 So welcome back after the brief tea break. So before we begin our tutorial session, let me solve just one problem, the ladder problem, the simplest problem using another approach which is typically done in textbooks. I will just take 5 minutes and then we will get on with our tutorial session. This is theta angle, say this is O the origin AB, this is weight W, this is force P. This problem can also be alternatively solved using another approach and another approach is as follows. We can say that with respect to a fixed coordinate system, it is very important to have a fixed coordinate system which is at the origin here intersection of these two sides. We can say that the y coordinate of point A is equal to 2L or this is L by 2, L by 2 is equal to L sin theta is the y coordinate. Now note that this is our y axis and this is our x axis. We can also say that x coordinate of point B is equal to L cos theta. Now this implies that delta yA is equal to L cos theta theta and delta xB is equal to minus L sin theta delta theta. Now note one thing that unlike the geometrical approach that we had used earlier, we do not need to take care of the directions up, down, sideways and so on. It just depends on the coordinate system that you are using. Think here the coordinate system here is y is up, x is to the right. So when you say delta yA, delta yA and delta theta. Now how is delta theta defined? It is defined like this. If you increase theta then this is what will happen. Now think about it that if delta theta is positive then what happens delta yA is L cos delta theta is also positive. As a result what we see is that delta yA go upwards and this is precisely what we will see from here that if delta theta is positive delta yA automatically becomes positive. What about delta xB? Now delta xB think about it. If delta theta is positive then this point moves inwards and as a result delta xB becomes minus L sin theta delta theta which is negative so that negative sin is automatically taken care of. We do not have to worry about what is positive, what is negative and then using principle of virtual work what we can say is that the w y is upwards is positive so w downwards is negative so minus w delta yA p negative because it is on the other side of x minus p delta xB is equal to 0 or in other words what we will get is that that minus w L w yA by 2 because this is at the center the load is at the center so it will be L cos theta by 2 delta theta this is minus this is also minus so we will get plus p L sin theta delta theta is equal to 0 and so we immediately get our answer that we had obtained early because this is true for any arbitrary delta theta. So this was another alternative method which is typically also done in the text books so I just wanted to mention that but I still like to go by the geometrical method which I had discussed in the morning because that gives a very very good feel for the problem whereas this becomes very mechanical and if you are not careful enough to choose the origin at a fixed point then the answer that you may get would be complete garbage but ultimately depends on what you are comfortable with and both are equally fine so long as you do them appropriately. So with this much of a preamble let us move on to tutorial problems a very very simple problem okay it will like not take more than 5 minutes okay we will discuss this problem I will give 5 minutes for you the 2 bar link is shown is supported by a pin okay this is a pin connection and a bracket connection at B and D determine the force P okay so this force P is applied upwards force E which is 100 pounds is applied downwards 150 pounds is applied downwards so what we are asked is that I determine the force P required to maintain equilibrium of the linkage very simple problem the overall virtual displacement is very easy what we are asked to find out is what is this force required to keep this system in equilibrium okay very simple problem think about it we will take 5 minutes to solve this problem any questions please direct I will be happy to answer via chat okay I can see that many centers have solved this problem so what we realize is this that this particular triangle okay you can clearly have a mechanism like this look at this look at this cylinder here okay this what can happen you have 2 members which are connected by a pin joint okay and add the collar okay this member is unable again connected by a pin joint now what can happen is this okay how do we give the virtual displacement to this structure look at member ABC suppose ABC and EDFC were not connected to each other okay then this clearly becomes a mechanism why a mechanism because free rotation for member ABC is possible about joint B if these 2 are not connected so what we do is that let us give us that rotation delta theta okay what possible mechanism that can happen but now note one thing we can do that but because these 2 are connected with each other what will happen is that this joint will open up if you give it a virtual rotation of delta theta so what we do is that this joint open up why how will it open up that this point BC C will move exactly downwards okay C will move exactly downwards okay and point A will move exactly upwards okay because it has to move perpendicular to the line joining A and B okay this is the line joining A and B we are rotating this about point A so this will move exactly upwards so this will go up this will go down and if we do not provide any moment to this member then the joint will open up like this so what we do that this delta C will be given by how much okay delta C will be given by this rotation multiplied by this distance rotation multiplied by 8 inches now what do we do this point moved by that much amount so the same displacement that comes here should be provided to this entire mechanism that this rod has to completely translate down by how much amount by this amount that is delta E delta F should be equal to delta C all of them move parallel to each other and this is the ultimate resulting virtual displacement which is completely compatible why because there is no virtual work done at point B in this mechanism there will no virtual work done at this joint there will be no virtual done at the collar why because we are not providing any displacements at this point only virtual work will be done by this and this and this and how do we get a relation between what should this delta theta and what should be this delta A and what should this delta E B the relation is straight forward that delta A because this is 16 this is 8 if this distance if this displacement is delta C delta A will only be twice of that distance and delta E and delta F okay if you look here they all of them are together so delta A will be 2 times delta C and all these displacements are equal to delta C so put everything in the principle of virtual work according to our virtual displacement this point okay moved up but the force applied is down so the virtual work done is minus so minus P delta A what about the other loads this is 100 into delta E plus 150 into delta F both are in the same direction now we substitute the relation between them that delta A is equal to 2 times delta C whereas delta E is equal to delta C and both delta F is equal to delta C put everything together we immediately see that P is equal to 125 pounds so the entire trick was in just recognizing that this undergoes some rotation this undergoes pure translation and to prevent this linkage from opening up we need to make sure that this relation is satisfied that delta A should be equal to 2 times delta C once we satisfy that we are done if you try solving this problem is in Newton's law also not a big deal but you have to draw two free body diagrams but in this case just this one recognition you immediately know okay what the answer should be for the P to maintain this body in equilibrium now let us move on to this second problem okay what do we have is that we have a mix of linkages okay so these are called as something like Nuremberg scissors so this is a demonstration that I have taken from demonstration.wolffrom.com and what you will see here is that this particular mechanism I can decrease the number of scissors okay increase the number of scissors okay and then we can change whatever size here is but ultimately what we want to do is we can extend the scissors like this or we can reduce the scissors like this so this precisely is the mechanism that I am showing here okay so AB, BG, GD, DE okay so we have members AB and BC and CD and DE and similarly we have members AH and HC and CF okay and FE okay so we have all these members they are connected okay to a hinge support here and they are all connected to each other by points note that this BC, BG is one complete member, HC is one complete member, CF is one complete member and GD is one complete member okay so these are not two force members now this are supported by at another place by another set of by another hinge or roller hinge or pin connection at point E now to this mechanism we are applying a load of P here we are applying a load of 2P here and force Q in the horizontal direction at point D and what we are asked to find out can we find out what is the horizontal reaction okay we have here in the question I have asked to find out what is the complete reaction but for the time being can we find out what is the horizontal reaction at point E that comes from the applications of these forces okay note one thing that this problem can be more easily solved okay using the coordinate approach if we put a at point A a coordinate axis in the x direction coordinate axis in the y direction and then write down appropriate y coordinate for HG x coordinate for D and release this support convert it into a roller and a horizontal reaction at E and correspondingly the x coordinate of point E then we do appropriate differentiations find out that if this angle changes okay think about it this is a one degree of freedom problem that if this angle theta changes by a small angle okay delta theta which means that this entire angle changes by a small value of delta theta then that change is reflected everywhere okay so it is a one degree of freedom system so if you put a coordinate system here and here in the x and y direction write the appropriate coordinates we can solve this problem quite easily and this is one particular example where we can see the beauty of principle of virtual world I can have 50 hundred linkages like this if you try solving this problem using Newton's laws moment balance force balance the number of free bodies that will be involved will be phenomenal whereas in this case it will be quite easy to do this problem okay so it will be a matter of like just like four or five lines okay why don't we solve this problem for around between five and ten minutes and then we will get back so let's get to the solution of the problem quite a few of centers gave the appropriate answer so what we did first release the support keep this roller here but the so what we did is that this was a pin connection or a hinge connection we replace this with a roller and a corresponding horizontal force now what do we do we put a coordinate axis at x and y this is only one degree of freedom system because if you provide a virtual displacement from theta to delta theta it changes uniformly everywhere so what we can do is that we can say what is the coordinate of point P it will be minus a sin theta with respect to this fixed coordinate system coordinate of point G again will be equal to minus a sin theta whereas the x coordinate of point D why I am interested in x coordinate because I need to find out that upon giving a small virtual rotation here okay changing this angle how much does the point move in the horizontal direction I don't care about a vertical direction why because this force will not do any work against the vertical direction so I am only interested is what in what the horizontal displacement is in this direction so we write down it is 1 2 3 4 5 5 a cos theta is the horizontal coordinate of point D now if this is yp delta yp is how much minus a cos theta delta theta let us pause to think about it what is happening here if delta theta is positive then what happens delta theta increases this point will move down even much even more so this quantity is what this quantity is become negative and clearly as expected positive displacement upward displacement is positive according to the sin convention downward displacement is negative so that negative is automatically taken care of here similarly delta yg same thing will happen if you increase this angle this point will go down and that is what is reflected in the negative sign here xd is 5 a cos theta so if you do delta xd it will become minus phi a sin theta delta theta now why this negative here if you increase this angle think about it this entire this point D will shift inwards what is our positive convention in this direction everything is positive in this direction everything is negative so that negative value is telling us that with increasing delta theta delta theta being positive this thing has a value that is negative which displacement is in the horizontal direction now we apply principle of virtual work what do we do this p downwards so according to sign convention minus p into delta yp g downwards minus 2p into delta yg okay and ultimately q is positive okay because it is in the same direction as x so q into delta xd and then without worrying about it is left or right we just put this value in with all the signs intact okay everything all the signs should be intact put it here and then we solve this we immediately get what is fex okay and one thing which I forgot to mention that xe what is xe xe is 1 2 3 4 5 6 6 cos theta so delta xe is equal to minus 6 a sin theta delta theta again when delta theta increases this point e tends to move inwards that is why this negative sign and the virtual work now is done by this this this and this force fex by our sin convention okay it is acting in this direction outward direction so we take it as positive I forgot to mention that here so it should be plus fex into delta xe write down all the equations here okay this is come here this comes here this is known this is known this is known we can solve for it because this is true for arbitrary delta theta and from that we get fex is this value which most of us have obtained okay so let us move on to this problem what do we have here okay little bit tricky not too difficult problem we have a mass which is lowered or raised okay by 4 identical links 1 2 and 3 and 4 on the other side so this mass has a dimension which is coming out of the plane so 2 is what we can see and 2 exact identical or on the other side okay so what we can do is that we can replace this mass by m by 2 and just consider only the front the front portion that we can see and what we are asked is that that the elevation of this mass is controlled by this hydraulic cylinders which are attached like this okay all the dimensions are given to you so what is asked is that that in this particular configuration when this length is B okay what should be the compression in each of the cylinders so how much force should the cylinders provide for the system to be in equilibrium okay so this is the problem so do that with principle of virtual work you can also do that with Newton's laws then these are two force members you have to solve everything but you will see that this particular problem is much more elegantly solved using principle of virtual work once you realize what the geometry is what the displacements what the angles are it is a matter of just three lines and you will get the answer okay so any questions about this problem or the previous problem please feel free to direct to me we will take 10 more minutes okay and then we will come back to this problem so we get back to the problem this solution is one of the older solutions that I had written it is not the most optimal but the point here is this okay first we have to figure out the full geometry okay many many colleges many centers have already solved it so this is B this is B so these two angles will be the same alpha so because some of all the angles is 180 degrees we immediately find out that alpha is equal to pi by 2 minus theta by 2 now note one thing that to raise or lower this mass what you have to do in this case the virtual displacement is provided in such a way that this mass is actually lowered okay and because of that what happens this hydraulic cylinder is replaced by the corresponding force that it is being that it is applying f of h now when we lower so once we remove the hydraulic cylinder note one thing that this entire assembly becomes a mechanism you remove this hydraulic cylinder you remove this hydraulic cylinder okay by symmetry okay we are again exploiting the symmetry because this assembly is symmetric we are saying that if the force in this is fh this also has to be fh so we are using only one half part of this okay that is one thing to note otherwise the number of unknowns present in the system are quite a bit okay but the only reason why we are using is is that there is a symmetry also about the central axis so we look only at one part now think about it when we release this hydraulic cylinder this assembly this entire mechanism this entire thing becomes a mechanism and as a result we can push it down and we keep that in mind while providing the virtual displacement we provide the virtual displacement the same as is there in this mechanism how do you what happens is that this point D moves to D prime okay if this angle is theta this angle is theta if the rotation in this is delta theta then again the rotation in this also has to be delta theta now note that this is a fixed point E and if the rotation in this case is delta theta which in this case is in the anticlockwise is in the clockwise direction then what we need to find out that what is the displacement of this point B along this direction okay what is the displacement of point B along the direction of the hydraulic cylinder why because only that component of the displacement does any work now we recall what we had done that we want to find out that what is in that along this direction how does it move so what do we do we draw this line and drop a perpendicular from point E okay where about which we are rotating this rod be draw a perpendicular like this drop it here so this distance is nothing but B sin alpha okay this is B sin alpha but what was alpha alpha was pi by 2 minus theta by 2 so this virtual displacement along this will be nothing but B sin alpha okay into delta theta but since alpha is equal to pi by 2 minus theta 2 that B sin alpha become B cos theta by 2 so immediately what do we say that the displacement along this direction is nothing but B cos theta by 2 into delta theta so now we know that what is the virtual displacement along this we also want to know what is the virtual displacement for this mass so it is very simple there are two ways of doing this one is we just write down the y coordinate for this so what is the y coordinate for this with respect to this fixed axis the y coordinate B B theta so for point D y D is equal to 2 B sin theta as a result delta y D is also equal to 2 B cos theta delta theta since we are decreasing the delta theta this becomes negative and why delta y D actually comes down so we can do it this way but delta y D is equal to 2 B cos delta theta downwards for the virtual displacement that we had provided so now we have delta y which is 2 cos theta B delta theta we have delta or the displacement virtual displacement along the hydraulic cylinder which is B cos theta by 2 delta theta put everything together use principle of virtual work you will see that force in the hydraulic cylinder will be W cos theta into cos theta by 2 okay so this problem becomes so much more simple when you use it solve it using principle of virtual work it will not be that difficult if you use Newton's Newton's laws also principles of equilibrium but then you have to draw multiple free body diagrams and do the balance whereas here you get the answer in one shot so there is one question from 1331 that the concept that we have used for E p is not convincing okay so let us see if I can convince them so as far as this is concerned this is a triangle okay E B so let us look only at this portion okay only at the portion which is below this this is the force in the hydraulic cylinder F H this angle is theta this angle is alpha this angle is alpha so what I had said is that that the displacement okay we have line E B and for line E B we are giving a virtual rotation okay along the direction in the clockwise direction now what we know clearly is that this point B if I draw it separately here because of this rotation it will have a displacement which is approximately perpendicular okay so this angle is 90 degrees this is the line of action of force so this data is equal to B times delta theta now what we want to know that this angle was alpha okay so this will be 90-alpha and this will be equal to alpha so what do we have here so we have here we can write it in a different way that if I magnify this portion you will see that this is this is delta which is B which is equal to B times delta theta and this length will be nothing but delta into sin alpha which will be equal to then the displacement along this line will be nothing but equal to B delta theta sin alpha or equivalently it is equal to B sin alpha delta theta but just note one thing what is this B sin alpha this B sin alpha is nothing but if this is B B sin alpha is essentially this distance and what this is telling you is that that if you had a rigid body like this then any point along this line okay if you have a big rigid body like this to which we are giving a small rotation in the clockwise direction then any point along this line will have a displacement in this direction only given by B sin alpha delta theta so if this is this geometry is not clear just draw it again you will see that there are two ways of doing this first it is to recognize that if the rotation for this E B happens about point E then the relative displacement will be perpendicular to the original line and then draw the angles and then find out that what is the corresponding component along this direction of FH that we want or the other way is that we know what is the line of action of this force so from point E drop a perpendicular here and this distance multiplied by delta theta you will see that it is also exactly equal to what is the displacement of this point along this direction so both are equivalent ways of doing it whatever you find convenient just think about it and it will become clear you have to just draw the appropriate figures and then interpret that accordingly so if that is clear then kindly send a chat message will have one last query before we move on to the next problem so this is a matter of geometry okay so if you are not comfortable in case right now it's just a matter of sitting down drawing the figures if you just stare at it for half an hour everything will become perfectly clear so it's just how much you stare at it okay I am almost sure that most of you know all these things but in case things are not clear or for example that there is some ambiguity the slides are there just stare at them just if you stare at them for half an hour everything will become clear the idea is that when you remove some support if you remove some hydraulic reaction the assembly becomes a mechanism and you want to provide virtual displacement to that assembly in such a way that the way the mechanism will deform that should be your virtual displacement and then you also don't want to change the length of any members neither do you want to open up any link okay and when you satisfy all these criteria you can move individual members and then deform and then give appropriate virtual displacement virtual angles in such a way that the criteria about no length no member should change its length any joints you don't open them if you satisfy that then you have your virtual displacement where the only work is done by the given forces and the force that you are interested in finding out so the idea is in recognizing that this becomes a mechanism and then figuring out how will that mechanism deform okay we will move on to this for this problem is a bit involved okay but I leave it to you for your own thinking when you go back not really involved but but it requires some time so what we do now is we look at this mechanism so the claw of a remote action actuator develops a clamping for C as a result of the tension P in the control rod so what we are doing is that that this is an assembly where in this configuration you try to pull on this rod P and when you try to pull on this rod P what will happen is that this will try to clamp in and if you keep some if you keep some material here for example then this will try to clamp in on that material and that material in turn will exert reaction C C symmetrically on the two jaws and what you are asked to find out that for for under the application of this force P what will be the clamping force at the jaws and all the assembly is given to you okay so think about this problem okay you can use some coordinate systems and so on or you can visualize this appropriately and get the answer okay it is it requires some thinking but not that much we will take ten minutes for this and after that we will discuss this problem if you have any questions about this problem or any other problems please send a request on chat and we can see if we can discuss there or we can discuss after after the end of ten minutes okay so I can see that many centers have solved a problem properly there is one simple way to solve this problem so what we do is that we know that the virtual work should be equal to 0 so what is this we give the virtual displacement okay I have not done a figure here but look at this top jaw okay if we pull on this okay what happens if you pull on this one okay suppose we put a virtual displacement to this entire assembly that let this be pushed outwards okay let this be pushed outwards or let this be put push inverse whatever it does not matter so if you take this push it inverse by little way little bit what happens this angle has to change as a result this will rotate and when this rotate to maintain all the compatibility at all the joints this member also has to rotate now if this member rotate this upper jaw by an angle delta theta then you will immediately see that this delta E okay this point E which is connected here will also move up okay by an amount how much C plus D delta theta whereas this jaw will move by an amount B times delta theta so what does that mean that would mean that delta A which is the vertical displacement here is equal to B divided by D plus C okay so B divided by D plus C why because this will be B delta theta this vertical displacement will be C plus D delta theta and so you take the ratio and what you will see is that delta A will be equal to B by D plus C into delta E so we have a relation that we want to find out what is the virtual work done by this force and this force so we need to find out what is delta A and that is what we get here so delta A we get in terms of delta E but now this is not enough why because ultimately the work done at this force P will be P times delta C what is delta C is how much this distance change by that will be this the change in this length that will be delta C so P times delta C will be the work done here so we need to actually express delta E which is this vertical virtual displacement in terms of the horizontal virtual displacement delta C to do that we recognize that point B can slide only upwards okay this can slide only sideways so this is very much like the ladder problem that this can only go upwards this can only go sideways so what we know is that that E square plus C square is equal to length of this link so we can use calculus and say that delta this is 0 because the length of the link should not change so 2 E delta E is equal to minus 2 C delta C or we then get a relation between delta E and delta C so now we know delta A in terms of delta E we know delta E in terms of delta C so what is the virtual work virtual work done will be 2 C into delta A plus P into delta C okay 2 C into delta A plus P into delta C substitute everything here okay because delta A is now expressible in terms of delta C since it is true for any arbitrary delta C you will see that compression will come out to be P by 2 into E D plus C divided by BC simple geometry okay so are there any questions on the chat about this so it seems that this problem is pretty clear to everyone now this problem okay is somewhat interesting problem so what we have is that we have an assembly what is reused to crush stones for example crush rocks now how is this assembly there we have one member two member three member four member they are pinned here and they are connected by hinge support at the top point now what is the mechanism that there is a hydraulic assembly which is pushing against this members what is the pressure okay P1 is 100 MPa here P2 is equal to 60 MPa here the diameters of the pistons here are given as 100 millimeters this angle is 30 degrees now what happens is that that when you apply this force this will try to extend okay this angle 30 will tend to decrease and as a result it will increase exert some force here may be horizontal and vertical both okay it is not clear immediately so it can exert a vertical and horizontal force here because of those forces this AB we will try to go down this angle 10 will tend to decrease and this point C will tend to I am not singing it will it will tend to crush the rock okay so ultimately what do we want is that that there will be some force F that will be exerted here we replace this rock with that force F and we want to provide virtual displacements to this to this structure in such a way that the only work is done by the moment here delta D by the moment here delta E because these forces are known we only want the horizontal displacement here and here because these forces are known and what is a what is a displacement here why because we have replaced this rock and so in the absence of rock this is a mechanism and the way this mechanism will deform is the same way we want to give small virtual displacements and then use principle of virtual work where the work will be done here and here some should be 0 and that will give us force on the rock so we will take just 5 minutes then we will briefly discuss the solution now this problem just a little bit tricky not too much just think about it okay if you have your full screen put on okay I want to make some hand gestures I did not know I could make them but apparently I can so look at the top assembly FDE it will form one triangle like this and a bottom triangle like this if you give this assembly a virtual rotation okay a virtual displacement how by changing this angle okay just a bit if you change the angle just a bit what happens this point B will move down okay this point D this point B will move down okay how much amount you can calculate that because we have enough problems we can easily figure out how much distance it will move down by okay now what happens what will happen to point D and E when you put a virtual displacement point D and E both come inside in the horizontal direction okay both vertical and horizontal directions will be present but because this force is horizontal we only care about what is the horizontal displacement of D and E so what happens is that you have this rhombus okay if the point B goes down point D and E they come in and they also go down we worry only about the horizontal displacement of D and E but now what happens if point B goes down point B is common to AB and BC also so we cannot do that we cannot provide a virtual displacement where only point E moves down because then there will be a gap between these two assembly so how do we clear that gap we clear that gap by also providing a virtual displacement in such a way that for the bottom assembly point B also moves down okay point B also moves down how much does it moves down by that how much distance this point B moves down from the top assembly is the same distance this bottom point B should move down by but then what happens just note our ladder problem that for the virtual work for the virtual work problem we had done previously if this point is free to move horizontally then this top point B will not only move downwards but it will also move sideways just look at the figure we had previously for the problem where there was a piston here that the point B will not only move down it will also move sideways then what happens then we can match the vertical displacement of point B coming from the top assembly and a vertical displacement of point B coming from the bottom assembly but the horizontal gap still remains and to take care of that horizontal gap what should we do now is that we should take this entire assembly okay and rotate it rigidly about point F okay the entire assembly rigid about point F how much does it rotate by by an amount so as to clear this gap but as a result what happens initially point D and point E both by symmetry had the same horizontal displacement but now because of this rotation you will see that one point D has more displacement other point D has less displacement and as a result the virtual work components that come from here will not have the same virtual displacements and they will differ you look at this figure okay when you go back you look at in leisure okay all the entire solution is given to you but if you use the coordinate system put a coordinate system here put this theta delta theta you will never realize that this rigid body rotation should also be present in order to make sure that there are no internal gaps and when you put that you will see that the virtual work equation will look like this and the p which is the force that is exerted at the rock will be equal to F1 plus F2 where F1 is the force exerted by one hydraulic cylinder F2 is the force exerted by the hydrolyzed cylinder tan theta tan alpha are all given this is theta this bottom is alpha okay plus F1 minus F2 by 4 and this term this rigid body rotation that is required for the virtual displacement becomes only important when F1 is not equal to F2 otherwise you will not even see this but when F1 is not equal to F2 that will contribute and the virtual displacement will be 3 components the top one has an angle change like this the bottom one has an angle change like this and a sliding and to match that gap this entire assembly should rotate and get fixed here okay these 3 components of virtual displacement when you put them together you will see a final assembly which looks like this green line okay and from that okay stare at this if you already have not figured it out you will see that this work will be done this work will be done and ultimately this point see also moves sideways so the unknown force here will also do work and all these 3 together will do some work and we get the final answer now a disclaimer here that this problem is much easier solved using Newton's law but this problem is given precisely to show that what possible virtual displacement can be present in the system if you want to solve a particular problem using this method so it is a very clear demonstration that how different geometry changes in a mechanism can happen when we remove this thing these are the mechanism kind of geometry changes that happen and how to incorporate them all together in order to visualize that in the absence of this what particular mechanism does this collapse to and then use that particular mechanistic collapse to find out what are the virtual displacements at various points use principle of virtual work to get our appropriate answers okay so I hope that in the lecture okay I have convert the beauty and power of virtual work at least to some extent to most of you okay if you still have some concerns to still have some confusions okay you can write to me on the chat you can write to me on Moodle okay and we will see if we can answer those questions and can answer those doubts okay so the entire next week we are going to do dynamics we will do kinetics kinematics of particles kinetics kinematics of rigid bodies okay and hopefully have a lot of fun doing that okay so have a good day.