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Engineering Statics: Open and Interactive

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 call its coordinates \((\bar{x}, \bar{y})\) or \((\bar{x}, \bar{y}, \bar{z})\text{.}\) This point represents the average location of all the particles which make up the body.
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.
Square shape shown on x-y plane in Quadrant I. x-bar is distance of centroid from y axis, and y-bar is distance of centroid from x-axis.
Figure 7.2.1. Location of the centroid, measured from the origin.
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 4.8 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 \(W_2\text{.}\)
A pen with split into two parts with two centers of gravity. The distances from the tip of the pen to the CG are x1 and x2. Lastly, an equivalent system consisting of a single weight acting at the pencil’s center of gravity x-bar.
Figure 7.2.2. (top) Side view of a pencil representing each half as a particle. (middle) A force diagram showing the weights of the two particles. (bottom) An equivalent system consisting of a single weight acting at the pencil’s center of gravity.
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:
\begin{equation*} \sum M_O=-x_1 W_1 -x_2 W_2 \end{equation*}
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 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:
\begin{equation*} \sum M_O=-\bar{x} W \end{equation*}
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 \(\bar{x}\text{.}\)
\begin{align*} -\bar{x} W \amp=-x_1 W_1 -x_2 W_2\\ \bar{x} \amp =\frac{x_1 W_1 +x_2 W_2}{W_1 + W_2} \end{align*}
This result is exactly in the form of (7.1.2) where the value being averaged is distance \(x\) and the weighting factor 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:
\begin{equation} \bar{x}=\frac{\sum \bar{x}_{i} W_i}{\sum W_i}\tag{7.2.1} \end{equation}
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
\begin{equation} \bar{x}=\frac{\sum \bar{x}_{i} W_i}{\sum W_i} \quad \bar{y}=\frac{\sum \bar{y}_{i} W_i}{\sum W_i} \quad \bar{z}=\frac{\sum \bar{z}_{i} W_i}{\sum W_i}\text{.}\tag{7.2.2} \end{equation}
In words, these equations say
\begin{equation*} \text{distance to CG}=\frac{\text{sum of first moments of weight}}{\text{total weight}} \end{equation*}
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?
Answer.
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
\begin{equation*} \bar{x} = \frac{\sum \bar{x}_{i} W_i}{\sum W_i} = \frac{0}{W} = 0\text{.} \end{equation*}
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 boxes are distributed along the \(x\) axis as shown.
  1. Find the distance from the origin to the center of gravity of the system.
  2. How would the center of gravity change if the origin was moved to the right-hand box?
Answer.
  1. \(\displaystyle \bar{x}=\m{0.8}\)
  2. The center of gravity does not change, however when measured from the new origin, \(\bar{x} = \m{-1.2}\text{.}\)
Solution.
The center of gravity of the system is the point where the total weight can be considered to act.
The total weight is \(W = \Sigma W_i = \N{90}\text{,}\) and it acts at the point where it produces the same total moment as the individual weights. In other words, it is an equivalent system. The moment comparison may be taken about any point, so for convenience we choose the origin.
Let \(x_i\) represent the \(x\)-coordinates of the individual weights, and \(\bar{x}\) represent the \(x\)-coordinate of the center of gravity. For both systems to be equivalent:
\begin{align*} W\ \bar{x} \amp= \sum (W_i\ x_i)\\ (\N{90})\ \bar{x} \amp = (\N{50})(0) + (\N{10})(\m{1.2}) + (\N{30})(\m{2.0}) \\ \bar{x} \amp= \frac{\Nm{72}}{\N{90}}\\ \amp= \m{0.8} \end{align*}
If the origin was moved to the right-hand box, the center of gravity would not change; however, the coordinate of the center of gravity would be different. The distance to the center of gravity from the new origin would be:
\begin{equation*} \bar{x}= \frac{(\N{10})(\m{-0.8}) + (\N{50})(\m{-2.0})}{\N{90}} = \m{-1.2} \end{equation*}