| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard mechanics ladder problem requiring moments, resolving forces, and friction at limiting equilibrium. Parts (a)-(d) follow routine procedures (taking moments about A, resolving horizontally/vertically, using F=μR), with straightforward arithmetic given tan θ = 3/4. Part (e) tests conceptual understanding of how moving the center of mass affects forces. While multi-part with several marks, all techniques are standard A-level mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Normal reaction at \(B\) acts left, so friction at \(A\) acts right to balance/oppose; weight causes rod to tend to slip left, so friction opposes that. Accept "towards the wall". B0 if rod said to be moving. B0 if mass/weight mentioned as cause | B1 | 2.4 |
| (1 mark total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Take moments about \(A\) | M1 | 3.4 |
| \(S \times 2a\sin\theta = Mga\cos\theta\) | A1 | 1.1b |
| \(S = \dfrac{1}{2}Mg\cot\theta\) * | A1* | 2.2a |
| (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve vertically: \(R = Mg\) | B1 | 3.3 |
| Resolve horizontally: \(F = S\) | B1 | 3.3 |
| Other possible equations listed (moments about \(B\), \(G\); resolve along/perp to rod) | N.B. on ePEN: first B1 = vertical resolution, second B1 = horizontal resolution | |
| \(F = \mu R\) | B1 | 1.2 |
| \(\dfrac{1}{2}Mg\times\dfrac{4}{3} = \mu Mg\) | dM1 | 2.1 |
| \(\mu = \dfrac{2}{3}\) (accept 0.67 or better) | A1 | 2.2a |
| S.C. for \(F \neq \mu R\): B0 then \(\frac{1}{2}Mg\times\frac{4}{3} \approx \mu Mg\) scores M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Any equivalent appropriate statement | B1 | Any equivalent appropriate statement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct number of terms, dimensionally correct | M1 | Condone sin/cos confusion and sign errors. If \(a\)'s never appear, M0 |
| Correct equation | A1 | — |
| Correct given answer correctly obtained, with no wrong working seen. Allow \(\frac{1}{2}Mg\cot\theta = S\) or \(S = \frac{Mg\cot\theta}{2}\) or \(\frac{Mg\cot\theta}{2} = S\) or \(S = \frac{Mg}{2}\cot\theta\) or similar, but NOT \(S = \frac{1}{2}\cot\theta\ Mg\) or similar. Allow \(m\) instead of \(M\). Must be \(\theta\) in final answer but allow a different angle in the working. | A1* | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| cao | B1 | — |
| cao | B1 | — |
| Seen anywhere, e.g. on the diagram | B1 | — |
| Using \(F = \mu R\), their two equations and substitute for trig (not necessarily correctly) to produce an equation in \(\mu\) only. This mark is dependent on the 3 previous B marks. | dM1 | — |
| \(\frac{2}{3}\), \(\mu\) | A0 | Accept 0.67 or better. N.B. If \(\mu = \frac{2}{3}\) follows this, they could score all the marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sqrt{F^2 + R^2}\) | M1 | Use of Pythagoras with square root to find the required magnitude, but \(F\) and \(R\) do not need to be substituted |
| \(\sqrt{\left(\frac{2}{3}Mg\right)^2 + (Mg)^2}\) | M1 | Substitute for their \(F\) and their \(R\) in terms of \(Mg\) and take square root to obtain magnitude in terms of \(M\) and \(g\) only. N.B. Must be using Pythagoras |
| \(\frac{1}{3}Mg\sqrt{13}\) or \(1.2Mg\) or better | A1 | Any equivalent surd form or \(1.2Mg\) or better. Must be in terms of \(M\) and \(g\) |
| ALTERNATIVE using trig on triangle of forces: \(X = \frac{Mg}{\sin\alpha}\) or \(\frac{S}{\cos\alpha}\) | M1 | — |
| Substitute for \(\sin\alpha\) or \(\cos\alpha\) and \(S\), where \(\tan\alpha = \frac{Mg}{S}\left(=\frac{3}{2}\right)\), to obtain \(X\) in terms of \(M\) and \(g\) only | M1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| New value of \(S\) would be larger as the moment of the weight about \(A\) would be larger | B1 | Correct answer and any equivalent appropriate statement |
# Question 6:
## Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal reaction at $B$ acts left, so friction at $A$ acts right to balance/oppose; weight causes rod to tend to slip left, so friction opposes that. Accept "towards the wall". B0 if rod said to be moving. B0 if mass/weight mentioned as cause | B1 | 2.4 |
| **(1 mark total)** | | |
## Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Take moments about $A$ | M1 | 3.4 |
| $S \times 2a\sin\theta = Mga\cos\theta$ | A1 | 1.1b |
| $S = \dfrac{1}{2}Mg\cot\theta$ * | A1* | 2.2a |
| **(3 marks total)** | | |
## Part 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically: $R = Mg$ | B1 | 3.3 |
| Resolve horizontally: $F = S$ | B1 | 3.3 |
| Other possible equations listed (moments about $B$, $G$; resolve along/perp to rod) | | N.B. on ePEN: first B1 = vertical resolution, second B1 = horizontal resolution |
| $F = \mu R$ | B1 | 1.2 |
| $\dfrac{1}{2}Mg\times\dfrac{4}{3} = \mu Mg$ | dM1 | 2.1 |
| $\mu = \dfrac{2}{3}$ (accept 0.67 or better) | A1 | 2.2a |
| S.C. for $F \neq \mu R$: B0 then $\frac{1}{2}Mg\times\frac{4}{3} \approx \mu Mg$ scores M1 | | |
## Question 6:
### Part 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Any equivalent appropriate statement | B1 | Any equivalent appropriate statement |
### Part 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct number of terms, dimensionally correct | M1 | Condone sin/cos confusion and sign errors. If $a$'s never appear, M0 |
| Correct equation | A1 | — |
| Correct given answer correctly obtained, **with no wrong working seen**. Allow $\frac{1}{2}Mg\cot\theta = S$ or $S = \frac{Mg\cot\theta}{2}$ or $\frac{Mg\cot\theta}{2} = S$ or $S = \frac{Mg}{2}\cot\theta$ or similar, but NOT $S = \frac{1}{2}\cot\theta\ Mg$ or similar. Allow $m$ instead of $M$. Must be $\theta$ in final answer but allow a different angle in the working. | A1* | — |
### Part 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| cao | B1 | — |
| cao | B1 | — |
| Seen anywhere, e.g. on the diagram | B1 | — |
| Using $F = \mu R$, their two equations and substitute for trig (not necessarily correctly) to produce an equation in $\mu$ only. This mark is **dependent** on the 3 previous B marks. | dM1 | — |
| $\frac{2}{3}$, $\mu$ | A0 | Accept 0.67 or better. **N.B.** If $\mu = \frac{2}{3}$ follows this, they could score all the marks. |
### Part 6(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sqrt{F^2 + R^2}$ | M1 | Use of Pythagoras with square root to find the required magnitude, but $F$ and $R$ do not need to be substituted |
| $\sqrt{\left(\frac{2}{3}Mg\right)^2 + (Mg)^2}$ | M1 | Substitute for their $F$ and their $R$ in terms of $Mg$ and take square root to obtain magnitude in terms of $M$ and $g$ only. **N.B.** Must be using Pythagoras |
| $\frac{1}{3}Mg\sqrt{13}$ or $1.2Mg$ or better | A1 | Any equivalent surd form or $1.2Mg$ or better. Must be in terms of $M$ and $g$ |
| **ALTERNATIVE** using trig on triangle of forces: $X = \frac{Mg}{\sin\alpha}$ or $\frac{S}{\cos\alpha}$ | M1 | — |
| Substitute for $\sin\alpha$ or $\cos\alpha$ and $S$, where $\tan\alpha = \frac{Mg}{S}\left(=\frac{3}{2}\right)$, to obtain $X$ in terms of $M$ and $g$ only | M1 | — |
### Part 6(e):
| Answer | Mark | Guidance |
|--------|------|----------|
| New value of $S$ would be **larger** as the **moment** of the **weight** about $A$ would be larger | B1 | Correct answer and any equivalent appropriate statement |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f9dc8158-8ed8-4138-9c75-050cf52e6f7e-16_408_967_246_539}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A $\operatorname { rod } A B$ has mass $M$ and length $2 a$.\\
The rod has its end $A$ on rough horizontal ground and its end $B$ against a smooth vertical wall.
The rod makes an angle $\theta$ with the ground, as shown in Figure 3.\\
The rod is at rest in limiting equilibrium.
\begin{enumerate}[label=(\alph*)]
\item State the direction (left or right on Figure 3 above) of the frictional force acting on the $\operatorname { rod }$ at $A$. Give a reason for your answer.
The magnitude of the normal reaction of the wall on the rod at $B$ is $S$.\\
In an initial model, the rod is modelled as being uniform.\\
Use this initial model to answer parts (b), (c) and (d).
\item By taking moments about $A$, show that
$$S = \frac { 1 } { 2 } M g \cot \theta$$
The coefficient of friction between the rod and the ground is $\mu$\\
Given that $\tan \theta = \frac { 3 } { 4 }$
\item find the value of $\mu$
\item find, in terms of $M$ and $g$, the magnitude of the resultant force acting on the rod at $A$.
In a new model, the rod is modelled as being non-uniform, with its centre of mass closer to $B$ than it is to $A$.
A new value for $S$ is calculated using this new model, with $\tan \theta = \frac { 3 } { 4 }$
\item State whether this new value for $S$ is larger, smaller or equal to the value that $S$ would take using the initial model. Give a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2023 Q6 [13]}}