| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Standard +0.3 This is a standard M1 moments problem requiring equilibrium equations (sum of forces = 0, sum of moments = 0) and algebraic manipulation. The setup is straightforward with clearly defined positions, and the 'show that' format guides students to the answer. Part (b) requires recognizing physical constraints (W > 0, x must be between supports), which is routine for this topic. Slightly above average difficulty due to the algebraic manipulation required, but well within typical M1 scope. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(M(A)\): \(Wx + 120 \times 1.5 = R \times 2 + 2R \times 1\) | M1 A2, 1, 0 | |
| \(R(\uparrow)\) \(3R = W + 120\) | M1 A1 | |
| Hence \(Wx + 180 = 3R = W + 120\) | M1 | |
| \(W(1 - x) = 60\) | A1 | |
| \(W = \frac{60}{1 - x}\) | M1 A1 so | (8 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(W > 0 \Rightarrow x < 1\) | M1 A1 | (2 marks) |
## Part (a)
$M(A)$: $Wx + 120 \times 1.5 = R \times 2 + 2R \times 1$ | M1 A2, 1, 0 |
$R(\uparrow)$ $3R = W + 120$ | M1 A1 |
Hence $Wx + 180 = 3R = W + 120$ | M1 |
$W(1 - x) = 60$ | A1 |
$W = \frac{60}{1 - x}$ | M1 A1 so | (8 marks)
## Part (b)
$W > 0 \Rightarrow x < 1$ | M1 A1 | (2 marks)
**Total: 10 marks**
---\includegraphics{figure_3}
A uniform rod $AB$ has length 3 m and weight 120 N. The rod rests in equilibrium in a horizontal position, smoothly supported at points $C$ and $D$, where $AC = 0.5$ m and $AD = 2$ m, as shown in Fig. 3. A particle of weight $W$ newtons is attached to the rod at a point $E$ where $AE = x$ metres. The rod remains in equilibrium and the magnitude of the reaction at $C$ is now twice the magnitude of the reaction at $D$.
\begin{enumerate}[label=(\alph*)]
\item Show that $W = \frac{60}{1-x}$. [8]
\item Hence deduce the range of possible values of $x$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2003 Q6 [10]}}