| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Resultant force on lamina |
| Difficulty | Standard +0.3 This is a straightforward application of the center of mass formula for a uniform triangular lamina (centroid at 1/3 from each side) followed by basic equilibrium geometry. The question requires knowing the standard result for a triangle's centroid and applying simple trigonometry to find the angle when suspended, which is slightly above average difficulty due to the mechanics context but remains a standard textbook exercise. |
| Spec | 6.04b Find centre of mass: using symmetry6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| 2 | Suppose X is the midpoint of BC |
| Answer | Marks |
|---|---|
| = 38.6598… so angle is 38.7° (3 s. f.) | B1 |
| Answer | Marks |
|---|---|
| [5] | m3.1b |
| Answer | Marks |
|---|---|
| 1.1 | n |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 3 | 1 |
| 3i | 2 | 0 |
| 3ii | 1 | 1 |
| 3iiiA | 2 | 0 |
| 3iiiB | 0 | 1 |
| 4i | 1 | 0 |
| 4ii | 3 | 0 |
| 4iii | 1 | 1 |
| 5i | 4 | 0 |
| 5ii | 2 | 1 |
| 5iii | 1 | 0 |
| 6i | 1 | 0 |
| 6ii | 2 | 0 |
| 6iii | 5 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 6ivA | 0 | 0 |
| 6ivB | 0 | 0 |
| 1 | 1 | |
| 7i | 1 | 0 |
| 7ii | 1 | 0 |
| 7iii | 1 | 0 |
| 0 | 1 | 2 |
| 7iv | 2 | 0 |
| 1 | 0 | 3 |
| 7v | 0 | 0 |
| Totals | 36 | c |
| 5 | 4 | 15 |
Question 2:
2 | Suppose X is the midpoint of BC
XC = 5 and AX = 12
CoM, G, is on AX, so XG = 4cm
CG is vertical and required angle is GCX
= arctan(4 )
e
5
= 38.6598… so angle is 38.7° (3 s. f.) | B1
M1
A1
c
M1
A1
[5] | m3.1b
1.1
i
1.1
2.1
1.1 | n
e
Finding length of the median
CoM 1 along median from the
3
base
GX = 4
soi
2 | 3 | 1 | 1 | 0 | 5
3i | 2 | 0 | 0 | 1 | 3
3ii | 1 | 1 | 0 | 1 | 3
3iiiA | 2 | 0 | 0 | 0 | 2
3iiiB | 0 | 1 | 0 | 0 | 1
4i | 1 | 0 | 1 | 0 | 2
4ii | 3 | 0 | 0 | 1 | 4
4iii | 1 | 1 | 0 | 0 | 2
5i | 4 | 0 | 0 | 0 | 4
5ii | 2 | 1 | 0 | 1 | 4
5iii | 1 | 0 | 1 | 0 | 2
6i | 1 | 0 | 0 | 0 | 1
6ii | 2 | 0 | 0 | 1 | 3
6iii | 5 | 0 | 0 | 2 | n
7
6ivA | 0 | 0 | 0 | 1 | 1
6ivB | 0 | 0 | 0 | e
1 | 1
7i | 1 | 0 | 0 | 2 | 3
7ii | 1 | 0 | 0 | 1 | 2
7iii | 1 | 0 | m
0 | 1 | 2
7iv | 2 | 0 | i
1 | 0 | 3
7v | 0 | 0 | 0 | 1 | 1
Totals | 36 | c
5 | 4 | 15 | 60
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S
122 A triangular lamina, ABC , is cut from a piece of thin uniform plane sheet metal. The dimensions of ABC are shown in Fig. 2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be1851d6-af11-40e1-8a36-5938ee7864d4-2_410_572_689_792}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
This piece of metal is freely suspended from a string attached to C and hangs in equilibrium.
Calculate the angle of BC with the downward vertical, giving your answer in degrees.
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS Q2 [5]}}