| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: reduction to single force |
| Difficulty | Standard +0.8 This M5 question requires finding unit vectors along given directions, computing a resultant force, and determining the line of action using moments—a multi-step problem involving 3D vector manipulation. While the techniques are standard for Further Maths Mechanics, the 3D context, multiple force vectors, and the need to apply moment principles to find the line of action elevate it above routine exercises. It's moderately challenging but follows established methods. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03a Force: vector nature and diagrams3.04a Calculate moments: about a point |
| Answer | Marks |
|---|---|
| \(F_1 = 7 \cdot \frac{1}{\sqrt{4^2 + (-6)^2 + (-12)^2}} \begin{pmatrix} 4 \\ -6 \\ -12 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}\) | B1 |
| \(F_2 = 3 \cdot \frac{1}{\sqrt{2^2 + 4^2 + 4^2}} \begin{pmatrix} -2 \\ -4 \\ -4 \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \\ -2 \end{pmatrix}\) | B1 |
| \(F_3 = 3\sqrt{10} \cdot \frac{1}{\sqrt{2^2 + (-10)^2 + (-16)^2}} \begin{pmatrix} 2 \\ -10 \\ -16 \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -8 \end{pmatrix}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum F_i = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix} + \begin{pmatrix} -1 \\ -2 \\ -2 \end{pmatrix} + \begin{pmatrix} 1 \\ -5 \\ -8 \end{pmatrix} = \begin{pmatrix} 2 \\ -10 \\ -16 \end{pmatrix}\) | M1 A1 | PRINTED ANSWER |
| Answer | Marks |
|---|---|
| Taking moments about O, | M1 |
| \(\begin{pmatrix} 4 \\ -6 \\ -12 \end{pmatrix} \times \begin{pmatrix} 1 \\ -5 \\ -8 \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} 2 \\ -10 \\ -16 \end{pmatrix}\) | |
| \(\begin{pmatrix} -12 \\ 20 \\ -14 \end{pmatrix} = \begin{pmatrix} -16y + 10z \\ 2z + 16x \\ -10x - 2y \end{pmatrix}\) | A1 A1 M1 |
| put \(x = 0 \Rightarrow z = 10 \Rightarrow y = 7\) | |
| so, \(r = \begin{pmatrix} 0 \\ 7 \\ 10 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -5 \\ -8 \end{pmatrix}\) is a vector equation. | A1 A1 |
## (a)
$F_1 = 7 \cdot \frac{1}{\sqrt{4^2 + (-6)^2 + (-12)^2}} \begin{pmatrix} 4 \\ -6 \\ -12 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}$ | B1 |
$F_2 = 3 \cdot \frac{1}{\sqrt{2^2 + 4^2 + 4^2}} \begin{pmatrix} -2 \\ -4 \\ -4 \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \\ -2 \end{pmatrix}$ | B1 |
$F_3 = 3\sqrt{10} \cdot \frac{1}{\sqrt{2^2 + (-10)^2 + (-16)^2}} \begin{pmatrix} 2 \\ -10 \\ -16 \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -8 \end{pmatrix}$ | B1 |
## (b)
$\sum F_i = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix} + \begin{pmatrix} -1 \\ -2 \\ -2 \end{pmatrix} + \begin{pmatrix} 1 \\ -5 \\ -8 \end{pmatrix} = \begin{pmatrix} 2 \\ -10 \\ -16 \end{pmatrix}$ | M1 A1 | PRINTED ANSWER |
## (c)
Taking moments about O, | M1 |
$\begin{pmatrix} 4 \\ -6 \\ -12 \end{pmatrix} \times \begin{pmatrix} 1 \\ -5 \\ -8 \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} 2 \\ -10 \\ -16 \end{pmatrix}$ |
$\begin{pmatrix} -12 \\ 20 \\ -14 \end{pmatrix} = \begin{pmatrix} -16y + 10z \\ 2z + 16x \\ -10x - 2y \end{pmatrix}$ | A1 A1 M1 |
put $x = 0 \Rightarrow z = 10 \Rightarrow y = 7$ |
so, $r = \begin{pmatrix} 0 \\ 7 \\ 10 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -5 \\ -8 \end{pmatrix}$ is a vector equation. | A1 A1 |The points $P$ and $Q$ have position vectors $4\mathbf{i} - 6\mathbf{j} - 12\mathbf{k}$ and $2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}$ respectively, relative to a fixed origin $O$.
Three forces, $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$, act along $\overrightarrow{OP}$, $\overrightarrow{OQ}$ and $\overrightarrow{QP}$ respectively, and have magnitudes 7 N, 3 N and $3\sqrt{10}$ N respectively.
\begin{enumerate}[label=(\alph*)]
\item Express $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ in vector form.
[3]
\item Show that the resultant of $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ is $(2\mathbf{i} - 10\mathbf{j} - 16\mathbf{k})$ N.
[2]
\item Find a vector equation of the line of action of this resultant, giving your answer in the form $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$, where $\mathbf{a}$ and $\mathbf{b}$ are constant vectors and $\lambda$ is a parameter.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2012 Q5 [10]}}