Problem Solving Strategies

harambe

05:20

@JohnRennie good morning

John Rennie

@harambe morning :-)

Nobody recognizeable

Hey guys how did jee go for all?

harambe

@JohnRennie can you help me with some doubts on communication systems

Nobody recognizeable

@sammygerbil are you there ?

harambe

@Nobodyrecognizeable it went okay for me. Found chemistry kinda difficult because I was not prepared for it

John Rennie

05:22

@Nobodyrecognizeable Sammy is normally here in the afternoo UK time. Right now it's 05:22 a.m.

@harambe I can try...

harambe

@JohnRennie okay

sammy gerbil

@Nobodyrecognizeable hello. My sleeping pattern is erratic. I didn't sleep at all on Friday night, and slept from 9pm to 3am on Saturday night.

Nobody recognizeable

@JohnRennie this guy always forgets to turn off the internet. So sometimes... i ping him in morning.

harambe

@JohnRennie imgur.com/a/iyn82mQ

What is the meaning of decimal system with system base 10 in Analog signal

Nobody recognizeable

@sammygerbil anyway try to have a definite pattern. And try to stick with it anyway. Although i know extreme pressure breaks those patterns.

harambe

05:26

Similarly a base 2 to deal with digital signals

sammy gerbil

@Nobodyrecognizeable Better than an alarm clock.

Nobody recognizeable

@sammygerbil are you ok if i ask you a question (if you don't feel sleepy now)?

@sammygerbil :) : xD

sammy gerbil

@Nobodyrecognizeable Better to ask John Rennie when done with harambe. I came in to look up xcoder's problem with the U tube.

Nobody recognizeable

@sammygerbil can a particle have momentum without energy?

@sammygerbil anyway i am also interested in the U tube problem. So one bookmark is pending: p

harambe

@Nobodyrecognizeable physics.stackexchange.com/questions/90231/…

John Rennie

05:31

Ah, I have some servers down. I'm going to have to work for a while. Sorry.

Nobody recognizeable

@harambe thanks. @sammygerbil so my problem is resolved.

harambe

@John Rennie Okay. Ping me when free

John Rennie

@harambe will do

Nobody recognizeable

@harambe you have 2 more months for chemistry.

harambe

@Nobodyrecognizeable yes. I am betting on that only xd

But first boards xD

Abcd

05:37

Why boards? Arent you a dropper?

Jul 8 '18 at 10:04, by harambe

@Blue As a dropper,That's gonna be my goal

Sid

Today is JEE Exam,no?

Abcd

No, it ended yesterday

Nobody recognizeable

@Abcd how did it go ?

Sid

Oh. Okay.

harambe

@Abcd had to miss boards because I got Ill. So I am repearting 12

Abcd

05:41

@Nobodyrecognizeable Not v good but not bad. Moderate.

@harambe Oh I see.

Sid

@harambe It means the normal signal that you have. Like 5V or 8V and so on(Analog signal)

Nobody recognizeable

@harambe that's a pain.

Sid

In Digital, there are only two values, 0 and 1. Hence, binary number system is used to represent it

0 and 1 usually means, "Either it's ON or it's OFF"

Nobody recognizeable

@Abcd anyway you have 2 months ahead analyze mistakes and itll very good next time .

Abcd

@Nobodyrecognizeable yes but focusing more on Advanced and Boards atm.

harambe

05:43

@Sid okay. Got it

@Sid what about in Analog

What does a system with base 10 mean

Nobody recognizeable

@Abcd yep best of luck. After all advanced is the big deal. Anyway gotta go. Cya guys.

harambe

@Nobodyrecognizeable cya.

Sid

@harambe Basically, it means, in an analog signal, you represent the values as you see them. Like, say a Sine wave. That's an analog signal. Its values will be, as you measure them

That is, if its peak is 5V, you say it is 5V. Not, something like 0 or 1

harambe

Okay

@Abcd are you done with semiconductors? How long does it take to finish it

Abcd

@harambe takes lot of time to understand but 1 week to finish.

harambe

05:55

Okay. Thanks for the info

@Sid free?

Sid

For around 15-30 minutes, yes

harambe

@Sid imgur.com/a/oeRIV4W

Can you explain point (b) to me

Sid

is that Communication System?

I am not too fond of that chapter and neither do I know much of their concepts. Sorry.

harambe

Yes it's communication system.

No problem.

sammy gerbil

06:23

@Mr.Xcoder Accelerating U tube problem : The solution suggested by the diagram seems unnecessarily complicated. Like $g$, the vertical component of applied acceleration $a$ has the same effect on both limbs, so we should make this component zero to minimize the magnitude of $a$.

We accelerate the tube horizontally to the right. Then the pressure at point A at the bottom of the column of oil in the leading limb is $P_A=\rho_u gH$ and that at point B at the same height in the trailing limb is $P_B=P_A+\rho_w a L$. The reason is that point B is at "depth" of $L$ “below” point A in water in a horizontal gravity field of strength $a$.

The pressure at B is also $P_B=\rho_w gH$ because the fluids are required to be level in both limbs. Combining expressions for $P_A, P_B$ gives $$\rho_w aL=(\rho_w-\rho_u)gH$$

@Nobodyrecognizeable ^^^

Abcd

@JohnRennie Are you free

John Rennie

@Abcd working I'm afraid

Abcd

06:51

@JohnRennie how long would it take?

John Rennie

@Abcd another hour I'm afraid

Abcd

ohkay :(

Nobody recognizeable

07:26

@sammygerbil are you free now?

tarit goswami

@JohnRennie Hi :) I have a question

John Rennie

@taritgoswami I'm working at the moment I'm afraid.

tarit goswami

Actually I am facing problem to properly understand wave particle duality..

If we think light as Photon particle, how we will explain it's polarization

@JohnRennie Sorry I'm not a native English speaker, not getting what it means :p

Abcd

@taritgoswami He means that he is busy with his job

tarit goswami

Ooh ok

John Rennie

07:57

@harambe @Abcd I'm done with work now

blue_eyed_...

08:13

@JohnRennie

John Rennie

@blue_eyed_... hi

blue_eyed_...

@JohnRennie, An electric lamp which runs at 100V DC and 10A current is connected to 220V, 50Hz ac mains. Determine the inductance of choke coil.

John Rennie

@blue_eyed_... Didn't you ask this before?

Nobody recognizeable

@JohnRennie please ping me when you're free.

blue_eyed_...

@JohnRennie, I can't remember right now if I've already asked.

John Rennie

08:22

@blue_eyed_... let me draw a diagram

harambe

@JohnRennie hi. I will ask after @Nobodyrecognizeable

Nobody recognizeable

@harambe from where are you asking?

harambe

@Nobodyrecognizeable communication

John Rennie

Nobody recognizeable

@harambe then i can't help. Can you do a pulley problem for me ?

harambe

08:27

@Nobodyrecognizeable I can try

John Rennie

@blue_eyed_... that's your circuit. I've drawn the lamp in series with the inductor. We know the resistance of the lamp is ten ohms because we know it draws tem amps from a 100V source. OK so far?

blue_eyed_...

@JohnRennie, yes!

Nobody recognizeable

John Rennie

@blue_eyed_... What we need is for the combined impedance of the lamp and the inductor to be 22 ohms so it draws 10 amps from the 220V source. Yes?

Nobody recognizeable

@harambe is it understandable? I get $Tcos\alpha = mgcos\theta$

blue_eyed_...

08:31

@JohnRennie, yes. I can carry on with that

John Rennie

@blue_eyed_... OK, ping me if you need anything more

2

Nobody recognizeable

@JohnRennie are you free now?

John Rennie

@Nobodyrecognizeable yes

harambe

@Nobodyrecognizeable it's understandable. Let me work on it

Nobody recognizeable

@JohnRennie carrying on above problem : i get $mgsin\theta -Tsin\alpha = mgsin\theta$

@JohnRennie there's bound to be something wrong in the above equation.

@harambe i was thinking of working on the inclined plane axis and balancing forces. Won't that help?

Abcd

08:37

@JohnRennie Do we need parametrised constructor when we have setters?

Nobody recognizeable

@JohnRennie are you still here?

@harambe did you get any answer?

harambe

@Nobodyrecognizeable I get (d)

Nobody recognizeable

@harambe they say c.

John Rennie

@Nobodyrecognizeable hi, yes, I was doing something lese.

Nobody recognizeable

@JohnRennie are my two equations correct?

harambe

08:42

@Nobodyrecognizeable I get (c)

Nobody recognizeable

@harambe how did you get that?

harambe

I ignored acveleeation of incline... Thought gsintheta was of that only

John Rennie

@harambe I agree

harambe

@Nobodyrecognizeable pseudo forces

Nobody recognizeable

@harambe please write the two equations.

A picture of equations would be helpful as well.

@JohnRennie how ?

John Rennie

08:48

Suppose you're standing on the trolley next to the pendulum. What acceleration would you feel?

Nobody recognizeable

@JohnRennie gcos,

harambe

The component of the gravity acting along the string is in equilibrium

John Rennie

@Nobodyrecognizeable Yes, $g\cos\theta$. So that's the acceleration the pendulum feels.

And we know $F=ma$, so the force on the pendulum must be $F = mg\cos\theta$.

And since the length of the string isn't changing, the tension in the string must be equal to the force on the pendulum.

So the answer is (3)

harambe

@JohnRennie yes.. Is my explanation correct

John Rennie

@harambe yes

Nobody recognizeable

08:51

@JohnRennie shouldn't $Tcos\alpha = mgcos\theta $

John Rennie

@Nobodyrecognizeable a bit of quick geometry should convince you that $\alpha = \theta$

Nobody recognizeable

@JohnRennie then my above equation shouldn't be true.

John Rennie

@Nobodyrecognizeable oh, hang on ...

Nobody recognizeable

@harambe ok please ask your question. Gotta go for a while. @JohnRennie please carry on with @harambe

harambe

@JohnRennie this question isn't being solved by resolving and then equating all forces

John Rennie

08:59

@harambe I think I've managed to confuse myself by trying to answer it while I was also doing something else. I propose to think no more about it for now.

harambe

@JohnRennie okay

John Rennie

In any case I need to get back to work for a bit. Did you have a quick question before I restart work?

harambe

@JohnRennie yes

John Rennie

Go on then ...

harambe

imgur.com/a/cDNzxlO

@JohnRennie Q11

How is statement 2 explanation of 1

@JohnRennie you can answer when you get back too. I will be here

John Rennie

09:04

Ground wave propagation relies on the fact that the radio waves interacts with the Earth.

@harambe Specifically the electric field of the radio wave induces current in the Earth. These currents then act as secondary sources i.e. they retransmit waves that interact with the original wave. The result is that the original wave is refracted downwards towards the Earth. OK so far?

harambe

@JohnRennie ok

John Rennie

But the Earth has a resistance so some of the energy of the original wave is dissipated as heat instead of being reradiated as a secondary wave. This reduces the amount that the original wave gets refracted downwards towards the Earth. Still clear?

harambe

Yes

John Rennie

The amount of energy that is lost as heat increases with increasing frequency, and that means the refraction of the original wave towards the ground decreases with increasing frequency i.e. the ground propagation is less effective as we increase the frequency.

harambe

Ohkay. Makes sense

Got it

John Rennie

09:11

Cool :-)

3

Now back to work. I'll be finished in an hour or so, possibly sooner.

Nobody recognizeable

@JohnRennie are you free now?

@JohnRennie whenever you come back please ping me.

Mr. Xcoder

@sammygerbil There are a couple of issues with your approach, in my opinion: 1. The problem asks you to find the minimum non-horizontal acceleration. What you have calculated is the minimum horizontal acceleration, but anyway: 2) it doesn't give a correct result, as it is still higher than the minimum required acceleration. Finishing the calculations, you get $a=\frac{(\rho_w-\rho_u)H}{\rho_wL}$. With the data given, this gives $a=g$.

But the minimum possible value for $a$ is $g/\sqrt{2}$ for an angle of $45^\circ$. I will show you my approach shortly. While my solution is completely different from the one in the grading sheet, it does give the correct result.

Nobody recognizeable

@Mr.Xcoder can you post your solution because i am interested too?

Mr. Xcoder

Your claim the vertical component of applied acceleration $a$ has the same effect on both limbs, so we should make this component zero to minimize the magnitude of $a$ is not necessarily correct, as its vertical component aims to reduce the pressure difference

@Nobodyrecognizeable Sure, I'm writing it down right now.

Nobody recognizeable

@Mr.Xcoder can i repost your problem For bookmark purposes?

Mr. Xcoder

09:22

Jan 9 at 13:48, by Mr. Xcoder

Hi all, I have this problem, The moving tube which I really don't know how to approach. In a tall and thin U-shaped tube of constant cross-sectional area and open at both ends lies a quantity of water. Through the upper part of one of half of the tube oil is poured, such that the height of the oil is $H=25cm$ and $\rho_u=0.8g/cm^3$. Given $g=10m/s^2$ and $\rho_w=1g/cm^3$, calculate the minimal non-horizontal acceleration $\vec{a}$ such that the liquids in both halves become the same height..

@Nobodyrecognizeable Done :)

Nobody recognizeable

Mr. Xcoder

So first of all, let $\phi$ be the angle $a$ forms with the horizontal. Then, he have that $\Delta P=\rho_w (g-a\sin\phi)H-\rho_o (g-a\sin\phi)H=(\rho_a-\rho_o)(g-a\sin\phi)H$.

On the other hand, the horizontal component of the acceleration also creates a pressure difference, given by: $\Delta P'=\rho_w aL\cos\phi$ (proof: $F_x=ma_x=m\rho_w\cdot SL$, where $S$ is the cross-sectional area of the tube, and $\Delta P'=F_x/S=\rho_w La_x$). From the equilibrium of the liquid columns, we deduce that $$\Delta P=\Delta P'\implies a=\dfrac{g}{\sin\phi+\frac{\rho_wL\cos\phi}{(\rho_w-\rho_o)H}}$$

Nobody recognizeable

@Mr.Xcoder great!

Mr. Xcoder

09:40

Wait, something must have gone wrong here.

Nobody recognizeable

@Mr.Xcoder hi how do you get the L?

Mr. Xcoder

@Nobodyrecognizeable L is given, 5cm.

Nobody recognizeable

@Mr.Xcoder you didn't write that on the question.

Mr. Xcoder

Jan 9 at 13:49, by Mr. Xcoder

The distance between the two arms of the U-shape is $L=5cm$. Neglige the adherence of water to the walls. Here is a picture of the initial situation:

Nobody recognizeable

@Mr.Xcoder in your last calculations you didn't put the value of H .

Mr. Xcoder

09:43

Yes, I've noticed now

Nobody recognizeable

@Mr.Xcoder you are right $\phi = 45°$

Mr. Xcoder

@Mr.Xcoder Now to evaluate $\dfrac{da}{d\phi}=0$: Let $A\equiv \dfrac{\rho_wL}{(\rho_w-\rho_o)H}$, then $\dfrac{d}{d\phi}(\sin\phi+A\cos\phi)=0\implies A=1\cot\phi\implies \phi=45^\circ$. Plugging into the initial expression of $a$: $$\boxed{a=\dfrac{g}{\sqrt{2}}}\approx 7m/s^2$$

Nobody recognizeable

@Mr.Xcoder great approach. Thanks for posting.

Mr. Xcoder

No problem, I hope it is also correct.

Nobody recognizeable

@Mr.Xcoder if you're in doubt we can ask jr, if the approach is correct or not.

@Mr.Xcoder basically i am asking for permission.

@Mr.Xcoder putting $\phi =0$ ie horizontal case you get @sammygerbil 's answer. I think this is another evidence you are correct.

Mr. Xcoder

09:57

Don't bother JR with this question again, I think this approach is right.

Plus, we will receive sammygerbil's feedback once they come online and see the pings.

harambe

@Mr.Xcoder your olympaids are coming near, right?

Nobody recognizeable

@Mr.Xcoder ok. This approach is right. And your equation is right for several cases we don't need to bother. Will see what @sammygerbil says. But greatly done.

Mr. Xcoder

The local phase was on December 8th and the next one is on February 23rd.

2

Nobody recognizeable

@Mr.Xcoder how did you do ?

Mr. Xcoder

26.25/30, second place

Nobody recognizeable

10:00

@Mr.Xcoder great!

harambe

@Mr.Xcoder All the best then!

Mr. Xcoder

Thanks to you both :)

Nobody recognizeable

@harambe @Mr.Xcoder if you have any mechanics problem then please ping me. Ill like to jump on that.

Mr. Xcoder

So you're interested in mechanics problems?

harambe

@Nobodyrecognizeable I have a rotation question. If you are up then I can post

Nobody recognizeable

10:03

@Mr.Xcoder best of luck for second phase then. Don't wanna waste more time of you.

@Mr.Xcoder yes in 1st sem we had mechanics , gpm, waves and oscillations.

@harambe yep. Sure.

Mr. Xcoder

@Nobodyrecognizeable You're not wasting my time at all! I enjoy spending time here for a reason :) You guys are all great and find very interesting problems I sometimes solve after JR or sammy reply :)

Nobody recognizeable

@Mr.Xcoder same for me too. As a bs student. I think we have a problem from harambe now.

Mr. Xcoder

I'll attempt the problem once posted, but I cannot guarantee I also know how to solve it, because I don't have the necessary knowledge to attempt all of those.

For instance I didn't learn oscillations, moments of intertia/angular momentum etc.

Nobody recognizeable

@Mr.Xcoder its same for me too. But by jumping on the problems we learn new techniques.

blue_eyed_...

@JohnRennie

Mr. Xcoder

10:08

@Nobodyrecognizeable Any concept in particular you would like to focus on? If encounter difficulties with that concept in the months to follow, I'll know who to ping, then :)

Nobody recognizeable

@blue_eyed_... he has gone for working a while.

blue_eyed_...

@Nobodyrecognizeable, oh

Mr. Xcoder

I'll be back in about half an hour, I think, so if I don't respond to pings sorry (in advance) for the late replies

harambe

imgur.com/a/xPVVHPD

Nobody recognizeable

@Mr.Xcoder mechanics, gpm, oscillations were on this sem i will like em theory after a few months.

@harambe ill try this. If i get it ill answer.

1

Q: Finding velocity of a rolling disc down a slope

When finding the velocity of a rolling disc using newton's second law method we need to take into account the friction. We use $$ mg \sin(\theta) - F = ma $$ and $$ Fr = I \alpha$$ $$I = \dfrac{mr^2}{2}$$ However, if we are finding the velocity using the energy method. We use $ T = \dfrac{...

@harambe this is related.

@JohnRennie are you free now?

John Rennie

10:25

@Nobodyrecognizeable free ish. I'm around for a few hours but have bits and pieces of other stuff to do.

Nobody recognizeable

@JohnRennie whats the problem in my question when i break components of force?

@JohnRennie in my knowledge if a block has less velocity then it slides after reaching a certain velocity it starts rolling.

-1

Q: What is the minimal friction force required in order for a ball on an incline to roll without slipping?

I read an article now and it confused me completely. Since I thought I have read somewhere else a completely different solution. You put a ball on an incline at rest. What is the minimal friction force so that the ball rolls without slipping? I thought there is a simple and easy solution. ...

newtonian-mechanics forces friction

John Rennie

on phone

Nobody recognizeable

@harambe

@JohnRennie no problem.

Mr. Xcoder

I don't know how to solve such questions, sorry.

John Rennie

10:46

Ok, the phone call is over. Where were we?

Nobody recognizeable

Rolling Without Slipping Friction

►

@JohnRennie on the mgcos\theta problem resolving the forces.

@harambe ^^ @Mr.Xcoder

John Rennie

You do this by calculating $\alpha$ and using $T = I\alpha$ to calculate the the frictional force.

Nobody recognizeable

@JohnRennie ok thats fine . But would the wedges movement disturb with the above solution anyway?

@JohnRennie are you still here?

John Rennie

Ah, I assumed the wedge was fixed ...

Nobody recognizeable

@JohnRennie then what to do now ?

John Rennie

10:59

So is the wedge free to move?

Nobody recognizeable

@JohnRennie ^^

@JohnRennie i needed an error analysis book for physics practicals. Can you recommend me one ?

John Rennie

@Nobodyrecognizeable I'm afraid I'm out of touch with current physics books.

@Nobodyrecognizeable I don't really have the time for detailed analysis of problems this morning. Sorry.

Nobody recognizeable

@JohnRennie i don't need current books. I just wanna learn that anyway. I think you've read something in error analysis. That would serve good for me.

@JohnRennie ok no problem.

@JohnRennie if you recall any book. Then please suggest me later this day. Or tommorow.

@JohnRennie have a nice day professor, goodbye.

John Rennie

Bye

blue_eyed_...

@JohnRennie, 46

2

Dante

15:16

@harambe How was your JEE?

harambe

15:37

@Dante went kinda average. I found chemistry tough though it was easiest and calculus was hard too

Dante

@harambe We felt chem was tough because we didn't study it (or I didn't study it).

I kept ignoring it till the last, though it was scoring :(

What are your expected marks?

harambe

Around 160-170.Kinda low

Dante

@harambe same, 10-20 marks more.

I want to go back in time man :(

harambe

XD

Dante

And start my 11th again

Come on, I'm feeling low :/

Nobody recognizeable

15:41

you all got one more time guys. To rectify the mistakes don't redo the mistakes.

harambe

Everyone gets that feeling

Dante

Hmmm...

harambe

@Nobodyrecognizeable yeah

@Dante gotta go now. See you tomorrow

@Nobodyrecognizeable see you tomorrow

Nobody recognizeable

@harambe anyhow never feel that you have excessive time nor feel less time just go on seeing what you left. And rectifying them.

@harambe cya.

Dante

Cya

Nobody recognizeable

15:44

@Dante any idea when @sammygerbil will come ?

Dante

@Nobodyrecognizeable No idea dude.Haven't asked any doubt since a week here.

Nobody recognizeable

@Dante i see. Out of doubts isn't a good thing anyway. Although it depends on situations.

Dante

Out of doubts because I was revising the last week.

the formulae

And we have practicals and pre-boards

2

Competitive study is paused for a while.

2

Nobody recognizeable

@Dante its ok then. Whats the formulae for kinetic energy lost in in an inelastic collision in terms of velocity and masses of the particle?

Dante

@Nobodyrecognizeable Idk if there is any 'formula' for that, we can find it generally according to the question.

For JEE level collisions, one formula does all the job, final velocity of both the bodies in terms of their masses,velocities and coefficient of restitution. This formula can suffice most problems.

Nobody recognizeable

15:52

@Dante ok anyway, sometimes quick formulas help you do the question in a flash rather than deriving from scratch. Although this gives you pressure as well. But one can go oneway .

@Dante anyway gotta go. Cya.

Dante

Yeah, cya!

wait

@Nobodyrecognizeable tell this answer before going ^^

5 hours ago, by Nobody recognizeable

Nobody recognizeable

16:12

@Dante thats harambes problem. Unfortunately i don't know the answer.

Dante

Oh, ok

Looks like AITS problem

Abcd

16:35

hello @JohnRennie ??

John Rennie

@Abcd hi

Abcd

@JohnRennie can you please check 1 code using tests?

John Rennie

@Abcd I'm eating my lunch at the moment. Tomorrow morning would be a better time.

Abcd

@JohnRennie Oh I see professor.

Mr. Xcoder

23:26

The most unexpected answer to a counting puzzle

►

I think that this video is definitely worth watching

Transcript for

Problem Solving Strategies