| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2009 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: equilibrium conditions |
| Difficulty | Challenging +1.2 This M5 question requires vector cross products for moments and equilibrium conditions in 3D, which are advanced mechanics topics. However, the solution is methodical: (a) uses force equilibrium (straightforward vector addition), (b) requires moment equilibrium about a point to find the line of action (standard technique), and (c) involves calculating the moment of a couple. While the 3D vector manipulation is more sophisticated than basic mechanics, this follows standard M5 procedures without requiring novel insight or particularly complex reasoning. |
| 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 | Guidance |
| \(\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2) = -(2\mathbf{i}+\mathbf{j}-2\mathbf{j}-\mathbf{k}) = -(2\mathbf{i}-\mathbf{j}-\mathbf{k})\) | M1 | Equilibrium condition |
| \(\mathbf{F}_3 = -2\mathbf{i}+\mathbf{j}+\mathbf{k}\) | A1 | Correct vector |
| \( | \mathbf{F}_3 | = \sqrt{4+1+1} = \sqrt{6}\) N |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Total moment about \(O\): \(\mathbf{r}_1 \times \mathbf{F}_1 + \mathbf{r}_2 \times \mathbf{F}_2\) | M1 | Summing moments |
| \(\mathbf{r}_1 \times \mathbf{F}_1 = (3\mathbf{i}+\mathbf{j}+\mathbf{k})\times(2\mathbf{i}+\mathbf{j}) = -\mathbf{i}+2\mathbf{j}-\mathbf{k}+\mathbf{k}-2\mathbf{j}... \) | M1 | Correct cross product method |
| \(\mathbf{r}_1 \times \mathbf{F}_1 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\3&1&1\\2&1&0\end{vmatrix} = -\mathbf{i}+2\mathbf{j}+\mathbf{k}\) | A1 | Correct first moment |
| \(\mathbf{r}_2 \times \mathbf{F}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-2&0\\0&-2&-1\end{vmatrix} = 2\mathbf{i}-(-1)\mathbf{j}+(-2)\mathbf{k} = 2\mathbf{i}-\mathbf{j}-2\mathbf{k}\) | A1 | Correct second moment |
| Total moment \(= \mathbf{i}+\mathbf{j}-\mathbf{k}\) | A1 | Correct total moment |
| Line of action: \(\mathbf{r} \times \mathbf{F}_3 = \mathbf{i}+\mathbf{j}-\mathbf{k}\) | M1 | Setting up equation for line |
| \(\mathbf{r} = \lambda(-2\mathbf{i}+\mathbf{j}+\mathbf{k}) + \mathbf{a}\) where \(\mathbf{a}\) satisfies moment condition | M1 | Parametric line through correct direction |
| \(\mathbf{r} = (t)(-2\mathbf{i}+\mathbf{j}+\mathbf{k}) + (\mathbf{i})\) or equivalent correct form | A1A1 | Correct line equation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mathbf{F}_4 = \mathbf{F}_3 = -2\mathbf{i}+\mathbf{j}+\mathbf{k}\) acting through \(O\) | M1 | Recognising couple = moment difference |
| Couple \(= \mathbf{i}+\mathbf{j}-\mathbf{k}\) (the total moment computed above) | M1A1 | Using total moment as couple |
| \( | \text{couple} | = \sqrt{1+1+1} = \sqrt{3}\) N m |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2) = -(2\mathbf{i}+\mathbf{j}-2\mathbf{j}-\mathbf{k}) = -(2\mathbf{i}-\mathbf{j}-\mathbf{k})$ | M1 | Equilibrium condition |
| $\mathbf{F}_3 = -2\mathbf{i}+\mathbf{j}+\mathbf{k}$ | A1 | Correct vector |
| $|\mathbf{F}_3| = \sqrt{4+1+1} = \sqrt{6}$ N | M1A1 | Correct magnitude |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Total moment about $O$: $\mathbf{r}_1 \times \mathbf{F}_1 + \mathbf{r}_2 \times \mathbf{F}_2$ | M1 | Summing moments |
| $\mathbf{r}_1 \times \mathbf{F}_1 = (3\mathbf{i}+\mathbf{j}+\mathbf{k})\times(2\mathbf{i}+\mathbf{j}) = -\mathbf{i}+2\mathbf{j}-\mathbf{k}+\mathbf{k}-2\mathbf{j}... $ | M1 | Correct cross product method |
| $\mathbf{r}_1 \times \mathbf{F}_1 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\3&1&1\\2&1&0\end{vmatrix} = -\mathbf{i}+2\mathbf{j}+\mathbf{k}$ | A1 | Correct first moment |
| $\mathbf{r}_2 \times \mathbf{F}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-2&0\\0&-2&-1\end{vmatrix} = 2\mathbf{i}-(-1)\mathbf{j}+(-2)\mathbf{k} = 2\mathbf{i}-\mathbf{j}-2\mathbf{k}$ | A1 | Correct second moment |
| Total moment $= \mathbf{i}+\mathbf{j}-\mathbf{k}$ | A1 | Correct total moment |
| Line of action: $\mathbf{r} \times \mathbf{F}_3 = \mathbf{i}+\mathbf{j}-\mathbf{k}$ | M1 | Setting up equation for line |
| $\mathbf{r} = \lambda(-2\mathbf{i}+\mathbf{j}+\mathbf{k}) + \mathbf{a}$ where $\mathbf{a}$ satisfies moment condition | M1 | Parametric line through correct direction |
| $\mathbf{r} = (t)(-2\mathbf{i}+\mathbf{j}+\mathbf{k}) + (\mathbf{i})$ or equivalent correct form | A1A1 | Correct line equation |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{F}_4 = \mathbf{F}_3 = -2\mathbf{i}+\mathbf{j}+\mathbf{k}$ acting through $O$ | M1 | Recognising couple = moment difference |
| Couple $= \mathbf{i}+\mathbf{j}-\mathbf{k}$ (the total moment computed above) | M1A1 | Using total moment as couple |
| $|\text{couple}| = \sqrt{1+1+1} = \sqrt{3}$ N m | A1 | Correct magnitude |
---\begin{enumerate}
\item Two forces $\mathbf { F } _ { 1 } = ( 2 \mathbf { i } + \mathbf { j } ) \mathrm { N }$ and $\mathbf { F } _ { 2 } = ( - 2 \mathbf { j } - \mathbf { k } ) \mathrm { N }$ act on a rigid body. The force $\mathbf { F } _ { 1 }$ acts at the point with position vector $\mathbf { r } _ { 1 } = ( 3 \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }$ and the force $\mathbf { F } _ { 2 }$ acts at the point with position vector $\mathbf { r } _ { 2 } = ( \mathbf { i } - 2 \mathbf { j } ) \mathrm { m }$. A third force $\mathbf { F } _ { 3 }$ acts on the body such that $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 }$ are in equilibrium.\\
(a) Find the magnitude of $\mathbf { F } _ { 3 }$.\\
(b) Find a vector equation of the line of action of $\mathbf { F } _ { 3 }$.
\end{enumerate}
The force $\mathbf { F } _ { 3 }$ is replaced by a fourth force $\mathbf { F } _ { 4 }$, acting through the origin $O$, such that $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 4 }$ are equivalent to a couple.\\
(c) Find the magnitude of this couple.
\hfill \mbox{\textit{Edexcel M5 2009 Q5 [16]}}