| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2013 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: equilibrium conditions |
| Difficulty | Challenging +1.2 This is a Further Maths M5 question on 3D equilibrium and couples requiring vector cross products and moment calculations. Part (a) is straightforward force resolution, part (b) requires finding a line of action using moments (standard technique), and part (c) involves calculating a couple from a non-equilibrium system. While it requires multiple steps and 3D vector manipulation, these are well-practiced techniques for M5 students with no novel insight needed. |
| Spec | 3.04b Equilibrium: zero resultant moment and force4.04a Line equations: 2D and 3D, cartesian and vector forms |
Three forces $F_1$, $F_2$ and $F_3$ act on a rigid body. The forces $F_1$ and $F_2$ act through the points with position vectors $r_1$ and $r_2$ respectively.
$r_1 = (-2i + 3j)$ m, $F_1 = (3i - 2j + k)$ N
$r_2 = (3i + 2k)$ m, $F_2 = (-2i + j - k)$ N
Given that the system $F_1$, $F_2$ and $F_3$ is in equilibrium,
**(a)** Find $F_3$.
(2 marks)
**(b)** Find a vector equation of the line of action of $F_3$, giving your answer in the form $r = a + tb$.
(5 marks)
The force $F_3$ is replaced by a force $F_4$ acting through the point with position vector $(i - 2j + 3k)$ m. The system $F_1$, $F_2$ and $F_4$ is equivalent to a single force $(3i + j + k)$ N acting through the point with position vector $(i + j + k)$ m together with a couple.
**(c)** Find the magnitude of this couple.
(8 marks)
---\begin{enumerate}
\item Three forces $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 }$ act on a rigid body. The forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ act through the points with position vectors $\mathbf { r } _ { 1 }$ and $\mathbf { r } _ { 2 }$ respectively.\\
$\mathbf { r } _ { 1 } = ( - 2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { m }$,\\
$\mathbf { F } _ { 1 } = ( 3 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } ) \mathrm { N }$\\
$\mathbf { r } _ { 2 } = ( 3 \mathbf { i } + 2 \mathbf { k } ) \mathrm { m }$,\\
$\mathbf { F } _ { 2 } = ( - 2 \mathbf { i } + \mathbf { j } - \mathbf { k } ) \mathrm { N }$\\
Given that the system $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 }$ is in equilibrium,\\
(a) find $\mathbf { F } _ { 3 }$,\\
(b) find a vector equation of the line of action of $\mathbf { F } _ { 3 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + t \mathbf { b }$.
\end{enumerate}
The force $\mathbf { F } _ { 3 }$ is replaced by a force $\mathbf { F } _ { 4 }$ acting through the point with position vector $( \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } ) \mathrm { m }$. The system $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 4 }$ is equivalent to a single force ( $3 \mathbf { i } + \mathbf { j } + \mathbf { k }$ ) N acting through the point with position vector $( \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }$ together with a couple.\\
(c) Find the magnitude of this couple.\\
\hfill \mbox{\textit{Edexcel M5 2013 Q4 [15]}}