| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: reduction to single force |
| Difficulty | Standard +0.3 Part (a) requires simple vector addition of forces. Part (b) involves finding the moment about the origin, setting it equal to the moment of R about a general point, and solving for the line of action—a standard M5 technique with straightforward vector calculations. This is slightly easier than average due to the routine nature of the procedures involved. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors4.04g Vector product: a x b perpendicular vector6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 = (\mathbf{i}+2\mathbf{j}-\mathbf{k}) + (-2\mathbf{i}+\mathbf{j}+3\mathbf{k}) = -\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Total moment about origin \(= \mathbf{G} + \mathbf{a} \times \mathbf{R}\) must equal moment of \(\mathbf{R}\) about a point on line of action | M1 | Setting up moment equation |
| \(\mathbf{a} \times (\mathbf{F}_1 + \mathbf{F}_2) + \mathbf{G}\) where \(\mathbf{a} = \mathbf{i}+\mathbf{j}+\mathbf{k}\) | M1 | |
| \(\mathbf{a} \times \mathbf{R} = (\mathbf{i}+\mathbf{j}+\mathbf{k}) \times (-\mathbf{i}+3\mathbf{j}+2\mathbf{k})\) | M1 | Computing cross product |
| \(= \mathbf{i}(2-3) - \mathbf{j}(2+1) + \mathbf{k}(3+1) = -\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}\) | A1 | |
| Total moment \(= (-\mathbf{i}-3\mathbf{j}+4\mathbf{k}) + (7\mathbf{i}-3\mathbf{j}+8\mathbf{k}) = 6\mathbf{i} - 6\mathbf{j} + 12\mathbf{k}\) | A1 | |
| For line of action: \(\mathbf{r} \times \mathbf{R} = 6\mathbf{i} - 6\mathbf{j} + 12\mathbf{k}\) | M1 | |
| \(\mathbf{r} = \mathbf{i}+\mathbf{j}+\mathbf{k} + t(-\mathbf{i}+3\mathbf{j}+2\mathbf{k})\) | A1 | Accept equivalent vector equations |
# Question 3:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 = (\mathbf{i}+2\mathbf{j}-\mathbf{k}) + (-2\mathbf{i}+\mathbf{j}+3\mathbf{k}) = -\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$ | M1 A1 | |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Total moment about origin $= \mathbf{G} + \mathbf{a} \times \mathbf{R}$ must equal moment of $\mathbf{R}$ about a point on line of action | M1 | Setting up moment equation |
| $\mathbf{a} \times (\mathbf{F}_1 + \mathbf{F}_2) + \mathbf{G}$ where $\mathbf{a} = \mathbf{i}+\mathbf{j}+\mathbf{k}$ | M1 | |
| $\mathbf{a} \times \mathbf{R} = (\mathbf{i}+\mathbf{j}+\mathbf{k}) \times (-\mathbf{i}+3\mathbf{j}+2\mathbf{k})$ | M1 | Computing cross product |
| $= \mathbf{i}(2-3) - \mathbf{j}(2+1) + \mathbf{k}(3+1) = -\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ | A1 | |
| Total moment $= (-\mathbf{i}-3\mathbf{j}+4\mathbf{k}) + (7\mathbf{i}-3\mathbf{j}+8\mathbf{k}) = 6\mathbf{i} - 6\mathbf{j} + 12\mathbf{k}$ | A1 | |
| For line of action: $\mathbf{r} \times \mathbf{R} = 6\mathbf{i} - 6\mathbf{j} + 12\mathbf{k}$ | M1 | |
| $\mathbf{r} = \mathbf{i}+\mathbf{j}+\mathbf{k} + t(-\mathbf{i}+3\mathbf{j}+2\mathbf{k})$ | A1 | Accept equivalent vector equations |
---3. A system of forces acting on a rigid body consists of two forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ acting at a point $A$ of the body, together with a couple of moment $\mathbf { G } . \mathbf { F } _ { 1 } = ( \mathbf { i } + 2 \mathbf { j } - \mathbf { k } ) \mathrm { N }$ and $\mathbf { F } _ { 2 } = ( - 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } ) N$. The position vector of the point $A$ is $( \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }$ and $\mathbf { G } = ( 7 \mathbf { i } - 3 \mathbf { j } + 8 \mathbf { k } ) \mathrm { Nm }$.
Given that the system is equivalent to a single force $\mathbf { R }$,
\begin{enumerate}[label=(\alph*)]
\item find $\mathbf { R }$,
\item find a vector equation for the line of action of $\mathbf { R }$.\\
(Total 9 marks)
\section*{4.}
\section*{Figure 1}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{43ce237f-c8e4-428a-b8cd-04673e62abb9-3_896_515_276_772}
\end{center}
A thin uniform rod $P Q$ has mass $m$ and length $3 a$. A thin uniform circular disc, of mass $m$ and radius $a$, is attached to the rod at $Q$ in such a way that the rod and the diameter $Q R$ of the disc are in a straight line with $P R = 5 a$. The rod together with the disc form a composite body, as shown in Figure 1. The body is free to rotate about a fixed smooth horizontal axis $L$ through $P$, perpendicular to $P Q$ and in the plane of the disc.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2005 Q3 [9]}}