Hydrostatics

[MartaBraga]

How do I calculate the pressure that a certain amount of water applies on the wall of a dem(example)?

Is the equation Pressure = P(atmosphere) + (density(water).hight.gravity) correct? Or does that apply only for the pressure aplied on the bottom of the system?

(I'm sorry about the bad english)

[Doc Al]

That equation is correct. Realize that "height" is really depth below the water surface.

[Cyrus]

.You have to integrate the pressure distribution going up the wall...or you could be clever and just find the area of the triangle and do it geometrically.

[mgb_phys]

Thats the correct formula for the pressure at that depth.
(the force sideways is the same as the force downwards because it is a fluid, so pushes in all directions)

Cyrus's reply is about calculating the TOTAL force on the wall.
At the bottom is the pressure you calculated, at the surface there is obviously no pressure and half way down is half the pressure.

[MartaBraga]

.You have to integrate the pressure distribution going up the wall...or you could be clever and just find the area of the triangle and do it geometrically.


How do I do that, I mean, calculate it through the area of the triangle? (I rather do anything than integrate!)

[Cyrus]

The pressuer at the bottom is: P=\rho * g*h. The pressure at the top is zero. It varies linearly. I'll leave it to you to show us the rest of the derivation and logic. :smile:

[MartaBraga]

Ok, so is it that simple?
I calculated the pressure through that formula and then divided it by 2 and multiplied it by the area... and the result is correct... But I thought I had to use the atmospheric pressure in the formula...

Or is this pure luck?

[Cyrus]

Can you show your steps. It would be more insightful if you knew why your answer is correct. Right now, you are just blindly plugging away and hoping for the best to come out.

[MartaBraga]

Ok, the problem is to calculate the pressure the water applies on a dem, with 40m of heigth and 150m of lenght.
I calculated the pressure = 1000 (the density) x 40 (the height) x 9,8 (the aceleration of gravity. The result is 392000, which I divided by 2 = 196000. I multiplied this value by the area of the dem in contact with the water, 6000m2. The result is 1,176x10^9.
This was the result I was looking for.

[Cyrus]

Ok, first of all the water does not exert a 'pressure' on the dam. It has a pressure distribution. This causes a net force acting at the center of pressure. (Think center of gravity, same concept different name).

Yes that exactly right. Do you know what you are doing though?

[MartaBraga]

That's the problem. I just thought, well, if the pressure on the top is 0, and on the bottom is, in this case, 392000, lets find the average pressure, and that's why I divided it by 2. and then, I multiplied it by the area.

(I'm really sorry about the grammar errors, but I'm Portuguese, so there are some words and expressions I don't fully know how to spell or use)

[gmax137]

"But I thought I had to use the atmospheric pressure in the formula..."

You can leave out the atmospheric pressure because it is applied to the downstream side of the dam also - hence it cancels out. If there were no water in the lake at all, just the air on both sides, there would be no net force on the dam.

[Doc Al]

If the problem is to calculate the net force that the water exerts on the dam, then you must include atmospheric pressure. (But as gmax137 states, for some purposes you can forget atmospheric pressure when it acts on both sides of the dam.)

[Cyrus]

That's the problem. I just thought, well, if the pressure on the top is 0, and on the bottom is, in this case, 392000, lets find the average pressure, and that's why I divided it by 2. and then, I multiplied it by the area.

(I'm really sorry about the grammar errors, but I'm Portuguese, so there are some words and expressions I don't fully know how to spell or use)

Hint: The area of a triangle is 1/2*(base)*(height)!