| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard M2 ladder equilibrium problem requiring resolution of forces, taking moments, and applying friction laws. While it involves multiple parts and algebraic manipulation, the techniques are routine for M2 students: resolve horizontally/vertically, take moments about a point, apply F ≤ μR, and solve. The given result in part (i) guides students through the setup, and parts (ii)-(iii) follow standard patterns. Slightly easier than average due to the structured guidance. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(Fr = R_w\) | B1 | Resolving horizontally |
| \(M1\) | Moments about \(A\) all forces accounted for | |
| \(Wa \cos \theta + 6Wx \cos \theta = 2aR_w \sin \theta\) | A1 | A1 for at least two terms correct, may involve \(\theta\) |
| A1 | Fully correct equation without \(\theta\) | |
| \(Fr = \frac{5W}{24}\left(1 + \frac{6x}{a}\right)\) | A1 [5] | AG Correctly shown |
| OR | ||
| \(R_g = 7W\) | B1 | Resolving vertically |
| M1 | Moments about \(B\) all forces accounted for | |
| \(Wa \cos \theta + 6W(2a - x) \cos \theta + 2aFr \sin \theta = 2aR_g \cos \theta\) | A1 A1 | A1 for two terms correct |
| \(Fr = \frac{5W}{24}\left(1 + \frac{6x}{a}\right)\) | A1 | AG Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(R_g = 7W\) | B1 | Resolving vertically (maybe seen in (i) and used in (ii)) |
| \(\frac{5W}{24}\left(1 + \frac{6x}{a}\right) = \frac{1}{3}(7W)\) | M1 | Use of \(Fr = \mu R\) with candidates \(R_g\); allow \(\leq\), \(<\), \(>\), \(\geq\) |
| \(x = 1.7a\) | A1 [3] | \(17a/10\), \(51a/30\); must be \(=\) or \(\leq\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{Wa\cos\alpha + 6W(2a)\cos\alpha}{2a\sin\alpha} = \frac{1}{3}(7W)\) (oe for moments about \(B\)) | M1 | Setting \(x = 2a\) in their dimensionally correct moments equation and use of \(Fr = \mu R\) allow \(\leq\), \(<\), \(>\), \(\geq\) |
| \(\tan \alpha = \frac{39}{14}\) | A1 | \(\tan \alpha = 2.7857142...\); must be \(=\) or \(\geq\) |
| [3] |
## (i)
$Fr = R_w$ | B1 | Resolving horizontally
$M1$ | | Moments about $A$ all forces accounted for
**$Wa \cos \theta + 6Wx \cos \theta = 2aR_w \sin \theta$** | A1 | A1 for at least two terms correct, may involve $\theta$
| A1 | Fully correct equation without $\theta$
$Fr = \frac{5W}{24}\left(1 + \frac{6x}{a}\right)$ | A1 [5] | AG Correctly shown
OR | |
$R_g = 7W$ | B1 | Resolving vertically
| M1 | Moments about $B$ all forces accounted for
**$Wa \cos \theta + 6W(2a - x) \cos \theta + 2aFr \sin \theta = 2aR_g \cos \theta$** | A1 A1 | A1 for two terms correct
$Fr = \frac{5W}{24}\left(1 + \frac{6x}{a}\right)$ | A1 | AG Correctly shown
## (ii)
$R_g = 7W$ | B1 | Resolving vertically (maybe seen in (i) and used in (ii))
$\frac{5W}{24}\left(1 + \frac{6x}{a}\right) = \frac{1}{3}(7W)$ | M1 | Use of $Fr = \mu R$ with candidates $R_g$; allow $\leq$, $<$, $>$, $\geq$
$x = 1.7a$ | A1 [3] | $17a/10$, $51a/30$; must be $=$ or $\leq$
## (iii)
$\frac{Wa\cos\alpha + 6W(2a)\cos\alpha}{2a\sin\alpha} = \frac{1}{3}(7W)$ (oe for moments about $B$) | M1 | Setting $x = 2a$ in their dimensionally correct moments equation and use of $Fr = \mu R$ allow $\leq$, $<$, $>$, $\geq$
$\tan \alpha = \frac{39}{14}$ | A1 | $\tan \alpha = 2.7857142...$; must be $=$ or $\geq$
[3] | |
---A uniform ladder $AB$, of weight $W$ and length $2a$, rests 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 $\sin \theta = \frac{12}{13}$. A man of weight $6W$ is standing on the ladder at a distance $x$ from $A$ and the system is in equilibrium.
\begin{enumerate}[label=(\roman*)]
\item Show that the magnitude of the frictional force exerted by the ground on the ladder is $\frac{5W}{24}\left(1 + \frac{6x}{a}\right)$. [5]
\end{enumerate}
The coefficient of friction between the ladder and the ground is $\frac{1}{3}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find, in terms of $a$, the greatest value of $x$ for which the system is in equilibrium. [3]
\end{enumerate}
The bottom of the ladder $A$ is moved closer to the wall so that the ladder is now inclined at an angle $\alpha$ to the horizontal. The man of weight $6W$ can now stand at the top of the ladder $B$ without the ladder slipping.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the least possible value of $\tan \alpha$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2016 Q5 [11]}}