I Water tension of hemisphere

1. Mar 15, 2017

Just_enough

I have a 8mm diameter glass tube willed with water that have a bulge outward above the top.
I know that the preasure inside the bulge is higher than the outside (not postive as to why, probably due to it being in a liquid compare to the atmophere)
the question is:
1. how do I calculate the force due surface tension when this is all the info I was given?

I feel like I have to use BERNOULLI’S equation, but I think im missing too much info to do this.

2. Mar 15, 2017

Staff: Mentor

Are you familiar with the concept of surface tension?

3. Mar 15, 2017

Just_enough

yes

4. Mar 15, 2017

Staff: Mentor

Surface tension works the same as if there is a stretched membrane at the interface between the liquid and the gas. The tension in the membrane per unit length within the surface is equal to the surface tension. What is the downward surface tension force around the circumference of the tube exit.

5. Mar 15, 2017

Just_enough

I tried to go back and read on "fluids" chapter in my boo (Principle of physics by eric mazur) to try to understand what you mean, but I dont get it

6. Mar 16, 2017

Andy Resnick

This question is ill-posed, since the contact line is pinned at the tube edge. The pressure jump across the water-air interface is proportional to the curvature of the interface, and a range of curvatures are possible. Similarly, using Young's equation is complicated as a range of contact angles are possible.

7. Mar 16, 2017

Staff: Mentor

Hi Andy,
The problem title calls the surface hemispherical, so the contact angle is specified. I have trouble with the specification that the tube diameter is 0.8 cm and the surface could be hemispherical. I don't think this could be stable. I could better believe a 0.8 mm tube diameter.

Chet

8. Mar 16, 2017

lychette

In what way would it not be stable. Do you have some mathematical reason? Genuinely interested in your statement

9. Mar 16, 2017

Staff: Mentor

It must seems to me that a hemispherical bulge 1 cm in diameter would not stay stable under the action of surface tension.

10. Mar 17, 2017

256bits

I tried it out with a plastic straw - 07mm OD, maybe .065 cm ID, so a very thin wall.
What looks like a hemisphere of water does form above by squeezing the straw.
Any more squeezing and there is a collapse.

Also tried a coke bottle - approx. 2 cm ID. The wall is 1.5 mm, somewhat flat.
The water rises up to what looks to be about 2 to 3 mm, and then collapses.

Now that is with plastic tubing.
Glass and water would ( should ) behave differentially.
And curvature of the rim would have an effect also.

Maximum radius for a hemispherical bubble?
Seems to be just what you had stated.

11. Mar 17, 2017

Andy Resnick

Yes, that's the title of this thread. But it's not in the stated problem ("a bulge outward").

12. Mar 17, 2017

Khashishi

13. Mar 17, 2017

Staff: Mentor

I get $$2\pi r \sigma=\pi r^2 \Delta p$$