If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you *and* that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense.

Consider the container in [link]. Its bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure is the weight of the fluid $\text{mg}$ divided by the area $A$ supporting it (the area of the bottom of the container):

We can find the mass of the fluid from its volume and density:

The volume of the fluid $V$ is related to the dimensions of the container. It is

where $A$ is the cross-sectional area and $h$ is the depth. Combining the last two equations gives

If we enter this into the expression for pressure, we obtain

The area cancels, and rearranging the variables yields

This value is the *pressure due to the weight of a fluid*. The equation has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus the equation $P=\mathrm{h\rho g}$ represents the pressure due to the weight of any fluid of *average density* $\rho $ at any depth $h$ below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered. [link] illustrates this situation.

In [link], we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See [link].) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be $1.96\times {10}^{13}\phantom{\rule{0.25em}{0ex}}\text{N}$).

**Strategy for (a)**

The average pressure $\overline{P}$ due to the weight of the water is the pressure at the average depth $\overline{h}$ of 40.0 m, since pressure increases linearly with depth.

**Solution for (a)**

The average pressure due to the weight of a fluid is

Entering the density of water from [link] and taking $\overline{h}$ to be the average depth of 40.0 m, we obtain

**Strategy for (b)**

The force exerted on the dam by the water is the average pressure times the area of contact:

**Solution for (b)**

We have already found the value for $\overline{P}$. The area of the dam is $A=\text{80.0 m}\times \text{500 m}=4.00\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$, so that

**Discussion**

Although this force seems large, it is small compared with the $1.96\times {\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{N}$ weight of the water in the reservoir—in fact, it is only
$\mathrm{0.0800\%}$ of the weight. Note that the pressure found in part (a) is completely independent of the width and length of the lake—it depends only on its average depth at the dam. Thus the force depends only on the water’s average depth and the dimensions of the dam, *not* on the horizontal extent of the reservoir. In the diagram, the thickness of the dam increases with depth to balance the increasing force due to the increasing pressure.epth to balance the increasing force due to the increasing pressure.

*Atmospheric pressure* is another example of pressure due to the weight of a fluid, in this case due to the weight of *air* above a given height. The atmospheric pressure at the Earth’s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth’s rotation (this creates weather “highs” and “lows”). However, the average pressure at sea level is given by the *standard atmospheric pressure* ${P}_{\text{atm}}$, measured to be

This relationship means that, on average, at sea level, a column of air above $1.00\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ of the Earth’s surface has a weight of $1.01\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}$, equivalent to $\text{1 atm}$. (See [link].)

Calculate the average density of the atmosphere, given that it extends to an altitude of 120 km. Compare this density with that of air listed in [link].

**Strategy**

If we solve $P=\mathrm{h\rho g}$ for density, we see that

We then take $P$ to be atmospheric pressure, $h$ is given, and $g$ is known, and so we can use this to calculate $\overline{\rho}$.

**Solution**

Entering known values into the expression for $\overline{\rho}$ yields

**Discussion**

This result is the average density of air between the Earth’s surface and the top of the Earth’s atmosphere, which essentially ends at 120 km. The density of air at sea level is given in [link] as $1\text{.}\text{29}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$ —about 15 times its average value. Because air is so compressible, its density has its highest value near the Earth’s surface and declines rapidly with altitude.

Calculate the depth below the surface of water at which the pressure due to the weight of the water equals 1.00 atm.

**Strategy**

We begin by solving the equation $P=\mathrm{h\rho g}$ for depth $h$:

Then we take $P$ to be 1.00 atm and $\rho $ to be the density of the water that creates the pressure.

**Solution**

Entering the known values into the expression for $h$ gives

**Discussion**

Just 10.3 m of water creates the same pressure as 120 km of air. Since water is nearly incompressible, we can neglect any change in its density over this depth.

What do you suppose is the *total* pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water’s surface affect the pressure below? The answer is yes. This seems only logical, since both the water’s weight and the atmosphere’s weight must be supported. So the *total* pressure at a depth of 10.3 m is 2 atm—half from the water above and half from the air above. We shall see in Pascal’s Principle that fluid pressures always add in this way.

# Section Summary

- Pressure is the weight of the fluid $\text{mg}$ divided by the area $A$ supporting it (the area of the bottom of the container):
$P=\frac{\text{mg}}{A}.$
- Pressure due to the weight of a liquid is given by
$P=\mathrm{h\rho g},$
where $P$ is the pressure, $h$ is the height of the liquid, $\rho $ is the density of the liquid, and $g$ is the acceleration due to gravity.

# Conceptual Questions

Atmospheric pressure exerts a large force (equal to the weight of the atmosphere above your body—about 10 tons) on the top of your body when you are lying on the beach sunbathing. Why are you able to get up?

Why does atmospheric pressure decrease more rapidly than linearly with altitude?

What are two reasons why mercury rather than water is used in barometers?

[link] shows how sandbags placed around a leak outside a river levee can effectively stop the flow of water under the levee. Explain how the small amount of water inside the column formed by the sandbags is able to balance the much larger body of water behind the levee.

Why is it difficult to swim under water in the Great Salt Lake?

Is there a net force on a dam due to atmospheric pressure? Explain your answer.

Does atmospheric pressure add to the gas pressure in a rigid tank? In a toy balloon? When, in general, does atmospheric pressure *not* affect the total pressure in a fluid?

You can break a strong wine bottle by pounding a cork into it with your fist, but the cork must press directly against the liquid filling the bottle—there can be no air between the cork and liquid. Explain why the bottle breaks, and why it will not if there is air between the cork and liquid.

# Problems & Exercises

What depth of mercury creates a pressure of 1.00 atm?

0.760 m

The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down.

Verify that the SI unit of $\mathrm{h\rho g}$ is ${\text{N/m}}^{2}$.

Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How high above a user must the water level be to create a gauge pressure of $3\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$?

The aqueous humor in a person’s eye is exerting a force of 0.300 N on the $1\text{.}\text{10}{\text{-cm}}^{2}$ area of the cornea. (a) What pressure is this in mm Hg? (b) Is this value within the normal range for pressures in the eye?

(a) 20.5 mm Hg

(b) The range of pressures in the eye is 12–24 mm Hg, so the result in part (a) is within that range

How much force is exerted on one side of an 8.50 cm by 11.0 cm sheet of paper by the atmosphere? How can the paper withstand such a force?

What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?

$1\text{.}\text{09}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$

Calculate the average pressure exerted on the palm of a shot-putter’s hand by the shot if the area of contact is $\text{50}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{2}$ and he exerts a force of 800 N on it. Express the pressure in ${\text{N/m}}^{2}$ and compare it with the $1\text{.}\text{00}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{Pa}$ pressures sometimes encountered in the skeletal system.

The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of $\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{2}.$ What force does it exert to accomplish this?

24.0 N

Show that the total force on a rectangular dam due to the water behind it increases with the *square* of the water depth. In particular, show that this force is given by $F=\rho {\mathrm{gh}}^{2}L/2$, where $\rho $ is the density of water, $h$ is its depth at the dam, and $L$ is the length of the dam. You may assume the face of the dam is vertical. (Hint: Calculate the average pressure exerted and multiply this by the area in contact with the water. (See [link].)

- College Physics
- Preface
- Introduction: The Nature of Science and Physics
- Kinematics
- Introduction to One-Dimensional Kinematics
- Displacement
- Vectors, Scalars, and Coordinate Systems
- Time, Velocity, and Speed
- Acceleration
- Motion Equations for Constant Acceleration in One Dimension
- Problem-Solving Basics for One-Dimensional Kinematics
- Falling Objects
- Graphical Analysis of One-Dimensional Motion

- Two-Dimensional Kinematics
- Dynamics: Force and Newton's Laws of Motion
- Introduction to Dynamics: Newton’s Laws of Motion
- Development of Force Concept
- Newton’s First Law of Motion: Inertia
- Newton’s Second Law of Motion: Concept of a System
- Newton’s Third Law of Motion: Symmetry in Forces
- Normal, Tension, and Other Examples of Forces
- Problem-Solving Strategies
- Further Applications of Newton’s Laws of Motion
- Extended Topic: The Four Basic Forces—An Introduction

- Further Applications of Newton's Laws: Friction, Drag, and Elasticity
- Uniform Circular Motion and Gravitation
- Work, Energy, and Energy Resources
- Introduction to Work, Energy, and Energy Resources
- Work: The Scientific Definition
- Kinetic Energy and the Work-Energy Theorem
- Gravitational Potential Energy
- Conservative Forces and Potential Energy
- Nonconservative Forces
- Conservation of Energy
- Power
- Work, Energy, and Power in Humans
- World Energy Use

- Linear Momentum and Collisions
- Statics and Torque
- Rotational Motion and Angular Momentum
- Introduction to Rotational Motion and Angular Momentum
- Angular Acceleration
- Kinematics of Rotational Motion
- Dynamics of Rotational Motion: Rotational Inertia
- Rotational Kinetic Energy: Work and Energy Revisited
- Angular Momentum and Its Conservation
- Collisions of Extended Bodies in Two Dimensions
- Gyroscopic Effects: Vector Aspects of Angular Momentum

- Fluid Statics
- Introduction to Fluid Statics
- What Is a Fluid?
- Density
- Pressure
- Variation of Pressure with Depth in a Fluid
- Pascal’s Principle
- Gauge Pressure, Absolute Pressure, and Pressure Measurement
- Archimedes’ Principle
- Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
- Pressures in the Body

- Fluid Dynamics and Its Biological and Medical Applications
- Introduction to Fluid Dynamics and Its Biological and Medical Applications
- Flow Rate and Its Relation to Velocity
- Bernoulli’s Equation
- The Most General Applications of Bernoulli’s Equation
- Viscosity and Laminar Flow; Poiseuille’s Law
- The Onset of Turbulence
- Motion of an Object in a Viscous Fluid
- Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

- Temperature, Kinetic Theory, and the Gas Laws
- Heat and Heat Transfer Methods
- Thermodynamics
- Introduction to Thermodynamics
- The First Law of Thermodynamics
- The First Law of Thermodynamics and Some Simple Processes
- Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
- Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
- Applications of Thermodynamics: Heat Pumps and Refrigerators
- Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
- Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation

- Oscillatory Motion and Waves
- Introduction to Oscillatory Motion and Waves
- Hooke’s Law: Stress and Strain Revisited
- Period and Frequency in Oscillations
- Simple Harmonic Motion: A Special Periodic Motion
- The Simple Pendulum
- Energy and the Simple Harmonic Oscillator
- Uniform Circular Motion and Simple Harmonic Motion
- Damped Harmonic Motion
- Forced Oscillations and Resonance
- Waves
- Superposition and Interference
- Energy in Waves: Intensity

- Physics of Hearing
- Electric Charge and Electric Field
- Introduction to Electric Charge and Electric Field
- Static Electricity and Charge: Conservation of Charge
- Conductors and Insulators
- Coulomb’s Law
- Electric Field: Concept of a Field Revisited
- Electric Field Lines: Multiple Charges
- Electric Forces in Biology
- Conductors and Electric Fields in Static Equilibrium
- Applications of Electrostatics

- Electric Potential and Electric Field
- Introduction to Electric Potential and Electric Energy
- Electric Potential Energy: Potential Difference
- Electric Potential in a Uniform Electric Field
- Electrical Potential Due to a Point Charge
- Equipotential Lines
- Capacitors and Dielectrics
- Capacitors in Series and Parallel
- Energy Stored in Capacitors

- Electric Current, Resistance, and Ohm's Law
- Circuits, Bioelectricity, and DC Instruments
- Magnetism
- Introduction to Magnetism
- Magnets
- Ferromagnets and Electromagnets
- Magnetic Fields and Magnetic Field Lines
- Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
- Force on a Moving Charge in a Magnetic Field: Examples and Applications
- The Hall Effect
- Magnetic Force on a Current-Carrying Conductor
- Torque on a Current Loop: Motors and Meters
- Magnetic Fields Produced by Currents: Ampere’s Law
- Magnetic Force between Two Parallel Conductors
- More Applications of Magnetism

- Electromagnetic Induction, AC Circuits, and Electrical Technologies
- Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
- Induced Emf and Magnetic Flux
- Faraday’s Law of Induction: Lenz’s Law
- Motional Emf
- Eddy Currents and Magnetic Damping
- Electric Generators
- Back Emf
- Transformers
- Electrical Safety: Systems and Devices
- Inductance
- RL Circuits
- Reactance, Inductive and Capacitive
- RLC Series AC Circuits

- Electromagnetic Waves
- Geometric Optics
- Vision and Optical Instruments
- Wave Optics
- Introduction to Wave Optics
- The Wave Aspect of Light: Interference
- Huygens's Principle: Diffraction
- Young’s Double Slit Experiment
- Multiple Slit Diffraction
- Single Slit Diffraction
- Limits of Resolution: The Rayleigh Criterion
- Thin Film Interference
- Polarization
- *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light

- Special Relativity
- Introduction to Quantum Physics
- Atomic Physics
- Introduction to Atomic Physics
- Discovery of the Atom
- Discovery of the Parts of the Atom: Electrons and Nuclei
- Bohr’s Theory of the Hydrogen Atom
- X Rays: Atomic Origins and Applications
- Applications of Atomic Excitations and De-Excitations
- The Wave Nature of Matter Causes Quantization
- Patterns in Spectra Reveal More Quantization
- Quantum Numbers and Rules
- The Pauli Exclusion Principle

- Radioactivity and Nuclear Physics
- Medical Applications of Nuclear Physics
- Particle Physics
- Frontiers of Physics
- Atomic Masses
- Selected Radioactive Isotopes
- Useful Information
- Glossary of Key Symbols and Notation