| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Prism or block on inclined plane |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring force resolution, friction at two surfaces, moment equilibrium about a strategic point, and algebraic manipulation to derive a specific expression. The multi-surface friction and applied force geometry make it substantially harder than typical A-level mechanics, though the mathematical techniques (resolving forces, taking moments) are standard for Further Maths students. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (b) | P 0 1 6 s i n 1 3 c o s 0 − |
| Answer | Marks | Guidance |
|---|---|---|
| 1 6 | A1 | 1.1 |
| For the prism to be in equilibrium t a n 4 | M1 | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 75.96... | A1 | 1.1 |
Question 11:
11 | (b) | P 0 1 6 s i n 1 3 c o s 0 − | M1 | 3.4 | Considers denominator of given answer in (a) either >
0 or = 0 or 0
1 3
t a n 3 9 .0 9 ...
1 6 | A1 | 1.1 | awrt 39
For the prism to be in equilibrium t a n 4 | M1 | 3.1b | y
Considers tan= (or reciprocal) for triangular
x
lamina
75.96... | A1 | 1.1 | awrt 76 (For full marks must imply an interval
between these two values)
[4]\includegraphics{figure_11}
The diagram shows the cross-section through the centre of mass of a uniform solid prism. The cross-section is a right-angled triangle ABC, with AB perpendicular to AC, which lies in a vertical plane. The length of AB is 3 cm, and the length of AC is 12 cm.
The prism is resting in equilibrium on a horizontal surface and against a vertical wall. The side AC of the prism makes an angle $\theta$ with the horizontal.
A horizontal force of magnitude $P$ N is now applied to the prism at B. This force acts towards the wall in the vertical plane which passes through the centre of mass G of the prism and is perpendicular to the wall.
The weight of the prism is 15 N and the coefficients of friction between the prism and the surface, and between the prism and the wall, are each $\frac{1}{2}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the least value of $P$ needed to move the prism is given by
$$P = \frac{40 \cos \theta + 95 \sin \theta}{16 \sin \theta - 13 \cos \theta}.$$ [8]
\item Determine the range in which the value of $\theta$ must lie. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q11 [12]}}