| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on peg or cylinder |
| Difficulty | Standard +0.3 This is a standard moments problem requiring taking moments about a point, resolving forces, and using friction in limiting equilibrium. The geometry is straightforward (given tan θ = 4/3), and the method is routine for A-level mechanics: moments about A to find S, then resolve horizontally/vertically to find friction coefficient. Slightly easier than average due to clear setup and standard technique. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Take moments about \(A\) | M1 | 3.1a — Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors. Allow different letter for angle. N.B. M0 if one or both \(a\)'s not present originally. |
| \(S \times 1.5a = Mga\cos\theta = \left(Mga \times \frac{3}{5}\right)\) | A1 | 1.1b — Correct equation; \(\cos\theta\) may or may not be replaced by \(\frac{3}{5}\). N.B. may see \(S \times 1.5a = Mg\cos\theta \times a\) |
| \(S = \dfrac{2}{5}Mg\) * | A1* | 2.2a — Given answer correctly obtained; need to see \(\cos\theta = \frac{3}{5}\) used. Allow \(S = \frac{2Mg}{5}\) or \(\frac{2Mg}{5} = S\). A0 if \(S\) is missing. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve horizontally: \(F = S\sin\theta\) | M1, A1 | 3.4, 1.1b — Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors. \(S\) does not need to be substituted. |
| Resolve vertically: \(R = Mg - S\cos\theta\) | M1, A1 | 3.3, 1.1b — Correct no. of terms, dimensionally correct. \(S\) does not need to be substituted. |
| Other possible equations (max M1A1 each): (parallel to rod): \(F\cos\theta + R\sin\theta = Mg\sin\theta\); (perp to rod): \(F\sin\theta + Mg\cos\theta = S + R\cos\theta\); \(M(B)\): \((S \times 0.5a)+(R \times 2a\cos\theta) = (Mg \times a\cos\theta)+(F \times 2a\sin\theta)\); \(M(C)\): \((R \times 1.5a\cos\theta) = (Mg \times 0.5a\cos\theta)+(F \times 1.5a\sin\theta)\); \(M(G)\): \((R \times a\cos\theta) = (S \times 0.5a)+(F \times a\sin\theta)\) | N.B. If more than two equations, mark only those used to find \(\mu\). | |
| \(F = \mu R\) and two equations used to solve for \(\mu\) | DM1 | 3.1a — Dependent on previous two M marks. |
| \(\mu = \dfrac{8}{19} = 0.42105\ldots\) | A1 | 2.2a — Accept \(0.42\) or better (as \(g\) cancels). |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Take moments about $A$ | M1 | 3.1a — Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors. Allow different letter for angle. N.B. M0 if one or both $a$'s not present originally. |
| $S \times 1.5a = Mga\cos\theta = \left(Mga \times \frac{3}{5}\right)$ | A1 | 1.1b — Correct equation; $\cos\theta$ may or may not be replaced by $\frac{3}{5}$. N.B. may see $S \times 1.5a = Mg\cos\theta \times a$ |
| $S = \dfrac{2}{5}Mg$ * | A1* | 2.2a — Given answer correctly obtained; need to see $\cos\theta = \frac{3}{5}$ used. Allow $S = \frac{2Mg}{5}$ or $\frac{2Mg}{5} = S$. A0 if $S$ is missing. |
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## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve horizontally: $F = S\sin\theta$ | M1, A1 | 3.4, 1.1b — Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors. $S$ does not need to be substituted. |
| Resolve vertically: $R = Mg - S\cos\theta$ | M1, A1 | 3.3, 1.1b — Correct no. of terms, dimensionally correct. $S$ does not need to be substituted. |
| Other possible equations (max M1A1 each): (parallel to rod): $F\cos\theta + R\sin\theta = Mg\sin\theta$; (perp to rod): $F\sin\theta + Mg\cos\theta = S + R\cos\theta$; $M(B)$: $(S \times 0.5a)+(R \times 2a\cos\theta) = (Mg \times a\cos\theta)+(F \times 2a\sin\theta)$; $M(C)$: $(R \times 1.5a\cos\theta) = (Mg \times 0.5a\cos\theta)+(F \times 1.5a\sin\theta)$; $M(G)$: $(R \times a\cos\theta) = (S \times 0.5a)+(F \times a\sin\theta)$ | | N.B. If more than two equations, mark only those used to find $\mu$. |
| $F = \mu R$ and two equations used to solve for $\mu$ | DM1 | 3.1a — Dependent on previous two M marks. |
| $\mu = \dfrac{8}{19} = 0.42105\ldots$ | A1 | 2.2a — Accept $0.42$ or better (as $g$ cancels). |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{184043b7-1222-44fb-bc9f-3f484f72147b-16_458_798_258_630}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a uniform rod $A B$ of mass $M$ and length $2 a$.
\begin{itemize}
\item the rod has its end $A$ on rough horizontal ground
\item the rod rests in equilibrium against a small smooth fixed horizontal peg $P$
\item the point $C$ on the rod, where $A C = 1.5 a$, is the point of contact between the rod and the peg
\item the rod is at an angle $\theta$ to the ground, where $\tan \theta = \frac { 4 } { 3 }$
\end{itemize}
The rod lies in a vertical plane perpendicular to the peg.\\
The magnitude of the normal reaction of the peg on the rod at $C$ is $S$.
\begin{enumerate}[label=(\alph*)]
\item Show that $S = \frac { 2 } { 5 } M g$
The coefficient of friction between the rod and the ground is $\mu$.\\
Given that the rod is in limiting equilibrium,
\item find the value of $\mu$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2024 Q6 [9]}}