| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2003 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Three-dimensional force systems: finding resultant and couple |
| Difficulty | Challenging +1.2 This is a systematic 3D moments problem requiring vector cross products and equilibrium conditions, but follows a standard M5 template. Parts (a)-(b) involve routine vector addition and moment calculations about a point. Parts (c)-(d) require applying equilibrium conditions (sum of forces = 0, sum of moments = 0) to find an unknown force and position, which is methodical rather than requiring novel insight. The 3D nature and multiple steps elevate it above average difficulty, but it's a textbook M5 question type. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| (a) \(\mathbf{R} = \begin{pmatrix} 4 \\ -2 \\ 0 \end{pmatrix} + \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 8 \\ 0 \\ 2 \end{pmatrix}\) or \(8i + 2k\) | M1 A1 (2) |
| Answer | Marks |
|---|---|
| Results: \(\begin{pmatrix} 0 \\ 0 \\ -6 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ -10 \end{pmatrix}, \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -2 \\ 0 \\ 0 \end{pmatrix}\) | M1 |
| [A1 one correct, A2 at least three correct] | A2, 1, 0 |
| Resultant = \(\begin{pmatrix} 4 \\ 0 \\ -16 \end{pmatrix}\) in any form | M1 A1 (5) |
| (c) \(\mathbf{F} = -8i - 2k\) | B1 ft (1) |
| (d) For equilibrium: \(\mathbf{r} \times \begin{pmatrix} -8 \\ 0 \\ -2 \end{pmatrix} = -\begin{pmatrix} 4 \\ 0 \\ -16 \end{pmatrix}\) or equivalent | M1 |
| \(\mathbf{PX} = \begin{pmatrix} 0 \\ \lambda \\ 0 \end{pmatrix} \Rightarrow \mathbf{r} \times \begin{pmatrix} -8 \\ 0 \\ -2 \end{pmatrix} = \begin{pmatrix} -2\lambda \\ 0 \\ 8\lambda \end{pmatrix}\) | M1 A1 ft |
| Finding \(\lambda\): \(PX = 2\) | M1; A1 (5) |
**(a)** $\mathbf{R} = \begin{pmatrix} 4 \\ -2 \\ 0 \end{pmatrix} + \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 8 \\ 0 \\ 2 \end{pmatrix}$ or $8i + 2k$ | M1 A1 (2) |
**(b)** Finding one of:
- $\begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 4 \\ -2 \\ 0 \end{pmatrix}$
- $\begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix} \times \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix}$
- $\begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \end{pmatrix}$
- $\begin{pmatrix} 0 \\ 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
Results: $\begin{pmatrix} 0 \\ 0 \\ -6 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ -10 \end{pmatrix}, \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -2 \\ 0 \\ 0 \end{pmatrix}$ | M1 |
[A1 one correct, A2 at least three correct] | A2, 1, 0 |
Resultant = $\begin{pmatrix} 4 \\ 0 \\ -16 \end{pmatrix}$ in any form | M1 A1 (5) |
**(c)** $\mathbf{F} = -8i - 2k$ | B1 ft (1) |
**(d)** For equilibrium: $\mathbf{r} \times \begin{pmatrix} -8 \\ 0 \\ -2 \end{pmatrix} = -\begin{pmatrix} 4 \\ 0 \\ -16 \end{pmatrix}$ or equivalent | M1 |
$\mathbf{PX} = \begin{pmatrix} 0 \\ \lambda \\ 0 \end{pmatrix} \Rightarrow \mathbf{r} \times \begin{pmatrix} -8 \\ 0 \\ -2 \end{pmatrix} = \begin{pmatrix} -2\lambda \\ 0 \\ 8\lambda \end{pmatrix}$ | M1 A1 ft |
Finding $\lambda$: $PX = 2$ | M1; A1 (5) |
**Total: (13 marks)**
---3.
\section*{Figure 1}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{1fa7336c-20aa-45c1-b685-d8e205367227-3_528_755_317_618}
\end{center}
Figure 1 shows a box in the shape of a cuboid $P Q R S T U V W$ where $\overrightarrow { P Q } = 3 \mathbf { i }$ metres, $\overrightarrow { P S } = 4 \mathbf { j }$ metres and $\overrightarrow { P T } = 3 \mathbf { k }$ metres. A force $( 4 \mathbf { i } - 2 \mathbf { j } ) \mathrm { N }$ acts at $Q$, a force $( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { N }$ acts at $R$, a force $( - 2 \mathbf { j } + \mathbf { k } ) \mathrm { N }$ acts at $T$, and a force $( 2 \mathbf { j } + \mathbf { k } ) \mathrm { N }$ acts at $W$. Given that these are the only forces acting on the box, find
\begin{enumerate}[label=(\alph*)]
\item the resultant force acting on the box,
\item the resultant vector moment about $P$ of the four forces acting on the box.
When an additional force $\mathbf { F }$ acts on the box at a point $X$ on the edge $P S$, the box is in equilibrium.
\item Find $\mathbf { F }$.
\item Find the length of $P X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2003 Q3 [13]}}