| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Moderate -0.3 This is a standard M2 moments question requiring taking moments about point A to find tension, then resolving forces for the reaction. The setup is straightforward with given angle information (tan α = 4/3 means sin α = 4/5, cos α = 3/5), and the multi-part structure guides students through the solution methodically. While it requires careful bookkeeping of forces and distances, it involves routine application of equilibrium principles without novel problem-solving insight. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow T = 100g = 980 \text{ N}\) | M1 A2, B1; M1 A1 | (5) |
| Answer | Marks |
|---|---|
| \((\rightarrow): X = T \cos \alpha\) | B1 |
| \((\uparrow) Y + T \sin \alpha = 100g\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 877 \text{ N}\) (3 sf) | M1 A1; M1 A1; A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Cable light \(\Rightarrow\) tension same throughout \(\Rightarrow\) force on rod at \(D\) is \(60g\) | B1 | (1) |
## Part (a)
Taking moments about A: $40g \times \frac{3}{2} + 60g \times 2 = T \sin \alpha \times 3$
Using $\sin \alpha = \frac{3}{5}$:
$60g + 120g = \frac{9T}{5}$
$\Rightarrow T = 100g = 980 \text{ N}$ | M1 A2, B1; M1 A1 | (5)
## Part (b)
$(\rightarrow): X = T \cos \alpha$ | B1
$(\uparrow) Y + T \sin \alpha = 100g$ | M1 A1
$R = \sqrt{X^2 + Y^2} = \sqrt{(784^2 + 392^2)}$
$= 877 \text{ N}$ (3 sf) | M1 A1; M1 A1; A1 | (6)
## Part (c)
Cable light $\Rightarrow$ tension same throughout $\Rightarrow$ force on rod at $D$ is $60g$ | B1 | (1)
**Total: (12 marks)**
---\includegraphics{figure_2}
A uniform steel girder $AB$, of mass 40 kg and length 3 m, is freely hinged at $A$ to a vertical wall. The girder is supported in a horizontal position by a steel cable attached to the girder at $B$. The other end of the cable is attached to the point $C$ vertically above $A$ on the wall, with $\angle ABC = \alpha$, where $\tan \alpha = \frac{4}{3}$. A load of mass 60 kg is suspended by another cable from the girder at the point $D$, where $AD = 2$ m, as shown in Fig. 2. The girder remains horizontal and in equilibrium. The girder is modelled as a rod, and the cables as light inextensible strings.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the cable $BC$ is 980 N.
[5]
\item Find the magnitude of the reaction on the girder at $A$.
[6]
\item Explain how you have used the modelling assumption that the cable at $D$ is light.
[1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2003 Q4 [12]}}