| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Prism or block on inclined plane |
| Difficulty | Challenging +1.2 This is a multi-part moments question requiring force diagrams, equilibrium verification, couple analysis, and comparison of toppling vs sliding conditions. While it involves several steps and the toppling analysis requires taking moments about an edge, the techniques are standard for Further Mechanics and the cylinder geometry is straightforward. The question guides students through each stage systematically, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | Forces W, F and R marked correctly |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | tan15° = 0.268 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (𝜇 =)tan15° < 0.3 so does not slip | E1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | E1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | 𝐶 +5𝑔(12sin15°) > 5𝑔(5cos15°) | M1 |
| cylinder; C is mag of couple | Allow = or any inequality |
| Answer | Marks | Guidance |
|---|---|---|
| 𝐶 > 84.5 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝐶 ≤ 5𝑔(12sin15°)+5𝑔(5cos15°) | M1 | 3.3 |
| 𝐶 ≤ 389 | 388.837… accept 39.7g | |
| 84.5 < 𝐶 ≤ 389 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | Will slide when tan𝛼 = 0.3 | M1 |
| α = 16.7° | 16.699 |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | M1 | 3.4 |
| α = 22.6° | For both angles correct 22.61986 | |
| Will slide first | A1 | 2.2a |
Question 6:
6 | (a) | Forces W, F and R marked correctly | B1 | 2.5 | And no others –R within base
[1]
(b) | tan15° = 0.268 | M1 | 1.1 | May see 𝐹 = 𝑊sin15° and / or
𝑅 = 𝑊cos15°
(𝜇 =)tan15° < 0.3 so does not slip | E1 | 2.4 | Accept in equilibrium
5
tan−1 = 22.6(1986…)°
12 | M1 | 1.1 | Or equivalent method
tan−1 5 > 15° so does not topple
12 | E1 | 2.4 | Accept in equilibrium
[4]
(c) | 𝐶 +5𝑔(12sin15°) > 5𝑔(5cos15°) | M1 | 3.3 | Moment about centre of base of
cylinder; C is mag of couple | Allow = or any inequality
for M1A1 (X2)
𝐶 > 84.5 | A1 | 1.1 | 84.466… accept 8.62g | Could take moments
about other points.
𝐶 ≤ 5𝑔(12sin15°)+5𝑔(5cos15°) | M1 | 3.3
𝐶 ≤ 389 | 388.837… accept 39.7g
84.5 < 𝐶 ≤ 389 | A1 | 1.1 | cao
[4]
(d) | Will slide when tan𝛼 = 0.3 | M1 | 3.4
α = 16.7° | 16.699
Will topple when 𝛼 = tan−1 5
12 | M1 | 3.4
α = 22.6° | For both angles correct 22.61986
Will slide first | A1 | 2.2a | www
[3]
PPMMTT
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© OCR 20196 A uniform solid cylinder, L, has base radius 5 cm , height 24 cm and mass 5 kg . L is placed on a rough plane inclined at an angle $\alpha$ to the horizontal, as shown in Fig. 6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b808042-95b8-4862-8355-3979c1981089-5_431_951_351_242}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item On the copy of Fig. 6 in the Printed Answer Booklet mark the forces acting on L .
The coefficient of friction between L and the plane is 0.3 . Initially $\alpha$ is $15 ^ { \circ }$.
\item Show that L rests in equilibrium on the plane.
A couple is applied to L . It is given that L will topple if the couple is applied in an anticlockwise sense, but L will not topple if the couple is applied in a clockwise sense.
\item Find the range of possible values of the magnitude of the couple.
The couple is now removed and the plane is slowly tilted so that $\alpha$ increases.
\item Determine whether L topples first without sliding or slides first without toppling.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2019 Q6 [12]}}