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Section 4.3 Moment Cross Products

You may have noticed in Figure 4.2.1 that there were two different position vectors shown from bolt \(A\) to the line of action of the force \(\vec{F}\text{.}\) The more general of the two is \(\vec{r}\) which, by definition, goes from the point of interest to any point on the line of action of the force. You can use any \(\vec{r }\) that originates at A and ends at a point on the line of action of \(\vec{F}\text{,}\) but the one down the handle of the wrench is probably the easiest to construct.

Although we refer to the moment about a point (on the two-dimensional body), the moment is actually rotating about an axis. Using our standard \(x\)-\(y\) axes in the plane of the screen/paper, all two-dimensional moments rotate about a \(z\) axis which is perpendicular to the page/screen.

Finding a cross product is a vector operation where you feed the cross product two vectors and it gives you a vector answer that is perpendicular to both vectors you crossed. The mathematics of cross products were discussed in Cross Products where the basic equation was given as (2.8.1) where \(\theta\) is the angle between the two vectors as shown in Figure 4.2.1 above, and \(\hat{\vec{u}}\) is the unit vector perpendicular to both \(\vec{A}\) and \(\vec{B}\) with the direction coming from the right-hand rule.

Evaluating a moment using this method,

\begin{equation*} \vec{M} = \vec{r} \times \vec{F} = \begin{vmatrix} \ihat \amp \jhat \amp \khat\\ {r}_{x} \amp {r}_{y} \amp 0\\ {F}_{x} \amp {F}_{y} \amp 0 \end{vmatrix} \color{Peach}{ \begin{matrix} \ihat \amp \jhat\\ {r}_{x} \amp {r}_{y} \\ {F}_{x} \amp {F}_{y} \end{matrix}} =( {r_x}{F_y} - {r_y}{F_x})\khat \end{equation*}

Note that it is the perpendicular components that cause moments.

It is important to avoid three common mistakes in setting up the cross product.

  • The order must always be \(\vec{r}\times\vec{F}\text{,}\) never \(\vec{F}\times\vec{r}\text{.}\) Thus 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 point of interest to 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 (covered in Ch 2). The most common right-hand coordinate system has the x-axis positive to the right, the y-axis positive upward, and the z-axis positive coming out of the screen/page.

Often the most robust approach to finding the moment of a force is using the cross product of \(\vec{r}\) and \(\vec{F}\) in the equation:

\begin{equation} \vec{M} = \vec{r} \times \vec{F}\tag{4.3.1} \end{equation}

where \(\vec{r}\) is a position vector from the point of interest to the line of action of the force, and \(\vec{F}\) is a force vector.

As introduced in the previous section, if you know the angle \(\theta\) between \(\vec{r}\) and \(\vec{F}\) you can use

\begin{align*} \vec{r} \times \vec{F}\amp =\left |\vec{r} \right | \left |\vec{F} \right | \sin \theta \, \hat{\vec{u}} \end{align*}

However if you know the components of the position \(\vec{r}\) and force \(\vec{F}\) vectors, it is typically most efficient to express the cross product as a determinant which includes both \(\vec{r}\) and \(\vec{F}\) in the form:

\begin{equation*} \vec{M} = \vec{r} \times \vec{F} = \begin{bmatrix} \ihat \amp \jhat \amp \khat\\ {r}_{x} \amp {r}_{y} \amp 0\\ {F}_{x} \amp {F}_{y} \amp 0 \end{bmatrix} \end{equation*}

The interactive below will allow you to observe the multiple ways to solve for moments that were covered in the previous sections and also let you practice finding moments using the various techniques (by hiding the solutions until after you have tried yourself).

Interactive allowing you to see and practice (by showing the solutions after you try the computations) three different methods to compute two-dimensional moments. Every time you open the interactive it will give you a new set of values.

Figure 4.3.1.