 Let us begin with today's lecture which involves equilibrium of rigid bodies and particularly we will discuss about 2D equilibrium. Now let me first tell that the main text that I am using for this course is Vector Mechanics for Engineers by Bear and Johnston. The most recent edition which is out just a few months ago is the 10th edition. I will refer to that book as BJ 10 all throughout and Indian edition of this book is available and highly recommended. In this course I also have used problems from the previous versions of Bear and Johnston notably 3rd edition and 8th edition. And note that due to the request from the coordinators during the coordinators workshop we are also including an extensive session like around 5 to 6 lectures on dynamics and I wish to point out that dynamics will be exclusively taught from some of the topics covered and the material covered in BJ 10 okay. So there are many new resources that are available in BJ 10 and it will become more clear as we proceed further. So many of the slide contents in our lectures is from BJ 10 from the instructor resources and many features are available from the instructors at this website. If you are interested in more about the book purchase and other details okay these are available to you you can act on that sometime later. Now let us go to the next steps that I have used so there are many interesting and challenging problems in engineering mechanics, tactics and dynamics and Miriam and Kray. So there is a older edition, editions 2, 5 and new edition 7 I refer to them as MK3, MK5, MK7 and a lot of interesting problems are taken from this book. And there are a lot of online resources okay there are a lot of demonstrations at Mathematica s website demonstrations.pollform.com. So Mathematica is a software which is developed by Wolfram and there are excellent lectures by Professor Alan Borat Brown University and there are general lectures on dynamics on YouTube and nice animations okay so all these links will be provided to you. So I really urge you to go and have a look at these resources okay. Now the idea is that that what is the purpose of learning 2D equilibrium okay the real world okay as we see around us is mostly a moving world it is this dynamic world and those topics are definitely covered will be covered in next week. Then the question arises that what is the big deal that why do we want to learn about equilibrium at all? So the point is that even though equilibrium is boring it is great to know about equilibrium because if you are sleeping peacefully on our bed we do not want that bed to collapse okay so we want to make sure that we know what are the forces acting in each and every component so that everything is stable. So as I pointed out that even though equilibrium systems which are stable are boring but boring is useful okay and in fact you will see I hope that even this boring is actually interesting so with that in mind okay. So let us briefly have a look at what you did previously so Professor Shovik Banerjee taught you vector mechanics or more than taught you he discussed vector mechanics with you defined what are forces, moment torque, he discussed equivalent systems, he discussed distributed loads, centroid moment of inertia and so on okay. Now what we are going to do is we are going to do use those concepts which he had discussed and then go on to what is called as rigid body equilibrium and before we go on to that concept let us just briefly remind ourselves about what we mean by rigid body. So strictly speaking no object in the world is rigid but mathematical definition of a rigid body is that for example if you take a nice hard bound textbook you draw two points on that join them by line measure the length of the line. Now you take the book okay turn it around flip it spin in okay do all manner of things with it what happens is that at a distance between the two points or the length of that line does not change so that is a mathematical definition of a rigid body even though in reality nobody is fully rigid, rigidity only implies that the deformations are very small when compared with the body dimensions and as a result many real life structures can really be idealized as rigid and as far as finding out the forces for the statically determinate structures which we will come to very soon okay it is a very good approximation to model those structures as rigid okay a few example okay this is a building okay typically all the forces that are acting on it you do not want the building to deform so much that the people inside are totally scared about staying inside okay. So we design a building in such a way that the components typically do not change the distances between each other a room which is like almost like a cube remains almost like a cube as opposed to being deformed to a parallel pipette. Another simple example is this cranes which for example you see around many construction sites okay so even though this crane is actually moving because of the help of this hydraulic cylinder you will see that there are many instances when the crane has to remain stable okay and at that instant we can say that the crane is actually a rigid body and when we are talking about the equilibrium of that crane we can talk about equilibrium of rigid bodies okay now let us come to equilibrium so let me just tell you that we will go on now for another hour or so and at around say 3, 3, 15 we will start with discussion and at that time we can have like have a transfer of information and knowledge but till then let me just discuss what this topic is now what do we mean by a body in equilibrium. So what we had seen earlier is that you take a rigid body 2D, 3D does not matter now that rigid body as far as mechanics is concerned okay is subjected to various forces if it is a structure then it is subjected to wind loads it is subjected to self loads if it is a machine it is subjected to the internal torques that are created from the pistons of the engine and so on there can also be internal couples or there can also be external couples that can act on the structure. Now a structure is in equilibrium okay by definition okay so these are the Newton's laws and extended Newton's laws if sum of all the forces that act on the structure okay is equal to 0 now what do forces do we will see that even more clearly when we do dynamics that all the forces acting on the body they end up in translating the body. So if the body is here if the sum of all the forces is not 0 then what will happen is that the center of mass of the structure will start translating and will have some acceleration okay and when the sum of all the forces is 0 then we know that there will be no acceleration the body will either stay uniform or will move at a constant velocity and we say that body is in equilibrium as far as the forces or translation is concerned. Now in addition to forces we also have external moments acting on the structure these are the external couples M i on top of that the forces acting on the body can also cause torque or moment given by R i cross F i and the total torque that is acting on the structure about some point O let us say it is the origin of the structure is sigma M j and sum of all the torques or the moments that are caused by these forces and if the body is supposed to be moment equilibrium this should be equal to 0. Now let us briefly remind ourselves what does that mean that if the moment or the torque about the structure is not 0 then the structure undergoes rotation and in fact to be precise okay all rotational moments okay they involve acceleration okay we will see that more when we discuss about kinetics of and kinetics and kinematics of rigid bodies. So if the sum of the moments acting on a structure is not 0 then the body will start rotating and the body in equilibrium neither translates nor rotates okay and as a result we say that this structure is purely in equilibrium okay. Now second topic that we come to is supports and equilibrium now what we see is that take for example a building a building has a nice deep foundation and that acts as a support for the building. So if the foundation for example is not firm then the building will collapse or it will not stand wind loads or lateral loads okay and fall over okay. Now we will discuss in the next class that many structures in real life is not just one rigid body but it is a combination of various rigid bodies that are connected by linkages okay and as we will proceed further you will see that different linkages and supports keep the body in equilibrium okay and they provide enough reactions that whatever forces are acting on it corresponding reactions are provided by the linkages and supports to keep the body stable and prevent it from both translating as well as rotating that is force equilibrium and moment equilibrium okay. So what are different possible structural supports okay we will have a look at that okay. So one thing that we will mention now is so there are two complementary things which are called as constraints and reactions okay. So there is a intricate relation between kinematics what is kinematics that something is either rotating or translating or something which are parameters of motion which we can see by our naked eyes or which we can for example use a camera and see the motion that is something that we can see and the next component are the forces of the reactions okay. So kinematics and reactions as we will see as we proceed further are intricately linked with each other and what we will see is that that if a support rigidly constraints a given degree of freedom for a rigid body then it gives rise to reaction corresponding to that degree of freedom. Similarly okay I want to emphasize this point because many a times I have seen that for example when you teach your students I have particularly seen that if this point is emphasized properly then students are much more clear about what they expect from the support reactions. Say if a support okay freely allows motion of particular degree of freedom then there is no reaction from that support in that particular direction. Now this point we will emphasize okay when we go on to what are different types of supports that are present in engineering mechanics okay and see that what are the kinematics degrees of freedom of that structure and correspondingly what are the reactions that the support can produce or provide. Now the one of the questions that the student ask a lot of times is that what are 2D structures no real life structure is 2D okay we live in a 3 dimensional world and no structure can really be a 2D structure. So what is this entire deal about doing this a big topic and writing books after books and discussing sessions after sessions about 2D equilibrium, 2D motion and so on. So for example let us take example of this 3 dimensional structure this is a 3D truss which professor Shobhik Banerjee will teach when he goes to this section on trusses you will see that this is a 3D structure but it is symmetric about this central axis. So if you draw an imaginary plane passing through the center line a vertical plane you will see that about that plane you have one part of the structure here and another part of the structure on the other side and they are essentially mirror images of each other and look at the loading acting on the structure for example if a train okay passes or a car or a truck any vehicle passes depending on what kind of bridge this is then you will see that it is designed in such a way typically that the forces are also uniformly distributed about the central line or the central plane. As a result what we have is we have a center line we have a center plane and the structure is symmetric about this side and this side as well as the loading is symmetric about the center plane. Now in that case what will happen is that because of the inherent symmetry you can just look at the structure from the side and whatever one section that you have and the corresponding load you can say that I am reducing that effective 3 dimensional structure to a planar structure by exploiting the symmetry that is present in the structure and the load that acts on the structure. So these kind of 3D structures can typically be modeled as 2D structures. Other type of structures that can also be modeled as 2D structures are when the third dimension of the structure is very small as compared to the other 2 dimensions and the load is coplanar what is coplanar is that the load lies in the same plane as the structure. For example if you look at a pair of scissors if you look from the top what you will see is that the thickness of the scissor is very small okay as compared to the other 2 dimensions in this plane and so if you look from the top it is as if we are dealing with the 2 dimensional objects. So these are the 2 typical cases where a really 3 dimensional objects can be modeled as effectively 2 dimensional objects. Now let us come to the most important part of 2D equilibrium and that is reactions and supports reactions at supports and connections for 2 dimensional structures. So we discussed what do we mean by a 2 dimensional structure. Now what we will discuss here is that what are the various supports that are used in engineering and what kind of reactions that they provide. So let us take the simplest example which is a roller. So let me spend a few minutes on this because it is a very important topic and I just want to ensure that all of us okay are on the same page okay effectively because when we teach our students this is something that if they master this topic they understand what are the support what are the different supports that are possible and what are the different reactions that they provide then it becomes much more easier for them to draw the free body diagrams and solve the problems effectively okay. So let us look at these structures which are roller structures. Now what do we mean by rollers? Now look here look at this particular structure we have a main structure which is some beam element or some truss element whatever is there that it is connected by a pin okay to this triangle and these wheels. Now what happens here? Now what are the possible kinematic degrees of freedom at this support? The possible kinematic degrees of freedom now what is the degree of freedom is what possible motions okay are available at this structure. You will see that the possible motions okay that are that are allowed is translation at this point okay and rotation at this point okay because this is connected by a pin so this top portion is free to rotate and so there are 2 degrees of freedom that are free here but the vertical motion is completely constrained and as a result what happens is that. So we can see that since this degree of freedom is not constrained this support is will be unable to provide any reaction in this direction since the rotation is not constrained. So rotation is another degree of freedom which is not constrained. So this structure at this point will not be able to support any torque about this point but whereas because we are preventing the motion in the vertical direction this support okay is possible okay it is possible at this support to exert a reaction in the vertical direction. Now let us take this support which is a guided roller in this support if the force is such that is roller tries to move upwards then there is no support for the roller and this support will be unable to provide any reaction in the vertical direction. Whereas if you come to this structure which is a guided roller the motion is constrained both in the downward direction and in the upward direction and it can provide a reaction which can be both downwards as well as upward. Now you also have a rocker support a rocker support is also equivalent to this and a frictionless surface now in a frictionless surface only possible reaction is a normal reaction but there is no reaction possible in the horizontal direction and the biggest example of this particular type of structure is the way we stand on the ground okay. If we stand on the ground okay if the ground is perfectly frictionless it will be almost impossible for us to keep balance although for example the ground is providing a support to preventing us from sinking down okay the lateral support or the support in the horizontal direction will not be possible without any friction. Now the second type of support we have is a short cable or a short link now what kind of reaction it will provide note one thing is that the short cable is supposed to be inextensible and so is the short link and as a result if you take this point and try to move in the perpendicular direction to the string or to the short link then that particular infinitesimal motion is possible but when you try to pull this along the direction of the string that motion is prevented because of the inextensibility of the string and as a result we can replace when we draw a free body diagram this string with a corresponding force in the direction of the string then we have a collar on a frictionless rod note one thing that at this point this is connected okay it is a short collar so it is connected by a pin so this degree of freedom the rotation is freely possible if this is frictionless the motion along the rod is also freely possible but we will not be able to move this collar perpendicular to the rod and as a result the only degree of freedom at this joint that is constrained is the motion in the vertical direction and when we draw a free body diagram we can see that there can be a corresponding reaction that can act on it perpendicular to the rod and same goes for a frictionless pin pin slot then let us come to what is a frictionless pin or a hinge now what do we mean by frictionless pin frictionless pin means that you have a rod which is connected by a pin and this rod for example is freely free to rotate about this because of because it is it is frictionless but what is not possible is this joint is perfectly connected okay to the structure as a result a translational degree of freedom in the vertical direction is not possible translational degree of freedom in the horizontal direction is constrained and because these two degrees of freedom are constrained what we have is we can either say that this particular support can provide two unknown reactions. Okay because this degree of freedom is constrained so it can provide a vertical reaction this degree of freedom is constrained so it can provide a horizontal reaction the rotation is not constrained because it's a frictionless pin and so we can have two possible support reactions a vertical support reaction I need a horizontal support reaction and this can also be equivalently represented as a force acting at an unknown angle alpha. And ultimately what do we have, ultimately we have a fixed support, this is for example what happens when we have foundations or when we have huge pillars which are or huge towers which are connected to wires or cables and so on. Now those of them what do we have is that because the rod is very deep inside at this particular connection the degree of freedom in the horizontal direction is also constrained, the degree of freedom in the vertical direction is also constrained and the corresponding rotation is also constrained. And so in three dimensions, there are three degrees of freedom two translations one rotation and this particular support what it does is that it constraints all those degrees of freedom and as a result it can provide three possible reactions, reaction in the x direction corresponding to constraining the degree of freedom in the x direction, reaction in the y direction corresponding to the degree of freedom in the y direction and a torque counter clockwise clockwise corresponding to constraining the rotation about the axis which is coming out of the plane. So this rotation either clockwise or anticlockwise. So if you want to convince yourself have a look at various of the supports this figure is a very nice figure which demonstrates how various connections what are the degrees of freedom that are constrained and what are the particular support reactions that can be provided. So this is a roller and this slide if you want to for example go into greater details with your students on what this support means you can have a look at all this particular supports and all of them act as rollers see how are the forces transmitted and for the structure to be everything to be in equilibrium what will be the corresponding supports support reactions that will be provided by these supports okay slot connection okay all these slides okay you have you will have access to these slides so you can think about this in greater details if you want to explain to your students what this particular connections mean okay. So these are some very simple examples for roller support and a fixed support. Now for example somebody would ask a question that in equilibrium we are interested that the structure should be in equilibrium okay. So why do not we just go ahead and provide this best constraint to the structure okay let us say that the structure should be in equilibrium let us not worry about anything we will only provide this support let us not bother about any other support reactions at all and it is very clear that we cannot do that is for example we need to have the structure okay for different purposes say for example this carbonate if we open the carbonate it is free to rotate about this hinge and when we want to prop it up then we support it like this okay and keep it in equilibrium in this direction okay so we keep it supported in this direction. So what we have is that that we are keeping this as a hinge because if this were a fixed support to begin with then there is no way we can move this flap up and down okay and adjust it in whatever position we want similarly for a roller okay we have a trolley which has rollers now what do you want to do we want to move this trolley from one place to other place depending on we want to carry things around and then lock it and convert it into roller on one place and hinge on other place. So different supports are required depending on what are the requirements for example if you want to build a cable bridge which for example is much more convenient to build much more it has many more advantages can take much more load thus the amount of material required is much less okay so but the support is now provided by many of the strings now we can as well provide a completely rigid structure but that will be counter productive. So depending on whatever structures that we are interested in we provide different particular types of supports and that is why all these supports okay even though we want to keep the structure in equilibrium depending on what is our requirement from the structure we end up using various linkages and various connections. Now with this much of introduction let us come to the main part of this 2D equilibrium. So free body diagram is essentially the heart of mechanics okay and we all know okay we have all been teaching this course for quite some time and we know that free body diagram is essentially the basis of anything and everything we do in engineering mechanics and just to remind ourselves what free body diagram means. So what do we do is we zoom into components of a structure what does the free body diagram means that we replace supports or connections with the corresponding reactions or to put it compactly what we do is that we replace kinematic constraints with corresponding reactions now this point will become more clear when we solve a few problems simple examples of free body diagrams. So let us take simple example of a beam this particular triangle and a circle it is a typical notation that is used for a hinge support same here. Now we saw a few moments ago that what is done by a hinge support and the hinge support look at the joint A at point A what is what are the kinematic degrees of freedom there are 3 kinematic degrees of possible freedom one is translation in the horizontal direction one is translation in the vertical direction and third is rotation. Now what is this joint doing it is preventing translation in the horizontal direction and as a result it can provide a reaction AX in the horizontal direction at this joint we are also preventing a kinematic motion in the vertical direction and as a result this support is capable of providing a reaction in the world y direction similarly the same logic works at point B where point B is also a hinge joint that because we are constraining 2 degrees of freedom horizontal and vertical translation we can have 2 possible reactions a vertical reaction and a horizontal reaction at point B and now this continuous this thing where we have external loads forces that can be provided by these supports where all the kinematic constraints are now replaced by forces and whatever the resulting diagram we get is what we call typically as the free body diagram okay I am just reminding ourselves that what is the definition of a free body diagram. Now let us look at another structure quickly what do we have here we have a fixed support and we saw that at a fixed support what are the kinematics degree of freedom that are constrained. So the kinematic degree of freedom constrained is a translation in the x direction translation in the y direction but in this case because the rotation is also fixed because it is a fixed support we are also constraining rotation and as a result when we replace this support we have to replace the kinematic degrees of freedom with the corresponding reactions which will now be A x A y and the corresponding torque and now point A this is one way of representing a roller that note that the gap here which means that this is free to slide in the horizontal direction okay if left alone. Now as a result of this the degrees of freedom that is constrained is only the translation in the y direction and as a result this point B in this case is capable of providing only a reaction in the vertical direction and this now is the main structure with the supports and applied load and what we are doing here is that we are replacing that with the corresponding reactions that can be provided by the supports and this is now our free body diagram there is absolutely no kinematic constraint here only thing we have are the reactions and applied forces okay so force results in the reactions because of the presence of the supports and in a free body diagram we replace these kinematic constraints purely by the corresponding reactions that are available here. So these are various examples okay you can have a look at that okay I am sure for example we have discussed many of these things at great leisure but it is one point I have seen that if you emphasize free body diagram more and more with the students with many many examples then that concept become much more clear to them and they feel much more comfortable solving simple problem to begin with and graduating to more complex problems okay. So concept of free body diagram okay just between ourselves like we cannot emphasize this enough to the students so whenever students are writing solving any problem we have to like it is our experience that we just keep hammering this concept on them again and again that you want to find out forces you have to talk about free body diagram okay draw free body diagram and then and only then talk about equilibrium and any other thing. So free body diagram is the single most important concept okay and we will all agree on this in engineering mechanics. Let us take a simple problem this problem is taken from Bir and Johnston 10 from the instructor slides what we have here okay is we have a structure okay the frames shown supports part of the roof of a small building okay so we have to give this problem to the students saying for example that their goal is to draw a free body diagram for the problem what is given to us is a tension in the cable BDF which is 150 kilo Newton. Now what we do to draw the free body diagram of this structure what are the kinematic degrees of freedom what we have here is that this tension is known to us which is 150 kilo Newton okay so we replace we make a cut in the string here okay and we can replace that with the corresponding force at this joint E it is a fixed joint what are the kinematic degrees of freedom that are constrained here so one is a translation at point E in the horizontal direction is completely constrained not possible at all okay vertical translation is completely constrained at joint E and the rotation is also constrained. So we have constrained 3 kinematic degrees of freedom so we replace that okay so when we go on to the next slide we see that we can have various ways in which we can draw the free body diagram at this point we have moment vertical force horizontal force why because all this kinematic degrees of freedom are constrained now one of the ways for example we can tell the students so these are the common mistakes that students make while drawing while having the drawing the free body diagram. So there are various ways in which we can draw the purported free body diagram. So if we go to point A okay if you go to sub figure A we will see that we have replaced this cable with a tension of 150 kilo Newton okay we have make a cut here and say that the force acting on it is acting directly at point B and say that we replace the cable with a force here we replace the support at point E with respect to these 3 support reactions okay that is clearly not a proper free body diagram why because this string is not only exerting force at joint B but it is also exerting force at joint D from this side as well as this side so drawing this is not at all right if we take this and add 150 150 and 150 so because this string is applying forces both at joint D and joint B this is fine okay this is like we can emphasize to the students that this is the common mistake that they can make but this is an appropriate way of drawing the free body diagram why because this string is connected to the structure not only at point B but also at point D. So when we replace this string with the corresponding tension we have to make sure that the points where the tension acts are all taken care of okay so this is also valid free body diagram this is the best free body diagram why because if you want to find out what are the reactions at this point we do not need to go into all these details because these 2 are forces which are acting along the same line and their equivalent effect on the structure will cancel out why because they are equal and opposite and they are acting along the same line so they cannot produce a torque and the sum of 150 and minus 150 will cancel each other so effectively this we have emphasized to the students is the best way of drawing the free body diagram and this D okay many students actually do this but this is completely wrong why because we have not released this kinematic constraint what is the kinematic constraint here that is string okay you cannot move the string in this direction so there can be a possible reaction which in this case is given to us 150 kilo Newton but that is possible and since it is not replaced here you may end up getting some complete garbage answer okay the students if for example you do not draw the appropriate free body diagram so one of the things for example which is recommended is that we expose students to all possible free body diagrams we tell them that what are the possible mistakes that they can make you by inventing situations ourselves okay that these are the possible mistakes they make and what is the reason why a particular free body diagram is right or it is not right and once this concept becomes very clear we have seen that the students become very very comfortable solving simple to complicated okay all possible problems okay they can solve them much properly once they understand that what a particular what is a free body diagram and that the equations of equilibrium are to be applied only to a free body diagram and not to a structure like this okay now we come to actually that we have a free body diagram and what is our interest that we have applied some forces we have applied some talks and we want to find out what are the reactions what are the reactions at the supports what are the reactions at the linkages that is the main goal of this course okay and with this mind what do we need to do we need to write equations of equilibrium now we saw that in three dimensions if you look at this full body and this problem this particular concept will become even more clear okay for all of us means like we will remind ourselves next week that what this concept is and so on when we do dynamics of rigid bodies now a rigid body in up in a plane has three degrees of freedom okay take this full rigid body AB this rigid body AB can translate in the horizontal direction it can translate in the vertical direction and it can rotate now because of this okay this simple idea in terms of equilibrium what does it translate to that when this is in equilibrium we are preventing the full translation of this previously what we had looked at is we had looked at only one joint for us to understand that what the support reaction should be now once we decided that what particular kinematic degrees of freedom are constrained here we found out that these are the possible reactions that are possible at the supports now we are looking at a full free body diagram and we know that for this entire system to be in equilibrium we should have translation in the horizontal direction that is prevented rotation in the vertical translation the vertical direction is prevented as well as the rotation is to be prevented and the corresponding equilibrium are FX or the equilibrium in the horizontal direction some of all the forces in the horizontal direction should be 0 FY some of all the forces in the vertical direction is 0 to prevent motion in the y direction and some of all the talks about some point it can be AB or C whatever it should be equal to 0 you can also convince the students that these are not the only way it can happen for example we can also use equation like sigma FX equal to 0 sigma MB and sigma MA equal to 0 so writing 3 conditions like this is also fine because we can see in dynamics that any motion of a body 2 translations and one rotation can also be thought of as one translation and 2 rotations about 2 different points you can think about a motion like this or furthermore any motion a translation 2 translations and one rotation can also be if you think about it carefully we can convince ourselves that that particular motion can also be replaced by 3 rotations 1 about point A 1 about point B and 1 about point C and all these 3 points do not lie in a straight line they form a triangle and any rigid motion can also be written to be a combination of these 3 and as a result we can also replace that these 3 conditions are all equivalent to each other that we can say that if we say that some of all the moments of point A some of all the moments of point B and some of all the moments or talks about point C is equal to 0 all these 3 conditions are equivalent to each other and there is one thing which we for example should emphasize students all the time is that these 3 are not independent equations of motion we write these 3 they will imply these 3 they will imply these 3 all these are equivalent to each other there is no new information that is given here so writing more than 3 equations per free body okay it is clearly should be punishable by law okay if it is not already okay so it is definitely not allowed to do this thing so only in 3 in 2 dimensions the number of equations per free body will be 3 and only 3 no more than 3 okay and as we see further that the equations can be trivially satisfied in some cases okay but you cannot write equations number of equations of equilibrium more than 3 in 2d equilibrium now with this much of a preamble okay let us solve a very simple problem okay so I had discussed this problem in the coordinators workshop and it is a very simple problem but why I am doing this is that so that everybody is on the same page okay so what we have here is that we have a rigid structure ACD which is pinned or connected at point C with a hinge joint further what we have is that we have a cable ABC which is connecting point D through a roller okay which is which has a frictionless hinge connection here okay the roller is free to rotate there is no friction at the hinge for this roller it goes over this roller and then it is connected at point A now this structure is subjected to a load of 150 Newton at this point and what we are asked to do is we are asked to find out what is the tension in this cable ABD and what are the support reactions at point C now this is a very simple problem but there is a reason why we took this simple problem is we want to understand what is the role played by the cable and what is the role played by this support to keep this system in equilibrium okay so let us think about it let us ask ourselves a question what is the role played by this cable ABD now if you cut this cable if the cable were not there then what happens is that we all know that this system now is free to rotate about point C so the cable what it is doing is preventing the rotation of the body ACD about point C so as a result we immediately know that what should be our equation of equilibrium that this is the tension in the cable this is the tension in the cable and what is the cable doing for this free body diagram that the support C is preventing two translational degrees of freedom okay motion in the x direction motion in the x in the y direction so that will lead to two reactions FCX in the horizontal FCY in the vertical direction the cable what is it doing that it is preventing motion of point A along direction AB that will lead to tension along AB what happens at point D this cable is preventing any motion of point D okay in the vertical direction so that should lead to or that should lead to a possible tension in the cable in the vertical direction now the question we ask ourselves is that that there can be some tension here T1 there can be some tension here T2 what can we say about the relation between T1 and T2 note here that we have mentioned very clearly okay or typically it is okay to assume in this course okay as a purpose of idealization that if this pulley is well oiled if you have put enough lubrication in the pulley then this pulley is free to rotate and as a result if you draw the free body diagram for the pulley itself at the joint B okay this joint B is where the pulley is connected to this support what do we have rotation is free which means that there is no possibility of an internal torque that is provided at joint B translation of the pulley is prevented by the hinge there so there is a possibility at a vertical reaction horizontal translation is prevented at joint B as a result there is a possibility of having a vertical there is a possibility of having reaction in the horizontal direction now for this pulley if we take moment or torque about point B you will see that the only torque is provided by this T1 into R T2 into R and because now T1 minus T1 into R minus T2 into R should be equal to 0 because this pulley also is in equilibrium will immediately see that T should be equal to T1 should be equal to T2 and it is a commutation all throughout the string and now we come to this free body diagram and what we will see is that the string here or the tension provided by the string is preventing the rotation of this free body diagram ACD about point C and so we immediately know that the equation of equilibrium that we should write in order to obtain this tension why because what is this tension doing it is preventing the rotation about point C and so the corresponding equation of equilibrium is some of all the torques or moments about point C what are they moment provided by this 150 kilo Newton what direction moment will it provide just think about it assume as if this is a door which is hinged about point C now if you apply a force on this door like this the door will tend to rotate in the clockwise direction anticlockwise direction so that is a direction of the moment we have written all the moments in the clockwise direction so 150 into 225 anticlockwise moment so this is clockwise positive so minus 150 into 225 okay and then all the torques so this tension if you say look at this CD assume this to be a door CD you apply a tension in the vertical direction how will the door tend to rotate we emphasize due to these to the students the door will tend to rotate in the anticlockwise direction so the torque sorry and the anticlockwise direction so the torque produced by this tension will be minus T into this 75 this is vertical so T into 75 is the torque provided by this thing and then coming to this tension now this tension can be broken into two components and then as professor Banerjee discussed we can use the very known theorem that the moment or the torque which is provided by this tension about point C will be nothing but the torque provided by the horizontal component plus the torque provided by the vertical component about point C look here the vertical component if this is an imaginary door how will it tend to rotate about point C because of this force it will tend to rotate in the clockwise direction so we have this particular positive sign 5 by 13 into 225 is the torque produced by this vertical force about point C 12 by 13 T is the horizontal force think about it what is the vertical distance is 175 so the torque provided by this force about axis C okay or about joint C will be simply 12 by 13 T into 175 should it be clockwise or anticlockwise just imagine that door again if this is a door if you apply a force like this how will it tend to rotate about point C it will tend to rotate in the clockwise direction and that is why it is 12 by 13 plus okay into 175 so these are the all possible torques that equal to 0 for equilibrium and we immediately get what the tension is now we ask ourselves okay we got the tension we are also asked to find out what are the displays what are the reactions at point C note one thing now we are again ask ourselves a question okay what is joint C doing here suppose what happens if this joint C okay will wear a roller joint okay not a hinge joint so it was free to translate in the horizontal direction such that no reaction can be provided so what you will see is that if you prevent this if you put that a roller instead of a hinge then because of the presence of the tension in this direction look at a free body diagram okay this tension now we tend to translate this in the horizontal direction and there will be absolutely no resistance to that motion so what is joint C doing is against this tension T it is preventing the translation in the horizontal direction so we write down the equation of equilibrium for this free body diagram in the horizontal direction and we immediately get what is FCX and correspondingly we ask the same question why should there be a vertical reaction what if we put a roller here in such a way that the only reaction produced in the horizontal direction and not in the vertical direction then the answer is again clear this tension T this tension T the vertical component these two vertical components and this vertical load they will not balance each other okay without the support and as a result without the present of this vertical constraint or the corresponding vertical reaction FCY it is not possible to prevent a vertical translation of this particular free body and so the immediate equation that comes to our mind is that sigma FY Y is the vertical direction we are taking should be equal to 0 or sigma F we write all the forces 13 5 by 13 T plus plus T minus 150 TV already obtained plus FCY equal to 0 and we see that FCY is equal to minus 120 and what we have done here arbitrarily we have assumed that the direction here is in this direction it is upwards for FCY but what we are getting here is we are getting minus 180 here we are getting minus 120 here it only means we all know and we should emphasize this to the students that minus just means that the direction that we have assumed before was not right actually the reaction is provided in the other direction or not in the to the right but to the left so that is a reaction and 120 Newton minus means it is not upwards but it is downwards and we emphasize this point to our students so that this point becomes very clear that a priori if you have enough intuition then we are good we can choose the direction but somehow if you are not completely sure to begin with what direction should we choose then the best choice is without worrying too much without lot of us just to the direction and if the force comes out to be negative it just means that whatever direction you have chosen the actual direction is just opposite of that okay now let us discuss this brief point okay this is taken from Maryam and Craig fifth edition this particular figure we note one thing that there are various force systems so a few slides ago we emphasized that the maximum number of equations that you can write for a free body diagram in 2d in equilibrium is 3 but that is the upper limit now say if you have forces which are all collinear to each other they all act in a straight line then we can simplify our problem greatly by choosing x direction along the line of action of the forces take the x as a line on which the forces act then what happens is that 3 collinear forces the only equation of equilibrium we need is sigma of all the forces along the x direction should be 0 because there is no force component in the y direction which is now perpendicular to this that is automatically satisfied moment equilibrium is automatically satisfied second special case is when all the forces meet at one point then what we see is that we can write 2 independent equations one is some of all the forces acting on this free body in the y direction is 0 some of all the forces acting on this free body in the x direction is 0 but because the forces are all concurrent about point O if you take torque about point O we will see that the moment balance is automatically satisfied ok we can convince ourselves by just little bit of a thought that they are concurrent so we can take torque about point O and moment balance is automatically satisfied we can also convince ourselves that if you take torque about some other point also the fact that the forces are concurrent and the sum of forces in the x and y direction is 0 will automatically ensure that the torque about some other arbitrary point will also be equal to 0 another special case is when all the forces are parallel to each other then what we can do we can say x axis is along the direction of which this force is at y axis is perpendicular now note one thing that we need an equation sigma fx is equal to 0 in the horizontal direction clearly that for the body to be in equilibrium these forces should be 0 but the equation of equilibrium in the y direction is automatically satisfied because there is no force component in the y direction and the only extra equation now you need is that these forces can cause a torque or moment about some chosen point and so we need an extra constraint that sigma m about z so z axis is coming out so all the forces will be about the axis which is coming out should also be 0 so instead of having three equations of equilibrium we have for this special case only one equation this special case only two equations this special case for parallel forces only two equations and ultimately a most general case where we have forces which did not be concurrent plus we can have couples acting on the free body diagram then all three equations should be written in order to solve what are the support reactions what are the link reactions for a free body diagram now this one important point which I want to emphasize is how do we know that a structure is properly constrained because we said that the support reactions can be provided by various supports okay but ultimately what we want to know is for a given structures when we are providing the supports are they capable of providing reactions in such a way that under any possible loading acting on the structure this support this structure is in equilibrium so let us look at three special cases let us look at this some four cases let us look at this structure this figure is taken from Merrim and Craig fifth edition we have a structure okay let us call this body which is constrained by a link in the horizontal direction so it can provide a reaction only in the horizontal direction second link this link is preventing motion of point O in the vertical direction so it can provide a reaction in the vertical direction and a third support at this point which prevents translation in the vertical direction so what we have prevented is that that with this pin we are preventing translation in the horizontal direction with this support we are preventing translation of this body in the vertical direction and with this support we are preventing the translation at this point for this body in the vertical direction and then you can convince yourself that with these constraints it is not possible to translate or rotate this body without deforming the body and so this is an adequately constrained system you can also convince yourself that whatever load okay or you can also ask the students to convince themselves or we can have discussions with them to make them sure that in one way neither is it possible to take this body and either translate or rotate it without deforming the body or we can apply any forces or any set of forces on it the reactions provided by these three supports will be enough to keep this body in equilibrium so this is an adequately constrained system but now what we do is that we play some tricks let us say we remove the structure sorry we remove the support and connect it at this point now the way we connect it is that that the line of the support reaction that the support reaction that can now be provided only in this direction and the line is such that it intersects at this point now this is clearly not a system in equilibrium why because if I apply any vertical load at this point because all the support reactions because they are intersecting at one point if I apply a force which does not pass through that point A then it can create a torque about point A and this support reactions will be incapable of supporting the structure under a loading which does not pass through the point through which all the support reactions pass so in simple words if all the support reactions intersect at a point then that is not an adequately constrained system why because we can apply a force which does not pass whose line of action does not pass through the point of intersection of all the supports and there is nothing to balance the torque created by that force about the point through which all these reactions pass through now let us example is that we take this free body take this support structure we have three supports now okay 1 2 3 now we have three support all of them are parallel to each other so one thing is support produce a reaction in the horizontal direction this support can provide a reaction in the horizontal direction and this support can provide a reaction in the horizontal direction now what is the problem in this structure clearly if I apply a vertical force then support reactions will be able to provide a reaction to a vertical and the system will be in not in equilibrium with respect to force balance in the vertical reaction in the vertical direction so as a result we can make another statement that if all the reactions that are provided by the support are parallel to each other then that system is not adequately constrained why because any force that is then applied perpendicular to the directions in which the reactions are acting then there is nothing to balance the force in that particular direction and then ultimately let us take an extra case where while removing three supports we remove without removing any support we just add an external support here now what we have we have particular support reactions but we had emphasized okay we had emphasized previously that we can write only three equations of equilibrium for a free body diagram and so for a free body diagram the equations that are allowed are only three but the constraints or the unknowns that are present here are four and as a result the number of unknowns is more than the number of equations that are present in the system and such kind of systems are called as statically indeterminate systems and that is not particularly topic of engineering mechanics these kind of structures I am sure for example when we teach a course on solid mechanics structural mechanics strength of material okay so these are the kind of problems that are dealt with in this course you know in those courses whereas in this particular course we only deal with statically determinate systems in which the total number of unknowns for for the system is equal to the total number of equations that we can write for the system and for this simple free body diagram even the sample problem which we had saw solved earlier the number of unknowns we had was one okay and two three so three were the number of unknowns but we could write three equations of equilibrium and could obtain the values of all the unknowns if we had some extra constraint provided somewhere then that structure would be statically indeterminate and that is beyond the the scope of this course okay so let us have a brief look at this at this simple problem taken from beer and Johnston so what do we have here we have this is a particular mechanism okay this is a door okay we are looking it only from one side the door you can imagine that the door is coming out of the plane of EDC and what we are looking at is that a door is supported on two sides this there is a pin or a roller at point A there is a roller at point B now what is happening here is that that we apply some tension on the on the door just think about it that this is the door imagine it coming out in the third direction on one side of the door we are providing a tension T on the other line of support we are also providing a tension T and what we are asked is that that if we make the tension to be equal to 0 then what happens is that we will immediately we can imagine that this force W okay will be unbalanced okay the only support reactions provided at A okay is the vertical reaction in the y direction why because it is a free roller this degree of freedom is not constrained this degree of freedom is constrained as a result this dough this point A is free to slide in the horizontal direction point B is free to slide in the vertical direction this degree of freedom is constrained the rotation is also free as a result if there is no tension that is acting on this door in the horizontal direction then this door will completely collapse under its own weight and become straight but now if you want to open this garage door what we do is that we apply some tension T from both sides on this side of the door and on the other side of the door and depending on how much tension we apply we can keep this door to be in equilibrium at some certain angle theta. Now what we are asked to find out is we are given all the parameters we are given this dimension A we are given this dimension B okay we are also told that the total distance that the door is hold in this particular position for which B D okay is equal to 1 0 5 0 so we know angle theta AC is equal to 2100 so we also know that G or the center of gravity through which the entire load effectively of the gate acts okay passes through the center and now what do we have is if we draw the free body diagram of this entire mechanism think about it what do we have this is the weight of the door okay the door is symmetry even though the door is three dimensional it has this support on one side the same support on the other side so what we do is that we just take one half of the door and take it as if it is a planar problem at point A we have a reaction in the vertical direction why because this degree of freedom is free rotation is free only degree of freedom constraint is the motion in the vertical direction this is the reaction tension T is the applied tension which is an unknown this reaction also is an unknown look at support B what do we have at support B this motion is free this degree of freedom is constraint so what do we have only possible reaction that can act will be like this now we look at this free body diagram for the entire door this angle theta is given to us the position through which the load act is given to us so all the dimensions all the geometry everything is given to us the known weight is W and the three unknowns in this problem are R A R B and T and what we are asked to find out you are asked to find out all these three quantities now the number of equation that we can write is 3 and the number of unknowns present is also 3 T R A and R B so this is a statically determinate problem which can be solved using standard techniques that we have discussed just a few moments ago now the idea is that what is the appropriate equation of equilibrium that we should look at now let us think about it that if this tension they are not there then what will happen is that there is a concept called as instantaneous center of rotation which we will learn or which we will brush our concept of on when we discuss this topic next week so this point A will try to slide in the horizontal direction point B will try to slide in the horizontal direction and you will see that the line joining these two will be the point okay where this entire structure will tend to instantaneously rotate about or you can also think about it this way that the reaction will pass through O here the reaction from B will pass through O here it will meet at this point and so if we take the for this free body diagram the moment balance about point O then the contribution of this goes away the contribution of this goes away and the only contribution remains is from the weight which is a known quantity and from the tension which is also a known quantity and we can write down a simple equation of equilibrium this is A this is B so A plus B sin theta is this vertical distance minus t times sin theta why minus because the convention for torque we have taken to be clockwise look at this if I apply a force like this about apply a force in the horizontal direction if you want to take torque about point O imagine that there is some imaginary door O A which is his about point O upon application of this force the door will tend to rotate in the anticlockwise direction that is why this minus and this W imagine that same door about point O you apply this force W that imaginary door will tend to rotate in the clockwise direction that is why plus and what is the horizontal direction here it is W A cos theta so this is the torque created by this this is the torque created by this one is anticlockwise one is clockwise sum equal to 0 we will see that we will immediately get what is t and then correspondingly we do equilibrium of all the forces for this 3 body diagram in x direction and y direction and then what do we get we can immediately find out what is the reaction at point A and what is the reaction at point B and here we are dividing it by 2 y because this load is equally shared on both sides that supports are here and then supports are on the other side of the door okay.