| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod or block on rough surface in limiting equilibrium (no wall) |
| Difficulty | Standard +0.8 This is a challenging M2 statics problem requiring resolution of forces in two directions, taking moments about a point, and applying limiting equilibrium conditions with friction. The geometry involves non-standard angles (tan α = 5/12) and an unknown cable angle β. Part (a) requires careful moment calculation (6 marks suggests multiple steps), part (b) needs simultaneous equations from force resolution, and the multi-part structure with interdependent solutions elevates this above routine mechanics questions. |
| Spec | 3.03v Motion on rough surface: including inclined planes3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 0.6R\) (seen anywhere) | ||
| Moments about \(B\): \(R \times 2a \cos \alpha + F \times 2a \sin \alpha = W \times a \cos \alpha\) | ||
| Using \(\cos \alpha = \frac{12}{13}\) and \(\sin \alpha = \frac{5}{13}\) | ||
| Solving for \(R\): \(\frac{24}{13}R + \frac{6}{13}R = \frac{12}{13}W \Rightarrow 30R = 12\) | ||
| \(\Rightarrow R = \frac{2}{5}W\) * |
| Answer | Marks | Guidance |
|---|---|---|
| Resolve \(\leftrightarrow\): \(T \cos \beta = F\); \(0.6R = \frac{6}{25}W\) | ||
| Resolve \(\uparrow\): \(T \sin \beta + R = W\) | ||
| \(T \sin \beta = \frac{3}{5}W\) | ||
| Complete method for \(\beta\) [e.g \(\tan \beta = 2.5\)]; \(\beta = 68.2°\) | ||
| Complete method for \(T\): substitute for \(\beta\) or \(\sqrt{(0.6W)^2 + (0.24W)^2}\) | ||
| \(T = 0.646...W \Rightarrow k = 0.65\) or \(0.646\) |
## Part (a)
| $F = 0.6R$ (seen anywhere) | | | | M1 |
| Moments about $B$: $R \times 2a \cos \alpha + F \times 2a \sin \alpha = W \times a \cos \alpha$ | | | | M1 A1 |
| Using $\cos \alpha = \frac{12}{13}$ and $\sin \alpha = \frac{5}{13}$ | | | | M1 |
| Solving for $R$: $\frac{24}{13}R + \frac{6}{13}R = \frac{12}{13}W \Rightarrow 30R = 12$ | | | | M1 |
| $\Rightarrow R = \frac{2}{5}W$ * | | | | A1 (6) |
## Part (b)
| Resolve $\leftrightarrow$: $T \cos \beta = F$; $0.6R = \frac{6}{25}W$ | | | | M1 A1 |
| Resolve $\uparrow$: $T \sin \beta + R = W$ | | | | M1 A1 |
| | $T \sin \beta = \frac{3}{5}W$ | | | | |
| Complete method for $\beta$ [e.g $\tan \beta = 2.5$]; $\beta = 68.2°$ | | | | M1; A1 (6) |
| Complete method for $T$: substitute for $\beta$ or $\sqrt{(0.6W)^2 + (0.24W)^2}$ | | | | M1 |
| $T = 0.646...W \Rightarrow k = 0.65$ or $0.646$ | | | | A1 (2) |
**Total: 14 marks**\includegraphics{figure_3}
A straight log $AB$ has weight $W$ and length $2a$. A cable is attached to one end $B$ of the log. The cable lifts the end $B$ off the ground. The end $A$ remains in contact with the ground, which is rough and horizontal. The log is in limiting equilibrium. The log makes an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac{5}{12}$. The cable makes an angle $\beta$ to the horizontal, as shown in Fig. 3. The coefficient of friction between the log and the ground is 0.6. The log is modelled as a uniform rod and the cable as light.
\begin{enumerate}[label=(\alph*)]
\item Show that the normal reaction on the log at $A$ is $\frac{5}{8}W$. [6]
\item Find the value of $\beta$. [6]
\end{enumerate}
The tension in the cable is $kW$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $k$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2002 Q7 [14]}}