| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: equilibrium conditions |
| Difficulty | Challenging +1.2 This is a standard M5 three-dimensional equilibrium problem requiring force resolution and moment equations. Part (a) uses straightforward vector addition for equilibrium of forces. Part (b) requires taking moments about a point and solving for the line of action, which is a routine M5 technique but involves more algebraic manipulation with cross products. The question is methodical rather than requiring insight, making it moderately above average difficulty for A-level but standard for Further Maths M5. |
| Spec | 3.04b Equilibrium: zero resultant moment and force4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For equilibrium: \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}\) | M1 | Resolving forces condition |
| \(\mathbf{F}_3 = -\mathbf{F}_1 - \mathbf{F}_2 = -(2\mathbf{j} + \mathbf{k}) - (-2\mathbf{i} - \mathbf{j})\) | M1 | Substituting values |
| \(\mathbf{F}_3 = 2\mathbf{i} - \mathbf{j} - \mathbf{k}\) | A1 | Correct vector |
| \( | \mathbf{F}_3 | = \sqrt{4 + 1 + 1} = \sqrt{6}\) N |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Taking moments about origin: \(\mathbf{r}_1 \times \mathbf{F}_1 + \mathbf{r}_2 \times \mathbf{F}_2 + \mathbf{r}_3 \times \mathbf{F}_3 = \mathbf{0}\) | M1 | Moment equilibrium condition |
| \(\mathbf{r}_1 \times \mathbf{F}_1 = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \times (2\mathbf{j} + \mathbf{k})\) | M1 | Setting up cross product |
| \(= \mathbf{i}(2\cdot1 - 2\cdot2) - \mathbf{j}(1\cdot1 - 2\cdot0) + \mathbf{k}(1\cdot2 - 2\cdot0) = -2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) | A1 | Correct first cross product |
| \(\mathbf{r}_2 \times \mathbf{F}_2 = (-\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (-2\mathbf{i} - \mathbf{j})\) | M1 | Setting up second cross product |
| \(= \mathbf{i}((-1)(0) - (1)(-1)) - \mathbf{j}((-1)(0) - (1)(-2)) + \mathbf{k}((-1)(-1) - (-1)(-2)) = \mathbf{i} - 2\mathbf{j} - \mathbf{k}\) | A1 | Correct second cross product |
| Total moment \(= (-2+1)\mathbf{i} + (-1-2)\mathbf{j} + (2-1)\mathbf{k} = -\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) | A1 | Correct total moment |
| \(\mathbf{a} \times \mathbf{F}_3 = -(-\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) so \(\mathbf{a} \times (2\mathbf{i} - \mathbf{j} - \mathbf{k}) = \mathbf{i} + 3\mathbf{j} - \mathbf{k}\) | M1 | Setting up equation for point on line |
| Point on line of action found, e.g. \(\mathbf{a} = (0\mathbf{i} + \mathbf{j} - 2\mathbf{k})\) or equivalent | A1 | Correct position vector |
| \(\mathbf{r} = (\mathbf{j} - 2\mathbf{k}) + t(2\mathbf{i} - \mathbf{j} - \mathbf{k})\) | A1 | Correct final equation |
## Question 3:
### Part (a): Find the magnitude of **F**₃
| Answer/Working | Mark | Guidance |
|---|---|---|
| For equilibrium: $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}$ | M1 | Resolving forces condition |
| $\mathbf{F}_3 = -\mathbf{F}_1 - \mathbf{F}_2 = -(2\mathbf{j} + \mathbf{k}) - (-2\mathbf{i} - \mathbf{j})$ | M1 | Substituting values |
| $\mathbf{F}_3 = 2\mathbf{i} - \mathbf{j} - \mathbf{k}$ | A1 | Correct vector |
| $|\mathbf{F}_3| = \sqrt{4 + 1 + 1} = \sqrt{6}$ N | A1 | Correct magnitude |
### Part (b): Find the vector equation of the line of action of **F**₃
| Answer/Working | Mark | Guidance |
|---|---|---|
| Taking moments about origin: $\mathbf{r}_1 \times \mathbf{F}_1 + \mathbf{r}_2 \times \mathbf{F}_2 + \mathbf{r}_3 \times \mathbf{F}_3 = \mathbf{0}$ | M1 | Moment equilibrium condition |
| $\mathbf{r}_1 \times \mathbf{F}_1 = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \times (2\mathbf{j} + \mathbf{k})$ | M1 | Setting up cross product |
| $= \mathbf{i}(2\cdot1 - 2\cdot2) - \mathbf{j}(1\cdot1 - 2\cdot0) + \mathbf{k}(1\cdot2 - 2\cdot0) = -2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ | A1 | Correct first cross product |
| $\mathbf{r}_2 \times \mathbf{F}_2 = (-\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (-2\mathbf{i} - \mathbf{j})$ | M1 | Setting up second cross product |
| $= \mathbf{i}((-1)(0) - (1)(-1)) - \mathbf{j}((-1)(0) - (1)(-2)) + \mathbf{k}((-1)(-1) - (-1)(-2)) = \mathbf{i} - 2\mathbf{j} - \mathbf{k}$ | A1 | Correct second cross product |
| Total moment $= (-2+1)\mathbf{i} + (-1-2)\mathbf{j} + (2-1)\mathbf{k} = -\mathbf{i} - 3\mathbf{j} + \mathbf{k}$ | A1 | Correct total moment |
| $\mathbf{a} \times \mathbf{F}_3 = -(-\mathbf{i} - 3\mathbf{j} + \mathbf{k})$ so $\mathbf{a} \times (2\mathbf{i} - \mathbf{j} - \mathbf{k}) = \mathbf{i} + 3\mathbf{j} - \mathbf{k}$ | M1 | Setting up equation for point on line |
| Point on line of action found, e.g. $\mathbf{a} = (0\mathbf{i} + \mathbf{j} - 2\mathbf{k})$ or equivalent | A1 | Correct position vector |
| $\mathbf{r} = (\mathbf{j} - 2\mathbf{k}) + t(2\mathbf{i} - \mathbf{j} - \mathbf{k})$ | A1 | Correct final equation |\begin{enumerate}
\item A rigid body is in equilibrium under the action of three forces $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 } \mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ act at the points with position vectors $\mathbf { r } _ { 1 }$ and $\mathbf { r } _ { 2 }$ respectively, where $\mathbf { F } _ { 1 } = ( 2 \mathbf { j } + \mathbf { k } ) \mathrm { N } \quad \mathbf { r } _ { 1 } = ( \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k } ) \mathrm { m } \mathbf { F } _ { 2 } = ( - 2 \mathbf { i } - \mathbf { j } ) \mathrm { N } \quad \mathbf { r } _ { 2 } = ( - \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { m }$\\
(a) Find the magnitude of $\mathbf { F } _ { 3 }$\\
(b) Find a vector equation of the line of action of $\mathbf { F } _ { 3 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + t \mathbf { b }$, where $\mathbf { a }$ and $\mathbf { b }$ are constant vectors and $t$ is a scalar parameter.\\
\end{enumerate}
\includegraphics[max width=\textwidth, alt={}, center]{cac4dd38-796c-414b-9b80-fe39ab12d41b-11_62_49_2643_1886}\\
\hfill \mbox{\textit{Edexcel M5 2015 Q3 [12]}}