8.2 Pressure

Learning objectives

8.2A—Describe the pressure exerted on a surface by a given forceDescribe the pressure exerted on a surface by a given force.• Pressure is defined as the magnitude of the perpendicular force component exerted per unit area over a given surface area, as described by the equation• Pressure is a scalar quantity.• The volume and density of a given amount of an incompressible fluid is constant regardless of the pressure exerted on that fluid.8.2B—Describe the pressure exerted by a fluidDescribe the pressure exerted by a fluid.• The pressure exerted by a fluid is the result of the entirety of the interactions between the fluid’s constituent particles and the surface with which those particles interact.• The absolute pressure of a fluid at a given point is equal to the sum of a reference pressure P0, such as the atmospheric pressure Patm, and the gauge pressure Pgauge. Relevant equation:• The gauge pressure of a vertical column of fluid is described by the equation TOPIC 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.

Next Topic

—8.3 Fluids and Newton’s LawsRead