"A square plate of side a feet is dipped in a liquid of weight density ρ lb/ft3. Find the fluid force on the plate if a vertex is at the surface and a diagonal is perpendicular to the surface."
Assuming that we have a function that returns the width of the object at any point x, w(x), and the depth in-water at any x, h(x), shouldn't we be able to find that for every layer of the object
Work(xi) = p * h(xi) // where p is weight density
F(xi) = Work(xi) * w(xi) * dx // where w(xi) * dx represents Area
which should in the end be
F(x) = p * integral(Work(x) * w(x) dx) ?

- LanHikari22

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- Anaise

hi

- anonymous

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## More answers

- anonymous

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- ganeshie8

The key idea here is that the fluid pressure on the plate changes with depth

- ganeshie8

\[p = \dfrac{F}{A} \implies F = pA\]

- ganeshie8

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- LanHikari22

Oh my god, I didn't even interpret it that way, I assumed it meant that half the square was inside the liquid!

- ganeshie8

Consider a small area element of thickness \(\Delta x\) :
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- ganeshie8

Let try and find its area in terms of the depth \(x\) first

- ganeshie8

Would you agree that the width of that area element is \(w(x) = 2x\) ?

- ganeshie8

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- LanHikari22

is it because of how the vertex cut the angle into 45 45 identity triangle?

- ganeshie8

Exactly! those two triangles are 45-45-90 isosceles triangles

- LanHikari22

what about the other half below the horizontal vertex? w(x) decreases as x increases instead, it's the opposite of the one above it

- ganeshie8

Very good observation! so we need to consider two integrals. One for each half

- ganeshie8

Lets work the top half first

- LanHikari22

yessir!

- ganeshie8

\(A = w(x)*dx = 2x*dx\)
\(F=pA = p*2x*dx \)

- ganeshie8

do you know how the pressure \(p\) changes with depth ?

- LanHikari22

Yeah! we can derive pressure from the weight density of the liquid, mg / xyz * z = mg/xy = F/A = Pressure.
pressure increases as depth increases.

- ganeshie8

\(p = \rho x\) ?

- LanHikari22

yeah, in this coordinate.

- ganeshie8

\(A = w(x)*dx = 2x*dx\)
\(F=pA = p*2x*dx = \rho *2x^2\, dx\)

- ganeshie8

integrate that between \(x=0\) and \(x=\dfrac{a\sqrt{2}}{2}\) to get the Force of liquid on the top half of the plate

- ganeshie8

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- ganeshie8

Do we get
\[F_{\text{top half}} = \int\limits_{x=0}^{x=a\sqrt{2}/2}\rho *2x^2\, dx\]
?

- LanHikari22

hmmm, I wonder if rotating this, and saying the lengths are asqrt(2)-x instead of x would work, that sounds like a complicated way of doing it though

- LanHikari22

and, yeah!

- ganeshie8

lets finish the top half first, one thing at a time :)

- LanHikari22

so, I believe I understand riemann sum in theory, and how intergrals are like the function level of multiplication but, quick question:
can I turn any multiplication statement into an integral if I have delta x there?
Like how we turned this F=pA=p∗2x∗dx=ρ∗2x2dx into an integral just now

- ganeshie8

For sure yes, lets consider a quick example

- ganeshie8

Maybe just take our case

- ganeshie8

Think about different ways to find below shaded region
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- ganeshie8

Since it is a familiar shape, finding the area is easy : 1/2*b*h = 1/2*a*a = 1/2a^2
yes ?

- LanHikari22

yeah, but that's independent of x

- ganeshie8

we don't care, all we want is area for now..
lets find the area using another method

- ganeshie8

Lets pretend that we know only the formula for area of rectangle and we don't remember any other formulas

- ganeshie8

How to find area of above triangular region using just the area of rectangle formula ?

- ganeshie8

Lets divide that triangle into rectangles and see

- LanHikari22

divide the triangle into rectangles?

- ganeshie8

Yes, lets approximate the area of that triangle with 2 rectangels of equal height :
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- LanHikari22

w(x) * a - a^2

- ganeshie8

Can you write an expression for the area of those two rectangles ?

- LanHikari22

Wouldn't that be w(x) * a ?

- ganeshie8

what are the heights of each rectangle ?
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- LanHikari22

2x^2, in other words

- LanHikari22

depending on how we're breaking the rectangles, if the index increases by a, it works.

- LanHikari22

oh wait, 4 rectangles, halfpoint is sqrt/2, is this sqrt/4?

- ganeshie8

the height of each rectangle is simply \(a\sqrt{2}/4\) right ?

- LanHikari22

I mean a * sqrt(2) / 4

- ganeshie8

Yep!

- ganeshie8

therefore the area of rectangles is given by :
\(A = \text{height}(w_1 + w_2)\)
\(A=\dfrac{a\sqrt{2}/2}{2}\left( 1*\dfrac{a\sqrt{2}
}{2} + 2*\dfrac{a\sqrt{2}}{2}\right)\)
yes ?

- ganeshie8

If we split into \(n\) rectangles instead of \(2\), we get :
\(A = \text{height}(w_1 + w_2+\cdots+w_n)\)
\(A=\dfrac{a\sqrt{2}/2}{n}\left( 1*\dfrac{a\sqrt{2}
}{n} + 2*\dfrac{a\sqrt{2}}{n}+\cdots + n*\dfrac{a\sqrt{2}}{n} \right)\)

- ganeshie8

see if that makes sense more or less

- ganeshie8

Notice that the approximation gets better as we increase the number of rectangles

- ganeshie8

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