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Engineering Statics: Open and Interactive

Daniel W. Baker, William Haynes

Section 4.5 3D Moments

The circular arrows we used to represent vectors in two dimensions are unclear in three dimensions, so instead, moments are drawn as arrows and represented by \(x\text{,}\) \(y\) 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 find the moment arm and use equation (4.2.1), so instead we will use the vector cross product, which is easier to apply but less intuitive.

Subsection 4.5.1 Moment Cross Products

The most robust and general method to find the moment of a force is to use the vector cross product
\begin{equation} \vec{M} = \vec{r} \times \vec{F}\text{,}\tag{4.5.1} \end{equation}
where \(\vec{F}\) is the force creating the moment, and \(\vec{r}\) is a position vector from the moment center to the line of action of the force. The cross product is a vector multiplication operation and the product is a vector perpendicular to the vectors you multiplied.

Instructions.

This interactive shows the force \(\vec{F}\) and position vector \(\vec{r}\) for use in the moment cross product. The position vector is a vector from the moment center to any point on the line of action of the force.
Figure 4.5.1. Moment cross product. \(\vec{M} = \vec{r} \times \vec{F}\)
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{4.5.2} \end{align}
Here, \(\theta\) is the angle between the two vectors as shown in Figure 4.5.1 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 (4.4.2).
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{4.5.3} \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.
\begin{equation} M = |\vec{M}| = \sqrt{ {M_x}^2 +{M_y}^2+{M_z}^2 }\tag{4.5.4} \end{equation}
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 (4.5.3) 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{4.5.5} \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 \(\vec{r}\text{.}\)

Subsection 4.5.2 Moment about a Point

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.

Instructions.

Interactive shows \(\vec{M} = \vec{r} \times \vec{F}\text{.}\) Vectors \(\vec{F}\) and \(\vec{r}\) can be adjusted interactively by dragging or changing the values in the table. Moment \(\vec{M}\) is calculated using the cross product.
Figure 4.5.2. Moment about the origin.
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.

Instructions.

Interactive shows the moment
\begin{equation*} \vec{M} = \vec{r} \times \vec{F} \end{equation*}
about an arbitrary point \(A\text{.}\) Force \(\vec{F}\) is defined by a scalar magnitude \(F\) and two points on its line of action, \(P_1\) and \(P_2\text{.}\) Position vector \(\vec{r}\) is the displacement from point \(A\) to a point on the line of action.
The locations of \(A\text{,}\) \(P_1\text{,}\) \(P_2\) and magnitude \(F\) can be set interactively. The interactive determines the components of \(\vec{r}\text{,}\) the unit vector \(\lambda\) of the line of action, and the components of \(\vec{F}\) and \(\vec{M}\text{.}\)
Note that sliding vector \(\vec{F}\) or the tip of vector \(r\) along the line of action has no effect on the resulting moment \(\vec{M}\text{.}\)
The magnitude of the moment can be visualized as the area of a parallelogram with sides \(r\) and \(F\text{.}\)
Figure 4.5.3. Moment about an arbitrary point.

Example 4.5.4. 3D Moment about a Point.

In an axis system with x to the right, y up on the page, and z coming A thin plate OABC sits in the xy plane. Cable BD pulls with a tension of 2 kN down to anchor point D. Find the moment of the tension force around point O.
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 \(O\text{.}\)
Answer.
\(\vec{M_O}=\kNm{\left\langle 1.114, 0.911, -0.446 \right\rangle}\)
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 \(\vec{r}_{OB}\)
described in detail following the image
A thin plate \(OABC\) sits in the xy plane. Cable BD pulls with a tension of 2 kN down to anchor point D. A position vector r_OB goes from the origin to the tail of the 2 kN force vector.
  • 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 \(\vec{r}_{OD}:\)
described in detail following the image
A thin plate \(OABC\) sits in the xy plane. Cable BD pulls with a tension of 2 kN down to anchor point D. A position vector r_OB goes from the origin to the tail of the 2 kN force vector.
  • 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.
Solution drawing to moment form force F_BD on the upper left corner of thin plate OABC.
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.

Subsection 4.5.3 Moment about a Line

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 2.7.3.

Instructions.

This interactive shows the moment produced by pushing on a door handle with a force \(\vec{F}\text{,}\) and the component of that moment along the axis of the door hinge axis, \(\vec{M}_z\text{.}\)
Figure 4.5.5. Moment on a hinge.
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 \(B\text{.}\)

Instructions.

This interactive shows the projection of moment \(\vec{M}\) on line through \(A\) and \(B\) with unit direction vector \(\hat{u}\text{.}\) The projection of \(M\) on line \(AB\) is (\(M \cdot \hat{u}) \hat{u}\text{.}\)
Figure 4.5.6. Moment of a force about a line
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.
With these vectors known, calculating the moment combines skills you already have learned:
  • finding the moment of a force about a point using the cross product, (4.5.1).
    \begin{equation*} \vec{M} = (\vec{r} \times \vec{F}) \text{.} \end{equation*}
  • finding the scalar projection of one vector onto another vector using the dot product, (2.7.9)
    \begin{equation*} \|\proj_{\vec{u}} \vec{M}\| = \hat{\vec{u}} \cdot \vec{M} \end{equation*}
    This combined dot and cross product is a signed scalar value called the scalar triple product. A positive sign indicates that the moment vector points in the positive \(\hat{\vec{u}}\) direction.
  • and multiplying a scalar projection by a unit vector to find the vector projection, (2.7.10)
    \begin{equation} \vec{M}_{\hat{\vec{u}}} = \|\proj_{\vec{u}} \vec{M}\|\ \hat{\vec{u}}\text{.}\tag{4.5.6} \end{equation}
Carrying these three operations out produces a vector \(\vec{M}_{\hat{\vec{u}}}\) that is the component of moment \(\vec{M}\) along a line in the \(\hat{\vec{u}}\) direction.
The scalar triple product can be calculated efficiently in a single step by evaluating a 3 \(\times\) 3 determinant consisting of the components of \(\hat{\vec{u}}\) 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 using the augmented determinant method Figure 2.8.6.
\begin{align*} \| \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*}
To find the vector projection along the selected axis, multiply this value by the unit vector for the axis, equation (4.5.6).
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