Integral Applications - Hydrostatic Pressure + Force

[MermaidWonders]

Question #1 - A lobster tank in a restaurant is 1 m long by 0.75 m wide by 60 cm deep. Taking the density of water to be 1000 kg/m$^3$, find the water forces
(a) on each of the larger sides of the tank;
(b) on each of the smaller sides of the tank.

 

[skeeter]

Question #1 - A lobster tank in a restaurant is 1 m long by 0.75 m wide by 60 cm deep. Taking the density of water to be 1000 kg/m$^3$, find the water forces
(a) on each of the larger sides of the tank;
(b) on each of the smaller sides of the tank.

for constant density fluids ...

$\displaystyle F = \rho \cdot g \int_c^d h(y) \cdot L(y) \, dy$

larger side ...

$\displaystyle F = 9800 \int_0^{0.6} y \cdot 1.0 \, dy$

shorter side ...

$\displaystyle F = 9800 \int_0^{0.6} y \cdot 0.75 \, dy$

 

[MermaidWonders]

The above worked for me, but I'm still finding applications like these quite intuitive. Would you mind explaining the process behind it?

 

[MermaidWonders]

Here's another one...

Question #2 - On August 12, 2000, the Russian submarine Kursk sank to the bottom of the sea, approximately 95 meters below the surface. Find the following at the depth of the Kursk. (Use g=9.8 m/s$^2$.)

The water pressure: pressure = _________

The force on a 5 meter square metal sheet held
Horizontally 95 meters below the surface:
force = __________

Vertically with its bottom 95 meters below the surface:
force = __________

 

[skeeter]

Go to this link ...

https://www.ck12.org/calculus/fluid-pressure/lesson/Applications:-Fluid-Forces-CALC/

... scroll down to
The Fluid Force on a Vertical Surface

 

[MermaidWonders]

Thanks for the link! Now I'm wondering, when would $h(y)$ not equal to $y$? Is it when we choose a vertical y-axis that doesn't have its origin at the water's surface? Like if the water surface is to be -5, then the depth y metres below would be obtained with the function -5 - y?

 

[MarkFL]

Let's find the hydrostatic force on a vertically submerged rectangle, whose width is $w$ and height is $h$. If we orient a vertical $x$-axis along one side of the rectangle, with the origin at the surface and the positive direction down, we may compute the force along one horizontal elemental rectangle having area $A(x)$ making up the entire rectangle as:

$$dF=\rho xA(x)$$

The area of this elemental rectangle is its width at $x$ times its height $dx$, hence:

$$dF=\rho xw(x)\,dx$$

where:

$$w(x)=w$$ (the width is constant for a rectangle)

And so we have:

$$dF=\rho w x\,dx$$

Now, if the top edge of the rectangle is at $x=x_1$ and the bottom edge is at $x=x_2$, then by summing all of the elements of the force, we find:

$$F=\rho w\int_{x_1}^{x_2}x\,dx$$

Applying the FTOC, we obtain:

$$F=\frac{\rho w}{2}\left[x^2 \right]_{x_1}^{x_2}=\frac{\rho w}{2}\left(x_2^2-x_1^2 \right)$$

Since the height $h$ of the rectangle is $$h=x_2-x_1$$, we may write:

$$F=\frac{\rho wh}{2}\left(2x_1+h \right)$$

 

[skeeter]

... when would $h(y)$ not equal to $y$?

what if the submerged planar surface is not vertical?

 

[MermaidWonders]

Oh, true...

 

[MarkFL]

Let's consider a rectangle of width $w$ and height $h$ submerged at an angle such that the rectangle is an inclined plane. We may orient an $x$-axis that runs down the surface of the plane, parallel to its height, and whose origin is at the upper edge. Let $d(x)$ be the depth of the fluid, where $d(0)=d_1$ is the depth of the upper edge and $d\left(h \right)=d_2$ is the depth of the lower edge. We may then state:

$$d(x)=\frac{d_2-d_1}{h}x+d_1$$

We may begin as before, computing the elemental force:

$$dF=\rho w \left(\frac{d_2-d_1}{h}x+d_1 \right)\,dx$$

Summing the elements, we obtain:

$$F=\rho w\int_0^{h} \frac{d_2-d_1}{h}x+d_1\,dx$$

Application of the FTOC gives us:

$$F=\rho w\left[\frac{d_2-d_1}{2h}x^2+d_1x \right]_0^{h}=\rho w\left(\frac{d_2-d_1}{2h}h^2+d_1h \right)=\rho w\left(\frac{d_2h+d_1h}{2} \right)=\frac{\rho wh\left(d_1+d_2 \right)}{2}$$

If $\ell$ is the horizontal distance through which the plane occupies, then by Pythagoras, we may state:

$$h=\sqrt{\ell^2+\left(d_1-d_1 \right)^2}$$

And so we may state:

$$F=\frac{\rho w\sqrt{\ell^2+\left(d_1-d_1 \right)^2}\left(d_1+d_2 \right)}{2}$$

 

[MermaidWonders]

Hmmm... let me take some time to digest this here. What is $dF$?

Let's consider a rectangle of width $w$ and height $h$ submerged at an angle such that the rectangle is an inclined plane. We may orient an $x$-axis that runs down the surface of the plane, parallel to its height, and whose origin is at the upper edge. Let $d(x)$ be the depth of the fluid, where $d(0)=d_1$ is the depth of the upper edge and $d\left(h \right)=d_2$ is the depth of the lower edge. We may then state:

$$d(x)=\frac{d_2-d_1}{h}x+d_1$$

We may begin as before, computing the elemental force:

$$dF=\rho w \left(\frac{d_2-d_1}{h}x+d_1 \right)\,dx$$

Summing the elements, we obtain:

$$F=\rho w\int_0^{h} \frac{d_2-d_1}{h}x+d_1\,dx$$

Application of the FTOC gives us:

$$F=\rho w\left[\frac{d_2-d_1}{2h}x^2+d_1x \right]_0^{h}=\rho w\left(\frac{d_2-d_1}{2h}h^2+d_1h \right)=\rho w\left(\frac{d_2h+d_1h}{2} \right)=\frac{\rho wh\left(d_1+d_2 \right)}{2}$$

If $\ell$ is the horizontal distance through which the plane occupies, then by Pythagoras, we may state:

$$h=\sqrt{\ell^2+\left(d_1-d_1 \right)^2}$$

And so we may state:

$$F=\frac{\rho w\sqrt{\ell^2+\left(d_1-d_1 \right)^2}\left(d_1+d_2 \right)}{2}$$

 

[MarkFL]

Hmmm... let me take some time to digest this here. What is $dF$?

$dF$ is the force differential. :)

 

[MermaidWonders]

Oh, I see. :)

 

[MermaidWonders]

Another question... here we go...Suppose that a cubic container measuring 4 m on a side has been suspended in the ocean from a ship by a 20 m cable attached to its top. Assume that the top of the cable is exactly at water level...

(The entire question is actually quite long, and so, I'm only going to type in the part of this question that I don't get.)

... Consider the hydrostatic force on the bottom face of the container vs. that on one vertical side of the container. Which one is greater?

Can someone explain why the force on one vertical side of the container is greater than that on the bottom face only? I thought it's the other way around, since if you were to compute the force on one vertical side of the container, you'll have to integrate for pressure due to the fact that the depth varies along the vertical side, so when this "integrated" pressure is multiplied by area to get force, the value would be larger than that for just the bottom face of the container... (Random logic used here, I think.)

 

[MarkFL]

The force $F_v$ on a square sheet held vertically in the water, where the square measures $x$ units on a side is given by:

$$F_v=\frac{\rho x^3}{2}$$

The force $F_x$ on a square sheet held horizontally in the water at a depth of $x$ units, where the square measures $x$ units on a side is given by:

$$F_h=\rho x^3$$

So, we see the force on the bottom is double that on anyone of the sides.

 

[MermaidWonders]

Sorry, but could you explain how you got these 2 formulas? I'm lost. :(

The force $F_v$ on a square sheet held vertically in the water, where the square measures $x$ units on a side is given by:

$$F_v=\frac{\rho x^3}{2}$$

The force $F_x$ on a square sheet held horizontally in the water at a depth of $x$ units, where the square measures $x$ units on a side is given by:

$$F_h=\rho x^3$$

So, we see the force on the bottom is double that on anyone of the sides.

 

[MarkFL]

Sorry, but could you explain how you got these 2 formulas? I'm lost. :(

I got them from my posts in this thread. However, we could just use the last one I posted (which should read)

$$F=\frac{\rho w\sqrt{\ell^2+\left(d_2-d_1 \right)^2}\left(d_1+d_2 \right)}{2}$$

Sorry for the typo in the original post.

For the vertical sheet, one of the sides of the cube, we have:

$$w=x,\,\ell=0,\,d_1=0,\,d_2=x$$

And so we find:

$$F_v=\frac{\rho x\sqrt{0^2+x^2}(0+x)}{2}=\frac{\rho x^3}{2}$$

For the bottom of the cube, we have:

$$w=x,\,\ell=x,\,d_1=x,\,d_2=x$$

And so we find:

$$F_h=\frac{\rho x\sqrt{x^2+(x-x)^2}(x+x)}{2}=\rho x^3$$

 

[MermaidWonders]

OK. But if we were to use slicing to get the total hydrostatic force on the vertical side and then compare that to the total hydrostatic force on the bottom face, I do get a larger value for the bottom face, which is correct, but the force on the bottom face is not exactly double that on any of the sides... :(

I got them from my posts in this thread. However, we could just use the last one I posted (which should read)

$$F=\frac{\rho w\sqrt{\ell^2+\left(d_2-d_1 \right)^2}\left(d_1+d_2 \right)}{2}$$

Sorry for the typo in the original post.

For the vertical sheet, one of the sides of the cube, we have:

$$w=x,\,\ell=0,\,d_1=0,\,d_2=x$$

And so we find:

$$F_v=\frac{\rho x\sqrt{0^2+x^2}(0+x)}{2}=\frac{\rho x^3}{2}$$

For the bottom of the cube, we have:

$$w=x,\,\ell=x,\,d_1=x,\,d_2=x$$

And so we find:

$$F_h=\frac{\rho x\sqrt{x^2+(x-x)^2}(x+x)}{2}=\rho x^3$$

 

[MarkFL]

OK. But if we were to use slicing to get the total hydrostatic force on the vertical side and then compare that to the total hydrostatic force on the bottom face, I do get a larger value for the bottom face, which is correct, but the force on the bottom face is not exactly double that on any of the sides... :(

I used the calculus to derive a general formula, and then applied it to both cases. Can you post your work?

 

[MermaidWonders]

View attachment 7899

I was going to attach photos of my solution, but MATHHELPBOARDS said that each of my files exceeded the max size for an attachment, so I'm just going to type out what I have on paper.${Force}_{bottom}$ = ${Pressure}_{bottom}$ * ${Area}_{bottom}$
= (1000 kg*m$^3$)(9.81 m/s$^2$)(24 m)(16 m$^2$)
= 3 767 040 N
**NOTE: 1000 kg*m$^3$ is the water density for the ocean.

As for the vertical side, we must integrate to get the force...

Slice so pieces have approximately the same hydrostatic pressure. Therefore, one must slice horizontally.

${Force}_{slice}$ $\approx$ ${Pressure}_{slice}$ * Area
$\approx$ 1000(9.81)${h}_{i}$ * 4 $\Delta$h

Adding slices, ${F}_{TOTAL}$ $\approx$ $\sum_{i = 1}^{n} 1000(9.81)$${h}_{i}$ * 4 $\Delta$h

As $\Delta$h $\to$ 0,
${F}_{TOTAL}$ $\approx$ $\lim_{{\Delta h}\to{0}}$ $\sum_{i = 1}^{n}$ 1000(9.81)${h}_{i}$ * 4 $\Delta$h
$\approx$ $\int_{20}^{24} 4000g{h}_{i}\,dh$
$\approx$ 3 453 120 N

 

[MarkFL]

Ugh...I misread the problem, and thought the top of the cube is at the water's surface, not the top of the 20 m cable. So, if we let $c$ be the length of the cable, we find:

$$F_v=\frac{\rho x^2(2c+x)}{2}$$

$$F_h=\rho x^2(c+x)$$

 

[MermaidWonders]

So does this mean that the larger hydrostatic force isn't necessary double that of the smaller force?

 

[MarkFL]

So does this mean that the larger hydrostatic force isn't necessary double that of the smaller force?

Only when $c=0$ is the force on the bottom double that on one of the sides.

 

[MermaidWonders]

Yeah, OK, but does my solution make sense? I don't even know if it's right or not with these problems anymore... :(

 

[MarkFL]

Yeah, OK, but does my solution make sense? I don't even know if it's right or not with these problems anymore... :(

Okay, if we plug in the given data to the formulas I posted, we get:

$$F_v=\frac{\left(9.81\frac{\text{m}}{\text{s}^2}\cdot1000\frac{\text{kg}}{\text{m}^3}\right)(4\text{ m})^2(2\cdot20+4)\text{ m}}{2}=3453120\text{ N}$$

$$F_h=\left(9.81\frac{\text{m}}{\text{s}^2}\cdot1000\frac{\text{kg}}{\text{m}^3}\right)(4\text{ m})^2(20+4)\text{ m}=3767040\text{ N}$$

Yes, our results agree. :)

 

[MermaidWonders]

Yay! :)