| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Ring on vertical rod equilibrium |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem with friction requiring resolution of forces in two directions and application of F ≤ μR. Parts (a)-(b) involve straightforward force resolution with given values. Parts (c)(i)-(ii) require understanding that friction can act in either direction and setting F = μR at limiting equilibrium, which is a common textbook exercise. The trigonometry is given explicitly (tan θ = 12/5), removing any geometric challenge. Slightly easier than average due to its routine nature and clear structure. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = 10\cos\theta - 0.2g\) or \(F = 0.2g - 10\cos\theta\) | M1 A1 | Resolve vertically, dimensionally correct, condone sin/cos confusion and sign errors |
| \( | F | = 1.9\) or \(1.89\) (N) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Friction acts downwards or down | A1 | Correct direction from a correct, but possibly unrounded, answer to part (a) |
| A0 for anything else | — | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = T\sin\theta\) \(\left(R = \frac{12T}{13}\right)\) | M1 A1 | Resolve perpendicular to rod, dimensionally correct, correct no. of terms, condone sin/cos confusion. M0 if \(T=10\) used |
| \(F = \frac{1}{4}R\) | B1 | \(F = \frac{1}{4}R\) seen or implied |
| Resolve vertically, for min: \(T\cos\theta = 0.2g - F\) | M1 A1 | Resolve parallel to rod, dimensionally correct, correct no. of terms. M0 if \(T=10\) or \(F\) from part (a) used |
| For max: \(T\cos\theta = 0.2g + F\) | A1 | Correct maximum case equation |
| (i) Min \(T\): \(3.2\) or \(3.19\) (N) | A1 | Allow \(0.325g\) |
| (ii) Max \(T\): \(13\) or \(12.7\) (N) | A1 | Allow \(1.3g\). If both found with no labels, allow all marks. If labelled wrongly, lose final two A marks |
## Question 5:
### Part 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = 10\cos\theta - 0.2g$ **or** $F = 0.2g - 10\cos\theta$ | M1 A1 | Resolve vertically, dimensionally correct, condone sin/cos confusion and sign errors |
| $|F| = 1.9$ or $1.89$ (N) | A1 | Correct value for friction, must be positive. If $\mu R$ used as notation for $F$ and never separated, allow M1A1A1 |
### Part 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Friction acts downwards or down | A1 | Correct direction from a correct, but possibly unrounded, answer to part (a) |
| A0 for anything else | — | — |
### Part 5(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = T\sin\theta$ $\left(R = \frac{12T}{13}\right)$ | M1 A1 | Resolve perpendicular to rod, dimensionally correct, correct no. of terms, condone sin/cos confusion. M0 if $T=10$ used |
| $F = \frac{1}{4}R$ | B1 | $F = \frac{1}{4}R$ seen or implied |
| Resolve vertically, for min: $T\cos\theta = 0.2g - F$ | M1 A1 | Resolve parallel to rod, dimensionally correct, correct no. of terms. M0 if $T=10$ or $F$ from part (a) used |
| For max: $T\cos\theta = 0.2g + F$ | A1 | Correct maximum case equation |
| **(i)** Min $T$: $3.2$ or $3.19$ (N) | A1 | Allow $0.325g$ |
| **(ii)** Max $T$: $13$ or $12.7$ (N) | A1 | Allow $1.3g$. If both found with no labels, allow all marks. If labelled wrongly, lose final two A marks |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{017cc2b0-9ec3-45ff-94c0-9d989badfd5d-16_757_460_246_804}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A small ring of mass 0.2 kg is attached to one end of a light inextensible string.\\
The ring is threaded onto a fixed rough vertical rod.\\
The string is taut and makes an angle $\theta$ with the rod, as shown in Figure 3, where $\tan \theta = \frac { 12 } { 5 }$
Given that the ring is in equilibrium and that the tension in the string is 10 N ,
\begin{enumerate}[label=(\alph*)]
\item find the magnitude of the frictional force acting on the ring,
\item state the direction of the frictional force acting on the ring.
The coefficient of friction between the ring and the rod is $\frac { 1 } { 4 }$\\
Given that the ring is in equilibrium, and that the tension in the string, $T$ newtons, can now vary,
\item \begin{enumerate}[label=(\roman*)]
\item find the minimum value of $T$
\item find the maximum value of $T$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2023 Q5 [12]}}