# Is the Force on a Dam dependant on the height of the water or the amount of water?

## Main Question or Discussion Point

Just had this pondering: Does the force on a water dam depend merely on the height of the water column behind it or does it depend on the total amount of water behind it? For the same dam if I have a lake behind it that is just a few metres wide vs. having a vastly large lake behind it with the same height.

Initially I want to say that the pressure at any given depth is dependant on water height/column only, so the force on the dam would be the same no matter the size of the lake, But that doesn't feel right somehow....

## Answers and Replies

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Baluncore
2019 Award
Only the hydrostatic pressure due to the height is important.

berkeman
Mentor
Initially I want to say that the pressure at any given depth is dependant on water height/column only, so the force on the dam would be the same no matter the size of the lake, But that doesn't feel right somehow....
Remember that Force is Pressure x Area. In your question you ask about "Force", but I think you really mean Pressure.

The length of the lake behind the dam would not matter, but the width of the lake at the dam (and hence the width of the dam) would change the total "Force" on the dam, but not the max pressure on the dam base. Does that help?

hutchphd and russ_watters
Yes, thanks all.

berkeman
I just read through an exhausting 11-page thread on another forum arguing this same question. There are a few participants who "just can't believe" a thin column of water will exert the same total force on the dam wall. The thread over there just bounces back and forth; "does..." "does not..."

So I came here looking for a convincing argument or approach to explain "how can so little water apply tons of force." Anyone have good pedagogical approach? Thanks!

russ_watters and Lnewqban
etotheipi
Gold Member
2019 Award
@gmax137 Surely they must agree that the force the little bit of water at the very bottom exerts on the column above is ##A h \rho g##, and that the pressure at this depth is then ##h \rho g##. All that remains is to note that hydrostatic pressure is isotropic.

Lnewqban
Surely they must agree
Yes... "but still, how can so little water..."

There is some confusion between pressure, force, total force on the wall.. Like I said, it is exhausting. I was hoping someone here had a good analogy, that would prompt a "flash of insight" for the guys on the other forum. Or maybe a simple demonstration used in a high-school physics class to show (seeing is believing).

etotheipi
etotheipi
Gold Member
2019 Award
I see. Maybe you could try poking holes in straws of different diameter and showing that the speed of efflux of water at a specific depth of the hole (measured by lateral distance travelled before hitting the table) is unchanged.

Of course, the most unambiguous thing to do in the case of the dam of depth ##h## and width ##w## is just to compute the total force explicitly,$$\vec{F} = \iint_S p d\vec{A} = \left( \int_{0}^{w} \int_{0}^{h} \rho g y \, dy dx \right) \hat{n} = \frac{1}{2} \rho g h^2 w \hat{n}$$and noting that this is independent of how far the water extends backwards from the damn in the ##z## direction.

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... There are a few participants who "just can't believe" a thin column of water will exert the same total force on the dam wall.

So I came here looking for a convincing argument or approach to explain "how can so little water apply tons of force."...
What do you mean by "thin column of water" and by "so little water"?

pbuk
Gold Member
Imagine that you have a lake 10 km long exerting a total force F on a dam. Build a new dam 1 km upstream of the existing dam (without changing the water level); does the force F change? Take all the water out of the 9 km long upstream lake, does the force F change?

sysprog, gmax137 and etotheipi
What do you mean by "thin column of water" and by "so little water"?
Suppose the "lake" extends only 6 inches upstream.

Actually the other thread is about aquariums; they were discussing and aquarium 8 feet wide, 3 feet deep, and 6 feet "thick." So it holds 8*3*6= 144 cubic feet of water; about 9000 pounds. The pressure at the bottom is 3 * 62 186 pounds/ft^2 or 1.3 psi. The force on the 8 x 3 wall is 8*3*186/2 =2200 pounds.

Now imagine the "thickness" is 1/2 inch (instead of 6 feet). It now holds only 8*3*(0.5/12) = 1 cubic foot; 62 pounds of water. The "I don't believe it" guys can't mentally get over 62 pounds of water ("so little") exerting 2200 pounds of force on the wall.

For what it's worth, they have gotten past the complication of surface tension and meniscus, etc. that would come into it if the "thickness" gets too thin. The conversation there agrees to leave that out as a complicating factor unrelated to the "I don't believe it" problem.

pbuk
Gold Member
Actually the other thread is about aquariums; they were discussing and aquarium 8 feet wide, 3 feet deep, and 6 feet "thick." So it holds 8*3*6= 144 cubic feet of water; about 9000 pounds. The pressure at the bottom is 3 * 62 186 pounds/ft^2 or 1.3 psi. The force on the 8 x 3 wall is 8*3*186/2 =2200 pounds.
Even easier to visualise with an aquarium: divide it in two with a sheet of glass, does the force on the walls magically halve?

sysprog, Ibix, russ_watters and 1 other person
Baluncore
2019 Award
Place two different diameter vertical tubes next to each other. Join them at the bottom.
Pour water into one tube and notice the water settles to the same level in the two tubes.

russ_watters and pbuk
Thank you.
Have those poor souls heard about the Pascal's barrel experiment?

Either you dive 10 meters below the surface of Lake Ontario or 10 meters below the surface of a small water well, each one of your tympanic membranes will feel a pressure that is equivalent to 2 atmospheres.

Multiply that value by the area of that membrane that is in contact with water (or trapped air), and you will have the value of the uniformly distributed force that that column of water is exerting over each membrane.

Now, if you need to generate electricity, you want huge amounts of water to be available, because the power you can expect to obtain depends on the flow rate of water, besides the difference of heights between upstream and downstream the generator.

https://en.wikipedia.org/wiki/Hydroelectricity#Calculating_available_power

sysprog, russ_watters and pbuk
pbuk
Gold Member
So it's pretty easy to come up with both thought experiments and practical demonstrations of hydrostatic pressure (or insert any other basic scientific concept), unfortunately there will always be people participating in exhausting 11 page threads on other forums for whom evidence is not enough.

They are beyond help, sometimes you have to just walk away.

Tom.G and Lnewqban
jbriggs444
Homework Helper
2019 Award
Stretch a sheet of cellophane across a reservoir just upstream of the dam. If the know-nothings are to be believed, the force from the lake side will vastly exceed that from the dam side and the cellophane will be pushed into the dam.

One wonders where the trapped water would go -- squirted skyward maybe?

Thank you all for your thoughtful suggestions. Turns out, @pbuk is right about this. (Except that other thread is now up to 16 pages...)

unfortunately there will always be people participating in exhausting 11 page threads on other forums for whom evidence is not enough.

They are beyond help, sometimes you have to just walk away.

Lnewqban
russ_watters
Mentor
Stretch a sheet of cellophane across a reservoir just upstream of the dam. If the know-nothings are to be believed, the force from the lake side will vastly exceed that from the dam side and the cellophane will be pushed into the dam.
And for the love of god, don't go swimming in a large pool -- you'll be crushed against the side of the pool!

Bystander, jbriggs444 and gmax137
rbelli1
Gold Member
you'll be crushed against the side of the pool!
Unless you stand exactly in the center!

BoB

russ_watters
sophiecentaur