| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: equilibrium conditions |
| Difficulty | Challenging +1.2 This M5 question requires computing moments using cross products and understanding couple conditions (zero resultant force). Part (a) involves standard moment calculations about a point, while part (b) requires finding a line of action given a couple constraint. Though it involves 3D vectors and multiple steps, the techniques are routine for Further Maths mechanics students with no novel problem-solving insight required. |
| Spec | 3.04a Calculate moments: about a point4.04a Line equations: 2D and 3D, cartesian and vector forms |
Three forces $\mathbf{F}_1 = (3\mathbf{i} - \mathbf{j} + \mathbf{k})$ N, $\mathbf{F}_2 = (2\mathbf{i} - \mathbf{k})$ N, and $\mathbf{F}_3$ act on a rigid body.
The force $\mathbf{F}_1$ acts through the point with position vector $(\mathbf{i} + 2\mathbf{j} + \mathbf{k})$ m, the force $\mathbf{F}_2$ acts through the point with position vector $(\mathbf{i} - 2\mathbf{j})$ m and the force $\mathbf{F}_3$ acts through the point with position vector $(\mathbf{i} + \mathbf{j} + \mathbf{k})$ m.
Given that the system $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ reduces to a couple $\mathbf{G}$,
**(a)** find $\mathbf{G}$.
(6 marks)
The line of action of $\mathbf{F}_3$ is changed so that the system $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ now reduces to a couple $(6\mathbf{i} + 8\mathbf{j} + 2\mathbf{k})$ N m.
**(b)** Find an equation of the new line of action of $\mathbf{F}_3$, giving your answer in the form $\mathbf{r} = \mathbf{a} + t\mathbf{b}$, where $\mathbf{a}$ and $\mathbf{b}$ are constant vectors.
(5 marks)
---2. Three forces $\mathbf { F } _ { 1 } = ( 3 \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { N } , \mathbf { F } _ { 2 } = ( 2 \mathbf { i } - \mathbf { k } ) \mathrm { N }$, and $\mathbf { F } _ { 3 }$ act on a rigid body.
The force $\mathbf { F } _ { 1 }$ acts through the point with position vector $( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } ) \mathrm { m }$, the force $\mathbf { F } _ { 2 }$ acts through the point with position vector $( \mathbf { i } - 2 \mathbf { j } ) \mathrm { m }$ and the force $\mathbf { F } _ { 3 }$ acts through the point with position vector $( \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }$.
Given that the system $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 }$ reduces to a couple $\mathbf { G }$,
\begin{enumerate}[label=(\alph*)]
\item find $\mathbf { G }$.
The line of action of $\mathbf { F } _ { 3 }$ is changed so that the system $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 }$ now reduces to a couple $( 6 \mathbf { i } + 8 \mathbf { j } + 2 \mathbf { k } ) \mathrm { N }$ m.
\item Find an equation of the new line of action of $\mathbf { F } _ { 3 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + t \mathbf { b }$, where $\mathbf { a }$ and $\mathbf { b }$ are constant vectors.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2013 Q2 [11]}}