| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against smooth wall in limiting equilibrium |
| Difficulty | Standard +0.3 This is a standard M2 statics problem requiring resolution of forces and taking moments about a point. While it involves multiple steps (resolving horizontally/vertically, taking moments, using friction law), these are routine techniques practiced extensively in M2. The 'show that' in part (a) guides students to the answer, and part (b) follows directly from standard equilibrium equations. Slightly easier than average due to its predictable structure. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force |
3 A uniform ladder, of length 6 metres and mass 22 kg , rests with its foot, $A$, on a rough horizontal floor and its top, $B$, leaning against a smooth vertical wall. The vertical plane containing the ladder is perpendicular to the wall, and the angle between the ladder and the floor is $\theta$.
A man, of mass 90 kg , is standing at point $C$ on the ladder so that the distance $A C$ is 5 metres. With the man in this position, the ladder is on the point of slipping. The coefficient of friction between the ladder and the horizontal floor is 0.6 . The man may be modelled as a particle at $C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{9cfa110c-ee11-447a-b21a-3f436432e27d-3_707_702_742_646}
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the frictional force between the ladder and the horizontal floor is 659 N , correct to three significant figures.
\item Find the angle $\theta$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q3 [9]}}