| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Prism or block on inclined plane |
| Difficulty | Standard +0.8 This is a Further Mechanics question requiring understanding of limiting equilibrium, triangle of forces, and toppling vs sliding analysis. Part (d) requires comparing critical angles for two failure modes—a non-routine problem-solving step beyond standard textbook exercises, though the individual techniques (moments about edge, friction conditions) are well-established. The multi-part structure and conceptual depth place it moderately above average difficulty. |
| Spec | 3.03a Force: vector nature and diagrams3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | Block is in equilibrium so the forces sum to zero. |
| Hence the vectors can be joined to form a triangle. | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Sides F and R perpendicular. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| sides R and W. | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | If block is in limiting equilibrium, F R = . | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| R | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (d) | Let and be the angles at which the block first slides |
| Answer | Marks | Guidance |
|---|---|---|
| T 8 .9 T | M1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| T S | A1 | 2.2a |
Question 2:
2 | (a) | Block is in equilibrium so the forces sum to zero. | B1 | 2.4
Hence the vectors can be joined to form a triangle. | B1 | 1.1
[2]
(b) | Sides F and R perpendicular. | B1 | 1.1 | Either explicitly marked, or
convincingly drawn.
Side lengths labelled F, R and W, with included between
sides R and W. | B1 | 1.1
[2]
(c) | If block is in limiting equilibrium, F R = . | M1 | 3.4 | F R = must be stated or implied
by substitution for F or seen in
diagram in part (b).
So tan=R =
R | A1 | 1.1 | AG
.
[2]
(d) | Let and be the angles at which the block first slides
S T
and topples, respectively.
From (c), t a n S 1 .3 5 = ( S 5 3 .4 7 1 ) =
t a n 1 1 .6 1 .3 0 3 = = ( 5 2 .5 0 3 ) =
T 8 .9 T | M1 | 2.1 | Angle needn’t be calculated
explicitly.
( ) so the block will topple first.
T S | A1 | 2.2a | Conclusion clearly stated.
[2]2 The diagram below shows the cross-section through the centre of mass of a uniform block of weight $W \mathrm {~N}$, resting on a slope inclined at an angle $\alpha$ to the horizontal. The cross-section is a rectangle ABCD . The slope exerts a frictional force of magnitude $F \mathrm {~N}$ and a normal contact force of magnitude $R \mathrm {~N}$.\\
\includegraphics[max width=\textwidth, alt={}, center]{9b624694-edb6-4000-838f-3557e078952d-3_546_940_450_242}
\begin{enumerate}[label=(\alph*)]
\item Explain why a triangle of forces may be used to model the scenario.
\item In the space provided in the Printed Answer Booklet, draw such a triangle, fully annotated, including the angle $\alpha$ in the correct position.
The coefficient of friction between the block and the slope is $\mu$.
\item Given that the block is in limiting equilibrium, use your diagram in part (b) to show that $\mu = \tan \alpha$.
It is given that $\mathrm { AB } = 8.9 \mathrm {~cm}$ and $\mathrm { AD } = 11.6 \mathrm {~cm}$. The coefficient of friction between the slope and the block is 1.35 . The slope is slowly tilted so that $\alpha$ increases.
\item Determine whether the block topples first without sliding or slides first without toppling.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2022 Q2 [8]}}