Question

A plate shaped like an equilateral triangle 1 m on a side is placed on a vertical wall 1 m below the
surface of a pool filled with water. On which plate in the figure is the force greater? Compute the force
on the two plates

Answer 1

Hi! I'll help if I can! For starters, would you be able to attach the diagram? You can take a picture screenshot, crop it, and attach it. Or you can provide a link to it, if it's publicly available!

This sounds like you will have a gradient of force due to the pressure. So, force increases with the greater depth. You will have a function for this force that varies vertically, and you'll integrate it over the shape of the equilateral triangle.

I think that I'd need more information about the force due to the water, though!

Answer 2

Thanks! @theEric im not sure how I must find the function. but heres the screenshot!

Answer 3

Oh, I see!

So you know, the pressure increases with depth. That's why submarines can go only so far before they have issues.

The pressure increases with the weight of the water above any spot, and this pressure is usually in all directions.

We want the formula for the weight of the water to a specific depth. We will use that to integrate over each triangle's varying depths.

Pressure can also be seen to be a "force per area." Considering depth, you can imagine infinitesimal depth levels, and each level corresponds to a "force per area." The force per area is the same on the entire strip of the triangle at that depth, so more area = more force.

So, that's the background.
The pressure of the water above a certain point can be given by
\(p=\rho g h\) where \(\rho\) is the mass/volume density, \(g\) is gravitational acceleration, and \(h\) is the depth.

So, now you have a pressure function that varies with the height (along your triangle).

By now, do you have a guess as to which triangle has a greater force on it?

Hint: more pressure at lower level. More force from more area.

Answer 4

If you don't know yet, you will after calculating!

So, you have to find a way to approach integrating over these triangles conveniently, knowing that it will be with respect to both dimensions of your 2D shape.

So, vertical and horizontal, but how to you want to do this double integral?

Answer 5

Note, the pressure varies with depth, and the force varies with area (hint: area at a depth?)

For the pressure, it's
\(p=\rho g h\) where \(\rho\) and \(g\) are constants that you can look up later. I wonder what you'll integrate, and through what range?
(This confirms the equation: http://en.wikipedia.org/wiki/Fluid_statics#Hydrostatic_pressure)

For the width of the triangle (and you might guess why I'm just caring about the width for my method), we can use bounds that are dependent upon the depth. That is, each side of the triangle (boundary of integral?) can be modeled as a function of height.

Answer 6

I eventually got the answer in class thanks @theEric!!!!

Answer 7

You're welcome! I'm glad you got it! :)

Answer 8

That was sort of a fun one. Simple shape and possibly applicable!!