| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder or rod with friction at both contacts |
| Difficulty | Standard +0.8 This is a standard ladder equilibrium problem requiring resolution of forces in two directions, taking moments about a point, and applying friction laws at two surfaces simultaneously. While it involves multiple steps and careful algebraic manipulation to reach the given result, it follows a well-established method taught in M2. The presence of friction at both contacts and the need to coordinate limiting equilibrium conditions at both surfaces elevates it slightly above average M2 difficulty, but it remains a recognizable textbook-style problem. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| resolve \(\uparrow\): \(\frac{1}{3}S + R - mg = 0\) | M1 | |
| resolve \(\rightarrow\): \(\frac{1}{3}R - S = 0\) | M1 | |
| solve simul. giving \(S = \frac{4}{9}R\); \(R = \frac{9}{7}mg\) | M1 A1 | |
| mom. about top of ladder: \(R.2a\cos\theta - \frac{1}{3}R.2a\sin\theta - mg.a\cos\theta = 0\) | M1 A1 | |
| \(\therefore \tan\theta = \frac{2R-mg}{\frac{1}{3}R} = \frac{5}{4}\) | M2 A1 | (9) |
resolve $\uparrow$: $\frac{1}{3}S + R - mg = 0$ | M1 |
resolve $\rightarrow$: $\frac{1}{3}R - S = 0$ | M1 |
solve simul. giving $S = \frac{4}{9}R$; $R = \frac{9}{7}mg$ | M1 A1 |
mom. about top of ladder: $R.2a\cos\theta - \frac{1}{3}R.2a\sin\theta - mg.a\cos\theta = 0$ | M1 A1 |
$\therefore \tan\theta = \frac{2R-mg}{\frac{1}{3}R} = \frac{5}{4}$ | M2 A1 | (9)
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f8ece90a-5042-4db1-9855-ffe74333a899-3_407_341_201_635}
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\caption{Fig. 1}
\end{center}
\end{figure}
Figure 1 shows a uniform ladder of mass $m$ and length $2 a$ resting against a rough vertical wall with its lower end on rough horizontal ground. The coefficient of friction between the ladder and the wall is $\frac { 1 } { 2 }$ and the coefficient of friction between the ladder and the ground is $\frac { 1 } { 3 }$.
Given that the ladder is in limiting equilibrium when it is inclined at an angle $\theta$ to the horizontal, show that $\tan \theta = \frac { 5 } { 4 }$.\\
(9 marks)\\
\hfill \mbox{\textit{Edexcel M2 Q4 [9]}}