OCR MEI Further Mechanics Minor 2024 June — Question 6 11 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
Year2024
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRange of equilibrium positions
DifficultyChallenging +1.2 Part (a) is a straightforward couple calculation requiring one equation. Parts (b-c) involve standard limiting equilibrium with friction at two contacts, requiring moment and force resolution equations, but the algebra is guided ('show that') and the setup is conventional for Further Mechanics. More demanding than typical A-level mechanics but standard for Further Maths.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems
6 Fig. 6.1 shows three forces of magnitude \(15 \mathrm {~N} , 15 \mathrm {~N}\) and 30 N acting on a rigid beam AB of length 6 m . One of the forces of magnitude 15 N acts at A, and the other force of magnitude 15 N acts at B. The force of magnitude 30 N acts at distance of \(x \mathrm {~m}\) from B. All three forces act in a direction perpendicular to the beam as shown in Fig. 6.1. The beam and the three forces all lie in the same horizontal plane. The three forces form a couple of magnitude 42 Nm in the clockwise direction. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Fig. 6.1} \includegraphics[alt={},max width=\textwidth]{0a790ad0-7eda-40f1-9894-f156766ae46f-6_504_433_591_246}
\end{figure}
  1. Determine the value of \(x\). Fig. 6.2 shows the same beam, without the three forces from Fig. 6.1, resting in limiting equilibrium against a step. The point of contact, C , between the beam and the edge of the step lies 1.5 m from A. The other end of the beam rests on a horizontal floor. The contacts between the beam and both the step and the floor are rough. \begin{figure}[h]
    \captionsetup{labelformat=empty} \caption{Fig. 6.2} \includegraphics[alt={},max width=\textwidth]{0a790ad0-7eda-40f1-9894-f156766ae46f-6_348_412_1633_244}
    \end{figure} It is given that the beam is non-uniform, and that its centre of mass lies \(\sqrt { 3 } \mathrm {~m}\) from B .
  2. Draw a diagram to show all the forces acting on the beam. The coefficient of friction between the beam and the step and the coefficient of friction between the beam and the floor are the same, and are denoted by \(\mu\).
    1. Show that \(\mu ^ { 2 } - 6 \mu + 1 = 0\).
    2. Hence determine the value of \(\mu\).
Question 6:
AnswerMarks Guidance
6(a) Take moments about eg. B:
30x−156=42
AnswerMarks
 x = 4 .4M1
A1
AnswerMarks
[2]1.1
1.1Taking moments about a
suitable point and equating this
to the magnitude of the couple.
AnswerMarks
(b)R
F C
C
R
B
W
F
AnswerMarks
BB1
B1
AnswerMarks
[2]1.1
3.3Accept different labels in this
question. Dimensions need not
be shown.
Any 3 correct forces shown.
AnswerMarks
All forces correctly shown.Normal contact
force at C
perpendicular to
beam, acting away
from step, in
roughly the correct
position. Frictional
contact force at C
acting towards A
parallel to the
beam.
Weight vertically
downwards, in
roughly the correct
position (must be
clearly below the
halfway point).
Both frictional and
normal components
of the contact force
at B.
AnswerMarks
(i)R 4.5=W 3cos60
Taking moments about eg. B:
C
R + R s i n 3 0 + F s i n 6 0 = W
B C C
R c o s 3 0 = F c o s 6 0 + F
C C B
AnswerMarks
OrM1*
M1*
AnswerMarks
A13.1b
3.1b
AnswerMarks
1.1Taking moments about a
suitable point.
Attempt at N2L both
horizontally and vertically.
Must have correct number of
terms in each. Condone sin/cos
transposition.
Taking moments about CoM:
𝑅 √3cos60 ° = 𝑅 (4.5−√3)+𝐹 √3sin60
𝐵 𝑐 𝐵
R c o s 3 0 = F c o s 6 0 + F
AnswerMarks Guidance
C C BM1* 3.1b
M1*3.1b Resolving horizontally.
Condone sin/cos transposition.
AnswerMarks
A11.1
( )
32 R 12 F W 12 R 32 F  − = − −
C C C C
( )
32 R R W 12 R 32 R    − = − −
C 2 C C C
( ) ( ) ( ( ) ( ) )
39 32 39 1 12 39 32 39     − = − −
2
31 38 38 31 2     − = − −
8 1 1 8
AnswerMarks
2 6 1 0    − + =M1dep*
A1
AnswerMarks
[5]3.4
2.1Using both F =R and
B B
F R  = to obtain an equation
C C
in only
AG – so sufficient working
must be shown
Taking moments about CoM:
𝑅 √3cos60 ° = 𝑅 (4.5−√3)+𝐹 √3sin60
𝐵 𝑐 𝐵
R c o s 3 0 = F c o s 6 0 + F
C C B
The other root (3+2 2)cannot be the coefficient of friction
as e.g. when takes this value R  0 which is not
B
 (4.5− 3)W 
possible R =  as the direction of R must be
 B   4.5(1− 3) B
AnswerMarks Guidance
vertically upwards.B1
[2]2.3 Any valid reason for the
rejection of the larger of the
two roots – note that ‘because
1   ’ only is B0
PMT
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Question 6:
6 | (a) | Take moments about eg. B:
30x−156=42
 x = 4 .4 | M1
A1
[2] | 1.1
1.1 | Taking moments about a
suitable point and equating this
to the magnitude of the couple.
(b) | R
F C
C
R
B
W
F
B | B1
B1
[2] | 1.1
3.3 | Accept different labels in this
question. Dimensions need not
be shown.
Any 3 correct forces shown.
All forces correctly shown. | Normal contact
force at C
perpendicular to
beam, acting away
from step, in
roughly the correct
position. Frictional
contact force at C
acting towards A
parallel to the
beam.
Weight vertically
downwards, in
roughly the correct
position (must be
clearly below the
halfway point).
Both frictional and
normal components
of the contact force
at B.
(i) | R 4.5=W 3cos60
Taking moments about eg. B:
C
R + R s i n 3 0 + F s i n 6 0 = W
B C C
R c o s 3 0 = F c o s 6 0 + F
C C B
Or | M1*
M1*
A1 | 3.1b
3.1b
1.1 | Taking moments about a
suitable point.
Attempt at N2L both
horizontally and vertically.
Must have correct number of
terms in each. Condone sin/cos
transposition.
Taking moments about CoM:
𝑅 √3cos60 ° = 𝑅 (4.5−√3)+𝐹 √3sin60
𝐵 𝑐 𝐵
R c o s 3 0 = F c o s 6 0 + F
C C B | M1* | 3.1b
M1* | 3.1b | Resolving horizontally.
Condone sin/cos transposition.
A1 | 1.1
( )
32 R 12 F W 12 R 32 F  − = − −
C C C C
( )
32 R R W 12 R 32 R    − = − −
C 2 C C C
( ) ( ) ( ( ) ( ) )
39 32 39 1 12 39 32 39     − = − −
2
31 38 38 31 2     − = − −
8 1 1 8
2 6 1 0    − + = | M1dep*
A1
[5] | 3.4
2.1 | Using both F =R and
B B
F R  = to obtain an equation
C C

in only
AG – so sufficient working
must be shown
Taking moments about CoM:
𝑅 √3cos60 ° = 𝑅 (4.5−√3)+𝐹 √3sin60
𝐵 𝑐 𝐵
R c o s 3 0 = F c o s 6 0 + F
C C B
The other root (3+2 2)cannot be the coefficient of friction

as e.g. when takes this value R  0 which is not
B
 (4.5− 3)W 
possible R =  as the direction of R must be
 B   4.5(1− 3) B
vertically upwards. | B1
[2] | 2.3 | Any valid reason for the
rejection of the larger of the
two roots – note that ‘because
1   ’ only is B0
PMT
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If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
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Alternatively, you can email us on
support@ocr.org.uk
For more information visit
ocr.org.uk/qualifications/resource-finder
ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2024 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
6 Fig. 6.1 shows three forces of magnitude $15 \mathrm {~N} , 15 \mathrm {~N}$ and 30 N acting on a rigid beam AB of length 6 m . One of the forces of magnitude 15 N acts at A, and the other force of magnitude 15 N acts at B. The force of magnitude 30 N acts at distance of $x \mathrm {~m}$ from B. All three forces act in a direction perpendicular to the beam as shown in Fig. 6.1. The beam and the three forces all lie in the same horizontal plane. The three forces form a couple of magnitude 42 Nm in the clockwise direction.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
  \includegraphics[alt={},max width=\textwidth]{0a790ad0-7eda-40f1-9894-f156766ae46f-6_504_433_591_246}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $x$.

Fig. 6.2 shows the same beam, without the three forces from Fig. 6.1, resting in limiting equilibrium against a step. The point of contact, C , between the beam and the edge of the step lies 1.5 m from A. The other end of the beam rests on a horizontal floor. The contacts between the beam and both the step and the floor are rough.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
  \includegraphics[alt={},max width=\textwidth]{0a790ad0-7eda-40f1-9894-f156766ae46f-6_348_412_1633_244}
\end{center}
\end{figure}

It is given that the beam is non-uniform, and that its centre of mass lies $\sqrt { 3 } \mathrm {~m}$ from B .
\item Draw a diagram to show all the forces acting on the beam.

The coefficient of friction between the beam and the step and the coefficient of friction between the beam and the floor are the same, and are denoted by $\mu$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\mu ^ { 2 } - 6 \mu + 1 = 0$.
\item Hence determine the value of $\mu$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2024 Q6 [11]}}
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