Gravity causes a liquid to fill the bottom of a container. Thus it’s not surprising that the pressure in a liquid is due almost entirely to the gravitational contribution. We’d like to determine the pressure at depth $d$ below the surface of the liquid. We will assume that the liquid is at rest; flowing liquids will be considered later in this chapter.

The shaded cylinder of liquid in FIGURE 14.10 extends from the surface to depth d. This cylinder, like the rest of the liquid, is in equilibrium with $\vec{F}{}_{\text{net}}=\vec{0}$. Three forces act on this cylinder: the gravitational force mg on the liquid in the cylinder, a downward force $p_{0}A$ due to the pressure $p_0$ at the surface of the liquid, and an upward force pA due to the liquid beneath the cylinder pushing up on the bottom of the cylinder. This third force is a consequence of our earlier observation that different parts of a fluid push against each other. Pressure p, which is what we’re trying to find, is the pressure at the bottom of the cylinder.

FIGURE 14.10 Measuring the pressure at depth $d$ in a liquid.

The upward force balances the two downward forces, so

The liquid is a cylinder of cross-section area $A$ and height d. Its volume is $V=Ad$ and its mass is $m=\rho V=\rho Ad$. Substituting this expression for the mass of the liquid into Equation 14.4, we find that the area $A$ cancels from all terms. The pressure at depth $d$ in a liquid is

where $\rho$ is the liquid’s density. Because the fluid is at rest, the pressure given by Equation 14.5 is called the hydrostatic pressure. The fact that $g$ appears in Equation 14.5 reminds us that this is a gravitational contribution to the pressure.

As expected, $p=p_0$ at the surface, where $d=0$. Pressure $p_0$ is often due to the air or other gas above the liquid. $p_0=1\text{atm}=101.3\text{kPa}$ for a liquid that is open to the air. However, $p_0$ can also be the pressure due to a piston or a closed surface pushing down on the top of the liquid.

The hydrostatic pressure in a liquid depends only on the depth and the pressure at the surface. This observation has some important implications. FIGURE 14.11a shows two connected tubes. It’s certainly true that the larger volume of liquid in the wide tube weighs more than the liquid in the narrow tube. You might think that this extra weight would push the liquid in the narrow tube higher than in the wide tube. But it doesn’t. If $d_1$ were larger than $d_2$, then, according to the hydrostatic pressure equation, the pressure at the bottom of the narrow tube would be higher than the pressure at the bottom of the wide tube. This pressure difference would cause the liquid to flow from right to left until the heights were equal.

FIGURE 14.11 Some properties of a liquid in hydrostatic equilibrium are not what you might expect.

Thus a first conclusion: A connected liquid in hydrostatic equilibrium rises to the same height in all open regions of the container.

FIGURE 14.11b shows two connected tubes of different shape. The conical tube holds more liquid above the dotted line, so you might think that $p_1>p_2$. But it isn’t. Both points are at the same depth, thus $p_1=p_2$. If $p_1$ were larger than $p_2$, the pressure at the bottom of the left tube would be larger than the pressure at the bottom of the right tube. This would cause the liquid to flow until the pressures were equal.

Thus a second conclusion: The pressure is the same at all points on a horizontal line through a connected liquid in hydrostatic equilibrium.

EXAMPLE 14.4 Pressure in a closed tube

Water fills the tube shown in FIGURE 14.12. What is the pressure at the top of the closed tube?

FIGURE 14.12 A water-filled tube.

FIGURE 14.12 Full Alternative Text

MODEL This is a liquid in hydrostatic equilibrium. The closed tube is not an open region of the container, so the water cannot rise to an equal height. Nevertheless, the pressure is still the same at all points on a horizontal line. In particular, the pressure at the top of the closed tube equals the pressure in the open tube at the height of the dashed line. Assume $p_0=1.00\text{atm}$.

SOLVE A point 40 cm above the bottom of the open tube is at a depth of 60 cm. The pressure at this depth is

$$\begin{array}{ccc}\hfill p& \hfill =\hfill & p_0+\rho gd\hfill \\ \hfill & \hfill =\hfill & 1.013\times {10}^5\text{Pa}+(1000{\text{kg/m}}^3)(9.80{\text{m/s}}^2)(0.60\text{m})\hfill \\ \hfill & \hfill =\hfill & 1.072\times {10}^5\text{Pa}=1.06\text{atm}\hfill \end{array}$$

This is the pressure at the top of the closed tube.

REVIEW The water in the open tube pushes the water in the closed tube up against the top of the tube, which is why the pressure is greater than 1 atm.

We can draw one more conclusion from the hydrostatic pressure equation $p=p_0+\rho gd$. If we change the pressure $p_0$ at the surface to $p_1$, the pressure at depth $d$ becomes $p\prime =p_1+\rho gd$. The change in pressure $\text{Δ}p=p_1-p_0$ is the same at all points in the fluid, independent of the size or shape of the container. This idea, that a change in the pressure at one point in an incompressible fluid appears undiminished at all points in the fluid, was first recognized by Blaise Pascal and is called Pascal’s principle.

For example, if we compressed the air above the open tube in Example 14.4 to a pressure of 1.5 atm, an increase of 0.5 atm, the pressure at the top of the closed tube would increase to 1.56 atm. Pascal’s principle is the basis for hydraulic systems, as we’ll see in the next section.