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Section 4.5 Moment about a Line

While finding the moment about a point is useful, often in mechanical systems, we are looking to compute the amount of moment caused by a force around an axis. An example of this would be the moment that an accessible door opener causes around the hinge axis of a door. To compute the moment around a line we can combine two skills you already have learned:

  • finding the moment about a point (from the previous section) and
  • using a dot product to determine how much of one vector is along another vector.

The combined computation of the dot and cross product can either be done in two steps, or in a single step. Both techniques require three vectors which are:

  • Unit vector along the line of interest: If your line of interest is \(DE\text{,}\) then you will recall that a unit vector \(\hat{\vec{u}}_{DE}\) is the pure direction of line \(DE\text{.}\)
  • Position vector: The position vector \(\vec{r}\) goes from any point along your line of interest to any point on the line of action of the force. Any one of the infinite vectors connecting the line of interest to the line of action will give you the same answer, but practically there are just a few which line up with the problem geometry
  • Force vector: This technique only allows you to find the moment for one force at a time. If you had multiple concurrent forces, you could find the resultant force first before taking the moment about the line of interest.

The techniques covered below require that you find the components of these three vectors prior to starting the cross and dot products. Where we always label a moment about a point with a subscript of the point you are taking the moment around, we label a moment about a line with a subscript which is the line you are taking the moment around. Thus:

\begin{gather*} \vec{M_{D}} \equiv \text{moment of force } \vec{F} \text{ around point D}\\ \vec{M_{DE}} \equiv \text{moment of force } \vec{F} \text{ around line DE} \end{gather*}

In some books, you will find a moment about a line labeled as \(\vec{M_{\lambda}}\) instead of \(\vec{M}_{DE}\text{.}\) The vector equation for the magnitude of this moment about line \(DE\) using the two-step cross and dot product steps can be written as: \(\vec{M}_{DE} = ( \vec{r}\times \vec{F} )\, \hat{\vec{DE}}\)

To streamline your calculations even further, mathematically we can combine the cross and dot product into what is known as a mixed triple product. In the mixed triple product we write the cross product as a determinate in the form:

\begin{equation*} {M_{DE}} = \hat{\vec{u}}_{DE}\ \cdot \left ( \vec{r}\; \times \vec{F}\right ) = \begin{bmatrix} {u}_{DE_x} \amp {u}_{DE_y} \amp {u}_{DE_z}\\ \mathit{r}_{x} \amp \mathit{r}_{y} \amp \mathit{r}_{z}\\ \mathit{F}_{x} \amp \mathit{F}_{y} \amp \mathit{F}_{z} \end{bmatrix} \end{equation*}

To calculate the value of the mixed triple product, evaluate the determinant of a matrix consisting of the components of the unit vector \(\hat{\vec{u}}_{DE}\) along the line of interest \(DE\) in the top row, the components of a position vector \(\vec{r}\) from line-of-interest \(DE\) to the line-of-action of force \(\vec{F}\) in the middle row, components of the force vector \(\vec{F}\) in the bottom row.

You may be wondering a couple things:

  • Why are we using the unit vector of DE instead of the full vector DE when taking the dot product? Recall that when taking the dot product that you end up with the projection of the first vector onto a second vector times the magnitude of the second vector. If you only want the projection of the first onto the second vector then set the second vector as a unit vector which has a length of one.
  • Why do we get a magnitude out of an equation which contains a cross product? You are correct that the cross product will produce a vector. But, recall that all dot products produce a scalar, hence the result of the two-step cross and dot product is a scalar.
  • What does it mean if my scalar value is negative? Recall this dot product with a unit vector tell you how much of the moment vector is in the direction of the dot product and negative scalar values tell you that the direction of the moment right-hand rotational axis (analogous to your right-hand thumb) is opposite the unit vector’s direction.

[If you do need to find the vector representation of the moment about a line, then you simply need to multiply your scalar magnitude times...]

[Develop interactive of a moment about a point (restricting the point to the line of interest) and then a variable location/ magnitude single force in Three-dimensional space. Checkboxes for show moment about a point result and show moment about a line result (both magnitude + vector)]