| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform beam on supports |
| Difficulty | Standard +0.3 This is a standard M1 moments problem requiring taking moments about a point and applying equilibrium conditions. Part (a) involves straightforward moment calculations to reach a given result. Parts (b) and (c) repeat the process with changed conditions and solve simultaneous equations. While multi-step, it follows a routine textbook approach with no novel insight required, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| M(D): \(160 \times 2.5 = W \times 4 + 200(4 - x)\) | M1 | A2, 1, 0 |
| \(400 = 4W + 800 - 200x\) | M1 | A1 |
| \(200x - 4W = 400 \Rightarrow 50x - W = 100\) ★ | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| M(D): \(50 \times 2.5 + W \times 1 = 200(4 - x)\) | M1 | A2, 1, 0 |
| \(200x + W = 675\) | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Solving \(\rightarrow x = 3.1\) m | M1 | A1 |
| \(: \quad W = \underline{55 \text{ N}}\) | M1 | A1 |
| (4) |
## (a)
M(D): $160 \times 2.5 = W \times 4 + 200(4 - x)$ | M1 | A2, 1, 0
$400 = 4W + 800 - 200x$ | M1 | A1
$200x - 4W = 400 \Rightarrow 50x - W = 100$ ★ | | (5)
## (b)
M(D): $50 \times 2.5 + W \times 1 = 200(4 - x)$ | M1 | A2, 1, 0
$200x + W = 675$ | | (3)
## (c)
Solving $\rightarrow x = 3.1$ m | M1 | A1
$: \quad W = \underline{55 \text{ N}}$ | M1 | A1
| | (4)
**Total: 12 marks**
---\includegraphics{figure_2}
A non-uniform rod $AB$ has length 5 m and weight 200 N. The rod rests horizontally in equilibrium on two smooth supports $C$ and $D$, where $AC = 1.5$ m and $DB = 1$ m, as shown in Fig. 2. The centre of mass of $AB$ is $x$ metres from $A$. A particle of weight $W$ newtons is placed on the rod at $A$. The rod remains in equilibrium and the magnitude of the reaction of $C$ on the rod is 160 N.
\begin{enumerate}[label=(\alph*)]
\item Show that $50x - W = 100$. [5]
\end{enumerate}
The particle is now removed from $A$ and placed on the rod at $B$. The rod remains in equilibrium and the reaction of $C$ on the rod now has magnitude 50 N.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Obtain another equation connecting $W$ and $x$. [3]
\item Calculate the value of $x$ and the value of $W$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2003 Q6 [12]}}