 So, let us move on to our first problem in the tutorial. What we have is we have a cylinder and the task that we want to do is this is the step and we want to pull the cylinder over the step that is what we want to do. And what is asked of us is this that the weight of cylinder 25 kg is given to us and we are told that at the point of contact A the coefficient of friction which in this case clearly is static friction is equal to 0.3. And what you are asked to find out that what is the maximum step size or what is the maximum height of the step edge which over which you can roll the cylinder over. So, we pull on this and we want to roll the cylinder over and we are asked to find out what is the maximum possible height given the coefficient of static friction between these two is 0.3 is possible for this problem. So, why do not we take around 10 minutes if you have any questions post on chat I will be available there and then after the end of 10 to 15 minutes we will go into the details of the problem what are the ideas what are the techniques that we are using. So, I could see from the chat that many colleges have done this problem properly. So, let me go over the problem so, what we have asked is that what is the maximum height at which this can be rolled over now the question is that if you try to exert force on this the cylinder will try to like roll about point A it may also try to slip both things can happen and it will lose this contact here. Now, the question is that that when it when it loses its contact here the cylinder is in somewhat hanging position that this will have the free body diagram of this cylinder will look like that this is the contact point this is the applied force and the cylinder has to be lifted off before you can tip it over now think about it the weight of the cylinder acts in the vertical direction as the horizontal force P which we have applied that and the weight by geometry they meet at point A. Now, this clearly is a 3 force body or is a 3 force member why because the force acts precisely at 3 points 1, 2 and 3 furthermore we also know the line of actions of W as well as P and when they meet at A for this entire body to be in equilibrium for this cylinder to be in equilibrium the reaction should clearly pass through A now what happens suppose this angle is 5 then this is an isosceles angle this angle should also be 5 and this is the outside angle which should be 2 5 then what do we have we have that for some given height h this is r-h okay this is r-h is equal to r cos 2 5 or we can rewrite this equation simple geometry okay this is r this is r-h this gap and r-h is nothing but this r times cos 2 5 or h is equal to r times 1-cos 2 5 now note one thing that to increase the height h what do we need to do the geometry dictates that to increase the height h this angle 5 has to keep increasing why because when angle 5 increases cos 2 5 decreases and this value keeps on increasing and h max when this 5 is as large as possible but now for this contacting surface the contact surface between the cylinder and this step what we know we know that the coefficient of friction is equal to 0.3 now since the coefficient of friction is equal to 0.3 the maximum angle and also note that the normal reaction between this as we had discussed earlier that this is the tangent okay at this point we draw a tangent and this line connecting the point of contact with the center of the cylinder is really the direction of the normal why because the tangent plane is this and the corresponding normal reaction or the normal direction will act in the direction connecting this point h and the center of the cylinder and now the friction will act tangential to this and this direction of r okay the angle this reaction r makes with this radius is nothing but the angle that the effective reaction r makes with the normal reaction because the normal reaction has to be along this is the effective reaction and that angle okay is the corresponding angle which cannot become more than the angle of friction which is tan inverse of mu s the maximum possible value of that angle okay we had discussed cone of friction okay the cone of friction in 2d is nothing but this effective angle this is essentially a 2 dimensional cone of friction and in this 2 dimensional cone of friction the reaction should can at most lie at the periphery of that cone of friction this phi becomes equal to phi s and that is the maximum angle that this reaction can make with the normal and so our h max will happen when phi is equal to phi max we substitute that value and we will see that height max or the maximum height of the step is equal to 0.248 now note there is one very interesting about this thing about this problem suppose okay that we have enough friction here we have some height we satisfy this criteria that the height is less than 0.248 when that happens okay the cylinder loses the contact and it is stable in this horizontal position exactly when height max becomes equal to 0.248 this angle phi becomes equal to phi s or this angle phi becomes equal to tan inverse of 0.3 then in that case there is an impending slippage that this is just about to slip any friction okay any value of friction or any value of lesser than 0.3 or any value of height more than this height will cause a slip here and the body will lose equilibrium now the question is that if we satisfy all these rules that h max is maximum equal to 0.248 then in that case can we say that we can roll this cylinder over afterwards if we can make it stable like this then can we just keep on going further and roll it over or will it slip at some intermediate position the answer is that that when we start pulling it just note that when you start raising the cylinder up you will see that the effective height of the step that the cylinder sees will keep decreasing because this point goes up and the effective height of the step is only this and as a result if you can keep the cylinder in equilibrium in this position you can just safely take it over and make it go and cross over the step okay. So we can take now a couple of questions 1 to 0 5 yes sir I want to ask this is friction depends on the velocity of body during motion. So I had discussed this earlier right briefly there was one question which was asked before that for dry friction or coulombic friction the frictional force between two moving surfaces does not depend on velocity but there are some other models which work for certain contacting surfaces in which the friction can depend on velocity but those kind of frictions are called as fluid friction but in the dry friction or the coulomb friction model. So this is one thing which I want to emphasize that this study about friction is an extremely complicated it is an extremely complicated problem and the model that we are using in this course and the model we are is typically used in engineering mechanics is one of the simpler models that does not mean it is a reality and it is true for any problem but for a large cases therefore a large number of solid contact problems this model where the contact force depend does not on the velocity but only on the normal reaction is a reasonable model and that is what model we are following of course there are some model models in which and there are certain problems in which the frictional force would clearly be proportional to velocity and maybe even be complicated dependent on velocity in some very complicated manners okay there are clearly instances of that but for this course we are worried only about the frictional force is proportional to the normal reaction 1071 I have a doubt about the friction okay consider two metallic surfaces okay while considering the surfaces the rough and smooth is based on coefficient of friction okay suppose the smooth surface having less friction compared to rough surface so friction that one thing I want to emphasize that friction is not property of an individual surface it is a property combined of the two surfaces okay so friction cannot be defined only for one material it has to define for a combined property it is like depends on two surfaces you cannot separate those two surfaces it is always glass on glass or metal on metal it is rubber on concrete it is like that it is never independently rubber or independently concrete or independently metal and other metal it is always a combined property so there is one okay nice question that is asked is that why do this is a question by center 1080 and I have asked that why do not we consider W in this problem it is not that we not consider okay let us go to the slide we clearly consider that the value of the P that you require okay how much P you need to apply depends on the weight but the beauty of this problem is that what it is telling you is that the criteria for slippage is completely independent of W it is purely dependent on the radius of the wheel and the height okay clearly if you solve this problem fully if you do equilibrium in the y direction in the x direction or for example if you take for this free body diagram moment balance about point H you will see that clearly that the value of P that is required will be equal to W so I am not taking W why because it is not relevant for the question asked that it the height does not depend on the weight it is independent but the amount of force I require if I want to lift a simple small cylinder which is of wood it is fine but if I want to lift a cylinder okay which is as big as the wheel of a road roller of course I have to apply a lot of load but the step size depending on the friction coefficient is the same it does not depend on the load it just depends on R and the coefficient of friction okay that is the answer to that question. The second question related to the problem is why reaction force passes through A okay think about it the force on this body acts at three points one is through the center which is the weight okay the body is already lifted up so the normal reaction we are completely disregarding then second is the horizontal force that we have applied and third is the reaction that passes from this point okay now the question is that why does this reaction passes through A now the question is the answer to that question is that this wheel has to be in equilibrium now this weight and this P they intersect at point A these two directions are given to us now if this reaction does not pass through A then what we can do is it has a line of action not passing through A so the moment that this force creates about A cannot be balanced so if the system has to be in equilibrium then the necessary and sufficient condition is that this reaction should pass through A okay so with this okay let us stop the discussion for this problem I can answer some questions on chat let us move on a problem number 3 okay this is a very very interesting problem so what we have is that we have almost massless Lender rod which is connected which is supported at peg C okay this peg C and it is supported on the other side of the wall at point B the coefficient of friction between both this and this is point 2 and this angle theta okay I have not mentioned here note that angle theta is 35 degrees okay angle theta here is 35 degrees please note that okay now when this angle theta is 35 degree what we are asked is if we apply a vertical load P to this rod then what is the what are the ranges of value of this ratio L by A that what is the upper limit and what is the lower limit on this ratio L which is the length of the rod and A which is the horizontal separation of this peg from the wall okay given that this angle theta is equal to 35 degrees which sorry I forgot to mention here and the coefficient of friction between C and the rod and between the B and the wall okay between between the rod and the wall is point 2 find out what is the upper limit on L by A what is the lower limit on L by A such that this rod remains in equilibrium when you apply a load P now the question is okay so let us take 10 minutes on this problem okay it may take more than that but there is lot of thinking that is involved in this problem that what can happen that why is there a range that is coming into picture think about it discuss with your colleagues you can put messages on the chat okay and think through it and then find out what is the range in which range of L by A upper limit and lower limit within which the rod remains stable okay so welcome back I could see that few colleges got the right answer but before I go to the solution of this problem let me discuss what is happening here okay what kind of mechanics is happening here so this was our rod that why is there a range okay this was a rod this is the wall this was the peg this is point A this point B and peg C now the question is why is there this ratio L by A why is there a range that it should be less than or equal to and greater than or equal to such that this system or this entire assembly is in equilibrium what is happening here and a simple thing one simple way to look about it very similar to the ladder problem that we had done earlier okay so think about this suppose what we did there is that under the application of this force P what if there were no friction at this point and this point and in that case clearly you will have a normal reaction only normal reaction coming from here and only normal reaction coming from here now think about it what happens these two normal reactions will interact we will intersect at this point O okay and this point O will lie on this side of the rod with we will lie within this range if A is reasonably small okay if A is too large then it will go out if A is small it will lie here now take the case where A is reasonably small now these two points intersect here now as a result what happens as a result note that this force P creates a counter clock a counter clockwise torque okay on this rod and when this happens the rod will have a tendency okay that when A is very small and the friction is low or friction is 0 then this rod has a tendency to rotate in the counter clockwise direction but now what happens okay this friction that comes here would prevent that motion why because it can exert appropriate torques and balance this force P and it will prevent that motion but what we see here is that that way when A is too small then even the friction that acts there may not be enough to prevent that rotation okay so let us take the first case when A is reasonably small then this can happen and because of this what will be the impending motion of the rod okay when this L by A ratio reaches some crucial value now think about it this if it rotation happens about this point let us say that the rotation happens about this point you will see that this point will have a tendency to move upwards here okay and also inverts but then this inward motion is not allowed okay so what do we do since this inward motion is not allowed what we do is that we take this rod and slide it upwards okay take this slide it upwards and then bring it down such that it will again remain in contact with this peg now as a result what has happened that this contact point has moved a little bit upwards and this contact point has moved a little bit upwards so the slippage okay so this relative slip happening here this point goes up so the friction acting on this will be downwards since this point is forced to move up the friction acting on this is downwards okay now so in the first case when A by A is small okay when the A by A is a reasonably small value then the impending slippage can happen like this where this point tries to move up this point tries to move up and so the friction acts like this in this direction downwards friction act like this in the downward direction so any A when the A becomes smaller than this the system cannot be in equilibrium and then let us say that what is the impending case that why is the impending motion happening so when the impending motion happens then what is the corresponding value of A by L now this I leave it to you look at the equations that are written here you can write down three equations of equilibrium for this now note one thing how many unknowns are present here 1 2 3 4 okay and this value of A phi phi unknowns how many equations of equilibrium we can write 3 in this case moment about b equal to 0 sigma fx equal to 0 and sigma fy equal to 0 so we can write down 3 equations okay but we have phi unknowns so we need slippage at 2 points to understand what is the impending motion and that slippage okay is happening here and simultaneously it is happening here and that is the reason why we should be careful about the direction of the forces because when A is small okay then the tendency of the rod okay will be to rotate in the anticlockwise direction and overall motion will be such the impending moment will be such that this point b moves up this point c moves up and as a result the frictional force here acts downwards frictional force here act downwards and that is the consistent way to choose what are the sign of the friction forces and when we do that we write down these three equations of equilibrium fc is equal to n times mu times nc fb is equal to mu times nb so the only unknowns are A nb nc we solve these three equations we will get an equation of this form okay in terms of mu sin theta cos theta theta equal to 35 degrees and A by L will be like this we substitute all the values and we will see that A by L comes out to be 0.07335 or what is asked that L by A is equal to 13.6 degrees now the question is that that this is one value we obtained why can't we just stop that this is the value of A by L any value of A by L when it becomes lesser sorry this is the value of A L by A if your L by A for example becomes or your A by L becomes smaller than this value or L by becomes larger than 3.46 you are done why is there a necessity of the second limit and the necessity of the second limit comes because if you increase this A if you increase this A then what happens is that just think about it if you increase A this is the vertical wall okay then what can happen is that if the peg is moved really upwards here then the normal assume then for example the friction is not there then what happens the normal reaction will have a line of action like this this normal reaction will have a line of action like this and this will be the instantaneous center of rotation for this slippage now think about it if you apply a force P like this then about this point what is the moment that it will create the moment will be a clockwise moment okay and then consistently what will happen is that that because of this clockwise moment if the friction is not enough then the rod will have a tendency to rotate in the clockwise direction like this and in order to cover this gap okay this point moved downwards in order to cover this gap what do we need to do we need to bring this downwards okay this entire rod now is slip downwards and then taken here perpendicularly so what you see is that that this entire infinitesimal movement what it does is that that this point C okay this point C it tends to slip downwards and this point B it tends to slip downwards what does that mean that previously because both of these points had a tendency to slip upwards at impending motion the resulting friction force acted downwards and here also downwards but in this case if you increase a then the tendency of the rod at some critical point is to rotate overall in the clockwise direction and in order to maintain compatibility which means contact at all the points this point C will tend to move down this point B will tend to move down and as a result the friction force will act upwards this friction force will act upwards and in other words what we can say is that because we had taken that the friction force is nothing but mu times NB we do not need to re-solve the problem again with plus and minus mu what with by reversing the directions and solving it again we can just note that since mu s times NA NB is equal to FB and mu s times NC is equal to FC since the direction of FB previously was up was downwards and this one okay was also downwards in the second case when l by when a becomes larger then this value this just become upward and this becomes upward so rather than re-solving the problem again okay so mu s NB is equal to FB previously it was downwards mu s NC is equal to FC previously it was downward in this case it becomes upwards so rather than solving it I can just say that we replace mu s with minus mu s and so the direction of friction forces are automatically so we just replace mu s with minus mu s and so that will switch the directions automatically okay and put mu equal to minus mu and we will get a second value of a by l which is larger or l by a which is 3.46 and what we see is that that that is the second limit that if you go beyond a which is larger than this value 0.29 okay or ratio l by a which is smaller than this value 3.46 then you will have an impending slippage which is in the anticlockwise direction and that will make the rod unstable so that is the reason that why we have this limit that at this point okay 13.6 the impending motion of the rod is such that is overall clockwise and downwards so and upwards whereas at this point the overall impending motion is in the clockwise sense and upwards and between these the rod will be in stable equilibrium under the application of force P you go beyond this and beyond this the rod will be unstable okay so what we will quickly do is there is another problem which I wanted to discuss okay is this so we will discuss it for 5 minutes and then we will move on to the quiz so this is a wedge problem so what do we have here we have a cylinder we have a wedge the coefficient of friction is given as 1 by 4 between this contact is given as 1 by 4 between this contact and what we are asked to find out is if the coefficient of friction is 1 by 4 for all surfaces determine the force P required to move the wedge that when you apply some force P what is the force minimum force that you require to move this wedge now this problem just remember that how many unknowns we have here we have 2 unknowns here normal reaction friction we have 2 unknowns here normal reaction and friction 4 and 1 P 5 how many equations of equilibrium we can draw we can write equations of equilibrium 3 for this 2 for this so we need 1 extra equation and that 1 extra equations so we have 3 plus 2 5 okay so we should have slippage exactly at one point and these are the different modes in which the slippage can happen okay you can convince yourself that what will happen is that when this point is moved inside this cylinder will tend to slip here you put the direction of slippage downwards and then put the appropriate forces and then you can solve this problem okay there isn't enough time to do this problem fully if you have any questions you can ask I can pose the answers on the moodle