- #1

peanutaxis

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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...

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- Thread starter peanutaxis
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- #1

peanutaxis

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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...

- #2

Baluncore

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Only the hydrostatic pressure due to the height is important.

- #3

berkeman

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Remember that Force is Pressure x Area. In your question you ask about "Force", but I think you really mean Pressure.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...

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?

- #4

peanutaxis

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Yes, thanks all.

- #5

gmax137

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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!

- #6

- #7

gmax137

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

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).

- #8

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 traveled 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.

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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- #9

Lnewqban

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What do you mean by "thin column of water" and by "so little water"?... 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."...

- #10

pbuk

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- #11

gmax137

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

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.

- #12

pbuk

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Even easier to visualise with an aquarium: divide it in two with a sheet of glass, does the force on the walls magically halve?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.

- #13

Baluncore

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Pour water into one tube and notice the water settles to the same level in the two tubes.

- #14

Lnewqban

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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.

Please, see:

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

- #15

pbuk

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

- #16

jbriggs444

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

- #17

gmax137

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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.

- #18

russ_watters

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And for the love of god, don't go swimming in a large pool -- you'll be crushed against the side of the pool!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.

- #19

rbelli1

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you'll be crushed against the side of the pool!

Unless you stand exactly in the center!

BoB

- #20

sophiecentaur

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