| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| 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 with a horizontal rod supported by an angled strut. Part (a) requires taking moments about the hinge, part (b) needs resolving forces, and part (c) involves showing a specific geometric result. While it requires multiple steps and careful force resolution, it follows a very standard template for this topic with no novel insights required—slightly easier than average for M2. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Mass ratios: \(ABC = M\), \(ADE = \frac{4M}{9}\), \(BCED = \frac{5M}{9}\) | B1 | Correct mass ratios |
| Distance ratios: \(ABC = \frac{h}{3}\), \(ADE = \left(\frac{h}{3} + \frac{1}{3}\cdot\frac{2h}{3}\right)\), \(BCED = \bar{y}\) | B1 | Correct distance ratios |
| Moments equation | M1 | Condone sign slip |
| \(M\frac{h}{3} - \frac{4M}{9}\cdot\frac{5h}{9} = \frac{5M}{9}\bar{y}\) | A1 | |
| \(\bar{y} = \frac{7h}{45}\) Answer Given | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Moments equation for folded shape | M1 | Requires correct mass ratios and terms of correct structure |
| \(\frac{5M}{9}\cdot\frac{7h}{45} + \frac{4M}{9}\left(\frac{h}{3} - \frac{1}{3}\times\frac{2h}{3}\right) = M\bar{x}\) | A1 A1 | \(-1\) each error; \(\frac{h}{9}\) |
| \(\bar{x} = \frac{11h}{81}\) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\tan\alpha = \dfrac{\frac{h}{3} - \bar{x}}{\frac{2a}{3}}\) | M1 | Use of tan in correct triangle. Allow reciprocal |
| A1 ft | Correct unsimplified for their \(\bar{x}\) | |
| \(= \dfrac{8h}{27a}\) | DM1, A1 (4) | Substitute and simplify |
# Question 6:
## Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| Mass ratios: $ABC = M$, $ADE = \frac{4M}{9}$, $BCED = \frac{5M}{9}$ | B1 | Correct mass ratios |
| Distance ratios: $ABC = \frac{h}{3}$, $ADE = \left(\frac{h}{3} + \frac{1}{3}\cdot\frac{2h}{3}\right)$, $BCED = \bar{y}$ | B1 | Correct distance ratios |
| Moments equation | M1 | Condone sign slip |
| $M\frac{h}{3} - \frac{4M}{9}\cdot\frac{5h}{9} = \frac{5M}{9}\bar{y}$ | A1 | |
| $\bar{y} = \frac{7h}{45}$ **Answer Given** | A1 **(5)** | |
## Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| Moments equation for folded shape | M1 | Requires correct mass ratios and terms of correct structure |
| $\frac{5M}{9}\cdot\frac{7h}{45} + \frac{4M}{9}\left(\frac{h}{3} - \frac{1}{3}\times\frac{2h}{3}\right) = M\bar{x}$ | A1 A1 | $-1$ each error; $\frac{h}{9}$ |
| $\bar{x} = \frac{11h}{81}$ | A1 **(4)** | |
## Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\tan\alpha = \dfrac{\frac{h}{3} - \bar{x}}{\frac{2a}{3}}$ | M1 | Use of tan in correct triangle. Allow reciprocal |
| | A1 ft | Correct unsimplified for their $\bar{x}$ |
| $= \dfrac{8h}{27a}$ | DM1, A1 **(4)** | Substitute and simplify |
---6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{173a2029-a0b8-437f-9339-5a1b6f30a8e3-010_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 Q6}}