Pressure in Fluids

What pressure is and why it is scalar

A fluid at rest cannot sustain a shearing (tangential) force — it would simply flow. The only force it exerts on a surface in contact with it is therefore normal to that surface, whether the surface is a container wall or an immersed object. Pressure is the magnitude of that normal force divided by the area over which it acts. For a force $\Delta F_\perp$ on a small area $\Delta A$,

$$P = \frac{\Delta F_\perp}{\Delta A}, \qquad P = \lim_{\Delta A \to 0}\frac{\Delta F_\perp}{\Delta A}.$$

NCERT is explicit that the numerator is "the component of the force normal to the area under consideration and not the (vector) force." Pressure therefore has dimensions $[\mathrm{ML^{-1}T^{-2}}]$ and is a scalar quantity — no direction can be assigned to it. The proof of Pascal's law (the prism element ABC–DEF in NCERT Fig. 9.2) shows the pressure on the three faces is equal, $P_a = P_b = P_c$, so pressure at a point is the same in every direction. The force on a chosen surface is a vector and is always perpendicular to that surface, but the pressure itself is not.

Units of pressure

The SI unit is $\mathrm{N\,m^{-2}}$, named the pascal (Pa) after Blaise Pascal. Several legacy and field-specific units survive because the pascal is inconveniently small for atmospheric and physiological work.

Two anchors are worth memorising: $1~\text{atm} = 1.013 \times 10^{5}~\text{Pa} = 76~\text{cm of Hg} = 760~\text{torr}$, and $1~\text{bar} = 10^{5}~\text{Pa}$, slightly below one atmosphere.

Density and relative density

The quantity that converts a depth into a pressure is the fluid's density. For a fluid of mass $m$ occupying volume $V$,

$$\rho = \frac{m}{V}, \qquad [\rho] = \mathrm{ML^{-3}}, \qquad \text{SI unit } \mathrm{kg\,m^{-3}}.$$

Density is a positive scalar. A liquid is largely incompressible, so its density is nearly constant at all pressures; a gas, by contrast, shows a large variation of density with pressure and temperature — the reason liquids are treated as incompressible and gases are not. The relative density (or specific gravity) of a substance is the ratio of its density to that of water at $4\,^{\circ}\mathrm{C}$, where $\rho_{\text{water}} = 1.0 \times 10^{3}~\mathrm{kg\,m^{-3}}$.

$$\text{Relative density} = \frac{\rho_{\text{substance}}}{\rho_{\text{water at } 4\,^{\circ}\mathrm{C}}}.$$

It is dimensionless. For example, aluminium has a relative density of $2.7$, so its density is $2.7 \times 10^{3}~\mathrm{kg\,m^{-3}}$.

Variation of pressure with depth

Consider a cylindrical column of fluid of base area $A$ between a point 1 at height $h$ above a point 2 (NCERT Fig. 9.3). Horizontal forces cancel; vertically, the pressure forces and the column's weight must balance. With $P_1 A$ acting down on the top and $P_2 A$ acting up on the bottom,

$$(P_2 - P_1)A = mg = (\rho h A)g \;\;\Rightarrow\;\; P_2 - P_1 = \rho g h.$$

Taking point 1 at the free surface open to the atmosphere ($P_1 = P_0$) and point 2 at depth $h$ ($P_2 = P$):

$$\boxed{P = P_0 + \rho g h}$$

The pressure at depth $h$ exceeds atmospheric pressure by $\rho g h$. Notably, the base area $A$ cancels: the result depends only on the vertical depth $h$, the density $\rho$ and $g$ — never on the cross-sectional area or the shape of the vessel.

Gauge versus absolute pressure

The excess of pressure over atmospheric, $P - P_0 = \rho g h$, is the gauge pressure. The total pressure $P$ measured against a perfect vacuum is the absolute pressure. Most instruments — tyre gauges, the open-tube manometer — read gauge pressure, because they compare the system against the atmosphere on their other side.

Same level, same pressure

For a fluid in equilibrium, the pressure is the same at all points on a horizontal plane. A horizontal bar of fluid is in equilibrium only if the pressures at its two ends are equal; were they unequal, a net horizontal force would drive a flow. Since the fluid is at rest, the pressure must be identical everywhere on a horizontal level — and this holds across connected regions of the same fluid.

This is the principle behind the open-tube manometer: pressure is equal at the same level on both arms of a U-tube, so the difference in liquid heights between the two arms reads the gauge pressure of the connected system, $P - P_0 = \rho g h$. A low-density liquid (oil) is used for small differences and a high-density liquid (mercury) for large ones.

Atmospheric pressure and the barometer

Atmospheric pressure at any point equals the weight of a column of air of unit cross-section extending from that point to the top of the atmosphere; at sea level it is $1.013\times10^{5}~\text{Pa}$. Torricelli measured it with the mercury barometer: a glass tube closed at one end, filled with mercury, and inverted into a mercury trough. The space above the column holds only mercury vapour at negligible pressure, so the pressure at the top of the column (point A) is essentially zero.

The pressure at B inside the column at trough level equals the pressure at C on the open surface, which is atmospheric. Therefore

$$P_0 = \rho g h,$$

with $\rho$ the density of mercury and $h$ the column height. At sea level $h \approx 76~\text{cm}$, equal to one atmosphere — which is why pressures are quoted in "cm" or "mm of Hg," and why a pressure equivalent to $1~\text{mm of Hg}$ is named one torr ($1~\text{torr} = 133~\text{Pa}$). Mercury's high density is the reason the instrument is compact: a water barometer would need a column roughly $10.3$ m tall to balance the same atmospheric pressure.

Buoyancy and Archimedes' principle

Because pressure rises with depth, the fluid pushes harder on the bottom of an immersed body than on its top. The net upward force is the buoyant force (upthrust). Archimedes' principle states that the buoyant force on a body partially or fully immersed in a fluid equals the weight of fluid the body displaces:

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

The apparent weight of a body hanging in a fluid is its true weight reduced by this upthrust, $W_{\text{app}} = W - F_B$, which is why an object feels lighter in water than in air. The fate of a fully submerged body follows directly by comparing densities.

For a floating body only part of the volume is submerged, and the equilibrium condition $F_B = W$ becomes the law of flotation: a floating body displaces a weight of fluid exactly equal to its own weight. The fraction submerged equals the ratio of the body's density to the fluid's density, $V_{\text{sub}}/V_{\text{body}} = \rho_{\text{body}}/\rho_{\text{fluid}}$.