 So, let us start with problem number 1, ok. So, we will do it for around 15 minutes. So, we have a simple mechanism in this problem. What we have is this is a device which is used to supposedly trim various things. So, this is where this is the blade this vertical portion this is a frictionless guide in this problem it is frictionless. And what we are asked to do is that this knife is connected by a 2 force member db to this to this full member abc. And what we do is that if you apply a force here then that force get transmitted here. And in the end at point e the force gets transmitted to the plate which then get sheared. So, what we are asked in this question is it for this very simple system we are asked to find out what is the vertical component of force exerted by the blade at d, ok. So, what is the vertical force here and because there is no friction between these 2 rigid supports this vertical force automatically imply what is the shearing force. And second we want what are the reactions at c, ok is the problem clear, ok. So, let us solve this problem, ok. So, this will be here, but what is the logic, ok. For example, what then in the yesterday's class is we ask ourselves a question if you want to know what is the force in member db the simple question we ask is what if we remove db. So, we know if we remove db then whatever force you apply here, ok there is nothing balanced about c. So, what we know is that this db is providing a counter balance for this force about point c. So, immediately we know that the thing we want to do is take this free body diagram and take moment about point c in order to find out what is the force in this 2 force member. We know what is the orientation of this member why because we know this 2 distances. So, we know what is angle. So, the moment we know this angle we immediately know if the force in db is given vertical and horizontal force is immediately obtained, ok. So, I think I forgot time p let me p is not unknown p is missing let me tell you what p is, but even then you can obtain the answer in terms of p, but let me quickly tell what the value of p is, ok get the answer in terms of p and then report what the value is. p is not a it is a it is a hinge between d and b, but member abc is a full rigid body, abc is a rigid body and it is connected by pin joint to member db at point b. At point b. Point b. Point b pin bb, ok. Right. So, abc is a whole rigid body it is pinned about point b with respect to member db. So, member db is a 2 force member member abc is north, ok. So, load is equal to 400, load is 400. So, use the load as 400 and tell what the answer is. 26. Yeah. It looks fine dy, right 2675. 2534. 2534 Newtons. So, between those like approximately similar value, but not exactly load is p is 400, ok please take p is equal to 400 I am sorry. Answer 2.7 kilo Newton for dy 2856 yeah 2.8 kilo Newton, 2.8 kilo Newton yes. Yeah, yeah, yeah. That is vertical reaction at d, yeah 2.86, 2.86 is the vertical reaction at d, good. 2 force member. I have found that as a db that rdb member comes 3.076 Newton. Ok maybe, but we are asked. Yeah, but what is the vertical reaction? Because that is what is asked in the question and it has miraculously disappeared. You may be right, but what I want is what is the vertical reaction at d? That is what the problem demands, ok at vertical reaction at d. Your like force in member db is fine, ok. So, do you think I can just briefly show the solution and we move on, ok. So, let me just quickly show the solution and we can move on, yes please. Then I got answer 2.83. 8386, ok that is fine. 2.8. Yeah, depends on how, where did you truncate decimal places is fine. 2.8 kilo Newton approximately is the right answer, ok. So, just simple, ok very straight forward problem we begin with a simple problem. We will move on to little bit more difficult problems. So, this is the entire free body diagram. We have replaced the force in member db, ok, along the because db is a 2 force member. This is a line of action of the member. Take moment about point c, ok for the entire free body diagram. You will get an equation like this. Note one thing that the inclination is such that dy by dx, ok is equal to 60 divided by 25. The dy by dx is equal to 60 divided by 25 and when we say that this is the direction of the force which is provided by member db, we are assuming implicitly that db may there is a compression, ok. There is a compression in db is what we are implicitly assuming. So, it will push the other member, ok. We solve this, ok because we are interested in dy what we can do is that while taking the torque of member d, of horizontal force dx about point c, we just replace dx in terms of dy here and we solve this equation we get dy. Then what we do simply that for this complete free body diagram abc, take equilibrium of the force in the horizontal direction. We can immediately then find out what is fcx reaction at point c in the horizontal direction. Similarly taking equilibrium in the y direction, ok. We can immediately find out what is the vertical reaction, ok at point c. The total effective reaction is sigma fx square plus fy square and the angle will be just tan inverse of fcy by fcx, ok because both are negative means we have taken the direction fcx to be upwards sorry fcy to be upwards fcx to be in the horizontal direction, ok and both come out to be negative which means that the direction that we had assumed are exactly opposite. So fcy is downwards fcx is to the left and so the effective angle will be 68.5 in this direction for the force. Any doubts can we move on to the next problem? So what we have is the same kind of a plier what we are asked to find out that what is the clamping force q in terms of the parameters what are the parameters in this system? The parameters in this system are c, e, alpha, b, a. There are 4 dimensions of the system and for this system, ok and there is an angle alpha. So for these dimensions what we are asked to find out is given this p what is the clamping force produced at the clamping points q. Question straight forward just 2 lines question, ok. So let us solve this problem, ok using the symmetry arguments which we had done yesterday little bit difficult problem not too difficult. Which one is alpha? Ok so let me mention that properly here. What is alpha is if you look here focus on these 2 pins these 2 if you take a line joining the centers of these 2 it is not perfectly vertical it makes an angle alpha with respect to the vertical, ok. So that line is at some inclination and you see because of that that angle is also reflected like this because this dotted dashed line is perpendicular to this so that also becomes alpha, ok. Is the geometry clear? What alpha is? Yeah so let us solve this problem. Any question, ok. So you can ask me just note one thing is that even though the geometry looks all curved bars and so on if you can just replace that with straight lines, ok. That will really simplify the problem because all the intricate details of the geometry handles being curved and everything is completely irrelevant to the problem. If you replace it with a nice straight lines problem it will become very straight forward and use little bit of symmetry arguments not much of what we did yesterday. This is even simpler than what we did yesterday. In this problem, ok a simple hint is that, a simple hint is that take this line this horizontal line is actually the line of symmetry in the sense that you take the plier and turn it around about this axis you will see exactly the same thing, right. So what does that mean that at this point of intersection two of these bodies are meeting, ok. This top handle and the bottom handle. They are intersecting at this center point let us call it O and in that case because of Newton's law as we saw yesterday the reactions will be equal and opposite but by the symmetry in the problem about this axis they should also be equal and mirror images of each other. So that will dictate as yesterday that the horizontal force, ok when you remove those two will be 0. You will see that and that is the only symmetry you need to use in this problem. Why? Because then you can just disassemble this portion, disassemble the top portion and we know from symmetry that the horizontal reaction here is 0. Same goes for this one that this pin is common to the top jaw and the bottom jaw, ok. So by Newton's third law when you disassemble these two the reactions from one to the other should be equal and opposite and it is a pin joint, ok. So these two degrees of freedom are restricted rotation is free. So there will be vertical reaction there is horizontal reaction Newton's third law will dictate that they will be equal and opposite but the mirror symmetry about this axis will dictate that they are mirror images of each other and if you just draw those arrows you will immediately see that even at this point the horizontal reaction is 0. And then you are left with two very simple problems you have to analyze this jaw sorry this handle and this jaw, ok. And you will immediately realize that what is this handle doing? This handle is trying to after applying these two forces P these two handles tend to rotate about these two points both the bottom one and the top one, ok. So we know that somehow the torque about this point has to be involved for this free body diagram. Similarly what about the jaw? Even in the jaw you can see that the jaw tends to rotate about this point. When we do principle of virtual work we will solve this exact problem using principle of virtual work and you can actually visualize what the displacements are but here in very hand waving or qualitative terms you will see that the top jaw and the bottom jaw tend to rotate about this point. So the immediate equation that would come to your mind is for this free body diagram torque about this point, ok. So that is the overall idea and by symmetry what did we achieve? We made sure that for the top free body diagram at this point of intersection there is no horizontal reaction. For the jaw, the top jaw if you draw the free body diagram then at the point of intersection there is no horizontal reaction and we simplified the problem we got rid of two unknowns immediately, ok by using symmetry arguments. You do not need to use symmetry argument but then you will end up solving the complete all four free body diagrams and that is not desirable if you can use the symmetry in the problem simplify the problem why not, ok. So that is the idea, ok. So if you can work on it for other five minutes, ok. Now with this extra piece of information if you have not already used it, ok. So what is A? Ok maybe I, this line we saw is the line joining the centers of these two, ok. Now what you do is that you draw this line, this line is parallel to this line. So these two lines are parallel to each other, ok and where these two lines intersect like at this point, ok. So this angle is also alpha, the angle that P makes with this line here is also alpha, ok. And from that point, ok, this distance is A. So for example I can say that P cos alpha into A is the torque of produced by this force P about this inch, ok. This angle here is also alpha. So P cos alpha into A is the torque for this free body diagram, ok of this force about this because the other component will just pass through this. So that is the reason that for example if this entire curved assembly if we reduce it into just straight lines, then the problem will be greatly simplified because all those details are extraneous. Sir I have a doubt. Yes. When we apply the load, we have two hinges in the line of symmetry. Two hinges in the line of symmetry but we, yes. The distance between that reduce when we apply the load. No, no, no, see this right now we are not looking at a mechanism. So for example if there were no material to clamp, then you apply the load of course the distance will change. But in this configuration when you are about to apply load, ok and the presence of the material between the clamps is keeping the system in equilibrium, ok. So we have put some rigid clamps, ok or whatever some material at the clamps. So the system currently is in equilibrium in that configuration and what we want to know is that in that configuration, ok what is the force that is acting at the clamp given the force that handles it. In the diagram of any of that. Yes. Component, we still have a horizontal component at the hinge. At which hinge? The symmetric hinge, the hinge on the line of symmetry. There can be horizontal component if you disassemble those two, ok but only thing that by symmetry. What is symmetry that you take this here, you rotate about the center line, ok. It will look the same. Let us just think about it. You just flip it, you will exactly get the same figure. So there is a symmetry about the center line joining the two center hinges. And that symmetry ensures that there is no horizontal reaction, ok that will add those two, both connection points. And now on watch free world diagram, both top free world diagram and the bottom free world diagram. Can you see this, ok. So what we have done is we have just made a simple caricature. This is the center point, ok. Back to this. This point is this. This center point is this, ok. So all the extraneous details we got rid of. And what we did is that, that this entire handle, we can just replace it by a nice straight of, a pair of straight lines. So this is one line, ok. This is this particular line EB, ok. This line EB is just this. Line joining the center of this and the center of this, ok. I will pose this, ok. You can use these figures if you wish to. So this is EB. And from the dimensions given in this figure, what we have seen is that this line, ok, is perpendicular to EB. Similarly for the top figure also, this line will be perpendicular to EB. So this distance is A. This is the right angle. Now from symmetry we saw that there is only vertical force, no horizontal force. Can everybody see this figure? Is it clear? Ok, no horizontal force. So there is only one vertical force. At point B, there is a possibility of having two forces, ok. There is a possibility of getting two forces. But for this free body diagram, if you take sigma FX equal to 0, you immediately see that FBX has to be 0. And then for this free body diagram, if you take moment about point E to be equal to 0, ok. Then you immediately get what is the value. But what is the moment about point A? So it will be P A cos alpha, ok. Because we want this horizontal distance which is A cos alpha. So P A cos alpha will give you distance, horizontal distance between this point and B. Minus B sin alpha will essentially give you that what is the horizontal distance of the line of action of P from E, ok. So P A cos alpha minus B sin alpha, ok, should be equal to FBY times B sin alpha. So that is an equation. You can immediately find out what is FBY. Now come to this free body diagram. What we know is that, that what is happening is the clamping force, ok, is trying to rotate this free body diagram, from the jaw about point C. And if you then immediately see that, if you take torque about point C to be equal to 0, or the moment balance about point C, similarly since FBX is 0, this goes away. So FBY into E, this is E, is equal to clamping force into C. So you just use these two equations and you will immediately see that Q will be given by P E by C A by B cos alpha minus 1, ok. Which I think very close some answer, some people got it, because we are doing this problem in a hurry. I am sure that if you do it at some leisure, then you will exactly get this answer, ok. It is ok, because there are so many distances and we are doing it in a hurry, ok. If you do it at leisure, ok, everything will work out fine. Is the overall idea clear to most of you? Q is equal to P or less than or greater than? It will have to be more than P. More than P. Yes, because it is a machine. So the advantage is that, why are we doing this? Why don't we just go there and hold it like this? We don't want to just hold it in our hands because too much load is required, whereas with a machine, a tiny load here, you can come, because A is very large. A is very large, ok. So what you can say that A is extremely large, you can throw away that 1, ok. And then A will essentially win over everything. Take some extreme limit. So that is the mechanical advantage you have. Otherwise I can just go and hold it like this, right. But that is too much force. So there is one thing that I can tell immediately is this. The solution looks like this. E by A, A by B, ok. So what you can see here is that, that you can have double advantage is what I can tell you. That A by B is a number which is more than 1. E by C also if you ensure is a number more than 1, then you are getting more advantage than just having one plan, ok. Less force will be applied. So for less force because we are multiplying 2 numbers which are more than 1, ok. So that may be one of the reasons, ok. But again that is speculation. Because if you look here there are 2 ratios involved. E by C, ok. And it is A by B, ok. And alpha, if you keep to be very small, then just note that cot alpha will be very large when alpha is very small. So you are getting a significant amount of advantage, ok. Mechanical advantage with this one. Because when alpha becomes small, ok, cot alpha becomes large. When alpha becomes close to pi by 2 and becomes very large, cot is just inverse of alpha. So when tan is small, cot is large. So it is a huge advantage that we get here. E by C greater than 1. A by B greater than 1. Cot alpha very large. One you can almost forget actually, ok. So one would be very negligible compared to A by B cot alpha. Actually there are numbers for this problem, ok. So this is, I got it in symbols because it is very easy to display the solution. Because numbers what happens is that like you do something, you may do two mistakes and you get some number which is very close to the actual value. With symbols you cannot lie with the symbols. Numbers you can always fudge. There is idea clear but there is significant amount of mechanical advantage that is coming in this one. Can we move on to the next problem before we start the quiz? This is from the older version of Bear and Johnston, ok. So if you get hold on older version of Bear and Johnston, it has some really interesting problems, ok. So what we have here is an assembly. So like in many problems which we had done yesterday, we have a rigid post which is nicely bolted. So for all practical purposes we can take it to be a rigid body. Nothing to do with no free world diagram that is involved. But what is happening here is that we have a hoisting mechanism to hoist this truss. Now all these details of the truss are extraneous. We do not really need them. How are we hoisting it? We have two parallel members ABC, DEF. We pin this member DEF at point F to the truss. We pin ABC to point C, ok, to the truss. And for simplicity what the textbook tells us is you assume, ok, that the pin at C can only transmit horizontal force, no vertical force. This is only to make the structure statically determinate. That at C there is no vertical reaction even though it is represented as a pin, ok. But what I am telling you is that if you think about it that restriction is not required, but it simplifies the problem, ok, that at point C, ok, even though it is a pin, ok, only horizontal reaction, no vertical reaction can be transmitted. It is written in the statement of the problem. Now this assembly, what are we doing is for this assembly we are hoisting it using this hydraulic cylinder which is attached to this main portion at point H. Now how are we raising or lowering? Look in this position, if you increase the length of this arm, this parallelogram can deform go up and the assembly will go up. If you reduce the length of this arm then the parallelogram which is like this will come back and the entire assembly can be lowered. This 5 kilo Newton, ok, is the weight that we are lowering or raising and what we are asked that for this particular configuration, ok. So for example what I have seen is that some of you are asking that the system will move. So what I have to tell you again is that in this particular course what we are interested is if you are cosy statically lowering or raising or if you are holding this assembly in a particular position then for that position what are the forces and what we are asked here is that given this is the geometry of the problem what is the force that is exerted by this hydraulic cylinder at point H, ok and all the dimensions are given to you. So there is one, a naive way of solving this problem, there is an elegant way of solving this problem, ok. So why do not you have a go at it? Any questions you can ask me. Towards then I will flash the solution, ok. What we are asked, ok, just think about the problem. Is the hydraulic cylinder is attached to the main vertical wall or the main vertical pillar at point H. So what are the forces exerted by the hydraulic cylinder at point H? So we are going to find out FHX and FHY for simplicity and all the dimensions are given. The problem from old Beer and Johnson, get that book if you have, there are many, many nice problems. The new one does not have that many good problems. Problem clear, is there any issue with the problem statement? Only thing that is not clear from the figure is that at point C which is at the top even though it looks a pin joint just for simplicity in the textbook it is given that point C can transmit only a horizontal force, no vertical force. But I assure you that even if that restriction is not there you can still solve this problem. A B is not a two force member. Again see, A B C is a whole rigid member. A B is not a two force member. Similarly D E F is a full member. But B E which is a vertical link, that is a two force member and of course the hydraulic cylinder is also a two force member. Strictly speaking it is not because of the weight but the weight is generally smaller as compared to the other quantity, other forces involved. No H X is not 5. Okay, H Y is how much? 5. 5, H Y is not 5. H X is 5, H Y is 5. No. See note one thing, there is a consistency in H X, H X and H Y both are 5. Then the inclination of that member should be 45 degrees which is clearly it is not. So you know that H X and H Y cannot be both equal. Okay, but the member in the middle portion is 12.5, that is fine. Also for example, if you report that, it is a two force member you should in principle also report its compressive or tensile. But in this case it does not matter because that is not the final goal of this problem. So the consistency check even without going into any details is that H X and X Y should not have the same values because angle is tan inverse of 1.2 by 0.8 with respect to horizontal and not 45. So let me quickly go over okay because we are running out of time. So the way, so everybody can find out force in link B. Everybody can right? Or is there any doubt about finding force in link B? Yeah, you can say something. Okay, let me put it this way. Most of you, can you not solve force in member B? You can solve? Yeah, yeah please. Okay, let me do that okay. So the way I can explain okay. So let me just flash this thing here. Okay, you can see that if you draw a free body diagram okay, for the truss okay, free body diagram for the top member what you will see is at point C according to the problem there is only horizontal force okay. Let me go again here. So point C there is only horizontal reaction. Point F, there can be two reactions both vertical and horizontal. Then by drawing the free body diagram of this okay, we can find out all these three forces fcx, ffy, ffx. You agree with me? Okay, you can find all the three forces. Find simple equilibrium. You can talk about point F, find out fcx, equilibrium in the horizontal direction, find ffx and p will be equal to f of fy, straight up, straight forward. So all three are obtained. Now what we do okay is coming back to this problem again. We know what are the reactions at point C and point F. One horizontal reaction, horizontal and vertical. So we look at this member okay, member ac. We look at member ac. Then what you see is that there is an internal member be. What is that member doing? If the be were not present then what will happen is that this fcx will have nothing to balance its torque for member abc about point A, right? So from that as a few of you obtained, you can obtain what is a member force in member be. Note that we have taken the line of action from member be on to rod abc downwards. So what we have implicitly assumed that member be is in tension. Okay, member be is in tension, it is downwards. So we can solve for this. And as few of you obtained that fbe will be in 12.5 kilo Newton in tension. Now what we do is look at the bottom member bef. Now this fbe will act equal and opposite. Okay, it is a tensile force. It will act, provide a pulling force at point B. There is ffy, ffx. If this force were not present then there will be nothing to balance the moment for this free body diagram about point D from this okay. But what is additional constraint that we have here? Okay, we have the additional constraint we have is that there is a hydraulic cylinder which is attached at point E. So now how many unknowns do we have? We have one unknown. The direction of that unknown is known. Okay, so we can have the horizontal and vertical components easily. ffx, ffy are known. So for this free body diagram, taking moment about point D, we can immediately find out what is fhx and what is fhy. Okay, and there is an even more elegant method to solve this problem. Okay, I will quickly flash that. And this follows. What we do is this, like using the trick which we used yesterday for the top truss. Okay, for the top truss, take your x and y axis which are not straight but which which are inclined. Because then the y axis is along, since then the x axis is along the member ABC, the inclined axis. And this is the y axis. So what we will do is that, that for this top free body diagram, take equilibrium in this inclined direction. Okay, you can immediately say that you have fxy. Okay, this is y. Plus fy is equal to p cos theta. Immediately, p can be written as p cos theta. Now for these two, we just have equal and opposite reactions when you just go along. And you can take for this free body diagram you will get one equation like this. Second free body diagram, take moment about point D. You will again get an equation like this, both in terms of fbe, fcy, ffy. Now just add these two equations. What you will get is a simple equation of this form, ffy plus fcy which is nothing but p cos theta into all the momentum will be equal to the force in the hydraulic cylinder. So in four steps, you can solve this problem. Okay, and get the answer to that. Think about it. That instead of resolving the forces in x and y direction as horizontal, you keep your x and y axis which is tilted. How much is the tilt? The tilt is along the direction abc which is the same as the direction of the ef. Think about it, try it. If you got it, it's great. If not, you can talk me afterwards or just think about it. It will come over. What you will realize is that that we can just write three and four equations and without solving anything, it will be a very transparent answer and it will be extremely transparent answer because you immediately know a simple relation between the applied force, angle of this link, applied force, and all the distances. Very transparent answer will come which will not be clear from the earlier procedure which is simple but a bit convoluted. So idea is that again by choosing appropriate coordinate axes you can really make the problem much simpler than it was.