| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on inclined plane |
| Difficulty | Standard +0.8 This is a non-trivial statics problem requiring moments about a point, resolution of forces in two directions, and friction analysis. The non-uniform rod with specified centre of mass, combined with the angled applied force and inclined plank, requires careful geometric reasoning and systematic application of equilibrium conditions across multiple parts. While the techniques are standard A-level mechanics, the multi-step nature and geometric complexity place it above average difficulty. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) | a.c moments about A |
| Answer | Marks |
|---|---|
| T(cid:117)2.5sin(55(cid:14)20)(cid:16)900(cid:117)2cos20(cid:32)0 | A1 |
| A1 | 1.1 |
| 2.1 | 55 + 20 |
| Answer | Marks | Guidance |
|---|---|---|
| Tcos55(cid:117)2.5sin20(cid:14)Tsin55(cid:117)2.5cos20... | A1 | A1 |
| Answer | Marks |
|---|---|
| of T both correct | N |
| Answer | Marks |
|---|---|
| (cid:16)900(cid:117)2cos20(cid:32)0 | of T both correct |
| A1 | E |
| Answer | Marks |
|---|---|
| so T = 700.4457… so 700N (3 s. f.) AG | A1 |
| [4] | M |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (ii) | At A, take friction F → and normal reaction R ↑ |
| Answer | Marks |
|---|---|
| so (cid:80)(cid:116) 1.2315… so (cid:80)(cid:116) 1.23 (3 s. f.) | I |
| Answer | Marks |
|---|---|
| [5] | 3.1b |
| Answer | Marks |
|---|---|
| 3.2a | Use of Coulomb’s law. Allow use of |
Question 4:
4 | (i) | a.c moments about A | M1 | 3.1b | All required forces present and
resolved
either
T(cid:117)2.5sin(55(cid:14)20)(cid:16)900(cid:117)2cos20(cid:32)0 | A1
A1 | 1.1
2.1 | 55 + 20
All correct
or
Tcos55(cid:117)2.5sin20(cid:14)Tsin55(cid:117)2.5cos20... | A1 | A1 | N
Horizontal and vertical components
of T both correct | N
Horizontal and vertical components
(cid:16)900(cid:117)2cos20(cid:32)0 | of T both correct
A1 | E
All correct
so T = 700.4457… so 700N (3 s. f.) AG | A1
[4] | M
1.1
4 | (ii) | At A, take friction F → and normal reaction R ↑
C
Resolving → and ↑
F – Tcos55 = 0 E
R + Tsin55 – 900 = 0
P
F(cid:100)(cid:80)R
so (cid:80)(cid:116) 1.2315… so (cid:80)(cid:116) 1.23 (3 s. f.) | I
M1
A1
A1
M1
A1
[5] | 3.1b
1.1
1.1
3.4
3.2a | Use of Coulomb’s law. Allow use of
=
(cid:116)
Must have4 Fig. 4 shows a non-uniform rigid plank AB of weight 900 N and length 2.5 m . The centre of mass of the plank is at G which is 2 m from A . The end A rests on rough horizontal ground and does not slip. The plank is held in equilibrium at $20 ^ { \circ }$ above the horizontal by a force of $T \mathrm {~N}$ applied at B at an angle of $55 ^ { \circ }$ above the horizontal as shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{54711a46-83ce-4fb9-b6d3-53b264725c74-3_426_672_539_605}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Show that $T = 700$ (correct to 3 significant figures).\\
(ii) Determine the possible values of the coefficient of friction between the plank and the ground.
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor Q4 [9]}}