| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| 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 resolution of forces and taking moments about a point. Part (a) and (b) involve straightforward equilibrium equations with one geometric calculation (finding the perpendicular distance). Part (c) requires recognizing that when the rod is at the midpoint, the vertical component of the reaction must equal the weight, making the wall force horizontal. While it requires careful setup and multiple steps, it follows standard textbook methods without requiring novel insight, making it slightly easier than average. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(M(A): P \times 0.5\sin 60 = 30g \times 1.5\) | M1 A2 | |
| \(P = 90g \cdot \frac{2}{\sqrt{3}} \approx 1020 \text{ N (1000 N)}\) | A1 | |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\rightarrow \quad X = P\cos 60 = \frac{1}{2}P\) | M1 A1 | |
| \((\approx 509 \text{ N (510 N)})\) | ||
| \(\uparrow \quad Y + P\cos 30 = 30g\) | M1 A1 | |
| \((\Rightarrow Y = -588 \text{ N})\) | ||
| resultant \(= \sqrt{X^2 + Y^2} = \sqrt{509^2 + 588^2} \approx 778 \text{ N or } 780 \text{ N}\) | M1 A1 | |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| In equilibrium all forces act through a point; \(P\) and weight meet at mid-point; hence reaction also acts through mid-point so reaction is horizontal | M1 A1 cso | |
| OR \(M(\text{mid-point}): Y \times 1.5 = 0 \Rightarrow Y = 0\), hence reaction is horizontal | M1 A1 | |
| (2) |
## Question 6:
### Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $M(A): P \times 0.5\sin 60 = 30g \times 1.5$ | M1 A2 | |
| $P = 90g \cdot \frac{2}{\sqrt{3}} \approx 1020 \text{ N (1000 N)}$ | A1 | |
| | **(4)** | |
### Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\rightarrow \quad X = P\cos 60 = \frac{1}{2}P$ | M1 A1 | |
| $(\approx 509 \text{ N (510 N)})$ | | |
| $\uparrow \quad Y + P\cos 30 = 30g$ | M1 A1 | |
| $(\Rightarrow Y = -588 \text{ N})$ | | |
| resultant $= \sqrt{X^2 + Y^2} = \sqrt{509^2 + 588^2} \approx 778 \text{ N or } 780 \text{ N}$ | M1 A1 | |
| | **(6)** | |
### Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| In equilibrium all forces act through a point; $P$ and weight meet at mid-point; hence reaction also acts through mid-point so reaction is horizontal | M1 A1 cso | |
| OR $M(\text{mid-point}): Y \times 1.5 = 0 \Rightarrow Y = 0$, hence reaction is horizontal | M1 A1 | |
| | **(2)** | |
---6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{3847deb8-d86e-4254-828f-5d62f20c186f-09_442_689_292_632}
\end{center}
\end{figure}
A uniform pole $A B$, of mass 30 kg and length 3 m , is smoothly hinged to a vertical wall at one end $A$. The pole is held in equilibrium in a horizontal position by a light rod CD. One end $C$ of the rod is fixed to the wall vertically below $A$. The other end $D$ is freely jointed to the pole so that $\angle A C D = 30 ^ { \circ }$ and $A D = 0.5 \mathrm {~m}$, as shown in Figure 2. Find
\begin{enumerate}[label=(\alph*)]
\item the thrust in the rod $C D$,
\item the magnitude of the force exerted by the wall on the pole at $A$.
The rod $C D$ is removed and replaced by a longer light rod $C M$, where $M$ is the mid-point of $A B$. The rod is freely jointed to the pole at $M$. The pole $A B$ remains in equilibrium in a horizontal position.
\item Show that the force exerted by the wall on the pole at $A$ now acts horizontally.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2005 Q6 [12]}}