Hydrostatic pressure in a narrow container

In summary, pressure increases with depth regardless of the shape of the container. However, this doesn't sit well with me.
  • #1
RFreund
5
0
TL;DR Summary
Hydrostatic pressure just doesn't sit well with me and I'm hoping someone can help me understand it better.
I understand that pressure increases with depth regardless of the shape of the container. However, this doesn't sit well with me. Imagine a container: 8' tall x 1" wide x 1' long. The pressure at the bottom of the container is 8' x 62.4 pcf = 499.2psf. However, the weight of the water in the container is only 8' x 1/12 x 1' x 62.4 = 41.6 lbs.
Can someone help me rationalize this?
It's hard to imagine pouring 5 gallons of water into a container and generating this type of force.
Or another crazy thought experiment is starting with gallon of water in a box with a closed top and a tall, very narrow tube on top. Now imagine the sides can slide inward to squeeze the water up. At some point the force becomes too large for you to squeeze all the water out. But it's only 1 gallon!
 
Engineering news on Phys.org
  • #2
The "psf" is pounds per square foot. If you put 1 pound on a square inch surface, the pressure will be 144 psf.
So the "pounds" value can readily be less than the "pounds per square foot" value.

If I had a 1-gallon contained that was 7 miles deep, the pressure at the bottom would be the same as the deepest spot on the ocean floor. If that seems unrealistic, stretch a thin tube all the way from the ocean surface to the bottom of Marianna Trench - then let it fill with water.
 
  • Like
Likes berkeman
  • #3
Thanks for the reply. I guess my "issue" with it, is just, how can something with such little weight generate such a large force. I suppose you could liken it to air pressure, but in that case, something is "forcing" air into a container. In the hydrostatic pressure case, there is only gravity acting on the fluid. Maybe there is a better mathematical way to express the hydrostatic pressure other than depth x density that could help make me more at peace with this phenomenon?
 
  • #4
RFreund said:
Thanks for the reply. I guess my "issue" with it, is just, how can something with such little weight generate such a large force.
 
Last edited:
  • Like
Likes RFreund and DrClaude
  • #5
Do you understand why a lever "multiplies" force? How about an old fashioned bumper jack? What about an hydraulic jack?
The work done is the same at each end (small force)x(large distance)=(large force)x(small distance) and energy is conserved.
 
  • Like
Likes RFreund and russ_watters
  • #6
The molecules making up the water don't like being pushed together, but as the force around them grows they don't have a choice, they move imperceptibly (at our scales) closer to each other simultaneously pushing away from each other with greater force. Gravitational forces squeezing them together, and Columbic forces are keeping them apart. The change in pressure as you descend is a measure of change in stored energy per unit volume; the volume is being squeezed like a really stiff linear spring.

I'm just speculating, but I guess the distances between fluid molecules is relatively large such that the columbic force ( which goes as ##\propto \frac{1}{r^2}##) is approximately linear? We measure a linear increase in force with depth (in hydrostatics), but I would imagine at some depth the pressure as a function of depth will become apparently non-linear as the space between molecules becomes diminished?
 
  • #7
How much do you think 5 gallons weighs?
 
  • #8
RFreund said:
Imagine a container: 8' tall x 1" wide x 1' long. The pressure at the bottom of the container is 8' x 62.4 pcf = 499.2psf. However, the weight of the water in the container is only 8' x 1/12 x 1' x 62.4 = 41.6 lbs.
Can someone help me rationalize this?
It's hard to imagine pouring 5 gallons of water into a container and generating this type of force.

Pressure isn't force.
 
  • Like
Likes jbriggs444, gmax137 and hutchphd
  • #9
RFreund said:
The pressure at the bottom of the container is 8' x 62.4 pcf = 499.2psf. However, the weight of the water in the container is only 8' x 1/12 x 1' x 62.4 = 41.6 lbs.
Can someone help me rationalize this?
It's hard to imagine pouring 5 gallons of water into a container and generating this type of force.
RFreund said:
I guess my "issue" with it, is just, how can something with such little weight generate such a large force.
When you say "force", you actually mean "pressure". The force acting on the ground from 5 gallons of water is always 41 lb no matter the size of the container. The pressure is the force (weight in this case) divided by the area. The smaller the area on the ground, the greater the pressure.

Put a 5-gallon can of water on a nail and the nail will sink into the ground. Put the same can directly on the ground and it doesn't sink. Same weight, different area, different pressure.

RFreund said:
I suppose you could liken it to air pressure, but in that case, something is "forcing" air into a container. In the hydrostatic pressure case, there is only gravity acting on the fluid.
Do you realize that having a standard atmospheric pressure of 15 psi means that over a 1" X 1" square over your head, there is actually 15 lb of air resting on it? This means, depending on your size, that you have between 500 and 1000 lb of air resting on your head and shoulders as we speak! Just gravity acting on the fluid.

You can find more info in this post.
 
  • #10
JT Smith said:
Pressure isn't force.
@RFreund , re-read your original post in this thread, and then say, "pressure is not force... force is not pressure" do this for each of your posts.
 
  • #11
Note, the total force (pressure times area) is not the same between the bottom of the container and the water; and the bottom of the container and the ground, unless the container is a consistent cross section (a rectangle or cylinder, etc.).

Pressure is independent of container shape, and it is also independent of container size. This may be hard to visualize with hydrostatic pressure, but it should be a lot easier with pump-generated static pressure. When blowing up a balloon, for example, your ability to pressurize it only depends on how much pressure your lungs can generate, not on the size of the balloon. A bigger balloon just takes longer. A typical male can generate 1psi by exhaling, which is enough to lift a car with a queen sized air mattress. I feel like this is something mythbusters should have tried...
 
  • #12
Compare bursting a narrow tube with a liquid vs. bursting the tube with a narrow solid wedge shoved in.
In both cases, a small force exerted over a long distance (shoving the liquid/wedge into the tube) gets leveraged into a large force over a short distance (bursting the tube).
In case of the solid wedge, a lot of the force exerted along the wedge gets wasted on the friction pushing the walls of the tube along at the spot of contact. A liquid pressed into the tube is perfectly lubricated, so the whole force exerted along the tube is available to be leveraged into bursting the tube.
 
  • #13
Thanks for the replies. A great video suggestion by erobz and I kinda like the "lever" comparison by Hutchphd.

A few comments on pressure vs force. This seems to be side tracking my question a little bit. Force is simply pressure times an area. When I think about pressure problems, I like to eventually get into a force. For example, when I say "squeeze the container". I don't imagine squeezing "5 psi", but I can imagine what it feels like to squeeze at a force of 50lbs.
 
  • #14
RFreund said:
A few comments on pressure vs force. This seems to be side tracking my question a little bit. Force is simply pressure times an area. When I think about pressure problems, I like to eventually get into a force.
It's great that you mathematically understand the difference/relationship, but in my opinion it's part and parcel of the issue of what's up with a narrow container -- I feel like you haven't transitioned that to intuition. You could rupture a tanker truck with a tall metal straw. Making sure you keep them separate but squared really is how you understand the answer the question you asked.
For example, when I say "squeeze the container". I don't imagine squeezing "5 psi", but I can imagine what it feels like to squeeze at a force of 50lbs.
Ok, but please recognize that squeezing a force of 50lb doesn't directly translate into pressure. 5 psi lacks ambition; how about 500psi?
 
  • Like
Likes RFreund
  • #15
russ_watters said:
You could rupture a tanker truck with a tall metal straw. Making sure you keep them separate but squared really is how you understand the answer the question you asked.
That's a fair point, but my 'bamboozlement' is that the pressure on the tanker truck is the same whether that tall metal straw is 1 inch in diameter or 100 feet in diameter.
If we imagine the straw as rectangular (not necessary, but helpful for the transition), I do understand the "force" on the walls of a thin straw would be less than that of a wide straw due to the difference in force and pressure. The pressure on the walls is equal in both cases, but because the area of the wider straw is greater, the force is greater.
 
  • #16
It isn't intuitive, at least not at first.

Think of it this way. If it were a tall solid straw, instead of one filled with a liquid, then a wider straw would weigh more than a thin one. The total force on the ground would be greater with the wider, solid straw. That's easy to understand. But the pressure on the ground would be the same. Make sense?

With a liquid the pressure includes the sides as well as the bottom, but the idea is basically the same.
 
  • Like
Likes russ_watters
  • #17
RFreund said:
That's a fair point, but my 'bamboozlement' is that the pressure on the tanker truck is the same whether that tall metal straw is 1 inch in diameter or 100 feet in diameter.
If we imagine the straw as rectangular (not necessary, but helpful for the transition), I do understand the "force" on the walls of a thin straw would be less than that of a wide straw due to the difference in force and pressure. The pressure on the walls is equal in both cases, but because the area of the wider straw is greater, the force is greater.
Consider the comparison between the column of water and a solid wedge. A thick water column works like a blunt wedge. A thin water column works like a sharp wedge. So the sharp wedge exerts a lot of sideward force, but over a shorter distance.
The wedge has some side friction so some of the force is wasted on side friction - especially in the case of the sharp wedge. Whereas the water column is perfectly lubricated. A thinner water column is a sharper wedge, so its smaller weight is leveraged into same side force.
 

FAQ: Hydrostatic pressure in a narrow container

What is hydrostatic pressure?

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases with depth in the fluid and is calculated using the formula P = ρgh, where P is the pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column.

Does the shape of the container affect hydrostatic pressure?

No, the shape of the container does not affect hydrostatic pressure. Hydrostatic pressure depends solely on the height of the fluid column and the density of the fluid, not on the width or shape of the container.

How does the width of a narrow container influence hydrostatic pressure?

The width of a narrow container does not influence hydrostatic pressure. Hydrostatic pressure is determined by the height of the fluid and its density. Therefore, a narrow container will have the same hydrostatic pressure at a given depth as a wider container with the same fluid height.

What role does fluid density play in hydrostatic pressure?

Fluid density is a crucial factor in determining hydrostatic pressure. Higher density fluids exert greater pressure at a given depth. The relationship is linear, meaning that if the density of the fluid doubles, the hydrostatic pressure at any given depth will also double.

How can you measure hydrostatic pressure in a narrow container?

Hydrostatic pressure in a narrow container can be measured using a pressure sensor or a manometer placed at the desired depth. These instruments will provide a reading of the pressure, which can be compared to the theoretical value calculated using the formula P = ρgh.

Similar threads

Replies
12
Views
1K
Replies
5
Views
6K
Replies
4
Views
6K
Replies
7
Views
4K
Replies
1
Views
2K
Back
Top