| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Three-dimensional force systems: finding resultant and couple |
| Difficulty | Standard +0.3 This is a standard M5 three-dimensional moments question requiring routine vector addition for the resultant force and systematic calculation of moments about a point using cross products. While it involves multiple 3D vector calculations, the method is entirely procedural with no problem-solving insight required—slightly easier than average due to its mechanical nature. |
| Spec | 3.04a Calculate moments: about a point |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = (2-1+3)\mathbf{i} + (-1+1+2)\mathbf{j} + (1-2+2)\mathbf{k}\) | M1 | Sum of forces |
| \(\mathbf{R} = 4\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mathbf{G} = \sum \mathbf{r}_i \times \mathbf{F}_i - \mathbf{a} \times \mathbf{R}\) where \(\mathbf{a} = 3\mathbf{i} - \mathbf{j} + \mathbf{k}\) | M1 | Correct method for couple |
| \(\mathbf{r}_1 \times \mathbf{F}_1\): \(({\bf i}-{\bf k})\times(2{\bf i}-{\bf j}+{\bf k})\) | M1 | Attempt cross product |
| \(= \mathbf{i}(-1\cdot1 - 0\cdot(-1)) - \mathbf{j}(1\cdot1-(-1)\cdot2)+\mathbf{k}(1\cdot(-1)-0\cdot 2) = -\mathbf{i} - 3\mathbf{j} - \mathbf{k}\) | A1 | Correct \(\mathbf{r}_1\times\mathbf{F}_1\) |
| \(\mathbf{r}_2 \times \mathbf{F}_2\): \((2\mathbf{i}-\mathbf{j}+\mathbf{k})\times(3\mathbf{i}+\mathbf{j}-2\mathbf{k}) = \mathbf{i}(2-1)-\mathbf{j}(-4-3)+\mathbf{k}(2+3) = \mathbf{i}+7\mathbf{j}+5\mathbf{k}\) | A1 | Correct \(\mathbf{r}_2\times\mathbf{F}_2\) |
| \(\mathbf{r}_3 \times \mathbf{F}_3\): \((\mathbf{i}+\mathbf{j}-\mathbf{k})\times(-\mathbf{i}+2\mathbf{j}+2\mathbf{k}) = \mathbf{i}(2+2)-\mathbf{j}(2-1)+\mathbf{k}(2+1) = 4\mathbf{i}-\mathbf{j}+3\mathbf{k}\) | A1 | Correct \(\mathbf{r}_3\times\mathbf{F}_3\) |
| \(\sum \mathbf{r}_i\times\mathbf{F}_i = 4\mathbf{i}+3\mathbf{j}+7\mathbf{k}\) | A1 | Correct sum |
| \(\mathbf{a}\times\mathbf{R} = (3\mathbf{i}-\mathbf{j}+\mathbf{k})\times(4\mathbf{i}+2\mathbf{j}+\mathbf{k}) = \mathbf{i}(-1-2)-\mathbf{j}(3-4)+\mathbf{k}(6+4) = -3\mathbf{i}+\mathbf{j}+10\mathbf{k}\) | M1 A1 | Correct \(\mathbf{a}\times\mathbf{R}\) |
| \(\mathbf{G} = (4\mathbf{i}+3\mathbf{j}+7\mathbf{k})-(-3\mathbf{i}+\mathbf{j}+10\mathbf{k}) = 7\mathbf{i}+2\mathbf{j}-3\mathbf{k}\) | A1 | Correct final answer |
## Question 6:
**(a) Find R**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = (2-1+3)\mathbf{i} + (-1+1+2)\mathbf{j} + (1-2+2)\mathbf{k}$ | M1 | Sum of forces |
| $\mathbf{R} = 4\mathbf{i} + 2\mathbf{j} + \mathbf{k}$ | A1 | Correct answer |
**(b) Find G**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{G} = \sum \mathbf{r}_i \times \mathbf{F}_i - \mathbf{a} \times \mathbf{R}$ where $\mathbf{a} = 3\mathbf{i} - \mathbf{j} + \mathbf{k}$ | M1 | Correct method for couple |
| $\mathbf{r}_1 \times \mathbf{F}_1$: $({\bf i}-{\bf k})\times(2{\bf i}-{\bf j}+{\bf k})$ | M1 | Attempt cross product |
| $= \mathbf{i}(-1\cdot1 - 0\cdot(-1)) - \mathbf{j}(1\cdot1-(-1)\cdot2)+\mathbf{k}(1\cdot(-1)-0\cdot 2) = -\mathbf{i} - 3\mathbf{j} - \mathbf{k}$ | A1 | Correct $\mathbf{r}_1\times\mathbf{F}_1$ |
| $\mathbf{r}_2 \times \mathbf{F}_2$: $(2\mathbf{i}-\mathbf{j}+\mathbf{k})\times(3\mathbf{i}+\mathbf{j}-2\mathbf{k}) = \mathbf{i}(2-1)-\mathbf{j}(-4-3)+\mathbf{k}(2+3) = \mathbf{i}+7\mathbf{j}+5\mathbf{k}$ | A1 | Correct $\mathbf{r}_2\times\mathbf{F}_2$ |
| $\mathbf{r}_3 \times \mathbf{F}_3$: $(\mathbf{i}+\mathbf{j}-\mathbf{k})\times(-\mathbf{i}+2\mathbf{j}+2\mathbf{k}) = \mathbf{i}(2+2)-\mathbf{j}(2-1)+\mathbf{k}(2+1) = 4\mathbf{i}-\mathbf{j}+3\mathbf{k}$ | A1 | Correct $\mathbf{r}_3\times\mathbf{F}_3$ |
| $\sum \mathbf{r}_i\times\mathbf{F}_i = 4\mathbf{i}+3\mathbf{j}+7\mathbf{k}$ | A1 | Correct sum |
| $\mathbf{a}\times\mathbf{R} = (3\mathbf{i}-\mathbf{j}+\mathbf{k})\times(4\mathbf{i}+2\mathbf{j}+\mathbf{k}) = \mathbf{i}(-1-2)-\mathbf{j}(3-4)+\mathbf{k}(6+4) = -3\mathbf{i}+\mathbf{j}+10\mathbf{k}$ | M1 A1 | Correct $\mathbf{a}\times\mathbf{R}$ |
| $\mathbf{G} = (4\mathbf{i}+3\mathbf{j}+7\mathbf{k})-(-3\mathbf{i}+\mathbf{j}+10\mathbf{k}) = 7\mathbf{i}+2\mathbf{j}-3\mathbf{k}$ | A1 | Correct final answer |
---6. Three forces $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 }$ act on a rigid body at the points with position vectors, $\mathbf { r } _ { 1 } , \mathbf { r } _ { 2 }$ and $\mathbf { r } _ { 3 }$ respectively, where\\
$\mathbf { F } _ { 1 } = ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { N }$\\
$\mathbf { F } _ { 2 } = ( 3 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } ) \mathrm { N }$\\
$\mathbf { F } _ { 3 } = ( - \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k } ) \mathrm { N }$\\
$\mathbf { r } _ { 1 } = ( \mathbf { i } - \mathbf { k } ) \mathrm { m }$\\
$\mathbf { r } _ { 2 } = ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { m }$\\
$\mathbf { r } _ { 3 } = ( \mathbf { i } + \mathbf { j } - \mathbf { k } ) \mathrm { m }$
The system of the three forces is equivalent to a single force $\mathbf { R }$ acting at the point with position vector ( $\mathbf { 3 i } - \mathbf { j } + \mathbf { k } ) \mathrm { m }$, together with a couple of moment $\mathbf { G }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { R }$.
\item Find $\mathbf { G }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2014 Q6 [11]}}