## Section 7.2 Center of Gravity

So far in this book we have always taken the weight of an object to act at a point at its center. This is the center of gravity: the point where all of an object’s weight may be concentrated and still have the same *external* effect on the body. In this chapter we will learn to actually locate this point.

We will indicate the center of gravity with a circle with black and white quadrants, and we will use \((\bar{x}, \bar{y})\) as the coordinates of this point. This represents the average location of all the particles which make up the body. The same symbols are also used for centroids, and \(\bar{z}\) is used as well for three-dimensional problems.

The center of gravity of a body is fixed with respect to the body, but the coordinates depend on the choice of coordinate system. For example, in Figure 7.2.1 the center of gravity of the block is at its geometric center meaning that \(\bar{x}\) and \(\bar{y}\) are positive, but if the block is moved to the left of the \(y\) axis, or the coordinate system is translated to the right of the block, \(\bar{x}\) would then become negative.

Lets explore the center of gravity of a familiar object. Take a pencil and try to balance it on your finger. How do you decide where to place it? You likely supported it roughly in the middle, then adjusted it until it balanced. You found the point where the moments of the weights on either side of your finger were in equilibrium.

Let’s develop this balanced moment idea mathematically.

Assume that the two halves of the pencil have known weights acting at points 1 and 2. How could we replace the two weights with a single statically equivalent force? Recall from Section 4.7 that statically equivalent systems produce the same external effect on the object —the net force on the object, and the net moment about any point don't change. An upward force at this point will support the pencil without tipping.

To be equivalent, the total weight must equal the total weight of the parts. \(W = W_1 + W_2\text{.}\) Common sense also tells us that \(W\) will act somewhere between \(W_1\) and \(W_2\text{.}\)

Next, let’s do the mathematical equivalent of sliding your finger back and forth until a balance point is located. Pick any point \(O\) to be the origin, then calculate the total moment about \(O\) due to the two weights.

The sum of moments around point \(O\) can be written as:

Notice that the moment of both forces are clockwise around point \(O\text{,}\) so the signs are negative according to the right-hand rule. We want a single equivalent force acting at the (unknown) center of gravity. Call the distance from the origin to the the center of gravity \(\bar{x}\text{.}\)

\(\bar{x}\) represents the mean distance of the weight, mass, or area depending on the context of the problem. We are evaluating weights in this problem, so \(\bar{x}\) represents the distance from \(O\) to the center of gravity.

The sum of moments around point \(O\) for the equivalent system can be written as:

The moment of total weight \(W\) is also clockwise around point \(O\text{,}\) so the sign of moment will also be negative according to the right-hand rule. Since the two representation are equivalent we can equate them and solve for \(\bar{x}\text{.}\)

This result is exactly in the form of (7.1.2) where the value being averaged is distance \(x\) and the weighting factor is is the weight of part \(W_i\) and the result is the mean distance \(\bar{x}\text{.}\)

The pencil was made up of two halves, but this equation can easily be extended \(n\) discrete parts. The resulting general definition of the centroidal coordinate \(\bar{x}\) is:

where:

\(W_i\) is the weight of part \(i\text{,}\)

\(\bar{x}_{i}\) is the \(x\) coordinate of the center of gravity of element \(i\text{,}\) and

\(\sum\) is understood to mean “sum all parts” so there is no need to write \(\sum\limits_{i = 1}^n\text{.}\)

The numerator is the first moment of force which is literally a moment of force as we defined it in Chapter 3. The denominator is the sum of the weights of the pieces, which is the weight of the whole object. We will soon also see first moments of mass and first moments of area and in Chapter 10, we will introduce second moments, which are the integral of some quantity like area, multiplied by a distance *squared*.

We treated the pencil as a one-dimensional object, so this discussion focused on \(\bar{x}\text{.}\) There are similar formula for the other dimensions as well

In words, these equations say

They apply to any object which can be divided into discrete parts, and they produce the coordinates of the object’s center of gravity.

###### Question 7.2.3.

Can you explain why the center of gravity of a symmetrical object will always fall on the axis of symmetry?

If the object is symmetrical, every subpart on the positive side of the axis of symmetry will be balanced by an identical part on the negative side. The first moment for the entire shape about the axis will sum to zero, meaning that

In other words, the distance from the axis of symmetry of the shape to the centroid is zero.

###### Example 7.2.4. Simple Center of Gravity.

Three \(\lb{10}\) boxes are distributed along the \(x\) axis as shown.

Find the total weight and the distance from the origin to the center of gravity of the three boxes.

How would the center of gravity change if the right-most box weighed \(\lb{20}\) instead of \(\lb{10}\text{?}\)

a) \(W = \lb{30} \qquad \bar{x}=\ft{2.5}\)

b) \(W = \lb{40} \qquad \bar{x}=\ft{3.25}\)

The total weight increases by \(\lb{10}\) and the center of gravity shifts to the right by \(\ft{0.75}\text{.}\) Also, if the weights of box three doubles, the first moment of weight with respect to the origin of the third box would also double.