Integral Applications - Hydrostatic Pressure + Force

In summary: F is the elemental force, or the force exerted by a small section of the submerged rectangle on the fluid. We are calculating the total force by summing all of these elemental forces. Does that make sense?
  • #1
MermaidWonders
112
0
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.
 
Physics news on Phys.org
  • #2
MermaidWonders said:
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 ...

$\displaystyle F = 9800 \int_0^{0.6} y \cdot 0.75 \, dy$
 
  • #3
The above worked for me, but I'm still finding applications like these quite intuitive. Would you mind explaining the process behind it?
 
Last edited:
  • #4
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 = __________
 
  • #6
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?
 
Last edited:
  • #7
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:

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

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

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

where:

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

And so we have:

\(\displaystyle dF=\rho w x\,dx\)

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:

\(\displaystyle F=\rho w\int_{x_1}^{x_2}x\,dx\)

Applying the FTOC, we obtain:

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

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

\(\displaystyle F=\frac{\rho wh}{2}\left(2x_1+h \right)\)
 
  • #8
MermaidWonders said:
... when would $h(y)$ not equal to $y$?

what if the submerged planar surface is not vertical?
 
  • #9
Oh, true...
 
  • #10
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:

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

We may begin as before, computing the elemental force:

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

Summing the elements, we obtain:

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

Application of the FTOC gives us:

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

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

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

And so we may state:

\(\displaystyle F=\frac{\rho w\sqrt{\ell^2+\left(d_1-d_1 \right)^2}\left(d_1+d_2 \right)}{2}\)
 
  • #11
Hmmm... let me take some time to digest this here. What is $dF$?
MarkFL said:
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:

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

We may begin as before, computing the elemental force:

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

Summing the elements, we obtain:

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

Application of the FTOC gives us:

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

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

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

And so we may state:

\(\displaystyle F=\frac{\rho w\sqrt{\ell^2+\left(d_1-d_1 \right)^2}\left(d_1+d_2 \right)}{2}\)
 
  • #12
MermaidWonders said:
Hmmm... let me take some time to digest this here. What is $dF$?

$dF$ is the force differential. :)
 
  • #13
Oh, I see. :)
 
  • #14
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.)
 
  • #15
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:

\(\displaystyle F_v=\frac{\rho x^3}{2}\)

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:

\(\displaystyle F_h=\rho x^3\)

So, we see the force on the bottom is double that on anyone of the sides.
 
  • #16
Sorry, but could you explain how you got these 2 formulas? I'm lost. :(

MarkFL said:
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:

\(\displaystyle F_v=\frac{\rho x^3}{2}\)

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:

\(\displaystyle F_h=\rho x^3\)

So, we see the force on the bottom is double that on anyone of the sides.
 
  • #17
MermaidWonders said:
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)

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

Sorry for the typo in the original post.

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

\(\displaystyle w=x,\,\ell=0,\,d_1=0,\,d_2=x\)

And so we find:

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

For the bottom of the cube, we have:

\(\displaystyle w=x,\,\ell=x,\,d_1=x,\,d_2=x\)

And so we find:

\(\displaystyle F_h=\frac{\rho x\sqrt{x^2+(x-x)^2}(x+x)}{2}=\rho x^3\)
 
  • #18
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... :(

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

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

Sorry for the typo in the original post.

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

\(\displaystyle w=x,\,\ell=0,\,d_1=0,\,d_2=x\)

And so we find:

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

For the bottom of the cube, we have:

\(\displaystyle w=x,\,\ell=x,\,d_1=x,\,d_2=x\)

And so we find:

\(\displaystyle F_h=\frac{\rho x\sqrt{x^2+(x-x)^2}(x+x)}{2}=\rho x^3\)
 
  • #19
MermaidWonders said:
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?
 
  • #20
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
 

Attachments

  • FullSizeRender.jpg
    FullSizeRender.jpg
    15.8 KB · Views: 68
Last edited:
  • #21
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:

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

\(\displaystyle F_h=\rho x^2(c+x)\)
 
  • #22
So does this mean that the larger hydrostatic force isn't necessary double that of the smaller force?
 
  • #23
MermaidWonders said:
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.
 
  • #24
Yeah, OK, but does my solution make sense? I don't even know if it's right or not with these problems anymore... :(
 
  • #25
MermaidWonders said:
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:

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

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

Yes, our results agree. :)
 
  • #26
Yay! :)
 

FAQ: Integral Applications - Hydrostatic Pressure + Force

What is hydrostatic pressure?

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the weight of the fluid above it. It is dependent on the density of the fluid and the depth of the fluid.

How is hydrostatic pressure calculated?

Hydrostatic pressure can be calculated using the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the gravitational acceleration, and h is the depth of the fluid.

What are some applications of hydrostatic pressure?

Hydrostatic pressure has various applications, including measuring blood pressure, testing water pipes for leaks, and calculating buoyancy forces in ships and submarines.

What is the difference between hydrostatic pressure and hydrodynamic pressure?

Hydrostatic pressure is the pressure exerted by a fluid at rest, while hydrodynamic pressure is the pressure exerted by a fluid in motion. Hydrodynamic pressure is dependent on the velocity and direction of the fluid flow, while hydrostatic pressure is only dependent on the depth of the fluid.

How is hydrostatic pressure related to force?

Hydrostatic pressure is directly related to force, as an increase in pressure will result in an increase in the force exerted by the fluid. This is why hydrostatic pressure is often used to calculate the force on objects submerged in a fluid.

Back
Top