| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Hemisphere or sphere resting on plane or wall |
| Difficulty | Challenging +1.8 This is a challenging Further Mechanics question requiring 3D visualization of a composite body, careful force analysis with friction at limiting equilibrium at two contact points, resolution in two directions, and a moments equation about a strategic point. The final part requires algebraic manipulation to derive an inequality. While systematic, it demands strong spatial reasoning and multi-step problem-solving beyond standard A-level mechanics. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | B1 |
| [1] | 1.2 | All remaining forces adding on correctly |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (b) | F + R =W |
| Answer | Marks |
|---|---|
| C 10 | M1* |
| Answer | Marks |
|---|---|
| [5] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Resolve horizontally and vertically |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (c) | (r + hsinθ )W + (r + 2h cosθ ) F = (r + 2hsinθ )R |
| Answer | Marks |
|---|---|
| 4 | M1* |
| Answer | Marks |
|---|---|
| [6] | 3.1b |
| Answer | Marks |
|---|---|
| 2.2a | Taking moments about D (or any other |
Question 8:
8 | (a) | B1
[1] | 1.2 | All remaining forces adding on correctly
(with arrows to indicate directions) to the
figure in the Printed Answer Booklet
8 | (b) | F + R =W
D C
R = F
D C
1 1
F = R and F = R
D 3 D C 3 C
1 1
F + R = W ⇒ R + R = W
3 C C 9 C C
9
R = W
C 10 | M1*
A1
B1
M1dep*
A1
[5] | 3.3
1.1
3.4
3.4
1.1 | Resolve horizontally and vertically
(correct number of terms in both
equations)
Where R is the normal contact force at
C
C, etc.
Correct use of F = µ R at C and D
Combine results to get an equation in R
C
only
8 | (c) | (r + hsinθ )W + (r + 2h cosθ ) F = (r + 2hsinθ )R
C C
3
(r + h sinθ )W + (r + 2h cosθ ) W
10
9
= (r + 2h sinθ ) W
10
r = h(2sinθ −1.5cosθ )
2hsinθ −1.5hcosθ > 0
3
4sinθ − 3cosθ > 0 ⇒ tanθ >
4 | M1*
A1
M1dep*
A1
M1
A1
[6] | 3.1b
1.1
3.4
1.1
2.3
2.2a | Taking moments about D (or any other
equivalent point) – correct number of
terms
oe
Substitute expressions for F and R
C C
Setting their expression for r > 0
AG8 A capsule consists of a uniform hollow right circular cylinder of radius $r$ and length $2 h$ attached to two uniform hollow hemispheres of radius $r$.\\
The centres of the plane faces of the hemispheres coincide with the centres, A and B , of the ends of the cylinder.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{17e92314-d7df-49b8-a441-8d18c91dbbb0-06_702_684_445_244}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
Fig. 8 represents a vertical cross-section through a plane of symmetry of the capsule as it rests in limiting equilibrium with a point C of one hemisphere on a rough horizontal floor and a point D of the other hemisphere against a rough vertical wall.
The total weight of the capsule is $W$ and acts at a point midway between A and B . The plane containing $\mathrm { A } , \mathrm { B } , \mathrm { C }$ and D is vertical, with AB making an acute angle $\theta$ with the downward vertical.
\begin{enumerate}[label=(\alph*)]
\item Complete the copy of Fig. 8 in the Printed Answer Booklet to show all the remaining forces acting on the capsule.
The coefficient of friction at each point of contact is $\frac { 1 } { 3 }$.
\item By resolving vertically and horizontally, determine the magnitude of the normal contact force between the floor and the capsule in terms of $W$.
\item By determining an expression for $r$ in terms of $h$ and $\theta$, show that $\tan \theta > \frac { 3 } { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2021 Q8 [12]}}