# What is pressure?

Pressure is kind of like force, but not quite.

## What does pressure mean?

If you tried to hammer a bowling pin into the wall, nothing would probably happen except for people deciding to no longer lend you their bowling pins. However, if you hammer with the same force on a nail, the nail would be a lot more likely to penetrate the wall. This shows that sometimes just knowing the magnitude of the force isn't enough: you also have to know how that force is distributed on the surface of impact. For the nail, all the force between the wall and the nail was concentrated into the very small area on the sharp tip of the nail. However, for the bowling pin the area touching the wall was much larger, and therefore the force was much less concentrated.
Person hitting a bowling pin and a nail with a hammer.
To make this concept precise, we use the idea of pressure. Pressure is defined to be the amount of force exerted per area.
${\Large P=\dfrac{F}{A}}$
So to create a large amount of pressure, you can either exert a large force or exert a force over a small area (or do both). In other words, you might be safe lying on a bed of nails if the total surface area of all the nail tips together is large enough.
Yeah, people do this. Since it's the total area over which the force is distributed that counts, the total surface area of all the nails can reduce the pressure that's created by your weight downward. But there has to be a huge number of nails for this to work.
This definition also means that the units of pressure are newtons per square meter $\dfrac{\text{N}}{\text{m}^2}$ which are also called pascals or abbreviated as $\text{Pa}$.
Blaise Pascal was a 17th century scientist, mathematician, and philosopher. Not only did he contribute the the understanding of fluid pressure, but he is also noted for "Pascal's wager", "Pascal's triangle", and "Pascal's theorem".

## How do you find the pressure in a fluid?

A solid surface can exert pressure, but fluids (i.e. liquids or gases) can also exert pressure. This might seem strange if you think about it because it's hard to imagine hammering in a nail with liquid. To make sense of this, imagine being submerged to some depth in water. The water above you would be pushing down on you because of the force of gravity and would therefore be exerting pressure on you. If you go deeper, there will be more water above you, so the weight and pressure from the water would increase too.
Not only can the weight of liquids exert pressure, but the weight of gases can as well. For instance, the weight of the air in our atmosphere is substantial and we're almost always at the bottom of it. The pressure exerted on your body by the weight of the atmosphere is surprisingly large. The reason you don't notice it is because the atmospheric pressure is always there. We only notice a change in pressure above or below normal atmospheric pressure (like when we fly in an airplane or go underwater in a pool). We aren't harmed by the large atmospheric pressure because our body is able to exert a force outward to balance the air pressure inward. But this means that if you were to be thrown into the vacuum of outer space by space pirates, your body pressure would continue pushing out with a large force, yet no air would be pushing in.
You probably wouldn't blow up since your body/skin/bones are strong enough to hold you together. Still, it would be really, really uncomfortable. Besides the lack of oxygen and possible direct radiation exposure from the sun, your eyes would bulge, your eardrums could pop, and the saliva on your tongue would probably boil since the boiling point of water decreases as pressure goes down. At zero pressure your body temperature is enough to boil the water on your tongue as well as the fluid in your eyes. So basically, don't ever get caught by space pirates.
Okay, so the weight of a fluid can exert pressure on objects submerged in it, but how can we determine exactly how much pressure a fluid will exert? Consider a can of beans that got dropped in a pool as seen in the following diagram.
This is one of the great mysteries of the universe. I doubt we will ever know. If you find out, please contact the Department of Physics Mysteries immediately.
A can of beans submerged below the water to a depth h.
The weight of the column of water above the can of beans is creating pressure at the top of the can. To figure out an expression for the pressure we'll start with the definition of pressure.
$\Large P=\dfrac{F}{A}$
For the force $F$ we should plug in the weight of the column of water above the can of beans. The weight is always found with $W=mg$, so the weight of the column of water can be written as $W=m_{w}g$ where $m_w$ is the mass of the water column above the beans. We'll plug this into the equation for pressure above and get,
$P=\dfrac{m_w g}{A}$
At this point it might not be obvious what to do, but we can simplify this expression by writing $m_w$ in terms of the density and volume of the water. Since density equals mass per unit of volume $\rho=\dfrac{m}{V}$ , we can solve this for the mass of the water column and write $m_w=\rho_w V_w$ where $\rho_w$ is the density of the water and $V_w$ is the volume of the water column above the can (not the entire volume of the pool). Plugging in $m_w=\rho_w V_w$ for the mass of the water column into the previous equation we get,
$P=\dfrac{\rho_w V_wg}{A}$
At first glance this appears to have only made the formula more complex, but something magical is about to happen. We have volume in the numerator and area in the denominator, so we're going to try to cancel something here to simplify things. We know that the volume of a cylinder is $V_w=Ah$ where $A$ is the area of the base of the cylinder and $h$ is the height of the cylinder. We can plug in $V_w=Ah$ for the volume of water into the previous equation and cancel the areas to get:
$P=\dfrac{\rho_w (Ah)g}{A} = \rho_w h g$
Good question. The original area $A$ in the denominator was the area upon which the force is exerted, which was the area of the top of the can. The area $A$ in the numerator refers to the area of the column of water. Since the area of the column of water is equal to the area of the top of the can, these areas do in fact cancel.
Not only did we cancel the areas, but we also created a formula that only depends on the density of the water $\rho_w$, the depth below the water $h$, and the magnitude of the acceleration due to gravity $g$. This is really nice since nowhere does it depend on the area, volume, or mass of the can of beans. In fact, this formula doesn't depend on anything about the can of beans other than the depth it is below the surface of the fluid. So this formula would work equally well for any object in any liquid. Or, you could use it to find the pressure at a specific depth in a liquid without speaking of any object being submerged at all. You'll often see this formula with the $h$ and the $g$ swapping places like this,
${\Large P= \rho gh}$
Just to be clear here, $\rho$ is always talking about the density of the fluid causing the pressure, not the density of the object submerged in the fluid. The $h$ is talking about the depth in the fluid, so even though it will be "below" the surface of fluid we plug in a positive number. And the $g$ is the magnitude of the acceleration due to gravity which is $+9.8 \dfrac{\text{m}}{\text{s}^2}$ .
Now you might think, "OK, so the weight of the water and pressure on the top of the can of beans will push the can downward right?" That's true, but it's only a half truth. It turns out that not only does the force from water pressure push down on the top of the can, the water pressure actually causes a force that pushes inward on the can from all directions. The overall effect of the water pressure is not to force the can downward. The water pressure actually tries to crush the can from all directions as seen in the diagram below.
OK, if you are really clever you might have realized that the bottom of the can is slightly lower in the fluid than the top of the can, and since the pressure gets larger the deeper you go ($P_{gauge}=\rho g h$) the upward pressure on the bottom of the can should be slightly larger than the downward pressure on the top of the can. This means that the overall effect of the pressure from the water is to crush the can and to exert a net upward force on it. This net upward force from the difference in pressure is the reason why there's a buoyant force on objects submerged in a fluid! But...we're getting a little ahead of ourselves so let's hold this thought for now.
A can of beans being squeezed by water pressure.
If it helps, you can think about it this way. When the can of beans fell into the water, it quite rudely displaced a large amount of water molecules from the region where the can is now. This caused the entire water level to rise. But water is pulled down by gravity which makes it want to try and find the lowest level possible. So the water tries to force itself back into the region of volume that it was displaced from in an effort to try and lower the overall height of the body of water. So, whether a can of beans (or any other object) is in the water or not, the water molecules are always being squashed into each other from the force of gravity as they try to lower the water level to the lowest point possible. The pressure $P$ in the formula $\rho gh$ is a scalar that tells you the amount of this squashing force per unit area in a fluid.
OK, so here is a subtle fact about pressure; it's defined to be a scalar, not a vector. So why do people seem to represent pressure in diagrams with arrows as if it were a vector with a particular direction?
Even though pressure is not a vector and has no direction in and of itself, the force exerted by the pressure on the surface of a particular object is a vector. So when people draw diagrams with pressure pointing in specific directions, those arrows can be thought of as representative of the direction of the forces on those surfaces exerted by the pressure from the fluid.
If there were no surface upon which the pressure could exert a force, it would make no sense to draw a direction for the force at that point inside the water. On the left hand side of the diagram below there are water molecules and pressure, but no well defined direction of force. The right hand side of the diagram below shows the well defined directions of forces on an ice cream cone submerged in the water.
While we're on the topic, we might as well make it clear that the force exerted on a surface by fluid pressure is always directed inwards and perpendicular (at a right angle) to the surface.
At this point, if you've been paying close attention you might wonder "Hey, there's air above the water right? Shouldn't the weight of the column of air above the column of water also contribute to the total pressure at the top of the can of beans?" And you would be correct. The air above the column of water is also pushing down and its weight is surprisingly large.
Many people think air has no mass and no weight, but that's not true. The narrow column of air with the same radius as a typical can of beans that stretches from sea level to the top of the atmosphere has a mass of around $30 \text{ kg}$ (that's like the weight of 30 pineapples). The force from atmospheric pressure on the top of a chessboard would be comparable to the weight of a car.
You might wonder how we can pick up the chessboard so easily if the weight of a car is pushing down on it, but it's because the weight of a car is also pushing up on it. Remember that the force from fluid pressure does not just push down, it pushes inwards perpendicular to the surface from every direction. It may not seem like there is any air under the chessboard when placed on the table but the roughness and cracks of the chess board are enough to allow air underneath. If you could get rid of all the air underneath the chessboard and prevent air from being allowed to sneak back in, that board would be stuck to the table like a suction cup. In fact, that's how suction cups work. They push the air out to create less pressure inside than out. The smooth plastic of the suction cup prevents air from sneaking back in. The higher pressure outside air pushes the suction cup into the surface. (see the diagram below)
Once air sneaks back in, the inside pressure becomes the same as the outside pressure and the cup can easily be taken off the surface.
If you wanted a formula for the total pressure (also called absolute pressure) at the top of the can of beans you would have to add the pressure from the Earth's atmosphere $P_{atm}$ to the pressure from the liquid $\rho gh$.
${\Large P_{total}=\rho gh +P_{atm}}$
We typically don't try to derive a fancy term like $\rho_{air} g h$ for the atmospheric pressure $P_{atm}$ since our depth in the Earth's atmosphere is pretty much constant for any measurements made near land.
A problem with trying to use $\rho_{air} gh$ to find the pressure at a certain depth in the atmosphere is that unlike the water example, the density of the air in the atmosphere is not the same at all altitudes. As you go higher in the atmosphere the density of air decreases so we can't treat $\rho_{air}$ as a constant.
This means that the atmospheric pressure at the surface of the Earth stays relatively constant. The value of the atmospheric pressure at the surface of the Earth is stuck right around $1.01 \times10^5 Pa$. There are small fluctuations around this number caused by variations in weather patterns, humidity, altitude, etc., but for the most part when doing physics calculations we just assume that this number is a constant and stays fixed. This means, as long as the fluid you're finding the pressure for is near the surface of the Earth and exposed to the atmosphere (not in some sort of vacuum chamber) you can find the total pressure (also called absolute pressure) with this formula.
$P_{total}= \rho gh +1.01 \times 10^5 Pa$
The $\rho gh$ corresponds to the pressure created by the weight of a liquid, and the $1.01 \times 10^5 \text{ Pa}$ corresponds to the pressure of the Earth's atmosphere near sea level.

## What's the difference between absolute pressure and gauge pressure?

When measuring pressure, people often don't want to know the total pressure (which includes atmospheric pressure). People typically want to know the difference in some pressure from atmospheric pressure. The reason is that atmospheric pressure doesn't change much and it's almost always present. So including it in your measurements can feel a bit pointless at times. In other words, knowing that the air inside of your flat tire is at an absolute pressure of $1.01 \times 10^5 Pa$ isn't really all that useful (since being at atmospheric pressure means your tire's flat). The extra pressure in the tire above atmospheric pressure is what will allow the tire to inflate and perform properly.
Because of this, most gauges and monitoring equipment use what is defined to be the gauge pressure $P_{gauge}$ . Gauge pressure is the pressure measured relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, zero at atmospheric pressure, and negative for pressures below atmospheric pressure.
The total pressure is commonly referred to as the absolute pressure $P_{absolute}$. Absolute pressure measures the pressure relative to a complete vacuum. So absolute pressure is positive for all pressures above a complete vacuum, zero for a complete vacuum, and never negative.
This can all be summed up in the relationship between the absolute pressure $P_{absolute}$, gauge pressure $P_{gauge}$, and atmospheric pressure $P_{atm}$ which looks like this,
$\Large P_{absolute} = P_{gauge} + P_{atm}$
For the case of finding the pressure at a depth $h$ in a non-moving liquid exposed to the air near the surface of the Earth, the gauge pressure and absolute pressure can found with,
$P_{gauge}=\rho gh$
$P_{absolute} = \rho g h + 1.01 \times 10^5\text{ Pa}$
Because the only difference between absolute pressure and gauge pressure is the addition of the constant value of atmospheric pressure, the percent difference between absolute and gauge pressures become less and less important as the pressures increase to very large values. (see the diagram below)
Diagram showing the values of various gauge and absolute pressures.

People often want to plug in the density of the object submerged $\rho_{object}$ into the formula for gauge pressure within a fluid $P=\rho g h$, but the density in this formula is specifically referring to the density of the fluid $\rho_{fluid}$ causing the pressure.
People often mix up absolute pressure and gauge pressure. Remember that absolute pressure is the gauge pressure plus atmospheric pressure.
Also, there are unfortunately at least 5 different commonly used units for measuring pressure (pascals, atmospheres, millimeters of mercury, etc). In physics the conventional SI unit is the pascal Pa, but pressure is also commonly measured in "atmospheres" which is abbreviated as $\text atm$. The conversion between pascals and atmospheres is, not surprisingly, $1 \text{atm} = 1.01 \times10^5 \text{ Pa}$ since one atmosphere is defined to be the pressure of the Earth's atmosphere.

## What do solved examples involving pressure look like?

### Example 1: Finding the pressure from the feet of a chair

A $7.20 \text{ kg}$ fuchsia colored four legged chair sits at rest on the floor. Each leg of the chair has a circular foot with a radius of $1.30\text{cm}$. The well engineered design of the chair is such that the weight of the chair is equally distributed on the four feet.
Find the pressure in pascals between the feet of the chair and the floor.
$P=\dfrac{F}{A} \quad \text{(Use definition of pressure. Gauge pressure isn't applicable here since there's no fluid.)}$
$P=\dfrac{mg}{A} \quad \text{(Plug in formula for weight of the chair } W=mg \text{ for the force F)}$
$P=\dfrac{mg}{4\times \pi r^2} \quad \text{(Plug in the total area of the feet of the chair 4\times \pi r^2 for the area A.)}$
$P=\dfrac{(7.20\text{ kg})(9.8\dfrac{\text{m}}{\text{s}^2})}{4\times \pi (0.013\text{ m})^2} \quad \text{(Plug in numbers, making sure to convert from cm to m)}$
$P=\dfrac{70.56 \text{ N}}{0.002124 \text{ m}^2}=33,200 \text{ Pa} \quad \text{(Calculate, celebrate!)}$

### Example 2: Force on a submarine porthole

A curious seahorse is looking into the circular window of a submarine that is sitting at a depth of $63.0 \text{ m}$ underneath the Mediterranean sea. The density of the seawater is $1025\dfrac{\text{kg}}{m^3}$. The window is circular with a radius of $5.60 \text{ cm}$. The seahorse is impressed that the window does not break from the pressure caused by the weight of the seawater.
What is the magnitude of the force exerted on the surface of the circular submarine window from the weight of the water?
$P=\dfrac{F}{A} \quad \text{(Use the definition of pressure to relate pressure to force)}$
$F=PA \quad \text{(Solve the formula symbolically for the force)}$
$F=(\rho gh)A \quad \text{(Plug in the formula for gauge pressure } P_{gauge}=\rho gh \text{ for the pressure P)}$
$F=(1025\dfrac{\text{kg}}{m^3})(9.8\dfrac{m}{s^2})(63.0\text{ m})(\pi \times [0.056 \text{ m}]^2) \quad \text{(Plug in numbers for } \rho, g, h, \text{ and } A)$
Since the window is circular we are going to use the formula for the area of a circle $A=\pi r^2$.
$F=6,230 \text{ N} \quad \text{ (Calculate, and celebrate!)}$
Note: We used the gauge pressure in this problem since the question asked for the force caused from "the weight of the water", whereas the absolute pressure would yield a force caused by the weight of the water and the weight of the air above the water.