| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| 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/Further Mechanics question on 3D force systems requiring vector addition, moment calculations using cross products, and understanding of resultant forces and couples. While it involves multiple steps and 3D vectors, the techniques are routine for Further Maths students: (a)(i) is simple vector addition, (a)(ii) requires finding moments about a point and using r×R=M, and (b) involves calculating the moment of the couple. The calculations are methodical rather than requiring novel insight, placing it moderately above average difficulty. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10g Problem solving with vectors: in geometry3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(\mathbf{R} = (\mathbf{i}+2\mathbf{j}+3\mathbf{k}) + (3\mathbf{i}+\mathbf{j}+2\mathbf{k})\) | M1 | |
| \(= (4\mathbf{i}+3\mathbf{j}+5\mathbf{k})\) | A1 | Total: (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \((x\mathbf{i}+y\mathbf{j}+z\mathbf{k})\times(4\mathbf{i}+3\mathbf{j}+5\mathbf{k}) = (2\mathbf{i}+\mathbf{k})\times(\mathbf{i}+2\mathbf{j}+3\mathbf{k})+(\mathbf{j}+2\mathbf{k})\times(3\mathbf{i}+\mathbf{j}+2\mathbf{k})\) | M1 | |
| \((5y-3z)\mathbf{i}+(4z-5x)\mathbf{j}+(3x-4y)\mathbf{k} = (-2\mathbf{i}-5\mathbf{j}+4\mathbf{k})+(6\mathbf{j}-3\mathbf{k})\) | A2 | |
| \((5y-3z)\mathbf{i}+(4z-5x)\mathbf{j}+(3x-4y)\mathbf{k} = (-2\mathbf{i}+\mathbf{j}+\mathbf{k})\) | ||
| a solution: \(x=0, y=-\frac{1}{4}, z=\frac{1}{4}\); \(x=\frac{1}{3}, y=0, z=\frac{2}{3}\); \(x=-\frac{1}{5}, y=-\frac{2}{5}, z=0\) | B1 | |
| \(\mathbf{r} = -\frac{1}{4}\mathbf{j}+\frac{1}{4}\mathbf{k} + \lambda(4\mathbf{i}+3\mathbf{j}+5\mathbf{k})\) | M1 A1 ft | Total: (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \((\mathbf{i}+2\mathbf{j}+\mathbf{k})\times(4\mathbf{i}+3\mathbf{j}+5\mathbf{k})+\mathbf{G} = (-2\mathbf{i}+\mathbf{j}+\mathbf{k})\) | M1 A1 | |
| \((7\mathbf{i}-\mathbf{j}-5\mathbf{k})+\mathbf{G} = (-2\mathbf{i}+\mathbf{j}+\mathbf{k})\) | ||
| \(\mathbf{G} = (-9\mathbf{i}+2\mathbf{j}+6\mathbf{k})\) | A1 | |
| \( | \mathbf{G} | = \sqrt{(-9)^2+2^2+6^2}\) |
| \(= 11\) (Nm) | A1 ft | Total: (5), Q Total: 13 |
## Question 4:
### Part (a)(i):
| Working/Answer | Mark | Notes |
|---|---|---|
| $\mathbf{R} = (\mathbf{i}+2\mathbf{j}+3\mathbf{k}) + (3\mathbf{i}+\mathbf{j}+2\mathbf{k})$ | M1 | |
| $= (4\mathbf{i}+3\mathbf{j}+5\mathbf{k})$ | A1 | **Total: (2)** |
### Part (a)(ii):
| Working/Answer | Mark | Notes |
|---|---|---|
| $(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})\times(4\mathbf{i}+3\mathbf{j}+5\mathbf{k}) = (2\mathbf{i}+\mathbf{k})\times(\mathbf{i}+2\mathbf{j}+3\mathbf{k})+(\mathbf{j}+2\mathbf{k})\times(3\mathbf{i}+\mathbf{j}+2\mathbf{k})$ | M1 | |
| $(5y-3z)\mathbf{i}+(4z-5x)\mathbf{j}+(3x-4y)\mathbf{k} = (-2\mathbf{i}-5\mathbf{j}+4\mathbf{k})+(6\mathbf{j}-3\mathbf{k})$ | A2 | |
| $(5y-3z)\mathbf{i}+(4z-5x)\mathbf{j}+(3x-4y)\mathbf{k} = (-2\mathbf{i}+\mathbf{j}+\mathbf{k})$ | | |
| a solution: $x=0, y=-\frac{1}{4}, z=\frac{1}{4}$; $x=\frac{1}{3}, y=0, z=\frac{2}{3}$; $x=-\frac{1}{5}, y=-\frac{2}{5}, z=0$ | B1 | |
| $\mathbf{r} = -\frac{1}{4}\mathbf{j}+\frac{1}{4}\mathbf{k} + \lambda(4\mathbf{i}+3\mathbf{j}+5\mathbf{k})$ | M1 A1 ft | **Total: (6)** |
### Part (b):
| Working/Answer | Mark | Notes |
|---|---|---|
| $(\mathbf{i}+2\mathbf{j}+\mathbf{k})\times(4\mathbf{i}+3\mathbf{j}+5\mathbf{k})+\mathbf{G} = (-2\mathbf{i}+\mathbf{j}+\mathbf{k})$ | M1 A1 | |
| $(7\mathbf{i}-\mathbf{j}-5\mathbf{k})+\mathbf{G} = (-2\mathbf{i}+\mathbf{j}+\mathbf{k})$ | | |
| $\mathbf{G} = (-9\mathbf{i}+2\mathbf{j}+6\mathbf{k})$ | A1 | |
| $|\mathbf{G}| = \sqrt{(-9)^2+2^2+6^2}$ | M1 | |
| $= 11$ (Nm) | A1 ft | **Total: (5), Q Total: 13** |
---\begin{enumerate}
\item Two forces $\mathbf { F } _ { 1 } = ( \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } ) \mathrm { N }$ and $\mathbf { F } _ { 2 } = ( 3 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } ) \mathrm { N }$ act on a rigid body.
\end{enumerate}
The force $\mathbf { F } _ { 1 }$ acts through the point with position vector ( $2 \mathbf { i } + \mathbf { k }$ ) m and the force $\mathbf { F } _ { 2 }$ acts through the point with position vector $( \mathbf { j } + 2 \mathbf { k } ) \mathrm { m }$.\\
(a) If the two forces are equivalent to a single force $\mathbf { R }$, find\\
(i) $\mathbf { R }$,\\
(ii) a vector equation of the line of action of $\mathbf { R }$, in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }$.\\
(b) If the two forces are equivalent to a single force acting through the point with position vector $( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } ) \mathrm { m }$ together with a couple of moment $\mathbf { G }$, find the magnitude of $\mathbf { G }$.
\hfill \mbox{\textit{Edexcel M5 2010 Q4 [13]}}