| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.3 This is a standard M2 moments problem requiring taking moments about the hinge and resolving forces. The geometry is straightforward (30° angle given), and the method is routine: moments about A to find length, then resolve horizontally and vertically to find reaction components. Slightly easier than average due to clean numbers and standard setup. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(M(A): \quad 63\sin 30 \cdot 14 = 2g \cdot d\) | M1 A1 A1 | M1: take moments about A; 2 recognisable force × distance terms involving 63 and 2(g). A1: 63N term correct. A1: 2g term correct |
| Solve: \(d = 0.225\) m | ||
| Hence \(AB = 45\) cm | A1 | \(AB = 0.45\)(m) or 45(cm); no more than 2sf due to use of \(g\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R(\rightarrow): \quad X = 63\cos 30 \approx 54.56\) | B1 | Horizontal component (correct expression – no need to evaluate) |
| \(R(\uparrow): \quad Y = 63\sin 30 - 2g \approx 11.9\) | M1 A1 | Resolve vertically – 3 terms needed; condone sign errors; could have cos for sin. A1: correct expression (not necessarily evaluated); direction of Y does not matter. Alternatively, take moments about B or C |
| \(R = \sqrt{X^2 + Y^2} \approx 55.8, 55.9\) or \(56\) N | M1 A1 | M1: correct use of Pythagoras. A1: 55.8(N), 55.9(N) or 56(N). OR for X and Y expressed as \(F\cos\theta\) and \(F\sin\theta\): M1 square and add, or find \(\tan\theta\) and substitute |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $M(A): \quad 63\sin 30 \cdot 14 = 2g \cdot d$ | M1 A1 A1 | M1: take moments about A; 2 recognisable force × distance terms involving 63 and 2(g). A1: 63N term correct. A1: 2g term correct |
| Solve: $d = 0.225$ m | | |
| Hence $AB = 45$ cm | A1 | $AB = 0.45$(m) or 45(cm); no more than 2sf due to use of $g$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R(\rightarrow): \quad X = 63\cos 30 \approx 54.56$ | B1 | Horizontal component (correct expression – no need to evaluate) |
| $R(\uparrow): \quad Y = 63\sin 30 - 2g \approx 11.9$ | M1 A1 | Resolve vertically – 3 terms needed; condone sign errors; could have cos for sin. A1: correct expression (not necessarily evaluated); direction of Y does not matter. Alternatively, take moments about B or C |
| $R = \sqrt{X^2 + Y^2} \approx 55.8, 55.9$ or $56$ N | M1 A1 | M1: correct use of Pythagoras. A1: 55.8(N), 55.9(N) or 56(N). OR for X and Y expressed as $F\cos\theta$ and $F\sin\theta$: M1 square and add, or find $\tan\theta$ and substitute |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{778a0276-6738-40e6-90b2-a536ce5abe6a-08_376_874_205_525}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform beam $A B$ of mass 2 kg is freely hinged at one end $A$ to a vertical wall. The beam is held in equilibrium in a horizontal position by a rope which is attached to a point $C$ on the beam, where $A C = 0.14 \mathrm {~m}$. The rope is attached to the point $D$ on the wall vertically above $A$, where $\angle A C D = 30 ^ { \circ }$, as shown in Figure 3. The beam is modelled as a uniform rod and the rope as a light inextensible string. The tension in the rope is 63 N .
Find
\begin{enumerate}[label=(\alph*)]
\item the length of $A B$,
\item the magnitude of the resultant reaction of the hinge on the beam at $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q5 [9]}}