| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Resultant force on lamina |
| Difficulty | Standard +0.3 This is a standard A-level mechanics question on centres of mass and equilibrium. Parts (a) and (b) involve routine application of centre of mass formulas with given numerical values. Parts (c) and (d) require taking moments and resolving forces, which are standard techniques. The question is slightly easier than average because it's highly structured with clear guidance ('show that', specific values given), requires only straightforward algebraic manipulation, and involves no novel problem-solving insight. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | Let A be the origin of a coordinate system with x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Centre of mass of lamina lies 10 cm above AB | M1 | 1.2 |
| (π+1.7)Γ11.7 = 1.7Γ15+πΓ10 | M1 | 1.1 |
| ο m = 3 .3 | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | 3 .3 ο΄ 13 x + 1 .7 ο΄ 12 x = 5 ο΄ 1 6 .5 | M1 |
| Answer | Marks |
|---|---|
| M0 for β x | Can award |
| Answer | Marks | Guidance |
|---|---|---|
| ο x ο» 4 2 .3 (3 sf) | A1 | 1.1 |
| Alternatively, M lies on AC, so x / 30 = 11.7 / 16.5 | M1 | |
| οxο»42.3 (3 sf) | A1 | Accept a value rounding to 42.3 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | Q s i n 3 0 ο° ο΄ 3 2 = 5 g ο΄ 1 6 .5 | M1 |
| ο Q = 5 0 .5 3 1 2 5 ο» 5 0 .5 | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (d) | Let the tension in the string be T N. |
| Answer | Marks | Guidance |
|---|---|---|
| (πsinπ =) πcos30Β° (= 43.76) | M1 | 3.3 |
| Answer | Marks |
|---|---|
| cos switched. | May take |
| Answer | Marks | Guidance |
|---|---|---|
| t a n 1 .8 4 3 7 ο¦ = | M1 | 3.1a |
| 6 1 .5 2 6 3 ο¦ ο = ο° | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| XN = 15.5tan30Β° | M1 | Attempt to find XN |
| tanπ = 16.5/XN | tanπ = 16.5/XN | M1 |
| οο¦=61.5263 ο° | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| T2 = (5g)2 + 50.532 β 2(5g)(50.53)cos60 (T = 49.78) | M1 | Equation involving T |
| sinΟ / 50.53 = sin60 / T | M1 | |
| Ο = 61.5 | A1 |
Question 3:
3 | (a) | Let A be the origin of a coordinate system with x and y axes
pointing in directions A B and A D respectively. Let the
lamina have mass m kg.
Centre of mass of lamina lies 10 cm above AB | M1 | 1.2 | soi
(π+1.7)Γ11.7 = 1.7Γ15+πΓ10 | M1 | 1.1 | M0 if 16.5 used instead of 11.7 | Using their 10
ο m = 3 .3 | A1 | 2.2a | AG
[3]
(b) | 3 .3 ο΄ 13 x + 1 .7 ο΄ 12 x = 5 ο΄ 1 6 .5 | M1 | 1.1 | Allow equation involving m
Allow 11.7 if 16.5 used in (a)
M0 for β
x | Can award
both marks for
work in (a)
ο x ο» 4 2 .3 (3 sf) | A1 | 1.1 | Accept a value rounding to 42.3
Alternatively, M lies on AC, so x / 30 = 11.7 / 16.5 | M1
οxο»42.3 (3 sf) | A1 | Accept a value rounding to 42.3
[2]
(c) | Q s i n 3 0 ο° ο΄ 3 2 = 5 g ο΄ 1 6 .5 | M1 | 1.1 | Allow omission of g
ο Q = 5 0 .5 3 1 2 5 ο» 5 0 .5 | A1 | 1.1 | AG
[2]
(d) | Let the tension in the string be T N.
(πcosπ =) 5πβπsin30Β° (= 23.73)
(πsinπ =) πcos30Β° (= 43.76) | M1 | 3.3 | Resolving into horizontal and
vertical components. Condone sin /
cos switched. | May take
moments
t a n 1 .8 4 3 7 ο¦ = | M1 | 3.1a | Using tan. Allow tanΛ1(23.7/43.8)
6 1 .5 2 6 3 ο¦ ο = ο° | A1 | 1.1 | Allow 1.073838β¦ rad | Accept 61 β 62
Alternatively: all three lines of action must concur, so β¦
16.5 cm N 15.5 cm
30Β°
X
XN = 15.5tan30Β° | M1 | Attempt to find XN
tanπ = 16.5/XN | tanπ = 16.5/XN | M1 | M1 | Using tan in triangle including Ο | Using tan in triangle including Ο
οο¦=61.5263 ο° | A1
Alternatively, using triangle of forces
T2 = (5g)2 + 50.532 β 2(5g)(50.53)cos60 (T = 49.78) | M1 | Equation involving T
sinΟ / 50.53 = sin60 / T | M1
Ο = 61.5 | A1
[3]
3 Fig. 3.1 shows a thin rectangular frame ABCD , with part of it filled by a triangular lamina ABD . $\mathrm { AD } = 30 \mathrm {~cm}$ and $\mathrm { AB } = x \mathrm {~cm}$. Together they form the composite structure S .
The centre of mass of $S$ lies at a point $M , 16.5 \mathrm {~cm}$ from $A D$ and 11.7 cm from $A B$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d1ec7861-dc8b-450b-8e05-c70479ab0dc2-4_572_953_450_242}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
The frame and the triangular lamina are both uniform but made of different materials. The mass of the frame is 1.7 kg .
\begin{enumerate}[label=(\alph*)]
\item Show that the triangular lamina has a mass of 3.3 kg .
\item Determine the value of $x$, correct to $\mathbf { 3 }$ significant figures.
One end of a light inextensible string is attached to S at D . The other end is attached to a fixed point on a vertical wall. For S to hang in equilibrium with AD vertical, a force of magnitude $Q N$ is applied to S as shown in Fig. 3.2. The line of action of this force lies in the same plane as S . The string is taut and lies in the same plane as S at an angle $\phi$ to the downward vertical.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d1ec7861-dc8b-450b-8e05-c70479ab0dc2-4_611_994_1756_242}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
\item By taking moments about D , show that $Q = 50.5$, correct to 3 significant figures.
\item Determine, in degrees, the value of $\phi$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2022 Q3 [10]}}