Good evening, good evening, good evening my 11-sided warriors. Today's session is an extension of the previous session of Newton's laws where we had studied about pulleys, blocks and strings. Today you're going to have a session on constrained motion where you're going to have multiple pulleys which could be moving and you will see the advantage of having movable pulleys. And you will see that constrained motion is a part of you know all the machinery which is there around you so many machines involve in some or the other way constrained motion and you can think of many machines like that and probably you can put it up in the chat box so all of this with me Shreyas your physics master teacher at Vedantu going live hello Jai Dev, Kalyani, Ashoka, Sharma, Jai Telugu hello Archana, Jai, Cry you CR7 oh it's not CR7. Hi Monica, hello Aseel.
Long time guys. No C. Hi Ankit, hi Arun Kamwar. Hellollo Karti, hi Suchi Mama.
Hellollo, hello, hello. So nice to see all of you. I'm Shreyas, I'm the physics master teacher at Vedantu. I'm also the HOD for J Physics division. I've been teaching for a very very long time and I have produced amazing ranks in my teaching career.
That's what motivates me to you know come over here and produce even more ranks and not just limited to a small place or a town or a coaching institute but throughout India. So we started this channel WeEnthuse which is dedicated to all the English medium students who are aspiring to be amazing scientists, engineers and researchers. So welcome aboard and if you have forgotten to smash the subscribe button.
in case you're not a subscriber yet please do that right now so that you can get our sessions every day and if you are already a subscriber if you're already my warrior you know what to do go smash that like button yeah i hope you have done that already hello fatima hi make for seven hi samuel hi harini hello tanu so i was missing you guys i mean like hi sandhya hello kartish ran hi sajja and you should blame you know who who is to be blamed yeah your seniors super seniors 2021 they are not ready to write the exams they want to write many or rather they want to keep postponing their exams every time there's a new date there is a postponement and it's not just one mains exam they want to write four exams or do like okay i want to do my best and that's why we need to support them you know so you're super seniors all right so let's get going guys and let me show you the schedule as well for this week And there it is. So constraint motion wedges movable pulleys that's for you guys today and Friday there is going to be one more session that's non-inertial frames and pseudo forces and if you are a 12th standard student the ones in white is for you. So self induction and mutual induction to the EMI chapter it's going to continue and special session I'm not sure exactly for who I'll let you know very soon.
So that's coming up on Saturday. Remember my lecture timings are always at 7.30 also it's a good time that we congratulate all the NTSE cleared students from Vedantu and it's amazing to see more than 100 students from Vedantu who have cleared NTSE and it just doesn't stop there some of you probably might know that the J-Mains results were declared recently for the July attempt and guess what look at these numbers just look at this look at what your seniors are doing and that's why you know we are so much involved look at these guys it's amazing to see these improvements you know in first attempt it was 45 then 307 now 694 students from Vedantu are above 99 percentile in JEE mains many of them have even got 100 percentile by the way in individual subjects above 90 percentile look at the graph how it's been increasing now almost 3,000 students so basically 7.5 5 times more students from the first attempt that's how it is more than three almost like 3000 students already qualified for GE advance so you know gradual improvement is there and I'm pretty sure the fourth attempt will be even higher than this so you should definitely wish all your seniors you know all the best luck and you know wish all good for them and guys if you want to be there all right if you want to qualify for GE advance if you want to get into IIT start your serious preparation. I know you are preparing because you are there over here but please start your serious preparation right now so that you don't have to struggle in the end.
And if you're looking for a coaching who will give you guarantee who will tell you yes you will succeed better you will improve better okay so Vedantu is the one which you should consider we give you VIP which is Vedantu improvement promise and if you do not improve the money is returned back. Now the course prices you might see in the description link all right they are very cheap and there are three different packages depending on what exactly you're looking for but. even the most basic the light version covers all the essential things the classic one additionally has the doubt solving and if you want personal mentorship you need to go for the plus version the batches are starting just after the independence day so you should consider joining really fast because after this the portion you know will be squeezed a little bit because you will be joining in late into the batches so remember all the coaching institutes generally stop their admissions or if they put you in some previous batch once you know july or august is over so be careful guys you know vedantu is starting fresh batches you will not be put into previous batches you will be put into fresh batches and they are also in english medium so consider joining quickly the course prices are as low as 2700 rupees when you use my coupon code shhpro all right so that's what you should look for yes sarun 10 discount bacha hi ganeshwari i'm just about to start as well Yes, Archana for 2022 aspirants are also there by the way. I think Ankit is there. Look at that.
Hi Priyanka. Hellollo Siva. So up to new, how many months will the new batches start Siva?
There will be new batches starting even later on, but the time duration of those batches will be squeezed. Because if you join early, you end early. You join little late, you end little late.
Join little late, you end little more late. But you cannot keep pushing it, correct? Because then you'll reach 12th standard.
So After a point of time the batch duration will get squeezed, it will become more towards the crash course. So that's why I think you should join in August, that's the last time you should join. After that it will become condensed, I mean that will be like a late entry kind of thing. Okay, cool. Hellollo Remo, let's begin.
Now you might be wondering what is constrained motion. How many of you have seen this simulation somewhere and you know what it is? And look at this animation out here. This mass is sliding left and right.
It's connected by a thread over a pulley and a movable pulley and this block is going up and down and you can see the motion of this and this is related. If this moves the string moves, the pulley moves, the block moves. Okay, some of you have seen it.
Charita, I'll be conducting a lot of micro courses but think about it this way that as if my replicas or teachers better than me or as good as me only are teaching. Okay, so we do not differentiate. Alright, some of you know this.
So this is a piston cylinder arrangement which is there in your engine. So your engine, car engine, bike engine, all of this has this arrangement. This is constrained motion.
This is an example of constrained motion. Thank you Madhula Bachcha. Hi Kunj, welcome aboard.
So what exactly is constrained? Are you getting a sense or a feeling of what constrained is? Constrained means, you know, restriction. I say, Madhula moves 2 meters or 2 times more than Ankit. Madhula moves 2 times more than Ankit.
So, if Ankit moves 1 meter, Madhula has to move 2 times more. That's a constrained motion. The motion is restricted to some other person's motion.
That is constrained motion. So, you can see the motions are restricted, constrained in a way. If this rotates, this has to go up and down. If this moves, that has to go up.
or down that's what it is all right yes so this is what you should understand constrained motion is nothing but motion which is restricted in some or the other way okay look at this simulation look at this definition by the way you will be getting the pdf you know after the session is over as well and it you should consider joining the telegram group over there so that you will get this pdf as early as possible. Generally, you will see the PDF in a day or so. Now, there are different kinds of constraint motion.
There are different kinds of constraint motion. Now what are these kinds of constraints motion? First of all, some constraints involve pulleys and string, some will involve rods, some will involve wedges or you know, incline planes.
We are going to do all kinds of problems today. First, let's start with pulley related problem. Now you might have seen a crane or a lift where you know there is a pulley which moves and it lifts heavy load and there are multiple pulleys arranged in some way.
What's the reason for that? Does that make the job easy? Yes, it does make the job easy of lifting that load.
How exactly will come to that? But look at this, this bucket of water is being lifted by some pulley arrangement out here and when this block moves down or up, you will see that bucket also moves up. or down. Now there is one thing which is very interesting.
If you observe this was okay this was the original position or the initial position of this block. This was the initial position of that block. This block and this pulley moved down somehow okay it moved down. This bucket has moved up.
This bucket has moved up from its initial position but the amount by which it moves is different. than the amount by which this block comes down. There is some relationship between them.
Constraint means restriction Madhula. Okay, it's some kind of restriction. That's all.
Okay, so you can see the motion of this bucket is restricted or related to the motion of other things involved in the system. It could be any kind of restriction or relationship. So, we are here to study. how to figure out the relationship between you know different objects, pulleys, strings, blocks moving relative to each other.
That's what we are trying to study. Now whenever you have pulley block problems which involve constraint motion basically understand the pulley is moving then remember one simple rule okay and the rule is the length of the string remains constant. the length of the string remains constant. GTR and pseudo force will be taught next week.
Concentrate on the class right now. Alright, so think about it this way. If I have a string, alright, if I pull it here, this end will also move because the string's length will roughly remain constant. And in J, in IIT, in J-e-main, J-e-advance, with SAT, CET, we are going to assume our strings are, you know, not flexible. They...
cannot extend they cannot extend themselves so the length of the string is constant that's our assumption which is very fair in daily life also when you pull a string it will hardly elongate by a millimeter or so which is negligible considering the length of the string okay now knowing this this rule follows the first statement automatically and this will help you solve the problems Read this rule carefully because we are going to use this rule for solving the problems. Okay, the magnitude of the velocity or the displacement along the length of the string at the ends are same. The magnitude of the velocity or the displacement, all right, of the string towards the end along the string are same.
What does it mean? Let's take an example. Let's say I have a string.
Okay, look at this everybody. This is a string, it could be anything. It could be of any shape. It could be over a pulley.
I've just taken an example. I've just taken a straight string. I could have wrapped it over a pulley multiple times as well.
Okay, the length of the string is constant. So I know if I pull the string here, that end should also move. Now, if somebody pulls the string somewhere here and pulls this end somewhere here, it's possible. It's possible, right? I imagine this is a string.
I pull this point here, I pull this point here, I rotate it or I pull it in some way, weird way. So these two ends have these two velocities. Now what you should do, take the component of these velocities along the string. You have learned vector splitting, vector resolution in vectors.
So this velocity has two components, one here, one here and the same one with this guy. Look at this, this is the velocity component here. There will be one more component that way. But do not worry about it. Same way this velocity has two components.
I'm not worried about the perpendicular component. I'm only worried about the component which is along the string. That's what is important. The component which is along the length of the string at the ends.
These are the two ends. Now just let me tell you one thing. If this is v1, if this is theta1, If this is v2 and if this is theta2 then think about it the resolution of this vector all right in this direction will give me the magnitude because this is closer so this component will be v1.
cos theta 1 correct that would have been v1 sine theta 1 this is cos this is closer cos is adjacent all right now this component over here will be nothing but v2 this is adjacent so cos theta 2 i am not worried about the other component now these two components have to be same so this will be equal to this as simple as that i hope this is clear The string should move alright with the same velocity along its length because if one moves more and the string will break understand or if one moves more the string could even slack that means it will become loose so if the string is tight if this moves one meter this has to move one meter it's non-negotiable understand that. I hope this is clear yes everybody clear very good so that was cost pass csv1 No, v1 is not equal to v2. v1's cos component is v2's cos component.
Don't get me wrong. I never said v1 is equal to v2. No, v1's component is v2's component along the string. That's why I have highlighted this word along. Okay.
Yes, Sharma ji ka beta. We enthuse J advanced is also taught. Like you can see every nurture class or rather. After every chapter, I am doing you know, advanced level problem solving special session only for advanced.
But even in our regular sessions, I do involve advanced problems without telling you this is advanced or else you will develop a mentality. Oh, this is advanced problem. Maybe I will not be able to do it. But we are taking up all the concepts.
And by the way, mains and advanced syllabus is roughly the same. So it's not like, oh for advanced I have to join something else, for mains I have to join something else. Only the level of questions, pattern. and thinking is slightly different that's all okay cool let's move ahead and let's use this concept in this question let's see how do you do it yes yes dinesh i have taught resolution of vectors previously check it out in the playlist bacha you will see nurture series there is a chapter called vectors in the playlist watch it i have taught it okay so right now we are in newton's laws i've already completed units and dimensions vectors basic maths derivatives integration 1D, 2D, relative motion.
Now it's Newton's laws lecture number 3. Okay, so going ahead, here is the question. This pulley, this pulley, this pulley, all are fixed. This mass has a thread attached.
It goes over these pulleys and attaches to M. Question is, what is the relationship between the velocities and accelerations of these two people? Okay, now here is the thing.
First of all, think this way. Where will M move or where will M2 move? They have only one way to move, they can go up or down. M also can move either up or down, it's obvious. Now I don't know which one will go up, which one will go down because the masses are not given.
So assume something. Anyways we just have to find relationship between them, we don't have to find the actual values. So just assume something.
Let's say M, okay let's just assume M goes down by let's say v1 velocity. If this goes down, the string will be pulled. The string will go over it.
If it gets pulled from here, the string will come like this. And then the string will go over this like this. So M2 has no other option. It's related.
It's constrained. It's restricted. Because it's connected via a string over the pulleys.
So M goes down. M2 has to go up. Undoubtedly.
This is non-negotiable again. So this is V2 guys. This is V2.
Now tell me one thing. Yeah, tell me one thing guys. Is, is you know there is there one string or two strings?
Look at the diagram carefully. How many strings are there? Only one string is there and what was the rule I told you? The velocity of the endpoints. The velocity of the endpoints along the string should be same.
Now this velocity is completely along the string. Even this velocity is completely along the string. There is no another component or there is no separate component along the direction of the string.
I hope this is clear. So, hence can I say since velocity of the end points of the end points all right along the string are basically same along the string are same. Hellonce, can I say V2 should be equal to V1.
Everybody getting this point? Yes, V1 should be equal to V2. As simple as that. Yes, Gaurav, this is an English medium channel, bacha.
Like you can see the channel name itself. is Vedantu JEE Enthuse English. Okay. Arun, there will be a mock test for JEE 2021 on Thursday morning, 9 o'clock. Okay.
Now, next part of the problem. We have found out relationship between velocity. How about acceleration?
We all know acceleration is rate of change of velocity. So, differentiate both sides. Okay.
Differentiate both sides, guys. with respect to time. So dv2 by dt will be dv1 by dt. Now what is derivative of velocity? That's acceleration.
So a2 will be equal to a1. As simple as that. If velocities are equal, magnitude wise, accelerations also have to be equal.
And every time the velocities are same. So their acceleration will also be same. Is that clear?
Okay, so Kalyani looks like you have missed a lot. but it's okay you can watch it from lecture one but trust me attend today's class live you will slowly get some idea about it and probably that will help you in lecture one and lecture two as well cool let's get going kishore bacha leave your comment after the session is over that i'll definitely reply to your comments whatever counseling you need okay now let's concentrate on the current session okay so the answer was correct v1 is v2 a1 was a2 this was simple problem let's take a slightly more trickier problem and see what happens. Oh by the way there is one thing which I wanted to tell you and lot of coaching institutes teach that.
I'm not saying that's wrong. I'm just saying it's not needed. Now what is not needed?
Listen to me. This is called the derivative method. Derivative method.
Now whatever method I told you remember I just gave you a theory slide. Just take the velocity of the endpoints along the string, equate it, done, done, done. It's simple visualization, nothing else. Now there is a derivative method also. Now what is the derivative method?
It says write the length of the string in terms of variables by assuming some variables of, you know, one block to the pulley or pulley to pulley or whatever, you know, dimensions you can think of. Like from mass to this pulley, I can assume this as x2, example. I can assume this as let's say x3, I can assume this as x4.
So what is x3? x3 is this much, okay, x2 is basically this much. And then x4 is nothing but this much. And then x1 is nothing but this much.
Now, the reason why I've done it is because I can write the length of the string as x2 plus x3 plus, what is that? x4 plus x1. Correct?
This plus this plus this plus this. You might neglect that small length or even if you take it into account, it's I mean it won't matter you will see because we're going to differentiate it anyways. Now look at this differentiate, differentiate with respect to time okay on both sides.
We can take derivative on both sides. Now observe what will happen. dl by dt will be dx2 by dt plus dx3 by dt plus dx4 by dt plus dx1 by dt.
Now, think carefully what is dl by dt? What is the total length rate change? What is dl by dt? Thank you Nikhil Krishna. What is the derivative of the length of the string?
What is the rate of change of length of the string? Come on my warriors figure this out. Think about it. Length is constant. Did I not tell you just some time back the length of the string never changes.
So the derivative should be 0. So LHS is 0. Okay beautiful. X2 rate of change of X2. This position rate of change will give me velocity of 2. Because this block will go down or up. So rate of change of X coordinate or whatever that coordinate.
Will give me velocity of 2. Rate of change of position. Isn't that right? No problem, Suchi.
X3, the distance from here to here will never change. No matter how much the string goes over these pulleys, this distance between the pulleys, remember they are fixed, so that will not change. So, that should be 0. Same thing applies for X4.
That's constant. The distance between the pulleys never changes. The length of the string between these two pulleys never changes. So that's also 0. And the last term, dx1 by dt. change of x coordinate.
x1 is not constant as this block goes up and down. Hellonce, that should be the velocity of object 1. Now rearrange and you will get v2 is equal to negative v1. Now it just means that if one is moving in such a way that the length is increasing, the other is moving in such a way the length is decreasing. So if one goes down, the other goes up, but the magnitudes are same.
That's all. The magnitudes are same. Do you Think you will be using this method in the exam?
What do you guys say? Would you prefer my method or would you prefer this method? Because I want to tell you about all the methods undoubtedly because I'm a physics teacher I'm supposed to tell you all the different things but I'll tell you what will best work in the examination and which will always work in the examination.
So, that is the method which I taught you okay cool. So, this method can also be used in case you want to try out different approaches obviously. Alright so anyways you're going to get the same answer is just that you might get some plus and minus signs and you might get confused what does it mean but just understand if it is plus it just means that the length is increasing if it is minus and it means that the length of the string is decreasing that's all it says.
Okay so canal circular motion is a separate lecture separate session separate chapter but in your ncrt is divided into multiple chapters So that's why the confusion if you refer to HC Varma, you will not get confused So guys, let me tell you one hack whenever you are attending my classes or Vedantu classes Just read HC Varma before and come and even after the class read HC Varma just follow the flow of HC Varma That's the best book for theory and your starting level problems for J mains and advanced and even NEET so HC Varma is the Bible or it's the Bhagavad Gita or it's the Quran for all of you you know who are studying physics for all the competitive exams that's the book you should refer don't refer anything else for you know the basics the understanding basic problems medium problems and to develop strong concepts that's what you should do okay now let's look at this question so here you have to find the acceleration between the blocks m and 3m and now let's use my method and see what do I get. Okay, let's do this. First of all, the length of the string should remain constant.
Okay, now there are two strings. One is over here. The second one is over here. Now, just visualize this. Let's say this 3m moves up.
If 3m moves up, the string will go over the pulley. It will go down and as it goes down this pulley will also go down. Correct? This pulley will also go down. If this pulley goes down this string along with this mass will also go down.
Let me draw the new position of the entire setup or arrangement. So let's say the pulley goes down like this. So the string comes down like this.
Okay maybe I'll show it a little bit more. so that you get a feel of it. So there is some extra string coming over here. There is some extra string coming over here. Okay, this pulley is attached to this mass.
So it goes like this, maybe and this mass has now come over here. Okay, this mass has now come over here. Now, think this way. If this mass, if this mass goes down, Okay, it goes down by let's say x.
It's just all about keeping the length constant. That's it that I'm doing. Observe, if this mass goes down by x, how much did the pulley go down by? Quickly let me know. If this mass has gone down by x, how much did the pulley go down by?
Quickly let me know in the chat box. How much did the pulley go down by? Obviously x because the pulley and this mass are directly connected to each other. So the pulley has no other option but to go down by x. Now think there is a requirement of string at the dotted positions which I have shown.
Earlier there was no string in those dotted positions. Once the pulley comes down you need string here and here. How much is your requirement demand like I want the string. How much is the requirement? How much string do you require?
Quickly let me know in the chat box. Let me know. You need string here and string here as well.
Excellent Madhula, very good Dravid. So Siva, it doesn't matter. Even if 3M is more, M is more, it doesn't matter. Even if this goes up, this goes down, the relationship will hold true. If A1 is 5 times A2, it will still hold true whether it goes down or goes up.
It's independent of these masses. So it doesn't matter. Yes, so think about it. This pulley with this mass came down. So the moment it came down, understand you need extra string here as well as here.
So this pulley has come down, think about it. So you need string here and here. So the pulley comes down by x, you need x and x string over here.
So that means the requirement is 2x. I hope Suchi Mama you got it. Now, here is the thing.
Now here is the thing. If that is what the amount of string you require, if that's the amount of string you require, where will you get it from? This point won't move. This point is not going to move.
So in order to meet that requirement, understand you need it from here. So this 3M will have to go up. This 3M has no other job or it has no other way but to move up by 2X.
Understand? Yes. Okay.
So Charita and Suchi and others observe this again. Okay, this block, okay, was put over this pulley, this string and this roof over here, the moment this pulley goes down a little bit, you need string here and here, right? Because understand the string is wrapped over it.
So the moment this goes down, you need string here and here, where is this string going to come from, you need x string here, x string here. So that two x's are going to be demanded from here. So this needs to move up by 2x.
So that 2x is going to flow and that 2x will get distributed x here x here. So that comes down by x. That's what happens. Yes.
So guys, the final answer is look at this. If the mass m comes down by x, 3m moves up by 2x in the same time. So hence, if Velocity of m is v, the velocity of 3m will be how much guys? Velocity of m is v, how much will be the velocity of 3m? It will be twice the amount in the same time, so it will be 2v.
Isn't that true? I hope this is clear. Similarly, similarly, think this way, if acceleration, if acceleration of m is a, what will be the acceleration of 3m guys because the velocity of 3m is double so hence acceleration of 3m will also be double of a that's what it is a and 2a so i have got the relationship from here then so v3m is 2vm and from here a3m is 2 times of am so this and this is the answer by the way Let me also tell you other way. See, lot of you are saying, Siva, 3M will go down, M will go up.
It doesn't matter. Think this way. If M goes up by X, think the position which I have shown here, which I have drawn here is the original position. If this block goes up by X, don't you see two new strings on both sides will be free loose.
Imagine this pulley goes up. This string and this string will suddenly be loose. So, in order to tighten the string, you need to pull up. You need to pull this string up and you need to pull this down.
So, 3M will go down by 2X if this pulley goes up by X. That's another way of looking at it. Yes, I hope this makes sense.
So, if this pulley goes up, you will see strings will be freed on both sides. This goes up, so you have two new strings. And where will it go?
It will go up. and it will come down. So anyways the answer will remain same whether it goes up or whether it goes down.
Okay is that clear? Excellent. Yes you can do whichever way it doesn't matter.
Remember constrained motion has got nothing to do with these masses. Their relationship if this is twice of that their relationship is not dependent on the mass. When did we even consider this is 3m and this is m?
Did we even consider that? Never. So it's not even dependent on the mass. So constraint motion is based on the geometry, the arrangement, the pulleys, which one moves, which one doesn't move.
So that's what decides the relationship. Now look at this. This simulation makes it very very clear. Look at this. You have basically taken the same arrangement.
It is a mirror image. So if you pull this down, that goes up. That's what is happening. Okay, as simple as that.
As simple as that. So observe this. Observe this over here.
Oops, once again, I'll pause it at the right moment. Okay, so you can see this is like a block and If this block goes down, this has to go, this string has to go like this and this can only happen if this pulley goes up. Now the moment this pulley goes up, do you see the string here and the string here will be free.
Understand this. This part is the same, okay. Just that this part and this part will be free.
So if the pulley over here goes up by x, you have 2x string which is free. So that 2x will flow over this and this will come down by 2x as simple as that okay i hope arun kumar now it's clear okay this is not so easy you have to have some visualization bacha and you need to think also on your own you have to visualize if you don't visualize then you will not get it i mean you have to visualize put a effort sincere effort to visualize this okay cool so let's get going ahead into the next question okay Clear says Siva, very good Madhula, very good. Samaja, alright. Arun, clear oh, Roshni, very good.
Don't worry, you will get hang of it when we solve some more questions. Look at this question. Now you will be like, sir, you are going to kill us today or what?
Okay, so, by the way, you might be wondering, sir, what is the use of these multiple pulleys? You might be wondering, what is the use of that? Okay, you will come to know about it. First look at this, look at this arrangement and these three pulleys are fixed.
These three pulleys are free to move. Okay, and as this guy pulls, this entire arrangement goes up and there is some mass which is being lifted up. Usually, if there is let's say 100 kg mass, you need to apply 100 kg force to pull it up.
But... with the use of multiple pulleys trust me you don't have to apply that much force with less force you can pull it up you don't believe me let me go take you back to the previous problem okay let me take you back to the previous problem just one second okay look at this this entire thing is one string right this entire thing is one single string cool now if the tension in this thing is t This will be T here, T here, T here, T here, everywhere it will be T, T, T, T, T, correct? But this string over here, this string over here, look at this.
If you look at this string, it's different from this red one. This string will not have the same tension. In fact, if you look at the free body diagram of this pulley, you will see there are two tensions pulling it up.
There are two tensions pulling it up so it should be balanced by a tension 2t from below so this string has twice the tension as the red string i hope this is clear now think about this guys if this green string has twice the tension of the red string that means by using multiple pulleys i can somehow create double triple four times the tension which means I can amplify the forces. I can amplify. Amplify means I can enlarge.
I can make it more. I can, if there is T force here, it suddenly becomes 2T because it has been wrapped over the pulley. That's the advantage.
So let's see how this concept comes to play over here. Okay, over here, what exactly is happening? And by the way, when I showed you this simulation, also the exact same thing was happening. Observe this. Observe this.
Just one second. Look at this. You are applying T force. How much force are you applying? T force.
The same string is continued over here. This is also T. This is also T. Now T and T will become how much? 2T.
So indirectly the force gets amplified and you can easily lift it. You can easily lift it because you are applying let's say 5N force. By the time it reaches here, it is becoming 10N.
That's the best part about it. Alright. you'll be like sir this is very good there must be some loss well there is some loss and what is that loss see everything you'd rather you'd You don't get everything in life.
When you gain something, you lose something. Or when you lose something, you gain something. Hellore, you got extra force.
You applied 5 newton, you got 10 newton. Okay, what did I lose? You lost displacement. Look at this. So, you know, if this goes up by x, this has to be pulled by 2x.
I just told you about that. This has to be pulled about 2x. And that's...
what you should understand is what you're losing so if this block goes by x you have to pull twice the amount but with lesser force isn't this something like your geared cycle example it's a loss in a way i mean not real real loss in terms of energy but you get something you lose something in a way so you have to apply more you know displacement in a geared cycle in first gear i hope you have ridden geared cycle some point of time in your life In the first gear you pedal more but the effort is less. You're okay with that because you have to apply less effort and you have to pedal more but when the road is fine and you're going down the slope or the road is flat, no traffic, you want to go with full speed, high speed, you want to pedal less, the effort will be slightly more. So that's what happens. So this is how you know things work.
So this is like your lever principle in a way. you might have talked about or heard about the lever principle lever arm momentum sorry effort arm so you know that effort arm is large the effort is less and all of that in your seventh grade or eighth grade long long time back okay anyways now let's see what is happening over here let's see what is happening over here so this spandu is applying some force and Look at the strings, there is only actually one string over here. You can lift this heavy load by applying less force, believe me.
And let's do this question. Let's do this question. Okay, let's figure this out. If this Pandu applies some force, that means that force which he applies will be translated to tension.
Whatever force you apply on the string, that would be the tension in the string. That same tension is there throughout this string. This is one string.
So t t t t t t. Count the number of t's over here. Just count the number of t's over here. Charita Jai already said something.
Come on, come on, come on. No, if you connect the string over here. then I mean you can do that obviously probably that will give you a different answer that's it 60 very good now if there is 60 now just observe this system just observe this system over here okay this is my system okay so if there are six t's pulling it from the top how much t will pull from below obviously 60 again if there are six t's pulling from the top this is a massless pulley light pulley so the net force should be zero so there will be 60 over here and that same 60 will be pulling up this 120 kg the same 60 will be pulling up this 120 kg is that understood everybody clear this entire system the net force upwards is 60 so the force downwards will also be 60. that 60 is being transmitted to the 120 kg block hence Hellonce, 6T will be 120, this is the mass, so into G.
Therefore, T will be equal to 120 by 6, which is 20G. Therefore, the force which the Pandu is applying, force by Pandu, how much is it? It is 2, 0, 0 Newton, which is 20 kg force or kg weight. Look at this, this is one 20 kg object but Pandu is applying only how much force?
20 kg force. Hello feels as if he's pulling 20 kg block. Hello'll be singing a song, he'll be dancing, he'll be talking to people, he will be chilling out.
Hello won't even feel that he's lifting 120 kg. So that's the advantage. I hope you see that. Isn't this interesting guys? Now are you getting a hang of it?
Now are you getting a hang of it? Is this little bit simpler now? Shall we do the second part of this? So the question is you know what is the relationship between them?
What is the relationship between them? By now you should have guessed it. By now you should have guessed it.
How many of you can tell it? Quickly let me know. If the force multiplies n times the displacement decreases and times that's the logic yes you had another pulley probably it would become 8T so 120 by 8 yes think about it if force multiplies n times understand the displacement will decrease n times so over here look at it this way if the block of 120 kg gets six times more force when pandu pulls by let's say T force or some kg force so that means the displacement should be less. So if this block if this block goes up by x, Pandu has to pull by how much?
6x that's the answer. 6x. It's not directly it's inversely proportional actually okay it's inversely proportional.
If the force is six times more the displacement is six times less. So it's in a way inversely proportional yes. It's inversely proportional to the displacement.
So, Pandu has to pull 6 times more. But it's okay. Hello's listening to his songs and he's applying less force. So, he's okay with that. Hello's okay to pull it for a longer time with less force.
Hello's okay. Cool. So, if this block goes up with velocity v, this is going like this.
With how much velocity? 6v. That's what the answer is. Is that understood clear?
Now, I have told you this. using the logic of inverse of you know tension and displacement but even otherwise I can tell you even with visualization even using my technique I can tell you how look at this I just erase this now I don't need this anymore I just told you using the logic of forces amplification but you don't need that logic every time how do you do it observe this observe this let's say this pulley goes here Let's say this pulley goes here, this pulley goes here, this pulley goes here, like this center has gone here, this center you know has gone here, this center has gone here. So how much will the block go up by?
Think this way, how much will the block go up by? Let's assume the block goes up by x. The block goes up by x, the platform also goes up by x, the platform also will go up by x.
So think this way guys. The string is now here like this. The string is here. So this platform, the string, the block and the pulleys all are going to move up and down by the same amount.
So would this be also x, x, x, x, x and x. All these are x, x, x, x. How many x's you got?
6 x's you got. Everybody clear? So Understand earlier there was a string here completely.
Now the string is only here. Earlier the string was here. Now the string is only here.
So what happened to this part of the string? What happened to this string? What happened to this string? All these 6x strings are free.
And to make the string tight, pull this by 6x to make it tight so that the string gets wrapped or else it will hang over there. Correcto. you will anyways get the same answer.
So this is x, this is 6x. So if this is v, this is 6v. If this is a, this is 6a.
If this block is accelerating by a, you're pulling the string with an acceleration of 6a. That's what is the answer. Yes, the force gets divided equally among the string in a way. Yes, Suchi. Is that now clear?
Everybody getting a hang of it? Shall we go ahead? Uma you'll need to wait bacha first understand the concept just don't wait for menti you are now grown up i know till 10th standard every time you used to have menti and used to have fun but i know you're still going to have fun but you need to understand concepts j cracking j is not an easy task you need to have a lot of patience effort dedication focus on the concept and not just the menti quiz trust me we'll have menti quiz but hold on let's learn the concepts first all right okay so that was the answer we just got it 20 kg force or 200 newton and the Pandu has six times the velocity or pulls the string six times the velocity of the 120 kg block.
J and Ram we have not yet discussed power or the work method all right when I teach you work and all of that probably it will be a good time to talk about it. Right now we are in Newton's loss. Now let's look at this question coming up on your screen beautiful beautiful question coming up on your screen now this is slightly different there is a ring which can slide like this.
It can slide like this. And there is a thread runs over the pulley and this block goes up and down. This is similar to this.
Question somewhat similar okay somewhat similar this you can think of the ring and this thread goes over a pulley and that block goes up and down somewhat similar to this situation now try to visualize this particular problem in that manner so if the string goes here this sorry if the ring goes here the string will go over the pulley the block comes down but their velocities will not be equal now how do you figure this out Use my rule, trust me you will not get it wrong. So here is the logic. Let's assume let's assume that this guy the ring moves with a velocity of va.
Isn't this the end point? Isn't this the end point? What is the velocity of that end point along the string?
This is theta. So the velocity along the string won't it be va cos theta? Won't it be Va cos theta because that is what is important. Now the same string is going here.
This is Vb. Vb is going down and the component of Vb along the string is Vb only. It's not different.
Vb's component is Vb along the string. So no need to split that. And what did I just tell you?
The component of the velocities of the endpoints along the string are the same because the length of the string is the same. Hellonce, can I not say this that VA cos theta is equal to VB VA cos theta is equal to VB and differentiating it taking the derivative taking the derivative differentiating it I will just get a acceleration of a so a might be accelerating like this okay acceleration of a. and this is acceleration of B.
Cos theta is nothing but acceleration of B. But in the, okay the first question was done. Relationship between the velocities done. Relationship between acceleration when theta is 60. So put theta as 60. So cos 60. So acceleration of A cos 60 is equal to acceleration of B.
So acceleration of A by 2. acceleration of B. So this is one answer this is my another answer. So two answers I have got everybody fine till here.
Everybody fine till here shall we go ahead and solve the third part. Now the third part is the tricky part. Let's do this. Okay so you can also write it as A.
A is twice A you got it either ways. How to do the third part the tension in the string and the value of the acceleration. Please understand, I don't know what is the value, whether it is 3 meters per second square, 5 meters per second square, I don't know that.
I just know this, this guy's acceleration and this guy's acceleration are in the ratio of 1 is to 2 or 2 is to 1. Okay, how do I do that? Use Newton's laws, that's what I have been teaching you. Draw the free body diagram.
Okay, very simple. This mass will be, what is it? The ring A has a mass of 1 kg. So, this is 1 g here. There will be a normal force because it is resting on a slider.
Okay, that's a ring, right? And it can slide over it. So it's resting on it.
So there will be a normal force on the top. Okay. Also, there will be tension over here. Look at this, because it is pulled by a string. Apart from that, there is nothing, there is no friction and all of that.
The same string continues. So the tension should be same. And just that there will be an additional weight over here. Block B's mass is 2g. So 2G over here, everybody fine I guess.
Yes, Siva you can take a screenshot and I actually do not call this as a session PDF, I call it as a session workbook. You know whatever I write doesn't come in the PDF because of the functionality of Google Slides, there is no provision. If that was there I would have done it but these are more like workbooks.
You can always watch the recording and write down the steps plus it's better you write it down by your hand. because then you will remember it better. Okay, let's write the Newton's law equation for block B as well as block A. Let's write it down. Observe this now.
Okay, so for block A, will I write an equation in the x axis or should I write an equation in the y axis? In the y axis, there is no motion only. There is no motion.
So why would I waste an equation? Instead, because it is moving in the horizontal direction, I'll only worry about horizontal forces and accelerations. Okay, now observe.
Is normal in x? No. Is weight in x? No.
Is tension in x? A part of it. How much tension?
Tensions cos theta. So, for object A in the x direction, I can say tensions cos theta which is T cos 60 will be the mass which is m which is 1 into acceleration which is A. Therefore, I will get T by 2 is equal to A.
Everybody fine till here? Tension is nothing but you know tension by 2 is nothing but acceleration of a but hold on I hold on a a is twice a b I just got it over here a a is twice of a b I can use that substitution so therefore t by 2 a a is twice of a b so 2 a b therefore t is equal to 4 times of a b okay this is equation number one now let me write another equation but for this block because I cannot solve this any further. I do not know the value of tension. I know I have to find tension, but I can't say tension is 4ab.
What is ab? I cannot find it out so easily unless I write one more equation. So for B, so for B block in the y direction, okay, for the B block in the y direction, I'll write the equation.
It's going down, it's accelerating down. So bigger force minus smaller force is mass into acceleration, Newton's second law. So let's write it down.
So 2g minus tension is mass into acceleration of B block. Interesting part is tensions value I know in terms of B. So how about substituting that over here. So therefore 2g minus 4ab is equal to twice of ab.
Now going ahead, shift this guy over there, you will get 2g is equal to 4 plus 2 is 6 times of ab. Therefore, acceleration of B is nothing but g by 3 which is 10 by 3 meters per second square. So, this is the answer that I was looking for. I found out the value of acceleration of B. Done.
Now, in Suchi Newton's law equation, I hope you were there in the previous lecture. So, I have just used net force is equal to mass into acceleration. That's all I have done.
But, since this ring is accelerating, horizontally only consider the forces in horizontal direction. This block is going in vertical direction consider forces or their components only in vertical direction that's all I have done. Okay so that's all you need to do that's all you need to do.
Great so you can see t cos theta tensions cos theta is mass into acceleration. For block B weight minus tension that's the net force remember always on the LHS net force is there okay. So you need to account for the signs and the directions.
Net force is mass of B into acceleration of B. That's what I have done. Once I know A, I can easily find A.
So what is A guys? Okay. So A, acceleration of A from there will be acceleration of B into 2. That means acceleration of A will be, you know, 20 by 3 meters per second square. So that is the second answer that I was looking for.
Okay, the last thing, tension. Tension is 4 times A. Tension is 4 times of A. So, using that, I'll get 4 times A. What is A?
It is 10 by 3. So, tension will be 40 by 3 newton. So, that's my other or last answer. Okay, that's how you solve these questions.
Thank you, Mutu. Understood or clear or? I hope you guys got a hang of this.
Done and dusted. Yes Priyanka. So these are the ways in which you are supposed to solve the questions.
It's a beautiful problem. Trust me, if some You claims and tells you that watch my lecture on constraint motion or watch any physics lecture of mine and you will score full marks or you will be a topper, then he or she is lying. It's not possible. The only way, the only possible way to solve this problem is to watch my lecture on constraint motion.
You can score well, you can improve in physics is when you watch the lectures immediately after the lecture read through the theory, practice problems and when you get stuck also look at the problems which we have solved try to do it on your own and then again start doing the problems and ask your doubts you know have discussions with your friends and put a sincere effort in solving the problems that's a slow and steady process and that's you know the only way of getting good marks If somebody has told you there is a shortcut, you know, you just watch my lectures, that's it. And you can go ahead and sleep in your bed after my lectures. Hello or she is for surely lying.
okay so those are the answers like we were talking about it yeah yeah darshini you can use lamis theorem but it's of no use lamis theorem is only for equilibrium bacha this is not even in equilibrium so lamis theorem is of no use and by the way lamis theorem is a part of mathematics okay so some of the physics teachers teach you in 11 standard but it's of no use because you are anyways going to learn it in solutions of triangles and all of that in mathematics okay cool Now let's look at another question. Look at this. This is another beautiful example. There are two blocks tied to another block.
As this block goes down, these two blocks go up. Beautiful question and these kind of questions have been asked in J-Mains. Okay, look at this.
Look at this. So, N's P and Q are being pulled with speeds U and U. U, U. This block goes up. So, ends P and Q are pulled with speeds U and UH.
Okay, all the other pulleys are fixed. In case you wanted to see the visualization once more, look at this. So, in this case, it was going up, but imagine this going down.
If this goes down, it goes up. That's what happens. So, P and Q go down.
This block will go up. What's the speed of M? Let's see.
Etude machine Ayaman is, you know, when you have two blocks, one pulley. so one goes up one goes down that's eight one this is not exactly eight one okay let's see how many of you get this correct i'm pretty sure many of you are going to get it wrong i'm seeing how many of you get it correct guys come on my warriors figure this out all the four options are given to you figure it out which of the a b c d options are correct a says i remain interesting come on come on come on figure this out Madhula saying 2u by cos theta, Praneet saying c. Interesting.
Jaswant, I'll be there in the micro courses. In the long term batches, I'm not there but other teachers are there. Remember, I can't be there everywhere.
But think of it this way, replicas of me or teachers better than me only are there. So take up the long term batch, no matter which teacher is teaching, all are very well educated, very well experienced and amazing teachers. and I'll be there in the micro courses special lectures special topics special courses I'll be there and you will get all those things for free okay okay so people are saying B and C I'll tell you what observe this look at any one string there are two strings here look at any one string for example this one okay this string is being pulled down with you so this end should also have the same velocity along this direction.
Now, let's just assume that this M is going up, it's going up with velocity V. What did I tell you? Velocity along the string, along the string, okay, at the ends, at the ends are same. So what is the velocity at the end along the string?
That is u. What is the velocity along the string at this end? That's v's component. How much is v's component? That is v cos theta.
What is asked? v is asked. So v will be u by cos theta and hence the answer is b. The answer is b. That's it.
No, it is not 2u, Praneet. So this string is separate, this string is separate. So lot of people think it should be 2. No, there is no need of 2 because there are 2 different strings.
This string is different. This string also will have the same law. Think this way, for this string, you know, it's not, don't think that it is connected at the same point.
These are 2 independent strings. So if this goes up by v, v's component which is v cos theta will go here and that will be also equal to u. So v cos theta is again equal to u. so v is u by cos theta it will not get added it is supplying two strings at the same time but they are independent of each other they are two separate strings keep this in mind okay so lot of people mistake it for two and some people also mistake it for whether cos is on the top or whether it is in the denominator that's a common mistake you can see the answer is b now let's go to different kinds of questions but before that let me tell you if you need more problems for practice In the Vedantu Pro subscription, you will get all the assignments after every class, not chapter, every class based on what is taught.
And you can attempt this test or assignments. again and again for practice it is all recorded so whatever answers you give and let's say you solve the assignment let's say now in august and maybe in september you feel like okay maybe i don't know much about constraint motion maybe i have forgotten it so let's revise you can re-attempt it and all the previous answers will be erased or it will be kept out and you can see how you perform this time and you can compare your results so you can again and again revise all these assignments chapters concepts thoroughly and you will also get regular test series regular test series where you know the same way you can do analysis time analysis accuracy analysis strike rate analysis you know how many questions remain unattempted all this gets recorded in a beautiful way and all this can be analyzed for checking out your performance so guys if you are in the pro subscription please use these features and those of you are not in the pro subscription consider all these features because these are very important for additional practice you cannot just attend lectures you cannot just read random books you need to solve these assignments and give these tests and compete with real people who are aiming for j that's what is important now let's look at another type of question which is rod constraints now in rod constraints look at this it's the same way A rod will slide or move and the length of the rod will remain fixed. So same logic as that of the thread, the length of the rod will remain constant and you have to figure out what are the velocities at the end points. Okay now observe this. Do you think this point and this point always have the same speed or different speed?
Do you think their speeds are same every time or do you think this moves at a different speed than this speed, the ends? What do you think the actual speed is it same or is it different? I have a feeling I have a feeling that it is different and you can see that very clearly. Look at this when it is over here okay look at this this is hardly moving this is moving so much this is moving so much this is hardly moving. Now when it comes down you can see this is moving so much this is hardly moving.
So you can see their motion their amount of movement is different at different angles. So hence I need to find out is there any relationship for you know the velocities and the concept is again similar to that of the string. If there is a rod given and you know the velocities take the velocities component along the rod and these components should be equal.
The velocity components at the ends along the rod should be equal. Let's apply it. Let's apply this. There is this rod or ladder. It is sliding and let's assume that this is a top edge and this top edge is having velocity vt okay it is having velocity vt.
This bottom edge is having some velocity vb over here like this. As the rod slides down that point goes there this point comes down. What's the relationship between them as a function of theta?
Let's figure this out okay. Before that, let's extend the rod a little bit so that we are able to, you know, gauge where the components are going to be. Okay, this is along the rod, and this would be perpendicular to the rod.
These are components which are perpendicular to the rod. Okay. So what I'm going to do is split this velocity into two components. I will be worried about this component.
Okay, I won't care about this component. I don't care about this component velocity along the rod. Okay. Similarly, this velocity can be split into two components.
One is along one is perpendicular. I am interested in this component. I am not interested in this component.
Cool. Now this is theta. If this is theta, obviously this will be also theta.
If this is theta, this will be also theta. Okay, if this is theta, and if this is 90, obviously, this will be 90 minus theta. And if this is 90 minus theta, this angle over here will be theta.
Okay, I think everybody's getting it. Now, we all know the velocity at the ends along the rod should be same. Keep this in mind.
So what are the velocities along the rod? Vt is cos, no, Vt is sine component. Look at this, this is theta, this is adjacent, this is opposite or away from it. So this component is strictly speaking Vt sine.
sine theta and this component is VB cos theta. VB cos VB sine adjacent. So VT sine theta is VB cos theta therefore VT tan theta is equal to VB.
So that's the final answer. VT's tan theta is VB as simple as that. How many of you got it?
Did you understand how to do this question? So just take the components of the velocities along the rod rest all is geometry nothing else. Very good Samaja, very good Roshni, very good.
Excellente. Okay my god. Yes man you can get into IIT Bachcha, Sharma ji ka beta, you can definitely get it. I'm pretty sure other parents will give you give your example to their kids. Dekho Sharma ji ka beta dasvi me hi padhai chalo kara, IIT Delhi ke le tum kyun nahi padh rahe?
You will set an example I'm pretty sure Sharma ji ka beta. Okay so you got it correct let's get going to the next type of question and look at this look at this. This is wedge kind of constraints.
A lot of people get confused over here. I'm going to give you a simple trick by which you will not make mistakes. Yes, Ram Deshmukh, I'm from Pune, but now settled in Bangalore.
If there is some block, incline, which moves, wedge, wedge means like a triangular shape, okay, that's a wedge, okay, which moves, then something which is resting on it or something which is constrained to move along with it would also move. You can see If the yellow block moves with some speed, the red block also moves with some speed. Their speeds may not be the same. Okay, it depends on the inclination.
If the inclination is very small, then probably when the yellow block moves, and the yellow block moves, this red object will move very slowly if the inclination is small. So I need to figure out the relationship between them. Now, what is the way to do that?
It's very simple. All you need to do is find what is the point of contact. Okay, just check what is the point of contact and remember the velocity of the points of contact along the normal are same. What do I mean by that? Okay, let's take an example.
Look at this. Imagine the same situation. This is incline, this is that rod.
Okay, and look at the point of contact or points of contact. This is the region of contact over here. This blue object can move up and down.
This red object can move left and right. So this red object which moves that way okay which moves that way has two components of velocity. One is along the incline, one is perpendicular to the incline.
The red object which moves that way has two components one is along the incline or and the second one is perpendicular to the incline or perpendicular to the surface of contact. Similarly, the blue object is moving down. It's moving down. So, that velocity again has a component perpendicular to the incline and along the incline. Now, my point is this.
See, whenever you have two surfaces in contact, okay, whenever you have two surfaces in contact, the point on the red surface, for example, let's say it has two meters per second velocity. Then the point of contact on the black surface must also have 2 meters per second. Because if it doesn't then the surfaces will separate if their velocities are not same.
So they both must have the same velocity but perpendicular to the surface of contact. So the point over here is check the velocity which is perpendicular to the surface of contact. You'll be wondering sir why not along the contact? It's simple. See if the red one, if the red one, the red point on the red surface has some three meters per second here and the black one has some four meters per second here then it doesn't matter it will just slide over it.
They will not separate. So if they have to maintain contact so the sliding won't affect it. They should move together along that normal.
which is drawn along that perpendicular which is drawn. Is that clear? Thank you Empadu Babu.
See understand this. So whenever you have two surfaces, if both the surfaces have some velocity, they will just slide, they will not separate. So in order that both of them move together their velocities perpendicular to the surface should remain the same.
Keep this in mind. Now use this concept over here. The red point the red point on the wedge over here was moving with let's say velocity v1.
Okay let's assume this as angle theta. Let's just assume this as angle theta. So if this is theta over here.
Naturally, even this is theta. So that point on the red wedge will have a velocity here and a velocity component here. This component will be obviously v1 sine theta, the one which is perpendicular, okay, that will be v1 sine theta.
Similarly, the blue wedge or the blue object will have a velocity v2 over here. If this is V2, by the way, how much will that component be? That can be easily figured out. See guys, this will be 90 minus theta because this is a right angle triangle. So obviously, this is theta because this is again 90 degrees.
So that velocity which is perpendicular, the velocity which is perpendicular, how much it is? It is V2 cos theta. These two components will be same. That's what the concept is.
So hence V2 cos theta will be V1 sin theta. Therefore V2 will be V1 tan theta. So that's the relationship.
That's the relationship. Yes, Jessica, why won't you top it? Thank you, Ruti.
Okay, so did you guys get it? Did you guys understand it? How to do it? and the concept behind it. If the two surfaces should not separate or should not, you know, crumble onto each other, understand the velocity components perpendicular to the surface contact should be same.
Let's do a question based on this really quick. Cool. Let's do this. Look at this question. Look at this question.
Interesting question. Okay, observe this. There is a rod which can slide freely and because of its weight, visualize guys, try to simulate it in your mind.
Every time a simulation cannot pop up. So this rod will slide down like this. As it slides down, okay, it will push that wedge. This wedge is also having some mass, that wedge will move there. The question is what are the actual values of the acceleration?
Do you see that actually their accelerations won't be the same? Because this rod is not moving like this it's moving this way. So when the rod goes like this the wedge this triangular block moves that way.
So their accelerations might be related to each other. Okay observe this now. So let's take let's take a point let's take a point on the rod which is over here which is over here that point will be moving like this.
That point will be moving like this because it is going to slide between these two narrow gaps out here. So there is a gap over here and the rod is sliding in that freely. It's all smooth. It's given with what velocity?
Let's say V1. Is this V1 perpendicular to the surface? Yes, it is.
It is actually perpendicular to the surface. Everybody can see that. Now that is the velocity of the point on the rod. Look at the point. Look at the point on the wedge and see where it is going.
It is going in this direction because it is moving with the wedge. With what velocity? V2 velocity.
Now that velocity will have two components. Again it has two components and I am going to take components because remember the velocities along the normal, the perpendicular are same. So take the component in this direction.
What is that component? Well, that is not that difficult to see that if this is alpha, even this will be alpha guys. If this is alpha, okay, check this out. This will be 90 minus alpha and this angle will be alpha itself.
Okay, so this angle will be 90 minus alpha. So that will be alpha again. So what is this velocity then? Well, okay, oh, that is alpha. I need to draw one more thing.
Wait a minute. This is alpha, this is 90 minus alpha and this is alpha again. So v2's sine component, observe this. So this is actually speaking, this is v2, this is cos component, this is v2's sine alpha. Bonant.
Is that clear? So all you need to do is equate these two. So v1 is equal to v2 sine alpha.
Therefore, if v1 is v2 sine alpha, a1 will be equal to a2 sine alpha. That's it. So we have got the relationship between the acceleration.
We have got the relationship between the acceleration. Now once I get the relationship between the acceleration, I need to figure out what the acceleration is. Remember, I have not actually found the acceleration itself. I just found out the relationship.
Okay, I'll keep this aside. I don't need this anymore. I'll just remove this. The next step remember once I get a1 is a2 sine alpha. Okay, a1 is a2 sine alpha.
I need to apply Newton's laws. That is the only way. So draw the free body diagram. This rod will have its weight downwards.
and that weight will be mg. There will be some normal force between them. Okay, there will be some normal force between them. But this rod is accelerating that way. Okay, cool.
I'll keep it separate. Now what is mg's component in this direction? What is mg's component in this direction? Well, just draw this guys, just draw this perpendicular along and all that.
Remember, I had told you this rule, division of eight rule divided into eight parts. perpendicular, horizontal, vertical, along. So, we all know that this angle is alpha.
If this angle is alpha, this angle is alpha, this angle is alpha, this angle is alpha, this angle is alpha. So, what is the component of mg in this direction? It will be mg cos alpha.
Are you seeing it? This is mg that is alpha. So, mg cos alpha is in the downward direction.
Cool. There is one normal here also. So just put one normal.
And now what you will do is, so I will say for that rod of mass m, for that rod of mass m, this mg's cos component, because look at this, this is alpha mg's cos component. So mg cos alpha minus, why minus? Because the normal is opposing it.
So, minus n is equal to m into a1. Okay. Now, substituting a1 from there. So, mg cos alpha minus n is equal to small m into a2 sin alpha.
Cool. So, that was the equation I got for small m. Now, for capital M. For capital M, observe this.
The reaction of the normal will be this way. This is how the reaction of the normal will be. Okay.
There will be a weight of mass m downwards. There will be a normal reaction like this and again splitting these normal forces in different directions. Observe, there is one alpha over here.
This is another alpha. This is another alpha. This is also another alpha. So if this is alpha, this is alpha, this is alpha, this is alpha, this is alpha by our division of eighth method. Now this normal's sine component is giving that acceleration because these other forces are in the vertical direction.
So what will I say? Observe. Normals sine alpha component is the mass into acceleration. From this I'll get normal is m into acceleration by sine alpha.
Now I can take this normal and I will put it over here and let's see now what do we get. Observe this. So you will get mg cos alpha minus mass into acceleration by sine alpha is equal to m into a2 and there we have sine alpha over here. Now take this guy over there.
So what do we get? mg cos alpha is equal to a2 is common and I will get small m sine alpha plus capital M by sine alpha. Now, I want a2, right?
I want a2. So, what is a2? a2 will be this guy which is mg cos alpha divided by this whole thing which is small m sin alpha plus capital M by sin alpha.
So, that is the value of a2. Once I get a2, I'll also get a1. So, this is fine. Yes, yes, this is how it is. So, Observe what I have done.
I have drawn FBD of the rod and the wedge separately. I have written down Newton's laws equation for them separately and just solved them and I have got it. So once I get A2, I will also get A1.
A1 is A2 into sin alpha. So it will be mg cos alpha sin alpha divided by small m sin alpha plus m by sin alpha. So that's how it is. As simple as that. Okay, I hope this is clear guys what we have done.
Yes, we are correct. So I remember vaguely we had to get something like course over there. So solving it you should get the answer.
By the way, I've kept two questions for homework for all of you. Check out the homework questions. Okay, and please put it up in the comment section below.
These are the two homework questions for you. Two homework questions for you. Okay, so that's it from my side.
Thank you very much guys. And I'll see you in the next class which is non-inertial frames of reference and pseudo force. Okay so today's lecture was all about constraint motion, multiple pulleys, rod problems and wedge problems.
So please practice some questions on this and please do the homework and put it up in the comment section. Thank you very much. Hasta la vista. This was Shreyas here.