Problem Solving Strategies

harambe

12:01 AM

The arrow is towards the interface as surface tension is directed towards the interface and it is attractive?

sammy gerbil

@harambe Sorry I don't understand the question.

15 hours ago, by John Rennie

harambe

@sammygerbil here shouldn't the surface tension be downwards

It's due to imbalance of cohesive - adhesive force .

sammy gerbil

@harambe No. Imagine the surface as a sheet of stretched elastic or rubber. The tension is everywhere parallel to the surface, not perpendicular to it.

harambe

Okay

@sammygerbil physics.aps.org/assets/b4c2e0a3-7437-4592-be71-7a666a298652/…

At first the film was parallel so the surface tension would be parallel to surface

Like this

Now the air blower is trying to create a dimple in the film

What would be the direction of surface tension here

sammy gerbil

12:22 AM

@harambe The direction of surface tension is the same as before, parallel to the blue surface, as indicated by the double-headed arrows.

This does not seem to be a useful line of questioning. Is there some other question which you are trying to ask? What problem are you trying to solve?

Why pressure inside a bubble is higher than outside?

harambe

Trying to understand how surface tension works .i am just doing some research on it

Pressure is higher inside because it retains its shape against atmospheric pressure and surface tension

sammy gerbil

@harambe That's right. The surface tension is parallel to the surface of the bubble. It is always tangential. Consider an element in the surface. It is pulled left and right. But if the surface is curved these two forces don't quite line up. There is a small resultant which is directed towards the centre of curvature, radially inwards.

If the surface is a bubble, the resultant surface tension force is pulling the soap film inwards over the whole surface area of the bubble.

harambe

Got it

sammy gerbil

12:37 AM

The same effect happens in a guitar string which is displaced from its equilibrium position. The tension at every point of the string is longitudinal along the string, but because the string is curved the resultant tension force on any element of string is directed inwards towards the equilibrium position, pulling the string back towards the centre line.

animations.physics.unsw.edu.au/jw/…

harambe

@sammygerbil imgur.com/a/5TZr4JK

In the underlined sentence, it says each hemispherical surface pulls the other due to surface tension. Shouldn't it be cohesive forces

I don't see any interface there.. Both of them are liquid surface

The drop here is a liquid spherical drop

sammy gerbil

@harambe The surface of the drop is stretched like a sheet of elastic or rubber. There is a tension force everywhere in this sheet. If you draw any line in the sheet, there is a force pulling left and right on every element of liquid in the line.

Just as with a linear spring or string there are tension forces pulling left and right on every element of the spring/string.

harambe

12:56 AM

Yeah I get that. But is the diagram shows a net force towards right

Shouldn't it be zero

It's being stretched left and right

sammy gerbil

1:10 AM

@harambe The net force on the left hemisphere acts towards the right, the net force on the right hemisphere acts towards the left.

Like every force pair, the two forces act on different bodies, not the same body.

harambe

@sammygerbil yeah got it

harambe

5:25 AM

@JohnRennie hi

Good morning

John Rennie

@harambe morning :-)

harambe

@JohnRennie are you free

John Rennie

5:45 AM

@harambe yes, I'm around for a few hours this morning

harambe

@JohnRennie okay.

imgur.com/a/ztIwtzy

@JohnRennie is this asking for terminal velocity... If so the data is missing

John Rennie

Which question?

harambe

Q31

John Rennie

@harambe I think you have everything you need. I'm guessing you're supposed to assume Stokes drag otherwise they wouldn't have given you the viscosity of air.

(actually I suspect quadratic drag would apply rather than Stokes drag. Oh well.)

harambe

@JohnRennie the force acting on the raindrop is drag force by air, buoyant force in air and it's weight

John Rennie

6:00 AM

You can neglect the bouyant force because the density of air is only 0.1% of the density of water.

harambe

Ohkay

@JohnRennie imgur.com/a/VOyQwss

The direction of surface tension in first case is same as you have shown in second case, right

John Rennie

Your diagram isn't really the same as mine because in my diagram there are no walls - the dimple was being created by an air jet.

harambe

@JohnRennie how is the direction of surface tension in my pic determined

John Rennie

There are two slightly different questions there.

The force due to the surface tension always acts in the plane of the air-water interface.

So if the air-water interface makes an angle $\theta$ with the wall of the tube then the force due to the surface tension acts at the same angle $\theta$.

The second question is what determines that angle $\theta$ ...

@harambe Which question are you asking?

harambe

6:15 AM

First

John Rennie

The air-water interface behaves like an elastic sheet. It can only exert a force in the plane of the sheet.

If you zoom in on the region at the edge of the air-water interface then you'll see that angle $\theta$ is the contact angle.

harambe

@JohnRennie surface tension acts along tangent to the liquid surface

John Rennie

@harambe yes, that's what I mean by in the plane of the surface. The red line I've drawn above is the tangent to the surface at the point where it meets the walls of the tube.

harambe

@JohnRennie i.stack.imgur.com/gKar6.gif

Have you drawn surface tension along tangent here too

John Rennie

@harambe yes

The dashed lines are supposed to be parallel to the surface immediately under them

harambe

6:29 AM

Okay

@JohnRennie why is surface tension acting along 2πR in the capillary. The diagram doesn't look like the surface is a circle

John Rennie

It's a tube i.e. a cylinder with some radius $r$. So the contact line of the surface with the walls of the tube is a circle of radius $r$. The diagram shows a cross section view.

harambe

6:44 AM

@JohnRennie if I keep a thread inside a soap film and then I prick inside

Soap Film Loops

►

This happens

What is the mechanidm here

How to check direction of surface tension here

John Rennie

Once you burst the film inside the thread the thread becomes the edge of the soap file i.e. it becomes the line normal to which the surface tension exerts a force. OK so far?

harambe

Diagram please

John Rennie

OK, give me a moment ...

Suppose the thread starts off in some random shape like the squiggle I've drawn.

Before you burst the film there is film inside the thread, so if you zoom in on any part of the thread you have soap film on both sides and there is no net force. OK so far?

harambe

Ok

John Rennie

But as soon as you burst the film inside the thread there is now no soap film pulling on the inside of the thread. We only have the force due to the surface tension outside the thread pulling it outwards. Yes?

harambe

6:56 AM

Yes

John Rennie

Is it obvious this will result in the thread becoming a circle, or do I need to prove that?

harambe

@JohnRennie doubt

John Rennie

Yes?

harambe

The surface tension will pull the thread normally?

John Rennie

Suppose we zoom in on some small part of the thread. As we zoom in more and more the curvature of the thread becomes negligible i.e. the part of the thread we've zoomed in on looks approximately straight. OK so far?

harambe

7:00 AM

Ok

John Rennie

And if we have a straight line then the force exerted by the surface tension must be normal to the straight line by symmetry i.e. the system is symmetrical about a normal to the straight line.

harambe

OK

John Rennie

So that means the force exerted by the surface tension on the thread is always normal to the thread.

harambe

Okay. Got it

@JohnRennie how did it become circle. It's not obvious to me

John Rennie

The area of the soap film outside the thread is the original area minus the area of the part within the thread. The soap film wants to minimise its area because that minimises the energy, so that means it will maximise the area inside the thread. OK so far?

harambe

7:09 AM

Ok

John Rennie

We have a constraint because the length of the thread is fixed, so whatever shape the thread forms its perimeter must be constant. That means the thread will form the shape that has the largest area/perimeter ratio. Yes?

harambe

Ok

John Rennie

And the shape with the largest area/perimeter ratio is a circle.

harambe

Got it

Makes sense now

@JohnRennie what would the tension in the thread do after the thread forms a circle

Will it oppose the surface tension force to stop changing its shape

John Rennie

7:25 AM

Suppose we approximate the thread as a polygon, then consider the force at the vertices of the polygon. I'll knock up a quick drawing ...

@harambe The force in the surface is constant everywhere, because it's just the surface tension, so the net force on each vertex depends on the angle at that vertex. OK so far?

Actually I've just realised I've drawn the forces the wrong way round ...

Crap

there

Yes

If the forces at the vertices are different then the vertices will move relative to each other. So at equilibrium the forces must be the same at all the vertices. That means the angle must be the same at every vertex i.e. we have a regular polygon.

harambe

Okay

John Rennie

7:36 AM

And if let the number of sides go to infinity we get a circle.

harambe

Yes

John Rennie

So ... what were you asking about this?

harambe

I was asking that when the thread becomes a circle then would the tension in the thread be opposing the normal surface tension force on it to oppose change in surface of the thread from circle

John Rennie

@harambe Let's say the tension in the thread is $T$ and the radius of the circle is $r$. Now imagine increasing the radius of the circle by some infinitesimal amount $dr$. This means the length of the thread has to increase by $2\pi dr$. OK so far?

harambe

Ok

John Rennie

7:46 AM

To extend the thread by a distance $2\pi dr$ against the tension $T$ we have to do work $W = T 2\pi dr$.

But by extending the length of the thread we have increased the area of the circle by some amount $dA$, and that means we have decreased the area of the soap film by $dA$ and therefore decreased its energy by $\gamma dA$. Yes?

harambe

Ok

John Rennie

At equilibrium the energy we get back from decreasing the area of the soap film is equal to the work we have to put in to extending the thread i.e. the total energy change is zero. So we get $T 2\pi dr = \gamma dA$.

And $dA = 2\pi r dr$. Yes?

harambe

Yes

John Rennie

So we get $T 2\pi dr = \gamma 2\pi r dr$ or $T = \gamma r$

So the tension adjusts to balance out the force due to the surface tension.

Incidentally you can use the same argument to calculate the pressure in a soap bubble.

harambe

8:02 AM

@JohnRennie soap bubble has two interface so that would mean work done by surface tension in both surface would be equal to energy change due to surface

Conservation of energy is followed

John Rennie

Suppose we have a soap bubble of radius $r$ and the pressure inside it is $P$. The volume of the soap bubble is $\tfrac{4}{3}\pi r^3$ and the area is $4\pi r^2$.

If we let the bubble expand by a small amount $dr$ then this increases the volume by $dV$ and the area by $dA$.

harambe

Ok

John Rennie

The work done in the expansion is $PdV$ and the extra surface energy is $2\gamma dA$. As you say, there's a factor of 2 because there is both an outside and inside surface.

And at equilibrium $PdV = 2\gamma dA$

$$ P = 2\gamma\frac{dA}{dV} $$

harambe

@JohnRennie how did you write PdV... Like in thermodynamics

John Rennie

Yes, that's from thermodynamics.

harambe

8:09 AM

Okay

John Rennie

(we're assuming the gas is ideal)

harambe

Okay

@JohnRennie it's coming right

This is a cool. way to derive the expression too

John Rennie

@harambe naturally :-)

The method behind this approach is called the principle of virtual work

Virtual work

Virtual work arises in the application of the principle of least action to the study of forces and movement of a mechanical system. The work of a force acting on a particle as it moves along a displacement will be different for different displacements. Among all the possible displacements that a particle may follow, called virtual displacements, one will minimize the action. This displacement is therefore the displacement followed by the particle according to the principle of least action. The work of a force on a particle along a virtual displacement is known as the virtual work. Historically...

harambe

8:37 AM

@JohnRennie hi. You still free

John Rennie

@harambe yes

harambe

@JohnRennie imgur.com/a/ltqcUNJ

Q21

Using the capillary equation, I get height rise to be 0.003276m

It's not matching solution.... Do we have to do something else too

John Rennie

If the contact angle of the mercury was 90°, or its surface tension was zero, the height of the column would be 760mm i.e. just a standard barometer. Yes?

harambe

Yes

John Rennie

The effect of surface tension is to pull the column up a bit if the surface is concave or push it down if the surface is convex. Mercury forms a convex surface so the effect will be to decrease the height of the mercury column so it will be $h = 760 - x$ where $x$ is what you have to calculate.

harambe

8:44 AM

@JohnRennie on second thought, how did you conclude this

John Rennie

@harambe huh? If there are no surface tension effects the height is given by $\rho g h = 1 atm$

And it should be seared on your brain that 1 atm = 760mm of mercury

harambe

Yes

John Rennie

So the height will be 760mm plus or minus the effects of the surface tension

harambe

Ohhhh. Got it

Nobody recognizeable

9:04 AM

@JohnRennie are you free?

John Rennie

@Nobodyrecognizeable yes

Nobody recognizeable

@JohnRennie is it clear enough to understand?

John Rennie

@Nobodyrecognizeable yes, it's fine. When you extend a spiral spring you are actually twisting the wire that the spring is made from. That's what causes the force in the spring.

The point of this question is to calculate the spring constant $k$ from the torsional stiffness of the wire in the spring.

Nobody recognizeable

@JohnRennie rigidity modulus =$ F/A\theta$; mg=kx too

John Rennie

@Nobodyrecognizeable yes, that equation relates the extension x to the spring constant k and the force applied to the spring mg.

what you need to do is calculate the spring constant k from the rigidity modulus

Nobody recognizeable

9:16 AM

@JohnRennie i can find theta from first equation. Then what do i do next?

@JohnRennie are you still here?

John Rennie

I'm looking to see if I can find an article that describes the calculation

Nobody recognizeable

@JohnRennie i see. Are you going anywhere today?

John Rennie

@Nobodyrecognizeable I'll be going out in about half an hour. I'll be out for a couple of hours.

This gives the equation though not the derivation:

0

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