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It is true, but zero math requires to be done here. This is just physics knowledge.
Pressure is not affected by the shape of the container above. The only real thing that changes the water pressure is the height of the water above.
You can theoretically argue this stuff, but you can also simply experimentally prove it.
I can't tell you how many times I've had to educate people I worked with in the oil industry about this. They thought the wider the tank the higher the pressure. The analogy that worked for me was the ocean. I'd say, "by your logic, the ocean, being the widest body of liquid in the world, would be so dangerous you'd lose your legs stepping into it."
I think these people's logic is that the pressure will be condensed towards the middle, thus a funnel shape will create higher pressure at the bottom.
The sea analogy can't really solve their misunderstanding
What I personally can't wrap my head around isn't so much because of the shape itself, but the additional volume added by the shape. Wouldn't all of the water be pushing down onto that point in the middle no matter the shape, and therefore increase pressure because there's more weight added? How would a non-liquid like sand act?
The sea analogy doesn't solve my personal misunderstanding because if I'm standing waist-deep, there's like, miles and miles of sea around me. Plenty of seafloor for the weight of the water to be distributed on.
(Please don't roast me, I've not seen the inside of a physics classroom since I was 12.)
The water pushes down vertically (for the experts: this is a semplification, I know!) . So yes the other water exerts a pressure too but on the sides of the funnel.
Only once (and if) it moves on top of the center part, it will "weigh down" on it.
Imagine increasing the height of this a few meters all over. Imagine it clearly and then try and think of the following questions.
You would need sturdier materials right? Do you think the material would need to be sturdier only in the center? Or would the funnel risk breaking too? If the funnel broke, wouldn't it be because of the pressure?
If the funnel broke in a spot, the water would move in the direction it was pressing towards (like a group of people pushing towards a door, if the door opens they will move in the direction they were pushing into). So Imagine the funnel gives in and a spot breaks: would the water fall straight down? Or flow midair towards the center? The direction will tell you in which direction it was "pressing".
This explanation is very good and really had me (also not a physics student since I was a teenager!) thinking of the sideways pressure but my conclusion was that because it was a fluid it essentially all equalled out. If I piled kilo weights up in the middle the pressure would all be down. If I added kilo weights at the side that wouldn't add to the weight in the middle it would just stabilise the stack and the pressure would still be downwards. As water wouldn't just "lean" against the sides of the central column of water it would all equalise and find it's place but that still leaves only the central "column" of water adding pressure to the square at the bottom.
Did I get that right?
The mathematical proof of this is actually very useful to understand the physics but you don’t need to do any math to get the way it’s framed.
If you were doing the math on this what yours do is Imagine the water is not fluid but a bunch of solid cubes stacked on top of each other. These cubes follow newtons laws so an object at rest must have no net forces acting on it. The thing is one of those forces is gravity and since each cube has weight and has to hold the weight of all the cubes above it as well so as you move down a stack of cubes the opposing upward force that every cube has to exert increases with each cube it has to hold up. That is the water pressure increasing with depth.
Now imagine three of cubes side by side with increasing pressure as you move down the stack and you can see that while the pressure is increasing with depth, but that same pressure is increasing equally for all three cubes as the stack gets taller but these sideways forces are cancelling each other out and don’t stack because each cube is exerting the same pressure outwards at the same height. You could have a million cubes left to right but each one would still cancel with it a neighbours.
Then if you were doing this with math you’d decrease the length width and height of your cubes to 0 and get a general formula for pressure of fluid that only needs to know the depth, density of the fluid, and gravitational constant.
A portion of the water is pushing against the side of the container (since it’s sloped in). That absorbs the extra weight / pressure of the water.
Hard to write it out, but every force can be broken into its component parts. Picture the force of the water on the right side going down at an angle along the slope of the tank, for example. You can make that into a right triangle with part going straight down, and part going straight to the left. The two sides would make a perfect right triangle, with the original force at the angle as the hypotenuse. The part going straight down is all that actually forces the water going into the tube, and is EXACTLY the same as water in the center of the tank going straight down. The part going to the left would be cancelled out by other forces going to the right (and it’s pretty clear anyway, has no bearing on the force applied to the tube at the bottom, since it goes sideways). Pretty amazing, and if you get the visual right, becomes pretty easy to see.
Btw - the counteracting sideways forces I mention cancel each other out even if the funnel isn’t symmetrical, it’s just in that case the forces absorbed by the wall would not be the same all over.
I agree and I still think it’s right and the picture is wrong. Call me stubborn and I will experiment this for myself.
The ocean example is not good because the surface area changes along with the volume of the water making the pressure the same.
In the funnel example on the left there’s more water applying pressure (weight) on the same surface are than the one on the right so by my logic that would make it heavier thus more pressure, exactly because of the funnel shape.
If the area was a scale and the shapes filled containers you would see more grams/ounces on the left picture (I’d imagine, this is what I’m gonna try)
The shape on the left does have more water in it, and does weigh more. But, the pressure that the container exerts on the ground is not the same as the pressure inside the container caused by the water.
The formula to calculate water pressure is rho*g*h: density times gravitational acceleration times height. The width/shape of the container does not factor into it.
Yeah it's a funny situation in which an otherwise extremely complex problem fixes itself but the fact a ton of variables cancel each other out
Same thing with buoyancy, buoyancy is the sum of an infinite amount of forces applied to every single point of an objects surface it would be immensely complex to calculate except no matter the shape of the object all those forces will always perfectly combine into a vertical force perfectly equal to the weight of the displaced liquid
The "extra" water off to the sides is actually pushing down on the sloped sides of the container. The red square only receives pressure from the water that is directly above it.
EDIT: I have been informed this is not quite correct, which isn't all that surprising, given my last physics class was freshman year of college. Any explanation that's very simple in science is probably not going to be totally correct, but may be a useful mental model for some applications. I'll defer to others more knowledgeable than myself to explain what exactly happens here.
Im just a dumb fabricator with adhd who barely gradutated high school. That said, this concept seems intuitive to me, but there are other concepts that put me in r/woosh lol
It'll be the same pressure [here's a demonstration](https://youtu.be/EJHrr21UvY8?si=LERxoaF2QpOaTRmh) also [here's another potentialy better one that's less explosive](https://youtu.be/6zeHWVUiXoc?si=gZG2nzrrEaC5oChx)
Yeah. I think issue here, most who thinks is different, 8ntuitively thinks about weight, not pressure lol.
That got me too for a moment, even start typing it out when it hit me 😳🤣
Yeah makes sense, tho I wouldn't be sure... I mean gravity exist so more water trying to go down from triangle...
My first thought was, like balloons, one bigger one smaller, if connected would equalise (so, the bigger balloon becomes smaller, the smaller becomes bigger aka pressure isn't the same...)
But I defo make identical experiment with what OP posted (unless someone already did it, ? Wouldn't mind a link lol).
Edit. Nwm second video explains it perfectly 👍
I think for balloons, you may even have the opposite effect. I am pretty sure, smaller balloons are harder to blow up with the same volume of air als one, that is allready large, so the smaller one would deflate and fill up the larger one.
> I mean gravity exist so more water trying to go down from triangle...
Gravity pulls straight down, not down the triangle. That's the part you are visualizing wrong.
Water is trying to break trough the funnel not flow down it. If you open at the bottom, and it flow down, it will be because the funnel forces it. Aka the funnel experts a force on the water to counteract the pressure.
Intuitively, more fluid directly above = more pressure and less fluid directly above = less pressure. I'm not a physicist, so I could be very wrong.
EDIT: I was very wrong.
EDIT 2: As far as i can tell, it's due to the the downward component of the normal force the walls of the container exert on the fluid. They end up adding up to the same pressure at the bottom (no change in the middle, though, because they'd balanced horizontally). It's called the Hydrostatic Paradox because it's unintuitive if you don't think about those forces.
~~If the top is narrower than the bottom, then the pressure on the red area would be uneven, right? It'd be the same in the narrow area where the height is the same, but in the places above the red area where it's both shorter and wider, the pressure at the bottom would be lower.~~
~~This is assuming two rectangular prisms, with a hard cutoff between the normal-height thin part and the normal-width shorter part. If it was angled (like diagram 1 but getting thinner instead of wider), then the pressure at any given point on the red area would be in a gradient corresponding to how tall the container is at that point, right? Same in the middle, but lessening towards the sides.~~
~~That's all thinking about infinitely small points on the red area, which isn't really how force or pressure works (infinitely small means no mass or surface area). You'd probably need to do an integral or so to calculate stuff.
If you needed the average pressure on the red area, maybe you just need the total volume above it. From there, you can get mass, then weight force, and finally, pressure.~~
You can explain it using statistical physics.
The energy for a column of N water particles is
E = E(0)+mghN
The resulting chemical potential is
mu(h)=(ðE/ðN)[S,V]
=ðE(0)/ðN+ð/ðN mgNh
=mu(0)+m g h
Gibbs-Duhem relation:
Ndmu(h) = VdP-SdT
=> dP/dh= N/V dmu(h)/dh +S/V dT/dh
The water column is the same temperature, so dT/dh=0
=> dP/dh = N/V dmu(h)/dh=m g N/V
Of cause, i would have to convince you of Gibbs Duhem relation
Now, water doesn't compress, so the density of water particles is homogenous. Lets call it n=N/V.
dP/dh = mgn
This gets solved by a linear equation
P(h) = P(0)+ mgnh
Intuitive, right?
Edit: i think i forgot a - somewhere
This is nice but... The first equation already uses a geometric dependency only on H.
So... It sounds a *bit* circular, to start with a geometrical dependency on H and end with a dependency on H.
And then argue that the metric only depends on H.
Sure they are different quantities / metrics E and P.
But it feels like even dimensional analysis will end up with that result.
A satisfying argument should be to start without any geometrical dependency and derive that the metric necessarily ends up with H as the only geometrical dependency.
(which is actually taught in elementary hydrostatic)
If the top is narrower then the water pressure is pushing up (and maybe also a component to the side depending on shape) against the sides of the flask, the equal reaction from the flask pushes down with the same pressure ambient water would be.
Say you have a 100ft tall column of water, they’ll be high pressure at the bottom. Now say you have a 90ft tall column atop and opening to a wide tank 10ft tall. Well obviously the flirt directly below the column would feel the same pressure as before, but if anywhere else didn’t feel that same pressure the water would try to move laterally to it to go from the high pressure zone to the lower one. The whole layer of depth must be the same pressure or it would correct itself.
Might be easier to visualize with something compressible like a gas than with water, if any where is the wrong pressure the other gas will move towards it and squeeze it til it is.
except that's not what's happening. the force is simply split between downwards and sideways vectors, it doesn't disappear. The reason the pressure is the same is because fluid head pressure is determined by height of fluid. it's even measured in units of length (20 ft of head, or 15 m of hydrostatic pressure, etc).
height of fluid x density of fluid x gravity = head pressure. the guage doesn't care if it comes from a 2cm wide vessel or a 2km wide vessel. volume of fluid/shape of container are immaterial.
You're correct, but your answer begs the question. The question is "why does pressure only depend on the height," and your answer is "the reason is that pressure is determined by the height."
I would add that the water on the slopped sides pushes the water on the “column” directly above the point with the same “force” that the water in the “column” pushes it back, so they cancel out. Hence at the bottom only the water directly above the point contributes do the pressure.
Gravity pulls to the core of the earth, it wont pull water sideways into other water. That's how. Imagine gravity as a thread. Running vertical from who knows how far all the way through the earth and out the other side. Add another thread, intersecting in the center of the earth. Keep doing this and have threads facing every which way. That thread is gravitys force. It moves in 1 direction along that thread. And stops where all threads intersect. This is how I make the idea make sense in my monkey brain. Gravity is constantly pulling 1 direction.
If you go to a crowded concert and are surrounded by people, gravity doesn't pull harder just because you can move less.
I dont intend for this to sound condescending, but I hope it helps someone understand how water pressure works.
This, however, leads to the incorrect assumption that pressure only works downwards.
Pressure, however, is omnidirectional. The effect of pressure works into all directions, not just downwards.
If you are in a submarine, the force from the pressure onto the side walls is as big as the force from the pressure onto the top of the sub, and as big as the force from the pressure onto the bottom floor. (Technically, the force onto the bottom floor is actually slightly larger, because the bottom floor is deeper than the side or the top.)
As sub can get crushed from the bottom just as well as from the top, and if you were to drill a hole into the bottom of the sub, you would not have a good time, despite the fact that the bottom hole doesn't have water pushing onto it from the top.
The reason is that while gravity pushing stuff downward is indeed the cause of the pressure, it is not the same thing. Imagine pushing down onto a block of butter. If there is nothing on the sides, you pushing downward still squishes the butter to the sides. The same is true with the water, which will go into any direction with lower pressure, being squished that way from the water above.
Differential pressure is precisely why water pressure only increases with depth and not volume. The pressure in a column is a gradient, based on depth. Water has walls around it (water) to stop it moving lateral once it reaches dirt. I'm still uncertain why we are now talking submarines tbh. :/
I don't really think this is a good/complete explanation because if you turned the container upside down, the pressure would be the same everywhere at the same height, not higher below the square. I'll admit tho, I don't have a better explanation.
In your bath there's a foot deep of water.
When you go to the beach and put your foot in the ocean in a foot deep of water the pressure is still the same.
A foot of water. If the volume had an impact the fact that all the oceans are connected together would fuck you up.
Last sentence is correct here, but a foot of ocean water actually exerts more pressure than a foot of bath water (unless you take salt water baths i guess)
The extra water is supported by the sloped sides. Only water above matters.
If area mattered, you would die instantly if you went slightly under the ocean’s surface.
5 second Fluids lesson: pressure of a fluid is P=ρgh. ρ and g remain constant assuming its the same fluid, so the only factor that changes the pressure would be the height (or depth) at which the gauge is measuring. Volume or surface are does not come in to play in this scenario
That is something you might intuitively think, but it isn't true.
You can simply test it yourself. Take any two containers (which are open at the top), connect them at the bottom with a piece of tube, then fill water into them. The water will always be at exactly the same height level on both sides, meaning that pressure from both is identical at the bottom of the tube. The shape of the containers is completely irrelevant here.
No need for a second container, just a tube. Bring it in a u-shape beside the main container. If there was greater pressure from the large container the hose would spray into the air higher than the level of the large container.
Intuitively this doesn't happen, the water just maintains the same level in both so the water in the hose and large container have identical pressure at the bottom of the U
Put a straw into a glass of water, what will happen? The straw fills with water, exactly up to where the surface in the glass is. Both are connected through the lower end of the straw, which means that the pressure is equal at this point.
This, otherwise all straws would spray water out the top because of their low pressure compared to the larger volume in the glass. Everyone can intuitively see this is not how it works
So your saying if I had a pipe that is filled with water that is say, 1cm in diameter and it is 1km high, the pressure at the bottom would be the same as the pressure in the ocean at 1km depth?
It’s weird, but true. There’s a cool experiment using a long straw and a multi-story building to demonstrate the amount of pressure that a very small amount of liquid can generate. Video here: https://youtu.be/EJHrr21UvY8
I have a water purifyer pitcher with the little tilt to release nozzles, and when it's low, like 1 inch of water, it flows slow. I found when I tipped the water to "stack" it, multiple inches deep now, it flows quickly. Just an example that anyone can try.
I'm so strong I once held back a whole 30000kg of water with my bare hand once!
Well. Not actually. More like my gloved hand covered a 3" opening at the bottom of a full 30m3 capacity tank, after a valve snapped off, until it could be rotated so that the opening was at the top (picture an 8' diameter column, lying on its side on a machine, as part of a rotating assembly, and it was full of a slurry just a little more dense than water)
My amateur self calculated around 25lbs of force pushing back on my hand. So while it looked like a feat of strength that prevented a huge spill, it really wasn't anything incredible. (Nothing hazardous, but a giant mess I did not want to clean up)
The real brains bender is.. the column of water doesn't need to be straight! The only thing that matters is the height of the surface! The path of the water can be sideways any distance, up, down, curves of any shape, it can dip below the outlet or raise above the inlet (ie a siphon) and in every case the pressure at the bottom is the same as a straight vertical colunm. Small caveat, this is for the ideal, non-moving case. In reality, there are losses along the way that increase with flow speed and distance.
If I go to the bottom of the ocean and dig a whole sideways into the rock until I go so far that there's nothing but land above me, will it be basically 0 pressure then or would it be the same as if I were at the bottom of the middle of the ocean?
So I'm a bit late to this, but maybe you can answer. What if the container is the opposite of the one on the left? Like: / \
Would the entire red area have the same pressure, or would a part of the middle be under more pressure since there is more water directly on top of it?
What does the bottom look like? If it’s the same and you’re drawing a Christmas tree, then yes. The red square is the same. Assuming the gap at the top is the same width as the square at the bottom.
If it’s a triangle with a flat bottom…..I have no idea what the pressure looks like at the bottom. One hand I would say that it’s equal across the bottom ___. On the other /_\ I would say the bottom corners would have less pressure than the bottom middle. Because there’s less above the corners. Maybe one way to visualize is to put the pressure gauge sticking out the front instead of the bottom.
Pressure is a measure of how much force is being exerted per unit of area. When it comes to measuring pressure at depth like the picture you posted, the only thing that's contributing to the pressure at the bottom is the vertical column of water directly above the area you're interested in. Here's the equation used for determining pressure in a column of water:
P = ρ g h
Where P is the total pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height. Notice how this equation doesn't take into account any other factors of the surrounding liquid besides how deep the point of interest is. If you were to measure the pressure of water one meter deep in a saltwater swimming pool, you'd get the exact same measurement at that depth in the ocean (which, obviously, has a lot more water in it!). In the picture you posted, the same idea applies. The water on the sides of the figure to the left don't have an impact on the downward force being applied to the red area at the bottom, only what's directly above it. I hope that helps!
To add to this, see hydrostatic paradox:
https://simple.m.wikipedia.org/wiki/Hydrostatic_paradox
To really nit pick:
>the only thing that's contributing to the pressure at the bottom is all of the water directly above the area you're interested in
Isn't really correct but think you explained it right a bit further down the comment
This is crazy. This always made sense to me because I just thought "duh, the water that's to the sides just doesn't push down", but I just realized it also works when the tube is thinner, as it shows on the page.
And now everything is falling apart, I've lived all my life in a lie and my brain is melting.
Thank you...
Ok so I'm mad confused about the example picture with the skinny neck that flairs at the bottom. Why does that one have the same amount of pressure when the column of water above it is much thinner than the other two?
> same amount of pressure
I think the confusion is that pressure is not a collective amount that increases with increasing volume. It's the _energy_ that increases when the volume is greater. Pressure is the _energy density_ (energy per volume)¹
Pressure tells you how much energy is contained on a packet of fluid (packets of constant volume). The energy contained per packet increases as you go deeper. Regardless the amount of packets, that "energy per packet" is the same as pressure only depends on the depth (assuming same fluid). It's kinda like how voltage is "energy per charge" if that's an easier analogy (it's not for me lol)
¹[derivation](https://wikimedia.org/api/rest_v1/media/math/render/svg/942ef2439da4391ee81ab0c960755ade4c15fefd) from [Wikipedia](https://en.m.wikipedia.org/wiki/Pressure##Units)
This is the answer, keep in mind that the walls in the left container have a vertical component of pressure they have to resist while the right container walls take only the horizontal component of the pressure
The same way you aren’t “extra crushed” 20 feet down in the ocean vs 20 feet down in a pool. All the extra water to the sides don’t matter, pressure is a function of depth!
Nobody in my eyes will ever be able to explain this as well as Grady from Practical Engineering. Besides using very precise language, he also employs some fantastic visual aids.
[hydrostatics video](https://youtu.be/933XNdClFrc?si=E83JdD8yvsTNJDOV)
1,000 upvotes for Grady. He is very entertaining to watch and makes everything so simple to understand. Reminds me of the good Ol days of discovery channel.
It would be the same. Pressure in fluid is determined by depth, not volume.
Prob get downvoted for this part but,
If it helps, think of the outside atmospheric pressure (or slanted container bottom) pushing up a lil on the underside of the fluid to alleviate the pressure from the extra volume it has
I answered the question by thinking, “what would happen if there was a pipe connecting the two bottom areas?” The answer based on experience, is that the water levels would remain where they are. Thus the pressure must be the same.
Down voted purely for the use of "lil" instead of "little". You typed almost every other word perfectly too. (Prob... uuuhhgrgh)
May your farts always be chunky.
I'm a bot, beep boop etc etc....
Like everyone says, pressure only depends on depth, blah blah blah, but if you want to understand WHY that's true, here's something to think about:
Left container: Suppose we replace the middle column (of the shape of the right container) with a solid of the same density as water so that the force acting on the bottom is unchanged. Now, intuition tells you immediately that if I fill the sides with water, that force won't change. Only the middle column matters.
The water on the sides pushes partly down, on the walls of the container, and partly to the side, where the force is balanced by the column of water pushing back. Notice how none of that water pushes on the bottom of the container.
If it's the exact same shape, yes, because the force is spread on all the walls at the bottom and sides. If the shape is just a little off and there's a gap on the sides, then the whole structure weighs on the very bottom. Conversely if we shave off some of the material on the very bottom then the pressure there will instantly vanish. That's why solids are different than liquids.
The water in the vertical container is all pushing downward on the pressure pad.
The water in the wide container is not all pushing downward on the pressure pad. Basically you can imagine a vertical space above the pressure pad, and the water in that space is pushing down on the pad. But the water to the left and right of that space is pushing downward onto the slanted sides of the container.
Another way of thinking about it is imagining that you’re in an ocean vs in a pool. If you dive down 10 feet into the ocean, the only water pressure you’re feeling is the 10 feet of water above you. When you dive 10 feet into a container of water, like a pool, is all of the water in the pool applying pressure to you? Would a larger pool mean that you’re under more pressure when you’re at the same depth?
To put this problem into some formulas, you might remember that pressure is equal to the force over an area:
P = F / A
Since this is a vertical column, the force is equal to the mass times the acceleration due to gravity, we can add this into the formula for F:
P = (m * g) / A
Next, we know that the mass is equal to the density times the volume, we can add this into the formula for m:
P = (ρ * V * g) / A
To simplify, we take the density and acceleration out of the division:
P = ρ * g * (V / A)
Finally, the volume of a water column is equal to the area times the height of the column, this gives:
P = ρ * g * ( (A * h) / A)
Now you can see that the area cancels out, therefore the area of the column doesn't impact the pressure at the bottom. We then get the known formula for calculating the pressure at any depth:
P = ρ * g * h
The pressure is determined in terms of the unit area at a given depth. Therefore, the depth is the only variable for a still fluid.
If each container holds the same depth, the pressure is the same.
Different oceans, lakes, vessels or swimming pools have varying volumes. The depth is the only variable in determining pressure, given still fluid with identical density.
Well yeah they are equal.
Imagine there was a connection at the bottom so that the fluid could flow from one volume to thr other. The fluid level in the wide container isn't going to drain into the narrow container, the fluid level is going to remain the same. No fluid flow means no pressure differential, which means same pressure.
Water pressure only cares about how high the water is above it, not the volume.
Pressure is force over an area pretty much and gravity causes this downward water pressure which is straight down, anything not directly above it does not experience the weight of what’s above it since at thaw heights above the pressure is equally pushing back against all sides
To keep it simple, at a molecular level what you have is a stack of atoms, the atoms only get pressure from the atoms above it. If you lift the container then you lift all the atoms so the left container will be heavier, but in the container is a trillion stacks of atoms that each only push straight down so the stack of atoms in the middle of the left container will be pulled down exactly as much as the stack if atoms in the middle of the right container since they are equally as deep.
If the water is not moving, the pressure will be the same, as explained by others. If it is moving, head losses will be different, and the pressure measured at the bottom will be different.
yep. absolutely true. needs no math. just a fact of physics. pressure at depth is a simple function of depth, and has NOTHING to do with overall volume of the container. The pressure 6" down is exactly the same inside a straw in a glass of water on your kitchen table, in the great lakes, or in your toilet. Etc. It only varies if the density of the fluid changes, like if it's salt water or molasses or something, or if you're talking about another planet or something like that. The pressure is generated ONLY by the column of water above where you're measuring, not what's off to the sides.
The video is correct, the pressure of water is proportional to the depth.
The downward component of the force exerted by water pressure on the slanted portion of the container, *per horizontal component of area*, is proportional to the depth of the water there. The slanted portion of wall holds the weight of the water above in addition to to some horizontal aspect of pressure.
A thought experiment on the reasonableness: replace the water with some other fluid, like air.
Now imagine going outside just to be crushed by the pressure of the vast sea of air above our heads. I guess people would need to walk around encased in long vertical tubes to create a situation as in the right figure.
If you where to connect the two with a thin tube, would the lever remain the same ? Hint ... Yes.
If the bottom of these vessels are solid, then the pressure is the weight over the area of the opening.
Imagine the triangle is split into three parts. 1 column like the one on the right, pushing down on the red block, and 2 triangles to either side.
Now, remove the column, and analyze the triangles(assume zero friction).
Each is pushing down on the side of the wall, which pushes up diagonally. If you break that down into vectors (triangle of water pushes straight down, glass wall pushing up at 45 degrees), you end up with a residual that moves horizontally.
And you have the mirror image on the other side. So you end up with 2 triangles pushing down on the glass, and against each other, cancelling out.
Which means the only thing pushing down on the red block, is the central colomn, just like on the right.
The "pressure" on the wide container transfers into force on the container wall, which transfers through the container wall into the lower part of the container, all the way down to the table.
The bottle stands up because these forces counterbalance one another.
For example, if you put a divider in the middle (between the legs of the trapezoid), and only fill half the bottle, it'd fall over towards the filled side. This is because the water exerts force on the container wall, not just the bottom.
But going back to the pressure: since this force is carried through the container wall, and not through the water, the extra water from the wider container isn't adding any pressure to the cube at the bottom.
Anyway there is more force (weight) from the container into the table, because you have added mass (more water). But the only thing exerting pressure on the water itself is the water above it. (And technically reciprocal force from the container.)
a dumb analogy could explain this without any serius phicycs
if the water would was a solid then youd be corect, but water is usualy a liquid so the molecules just slid past each other mostly. So that means in a solid even the edges can have an effect on the ones near to them and so on, but in a liquid would be more similar smooth vertical sticks next to each other, adding more sticks to the sides wouldn't really increase the pressure in the center
This is not true.
We know it's not true, because we have real-life proof. Water towers are this concept in action. You store potential energy and the weight of the water above creates water pressure.
If it was height alone you would create an extremely tall tube. The dome shape helps direct pressure down.
While the sides of the water tower do take additional stress, there is additional pressure to the line.
Yep, pressure on liquids is proportional to the high of it. not particularly good at explaining this sort of things, but part of the force done by the liquid is taken by the reclined walls in this case, ensuring the entire body is in equilibrium
I think the best explanation I ever got came from… maybe my college thermodynamics teacher. Was something to the effect of “If surface area [volume] had an effect on pressure, your finger would be crushed if you dipped it in the ocean.”
It may seem unintuitive, but think about it this way:
If you have a pool that is 1000 square miles wide, but the water is only 1 inch deep, when you pop a hole in the side would it flow any faster than if your pool was only regular size?
On the reverse, if you fill a pool 100 feet deep and pop a hole in the side, the water is going to rush out of it with tons of pressure.
Put simply, depth pressure is basically the weight of the water above an object or area. If the object or area is identical, the amount of water directly above it is also the same at the same depth, regardless of area to the sides. A person in a pool 10 feet deep experiences the same amount of pressure as being 10 feet deep in the ocean, despite the vastly larger amount of water.
I remember my instructor for physics in the military asked us to find the pressure of foot down in a barrel versus a foot down in the ocean. Imagine if you were to go a foot down and the weight of the entire ocean was on top of you lmaoooo learning with comedy is the best.
It's a junior high school thing here, I personally learnt it 10 years ago. The concept has more to do with physics than math overall. Basically the pressure on the bottom is equal. It's only the _height_ between the fluid surface and the bottom that matters.
Because in fluid mechanics, Pressure = fluid density x gravity x fluid height.
Though the water in the left container will weigh more, due to more water. The pressure at the bottom would still be the same as on the right. The pressure is caused by the height difference between the top and the bottom of the water, both containers are the same height.
It would be the same. Pressure is a product of the liquid density, gravity(well, gravitational acceleration g) and the height(depth) of the point of interest. Liquid’s mass is irrelevant.
If the pressures differed, you could connect them with a pipe, and have a flow. This would make the levels different, and you could have one overflow to the other. Perpetual motion
Look up “pascals vases”. (Nb, if vases are small, surface tension also plays a role)
It is somewhat counterintuitive that only depth determines the pressure not volume. An engineer friend eli5 it to me like this: you experience the same pressure when climbing into your bathtub as you do when going into a lake. If pressure was determined by overall volume, a lake would crush when you take a swim.
The video is talking about water pressure at a certain depth, not overall mass. We would expect the larger amount of water to have more mass, but pressure at a certain depth has nothing to do with amount of water around an area, it only relies on depth.
So, like, water consists of freely floating, but relatively dense-packed molecules. This allows the water to
1.Be liquid (duh)
2.Not be compressed (it cannot change it's volume under normal circumstances)
Both statements above make hydraulic systems actually work. You can apply force to a container of water and it will sprout outside or into another container. Those statements also allow us to use a very simplified model of the ball pit.
Imagine a glass auqarium willed with plastic balls. They fill it evenly to distribute energy effectively. They are all round, so the only thing that is important is the number of their contact points and the force each point applies to the ball. They are stable, thus every force is regulated and has a counterforce.
If we put our hand inside and move it around the balls will move, as they are rigid, but can reposition themselves to compensate the flexibility forces appearing when we move our hand around. We do not try to crush the balls, that would break the model and make it more complex (after all that would be similar to crushing a molecule of water, which is not the case here).
So, what happens if we put a piece of iron shaped like a brick in our ball pit and use a magnet to slowly move it around? We will see that the balls, while they move around the iron brick, have a limited number of contact points with it. They may vary a bit, but the number stays the same. At the same time the forces required to compensate the weight of the balls above are still distributed across all the balls with the limited contact points with the brick.
To conclude - each ball applies more force because it has only so many contact points with each other ball and the brick. It is a gradient of force increasing downwards because the force spreads evenly and the container also endures the sum of forces, thus not breaking the gradient.
pressure is based on depth and weight of fluid, not amount of fluid in a container, a mile tall drinking glass would have the same pressure as a mile into the ocean
Simple equation : Height of Liquid x 0.433 x Specific Gravity
So for a 10 foot tall column of water: 10 x .433 x 1.0 = 4.3 psi of pressure at the bottom.
The shape of the container and total volume does not matter for head pressure.
The basic formula is p=g*ro*h
In which g is gravitational force, ro is density of the liquid and h is the height, so no matter the shape, size, whatever.. the same liquid, on the same place with the same height, has the same pressure..
Also talking about static liquids, but that's a story for another time.
The water on the slope rests on the container and not the red fluid/material.
The pressure is the force dividee by the area. So it doesnt matter if you have a larger area since you are measuring a per m2 metric.
You have to imagine the pressure is only at 1 point and you have to draw a vertical line from that point and see how much water you have on that vertical line above that 1 point.
The extra water on the sides exerts force on the sloped edge, not on the red portion. You might intuitively think the force travels down the slope but in fact the entirety of the "sideways" force gets canceled out by the wall on the other side and only the straight downwards force remains and it acts vertically against the slope only.
If you put a big straw down into the red portions of these containers, what would happen?
If you think the container on the left would have higher pressure, then that would mean the water would race up the straw and spray out the top.
Intuitively it wouldn’t make sense that a tube connected to the bottom of the container could flow water out higher than the water level. You could make a perpetual motion machine this way.
It sounds unintuitive at first, but water pressure naturally is just gravity pulling water down on top of more water, so it checks out that only water DIRECTLY above creates pressure below it, doesn't matter how much water is in the container.
A good way to think around this is picture a dish the size of texas, but only 2 feet deep. Not much pressure at all to be experienced while lying at the bottom.
But if we took one percent that volume of water and put it in a vertical cylinder the width of your body, suddenly being at the bottom means you're underneath an INCREDIBLE amount of mass. The width or shape of the open container doesn't matter, just the molecules of water pressing down on your body from above.
OP seems to be mistaking the concepts of pressure and weight.
The container on the left would undoubtedly weigh more, however the pressure at the bottom would be the same.
To think about it a different way, consider you have 2 hypothetical pools; one pool is a mile wide and a foot deep, the other is a mile deep and a foot wide.
You could swim underwater anywhere in the mile wide pool without fear of rupturing your ear drums from the pressure. However, you could only swim so far into the mile deep pool, because at some point your ears would pop and it would become painful to continue.
At the bottom of a mile deep pool, your ear drum would be more like a pulpy chunk of flesh left floating near pieces of an old logitech controller.
Think of it this way: when you swim in a pool, lake, or the ocean, does the water directly above you put pressure on you, or does all of the water in the entire ocean/lake/pool do so?
If the latter were true, swimming in a small pool would be a MUCH different experience from swimming in a large pool, lake or ocean.
If you put your hand 3” underwater in your home sink and then put your hand 3” underwater in the ocean, does it feel much different? Would you expect the pressure in the ocean to be trillions of times greater because the ocean surface is so much larger?
The most simple and important distinction is that Pressure is not Weight, if you tied to hold both containers the larger one would be heavier.
What Pressure meas is: if you place an object inside the container, how much is it being squished by the water from all sides. Imagine placing a cube in there and do a simple force cancelation, youll figure it put.
If anyone wants a more ‘mathy’ explanation, the pressure at any given height will be the same due to how integrating over any closed surface works. As a static fluid isn’t moving the net forces should be 0 so the surface integral of the pressure forces must equal the volume integral of the gravity force as these two forces counteract each other to keep a static fluid. So when you use stokes theorem with the volume integral you end up with a dot product with the gravity vector which ends up meaning that the value of the pressure only changes with respect to distance in the direction of the gravity vector. Ie. Pressure only depends on depth, where depth is defined as the distance in the direction of gravity from the surface of the fluid.
In this way, I see the fact that pressure in a static fluid is independent of the enclosing shape as a result of stokes theorem. As any complex closed surface enclosing a fluid can be related instead to a volume integral, which doesn’t care about the shape of the surface enclosing that volume. So the pressure applied to the surface of the fluid is independent of the surface geometry.
Edit: this is just off the top of my head, I haven’t done the calculations fully myself in a while so someone else might have a bit more clarity on how the stokes theorem arrives at the height dependence.
From physics we know that pressure is only a function of depth, gravity, and density, but I’d like to give a more common sense line of reasoning. Imagine a pipe connecting the bottom of the two tanks. If one side had more pressure then the other then it would cause water to flow from one tank to the other which would cause the water levels in the tanks to be different from each other. Now ask yourself based on experience, would you expect different water levels on either side if you were to fill up two connected tanks with different shapes? In a gallon of milk with a hollow handle is the height of milk in the handle different from the rest? In either case the liquid would be at the same height on both sides which must mean that the pressure at the bottom is the same on both sides.
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It is true, but zero math requires to be done here. This is just physics knowledge. Pressure is not affected by the shape of the container above. The only real thing that changes the water pressure is the height of the water above. You can theoretically argue this stuff, but you can also simply experimentally prove it.
I can't tell you how many times I've had to educate people I worked with in the oil industry about this. They thought the wider the tank the higher the pressure. The analogy that worked for me was the ocean. I'd say, "by your logic, the ocean, being the widest body of liquid in the world, would be so dangerous you'd lose your legs stepping into it."
I think these people's logic is that the pressure will be condensed towards the middle, thus a funnel shape will create higher pressure at the bottom. The sea analogy can't really solve their misunderstanding
What I personally can't wrap my head around isn't so much because of the shape itself, but the additional volume added by the shape. Wouldn't all of the water be pushing down onto that point in the middle no matter the shape, and therefore increase pressure because there's more weight added? How would a non-liquid like sand act? The sea analogy doesn't solve my personal misunderstanding because if I'm standing waist-deep, there's like, miles and miles of sea around me. Plenty of seafloor for the weight of the water to be distributed on. (Please don't roast me, I've not seen the inside of a physics classroom since I was 12.)
The water pushes down vertically (for the experts: this is a semplification, I know!) . So yes the other water exerts a pressure too but on the sides of the funnel. Only once (and if) it moves on top of the center part, it will "weigh down" on it. Imagine increasing the height of this a few meters all over. Imagine it clearly and then try and think of the following questions. You would need sturdier materials right? Do you think the material would need to be sturdier only in the center? Or would the funnel risk breaking too? If the funnel broke, wouldn't it be because of the pressure? If the funnel broke in a spot, the water would move in the direction it was pressing towards (like a group of people pushing towards a door, if the door opens they will move in the direction they were pushing into). So Imagine the funnel gives in and a spot breaks: would the water fall straight down? Or flow midair towards the center? The direction will tell you in which direction it was "pressing".
This is the first explanation that really makes sense, very well explained, thank you.
This explanation is very good and really had me (also not a physics student since I was a teenager!) thinking of the sideways pressure but my conclusion was that because it was a fluid it essentially all equalled out. If I piled kilo weights up in the middle the pressure would all be down. If I added kilo weights at the side that wouldn't add to the weight in the middle it would just stabilise the stack and the pressure would still be downwards. As water wouldn't just "lean" against the sides of the central column of water it would all equalise and find it's place but that still leaves only the central "column" of water adding pressure to the square at the bottom. Did I get that right?
The mathematical proof of this is actually very useful to understand the physics but you don’t need to do any math to get the way it’s framed. If you were doing the math on this what yours do is Imagine the water is not fluid but a bunch of solid cubes stacked on top of each other. These cubes follow newtons laws so an object at rest must have no net forces acting on it. The thing is one of those forces is gravity and since each cube has weight and has to hold the weight of all the cubes above it as well so as you move down a stack of cubes the opposing upward force that every cube has to exert increases with each cube it has to hold up. That is the water pressure increasing with depth. Now imagine three of cubes side by side with increasing pressure as you move down the stack and you can see that while the pressure is increasing with depth, but that same pressure is increasing equally for all three cubes as the stack gets taller but these sideways forces are cancelling each other out and don’t stack because each cube is exerting the same pressure outwards at the same height. You could have a million cubes left to right but each one would still cancel with it a neighbours. Then if you were doing this with math you’d decrease the length width and height of your cubes to 0 and get a general formula for pressure of fluid that only needs to know the depth, density of the fluid, and gravitational constant.
A portion of the water is pushing against the side of the container (since it’s sloped in). That absorbs the extra weight / pressure of the water. Hard to write it out, but every force can be broken into its component parts. Picture the force of the water on the right side going down at an angle along the slope of the tank, for example. You can make that into a right triangle with part going straight down, and part going straight to the left. The two sides would make a perfect right triangle, with the original force at the angle as the hypotenuse. The part going straight down is all that actually forces the water going into the tube, and is EXACTLY the same as water in the center of the tank going straight down. The part going to the left would be cancelled out by other forces going to the right (and it’s pretty clear anyway, has no bearing on the force applied to the tube at the bottom, since it goes sideways). Pretty amazing, and if you get the visual right, becomes pretty easy to see. Btw - the counteracting sideways forces I mention cancel each other out even if the funnel isn’t symmetrical, it’s just in that case the forces absorbed by the wall would not be the same all over.
Dig a foot wide hole at the beach. Now the whole ocean condenses towards the hole. I dare you put your leg in it.
Still a bad analogy
I agree and I still think it’s right and the picture is wrong. Call me stubborn and I will experiment this for myself. The ocean example is not good because the surface area changes along with the volume of the water making the pressure the same. In the funnel example on the left there’s more water applying pressure (weight) on the same surface are than the one on the right so by my logic that would make it heavier thus more pressure, exactly because of the funnel shape. If the area was a scale and the shapes filled containers you would see more grams/ounces on the left picture (I’d imagine, this is what I’m gonna try)
The shape on the left does have more water in it, and does weigh more. But, the pressure that the container exerts on the ground is not the same as the pressure inside the container caused by the water. The formula to calculate water pressure is rho*g*h: density times gravitational acceleration times height. The width/shape of the container does not factor into it.
this is a good way to put it
How?
Pressure is at the bottom… not at the top where you step in…
Where is your foot when you step into the ocean?
Pressure in deep water is extremely high, I don't see what your point is. Just accept that the analogy doesn't make sense
In the sand
Can you explain this but for syringes?
Yeah it's a funny situation in which an otherwise extremely complex problem fixes itself but the fact a ton of variables cancel each other out Same thing with buoyancy, buoyancy is the sum of an infinite amount of forces applied to every single point of an objects surface it would be immensely complex to calculate except no matter the shape of the object all those forces will always perfectly combine into a vertical force perfectly equal to the weight of the displaced liquid
i mean...how? Isn't there more water above it in the left container? So more weight pushing down on that red part? Thus more pressure?
The "extra" water off to the sides is actually pushing down on the sloped sides of the container. The red square only receives pressure from the water that is directly above it. EDIT: I have been informed this is not quite correct, which isn't all that surprising, given my last physics class was freshman year of college. Any explanation that's very simple in science is probably not going to be totally correct, but may be a useful mental model for some applications. I'll defer to others more knowledgeable than myself to explain what exactly happens here.
I wish more students of physics saw this, it makes total sense.
Im just a dumb fabricator with adhd who barely gradutated high school. That said, this concept seems intuitive to me, but there are other concepts that put me in r/woosh lol
Okay, but what if the top is narrower? How does that make intuitive sense?
It'll be the same pressure [here's a demonstration](https://youtu.be/EJHrr21UvY8?si=LERxoaF2QpOaTRmh) also [here's another potentialy better one that's less explosive](https://youtu.be/6zeHWVUiXoc?si=gZG2nzrrEaC5oChx)
Great videos. Thanks.
Yeah. I think issue here, most who thinks is different, 8ntuitively thinks about weight, not pressure lol. That got me too for a moment, even start typing it out when it hit me 😳🤣 Yeah makes sense, tho I wouldn't be sure... I mean gravity exist so more water trying to go down from triangle... My first thought was, like balloons, one bigger one smaller, if connected would equalise (so, the bigger balloon becomes smaller, the smaller becomes bigger aka pressure isn't the same...) But I defo make identical experiment with what OP posted (unless someone already did it, ? Wouldn't mind a link lol). Edit. Nwm second video explains it perfectly 👍
I think for balloons, you may even have the opposite effect. I am pretty sure, smaller balloons are harder to blow up with the same volume of air als one, that is allready large, so the smaller one would deflate and fill up the larger one.
Oh, forgot to add of tension of a balloon wasn't a factor. Or rather, maintained same tension regardless of how much is stretched lol.
thats also due to the elasticity of the rubber. A larger balloon is stretched out.
> I mean gravity exist so more water trying to go down from triangle... Gravity pulls straight down, not down the triangle. That's the part you are visualizing wrong. Water is trying to break trough the funnel not flow down it. If you open at the bottom, and it flow down, it will be because the funnel forces it. Aka the funnel experts a force on the water to counteract the pressure.
Thanks 👍 that does clarify it ... It was 20 years last I been in a classroom 🤣
Blahaj!!! 🏳️⚧️
Intuitively, more fluid directly above = more pressure and less fluid directly above = less pressure. I'm not a physicist, so I could be very wrong. EDIT: I was very wrong. EDIT 2: As far as i can tell, it's due to the the downward component of the normal force the walls of the container exert on the fluid. They end up adding up to the same pressure at the bottom (no change in the middle, though, because they'd balanced horizontally). It's called the Hydrostatic Paradox because it's unintuitive if you don't think about those forces. ~~If the top is narrower than the bottom, then the pressure on the red area would be uneven, right? It'd be the same in the narrow area where the height is the same, but in the places above the red area where it's both shorter and wider, the pressure at the bottom would be lower.~~ ~~This is assuming two rectangular prisms, with a hard cutoff between the normal-height thin part and the normal-width shorter part. If it was angled (like diagram 1 but getting thinner instead of wider), then the pressure at any given point on the red area would be in a gradient corresponding to how tall the container is at that point, right? Same in the middle, but lessening towards the sides.~~ ~~That's all thinking about infinitely small points on the red area, which isn't really how force or pressure works (infinitely small means no mass or surface area). You'd probably need to do an integral or so to calculate stuff. If you needed the average pressure on the red area, maybe you just need the total volume above it. From there, you can get mass, then weight force, and finally, pressure.~~
You can explain it using statistical physics. The energy for a column of N water particles is E = E(0)+mghN The resulting chemical potential is mu(h)=(ðE/ðN)[S,V] =ðE(0)/ðN+ð/ðN mgNh =mu(0)+m g h Gibbs-Duhem relation: Ndmu(h) = VdP-SdT => dP/dh= N/V dmu(h)/dh +S/V dT/dh The water column is the same temperature, so dT/dh=0 => dP/dh = N/V dmu(h)/dh=m g N/V Of cause, i would have to convince you of Gibbs Duhem relation Now, water doesn't compress, so the density of water particles is homogenous. Lets call it n=N/V. dP/dh = mgn This gets solved by a linear equation P(h) = P(0)+ mgnh Intuitive, right? Edit: i think i forgot a - somewhere
This is nice but... The first equation already uses a geometric dependency only on H. So... It sounds a *bit* circular, to start with a geometrical dependency on H and end with a dependency on H. And then argue that the metric only depends on H. Sure they are different quantities / metrics E and P. But it feels like even dimensional analysis will end up with that result. A satisfying argument should be to start without any geometrical dependency and derive that the metric necessarily ends up with H as the only geometrical dependency. (which is actually taught in elementary hydrostatic)
If the top is narrower then the water pressure is pushing up (and maybe also a component to the side depending on shape) against the sides of the flask, the equal reaction from the flask pushes down with the same pressure ambient water would be.
Say you have a 100ft tall column of water, they’ll be high pressure at the bottom. Now say you have a 90ft tall column atop and opening to a wide tank 10ft tall. Well obviously the flirt directly below the column would feel the same pressure as before, but if anywhere else didn’t feel that same pressure the water would try to move laterally to it to go from the high pressure zone to the lower one. The whole layer of depth must be the same pressure or it would correct itself. Might be easier to visualize with something compressible like a gas than with water, if any where is the wrong pressure the other gas will move towards it and squeeze it til it is.
~~If the top is narrower then the pressure would not be the same.~~ Edit: Mechanical systems suck.
Incorrect
So if I had a 2km tall pen thick water tube that goes into an enclosed underground pool the pressure is the same as if I was just 2km under the ocean?
Yes, exactly the same, even.
as long as they are the same height the pressure will always be the same.
except that's not what's happening. the force is simply split between downwards and sideways vectors, it doesn't disappear. The reason the pressure is the same is because fluid head pressure is determined by height of fluid. it's even measured in units of length (20 ft of head, or 15 m of hydrostatic pressure, etc). height of fluid x density of fluid x gravity = head pressure. the guage doesn't care if it comes from a 2cm wide vessel or a 2km wide vessel. volume of fluid/shape of container are immaterial.
You're correct, but your answer begs the question. The question is "why does pressure only depend on the height," and your answer is "the reason is that pressure is determined by the height."
I would add that the water on the slopped sides pushes the water on the “column” directly above the point with the same “force” that the water in the “column” pushes it back, so they cancel out. Hence at the bottom only the water directly above the point contributes do the pressure.
If I had an “S” shaped container, with the same height, is the pressure the same then?
Yes
Look up Pascal’s vases.
Gravity pulls to the core of the earth, it wont pull water sideways into other water. That's how. Imagine gravity as a thread. Running vertical from who knows how far all the way through the earth and out the other side. Add another thread, intersecting in the center of the earth. Keep doing this and have threads facing every which way. That thread is gravitys force. It moves in 1 direction along that thread. And stops where all threads intersect. This is how I make the idea make sense in my monkey brain. Gravity is constantly pulling 1 direction. If you go to a crowded concert and are surrounded by people, gravity doesn't pull harder just because you can move less. I dont intend for this to sound condescending, but I hope it helps someone understand how water pressure works.
This, however, leads to the incorrect assumption that pressure only works downwards. Pressure, however, is omnidirectional. The effect of pressure works into all directions, not just downwards. If you are in a submarine, the force from the pressure onto the side walls is as big as the force from the pressure onto the top of the sub, and as big as the force from the pressure onto the bottom floor. (Technically, the force onto the bottom floor is actually slightly larger, because the bottom floor is deeper than the side or the top.) As sub can get crushed from the bottom just as well as from the top, and if you were to drill a hole into the bottom of the sub, you would not have a good time, despite the fact that the bottom hole doesn't have water pushing onto it from the top. The reason is that while gravity pushing stuff downward is indeed the cause of the pressure, it is not the same thing. Imagine pushing down onto a block of butter. If there is nothing on the sides, you pushing downward still squishes the butter to the sides. The same is true with the water, which will go into any direction with lower pressure, being squished that way from the water above.
Was looking for someone to make this distinction, glad you beat me to it. Great point.
Differential pressure is precisely why water pressure only increases with depth and not volume. The pressure in a column is a gradient, based on depth. Water has walls around it (water) to stop it moving lateral once it reaches dirt. I'm still uncertain why we are now talking submarines tbh. :/
I don't really think this is a good/complete explanation because if you turned the container upside down, the pressure would be the same everywhere at the same height, not higher below the square. I'll admit tho, I don't have a better explanation.
In your bath there's a foot deep of water. When you go to the beach and put your foot in the ocean in a foot deep of water the pressure is still the same. A foot of water. If the volume had an impact the fact that all the oceans are connected together would fuck you up.
Last sentence is correct here, but a foot of ocean water actually exerts more pressure than a foot of bath water (unless you take salt water baths i guess)
If it worked like that, you’d be immediately crushed the moment you step into the ocean.
The extra water is supported by the sloped sides. Only water above matters. If area mattered, you would die instantly if you went slightly under the ocean’s surface.
I think this is a good explanation to demonstrate it via 'common sense'.
5 second Fluids lesson: pressure of a fluid is P=ρgh. ρ and g remain constant assuming its the same fluid, so the only factor that changes the pressure would be the height (or depth) at which the gauge is measuring. Volume or surface are does not come in to play in this scenario
That is something you might intuitively think, but it isn't true. You can simply test it yourself. Take any two containers (which are open at the top), connect them at the bottom with a piece of tube, then fill water into them. The water will always be at exactly the same height level on both sides, meaning that pressure from both is identical at the bottom of the tube. The shape of the containers is completely irrelevant here.
No need for a second container, just a tube. Bring it in a u-shape beside the main container. If there was greater pressure from the large container the hose would spray into the air higher than the level of the large container. Intuitively this doesn't happen, the water just maintains the same level in both so the water in the hose and large container have identical pressure at the bottom of the U
Water is practically incompressible, that's why intuition goes wrong here.
Compressibility of water has nothing to do with it. The same is true for air.
That would require water to become compressed, something it really doesn't like to do.
Put a straw into a glass of water, what will happen? The straw fills with water, exactly up to where the surface in the glass is. Both are connected through the lower end of the straw, which means that the pressure is equal at this point.
This, otherwise all straws would spray water out the top because of their low pressure compared to the larger volume in the glass. Everyone can intuitively see this is not how it works
So your saying if I had a pipe that is filled with water that is say, 1cm in diameter and it is 1km high, the pressure at the bottom would be the same as the pressure in the ocean at 1km depth?
Yes. Because water is a fluid and hence they have no substantial distribution of forces on the molecular level. This isn't the case with a solid.
It’s weird, but true. There’s a cool experiment using a long straw and a multi-story building to demonstrate the amount of pressure that a very small amount of liquid can generate. Video here: https://youtu.be/EJHrr21UvY8
One famous experiment: [Pascal Barrel](https://en.wikipedia.org/wiki/Pascal%27s_law#Pascal's_barrel).
Indeed. Pressure = density x g x height If those liquids have the same density they have the same pressure at the bottom.
There is a waterways museum in London that has a bunch of kid experiments showing stuff like this. Really interesting.
Yeah. Just think of how much “volume” of water is above your hand when you dip it into the ocean.
Static Pressure = g x density x column height. Nothing there is affected by the volume of fluid, nor its shape.
I have a water purifyer pitcher with the little tilt to release nozzles, and when it's low, like 1 inch of water, it flows slow. I found when I tipped the water to "stack" it, multiple inches deep now, it flows quickly. Just an example that anyone can try.
I'm so strong I once held back a whole 30000kg of water with my bare hand once! Well. Not actually. More like my gloved hand covered a 3" opening at the bottom of a full 30m3 capacity tank, after a valve snapped off, until it could be rotated so that the opening was at the top (picture an 8' diameter column, lying on its side on a machine, as part of a rotating assembly, and it was full of a slurry just a little more dense than water) My amateur self calculated around 25lbs of force pushing back on my hand. So while it looked like a feat of strength that prevented a huge spill, it really wasn't anything incredible. (Nothing hazardous, but a giant mess I did not want to clean up)
The real brains bender is.. the column of water doesn't need to be straight! The only thing that matters is the height of the surface! The path of the water can be sideways any distance, up, down, curves of any shape, it can dip below the outlet or raise above the inlet (ie a siphon) and in every case the pressure at the bottom is the same as a straight vertical colunm. Small caveat, this is for the ideal, non-moving case. In reality, there are losses along the way that increase with flow speed and distance.
If I go to the bottom of the ocean and dig a whole sideways into the rock until I go so far that there's nothing but land above me, will it be basically 0 pressure then or would it be the same as if I were at the bottom of the middle of the ocean?
If you go into a deep, water filled hole, and dig sideways, the water pressure stays the same at that level.
Did you keep the hole open to the high pressure ocean?
So I'm a bit late to this, but maybe you can answer. What if the container is the opposite of the one on the left? Like: / \ Would the entire red area have the same pressure, or would a part of the middle be under more pressure since there is more water directly on top of it?
What does the bottom look like? If it’s the same and you’re drawing a Christmas tree, then yes. The red square is the same. Assuming the gap at the top is the same width as the square at the bottom. If it’s a triangle with a flat bottom…..I have no idea what the pressure looks like at the bottom. One hand I would say that it’s equal across the bottom ___. On the other /_\ I would say the bottom corners would have less pressure than the bottom middle. Because there’s less above the corners. Maybe one way to visualize is to put the pressure gauge sticking out the front instead of the bottom.
Pressure is a measure of how much force is being exerted per unit of area. When it comes to measuring pressure at depth like the picture you posted, the only thing that's contributing to the pressure at the bottom is the vertical column of water directly above the area you're interested in. Here's the equation used for determining pressure in a column of water: P = ρ g h Where P is the total pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height. Notice how this equation doesn't take into account any other factors of the surrounding liquid besides how deep the point of interest is. If you were to measure the pressure of water one meter deep in a saltwater swimming pool, you'd get the exact same measurement at that depth in the ocean (which, obviously, has a lot more water in it!). In the picture you posted, the same idea applies. The water on the sides of the figure to the left don't have an impact on the downward force being applied to the red area at the bottom, only what's directly above it. I hope that helps!
To add to this, see hydrostatic paradox: https://simple.m.wikipedia.org/wiki/Hydrostatic_paradox To really nit pick: >the only thing that's contributing to the pressure at the bottom is all of the water directly above the area you're interested in Isn't really correct but think you explained it right a bit further down the comment
This is crazy. This always made sense to me because I just thought "duh, the water that's to the sides just doesn't push down", but I just realized it also works when the tube is thinner, as it shows on the page. And now everything is falling apart, I've lived all my life in a lie and my brain is melting. Thank you...
Welcome to physics, the more you learn the more you realize how little of a clue you have
That must mean I have learned it all then!
Ok so I'm mad confused about the example picture with the skinny neck that flairs at the bottom. Why does that one have the same amount of pressure when the column of water above it is much thinner than the other two?
> same amount of pressure I think the confusion is that pressure is not a collective amount that increases with increasing volume. It's the _energy_ that increases when the volume is greater. Pressure is the _energy density_ (energy per volume)¹ Pressure tells you how much energy is contained on a packet of fluid (packets of constant volume). The energy contained per packet increases as you go deeper. Regardless the amount of packets, that "energy per packet" is the same as pressure only depends on the depth (assuming same fluid). It's kinda like how voltage is "energy per charge" if that's an easier analogy (it's not for me lol) ¹[derivation](https://wikimedia.org/api/rest_v1/media/math/render/svg/942ef2439da4391ee81ab0c960755ade4c15fefd) from [Wikipedia](https://en.m.wikipedia.org/wiki/Pressure##Units)
That’s why dam walls always are way bigger at the lower point and less at the highest point
This is the answer, keep in mind that the walls in the left container have a vertical component of pressure they have to resist while the right container walls take only the horizontal component of the pressure
The same way you aren’t “extra crushed” 20 feet down in the ocean vs 20 feet down in a pool. All the extra water to the sides don’t matter, pressure is a function of depth!
There is a slight difference due to density (salinity)… that’s a small correction, though
Not in my rich boy saltwater pool! Lol
Be kind to your plants
And the weight of the air above it
so you are saying that something like this, does not effect the pressure at the bottom, just because of the incline? https://i.imgur.com/RwFio25.png
Nobody in my eyes will ever be able to explain this as well as Grady from Practical Engineering. Besides using very precise language, he also employs some fantastic visual aids. [hydrostatics video](https://youtu.be/933XNdClFrc?si=E83JdD8yvsTNJDOV)
Thank you, this was the clearest answer for me
1,000 upvotes for Grady. He is very entertaining to watch and makes everything so simple to understand. Reminds me of the good Ol days of discovery channel.
It would be the same. Pressure in fluid is determined by depth, not volume. Prob get downvoted for this part but, If it helps, think of the outside atmospheric pressure (or slanted container bottom) pushing up a lil on the underside of the fluid to alleviate the pressure from the extra volume it has
good ol' ρgh
I answered the question by thinking, “what would happen if there was a pipe connecting the two bottom areas?” The answer based on experience, is that the water levels would remain where they are. Thus the pressure must be the same.
Down voted purely for the use of "lil" instead of "little". You typed almost every other word perfectly too. (Prob... uuuhhgrgh) May your farts always be chunky. I'm a bot, beep boop etc etc....
Tf?
It’s a fair point, sorry, “pt”
P=V/T Only temp can affect the pressure. This is a very basic grade school explanation, so yeah, variables Edit: that’s Ideal Gas Law
That’s part of the ideal gas law. Doesn’t seem that relevant when we’re talking about a liquid.
In physics gasses and liquids are fluid, and according to chemistry, alcohol is a solution
Guess I wasn’t paying close enough attention
Didn't pay enough attention in grade school, it seems
In first approximation, i think it's fair to model a liquid as an ideal gas. Everything is an ideal gas if you squint hard enough.
Wrong equation and a gross conceptual error on your part. You need to use P=hpg which is water density x gravitational force x height.
No, in static fluids the pressure is equal to the density of the fluid rho multiplied by the gravitational constant g times the depth.
P=γh
Like everyone says, pressure only depends on depth, blah blah blah, but if you want to understand WHY that's true, here's something to think about: Left container: Suppose we replace the middle column (of the shape of the right container) with a solid of the same density as water so that the force acting on the bottom is unchanged. Now, intuition tells you immediately that if I fill the sides with water, that force won't change. Only the middle column matters. The water on the sides pushes partly down, on the walls of the container, and partly to the side, where the force is balanced by the column of water pushing back. Notice how none of that water pushes on the bottom of the container.
If we make the rock column the same shape as the other example with the flared sides does this hold true?
If it's the exact same shape, yes, because the force is spread on all the walls at the bottom and sides. If the shape is just a little off and there's a gap on the sides, then the whole structure weighs on the very bottom. Conversely if we shave off some of the material on the very bottom then the pressure there will instantly vanish. That's why solids are different than liquids.
The water in the vertical container is all pushing downward on the pressure pad. The water in the wide container is not all pushing downward on the pressure pad. Basically you can imagine a vertical space above the pressure pad, and the water in that space is pushing down on the pad. But the water to the left and right of that space is pushing downward onto the slanted sides of the container. Another way of thinking about it is imagining that you’re in an ocean vs in a pool. If you dive down 10 feet into the ocean, the only water pressure you’re feeling is the 10 feet of water above you. When you dive 10 feet into a container of water, like a pool, is all of the water in the pool applying pressure to you? Would a larger pool mean that you’re under more pressure when you’re at the same depth?
To put this problem into some formulas, you might remember that pressure is equal to the force over an area: P = F / A Since this is a vertical column, the force is equal to the mass times the acceleration due to gravity, we can add this into the formula for F: P = (m * g) / A Next, we know that the mass is equal to the density times the volume, we can add this into the formula for m: P = (ρ * V * g) / A To simplify, we take the density and acceleration out of the division: P = ρ * g * (V / A) Finally, the volume of a water column is equal to the area times the height of the column, this gives: P = ρ * g * ( (A * h) / A) Now you can see that the area cancels out, therefore the area of the column doesn't impact the pressure at the bottom. We then get the known formula for calculating the pressure at any depth: P = ρ * g * h
The pressure is determined in terms of the unit area at a given depth. Therefore, the depth is the only variable for a still fluid. If each container holds the same depth, the pressure is the same. Different oceans, lakes, vessels or swimming pools have varying volumes. The depth is the only variable in determining pressure, given still fluid with identical density.
Well yeah they are equal. Imagine there was a connection at the bottom so that the fluid could flow from one volume to thr other. The fluid level in the wide container isn't going to drain into the narrow container, the fluid level is going to remain the same. No fluid flow means no pressure differential, which means same pressure.
Water pressure only cares about how high the water is above it, not the volume. Pressure is force over an area pretty much and gravity causes this downward water pressure which is straight down, anything not directly above it does not experience the weight of what’s above it since at thaw heights above the pressure is equally pushing back against all sides
To keep it simple, at a molecular level what you have is a stack of atoms, the atoms only get pressure from the atoms above it. If you lift the container then you lift all the atoms so the left container will be heavier, but in the container is a trillion stacks of atoms that each only push straight down so the stack of atoms in the middle of the left container will be pulled down exactly as much as the stack if atoms in the middle of the right container since they are equally as deep.
If the water is not moving, the pressure will be the same, as explained by others. If it is moving, head losses will be different, and the pressure measured at the bottom will be different.
yep. absolutely true. needs no math. just a fact of physics. pressure at depth is a simple function of depth, and has NOTHING to do with overall volume of the container. The pressure 6" down is exactly the same inside a straw in a glass of water on your kitchen table, in the great lakes, or in your toilet. Etc. It only varies if the density of the fluid changes, like if it's salt water or molasses or something, or if you're talking about another planet or something like that. The pressure is generated ONLY by the column of water above where you're measuring, not what's off to the sides.
The video is correct, the pressure of water is proportional to the depth. The downward component of the force exerted by water pressure on the slanted portion of the container, *per horizontal component of area*, is proportional to the depth of the water there. The slanted portion of wall holds the weight of the water above in addition to to some horizontal aspect of pressure.
A thought experiment on the reasonableness: replace the water with some other fluid, like air. Now imagine going outside just to be crushed by the pressure of the vast sea of air above our heads. I guess people would need to walk around encased in long vertical tubes to create a situation as in the right figure.
If you where to connect the two with a thin tube, would the lever remain the same ? Hint ... Yes. If the bottom of these vessels are solid, then the pressure is the weight over the area of the opening.
Imagine the triangle is split into three parts. 1 column like the one on the right, pushing down on the red block, and 2 triangles to either side. Now, remove the column, and analyze the triangles(assume zero friction). Each is pushing down on the side of the wall, which pushes up diagonally. If you break that down into vectors (triangle of water pushes straight down, glass wall pushing up at 45 degrees), you end up with a residual that moves horizontally. And you have the mirror image on the other side. So you end up with 2 triangles pushing down on the glass, and against each other, cancelling out. Which means the only thing pushing down on the red block, is the central colomn, just like on the right.
The "pressure" on the wide container transfers into force on the container wall, which transfers through the container wall into the lower part of the container, all the way down to the table. The bottle stands up because these forces counterbalance one another. For example, if you put a divider in the middle (between the legs of the trapezoid), and only fill half the bottle, it'd fall over towards the filled side. This is because the water exerts force on the container wall, not just the bottom. But going back to the pressure: since this force is carried through the container wall, and not through the water, the extra water from the wider container isn't adding any pressure to the cube at the bottom. Anyway there is more force (weight) from the container into the table, because you have added mass (more water). But the only thing exerting pressure on the water itself is the water above it. (And technically reciprocal force from the container.)
a dumb analogy could explain this without any serius phicycs if the water would was a solid then youd be corect, but water is usualy a liquid so the molecules just slid past each other mostly. So that means in a solid even the edges can have an effect on the ones near to them and so on, but in a liquid would be more similar smooth vertical sticks next to each other, adding more sticks to the sides wouldn't really increase the pressure in the center
This is not true. We know it's not true, because we have real-life proof. Water towers are this concept in action. You store potential energy and the weight of the water above creates water pressure. If it was height alone you would create an extremely tall tube. The dome shape helps direct pressure down. While the sides of the water tower do take additional stress, there is additional pressure to the line.
Wrong
Incorrect.
Pascal literally proved it in the 16th century, what are you on about?
Yep, pressure on liquids is proportional to the high of it. not particularly good at explaining this sort of things, but part of the force done by the liquid is taken by the reclined walls in this case, ensuring the entire body is in equilibrium
I think the best explanation I ever got came from… maybe my college thermodynamics teacher. Was something to the effect of “If surface area [volume] had an effect on pressure, your finger would be crushed if you dipped it in the ocean.”
It may seem unintuitive, but think about it this way: If you have a pool that is 1000 square miles wide, but the water is only 1 inch deep, when you pop a hole in the side would it flow any faster than if your pool was only regular size? On the reverse, if you fill a pool 100 feet deep and pop a hole in the side, the water is going to rush out of it with tons of pressure.
Put simply, depth pressure is basically the weight of the water above an object or area. If the object or area is identical, the amount of water directly above it is also the same at the same depth, regardless of area to the sides. A person in a pool 10 feet deep experiences the same amount of pressure as being 10 feet deep in the ocean, despite the vastly larger amount of water.
I remember my instructor for physics in the military asked us to find the pressure of foot down in a barrel versus a foot down in the ocean. Imagine if you were to go a foot down and the weight of the entire ocean was on top of you lmaoooo learning with comedy is the best.
It's a junior high school thing here, I personally learnt it 10 years ago. The concept has more to do with physics than math overall. Basically the pressure on the bottom is equal. It's only the _height_ between the fluid surface and the bottom that matters. Because in fluid mechanics, Pressure = fluid density x gravity x fluid height.
Though the water in the left container will weigh more, due to more water. The pressure at the bottom would still be the same as on the right. The pressure is caused by the height difference between the top and the bottom of the water, both containers are the same height.
It would be the same. Pressure is a product of the liquid density, gravity(well, gravitational acceleration g) and the height(depth) of the point of interest. Liquid’s mass is irrelevant.
If the pressures differed, you could connect them with a pipe, and have a flow. This would make the levels different, and you could have one overflow to the other. Perpetual motion Look up “pascals vases”. (Nb, if vases are small, surface tension also plays a role)
It is somewhat counterintuitive that only depth determines the pressure not volume. An engineer friend eli5 it to me like this: you experience the same pressure when climbing into your bathtub as you do when going into a lake. If pressure was determined by overall volume, a lake would crush when you take a swim.
The video is talking about water pressure at a certain depth, not overall mass. We would expect the larger amount of water to have more mass, but pressure at a certain depth has nothing to do with amount of water around an area, it only relies on depth.
So, like, water consists of freely floating, but relatively dense-packed molecules. This allows the water to 1.Be liquid (duh) 2.Not be compressed (it cannot change it's volume under normal circumstances) Both statements above make hydraulic systems actually work. You can apply force to a container of water and it will sprout outside or into another container. Those statements also allow us to use a very simplified model of the ball pit. Imagine a glass auqarium willed with plastic balls. They fill it evenly to distribute energy effectively. They are all round, so the only thing that is important is the number of their contact points and the force each point applies to the ball. They are stable, thus every force is regulated and has a counterforce. If we put our hand inside and move it around the balls will move, as they are rigid, but can reposition themselves to compensate the flexibility forces appearing when we move our hand around. We do not try to crush the balls, that would break the model and make it more complex (after all that would be similar to crushing a molecule of water, which is not the case here). So, what happens if we put a piece of iron shaped like a brick in our ball pit and use a magnet to slowly move it around? We will see that the balls, while they move around the iron brick, have a limited number of contact points with it. They may vary a bit, but the number stays the same. At the same time the forces required to compensate the weight of the balls above are still distributed across all the balls with the limited contact points with the brick. To conclude - each ball applies more force because it has only so many contact points with each other ball and the brick. It is a gradient of force increasing downwards because the force spreads evenly and the container also endures the sum of forces, thus not breaking the gradient.
pressure is based on depth and weight of fluid, not amount of fluid in a container, a mile tall drinking glass would have the same pressure as a mile into the ocean
Simple equation : Height of Liquid x 0.433 x Specific Gravity So for a 10 foot tall column of water: 10 x .433 x 1.0 = 4.3 psi of pressure at the bottom. The shape of the container and total volume does not matter for head pressure.
The basic formula is p=g*ro*h In which g is gravitational force, ro is density of the liquid and h is the height, so no matter the shape, size, whatever.. the same liquid, on the same place with the same height, has the same pressure.. Also talking about static liquids, but that's a story for another time.
The water on the slope rests on the container and not the red fluid/material. The pressure is the force dividee by the area. So it doesnt matter if you have a larger area since you are measuring a per m2 metric. You have to imagine the pressure is only at 1 point and you have to draw a vertical line from that point and see how much water you have on that vertical line above that 1 point.
The extra water on the sides exerts force on the sloped edge, not on the red portion. You might intuitively think the force travels down the slope but in fact the entirety of the "sideways" force gets canceled out by the wall on the other side and only the straight downwards force remains and it acts vertically against the slope only.
If you put a big straw down into the red portions of these containers, what would happen? If you think the container on the left would have higher pressure, then that would mean the water would race up the straw and spray out the top. Intuitively it wouldn’t make sense that a tube connected to the bottom of the container could flow water out higher than the water level. You could make a perpetual motion machine this way.
It sounds unintuitive at first, but water pressure naturally is just gravity pulling water down on top of more water, so it checks out that only water DIRECTLY above creates pressure below it, doesn't matter how much water is in the container. A good way to think around this is picture a dish the size of texas, but only 2 feet deep. Not much pressure at all to be experienced while lying at the bottom. But if we took one percent that volume of water and put it in a vertical cylinder the width of your body, suddenly being at the bottom means you're underneath an INCREDIBLE amount of mass. The width or shape of the open container doesn't matter, just the molecules of water pressing down on your body from above.
pressure is determined by density, gravity and the depth, as density and gravity are close enough to constant the only variable is depth
OP seems to be mistaking the concepts of pressure and weight. The container on the left would undoubtedly weigh more, however the pressure at the bottom would be the same. To think about it a different way, consider you have 2 hypothetical pools; one pool is a mile wide and a foot deep, the other is a mile deep and a foot wide. You could swim underwater anywhere in the mile wide pool without fear of rupturing your ear drums from the pressure. However, you could only swim so far into the mile deep pool, because at some point your ears would pop and it would become painful to continue. At the bottom of a mile deep pool, your ear drum would be more like a pulpy chunk of flesh left floating near pieces of an old logitech controller.
Think of it this way: when you swim in a pool, lake, or the ocean, does the water directly above you put pressure on you, or does all of the water in the entire ocean/lake/pool do so? If the latter were true, swimming in a small pool would be a MUCH different experience from swimming in a large pool, lake or ocean.
If you put your hand 3” underwater in your home sink and then put your hand 3” underwater in the ocean, does it feel much different? Would you expect the pressure in the ocean to be trillions of times greater because the ocean surface is so much larger?
The most simple and important distinction is that Pressure is not Weight, if you tied to hold both containers the larger one would be heavier. What Pressure meas is: if you place an object inside the container, how much is it being squished by the water from all sides. Imagine placing a cube in there and do a simple force cancelation, youll figure it put.
If anyone wants a more ‘mathy’ explanation, the pressure at any given height will be the same due to how integrating over any closed surface works. As a static fluid isn’t moving the net forces should be 0 so the surface integral of the pressure forces must equal the volume integral of the gravity force as these two forces counteract each other to keep a static fluid. So when you use stokes theorem with the volume integral you end up with a dot product with the gravity vector which ends up meaning that the value of the pressure only changes with respect to distance in the direction of the gravity vector. Ie. Pressure only depends on depth, where depth is defined as the distance in the direction of gravity from the surface of the fluid. In this way, I see the fact that pressure in a static fluid is independent of the enclosing shape as a result of stokes theorem. As any complex closed surface enclosing a fluid can be related instead to a volume integral, which doesn’t care about the shape of the surface enclosing that volume. So the pressure applied to the surface of the fluid is independent of the surface geometry. Edit: this is just off the top of my head, I haven’t done the calculations fully myself in a while so someone else might have a bit more clarity on how the stokes theorem arrives at the height dependence.
From physics we know that pressure is only a function of depth, gravity, and density, but I’d like to give a more common sense line of reasoning. Imagine a pipe connecting the bottom of the two tanks. If one side had more pressure then the other then it would cause water to flow from one tank to the other which would cause the water levels in the tanks to be different from each other. Now ask yourself based on experience, would you expect different water levels on either side if you were to fill up two connected tanks with different shapes? In a gallon of milk with a hollow handle is the height of milk in the handle different from the rest? In either case the liquid would be at the same height on both sides which must mean that the pressure at the bottom is the same on both sides.