OCR MEI Further Mechanics A AS Specimen — Question 4 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics A AS (Further Mechanics A AS)
SessionSpecimen
Marks8
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Mark schemeDownload PDF ↗
TopicMoments
TypeRod on smooth peg or cylinder
DifficultyStandard +0.3 This is a straightforward moments equilibrium problem with standard steps: taking moments about P to find T's vertical component (already shown), then resolving forces vertically and horizontally to find components of F, and finally applying friction law. The setup is clear, calculations are routine, and it requires only standard A-level mechanics techniques with no novel insight.
Spec3.03r Friction: concept and vector form3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force
4 Fig. 4 shows a thin rigid non-uniform rod PQ of length 0.5 m . End P rests on a rough circular peg. A force of \(T \mathrm {~N}\) acts at the end Q at \(60 ^ { \circ }\) to QP . The weight of the rod is 40 N and its centre of mass is 0.3 m from P . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{be1851d6-af11-40e1-8a36-5938ee7864d4-4_506_960_977_605} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} The rod does not slip on the peg and is in equilibrium with PQ horizontal.
  1. Show that the vertical component of \(T\) is 24 N .
  2. \(F\) is the contact force at P between the rod and the peg. Find
    • the vertical component of \(F\),
    • the horizontal component of \(F\).
    • Given that the rod is about to slip on the peg, find the coefficient of friction between the rod and the peg.
Question 4:
AnswerMarks Guidance
4(i) Let the vertical cpt be U N
a.c. moments about P
0.5U – 0.3×40 = 0
AnswerMarks
so U = 24M1
E1
AnswerMarks
[2]3.1b
1.1Moments with only and all
appropriate forces
Clearly shown
AnswerMarks Guidance
4(iii) As on the point of slipping,F (cid:32)F (cid:32)(cid:80)R
max
8 3 3
So (cid:80)(cid:32) (cid:32)
AnswerMarks
16 2M1
A1
AnswerMarks
[2]m2.2a
1.1
AnswerMarks
ie
UsingF (cid:32)(cid:80)Rand substituting
max
their X and Y
BC
AnswerMarks
FT their (ii)Accept 0.866 or 0.8660 
Question 4:
4 | (i) | Let the vertical cpt be U N
a.c. moments about P
0.5U – 0.3×40 = 0
so U = 24 | M1
E1
[2] | 3.1b
1.1 | Moments with only and all
appropriate forces
Clearly shown
4 | (iii) | As on the point of slipping,F (cid:32)F (cid:32)(cid:80)R
max
8 3 3
So (cid:80)(cid:32) (cid:32)
16 2 | M1
A1
[2] | m2.2a
1.1
i | e
UsingF (cid:32)(cid:80)Rand substituting
max
their X and Y
BC
FT their (ii) | Accept 0.866 or 0.8660
4 Fig. 4 shows a thin rigid non-uniform rod PQ of length 0.5 m . End P rests on a rough circular peg. A force of $T \mathrm {~N}$ acts at the end Q at $60 ^ { \circ }$ to QP . The weight of the rod is 40 N and its centre of mass is 0.3 m from P .

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{be1851d6-af11-40e1-8a36-5938ee7864d4-4_506_960_977_605}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

The rod does not slip on the peg and is in equilibrium with PQ horizontal.\\
(i) Show that the vertical component of $T$ is 24 N .\\
(ii) $F$ is the contact force at P between the rod and the peg.

Find

\begin{itemize}
  \item the vertical component of $F$,
  \item the horizontal component of $F$.\\
(iii) Given that the rod is about to slip on the peg, find the coefficient of friction between the rod and the peg.
\end{itemize}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS  Q4 [8]}}
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