Next: Exercises Up: Centroids and Centers Previous: Purpose

## Background

In designing mechanisms or structures, one often has to deal with distributed forces, that is, forces that do not act at a discrete, finite set of points. The most common example of a distributed force is the force of gravity, which acts on all parts of any body of matter. Other examples are pressure in fluids and electrostatic forces, though there are many others.

One of the basic useful principles of analyzing distributed forces is the idea of replacing them with a single, aggregate force that acts at a single point and is somehow equivalent to the original distributed force. This may not always be possible, but this technique has found great use in engineering and science. As a simple example, suppose we have gravity acting on a solid plate of uniform thickness and density, but irregular shape. Finding the equivalent force is really the problem of finding the point where we could exactly balance the plate. This balance point is often called the center of mass of the body.

For symmetric objects, the balance point is usually easy to find. For example, the balance point of a see-saw is the exact center. Similarly, the balance points for rectangles or circles are the just the geometrical centers. For non-symmetric objects, the answer is not so clear, but it turns out that there is a fairly simple algorithm involving integrals for determining balance points.

To describe the algorithm abstractly, suppose that we have a three-dimensional body occupying a region D in 3. Suppose further, that the density of the body is described by a function defined on D. Then the mass m of our body can be computed by the triple integral

and the coordinates, written , of the center of mass can be shown to be given by the three equations

As an example, consider the unit cube, and suppose, for the sake of simplicity, that is constant and equal to 1. Then the mass is also unity and are given by

Performing the three integrals gives , as expected from symmetry. The commands below show how Maple can be used to perform this computation.

```  > mass := int(int(int(1,x=0..1),y=0..1),z=0..1);
```

```  > x_bar := int(int(int(x,x=0..1),y=0..1),z=0..1)/mass;
```

```  > y_bar := int(int(int(y,x=0..1),y=0..1),z=0..1)/mass;
```

```  > z_bar := int(int(int(z,x=0..1),y=0..1),z=0..1)/mass;
```

Next: Exercises Up: Centroids and Centers Previous: Purpose

William W. Farr
Sun Sep 17 18:36:55 EDT 1995