Edexcel M5 2018 June — Question 2 11 marks

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2018
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
Type3D force systems: reduction to single force
DifficultyChallenging +1.2 This M5 question requires calculating resultant force and total moment about origin, then applying the parallel condition via dot product equals zero. While it involves 3D vectors and multiple steps, the techniques are standard for Further Maths mechanics: vector addition, cross products for moments, and using perpendicularity conditions. The algebraic manipulation is straightforward once the method is identified.
Spec1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force
2. Three forces \(\mathbf { F } _ { 1 } = ( a \mathbf { i } + b \mathbf { j } - 2 \mathbf { k } ) \mathrm { N } , \mathbf { F } _ { 2 } = ( - \mathbf { i } + \mathbf { j } - 2 \mathbf { k } ) \mathrm { N }\) and \(\mathbf { F } _ { 3 } = ( - \mathbf { i } - 3 \mathbf { j } + \mathbf { k } ) \mathrm { N }\), where \(a\) and \(b\) are constants, act on a rigid body. The force \(\mathbf { F } _ { 1 }\) acts through the point with position vector \(\mathbf { k } \mathrm { m }\), the force \(\mathbf { F } _ { 2 }\) acts through the point with position vector \(( 3 \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { m }\) and the force \(\mathbf { F } _ { 3 }\) acts through the point with position vector \(( \mathbf { j } + 2 \mathbf { k } ) \mathrm { m }\). The system of three forces is equivalent to a single force \(\mathbf { R }\) acting through the origin together with a couple of moment \(\mathbf { G }\). The direction of \(\mathbf { R }\) is parallel to the direction of \(\mathbf { G }\). Find the value of \(a\) and the value of \(b\).
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mathbf{R} = \sum\mathbf{F}_i = \begin{pmatrix}a\\b\\-2\end{pmatrix}+\begin{pmatrix}-1\\1\\-2\end{pmatrix}+\begin{pmatrix}-1\\-3\\1\end{pmatrix} = \begin{pmatrix}a-2\\b-2\\-3\end{pmatrix}\)M1 A1 M1 for adding 3 forces; A1 for correct R
\(\mathbf{G} = \begin{pmatrix}0\\0\\1\end{pmatrix}\times\begin{pmatrix}a\\b\\-2\end{pmatrix}+\begin{pmatrix}3\\-1\\1\end{pmatrix}\times\begin{pmatrix}-1\\1\\-2\end{pmatrix}+\begin{pmatrix}0\\1\\2\end{pmatrix}\times\begin{pmatrix}-1\\-3\\1\end{pmatrix}\)M1 M1 for moments about \(O\); allow \(\mathbf{F}\times\mathbf{r}\)
\(= \begin{pmatrix}-b\\a\\0\end{pmatrix}+\begin{pmatrix}1\\5\\2\end{pmatrix}+\begin{pmatrix}7\\-2\\1\end{pmatrix}\)A3 -1 each error or omission (A1 for each product); allow consistent negatives
\(= \begin{pmatrix}8-b\\a+3\\3\end{pmatrix}\)A1 cao
\(8-b = \lambda(a-2)\)M1 A1ft M1 for using G parallel with R; A1ft for correct ft equations
\(a+3 = \lambda(b-2)\)
\(3 = \lambda\times -3\)
\(a = \frac{-7}{2};\; b = \frac{5}{2}\)A1; A1 cao A1 for \(-3.5\); A1 for \(2.5\)
Total: [11]
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{R} = \sum\mathbf{F}_i = \begin{pmatrix}a\\b\\-2\end{pmatrix}+\begin{pmatrix}-1\\1\\-2\end{pmatrix}+\begin{pmatrix}-1\\-3\\1\end{pmatrix} = \begin{pmatrix}a-2\\b-2\\-3\end{pmatrix}$ | M1 A1 | M1 for adding 3 forces; A1 for correct **R** |
| $\mathbf{G} = \begin{pmatrix}0\\0\\1\end{pmatrix}\times\begin{pmatrix}a\\b\\-2\end{pmatrix}+\begin{pmatrix}3\\-1\\1\end{pmatrix}\times\begin{pmatrix}-1\\1\\-2\end{pmatrix}+\begin{pmatrix}0\\1\\2\end{pmatrix}\times\begin{pmatrix}-1\\-3\\1\end{pmatrix}$ | M1 | M1 for moments about $O$; allow $\mathbf{F}\times\mathbf{r}$ |
| $= \begin{pmatrix}-b\\a\\0\end{pmatrix}+\begin{pmatrix}1\\5\\2\end{pmatrix}+\begin{pmatrix}7\\-2\\1\end{pmatrix}$ | A3 | -1 each error or omission (A1 for each product); allow consistent negatives |
| $= \begin{pmatrix}8-b\\a+3\\3\end{pmatrix}$ | A1 cao | |
| $8-b = \lambda(a-2)$ | M1 A1ft | M1 for using **G** parallel with **R**; A1ft for correct ft equations |
| $a+3 = \lambda(b-2)$ | | |
| $3 = \lambda\times -3$ | | |
| $a = \frac{-7}{2};\; b = \frac{5}{2}$ | A1; A1 cao | A1 for $-3.5$; A1 for $2.5$ |

**Total: [11]**

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2. Three forces $\mathbf { F } _ { 1 } = ( a \mathbf { i } + b \mathbf { j } - 2 \mathbf { k } ) \mathrm { N } , \mathbf { F } _ { 2 } = ( - \mathbf { i } + \mathbf { j } - 2 \mathbf { k } ) \mathrm { N }$ and $\mathbf { F } _ { 3 } = ( - \mathbf { i } - 3 \mathbf { j } + \mathbf { k } ) \mathrm { N }$, where $a$ and $b$ are constants, act on a rigid body.

The force $\mathbf { F } _ { 1 }$ acts through the point with position vector $\mathbf { k } \mathrm { m }$, the force $\mathbf { F } _ { 2 }$ acts through the point with position vector $( 3 \mathbf { i } - \mathbf { j } + \mathbf { k } ) \mathrm { m }$ and the force $\mathbf { F } _ { 3 }$ acts through the point with position vector $( \mathbf { j } + 2 \mathbf { k } ) \mathrm { m }$.

The system of three forces is equivalent to a single force $\mathbf { R }$ acting through the origin together with a couple of moment $\mathbf { G }$. The direction of $\mathbf { R }$ is parallel to the direction of $\mathbf { G }$.

Find the value of $a$ and the value of $b$.

\hfill \mbox{\textit{Edexcel M5 2018 Q2 [11]}}
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