| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.8 This is a multi-part statics problem requiring moment equilibrium about the hinge, resolution of forces in two directions, and a constraint analysis. Part (a) is standard moments application, but parts (b) and (c) require careful component resolution and the geometric constraint φ > θ to find the range of b, which demands insight beyond routine application. |
| Spec | 6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Taking moments about \(A\) | M1 | Requires all terms; condone trig confusion and sign errors |
| \(bF=3mga\cos\theta+mg\times 2a\cos\theta\) | A2 | -1 each error |
| \(bF=5mga\cos\theta\) | A1 | Given answer |
| \(F=\frac{5mga}{b}\cos\theta\) | — | — |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Component of \(\mathbf{R}\) parallel to \(AB\): \(R\cos(\phi-\theta)\) | M1 | Requires all terms; condone trig confusion |
| \(=3mg\sin\theta+mg\sin\theta=4mg\sin\theta\) | A1 | Correct unsimplified |
| Component of \(\mathbf{R}\) perpendicular to \(AB\): \(R\sin(\phi-\theta)+F=4mg\cos\theta\) | M1 | Requires all terms; condone consistent trig confusion and sign errors |
| Correct unsimplified | A1 | — |
| \(\left(R\sin(\phi-\theta)\right)=4mg\cos\theta-\frac{5mga}{b}\cos\theta\) | A1 | Correct with \(F\) substituted |
| ISW for incorrect work after correct components seen | (5) | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(X=F\sin\theta=\frac{5mga}{b}\cos\theta\sin\theta\) | M1 | Allow with \(F\); requires all terms; condone trig confusion |
| \(F\) substituted | A1 | — |
| \(Y=4mg-F\cos\theta=4mg-\frac{5mga}{b}\cos^2\theta\) | M1 | Allow with \(F\); requires all terms; condone trig confusion and sign errors |
| Correct unsimplified | A1 | — |
| Correct substituted | A1 | — |
| (5) | ||
| Alternatives for \(M(B)\): \(2aR\sin(\phi-\theta)+3mga\cos\theta=F(2a-b)\) | M1A1 | |
| \(M(C)\): \(bR\sin(\phi-\theta)+(2a-b)mg\cos\theta=3mg(b-a)\cos\theta\) | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Use of \(R\sin(\phi-\theta)>0\) | M1 | — |
| Solve for \(b\) in terms of \(a\): \(4>\frac{5a}{b}\), \((2a\geq)b>\frac{5}{4}a\) | A1 | \(2a\) not required; CSO |
| (2) | ||
| Special case: Misread of directions in (b) — NB: This MR can score full marks | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| For \(\varphi>\theta\), \(\tan\varphi>\tan\theta\): \(\tan\varphi=\frac{Y}{X}=\frac{4-\frac{5a}{b}\cos^2\theta}{\frac{5a}{b}\cos\theta\sin\theta}>\tan\theta\) | M1 | — |
| \(4-\frac{5a}{b}\cos^2\theta>\frac{5a}{b}\sin^2\theta\) | — | — |
| \(4>\frac{5a}{b}(\cos^2\theta+\sin^2\theta)\Rightarrow b>\frac{5}{4}a\) | A1 | cso |
| (2) | ||
| (11 marks) |
## Question 6(a):
| Working | Mark | Guidance |
|---------|------|----------|
| Taking moments about $A$ | M1 | Requires all terms; condone trig confusion and sign errors |
| $bF=3mga\cos\theta+mg\times 2a\cos\theta$ | A2 | -1 each error |
| $bF=5mga\cos\theta$ | A1 | **Given answer** |
| $F=\frac{5mga}{b}\cos\theta$ | — | — |
| | **(4)** | |
---
## Question 6(b):
| Working | Mark | Guidance |
|---------|------|----------|
| Component of $\mathbf{R}$ parallel to $AB$: $R\cos(\phi-\theta)$ | M1 | Requires all terms; condone trig confusion |
| $=3mg\sin\theta+mg\sin\theta=4mg\sin\theta$ | A1 | Correct unsimplified |
| Component of $\mathbf{R}$ perpendicular to $AB$: $R\sin(\phi-\theta)+F=4mg\cos\theta$ | M1 | Requires all terms; condone consistent trig confusion and sign errors |
| Correct unsimplified | A1 | — |
| $\left(R\sin(\phi-\theta)\right)=4mg\cos\theta-\frac{5mga}{b}\cos\theta$ | A1 | Correct with $F$ substituted |
| ISW for incorrect work after correct components seen | **(5)** | — |
**Alternative:**
| Working | Mark | Guidance |
|---------|------|----------|
| $X=F\sin\theta=\frac{5mga}{b}\cos\theta\sin\theta$ | M1 | Allow with $F$; requires all terms; condone trig confusion |
| $F$ substituted | A1 | — |
| $Y=4mg-F\cos\theta=4mg-\frac{5mga}{b}\cos^2\theta$ | M1 | Allow with $F$; requires all terms; condone trig confusion and sign errors |
| Correct unsimplified | A1 | — |
| Correct substituted | A1 | — |
| | **(5)** | |
**Alternatives for $M(B)$:** $2aR\sin(\phi-\theta)+3mga\cos\theta=F(2a-b)$ | M1A1
**$M(C)$:** $bR\sin(\phi-\theta)+(2a-b)mg\cos\theta=3mg(b-a)\cos\theta$ | —
---
## Question 6(c):
| Working | Mark | Guidance |
|---------|------|----------|
| Use of $R\sin(\phi-\theta)>0$ | M1 | — |
| Solve for $b$ in terms of $a$: $4>\frac{5a}{b}$, $(2a\geq)b>\frac{5}{4}a$ | A1 | $2a$ not required; CSO |
| | **(2)** | |
**Special case:** Misread of directions in (b) — NB: This MR can score full marks | **(2)**
**Alternative for 6(c):**
| Working | Mark | Guidance |
|---------|------|----------|
| For $\varphi>\theta$, $\tan\varphi>\tan\theta$: $\tan\varphi=\frac{Y}{X}=\frac{4-\frac{5a}{b}\cos^2\theta}{\frac{5a}{b}\cos\theta\sin\theta}>\tan\theta$ | M1 | — |
| $4-\frac{5a}{b}\cos^2\theta>\frac{5a}{b}\sin^2\theta$ | — | — |
| $4>\frac{5a}{b}(\cos^2\theta+\sin^2\theta)\Rightarrow b>\frac{5}{4}a$ | A1 | cso |
| | **(2)** | |
| | **(11 marks)** | |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f30ed5b8-880e-42de-860e-d1538fa68f11-20_757_1264_233_333}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform rod $A B$, of mass $3 m$ and length $2 a$, is freely hinged at $A$ to a fixed point on horizontal ground. A particle of mass $m$ is attached to the rod at the end $B$. The system is held in equilibrium by a force $\mathbf { F }$ acting at the point $C$, where $A C = b$. The rod makes an acute angle $\theta$ with the ground, as shown in Figure 3. The line of action of $\mathbf { F }$ is perpendicular to the rod and in the same vertical plane as the rod.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of $\mathbf { F }$ is $\frac { 5 m g a } { b } \cos \theta$
The force exerted on the rod by the hinge at $A$ is $\mathbf { \mathbf { R } }$, which acts upwards at an angle $\phi$ above the horizontal, where $\phi > \theta$.
\item Find
\begin{enumerate}[label=(\roman*)]
\item the component of $\mathbf { R }$ parallel to the rod, in terms of $m , g$ and $\theta$,
\item the component of $\mathbf { R }$ perpendicular to the rod, in terms of $a , b , m , g$ and $\theta$.
\end{enumerate}\item Hence, or otherwise, find the range of possible values of $b$, giving your answer in terms of $a$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q6 [11]}}