 Fine, let us solve a few questions all of you promise The ladder has mass m and length m It is placed like that There's a wall and this is floor wall is smooth floor has coefficient of friction mu Okay Right if I decrease the angle theta, what will happen if I keep on decreasing? There'll be a point at which will not be the degree of gravity just like so you need to find minimum theta for which it will not slide As in it is about to slide step number one draw the paper diagram Theta is always dimensionless when you get sign theta to be equal to something it has to be dimensionless if it has dimension Your answer is wrong You can check your answer that is right or not if mu increases theta should Decrease further It should allow lesser angle if theta is if mu is large There will be a normalization from here or not this will be normalization. Let's say n1 There will be friction Which direction It is about to slide so this way water magnitude Since it is about to slide mu n1 There will be gravity mg What else position from here? This is let's say n2 Okay You have drawn the favorite. I there's no friction at the top Okay, so next we can write net force along x axis to be 0 Then along y axis to be 0 my x and y axis These Along x axis. What are the forces mu n1? minus n2 This will be equal to 0 along y axis. We have n1 this And then I can write net torque is 0 about any point about which point should I write Why If I write about this torque because of these two will become zero Okay, so this is more sensible to write about this point Okay, so what is it talk about this point because of mg? What is this perpendicular sense? L L by 2 cost theta So mg into L by 2 cos theta is a top because of mg. What about talk into n2? You have to drop a pop into this. This is that is what? I'll sign I'll sign so minus of Why minus because sense of rotation is different and to into L Sign theta is equal to 0 now and to is what mu times n1 n1 is mg So n2 is L by 2 cos of theta minus mg L Sign theta equal to 0 mg L mg L cover So 10 theta is what is tan inverse 1 by Okay Any doubt Super job here are so on of mass M and length L this fulcrum. It can rotate this distance is L by 5 L by 5 a point masses get Small m you need to find out how much mass you should keep here. What is the point mass? You should keep here Simple simple question M by 4 which is wrong what is small m and three is capital M which M by 4 Okay, so So there is capital M You probably note the gravity force on the road Nobody is getting to and by Capital M by so one always getting that So you know also draw Capital mg force from where? L by 2 distance This is your normal reaction. Let's say n1 and actually can assume that mg force is acting on the rod here Actually, what is acting normal reaction between small m and the rod and since small in the rest Normalization is equal to mg Okay, so that is why this force can be considered as mg But it will not be equal to mg if rod starts rotating with angular acceleration Are you getting it? Then it's right n minus mg is equal to m into a and from there You get the normal reaction between small m and the rod right now rod is fixed Okay, let's say this mass is m1 so m1 g so about which About which I should equate talk to busy About this point Then anyone will not come in the equation itself Right, so I'll be able to write it as mg into l by 5 minus Capital mg into by 2 minus l by 5 then m1 g into What is like this? One last question a uniform rod of length l rest again the smooth roller This is roller, okay, and it is smooth I find coefficient of friction between the ground at the lower end Theta between the ground mu is what if you decrease that theta, it is like Okay, how does it make a difference whether it is a roller or not? Because of the roller what will happen? What is the normalization at this point? Many kilo to the rod Okay, that is the difference from the previous question. This is n2. This is n1 Cos per theta by No, it is a complicated expression The rod has mass Oh If you want to have master I'll write the answer first Okay, let's solve this there will be mg force There will be new times now the reaction I can take component of n2 Holiday vertical, this is theta and that is theta. So this is n2 Stop talking So n2 sin theta Minus mu n is equal to 0 Vertical direction n1 plus Both these equations you got still you are getting it wrong Then talk about this point. Let's equate that to 0. What is the torque because of mg? mg into What? mg into l by 2 No, it's a weird expression Then l by 2 cos theta How do you know what is this length? l is still here l minus h cos theta See this entire force is perpendicular to the rod So you don't need to worry about taking its component when you find in the torque Okay, so n2 into what? l minus h cos theta h by sin theta These are equations How many variables? n1, n2, mu Three variables, three equations We take mg into l by 2 cos theta We assume that mg is being applied at one end of the rod Here, center of the rod Yeah, and then it will rotate around the center of mass, right? What? The rod There is no fixed axis It depends on suppose this rod I fixed one end It will rotate about from its top It doesn't rotate from center of mass It doesn't seem to rotate about center of mass So sir, how can we calculate torque if we don't know what the axis is? There is no rotation at all It is an equilibrium So net torque about any end point is zero Yeah, but for the equation we still need force and distance from the How you write See, I have written torque about this point zero You can write torque about any point to be zero You get the same answer But if I write about this point, my equation will be simpler Because of these two, torque into zero Sir, why? There, we know from n to y to b Right, l minus h by sin theta So it is h by sin theta Yeah, so why is it not h by sin theta? Correct Yeah, why it is not h by sin theta only? It is No, I just wrote it This distance is h by sin theta only, right? So simply n to n to h by sin theta Sir, why does the roller make it perpendicular to the normal reaction? See, perpendicular to a point is not defined So roller makes it a smooth contact That's it And suppose it is not given it is a roller or not Still you take perpendicular to rod Okay, it's a reasonable assumption But then in reality it may not be perpendicular to rod Because the point will not only apply a normal force It will also be little bit clear inside the rod And then it will try to apply along the rod also some force Any other doubt? Sir, we take it perpendicular to the rod Only when it extends beyond that wall The rod See, if the normal reaction we have discussed in the roller motion Normal reaction force should be perpendicular to both the surfaces of contact Understood? If both surfaces are plane Multiple points of contact will be there And perpendicular plane is same anywhere you go And if the shape changes, there will be a point of contact They cannot be a surface of contact One sphere and the surface also has a simple point of contact If it has more than one point of contact It is not a sphere It may look like a sphere It has to be a point of contact No doubts, right?