| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2008 |
| 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 question on reducing a force system to a resultant force and couple. It requires routine vector addition for part (a) and moment calculations using cross products for part (b), but follows a well-practiced procedure with no novel insight needed. Slightly easier than average due to straightforward computational steps. |
| 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 | Marks | Notes |
| \(\mathbf{R} = (-2\mathbf{i} + \mathbf{j} - \mathbf{k}) + (3\mathbf{i} - \mathbf{j} + 2\mathbf{k})\) | M1 | |
| \(= (\mathbf{i} + \mathbf{k})\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\mathbf{G} + (5\mathbf{i}+\mathbf{j}-\mathbf{k}) \times (\mathbf{i}+\mathbf{k}) = (\mathbf{i}-\mathbf{j}+\mathbf{k}) \times (-2\mathbf{i}+\mathbf{j}-\mathbf{k}) + (4\mathbf{i}-\mathbf{j}-2\mathbf{k}) \times (3\mathbf{i}-\mathbf{j}+2\mathbf{k})\) | M1 A2 ft | |
| \(\mathbf{G} + (\mathbf{i} - 6\mathbf{j} - \mathbf{k}) = (-\mathbf{j} - \mathbf{k}) + (-4\mathbf{i} - 14\mathbf{j} - \mathbf{k})\) | A3 ft | |
| \(\mathbf{G} = (-5\mathbf{i} - 9\mathbf{j} - \mathbf{k})\) | A1 | |
| \( | \mathbf{G} | = \sqrt{(-5)^2 + (-9)^2 + (-1)^2} = \sqrt{107}\) Nm |
## Question 3:
### Part (a):
| Working | Marks | Notes |
|---|---|---|
| $\mathbf{R} = (-2\mathbf{i} + \mathbf{j} - \mathbf{k}) + (3\mathbf{i} - \mathbf{j} + 2\mathbf{k})$ | M1 | |
| $= (\mathbf{i} + \mathbf{k})$ | A1 | **(2)** |
### Part (b):
| Working | Marks | Notes |
|---|---|---|
| $\mathbf{G} + (5\mathbf{i}+\mathbf{j}-\mathbf{k}) \times (\mathbf{i}+\mathbf{k}) = (\mathbf{i}-\mathbf{j}+\mathbf{k}) \times (-2\mathbf{i}+\mathbf{j}-\mathbf{k}) + (4\mathbf{i}-\mathbf{j}-2\mathbf{k}) \times (3\mathbf{i}-\mathbf{j}+2\mathbf{k})$ | M1 A2 ft | |
| $\mathbf{G} + (\mathbf{i} - 6\mathbf{j} - \mathbf{k}) = (-\mathbf{j} - \mathbf{k}) + (-4\mathbf{i} - 14\mathbf{j} - \mathbf{k})$ | A3 ft | |
| $\mathbf{G} = (-5\mathbf{i} - 9\mathbf{j} - \mathbf{k})$ | A1 | |
| $|\mathbf{G}| = \sqrt{(-5)^2 + (-9)^2 + (-1)^2} = \sqrt{107}$ Nm | M1 A1 | **Total: 11** |
---3. A system of forces consists of two forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ acting on a rigid body.\\
$\mathbf { F } _ { 1 } = ( - 2 \mathbf { i } + \mathbf { j } - \mathbf { k } ) \mathrm { N }$ and acts at the point with position vector $\mathbf { r } _ { 1 } = ( \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { m }$.\\
$\mathbf { F } _ { 2 } = ( 3 \mathbf { i } - \mathbf { j } + 2 \mathbf { k } ) \mathrm { N }$ and acts at the point with position vector $\mathbf { r } _ { 2 } = ( 4 \mathbf { i } - \mathbf { j } - 2 \mathbf { k } ) \mathrm { m }$.\\
Given that the system is equivalent to a single force $\mathbf { R } \mathrm { N }$, acting at the point with position vector $( 5 \mathbf { i } + \mathbf { j } - \mathbf { k } ) \mathrm { m }$, together with a couple $\mathbf { G N m }$, find
\begin{enumerate}[label=(\alph*)]
\item $\mathbf { R }$,
\item the magnitude of $\mathbf { G }$.\\
(9)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2008 Q3 [11]}}