Please find the force Fz which the hemisphere endures

In summary, the conversation discusses a problem involving finding the force Fz that a hemisphere endures when submerged in a river at an angle. The individual asking the question is not confident in their calculation method and asks for help. The conversation delves into the concepts of Archimedes' principle and the displacement of fluid by a submerged object. The final summary concludes that the individual is unsure of the necessary information to solve the problem and is seeking clarification.
  • #1
dodonaomik
17
0

Homework Statement


My English of influid meachanics is poor, so I only can hope that
you can understand the figure (please forgive me!)

1111.jpg

Please find the force Fz which the hemisphere endures

Homework Equations



My way: "on" the hemisphere, I want to use triple integral(but I'm not sure the way's right) and cut into the part into many innumerable tiny 3D right angled trapezoid
(i.g. KJPQ-K'J'P'Q')

2222.jpg


Then I think that:
ρgV( the part + the hemisphere)=Fz

The Attempt at a Solution


Because my math is weak, I hope somebody can help me.
If you can understand the problem, please tell me whther my way or my calculation is right or not, thank you!

The following is my calculation.
3333.jpg

ps: one integral
4444.jpg
 
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  • #2
Is this force due to the internal pressure in the drop?
If so isn't this pressure generated by the surface tension?
I assume the drop of liquid is in air, maybe hanging or falling?
Not sure we are able to solve this, just trying to understand the
problem that you are dealing with.
 
Last edited:
  • #3
andrevdh, 1st thank you for your replying.
And you can imagine that one hollow hemisphere is located under one still river by 1 person.
Of course the hemisphere is fixed obliquely as in the fig( dip angle=30°).
 
  • #4
dodonaomik said:
andrevdh, 1st thank you for your replying.
And you can imagine that one hollow hemisphere is located under one still river by 1 person.
Of course the hemisphere is fixed obliquely as in the fig( dip angle=30°).
Is this right...
This is a hollow sealed hemisphere being held immersed in a liquid. It is held at an angle. You want the force required to hold it down?
 
  • #5
andrevdh said:
Is this force due to the internal pressure in the drop? ___NO____the water pressure(---->the hollow hemisphere) in a river
If so isn't this pressure generated by the surface tension?___NO____by the force of gravity of the water in a river
I assume the drop of liquid is in air, maybe hanging or falling?___NO____not in air but in a river
Not sure we are able to solve this, just trying to understand the
problem that you are dealing with.

7777.jpg
 
  • #6
haruspex said:
Is this right...
This is a hollow sealed hemisphere being held immersed in a liquid. It is held at an angle. You want the force required to hold it down?
Thanks for you replying.
The hemisphere is not sealed, :smile:in other words, the hemisphere isn't closed.And I want to know the force which the river water exerts to the hemisphere.
 
  • #7
Additionaly I don't know whether my method is right or not.
(PP'Q'Q is an right angled trapezoid just the same as JJ'K'K)
 
  • #8
Are you familiar with Archimedes principle? Does the vertical force exerted by the surrounding fluid on a submerged object depend on its depth or orientation?

Chet
 
  • #9
Chestermiller said:
Are you familiar with Archimedes principle? Does the vertical force exerted by the surrounding fluid on a submerged object depend on its depth or orientation?

Chet
Thank you! I am not familiar with Archimedes' spiral, and I olny know it's equation.

Additionaly your reply is irrelevant to this post.
 
  • #10
dodonaomik said:
Thank you! I am not familiar with Archimedes' spiral, and I olny know it's equation.

Additionaly your reply is irrelevant to this post.
I don't think so. The upward force is equal to the weight of the fluid that has been displaced by the hemisphere.

chet
 
  • #11
Chestermiller said:
I don't think so. The upward force is equal to the weight of the fluid that has been displaced by the hemisphere.

chet
Sorry, I mixed up 2 concepts: Archimedes principle and Archimedes' spiral.

Now I really cannot understand it_____How the fluid can be displaced by the hemisphere?
 
  • #12
dodonaomik said:
Thanks for you replying.
The hemisphere is not sealed, :smile:in other words, the hemisphere isn't closed.And I want to know the force which the river water exerts to the hemisphere.
If it's hollow and not closed, do we know the shell's thickness? If it has no thickness it displaces no volume, so there is no buoyancy.
Or... is this a question about hydrodynamics... force due to fluid flow over a surface?
 
  • #13
dodonaomik said:
Sorry, I mixed up 2 concepts: Archimedes principle and Archimedes' spiral.

Now I really cannot understand it_____How the fluid can be displaced by the hemisphere?
The displaced volume of fluid is the same thing as the submerged volume of the body. So the force exerted by the fluid on the submerged body in the vertical direction is given by:
$$F=\rho g V$$
where ##\rho## is the density of the fluid, g is the gravitational acceleration, and V is the submerged volume of the body.

Chet
 
  • #14
haruspex said:
If it's hollow and not closed, do we know the shell's thickness?
-----No, I feel we are not necessary to know the thicness. In other words, we in fact shouldn't concer the hemisphere shell's thicknessIf it has no thickness it displaces no volume, so there is no buoyancy.
----------I think we only know Fz(as in the fig.) and are unnecessary to know the buoyancy.Or... is this a question about hydrodynamics... force due to fluid flow over a surface?
_________NO, not hydrodynamics but hydraulics... force due to the still river water over the hemisphere shell

MY English about influid mechanics is weak, so please don't mind!
 
  • #15
Chestermiller said:
The displaced volume of fluid is the same thing as the submerged volume of the body. So the force exerted by the fluid on the submerged body in the vertical direction is given by:
$$F=\rho g V$$
where ##\rho## is the density of the fluid, g is the gravitational acceleration, and V is the submerged volume of the body.

Chet
First thank you so much,
but I don't want to know the buoyancy(vertical upward), and I want to know Fz(vertical downward, as in the 1st fig.).
 
  • #16
dodonaomik said:
we only know Fz(as in the fig.) and are unnecessary to know the buoyancy.
dodonaomik said:
ρgV( the part + the hemisphere)=Fz
Fz is what you are trying to find, no? So you do not know it.
ρgV is the buoyancy, but if the shell has no thickness and is full of water, V = 0.
 
  • #17
dodonaomik said:
First thank you so much,
but I don't want to know the buoyancy(vertical upward), and I want to know Fz(vertical downward, as in the 1st fig.).
To get that, you need to know how much the hemisphere weighs. Is the hemisphere solid, or is it just a thin shell? If it is just a thin shell, then your downward force is equal to the upward force from this equation.

Chet
 
  • #18
haruspex said:
Fz is what you are trying to find, no? So you do not know it.
ρgV is the buoyancy, but if the shell has no thickness and is full of water, V = 0.

I'm having trouble understanding what she's saying. I didn't think that there was water inside the body or shell. If there is, then the buoyant force is zero for a zero thickness shell. I feel like I'm preaching to the choir.

Chet
 
  • #19
haruspex said:
Fz is what you are trying to find, no? So you do not know it.
ρgV is the buoyancy, but if the shell has no thickness and is full of water, V = 0.
IT'S true that we are unnecessary to know the shell's thickness.ρgV( the part + the hemisphere)=Fz
the part=the water on the hemisphere(at the moment, you can imagine that the hemisphere is sealed or closed.)
the hemisphere= the water in the hemisphere(at the moment, you can imagine that the hemisphere is sealed or closed.)
 
  • #20
Chestermiller said:
To get that, you need to know how much the hemisphere weighs. Is the hemisphere solid, or is it just a thin shell? If it is just a thin shell, then your downward force is equal to the upward force from this equation.

Chet
In my opinion~~~~in my understanding, I really feel that we are unnecessary to know the weight of hemisphere and the thickness of shell of the hemisphere.And...yes! It's just a thin shell...a very thin shell... and we should ignore the thickness of the shell.

In fact I feel that Fz= the force of gravity of water "upward" and in the hemisphere
 
  • #21
dodonaomik said:
In my opinion~~~~in my understanding, I really feel that we are unnecessary to know the weight of hemisphere and the thickness of shell of the hemisphere.And...yes! It's just a thin shell...a very thin shell... and we should ignore the thickness of the shell.

In fact I feel that Fz= the force of gravity of water "upward" and in the hemisphere
Neither Chet nor I can make any sense of the question. We see no reason why there should be a net force.
There must be some text associated with the question as presented to you. If it is not in English, please provide the best translation you can make yourself, as well as a translation through an online utility such as Google or Babelfish.
 
  • #22
haruspex said:
Neither Chet nor I can make any sense of the question. We see no reason why there should be a net force.
There must be some text associated with the question as presented to you. If it is not in English, please provide the best translation you can make yourself, as well as a translation through an online utility such as Google or Babelfish.
Equipped with a D=1.2m diameter hemispherical cap in the circular oil depots inclined wall, hemispherical cap center point O, depth of H=4m in the page. Inclined wall and thelevel angle of α=30 degrees, there is the density of ρ=900kg/m^3. and total tensionbolts a, b.

111.jpg

222.jpg

333.jpg

444.jpg
All above calculations are made my teacher!
But I cannot agree with him( I feel very puzzled with R circled by me)
 
  • #23
kkkk.JPG


Maybe the above fig is more helpful to calculate Fz
 

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  • #24
dodonaomik said:
Maybe the above fig is more helpful to calculate Fz
Indeed it is, particularly the label "air".
Your challenge to the 'R' is in regard to Fx, not Fz. The part in square brackets (divided by 2) is the pressure at O, and by symmetry it is the mean pressure acting over the circular opening to the hemisphere. To get Fx, you want to find the effective area in a vertical plane. That will be sin(alpha) multiplied by the complete area. But what I see as wrong is that the complete area is written as 2R2 instead of ##\pi R^2##.
Your Fz corresponds to Fy in the later calculation. This calculation is the same, except that it's cos instead of sin, and you have to add the weight of the water in the hemisphere.
 

Related to Please find the force Fz which the hemisphere endures

What is the formula for calculating the force Fz on a hemisphere?

The formula for calculating the force Fz on a hemisphere is Fz = 2πRH, where R is the radius of the hemisphere and H is the height of the fluid above the hemisphere.

How do the properties of the fluid affect the force Fz on the hemisphere?

The force Fz on a hemisphere is affected by the density and viscosity of the fluid. A denser fluid will exert a greater force on the hemisphere, while a more viscous fluid will exert a lower force.

What factors can cause the force Fz on a hemisphere to change?

The force Fz on a hemisphere can change due to changes in the height of the fluid above the hemisphere, changes in the properties of the fluid, and changes in the shape or orientation of the hemisphere.

Is the force Fz on a hemisphere always directed vertically?

No, the force Fz on a hemisphere can be directed at different angles depending on the shape and orientation of the hemisphere. However, it is typically directed in the same direction as the gravitational force.

How is the force Fz on a hemisphere used in real-world applications?

The force Fz on a hemisphere is used in various applications in fluid mechanics, such as calculating the buoyant force on submerged objects or determining the stability of floating structures. It is also used in the design of ship hulls and other structures that interact with fluids.

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