Suppose the pressure at the Earth's surface is $P$.
Consider an air column of cross-sectional area $A$.
The upward force on the column is $F_{\text{up}}=PA$.
Denote the weight of the column as $W$.
By definition of "weight", the downward force on the column is $F_{\text{down}}=W$.
Suppose the pressure is too low, such that $F_{\text{up}}<F_{\text{down}}.$
The column of air will then fall downward.
As it does, it more air molecules are arriving at the surface of the Earth, increasing the density of air and therefore also increasing the pressure.
Since the pressure increases, so does $F_{\text{up}}$.
This will continue until $F_{\text{up}} = F_{\text{down}}$, at which time the system is in equilibrium and stays the same.
In other words, the pressure is such as to balance the weight of the column because that's the only situation which won't immediately change.
As John Rennie explained, in the American Engineering System, force is expressed in lb$_f$, mass in lb$_m$, and acceleration in ft/sec$^2$. This system is not coherent. Hence, a conversion factor other than one must be used in the equation for force; that is, $F=\frac{ma}{g_c}$, where $g_c=32.174 \frac{lbm \cdot ft}{sec^2 \cdot lbf}$ is a constant, known as the gravitational conversion constant.
Derivation of $g_c$
The principle of conservation of units is used to derive $g_c$. We wish to convert $F=ma$ from SI to AES.
First, find the dimension of the "hidden constant" in the given equation. In other words, $1=\frac{F}{ma}$ has dimension force divided by the product of mass and acceleration. Next, consider the units of the hidden constant -- the given units are $\frac{N \cdot s^2}{kg \cdot m}$ and the required units are $\frac{lbf \cdot sec^2}{lbm \cdot ft}$.
Now, convert the given units of the hidden constant to the required units. Thus, $$\left(\frac{1 N \cdot s^2}{kg \cdot m}\right) \left(\frac{0.45359 kg}{lbm}\right) \left(\frac{0.3048 m}{ft}\right)\left(\frac{lbf}{4.4482 N}\right)=\frac{1}{32.174}\frac{lbf \cdot sec^2}{lbm \cdot ft}$$
The hydrostatic pressure is calculated as follows:
$$p=D\frac{(\rho g_c)}{144g} \ ft \frac{ft^2 \ lbm \ ft \ sec^2 \ lbf}{in^2 \ ft^3 \ lbm \ ft \ sec^2}$$
where $p$ is hydrostatic pressure in psig, $D$ is depth (or height) of the fluid in ft, $\rho$ is the fluid density (lbm/ft^3). Note that the density of fresh water is 62.4 lbm/ft^3, $g_c$ is the gravitational constant of acceleration (32.2 ft/sec^2), and $g$ is units conversion to lbf (32.2 $\frac{lbm \ ft/sec^2}{lbf}$)
In AES the fluid density is often express in lbm/gal. There is 7.48 gal/ft^3, therefore $p=0.05194D\rho$.
It is often convenient to express the hydrostatic pressure as a fluid pressure gradient or hydrostatic pressure developed per unit height of fluid.
$\nabla p =0.052\rho$, where $\nabla p$ is the hydrostatic pressure gradient (psig/ft). For fresh water the pressure gradient is: $$\nabla p=0.052(8.33)=0.433\ \text{psi/ft}$$
Best Answer
The analogy is good, but you appear to misunderstand it. Water pressure is not a function of the surface area of the piping, any more so than voltage is a function of the surface area of the wiring.
The water pressure is a function of the strength of the pump which is charging the pipework (assuming the quantity of water available to charge the pipework in the first place, exceeds the capacity of the pipework). And in a situation where the water is flowing freely out after the pump, it is also a function of the resistance of the pipework that comes after the pump.
A static pressure of 1bar in a small diameter pipe is the same as a static pressure of 1bar in a large diameter pipe. Drill a small hole in either pipe, as if applying a multimeter to a wire, and you will measure the same 1bar of pressure in both.
The difference is when the water is flowing. If the water is at a pressure of 1bar leaving the pump at the pump's outlet, then there will be a pressure drop along the pipe according to the cross-section (in the ideal case of a circular pipe, the diameter) of the pipe and the distance from the pump outlet. The pressure at the open end of the pipe will not be 1bar, but something less depending on the diameter and the length from the pump.
The drop will be larger for smaller pipes than larger ones.
However, if the open end of the pipe is then closed, the pressure will charge up to the same pressure as at the outlet of the pump, and the final static pressure will be the same regardless of the diameter of the pipe.
This is why "voltage drop" has to be measured with the relevant load switched on, just as pressure drop has to be measured with the outlet open.
As to units, the electrical analogue unit for water volume - often measured in litres - is the coulomb. The electrical analogue unit for water flow - not given a first class unit, but often measured in litres/second - is amps.
Pounds per square inch - psi - is slightly confusing in this context. It does not mean the pressure is a given number of pounds distributed over the available square-inchage of the pipework (so that increasing the square-inchage of the pipework leads to a reduction in the number of pounds applied to each square inch). The psi is the amount of pressure being applied to every square inch of the pipework, howsoever many square inches there may in fact be.
Increasing the internal surface area of a solid container does not reduce the pressure of the contents. But increasing the volume of a container requires more litres to be added to charge the container to a given pressure in the first place.
Similarly, the surface area of a wire does not change the voltage - the voltage is the pressure that is apparent across all surface of the wire, and increasing the surface area of the wire will simply mean the exact same voltage level being apparent across a larger surface than before.
What a wire of larger volume does do, is require more coulombs of electricity to charge the wire to that voltage in the first place. The eletrical analogue unit of volumetric capacity - also often measured in litres, confusingly (mainly because liquids are conventionally treated as incompressible) - is the farad.
Edit: I thought I'd add further to this as it provoked my own interest about the water analogy. I hope I haven't spoiled a good start.
I was wondering how to analogise watts and joules. I've already said that coulombs are the analog unit of water quantity in litres. We would usually measure water transfer or consumption in terms of such litres. But why do we very rarely talk of electrical quantities in terms of coulombs?
We more often talk of everyday electricity consumption in terms of watt-hours - which is an equivalent to joules (1 watt-hour = 3,600 joules). What then is the water analog unit for joules?
The answer is that it is actually joules for both. Joules (and by implication watt-hours) is a unit of energy or work quantity.
"Work" here is not a description of a physical quantity. It is a description of a purely abstract mechanical quantity, based on the ability of the system in question to bring about change in the physical properties or spatial arrangements of physical things. When a quantity of work is transferred in a system, we are describing what quantity of change has occurred in that system.
The word "change" here is intentionally elusive. Specific kinds of physical change can be measured in different units - a litre of water is raised off the ground by 1 meter, or a litre of water is heated by 100 degrees. But each of these specific changes can be reduced (by applying the laws of physics) to a general description of the quantity of work transferred.
Note also that I talk of "work transferred", not "work done". Because this quantity of work is neither created nor destroyed, nor produced or consumed, only transferred. And it can be transferred on (so as to cause consequential change), or transferred back (so as to undo the change), and the laws of physics describe the terms on which this can be done. And there are always two sides to the equation - if work is transferred to some change, there must be another change elsewhere of opposite work quantity.
Nowadays we rarely use transfers of water through pipework to do work in this mechanical sense. We pipe water into the home as a solvent for washing, as a carrier for sewage, or for human consumption, and it is the water itself (i.e. it's intrinsic physical or chemical properties) which is seen as the main feature for this purpose, not the work which the piping and the water transfer system is capable of performing.
Transferring water in and out of the home does perform work, but it is not of everyday interest to measure exactly how much work is performed in doing so.
The contexts in which water, or steam, are used specifically to do mechanical work, and where that work would be subject to measurement, are in the realm of mechanical engineering. A main application is in turbines (of which a simple waterwheel is but a simple example). Now we see an application in which the quantity of water in litres is no longer intrinsically relevant, but instead it's ability to do work by driving the turbine is the quantity of interest. An unpressured quantity of water is not capable of doing any work in driving the turbine. But a pressured quantity is. And the amount of work is a function of the pressure of the transfer, not just the volume.
And that brings us back around to electricity. We do not move eletrical charges around wires for the intrinsic properties conveyed by charge, or so that we can hold the charge itself in our hands. We do it so that the movement of electrical charge performs work - usually in the end, work of a non-electrical kind, such as heating things or driving mechanical appliances. And that is why we measure electrical consumption in terms of joules, or watt-hours, because the ability to transfer work (rather than to transfer electrical charge) is the main purpose of electrical systems.
And finally, what of the watt unit? That is a measure of power, or the rate of transfer. But crucially, it is not a measure of the rate of transfer of litres of water, or of coulombs of electrical charge. It is a measure of the rate of transfer of work (measured in joules), the abstract quantity which I've described above.
It is not, as might first be thought, like the transfer of litres of water. A transfer of a very small amount of water under very great pressure, is capable of doing the same amount of work, as the transfer of a very large amount of water under more modest pressure. And watts measures the rate of work transferred electrically, regardless of the voltage or amperage used to do it (and thus regardless of the amount of coulombs actually moved, which in AC circuits is a net of zero anyway - it's like the same water volume being pumped and sucked back and forth in rapid alternation).