| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with rough contact at free end |
| Difficulty | Standard +0.3 This is a standard M2 moments equilibrium problem with three parts requiring systematic application of resolving forces and taking moments. Part (a) uses moments about A to find x (routine), part (b) resolves horizontally (straightforward), and part (c) applies friction law F=μR (standard). All techniques are textbook exercises with no novel insight required, making it slightly easier than average for M2. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| M(A) \(3mg \times 2a + 3mgx = T \cos \theta \times 4a\) | M1 A2, 1.0 | |
| \(= \frac{12}{5}aT\) | ||
| \(\frac{12}{5}aT = 6mga + 3mgx\) | ||
| \(T = \frac{25}{4}mg\) \(\frac{12}{5} \times \frac{25}{4}mg = 6mga + 3mgx\) | M1 | |
| \(15a = 6a + 3x\) | ||
| \(x = 3a\) ** | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| R(\(\to\)) \(R = T \sin \theta\) | M1 | |
| \(= \frac{25}{4}mg \times \frac{4}{5}\) | A1 | |
| \(= 5mg\) ** | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| R(\(\uparrow\)) \(F + \frac{25}{4}mg \times \frac{3}{5} = 3mg + 3mg\) | M1 A2, 1.0 | |
| \(F = 6mg - \frac{15}{4}mg = -\frac{9}{4}mg\) | ||
| \(\mu = \frac{F}{R} = \frac{\frac{9}{4}mg}{5mg} = \frac{9}{20}\) | DM1 A1 |
## Part (a):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| M(A) $3mg \times 2a + 3mgx = T \cos \theta \times 4a$ | M1 A2, 1.0 | |
| | $= \frac{12}{5}aT$ | |
| $\frac{12}{5}aT = 6mga + 3mgx$ | | |
| $T = \frac{25}{4}mg$ $\frac{12}{5} \times \frac{25}{4}mg = 6mga + 3mgx$ | M1 | |
| $15a = 6a + 3x$ | | |
| $x = 3a$ ** | A1 | |
## Part (b):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| R($\to$) $R = T \sin \theta$ | M1 | |
| $= \frac{25}{4}mg \times \frac{4}{5}$ | A1 | |
| $= 5mg$ ** | A1 | |
## Part (c):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| R($\uparrow$) $F + \frac{25}{4}mg \times \frac{3}{5} = 3mg + 3mg$ | M1 A2, 1.0 | |
| $F = 6mg - \frac{15}{4}mg = -\frac{9}{4}mg$ | | |
| $\mu = \frac{F}{R} = \frac{\frac{9}{4}mg}{5mg} = \frac{9}{20}$ | DM1 A1 | |\includegraphics{figure_3}
A uniform rod $AB$, of mass $3m$ and length $4a$, is held in a horizontal position with the end $A$ against a rough vertical wall. One end of a light inextensible string $BD$ is attached to the rod at $B$ and the other end of the string is attached to the wall at the point $D$ vertically above $A$, where $AD = 3a$. A particle of mass $3m$ is attached to the rod at $C$, where $AC = x$. The rod is in equilibrium in a vertical plane perpendicular to the wall as shown in Figure 3. The tension in the string is $\frac{25}{4}mg$.
Show that
\begin{enumerate}[label=(\alph*)]
\item $x = 3a$, [5]
\item the horizontal component of the force exerted by the wall on the rod has magnitude $5mg$. [3]
\end{enumerate}
The coefficient of friction between the wall and the rod is $\mu$. Given that the rod is about to slip,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the value of $\mu$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2011 Q7 [13]}}