| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| 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 statics problem requiring moments about a point, resolution of forces, and friction in limiting equilibrium. The setup is clearly defined with given geometry (tan θ = 3/4), and the solution follows a routine three-step process: take moments about A to find tension, resolve horizontally to verify the given result, then use F = μR for limiting equilibrium. While it requires careful bookkeeping of forces and distances, it involves no novel insight beyond standard M2 techniques, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Accept 32.7m, 33m | M1*, A1=A1, DM1*, A1 | 5 marks |
| (b) \(\rightarrow\) \(R = T \cos \theta = \frac{10}{3}mg \times \frac{4}{5} = \frac{8}{3}mg\) ★ | M1 A1ft; A1 | cso, fit their T; 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = \mu R \Rightarrow \mu = \frac{3}{8}\) | M1, A1ft, M1 A1 | 4 marks; 12 marks total |
(a) $M(A)$ $T \sin \theta \times 4a = mg \times 2a + 2mg \times 3a$
$T = \frac{8mg}{4} \times \frac{5}{3} = \frac{10}{3}mg$
Accept 32.7m, 33m | M1*, A1=A1, DM1*, A1 | 5 marks
(b) $\rightarrow$ $R = T \cos \theta = \frac{10}{3}mg \times \frac{4}{5} = \frac{8}{3}mg$ ★ | M1 A1ft; A1 | cso, fit their T; 3 marks
(c) $\uparrow$ $F + T \sin \theta = 3mg \Rightarrow F = mg$ fit their T
Or: $M(B)$ $F \times 4a = mg \times 2a + 2mg \times a \Rightarrow F = mg$
$F = \mu R \Rightarrow \mu = \frac{3}{8}$ | M1, A1ft, M1 A1 | 4 marks; 12 marks total
**Alternative approach:**
$\rightarrow$ $R = T \cos \theta$
$\uparrow$ $F + T \sin \theta = 3mg$
$M(B)$ $F \times 4a = mg \times 2a + 2mg \times a$ ($\Rightarrow F = mg$)
$\Rightarrow mg + T \sin \theta = 3mg \Rightarrow T = \frac{10mg}{3}$
If they use this method, watch out for F=mg just quoted in (c): M1A1
---\includegraphics{figure_2}
A horizontal uniform rod $AB$ has mass $m$ and length $4a$. The end $A$ rests against a rough vertical wall. A particle of mass $2m$ is attached to the rod at the point $C$, where $AC = 3a$. One end of a light inextensible string $BD$ is attached to the rod at $B$ and the other end is attached to the wall at a point $D$, where $D$ is vertically above $A$. The rod is in equilibrium in a vertical plane perpendicular to the wall. The string is inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac{3}{4}$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string. [5]
\item Show that the horizontal component of the force exerted by the wall on the rod has magnitude $\frac{5}{8}mg$. [3]
\end{enumerate}
The coefficient of friction between the wall and the rod is $\mu$. Given that the rod is in limiting equilibrium,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the value of $\mu$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q5 [12]}}