Problem Solving Strategies - 2020-01-17

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1:59 AM

@JohnRennie Good morning sir! The barometer question that I posted there got two answers and the one by BillN seems to imply that (though I am not sure if my conclusion is right) that the tube may indeed jump up (if its mass is negligible). I would like to have your perspective on this sir. What do you think happens? Thanks!

Nobody recognizeable

2:28 AM

@JohanLiebert John rennie will be around from 11 India time. Now he may be in sleep.

3 hours later…

5:22 AM

Good morning Sir :-) @JohnRennie

5:43 AM

@user8718165 hi :-)

@JohanLiebert hi Johan

@JohnRennie hello sir.. could you please tell me what this means

@user8718165 I'll just be a moment. I'm busy in another room ...

@JohnRennie sure sir.

6:00 AM

@JohnRennie hi sir :-)

@JohanLiebert hi. Let me answer @user8718165 then I'll answer you.

@user8718165 hi

@JohnRennie hello sir... please tell me what he means. I didn't understand.

@user8718165 suppose you are pushing a block up an incline that is at 30°

@JohnRennie okay sir

The normal force between the block and the incline is at 30° to the vertical.

6:08 AM

yeah

The normal force is (obviously) normal to the displacement along the slope, so $F.ds = 0$

@JohnRennie yeah sir... its 0

That's what Aaron means when he says the normal force does no work.

@JohnRennie but he says there will be no net work but vertical work by normal force...

@JohnRennie How can normal force do vertical work sir?

It's just terminology. The normal force has a vertical component, and that is added to the other vertical forces like gravity to calculate the net vertical force.

6:13 AM

@JohnRennie so he means that the net vertical force does work.... right sir?

Yes

@JohnRennie Thank you very much sir...I was thinking this... but was unsure... thank you sir

@JohnRennie Today is better :-)

@user8718165 why is today better?

You mean you're feeling better? If so that's good :-)

@JohnRennie good morning sir! Now are you free?

@JohanLiebert yes.

Can you clarify exactly what your experiment is?

Are you holding an evacuated tube above the mercury then suddenly allowing the mercury to flow in?

6:23 AM

@JohnRennie sir the experiment goes like this: take a tube of height say 15cm and fill it with mercury and then place a cardboard at the bottom so that while inverting the tube the mercury does not fall. Now as soon as the mercury filled tube is placed in container we remove the cardboard. It is clear that at immediate bottom of the tube the pressure is lower that surrounding. So what happens then?

The pressure isn't lower at the bottom of the tube. The pressure at the bottom of the tube is the same as the pressure of the mercury outside the tube i.e. 1 atm.

@JohnRennie so sir what happens to the pressure applied by the the mercury column of that height(assume the tube is massless)?

user image

Those are the pressures.

@JohnRennie but if there isn't any vacuum or gas in the above part of the tube then where does the pressure $P_{\text {atm}} -\rho g h$ come from?

OK, I think I see what you mean.

6:35 AM

@JohanLiebert I'd say it's from the contact force of mercury with the topmost part of the test tube.

The mercury will not flow down it'll stay as such.

The atmosphere outside the tube is pushing on the surface of the mercury in the dish. Yes?

@JohnRennie yes sir!

@GuruVishnu yes i have performed the experiment with water and it looks like while pulling the bottle upward a downward force acts on the bottle filled with water.

And since the mercury is a fluid it transmits this pressure through the mercury. Let me draw this on your diagram ...

user image

So at the base of the tube there is an upwards pressure of 1 atm. Since the mercury isn't going anywhere this must be balanced by an equal and opposite downwards pressure of 1 atm. Yes?

@JohnRennie yes that's what i was thinking.

The downwards pressure comes from two sources:
1 the weight of the mercury in the tube
2. the atmosphere pressing down on the top surface of the tube.

user image

6:44 AM

@JohnRennie but sir wasn't the atmosphere applying force in the case of mercury barometer with the height larger than 76cm?

Suppose you have a tube just in air. Then the atmosphere presses on all surafces of the tube with a pressure of 1 atm.

That means the glass that the tube is made of is being compressed by the atmosphere.

Glass is very rigid, so it isn't compressed much, but it is being compressed by a tiny amount.

@JohnRennie ok sir

Now consider the usual barometer i.e. longer than 76cm and with vacuum at the top.

@JohnRennie ok....

The air still presses on the glass at the outside of the tube, but there is no force on the glass on the inside of the tube because there's a vacuum there.

So the glass will be expanded very slightly compared to the same glass with 1 atm on all sides.

6:49 AM

@JohnRennie yes sir

You can think of the glass as a (very stiff!) spring that expands or contracts slightly in response to the external pressure.

Your diagram is somewhere in between.

The pressure on the outside of the glass is 1 atm, and inside the glass it is 1 atm - $\rho g h$.

@JohnRennie so that pressure is applied by the glass then?

Yes. The glass will be behaving like a spring that expands and contracts in response to the pressure put on it.

With air on both side the glass is fully compressed. With air and vacuum the glass is fully expanded. In your experiment the glass is in between these two extremes.

So the atmosphere is pressing on the outside of the glass, and the glass is pressing on the mercury inside it, but with a reduced pressure.

@JohnRennie ok sir! I think the same thing happened when i used the bottle with water(i don't much mercury!)

@JohnRennie so sir would you post this as an answer there?

@JohnRennie or should i just refer the future visitors to this transcript?

It's quite a messy answer to write. I'll have a think about whether I can write a coherent summary of our discussion here and post it as an answer.

7:02 AM

@JohnRennie sir does this mean that if the glass isn't strong enough to resist this then it might break?

Yes, if you tried it with a really thin glass tube then the glass would break.

@JohnRennie thank you very much sir!

@JohnRennie yeah sir... Sorry I was a bit busy :-)

@user8718165 Cool :-)

7:17 AM

@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

@JohnRennie: Do you also say capacitance of any capacitor is increased by a factor of dielectric constant when the space between the plates is filled by a dielectric sir?

Reference:

0

Q: Is the capacitance of all types of capacitors increased by a factor of dielectric constant like parallel plate capacitors?

I've learnt that the capacitance $C$ of a parallel plate capacitor is given by: $$C=\frac{A\epsilon_0}{d}$$ where, $A$ is the area of cross section of the plates and $d$ is the separation between the two plates. When the space in between the two plates is filled by a dielectric of dielectric cons...

electrostatics electric-circuits capacitance dielectric

I need to work for a few minutes so I'll have to get back to this later ...

@JohnRennie Ok sir. No problem :-)

8:15 AM

@JohnRennie Sir I'm unable to find the mistake in this $$2\int \frac{\frac{u}{2\sqrt{2}}{\left(1+2u^2 - \sqrt 2 u\right)} - 2\int \frac{\frac{u}{2\sqrt{2}}{\left(1+2u^2 + \sqrt 2 u\right)} $$

Texmaker writes ! Paragraph ended before \frac was complete.

@JohnRennie Sir are you around?

Oh! I got it.

Sorry for pinging you.

$$
2
\int
\frac
{
  \frac{u}{2\sqrt{2}}
  {
    \left( 1+2u^2 - \sqrt 2 u \right)
  }
  -
  2
  \int
  \frac
  {
    \frac{u}{2\sqrt{2}
  }
  {
    \left(1+2u^2 + \sqrt 2 u\right)
  }
$$

@Knight I've used indents to mark where brackets are opened and closed. Looking at this it is immediately clear that you have an opening \frac{ that isn't closed.

Yes 6th line

ha?

If in doubt it's often useful to use indents. That's exactly what programmers do when writing code.

How to do indents?

Just type spaces ...

Your MathJax source is just text. Actually it's effectively a program that when run produces the image.

8:29 AM

Oh, means I should write my program with line breaks and spacings?

You can use carriage returns and spaces in your MathJax. It doesn't all have to be on one line.

@Knight if it's long and complicated you can use line breaks and spaces to help you get it right. If it's short and simple I wouldn't bother.

Sir how can we move on to next line in this chat, clicking enter results in sending the message.

Press shift-enter

Wow! It worked

Yes
it
does
:-)

8:54 AM

@JohnRennie Sir what to do in this problem

user image

@Knight Magnetic field lines never start or end. They only travel in closed loops. Yes?

Yes I agree

Imagine looking at the coil edge on (I can draw a diagram if necessary) and oriented so that the field lines travel upwards through the coil. Then every field line that travels upwards through the coil must bend round and travel downwards through the region outside the coil. Yes?

That needs a little rigorous explanation, why the field lines will go up from the ring and come down outside of it?

Field lines have to be loops.

If you draw a line upwards through the coil then it must come down again or it can't find a loop.

It can come down inside the loop or outside the loop, but if it comes down inside the loop then it cancels the line moving up and the net contribution to the flux through the loop is zero.

9:02 AM

Why can't we say that field lines goes and comes down all inside the loop?

Oh!

WoW!

After this what we should do?

So if we are counting the flux through the loop this can only be due to lines that go up through the loop and down outside it.

yes

So that means the number of lines going up through the loop must be equal to the number of lines going down outside the loop. Yes?

I doubt it. Field lines can originate from outside of the loop and since they were of lesser strength so they will not cancel the field inside the loop.

Field lines can't originate from anywhere. They are always loops and have no starting point or ending point.

So there can be only three types:
1. go up and down inside the loop
2. go up and down outside the loop
3. go up through the loop and down outside the loop

9:08 AM

yes, I agree.

1 and 2 do not contribute to the total flux inside the loop or the total flux outside the loop because the up and down parts of the line cancel each other.

Only 3 can produce a net flux inside the loop or a net flux outside the loop.

yes.

So for every field line that goes up through the loop a field line must go down through the area outside the loop.

And that means the number of lines going up through the loop must be equal to the number of lines going down outside the loop.

I tried to solve it like this $$ \textrm {Flux} = \int_S \mathbf{B} \cdot d\mathbf{S}$$

Since, total area of outside is larger therefore I concluded that $\phi_i$ must be greater than $\phi_0$

But you don't know how $\mathbf B$ varies with distance outside the loop, so how can you do the integral?

9:13 AM

Although B is smaller outside but area is much much bigger than the loop.

Suppose we write $\mathbf B(r)$ as a function of the distance $r$ from the centre of the loop, then $ds = 2\pi r dr$ and the integral is:

$$ \Phi = 2 \pi \int_{r=R}^{r=\infty} \mathbf B(r) r dr $$

Yes.

Okay for outside

where $R$ is the radius f the loop i.e. we're just calculating the flux outside.

@Knight Yes. for the inside integrate from zero to $R$

Yes, I agree

But you have no idea what the form of $\mathbf B(r)$ is. You're saying you think it will fall with distance slowly enough to give a bigger integral that the interior integral.

What I'm saying is that no it won't.

9:18 AM

Yes, I speculate that. Because B varies as $\frac{1}{r^2}$

No, for a dipole $B(r) \propto r^{-3}$

You only get an inverse square law for a monopole i.e. a charge, and there are no magnetic charges.

So, why my speculation won't work?

> So, why my speculation won't work?

@Knight So you've made a guess based on no evidence, and you're asking me why your guess is wrong?

Yes I'm getting you

What I'm telling you is that it's wrong because the field lines have to be loops, so the number of lines inside the coil must be equal to the number outside. There is no way they can be different.

9:26 AM

Yes sir. Thanks

10:02 AM

@JohnRennie: Hi sir :-)

@GuruVishnu hi :-)

@JohnRennie Are you free now? If yes shall we continue our discussion on Optics and then Capacitors, sir? If not, could you please ping me once you're free?

I'm free, but I'm not sure what more there is to say.

@JohnRennie: May be this could help:

in General chat, 1 hour ago, by John Rennie

Yes, and that's why it's more complicated than it seems at first glance.

(regarding capacitors)

I don't know a general proof that for an arbitrary shape of capacitor the capacitance is always increased by a factor of $\epsilon_r$. I suspect that would be a complicated proof.

10:10 AM

@JohnRennie Isn't the capacitance increased by $K$ (dielectric constant sir)?

Relative permittivity and dielectric constant are different names for the same thing.

@JohnRennie Ok sir. Thank you for remembering me. I have used only $K$ so far and so I got confused.

@JohnRennie: And the following regarding optics:

in General chat, 1 hour ago, by John Rennie

I'll have to think about this and I have other things to do right now. It isn't obvious why it's true for a plano lens.

I'll need to set aside a couple of hours to think about that. I don't have a couple of hours free any time today.

@JohnRennie Ok sir. And may I proceed with the bounty? BTW I had a doubt regarding the optical metamaterials from the Wiki page you linked there.

I think your bounty will be wasted because your question isn't clear enough to get a good answer.

You're welcome to ask about metamaterials, though it's a subject I don't know much about.

10:17 AM

@JohnRennie I don't know how to make the question clear sir. The one you suggested was a better concise question. I thought both of us meant the same thing. Did you see the original question, sir?

1

Q: Is real image always inverted and virtual image always erect?

Question: Is real image always inverted and virtual image always erect with respect to the object? Is there any single optical device (a mirror, or a lens, or anything else) which could produce a real and erect image or a virtual and inverted image? Till now, I haven't encountered such a device....

optics reflection geometric-optics lenses

The one I specified was only the Bounty message.

This and the rest of the messages that follows this one: chat.stackexchange.com/transcript/message/53257516#53257516

> any single optical device (a mirror, or a lens, or anything else)

That is so broad that it's difficult to answer.

@JohnRennie Ok sir. Will it suffice if I remove "or anything else"? I thought if it wasn't a mirror or a lens there must be something which satisfies those properties.

If the question is Is there a simple lens that creates a real erect image then the answer is no.

If the question is Is there a simple mirror that creates a real erect image then the answer is no.

But there can be things like Fresnel lenses that don't use refraction to form images.

If the question is Is there a (any device) that creates a real erect image then the answer is ________?

Maybe a Fresnel lens could create a real erect image - I don't know.

10:24 AM

@JohnRennie Did you mean the white dome of a PIR sensor sir?

A Fresnel lens ( FRAYN-, FREN-el, -⁠əl, fray-NEL or FREZ-nəl) is a type of composite compact lens originally developed by French physicist Augustin-Jean Fresnel for lighthouses. It has been called "the invention that saved a million ships."The design allows the construction of lenses of large aperture and short focal length without the mass and volume of material that would be required by a lens of conventional design. A Fresnel lens can be made much thinner than a comparable conventional lens, in some cases taking the form of a flat sheet. A Fresnel lens can capture more oblique light from...

Also:

A zone plate is a device used to focus light or other things exhibiting wave character. Unlike lenses or curved mirrors however, zone plates use diffraction instead of refraction or reflection. Based on analysis by Augustin-Jean Fresnel, they are sometimes called Fresnel zone plates in his honor. The zone plate's focusing ability is an extension of the Arago spot phenomenon caused by diffraction from an opaque disc.A zone plate consists of a set of radially symmetric rings, known as Fresnel zones, which alternate between opaque and transparent. Light hitting the zone plate will diffract around...

Maybe a sufficiently complicated zone plate could create an erect real image.

@JohnRennie: Ok sir. I don't know either of the two concepts and I have to learn about them. BTW I was replying to some comments below my question. You may wish to see them:

i.stack.imgur.com/cX9dc.jpg — Yuvraj Singh... 5 mins ago

And I suppose I have understood correctly:

@YuvrajSingh... : See, our eye lens (a convex lens) produces an inverted image on the retina and our brain perceives it as an erect image. It's an altogether different phenomenon. Convex lens (eye lens) produced real inverted image and not real and erect image. — Guru Vishnu 2 mins ago

Yes, you are correct.

The brain flips the image so you see it as erect even when it isn't.

@JohnRennie Ok sir. Thank you for clarifying :-)

@JohnRennie: Do you have any ray diagrams for Fresnel lens or Zone plates producing a real and erect image, sir? I hope I will be able to understand diagrams if it's based on simple reflection or refraction?

@GuruVishnu no.

Yuvraj Singh...

10:36 AM

@Guru vishnu I will give you a good answer.

@JohnRennie :-(

@YuvrajSingh... Thank you. But please not the eye lens :)

@JohnRennie May I give it a shot, sir?

@GuruVishnu give what a shot? Place the bounty?

@JohnRennie Yes sir :

@GuruVishnu yes, go ahead.

It's your points :-)

@JohnRennie :-)

@JohnRennie: Bounty is online! :-)

10:47 AM

@GuruVishnu good luck!

@YuvrajSingh... hi Yuv. I edited one of your gravity questions. It now looks much better :-)

@JohnRennie :-)

@JohnRennie: Even if you invent something new regarding this please post an answer sir :-)

@YuvrajSingh...: Hi! re:chat.stackexchange.com/transcript/message/53258835#53258835 = I've started a bounty for the same question.

:-)

@user8718165 really you have radically modified the question!

@JohanLiebert yeah ;-) Isn't it looking beautiful now?

11:03 AM

@user8718165 yes it does. Now it is much more easier to understand the question.

2 hours later…

12:33 PM

-2

Q: Physics olympiad. My question is about classical-mechanics

Help please i can't find $u2$. Two planes move perpendicular to each other with speeds $v$ = $1$m / s. A body flies at them with a velocity of $u$ = $1$ m / s at an angle $a$ = $45$ to the surfaces of the planes and is elastically reflected successively from each of the planes. Find the final vel...

homework-and-exercises newtonian-mechanics kinematics collision

@JohnRennie sir does this question have enough information to be answered? (Not considering the fact that it is a homework and exercise type question)

Yes I think that can be answered. I assume the planes are taken to be infinitely massive so their speed doesn't change when the object bounces off them.

You do it by working in the rest frame of the planes.

12:57 PM

Does anyone know where I can find a formulary (is that what they call a collection of formulas) online?

3 hours later…

3:48 PM

hello sir :) @JohnRennie

@JohnRennie I've a little question sir :-)

@Rainb On what topic?

@user8718165 hi :-)

@JohnRennie hello sir :-) Did you have your lunch :-)

Yes, toasted sandwiches!

@JohnRennie yeah sir... can I ask you a little question sir?

4:00 PM

@user8718165 yes

@JohnRennie I learnt a little bit about multiple summations.... could you please explain it sir just a bit please? I'll get it. :-)

With only two objects present there can only be two forces acting: $F_{12}$ and $F_{21}$. Yes?

@JohnRennie Are you free for a quick question in physics?

@JohnRennie yeah sir

@AjayMishra you're welcome to post the question but I need to finish answering user8718165's question first.

4:05 PM

@JohnRennie sir, I'll wait... you can help Ajay or let's do it (if you wish) :)

@user8718165 OK. @AjayMishra what did you want to ask?

Why we assume that in the normal mode the frequency of each element is same?

In coupled oscillations ^

If different elements were moving at different frequencies then they would get out of phase with each other. One of the key features of a normal mode is that every part of the object moves in phase.

Why? Is that a defining property? [Not always in phase, but I got your point]

I suspect normal modes have a technical definition to do with eigenvectors of a matrix, but it seems pretty obvious to me that in a normal mode all the elements move in phase.

4:10 PM

Okay. I'll think over it, @user8718165 I am done. Thanks.

@user8718165 hi again

@AjayMishra :)

@JohnRennie hello sir. Go on :)

7 mins ago, by John Rennie

With only two objects present there can only be two forces acting: $F_{12}$ and $F_{21}$. Yes?

@JohnRennie yeah sir

If the interaction lasts a time $dt$ then the impluses due to the two forces are $F_{12}dt$ and $F_{21}dt$.

4:13 PM

@JohnRennie yeah sir

Since momentum has to be conserved the total impulse has to sum to zero, and that gives us $F_{12}dt + F_{21}dt = 0$ and hence $F_{12}= -F_{21}$. All very straightforward.

@JohnRennie yeah sir

But suppose we add third body. Now we get $F_{12}dt + F_{21}dt + F_{13}dt + F_{31}dt + F_{23}dt + F_{32}dt = 0$

Now we have one equation is six variables, so we can no longer work out the values of all the forces.

So we cannot say that $F_{12}dt + F_{21}dt = 0$. That is, as soon as we add a third body conservation of momentum no longer implies the third law.

@JohnRennie okay sir...

That's the point being made in the answer you linked.

:53262858 yes, but any external bodies exerting a force can just be added to the number of bodies in the system, so we can always work with an isolated system if we make it big enough.

4:20 PM

@JohnRennie got it sir :-)

@JohnRennie thank you very much sir for your help :-)

:-)

@JohnRennie Good Night sir... We'll chat tomorrow :-)

@user8718165 bye :-)

5:09 PM

@JohnRennie could you please explain a physics question to me if you are free ?

@MKC hi, yes, what's the question?

Thanks for the response . It is regarding an acceleration vs time graph data which I recorded . Please have a look at it . physics.stackexchange.com/q/525519/218252

The x plot seems fairly clearly related to the two turns. I suspect the acceleration when you start pedalling is too small to be see though the noise, though I agree the braking towards the end of the trip seems clear.

I suspect the phone is being rattled around as you cycle and that's why there is so much noise in the traces.

I'm not sure there is much more to say about it.

No , actually what happened was another person was driving the bike and I was just sitting on the backseat recording the data with the phone in my hand( with the + Y direction in the direction of travel).

is it correct that the X acceleration directly gives me the lateral acceleration?

@MKC Yes.

Well, only if you kept the phone horizontal as the bike leaned into the turns.

If you allowed the phone to tilt the acceleration would be a combination of the x and z traces.

5:20 PM

@JohnRennie thanks. Actually I had to show these data in a report etc. Hence I was confirming. And yes the phone was perfectly horizontal.

If you were holding the phone then I suspect the rise at the end is you swaying forward immediately after the bike stops.

I just wanted the braking and acceleration to be near about 1 to 1.5 g , but I got less. I guess I will have to get the bike to travel faster.

And yes the human body sways forward on braking and back on acceleration, so maybe attaching the phone to the bike would be better.

I would attach the phone to the bike. That does mean the phone will tilt, but that's no problem. The lateral acceleration is then given by a^2 = a_x^2 + a_z^2

Yeah sure. But as you can see the x and z graph become negative and positive very quickly and are not smooth . Someone can raise a question regarding the direction of acceleration during turns, as the graph changes the directions of x and z quite frequently.

@MKC I think that's vibration from the bike.

5:31 PM

Interpreting that and filtering out the true acceleration of the bike is not easy. Anyways this data will somehow do my job I guess of showing the maximum acceleration experienced on the vehicle.

@MKC yes, I agree. You could try smoothing the data but that's not usually very successful.

Maybe around turns , mv^2/r will work better, as this is something where all data can be determined easily , if the average velocity around turns is recorded and radius of turn is defined.

You could try getting the bike to ride in a circle and gradually increase the speed. Use a datalogger to record the speed. I've used the GPS Logger app for this.

@JohnRennie Yeah you are correct, this can sure be done. I will surely try this later. Anyways thanks for the help and opinion.

Bye

1 hour later…

6:51 PM

@Knight Electrical Engineerung

something like this de.wikibooks.org/wiki/Formelsammlung_Physik/_Druckversion but in english

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Transcript for

Jan '2017

Problem Solving Strategies

General chat for high school physics. For MathJax see [here](m...

fluid-statics homework-and-exercises newtonian-mechanics

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