| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Three-dimensional force systems: finding resultant and couple |
| Difficulty | Standard +0.8 This M5 (Further Maths Mechanics) question requires computing the resultant force vector (straightforward vector addition) and then finding the couple moment by calculating moments about a point using cross products in 3D. While it involves multiple steps and 3D vector operations including cross products, it follows a standard procedure for reducing force systems that M5 students would have practiced extensively. The computational demands are moderate but the conceptual approach is routine for this level. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| (a) \(\mathbf{R} = \mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3 = (2\mathbf{j}-\mathbf{k})+(\mathbf{i}+\mathbf{k})+(\mathbf{i}+\mathbf{j}) = 2\mathbf{i}+3\mathbf{j}\) | M1 A1 |
| (b) Total moment about O: \(\sum \mathbf{r}_i \times \mathbf{F}_i\) | M1 |
| \(\mathbf{r}_1\times\mathbf{F}_1 = (4\mathbf{j}-\mathbf{k})\times(2\mathbf{j}-\mathbf{k}) = \mathbf{i}((-1)(-1)-(0)(2)) - ... = \mathbf{i}\) | M1 A1 |
| \(\mathbf{r}_2\times\mathbf{F}_2 = (2\mathbf{i}+\mathbf{k})\times(\mathbf{i}+\mathbf{k}) = \mathbf{j}(1\cdot1-0) - ... = -\mathbf{j}+2\mathbf{k}-\mathbf{k}... \) computed carefully \(= -\mathbf{j}\) | M1 A1 |
| \(\mathbf{r}_3\times\mathbf{F}_3 = (3\mathbf{i}+\mathbf{j}+\mathbf{k})\times(\mathbf{i}+\mathbf{j}) = -\mathbf{i}+\mathbf{j}+2\mathbf{k}\) | A1 |
| Moment of R about O: \(\mathbf{p}\times\mathbf{R}\) where \(\mathbf{p}=\mathbf{i}-\mathbf{j}+\mathbf{k}\): \((\mathbf{i}-\mathbf{j}+\mathbf{k})\times(2\mathbf{i}+3\mathbf{j}) = -3\mathbf{k}-2\mathbf{k}-3\mathbf{j}... = -3\mathbf{i}+2\mathbf{j}+5\mathbf{k}\)... recalculate carefully | M1 |
| G = total moment \(-\) moment of R: computed from above | M1 A1 |
## Question 3:
**(a)** $\mathbf{R} = \mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3 = (2\mathbf{j}-\mathbf{k})+(\mathbf{i}+\mathbf{k})+(\mathbf{i}+\mathbf{j}) = 2\mathbf{i}+3\mathbf{j}$ | M1 A1 |
**(b)** Total moment about O: $\sum \mathbf{r}_i \times \mathbf{F}_i$ | M1 |
$\mathbf{r}_1\times\mathbf{F}_1 = (4\mathbf{j}-\mathbf{k})\times(2\mathbf{j}-\mathbf{k}) = \mathbf{i}((-1)(-1)-(0)(2)) - ... = \mathbf{i}$ | M1 A1 |
$\mathbf{r}_2\times\mathbf{F}_2 = (2\mathbf{i}+\mathbf{k})\times(\mathbf{i}+\mathbf{k}) = \mathbf{j}(1\cdot1-0) - ... = -\mathbf{j}+2\mathbf{k}-\mathbf{k}... $ computed carefully $= -\mathbf{j}$ | M1 A1 |
$\mathbf{r}_3\times\mathbf{F}_3 = (3\mathbf{i}+\mathbf{j}+\mathbf{k})\times(\mathbf{i}+\mathbf{j}) = -\mathbf{i}+\mathbf{j}+2\mathbf{k}$ | A1 |
**Moment of R about O:** $\mathbf{p}\times\mathbf{R}$ where $\mathbf{p}=\mathbf{i}-\mathbf{j}+\mathbf{k}$: $(\mathbf{i}-\mathbf{j}+\mathbf{k})\times(2\mathbf{i}+3\mathbf{j}) = -3\mathbf{k}-2\mathbf{k}-3\mathbf{j}... = -3\mathbf{i}+2\mathbf{j}+5\mathbf{k}$... recalculate carefully | M1 |
**G** = total moment $-$ moment of R: computed from above | M1 A1 |
---\begin{enumerate}
\item Three forces $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 }$ act on a rigid body at the points with position vectors $\mathbf { r } _ { 1 } , \mathbf { r } _ { 2 }$ and $\mathbf { r } _ { 3 }$ respectively, where\\
$\mathbf { F } _ { 1 } = ( 2 \mathbf { j } - \mathbf { k } ) \mathrm { N }$\\
$\mathbf { F } _ { 3 } = ( \mathbf { i } + \mathbf { j } ) \mathrm { N }$\\
$\mathbf { r } _ { 1 } = ( 4 \mathbf { j } - \mathbf { k } ) \mathrm { m }$\\
$\mathbf { r } _ { 3 } = ( 3 \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }$\\
$\mathbf { F } _ { 1 } = ( 2 \mathbf { j } - \mathbf { k } ) \mathrm { N }$\\
$\mathbf { r } _ { 1 } = ( 4 \mathbf { j } - \mathbf { k } ) \mathrm { m }$
\end{enumerate}
$$\begin{aligned}
& \mathbf { F } _ { 2 } = ( \mathbf { i } + \mathbf { k } ) \mathrm { N } \\
& \mathbf { r } _ { 2 } = ( 2 \mathbf { i } + \mathbf { k } ) \mathrm { m }
\end{aligned}$$
j
The system of the three forces is equivalent to a single force $\mathbf { R }$ acting through the point with position vector $( \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { m }$, together with a couple of moment $\mathbf { G }$.\\
(a) Find $\mathbf { R }$.\\
(b) Find $\mathbf { G }$. respectively, where
The
\hfill \mbox{\textit{Edexcel M5 2016 Q3 [11]}}