| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: reduction to single force |
| Difficulty | Challenging +1.2 This is a standard M5 question on reducing a force system to a single resultant force and finding its line of action. Part (a) requires simple vector addition (2 marks). Part (b) involves computing moments about a point, setting the moment of R equal to the sum of individual moments, and solving for the position vector—a methodical but lengthy calculation requiring careful vector cross products and algebraic manipulation. While the computation is substantial (10 marks), it follows a well-established procedure taught explicitly in M5 with no novel insight required. |
| 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 |
|---|---|
| \(R = \left(\frac{2}{6}\right) + \left(\frac{0}{6}\right) + \left(\frac{3}{6}\right) = \left(\frac{4}{9}\right) = (4\dot{i} + 2\dot{k}) \sim\) | M1 A1(6) |
| Answer | Marks |
|---|---|
| \(= \left(\frac{3}{-6}\right)\) | M1 A1 A1 A1 |
| Answer | Marks |
|---|---|
| \(\left(\frac{4 \frac{23}{2}x}{4+y}\right) = \left(\frac{3}{-6}\right)\) | A1 FE B1 |
| Answer | Marks |
|---|---|
| \(r = \left(\begin{array}{c}-\frac{3}{2} \\ \frac{3}{2} \\ 0\end{array}\right) + \lambda \left(\begin{array}{c}0 \\ 1 \\ 1\end{array}\right)\) | M1 A1(w)(o) |
## Part (a)
$R = \left(\frac{2}{6}\right) + \left(\frac{0}{6}\right) + \left(\frac{3}{6}\right) = \left(\frac{4}{9}\right) = (4\dot{i} + 2\dot{k}) \sim$ | M1 A1(6) |
## Part (b)
$\left(\frac{1}{6}\right) \times \left(\frac{2}{6}\right) + \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) + \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right)$
$= \left(\frac{6}{-6}\right) + \left(\frac{-2}{2}\right) + \left(\frac{5}{2}\right)$
$= \left(\frac{3}{-6}\right)$ | M1 A1 A1 A1 |
$\left(\frac{3}{2}\right) \times \left(\frac{4}{2}\right) = \left(\frac{3}{-6}\right)$
$\left(\frac{4 \frac{23}{2}x}{4+y}\right) = \left(\frac{3}{-6}\right)$ | A1 FE B1 |
$<-3:$ $x = -\frac{3}{2}, y = \frac{3}{2}, z = 0$
$r = \left(\begin{array}{c}-\frac{3}{2} \\ \frac{3}{2} \\ 0\end{array}\right) + \lambda \left(\begin{array}{c}0 \\ 1 \\ 1\end{array}\right)$ | M1 A1(w)(o) |
---A force system consists of three forces $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ acting on a rigid body.
$\mathbf{F}_1 = (\mathbf{i} + 2\mathbf{j})$ N and acts at the point with position vector $(-\mathbf{i} + 4\mathbf{j})$ m.
$\mathbf{F}_2 = (-\mathbf{j} + \mathbf{k})$ N and acts at the point with position vector $(2\mathbf{i} + \mathbf{j} + \mathbf{k})$ m.
$\mathbf{F}_3 = (3\mathbf{i} - \mathbf{j} + \mathbf{k})$ N and acts at the point with position vector $(\mathbf{i} - \mathbf{j} + 2\mathbf{k})$ m.
It is given that this system can be reduced to a single force $\mathbf{R}$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf{R}$. [2]
\item Find a vector equation of the line of action of $\mathbf{R}$, 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. [10]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2006 Q4 [12]}}