| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2021 |
| Session | November |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Toppling on inclined plane |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass question requiring composite body calculations, toppling conditions on an inclined plane, and friction analysis. While it involves several steps and 3D visualization, the techniques are standard for Further Mechanics: finding COM by splitting into rectangles, applying toppling criteria (vertical line through COM must pass through base), and moments about pivot points. The calculations are methodical rather than requiring novel insight, making it moderately above average difficulty for A-level Further Maths. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | Let the coordinates of centre of mass be (x,y,z) |
| z =1.5 | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑦𝑦� 5 8.5 | M1 | 1.1 |
| Answer | Marks |
|---|---|
| parts. | e.g. could also |
| Answer | Marks | Guidance |
|---|---|---|
| x =5 | A1 | 1.1 |
| y =6.5 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | θ =arctan 1 or θ =arctan 5 | |
| min 6.5 max 6.5 | M1 | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| min | A1 | 1.1 |
| A1 | 1.1 | 37.568592… |
| Answer | Marks |
|---|---|
| (c) | 𝜃𝜃L 𝑚𝑚 et 𝑚𝑚 t 𝑚𝑚 he= th3r7e.s6holds for breaking equilibrium be sliding and |
| Answer | Marks | Guidance |
|---|---|---|
| s s | M1* | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| s max s | M1* | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| s cos30°+µsin30° | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| t t | M1* | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| t 14sin30°+10cos30° | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| µ(9sin30°+10cos30°)>5cos30° | M1dep* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝐿𝐿𝑜𝑜 𝜇𝜇𝑚𝑚𝑚𝑚𝑚𝑚= = | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| inequality is present. | 0.3290306… | |
| 3√3+10 73 | [7] | |
| (d) | Either | |
| If the angle. were smaller then would be smaller … | M1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| min | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| If the angle were smaller then P would have a larger | M1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| min | A1 | 2.2a |
Question 5:
5 | (a) | Let the coordinates of centre of mass be (x,y,z)
z =1.5 | B1 | 1.1
𝑥𝑥̅ 2 9
210 � � =120 � �+90 � �
𝑦𝑦� 5 8.5 | M1 | 1.1 | Any correct equation, using correct
ratio of masses of the constituent
parts. | e.g. could also
have
2 7
28 +42
3.5 8.5
x =5 | A1 | 1.1
y =6.5 | A1 | 1.1
[4]
(b) | θ =arctan 1 or θ =arctan 5
min 6.5 max 6.5 | M1 | 3.1b | 1 x
tanθ= or tanθ= - condone
y y
reciprocal fractions for this mark
θ =8.75
min | A1 | 1.1 | 8.746162…
A1 | 1.1 | 37.568592…
[3]
(c) | 𝜃𝜃L 𝑚𝑚 et 𝑚𝑚 t 𝑚𝑚 he= th3r7e.s6holds for breaking equilibrium be sliding and
toppling be P and P
s t
R+P sin30°=mg⇒R=mg−P sin30°
s s | M1* | 3.3 | Resolve vertically – correct number
of terms (allow sin/cos errors)
P cos30°=F =µ(mg−P sin30°)
s max s | M1* | 3.4 | Resolve horizontally and use of
F =µR
µmg
P =
s cos30°+µsin30° | A1 | 1.1 | oe
14Psin30°+10Pcos30°=5mg
t t | M1* | 3.1b | Moment – correct number of terms.
Condone sin/cos errors (and sign
errors)
5mg
P =
t 14sin30°+10cos30° | A1 | 1.1 | oe
µmg 5mg
P >P, so >
s t cos30°+µsin30° 14sin30°+10cos30°
µmg(14sin30°+10cos30°)>5mg(cos30°+µsin30°)
µ(9sin30°+10cos30°)>5cos30° | M1dep* | 2.1 | Dependent on all previous M marks
5
𝜇𝜇 >
9tan30°+10
5
𝜇𝜇 >
3√3+10
5 50−15√3
𝐿𝐿𝑜𝑜 𝜇𝜇𝑚𝑚𝑚𝑚𝑚𝑚= = | A1 | 2.2a | Accept any equivalent exact form,
or 0.329 (or better)
Need not be explicitly stated if
inequality is present. | 0.3290306…
3√3+10 73 | [7]
(d) | Either
If the angle. were smaller then would be smaller … | M1 | 2.1
… so µ would be larger.
min | A1 | 2.2a
𝜃𝜃, tan𝜃𝜃
Or
If the angle were smaller then P would have a larger | M1 | M1 | 2.1 | 2.1
horizontal component and a smaller anticlockwise turning
effect …
… so µ would be larger.
min | A1 | 2.2a
[2]5 Fig. 5.1 shows a solid L-shaped ornament, of uniform density. The ornament is 3 cm thick. The $x , y$ and $z$ axes are shown, along with the dimensions of the ornament. The measurements are in centimetres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b3e369f4-13f7-457b-9a43-04ed2e2a2bba-6_556_887_406_244}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Determine, with reference to the axes shown, the coordinates of the ornament's centre of mass.
Fig. 5.2 shows the ornament placed so that the shaded face (indicated in Fig. 5.1) is in contact with a plane inclined at $\theta ^ { \circ }$ to the horizontal, with the 4 cm edge parallel to a line of greatest slope. The surface of the plane is sufficiently rough so that the ornament will not slip down the plane.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b3e369f4-13f7-457b-9a43-04ed2e2a2bba-6_646_844_1452_242}
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
\end{center}
\end{figure}
\item Determine the minimum and maximum possible values of $\theta$ for which the ornament does not topple.
The ornament is now placed with its shaded face in contact with a rough horizontal surface. A force of magnitude $P$ N, acting parallel to the planes of the L -shaped faces, is applied to one of the edges of the ornament, as shown in Fig. 5.3. The force is inclined at an angle of $30 ^ { \circ }$ to the horizontal. The coefficient of friction between the ornament and the surface is $\mu$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b3e369f4-13f7-457b-9a43-04ed2e2a2bba-7_524_680_452_246}
\captionsetup{labelformat=empty}
\caption{Fig. 5.3}
\end{center}
\end{figure}
The value of $P$ is gradually increased until the ornament is on the point of toppling but does not slide.
\item Determine the minimum value of $\mu$.
\item Explain how your answer to part (c) would change if the angle between $P$ and the horizontal was less than $30 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2021 Q5 [16]}}