| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with strut or direct force support |
| Difficulty | Standard +0.3 This is a standard M2 moments problem with a rod in equilibrium supported by a strut. It requires taking moments about point A to find the thrust, then resolving forces to find the reaction at the hinge. The geometry is straightforward (3-4-5 triangle), and the method is a textbook application of equilibrium conditions. Slightly easier than average due to the simple geometry and clear structure. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Taking moments about A: \(3g \times 0.75 = \frac{T}{\sqrt{2}} \times 0.5\) | M1A1A1 | |
| \(T = 3\sqrt{2}g \times \frac{7.5}{5} = \frac{9\sqrt{2}g}{2}\ (= 62.4\text{N})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pm H = \frac{T}{\sqrt{2}}\left(= \frac{9g}{2} \approx 44.1\text{N}\right)\) | B1 | |
| \(\pm V + \frac{T}{\sqrt{2}} = 3g\ \left(\Rightarrow V = 3g - \frac{9g}{2} = \frac{-3g}{2} \approx -14.7\text{N}\right)\) | M1A1 | |
| \(\Rightarrow | R | = \sqrt{81 + 9} \times \frac{g}{2} \approx 46.5\text{(N)}\) |
| at angle \(\tan^{-1}\frac{1}{3} = 18.4°\) (0.322 radians) below line of BA | M1A1 | |
| \(161.6°\) (2.82 radians) below line of AB \((108.4°\) or 1.89 radians to upward vertical) |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Taking moments about A: $3g \times 0.75 = \frac{T}{\sqrt{2}} \times 0.5$ | M1A1A1 | |
| $T = 3\sqrt{2}g \times \frac{7.5}{5} = \frac{9\sqrt{2}g}{2}\ (= 62.4\text{N})$ | A1 | |
**Subtotal: (4)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pm H = \frac{T}{\sqrt{2}}\left(= \frac{9g}{2} \approx 44.1\text{N}\right)$ | B1 | |
| $\pm V + \frac{T}{\sqrt{2}} = 3g\ \left(\Rightarrow V = 3g - \frac{9g}{2} = \frac{-3g}{2} \approx -14.7\text{N}\right)$ | M1A1 | |
| $\Rightarrow |R| = \sqrt{81 + 9} \times \frac{g}{2} \approx 46.5\text{(N)}$ | M1A1 | |
| at angle $\tan^{-1}\frac{1}{3} = 18.4°$ (0.322 radians) below line of BA | M1A1 | |
| $161.6°$ (2.82 radians) below line of AB $(108.4°$ or 1.89 radians to upward vertical) | | |
**Subtotal: (7) Total: [11]**
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8e220b8a-46f1-4b9b-88a4-f032c7fbda50-05_568_956_205_516}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform rod $A B$, of length 1.5 m and mass 3 kg , is smoothly hinged to a vertical wall at $A$. The rod is held in equilibrium in a horizontal position by a light strut $C D$ as shown in Figure 1. The rod and the strut lie in the same vertical plane, which is perpendicular to the wall. The end $C$ of the strut is freely jointed to the wall at a point 0.5 m vertically below $A$. The end $D$ is freely joined to the rod so that $A D$ is 0.5 m .
\begin{enumerate}[label=(\alph*)]
\item Find the thrust in $C D$.
\item Find the magnitude and direction of the force exerted on the $\operatorname { rod } A B$ at $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q4 [11]}}