 Welcome back. So, let us start from where we had left. So, coming back to the concept of two force members, what we saw is that like there are these different ways in which two force can happen. Now, a two force member did not have a particular shape. In this case, it is a shape of a right angle. What we see is that the only requirement for something to be a two force member by definition is that the force that the member is connected at one end by a pin, at another end by another pin and there are no forces in between. No moment, no force, nothing. And the self weight which is also a force in between passing through the center of gravity is taken to be negligible. Why? Because the other forces involved are typically much larger when compared to the force from the weight. Now, note that in a two force member, suppose at the end, because it is a pin, you can have a force in any arbitrary direction. At this other end, because again it is a pin connection, the force can be in any arbitrary direction. But note that if the forces are in the arbitrary direction, equilibrium will not be satisfied. So, suppose the forces are in equal and opposite direction, then they will cause a couple because of which the link will not be in equilibrium. So, the only way in which when you have a link which has only forces at two ends, the only way this can be in equilibrium is the two forces act along the line joining the two ends. So, the force can be either inverse or it can be outwards. So, this is what we mean by a two force member and what does a two force member do that many times if you recognize the two force members, then the problems as we will see are drastically simplified. Now, this is the concept of hydraulic cylinder. So, most of us have seen hydraulic cylinders in many sites for example, when they are digging trenches, various places you will see the hydraulic cylinders. Now, what do hydraulic cylinders do is that typically they have an extendable arm and that extendable arm is connected to different parts of the structure and that arm can either extend or contract in order to get the parts of the structure in different configurations as per the requirement. Now, hydraulic cylinder also has some weight but the forces that are involved at the ends of the hydraulic cylinder are much larger when compared to the weight of the hydraulic cylinder. So, that can also be safely considered to be a two force member. So, with this let us move on to this problem. Pretty interesting problem. What we have here is we have an assembly which is taken to for example, raise or lower a car onto a platform. When this assembly is such that this link is completely inclined, what you can say that the car can move on to the ramp placed here and what you can do is that you can raise it up and for particular servicing and what we are asked here is that in this mechanism there are two components. We have this component, this horizontal component which is pinned at point A. We have a hydraulic cylinder which is pinned at point B and A and we have a link which is supported at the ground at D and C and B. And what we are asked is that that given that this point is where the effective weight of the car acts and the weight is given to us. All the necessary dimensions are given to us. We are asked to find out what should be the force in this hydraulic cylinder AB to keep this configuration in equilibrium in this given position is what we are asked to find out. Now the typical question, the question we will ask ourselves is this. How many free bodies do we have in this problem? How many unknowns? So we have one free body diagram, this horizontal member, two second free body diagram, DCB and third which is a two force member which is AB but two force member essentially just means one unknown. What is the force in that two force member? So the total number of unknowns present in this system is two unknowns here. It is a pin support. Horizontal, vertical both are constrained. So we have two unknowns. At point we have a roller support. So only degree of freedom constraint is the kinematic degree of freedom in the vertical direction. So that will give a reaction, vertical. Point C is a pin, two more and this is a two force member. So 1, 2, 3, 4, 5, 6. We have six unknowns. How many free body diagrams can we draw? How many free bodies we have this one in two? So we have two free bodies, six unknowns. So 3 into 2 it is a good. It is a statically determinate system. But now the question is that is there a nice way because for example we can draw various free body diagrams. I can draw the complete free body diagram. I can draw free body diagram only of this link. Then what is the appropriate free diagram to draw? Okay and how do we proceed about it? This is a very state forward problem. But let me ask you one question. I need to find out the force in the hydraulic cylinder AB. So if I just draw the free body diagram for DCB, will that be good? No. Why? Because we have unknowns. 1, 2, 3, 4. We have four unknowns, one free body diagram. Okay. So we can write only three equations for not a good idea. So we ask ourselves a question. Okay. What is this ground doing? If this ground were not present then what happens? Suppose there were no ground or suddenly it liquefies then what will happen is that this entire assembly rotates about point O and sinks down. So what is this ground doing is it is providing a vertical reaction and what is that vertical reaction doing? It is preventing rotation of the entire assembly about O. So that immediately tells us what is our free body diagram and what is our equation of equilibrium? So our free body diagram is this complete free body diagram of this entire assembly and the equation of equilibrium is torque of all the forces or moment of all the forces about point O. Immediately. And once we do that in one shot we can obtain what is the vertical reaction at D. Fair enough. Okay. We can obtain vertical reaction at D. But now the question is what is the hydraulic cylinder doing? Suppose we do not have the hydraulic cylinder. We have the ground but hydraulic cylinder is not present. Then what do you think will happen? Then this link DCB will rotate about point C because this reaction there is nothing to balance the torque provided by this reaction about point C. So what is hydraulic cylinder doing? That it is preventing the rotation of this link DCB about point C. So immediately then we know what is our next free body diagram. Our free body diagram is DCB. Okay. 2 unknowns 1, 2, 3. This we already found out vertical reaction. Okay. And we take the torque about point C. We immediately know what is the force in the hydraulic cylinder. So what we have done is that rather than brute forcing which free body diagram to draw and so on, we just went step by step that what is the purpose of every component in this structure? The ground is providing vertical reaction without which the entire system will rotate about point O. The hydraulic cylinder is providing horizontal force which is providing a counterbalance to prevent the torque to balance the torque by the ground reaction and prevent the link DCB from rotating about C. So once we do that logic then immediately what is the free body diagram that is required and what is the appropriate equation of equilibrium that immediately becomes apparent rather than just brute forcing the entire thing. Is that approach fine? Is the idea okay? So if you just look at the solution okay you can have a look at this and you can whatever I had just discussed in words essentially it is put into equations and from that we can find out what is the force in the hydraulic cylinder. Is the approach fine? Any question? So let us move on to the next problem. It's a very similar problem. This is an assembly okay. What does this assembly do? Okay by the way this is used in still this is a real life. I should have put the picture. I forgot to put it. It's still used at many small domestic airports to tow aircrafts. So what is done is that this hook is connected to the aircraft and from the hangar the aircraft is like tow to and fro. Now this mechanism in this particular position what is happening is that no aircraft connected just the mechanism just lying like this. What we have is we have a tractor okay which is pulling on this mechanism. How is this tractor connected to the mechanism that there is a slot here okay. So there is a big slot and this one so if I draw it here if I look vertically what I see is that at point E there will be a slot like this okay two of them and through this hole okay this there is a if I look from the top okay it will look like this. There is a hole here okay hole here another hole here. So what you do is that you put a link like this and so with that one you connect to this tractor this entire assembly. So it is acting essentially like a hinge joint why because there is enough gap such that the translation in the horizontal and vertical direction is prevented but the rotation is not but the rotation is still free at least infinitesimal rotation is still allowed. So it is essentially acting like a pin support or a hinge support and it is providing two reactions EX and EY. What we are asked to find out that in this particular position okay what is the force okay that is provided by this hydraulic cylinder to keep this entire assembly in equilibrium. Now what we note again is that that if there were no ground what will happen this entire assembly will try to rotate about E. So for this free body diagram okay if I draw this free body diagram here this is the weight this is the reactions at E. If we balance the moment produced by this W and FY about point E we will immediately get what is the reaction. But now what is happening once this reaction is being produced if there were no hydraulic cylinder like in the last problem this assembly this triangle ACB will tend to rotate about point A. So we immediately know that our next equation of equilibrium is just this assembly ACB forced by the hydraulic cylinder point A is a pin support you have two unknowns here. So two unknowns three unknowns this we already found. So one free body diagram three equations we can write three unknowns it is a solvable problem and we don't even bother about these two reactions because we are only interested in the force in the hydraulic cylinder. And what we know is from the geometry we know this geometry and we know that hydraulic cylinder is a two force member so the four should act along this. So we know also what is the line of action of the force okay. So we can just resolve it into FHX FHY okay and what we know is that FHY by FHX is equal to 20 divided by 80 from all the dimensions we are done okay. We solve this entire problem we can find out what is the force in the hydraulic cylinder okay. Now just briefly what I want to discuss in this slide is just an overall summary that how do we decide what is happening to the structure is it a well constrained structure is it a jambed structure or a highly constrained structure over constrained structure or is it a mechanism or a wobbly structure. You can see from here that if you have a pin connection a beam and a roller okay this is a roller then two reactions third reaction is in the horizontal direction all meeting at a point so this is a wobbly structure okay wobbly or a statically it is not a well constrained structure okay same for this it is a this is a pin connection okay this is another pin connection so what you can see is that the this structure is free to rotate about this point okay so it is not a proper structure. So okay so I think I don't need to go through all these things right you are okay with this this is one different problem which I want to discuss in good details okay now in this problem what we have okay we have a mechanism of a paper punch so this is a bunch of papers and you are using this mechanism to apply some force P here and putting the force at the ends you want to exert this punching force which will go and punch the papers okay this is an entire assembly how many arms do we have one free body two three four so we have four free bodies okay simply complex linkage how many unknowns we can have at this point how many unknowns possible at this contact point two actually so two possible why because this when you are jamming the paper the papers need not come out okay so if it is frictionless then only one okay but otherwise if the there is a friction between these two then there can be in principle two okay so two here two here four how many unknowns here two six two unknowns okay seven eight nine ten this is a slot so there is no reaction in the horizontal direction only vertical okay eleven twelve so we have twelve unknowns how many free bodies we had four so four into three is the number of equations we can write max twelve so it seems that everything is fine but the point is that there are twelve free bodies twelve unknowns okay it's a huge pain okay the problem can get very messy so is there any way do you think that this problem can be simplified a bit what do you think so what are the appropriate free body diagrams that you think we can use if you take the combined free body diagram okay then what do we learn from that what will we learn is something like this that this is a free body diagram for all the linkage okay just brooding it out okay so this free body is the top link this one this one is the bottom link okay the bottom bottom part of the plier or the punch this component okay this component is this okay is this part the top one is this okay now note one thing that if I take the combined free body diagram then what we know immediately is that if this is Q this also has to be Q because a total vertical force on the full free body is 0 so Q Q 0 P P 0 so we know that this this is Q this has to be Q similarly what we know that the horizontal forces should also be 0 why because if the horizontal forces are not 0 then they should be equal and opposite for balancing in the horizontal direction for the complete free body but if they are not they are equal and opposite then they will produce a couple but there is no couple acting here so the horizontal forces have to be 0 so what we know now is that that we have gotten rid of like two extra unknowns so these horizontal reactions go away the vertical force are equal and opposite on the two parts of the clamp now let us ask ourselves a question okay that how many unknowns in principle we can have just note this is at point a there is a Q force Q force downwards coming to point C they're coming to point B here we have two possible forces the direction can be upward or downwards we don't care okay it will come automatically so f b y f b x okay note one thing that this point B is also common to this link okay it's also common to this link so what do we do this is that link so by Newton's third law if this is f b here it will be equal and opposite f b x equal and opposite you agree with me fine similarly for the bottom link you have a vertical reaction two reactions here one two but this point E is also connected at this one so we see here that if this is f e y this is that point okay f e y equal and opposite f e x is horizontal f e x is equal and opposite and now at this connection point note okay look at this connection point this point is common to this horizontal link as well as this arm so what do we have vertical force f f y equal and opposite there are these many number of forces and of course p and p at the end so there are so many forces that are acting here now note one thing do you see anything is this clear there is a reason why I'm doing this in so many gory details okay there is a reason which will come become clear in a few moments but anything peculiar about this structure it's a 2d planar structure assuming that the third dimension is very small 2d planar structure we can use 2d equilibrium concepts is there any additional thing that is happening here is the does it have any extra symmetry it's a symmetric structures means what does it happen is that but if you take this structure if I flip it around like this I will see the same structure okay or in other words that if I draw a center line then there is a mirror symmetry about these two now that mirror symmetry what does that tell us that mirror symmetry tells us very important things what it tells us is that come to point O which is common to this part and this part you agree with me that point O is common to this one arm and the other arm okay just for simplicity I have drawn this separately but they are actually together okay I didn't want them to overlap look here Newton's third law will dictate for the top member the Newton's third law equal and opposite reaction will guarantee that on the other member if this direction is vertical it should be up like this similarly this is horizontal the right it should be horizontal in the left but what the symmetry tell you if you want to have a mirror image about this line is this line a mirror image of this this vertical line pointing like this pointing like this they are mirror images of each other okay this arrow is a mirror image of this arrow about this but what about this f o x is it a mirror image of this no it is not a mirror image it one is looking like this other one is looking like this but Newton's law has to tell you that they should be equal and opposite mirror symmetry because of that this symmetry in the problem it is it has to guarantee that they should also be symmetric about the center axis what is the only conclusion they should also be equal and opposite okay and they should also have the same direction so two contradictory things how can that be satisfied only satisfied if that reaction in the horizontal direction itself is 0 okay so symmetry guarantees that f o x is 0 straight away now note one thing what do you immediately see that this f e y should also be equal to f b y just my mirror symmetry f b y equal to f b y f e x is equal to f b x and by mirror symmetry again f c y is equal to f f y so many symmetries are happening here okay you is this fine so all these symmetries are essentially reducing the number of unknowns in the problem now how is the how is that good for us if I decide to draw the free body diagram only for this top part okay just 1 and 2 okay how many unknowns are we exposing 1 here 2 here 3 here 4 by symmetry just one extra on 4 okay but then we are exposing this point but whereas when you this link this when you remove this connection there were two reactions here but when you remove this top part you also expose this reaction but fortunately by symmetry what happened is that that the forces acting here will be equal and opposite to forces acting here okay any questions sir f o x of upper link yes there are two links okay so these are separated only like this these two links are crossing like this yes so that f o x is to us right and lower one is left one by Newton's third law yes and when you assembled it plus minus get cancelled remains in equilibrium yes right that's why those forces are equal they have to be okay now we have shown same forces at all on upper link as well as same forces on lower link yes then why it was not there in earlier case of the pulley which one that pulley wall awesome before yeah so you said give 4 unknowns because because in this case because there are only two things coming together here automatically no in the pulley case there are 4 but now if you write the equation of equilibrium for the pulley then you will see that they are not independent there is a relation between them here here also two bodies are there by the same but what we do is that the top body okay there is like this there is a link like this yes we remove this and the screw yes forces from upper as well as lower coming on the same axle but this just as earlier yes listen so we take this and the screw out one free body so what we have essentially is that essentially we have removed only one portion okay we have received removed only one portion so because of which whatever acts on one is equal and opposite to the other one but in the previous portion what we had discussed there were three things which are meeting at one point and when three of them meet at one screw then also just equal there are two of them here there are three third one pin screw actually I take pin as a part of that thing the same way I had done in the previous one pin was a part of the pulley okay sir we'll discuss it separately yeah but you can discuss that okay but this is a that's why when three members come together it's a huge pain two members is good okay two members is fine we'll discuss that after 530 if you are okay thank you now note okay so far so good everybody is okay with the symmetry arguments okay now just go on to the next part we just look at the top one we took all this free body diagram just exported it here look at this one now what we just noted that f e y is equal to f b y by symmetry and f e x okay is equal to f b x by symmetry similarly okay so this is what we have done here so essentially now what we are exposed this is f b y f b y f b x f b x so we by symmetry we were saved because otherwise if there were no symmetry then if we just take the top just the top two free body diagrams then this will be e this is b but by symmetry what happened is that that f b y is equal to f e y f b x is equal to f e x and we are totally fine now with these two free body diagrams what are the appropriate equations of equilibrium that we can write okay just look at the top free body diagram okay there is a there is one small logic here which I'm going to explain in a few moments but just bear with me if I do equilibrium in the vertical direction what do I get that q is equal to f b y plus f c y now note one thing that how is this force transmitted from this arms all the way to the clamp the way it is transmitted is that that this assembly okay this top portion tries to rotate about this hinge this point will try to move like this this point will try to come down okay and because of this there is a force exerted here which will lead to this getting clamped so what we do here is that we take by noting that the center is the pivotal point that is the center o about which the assembly is rotating okay this top one is rotating about this bottom one also is rotating about this in order to create the desired effect of producing the clamp of the clamping for this free body diagram if we take torque about o look at the dimensions this is a this is a you will immediately see that f c y into a plus f b y into a this distance is a will be equal to p into b and we just note that f b y plus f c y equal to q and we immediately note from these two equations that the clamping force q is equal to p times b by a now there are two that you can take from here that a is very small compared to b so what is this telling you that this mechanism in design in such a way that a small force p here can be converted into large force q at the other end this is called as mechanical advantage okay as you would know okay so is this problem reasonably clear find any question yes not in this problem if these two are a very good question if both of them are a c does not matter what will that change is the reaction at b but the clamping force is independent of c if both of them are equal a and a now can somebody tell me what may be the purpose behind having both of them to be equal just note these distance just see in this figure that both are a what do you think maybe the reason that they are a why is not one a1 and other a2 there is no symmetry there because symmetry is there about this that is fine what does having both of them a due to this problem moment is fine I can always balance only thing I would need to do is that I will need to write one extra equation as they as here sir yes here where as the as the lever moves away the point on the left hand side slides by the similar distance point on which point this point yeah the roller roller it will slide yes by the equal amount yes yes so both will always remain equal that is fine but no but a is a distance okay like in this configuration okay in the horizontal configuration okay what is that having but just this for example here both when it is about to be clamped yeah okay about to be clamped for example if it is not about to be clamped then a can be different because it can slide in and in but in this configuration when you are about to punch okay this a remaining same what can it achieve what it will do you are right what it will do is that that if you now try to press the punch if this a and b were not equal then after you press it infinitesimally little bit this will take okay this will take and so the point is that when you are punching you do not want your thing you wanted to go exactly horizontal okay and not these two being equal will ensure that in this position when you are about to press the punch this entire assembly will just go down and not it and like make a nice clamp on the paper sir yes can we state like this whatever forces and reactions are occurring on that joint by keeping the same distance both the things will be distributed equally that is fine but the point is that if they are not distributed equally so what okay so what but the purpose of this thing is to create a nice punch like this and if these two are not equal okay then the point is that while punching there will be a tilt it will not just go like this okay so that is for example one way to think about it in language of forces forces will be different but geometrically that's what I think is happening here and because of this those two way is just to ensure that this entire assembly move exactly vertically and this points will become more clear when we do principle of virtual work we will actually see that how does different components of the structure move when we try to create a mechanism good fine and because of that that C is independent but if a and both of them are not a C will be required to solve this problem okay any questions sir our aim is to find out force Q okay so can I write the simple equation moment produced by Q about that oh yeah what will happen is that that is not good because if you take the full free body diagram then there will be Q here they'll be Q here okay so if you take moment about this this will become positive clockwise this will become anti-clockwise and they will cancel each other out no so the full free body diagram moment is no good as far as finding Q is concerned because there is a Q here there is a Q also here so if you take the whole thing as a free body take torque about P this is a P it will produce a clockwise torque anti-clockwise cancel clockwise anti-clockwise cancel so 0 equal to 0 we don't get anything and that makes sense why because this full full free body means that internally this can be all welded but full free body still remains the same but think about it that if all of them are internally welded there is no mechanism so that is one way for us to know that the complete free world diagram is not a good thing because if you weld everything feel free world diagram for the complete thing will not change okay okay but we know that if you weld everything then there's not mechanism okay that's the reason we know that we have to go to internal details and just the full free world diagram is not good enough okay sir cancer this answer could be like this Q is equals to P into B upon C plus a no it is only a C C plus a no it is not because movement produce about that it is not if you do this problem it will be PB divided by a you solve it okay and then you tell me if I am wrong I will correct it correct myself in the next class I don't think so okay so any question fine okay this is very interesting problem okay what we have okay so these are one of the few other problems we had given in our exams okay final exam is a very nice problem it's from Merriam statics okay if you want to solve really interesting problems in you make but get older version of Merriam not the newer version that version doesn't have that group problems the older version 1972 edition 1978 edition that has fantastic problems so what this problem is as follows so we have a beam AB weight is given 300 kg total length of the beam is 6 meter the entire assembly now what is being done is suppose we want to hoist this beam up take this beam up initially it is on these two supports you want to take this up now how are we taking this up we are having an assembly where there is a there is a ring C through which we are passing ropes okay and we form a nice triangle ACB the string has a length of 4 meter 4 meter and length of AB also is 4 meter now what we do is that we apply a force on the top and when the force at the top okay becomes equal to W you can lift this beam of the two supports you agree with me there are two supports are providing reactions which are balancing weight but if you apply a load which is equal to P becomes equal to W then you can lift the beam of the two supports okay now what you're asked to find in this problem that if you if you apply a load W and hoist the entire assembly upwards then upwards then what is the tension that will be created in string AC and string BC is the problem is the problem clear you have this you want to hoist it up pull on it okay it will lift up the supports when the weight the pool pulling force becomes equal to W and now you are asked to find out that when it is in this hanging or a suspended position what is the tension in AC and tension in BC is the question can you tell me what happens somebody tell that if you hoist the beam of the supports what will happen yes somebody said we want it to be cleared of both supports no supports then what will happen somebody said something inclination of AB okay so AB will incline I'm saying that this is this will become a triangle what happened force in one direction W will be known because P will be equal to weight of the beam okay so one is known other three angles are known which angles are known this is 444 angle at sea will be known angle at sea AC but but but what what will rotate you are saying yes sir what somebody said something will rotate what is that beam will rotate beam is rotating say we will rotate about me not necessarily not necessarily not necessarily not necessarily I agree with the statement that the beam will rotate first of all somebody tell me that why should the beam rotate couple produces yes the beam does not rotate then the center of mass is at 3 meters this is at 2 meters so there is an offset between the center of mass okay there is an offset between the center of mass and the load applied so that will produce a torque it will rotate the beam what that what will it rotate the beam in the clockwise direction or anti-clockwise direction clockwise direction how much will it rotate till at the top and the bottom that will W and the top align with each other okay so then now essentially how many unknowns to be have and how many equations can be right okay so we have a inclined position okay what we know is that after the beam is lifted of the supports when these strings become out okay the angle this angle will become 60 60 60 and the beam will rotate okay like this okay so unknowns are how many and what is the free body diagram and how many unknowns do we have two unknowns are the tensions two tensions are unknown is there any other unknown the rotation because support actions are zero we want it to take all the support so we have three unknowns but we treat as the unknown but we can write three equations for one equilibrium okay we will just one beam okay so we can write three equations three unknowns and can effectively solve this problem so the only trick as you rightly pointed out is that when you lift it off the beam will not even horizontal it will tilt okay and a tilt will create an additional unknown okay and two unknowns are the tensions so we have a properly constrained system with three equations unknowns which can be solved and if you just look at it okay as you rightly pointed out you will get a configuration like this p will be equal to w both will be in the same line and this angle if it is theta ta will make an angle of 60 minus theta with respect to horizontal tb will make an angle of 60 plus theta with respect to horizontal and we can write two equations three equations first is ta and tb horizontal components should match with each other second equation of equilibrium is for this three body diagram okay if you take torque about point b then you will get one simple equation tb take torque about point a we will get another equation and from these two equations we will get tb equal to 3 ta and these two equations you combine them you will see that there will be a relation like this from which we can get theta and once we get theta immediately we know what is ta and tb so it is not a mechanism but only small trick which I think all of you immediately recognize is that the tilt will happen okay that is the only trick in this problem any question about this so we can get theta once we get theta you can immediately obtain what is ta and what is tb any questions any comments 3 plus 1 3 is where the center of mass okay 3 is where the center of mass is plus 1 is from center of mass to be 3 plus 1 4 okay so ultimately it is forming equilateral triangle yeah because all the links are given the slack length is 4 so when you lift it up the slack will become taught and so it will become a nice equilateral triangle now this is another looks like a very complex problem okay but I am almost sure that you are going to figure it out in 10 minutes okay not 10 minutes 10 seconds so what the assembly have is as follows but there is an important point which I want to make with this problem so what we have is a mechanism which is lift used to lift this like there is a there is a figure showing a special before electing vertical sections of a construction tower it can be a construction tower it can be TV tower or mobile tower it doesn't matter as far as this problem is concerned but what we have is we want to lift this make this entire assembly okay what assembly do we have we have this assembly this pink portion using this assembly we lift this portion which is the part of a tower so called tower and how are we lifting it that there is a vertical pillar rigid vertical pillar which is clamped here okay so it is not going to be disturbed it is perfectly rigid and there are wheels on four sides and what we are doing is that that this assembly can slide up or down but we are not going to sit down and make it go up and down what are you doing is that that there is a link FH okay this link FH is used to exert force on this tower to go up or down okay take take the load from the tower but this in turn is connected to a arm EDF and this arm EDF is hinged at point E and there is a hydraulic jack or a hydraulic cylinder which is connected like this at point D now if this hydraulic cylinder extends this assembly will go up if the hydraulic cylinder reduces this assembly will come down and what we are asked to find out in this problem is what is the force in this hydraulic cylinder is the question that we are asked is the idea clear that a hydraulic cylinder of the arm increases this assembly goes up arm decreases assembly come down okay question is what is the force in the hydraulic cylinder in this configuration it can have many configuration but given the entire dimensions configurations everything what how do we find out what is the force some suggestions first of all it looks like a reasonably complicated problem because there are four wheels okay and so many things are happening but what do you think is the crux of this problem hinge what hinge F okay so what do you say should be what is the free body diagram that we should be on the lookout for and what are the corresponding equations of equilibrium and what is our motivation to write those equations of equilibrium for those free body diagrams let us go step by step let us let us answer this one let us like turn the argument around and say we are asked to find out what is the force in the hydraulic cylinder so let us use our normal strategy and say what if the hydraulic cylinder were not there okay what will happen why will it come down why because this assembly EDF okay the assembly EDF will try to rotate okay about this point why because the force coming from the top link there is no counter counterbalance for it so if you remove the hydraulic cylinder this assembly will rotate about point E and the entire assembly will come come down okay but why is this link E FH required what is that link FH doing if link FH were not because what we are learning now is if we remove this then link FH will make sure that the assembly collapses but then let us ask another question that okay let us remove the link FH that is a problem creator but if you remove FH then what can happen what will happen nothing what is link FH doing so that is the question what is link FH doing how is it supporting the top one if this link FH were not there what will happen to this assembly will it topple will it will it rotate it will not rotate but it will just slide down okay so what we know is that link FH is preventing the assembly from coming down and in turn the hydraulic cylinder what is it doing it is preventing the rotation of E D F about point E so we know the entire mechanism and once we know the mechanism we immediately know what are the appropriate free body diagrams and what are the appropriate equations of equilibrium so we immediately know that even though the link looks complicated instead of having four wheels if there are 15 wheels it doesn't matter the assembly will still topple if there is no link FH so we immediately know is that for the top free body diagram okay the top part what we know is we need to take that free body diagram these two rollers they can only provide reaction in the horizontal direction but there is nothing to balance equilibrium in the vertical direction so the only equation of equilibrium we need to write is equation or in the vertical direction and from that we can immediately find out what is force in FH you agree with me okay so that's we immediately know so one line we can immediately find out what is the force in FH all the dimensions are given to you 1.753 so we can know what is inclination of F okay everything is given to you we know what is the angle we know what is the force in FH and now once we know FH we know that FH is trying to rotate the free body diagram EDF about D sorry about E so we need to take that free body diagram and write down the equation of equilibrium moment equilibrium about point E and then we can immediately find out what is the force in the hydraulic cylinder is this okay the procedure so the idea is this that the problem you look for for example we can make this problem even more complicated by having instead of having four wheels I have like eight wheels ten wheels okay a huge assembly but that doesn't change the problem at all as far as finding the force in the hydraulic cylinder is concerned the structure may then become statically indeterminate but for our purposes we exactly recognize what are the appropriate free body diagrams and what are the appropriate equations of equilibrium procedure clear okay the logic clear that even think in terms of mechanisms then all the entire assembly even it may look complicated immediately becomes apparent okay so this is the solution you can have a look at this okay just simple whatever we talked in words is essentially discussed in simple algebra which you can definitely do it's not an issue any questions any doubts which one huh so that's one good thing okay so note that FH we are neglecting the weights of the assembly so FH is connected only at ends and if you read the problem statement we have to neglect the forces weight of the link okay so only forces acting at the ends so it's a two force member and in a two force member the force will act along the two inches which in this case because the member is straight is along FH so it's a two force member and hydraulic cylinder also is a two force member yes means the force required to lift the assembly will become less yes yes yes because as the movement of the torque is created about but then what happens is that I agree with you that it will do that way but then think about it from mechanism point of view that in this case if I want to lift this assembly now for example it's a question of what you want to do if you think in terms of displacements if I want to lift this assembly if I increase the length of this a tiny increase here will lead to a huge increase in F so by tiny movement of hydraulic cylinder you can lift this assembly by whole lot but as you said you are right that if you move it here then what will happen is that the cylinder arm can only be of a certain length so you need to just to lift the assembly the amount of extension it need to have will be far more we are right the force may be low but then you have to look at this kinematic considerations that like how much you need to extend the arm in either to make it go up or down mega gram mega 10 power 6 gram okay 10 power 6 any other questions so I think with this okay so let us move on to the last problem of the day is a solved problem in beer and Johnson but we are solving it in a different way and we are giving it some kind of mechanistic understanding of what the assembly is doing and how can we think about that and simplify the way the problem is approached so what we have here so in plain English what the problem is is as follows so this is a mechanism used to lift some weight okay so what we have is that we have a platform here okay wide platform and a platform what it has is that on two sides we have two assemblies like this one assembly on one side on the other side there is another assembly okay so if the weight that we are lifting is w then if you recall our arguments about symmetry then one half of the side will have a weight w by two which it is sharing so essentially it's a 2d problem now in this 2d problem what we are doing is the assemblies like this there is a base which is rigid this platform is hollow from inside what we are doing is that there are these wheels two wheels connected by link and on this link we have e b and g c which are pinned so it is forming a parallel gram okay so e g c b is a parallel gram and all of them are pin connection pin pin pin pin roller roller now further what we are doing is that that we are adding a link which is connected to this platform and this member at d so it is a two force member finally we have a hydraulic cylinder which is connected at point d and at the other end attached at point h and all the dimensions are appropriately given a a theta okay everything is given and what we are asked to find out that to keep this assembly in equilibrium in this configuration what should be the force in the hydraulic cylinder okay the reasonably complex assembly and what we are asked to find out ultimately is what is the force for this assembly in the hydraulic cylinder so let us go step by step okay is the question clear is the problem clear what is the mechanism just note one thing that this assembly everything is pinned so the angle can change so if you want to lift the assembly you increase theta if you want to make the assembly go down you decrease the theta okay now the question we want to ask is what is the force in the hydraulic cylinder is the question so let us think about the problem in different ways so can somebody tell me what may be the purpose of having this link ad so let us always ask that what is the purpose of putting some links what is ad doing you think it's a two force member it rises the load set when the suppose suppose ad is not there suppose ad is not there okay version of bc in but horizontal life for example I don't need to resist anything right horizontal is fine because if I want to raise the assembly I will take it up lower that internal mechanism can slide it's fine horizontal motion of the platform okay so what it is doing is this that if you look at this platform if you look at the top platform what are the reactions acting on this platform if I just draw the free body of the platform b will provide a vertical reaction c will provide a vertical reaction okay w by 2 acting from the top and this ad so if I look at the free body diagram it will look like this if this link were not present then what happens that any tiny force okay any wind loading or anything that acts on the top platform it will just slide it out so to maintain that top platform in equilibrium we need the presence of this link because if it the platform tries to shift horizontally because of some external load some noise or anything then this link will prevent it from going and getting out of equilibrium so that is the purpose of that link okay as you rightly pointed out it will prevent this equilibrium or it will keep the system in equilibrium the platform in the vertical in the horizontal direction is this fine argument fine that's what it is doing now let us look at this free body diagram in the current configuration when there is no horizontal force yes please which one for the parallel link if we make one of the giant giant b has been designed and it's a pin joint no but if you make it a pin joint if you don't know now if you make it I agree instead of roller you are saying you to make it pin right yeah so in this case the purpose of that assembly because now it's in theta position if I want to raise it then people will make sure that you cannot like change the configuration so we can't keep that as a pin it has to be a roller because only then can that theta can be changed and assembly can be lowered or raised okay so pin is not possible so but what we can do is that instead we have this link and if there is any horizontal force that acts on the body then this link will balance that but in the current configuration there is no horizontal position that is force acting on this platform what does that mean that the link force is zero so as far as the current problem is concerned it it will as well be good if the link were not there okay so we can keep that link to be equal to zero now second question the main question what is a hydraulic cylinder doing if you don't put the hydraulic cylinder what will happen it will just for example this it is like this it will just keep shifting like this and just go become flat okay it will not be able to stay in that configuration now how do we translate that because ultimately we want to find out the force in the hydraulic cylinders there are way too many unknowns here okay so we know in some kind of course terms that what is hydraulic cylinder doing but in terms of more free body diagrams and equilibrium what what is happening to the to the entire assembly because of just tell me if the hydraulic cylinder were not there what happens to FB and GC forget about the toppling let us think in terms of only only force equilibrium without the hydraulic cylinder what can we say about force in BE and force in CG no hydraulic cylinder no no theta is given suppose given theta okay let us not talk about mechanics because mechanism is like you are trying to look at kinematics let us talk about force equilibrium if there are no hydraulic cylinder just two parallel limbs and assume that like suppose the system by miracle is in equilibrium what happens to those two systems to what can we say about those two links about the forces in those two links no no no forget about coming down or no we realize that hydraulic cylinder is preventing for coming down but now let us talk another language this is a language of kinematics where we are looking at the motions and so on but now we are doing equilibrium so let us come to that language and what I am saying is that suppose if hydraulic cylinder were not there and still if you want to believe that the system assembly still in equilibrium what what can we say about the forces in those two links action force there are two force members there are two force members now if we draw the free body diagram of this inner wheel like this okay there are two force members acting on this but there is a load acting on it in the vertical direction now when the support reactions are parallel what is the issue that we saw both are is parallel so if there is any force acting on the system which is perpendicular to them you are destroyed that's what is happening here that if there were no hydraulic cylinder those two links will be parallel there will be two force members but there is still a vertical force acting on it which has a component in a direction perpendicular to those two so the system will not say in equilibrium so what is that thing doing is because by providing a force in the middle this this link what it is doing hydraulic cylinder it is doing it is making sure that one of the members is not a two force member okay and when it is not a two force member what does that mean that it can have two unknowns it can have one reaction two reaction but now what we do is that instead of resolving the reactions in vertical and horizontal direction let us say that the two reactions are provided along and perpendicular to the length of the link just for our convenience because that's what we saw that without hydraulic cylinder the force will be along the member because of the hydraulic cylinder the member is no longer two force so it will have another component we which it can which it can generate you see the point see the point any question any doubt yes please if I go to the previous one yes joint what joint a a b and a b and d d yeah no no but the point is this what it can do is this that for example that a d now it is not fixed it's hinged that is fine but that angles can change because of that it can accommodate that motion but but the angle can still change the distance is that the length of the triangle are constant but the angles between those three connections the length of that the triangle you can have a triangle with three sides but I can always deform the angles what no but he is no so big distance can change now a b because it's a roller support that angle can change that distance can change right so both the distance can change and the angles can change fine no it's a rule because without roller support you can't even move okay now so so is the entire argument clear what we are trying to do at so what is this hydraulic cylinder doing the hydraulic cylinder is ensuring that this member doesn't remain the member e db does not remain a two-force member and now it is straightforward what we do is that for this assembly of wheels we have two unknowns 1 2 3 and rb and rc are also unknown but what we know is the sum of rb and rc by equilibrium for the top body should be equal to w by 2 pardon okay I have written w say it's w by 2 so how do we find out what is fb2 just take equation of equilibrium in the inclined direction this direction so what you will see is that if this angle is theta this angle is also theta so for this free body diagram equation of equilibrium sum of all the forces along this direction being equal to 0 will immediately tell you that fb2 is equal to rb plus rc into cos theta but rb plus rc is just w so fb2 or the normal component of the force is w cos theta immediately yes please you have shown in the fbd we have the force fad right huh so that has to be equivalent opposite has to be there in the freeboard diagram 2 as well yeah so freeboard diagram on the top is rb rc upwards look there and rb rc will be equal and opposite on this one no no I'm asking about the freeboard diagram 2 yeah so freeboard diagram 2 what are you asking I don't understand no the you have shown fb2 right which is perpendicular to that link yeah but that is that is the resolved component here I should show it the other way you are going here I should show you the other way there I can choose whatever direction I want to write that direction is given to me if it is opposite it will come out to be negative I have made a mistake in the third free body diagram okay that I realize now okay but as far as that one is concerned what I'm saying that this this member is connected to the pin if you remove this to draw the free body diagram instead of writing fx and fy I'm writing at fb1 fb2 and fb1 two perpendicular direction what is so special about horizontal and vertical I can take my x and y axis in the inclined way and the reason the reason we did that was for the simple logic that if there were no hydraulic cylinder see our entire motivation is to find out what is the force in the hydraulic cylinder if there were no hydraulic cylinder okay if there were no hydraulic cylinder then there will be only fb1 there is no fb2 because this becomes a two force member but the presence of the cylinder made this present also created this fb2 and that fb2 is now ensuring that for this free body diagram fb2 okay the reactions are not just parallel but there is also a perpendicular component and we can write down equation of equilibrium sigma f in that inclined direction to be equal to 0 which will immediately give you what is fb2 what is fb2 is the logic clear that how did we build on it that for example the way we motivated the entire thing because otherwise it will get lost like for example there will be so many unknowns it is not immediately clear that what you do but the moment we think about in terms of mechanisms that what is every component doing then we can logically come up to these steps okay now if you get fb2 then I made a mistake okay this I have to admit that Newton's law will guarantee this I made the other way this I forgot okay it should be downwards this arrow should be flipped it should be the other way okay I made the mistake there okay now what is happening finally that on the member edb there is a force coming from the hydraulic cylinder this coming downwards fb2 this doesn't create any torque about e so for this free body diagram ultimately what is hydraulic cylinder doing it is preventing the rotation so take torque on this free body diagram of all the forces about point e you will immediately see that you can find out fb2 times 2a will be equal to fh perpendicular component into a and we know what is fb2 w cos theta so then immediately from all the dimensions we can immediately find out what is the force in the hydraulic cylinder problem fine so by using all these different ways in for example what is every component doing we could reduce our structure into small small components we could use logic use appropriate free bodies use appropriate equations of equilibrium and finally could solve the full problem is the idea clear so the idea what I'm trying just trying to say that's what we emphasize to the students is that just ask what is the quantity that you are looking for and then go from there rather than like just to begin with just destroy everything and try to write some equations of equilibrium ok just write arbitrary free world diagrams ask what is required and then build up from there logically and it may not always happen there are some problems which are very complicated in the sense like you cannot use some logic to get nice equations of equilibrium but more often than not this approach works quite well that you find out what you need and then go upwards ok by ok what if that were not there what will happen and then that immediately gives rise to the logic that what will be the free body diagram and what will be the equation of equilibrium ok so with this I think we have our time so any questions any discussions ok we can have now I want to know one direction of normal reaction of the free body diagram if larger sphere is there and smaller hemisphere small hemisphere and a larger sphere these two are connected along the line joining the centers of two that will be normal reaction no that is not the point of my question is hemisphere is smaller and the sphere is larger that doesn't matter the reaction point of contact will have a tangent ok so the reaction will always be perpendicular to the tangent no at the point of contact actually that's horizontal surface another one is curve surface so you cannot draw a tangent at that point now what are you what do we want ok so do we want this hemisphere hemisphere is smaller one ok this is inverted inverted hemisphere no not like that inverted hemisphere like this ok yes and and the largest we are touching it how does it left from the bottom no no not the bottom over the ground they are side by side can you draw it here come and draw this is a normal reaction is good this one is fine and this one is this is this one is deadly ok so it is not clear at all but for this one ok what you can do is that for this particular sphere this is the point of contact ok let us zoom in here let us say this is a this is a and if there is no friction it has to be normal so it should just act like this to the center that is the only way the normal reaction can act in this case even the other one is an edge you can just replace that with a like any edge you can say that it is a small radius of curvature so there will be some kind of a tangent but on the main sphere there cannot be any tangential force because tangential force by definition is friction by definition so the reaction can only be like this at the point of contact if this is the point of contact we note that down and from the point of contact going into the center is the normal reaction along just it will be equal and opposite just equal and yes but this one is dicey because there is no nice normal to define here this point of contact they are edges this one is very dicey there is no clear cut way to define normal reaction there not that I know of anybody has any idea they can say but because here there is at least one nice surface here both surfaces are discontinuous that but those are all approximations you can make an approximate but here there is no approximation it's exact okay so thank you