| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| 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 moments equilibrium problem requiring taking moments about a point and resolving forces. Part (a) involves routine moment calculation about point A (5 marks for a 'show that' is generous). Parts (b) and (c) use vertical force equilibrium and horizontal force equilibrium respectively. All steps follow standard mechanics procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| - \(T = \frac{5(5a-2x)}{6a}W\) (cso) | M1, A2(1,0) | M1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\frac{5}{6}W + 2a = Wa + 2Wx\), \(x = \frac{2}{3}a\) | M1, B1 | O.E. |
| Answer | Marks | Guidance |
|---|---|---|
| - \(x = T\cos\theta = \frac{5}{6}(5-\frac{4}{5})W \times \frac{7}{5} = \frac{22}{9}W\) | M1, B1, M1, B1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| - \(x = \frac{2}{3}a\) | M1, B1 | A1 |
## (a)
**Answer/Working:**
- Moments about D: $T \times 2a\sin\theta = Wa + 2W(2a-x)$
- Moments about C: $T \times \frac{5}{6}a = 5Wa - 2Wx$
- $T = \frac{5(5a-2x)}{6a}W$ (cso) | M1, A2(1,0) | M1, B1 | 5
## (b)
**Answer/Working:** $\frac{5}{6}W + 2a = Wa + 2Wx$, $x = \frac{2}{3}a$ | M1, B1 | O.E. | A1 | 3
## (c)
**Answer/Working:**
- $x = T\cos\theta = \frac{5}{6}(5-\frac{4}{5})W \times \frac{7}{5} = \frac{22}{9}W$ | M1, B1, M1, B1 | A1 | 4
**Alternative to (c):**
- **Working:** $R(→)$: $\frac{7}{6}W + T\sin\theta = 3W$
- $\frac{7}{6}W + \frac{5(5a-2x)}{6a} \times \frac{3}{5} = 3W$
- $x = \frac{2}{3}a$ | M1, B1 | A1 | 3
---\includegraphics{figure_2}
Figure 2 shows a horizontal uniform pole $AB$, of weight $W$ and length $2a$. The end $A$ of the pole rests against a rough vertical wall. One end of a light inextensible string $BD$ is attached to the pole at $B$ and the other end is attached to the wall at $D$. A particle of weight $2W$ is attached to the pole at $C$, where $BC = x$. The pole is in equilibrium in a vertical plane perpendicular to the wall. The string $BD$ is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{3}{5}$. The pole is modelled as a uniform rod.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in $BD$ is $\frac{5(5a - 2x)}{6a}W$.
[5]
\end{enumerate}
The vertical component of the force exerted by the wall on the pole is $\frac{7}{4}W$. Find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item $x$ in terms of $a$,
[3]
\item the horizontal component, in terms of $W$, of the force exerted by the wall on the pole.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2002 Q5 [12]}}