 So, now let us have a discussion session if you have any suggestions, any clarifications. So, let us discuss about that for the next 20 minutes or so. 1, 2, 5, 6, yes you are audible clearly, please go ahead with your question. Sir, the difference between ladder friction, is there any ladder friction, friction lag just now in this problem, you have explained, sliding force will be there in bottom. Okay, sure, so if we look at the ladder problem, okay, so let me use this here, so if you think about the ladder problem, okay in this problem what do we have, we had a free roller, okay there is lot of lubrication, we said that this is properly lubricated, so this point is free to move in the horizontal direction and there was a slot like this, so these two are free to move and as a result we said that because they are free to move these joints are well lubricated and as a result we said that only reactions that are possible are in this direction and in this direction, okay plus or minus. But if you think about the ladder problem and we are going to discuss that it is a very famous problem, okay amongst all the colleges that if you take a ladder like this, then you can have a friction at this point, you can also have a friction at this point and as a result as you rightly pointed out we do not need to have a tension like this in order to keep the system in equilibrium because the friction can provide a lateral force like this to keep the system in equilibrium. So as you rightly pointed out that if you have a ladder problem where you have friction at this point as well as this point then you can have possible forces from the friction coming here and then the requirement of this T is not there. So you rightly pointed out that if it is a ladder problem with friction then right then the lateral forces will be present. 1, 2, 0, 3 you are audible. Sir I have no doubt if it is a 2D problem Sandal we are using only three equations sigma f is equal to 0, sigma f y is equal to 0, sigma m is equal to 0. Yes. Sir how do you identify the nature of the support sir? If it is a roller supporter is placed with any inclination, how do you solve those kinds of problems sir? The load is acting irrespective of any problems load is always acting towards the earth, roller supporter is placed with respect to some inclination how do you solve those kinds of problems? Okay, so what you are saying is that if I understand you correctly that suppose you have a roller which is like this, this right is that what you have in mind if you have something like this I can simply draw a structure like this okay. So we have a pin support here, we have a roller support here and we can have any manner of load acting on this structure we can have a vertical load like this, we can have an inclined load like this but the way to think about the support reaction is that at this point okay what can happen here that this roller itself okay you have to draw the free body diagram for the roller itself think about the roller support itself. Now what can happen is that that this is the main part of the structure and we want to understand that what if we draw the equilibrium for the roller itself now this roller itself is constrained it is free to move in this direction okay but it is constrained to move in the vertical direction and as a result that this particular degree of freedom along this direction is constrained so when we remove this roller we replace that now okay equivalently with a force which is acting in this direction which is perpendicular to the incline this is the point clear okay. Yes sir thank you very much sir okay 1, 2, 9, 8. Professor in most of the expressions you are showing division by 2 in blue ink that is not clear. Yes I will show you why that it because of the gestures for example the way I want to show you is like this okay. So the garage door actually is a three-dimensional structure okay so let me try to make an attempt to draw the three-dimensional structure okay so this is how the door looks like so this is our Z axis okay and our X and Y are the planar axis X and Y. Now when I said that the door is supported from two sides so there are two slots running like this there is one slot like this one vertical slot like this another slot like this another vertical slot like this okay so my drawing is very poor and now so there are the door is supported equally from one side okay so this is one side so this is side 1 and this is side 2 and the way it acts through the center of the door like this. Now if I look at this door from this direction okay from the side then what we see is our problem this this and this so this is support 1 okay so this is 1 2 this is A B this is C D so A and C are mirror images of each other D and B are mirror images of each other about the central axis and if I look from the side this is A and this is B. So what we are saying is that because the door is symmetric about this we effectively consider this 3D problem as actually a 2D problem with half of the load supported by this side half of the load supported by the other side. So in principle you can say that W to begin with was equal to W by 2 and then you do not need to divide everything by 2 but because I had forgotten to do that at the last when I solved everything I just said that the W was the total weight of the door but because now the the reactions are shared with between 1 and 2 and between A and C and it and between D and B after solving everything I divided by 2 that is just to say that that everything was shared between these two and between these two is that point clear thank you for this okay yeah 1 2 1 5 please explain me the principle of transmissibility with the example principle of transmissibility okay. So okay one simple thing is like this okay suppose you have a rigid body like this okay so if you apply a force like this okay and I want to take a moment of that force about this point A now what do we have I can join these two this is F this is R okay so we say that the torque of this force about A is equal to R cross F but as far as rigid bodies are concerned we can just take the line of this and instead of applying this force here we just move this force all the way along this line and apply it here. So as far as the force and moment balance are concerned you will see that there is absolutely no difference so this is R bar so it is the same vector it is the same vector rather than having it here what we are doing is that we are taking it and putting it here on the same line and what you will see is that this is R bar prime so the new torque the torque of this force now should be MA prime which should now be equal to R prime cross F bar and you will see you can convince yourself that this will be equal to MA prime just do simple algebra just note that this is R bar okay and this is delta R bar so R bar plus delta R bar is equal to R bar prime just substitute these expression and you will see that the effect or the force direction remains the same as well as the turning capacity of the force which means the torque or the moment that the force produces about any point if you keep that force and move it along that line then that effect does not change and this I believe is the principle of transmissibility that you can either if you want to take the if you if the force is given the line of action of force is given then if you want to find out the torque I can either take the torque like this or I can just move the force along that line and then take the torque about that point like this both will give the same effect is that okay okay thank you sir 1 2 7 1 sir while presenting the result for hinge we used to write one horizontal component and vertical component and some people used to write an inclined support reaction as a result it is the correct way to present it both are equivalent depending on what you are comfortable with and sometimes it depends on the problem okay so we will solve one problem okay sometime later okay so what you are saying what I had said is that that suppose you have a roller a hinge acting on a member what we say is that we replace that this is by a force in the y direction and a force in the x direction so how many unknowns do we have we have two unknowns but we can also say that iska effect yes yes but I can as well say that this is using parallelogram law okay that these are the two forces okay this effect is like this that it is acting the resultant force F acts at some angle alpha so these two are completely equivalent why because I can as well say that my FX is equal to F sin alpha and my FY is equal to F cos alpha so you can do it in this way that have F and alpha you will get FX and FY or vice versa if I give you FX and FY I can also find what is the capital F and what is alpha so both ways are perfectly equivalent what way you choose it completely depends on what you are comfortable with and sometimes it depends on what problem you are solving that for particular problems it may be more comfort for you it may be a better idea we will solve a few problems in the tutorial it may be a better idea to use F and alpha and for certain type of problems it may be a better idea to use FX and FY if you are using geometrical methods for example Lame's theorem and other things then a good idea used to use F and alpha but if you are using moment equilibrium force equilibrium which are equivalent then for example it is a good idea to use FX and FY is that okay but they are both are equivalent there is nothing that one is right and other is wrong both are perfectly equivalent but what you use depends on the problem and depends on how you are comfortable with solving that problem hello 1, 2, 7, 5 go ahead we can hear you sir my question is you always use this word torque and we understand moment so can you emphasize difference between torque and moment there is absolutely no difference as far as engineering mechanics is concerned okay as far as e make is concerned there is no difference but if you go to structural mechanics for example or solid mechanics then torque is typically referred to for a torsional torque so any kind of moment of a force okay that is relative to torsion you use as torque whereas moment is typically used for bending moment but as far as e make is concerned okay torque and moment okay they are just interchangeably used and both are perfectly logical and perfectly right way of using that term so this term R cross F okay this is called as moment of a force typically but this is just semantics okay let us not get embroiled in that and at moment of a force you just use a shortcut moment or you also call it as torque and many a times what I have noted is that that physicist typically use this word torque and engineers typically use this word moment okay but both are equivalent okay thank you sir one more question sir how many equilibrium equation for a in a three-dimensional structure three-dimensional structure six maximum and professor Shobik Banerjee will deal with 3d equilibrium in another few classes so for 3d equilibrium the number of equations max you can write is six three forces and three moments okay it was not a slide actually because I had to write on this so what is the question can you quickly ask I can go over it again sir exterior force is applied from bottom in that particular signal if we consider that force at that bottom oh if you take the force like the bottom okay yes sir so as an actually if we consider this force as an axial it will be compression force F yes so you are totally right so as far as the engineering effect is concerned okay as far as a deformation of structures are concerned these are not equivalent okay this F bar okay this and this are not equivalent but as far as the force and moment balance are concerned they are equivalent these are not equivalent for deformations okay okay but they are equivalent for equilibrium okay so there is one question for example I think I should answer this definitely and which comes for example because many students for ask these questions is that why do we neglect the friction for the belt pulley arrangement so first of all it is not that we are neglecting the friction between the belt and the pulley but what we are really saying is that see if you think about it that this is a pulley and that pulley as a hole and through this hole there is a screw that goes through that hole and then the pulley rotates about that screw okay now what we say is that that typically there is a contact between this screw okay so if I magnify this what do we say is that that that this is the screw and this is the internal hole and there is a contact between these two now if you lubricate this joint properly then this friction which is called as journal friction or if you use ball bearings for example which are well lubricated then if this is journal friction and upon lubrication this can be neglected so as a result if you look at that what are the contact forces if you draw the free body diagram of this pulley then what do I need to do when I draw free body diagram I remove this screw and I just draw the pulley like this and these are the tensions that are acting on it so this is the belt and between belt and the pulley there can be friction okay there can be friction here but we neglect the friction between the pulley and the screw and so when we draw the free body diagram of the pulley by removing the internal screw this friction which can provide a torque is completely missing and the corresponding equivalent reactions are only this and this and as a result even though there is friction between belt and the pulley if this is T1 and this is T2 for the pulley to be in equilibrium T1R should be equal to T2R that means T1 is equal to T2 and it so happens that even though there is a possibility of friction between the pulley and the belt it is not activated so the friction is present between the pulley and the belt does not mean that it will always be activated it will only be activated if it is required by the external forces but the external forces are such that if you take torque of this free body diagram for the pulley about the center then the only balance equation you can write is this because this joint is well lubricated so there is no internal friction no internal torque okay as opposed to if you have a pulley which is not properly lubricated at this joint then you see that it is not easy to turn that pulley you have to put lot of force just to move that pulley and so in that case at this contact you can have an internal torque in addition to having this forces and in that case T1 will not be equal to T2 and there will be a friction between the belt and the pulley will be activated so just because there is friction does not mean it has to be activated it depends on what the loading is okay so this is one of the things which I wanted to emphasize that it is not at the belt and the friction does not the belt and the pulley does not have friction it means that the screw around which the pulley is hinged that and the pulley it does not have friction it is properly lubricated okay so we will take a short break