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 [latex]\text{mg}[/latex] divided by the area [latex]A[/latex] 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 [latex]V[/latex] is related to the dimensions of the container. It is

where [latex]A[/latex] is the cross-sectional area and [latex]h[/latex] 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 [latex]P=\mathrm{h\rho g}[/latex] represents the pressure due to the weight of any fluid of average density [latex]\rho[/latex] at any depth [latex]h[/latex] 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.

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 [latex]{P}_{\text{atm}}[/latex], measured to be

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

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.

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 [latex]\mathrm{h\rho g}[/latex] is [latex]{\text{N/m}}^{2}[/latex].

[latex]\begin{array}{lll}{\left(h\rho g\right)}_{\text{units}}& =& \left(\text{m}\right)\left({\text{kg/m}}^{3}\right)\left({\text{m/s}}^{2}\right)=\left(\text{kg}\cdot {\text{m}}^{2}\right)/\left({\text{m}}^{3}\cdot {\text{s}}^{2}\right)\\ & =& \left(\text{kg}\cdot {\text{m/s}}^{2}\right)\left({\text{1/m}}^{2}\right)\\ & =& {\text{N/m}}^{2}\end{array}[/latex]

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 [latex]3\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}[/latex]?

The aqueous humor in a person’s eye is exerting a force of 0.300 N on the [latex]1\text{.}\text{10}{\text{-cm}}^{2}[/latex] 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?

[latex]1\text{.}\text{09}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}[/latex]

Calculate the average pressure exerted on the palm of a shot-putter’s hand by the shot if the area of contact is [latex]\text{50}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{2}[/latex] and he exerts a force of 800 N on it. Express the pressure in [latex]{\text{N/m}}^{2}[/latex] and compare it with the [latex]1\text{.}\text{00}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{Pa}[/latex] 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 [latex]\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{2}.[/latex] 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 [latex]F=\rho {\mathrm{gh}}^{2}L/2[/latex], where [latex]\rho[/latex] is the density of water, [latex]h[/latex] is its depth at the dam, and [latex]L[/latex] 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].)