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8.1 Internal Structure and Density

Identify a fluid by shape behavior

A fluid is a substance with no fixed shape, so it flows and takes the shape allowed by its container. Both liquids and gases are fluids; a solid retains a fixed shape.

Connect particles to macroscopic properties

State	Particle interaction and arrangement	Shape behavior
Solid	Strong enough to maintain an organized structure	Fixed shape
Liquid	Particles remain close but rearrange	No fixed shape; nearly fixed volume
Gas	Particles are widely separated and interact weakly except during collisions	No fixed shape; expands to fill available volume

Calculate mass per volume

$\rho=\frac{m}{V}$

Example: a fluid sample has mass $1.20\,\text{kg}$ and volume $5.00\times10^{-4}\,\text{m}^3$ .

$\rho=\dfrac{1.20}{5.00\times10^{-4}}=2.40\times10^3\,\text{kg}\,\text{m}^{-3}$ .

Density is mass per unit volume; doubling both mass and volume leaves the density unchanged.

State the ideal-fluid assumptions

An ideal fluid is modeled as incompressible, so its density does not change, and as having no viscosity, so there is no internal resistance to flow.

Use the model as an approximation

Ideal-fluid assumptions simplify later pressure and flow models. They are not claims that every real fluid is perfectly incompressible or frictionless; check whether the model is appropriate for the situation.

8.2 Pressure

Connect force and area

Pressure measures how much force acts perpendicular to each unit of surface area. Pressure is a scalar: it has magnitude but no direction.

$P=\frac{F_{\perp}}{A}$

Choose the correct quantities

Quantity	Use in the equation	SI unit
$F_{\perp}$	Component of force perpendicular to the surface	newton (N)
$A$	Area over which that force acts	square metre ( $\text{m}^2$ )
$P$	Perpendicular force per unit area	pascal ( $\text{Pa}=\text{N}/\text{m}^2$ )

Substitute with units

Example: a perpendicular force of $600\,\text{N}$ acts over $0.030\,\text{m}^2$ .

$P=\dfrac{F_{\perp}}{A}=\dfrac{600}{0.030}=2.0\times10^4\,\text{Pa}$ .

This value means each square metre would experience $2.0\times10^4\,\text{N}$ of perpendicular force at the same pressure.

Predict the effect of area

For the same perpendicular force, pressure is inversely proportional to contact area. Halving the area doubles the pressure; doubling the area halves it.

Control two common errors

Use only the force component perpendicular to the surface, not the full magnitude of an angled force. In an incompressible-fluid model, changing pressure does not change the fluid's density or volume.

Explain where fluid pressure comes from

A fluid exerts pressure on a surface through the collective interactions of many fluid particles with that surface. At rest, greater depth means more fluid above the point and therefore greater pressure.

Separate gauge and absolute pressure

$P_{\text{gauge}}=\rho g h$

$P=P_0+P_{\text{gauge}}=P_0+\rho g h$

Location	Gauge pressure	Absolute pressure
Reference surface, $h=0$	$0$	$P_0$
Vertical depth $h$	$\rho g h$	$P_0+\rho g h$

Here $\rho$ is fluid density, $g$ is gravitational field strength, $h$ is vertical depth below the reference surface, and $P_0$ is the pressure at that surface.

Calculate pressure at a depth

Example: water has $\rho=1000\,\text{kg m}^{-3}$ . At $h=2.0\,\text{m}$ , with $g=9.8\,\text{N kg}^{-1}$ and surface pressure $P_0=1.01\times10^5\,\text{Pa}$ :

$P_{\text{gauge}}=(1000)(9.8)(2.0)=1.96\times10^4\,\text{Pa}$

$P=1.01\times10^5+1.96\times10^4=1.206\times10^5\,\text{Pa}\approx1.21\times10^5\,\text{Pa}$ .

Use the correct reference and depth

Gauge pressure excludes the reference pressure; absolute pressure includes it. Use vertical depth, not the distance traveled through the fluid. In the same fluid at rest, points at the same depth have the same pressure regardless of container shape.

8.3 Fluids and Newton’s Laws

1

Apply Newton’s second law

A fluid system changes its velocity when a net external force changes its momentum. For a constant-mass fluid parcel, a nonzero net force produces acceleration.

$\sum \vec F_{\text{external}}=\frac{d\vec p}{dt}=m\vec a$

2

Classify forces using the system boundary

Interaction	Role for the chosen fluid system
Collisions and forces between particles inside the fluid	Internal: transfer momentum between parts of the fluid
Gravity, a container wall, a piston, or another object acting on the fluid	External: can change the total momentum of the fluid system

3

Connect particles to fluid behavior

Microscopic particle interactions transmit forces through the fluid. Together with external forces, these interactions create the macroscopic patterns we observe, such as acceleration or changing flow speed.

4

Predict the velocity change

Example: model a $2.0\,\text{kg}$ parcel of fluid as having constant mass. If the net external force is $6.0\,\text{N}$ to the right,

$a=\dfrac{F_{\text{net}}}{m}=\dfrac{6.0}{2.0}=3.0\,\text{m s}^{-2}$ to the right.

Its velocity therefore changes by $3.0\,\text{m s}^{-1}$ each second while that net force acts.

Keep internal forces in their proper role

Internal forces do not vanish: they redistribute momentum within the fluid. But for the whole chosen system, internal force pairs cannot by themselves change total momentum; identify the system boundary before deciding which forces are external.

Combine distributed fluid forces

The buoyant force is the net upward force that a fluid exerts on an object. It is the combined result of forces from many fluid particles over the object’s surface.

Fluid pressure is greater on the deeper parts of the object than on the shallower parts. When all the distributed fluid forces are combined, their vertical components produce a net upward force.

Use the displaced-fluid weight

$F_B=\rho_{\text{fluid}}V_{\text{displaced}}g$

Quantity	Meaning	SI unit
$\rho_{\text{fluid}}$	Density of the fluid, not the object	$\text{kg m}^{-3}$
$V_{\text{displaced}}$	Volume of fluid displaced by the submerged part	$\text{m}^3$
$F_B$	Weight of that displaced fluid	N

Calculate the upward force

Example: an object displaces $0.020\,\text{m}^3$ of water with $\rho=1000\,\text{kg m}^{-3}$ . Using $g=9.8\,\text{N kg}^{-1}$ ,

$F_B=(1000)(0.020)(9.8)=196\,\text{N}$ upward.

This equals the weight of the displaced water.

Compare buoyancy with weight

Buoyant force is upward, but an object does not necessarily accelerate upward. Compare $F_B$ with the object’s weight: greater buoyant force gives upward acceleration, equal forces give zero vertical acceleration, and smaller buoyant force gives downward acceleration. Use only the displaced volume—not automatically the object’s full volume.

8.4 Fluids and Conservation Laws

Connect pressure difference to flow

A pressure difference between two locations can drive fluid flow. In a completely filled tube carrying steady incompressible flow, matter cannot accumulate, so the rate entering equals the rate leaving.

Calculate volume flow rate

$\frac{V}{t}=Av$

Quantity	Meaning	SI unit
$V/t$	Volume flow rate through a cross section	$\text{m}^3\,\text{s}^{-1}$
$A$	Cross-sectional area perpendicular to the flow	$\text{m}^2$
$v$	Average fluid speed through that cross section	$\text{m s}^{-1}$

Apply mass conservation

$A_1v_1=A_2v_2$

For an incompressible fluid, density is constant. Equal mass flow rates therefore mean equal volume flow rates: $A_1v_1=A_2v_2$ . A smaller cross-sectional area requires a greater speed to carry the same volume each second.

Relate area and speed

Example: water moves at $2.0\,\text{m s}^{-1}$ through area $A_1=6.0\,\text{cm}^2$ and enters a section with $A_2=3.0\,\text{cm}^2$ .

$v_2=\dfrac{A_1v_1}{A_2}=\dfrac{(6.0)(2.0)}{3.0}=4.0\,\text{m s}^{-1}$ .

The area halves, so the speed doubles; the volume flow rate is unchanged.

Use the incompressible-flow condition

Continuity does not say pressure stays constant. It conserves mass flow. The simplified form $A_1v_1=A_2v_2$ requires incompressible flow; if density changes, conserve $\rho Av$ instead.

Conserve mechanical energy

In ideal fluid flow, pressure, kinetic, and gravitational energy can change between two locations while their total mechanical energy is conserved.

$P_1+\rho gy_1+\frac12\rho v_1^2=P_2+\rho gy_2+\frac12\rho v_2^2$

Track three energy stores

Bernoulli term	Energy represented per unit volume
$P$	Pressure energy
$\rho gy$	Gravitational potential energy
$\tfrac12\rho v^2$	Kinetic energy

Compare the same three terms at locations 1 and 2. At equal height, a larger kinetic term must be balanced by a smaller pressure term. When height changes, keep all three terms.

Derive the exit-speed relationship

$v=\sqrt{2g\,\Delta y}$

Torricelli example: a small opening is $1.25\,\text{m}$ below the liquid surface. The surface and opening are both exposed to the same pressure, and the wide surface moves negligibly. Bernoulli’s equation reduces to

$\rho g\Delta y=\tfrac12\rho v^2$ , so $v=\sqrt{2g\Delta y}$ .

$v=\sqrt{2(9.8)(1.25)}=4.95\,\text{m s}^{-1}\approx5.0\,\text{m s}^{-1}$ .

Check the model conditions

Do not use ‘faster means lower pressure’ without checking height and Bernoulli’s assumptions. Unless stated otherwise here, treat the fluid as ideal and the pipe as completely filled. In Torricelli’s theorem, $\Delta y$ is the vertical height difference, not the path length.

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