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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 an empty see-saw is the exact center. Similarly, the balance points for rectangles or circles are 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.

We begin by restricting our attention to thin plates of uniform
density. In Engineering and Science, this type of object is called a
lamina. For mathematical purposes, we assume that the lamina is
bounded by *x*=*a*, *x*=*b*, *y*=*f*(*x*), and *y*=*g*(*x*), with . Then the book gives the following formulas for the coordinates
of the center of mass.

To go from these formulas to the integral formulas presented earlier,
we first partition the interval [*a*,*b*] into *n* subintervals

The right hand sides of these equations should be easy to recognize as Riemann sums, so that when we take limits as

We end this section with an example, including Maple commands, for
computing the center of mass. Suppose you have a lamina bounded by the
curves *f*(*x*)=*x ^{3}*-3*

> f := x -> x^3-3*x^2-x+3;

> g := x -> 3-x;

If you are not sure about the relative positions of the two curves, it is a good idea to plot them both.

> plot({g(x),f(x}),x=0..3);Notice that the graph of

Now we are ready to compute the center of mass. Using labels, as shown below, can help you organize your calculations and avoid mistakes. Computing the mass separately also lets you check it. If you get a negative value for the mass, something is wrong and you have to check what you have done. A common mistake is reversing the order of the functions.

> mass := int(g(x)-f(x),x=0..3);

> x_bar := int(x*(g(x)-f(x)),x=0..3)/mass;

> y_bar := 1/2*int(g(x)^2-f(x)^2,x=0..3)/mass;

- 1.
- Find the coordinates of the center of mass of the lamina bounded by the curves and for . Make sure that you plot both functions on the same graph and check to see if your center of mass coordinates make sense.
- 2.
- Find the coordinates of the center of mass of the lamina bounded
by
*x*=0,*x*=*a*, and*y*=*k x*, where^{2}*k*and*a*are positive constants. In engineering texts, this is called a parabolic spandrel. Note that your answer will depend on the parameters*k*and*a*. Make sure that you plot both functions on the same graph and check to see if your center of mass coordinates make sense. (To do this, just pick values for the parameters*a*and*k*.) - 3.
- Sometimes you have to find the center of mass of a composite
lamina, that is, a lamina made up of 2 or more parts. For
definiteness, suppose that you have a composite lamina made of two
parts. Part 1 is bounded by
*x*=*a*,*x*=*b*,*y*=0 and*y*=*f*(*x*), where*b*>*a*. Part 2 is bounded by*x*=*b*,*x*=*c*,*y*=0 and*y*=*g*(*x*), where*c*>*b*. Show that the coordinates of the center of mass of the composite lamina satisfy the following equations. - 4.
- Find the coordinates of the center of mass of a triangle whose
vertices are located at the points (0,0), (0,
*c*), and (*b*,0), where*b*> 0.

9/21/1999