| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.8 This is a multi-part statics problem requiring resolution of forces, moments about a point, and analysis of limiting equilibrium. While it involves standard M2 techniques (resolving forces, taking moments, finding resultants), the three-part structure with the painter climbing adds complexity. The calculation requires careful handling of the geometry (tan α = 3/4 means sin α = 3/5, cos α = 4/5), and part (c) involves setting up a moment equation with a variable distance. This is more demanding than a routine statics question but doesn't require exceptional insight—it's a solid exam question testing multiple skills in sequence. |
| 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 |
|---|---|---|
| (a) \(R = mg\), \(F = S\) | B1 B1 M1 A1 | |
| \(M(B): mga \cos \alpha = 2a S \sin \alpha\) | M1 | |
| \(S = \frac{mg}{2 \tan \alpha} = \frac{2mg}{3}\) | A1 | |
| \(F = S\) | A1 | |
| Resultant force at \(B = \sqrt{(mg)^2 + \left(\frac{2mg}{3}\right)^2} = \frac{\sqrt{13}}{3}mg\) | M1 A1 | |
| (b) Angle \(= \tan^{-1}(3/2) = 56°\) to horizontal | M1 A1 A1 | |
| (c) \(M(B): mga \cos \alpha + 6mgx \cos \alpha = 2a S \sin \alpha\) | M1 A1 A1 | |
| \(S = \frac{2mg(a + 6x)}{3a}\) | A1 M1 A1 | |
| When \(S = 2mg\), \(a + 6x = 3a\) | A1 M1 A1 | |
| \(6x = 2a\) | A1 | |
| \(x = \frac{a}{3}\) | A1 | (17 marks) |
(a) $R = mg$, $F = S$ | B1 B1 M1 A1 |
$M(B): mga \cos \alpha = 2a S \sin \alpha$ | M1 |
$S = \frac{mg}{2 \tan \alpha} = \frac{2mg}{3}$ | A1 |
$F = S$ | A1 |
Resultant force at $B = \sqrt{(mg)^2 + \left(\frac{2mg}{3}\right)^2} = \frac{\sqrt{13}}{3}mg$ | M1 A1 |
(b) Angle $= \tan^{-1}(3/2) = 56°$ to horizontal | M1 A1 A1 |
(c) $M(B): mga \cos \alpha + 6mgx \cos \alpha = 2a S \sin \alpha$ | M1 A1 A1 |
$S = \frac{2mg(a + 6x)}{3a}$ | A1 M1 A1 |
When $S = 2mg$, $a + 6x = 3a$ | A1 M1 A1 |
$6x = 2a$ | A1 |
$x = \frac{a}{3}$ | A1 | (17 marks)\includegraphics{figure_7}
A uniform ladder $AB$, of mass $m$ kg and length $2a$ m, rests with its upper end $A$ in contact with a smooth vertical wall and its lower end $B$ in contact with a fixed peg on horizontal ground. The ladder makes an angle $\alpha$ with the ground, where $\tan \alpha = \frac{3}{4}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the resultant force acting on the ladder at $B$ is $\frac{\sqrt{13}}{3}mg$. [7 marks]
\item Find, to the nearest degree, the direction of this resultant force at $B$. [3 marks]
\end{enumerate}
The peg will break when the horizontal force acting on it exceeds $2mg$ N.
A painter of mass $6m$ kg starts to climb the ladder from $B$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find, in terms of $a$, the greatest distance up the ladder that the painter can safely climb. [7 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [17]}}