LanHikari22
  • LanHikari22
"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) ?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
\(\star\)
Anaise
  • Anaise
hi
anonymous
  • anonymous
|dw:1450407935797:dw|

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anonymous
  • anonymous
|dw:1450408153307:dw|
ganeshie8
  • ganeshie8
The key idea here is that the fluid pressure on the plate changes with depth
ganeshie8
  • ganeshie8
\[p = \dfrac{F}{A} \implies F = pA\]
ganeshie8
  • ganeshie8
|dw:1450410116289:dw|
LanHikari22
  • 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
  • ganeshie8
Consider a small area element of thickness \(\Delta x\) : |dw:1450410291519:dw|
ganeshie8
  • ganeshie8
Let try and find its area in terms of the depth \(x\) first
ganeshie8
  • ganeshie8
Would you agree that the width of that area element is \(w(x) = 2x\) ?
ganeshie8
  • ganeshie8
|dw:1450410579411:dw|
LanHikari22
  • LanHikari22
is it because of how the vertex cut the angle into 45 45 identity triangle?
ganeshie8
  • ganeshie8
Exactly! those two triangles are 45-45-90 isosceles triangles
LanHikari22
  • 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
  • ganeshie8
Very good observation! so we need to consider two integrals. One for each half
ganeshie8
  • ganeshie8
Lets work the top half first
LanHikari22
  • LanHikari22
yessir!
ganeshie8
  • ganeshie8
\(A = w(x)*dx = 2x*dx\) \(F=pA = p*2x*dx \)
ganeshie8
  • ganeshie8
do you know how the pressure \(p\) changes with depth ?
LanHikari22
  • 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
  • ganeshie8
\(p = \rho x\) ?
LanHikari22
  • LanHikari22
yeah, in this coordinate.
ganeshie8
  • ganeshie8
\(A = w(x)*dx = 2x*dx\) \(F=pA = p*2x*dx = \rho *2x^2\, dx\)
ganeshie8
  • 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
  • ganeshie8
|dw:1450411244834:dw|
ganeshie8
  • ganeshie8
Do we get \[F_{\text{top half}} = \int\limits_{x=0}^{x=a\sqrt{2}/2}\rho *2x^2\, dx\] ?
LanHikari22
  • 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
  • LanHikari22
and, yeah!
ganeshie8
  • ganeshie8
lets finish the top half first, one thing at a time :)
LanHikari22
  • 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
  • ganeshie8
For sure yes, lets consider a quick example
ganeshie8
  • ganeshie8
Maybe just take our case
ganeshie8
  • ganeshie8
Think about different ways to find below shaded region |dw:1450412134674:dw|
ganeshie8
  • 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
  • LanHikari22
yeah, but that's independent of x
ganeshie8
  • ganeshie8
we don't care, all we want is area for now.. lets find the area using another method
ganeshie8
  • ganeshie8
Lets pretend that we know only the formula for area of rectangle and we don't remember any other formulas
ganeshie8
  • ganeshie8
How to find area of above triangular region using just the area of rectangle formula ?
ganeshie8
  • ganeshie8
Lets divide that triangle into rectangles and see
LanHikari22
  • LanHikari22
divide the triangle into rectangles?
ganeshie8
  • ganeshie8
Yes, lets approximate the area of that triangle with 2 rectangels of equal height : |dw:1450412904544:dw|
LanHikari22
  • LanHikari22
w(x) * a - a^2
ganeshie8
  • ganeshie8
Can you write an expression for the area of those two rectangles ?
LanHikari22
  • LanHikari22
Wouldn't that be w(x) * a ?
ganeshie8
  • ganeshie8
what are the heights of each rectangle ? |dw:1450412974796:dw|
LanHikari22
  • LanHikari22
2x^2, in other words
LanHikari22
  • LanHikari22
depending on how we're breaking the rectangles, if the index increases by a, it works.
LanHikari22
  • LanHikari22
oh wait, 4 rectangles, halfpoint is sqrt/2, is this sqrt/4?
ganeshie8
  • ganeshie8
the height of each rectangle is simply \(a\sqrt{2}/4\) right ?
LanHikari22
  • LanHikari22
I mean a * sqrt(2) / 4
ganeshie8
  • ganeshie8
Yep!
ganeshie8
  • 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
  • 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
  • ganeshie8
see if that makes sense more or less
ganeshie8
  • ganeshie8
Notice that the approximation gets better as we increase the number of rectangles
ganeshie8
  • ganeshie8
|dw:1450413837041:dw|