Fluids and Newton's Laws - AP Physics 1

SECTION 1

Historical Context & Motivation

The study of fluids is one of the oldest branches of physics, predating even the formalization of mechanics. Ancient civilizations in Mesopotamia, Egypt, and Rome developed sophisticated hydraulic engineering—aqueducts, irrigation canals, and dams—long before anyone articulated the underlying principles. Yet these accomplishments were empirical; builders worked from experience rather than theory. The breakthrough came when natural philosophers began asking why a submerged object feels lighter, or why water rises in a pump. The answers, it turned out, lie in the same force and equilibrium concepts that govern blocks on ramps and planets in orbit: Newton's laws of motion applied to continuous media rather than rigid bodies.

c. 250 BCE • Archimedes' Principle

Archimedes of Syracuse discovers that a body immersed in a fluid experiences an upward buoyant force equal to the weight of the displaced fluid—arguably the first quantitative law of fluid statics.

1586 • Stevin's Law of Hydrostatic Pressure

Simon Stevin demonstrates that the pressure exerted by a liquid depends only on its depth and density, not on the shape of its container—foreshadowing Pascal's later work.

1653 • Pascal's Law

Blaise Pascal formalizes that pressure applied to a confined fluid is transmitted undiminished in every direction, laying the theoretical foundation for hydraulic systems.

1687 • Newton's Principia

Isaac Newton publishes the three laws of motion and universal gravitation, providing the general framework within which all static and dynamic fluid behavior can be derived.

1738 • Bernoulli's Hydrodynamica

Daniel Bernoulli applies energy conservation to flowing fluids, linking pressure, velocity, and elevation in what becomes Bernoulli's equation—an extension of Newtonian mechanics to fluid dynamics.

The central question this lesson addresses is straightforward yet profound: how do Newton's laws—originally formulated for point masses and rigid bodies—apply to substances that flow and deform? When you draw a free-body diagram for a small parcel of fluid, you discover that familiar ideas like net force, equilibrium, and acceleration yield powerful results including pressure-depth relations, buoyant forces, and constraints on fluid flow. These ideas are not separate from mechanics—they are mechanics, applied to a new and beautifully deformable medium.

SECTION 2

Core Principles & Definitions

Before applying Newton's laws to fluids, you need a vocabulary shift. In rigid-body mechanics, you track individual objects with definite shapes and masses. In fluid mechanics, the medium is continuous and deformable, so the relevant quantities become density and pressure rather than mass and force alone. A fluid is any substance—liquid or gas—that cannot sustain a shear stress at rest; it yields and flows instead. This single property explains why fluids conform to their containers and transmit forces in all directions.

Density (ρ)

Mass per unit volume, ρ = m / V (SI unit: kg/m³). Density is the intensive property that lets you compare how 'concentrated' different fluids are. Water has ρ ≈ 1000 kg/m³; air at sea level has ρ ≈ 1.2 kg/m³.

Pressure (P)

Force per unit area perpendicular to a surface, P = F⊥ / A (SI unit: Pa = N/m²). Pressure is a scalar; at a point inside a fluid at rest, it acts equally in all directions. Atmospheric pressure at sea level is approximately 1.013 × 10⁵ Pa.

Newton's Second Law for Fluids

Isolate an imaginary fluid parcel and apply ΣF = ma. In static equilibrium (a = 0), the net force on the parcel—from surrounding pressure and gravity—is zero. This reasoning directly produces the hydrostatic pressure equation.

Buoyancy (Archimedes' Principle)

The net upward pressure force on a submerged object equals the weight of fluid displaced:

F_b

= ρ_fluid × V_displaced × g. This is not a new law but a direct consequence of Newton's second law applied to the surrounding fluid.

Pascal's Law

A change in pressure applied to a confined, incompressible fluid is transmitted undiminished throughout the fluid. This follows from Newton's laws: if pressure were not transmitted equally, unbalanced forces would accelerate fluid parcels until equilibrium is restored.

✦

KEY TAKEAWAY

Think of a fluid at rest like a crowd of people in a packed elevator. Every person (fluid parcel) is pushed from all sides by their neighbors. When the elevator is stationary, the pushes balance perfectly—that is static equilibrium. If someone at the back pushes harder, everyone in the crowd feels the extra push equally (Pascal's law). And if someone is standing on a raised platform (deeper in the fluid), the people below shoulder more weight, so the pressure they feel is greater—exactly as the hydrostatic equation predicts.

SECTION 3

Visual Explanation: Free-Body Diagram of a Fluid Parcel

The single most illuminating exercise in fluid statics is drawing a free-body diagram for an imaginary rectangular parcel of fluid inside a larger body of liquid at rest. The diagram below shows a small parcel at depth h beneath the surface. The parcel has cross-sectional area A and height Δy. Three forces act on it: the downward pressure from fluid above, the upward pressure from fluid below, and the downward gravitational force (weight). Because the parcel is in static equilibrium, Newton's second law requires the vector sum of these forces to be zero, which leads directly to the hydrostatic pressure equation.

A rectangular fluid parcel (violet) at depth h below the surface experiences three forces: pressure from above (red arrow, downward), pressure from below (green arrow, upward), and its own weight (amber arrow, downward). Setting ΣF = 0 yields the hydrostatic pressure equation P = P₀ + ρgh.

Notice how the derivation requires nothing beyond Newton's second law in the y-direction and the definition of pressure. The green upward arrow represents the force exerted by the fluid below the parcel, which is larger than the red downward force from the fluid above because the lower face sits at greater depth and therefore higher pressure. The difference between these two pressure forces exactly balances the parcel's weight—this is the origin of buoyancy. If you replace the fluid parcel with a solid object of the same dimensions, the surrounding fluid still pushes with the same pressure distribution, so the net upward force remains ρ_fluidgV—Archimedes' principle.

SECTION 4

Mathematical Framework

The equations of fluid statics and buoyancy follow directly from Newton's second law applied to carefully chosen fluid parcels. Each equation below can be derived by drawing a free-body diagram, summing forces, and applying the equilibrium condition ΣF = 0 (for statics) or ΣF = ma (for dynamics). Mastering these derivations—not just memorizing the results—is essential for the AP Physics 1 exam.

HYDROSTATIC PRESSURE

P = P₀ + ρgh

P = absolute pressure at depth h, P₀ = pressure at the surface (often atmospheric), ρ = fluid density (kg/m³), g = gravitational acceleration (9.8 m/s²), h = depth below the surface (m). This follows from ΣF_y = 0 on a horizontal fluid parcel.

GAUGE PRESSURE

P_gauge = P − P₀ = ρgh

Gauge pressure is the pressure relative to atmospheric pressure. Many instruments (tire gauges, blood pressure cuffs) measure gauge pressure. The absolute pressure at a point is the gauge pressure plus atmospheric pressure.

BUOYANT FORCE (ARCHIMEDES' PRINCIPLE)

F_b = ρ_fluid × g × V_displaced

F_b = buoyant force (N, directed upward), ρ_fluid = density of the surrounding fluid, V_displaced = volume of fluid displaced by the object. Derived by integrating the pressure difference between the top and bottom surfaces of the submerged object.

PASCAL'S LAW (HYDRAULIC SYSTEMS)

F₁ / A₁ = F₂ / A₂

In a confined incompressible fluid, pressure changes are transmitted equally. A small force F₁ on a small piston of area A₁ creates the same pressure as a large force F₂ on a large piston of area A₂. This is a direct application of Newton's third law across the fluid medium.

💡

Connecting to Newton's Laws

Every fluid equation on the AP exam can be traced to Newton's second or third law. The hydrostatic equation comes from ΣF = 0 on a fluid column. Archimedes' principle is the net pressure force on a submerged object—still ΣF analysis. Pascal's law invokes Newton's third law: the fluid pushes back on every surface with equal pressure. When you face an unfamiliar fluid problem, start with a free-body diagram of a fluid element or submerged object, just as you would for any mechanics problem.

SECTION 5

Buoyancy and Floating: A Detailed Breakdown

Whether an object sinks, floats, or hovers in a fluid depends on the relationship between the buoyant force and the object's weight. Because both forces depend on volume and density, the outcome reduces to a simple density comparison. An object denser than the fluid sinks because its weight exceeds the maximum buoyant force (achieved when fully submerged). An object less dense than the fluid floats at the surface, submerging just enough volume to generate a buoyant force equal to its weight. An object whose density equals the fluid's density is neutrally buoyant and will remain at rest wherever you place it—a condition exploited by submarines adjusting their ballast.

Three scenarios for an object in a fluid. Left: Object less dense than the fluid floats, partially submerged, with F_b = W. Center: Object density equals fluid density: neutral buoyancy. Right: Object denser than the fluid sinks; W > F_b, producing a net downward force and downward acceleration.

Condition	Density Relation	Net Force	Behavior
Floating	ρ_obj < ρ_fluid	ΣF = 0 (equilibrium at surface)	Partially submerged; fraction submerged = ρ_obj / ρ_fluid
Neutrally buoyant	ρ_obj = ρ_fluid	ΣF = 0 (equilibrium anywhere)	Fully submerged, remains at any depth
Sinking	ρ_obj > ρ_fluid	ΣF ≠ 0 (net downward)	Fully submerged, accelerates downward until reaching the bottom or a denser layer

Summary of buoyancy conditions determined by comparing object and fluid densities

SECTION 6

Worked Example: Buoyancy and Apparent Weight

A solid aluminum cube with side length 0.10 m is suspended by a string and fully submerged in water. Determine (a) the buoyant force on the cube, (b) the tension in the string, and (c) the apparent weight of the cube. Use ρ_Al = 2700 kg/m³, ρ_water = 1000 kg/m³, and g = 9.8 m/s².

Aluminum Cube in Water

1
Step 1 — Identify the Given Values

Side length s = 0.10 m, so V = s³ = (0.10)³ = 1.0 × 10⁻³ m³. The cube is fully submerged, so V_displaced = V = 1.0 × 10⁻³ m³. We also know ρ_Al = 2700 kg/m³ and ρ_water = 1000 kg/m³.
2
Step 2 — Calculate the Buoyant Force

Apply Archimedes' principle: F_b = ρ_water × g × V_displaced = (1000)(9.8)(1.0 × 10⁻³).

$F_b$ = 9.8 N
3
Step 3 — Calculate the Weight of the Cube

W = m × g = ρ_Al × V × g = (2700)(1.0 × 10⁻³)(9.8).

W = 26.46 N
4
Step 4 — Apply Newton's Second Law (Equilibrium)

The cube is stationary (a = 0), so ΣF_y = 0. Three forces act on the cube: tension T (up), buoyant force F_b (up), and weight W (down). Therefore T + F_b − W = 0, giving T = W − F_b = 26.46 − 9.8.

T = 16.66 N ≈ 16.7 N
5
Step 5 — Determine Apparent Weight

The apparent weight is the reading a scale or string tension would show—effectively the true weight minus the buoyant force. Since the string tension supports the cube against both gravity and the buoyant force, the apparent weight equals the tension.

Apparent weight = T = 16.7 N (compared to true weight of 26.5 N)

A fully submerged 0.10 m aluminum cube is supported without touching the vessel: weight 26.46 N downward, buoyancy 9.80 N upward and tension 16.66 N upward. T + B = W at equilibrium; arrow lengths are schematic.

Matched spring scales suspend an aluminum cube in air and fully underwater, with a taut thread and clear gaps around the submerged cube. Submersion changes a supporting scale’s tension without changing the object’s gravitational force. At rest, upward tension plus buoyancy balances the weight.

Apparent weight is the supporting tension. Submersion changes a supporting scale’s tension without changing the object’s gravitational force. At rest, upward tension plus buoyancy balances the weight. Submersion changes a supporting scale’s tension without changing the object’s gravitational force. At rest, upward tension plus buoyancy balances the weight.

Buoyancy reduces the scale reading, not gravity. Submersion changes a supporting scale’s tension without changing the object’s gravitational force. At rest, upward tension plus buoyancy balances the weight. Submersion changes a supporting scale’s tension without changing the object’s gravitational force. At rest, upward tension plus buoyancy balances the weight.

SECTION 7

Strengths, Limitations & Common Pitfalls

The Newtonian approach to fluids is elegant and powerful, but it comes with assumptions and common student errors that are worth cataloging. The AP exam frequently tests your ability to recognize when these assumptions hold and when they break down.

Strength / Feature	Limitation / Pitfall
Hydrostatic equation P = P₀ + ρgh is simple and widely applicable to any static fluid of uniform density.	Assumes constant density ρ. In gases, density changes with altitude; the equation is approximate for tall gas columns.
Archimedes' principle applies to objects of any shape—only the displaced volume matters.	Common error: using the object's total volume instead of the displaced volume when the object is only partially submerged.
Free-body diagram approach transfers seamlessly from rigid-body mechanics to fluid problems.	Common error: forgetting that the buoyant force uses the fluid's density, not the object's density. Many students accidentally substitute ρ_obj for ρ_fluid.
Pascal's law allows enormous mechanical advantage in hydraulic systems.	Pascal's law assumes an incompressible fluid and a closed system. It breaks down for gases under large pressure changes.
Pressure at a given depth is independent of container shape (the hydrostatic paradox).	Students often think a wider container creates more pressure at the same depth. The extra weight is supported by the walls, not the fluid at the bottom.

Key strengths and common pitfalls when applying Newton's laws to fluids

⚠️

EXAM TIP

On the AP exam, many fluid questions are disguised Newton's second law problems. If you see an object in a fluid, your first instinct should be to draw a free-body diagram with three forces: weight (down), buoyant force (up), and any applied or contact force (string tension, normal force, etc.). Then write ΣF = ma. This strategy works whether the object is floating, sinking, submerged on a spring, or sitting on the bottom of a container.

SECTION 8

Connection to Advanced Fluid Theory

The static fluid concepts covered in AP Physics 1 are the foundation upon which the full edifice of fluid dynamics is built. While the AP Physics 1 curriculum treats fluids primarily at rest or in simplified steady flow, the underlying Newtonian reasoning extends directly to more advanced formulations. Understanding where AP-level concepts fit within this larger picture can deepen your intuition and prepare you for further study in physics or engineering.

AP Physics 1 Concept	Advanced Extension	Where You'll See It
P = P₀ + ρgh (constant ρ)	General hydrostatic equation dP/dy = −ρg, allowing variable density (e.g., atmospheric pressure vs. altitude)	AP Physics 2, introductory engineering courses
Archimedes' principle (static buoyancy)	Dynamic lift and drag forces on objects moving through fluids; Navier-Stokes equations	College-level fluid mechanics, aerospace engineering
ΣF = 0 on a fluid parcel	ΣF = ma for accelerating fluid parcels → Euler's equation, then add viscosity → Navier-Stokes	Graduate-level fluid dynamics, computational fluid dynamics (CFD)
Pascal's law (hydraulics)	Compressible fluid mechanics; thermodynamic equations of state relating P, ρ, and T	Thermodynamics, chemical engineering

How AP Physics 1 fluid concepts extend into more advanced physics and engineering courses

The key insight is that the Navier-Stokes equations—the governing equations for all fluid motion—are simply Newton's second law (ΣF = ma) applied to an infinitesimally small fluid element, with forces from pressure gradients, viscosity, and gravity. Every fluid phenomenon from ocean currents to airplane lift ultimately traces back to the same free-body diagram reasoning you practice in AP Physics 1. The equations become mathematically more demanding (they involve partial differential equations), but the physical idea never changes: forces cause acceleration, and equilibrium means zero net force.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A solid steel ball is held at rest by a string while fully submerged in a tank of water. The string is then cut. Which of the following correctly describes all the forces acting on the ball at the instant after the string is cut?

A. Only the gravitational force acts downward on the ball.
B. The gravitational force acts downward and a buoyant force acts upward, but the gravitational force is greater in magnitude.
C. The gravitational force acts downward and a buoyant force acts upward, and they are equal in magnitude.
D. The gravitational force acts downward and a buoyant force acts upward, but the buoyant force is greater in magnitude.

Show answer (B)

The correct answer is B. After the string is cut, only two forces act on the ball: gravity (W = ρ_steel × V × g, downward) and the buoyant force (

F_b

= ρ_water × V × g, upward). Since steel (ρ ≈ 7800 kg/m³) is denser than water (ρ = 1000 kg/m³), the weight exceeds the buoyant force, producing a net downward force and downward acceleration. Choice A is incorrect because the buoyant force from the surrounding water does not vanish when the string is cut—it depends on the fluid pressure, not the string. Choice C is incorrect because equal forces would mean neutral buoyancy, which requires the object and fluid to have the same density. Choice D is incorrect because a buoyant force exceeding the weight would cause the ball to accelerate upward—impossible for steel in water.

Question 2: BASIC CALCULATION

A diver is swimming at a depth of 15 m in a freshwater lake. Atmospheric pressure is 1.013 × 10⁵ Pa and the density of freshwater is 1000 kg/m³. What is the absolute pressure at the diver's depth?

A. 1.47 × 10⁵ Pa
B. 2.48 × 10⁵ Pa
C. 1.013 × 10⁵ Pa
D. 2.48 × 10⁶ Pa

Show answer (B)

The correct answer is B. Using the hydrostatic pressure equation P = P₀ + ρgh: P = 1.013 × 10⁵ + (1000)(9.8)(15) = 1.013 × 10⁵ + 1.47 × 10⁵ = 2.483 × 10⁵ Pa ≈ 2.48 × 10⁵ Pa. Choice A (1.47 × 10⁵ Pa) represents only the gauge pressure ρgh, forgetting to add atmospheric pressure P₀. Choice C (1.013 × 10⁵ Pa) is just atmospheric pressure, ignoring the water column entirely. Choice D (2.48 × 10⁶ Pa) is off by a factor of 10, likely from a decimal error in the calculation.

Question 3: INTERMEDIATE

A wooden block of density 600 kg/m³ and volume 0.050 m³ is floating in a tank of oil with density 800 kg/m³. What fraction of the block's volume is submerged, and what is the buoyant force on the block?

A. 75% submerged; $F_b$ = 294 N
B. 60% submerged; $F_b$ = 235 N
C. 75% submerged; $F_b$ = 392 N
D. 80% submerged; $F_b$ = 294 N

Show answer (A)

The correct answer is A. For a floating object in equilibrium,

F_b

= W. The weight of the block is W = ρ_block × V × g = (600)(0.050)(9.8) = 294 N, so

F_b

= 294 N. The fraction submerged equals ρ_block / ρ_fluid = 600 / 800 = 0.75, or 75%. Choice B incorrectly reverses the density ratio (800/600 inverted to get 60%, which is wrong) and miscalculates the force. Choice C has the correct submerged fraction but incorrectly calculates the buoyant force using the oil's density times the total volume (

F_b

= 800 × 0.050 × 9.8 = 392 N), which would be the buoyant force if the block were fully submerged—but it floats. Choice D uses an incorrect fraction (80%) which doesn't correspond to the given densities.

Question 4: APPLIED

5 pts

A research team wants to determine the density of an irregularly shaped rock sample. They propose using a spring scale and a beaker of water. Design an experimental procedure to determine the rock's density. Include: (a) A list of quantities to be measured and equipment needed. (b) A step-by-step procedure. (c) An explanation of how to analyze the data to calculate the rock's density. (d) One significant source of experimental uncertainty and how it affects the result.

Show answer

Point 1 (Equipment & Quantities — 1 pt): Measure the weight of the rock in air (W_air) using a spring scale, and the apparent weight of the rock fully submerged in water (W_water) using the same spring scale. Equipment: spring scale (or force sensor), beaker of water large enough to submerge the rock, string to suspend the rock, known density of water (1000 kg/m³). Point 2 (Procedure — 1 pt): (1) Hang the rock from the spring scale in air and record the reading W_air. (2) Lower the rock, still attached to the spring scale, until it is fully submerged in the water, ensuring no part touches the beaker walls or bottom and no air bubbles cling to the surface. (3) Record the submerged reading W_water. Point 3 (Data Analysis — 2 pts): The buoyant force is

F_b

= W_air − W_water. By Archimedes' principle,

F_b

= ρ_water × g × V_rock, so V_rock = (W_air − W_water) / (ρ_water × g). The mass of the rock is m = W_air / g. Therefore, ρ_rock = m / V_rock = W_air / (W_air − W_water) × ρ_water. This eliminates the need to directly measure the irregular volume. Point 4 (Uncertainty — 1 pt): A significant source of error is air bubbles clinging to the rock's surface when submerged, which would increase the displaced volume, making

F_b

appear larger than it should be. This would lead to an overestimate of V_rock and consequently an underestimate of ρ_rock. This can be minimized by gently tapping the rock or submerging it slowly and allowing bubbles to escape before recording data.

Question 5: CRITICAL THINKING

5 pts

A hollow metal sphere has an outer radius R and total mass M. It is placed in a fluid of density ρ_f. (a) Derive an expression for the minimum fluid density ρ_f,min required for the sphere to float. (b) If the sphere floats, derive an expression for the fraction of its volume that is submerged. (c) Explain, using Newton's second law, what would happen if the floating sphere were pushed slightly deeper into the fluid and then released. Would it return to equilibrium? Justify your answer in terms of forces.

Show answer

Point 1 (Part a — Derivation of ρ_f,min — 1 pt): For the sphere to float, the maximum buoyant force (when fully submerged) must at least equal the weight: ρ_f × g × V_total ≥ Mg. The total volume is V_total = (4/3)πR³. Therefore ρ_f,min = M / [(4/3)πR³] = 3M / (4πR³). This is simply the average density of the hollow sphere. Point 2 (Part b — Fraction submerged — 2 pts): For a floating sphere in equilibrium, ΣF = 0 means

F_b

= W, so ρ_f × g × V_sub = Mg. The fraction submerged is f = V_sub / V_total = M / (ρ_f × V_total) = M / [ρ_f × (4/3)πR³] = 3M / (4πR³ρ_f). Equivalently, f = ρ_avg / ρ_f where ρ_avg is the average density of the hollow sphere. Point 3 (Part c — Stability analysis using Newton's second law — 2 pts): If the sphere is pushed slightly deeper, the submerged volume increases, which increases the buoyant force

F_b

= ρ_f × g × V_sub. However, the weight Mg remains constant. Therefore the net upward force

F_b

− Mg > 0, producing an upward acceleration by Newton's second law (ΣF = ma, with a directed upward). The sphere will accelerate back toward its original equilibrium position. Similarly, if lifted slightly above equilibrium, V_sub decreases,

F_b

decreases, and the net force becomes downward, pushing it back down. This is a restoring force—the sphere undergoes stable equilibrium and would oscillate about its equilibrium position (simple harmonic motion if the restoring force is approximately linear in displacement). The key physics: the buoyant force is depth-dependent because V_sub changes, while the weight is constant, creating the restoring mechanism.

SUMMARY

Lesson Summary

Summary

Fluid statics is not a separate branch of physics but a direct application of Newton's laws of motion to continuous, deformable media. By isolating an imaginary fluid parcel and applying ΣF = ma in the vertical direction, we derived the hydrostatic pressure equation P = P₀ + ρgh, which states that pressure increases linearly with depth in a fluid of uniform density. The net upward pressure force on a submerged object yields Archimedes' principle: $F_b$ = ρ_fluid × g × V_displaced. Whether an object floats, sinks, or is neutrally buoyant depends on comparing the object's average density to the fluid's density.

Pascal's law tells us that pressure changes are transmitted undiminished through a confined incompressible fluid, enabling hydraulic force multiplication. For every AP-style fluid problem, the winning strategy is the same one you use in mechanics: draw a free-body diagram, identify all forces (weight, buoyant force, tension, normal force), set up Newton's second law, and solve. The concepts here form the foundation for more advanced topics such as fluid dynamics, Bernoulli's equation, and the continuity equation, which you will encounter in AP Physics 2 and college-level courses.