| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 7 |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard ladder equilibrium problem requiring resolution of forces and taking moments, which are core M1 techniques. Part (a) is a 'show that' requiring straightforward moment calculation about point B. Part (b) applies the friction law F ≤ μR. While it involves multiple steps and careful algebra with the given tan θ = 3, it follows a well-established method with no novel insight required, making it slightly easier than average. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(2(150\cos\theta) + x(750\cos\theta) = 4(R_B \sin\theta)\) | M1 | Attempt moments e.g. about A |
| \(A_2\) | A1 for any two terms correct | |
| \(R_B = F_A \Rightarrow F_A = \frac{25}{2}(2 + 5x)\) | A1 | AG – sufficient working must be shown to justify the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(R_A = 150 + 750\) | B1 | Resolving vertically |
| \(\frac{25}{2}(2 + 5x) \leq \frac{1}{4}(900)\) | M1 | Correct use of \(F_A \leq \mu R_A\) |
| \(x \leq 3.2 \text{ so the greatest value of } x \text{ is } 3.2\) | A1 | Must have maximum value of \(x\) explicitly stated. |
## Part (a)
$2(150\cos\theta) + x(750\cos\theta) = 4(R_B \sin\theta)$ | M1 | Attempt moments e.g. about A
$A_2$ | A1 for any two terms correct
$R_B = F_A \Rightarrow F_A = \frac{25}{2}(2 + 5x)$ | A1 | AG – sufficient working must be shown to justify the given answer
## Part (b)
$R_A = 150 + 750$ | B1 | Resolving vertically
$\frac{25}{2}(2 + 5x) \leq \frac{1}{4}(900)$ | M1 | Correct use of $F_A \leq \mu R_A$ | Allow equals throughout
$x \leq 3.2 \text{ so the greatest value of } x \text{ is } 3.2$ | A1 | Must have maximum value of $x$ explicitly stated.A uniform ladder $AB$, of weight $150\text{N}$ and length $4\text{m}$, rests in equilibrium with the end $A$ in contact with rough horizontal ground and the end $B$ resting against a smooth vertical wall. The ladder is inclined at an angle $\theta$ to the horizontal, where $\tan \theta = 3$. A man of weight $750\text{N}$ is standing on the ladder at a distance $x\text{m}$ from $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the frictional force exerted by the ground on the ladder is $\frac{75}{2}(2 + 5x)\text{N}$. [4]
\end{enumerate}
The coefficient of friction between the ladder and the ground is $\frac{1}{4}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the greatest value of $x$ for which equilibrium is possible. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q8 [7]}}