| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on peg or cylinder |
| Difficulty | Standard +0.8 This is a non-trivial ladder equilibrium problem requiring careful geometric analysis to find where the rail contacts the ladder (not at the top), followed by resolving forces and taking moments about a strategic point. The limiting equilibrium condition and the need to handle both smooth and rough contacts elevates this above standard ladder problems, but it follows established mechanics methods without requiring novel insight. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Take moments about \(A\) | M1 | |
| \(N \times \frac{4a}{\sin\alpha} = Mg \times 3a\cos\alpha\) | A1 | |
| \(\frac{9Mg}{25}\) * | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve horizontally: \((\rightarrow)\ F = \frac{9Mg}{25}\sin\alpha\) | M1, A1 | |
| Resolve vertically: \((\uparrow)\ R + \frac{9Mg}{25}\cos\alpha = Mg\) | M1, A1 | |
| \(F = \mu R\) used | M1 | |
| Eliminate \(R\) and \(F\) and solve for \(\mu\) | M1 | |
| \(\left(F = \frac{36Mg}{125},\ R = \frac{98Mg}{125}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(M(C)\): \(Mg \cdot 2a\cos\alpha + X \cdot 5a\sin\alpha = Y \cdot 5a\) | M1A1 | Correct no. of terms, dim correct, condone sin/cos confusion and sign errors |
| \(M(G)\): \(\frac{9Mg}{25} \cdot 2a + X \cdot 3a\sin\alpha = Y \cdot 3a\) | M1A1 | Correct no. of terms, dim correct, condone sin/cos confusion and sign errors |
| \(M(B)\): \(Mg \cdot 3a\cos\alpha + X \cdot 6a\sin\alpha = Y \cdot 6a + \frac{9Mg}{25}a\) | — | — |
| \(\left(X = \frac{4Mg}{3},\ Y = \frac{98Mg}{75}\right)\) | — | — |
| \(F = \mu R\) becomes: \(X - Y\sin\alpha = \mu Y\cos\alpha\) | M1 | Must be used; M0 if \(F = \mu \times \frac{9Mg}{25}\) used |
| Eliminate \(X\) and \(Y\) and solve for \(\mu\) | M1 | Must have 3 equations (and all 3 previous M marks) |
| \(\mu = \frac{18}{49}\) (0.3673…, accept 0.37 or better) | A1 | Accept 0.37 or better |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Take moments about $A$ | M1 | |
| $N \times \frac{4a}{\sin\alpha} = Mg \times 3a\cos\alpha$ | A1 | |
| $\frac{9Mg}{25}$ * | A1* | |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve horizontally: $(\rightarrow)\ F = \frac{9Mg}{25}\sin\alpha$ | M1, A1 | |
| Resolve vertically: $(\uparrow)\ R + \frac{9Mg}{25}\cos\alpha = Mg$ | M1, A1 | |
| $F = \mu R$ used | M1 | |
| Eliminate $R$ and $F$ and solve for $\mu$ | M1 | |
| $\left(F = \frac{36Mg}{125},\ R = \frac{98Mg}{125}\right)$ | | |
## Question 4 (part b continued):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $M(C)$: $Mg \cdot 2a\cos\alpha + X \cdot 5a\sin\alpha = Y \cdot 5a$ | M1A1 | Correct no. of terms, dim correct, condone sin/cos confusion and sign errors |
| $M(G)$: $\frac{9Mg}{25} \cdot 2a + X \cdot 3a\sin\alpha = Y \cdot 3a$ | M1A1 | Correct no. of terms, dim correct, condone sin/cos confusion and sign errors |
| $M(B)$: $Mg \cdot 3a\cos\alpha + X \cdot 6a\sin\alpha = Y \cdot 6a + \frac{9Mg}{25}a$ | — | — |
| $\left(X = \frac{4Mg}{3},\ Y = \frac{98Mg}{75}\right)$ | — | — |
| $F = \mu R$ becomes: $X - Y\sin\alpha = \mu Y\cos\alpha$ | M1 | Must be used; M0 if $F = \mu \times \frac{9Mg}{25}$ used |
| Eliminate $X$ and $Y$ and solve for $\mu$ | M1 | Must have 3 equations (and all 3 previous M marks) |
| $\mu = \frac{18}{49}$ (0.3673…, accept 0.37 or better) | A1 | Accept 0.37 or better |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d1989e18-1a4a-47e9-9f12-3beb8985ed87-12_803_767_239_647}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A ladder $A B$ has mass $M$ and length $6 a$.\\
The end $A$ of the ladder is on rough horizontal ground.\\
The ladder rests against a fixed smooth horizontal rail at the point $C$.\\
The point $C$ is at a vertical height $4 a$ above the ground.\\
The vertical plane containing $A B$ is perpendicular to the rail.\\
The ladder is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 4 } { 5 }$, as shown in Figure 1.\\
The coefficient of friction between the ladder and the ground is $\mu$.\\
The ladder rests in limiting equilibrium.\\
The ladder is modelled as a uniform rod.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item show that the magnitude of the force exerted on the ladder by the rail at $C$ is $\frac { 9 M g } { 25 }$
\item Hence, or otherwise, find the value of $\mu$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2020 Q4 [10]}}