 So next force is spring force, write out spring, now there can be many kinds of spring, right? The most common form of the spring is looks like this, this is a spring, right? The spring will have its natural length, okay? I say the length of the spring which is L0 is its natural length, okay? The behavior of the spring is such that it tries to regain its original length, fine? It will make an effort to regain its original length, fine? So if I change its length, it will try to come back to its original length, that is what it means, alright? So if I compress it like this, what will it try to do? It will try to expand, so it will push both sides with some force, okay? And if I stretch it, it tries to contract, fine? And if it is at its natural length, if it at its natural length, it will not apply any force or effort, fine? This is what it means, okay? So if you keep the spring on your hand like this, the small spring, it will not apply any force. But if I compress it, it will try to expand, if you expand it will try to compress, okay? So whatever is connected at the two ends will also feel the force, yes or no? Yeah. Okay? So suppose it is like this, if I pull the block like this, spring will try to pull it like that, okay? Yeah. And if I compress the block, it try to push it, fine? Fine, so if a mass is compressed, then it will get a force this way, because it is connected to the spring and spring is trying to regain its original length, fine? So this is the property of the spring and the spring can apply push or pull both. It can be both push or pull depending on whether it is stretched or compressed, okay? Now the force of the spring, write down the force of the spring acts along the spring, okay? It acts along the spring and the magnitude of force, magnitude of the force of the spring is proportional to what? What do you think it is proportional to? Length of the spring. Delta Length of the spring. Change in the length of the spring, delta x, okay? Or if you represent delta x, bring just x, you can say it is proportional to x, where x represents the change in the length, change of the length of the spring. And the force also should depend on the material and various other things, how it is packed and everything else, okay? So these are two things what depends on, so if you combine these two, you will get an equation like this, f is equal to k times x, fine? This is the magnitude of the force, k is called the spring constant. This is the spring constant, okay? So if k is high, it means it is a stiff spring, a very strong spring and this is about the spring force, any doubts? So these are the five forces that I have introduced to you, okay? So now we can apply these knowledge in solving few questions. When I say solve few questions, today's session is restricted on drawing the free body diagram. So I will quickly introduce to you what is the free body diagram, then we will start solving the questions, okay? Write down the free body diagram, so referred as FBD, as is moving with the excitation of 2 meter per second square, that excitation is because of the force only, right? Does it depend on who is creating that force, how it is getting created, the force, okay? Acting on the block, okay? We are taking one body at a time, one mass at a time and drawing all the forces acting on it. That's it. This is the free body diagram, okay? Touch the mass, okay? Write down, tail will touch the mass and the head will show the direction of the force. You have this situation, okay? Mass is kept over here, you take the mass in air like this and just show all the forces acting on it. Like this, this can be friction, this is normal reaction, this is mg, okay? So this is the free body diagram of this mass, this mass. Getting it? This is what I mean by free body diagram. We will be drawing free body diagram of various cases, fine? Once you draw free body diagram, then you can apply Newton's second law of equation, force equal to mass immigration, okay? Let's see how. So what do you think the first step in drawing free body diagram is? First step. Find out the object you are drawing free body diagram. Find out which object you are drawing free body diagram, okay? Shortlist or identify the object for which you are drawing free body diagram. That is step number one. If you don't know which object you are drawing free body diagram, you will not, I mean what is the, there is nothing like drawing free body diagram, right? So step number one, find out one object on which you are drawing the forces, okay? That is step number one. Step number two, write it down. Step two is take that object only and redraw it without changing its orientation. If object is kept like this, you know, it is kept at a slant like that, you should not rotate and make it straight. It should be like that only, okay? Redraw that object only, only that object, nothing else, okay? Suppose object is kept on the inclined plane, do only draw that mass without changing its orientation, okay? That is step number two. Step number three, three is draw your x and y axis. Choose x and y axis. Choose x and y axis. Fine? The chosen x and y axis, that is step number three. See, what I am telling you is not something to memorize. But I am trying to structure your thought in the right direction. Once you solve, let's say, 10, 15 questions, you don't need to think about these steps. They will automatically come. But if you don't write these step-by-step manner and think in this direction, you will still get there, but right now you may get there by solving 10 questions. If you don't think in a systematic manner, you may get there after solving 50 to 100 questions. That's all, okay? So that is step number three. What is that? Choose x and y axis. So you can choose x and y axis smartly. How should you choose x and y axis? So that most of the vectors are already along x and y axis. You have to take least number of components. Getting it? Yes or no? Right? Step number four. Step number four, draw all the forces as it is. Draw all the forces as it is and show the angles. Show the angles the forces are making with x or y axis. Okay? Just ensure the angles made by the forces with x and y axis. You don't need to show the angle if forces already along x and y axis. Fine? If it makes some angle, then you show the angle. That is 30 degrees, 40 degrees, whatever it is. Okay? That is step number four. Step number five, what do you think will be? Components. Take components. Good. Take components of all the forces along x and y axis. Take components of all the forces along x and along y axis. Okay? That is step five. Step number six. What do you think it will be? Add the x and y axis. Add all the forces along x axis and equate it to mass and maximization along x axis. Write down. Add all the forces along x axis and equate that to mass and maximization along x axis. That is step six. And similarly for y axis, under step six only you write. Okay? So now you get two equations. One along x and one along y axis. Will you get only two equations? In every question? You can also have this. What? You can also have this. Suppose z axis is not there. Only two equations? x component y component. There can be more masses also. Two equations per mass. Are you getting it? You have to do these steps for other masses also. Fine? You have to do these for the other masses also. Getting it? Identify the constraints of the motion. Constraints of the motion. For constraint there will be separate class of four hours just talking about constraint. Okay? So we are skipping the constraint right now. We will be talking about it later. Identify the constraints of the motion of the objects. Okay? Give me an example of constraints. One object is tied with the other object like this. If the restriction of this object this way is A only. That is constraint. If you say this is A1, this is A2, number of equations will be less than number of variables. Okay? You have to identify the very simple constraint. There can be, you know, little tricky constraint also. For example, various pulleys are there. So many pulleys are there. And one mass is coming down. One mass as a restriction is related to the other mass as a restriction or not? Yes. So what is the relation between them? That when you find, that is also constraint relation. Okay? When I am moving on a floor, I am constrained to move in forward direction only. I will not assume acceleration at some angle with the floor. Is that all? Yes. So that all kind of constraint without even, I mean, realizing that you are using constraint equation. Okay? And similarly, a mass is kept on an incline. It slides down the incline like that. So, acceleration is parallel to the incline plane. That is again a constraint. So like that, there can be several constraints. You have to identify, if you have not used it, identify one by one and write down equation related to it. For example, if you assume A1 and A2 here, the constraint equation relation will be A1 is equal to A2. Simple. Okay? There is nothing, there is no physics about it. It is mathematical. Okay? And at times, the most tricky part of a question is identifying constraint relation only. Everything else is straightforward. Identifying constraint equation is where the people struggle. Okay? Just to give you an example. I am not asking you to solve it. I am just giving you an example like this. Suppose this wedge can move on the surface. There is this rod which comes out from the two supports. Okay? Then this rod can move down. And as this rod moves down, this wedge will move forward. There is a relation between this acceleration and that acceleration. The equation of motion, by assuming this has A1 and this has A2, but number of equations will be less than the variable. So you have to get a relation between A1 and A2 also. Okay? Another example, let me tell you, is this. Acceleration of this and acceleration of these two. This pulley can move up and down. Okay? So there is a relation among these three accelerations. Okay? Another example could be like this. You know, this wedge can move. Let's focus here. This wedge can move. Okay? So what will happen to this mass? This mass not only moves along the string. As the wedge moves, this moves with the wedge also. So there is a relation among this acceleration and that acceleration also. Okay? So only one thing in this laws of motion which we cannot put in some sequence or I can't tell you a systematic way of identifying the constraint relation. It's more like a solving puzzle. Getting it. Though there are systematic methods to identify the constraint relation, but many a times it is your own visualization and it's, I mean, whether you get the trick at that moment or not. Okay? So this is the most tricky part of the problem. That we'll be not taking it right now because this has nothing to do with vectors. Okay? So we'll be taking up simple question where you don't have to break your head on identifying the constraint relation. But they do exist. That's why there's this step of identifying the constraint relation. All right? Then step number nine is solve these equations and get the answer. All right, guys. So we are all set. Finally, should take up few numerals. Fine. Tell me this one. There is this mass m. Okay? This surface is rough. Coefficient of friction is mu. Okay? There is this force f applied on it. F. You have to find out the acceleration of mass m. Follow all those steps one by one, sequentially. You may be slow when you're learning but if you learn it systematically, your speed goes up. Okay? So one by one, all the steps you follow is all this. Have you drawn the free word diagram? Yes. I want to see free word diagram, not intersecting final answer, which might be wrong also. There is only one mass, so nothing there to shortlist. Right? The one of mass is this. And then, what are the forces? Mg? Yeah. Mg is down? Any doubt in that? Mg is down? Yeah. Then what else? No. F is there? Which is the external force. Then normal? How much? How will it be applied like this? Then what else? And then friction. Friction. Friction? Yeah. How much? Mu n. Yes. Why mu n? Mu is the rotation. But you are assuming it is moving. You are assuming it is moving. You are assuming it is moving. Suppose it is not moving. Then friction will be zero? And friction will be? F. Yes, you have ignored. Acceleration will be zero. The maximum value of friction is how much? Normal reaction? Yes or no? Okay. Now let's see here. I will be taking one axis like this. X axis and this is along Y axis. Okay. Net force along X axis is what? F minus friction. Yeah. F minus friction. Okay. This is net force. This will be equal to mass time acceleration. Along X axis. X axis. There is only one axis. Due to constraint it will move only along X axis. So Y axis you have. And minus mg. You see. How many of you written n is equal to mg directly? Don't do that. Okay. You may be doing well in this question. Okay. You confuse yourself later on. All the forces left hand side. Left hand side is equal to mass acceleration on the right hand side. Keep it simple. Okay. Leave your thought process for imagination. Don't assume the formula. Formula is what? Net force is equal to mass immigration. Formula is not force on the left hand side is equal to force on the right hand side. Which you have written. Fine. Although that may be mathematically equal. But physically it means nothing. Okay. Yeah. All right. The normal direction is mg. From this equation normal direction is mg. So friction, maximum friction value will be equal to mu times m into g. What is the maximum force for which a is 0? How do you find that? Put a equal to 0 here. F will be equal to F r. So F is, since F will be equal to F r. Because F minus F r is 0. So F, so maximum value of F is maximum value of friction. Which is equal to mu m into g. Right. It will not move. Understood. All of you understood this? Yes. Okay. Is your applying more force than mg? Then what is acceleration? F minus mu mg divided by m. If force is more than mg. Then it is accelerating forward. Okay. Then F minus mu times normal reaction which is again mg only will be equal to m into a. Yeah. So a will be equal to F by m minus mu g. Question of laser motion. Understood. Yes. Any doubts please ask. Next question. Be careful. So the first thing we have learned from this question is that you are ignoring cases. Okay. Acceleration may be 0. Okay. This question. Follow all these steps and get the value of acceleration for capital M. It's not as easy as it looks. That's the end. Have it run free by diagram? Yes. The force is? Normal. This mg? Yes. Then? Normal. Normal reaction? Normal reaction. There is this force. What else? Friction. Friction. Friction. This is theta. How will you take your x and y axis? Normal. Normal one. Normal. Okay. The usual one. Now what? You have to take component of all the forces along x and y. Capital F. Capital F. So you have to take component. So this component will be? F cos theta. Alright. So this is F cos theta. So this is F sin theta. And this is F sin theta. F sin theta. Then n doesn't matter. Then n doesn't matter. So it may not matter. So it may not matter. What next? Now we change. What next? Now we can come back to the student's class. So we change the order of x axis. What is the equation of long x axis? F sin theta minus F r. F minus F r. F minus F r. Is equal to m A x. m A x. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. R. then solve it. Can you do it yourself now? It may happen that it does not move at all. That is another case. Acceleration along x axis is 0, okay. If A x is 0 then what will be F cos theta minus F r? 0, right. This is the case. If A x is 0, can A y be 0 necessarily? No, no. It may not be 0. If A y is there, what will be friction? 0. So, F cos theta will be in the net force, it will create acceleration in x direction also. If A y is not 0, it is accelerating upwards. Friction is gone. So, can F cos theta be 0? F cos theta has 0. So, there will be acceleration in x direction also. Yes. Understood? If A x is 0 then you cannot have A y. Yes, no. If A x is 0, you cannot have A y also. Yes. What I am saying, if it lifts up, if it lifts up, there is nothing which opposes F cos theta. Yes. So, there will be A x. F cos theta will be equal to m into A x. So, but if A y is 0 then A x can also not be 0. Yes, that is not possible. But case number one, if A x is 0, A y is automatically 0. Yes. Understood? All of you understood this? Yes. Okay. What is the maximum value of force for which A x is 0? Friction divided by cos theta. F r max. What is F r max? Mu times? Yes. Normal reaction? Yes. Okay. Stop talking. N plus F sin theta minus mg is equal to 0. Right? So, normal reaction will be equal to mg minus F sin theta. Okay? So, maximum value of friction is mu times of that. Fine? So, F will become equal to mu times mg minus sin theta divided by cos theta. So, you can solve this equation to get the value of F. F is there on the right hand side also. So, you can rearrange the term to get the value of F. Fine? So, F will come out to be equal to mu mg divided by sin theta plus cos theta. Let's say the max force sin theta, no, mu sin theta. Mu sin theta plus cos theta. Understood? If you didn't get how, then you do it yourself. You will get it. Nothing but just simple cross multiplication. No, not that part. So, is this along the x-axis? This one? Oh, it is along the y-axis. This is along the y-axis. This is along the y-axis. This is along the y-axis. No. Now, for ax to be 0, ay also has to be 0. That is our net force along y-axis is 0. Understood? Next case is case number 2, ax is not equal to 0, but ay is 0. And then third case is both are not equal to 0. Sorry. Both are not equal to 0. Fine? Can you do it for case number 2 and then for case number 3? So case number two is AX and AY are both positive. Case number two, write down not equal to zero but AY is zero. Solve this. It is accelerating forward but not accelerating in Y direction. Any doubts here? How do you go to the last step over there? You can do it yourself. You can solve for F. Forget about this. Can you solve for F? Do it now. Solve it. Forget about this. Can you solve for F? You will get this. Sir, AX is zero while AY is zero. If AX is zero, that means friction must be present. But if it moves up, who will create friction? It loses contact. Usually students will solve only for this case. Case number two, they forget other two cases. If AX is not equal to zero, then what should I write? Cos theta minus nu times. Since AX is not equal to zero, friction will be maximum possible. Nu times N, it is constant. This should be equal to M into A plus F cos theta minus mg is equal to zero. Sign theta. Sign theta. You don't need to solve this. Just leave it till here. Solving this, you can get the value of A. Get N from here, substitute there, you get the value of A. Case number two, both are not equal to zero. Case number three, sorry. Case three. We got a really weird answer. Just one? No, no, no. Three is like a big thing. Well, big thing under the rule. Case number three, focus here. X not equal to zero and AY also not equal to zero. What will be the free bullet diagram for case number three? What all things will be zero? Friction will be there or not? Zero. It will not be there. It is moving up. No contact. No forces. Even normal force is gone. Only this much is the free bullet diagram. If it moves up. Getting it? Get it out. But you can divide the expression into two components. This is AY and this is AX. Next force horizontally is what now? F cos theta only? So F cos theta will be equal to M into AX and vertically. Sine theta minus mg is equal to M AY. These are two equations. That's it. Total expression will be root over AX square plus AY square. AX and AY are the component of expression along X and Y. Any doubts? So we have learned many things from just one simple question. People ignore case number one and case number three. They just focus on case two. Now, please relook it at again and let me know if you have any doubts. Read it again. It is important that you understand this. If you have not understood, have any doubts, please ask. What? AX is alpha star by M. AY is F cos theta minus mg by M. All of you do it. Like that. There is no X. No variable AX is equal to X cos theta by M, right? AY is F cos theta minus mg by M. See, I can understand. This is the beginning of the second law of motion chapter or laws of motion chapter. So hesitation or you will feel that very difficult questions are. These are difficult questions. But all it takes, just 10, 15 such questions and then you will be at ease with it. I can show you very, very simple question. You don't just directly substitute force and excretion. You will get an answer. But the reality is straight-forward questions are not asked. So let's not do something which is not asked. What is the point of I telling you 500 forces this way and mass is 5 kg, so what is the excretion? You find, excretion is 1 meter per second square. You feel happy about yourself that yes, I understood. That doesn't mean anything. So we have to go through with the phase of frustration and with the phase of solving a lot of difficult questions. Then only it will help you do something. It is square plus AY square. Excretion is AXI square plus AYJ square minus MG by M. So now I am going to reduce something which will come again and again in questions. That is inclined plane. Suppose this is angle theta. You have a mass kept on it. Mass is capital M, coefficient of friction is mu. You have to find excretion of capital M. No force applied. Gravity will be there. I don't need to specify. Show all the forces, only forces. Sir, now force will not be like that. Now force will push it up. Friction will expose the direction of motion. That is the case number one. Acceleration is not equal to zero. Don't worry about it. Take the mass like this as it is. Downward is MG. What is in vertical? There is a normal reaction. What else? Friction. Friction is like this or like that? Opposite to the direction of motion. This way. How much will be the friction? Mu times M. Because it is sliding. This is capital M. Now what is... Which direction you have to take X and Y axis? Along the axis. Along the axis. Even excretion is like this. So it makes sense to take X and Y axis like that. This is Y and that is X. Getting it? Now what? You have three vectors aligned in either X and Y axis. MG is not. They are compared of MG. What it will be? MG cos theta. What is this angle? This angle. 45. 90 minus theta. This is 90 minus theta. This is theta. This was MG. This was MG. So this one will be MG cos theta. You are down and down. Okay. So this direction of... This direction the compared of MG is MG cos theta MG sin theta. So this one to the right side is MG cos theta. Getting it? Any doubts? Any doubts here? Acceleration. Are aligned either in X or Y direction. So I can write down 4 is equal to Mach M acceleration along X and Y. Quickly tell me what it is. Acceleration is what? Acceleration along that. MG sin theta minus mu N will be MG. Next force along Y axis is what? N minus MG cos theta. This will be equal to 0. So normal reaction will be equal to MG cos theta. Along X axis. Guys one more thing. When we write 4 is equal to Mach M acceleration. Direction of acceleration we usually take positive. Direction of acceleration is positive. So all the forces acting in the direction of acceleration is positive. Opposite will be negative. Friction is negative. Log X is G sin theta minus mu N. So MG sin theta minus friction which is how much? Mu times? MG cos theta. Mu times N. Okay. Mu N. This is the net force is equal to Mach M acceleration. So we can't write N and MG part. We'll write it. Why there is a hurry? Understand this question. But whatever you learn here you have to use it in other questions also. Okay. I understand what is... I mean you think that is related to write like this. I have been writing it 100 times at least. Do I have acceleration along Y axis? It is not moving perpendicular to the plane. It cannot on its own. That's a constraint. That's a constraint. Any doubt? So when you solve this you get the value of acceleration. Fine.