to fluid mechanics what statics is to solid mechanics: the simplest case (a = 0) within a broader Newtonian framework. Just as a structural engineer must check whether a bridge member is truly in static equilibrium before applying statics, a fluid engineer must verify that the fluid is genuinely at rest—or can be treated as nearly so—before using P = P₀ + ρgh. When the fluid moves, Newton's second law still applies, but the mathematics becomes richer (and often nonlinear).

Connection to Advanced Fluid Dynamics

Everything developed in this lesson—pressure variation with depth, Pascal's law, and Archimedes' principle—represents the static limit of a much broader theory. When fluid elements accelerate, Newton's second law in its full generality must be applied. Leonhard Euler formalized this for inviscid (frictionless) fluids, and Claude-Louis Navier and George Gabriel Stokes extended it to viscous fluids. The resulting Navier–Stokes equations are among the most important—and most challenging—equations in all of physics, governing phenomena from blood flow in arteries to turbulence behind aircraft.

From hydrostatics to the Navier–Stokes equations: same physics, greater generality
FeatureHydrostatics (This Lesson)Full Fluid Dynamics (Advanced)
Newton's law formΣF = 0 (first law, equilibrium)ΣF = ma (second law, general)
Governing equation∇P = ρg (vector form)ρ(Dv/Dt) = −∇P + μ∇²v + ρg
Velocity fieldv = 0 everywherev(x, y, z, t) — spatially and temporally varying
Forces consideredPressure gradients and gravityPressure gradients, gravity, and viscous (shear) stresses
Mathematical difficultyOrdinary differential equation (1D integration)Coupled nonlinear partial differential equations — analytical solutions rare
Typical applicationsDams, manometers, hydraulic lifts, submarinesPipe flow, aerodynamics, weather modeling, ocean currents

An intermediate result you will encounter soon is Bernoulli's equation: P + ½ρv² + ρgy = constant along a streamline. This equation is derived by applying Newton's second law (in the form of the work–energy theorem) to a fluid element moving along a streamline in steady, inviscid flow. Notice that when v = 0, Bernoulli's equation reduces to P + ρgy = constant, which is exactly the hydrostatic result P = P₀ + ρgh. The static case is always embedded within the dynamic theory as a special limit, underscoring the unity of Newton's framework across all of fluid mechanics.

Practice Problems

PROBLEM 1CONCEPTUAL
A container of water is placed on a scale. A steel ball suspended from a string is lowered into the water but does not touch the bottom. The string is held by a stand that does not rest on the scale. Does the scale reading increase, decrease, or stay the same compared to the container of water alone? Explain your reasoning using Newton's third law.
PROBLEM 2BASIC CALCULATION
A scuba diver descends to a depth of 25 m in seawater (ρ = 1025 kg/m³). Atmospheric pressure at the surface is 1.013 × 10⁵ Pa. What is the absolute pressure at this depth, and what is the gauge pressure?
PROBLEM 3INTERMEDIATE
A hydraulic lift has a small piston of diameter 4.0 cm and a large piston of diameter 24 cm. A mechanic pushes down on the small piston with a force of 150 N. (a) What force is exerted by the large piston? (b) If the mechanic pushes the small piston down by 30 cm, how far does the large piston rise? (c) Verify that work input equals work output.
PROBLEM 4APPLIED
An ice cube of mass 50 g (density 917 kg/m³) floats in a glass of freshwater (density 1000 kg/m³). (a) What fraction of the ice cube is submerged? (b) As the ice melts, does the water level in the glass rise, fall, or remain the same? Justify your answer using Newton's first law and Archimedes' principle.
PROBLEM 5CRITICAL THINKING
A closed container is completely filled with water and placed on a scale reading W₀. A ping-pong ball (mass m, volume V, with ρ_ball < ρ_water) is attached to a string that is fixed to the bottom of the container; the ball is fully submerged and the string is taut. Now the string is cut. After the ball reaches equilibrium floating at the surface, has the scale reading changed? Provide a rigorous argument based on Newton's laws applied to the entire system.

Lesson Summary

This lesson demonstrated that the foundational results of fluid statics are not independent postulates but direct consequences of Newton's laws of motion applied to continuous media. By isolating an infinitesimal fluid element and demanding static equilibrium (ΣF = 0), we derived the hydrostatic pressure equation P = P₀ + ρgh, which shows that pressure increases linearly with depth in an incompressible fluid. Pascal's law follows from the isotropy of pressure in a static fluid, enabling hydraulic force multiplication. Archimedes' principle—which states that the buoyant force equals the weight of displaced fluid—emerges from integrating pressure over the surface of a submerged body.

When the fluid is no longer at rest, Newton's second law (ΣF = ma) must be retained in its full form, leading to Bernoulli's equation for steady inviscid flow and ultimately to the Euler and Navier–Stokes equations for general viscous flow. The static results of this lesson are always recovered as the v = 0 special case of these dynamic equations. The unifying theme is that every equation in fluid mechanics is Newton's second law, rewritten for a medium that flows.

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