| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder on smooth wall and rough ground |
| Difficulty | Standard +0.3 This is a standard M2 ladder problem requiring three equilibrium equations (horizontal/vertical forces and moments) with given coefficient of friction. The setup is straightforward with clear numerical values, and part (c) is a routine modelling statement. Slightly easier than average due to the direct application of standard methods without requiring novel insight or complex algebraic manipulation. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03u Static equilibrium: on rough surfaces6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R(\uparrow): R = 25g + 75g (= 100g)\) | B1 | |
| \(F = \mu R \Rightarrow F = \dfrac{11}{25} \times 100g\) | M1 | |
| \(= 44g\ (= 431)\) | A1 | |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(A): 25g \times 2\cos\beta + 75g \times 2.8\cos\beta = S \times 4\sin\beta\) | M1, A2,1,0 | |
| \(R(\leftrightarrow): F = S\) | ||
| \(176g\sin\beta = 260g\cos\beta\) | M1A1 | |
| \(\beta = 56(°)\) | A1 | |
| Total: (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| So that Reece's weight acts directly at the point \(C\). | B1 | |
| Total: [10] |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R(\uparrow): R = 25g + 75g (= 100g)$ | B1 | |
| $F = \mu R \Rightarrow F = \dfrac{11}{25} \times 100g$ | M1 | |
| $= 44g\ (= 431)$ | A1 | |
| **Total: (3)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(A): 25g \times 2\cos\beta + 75g \times 2.8\cos\beta = S \times 4\sin\beta$ | M1, A2,1,0 | |
| $R(\leftrightarrow): F = S$ | | |
| $176g\sin\beta = 260g\cos\beta$ | M1A1 | |
| $\beta = 56(°)$ | A1 | |
| **Total: (6)** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| So that Reece's weight acts directly at the point $C$. | B1 | |
| **Total: [10]** | | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c8ebad3-0ebb-4dfe-8036-54b651deb9cf-03_602_554_205_712}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a ladder $A B$, of mass 25 kg and length 4 m , resting in equilibrium with one end $A$ on rough horizontal ground and the other end $B$ against a smooth vertical wall. The ladder is in a vertical plane perpendicular to the wall. The coefficient of friction between the ladder and the ground is $\frac { 11 } { 25 }$. The ladder makes an angle $\beta$ with the ground. When Reece, who has mass 75 kg , stands at the point $C$ on the ladder, where $A C = 2.8 \mathrm {~m}$, the ladder is on the point of slipping. The ladder is modelled as a uniform rod and Reece is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the frictional force of the ground on the ladder.
\item Find, to the nearest degree, the value of $\beta$.
\item State how you have used the modelling assumption that Reece is a particle.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q2 [10]}}