| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod with end on ground or wall supported by string |
| Difficulty | Standard +0.3 This is a standard M2 ladder/rod equilibrium problem requiring resolution of forces and taking moments about a point. While it involves multiple steps (finding tension, horizontal force, then friction coefficient), the approach is methodical and follows standard textbook procedures. The perpendicular string simplifies the geometry, and limiting equilibrium is a routine concept. Slightly easier than average due to the straightforward setup. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(M(A)\) or alternative complete method to an equation in \(T\) only | M1 | Must have all terms. Terms must be dimensionally correct. Condone sign errors and sin/cos confusion. |
| \(T\times 2a = mg\times 3a\sin 60° + mg\times 6a\sin 60°\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation | |
| \(T = 9mg\dfrac{\sqrt{3}}{4}\) | A1 (4) | With trig. substituted. 3.90\(mg\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(R(\rightarrow)\quad R = T\cos 60°\) | M1 | Resolve horizontally. Condone sin/cos confusion |
| \(\left(= 9mg\dfrac{\sqrt{3}}{4}\times\dfrac{1}{2}\right)\) | A1ft | Follow their \(T\). Allow with \(\cos 60°\) |
| \(R = \dfrac{9\sqrt{3}}{8}mg\) | A1ft (3) | 1.95\(mg\) or better. Follow their (a). |
| Alt 4(b): \(2mg\cos 60° = R\cos 30° - F\cos 60°\) and \(T - F\cos 30 = 2mg\cos 30° + R\cos 60°\) | (M1) | Resolve parallel and perpendicular to the rod and eliminate \(F\) |
| \(\dfrac{5mg\sqrt{3}}{4} - \dfrac{R}{2} = -\sqrt{3}mg + \dfrac{3R}{2}\) | (A1ft) | Equation in \(R\) only. Follow their \(T\) |
| \(R = \dfrac{9\sqrt{3}}{8}mg\) | (A1ft) | With trig. substituted. Follow their (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(R(\uparrow)\quad T\cos 30° - F = mg + mg\) | M1 | Resolve vertically. Need all terms. Condone sign errors and sin/cos confusion. Allow for \(\pm F\) |
| A1 | Unsimplified equation with at most one error. Allow for \(\pm F\) | |
| \(F = 9mg\dfrac{\sqrt{3}}{4}\times\dfrac{\sqrt{3}}{2} - 2mg\left(= \dfrac{11}{8}mg\right)\) | A1 | Correct unsimplified expression for \(F\), with trig. substituted. Allow for \(\pm F\). Seen or implied. |
| \(\mu = \dfrac{F}{R} = \dfrac{\dfrac{11}{8}mg}{\dfrac{9\sqrt{3}}{8}mg}\) | dM1 | Use of \(F = \mu R\). Dependent on the two previous M marks |
| \(= \dfrac{11}{9\sqrt{3}}\) \((= 0.71\) or \(0.706\) or better\()\) | A1 (5) | (\(g\) cancels) |
| Alt (c) 1st 3 marks: \(2mg\cos 60° = R\cos 30° - F\cos 60°\) | (M1) | Resolve parallel to the rod. Need all terms. Condone sign errors and sin/cos confusion. Allow for \(\pm F\) |
| \(mg = \dfrac{27}{16}mg - \dfrac{1}{2}F\) | (A1) | Unsimplified equation with at most one error. Allow for \(\pm F\). sin/cos confusion is one error |
| \(F = \dfrac{11}{8}mg\) | (A1) | Correct unsimplified expression for \(F\). Allow for \(\pm F\). Seen or implied. |
| Total: [12] |
# Question 4(a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $M(A)$ or alternative complete method to an equation in $T$ only | M1 | Must have all terms. Terms must be dimensionally correct. Condone sign errors and sin/cos confusion. |
| $T\times 2a = mg\times 3a\sin 60° + mg\times 6a\sin 60°$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $T = 9mg\dfrac{\sqrt{3}}{4}$ | A1 (4) | With trig. substituted. 3.90$mg$ or better |
# Question 4(b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $R(\rightarrow)\quad R = T\cos 60°$ | M1 | Resolve horizontally. Condone sin/cos confusion |
| $\left(= 9mg\dfrac{\sqrt{3}}{4}\times\dfrac{1}{2}\right)$ | A1ft | Follow their $T$. Allow with $\cos 60°$ |
| $R = \dfrac{9\sqrt{3}}{8}mg$ | A1ft (3) | 1.95$mg$ or better. Follow their (a). |
| **Alt 4(b):** $2mg\cos 60° = R\cos 30° - F\cos 60°$ and $T - F\cos 30 = 2mg\cos 30° + R\cos 60°$ | (M1) | Resolve parallel and perpendicular to the rod and eliminate $F$ |
| $\dfrac{5mg\sqrt{3}}{4} - \dfrac{R}{2} = -\sqrt{3}mg + \dfrac{3R}{2}$ | (A1ft) | Equation in $R$ only. Follow their $T$ |
| $R = \dfrac{9\sqrt{3}}{8}mg$ | (A1ft) | With trig. substituted. Follow their (a) |
# Question 4(c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $R(\uparrow)\quad T\cos 30° - F = mg + mg$ | M1 | Resolve vertically. Need all terms. Condone sign errors and sin/cos confusion. Allow for $\pm F$ |
| | A1 | Unsimplified equation with at most one error. Allow for $\pm F$ |
| $F = 9mg\dfrac{\sqrt{3}}{4}\times\dfrac{\sqrt{3}}{2} - 2mg\left(= \dfrac{11}{8}mg\right)$ | A1 | Correct unsimplified expression for $F$, with trig. substituted. Allow for $\pm F$. Seen or implied. |
| $\mu = \dfrac{F}{R} = \dfrac{\dfrac{11}{8}mg}{\dfrac{9\sqrt{3}}{8}mg}$ | dM1 | Use of $F = \mu R$. Dependent on the two previous M marks |
| $= \dfrac{11}{9\sqrt{3}}$ $(= 0.71$ or $0.706$ or better$)$ | A1 (5) | ($g$ cancels) |
| **Alt (c) 1st 3 marks:** $2mg\cos 60° = R\cos 30° - F\cos 60°$ | (M1) | Resolve parallel to the rod. Need all terms. Condone sign errors and sin/cos confusion. Allow for $\pm F$ |
| $mg = \dfrac{27}{16}mg - \dfrac{1}{2}F$ | (A1) | Unsimplified equation with at most one error. Allow for $\pm F$. sin/cos confusion is one error |
| $F = \dfrac{11}{8}mg$ | (A1) | Correct unsimplified expression for $F$. Allow for $\pm F$. Seen or implied. |
| **Total: [12]** | | |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ef8231d-5b95-4bbb-a8e2-788c708fa078-12_518_696_319_625}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform $\operatorname { rod } A B$ has mass $m$ and length $6 a$. The end $A$ rests against a rough vertical wall. One end of a light inextensible string is attached to the rod at the point $C$, where $A C = 2 a$. The other end of the string is attached to the wall at the point $D$, where $D$ is vertically above $A$, with the string perpendicular to the rod. A particle of mass $m$ is attached to the rod at the end $B$. The rod is in equilibrium in a vertical plane which is perpendicular to the wall. The rod is inclined at $60 ^ { \circ }$ to the wall, as shown in Figure 1.
Find, in terms of $m$ and $g$,
\begin{enumerate}[label=(\alph*)]
\item the tension in the string,
\item the magnitude of the horizontal component of the force exerted by the wall on the rod.
The coefficient of friction between the wall and the rod is $\mu$. Given that the rod is in limiting equilibrium,
\item find the value of $\mu$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2017 Q4 [12]}}