Problem Solving Strategies

napstablook

03:37

@IITM no it does work. you have to further integrate

i crosschecked the integration. about the answer, the solution takes SHM approach as both the systems are mathematically equivalent

which is way neater tbh. but since you said you couldn't understand I presume you have not read SHM?

anyways here is the integral prnt.sc/12sd3y5 you will need to know how def. integral sub works tho.

IITM

04:21

@napstablook Now I understood where I made mistake, I didn't change limits so I got something weird

@napstablook No,I actually understand SHM but,I just noticed that it is in SHM after JohnRennie told me when I was looking at the solution also,I didn't observe that

anyways,thanks for crosschecking btw,you are preparing for IIT?

napstablook

Yep

IITM

04:39

@JohnRennie No need to check my solution,napstablook crosschecked it,but Thanks for your attention

John Rennie

@IITM cool :-)

IITM

@napstablook you gave mains this year?

@JohnRennie :)

napstablook

@IITM I did you also final year?

IITM

@napstablook yep

@napstablook so what's the percentile?

napstablook

99.79 your?

IITM

04:45

99.76

napstablook

oh cool, I will be giving the next 2 attempts (whenever they come) are you going to?

IITM

@napstablook yes,ofcourse

@napstablook studying in any institute?

napstablook

allen, you?

IITM

@napstablook self-study

napstablook

05:02

@JohnRennie can you please check this question prnt.sc/12sg7kx

John Rennie

@napstablook I Googled to check what Searle's experiment was, and I found this solution.

napstablook

No my doubt is not related to searle's expt. I should have elaborated. my problem is with the vernier caliper here

can you tell me what the intitial and the final readings of the vernier caliper are? my calculation seems to show that the string contracted after a load of 2kg was added

John Rennie

I assume the Vernier scale has 50 divisions, since it covers 0.05cm i.e. from 3.20 to 3.25cm.

Before loading the wire the 20th division on the Vernier scale lines up so the reading is 3.220cm. OK so far?

napstablook

the 0 is somewhere between the 2 readings though

so I don't think we can presume that it covers the entire length between the 2 reading 3.2 and 3.25

John Rennie

The reading is greater than 3.20cm and less than 3.25cm. The way you find out the exact reading is to see where the Vernier scale aligns with a main scale marking and add that to 3.20. Yes?

napstablook

05:13

Im sorry but can you show me a diagram off how the scales are aligned?

John Rennie

I'll have a go, but that's going to take some drawing ...

napstablook

ok prnt.sc/12sgm67 check my drawing though

sry for the poor quality I rushed it a bit

John Rennie

@napstablook there, that's the scale before loading. The main scale is the top scale and Vernier is the bottom scale. The 20th division on the Vernier scale lines up.

napstablook

one thing that is weird is that they both have same spacings in the diagram, hence the lining up occurs at a rather far off distance

0.02 cannot line up with anything beyond 3.25 as the next reading is 3.3 which would require 50 divisions to line up at all

John Rennie

@napstablook 50 divisions on the Vernier scale correspond to 49 divisions on the main scale.

This is the scale after loading. Now the 45th division on the Vernier scale lines up:

napstablook

05:28

ya but this is not the traditional vernier scale, the least count is 0.001cm so 50 divisions should fit between a length of 0.05cm which is the LC spacing of main scale

ConGovDeIn

05:41

@JohnRennie Hi :-)

John Rennie

@ConGovDeIn hi :-)

ConGovDeIn

Mausa, will you suggest a song to me?

John Rennie

A song?

ConGovDeIn

You once suggested “animals” to me and I enjoyed it too much

John Rennie

Ok, how about Crime of the Century by Supertramp. It's very different from Animals, but it was one of my favourites when I was student.

ConGovDeIn

05:48

Okay.

When was the last time you played that thing ?

John Rennie

It's a Bluetooth audio system. I play stuff on it from my phone or laptop.

ConGovDeIn

:)

John Rennie

It's this one

ConGovDeIn

@JohnRennie OH! Mausa, it’s an awesome song! Just heard it! Music is so refined.

John Rennie

Glad you like it :-)

ConGovDeIn

06:00

And sir, can you suggest me a song where a lover (who is alone now) thinks about his past time with his/her lover and how happy the time was but there was knowing in the back of head that it would all end?

napstablook

erm.. @JohnRennie would it be okay to discuss my doubt at a later time?

John Rennie

@napstablook I thought we'd finished ...

napstablook

@JohnRennie after discussing I realized I have some major prob in understanding how vernier works, but this discussion has gone too long so I wanted to talk later about this

John Rennie

OK. I'm here until about 17:00 Indian time.

napstablook

thx

John Rennie

06:08

The Indian books I've seen seem to make a Vernier scales much more complicated than they really are. I learned Vernier scales by using them in a workshop, not from a book, and I think when you have used them to actually measure things they are really easy to understand.

napstablook

I have never seen an actual vernier scale, only screw gauge.

satan 29

@ConGovDeIn whats "Mausa" ?

in Hindi, its a term for an "uncle"

ConGovDeIn

It’s Maussa

satan 29

and what does that mean

ConGovDeIn

:) you can mean anything by it. However, I use it as a southern pronunciation of “Master”

Srijan M.T

07:13

@JohnRennie Hii sir

I was in quarantine for few days suspected to be COVID +ve since I had high fever.

John Rennie

@SrijanM.T hi :-)

Wow! Are you OK?

Srijan M.T

I’m better now. Will get reports by Tom evening.

@JohnRennie A lot of weakness

John Rennie

:-(

Srijan M.T

Alone in a room now. Separated from parents

@JohnRennie I hope I’m -ve

John Rennie

At least you can use the Internet!

Srijan M.T

07:15

@JohnRennie I know right.

But yeah , it feels really weak. My hands move when I write

Kind of like shake

That’s how it is. Bad. But I hope I’ll recover soon.

John Rennie

If it is covid at least it means you're now immune

2

Srijan M.T

@JohnRennie Let’s see. I still have fever but less. Just took medicine.

Works for few hours

I hope you’re safe and healthy sir. Keep taking care.

John Rennie

I guess you just have to wait. It should only be a couple of days before the fever passes.

Srijan M.T

@JohnRennie Yes sir.

John Rennie

@SrijanM.T I've been vaccinated now (both jabs) so I'm OK.

Srijan M.T

07:18

@JohnRennie Great. I’m 17 and age limit in India is 18

So , I have to rely on immunity capsules

And protein

, I have a Q sir. It is of friction between 2 blocks.

I wrote an explanation but still have some confusions.

This is what I wrote.

The real thing is to find direction of static friction between two blocks in which the lower one is moving ahead

So , pls check my explanation and let’s discuss if you’re free on where I’m wrong🙂

John Rennie

The way you approach these sort of problems is just to apply the 2nd law.

Srijan M.T

Ohk.

John Rennie

If you look at block A the only (horizontal) force acting on it is the frictional force - let's call this Fa.

So the acceleration of block A has to be given by:

Srijan M.T

@JohnRennie Ok. So , it is in left direction right ?

John Rennie

Fa = Ma x a

If A is accelerating right then the force acting on A must also be to the right. Yes?

Srijan M.T

07:24

Hmm K. So , when B accelerates to right. Then , A also accelerate along with it. Due to friction . Otherwise , A would have had gone back. So , when B goes right , A tens to go back. This is force. Now comes friction . So , friction is in direction opposite of motion. So , it is in right direction ?

If any confusion sir , I’ll make a drawing then.

John Rennie

If no force was acting on A then when you pulled the block B the block A wouldn't move.

Yes?

Bhavay

@SrijanM.T It will act in right direction for the block A for the reason you stated. By 3rd Law , it will act on left direction on block b.

Srijan M.T

@JohnRennie Yes

@Bhavay K.

John Rennie

@SrijanM.T so A wouldn't go backwards. It would just stay where it is. It does go backwards relative to block B but be careful not to get confused about this. It just means A stays where it is while B is going forwards.

Srijan M.T

@JohnRennie Yes. A wouldn’t go backwards when B doesn’t move. When B moves ahead , the block A if I’m sitting on block B.I see it goes back ?

Like sir if we go deep in picture and look at the asperities present between the two surfaces.Can I say that as B moves ahead by a force of F. Let’s say someone pulled the block. Then , B applies force on A towards right. In return , A also applies force on B towards left ?

John Rennie

07:39

Yes, relative to block B block A moves backwards, but block B's rest frame is a non-inertial frame and you need to be very careful with non-inrtial frames.

@SrijanM.T correct

Srijan M.T

@JohnRennie That’s what I wrote in explanation. Nice to hear that

Ok so , then.

Can I say that the asperities of block A slip on block B. Since now we are considering friction

Or they kind of push onto each other and make holes or apply pressure.

There are two scenarios

I guess

If this happens. The 1st one.Then , I’m confused what is friction. The asperity of block B will apply force on block A in right but block A applied in left.

Where is the friction ? What is it ?

Or this force that I just said is friction ?

John Rennie

The exact mechanism by which friction is created will vary. In any case you do not need to worry about it for the JEE. Just consider the friction as a force given by μN and leave it at that.

Srijan M.T

Ohk. So , what ever force I wrote is not friction. So , considering JEE. I can just say that block A applied force on B in right and block B applies force on block A in left.?

Opposite of what I wrote earlier.

So , which one is friction now I should write as?

I feel considering what textbook says. It says that direction of friction for block A is on right. So , I think this one is right. Like sir if we go deep in picture and look at the asperities present between the two surfaces.Can I say that as B moves ahead by a force of F. Let’s say someone pulled the block. Then , B applies force on A towards right. In return , A also applies force on B towards left ?

John Rennie

Block B applies a frictional; force to the right on block A, and block A applies an equal and opposite frictional force on block B to the left. Both forces are friction.

Srijan M.T

07:54

@JohnRennie Ok.

Now , one more thing. It is just to clear if I understand third law well.

Is this right sir ?

Regarding acceleration I wrote

John Rennie

I'm not 100% sure what you mean in that, but it looks wrong to me ...

Srijan M.T

Block B pulled by force of 40N

John Rennie

And what is μ?

Srijan M.T

Coefficient of friction but I didn’t write it here

John Rennie

I know that, I mean what is the value of μ? The calculation depends on the value of μ.

Srijan M.T

08:02

@JohnRennie I didn’t consider it here. I just equated it normally. Because of the way they look like

@JohnRennie Since a guy pulling applies a force of 40N on B . Then , B applied 40N on A?

This is what I meant.

John Rennie

The thing is that we can have two different behaviours:
1. if we pull B gently A will not slide and A and B will have the same acceleration
2. if we pull B hard then A will slide and the two blocks will have different accelerations

To find out whether 1 or 2 happens you need to do the calculation and for that you need the value of μ.

Srijan M.T

@JohnRennie That is I guess due to adhesion breaking up. 2nd one.

@JohnRennie Ok. Got it sir.

John Rennie

Take the situation you have drawn.

Srijan M.T

Let us say it is 0.5

John Rennie

Start by assuming that A does not slide. That means A and B accelerate togther.

The applied force is 40N, and the total mass is Ma + Mb = 6kg because they are accelerating together as a single mass.

Srijan M.T

08:06

@JohnRennie Yes. Two cases here too. Either B didn’t move at opal or B moved with very less acceleration

Such that adhesion didn’t break between the,

John Rennie

That means a = 40/6 = 6.66 m/s²

Srijan M.T

@JohnRennie Yes

John Rennie

OK so far?

Srijan M.T

Yes

Prateek Mourya

sir where is pole of equivalent lens of two lenses kept at a distance

John Rennie

08:07

Now consider just block A. We know its acceleration is 6.66 m/s² so the force on it must be:

Fa = Ma x a = 2 x 6.66 = 13.33N

Yes?

Srijan M.T

@JohnRennie Ohk.

@JohnRennie Is it by block B? Like is it friction ?

John Rennie

Yes. The friction between A and B must be applying a force on A of 13.33N.

Srijan M.T

Ok.

John Rennie

But, suppose we are told μ = 0.5

Srijan M.T

So , on A it is 13.33 by B

And on B=4*6.66?

John Rennie

08:10

@SrijanM.T Yes

@SrijanM.T Yes.

Srijan M.T

Sir , we didn’t even need coefficient of friction ?

Thats amazing.

John Rennie

The force on B is the 40N force to the right minus the 13.33N force to the left that A applies on B.

Srijan M.T

Did we need it still but sir ? Coefficient of friction ?

@JohnRennie ok.

John Rennie

@SrijanM.T Careful! We assumed μ is high enough for A not to slip. So we have actually considered the value of μ.

Now suppose μ= 0.5

Srijan M.T

@JohnRennie ok.

John Rennie

08:12

That means the static friction is Fs = Ma x g x 0.5. Yes?

Srijan M.T

@JohnRennie Yes.

John Rennie

So Fs = 10N, but that is less than the 13.33N we calculate if we assume that A doesn't slip. Yes?

Srijan M.T

@JohnRennie Ohk.

John Rennie

That means if μ = 0.5 our assumption that A doesn't slip must be wrong. For an applied force of 40N and μ = 0.5 A will slip. And that means the force on A is just the frictional force of 10N.

So the acceleration of A is now a = 10/2 = 5 m/s²

Srijan M.T

@JohnRennie Ok. Maximum value . Yes

@JohnRennie Ok.

John Rennie

08:15

Now, if B applies a force of 10N on A that means A applies an equal and opposite force of -10N on B.

Srijan M.T

@JohnRennie Yes sirl

John Rennie

So the net force on B is the applied force of 40N minus the 10N frictional force i.e. 30N.

OK so far?

Srijan M.T

@JohnRennie Yes sir.

John Rennie

So the acceleration of B is 30/4 = 7.5 m/s²

Srijan M.T

Yes sir.

Different acc ?

Does it imply Sth significant sir ?

John Rennie

08:17

So in this case both blocks accelerate but B accelerates faster than A does because now A slides on block B.

Srijan M.T

@JohnRennie Ok. So,it slides with 2.5m/s^2 acc

?

towards left ( Block A)

John Rennie

I recommend you only ever consider accelerations relative to the ground.

Srijan M.T

Ohk. Sir.

John Rennie

Relative to the ground we have worked out that the acceleration of A is 5 m/s² and the acceleration of B is 7.5 m/s². Yes?

Srijan M.T

@JohnRennie Yes sir.

John Rennie

08:20

It is true that relative to B block A accelerates left at -2.5 m/s², but I advise you do not take relative accelerations like this as it can get confusing.

Srijan M.T

Ok sir. It just looks so obvious. Can we find slipping acceleration value ?

Avantgarde

John sir :D

John Rennie

@Avantgarde hi :-)

Avantgarde

@JohnRennie Hello!

John Rennie

@SrijanM.T Yes we can.

The maximum acceleration that A can have is when Ma x a = the frictional force. Yes?

Srijan M.T

08:23

@JohnRennie Ok.

John Rennie

So in our example of μ = 0.5 we get Fs = 10N and therefore a = 5 m/s²

And if there is no slipping that means B must have the same acceleration a = 5 m/s². Yes?

Srijan M.T

@JohnRennie yes

John Rennie

So the force to make both A and B accelerate at a = 5 m/s² is F = (Ma + Mb) x 5 = 30N

So 30N is the maximum force we can exert on B before A slides. Yes?

Srijan M.T

@JohnRennie Ok,

Sir, next ?

John Rennie

That's it.

What do you think we need to calculate next?

Srijan M.T

08:32

Hmm. Originally. Acc of B is 7.5 and A is 5

John Rennie

That's what we get if we apply a 40N force.

Srijan M.T

Yes sir.

If we apply 30 , they don’t slip.

John Rennie

If we apply a 30N force we find A and B have the same acceleration of a = 5 m/s²

Srijan M.T

But when we apply 40 , then what happens . How to calculate slipping friction value

@JohnRennie Yes sir.

John Rennie

> How to calculate slipping friction value

Can you clarify what you are asking here?

Do you mean what is the force on A if it is slipping?

Srijan M.T

08:34

@JohnRennie Ok. The acceleration value with which A goes behind since it is slipping on the surface on B

John Rennie

> acceleration value with which A goes behind

Srijan M.T

We know 7.5 and 5.

John Rennie

What does "goes behind" mean?

Srijan M.T

@JohnRennie ok.So , I am thinking that B goes ahead with 7.5 and A goes ahead with 5.

John Rennie

Yes... ?

Srijan M.T

08:36

Then , originally. It is more like. A is slipping behind?

Like if I run on a train which is going at 7.5. Let’s say I move on the train with 5m/s^2. Then , is it like I originally go back with 2.5 or some value ?

Im confused since I have seen people running on train too.

It looks like acc of A wrt ground will be 12.5?

Is I right sir ?

John Rennie

I strongly recommend you do not consider these relative accelerations as it's very easy to get confused.

Srijan M.T

Ok. Where do we consider I am going wrong then

John Rennie

With a 40N force we calculated that A accelerates at 5 m/s² relative to the ground and B accelerates at 7.5 m/s² relative to the ground. Yes?

Srijan M.T

@JohnRennie ok.

Can’t I compare this with a car at 5 on the train with a train at 7.5.

John Rennie

I recommend you do not do this.

Srijan M.T

08:43

ok.

@JohnRennie Let’s go with this then sir.

John Rennie

Yes

Srijan M.T

Sir. Then, if we consider relative to ground. A moves with -2.5 acc towards right

John Rennie

6 mins ago, by John Rennie

With a 40N force we calculated that A accelerates at 5 m/s² relative to the ground and B accelerates at 7.5 m/s² relative to the ground. Yes?

@SrijanM.T No.

Srijan M.T

Ok. Then , what should we say sir. First thing. Only when 30N is applied A is at rest with respect to B. But when 40N , Sth different right

John Rennie

That is correct, but we are calculating the acceleration of block A relative to the ground not relative to block B.

Srijan M.T

08:49

@JohnRennie Ok. But we can calculate if we wish to relative to the block also.

John Rennie

I recommend you do not do that.

Srijan M.T

Like as we know sir if I take a block and keep it on paper. Then , quickly pull the paper. The block droves ahead but slips back right ?

@JohnRennie Ok sir.

Isn’t this what is happening here. This paper example

John Rennie

Relative to the ground both the block and the paper move when you pull the paper. The paper moves more than the block, but both move.

Srijan M.T

@JohnRennie Yes right.

But there is a time paper moves faster than the block.

John Rennie

Yes

Srijan M.T

08:53

Then , paper goes so ahead that block is no longer on the paper

So , there is slipping acc , relative acc or kind of like the value with which it couldn’t be same as the paper

That value , I just wish to know

John Rennie

Let's leave this for now.

Srijan M.T

ok sure sir.

I’ll go for lunch now. Talk to you later about some other Q sir.

Thank you a lot sir.

John Rennie

Bye :-)

Anusha

09:45

@JohnRennie hi

John Rennie

@Anusha hi :-)

Anusha

U free?

John Rennie

Yes

Anusha

I was thinking of a question

John Rennie

Yes ... ?

Anusha

09:47

We do general case of resistance in cube network I hope u know that

John Rennie

OK ... ?

Anusha

Where we apply battery across the face diagonal

Wait let me attach a pic

John Rennie

There has just been a question about this ...

4

Q: The Resistor Cube?

This is an oldie, but still a goodie . Twelve identical resistors of resistance $R\Omega$ are soldered together in a cubic network. I want to find , $\rightarrow$ What is the resistance measured with an ohmmeter between $A \to C$ , $A\to B$ & $A \to D$ ? $\Rightarrow$ One solution to this pro...

Anusha

Yeah

But in addition to this can we solve it if we add resistances across diangonals also

John Rennie

Obviously yes it can be solved, but I don't know how to do it.

Anusha

09:51

Haha ok

Srijan M.T

10:25

@JohnRennie Hii sir

In the adjoining figure, the coefficient of friction between wedge (of mass M) and block (of mass m) is μ. Find the minimum horizontal force F required to keep the block stationary with respect to wedge.[![enter image description here][1]][1]

1) For this Q , total acc of $M+m$= $F/M+m. $

2) Considering inertial frame of reference , for mass M and m.FBD is as follows :[![enter image description here][2]][2]

I haven’t marked N1 or N2 but can assume them according to FBD.

For mass M ,

1) μN1 on left means the friction from surface.

This is Q1

Example 6

here , if less tan 15N force. No slipping

Then , it means after 15 or at 15. Kinetic friction should occur right

Then , how is it that μ_s *N=F where F=15N?

Transcript for

Problem Solving Strategies