10.8: Derivation moment on a system
In this section we will derive and prove that the resultant moment of internal forces on a system of point masses is zero, such that we only need to add the moments of external forces to analyse the kinetics with Euler’s second law.
If many forces are acting on a system \(C\) consisting of many point masses, we can calculate the resultant moment \(\overrightarrow{\boldsymbol{M}}_{C, P}\) on the system with respect to a single reference point \(P\), by adding Equation 10.15 for all point masses.
\[\overrightarrow{\boldsymbol{M}}_{C, P}=\sum_{i} \overrightarrow{\boldsymbol{M}}_{i / P}=\sum_{i} \overrightarrow{\boldsymbol{r}}_{i / P} \times \overrightarrow{\boldsymbol{F}}_{i} \tag{10.58} \label{10.58}\]
Note that we always use the same reference point \(P\) for every moment in the sum. The forces acting can be separated into external and internal forces, where each point mass \(i\) experiences a moment due to the sum of external forces \(\overrightarrow{\boldsymbol{F}}_{i, \text { ext }}\) and due to the internal force vectors \(\overrightarrow{\boldsymbol{F}}_{i j}\) acting on it by the other point masses for which index \(j \neq i\) :
\[\overrightarrow{\boldsymbol{M}}_{C, P}=\sum_{i} \overrightarrow{\boldsymbol{M}}_{i / P}=\sum_{i} \overrightarrow{\boldsymbol{r}}_{i / P} \times \overrightarrow{\boldsymbol{F}}_{i, \mathrm{ext}}+\sum_{i} \sum_{j \neq i} \overrightarrow{\boldsymbol{r}}_{i / P} \times \overrightarrow{\boldsymbol{F}}_{i j} \tag{10.59} \label{10.59}\]
We know from Newton’s third law that the internal forces always come in pairs, such that we have both moments \(\overrightarrow{\boldsymbol{r}}_{i / P} \times \overrightarrow{\boldsymbol{F}}_{i j}\) and \(\overrightarrow{\boldsymbol{r}}_{j / P} \times \overrightarrow{\boldsymbol{F}}_{j i}\) contributing to \(\overrightarrow{\boldsymbol{M}}_{C, P}\). When adding these, as illustrated for \(\overrightarrow{\boldsymbol{F}}_{12, \text { int }}\) and \(\overrightarrow{\boldsymbol{F}}_{21, \text { int }}\) in Figure 10.3, we get:
\[\begin{align} \overrightarrow{\boldsymbol{r}}_{i / P} \times \overrightarrow{\boldsymbol{F}}_{i j}+\overrightarrow{\boldsymbol{r}}_{j / P} \times \overrightarrow{\boldsymbol{F}}_{j i} & = \tag{10.60} \label{10.60}\\[4pt] \overrightarrow{\boldsymbol{r}}_{i / P} \times \overrightarrow{\boldsymbol{F}}_{i j}+\overrightarrow{\boldsymbol{r}}_{j / P} \times\left(-\overrightarrow{\boldsymbol{F}}_{i j}\right) & = \tag{10.61} \label{10.61}\\[4pt] \left(\overrightarrow{\boldsymbol{r}}_{i / P}-\overrightarrow{\boldsymbol{r}}_{j / P}\right) \times \overrightarrow{\boldsymbol{F}}_{i j} & = \tag{10.62} \label{10.62}\\[4pt] \overrightarrow{\boldsymbol{r}}_{i / j} \times \overrightarrow{\boldsymbol{F}}_{i j} & =\overrightarrow{\mathbf{0}} \tag{10.63} \label{10.63}\end{align}\]
Where we used that \(\overrightarrow{\boldsymbol{F}}_{j i}=-\overrightarrow{\boldsymbol{F}}_{j i}\) from Newton’s third law and use that \(\overrightarrow{\boldsymbol{r}}_{i / j} \times \overrightarrow{\boldsymbol{F}}_{i j}=\overrightarrow{\mathbf{0}}\) because of the collinearity of the forces along the line connecting the point masses, that is parallel to \(\overrightarrow{\boldsymbol{r}}_{i / j}\). From Equation 10.63 we conclude that every contribution to the resultant moment by an internal force \(\overrightarrow{\boldsymbol{F}}_{i j}\) is cancelled by another contribution by a force \(\overrightarrow{\boldsymbol{F}}_{j i}\). Therefore, the internal forces do not contribute to the resultant moment acting on the system and Equation 10.59 can be simplified by only considering external forces.
The resultant moment vector \(\sum_{i} \overrightarrow{\boldsymbol{M}}_{C, i / P}\) acting on a system of point masses \(C\) with respect to a reference point \(P\) is the sum over the moments of the external forces.
\[\sum_{i} \overrightarrow{\boldsymbol{M}}_{C, i / P, \mathrm{ext}}=\sum_{i} \overrightarrow{\boldsymbol{r}}_{i / P} \times \overrightarrow{\boldsymbol{F}}_{i, \mathrm{ext}} \tag{10.64} \label{10.64}\]