| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Three-dimensional force systems: finding resultant and couple |
| Difficulty | Standard +0.8 This M5 question requires computing moments of forces in 3D using vector cross products, then finding the resultant force and couple. While the individual calculations are systematic (vector addition, cross products), the 3D nature and multi-step process involving moment equivalence makes this moderately challenging, above average for A-level but standard for Further Maths mechanics. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors4.04g Vector product: a x b perpendicular vector6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\mathbf{F} = \sum \mathbf{F}_i = (8\mathbf{i}+3\mathbf{j}-4\mathbf{k})\) N | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\sum \mathbf{r}_i \times \mathbf{F}_i = (\mathbf{i}-2\mathbf{j})\times(3\mathbf{i}+4\mathbf{j}-6\mathbf{k}) + (3\mathbf{i}-\mathbf{k})\times(5\mathbf{i}-\mathbf{j}+2\mathbf{k})\) | M1 | |
| \(= (12\mathbf{i}+6\mathbf{j}+10\mathbf{k}) + (-\mathbf{i}-11\mathbf{j}-3\mathbf{k}) \quad (= 11\mathbf{i}-5\mathbf{j}+7\mathbf{k})\) | A2,1,0 | |
| \(\mathbf{G} + (\mathbf{i}-\mathbf{k})\times(8\mathbf{i}+3\mathbf{j}-4\mathbf{k}) = (11\mathbf{i}-5\mathbf{j}+7\mathbf{k})\) | M1 | |
| \((3\mathbf{i}-4\mathbf{j}+3\mathbf{k})\), \(\mathbf{G} = (8\mathbf{i}-\mathbf{j}+4\mathbf{k})\) | B1, A1 | |
| \( | \mathbf{G} | = \sqrt{8^2+(-1)^2+4^2} = 9\) Nm |
## Question 5:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\mathbf{F} = \sum \mathbf{F}_i = (8\mathbf{i}+3\mathbf{j}-4\mathbf{k})$ N | B1 | **(1)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\sum \mathbf{r}_i \times \mathbf{F}_i = (\mathbf{i}-2\mathbf{j})\times(3\mathbf{i}+4\mathbf{j}-6\mathbf{k}) + (3\mathbf{i}-\mathbf{k})\times(5\mathbf{i}-\mathbf{j}+2\mathbf{k})$ | M1 | |
| $= (12\mathbf{i}+6\mathbf{j}+10\mathbf{k}) + (-\mathbf{i}-11\mathbf{j}-3\mathbf{k}) \quad (= 11\mathbf{i}-5\mathbf{j}+7\mathbf{k})$ | A2,1,0 | |
| $\mathbf{G} + (\mathbf{i}-\mathbf{k})\times(8\mathbf{i}+3\mathbf{j}-4\mathbf{k}) = (11\mathbf{i}-5\mathbf{j}+7\mathbf{k})$ | M1 | |
| $(3\mathbf{i}-4\mathbf{j}+3\mathbf{k})$, $\mathbf{G} = (8\mathbf{i}-\mathbf{j}+4\mathbf{k})$ | B1, A1 | |
| $|\mathbf{G}| = \sqrt{8^2+(-1)^2+4^2} = 9$ Nm | M1 A1 | **(8),(9)** |
---5. Two forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ act on a rigid body, where\\
$\mathbf { F } _ { 1 } = ( 3 \mathbf { i } + 4 \mathbf { j } - 6 \mathbf { k } ) \mathrm { N }$ and\\
$\mathbf { F } _ { 2 } = ( 5 \mathbf { i } - \mathbf { j } + 2 \mathbf { k } ) \mathrm { N }$.\\
The force $\mathbf { F } _ { 1 }$ acts at the point with position vector $( \mathbf { i } - 2 \mathbf { j } ) \mathrm { m }$, and the force $\mathbf { F } _ { 2 }$ acts at the point with position vector ( $3 \mathbf { i } - \mathbf { k }$ ) m. The two forces are equivalent to a single force $\mathbf { F }$ acting at the point with position vector $( \mathbf { i } - \mathbf { k } ) \mathrm { m }$, together with a couple $\mathbf { G }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { F }$.
\item Find the magnitude of $\mathbf { G }$.\\
(8)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2007 Q5 [9]}}