| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod with end on ground or wall supported by string |
| Difficulty | Standard +0.8 This is a multi-part ladder equilibrium problem requiring force diagrams, resolving forces in two directions, taking moments about a strategic point, and using the friction law. Part (b) involves algebraic manipulation with the given angle, and part (c) requires physical insight about optimal rope placement. More demanding than standard M2 questions due to the rope configuration and the need to work with tan θ = 5/2 throughout, but still follows established ladder problem methodology. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Diagram showing: \(\mu S\) (vertical), \(S\) (horizontal), \(R\) (vertical), \(T\) (horizontal at angle), \(20g\) (vertical downward) | B2 | |
| (b) resolve \(\uparrow\): \(R + \mu S - 20g = 0 \quad \therefore R = 20g - \mu S\) | M1 | |
| resolve \(\rightarrow\): \(T - S = 0 \quad \therefore S = T\) | M1 | |
| eliminating \(S\) gives: \(R = 20g - \frac{1}{4}T\) | A1 | |
| mom. about top of ladder: \(T(4\sin\theta) + 20g(3\cos\theta) - R(6\cos\theta) = 0\) | M1 A1 | |
| \(47\tan\theta + 60g - 6R = 0\) | M1 | |
| \(10T + 60g - 120g + 2T = 0 \quad \therefore 12T = 60g\) and \(T = 5g\) | A1 | |
| (c) attach rope lower down ladder/wall gives larger moment about top of ladder with same tension | B1 B1 | (11) |
**(a)** Diagram showing: $\mu S$ (vertical), $S$ (horizontal), $R$ (vertical), $T$ (horizontal at angle), $20g$ (vertical downward) | B2 |
**(b)** resolve $\uparrow$: $R + \mu S - 20g = 0 \quad \therefore R = 20g - \mu S$ | M1 |
resolve $\rightarrow$: $T - S = 0 \quad \therefore S = T$ | M1 |
eliminating $S$ gives: $R = 20g - \frac{1}{4}T$ | A1 |
mom. about top of ladder: $T(4\sin\theta) + 20g(3\cos\theta) - R(6\cos\theta) = 0$ | M1 A1 |
$47\tan\theta + 60g - 6R = 0$ | M1 |
$10T + 60g - 120g + 2T = 0 \quad \therefore 12T = 60g$ and $T = 5g$ | A1 |
**(c)** attach rope lower down ladder/wall gives larger moment about top of ladder with same tension | B1 B1 | (11)3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ef2dd10c-5a3c-4868-af00-aaf7f2937d7e-2_421_474_1080_664}
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\caption{Fig. 1}
\end{center}
\end{figure}
Figure 1 shows a ladder of mass 20 kg and length 6 m leaning against a rough vertical wall with its lower end on smooth horizontal ground. The ladder is prevented from slipping along the ground by a light rope which is attached to the ladder 2 m from its bottom end and fastened to the wall so that the rope is horizontal and perpendicular to the wall.
The ladder is at an angle $\theta$ to the horizontal where $\tan \theta = \frac { 5 } { 2 }$ and the coefficient of friction between the ladder and the wall is $\frac { 1 } { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram showing all the forces acting on the ladder.
\item Show that the magnitude of the tension in the rope is $5 g$.
A man wishes to use the ladder but fears the rope will snap as he climbs the ladder.
\item Suggest, giving a reason for your answer, a more suitable position for the rope.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [11]}}