Question 4 - A-Level Maths

OCR MEI Further Mechanics A AS 2018 June — Question 4 9 marks

Exam Board	OCR MEI
Module	Further Mechanics A AS (Further Mechanics A AS)
Year	2018
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Centre of Mass 1
Type	Suspended lamina equilibrium angle
Difficulty	Standard +0.3 This is a standard centre of mass question involving composite shapes (subtraction method) and suspended equilibrium. Part (i) is trivial geometry, part (ii) requires routine application of the composite body formula with area weighting, and part (iii) uses basic equilibrium (vertical through COM). All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average.
Spec	6.04e Rigid body equilibrium: coplanar forces

4 A uniform lamina ABDE is in the shape of an equilateral triangle ABC of side 12 cm from which an equilateral triangle of side 6 cm has been removed from corner \(C\). The lamina is situated on coordinate axes as shown in Fig. 4. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-4_501_753_406_646} \captionsetup{labelformat=empty} \caption{Fig. 4}

\end{figure}

Explain why angle \(\mathrm { BDA } = 90 ^ { \circ }\).
Find the coordinates of the centre of mass of the lamina ABDE . The lamina ABDE is now freely suspended from D and hangs in equilibrium.
Calculate the angle DE makes with the downward vertical.

Show mark scheme Show mark scheme source

Question 4:

Answer	Marks	Guidance
4	(i)	Equilateral triangle, plus symmetry or D is
midpoint of BC	B1
[1]	2.4	Oe, eg reference to median
(ii)	Masses in right ratio; 4 : 1 : 3

4×6−1×9 = 3×𝑥̅

Distance AD is √122 −62

4×2√3−1×√3 = 3×𝑦̅

7 (5, √3)

Answer	Marks
3	B1

M1

B1

M1

A1

Answer	Marks
[5]	3.1a

3.4

1.1

Answer	Marks
3.2a	May see 36√3ρ, 9√3ρ and 27√3ρ oe

Allow sign error; one mass and one

distance right, at least

AD = 6√3

Allow sign error; one mass and one

distance (ft AD) right, at least

Answer	Marks
Or (5,4.04(145))	Implied by COM 2√3 from D

or 4√3 from A

3 6

OR 𝑚( )+𝑚( )+

√3 4√3

6 𝑥̅

𝑚( )= 3𝑚( )

2√3 𝑦̅

Note: other methods may be

seen

Answer	Marks
(iii)	Use Centre of Mass vertically below D

3 angle EDG = tan−1 [+30]

7√3

Answer	Marks
Angle is 43.9°	M1*ft

*M1

A1

Answer	Marks
[3]	3.1b

1.1

Answer	Marks
2.2a	7

Must see 1 (or 6 – 5) and √3

3 Accept reciprocal; can be implied by

angle 13.9° or 76.1°

Answer	Marks
43.897… or 0.76616… rads	Ft from (ii)

Or find angle BDG (= 76.1°)

180 – 76.1 – 60 = 43.9°

Note: Can find DG and use sin

or cos

Question 4:
4 | (i) | Equilateral triangle, plus symmetry or D is
midpoint of BC | B1
[1] | 2.4 | Oe, eg reference to median
(ii) | Masses in right ratio; 4 : 1 : 3
4×6−1×9 = 3×𝑥̅
Distance AD is √122 −62
4×2√3−1×√3 = 3×𝑦̅
7
(5, √3)
3 | B1
M1
B1
M1
A1
[5] | 3.1a
3.4
1.1
1.1
3.2a | May see 36√3ρ, 9√3ρ and 27√3ρ oe
Allow sign error; one mass and one
distance right, at least
AD = 6√3
Allow sign error; one mass and one
distance (ft AD) right, at least
Or (5,4.04(145)) | Implied by COM 2√3 from D
or 4√3 from A
3 6
OR 𝑚( )+𝑚( )+
√3 4√3
6 𝑥̅
𝑚( )= 3𝑚( )
2√3 𝑦̅
Note: other methods may be
seen
(iii) | Use Centre of Mass vertically below D
3
angle EDG = tan−1 [+30]
7√3
Angle is 43.9° | M1*ft
*M1
A1
[3] | 3.1b
1.1
2.2a | 7
Must see 1 (or 6 – 5) and √3
3
Accept reciprocal; can be implied by
angle 13.9° or 76.1°
43.897… or 0.76616… rads | Ft from (ii)
Or find angle BDG (= 76.1°)
180 – 76.1 – 60 = 43.9°
Note: Can find DG and use sin
or cos

Show LaTeX source

4 A uniform lamina ABDE is in the shape of an equilateral triangle ABC of side 12 cm from which an equilateral triangle of side 6 cm has been removed from corner $C$. The lamina is situated on coordinate axes as shown in Fig. 4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-4_501_753_406_646}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

(i) Explain why angle $\mathrm { BDA } = 90 ^ { \circ }$.\\
(ii) Find the coordinates of the centre of mass of the lamina ABDE .

The lamina ABDE is now freely suspended from D and hangs in equilibrium.\\
(iii) Calculate the angle DE makes with the downward vertical.

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2018 Q4 [9]}}

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Q1 6 Q2 12 Q3 9 Q4 9 Q5 13 Q6 11