 Welcome to a day filled with energy methods in engineering mechanics. So in the morning I will discuss what is the method of virtual work, it is a part of energy methods and in the afternoon we will discuss about minimum potential energy theorem and how it is applied in engineering mechanics for finding out equilibrium of various systems and the corresponding stability of those systems. For example in a few places that I know of that these topics are not typically discussed but there is a reason why I want to discuss this. So first of all towards the end of this class it will become very clear that principle of virtual work that if you can get a understanding of how this principle is applied then even when you go back to our equilibrium 2D equilibrium we had done a couple of days ago it will be apparent that a good hold or a good mastery over how to apply principle of virtual work will automatically give a good hold or good mastery over how to apply or how to solve problems in 2D equilibrium. You bear with me for some time, I am sure you will find it interesting if you have not done this before and there will be a tutorial at around 11.30 the same time we had a tutorial yesterday and if required we can also let this session spill over in the afternoon period when we will take the class. But you will see you will definitely appreciate by the end of this class and that is what I hope to that how this simple idea will let you understand that upon releasing some constraints how does the remainder of the structure move and from that by looking at any problem you would immediately have an idea that what appropriate free body diagrams to draw what are the appropriate equations of equilibrium and so on. But first things first what we will do is let us see how to use what is this principle how to use this principle and in what problems or in what cases the application of this principle becomes really useful and makes the problem very straight forward as compared to solving it by equilibrium methods ok. So with this brief reamble let me discuss ok so we have discussed four still now Shobik had taught 3D equilibrium trusses we did friction 2D equilibrium and so on ok. So we are going in little bit of a different territory now we spoke the language of force and here we will speak the language of energy or work. Now we all know that if there is a force if there is a particle A on which we apply force F the force we all know is a vector in a particular direction the force acts then the work of the force F ok corresponding to a small displacement if the particle undergoes it goes from A to A prime and in the meanwhile undergoing a displacement of dr bar vectorially then the work done clearly is F dot dr ok we have learned that since our high school or in scalar notation F is equal to ds cos alpha the angle between these two straight forward ok. And special cases is that we know that when alpha is equal to 0 then work will be just F into ds how much displacement alpha is equal to pi work done will be negative because the displacement is in the direction of the force so if alpha is equal to pi what is pi means that the force and displacements are exactly opposite to each other the direction wise. So what we see is that that a work done will be negative and simple case if the displacement is perpendicular to the force then the work done we all know is 0 ok simple concept. Now coming little bit one step ahead suppose if we have a couple acting on a rigid body so that was a point which we discussed in the last slide if we have an extended rigid body and we have a couple F and minus F acting on it we want to find out what is the work done by the couple. So what we will realize is that if point A is where the force minus F acts point B is where the force F acts and at point B ok suppose we give this entire rigid body a small infinitesimal rotation of d theta ok then we can immediately find out what is dr 1 what is dr 2 and just little bit of manipulation if you will do you will realize that a total work done will be simply equal to FR where R is the perpendicular distance between these two points and F times R is the couple or the torque or the moment whatever the term you prefer and work will be simply equal to M times d theta. So simply if we have a point and a force acts on the point we know that work done is equal to F times ds into cos alpha alpha is the angle in similar vein if we have a rigid body an extended rigid body and if we apply a torque of M then M times d theta is the work done ok this simple things we just have to remember and then we are going to apply these things when we use the principle of virtual work ok so far so good. Now suppose that a particle A is subjected to multiple forces F1, F2, F3 and so on and to which we apply a virtual displacement. Now what is this thing about virtual displacement virtual displacement the reason we use this extra qualifier virtual is to emphasize that this is for example almost like a thought experiment you have a particle subjected to many forces it is not that the forces are giving the particle a displacement they may or may not but as far as our thought experiment goes what we are doing is that let us say that these forces are acting to which we imagine that suppose we move the particle or displace it by a certain amount delta R so that is why we use this a pre qualifier small delta so as to say ok so as to distinguish dR which is the actual displacement from the virtual displacement so we give it a virtual displacement of delta R and ask ourselves that subject to those forces what is the virtual work ok again virtual displacement producing virtual work so what is the corresponding virtual work and straight forward like our logic just extending from whatever we have done in the previous slides F1 dot dr F2 dot dr plus F3 dot dr which can be just written as F1 plus F2 plus F3 dot dr which is nothing but resultant dot dr simple logic vector logic. So delta U or the virtual work which is done by all these forces on this particle when we provide it a displacement of small delta R will be given as resultant of this dot dr. Now the question we ask ourselves is this suppose if all these forces ok F1 F2 F3 they are all concurrent forces in this case and suppose they are all in equilibrium which means that resultant R is equal to 0 what does that tell us about the work clearly if R is 0 then the virtual work done is 0 so what we know one way is that if you take a particle apply forces on it such that all the forces are in equilibrium in the sense that sum of all the forces is equal to 0 then the virtual work done on the particle for any virtual displacement is equal to 0 so there is one way result that we know but let us think in the other way round suppose we have a particle on which we apply any number of forces now if we provide any virtual displacement any direction and even with that suppose we get that the virtual work for any virtual displacement is 0 then we can say that the sum of force is also equal to 0 so it goes both ways that if the forces are in equilibrium virtual work is 0 and if virtual work is 0 for any virtual displacement then the forces the resultant should be 0 so what it means is that that force balance and saying that virtual work is equal to 0 for any arbitrary displacement are two equivalent principles that the principle of virtual work what it implies is the same thing which Newton's law imply nothing less nothing more and with that you have an alternative formalism to discuss equilibrium that either you say that Newton's law true all the force are 0 when they act on a particle it means that the particle is in equilibrium or you say that for any virtual displacement if the virtual work is 0 then also you say that the body or the particle is in equilibrium now like I am not going to prove it here but you can look it up in any textbook if not we can discuss after the class okay but suppose now this was a point so we are okay with the principle of virtual work on a point are there any questions about that so far so good very easy now suppose we have an extended body not a point then what the principle of virtual work says that if a rigid body is in equilibrium the total virtual work of external forces acting on the body is 0 for any virtual displacement of the body so saying that force balance and moment balance okay implies that the rigid body is in equilibrium is equivalent to saying that if on a rigid body I apply any combination of virtual displacement and then if the virtual work turns out to be 0 this means that the rigid body is in equilibrium okay simple principle of virtual work is what we have here we go one step ahead again this also we can easily show but we are not going to do it here for the lack of time if you have a system of rigid bodies which are connected by links okay and if we provided virtual work such that it is consistent with the linkages okay that is consistent that we don't break up the linkages or we don't disobey or we don't disobey the constraints provided by the linkages then for any virtual displacement if the entire set of rigid bodies in equilibrium then the virtual work which is provided by any set of virtual displacement is equal to 0 okay and this principle be your principle will become more clear as we proceed further and as we solve a few examples so let us take a very very simple example a simple rigid body okay which is hinged or pinned at point O and it has a roller support okay at the other point now suppose we want to find out that if on this rigid body we apply a force P what is the reaction produced at this point simple question we know that if you want to use the equilibrium approach 2d equilibrium which we had discussed before what we will do is we will take this free body out we release this kinematic constraint because at point O displacements in horizontal and vertical direction both are constrained the releasing of this constraint will give rise to possibly two reactions O y O x since this is a roller upon releasing this constraint okay this was already free so no reaction but a vertical constraint when it is removed it may lead to a reaction R and what we do is we take the moment about point A for this free body diagram and we can immediately find what is R now here what we use is a very different approach what we do is this we release this constraint okay remove this kinematic constraint so essentially what we have is that we have this triangle which is which is hinged about point O now we all know that if we have a rigid body just pinned about one point this is not a stable body it's a mechanism but what we are doing now is on this body we apply a virtual displacement okay we apply a virtual displacement how so that after providing the virtual displacement or virtual rotation whatever you want to call it the body in the final configuration of the body is in this dotted lines okay it is not a real displacement what we are just doing is this that suppose I just release the constraint on the one side and apply a small rotation about point O look at what is the resulting configuration now if I want to find out what is a virtual work done by all the forces okay what are the two forces now that can do work the reactions here cannot do work why because the point is stationary there is no displacement point O sorry sorry this point can do a work which is given by R times the vertical displacement here note that this is B so we will see later that a vertical displacement will be simply B times delta theta when delta theta is small and it is in the upward direction so the work done at point here will be equal to R times B delta theta let us look at the top point when we provide this virtual displacement this point will move sideways with a displacement of a times delta theta now this displacement is opposite to the direction of force I should put a negative sign here okay my mistake so this displacement is opposite to the direction of force so the work done okay will be negative so minus PA delta theta is the virtual work done at this point RB delta theta is the virtual work done at this point and the principle of virtual work says that if the rigid body is in equilibrium we provide it a virtual displacement okay then if you sum up all the virtual work done by the forces acting on the body then that will be equal to 0 you make it equal to 0 and what do we get we immediately see that R is equal to PA by B and that's exactly the same answer you would get okay if we use the equilibrium approach so what we have shown here is that is an alternate way of looking at the same problem you may say that we are not gaining much we write this equation we also write another single equation for 2d equilibrium so what is the big deal the big deal is that we will see that there are certain classes of problems which can be solved really really well by this second thing okay I will come to those points is if we can understand how do bodies move what are the mechanisms okay upon releasing some constraint how does each component more relative to the other then that will give us an intuition even for solving 2d equilibrium problems now suppose I want to find out I know R now I want to find out what is the vertical reaction at point O what we'll do is this we will release the vertical support here and give this entire body now of vertical virtual displacement so what we will have that the force here times delta y plus R times delta y will be equal to 0 then it immediately tells you that O y plus R is equal to 0 which is nothing but equilibrium in the y direction is this point clear or should I write it down on the page so let us take this rigid body here this was point O let us say that this was point B we found out that this was a reaction R acting at point B and at point O we can have two possible reactions because of pin support O y O x this force is P now suppose to this rigid body we give a virtual translation of delta y then what does the principle of virtual work tell you that given this virtual displacement what is the overall virtual work the overall virtual work now will be done by P delta y are perpendicular to each other no work O x delta y are perpendicular to each other work done is 0 what is a virtual work done here R times delta y note that the direction of virtual displacement okay so how will this final figure look like it will look like this upon giving the virtual displacement this is how the body will look like so these are the displacement no work done at point P no work done at point O due to force O x and a work done at point B is equal to R times delta y both are upwards so positive work what is the work done at O plus O y times delta y and principle of virtual work tells us that if this body is in equilibrium that the sum of the virtual work done should be equal to 0 but since this is true for any arbitrary delta y this immediately implies that R plus O y is equal to 0 but what is this equilibrium equation this equation is nothing but equilibrium in the y direction okay so is this point clear that applying the principle of virtual work by providing it virtual displacement in the y direction is nothing but equivalent to force balance in the y direction is this point clear that we can do the same thing by applying a virtual displacement for this body in the x direction and the only component that does work is what P times delta x plus O x times delta x equal to 0 so what do we see that P delta x plus O x delta x is equal to 0 and this simply means that P plus O x equal to 0 which is nothing but equilibrium equation in the x direction so that's why we know that principle of virtual work and the moment balance and force balance are two equivalent things we can use one way or the either okay and we will get the same answer only thing is that the philosophically they are drastically different and we will see that because they're philosophically different okay principles of virtual work have a great value is solid mechanics in structural mechanics and infinite element method okay we'll come that to come to that later okay so let us come here now there's a terminology that we use while doing principle of virtual work we say that forces that do the work okay as we saw previously that in this case R and P where the forces that did the virtual work and those forces are called as active force there are reactive and internal forces is for example look at this structure okay what we have here is that it's we have we really had here a vertical roller which we are released now okay vertical roller this is P this is F if we give it a virtual displacement then the reactions here okay will not do work the reactions here will not do work the internal linkages will not do work only work will be done by P and F and those are called as active forces and the work and the reactions of the forces we don't do the work are called as reactive forces now virtual displacements are to be given carefully so that only the active forces okay that only the active forces are only the forces that we know and the forces that we're interested in obtaining means for example let us look at this picture okay let me go back to the paper mode again suppose take this rigid body suppose I I give it a virtual displacement which is a combination of rotation and translation in the y direction okay or suppose I give it something like this let us move it upwards we give it a vertical translation in the y direction as well as a small rotation in that case what will happen is that that because there is a translation here work will be done by OI there will be no virtual work done by OX since this point moves upwards work will be done by R also and work will be done by P also so OI R and P all of them end up doing work and because of that what happens is that we get a coupled equation okay and we'll have P data X P plus R data Y plus data Y let me call this data Y B and O Y data Y simply okay only delta Y O see it equal to 0 so in that case what happens is that we will not get a clear cut answer to the problem that we desire why because we have exposed two unknown forces here so the idea is that there is nothing wrong with doing like this but what happens is that if you provide virtual displacements which are not judicious then you will get some equations it is like for example if you take moment for a free body diagram suppose this is the free body diagram in 2d equilibrium somebody may say why should not I take talk about this point of course you can do there is nothing wrong with that but what will happen that this will also contribute this will also contribute this will also contribute so we will not get a clear cut equation where you only get this unknown or this unknown so that in the same way when we are applying principle of virtual work the virtual displacement should be such okay they should be judiciously chosen that what we want to obtain and what we know should be the only quantities that do the virtual work and because of that in one shot can we get the unknown that we desire is the point clear okay let us move on and in free body diagram okay in 2d equilibrium we draw a free body diagram in virtual work method we draw what is called as active force diagram now what is the active force diagram that any constraint we release only the forces which will do the work are represented and then we provide a virtual displacement or a virtual rotation and a combination of them okay so what is that actual active force diagram will come that come to that in a few moments but as of now it is only a terminology is a process by enlarge is the principle of virtual work and how are we going to apply it is the is the idea clear let us move on okay so when we provide virtual displacements which are the forces that do not do work okay just some examples so reactions at a frictionless pin okay what are the internal reactions here due to rotation of a body around the pin don't do work okay these are all the details they will become clear when we solve the problem reaction at frictionless surfaces for example if I provide a virtual displacement along the plane then a vertical reaction will not do work weight of the body with CG moving horizontally if the CG has probing provided a horizontal displacement then the gravity acts downwards okay so that will not do a virtual work and the sum of the work done by several forces may be 0 okay for bodies connected by a frictionless pin bodies connected by an in extensible core internal forces holding together parts of a rigid body these are some there's a list shopping list we'll do a few problems and all these concepts will automatically become clear when we'll do those problems now the most important thing okay if you remember when we started with 2d equilibrium we had said that there are kinematic degrees of freedom and there are force degrees of freedom what happened is that in 2d equilibrium we are mainly interested in what are the force degrees of freedom the kinematic degrees of freedom were used only for example that if you release a constraint how many forces you will replace those constraints by that was the only thing we were using the kinematic constraints for but now in principle of virtual work what becomes extremely important is upon releasing a constraint upon providing a virtual displacement how is the body going to move okay so that becomes extremely important say what we will look in in this course in this lecture is only one degree of freedom the one degree of freedom systems so look at this system how many degrees of freedom do you think it has one why because if you provide because if I know the coordinate of this point all the coordinates or all the confirmation of this body is automatically fixed similarly here if I know this rotation if I know this angle then I immediately know what is the resulting configuration of this body similarly if I know this theta you can immediately find out what is this configuration but in two degree of freedom systems this is theta 1 this is theta 2 okay you cannot change theta 1 and expect theta 2 to appropriately change these two are independent quantities similarly here there is theta 1 and theta 2 both of them are independent and we need two variables in order to specify the complete configuration okay so to summarize principle of virtual work okay all this will become clear when we saw this all simple problems the principle of virtual work what does it tell you the virtual work done by an external by external forces on an ideal mechanical system what do you mean by ideal is there are no friction forces not that we cannot solve problems with friction but if friction is involved problems of virtual work becomes very messy we are not going to deal with that complication what we are going to do is that that for ideal systems which don't have friction or which have negligible friction then the principle of virtual work states that virtual work done by external active forces on an ideal mechanical system in equilibrium is 0 for all virtual displacements consistent with the constraints now what is this consistent with the constraints will again become clear when we solve a few problems okay as we discussed just a few moments ago ideal system is a system such that all surfaces joints are frictionless okay and consistent with constraints what does that mean that a virtual displacement should be such that they should not allow the non active forces to do any work okay let us see what all these things mean now coming back okay from where we started that why do we need principle of virtual you would say that we discussed 2d equilibrium in great details friction problems 3d equilibrium all things we did and why do we need to use principle of virtual work first thing we will see is that there is certain class of problems there are certain complex mechanisms for which if you do the same problem using equilibrium approach it will require huge amount of effort and even after putting in huge amount of effort you may not be sure that answer you have obtained is right second thing is we can obtain the active unknown force in just one shot a very very complicated mechanism for which you may have to draw let's say 3 4 free body diagrams by using principle of virtual work appropriately okay we can find out what is that active unknown force in just one equation and last and the most important point is that that such type of analysis okay when we add deformation component to this will be man it will be essential in solid mechanics structural mechanics and for example in the future when you may be teaching or students may be learning courses like the finite element method okay so with this much of a preamble okay with this much tantalizing introduction let us solve okay some problems but before I do that let me describe a generic observation that if I take a rigid body then what happens if I provide it a small rotation look at it we have a rigid body AB suppose now on this rigid body AB we provide a small virtual rotation about 0.0 of delta theta so this is AB a small rotation about delta theta such that this AB becomes AB prime okay now can you tell me what will be the distance BB prime okay when delta theta is very small just a AB AB times delta theta and since the length of AB is equal to L this distance BB prime will be approximately equal to L delta theta right everybody agrees with me L delta theta there will be of course some tiny error okay why because you are taking arc of a circle let me show it here that if you have this rod here this will trans transfers arc of a circle so strictly speaking there will be a tiny displacement also in this direction but that will be much smaller compared to this displacement which you say is of the first order so this is L this is delta theta this becomes L delta theta and we neglect this tiny displacement perpendicular in that direction now can we also say that what is this angle ABB prime approximately can we say what is this angle ABB prime approximately it has to be 90 degree why because some of all these three angles is 180 but because delta theta is very small okay you will immediately become it will immediately become clear that this angle is approximately 90 degrees clear so what do we know from here is that that if you take a rigid body AB and apply a virtual rotation or a small rotation about point 8 then the resulting displacement BB prime is perpendicular to AB and the magnitude of the displacement is given by to the first order L times delta theta okay so BB prime will be L times delta theta point clear so very simple point but very important point we are going to use it again and again in the course fine now do let us do one thing this BB prime is at some angle okay what is the angle if this angle is theta this angle also becomes theta we all know so BB prime is at an angle theta with respect to the vertical now what is the vertical component of this displacement BB prime just BB prime cos theta okay delta y is equal to BB prime cos theta which is nothing but L cos theta delta theta I can rewrite it like this what is the component of BB prime or the virtual displacement in the horizontal direction nothing but BB prime sin theta but what is BB prime L delta theta so this can be rewritten as L sin theta delta theta now note one thing there is a very important interpretation of this what is L sin theta L sin theta is this vertical distance BP what is L cos theta is this horizontal distance AP so what is the vertical displacement it is nothing but this distance AP okay times delta theta you understand that you get that point so we are just rewriting this L delta theta into sin theta as L sin theta into delta theta but what we immediately know that displacement of point B if you want to find out what do we do just draw a line okay perpendicular to the direction in which we want to find a displacement about because we want to find out displacement in the y direction what you draw is take do is that that through point B draw a vertical line okay draw a perpendicular line from A to P okay this angle is 90 degree and this AP times delta theta will be the displacement of point B in the y direction similarly if you want to find out what is the horizontal displacement of point B what do we do draw a horizontal line okay and from point A drop a perpendicular to this which length will be the same as BP and that distance into delta theta will be equal to the displacement in the horizontal direction now let me ask you one question suppose AB were not a rod okay suppose we had a triangle instead ABP let us say that we do not have a simple rod like this but we have a rigid triangle given by ABP now if I provide a virtual displacement or a small displacement of delta theta to this triangle okay if I provide a small virtual displacement to this triangle what will be the vertical displacement at any point along this line BP is a question clear same same it will be equal to AP times delta theta so it's a very important result what we see is that if we have a rigid body a full rigid body ABP now to that full rigid body if we give a small rotation data theta then what we know is that that along this line BP all the points have the same vertical displacement is the point clear appreciated okay and similarly what you will also see is that that for the same thing can be done if for example you draw a triangle like this let me draw it here okay if I draw a triangle like this this is perpendicular ABB1 now if about point O we give this entire assembly a virtual rotation of delta theta what do you think can be say about the horizontal displacement along all these points the horizontal displacement I am asking horizontal displacement what is the horizontal displacement of all the points along this line BB1 same and it is equal to B1A delta theta so this is a very important result what it tells us is that that as far as a horizontal displacement is concerned okay if you just draw this line all the points if you have a big body okay and I want to find out that I draw this line horizontal line and from point A I draw a perpendicular to that horizontal line then all the points on this line will have the same displacement in the horizontal direction and it will be given by B1A this distance multiplied by delta theta it's a very simple result but it's a very powerful result you will see that if you want to visualize some mechanism okay then this will become extremely useful and important and if you want to find out that along this vertical line what is a displacement of any point in the vertical direction along this line what we do we drop a perpendicular font point O because we are rotating the body about point O let us call this point as B2 then AB2 times delta theta will be the vertical displacement along all these points is this simple point clear okay and now let me ask you a third question something like this you have a big rigid body let us take point A here if we rotate this body about point A and suppose I ask that there is a direction which is inclined okay it's at some angle alpha and we give this system a small rotation of delta theta which means that the entire body gets a rotation of delta theta about point O then what can you say about a displacement in this direction on any of these points it will simply be equal to draw a perpendicular line let me call this AB3 then AB3 times delta theta will be the displacement in this direction for all the points okay is it simple point clear okay now this will become very important when we solve a number of problems you can always have a displacement take components of that and so on but with this it is immediately apparent that all these points along this line will have the same displacement in the inclined direction it will be equal to from point O about which we are doing the rotation for the body drop a perpendicular the distance is of that line is AB3 now AB3 into delta theta displacement of all the points in this direction okay so let us move on now now with this idea okay these are some mechanisms which you can solve easily using virtual work let us come to this very simple problem but very very illuminating about how principle of virtual work is actually applied okay we have a very simple problem we have a rod okay or for example many times you see that people when they want to climb up something there's a ladder which lean against one point and second point and people climb up the ladder so what we have here is that there's a equivalent of that we have an inclined rod which is leaning against the wall frictionless wall at this point it is resting at this point weight is acting through the center but in order to prop this rod up because if you don't apply this horizontal force if you remember the problem we did in 2d equilibrium this rod will just slide out so in order to prop this rod or keep it in this position we apply a force P and what we are asked okay is for this problem which we will call as a ladder problem what will be the force P required to keep this rod in equilibrium simple question we ask we can solve this problem in one shot using equilibrium approach how can we do we just draw two perpendicular line draw the free body diagram two reactions here meet at this point here so take moment about this point immediately get P so you say what is the big deal about virtual work the big deal is that that this is a stepping stone to solving some more complicated problems okay now how do we apply principle of virtual work here so before I go to the next slide and actually display the solution can you tell me that for this problem in order to obtain load P without bothering about the reactions at the top and the bottom point how do we apply principle of virtual work what virtual displacement do you think will be appropriate can we apply yes somebody right or less suppose suppose let me ask you one simple question okay let us draw it here okay before I go to the solution let us have some discussion no it doesn't depend okay we will see that okay but it's a valid concern that it may depend but it does not suppose this is the ladder will it be good if I apply it a virtual displacement like this a pure translation in the x direction is this a good virtual displacement to apply no why because this reaction will do work in addition to this P doing work so P is unknown reaction is unknown we don't gain anything okay so this is no good suppose I apply it a virtual displacement like this here take rotation about this point okay so give a delta theta rotation about the bottom point so this displacement will become how much L delta theta how much will be the horizontal displacement we know this is L this is theta so remember what we had done earlier the horizontal displacement will be nothing but this distance L sin theta into delta theta will be the horizontal displacement here vertical displacement will be this this horizontal distance L data cos theta or if you want to write it like this L cos theta delta theta sorry sorry upwards but what is the problem with with this virtual displacement that reaction and that this point will also do work okay so we are again like we are back to square one so both of these are not applicable so is there any other virtual displacement which I can provide or yes slide so what we can do is that we can bring this up and now entire assembly we can slide it by how much distance by a distance of a sin theta delta theta so that this gap is closed okay so that's precisely what we do is if you look at this okay if you look at the resultant diagram if you just take rotation about point B and apply this entire body AB a virtual rotation delta theta what do we get this gap opens up how much is the distance how much is the horizontal component of this gap it will be nothing but L sin theta L sin theta delta theta just note how do we get L sin theta delta theta is what we had seen is that just this distance multiplied by delta theta that is L sin theta delta theta so we want to close the gap what do we do we give a complete translation in this configuration such that the ultimate virtual configuration is this line a prime B prime which is such that point B has gone inwards look here by how much amount delta x because we are translating this entire assembly by delta x how much has point a gone upward by delta y right because whatever delta y was there it was equal to L cos theta which is this distance into delta theta that distance is not going to change after this translation fine okay so this is delta x which is given by L sin theta delta theta delta y will be equal to L cos theta delta theta and we are just now we have to apply principle of virtual work will the reaction at point a do any work with this virtual displacement no because point a has gone to a prime which has which has gone up vertically but reaction at a is horizontal so no work with the vertical reaction at point do any work after this virtual displacement no why because point B has moved only horizontally to point B prime whereas the reaction at B is vertical so no work done so only work done now is by what is by the weight here because if this displacement is delta y what is a virtual displacement of the center half of that just delta y by 2 what is a virtual displacement here delta x okay so p times delta x minus w times delta y by 2 should be equal to 0 but what we see from here if you take the ratio of dx by dy what do we get dx by dy is nothing but L tan theta L L cancels out says nothing but tan theta so if I rewrite this here what we have is that dx let me write on a fresh piece of paper dx by dy is equal to and theta principle of virtual work tells us that delta y is upwards so the work done by gravity is minus w delta y by 2 plus p times delta x positive because both p and delta x are in the same direction this should be equal to 0 but what do we know that delta y and delta x are not independent I can say that delta x is equal to delta y tan theta so it will be delta y by 2 plus p delta y tan theta is equal to 0 and since delta y is some arbitrary number we can just cancel out delta y from both sides and we know that p will be equal to w by 2 and theta okay which is the result which we had obtained for millions of times now so is the principle clear now very simple way of applying principle of virtual work the idea is this that the virtual displacement is how we know some basic moves basic moves are what rotation and translation by applying a combination of those basic moves the final virtual displacement should be such that the work done is only by the forces that we know which is w and the force that we desire which is p so according to this virtual displacement the only active forces were w and p and only though they did virtual work and their virtual displacement of the weight and p were related by the simple relation and so you could immediately get an equation relating p and w is the idea clear the very simple problem but it's very basic but very important because many problems that will come across will be a combination of this and some other problems simple find yes please rotation about a there are would not have done that work the p would rotation about a okay what things to do work okay suppose if there were a torque okay if there were but this body had a torque acting on it or a couple acting on it then the world then will be m times delta theta straight away okay if there were a torque acting on it but now as of now there is no torque acting so just this rotation is not going to do any work okay but indirectly indirectly how does the rotation lead to work is because of just this rotation look at the point a is going upwards is also going sideways at point a there is a reaction in the horizontal direction that multiplied by the horizontal displacement dx will do the work but what we want is that we don't want that quantity at all we are interested in p which is acting here so what we do we know that we can have a combination of transfer rotation and translation such that ultimately we ended up having the final configuration after the after providing the virtual work what was it a combination of rotation about b plus a translation and at the end of that this did not do work the top one did not do work only work was work was done by the horizontal force and the way and that's what we wanted any other question yes please sir my question is that yeah for calculation of delta x it was l sin theta d theta yes delta x was l sin theta d theta yes I means we can do this in this way also that x is equal we can get the formula for x x is equal to l cos theta and then we can differentiate it yes so this is one approach so what is suggesting is that okay this is a varied approach but what I will tell you is that that's a somewhat tricky approach and I will give you some problems okay I will show you in some problems where this approach can be very dicey okay you can do that but that's a very mechanical approach for example many textbooks use that approach so let me do that once so what he is suggesting is something like this okay it's a good point but I will tell you that why that approach is dicey there will be a few problems that will be discussed here where this approach can just give you garbage results if not used properly and also it is a very mechanical approach it doesn't give you proper insights so let us do it that way also let us say this is x this is y this is point a point b l theta so we can say that y a or the coordinate y coordinate of point a can be written as l sin theta x coordinate of b is equal to l cos theta what is delta y if I just differentiate use simple calculus you will see that delta y a is equal to l cos theta delta theta and delta x b will be equal to minus l sin theta delta theta now tell me if delta theta increases then what are you know what are you seeing is that delta y a is increasing is a l cos theta times delta theta is what we saw if delta theta increase delta x b decreases point b moves inverse is also what we saw it is l sin theta delta theta so this is a perfectly valid approach but what happens is that if you use this approach if you want to get a physical understanding that when we give all these mechanisms what is happening first of all you don't get that second of all I will show you a couple of problems where this approach okay if not is properly okay can give you absolute disaster results whereas the approach we are discussing now it will take some time to develop intuition for that but once you develop an intuition to do systems or do problems that way then that will be completely rock solid okay so the learning curve for the approach that I am discussing is a bit steep as compared to this calculus driven approach which is very mechanical and seemingly easy but the dividends are much bigger okay you take my word for for take my word for this for the time being and you will see that what I am saying is right when we solve a few more problems or for example when you go back you also solve a few problems try to attempt a few system you will realize the power of doing things in this way as opposed to just writing the coordinates and differentiating in order to mechanically get the coordinates but there are some cases where this is much better than the approach that I am discussing but by and large this is better for building intuition yeah so I am asking a simple clarification yeah that we are applying the load w o so that sorry can you repeat the question again please the load is w is vertically downwards so the ladder will slip towards left what is towards the left that endpoint B will move towards left for to prevent that we are applying force right no no no see see it's one thing virtual display I want to emphasize this point again that the virtual displacements that we provide are not dependent on the force the way we apply are such that those four should do the work that we want but the work but the forces themselves they do not influence virtual displacement they are completely in our control but we would provide it in such a way that only the forces which we are interested in should do work okay so your question is that later problem that it will slip downwards and it will move towards right in the load is applied to the center the point B the bottom most point will move towards yes then we will be applying force towards left yes yes that's why we are applying force P in the towards left direction that P yes surely surely surely that's the point okay so and also in the at any point the load is applied will not make any difference in this virtual we can come we can consider the load applied at any point any point yes it need not be at the center it can be at any point and the load can even be horizontal okay for example it need not be weight okay for example that somebody is pushing against it in the horizontal direction that is also fine or somebody is pushing against it in this direction okay both are also fine thank you sir okay but note one thing what principle of virtual work immediately tells you that if you don't apply this P if you don't apply this P then what does it tell you that the work done is only W times delta y by 2 and that can never be zero so whenever you encounter such a situation where the virtual work cannot be made to be equal to zero that's an indication that you don't have a stable system it's a mechanism okay so that's another duty of virtual work for a principle that you apply all virtual displacement right down the works and you see that that can never be made to be equal to zero you know that you have a mechanism at hand and you better apply some constraint so that the system is in proper equilibrium so you are right so in principle you may ask me that that I want to give the displacement such that it is consistent with the geometry as we said consistent with the constraints what does that mean that no work done here by the vertical reaction no work done at point A in the horizontal reaction that's the way we give the displacement you may ask me it's a virtual displacement everything is in my control why don't I give it some arbitrary displacement okay I take it rotate it okay suppose if I give the horizontal displacement to point P the point where obviously will move upwards yes yes that's what I'm saying yes so as you have taken that angular displacement but if point A move upwards okay going upwards is not a problem for us point A going sideways is a problem for us because reaction at point A is in the horizontal direction okay but if point A moves upwards that doesn't do any work so a going to a prime in the vertical direction here is good not an issue right but if a goes sideways as it is here that's a problem it's not really it's not really a philosophical problem it's a it's a practical problem philosophically there is nothing wrong with doing that but only thing that you get some answer which you have no use for you can as well even break the rod and give that as a virtual displacement then what will happen internal forces will do virtual work you don't want that okay so virtual displacement is completely in our control we give it judiciously in such a way that what we desire is are the only components and what we know are the only components that do work is this point clear to everyone can we move on okay is this too slow too fast or do I move bit faster how is it too slow okay good