| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Stacked blocks with friction |
| Difficulty | Standard +0.8 This is a multi-part stacked blocks problem requiring careful force resolution, friction analysis in limiting equilibrium, and then dynamics with different friction coefficients. Part (i) is standard M1 friction, but parts (ii) and (iii) require analyzing two separate bodies with friction between them and with the ground, involving multiple simultaneous equations and Newton's second law. The conceptual jump from equilibrium to motion with different accelerations elevates this above typical M1 questions. |
| Spec | 3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| \(Fr = 4 + 5\sin60\), \(Fr = 8.33\), \(R = 12 - 5\cos60\), \(R = 9.5\), \(\mu = (4 + 5\sin60)/(12 - 5\cos60)\), \(\mu = 0.877\) | M1, A1, M1, A1, M1, A1 | All 4 + component 5 (4 + 4.333(01)). May be implied. +/-(All 12 – component 5 (12 – 2.5)). May be implied, +ve from correct work. Friction/Reaction, Fr>4, R<12, both positive |
| Answer | Marks | Guidance |
|---|---|---|
| Upper block: \(\mu = 5\sin60/(9-5\cos60)\) (=4.3/6.5), \(\mu = 0.666\) | M1, A1 | (Component 5)/(9-component 5) |
| Answer | Marks | Guidance |
|---|---|---|
| Upper mass = 9/g, \((9/g)a = 5\sin60 - 0.1(9 - 5\cos60)\), \(a = 4.01\) | B1, M1, A1 | 0.918(36...), N2L: 0.918(36.)a = 4.33(01...) – 0.1x6.5 where friction = 0.1x(9-component 5) |
| Lower mass: Tractive force = \(4 + 0.1(9-5\cos60)\) (= 4.65), Max Friction = 0.877(3x(9-5\cos60)\( (= 8.33), Tractive force < Max Friction, \)a = 0$ | M1, A1, A1 | Compares TF (tractive force) and max friction |
| OR for Lower Mass: \(ma = 4+0.1(9-5\cos60)-0.877(3+9-5\cos60)\), -ve a caused by friction impossible, hence \(a = 0\) | M1, A1, A1 | N2L with 3 force terms: |
**Part i**
| $Fr = 4 + 5\sin60$, $Fr = 8.33$, $R = 12 - 5\cos60$, $R = 9.5$, $\mu = (4 + 5\sin60)/(12 - 5\cos60)$, $\mu = 0.877$ | M1, A1, M1, A1, M1, A1 | All 4 + component 5 (4 + 4.333(01)). May be implied. +/-(All 12 – component 5 (12 – 2.5)). May be implied, +ve from correct work. Friction/Reaction, Fr>4, R<12, both positive |
**Part ii**
| Upper block: $\mu = 5\sin60/(9-5\cos60)$ (=4.3/6.5), $\mu = 0.666$ | M1, A1 | (Component 5)/(9-component 5) |
**Part iii**
| Upper mass = 9/g, $(9/g)a = 5\sin60 - 0.1(9 - 5\cos60)$, $a = 4.01$ | B1, M1, A1 | 0.918(36...), N2L: 0.918(36.)a = 4.33(01...) – 0.1x6.5 where friction = 0.1x(9-component 5) |
|---|---|---|
| Lower mass: Tractive force = $4 + 0.1(9-5\cos60)$ (= 4.65), Max Friction = 0.877(3x(9-5\cos60)$ (= 8.33), Tractive force < Max Friction, $a = 0$ | M1, A1, A1 | Compares TF (tractive force) and max friction |
| OR for Lower Mass: $ma = 4+0.1(9-5\cos60)-0.877(3+9-5\cos60)$, -ve a caused by friction impossible, hence $a = 0$ | M1, A1, A1 | N2L with 3 force terms: |7
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4b703cf9-b3d3-4210-b57b-89136595f8a5-04_305_748_260_699}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A rectangular block $B$ of weight 12 N lies in limiting equilibrium on a horizontal surface. A horizontal force of 4 N and a coplanar force of 5 N inclined at $60 ^ { \circ }$ to the vertical act on $B$ (see Fig. 1).\\
(i) Find the coefficient of friction between $B$ and the surface.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4b703cf9-b3d3-4210-b57b-89136595f8a5-04_307_751_1000_696}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
$B$ is now cut horizontally into two smaller blocks. The upper block has weight 9 N and the lower block has weight 3 N . The 5 N force now acts on the upper block and the 4 N force now acts on the lower block (see Fig. 2). The coefficient of friction between the two blocks is $\mu$.\\
(ii) Given that the upper block is in limiting equilibrium, find $\mu$.\\
(iii) Given instead that $\mu = 0.1$, find the accelerations of the two blocks.
\hfill \mbox{\textit{OCR M1 2010 Q7 [14]}}