 Okay, so now let us move on to the tutorial, okay, we saw a bunch of problems. So first is a concept question, okay. So this concept question, I have taken in two concept questions both from BJ10, okay. BJ10 has fabulous concept questions, okay. I strongly urge you to go through those concept questions because once we go through them, we will get a reasonable feel for the problems. Now what concept question this tells us is a 1000 Newton Boulder B is resting on 200 Newton platform A. So Iska mass, Iska weight is 200 Newton, Boulder Kavit is 1000 Newton. When trucks see accelerate, okay and this entire assembly goes over a frictionless and massless pulley, okay, both frictionless pulley and massless pulley it goes over this. And now, okay, this truck to which it is attached it starts accelerating, okay in this direction with a constant acceleration. Now we are asked that which of the following statements are true. So let us examine these statements and just use simple logic, okay. Let us not write down too many equations, only we will write down equations, okay. When we are at loss, okay to understand that like simple intuition will not give us enough insights. Then and only then we will move on to writing any equations. So let us think about it. First question. First alternative, the tension in the cord connected to the truck is 300 Newton, okay. So this is 200 Newton, okay this is 1000 Newton. So the total weight, okay is 1200 and it is said that the tension in the cord connected to the truck is 200 Newton that clearly cannot be right, okay. This clearly cannot be right. Second one, the tension in the cord connected to the truck is 1200 Newton. Now that is we have to think that if the truck would have been stationary, okay the truck where I have been stationary then this is 1000, this is 200 and then overall how much is it? It is 1200, okay. But is this true or is it not true? Let us think about it, okay. Now what is happening? Okay. Let me use the whiteboard. This is the assembly of A and B. So A plus B, okay is equal to the weight of them is 1200 Newton. There is some tension acting on it. This is the truck, there is some tension acting on it. Same tension will act on both of them and this has some acceleration, okay. Now note one thing that the acceleration of the truck by the constraint will be the same as the acceleration of this B and A together. Now what we want to know is if this tension is equal to 1200 because the sum of the weights acting is 1200. But note one thing, without doing a single equation we realize that if this tension is equal to 1200, then this assembly cannot accelerate. Only way this assembly can accelerate upwards because this tension will then balance this and overall force acting on this assembly in the vertical direction just becomes 0. So no acceleration. Only way it can have an acceleration upwards is if this tension exerts a force over and above this weight 1200. So this tension has to be necessarily more than 1200 for this entire assembly to go upwards. So what we can clearly say from here is that the tension in the cord connected to the truck cannot be 1200. Next, the tension in the cord connected to the truck is greater than 1200. That is the right answer in that case that because of the logic that we had just discussed. Second one, the normal force between A and B is 1000 is the second thing. Now is that true? That between A and B, let us look here. So let us look at this B diagram B. What do we have? B has some weight which is how much which is 200. There is a normal reaction between A and B. We do not know what that is. But the same logic again this B is accelerating upwards with acceleration A. So what will happen that if this normal reaction is same as 200, then this acceleration will be 0. So in order to create an acceleration in the upward direction, this normal reaction has to be over and above this 200. So NAB has to be more than 200 and this is the same effect. Say for example if you are standing in an elevator and the elevator start accelerating upwards then we feel that for example we are pressing more against the ground. Means for example if we take a weighing scale and stand on it, then we will see that weighing scale will actually feel more normal reaction than our regular weight. Why? Because that extra acceleration it is over and above our weight and that in addition to the weight that we have this normal reaction should be more and above this 200 because if it is 200 no acceleration. So only when this NAB becomes more than 200 can we have some acceleration. So I can say that this is something like the elevator effect. And so the normal force between A and B is 1000 Newton, sorry I made a mistake here. This is not 200, this is 1000 okay. So the normal force between A and B cannot be 1000, it has to be more than 1000. And the last is the normal force between A and B is 1200, so that we cannot say how much it is that completely depends on the acceleration okay. So we cannot say anything about E. So essentially what we can say here is that that the only true statement here that the tension in the cord connected to the truck is greater than 1200 okay. So this simple concept question okay we immediately understand all concepts about free body diagrams and without explicitly solving any problem we get a reasonable understanding about all these qualitative statements that are made about this simple problem. Now let us come to the second concept question. Again the idea is not doing a detail calculation but just analyzing it in brief details okay just brief details so that we get a feel about the problem that we are not doing anything in great detail but just to qualitatively understand what is happening here. So we have two systems shown starting from rest this is 200 Newton okay. On the left okay so two systems shown start from rest on the left two 200 Newton weights are connected by an inextensible cords. So these are 200 Newton weights they are connected by an inextensible cord. On the right a constant 200 Newton force pull on the cord okay so these are two systems one is this 200 200 200 200 okay and what we want to know is neglecting all the frictional forces what are the following are these following statements true okay or what are the what statements are true. Let us look at these statements one after one now note one thing if this system is in equilibrium or not let us ask ourselves neglecting all the frictions in the pull is this is 200 so the tension if the system is at equilibrium or if the system wants to be in equilibrium this tension has to be 200 but because there is no friction between this and that bottom surface any force acting here will start accelerating the block downwards. So what will be the tendency of this entire assembly the tendency of this entire assembly will be to slide downwards look at this system 200 it goes over the pulley this 200 Newton force acts on this so this will have a tendency to slide it to the right. So both of the systems have the tendency to accelerate this downwards this to the right and overall downwards now the question is that what statements are true block A and C okay have the same acceleration that block A here and block C here have the same acceleration now let us analyze the situation if this is 200 in this case we clearly see that acceleration is downwards now since the acceleration is downwards okay if I draw the free body diagram for this bottom one is 200 this weight is 200 and this tension is T and this acceleration is happening downwards from whatever logic we have used okay we clearly see that acceleration has to go only downwards now when acceleration is downwards if T is equal to 200 then acceleration has to be 0 but because acceleration is downward the effective force okay has to act in a downward direction so the weight has to be more than tension so tension clearly in the first case is less than 200 okay now what happens because the tension clearly is less than 200 this acceleration depends only on the tension okay because the tension is the only horizontal force acting on it so mass times acceleration is equal to the tension. So here the acceleration the tension is 200 here the tension is less than 200 and as a result block A and C will have the same acceleration is not right block C will have a larger acceleration than block A is right why because this will be 200 but here the tension acting here will be less than 200. Third block A will have a larger acceleration block C is wrong block A will not move again from the logic that we have used block A has to move none of the above is again wrong so the only right statement here is that that block C will have a larger acceleration than block A clearly because the tension here is smaller than the tension here and that is the only force acting on block A and C which are of the same masses in the horizontal direction okay so these two are two simple concept questions okay now let us solve problem number one the two blocks are shown to be originally at rest neglecting the masses of the pulleys okay neglect the mass of the pulleys and the effect of the friction in the pulleys and between block A and the horizontal surface okay so no friction here no friction involved here no friction involved here okay masses are negligible okay determine the acceleration of block B and the tension in the cable okay so tension here tension here and tension here okay so let us solve this problem for the next 5 to 10 minutes okay and we will discuss if you have any questions send them my way okay on the chat I will try to answer them there okay so this problem okay really straight forward problem so just note here is I can draw a free body diagram for this and this so let us if I use the white board for this mass on the top no friction only force acts is the tension because all the pulleys are frictionless w n and this is equal to m a for the top mass for the bottom mass what are the forces that act the forces acting are these two forces okay and this force and we neglect the mass of the pulleys here but we can as well write of a draw free body diagram like this we can as well draw t t t this is weight okay acting downwards and what do we know that this should be equal to 1 a 1 this should be equal to m 2 a 2 done these are two simple equations we can write okay this weight this weight is given so there are one unknown is tension okay this is a 1 and this is a 2 so tension is one unknown a 1 is other unknown a 2 is other unknown how many equations we have we have one equation here sigma f in this direction is equal to m 1 a 1 here we have the equation sigma f in the y direction is equal to m 2 a 2 so we have two equations and three unknowns so how do we take care of that we realize that the length of the string should remain constant so what is the length of the string this is x is x 1 plus okay we can just write down these two quantities y 1 plus y 1 plus y plus y plus y and these two are two constant lengths okay so we have x 1 plus 3 y 1 is the length or we can say that x 1 double dot plus 3 y 1 double dot is equal to 0 or a 1 which is the acceleration of the top mass plus 3 acceleration of the bottom mass is equal to 0 and we can immediately find out what are the three equations three unknowns we can solve for them now there is one variation that one person asked a question is that what if there were additional okay so these are this is just one string and because this is one string I can write a relation between them now suppose instead of having one string we had something like this one more string has to be added suppose so we can say I add another string here then what happens just note that if I had another string here then this problem is fully constrained not a mechanism okay not a mechanism at all so we have to use ideas from static equilibrium that we had done before in order to solve this problem okay so this is problem number one so why do not you have a go at I had already told that who have finished problem number one to go for problem number two okay there are a and b parts okay the same problem okay there are two different parts of it have a go at problem number two and you can go for problem number two b also this is little bit of a difficult problem problem three okay you have to do some thinking if you are if you want to be challenged go for this problem and later we will solve problem four problem five and problem six are also there okay so just have a have a go at problem number two a and when you are done with problem number two a solve problem two a go for problem number four after this solve problem number four problem two a then solve problem four so in problem two a okay let us briefly discuss this a three kg collar rest on the friction less arm a in case you have finished this problem go on to other problems go to problem four if you have done problem four go to problem six and seven in case you have finished everything go to problem number three the three kg collar b okay so this is a collar rest on the friction less arm a okay so this is the friction less arm the collar is held in place by the rope attached to the drum okay so the collar is held in place attached to the drum and rotates about o so this is the point about which the rotation happens in the horizontal plane the linear velocity of the collar b okay the linear velocity of the collar b is increase okay according to v is equal to 0.2 e square where v is in meter per second okay and t is in second so find the tension in the rope and the force in the bar on the collar after four seconds if this r is equal to 0.4 meters okay so as state forward so what we have to do is that we have to draw the free body diagram and the kinetic diagram for the collar equations of motion for the collar and kinematics of the collar so we have to combine all these things together and solve so first is this okay we are given that v is equal to 0.2 t square r is equal to 0.4 meter so how what are the forces that act on the collar in this collar one force is the tension other force is the normal reaction that the rod exerts on the collar okay so this is the normal direction that we have chosen because this is moving in a circle okay we choose that this is the normal that this is the tangential direction this is the normal direction as expected what are the forces that act tension here and the normal reaction from here now what do we know that Newton's laws tell us that this normal reaction should be equal to m at and this tension should be equal to m a n okay straight forward that f n is equal to a n f t is equal to m at now n will be equal to m into v square by rho v is the velocity here what is that velocity the velocity in the tangential direction rho is the radius of curvature which here clearly is r and what is the this tension t okay m dv by dt why because we are looking at at is what at is the acceleration in the tangential direction now what is the acceleration in the tangential direction clearly dv by dt because this v that is given to us is the velocity in the tangential direction so this tension should be equal to m dv by dt so ultimately what do we want we want to find out tangential velocity okay a normal acceleration and tangential acceleration so what is the tangential velocity expression it is given to us that a tangential velocity at any instant is 0.2 t square so this is given to us so after 5 seconds what we are asked is that that find a tension in the rope and the force in the of the bar on the collar after 5 seconds if r is equal to 0.4 meters so we are given that after 5 seconds what happens so the tangential velocity is 0.2 t square straight away put the numbers will come out to be 5 meter per second what is the acceleration in the normal direction okay the acceleration in the normal direction okay what is the normal direction this along the along the centre of the radius along the centre of the circle which the which this mass traverses so acceleration is given by v square by rho comes out to be 62.5 meter per second square and what is the tangential acceleration is nothing but dv by dt okay simple just this is the speed tangential speed so dv by dt will be the tangential acceleration so what we have done is that that we have used the path coordinates in this case we have used the tangential coordinate and the normal coordinate found out what is the velocity in the tangential coordinate it is given we take the appropriate derivatives we find out the acceleration in the tangential direction and correspondingly using v square by r we found out what is the acceleration in the normal direction so substitute in the equations of motion what do we see that sum of all forces in the normal direction what is that the only force here is this tension t should be equal to m a n a n we have already found out so we can find out what is the tension sigma f t is equal to m at so we can find out okay so just note that this is there is a confusion here okay so this should be t okay just there is a mistake in this slide so this sum of all the forces in the normal direction is t here okay so this n should be replaced by t and here okay the confusion came because t capital T is for tension whereas small t is for tangent and capital N is for the normal reaction and small n is for the normal component of the acceleration and so this n got replaced here this t got replaced here so this is not n or the normal reaction this is the t tension because that is the one which is contributing to the acceleration in this direction along the normal direction and this is n which is the force which is contributing to acceleration in the tangential direction and we can find out what this forces and acceleration are okay so this is t and this is n okay please note that and then the second part of the question is how would the motion change so this is happening in the horizontal plane second part of the question is how would the problem change if the motion was in the vertical plane the only way it will change is that that we also need to apply an mg component here and so the mg will have one component along the t direction mg will have another component in the n direction and we have to just proceed with that okay are there any questions on that okay so everyone is okay now let us move on to problem number 4 okay so very simple problem what is asked is that at a small ball of mass m is supported okay by a wire here okay a wire in the inclined direction and a chord in the horizontal direction okay this entire assembly is in equilibrium now suddenly this chord is cut and what we are asked to find out is a determine the ratio k of the tension in the wire immediately after the chord is cut to that in the wire before the chord is cut this is the wire this is the mass this is mg or the weight and when the horizontal chord was attached it will exert okay some force f okay and this is the angle theta now when we want to we first want to find out tension 1 before the chord is cut and tension 1 we can easily find out what because this is in equilibrium so if we take equation of equilibrium in the vertical direction then what do we have we have T1 cos theta is equal to mg so T1 will be equal to mg by cos theta now if we cut this wire then what happens we get a situation like this let us say this is T2 this is theta this is mg now let us use the path coordinates this is the tangential direction this is the normal direction now what we realize is that here this angle is theta so mg sin theta should be equal to m acceleration in the tangential direction straight away okay why because mg sin theta is the only force acting on this body in this tangential direction whereas along the normal direction what do we have we have T2 okay minus mg cos theta should be equal to v square divided by rho where rho is the length of this wire but since in this position okay when you immediately cut it the velocity 0 this has to be equal to 0 so what will we get we will get T2 is equal to mg cos theta and then T1 by T2 or T2 by T1 if you do it that is a ratio k that is asked of us that is nothing but cos square of theta so note one thing that the tension okay cos theta is always less than 1 so the tension suddenly decreases and which makes sense that the tension what it is doing it is preventing the body from moving out okay so keeping the body in equilibrium and the moment you cut this cord okay then the system no longer has to be in equilibrium and there is a sudden jump in tension because it can have an acceleration now in this direction which was previously not allowed and the tension was balancing this component of the acceleration and some part was balanced by this so the tension immediately jumps after you cut it and T2 by T1 sorry the tension immediately decreases after you cut and T2 by T1 will come out to be cos square theta okay so there are couple of questions okay let me answer them before we move on to the quiz okay so it is by center 1, 2, 3, 4 they ask that if there is no gravity then do we how it tackle the problem okay so that is what we discussed right that in problem number 2a that arm was rotating in the horizontal plane so this is the drum okay this was the sleeve and this entire assembly was rotating about 0.0 this is x this is y and then what did we see is that that if this is the horizontal plane then the gravity acts inside the plane of this paper okay so if you again look from the zx plane you will see that this is 0.0 this is the drum this is the rod this is the sleeve okay this is the tension wire connecting them and this is gravity but as we had seen earlier because the rod cannot go downwards because the sleeve cannot go downwards because of the presence of this rod there is no acceleration in the downward direction so gravity is not doing anything interesting but on the other hand if we put it in such a way that gravity acts in this direction then the only change that happens is that that we had a tension acting on it from the wire connecting the drum and the sleeve and the rod and the sleeve there was a normal reaction that acted from the rod to the sleeve in addition we will also have mg here nothing will change only thing that will happen is that we can resolve this mg into 2 components one like this and while writing out man equal to fn and mat is equal to ft this normal force will also have a component coming from the gravity and this tension will also have a this tangential force will have a component from tension as well as gravity and accordingly the normal force and tension will be modified okay that will be the only difference that will happen if the gravity also acts on it in the in the sense that the rod is not rotating in the horizontal plane but in the vertical plane.