OCR MEI Further Mechanics A AS 2020 November — Question 3 9 marks

Exam BoardOCR MEI
ModuleFurther Mechanics A AS (Further Mechanics A AS)
Year2020
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeResultant force on lamina
DifficultyStandard +0.3 This is a standard equilibrium problem requiring resolution of forces and taking moments about a point. While it involves multiple parts and both force resolution and moment calculations, the techniques are straightforward applications of AS-level mechanics principles with no novel problem-solving required. The geometry is clearly defined and the solution path is routine for students who have practiced equilibrium problems.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces
3 Fig. 3 shows a light square lamina ABCD , of side length 0.75 m , suspended vertically by wires attached to A and B so that AB is horizontal. A particle P of mass \(m \mathrm {~kg}\) is attached to the edge DC . The lamina hangs in equilibrium. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b20e2254-955e-466c-8161-9614d8ccdba0-3_586_702_404_251} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure} The tension in the wire attached to A is 14 N and the tension in the wire attached to B is \(T \mathrm {~N}\). The wire at A makes an angle of \(25 ^ { \circ }\) with the horizontal and the wire at B makes an angle of \(60 ^ { \circ }\) with the horizontal.
  1. Determine the value of \(T\).
  2. Determine
    1. the value of \(m\),
    2. the distance of P from D . P is moved to the midpoint of CD . A couple is applied to the lamina so that it remains in equilibrium with AB horizontal and the tension in both wires unchanged.
  3. Determine
Question 3:
AnswerMarks Guidance
3(a) Tcos60=14cos25
– allow sign errors and cos/sin confusion
AnswerMarks Guidance
T =25.4(N)A1 1.1
[2]
AnswerMarks Guidance
(b)(i) Tsin60+14sin25=mg
allow sign errors and cos/sin confusion
AnswerMarks Guidance
mg =27.893⇒m=2.85A1 1.1
(ii)mgx=Tsin60°×0.75
or (0.75−x)mg =0.75(14sin25)M1 3.3
Taking moments about A (or B) – correct number of
terms and with tension at either B or A resolved
AnswerMarks Guidance
x=0.591(m)A1 2.2a
[4]
AnswerMarks
(c)e.g.
• Midpoint AB: 14sin25°×0.375−Tsin60°×0.375
(clockwise)
• A: mg×0.375−Tsin60°×0.75 (clockwise)
• B: 14sin25°×0.75−mg×0.375(clockwise)
• Midpoint CD:
14sin25°×0.375+Tcos60°×0.75−14cos25°×0.75−
AnswerMarks Guidance
clockwiseM1
T3.1b Plate remains in translational equilibrium so total
turning effect of the two tensions and the weight is the
same about any point
Correct number of terms required (dependent on
point at which moments is taken)
=−6.02 [accept (−)6.03, from method of considering change
in mom of P: ]
AnswerMarks Guidance
(0.591−0.375)×2.85𝑔𝑔A1ft 1.1
either positive or negative following through their
values from earlier parts
So the couple has magnitude 6.02 Nm in the clockwise
AnswerMarks Guidance
directionA1 1.2
[3]
Question 3:
3 | (a) | Tcos60=14cos25 | M1 | 3.3 | Resolving horizontally – correct number of terms
– allow sign errors and cos/sin confusion
T =25.4(N) | A1 | 1.1 | For reference: 25.37661804…
[2]
(b) | (i) | Tsin60+14sin25=mg | M1 | 3.3 | Resolving vertically – correct number of terms –
allow sign errors and cos/sin confusion
mg =27.893⇒m=2.85 | A1 | 1.1 | For reference: 2.846270566…
(ii) | mgx=Tsin60°×0.75
or (0.75−x)mg =0.75(14sin25) | M1 | 3.3 | Where x is the distance of the particle from D.
Taking moments about A (or B) – correct number of
terms and with tension at either B or A resolved
x=0.591(m) | A1 | 2.2a | For reference: 0.5909127769…
[4]
(c) | e.g.
• Midpoint AB: 14sin25°×0.375−Tsin60°×0.375
(clockwise)
• A: mg×0.375−Tsin60°×0.75 (clockwise)
• B: 14sin25°×0.75−mg×0.375(clockwise)
• Midpoint CD:
14sin25°×0.375+Tcos60°×0.75−14cos25°×0.75−
clockwise | M1
T | 3.1b | Plate remains in translational equilibrium so total
turning effect of the two tensions and the weight is the
same about any point
Correct number of terms required (dependent on
point at which moments is taken)
=−6.02 [accept (−)6.03, from method of considering change
in mom of P: ]
(0.591−0.375)×2.85𝑔𝑔 | A1ft | 1.1 | For reference: 6.022552582 – accept answer as
either positive or negative following through their
values from earlier parts
So the couple has magnitude 6.02 Nm in the clockwise
direction | A1 | 1.2 | cao
[3]
3 Fig. 3 shows a light square lamina ABCD , of side length 0.75 m , suspended vertically by wires attached to A and B so that AB is horizontal. A particle P of mass $m \mathrm {~kg}$ is attached to the edge DC . The lamina hangs in equilibrium.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{b20e2254-955e-466c-8161-9614d8ccdba0-3_586_702_404_251}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

The tension in the wire attached to A is 14 N and the tension in the wire attached to B is $T \mathrm {~N}$. The wire at A makes an angle of $25 ^ { \circ }$ with the horizontal and the wire at B makes an angle of $60 ^ { \circ }$ with the horizontal.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $T$.
\item Determine
\begin{enumerate}[label=(\roman*)]
\item the value of $m$,
\item the distance of P from D .

P is moved to the midpoint of CD . A couple is applied to the lamina so that it remains in equilibrium with AB horizontal and the tension in both wires unchanged.
\end{enumerate}\item Determine

\begin{itemize}
  \item the magnitude of the couple,
  \item the direction of the couple.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2020 Q3 [9]}}
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