 You may have realized by now that in statics we often reduce more complicated two or three-dimensional bodies down to single points. We have also simplified complicated distributed loads to single forces acting at a single point. But what is the significance of this single point and how does it affect our analysis? In other words, how do we take a body we want to analyze, whether it is an aircraft or an engineering potato, and reduce it to a single point? You might have some qualitative idea of where it should be, but we want to come up with a robust and quantitative method to locate that point. So let's start by thinking about this intuitively. Intuitively, we understand that the point should somehow be an average position of the body. The problem is, not all averages are created equally. You may have already encountered this reality in your first quarter of this academic year. In high school, you were likely well above average and near the top of your class. That is one of the factors that enabled you to be successful in being accepted into the competitive aerospace engineering program here at TU Delft. But amongst your peers here, you might be simply average or even below average. That doesn't mean that you are suddenly less smart or capable, just that a new basis for defining what is average has been applied. What is average is dependent on the basis for that averaging. In engineering problems, we tend to use three different bases for defining the average position of a body. The average geometrical position based on length, area, or volume. We call this average position the centroid. We can also define the average position based upon mass, which we call the center of mass. Finally, we can define the average position based upon weight, which we call the center of gravity. Let's start by looking at the average geometrical position, or the centroid, something you are already familiar with and have likely calculated many times before during your time in high school. Take for instance this L-shaped area with the bottom left corner of the area located at the origin of our coordinate frame. If we wanted to calculate the average geometrical position of this area, we could break it up into smaller shapes that we know the location of the center for, like squares and rectangles. We can then determine the centroid of the overall shape by performing a weighted average of the known centroids of these smaller shapes. As our basis for the averaging is area, these positions are weighted by the area of the smaller shapes they belong to. So the location of the centroid indicated here can be determined by the summation of the x-coordinates of the known centroids multiplied by the area of the shape they belong to, divided by the summation of all the areas. As well as the following similar expression for the y-coordinate of the centroid. This approach works, but we would like to make it more generalized. Here we had to break up the area into smaller areas with known centroid locations. But what if the area was something a little more organic looking? It is not really straightforward to fill this more complicated shape with squares and rectangles of a finite size. We may be able to get close and get an approximate value for the location of the centroid, but it will not be accurate. So how can we make this more accurate? Well, we can make the squares and rectangles smaller and add more of them. This would get us closer, but there are still gaps in the area, so to really make it general, we need to make our squares infinitely small. Here we see our shape with one infinitely small area dA on it. We can thus replace our summation equations with integral equations representing the infinite sum of the area dA multiplied by the location of the centroid of the area, which is the same as the location of the area as it is infinitely small. Please note that the subscript A next to the integral indicates that this integral needs to be performed over the entire domain of the area A of our shape. We will come back to this in a minute. The numerator of these equations is often referred to as the first moment of area. Why? Well, I guess someone looked at the equation and thought if the moment of a force is force times a distance, then an area times a distance is a moment of an area. But that is just a bit of nomenclature that will crop up later in this course, and in other courses during your studies. What does an integral over domain A actually mean? Well, it actually means that the integral shown is not a single integral. If we zoom in on our area, we can see that area element dA has an infinitesimal width of dx and an infinitesimal height of dy. So dA is equal to dx times dy. If we insert this result into our double integral equation, then we see that we have to integrate over both the domains of x and y. So these equations are actually double integral equations. I've replaced the denominator of these two equations by A because the integral of the area element dA over the entire domain is, in fact, just the area of that domain. That is how we would generalize the location of the centroid of an area, but sometimes we also deal with volumes. So how would this look in case of a volume? The centroid of a volume is still the average geometrical position of all the points within the shape itself. A volume is in three dimensions, however, so it will get three coordinates for the centroid, which can be obtained by integrating the position of an infinitesimal volume element times the volume of that element over the domain of the total volume of the body. This is then divided by the total volume of the body itself. As with area, this integral is not a single integral, but in this case it would be a triple integral as the volume dV would be equal to dx times dy times dz. So we are done then, right? We know how to reduce a body to a single point by replacing it by its centroid, and we can move on. Well, not exactly. If you think about it, the types of analysis we are performing in this course involve forces that act upon a body, and since force is equal to mass times acceleration, perhaps what we really want to know is where the center of mass is, not the centroid. The center of mass is the average point in the body in terms of its mass, so the weighting variable for averaging changes from volume to mass, where mass is simply the density times volume. Now you may ask yourself, isn't this the same as the centroid? And it is under very specific conditions, when the body has a uniform density. However, that is generally not the case. Take a rocket as an example. The density of a rocket is very different in different areas due to the use of various materials. Furthermore, a significant portion of the mass of the rocket is in the liquid propellant that is contained in the fuel tanks, and this is continuously being consumed, shifting the center of mass after the rocket launches. So in this case, the centroid would be a poor approximation of the relevant center of the rocket needed in our analysis of the forces acting on it. Okay, so the center of mass is what we really want, and we can approximate it as the centroid of a body if the density of the body is relatively constant. Well, again, not exactly. We are interested in analyzing forces, and again, forces equal to mass times acceleration. So what we often are interested in is where the resultant force of a body is acting, the mass times acceleration. We call this point the center of gravity, and it is the point at which all of the weight of a body can be considered to act. So the weighing factor for our averaging changes from mass to weight, where the weight is simply mass times gravitational acceleration. But isn't acceleration due to gravity just 9.81 meters per second squared? Well, if you were only studying mechanical or civil engineering, you could get away with this. But you decided to be a narrow space engineer, and in astronomical sense, we have to worry about varying gravitational acceleration fields. The classic example for this is a space elevator, a structure that would extend all the way from the surface of the Earth up to geosynchronous orbit. Here we need to recognize that the acceleration due to gravity, which we can extract from Newton's law of universal gravitation, is inversely proportional to the distance squared from the center of the mass of the Earth. In this equation, G is the gravitational constant, and M Earth is the mass of the Earth. The average radius of the Earth is about 6,378 kilometers, while the radius of geosynchronous orbit is about 42,000 kilometers. This means that our acceleration due to gravity is our familiar 9.81 meters per second squared at the surface of the Earth, but drops all the way down to 0.31 meters per second squared in geosynchronous orbit. If we simplify this problem and neglect the mass of the cable and only look at the mass of the counterweight that is fixed in orbit, M1, and the mass of the elevator car, M2, that travels up and down the cable, we can see as that car moves up and down the cable, its weight is changing drastically through the gravitational field. So there will be a big difference between the overall center of mass of the system and the center of gravity based upon that varying gravitational field. When it comes right down to it, technically, we are almost always interested in the center of gravity of a body when we are analyzing it as a point in engineering mechanics. But the center of mass and or the centroid can be a very good approximation of the location of the center of gravity. So we often use these equations as easier shortcuts to approximate our center of gravity. So how do we know when to use which approximation? To help us with this, we can look at the following table. When the density and gravitational fields can be approximated as being uniform, then the centroid, center of mass, and center of gravity are equivalent. If the gravitational field is not constant, but the density is, then the centroid and center of mass are equivalent, but the center of gravity is not. Conversely, if the density is not uniform in the gravitational field is, then the center of mass and center of gravity are equivalent, but the centroid is not. And finally, when neither the density nor the gravitational field can be approximated as uniform over the domain of the body, none of these centers are equivalent. I hope that gives you some insight into these different representations of the center of a body. Don't get too overwhelmed by all the equations. Rather, try to think about those subtle differences between them. You will see the equations can be considered to be the same general set of equations with different assumptions applied, which changes that weighting average. But they are all simply average positions within a body.