# Minimum force required to form sphere

1. May 27, 2015

### Tanya Sharma

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Doing a vertical force balance 2Fcosθ=mg ,where m is the mass of water .

Not sure how to proceed .

What role does the pin hole at the top play ?

I would be grateful if somebody could help me with the problem.

#### Attached Files:

• ###### sphere.PNG
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2. May 27, 2015

### Simon Bridge

... that is the key question.
Think about it - what does putting a pinhole in the top do that not having a hole there not do?
i.e. imagine there is also a pinhole at the bottom - what difference would a hole at the top make?

3. May 27, 2015

### haruspex

Pace Simon, but the pinhole at the top is only to get around a complication that the setter did not want to get tangled up with. It looks like you didn't notice the complication anyway, so what you have done so far is ok. But you need another equation. The answer to the question should be an actual number, not dependent on any variables.

4. May 27, 2015

### Staff: Mentor

Here's a hint Tanya: Think hydrostatic.

Chet

5. May 27, 2015

### Tanya Sharma

6. May 27, 2015

### Staff: Mentor

The pressure (force per unit area) on each hemisphere is going to increase with depth. To get the horizontal force on each hemisphere, you need to integrate. I haven't figured out yet whether you also need to do a moments balance.

Chet

7. May 27, 2015

### Tanya Sharma

Chet ,

I somehow feel there is a simpler , less mathematically rigorous solution to this problem .

Is the force across the interface between the spheres given by $$2 ρg\int_0^R \sqrt{R^2 - (R-h)^2}h dh$$ ? But then ρ comes in picture and the integral also doesn't look nice .

8. May 27, 2015

### TSny

The integrand looks good. Check the limits.

If you want to avoid the integration, consider a vertically oriented circular area submerged in a static fluid. Consider two, symmetrically placed horizontal strips of area as shown in blue. How do the pressures at each of these strips compare to the pressure at z = 0? Does the sum of the forces on the two strips simplify?

#### Attached Files:

• ###### pressure 1.png
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Last edited: May 27, 2015
9. May 27, 2015

### Tanya Sharma

It should be from 0 to 2R . Right?

Yes .

If we denote the pressure at z=0 to be P ,pressure at the upper blue strip is P-ρgz . Pressure at the lower blue strip is P+ρgz

2P . Is that so ?

Edit : 2P is sum of pressures .

10. May 27, 2015

### TSny

Yes.
Yes.
This is the total force on the two strips?

11. May 27, 2015

### Tanya Sharma

Total force on two strips is 4P$\sqrt{R^2-z^2}dz$ .

12. May 27, 2015

### TSny

Yikes. (Correct, but messy.) Let the area of both strips together be ΔA. Express the sum of the forces on the two strips in terms of P and ΔA.

13. May 27, 2015

### Tanya Sharma

2PΔA

14. May 27, 2015

### TSny

Well, I was taking ΔA as representing the sum of the areas of the two strips. If so, then your expression is not quite corrrect.

15. May 27, 2015

### Tanya Sharma

Total force on vertically oriented circular area should be $\pi ρgR^3$ . Is that correct ?

16. May 27, 2015

### TSny

Should atmospheric pressure be included?

17. May 27, 2015

### Tanya Sharma

It should be $(P_{atm}+ρgR)πR^2$ ?

18. May 27, 2015

### TSny

Yes, the total force on the circle is the pressure at the center times the total area of the circle. Do you see why they put the little hole at the top?

19. May 27, 2015

### Tanya Sharma

What should be the next step ?

20. May 27, 2015

### TSny

The next step should probably be to take a break and sleep on it. (Isn't it late there?)

Anyway, one thing to think about is how your result for the circle can be used to get the net horizontal force of the fluid on the inner surface of the hemisphere. Also think about the effect of external atmospheric pressure acting on the outer curved surface of the hemisphere.