| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2017 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: equilibrium conditions |
| Difficulty | Challenging +1.8 This M5 question requires multiple 3D vector operations (normalizing direction vectors, computing moments via cross products, finding line of action) across four connected parts. While the techniques are standard for Further Maths Mechanics, the multi-step nature, 3D geometry, and need to carefully track forces and moments throughout makes this significantly harder than typical A-level questions but not exceptionally challenging for M5 students who have practiced these methods. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{F}_1 = \frac{1}{3}(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \times 6 = (2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k})\) N | M1 A1 | First M1 for complete method to find \(\mathbf{F}_1\) or \(\mathbf{F}_2\); First A1 for \(\mathbf{F}_1\) |
| \(\mathbf{F}_2 = \frac{1}{7}(3\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}) \times 14 = (6\mathbf{i} - 12\mathbf{j} + 4\mathbf{k})\) N | A1 | Second A1 for \(\mathbf{F}_2\) |
| \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0} \Rightarrow \mathbf{F}_3 = (-8\mathbf{i} + 16\mathbf{j} - 8\mathbf{k})\) N | M1 A1 (5) | Second M1 for equating sum to zero; Third A1 for correct \(\mathbf{F}_3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{G} = (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) \times (2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}) + (\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (6\mathbf{i} - 12\mathbf{j} + 4\mathbf{k})\) | M1 | First M1 for taking moments about \(O\) |
| \(= (4\mathbf{i} + 2\mathbf{j}) + (8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k})\) | A1 either | First A1 for either correct cross-product |
| \(= (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\) Nm | A1 (3) | Second A1 for \((12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{R} = (\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\) N | B1 (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \times (\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) = (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\) | M1 A1 | First M1 for taking moments about \(O\); First A1 for correct equation |
| \((4y - 3z)\mathbf{i} + (z - 4x)\mathbf{j} + (3x - y)\mathbf{k} = (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})\) | A1 | Second A1 for 3 correct simultaneous equations |
| One solution: \(x = -1,\ y = 3,\ z = 0\) | B1 | B1 for a correct point on the line |
| \(\mathbf{r} = (-\mathbf{i} + 3\mathbf{j}) + t(\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\) | M1 A1 (6) | Second M1 for \(\mathbf{r} = \mathbf{a} + t\)(their \(\mathbf{R}\)); Second A1 for correct answer (need \(\mathbf{r} = \ldots\)) |
# Question 3:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{F}_1 = \frac{1}{3}(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \times 6 = (2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k})$ N | M1 A1 | First M1 for complete method to find $\mathbf{F}_1$ or $\mathbf{F}_2$; First A1 for $\mathbf{F}_1$ |
| $\mathbf{F}_2 = \frac{1}{7}(3\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}) \times 14 = (6\mathbf{i} - 12\mathbf{j} + 4\mathbf{k})$ N | A1 | Second A1 for $\mathbf{F}_2$ |
| $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0} \Rightarrow \mathbf{F}_3 = (-8\mathbf{i} + 16\mathbf{j} - 8\mathbf{k})$ N | M1 A1 (5) | Second M1 for equating sum to zero; Third A1 for correct $\mathbf{F}_3$ |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{G} = (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) \times (2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}) + (\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (6\mathbf{i} - 12\mathbf{j} + 4\mathbf{k})$ | M1 | First M1 for taking moments about $O$ |
| $= (4\mathbf{i} + 2\mathbf{j}) + (8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k})$ | A1 either | First A1 for either correct cross-product |
| $= (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})$ Nm | A1 (3) | Second A1 for $(12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})$ |
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{R} = (\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})$ N | B1 (1) | |
**Part (d):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \times (\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) = (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})$ | M1 A1 | First M1 for taking moments about $O$; First A1 for correct equation |
| $(4y - 3z)\mathbf{i} + (z - 4x)\mathbf{j} + (3x - y)\mathbf{k} = (12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k})$ | A1 | Second A1 for 3 correct simultaneous equations |
| One solution: $x = -1,\ y = 3,\ z = 0$ | B1 | B1 for a correct point on the line |
| $\mathbf{r} = (-\mathbf{i} + 3\mathbf{j}) + t(\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})$ | M1 A1 (6) | Second M1 for $\mathbf{r} = \mathbf{a} + t$(their $\mathbf{R}$); Second A1 for correct answer (need $\mathbf{r} = \ldots$) |
---\begin{enumerate}
\item The position vectors of the points $P$ and $Q$ on a rigid body are $( \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } ) \mathrm { m }$ and $( \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { m }$ respectively, relative to a fixed origin $O$. A force $\mathbf { F } _ { 1 }$ of magnitude 6 N acts at $P$ in the direction $( \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } )$. A force $\mathbf { F } _ { 2 }$ of magnitude 14 N acts at $Q$ in the direction $( 3 \mathbf { i } - 6 \mathbf { j } + 2 \mathbf { k } )$. When a force $\mathbf { F } _ { 3 }$ acts at $O$, which is also a point on the rigid body, the system of three forces is equivalent to a couple of moment $\mathbf { G }$\\
(a) Find $\mathbf { F } _ { 3 }$\\
(b) Find G
\end{enumerate}
When an additional force $\mathbf { F } _ { 4 } = ( \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k } ) \mathrm { N }$ also acts at $O$, the system of four forces is equivalent to a single force $\mathbf { R }$.\\
(c) Write down $\mathbf { R }$.\\
(d) Find an equation of the line of action of $\mathbf { R }$ in the form $\mathbf { r } = \mathbf { a } + t \mathbf { b }$, where $\mathbf { a }$ and $\mathbf { b }$ are constant vectors and $t$ is a parameter.
\hfill \mbox{\textit{Edexcel M5 2017 Q3 [15]}}