 Here we have taken a new question. In this question we will try to apply the basics and learn how we can apply it and solve the question like this. So the question goes like this. There is a wedge of capital mass M and there is a small mass M which is connected through the pulley and their motions are constrained. Now why this wedge will move? This wedge will move when this small M comes down. When the small M comes down the wedge will get pulled forward. Can you imagine this? Let us try to visualize the problem in a step-by-step manner. Let us first take the wedge. Now how the wedge will move? The wedge will move along the horizontal surface. So its acceleration will be horizontal and if you take a small M, can you visualize how the small M will move? Small M will definitely come down. Now isn't small M also moving with the wedge? For example, if you visualize this motion M, suppose you are sitting here and look at the small M, how the small M will move with respect to you? You are moving with the wedge and this small M will appear as if it is going down. Now suppose you come from there and stand on the ground. If you stand on the ground you will not move with the wedge. So the wedge along with small M will move on your right hand side. So entire mass M, capital M and small M both are moving on the right hand side. So they are moving together right hand side along with that small M is coming down. So small M will have two acceleration. Let us call the acceleration of wedge as capital A and then we can show acceleration of small M has one component downward. Let us call this as small A and one in right hand side. Let us call that capital A and if we take this wedge and block individually, we can show all the forces. For example, let us take the wedge. If we take the wedge we have definitely mass, it has mass so there will be gravitational force and it will be always vertical direction. So in mechanics problem generally we assume that whatever is happening is near the earth's surface. The earth is assumed to be flat and direction of Mg which is gravitational force will be downward direction. So that is why Mg is downward and then there will be normal reaction from the surface. Which direction will be normal reaction? Normal force always pushes. So the surface, the horizontal surface will push the wedge in the upward direction. Then we have tension T because of this string. Tension will always pull. Tension can never push. So tension will act away from the point of application. So that is why tension is down here. Now this string is wrapped around this pulley and then there is another tension force at the other end. Now these two tensions the value of them as you can see we have marked to be same. So this T will be equal to that tension force. So there is some assumption here. The assumption is the string is massless and it is inextensible. And we have assumed that the pulley is smooth. So if we have all these three assumptions then we can say that the tension throughout the string is same. So these two tensions are same and then we have of course this small m is moving right hand side. It has an acceleration capital A. Now acceleration cannot happen without any force. So what kind of force horizontally will be there on small m which will cause the acceleration capital A? That force must be normal reaction from the wedge on the small m. So according to Newton's third law if one object applies normal reaction suppose there is a force on small m, small m will apply and equal opposite force same normal reaction on capital M in this side. So there is another force N1 here. What other force comes in your mind? What is given in the problem is this surface is smooth the horizontal surface but the surface between capital N and small m they are not smooth. The coefficient of friction is k. And we have we learned that small m is moving relative to capital M which direction it is moving relative to capital M? It is moving downward right that is the only relative motion between small m and capital M because the other motion is capital A which is with capital M. So friction is because of the relative motion and relative motion is downward. So friction on small m is upward and since this friction force is applied by the wedge in upward direction on small m there will be a pair of this force according to again Newton's third law which will be downward direction on capital M. So you know one thing I wanted to tell you here is that whenever you see any force always ask yourself where is the pair of that force because universe is an isolated system. If there is a force there must be a pair of that force and equal and opposite force somewhere right. Can I ask you one thing where is a pair of this mg that we haven't shown anywhere this mg is applied by whom? This mg is applied by the earth right. So the pair of mg is the mg applied by the wedge on the earth. Now the earth is not the part of our system so that is why we are not showing it here but whatever force is applied by capital M on small m there is a pair because both small m and capital M are part of our system. So this is the free body diagram of small m and capital M. So let us write Newton's second law of equation for the wedge as well as for small m. First let us write for wedge. So along x axis what all forces you see? One force is n1 right like that. There is another force t this one any other force? Only two forces are there. So net force is along the direction of acceleration t minus n1 right. This must be equal to mass time acceleration of wedge fine and along y direction what are the forces? We have n2 mg tension and this friction force right. So the net force in vertical direction is n2 minus mg minus kn1 minus t. This is equal to 0 because there is no acceleration along y axis right. Now let us write down equations of motion for small m along x and along y direction. Along x direction what are forces you can see? You have n1 force right. So n1 force should be equal to mass time acceleration. What is the accession of small m along x axis? Accession is capital A right. Along y axis what are the forces? Along y axis we have mg, kn1 and tension right. So mg minus kn1 minus t should be equal to mass time acceleration right. Now let us count number of variables. Here we have n1 n2 t small a and capital A. We have five variables right and we have only four equations. So we need another equation. What could be that? At the start of the problem we have analyzed the situation and we have learned that there is a relation between the motion of small m and capital N right. So how their motions are related? So that motion you can see that the length of the string is fixed right. If it is l1 this is l2 and this is l3 right. So I can say that l1 plus l2 plus l3 is a constant right and if we differentiate this equation what I will get? Rate of change of l1 plus what is the rate of change of l2? l2 does not change right. So its length is fixed. So this is 0 plus dl3 by dt is equal to 0. Rate of change of constant is also 0 right. Now if we differentiate this again what we will get is d square l1 by dt square plus d square l3 by dt square is equal to 0 right. So there is this relation which equates the acceleration of l1 with acceleration of l3. Now what is the acceleration with which l1 is increasing? Acceleration of l1 is small a right. This point is moving with acceleration a1 whereas this point is not moving at all. So acceleration of increase of l1 is a1 sorry small a. So d square l1 by dt square is a1. What about d square l3 by dt square? This is the acceleration of l3. Now if we have these two points so this point is moving with acceleration capital A right whereas this point is fixed. So what is the acceleration of increase of l3? Capital A is creating a decrease in length right. So d square l3 by dt square will be written as minus a. This is equal to 0 right. So small a will be equal to capital A right. So this is a constraint relation. So now we have five equations we have this one equation 1, second, fourth and we have this one as fifth equation and now we can solve this problem by you know simultaneously solving this five equation. You can also get the constraint equation by using our one of those three techniques by using the work done by tension should be equal to 0. So t into x1 should be equal to t into minus x2. So you will get x1 is equal to x2 and when you differentiate it you will get a1 is equal to it or small a is equal to capital A. So like that also you can get right. One thing you should note here is that small m has two acceleration small a and capital A its net acceleration will be root over a square small a square plus capital A square okay. So this is how you solve a problem like this right. So thanks for watching the video. We will come up with another question another similar question and we will discuss in detail how you can apply your basics to solve a difficult looking problem like this okay. Thank you.