Equilibrium of a Rigid body, moments and Center of mass by Prof. M. V. Satyanarayana Equilibrium of rigid body: - So, the topic for today’s discussion is equilibrium of a rigid body, so here, after mostly our focus is going to be on rigid bodies and well, we have seen that the forces acting on rigid bodies can be divided into two groups: one external force, external force - this we generally call it as a force external or internal force. We have been denoting it as force internal. We had seen that these internal forces usually do not contribute either for translation or rotation. This we had seen unless otherwise we specify most of the forces. What we are going to consider external forces, so we will simply denote it by , vector and now remember we have dp by dt, rate of change of momentum is equal to force and we also have the rate of change of angular momentum what we call it as torque, which is responsible for the rotational motion of the object. So, whenever a body is subjected to force, then there is going to be momentum or acceleration. Translational motion is possible and due to torque, you will have the rotational motion. Now we have the concept of under what conditions the system is in mechanical equilibrium. Concept of Mechanical Equilibrium: - There are different kinds: a body may be under different kinds of equilibria. There is something else called thermodynamic equilibrium, chemical equilibrium. Right now, we are concerned about the concept of mechanical equilibrium. So, what is this concept? This is like this. If linear momentum is a constant of motion, this is that linear momentum is a constant of motion. That means linear momentum is conserved. It is not changing, then, in that case, what happens? On the other hand, if the angular momentum is a constant of motion, then what can we have and so what happens in this case? This implies the sum of all the forces are acting on the body. There may be several forces. Force is a vector so So, for mechanical equilibrium, system of all forces acting on the body must be equal to 0. This is what you called as the translational equilibrium. So, this is what you call it as this terminology translational equilibrium. Ok, then next if tau is 0. I may be written here. If L is a constant of motion, then this implies so various stocks acting on the system, depending on how many of them this is equal to zero. This is what you call it as rotational equilibrium. Ok, so what is the meaning of the first equation? The meaning of first equation, I will write here. Meaning of first equation is remember. It is a vector equation. Therefore, the sum of all the x component of all the forces is equal to zero. So, if is the ith force, the x component, and I am summing over all the forces, this is equal to 0. Similarly, all the y components is equal to 0 and all the z components is equal to zero. Now, what this means that torque is a vector, again sum of all the x components of the torques over all for all the torques acting is 0 and sum of all the y components of torques is equal to 0 and sum of all the z components of torques is zero- and here I am writing in component notation. These two equations are in component notation. Therefore, there is no need to write vectors. Time: -06:47 Coplanar forces: - We shouldn’t write suppose we have coplanar process, essentially a special situation, for case of coplanar forces, and let us say that the system is under equilibrium. It is some say essentially like a two-dimensional problem, two dimensional means all the forces are acting on, let us say x y plane and then what happens for the translational equilibrium. This means two conditions. What is it essentially? What is it sum of the x components, is equal to 0 and sum of y components of the forces equal to zero? So, this is essentially two conditions and, and you need to look for a direction which is perpendicular to this 2d plane. There is no rotational motion about an axis. Therefore, tau is going to be about an axis, tau about an axis perpendicular to perpendicular to F1 and F2 vanishes. So essentially, it is only three conditions, essentially okay, right now let us say that we consider an important case. Suppose sum of all the torques are acting on this body they vanish. Now you may say that I’m going to calculate this torques with respect to some other origin, is it possible that the body will have rotational motion? The answer is no. The reason that how I’ll consider I have sum of tau equal to 0. This is for I will call it as rotational equilibria. Is it going to remain valid? Does it remain valid? If I see, the origin is changed, if the origin is changed, you’ll see that because it is very easy simple calculation, I have the origin here and one of the forces. This is Fab. The two points here remember: it’s a couple, acting minus F and on this it will be F. Remember these two have to be parallel. I mean these two lines, but in opposite directions. This is B. I will join this. This is position vector .This OA vector is called position vector . Similarly, OB vector the position vector is , so these two forces +F and -F constitute a couple on a rigid body. So now let us calculate the moment of the couple. What is the moment of the couple? This is crossed with the force crossed with , then plus this crossed with. Time: - origin under so translational. Rotational equilibrium, remember what it is independent of the location of the origin. So, if a particular, if a body is under rotational equilibrium, with respect to a particular coordinate system and then what you do, you change the origin and look at it still it will remain the same. That is the message and okay. Now we will consider various cases as possible. The first case is like this. I will you can treat it as a kind of illustration, I will treat it as an illustration that means explaining the basic concepts. Case first: - I consider a rod AB. This is AB here and you have the center here at C. This is a. This a is the distance. This is a uniform rod of uniform cross section. Now we will say that there is a F acting and then there is a F acting here, so this particular rod is subjected to two forces F here of here. Is there, it will induce a torque like this? This force will induce a torque in this direction. This is going to be anti-clockwise. This is the clockwise both of them they cancel. Therefore, tau is equal to zero, but the tau is equal to 0 means it is under rotational equilibrium. Sorry so it is under rotational equilibrium and what about sigma forces? Total forces are not equal to 0. In fact, that is equal to . So obviously what will happen? This is a case. Where is it under rotational equilibrium? Yes, what about translational equilibrium? No, it is not on a transmission. Now we will consider the other case. This other case is like this, we’ll consider the same rod and there is a at this end. I have a force acting like this. There is another force acting like this. It’s a couple strictly speaking, that’s the definition has the center here now only thing I changed, reverse the direction of one of the forces. In this case. What will happen? They are in opposite direction. Therefore, it vanishes. However, the sum of torques is equal to the torque total. What is the total torque acting on it, so this will with respect to this, this will have a torque in this direction? It is in the anti-clockwise direction. It is 2 times the torque due to each force, so this is not equal to 0. Therefore, what about translational equilibrium of the system? It is under translational equilibrium, whereas what about the rotational equilibrium of the system? No, this will rotate, and so such a situation where the body does not have translational motion where I just under translational equilibrium. However, it rotates about a particular point and axis this is what is called as a pure rotation. Okay, we will one special case. We will see. This is called as a liver problem, common liver which we use. This is called principle of moments. Actually, this is start even at a very element, even at the early stage in school. However, we shall discuss this from the point of view of translational equilibrium and rotationally, so I have a simple lever like this.

Time: - 17:37 Lever (principle of Motion): - What you have, you have a fulcrum. This is what you call it as a pivot or fulcrum as it is known, and this is subjected to here. There is a force, this distance is and then this distance is , call this point as A, call this point as B: okay right now, this divide point: you can call it as O about this particular point now what happens now this body, this whole rod, will have ideally speaking, a liver should not have any mass, so it is of negligible mass. So, an ideal liver has negligible mass, so there is going to be two forces, one force is acting here, another force . It will have a moment like this. It will reduce a moment. This will induce a moment like this. Finally, it will be there so generally, what you do, this is the low dom. This section is known as ledum. This portion is known as ledum. This A to O is known as ledum, then O to B what to be known, as this effortum. See what you do. There is a weight here which requires to be lifted or moved. You have here effort on some force. You are going to apply okay. Now there is forces are acting here. Therefore, there will be a reaction at this particular point, reaction at the fulcrum. This is a vector quantity, so the reaction at the fulcrum is the support reaction of the foreground, okay, so for translational equilibrium what we need, this R must be equal to plus . Now taking moments, now there are three forces acting essentially , and then the reaction R. It is also a force variety now taking moments about this particular point for moments about O. What do you get here? So, this is for rotational equilibrium. You have this condition for translation equilibrium and so from this, you can see this object is not rotating at all. It is in equilibrium. It is neither rotating like this, nor it is rotating like this. Both these moments are cancelling with each other. So, from this I have we have So, this is known as mechanical advantage. See what we want is, ideally, if is much much larger than , let us say then, in order to maintain this equilibrium, we need to have this distance much much larger than That is the idea. So is much smaller and this is a common sense. Center of Gravity: - We will go into concept of center of gravity all the time, the concept of centre of gravity. It is a common experience that would have seen everybody can do it. If you have a notebook or a cardboard, it can be held at a particular point where one can vertically hold it so that this book or cardboard it is balanced. How does this happen so that there is a going to be reaction at the tip, what we call it as arm. This reaction at the tip is going to balance the total weight of the book mg of the material of the book, let us say so, the notebook is under translational equilibrium. It is under translational equilibrium, not only that it is also under rotational equilibrium. Why? Otherwise, if there are the different forces acting here, they can tilt like this or tilt like this. It is not happening to unbalanced torque. If there is going to be one, it will tilt. So, what happens the center of gravity? This is now we define what is known as the center of gravity C.G. The center of gravity is located such that the total torque on the body due to the forces, let us say that there is . There is something else: , etc. So, the total torque due to different forces, they cancel out and so torque is equal to So, the center of gravity of a body is that point where the total gravitational torque on the body is zero. So, this is this definition of center of gravity. The total gravitational torque acting on the body must be equal to 0, and so and are perpendicular to each other. Therefore, essentially, you are left with this, or a summation is equal to zero. Okay at this stage, one would think that this is same as the center of mass but is not. Remember the center of mass definition is this quantity divided by the total mass? However, it will turn out to be the same if the origin is the center of mass of the body. Time: -28:06 If origin is going to be the center of mass of the body, then it will turn out to be the same. So, what happens that the center of mass and center of gravity, both of them will turn out to be the same if the body is subjected in a uniform gravitational field, so center of mass is similar to center of gravity in a uniform gravitational field. Now, if g, on the other hand, if g varies from point to point, if the g is going to point to point, then center of mass and center of gravity do not go inside. How is center of gravity of a body determined? It is again a very standard question, one supposes I have a cardboard or something, and I want to find the center of gravity. So, this is a very standard procedure. What you do, you suspended it from a particular point A, so the whole weight is going to act along this, so it will be along this direction. Now you take some other point B and suspend it again then its weight will act along this. When you put it this way or when you fix this whole body about the point B, so similarly, you will find that various lines will intersect. Suppose I have another point C here and so this is the center of gravity, this O is the C.G. of the body. Okay, we will work out an illustration. Problem Based on center of gravity: - You will work out a simple problem and it will illustration you can treat it as an example which are involved with various concepts, so I have a rod. This end I will call it as A, this end I will call it as B, so there is a pivot here . There is a pivot here, . It is obvious that there will be a reaction here. This is . This is I have the C.G. of the body, so this is mg. The weight will act down right. Let us say that this weight is mass times acceleration due to gravity. This is 4g. Then I have another weight at this particular point P. Let us say that this is the point P. There is a weight . This is to be 6g okay, now k1 and k2 are pivots or knife edges. Whichever way you want to take, there are few dimensions given in the problem. I will specify that AB is 70 centimeters, namely the length of the rod. Then Ag, because this is the center of gravity, is 35 centimeters right and AP is given to be 30 centimeters, so obviously PG would be five centimeters. Ok, then we need certain distances. What is this AK1, KK1? AK1 is equal to BK2, the location of knife edges, is 10 centimeters, so we know what is K1G and K2G? K1G is equal to K2g is equal to so 35 minus 10 is 25 centimeters. Now we have, it is under translational equilibrium. Therefore, the forces acting upwards, and must be equal to forces acting downwards, so translational equilibrium implies So, we can have this so Now taking moments about G for rotational equilibrium delta, it is going to be in the clockwise, whereas this will be in the anti-clockwise and we are taking moments with respect to the point G, so this will rotate in this direction so The moments are equal to 0, so we can substitute the numbers. From this you will get an equation. Strictly speaking, I should put if I want to write the units. I should write newtons here and some little bit of arithmetic is involved, do it. Now from these two equations you can calculate and . If I had these two equations, you would get So, it is very easy to do these kinds of problems. All that you need to do, is not going to better. You need to do this kind of problems, to write the balancing equations for forces and write the balancing equations for, namely translational equilibrium condition and rotational equilibrium condition. When you write down the rotational equilibrium condition, you need to take talks about a suitable point where your algebra would be simpler. And now we will consider an illustration another problem. But there are different kinds of problems that are involved, which are repeatedly asked in examination. Various things can be asked. One is a ladder problem this I will call it as an illustration or a problem. Time: -36:55 Illustration: Ladder Problem: - This is what ladder problem led up right now, so the situation is like this. I have a wall; I have a ladder here. This is a ladder. Ladder is AB. The wall is smooth, whereas the floor is rough. I can have this. So this point, I will call it as D. This is C, so the weight will act downwards mg and now what are the various forces acting on this ladder? That is first, we should sketch very clearly, then indicate the directions of torques, also very clearly now, because the wall is smooth, there is going to be a reaction here this I will call it as I need to have a little longer one for reasons you will realize why this is C, then, because it is rough here. What happens you have a frictional force f, and this is what is going to be the normal reaction of the floor on the foot of the ladder? Now these two forces can be combined into one. I mean I need to very clearly. So, this is going to be the f. For equilibrium when I produce this, when this force is produced, they must meet at a particular point. Not a bad diagram, and this point I will call it as O: why should they meet there? In that case what happens if they don’t meet? What will happen because if they meet what will happen, all the torques about this particular point will be zero. Therefore, it will be under really rotational equilibrium and it’s also we will see. This AB length is L. We will call this angle as theta , made by the ladder with the floor. So, what is the f? actually, f is nothing but the reaction of the floor on the foot of the ladder. Now let us workout, it is under translational equilibrium and also under rotational equilibrium. So translational equilibrium implies sigma of all forces is equal to zero. What are the forces acting? There are two kinds of forces, one along horizontal direction, one along vertical direction. Therefore, we will have it as two equations, and sum of all is equal to zero, is equal to zero. There are only two forces, the forces along horizontal direction. is one force. It is acting at A, there is another force, namely the frictional force f, which is acting at B.

Therefore, f is equal to What about the forces acting along y direction? Only one is the weight of the ladder, which is acting at the center of gravity, and the reaction at this particular point this implies is equal to mg. Two important equations we have got. All that we have done is to make use of translational equilibrium nothing more now rotational equilibrium, so sum of all the torques must be equal to 0, about which point I would like to calculate, and I would like to calculate that talks about B, and so I will say that net torques about the point B, you can choose any point you want earlier. We already seen it then what will happen then. This will rotate in this direction. The weight of the body will rotate in the clockwise direction, whereas this will rotate it in the counterclockwise direction so mg into the foot of the perpendicular from this particular point that will be equal to Now I know what is . I know what is . Therefore, I can calculate what is the total force F is same as , therefore, what is the total force? Actually, it is the reaction of the floor on the foot of the ladder. This will be equal to square root, as from this particular figure, I have here, This is the magnitude of the force. You can also do a little bit of calculation to what is the direction of the force. The direction of the force is given by what, suppose this is going to meet at this particular point. I need to give some name so the direction of the force, the direction of the reaction of the flow reaction R is given by this angle that can also be calculated from little bit of geometry. Time: -47:38 Problem related to inclined plane: - Now we will consider one more example. Now this again, one of the typical problems of ladder. Another typical problem what we call it as you keep a heavy block on an inclined plane. This kind of problems are called as inclined plane. We will call inclined plane and blocks placed on some inclined plane. This another typical problem: I am not going to write down the problem, but I will describe the physics of it. I have an inclined plane. This is theta and I have a block which is placed on it. H is the height of the block, and so-called length of the block, or a side of a block is b. Okay, the weight will act from the through the center of gravity. This will be mg. I’ll reduce it a little bit, and this can be resolved along two directions. This is , and this will be . Suppose this particular inclined plane, you can rotate about this void. Let me say then, the this planes this particular top portion will coincide like this. Then the block is placed on this. Now I can rotate it. Let us say that I can rotate the inclined plane. That means the angle can be increased when the block is on the floor, mg will act downwards. The reaction will also act. The normal reaction will be here now. If I keep rotating what will happen at this, there is a situation when this block can topple. When the block topples, this normal reaction will no longer act through the center of gravity, so normal reaction will be in some other point. This is going to be the direction of normal, so let us say, as you keep rotating this N will move so this distance x, I will produce it so that you will know clearly this distance. I will call it as x, so the point of application of N will shift from this line towards this, and the body will topple when N exactly coincides with this particular side. Okay, now this kind of problems can be done this. I will call it as A this point. I will call it as B. That is all I would need now. I will write down the block has two tendencies: one the block can slide down. It is the translational motion. The block can slide down, so one can write down for translational equilibrium Then for rotational equilibrium, we don’t want this to topple, therefore sigma of all torques. I can take it about a particular point. I am going to take it about C. This must be equal to C. I am going to take talks about this now. What about the translational equilibrium? What does it mean? It means first horizontal forces. What are the various horizontal forces, horizontal in the sense horizontal to the side of this block? Here there is a frictional force f. Therefore, this f must be equal to this. This is the only force that is acting along this direction along this direction. Then N is equal . Now what about the torques about the C. This will have a torque about this point C, this implies It is also going to create a torque f into perpendicular distance from here h by two. Time: -54:16 So, from these three equations, we can discuss, when the top link will take place, whether the toppling can take place before sliding or sliding can take place without toppling etc. So various such situations will discuss the problem session. Thank you.