OCR MEI Further Mechanics Major 2019 June — Question 4 6 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2019
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeLamina with attached triangle
DifficultyStandard +0.8 This is a composite lamina centre of mass problem requiring geometric analysis to find coordinates, then application of equilibrium principles. The geometry setup (finding C's position from the isosceles triangle constraint with FC = a) requires careful coordinate work, followed by standard but multi-step centre of mass calculations for composite shapes, then a straightforward equilibrium angle calculation. The geometric reasoning and composite mass handling elevate this above routine questions, but it follows established Further Mechanics patterns without requiring exceptional insight.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_4} Fig. 4 shows a uniform lamina ABCDE such that ABDE is a rectangle and BCD is an isosceles triangle. AB = 5a, AE = 4a and BC = CD. The point F is the midpoint of BD and FC = a.
  1. Find, in terms of \(a\), the exact distance of the centre of mass of the lamina from AE. [4]
The lamina is freely suspended from B and hangs in equilibrium.
  1. Find the angle between AB and the downward vertical. [2]
Question 4:
AnswerMarks Guidance
4(a) 2.5a 20a2    5a 1 a     1 4aa  ...
 3 2 
 1 
x20a2  4aa 
 2 
91
x  a
AnswerMarks
33M1
A1
A1
A1
AnswerMarks
[4]2.1
1.1
1.1
AnswerMarks
2.2aTable of values idea – correct number
of terms, dimensionally consistent
Must be exact and must come from
AnswerMarks
exact workingAllow one a slip – be
on the look out for
moments about F and C
For reference only:
2.757575…
AnswerMarks Guidance
4(b) 2a
tan
5ax
AnswerMarks
41.7M1
A1
AnswerMarks
[2]1.1
1.1Tan of a relevant angle; allow
reciprocal
oe (radians: 0.728317…) – answer
AnswerMarks
correct to at least 3 sf2a
M0 for tan
x
41.729512…
Question 4:
4 | (a) | 2.5a 20a2    5a 1 a     1 4aa  ...
 3 2 
 1 
x20a2  4aa 
 2 
91
x  a
33 | M1
A1
A1
A1
[4] | 2.1
1.1
1.1
2.2a | Table of values idea – correct number
of terms, dimensionally consistent
Must be exact and must come from
exact working | Allow one a slip – be
on the look out for
moments about F and C
For reference only:
2.757575…
4 | (b) | 2a
tan
5ax
41.7 | M1
A1
[2] | 1.1
1.1 | Tan of a relevant angle; allow
reciprocal
oe (radians: 0.728317…) – answer
correct to at least 3 sf | 2a
M0 for tan
x
41.729512…
\includegraphics{figure_4}

Fig. 4 shows a uniform lamina ABCDE such that ABDE is a rectangle and BCD is an isosceles triangle. AB = 5a, AE = 4a and BC = CD. The point F is the midpoint of BD and FC = a.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the exact distance of the centre of mass of the lamina from AE. [4]
\end{enumerate}

The lamina is freely suspended from B and hangs in equilibrium.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the angle between AB and the downward vertical. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2019 Q4 [6]}}
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