Class 11 Chap 5 || Laws Of Motion 02 || Newton's Second Law Of Motion || NLM IIT JEE NEET NCERT

Hello friends, we are back after a long break of 6-7 days. Finally we are back. There were some problems, you can see there is no caller mic, the lighting is very low and the video recording application is also outdated. There are many problems, the location may be changing.

Still, we tried to start the video today because these mics are not available in the ordinary shops in the city where I live. So, when they come, they will come. Till then, we have to start studying. The pain you have is the same pain we have. Let's finish all these things and start the part 2 of Laws of Motion. So, today we will study Newton's Second Law. Newton's Second Law of Motion. Newton's Second Law.

Newton's second law. Let me tell you the truth. There is only one law of Newton which is to be studied well. That is second law. First law and third law are based on second law. If second law is understood then first and third will definitely come. Now second law basically says F = ma. If I say in common language then F = ma is second law.

But we don't study like this. We study in our syllabus like this that there is a pulley here is a mass here is a mass here. Ok. We ask what is acceleration and tension. Or we say there is an incline plane. There is a mass on the incline plane. We put a pulley here and then we put it here. What is acceleration? Or we say there is some string. Ok. There is some string. Ok.

in an elevator, and it is moving, what is the acceleration? what is the tension? these are the things, either he will tell us that there is a block, he has another block, we put the codes from here, so what is the acceleration? what is the normal reaction? etc. etc. these kind of questions he has to solve in his syllabus and before that, you should know what is Newton's second law?

So actually second law is a particular condition. We will reach here today. F is equal to m. But where did it come from? Let's understand it first. The second law of the needle basically says that the rate of change in momentum is equal to force applied

is equal to force applied. Basically, it is proportional to force, then a constant comes and the value of k is created. So, it is saying that the rate of change momentum is equal to force applied. Now, this is very basic. Some people are studying it. So, read the video quickly and see the numerical. If you haven't read it, then understand what is basic and what is differential form. Very important. Differential form. When is this true? When is it true? No. I will explain these things to you in the starting.

The rate of change in momentum means change in momentum upon change in time is equal to force. Now many of you are thinking what is momentum? Momentum is mass into velocity. So we have seen change in momentum upon change in time is equal to force. This is our second law of needles. What is the second law? Change in momentum upon change in time is equal to force. What is momentum? Mass into velocity.

So how did this formula come here? Let's start. We said Force = Delta P/Delta T Change in momentum/change in time Suppose there is an object of mass m Its initial speed is u and I have put force in it, F

Now what will happen with force? This motion starts and its velocity increases. The initial velocity will be like this. The velocity will change until the force keeps on hitting. I have taken the example from here to here. The force will keep on hitting till the time E. Listen carefully. The velocity changes until the force keeps on hitting.

Now see the initial momentum of this. How much will be the initial momentum? Mass into initial velocity MU Final momentum mass into final velocity MV Here is the change in momentum. The change in momentum is Final momentum minus initial momentum and the time below. So final momentum is MV and initial momentum is MU and time is T.

F is equal to m common, so we have b minus u times t. Now what is acceleration? Change in velocity of motion. b minus u times t is acceleration. So we can say F is equal to m. And we have reached here. This was Newton's second law. And this is a particular case which we have to study a lot.

So you will ask why is this a particular case? Because you have considered mass m here and here also. As you have taken m common here. Means you have considered mass is constant. So if mass is constant then this is correct. If mass is not constant then this is not correct. You will ask what I am talking about? See this, F is equal to delta P by delta T. Change in momentum upon change in time.

What is the formula of momentum? Mass into velocity. So how can momentum change? Because of mass, velocity or both? It can be that the mass changes, the velocity changes or both. In this example, we have considered only velocity as the change. What did we consider mass? Constant. Whereas Newton's second law does not say that mass is constant.

It says F = change in moment upon change in time If the mass constant is constant, then F = ma otherwise F = delta P/delta D Who knows, the mass is also changing Here M1, here M2, then M is not common and this acceleration term is not there So the thing is that the mass constant is not common

And when the mass constant is not there, then what happens? There are questions of questions and there is no answer to them. So to do mass constant, a very popular case of rocket propulsion is in your syllabus. Rocket. Ok, below the rocket, exhaust gas comes. From here, exhaust gas comes. You must have seen it in TV.

Where does this exhaust gas come from? From fuel. And what is fuel? It is mass. When the rocket goes up, its mass changes continuously. Because the fuel is getting exhausted, which is released in the form of gas. In such a case, if the mass is changing, then we cannot use F = ma. What can we use? F = delta P/delta E.

The correct way to get force is change in momentum upon changing time. This is second law, always valid. But this is not valid. When does it get valid? When the mass is constant. Another popular case is when the mass is not constant. I have shown this very simply. Another popular case is when the speed of an object is around the speed of light. Somewhere around the speed of light.

even then its mass keeps changing and if I say it correctly, it keeps increasing. This is what Einstein told us in theory of relativity. He said that the actual mass of any object is equal to rest mass divided by under root 1 minus b square by c square. The rest mass of the M0 is seen in the 9x9 book. Like the one which is written in a box and written in a line. We don't understand that.

M is actual mass, B is speed of body and C is speed of light. So, when any object is in motion, its mass increases a little. So, any object like car, car, or scooter, the moment any object comes in motion, its mass increases a little.

But this thing is very evident when the speed of the object is near the speed of light. For example, the speed of light is 10^1 meter per second. Suppose the speed of object is 10^6 meter per second. 10^7 meter per second. For example, the speed of electrons is 10^6, 10^7, etc.

In such cases, we see that the actual mass of the object is slightly higher than the rest mass. The body is in rest phase when it is locked, while the actual mass is in motion.

Take an example, let's say the speed of an object is c/2, the speed of light is half of the speed of an object. So if you put c/2 here, c^2/4 will be c^2c^2 cancel, 1/4, 1-1/4, 1-1/4, 3/4.

Under root 3 by 4, we will have m0. C by 2 is cancelled. 1 by 4, 1 minus 1 by 4, this is equal to 3 by 4. Under root 3 by 4, we will have root 3 and root 4. m is equal to 2. Now see, this quantity is bigger than 1. The upper one is big and the lower one is small.

Means the value of m will be more than m0. This is what I am telling you that whenever any body is in motion, its mass will be slightly higher than its actual rest mass. Now this is very evident when the speed is near the speed of light. For example, our speed is 100 m/s, 200 m/s, it is very high, 100 m/s is very high.

So here 100 square, here 10 to the power of 8 square. So the item becomes 0. The item becomes very small in this way. Normally day to day like this. So 1 minus something very small is 1. Under root 1 is 1. M is equal to M naught. So our mass is equal to our rest mass. Close to close. Actually our mass also increases a little.

So, any object in motion, its mass changes. But, in day to day life, we don't believe it. Here, we have to believe it when the object's speed is near the speed of light. In this case also, F = ma is not valid. Why it won't happen? Because mass is not constant. So, guys, you will get these cases in this syllabus in 5% time. In 95% time, you will get those cases where mass is constant. 95% time.

So, 95% of the time we will say that this is true. F is equal to m and we will do numerical for this. The question is what to do when 5% of the time is there? When mass is not constant, when mass is variable, when mass can change, then what will be the value of F? Delta P multiplied by delta. We will expand this further, children. And after expanding further, we will teach you the differential form of Newton's second law. Differential

form of Newton's second law. So, we have F = delta P divided by delta D. Change in momentum. Let's say the momentum is a little bit F. So, we will write delta P as dE.

Time also changes to delta t instead of dT. Turface is equal to dT by dT. The movement is a bit serious because the atmosphere has changed a lot. I am not going to talk much. I will adjust to the polar bear. It will take some time to adapt to it. I am adapting to it now. I don't even have a small gold lightning or a mic. I will talk a little today. Let's go. Serious, okay. I am not going to talk much.

So, dp is a little change in momentum and little change in timing. This is the definition form. In the definition form, you will say f = dp/dt. Wait, I will not say this. You will understand later. So, you said f = dp/dt. If you know mass variable, then how to deal with this?

The formula of momentum is mass into velocity. So, we will write d = mv/dt Now, we have to differentiate uv. 2 variables uv. m is taken out. dv/dt. v is taken out. dm/dt

Are you getting this point? First, m is common here, dv/dt, plus v is common here, dm/dt So, f is equal to m dv/dt plus v dm/dt This is the exact second law. This is the exact second law. Write it down. I will put the notes in the copy and put it in my heart. This is the exact Newton's law. m dv/dt plus v dm/dt, always valid. Always valid. Always valid.

First term is telling us that how does force is formed by the change of velocity? Second term is telling us that how does force is formed by the change of mass? Force is formed when there is a change in momentum. In which there is a change? In momentum. In momentum, there is a change in mass also but also in velocity. So first term is telling us that if there is a change in velocity then what is the force? And if there is a change in mass then what is the force? So what is mean this second law in differential form? F is equal to either you can say change in momentum upon change in value.

Then I said the formula of momentum is mass into velocity. The way to differentiate is to take the mass and differentiate the velocity. Take the velocity and differentiate the mass. So you can say Newton's second law. Mgv/gg + Vgm/gg. Understood? And how will you come here? Let us say that you will come to the point of x=ma.

If mass is constant, then change in mass is 0. If mass is constant, then change in mass is 0. So this term becomes 0. Now dv/dt is acceleration. So f is equal to... So the method to come from here is that if mass is constant,

So change in mass means dm is zero, so this is zero, dv/dt acceleration, F = ma. So how can you prove that F = ma when the mass is constant? Remember, when you study in class, F = ma is not Newton's second law. It is a very different concept. F = ma is not Newton's second law. Newton's second law is F = change in momentum upon change in time. Yes, F = ma is possible when the mass is constant.

Otherwise, F is equal to m dv by dt plus v dm by dt. Note this. This is a new equation. m dv by dt plus v dm by dt. Suppose the mass is changing. In a second, 5 kg mass is decreasing. So, dm by dt is minus 5.

how much mass is decreasing in a second? 5 for example, if in a second 2 kg mass is increasing then the value of dm/dt is 2 change in mass upon change in time increasing or decreasing so in this way you can get the question in which it will say that mass is increasing or decreasing and it will ask to find the force then we will have to use dv/dt + v/dt so in my opinion you have to clear the basic of Newton's second law what is Newton's second law? change in momentum upon change in time and it also becomes the opposite of the first law if mass is constant

So, let's start our today's lecture and start with F=ma. We have a lot of questions on F=ma. So, let's start with a simple question. If you haven't seen the first one, then watch it. I have taught you to make a pre-body diagram. Let's say there is a mass. I will put a force on it. This mass is known as m.

Now I have to find its acceleration. If I apply force, it will accelerate. I said A and I have to find it. So, first thing, you have to see from where. I am seeing it from the ground. Now, to see from the ground, how much force is applied on it. One is its weight, on this side, empty. One is the normal weight applied on it, which is the surface.

and one is force. I have made this style video and I have represented it with dots. Downward is mg, upward is normal reaction, and to the front is F and the acceleration is T. You know that the mass is neither going up nor down. These two forces are canceling each other. So I can say that N is equal to mg. This is the force and this is the force. So neither the motion is upward nor downward. The motion is this way. So the direction in which the motion is taking place,

The force of that is F = NA. From here you can get A. A is equal to F but A is equal to M. This is a very basic case. Let's move on. Let's move on to the next case. The force on the body is unique. You can see that the force is unique. One is F, one is N, one is G. There can be many number of forces that are acting on the same body. But there will be a unique acceleration.

The body's acceleration can't be in one direction only. It can be in multiple directions. Forces can be multiple but the acceleration can be one. So the exact formula for acceleration is F net force upon mass. Don't look at the force in the body, look at the net force in the body. What do you see? Net force and from that comes acceleration. Always look at the net force.

Let's see. Let's take a new example. Here we have a block. We can change it a little. And here a force will be applied from here. F1. Here a force will be applied from here. F2. Mass is m. Now we have given that F2 is greater than F1. So where will this mass go? It will go to the front. So we have given the expression of A here. Right? Okay. Now we have made its MVD. I have made this mass m by dot.

The lower side will be called MG. The surface will be like this. Normal reaction. One F2 will be like this. And one F1 will be like this. Now, see F is bigger in the body. It means the body will be bigger here and the exclamation will be here. So, we have to cancel these two. So, we can say N = MG. Let's go further.

Consider the direction of acceleration as positive direction. Let's see the next equation. F = m. Fnet = m. Let's see Fnet. The direction of acceleration is positive. So, F2 is positive. F1 is negative. So, F2 - F1 = m.

you can take out A from here A is equal to F2 minus F1 upon F1 you don't need to write net force how much is F2 in front? F1 in the back? F1 in the front is positive and F2 in the back is negative why did you take positive? because of acceleration ok? ok, now it will not take any more than 2 minutes let me take it from the other side ok, let's change the force suppose here the mass is m and the force is like this and suppose this angle is theta

Now I will tell you to find the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal reaction and acceleration of body by doing this. Now we will solve the normal and acceleration of body by doing this. Now we will solve the normal and acceleration of body by doing this. Now we will solve the normal and acceleration of body by doing this. Now we will solve the normal and acceleration of body by doing this. Now we will solve the normal and acceleration of body by doing this. Now we will solve the acceleration of body by doing this

Where theta is there is cos, there is a perpendicular sign Now remove this force When its component is done, remove this force What other force will be applied on it? mg, upward normal reaction So we know that the body will move up and down like this Where will the body move? up with acceleration a

Now, if the body is not moving upward, then the force is not the same.

plus L sin theta is equal to mg, the force of product. From here you get N, mg minus L sin theta. Whatever values you have, give it. F, M, G, you will get normal reaction. Now take the direction of the result. What is acceleration A and force? F cos theta.

We have to use f net = m. What is the force? f cos theta. This is the force. is equal to m * a. You can take a from here. f cos theta divided by m. Whatever is given, it is the value. Did you understand? Let's take a question with the values. Let's take a question with the values. Let's assume there is a mass here.

5 kgs, ahaa, now I can do anything. And here, suppose a force is moving, 20 degrees, I can do anything. And this angle is stethoscopic mass. One force is moving like this, and this angle is depending on this force, 10 degrees. Now we ask how the body will move, what will happen, what will be the normal reaction, we ask that. Find, find, normal reaction and acceleration.

Let's understand this We have to make two components of this force One will be horizontal here That will be 20 cos 37 One will be vertical 20 sin 37 What to say about this force? This force will go like this Angle is 53 degree So, in horizontal will be 10 cos 53

Vertical is 10 sin 53. The other two are normal reaction. Here is a 3D model of the mass of 5 kg. The lower side is 5G. The upper side is normal reaction.

Here it is 20 cos 37 and above it is 20 sin 37. Here the force was 3, so 10 here is 10 cos 33 and below it is 10 sin 53. These are all the forces.

Now, one force on the body will be like this and the other force will be like this. It will be like a discoloration. Now, look at this video. Forces are like this. The force above is cancelled. So, we have written the discoloration here. Now, because vertical is in the frame, equate the force above and below. n+20 sin = 5g + 10 sin

I can explain this diagram again. I can explain it very easily. This is the force 15, 10 nm. Angle is 3. To increase this 10 nm, what will be the angle? Angle is 53. The component of 10 is 10 cos 53 and the component of 10 is 10 sin 53.

From here you can take out n. I have already told you to take out sin 37 and sin 53. This is 37 degree arc angle. Here 3, 4, 5 and this is 53 degree arc angle. In 53, 4, 3, 5.

In 37 degree, 37 is a small angle. So, in front of it, a small side is 3. 53 is a big angle. So, in front of it, a big side is 4. So, the same 5 is multiplied by 3, 4, 5. From here, sin 37 can be taken. 3 multiplied by 5.

and sin can be derived from here 4/5 perpendicular to the reaction. So what can you derive from here? Normal reaction and G value is 10. So we will get N+20 sin will be 3/5

is equal to 5g + 10 sin 53 4 by 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal to 5 4 by 5 is equal

and this is -12 = 46 Nm so the normal reaction is 46 Nm so you can calculate it by weighting it then we have to calculate the acceleration of 5 kg mass in the acceleration you have to see Fnet = 20 cos 37 + 10 cos 37 + 10

cos = ma = ma = 5 * 8 20 cos = 4/5 10 cos = 3/5 = 5*8 5*10 3 5*10 4*16

2,3,6 is equal to 5 and you can calculate the value of a from here So, in this way we can calculate normal reaction and acceleration F_Net is equal to m_a F_Net is equal to m_a Let's see if it is blocked or not If it is blocked, suppose it is blocked

Let's take one block Suppose its mass is m1 and its mass is m2 and we have applied a force from here Now we asked you to find the normal reaction between blocks and the acceleration of both blocks Find the acceleration of both blocks and the normal reaction between them Now tell me one thing When I apply force

So, the speed at which M1 will move forward, will also move forward at the speed of M2. Because M1 and M2 are stuck together. When you hit it, it will move forward with it. Meaning, the acceleration will be of this acceleration, its acceleration will be of this acceleration. If it moves forward 5 meters in a second, then it will move forward 5 meters in a second. It will move with the stickiness.

Means, both are same as tradition Now, we have to make FBD of both

Let's make N1 FPD I am making FPD of N1 Free Body Diagram In which that force is represented Which is on the body Listen to me Which is on the body Not the force which is on the body The force is acting on the body See them So on N1, it will go down M1G On the top, it will have a counter reaction I will write N

If there is a force on the other side, there is a selection of this. Now tell me, will there be any other force on M1? Say it, say it, say it, say it. Will there be any other force on M1? Say it, say it, say it, say it. Yes, there will be. Normal reaction will be applied on M2. Because M1 and M2 are in contact, so normal reaction will be applied on M2 and normal reaction will be applied on M1 and M2.

Let us call this normal reaction N' and M1 is placed on this M' We have understood this because both are in contact M2 is placed on S and M1 is placed on S M1 is placed on the same force as N' A is placed on M2 and M1 is placed on M2 So there is a force in M2 which is placed on P

Now this is in the vertical equation so we will call it N = MNG. This is the end of the story. This is not even called normal reaction. This is not even called normal reaction. We have to find out this one. Normal reaction between bonds and ash. Now from here. Now exploration is this way. One force is front and one is back. Consider the direction of exploration as positive. And what is the equation? F net = MNG.

So, the force will be positive in the front side F is positive The force is negative in the back side So, -N' Because acceleration is in front side, so the direction is positive So, the direction in the back side is negative This is the net force What is the mass? M1 Acceleration is the same Same diagram equation is given F-N' = M1A FBD is clear according to you It is perfectly clear Come, let's make FBD of M2 Let us draw FBD of M2 This is very important

On the bottom, we know that it is M2. On the top, there is a normal reaction surface called N1. On the surface, there is a normal reaction and we call it N1. Here we have N, here N' and here N. Which force is there on the front? F is the rate of M2. Many people mistake the concept that they think of F after M2. F is only applied to M1. F is not applied to M2. So, why does M2 go ahead?

M2 is moving forward because of N' M1 is pushing M2 M1 is pushing M2 means this N' is moving forward because of M1 and M2 means the trade force is N' from the front side notice that F is moving towards M1 not towards M2 what is happening on M2? normal reaction so who is moving towards M1? it is a bit of a diastolic problem but the concept is the same concept is the same now we will tell you the shortcut

So the force was on M1 and the pressure was applied on M2. This is the normal reaction. The acceleration is also A. In vertical equilibrium, the force above is not equal to the force below. We don't have to apply this normal reaction which was applied on surface M2.

For this equation we will write Fnet = m2a. This is equation 2. Now you have to write n and a. So there are two variables and two equations. You can write them. Exam will give you F, m1 and m2.

Now add these two questions. Add -n -cancel f = m1a + m2a

F/M1+M2. Very good. Now, if you put the value of escalation in the second equation, then you will get the normal reaction. M2*A and that is the normal reaction between blocks. So, the question is done. Now, I will show you something amazing in the same question. You have taken out the normal reaction by taking out the escalation. For this, we will do a short video.

I will tell you the shortcut later, first let's see the mean. If I consider this whole system, M1 and M2, this whole system is for me. So, the mass of this system is M1 + M2. And what is the total force applied on this system? So, I can say the acceleration is net force divided by net mass.

F = F and M = M1+M2. You can see the similar answer. Still, it is necessary to make FDD because the normal reaction will not be obtained from the shortcut. Normal reaction will not be obtained from the equation. Normal reaction will be obtained only when the equation is formed. You can find the explanation. This happened because both the masses were moving together.

It is not necessary that both have same acceleration. Here, both have same acceleration and both are moving at the same time. So, consider both as a system. Always consider both as a system. It is not that easy. Then, keep the normal reaction of acceleration here. Now, I will give you a question. Show it to me.

So here we have to do normal reaction and excavation. I will increase the mass from 2 to 3. 1, 2, 3. These are the 3 masses. Assume this is 2 kg, this is 3 kg and this is 2 kg. Let's increase it. Let's give it 4 kg. Here the force is of 18 N. We are asked to find

Now friends, you understand that we don't need to use the vertical force to make the question because the force is being directed in the horizontal direction

So, we have shown the energy and the normal reaction here, which means we just consider the horizontal force and solve the question. So, we get 2 kg mass. One force on 2 kg and one on 3 kg. What will be applied to the normal reaction? I have applied N1 here and N1 will be applied to 2 kg and 3 kg. Similarly, normal reaction will be applied to 3 kg and 4 kg and N2 will be applied to 4 kg and 3 kg.

3 kg is applied on 4 kg and 4 kg is applied on 3 kg. Normal reaction perpendicular to surface. Let's make 2 kg RBD. We don't have to see 2g force on the bottom. We don't have to see normal reaction on the top. It is not a matter of vertical forces. This is the force of the cut. And a force is applied on the back which is applied on 3 kg and 2 kg. Which is called N1.

So, the acceleration of this is a, the acceleration of this is a, and the acceleration of this is a, all three are the acceleration because all three are moving together. We have said that the acceleration is a. So, this direction is positive.

So, 18 is positive and N1 is negative. So, equation will be 18 minus N1 is equal to... What does the equation mean in second law? Fnet is equal to M. So, net force is 18 minus N1 is equal to mass is 2 kg. 2 into A. Now, 3 kg of equity is made.

N1 is on front side of 3 kg and N2 is on back side of 3 kg and N2 is negative of N1 is equal to MA = 3 and 4 kg FPD

4 kg has only one force which is N2 and the acceleration is 1, we are not looking at the vertical forces. So N2 is here and that is equal to MA, M is 4*A. Done kids? Add these three equations and you will get the acceleration. Add these three equations. N2 cancels with N2, N1 cancels with N1.

18 is equal to add here 2 3 5 4 9 9 2 3 5 9 1 from here the value of a will be 2 take out the value of 18

Now we are asked about the normal reaction between 4 kg and 3 kg. So, what is the normal reaction between 4 kg and 3 kg? N2. So, we have to ask about N2. N2 is equal to 4A. We have got A, so we can subtract N2. N2 is equal to 4A. 4 into 2 is equal to 8. And this is the second square on 2 meters.

If we ask the normal reaction between 2 kg and 3 kg block, then we will find the value of N2 and we will put it here and we will find N1 which was the normal reaction between these two blocks. Here we can also use the short pad, look carefully. We can use the short pad also. We have to consider these three as a system because they are moving together, so we can consider the system. So, A is equal to F net upon net mass. The net force of this whole system is 18.

Don't look at these forces as they are joining. These are internal forces. When you have made this whole system, the internal one, the internal one, these are internal forces. Because of which there is no motion. The motion is of external force.

So you have taken the whole system of net force and net mass 2,3,5,4,9 So the simulation is 2 but if you want a normal reaction then you need to make a free body diagram. Let me show you another question. Suppose here is a block and its mass is m1 and here is a block and its mass is m2. Suppose a force is added from here to this, f

Now we are asked to find the acceleration of M1 and M2 and we are given that all surfaces are smooth. So what we will do? We are looking at the force applied on M1 and M2. So we will consider both as a system.

Both have a system. The mass of this system is m1+m2. Now how will you find the escalation? Fnet/net mass Total force/total mass Total force/F Total mass is m1+m2 So the escalation is the same. And this is a very common mistake of our students. Which we should not do. Read carefully. All surfaces are smooth. This surface is smooth.

The force you applied is on M1. That means, on applying force on M1, M1 will move but M2 will not move with it. M1 and M2 are not connected. There is no friction between them. When you apply force on M1, then M1 will move. No force will be acting on M2. Then M2 will fall down. That means, starting acceleration of M2 will be zero.

and this complete force is applied on M1 so the escalation of M1 will be force upon mass so the escalation of both will not be same so you cannot call it system let's do it with free point diagram so you will get more information when we will make FBD of it let's make FBD of different mass make FBD of M1 M1G normal reaction on ground is M

and one force M2 will be applied on the angle because both are in contact so what will be applied in contact? Normal reaction, so M2 will be applied like this let us call this N' now let us make M1's FD, M1G will be applied on the lower side and the normal reaction will be applied on the upper side one normal reaction will be applied on M2, the lower side will be applied on N' and one force will be applied on the upper side

all these forces are acting on M1 now M1 is not going up, it is going down means these forces will cancel these forces these forces will cancel these forces there is only one force in horizontal direction that is F is equal to M1 and its acceleration is F is equal to M1 or A1 I have said acceleration of M1 is A1 and this force will be equal to the force below and whatever happens now this is M2's FPD this is M2

M2 will have a force M2G which will pull it from gravity and this N dash will be applied on M1 not M2, on M1 M1 and M2 will have a normal reaction on N dash on the top equivalent opposite this N dash M2 applied on M1 this N dash M1 applied on M2 on the top so N dash on the top

Now tell me, is there any horizontal force on M2? No force is on this side of M2. That means M2 will not escalate this side. These two forces are equal to each other. And there is no force on this side of M2. So the acceleration is zero. So we can say that the acceleration of M2 is zero. The acceleration of M1 will be equal to A1. A1 will be equal to F and the acceleration of M1. I am saying this in the beginning.

that this force is acting on M1 only and not on M2. So, pay attention here. It was written that all surfaces are smooth. If there is friction behind them, then M2 would have caught M1. Then M2 would have moved the force on M1. But now this is smooth. If we hit it, it will slide down. What will happen? M1 will slide down and M2 will slide down after some time.

So, you understood this. Now, let's see the position. Suppose this is a mass of 1 kg. We are applying force on it. Say, 1 meter. Here is a small mass. Suppose it is 2 kg. And its length is 4 meters. Now, when you apply force on it, it will come only after shifting.

If you apply force on 5 kg, then as you heard in the previous example, 5 kg will be removed and 2 kg will be removed because all surfaces are smooth. Now, we will ask find the time in which 2 kg falls off 5 kg. So, 2 kg falls off 5 kg. There is a question in XC program also. Let's see how to do this.

So, this one has some acceleration. How much acceleration will it have? Force upon mass 10 to the power of 5, which is 2. And how much acceleration will it have? Zero. In the previous question, you saw that all the forces are acting on the recipe. No force is acting on the direction of the holder. So, due to its acceleration, it will move forward from here and 2 kg will move downwards. Now, how do we solve this? We do one thing. We move here.

we bring the observer to this frame. This frame is moving forward by 2. Now we sit in this frame and see 2 kg. If we sit in this frame, we will see 2 kg going backwards. 2 m/sq. sq. Because we are sitting in the frame. When our brain is running, we think the front frame is going backwards. Similarly, this observer will be asked that how much this mass is going backwards? 2 m/sq. sq.

Now, this mass is given by the observer. The square on the back is 2. The distance is 4. And the initial velocity is z. You could have understood this in a different way. For example, I call this a and this b.

So we are looking at everything from A So, the exclamation of B as seen by A You must have read this in my relative motion videos, you must have read this yourself The exclamation of B as seen by A When B is seen, what will we say? The exclamation of B minus the exclamation of A How much was the exclamation of B? Zero. How much was A? Two. How much was A? Minus two So when the exclamation of B is seen from A, then how will it be? Minus two

means 2 will come in the backward direction distance to cover B is related to A and B is related to A is 4B what is the initial velocity of B? 0 you have to find out the time you can find out or you can calculate using your equation S = U+1/2 AT^2 distance to cover is 4 initial velocity is 0

So you know that we will not take negative time. So after 2 seconds, this mass will come back and will be released from here. Clear? Now you understood how to do this?

So guys, in this way we could have done block over block type equations or we could have done block along block in which normal reaction is not there. The next question that comes is about string. What it does is, suppose I took one block M1, string, and another block M2 and from here I have taken a force. Now it asks us, "Pine tension in string, Pine escalation of M1 and M2". Let's start.

FPD will be made. Look at the vertical force vector. No, look at the horizontal. We know that this will work here and this will work here. Now if both are made of string and string means that the base is not slow. So the further it goes, the further it will go. The further it goes, the further it will go. Means both will have displacement, velocity, acceleration will be same. Both will have displacement, velocity, acceleration will be same. Will be made of same string.

So if the escalation of this is A, then the escalation of this will also be A. Now let's see the forces of M1. One is downward, M1G. One is upward, normal reaction, which we don't have to consider now. We have to see the force of the opposite side. Here in this stream, tension force will be applied. And tension force is away from the point. That is, it goes away from the point. If you see tension anywhere, it goes away from that mass, that point.

So, I am making equity of M1. I am not making the force at the bottom. Tension is at the Asclaration. So, what will be the equation of M1? Net force = Mass * Asclaration. Let's make equity of M2. M2 has one force at the top. And tension will be at the middle. Away from the point. So, tension is at the T.

acceleration is here, A if you consider the direction of acceleration as positive then what will be the equation of red force? F is positive, T is behind, negative mass is M2*A we can write F minus T is behind, is equal to M2A now add both of them, add will also be P, so P will be the merger F is equal to M1A plus M2A, from here A is common, A is equal to F, but A M1 plus

and if someone is asking you about tension, then you can take the value of A and put it here and you will get tension. If you take the value of A, what will you get? Tension. So here also you can see that we can consider the question as a system. What do we consider as a system? If we consider it as a system, then net mass m1+m2, net force F, the acceleration F upon m1+m2. Why did this happen? Because both the accelerations were the same. So it is not necessary that the acceleration is always the same as we saw in the previous example.

Let's see the next question. Let's increase the weight of the muscle.

and the course is 6 minutes long and we are pulling it so this is block A, block B, block C

Here tension will be on this side, I called it T1, here T1, till the time the string is same, the tension remains same. Tension is away from point, here away from point. Here tension will be on this side, T2, here T2. Wherever you see tension, away from point, wherever you see tension, away from point. This string was same, so tension is same. This and this string are different, so tension is different.

Now let's make the equity. All three will run together, there is no doubt in this. Because all three strings are connected. So this has acceleration A, this has acceleration A, this has acceleration A. So let's make the equity of A. T1 is the tension on this side. Asceleration A. What is the equation? Mass is 2 kg. So what is the equation? T1 is equal to force is equal to M into A. Force is equal to M into. Now let's make the equity of 1 kg. Block B is 1 kg.

I am not looking at the vertical forces, I don't need that. This side is T2 and this side is T1. Ascalaion is here. Equation capability is T2. On the front side is positive and T1. On the back side is negative. This equals to mass. 1 into A. Now let's see the 3 kg activity. Block C 3 kg. On the front side is force F. On the back side is T2.

Backward T2, forward T4, acceleration forward. So the equation will be F minus T2, because F is positive and T2 is negative, is equal to 3A. Now you have to do the acceleration, add all three equations. T1 to T1 cancel, T2 to T2 cancel, F is equal to 2A plus 1A, 3, 3, 6, 6A. Force value is 6, is equal to 6A, A value is 1. Now A value is 1,

If you keep it here, you will get P1. If you keep the value of P1 here, you will get P2. And in this way, you can remove tension in different strain and also in acceleration. So, I think you understood this block-string question. Now, what other variations can you do? Another variation can be done by giving you an angle. For example, you are given a block.

and then we have given the force as S suppose we have given the force as 30 Nm and this angle is 60 degree now this mass is 2 kg and this mass is 8 kg and this mass is 7 kg now we have asked them to find the tension between these two

or there can be more masses, I am telling you the method of doing it. Now two components of this force will be there, one is this side, F cos 60 and one is up side, F sin 60. Now F cos 60 is pulling forward, so the acceleration of A will be same as the acceleration of A. Here tension is this side, T, here tension is this side, T. F is made and this is made of A kg.

What is the weight of 8 kg? The force will be on the bottom side, 8g. On the top side, normal reaction. Right? 8g on the bottom side. What will be on the top side? Normal reaction. On the front side, t, the part of periscopation. Because there is no vertical motion, so m is equal to 8g. Normal reaction is after 8g. And t is equal to m. By tension, the force that is pulling us, what is the mass? 8 into 8. On which question are we doing all the questions? F is equal to m into second power.

T=8. Now, what is the location of 7 kg? 7 kg. Lot is 7g. Normal reaction is 7g. Normal reaction is 7g. And there is another force F sin 60. This is the movement of this force. F sin 60. This also goes up. F sin 60. There is another force F cos 60.

So vertically it is in the frame, F sin = 70 and the acceleration is this way. There is one more force, the tension is this way, so I wrote T. Now see the horizontal direction, what will we write? F cos is pulling forward, minus T is taking backward, is equal to M into A.

Now you have to find out the acceleration, which you add to it, TCT, cancel, F cos 60 is equal to 8*8, 15a, F value 30, cos 60 is 1/2, is equal to 15a, this 15, how much is the value of a? 1. Now if the value of a is here, then you get tension. Suppose I tell you to find out the normal reaction on this block, on this block, then this will look like normal reaction, after this I will call the normal reaction as n1, and the normal reaction as n2.

Now you can't do normal reaction. Here it will be 80, if you keep the value of G, it will be 10. And here if you keep the value of F, then you will get normal reaction time. So, you can do string block questions easily like this. After this, we have the pulley system, which we call as the output machine. In pulley system, we have questions like this.

In the next video, we will discuss about the pulleys. Now I will tell you the concept of the AdWord machine. This is what it tells us. Here is a pulley. Here is a mass m1. Here is a mass m2. And it is placed like this. And it tells us that m2 is greater than m1. If m2 is greater than m1, then this pulley will fall down.

Suppose the acceleration is coming from A, then the speed at which it will come down will be the same as the acceleration at which it will come up. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it will come down will be A. So, the acceleration at which it

Now if the string is the same, then tension can be applied here also. Why is it the same? Because the string is the same. And in our brain, the pulley is massless and it is like friction. Most of the time, always. Whenever it is like this, tension is the same.

M2 is at the bottom and M2G is at the top. Both are different. M1 is at the bottom and M1G is at the top. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the top and M2G is at the bottom. M2 is at the

will be positive and this will be negative. P minus M1g is equal to M1a. Net force is equal to mass into acceleration. Let's see the M2's angle. Here I am making M2's angle. Downward M1g, upward Tension T. Where is the acceleration? Downward. M2's acceleration is downward. Where will be the net force? Downward. What is the direction of acceleration?

So this will be positive this time. M2G+ - T = M2A. Do you understand? The acceleration is below, so the force is positive. The force of the opposite of the acceleration is negative. Now you have to calculate the acceleration of this. So add both the equations. What is the term of T? M2G - M1G = M1A + M2A. What is the value of A from here?

U-M1 G/U2+M1 You got the escalation value. If you put the escalation value in this or in this, then you will get the tension value. Now you can see that the concept has not worked. Net force upon net mass. Did it work? Did it not work? What do you say? Did the concept work? It did. Let me show you one thing. I will tell you. Everybody understood?

Let's keep this whole thing straight on the table. How much force is pulled on this side? M2G. How much force is pulled on this side? M1G. Which one is bigger? M2. Here is tension T. Here is tension T.

Now tell me how much is the net force? Look at it and tell me. Net force. This whole thing is considered a system. This whole thing is considered a system. When you consider it a system, then you don't have to look at the pension. Why? Because it is internal force.

The tension is in the system. So, what is the net force? M2G here and M1G here. How much will it be? M2G minus M1G. What is the net mass? What is the net mass in the system? M1 plus M2. Just like you have told, net force upon net mass is M2G minus M1G upon M1 plus. See, it is still coming.

But if you want to get rid of tension, then you have to make a free body diagram. So either you make a free body diagram or if you want to do a straight escalation, then you have to do net force upon net mass. The question will be over. So we will give this first class. Here our mass is 5 kg. Here our mass is 2 kg. Here our mass is 4 kg. We will do two mass.

We are asked to find acceleration of blocks. Now you can use your brain to find the acceleration of blocks. So, what you have to do is, you have to add net force upon net mass. Now, how much is the total force on this side? This is 4G and this is 2G. How much is it? If you add both, you get 6G on this side and 5G on this side. So, definitely the system will fall from here to here and from here to here.

Here, the net mass or net force is 6g. We have to do minus 5g. How much will be the net mass? 4, 2, 6, 5, 11. This is the net estimation. How much will it be? g/11. What will be the value of this? +g/11. Then we go up. This one? -g/11 because it comes down.

Now if you were asked about tension in this question, then you would have to find Free Body Diagram. The answer is same. Free Body Diagram. So guys, in the next video, we will look at the questions of pulley which you must be more interested in. Which are very common in LC Brahm. We will tell you how to do a question of pulley. Like you must have seen a question like this, which is a question of movable pulley.

You must have seen the question of movable pulley. Like this pulley is movable. You must have seen the question of movable pulley on inclined plane. You must have seen the question of movable pulley on horizontal plane. We will bring all these in the next video. We hope that the quality will improve. Keep studying. All the best.