| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Two jointed rods in equilibrium |
| Difficulty | Standard +0.3 This is a standard M2 statics problem requiring moment equilibrium about a point and resolution of forces. The geometry is given explicitly (angles and lengths), and part (a) guides students to the thrust value. The solution involves taking moments about A, basic trigonometry, and resolving forces—all routine techniques for M2 with no novel insight required, making it slightly easier than average. |
| Spec | 6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Moments about \(A\): \(W \times 2a\cos 30 = T\cos 60 \times 2.5a\) | M1A1 | M1 for producing equation in \(T\) and \(W\) only, usually by taking moments about \(A\); A1 for correct equation |
| \(T = \frac{2W\sqrt{3}/2}{2.5 \times 1/2} = \frac{4W\sqrt{3}}{5}\) ANSWER GIVEN | A1 | A1 for given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontally: \((H =) \pm T\cos 60\) | M1A1 | M1 for producing horizontal component in terms of \(T\) and/or \(W\) only |
| Vertically: \((V =) \pm(W - T\cos 30)\) | M1A1 | M1 for producing vertical component in terms of \(T\) and/or \(W\) only |
| \(\ | R\ | = \frac{W}{5}\sqrt{1+12} = \frac{\sqrt{13}}{5}W\) (0.72\(W\) or better) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pm X = T\cos 30 - W\cos 60\) \(\left(= \frac{7W}{10}\right)\) | M1A1 | |
| \(\pm Y = W\cos 30 - T\cos 60\) \(\left(= \frac{\sqrt{3}W}{10}\right)\) | M1A1 | |
| \(\ | R\ | = \frac{W}{10}\sqrt{49+3} = \frac{\sqrt{13}}{5}W\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R^2 = W^2 + T^2 - 2TW\cos 30°\) | M1A1 | M1 for attempt at cos rule |
| Substituting for \(T\) and \(\cos 30°\) | M1A1 | M1 for substituting (either surd or decimal) |
| \(\frac{\sqrt{13}}{5}W\) or \(0.72W\) or better | DM1A1 | DM1 dependent on previous two M's |
# Question 3:
## Part 3a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Moments about $A$: $W \times 2a\cos 30 = T\cos 60 \times 2.5a$ | M1A1 | M1 for producing equation in $T$ and $W$ only, usually by taking moments about $A$; A1 for correct equation |
| $T = \frac{2W\sqrt{3}/2}{2.5 \times 1/2} = \frac{4W\sqrt{3}}{5}$ **ANSWER GIVEN** | A1 | A1 for given answer correctly obtained |
## Part 3b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontally: $(H =) \pm T\cos 60$ | M1A1 | M1 for producing horizontal component in terms of $T$ and/or $W$ only |
| Vertically: $(V =) \pm(W - T\cos 30)$ | M1A1 | M1 for producing vertical component in terms of $T$ and/or $W$ only |
| $\|R\| = \frac{W}{5}\sqrt{1+12} = \frac{\sqrt{13}}{5}W$ (0.72$W$ or better) | DM1A1 | DM1 dependent on previous two M's, for solving for $R$ in terms of $W$ only, usually squaring, adding and square rooting |
**OR: Components along rod ($X$) and perpendicular to rod ($Y$):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pm X = T\cos 30 - W\cos 60$ $\left(= \frac{7W}{10}\right)$ | M1A1 | |
| $\pm Y = W\cos 30 - T\cos 60$ $\left(= \frac{\sqrt{3}W}{10}\right)$ | M1A1 | |
| $\|R\| = \frac{W}{10}\sqrt{49+3} = \frac{\sqrt{13}}{5}W$ oe | DM1A1 | |
**Alternative using Triangle of Forces:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R^2 = W^2 + T^2 - 2TW\cos 30°$ | M1A1 | M1 for attempt at cos rule |
| Substituting for $T$ and $\cos 30°$ | M1A1 | M1 for substituting (either surd or decimal) |
| $\frac{\sqrt{13}}{5}W$ or $0.72W$ or better | DM1A1 | DM1 dependent on previous two M's |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5f79f83-ccfb-47a5-8100-88db81fd0862-05_1102_732_118_651}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform $\operatorname { rod } A B$ of weight $W$ is freely hinged at end $A$ to a vertical wall. The rod is supported in equilibrium at an angle of $60 ^ { \circ }$ to the wall by a light rigid strut $C D$. The strut is freely hinged to the rod at the point $D$ and to the wall at the point $C$, which is vertically below $A$, as shown in Figure 1. The rod and the strut lie in the same vertical plane, which is perpendicular to the wall. The length of the rod is $4 a$ and $A C = A D = 2.5 a$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the thrust in the strut is $\frac { 4 \sqrt { 3 } } { 5 } W$.
\item Find the magnitude of the force acting on the $\operatorname { rod }$ at $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q3 [9]}}