# Homework Help: Please find the force Fz which the hemisphere endures

1. Mar 31, 2015

### dodonaomik

1. The problem statement, all variables and given/known data
My English of influid meachanics is poor, so I only can hope that
you can understand the figure (please forgive me!)

Please find the force Fz which the hemisphere endures

2. Relevant 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')

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

3. 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.

ps: one integral

2. Apr 1, 2015

### andrevdh

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: Apr 1, 2015
3. Apr 1, 2015

### dodonaomik

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. Apr 1, 2015

### haruspex

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. Apr 1, 2015

6. Apr 1, 2015

### dodonaomik

The hemisphere is not sealed, in other words, the hemisphere isn't closed.

And I want to know the force which the river water exerts to the hemisphere.

7. Apr 1, 2015

### dodonaomik

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. Apr 1, 2015

### Staff: Mentor

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. Apr 1, 2015

### dodonaomik

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

10. Apr 1, 2015

### Staff: Mentor

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. Apr 1, 2015

### dodonaomik

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. Apr 1, 2015

### haruspex

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. Apr 1, 2015

### Staff: Mentor

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. Apr 1, 2015

### dodonaomik

15. Apr 1, 2015

### dodonaomik

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. Apr 1, 2015

### haruspex

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. Apr 1, 2015

### Staff: Mentor

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. Apr 1, 2015

### Staff: Mentor

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. Apr 1, 2015

### dodonaomik

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. Apr 1, 2015

### dodonaomik

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. Apr 1, 2015

### haruspex

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. Apr 1, 2015

### dodonaomik

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.

All above calculations are made my teacher!
But I cannot agree with him( I feel very puzzled with R circled by me)

23. Apr 1, 2015

### dodonaomik

Maybe the above fig is more helpful to calculate Fz

#### Attached Files:

• ###### kkkk.JPG
File size:
35.6 KB
Views:
96
24. Apr 1, 2015

### haruspex

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.