| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on peg or cylinder |
| Difficulty | Challenging +1.2 This is a standard M2 moments problem requiring geometric analysis to find the perpendicular distance from the hinge, then applying moment equilibrium and resolving forces. The geometry involves a 5-12-13 Pythagorean triple which students should recognize, making calculations cleaner. While it requires multiple steps (geometry, moments, force resolution), the techniques are all standard M2 material with no novel insights needed. Slightly above average difficulty due to the geometric setup and two-part structure. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(A\) | M1 | Or a complete method to form an equation in \(R\) and \(W\) |
| \(W \times 8b\cos\theta = R \times 12b\) | A1 | Correct unsimplified equation |
| \(R = \frac{2W}{3}\cos\theta = \frac{2W}{3} \times \frac{12}{13}\) | DM1 | Substitute correctly for trig and solve for \(R\). Dependent on preceding M1 |
| \(R = \frac{8W}{13}\) | A1 | Allow \(R = 0.615W\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve horizontally | M1 | Form one equation in \(X\) and/or \(Y\) |
| \((\rightarrow)\ X = R\sin\theta \left(= \frac{40W}{169}\right)\) | A1 | Correct unsimplified equation |
| Resolve vertically | M1 | Form a second equation in \(X\) and/or \(Y\) |
| \((\uparrow)\ Y = W - R\cos\theta \left(= \frac{73W}{169}\right)\) | A1 | Correct unsimplified equation |
| \(\tan\alpha = \frac{X}{Y}\) | DM1 | Use their \(X\) and \(Y\) to find \(\tan\alpha\). Dependent on M marks for the two equations |
| \(\tan\alpha = \frac{40}{73}\) Given answer | A1 | Obtain given answer from correct work |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $A$ | M1 | Or a complete method to form an equation in $R$ and $W$ |
| $W \times 8b\cos\theta = R \times 12b$ | A1 | Correct unsimplified equation |
| $R = \frac{2W}{3}\cos\theta = \frac{2W}{3} \times \frac{12}{13}$ | DM1 | Substitute correctly for trig and solve for $R$. Dependent on preceding M1 |
| $R = \frac{8W}{13}$ | A1 | Allow $R = 0.615W$ |
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## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve horizontally | M1 | Form one equation in $X$ and/or $Y$ |
| $(\rightarrow)\ X = R\sin\theta \left(= \frac{40W}{169}\right)$ | A1 | Correct unsimplified equation |
| Resolve vertically | M1 | Form a second equation in $X$ and/or $Y$ |
| $(\uparrow)\ Y = W - R\cos\theta \left(= \frac{73W}{169}\right)$ | A1 | Correct unsimplified equation |
| $\tan\alpha = \frac{X}{Y}$ | DM1 | Use their $X$ and $Y$ to find $\tan\alpha$. Dependent on M marks for the two equations |
| $\tan\alpha = \frac{40}{73}$ **Given answer** | A1 | Obtain given answer from correct work |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{54112b4a-3727-4e5b-97e5-4291e7172438-14_472_789_253_575}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform rod, of weight $W$ and length $16 b$, has one end freely hinged to a fixed point $A$. The rod rests against a smooth circular cylinder, of radius $5 b$, fixed with its axis horizontal and at the same horizontal level as $A$. The distance of $A$ from the axis of the cylinder is 13b, as shown in Figure 2. The rod rests in a vertical plane which is perpendicular to the axis of the cylinder.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $W$, the magnitude of the reaction on the rod at its point of contact with the cylinder.
\item Show that the resultant force acting on the rod at $A$ is inclined to the vertical at an angle $\alpha$ where $\tan \alpha = \frac { 40 } { 73 }$
\begin{center}
\end{center}
5 continued\\
\includegraphics[max width=\textwidth, alt={}, center]{54112b4a-3727-4e5b-97e5-4291e7172438-17_81_72_2631_1873}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q5 [10]}}