# Engineering Statics: Open and Interactive

## Section4.43D Moments

The circular arrows we used to represent vectors in two dimensions are unclear in three dimensions, so instead, moments are represented as arrows with $$x\text{,}$$ $$y$$ and $$z$$ components, like force and position vectors. You will sometimes see moments indicated with double arrowheads to differentiate them from force vectors.
In three dimensions, it is usually not convenient to use geometry to find the moment arm for equation (4.1.1), so instead we will use the vector cross product method, which is easier to apply but less intuitive.

### Subsection4.4.13D Cross Products

The mathematics of cross products was discussed in Section 2.8, and equation (2.8.1) provides one method to calculate a moment cross products
\begin{align} \vec{M} \amp = |\vec{r}| |\vec{F}| \sin \theta \, \hat{\vec{u}} \text{.}\tag{4.4.1} \end{align}
Here, $$\theta$$ is the angle between the two vectors as shown in Figure 4.3.5 above, and $$\hat{\vec{u}}$$ is the unit vector perpendicular to both $$\vec{r}$$ and $$\vec{F}$$ with the direction coming from the right-hand rule. This equation is useful if you know or can find the magnitudes of $$\vec{r}$$ and $$\vec{F}$$ and the angle $$\theta$$ between them. This equation is the vector equivalent of (4.3.2).
Alternately, if you know or can find the components of the position $$\vec{r}$$ and force $$\vec{F}$$ vectors, it’s typically easiest to evaluate the moment cross product using the determinant form discussed in Subsection 2.8.3.
\begin{align} \vec{M} \amp = \vec{r} \times \vec{F} \notag\\ \amp= \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ r_x \amp r_y \amp r_z \\ F_x \amp F_y \amp F_z \end{vmatrix}\notag\\ \amp = (r_y F_z - r_z F_y)\ \ihat - (r_x F_z - r_z F_x)\ \jhat + (r_x F_y - r_y F_x)\ \khat \tag{4.4.2} \end{align}
Here, $$r_x\text{,}$$ $$r_y\text{,}$$ and $$r_z$$ are components of the vector describing the distance from the point of interest to the force. $$F_x\text{,}$$ $$F_y\text{,}$$ and $$F_z$$ are components of the force. The resulting moment has three components.
\begin{align*} M_x \amp = (r_y F_z - r_z F_y)\\ M_y \amp = (r_x F_z - r_z F_x)\\ M_z \amp = (r_x F_y - r_y F_x)\text{.} \end{align*}
These represent the component moments acting around each of the three coordinate axes. The magnitude of the resultant moment can be calculated using the three-dimensional Pythagorean Theorem.
$$M = |\vec{M}| = \sqrt{ {M_x}^2 +{M_y}^2+{M_z}^2 }\tag{4.4.3}$$
It is important to avoid three common mistakes when setting up the cross product.
• The order must always be $$\vec{r}\times\vec{F}\text{,}$$ never $$\vec{F}\times\vec{r}\text{.}$$ The moment arm $$\vec{r}$$ appears in the middle line of the determinant and the force $$\vec{F}$$ on the bottom line.
• The moment arm $$\vec{r}$$ must always be measured from the moment center to the line of action of the force. Never from the force to the point.
• The signs of the components of $$\vec{r}\text{ and }\vec{F}$$ must follow those of a right-hand coordinate system.
In two dimensions, $$r_z$$ and $$F_z$$ are zero, so (4.4.2) reduces to
\begin{align} \vec{M} \amp = \vec{r} \times \vec{F} \notag\\ \amp= \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ r_x \amp r_y \amp 0 \\ F_x \amp F_y \amp 0 \end{vmatrix}\notag\\ \amp = ( {r_x}{F_y} - {r_y}{F_x})\ \khat\text{.}\tag{4.4.4} \end{align}
This is just the vector equivalent of Varignon’s Theorem in two dimensions, with the correct signs automatically determined from the signs on the scalar components of $$\vec{F}$$ and $$\vec{r}\text{.}$$

The next two interactives should help you visualize moments in three dimensions.
The first shows the force vector, position vector and the resulting moment all placed at the origin for simplicity. The moment is perpendicular to the plane containing $$\vec{F}$$ and $$\vec{r}$$ and has a magnitude equal to the ‘area’ of the parallelogram with $$\vec{F}$$ and $$\vec{r}$$ for sides.
The second interactive shows a more realistic situation. The moment center is at arbitrary point $$A\text{,}$$ and the line of action of force $$\vec{F}$$ passes through arbitrary points $$P_1$$ and $$P_2\text{.}$$ The position vector $$\vec{r}$$ is the vector from $$A$$ to a point on the line of action, and the force $$\vec{F}$$ can be slid anywhere along that line.

#### Example4.4.3.3D Moment about a Point.

A thin plate $$OABC$$ sits in the $$xy$$ plane. Cable $$BD$$ pulls with a tension of $$\kN{2}$$ through a frictionless ring at point $$D\text{.}$$ Find the moment caused by the tension force around point $$O\text{.}$$
$$\vec{M_O}=\kNm{\left\langle 1.114, 0.911, -0.446 \right\rangle}$$
Solution.
Start the problem by using the position information and tension magnitude to find the force vector $$\vec{F}_{BD}\text{.}$$ This will be done in three steps:
1. Find the position vector $$\vec{BD}$$: Find position vectors by either subtracting the start-point coordinates from the end-point coordinates or focusing on the changes in the position components from $$B$$ to $$D\text{.}$$
\begin{align*} \vec{BD} \amp = D-B\\ \amp=\m{(-0.9, 1.1, 0)} - \m{(0.4, 0, 1.0)} \\ \amp = \m{\left\langle 1.3, -1.1, 1 \right\rangle} \end{align*}
2. Find the unit vector of $$\vec{BD}$$: Compute a unit vector by dividing $$\vec{BD}$$ by the total length of $$BD\text{.}$$
\begin{align*} BD \amp = |\vec{BD}|=\sqrt{1.3^2+(-1.1)^2+1.0^2}\\ \amp = \m{1.975}\\ \vec{\widehat{BD}} \amp = \frac{\vec{BD}}{BD}\\ \amp = \frac {\m{\left\langle 1.3, -1.1, 1 \right\rangle}}{\m{1.975}}\\ \amp = \left\langle 0.658 -0.557 0.506 \right\rangle \end{align*}
Note that $$\vec{\widehat{BD}}$$ is unitless and is the pure direction of $$\vec{BD}\text{.}$$
3. Multiply the unit vector by force magnitude: Now multiply $$\vec{\widehat{BD}}$$ by the $$\kN{2}$$ force magnitude to find the force components.
\begin{align*} \vec{F}_{BD} \amp = F_{BD}(\vec{\widehat{BD}})\\ \amp =\kN{2}\left\langle 0.658, -0.557, 0.506 \right\rangle\\ \amp = \kN{\left\langle 1.317, -1.114, 1.013 \right\rangle} \end{align*}
Next, find the moment arm from point $$O$$ to the line of action of the force. There are two obvious options for moment arms, either $$\vec{r}_{OB}$$ or $$\vec{r}_{OB}\text{.}$$ To demonstrate how both moment arms give the same answer, solutions for both moment arms will be shown.
Option 1: Moment using $$\vec{r}_{OB}$$
• Moment arm $$\vec{r}_{OB}$$ starts at the point we are taking the moment around, $$O\text{,}$$ and ends at the point $$B\text{.}$$
\begin{align*} \vec{OB}\amp =B-O\\ \amp =\m{(-0.9, 1.1, 0)} - \m{(0, 0, 0)}\\ \amp = \m{\left\langle -0.9, 1.1, 0 \right\rangle} \end{align*}
• Cross $$\vec{r}_{OB}$$ with $$\vec{F}_{BD}$$ to find the moment of $$\vec{F}_{BD}$$ about point $$O\text{.}$$
\begin{align*} \vec{M}_O \amp = \vec{r}_{OB} \times \vec{F}_{BD}\\ \amp = \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ -0.9 \amp 1.1 \amp 0 \\ 1.317 \amp -1.114 \amp 1.013 \end{vmatrix}\\ \amp = \kN{\left\langle 1.114, 0.911, -0.446 \right\rangle} \end{align*}
Option 2: Moment using $$\vec{r}_{OD}:$$
• Moment arm $$\vec{r}_{OD}$$ starts at the point we are taking the moment around, $$O\text{,}$$ and ends at the point $$D\text{.}$$
\begin{align*} \vec{OD}\amp=D-O\\ \amp =\m{(0.4, 0, 1.0)} - \m{(0, 0, 0)} \\ \amp = \m{\left\langle 0.4, 0, 1.0 \right\rangle} \end{align*}
• Cross $$\vec{r}_{OD}$$ with $$\vec{F}_{BD}$$ to find the moment of $$\vec{F}_{BD}$$ about point $$O\text{.}$$
\begin{align*} \vec{M_O} \amp = \vec{r}_{OD} \times \vec{F}_{BD}\\ \amp = \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ 0.4 \amp 0 \amp 1.0\\ 1.317 \amp -1.114 \amp 1.013 \end{vmatrix}\\ \amp = \kN{\left\langle 1.114, 0.911, -0.446 \right\rangle} \end{align*}
It is worth your effort to compute moments both ways for this example, or another problem, to prove to yourself that the answers work out exactly the same with different moment arms. Technically, you could select a position vector from anywhere on line $$\vec{BD}$$ and get the correct answer, but $$\vec{r}_{OB}$$ or $$\vec{r}_{OB}$$ are the only two between defined points in this problem.
Drawing $$\vec{M}_O\text{,}$$ demonstrates that a moment vector direction is both 1) the axis of rotation caused by $$\vec{T}_{BD}$$ around point $$O\text{,}$$ with the moment aligning to your thumb and the moment rotating around your fingers from the right-hand rule and 2) that $$\vec{M}_O$$ is perpendicular to the plane formed by $$\vec{T}_{BD}$$ and $$\vec{T}_{BD}\text{.}$$ Recall that all cross products result in vectors perpendicular to the two crossed vectors.

In three dimensions, the moment of a force about a point can be resolved into components about the $$x\text{,}$$ $$y$$ and $$z$$ axes. The moment produces a rotational tendency about all three axes simultaneously, but only a portion of the total moment acts about any particular axis.
We are often interested in finding the effect of a moment about a specific line or axis. For example, consider the moment created by a push on a door handle. Unless you push with a force exactly perpendicular to the hinge, only a portion of the total moment you produce will act around the hinge axis and be effective to open the door. The moment we are looking for is the vector projection of the moment onto the axis of interest. Vector projections were first discussed in Subsection 2.7.4.
The axis of interest does not need to be a coordinate axis. This interactive shows the projection of moment $$\vec{M}$$ on a line passing through points $$A$$ and $$B\text{.}$$
To find the moment of a force about a line, three vectors are required:
• $$\hat{\vec{u}}\text{,}$$ a unit vector pointing in the direction of the line or axis of interest.
• $$\vec{r}\text{,}$$ a position vector from any point on the line of interest to any point on the line of action of the force.
• $$\vec{F}\text{,}$$ the force vector. If you have multiple concurrent forces, you can treat them individually or add them together first and find the moment of the resultant.
Using these three vectors, we can compute the magnitude of the moment about a line in two related, but unique ways.
Two-Step Method: Separates the cross and dot products into two distinct steps.
• First, compute the moment of a the force about any point on the line of interest using the cross product, (4.3.4).
\begin{equation*} \vec{M} = (\vec{r} \times \vec{F}) \end{equation*}
• Next, find the scalar projection of the moment onto the line of interest using a dot product, (2.7.10).
\begin{equation*} M_{line}=\|\proj_{\vec{u}} \vec{M}\| = \vec{M} \cdot \hat{\vec{u}} \end{equation*}
Scalar Triple Product: Combines the cross and dot product into the determinant of a 3 $$\times$$ 3 matrix formed from the three vectors above.
• The scalar triple product consists of taking the determinant of a matrix consisting of the components of the line’s unit vector $$\hat{\vec{u}}$$ in the top row, the components of a position vector $$\vec{r}$$ in the middle row, and the components of the $$\vec{F}$$ in the bottom row. The determiant can be computed in multiple ways, but we suggest using the augmented determinant method Figure 2.8.6 or computing it directly in a calculator which allows 3 $$\times$$ 3 matrix computations.
• Here are the computations:
\begin{align*} M_{line}=\| \proj_u \vec{M} \| \amp = \hat{\vec{u}} \cdot ( \vec{r}\ \times \vec{F}) \\ \amp = \begin{vmatrix} u_x \amp u_y \amp u_z\\ r_x \amp r_y \amp r_z\\ F_x \amp F_y \amp F_z \end{vmatrix}\\ \amp = (r_y F_z - r_z F_y)\ u_x + (r_z F_x - r_x F_z)\ u_y + (r_x F_y - r_y F_x)\ u_z \end{align*}
• Note that the above determinant does not contain any unit vectors, like the $$\ihat,\ \jhat\text{,}$$ and $$\khat$$ vectors that show up in the top of cross-product determinants. Therefore, you are just multiplying the numbers in the diagonals and adding them up to a scalar value.
A positive value indicates that the moment vector points in the $$\hat{\vec{u}}$$ direction, while a negative value indicate the moment points in the opposite direction.
Finally if you need to find the vector components of this moment about a line, multiplying the scalar projection by the line’s unit vector will add direction to your scalar moment value, (2.7.11)
$$\vec{M_{line}}=\vec{M}_{\hat{\vec{u}}} = \|\proj_{\vec{u}} \vec{M}\|\ \hat{\vec{u}}\text{.}\tag{4.4.5}$$

#### Example4.4.6.3D Moment about a Line.

A $$\ft{2}$$ per side square box is attached to fixed pipe $$BG$$ by two frictionless hinges $$B$$ and $$G\text{.}$$ There are two forces applied to the box, $$\vec{P_1}=\lbf{30}$$ and $$\vec{P_2}=\lbf{25}\text{.}$$
For this two part example, first find the rotational moment from $$\vec{P_1}$$ about line $$\vec{BG}$$ and then find the moment of and $$\vec{P_2}$$ about line $$\vec{BG}\text{.}$$
##### (a)
Find the magnitude of the moment caused by force $$\vec{P_1}$$ about line $$\vec{BG}\text{.}$$
$$M_{P_1} \text{ about } BG=\ftlbf{0}$$
Solution.
The key observation to find the moment of $$\vec{P_1}$$ about line $$\vec{BG}$$ is that the line of action of $$\vec{P_1}$$ goes directly through line $$\vec{BG}\text{,}$$ thus force $$\vec{P_1}$$ produces no moment about line $$\vec{BG}\text{.}$$ Force $$\vec{P_1}$$ still pulls on the box and line $$\vec{BG}\text{,}$$ but causes no rotation about line $$\vec{BG}\text{.}$$ While you could set up a two-step or mixed triple to compute this moment, the simplest moment arm from line $$\vec{BG}$$ to force $$\vec{P_1}$$ would be $$\vec{r}=\ft{\left\langle 0,0,0 \right\rangle}$$ which results in a $$M_{P_1} \text{ about } BG=\ftlbf{0}\text{.}$$
##### (b)
Find the magnitude of the moment caused by force $$\vec{P_2}$$ about line $$\vec{BG}\text{.}$$
$$M_{P_2} \text{ about } BG=\ftlbf{35.355}$$
Solution.
Keep in mind that to compute the magnitude of a moment about a line, we need to compute a vector moment about a point on the line and then dot this vector moment onto the line’s unit vector. There are two options for this computatons:
Option 1: The Two-Step Method separates the cross and dot products into two distinct steps.
• Compute the vector moment of $$\vec{P_1}$$ about any point on line $$\vec{BG}\text{,}$$ including $$\vec{M_G}$$ or $$\vec{M_B}\text{,}$$ then
• Dot $$\vec{M}$$ onto the unit vector $$\widehat{BG}$$
Option 2: The Scalar Triple Product combines the cross and dot product into a single matrix determinant including the following vectors:
• Top row: Unit vector for the line of interest $$\widehat{BG}$$
• Middle row: Moment arm from line of interest to force vector. We’ll use $$\vec{r}_{HG}$$ as it has the most zeros, which will simplify the algebra when computing the determinant.
• Bottom row: Components of force vector $$\vec{P_2}.$$
Let’s set up the vectors for the scalar triple product starting with the unit vector $$\widehat{BG}\text{.}$$
The unit vector $$\widehat{BG}$$ is found by dividing the positon vector $$\vec{BG}$$ by its own magnitude.
\begin{gather*} \widehat{BG} = \frac{\vec{BG}}{BG}=\frac{\ft{\langle -2.0, 0.0, 2.0 \rangle}}{\ft{\sqrt{(-2)^2+0^2+2^2}}} = {\langle -0.707, 0.0, 0.707 \rangle} \end{gather*}
Next, the middle row of the scalar triple product is one of the moment arms from $$\vec{BG}$$ to the force $$\vec{P_2}\text{.}$$ There are four options which will all result in the same answer after you compute the scalar triple product.
\begin{align*} \vec{r}_{GH} \amp= \ft{\langle 0.0, -2.0, 0.0 \rangle} \\ \vec{r}_{GD} \amp= \ft{\langle 2.0, -2.0, 0.0 \rangle} \\ \vec{r}_{BH} \amp= \ft{\langle -2.0, -2.0, 2.0 \rangle} \\ \vec{r}_{BD} \amp= \ft{\langle 0.0, -2.0, 2.0 \rangle} \end{align*}
We’ll use $$\vec{r}_{GH}$$ going forward as it is a one-dimensional vector and the zeros will simplify the determinant.
Finally, the bottom row of the scalar triple product are the force vector components.
\begin{gather*} \vec{P_2} = \lbf{\langle 25, 0, 0 \rangle} \end{gather*}
Now, combining the three vectors, here is the triple scalar product matrix:
\begin{align*} M_{P_2} \text{ about } BG = \begin{vmatrix} {-0.707} \amp {0} \amp {0.707} \\ \ft{0.0} \amp \ft{-2.0} \amp \ft{0.0} \\ \lbf{25} \amp \lbf{0} \amp \lbf{0} \end{vmatrix} =\ftlbf{35.355} \end{align*}
Hence, the magnitude of the moment caused by force $$\vec{P_2}$$ about line $$\vec{BG}$$ is $$\ftlbf{35.355}\text{.}$$ As we found a positive answer, this confirms that the rotation caused by $$\vec{P_2}$$ is in the same direction as the vector $$\vec{BG}\text{,}$$ this is reinforced by the right-hand rule thumb direction in the diagram.
Validating the answer with the 2-step approach, we could first sum the moment of $$\vec{P_2}$$ about point G.
\begin{gather*} \vec{M_G}=(\ft{2})(\lbf{25})\ \hat{k} \end{gather*}
Given that the $$\vec{M_G}\ \hat{k}$$ moment and line $$\vec{BG}$$ lie in a 2-D plane, you could even use a \ang{45} right triangle instead of a dot product to find the amount of $$\vec{M_G}$$ which lies along line $$\vec{BG}\text{.}$$
\begin{align*} M_BG \amp = M_G (\cos\ \ang{45})\\ M_BG \amp = \ftlbf{50}(0.707)\\ M_BG \amp = \ftlbf{35.355} \end{align*}
Not all problems have simple geometry where you can form a right triangle to perform a dot product, but this process demonstrates the two-step combination of a cross and a dot product to find a moment about a line.