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## Physics library

### Course: Physics library>Unit 9

Lesson 1: Density and Pressure

# 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.
P, equals, start fraction, F, divided by, A, end fraction
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.
This definition also means that the units of pressure are newtons per square meter start fraction, start text, N, end text, divided by, start text, m, end text, squared, end fraction which are also called pascals or abbreviated as start text, P, a, end text.

## 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.
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.
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.
P, equals, start fraction, F, divided by, A, end fraction
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, equals, m, g, so the weight of the column of water can be written as W, equals, m, start subscript, w, end subscript, g where m, start subscript, w, end subscript is the mass of the water column above the beans. We'll plug this into the equation for pressure above and get,
P, equals, start fraction, m, start subscript, w, end subscript, g, divided by, A, end fraction
At this point it might not be obvious what to do, but we can simplify this expression by writing m, start subscript, w, end subscript in terms of the density and volume of the water. Since density equals mass per unit of volume rho, equals, start fraction, m, divided by, V, end fraction , we can solve this for the mass of the water column and write m, start subscript, w, end subscript, equals, rho, start subscript, w, end subscript, V, start subscript, w, end subscript where rho, start subscript, w, end subscript is the density of the water and V, start subscript, w, end subscript is the volume of the water column above the can (not the entire volume of the pool). Plugging in m, start subscript, w, end subscript, equals, rho, start subscript, w, end subscript, V, start subscript, w, end subscript for the mass of the water column into the previous equation we get,
P, equals, start fraction, rho, start subscript, w, end subscript, V, start subscript, w, end subscript, g, divided by, A, end fraction
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, start subscript, w, end subscript, equals, A, h where A is the area of the base of the cylinder and h is the height of the cylinder. We can plug in V, start subscript, w, end subscript, equals, A, h for the volume of water into the previous equation and cancel the areas to get:
P, equals, start fraction, rho, start subscript, w, end subscript, left parenthesis, A, h, right parenthesis, g, divided by, A, end fraction, equals, rho, start subscript, w, end subscript, h, g
Not only did we cancel the areas, but we also created a formula that only depends on the density of the water rho, start subscript, w, end subscript, 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,
P, equals, rho, g, h
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 plus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction .
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.
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, g, h is a scalar that tells you the amount of this squashing force per unit area in a fluid.
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.
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, start subscript, a, t, m, end subscript to the pressure from the liquid rho, g, h.
P, start subscript, t, o, t, a, l, end subscript, equals, rho, g, h, plus, P, start subscript, a, t, m, end subscript
We typically don't try to derive a fancy term like rho, start subscript, a, i, r, end subscript, g, h for the atmospheric pressure P, start subscript, a, t, m, end subscript since our depth in the Earth's atmosphere is pretty much constant for any measurements made near land.
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, point, 01, times, 10, start superscript, 5, end superscript, P, a. 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, start subscript, t, o, t, a, l, end subscript, equals, rho, g, h, plus, 1, point, 01, times, 10, start superscript, 5, end superscript, P, a

## 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, point, 01, times, 10, start superscript, 5, end superscript, P, a 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, start subscript, g, a, u, g, e, end subscript . 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, start subscript, a, b, s, o, l, u, t, e, end subscript. 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, start subscript, a, b, s, o, l, u, t, e, end subscript, gauge pressure P, start subscript, g, a, u, g, e, end subscript, and atmospheric pressure P, start subscript, a, t, m, end subscript which looks like this,
P, start subscript, a, b, s, o, l, u, t, e, end subscript, equals, P, start subscript, g, a, u, g, e, end subscript, plus, P, start subscript, a, t, m, end subscript
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, start subscript, g, a, u, g, e, end subscript, equals, rho, g, h
P, start subscript, a, b, s, o, l, u, t, e, end subscript, equals, rho, g, h, plus, 1, point, 01, times, 10, start superscript, 5, end superscript, start text, space, P, a, end text
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, start subscript, o, b, j, e, c, t, end subscript into the formula for gauge pressure within a fluid P, equals, rho, g, h, but the density in this formula is specifically referring to the density of the fluid rho, start subscript, f, l, u, i, d, end subscript 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 start text, a, end text, t, m. The conversion between pascals and atmospheres is, not surprisingly, 1, start text, a, t, m, end text, equals, 1, point, 01, times, 10, start superscript, 5, end superscript, start text, space, P, a, end text 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, point, 20, start text, space, k, g, end text 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, point, 30, start text, c, m, end text. 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, equals, start fraction, F, divided by, A, end fraction, start text, left parenthesis, U, s, e, space, d, e, f, i, n, i, t, i, o, n, space, o, f, space, p, r, e, s, s, u, r, e, point, space, G, a, u, g, e, space, p, r, e, s, s, u, r, e, space, i, s, n, apostrophe, t, space, a, p, p, l, i, c, a, b, l, e, space, h, e, r, e, space, s, i, n, c, e, space, t, h, e, r, e, apostrophe, s, space, n, o, space, f, l, u, i, d, point, right parenthesis, end text
P, equals, start fraction, m, g, divided by, A, end fraction, start text, left parenthesis, P, l, u, g, space, i, n, space, f, o, r, m, u, l, a, space, f, o, r, space, w, e, i, g, h, t, space, o, f, space, t, h, e, space, c, h, a, i, r, space, end text, W, equals, m, g, start text, space, f, o, r, space, t, h, e, space, f, o, r, c, e, space, F, right parenthesis, end text
P, equals, start fraction, m, g, divided by, 4, times, pi, r, squared, end fraction, start text, left parenthesis, P, l, u, g, space, i, n, space, t, h, e, space, t, o, t, a, l, space, a, r, e, a, space, o, f, space, t, h, e, space, f, e, e, t, space, o, f, space, t, h, e, space, c, h, a, i, r, space, 4, times, pi, r, squared, space, f, o, r, space, t, h, e, space, a, r, e, a, space, A, point, right parenthesis, end text
P, equals, start fraction, left parenthesis, 7, point, 20, start text, space, k, g, end text, right parenthesis, left parenthesis, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, divided by, 4, times, pi, left parenthesis, 0, point, 013, start text, space, m, end text, right parenthesis, squared, end fraction, start text, left parenthesis, P, l, u, g, space, i, n, space, n, u, m, b, e, r, s, comma, space, m, a, k, i, n, g, space, s, u, r, e, space, t, o, space, c, o, n, v, e, r, t, space, f, r, o, m, space, c, m, space, t, o, space, m, right parenthesis, end text
P, equals, start fraction, 70, point, 56, start text, space, N, end text, divided by, 0, point, 002124, start text, space, m, end text, squared, end fraction, equals, 33, comma, 200, start text, space, P, a, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, comma, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text

### 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, point, 0, start text, space, m, end text underneath the Mediterranean sea. The density of the seawater is 1025, start fraction, start text, k, g, end text, divided by, m, cubed, end fraction. The window is circular with a radius of 5, point, 60, start text, space, c, m, end text. 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, equals, start fraction, F, divided by, A, end fraction, start text, left parenthesis, U, s, e, space, t, h, e, space, d, e, f, i, n, i, t, i, o, n, space, o, f, space, p, r, e, s, s, u, r, e, space, t, o, space, r, e, l, a, t, e, space, p, r, e, s, s, u, r, e, space, t, o, space, f, o, r, c, e, right parenthesis, end text
F, equals, P, A, start text, left parenthesis, S, o, l, v, e, space, t, h, e, space, f, o, r, m, u, l, a, space, s, y, m, b, o, l, i, c, a, l, l, y, space, f, o, r, space, t, h, e, space, f, o, r, c, e, right parenthesis, end text
F, equals, left parenthesis, rho, g, h, right parenthesis, A, start text, left parenthesis, P, l, u, g, space, i, n, space, t, h, e, space, f, o, r, m, u, l, a, space, f, o, r, space, g, a, u, g, e, space, p, r, e, s, s, u, r, e, space, end text, P, start subscript, g, a, u, g, e, end subscript, equals, rho, g, h, start text, space, f, o, r, space, t, h, e, space, p, r, e, s, s, u, r, e, space, P, right parenthesis, end text
F, equals, left parenthesis, 1025, start fraction, start text, k, g, end text, divided by, m, cubed, end fraction, right parenthesis, left parenthesis, 9, point, 8, start fraction, m, divided by, s, squared, end fraction, right parenthesis, left parenthesis, 63, point, 0, start text, space, m, end text, right parenthesis, left parenthesis, pi, times, open bracket, 0, point, 056, start text, space, m, end text, close bracket, squared, right parenthesis, start text, left parenthesis, P, l, u, g, space, i, n, space, n, u, m, b, e, r, s, space, f, o, r, space, end text, rho, comma, g, comma, h, comma, start text, space, a, n, d, space, end text, A, right parenthesis
F, equals, 6, comma, 230, start text, space, N, end text, start text, space, left parenthesis, C, a, l, c, u, l, a, t, e, comma, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text
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.

## Want to join the conversation?

• Why is not possible to have negative absolute pressure? • why boiling point of water decreases as the pressure goes down? Is here a video that explain this phenomena? • What applies pressure to the surface of a boiling liquid? You may answer, the air molecules. That is absolutely correct. But what about the molecules of the liquid which are above the surface of the liquid, that have been converted into gaseous phase? It is this pressure that prevents additional molecules of the liquid from escaping from the surface of the liquid. So, greater amount of energy(heat) has to be provided to the liquid to make it boil. So, if the pressure is reduced, you are essentially reducing the amount of molecules above the surface of the liquid(like vacuuming them out of the container). Ah yes, this happens only in a closed container. Now, there is no opposition to the liquid molecules to prevent them from escaping from the liquid phase. Therefore the liquid vaporizes quickly, lowering the boiling point of water.
By the way, the pressure exerted by the gaseous molecules on the surface of the liquid in a closed container is called vapour pressure. So, boiling point is not the temperature at which the liquid boils as the temperature is dependent on the pressure. Therefore, a more precise definition of boiling point would be, 'The temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure.'

Hope this helps
• It is said in the article that the weight of air is actually not negligible. My question is: if you're on an open field you would have more air above you than if you were in, say, your house. How come we don't feel that difference at all? • Pressure acts in all directions. When you are in your house, the air is pushed through windows, through doors, through every little crack until the pressure inside is equal to the pressure outside. This is because air is a fluid which conforms to its environment. Solids, on the other hand, behave as you are suggesting.
• "unlike the water example, the density of the air in the atmosphere is not the same at all altitudes."
It seems to me that the density of water should increase, if only slightly, with the massive increases in pressure as one declines to the base of the ocean. (The text implies that there is one atmosphere of pressure increase with every 30 meters below water surface.)
Is it really the case that water's density does not change even under these massive pressures, and if so can you please explain that? • You are right.

Water is often considered "incompressible" to simplify physics calculations, because it is so hard to compress. This is the same kind of simplification as saying that the surface of the Earth is "flat" or that a ball rolling down a hill experiences no friction.

In fact the density of water does increase with pressure. For water initially at normal pressures, you need about 217 atmospheres of pressure per one percent increase in density.
• How can we differentiate Pressure from stress? • I have a question, if im in space am i exerting pressure on something? • so can we also say that the pressure of the air above us is counteracted by our blood pressure that is why we feel a bursting sensation from inside of our body when our blood pressure is high ? and feel a squeezing sensation from the atmosphere when our blood pressure is low ? • i understood that the fluid is pushing downwards on the top of the immersed object due to gravity but why does the fluid want to push onto the sides and the botttom of immersed object and create pressure? • when you go underwater, where do you feel the pressure? top of your head? or ears?

any deeper, you would feel it on your chest too.

its because you have 'displaced' some water and you could think of it like the water is trying to get back into the space that you are taking...

in fact, the upthrust is equal to the weight of that water that you are displacing

okay??  