| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| 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 Further Maths M5 question on force systems and couples requiring vector addition, moment calculations using cross products, and understanding of equivalent force systems. While it involves multiple steps and 3D vectors, the techniques are standard for M5: finding resultant forces, computing moments about a point, and applying the condition for equilibrium under a couple. The calculations are methodical rather than requiring novel insight, making it moderately above average difficulty for A-level but routine for Further Maths mechanics. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.04a Calculate moments: about a point |
| Answer | Marks |
|---|---|
| \(\mathbf{R} = (3\mathbf{j} + \mathbf{k}) + (4\mathbf{i} + \mathbf{j} - \mathbf{k})\) | M1 |
| \(= (4\mathbf{i} + 4\mathbf{j})\) (N) | A1 |
Total: 2 marks
| Answer | Marks |
|---|---|
| \((\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \times (4\mathbf{i} + 4\mathbf{j}) + \mathbf{G} = (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) \times (3\mathbf{j} + \mathbf{k}) + (-3\mathbf{i} + 2\mathbf{k}) \times (4\mathbf{i} + \mathbf{j} - \mathbf{k})\) | M1 |
| \((-4\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}) + \mathbf{G} = (-10\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}) + (-2\mathbf{i} + 5\mathbf{j} - 3\mathbf{k})\) | A2 |
| \(\mathbf{G} = (-8\mathbf{i} - \mathbf{j} + 7\mathbf{k})\) (N m) | A1 |
Total: 4 marks
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{F}_3 = -\mathbf{R} = (-4\mathbf{i} - 4\mathbf{j})\) | B1 | |
| \(\mathbf{G} = (2\mathbf{i} - \mathbf{k}) \times (-4\mathbf{i} - 4\mathbf{j}) + (-12\mathbf{i} + 3\mathbf{j} + 3\mathbf{k})\) | M1 A1 | |
| \(= (-16\mathbf{i} + 7\mathbf{j} - 5\mathbf{k})\) | A1 | |
| \( | \mathbf{G} | = \sqrt{(-16)^2 + 7^2 + (-5)^2}\) |
| \(= \sqrt{330}\) (N m) | A1 |
Total: 6 marks + 12 marks overall
## (a)
$\mathbf{R} = (3\mathbf{j} + \mathbf{k}) + (4\mathbf{i} + \mathbf{j} - \mathbf{k})$ | M1 |
$= (4\mathbf{i} + 4\mathbf{j})$ (N) | A1 |
Total: 2 marks
## (b)
$(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \times (4\mathbf{i} + 4\mathbf{j}) + \mathbf{G} = (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) \times (3\mathbf{j} + \mathbf{k}) + (-3\mathbf{i} + 2\mathbf{k}) \times (4\mathbf{i} + \mathbf{j} - \mathbf{k})$ | M1 |
$(-4\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}) + \mathbf{G} = (-10\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}) + (-2\mathbf{i} + 5\mathbf{j} - 3\mathbf{k})$ | A2 |
$\mathbf{G} = (-8\mathbf{i} - \mathbf{j} + 7\mathbf{k})$ (N m) | A1 |
Total: 4 marks
## (c)
$\mathbf{F}_3 = -\mathbf{R} = (-4\mathbf{i} - 4\mathbf{j})$ | B1 |
$\mathbf{G} = (2\mathbf{i} - \mathbf{k}) \times (-4\mathbf{i} - 4\mathbf{j}) + (-12\mathbf{i} + 3\mathbf{j} + 3\mathbf{k})$ | M1 A1 |
$= (-16\mathbf{i} + 7\mathbf{j} - 5\mathbf{k})$ | A1 |
$|\mathbf{G}| = \sqrt{(-16)^2 + 7^2 + (-5)^2}$ | M1 |
$= \sqrt{330}$ (N m) | A1 |
Total: 6 marks + 12 marks overall
---Two forces $\mathbf{F}_1 = (3\mathbf{i} + \mathbf{k})$ N and $\mathbf{F}_2 = (4\mathbf{i} + \mathbf{j} - \mathbf{k})$ N act on a rigid body.
The force $\mathbf{F}_1$ acts at the point with position vector $(2\mathbf{i} - \mathbf{j} + 3\mathbf{k})$ m and the force $\mathbf{F}_2$ acts at the point with position vector $(-3\mathbf{i} + 2\mathbf{k})$ m.
The two forces are equivalent to a single force $\mathbf{R}$ acting at the point with position vector $(\mathbf{i} + 2\mathbf{j} + \mathbf{k})$ m together with a couple of moment $\mathbf{G}$.
Find,
\begin{enumerate}[label=(\alph*)]
\item $\mathbf{R}$,
[2]
\item $\mathbf{G}$.
[4]
\end{enumerate}
A third force $\mathbf{F}_3$ is now added to the system. The force $\mathbf{F}_3$ acts at the point with position vector $(2\mathbf{i} - \mathbf{k})$ m and the three forces $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ are equivalent to a couple.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the magnitude of the couple.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2011 Q4 [12]}}