Edexcel Paper 3 2022 June — Question 4 11 marks

Exam BoardEdexcel
ModulePaper 3 (Paper 3)
Year2022
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod with end on ground or wall supported by string
DifficultyStandard +0.3 This is a standard mechanics equilibrium problem requiring resolution of forces and taking moments about a point. While it has multiple parts and involves a composite system (rod + particle), the techniques are routine for Further Maths students: resolving vertically/horizontally, taking moments about A, and applying limiting friction. The 'show that' format provides target answers to work towards, reducing problem-solving demand. Slightly easier than average due to its structured, methodical nature.
Spec3.04b Equilibrium: zero resultant moment and force
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{414946db-64d7-44b8-801d-2c7805ee9cc6-12_716_1191_246_438} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A uniform rod \(A B\) has mass \(M\) and length \(2 a\) A particle of mass \(2 M\) is attached to the rod at the point \(C\), where \(A C = 1.5 a\) The rod rests with its end \(A\) on rough horizontal ground.
The rod is held in equilibrium at an angle \(\theta\) to the ground by a light string that is attached to the end \(B\) of the rod. The string is perpendicular to the rod, as shown in Figure 2.
  1. Explain why the frictional force acting on the rod at \(A\) acts horizontally to the right on the diagram. The tension in the string is \(T\)
  2. Show that \(T = 2 M g \cos \theta\) Given that \(\cos \theta = \frac { 3 } { 5 }\)
  3. show that the magnitude of the vertical force exerted by the ground on the rod at \(A\) is \(\frac { 57 M g } { 25 }\) The coefficient of friction between the rod and the ground is \(\mu\) Given that the rod is in limiting equilibrium,
  4. show that \(\mu = \frac { 8 } { 19 }\)
Question 4(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Horizontal component of \(T\) acts to the left, friction must act to the rightB1
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Take moments about \(A\) or complete method for equation in \(T\), \(M\) and \(\theta\) onlyM1
\(T\cdot 2a = Mga\cos\theta + 2Mg\times 1.5a\cos\theta\)A1 A0 if \(a\)'s missing
\(T = 2Mg\cos\theta\)A1* Given answer correctly obtained
Question 4(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Resolve verticallyM1
\((\uparrow),\ R + T\cos\theta = Mg + 2Mg\)A1
\(R = \dfrac{57Mg}{25}\)A1*
Question 4(d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Find equation containing \(F\), e.g. resolve horizontallyM1
\((\rightarrow),\ F = T\sin\theta\)A1
Question 4 (continued):
Part 4d:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(F = \mu R\) used, i.e. both \(F\) and \(R\) are substitutedM1 3.1b
\(\mu = \dfrac{8}{19}\)A1* 2.2a
Notes 4d:
- M1: For any equation with \(F\) in it; correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved
- A1: Correct equation, trig does not need to be substituted
- M1: Must be used i.e. M0 if merely quoting it
- A1*: Given answer correctly obtained with no wrong working seen
## Question 4(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal component of $T$ acts to the left, friction must act to the right | B1 | |

## Question 4(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Take moments about $A$ or complete method for equation in $T$, $M$ and $\theta$ only | M1 | |
| $T\cdot 2a = Mga\cos\theta + 2Mg\times 1.5a\cos\theta$ | A1 | A0 if $a$'s missing |
| $T = 2Mg\cos\theta$ | A1* | Given answer correctly obtained |

## Question 4(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | |
| $(\uparrow),\ R + T\cos\theta = Mg + 2Mg$ | A1 | |
| $R = \dfrac{57Mg}{25}$ | A1* | |

## Question 4(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Find equation containing $F$, e.g. resolve horizontally | M1 | |
| $(\rightarrow),\ F = T\sin\theta$ | A1 | |

## Question 4 (continued):

**Part 4d:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \mu R$ used, i.e. both $F$ and $R$ are substituted | M1 | 3.1b |
| $\mu = \dfrac{8}{19}$ | A1* | 2.2a |

**Notes 4d:**
- M1: For any equation with $F$ in it; correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved
- A1: Correct equation, trig does not need to be substituted
- M1: Must be used i.e. M0 if merely quoting it
- A1*: Given answer correctly obtained with no wrong working seen

---
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{414946db-64d7-44b8-801d-2c7805ee9cc6-12_716_1191_246_438}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

A uniform rod $A B$ has mass $M$ and length $2 a$\\
A particle of mass $2 M$ is attached to the rod at the point $C$, where $A C = 1.5 a$\\
The rod rests with its end $A$ on rough horizontal ground.\\
The rod is held in equilibrium at an angle $\theta$ to the ground by a light string that is attached to the end $B$ of the rod.

The string is perpendicular to the rod, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Explain why the frictional force acting on the rod at $A$ acts horizontally to the right on the diagram.

The tension in the string is $T$
\item Show that $T = 2 M g \cos \theta$

Given that $\cos \theta = \frac { 3 } { 5 }$
\item show that the magnitude of the vertical force exerted by the ground on the rod at $A$ is $\frac { 57 M g } { 25 }$

The coefficient of friction between the rod and the ground is $\mu$\\
Given that the rod is in limiting equilibrium,
\item show that $\mu = \frac { 8 } { 19 }$
\end{enumerate}

\hfill \mbox{\textit{Edexcel Paper 3 2022 Q4 [11]}}
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