| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: reduction to single force |
| Difficulty | Standard +0.8 This M5 question requires computing moments of forces in 3D using vector cross products, then finding the line of action of the resultant force. While systematic, it demands confident vector manipulation, understanding of moment-position relationships, and multi-step reasoning beyond standard M1/M2 content, placing it moderately above average difficulty. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{F} = (2\mathbf{j} + 3\mathbf{k}) + (\mathbf{i} + 4\mathbf{k}) = \mathbf{i} + 2\mathbf{j} + 7\mathbf{k}\) | M1 | Adding force vectors |
| \( | \mathbf{F} | = \sqrt{1 + 4 + 49} = \sqrt{54}\) N |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{F}\) acts through point with p.v. \(\mathbf{r}\), setting up \(\begin{pmatrix}x\\y\\z\end{pmatrix} \times \begin{pmatrix}1\\2\\7\end{pmatrix} = \begin{pmatrix}1\\0\\1\end{pmatrix} \times \begin{pmatrix}0\\2\\3\end{pmatrix} + \begin{pmatrix}0\\2\\0\end{pmatrix} \times \begin{pmatrix}1\\0\\4\end{pmatrix}\) | M1 | |
| \(\begin{pmatrix}7y-2z\\z-7x\\2x-y\end{pmatrix} = \begin{pmatrix}-2\\-3\\2\end{pmatrix} + \begin{pmatrix}8\\0\\-2\end{pmatrix} = \begin{pmatrix}6\\-3\\0\end{pmatrix}\) | A1 A1, A1 | |
| So \(7y - 2z = 6\), \(z - 7x = -3\), \(2x - y = 0\) | ||
| \(\Rightarrow y = 2x \Rightarrow\) e.g. \(x = 0\), \(y = 0\), \(z = -3\) | M1 A1 | |
| \(\mathbf{r} = \begin{pmatrix}0\\0\\-3\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\7\end{pmatrix}\) | M1 A1 | (8) |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{F} = (2\mathbf{j} + 3\mathbf{k}) + (\mathbf{i} + 4\mathbf{k}) = \mathbf{i} + 2\mathbf{j} + 7\mathbf{k}$ | M1 | Adding force vectors |
| $|\mathbf{F}| = \sqrt{1 + 4 + 49} = \sqrt{54}$ N | M1 A1 | (3) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{F}$ acts through point with p.v. $\mathbf{r}$, setting up $\begin{pmatrix}x\\y\\z\end{pmatrix} \times \begin{pmatrix}1\\2\\7\end{pmatrix} = \begin{pmatrix}1\\0\\1\end{pmatrix} \times \begin{pmatrix}0\\2\\3\end{pmatrix} + \begin{pmatrix}0\\2\\0\end{pmatrix} \times \begin{pmatrix}1\\0\\4\end{pmatrix}$ | M1 | |
| $\begin{pmatrix}7y-2z\\z-7x\\2x-y\end{pmatrix} = \begin{pmatrix}-2\\-3\\2\end{pmatrix} + \begin{pmatrix}8\\0\\-2\end{pmatrix} = \begin{pmatrix}6\\-3\\0\end{pmatrix}$ | A1 A1, A1 | |
| So $7y - 2z = 6$, $z - 7x = -3$, $2x - y = 0$ | | |
| $\Rightarrow y = 2x \Rightarrow$ e.g. $x = 0$, $y = 0$, $z = -3$ | M1 A1 | |
| $\mathbf{r} = \begin{pmatrix}0\\0\\-3\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\7\end{pmatrix}$ | M1 A1 | (8) |
---4. Two forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ act on a rigid body, where $\mathbf { F } _ { 1 } = ( 2 \mathbf { j } + 3 \mathbf { k } ) \mathrm { N }$ and $\mathbf { F } _ { 2 } = ( \mathbf { i } + 4 \mathbf { k } ) \mathrm { N }$. The force $\mathbf { F } _ { 1 }$ acts through the point with position vector $( \mathbf { i } + \mathbf { k } ) \mathrm { m }$ relative to a fixed origin $O$. The force $\mathbf { F } _ { 2 }$ acts through the point with position vector ( $2 \mathbf { j }$ ) m . The two forces are equivalent to a single force $\mathbf { F }$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of $\mathbf { F }$.
\item Find, in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }$, a vector equation of the line of action of $\mathbf { F }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q4 [11]}}