Standard +0.3 This is a straightforward centre of mass problem for composite shapes. Students need to find the centroids of two triangles (each at 1/3 of the median from the base), calculate their areas as weights, and apply the standard formula for combined centre of mass. The geometry is clearly defined with no tricky setup, making this easier than average despite being Further Maths content.
\includegraphics{figure_1}
A uniform lamina \(ABCD\) consists of two isosceles triangles \(ABD\) and \(BCD\). The diagonals of \(ABCD\) meet at the point \(O\). The length of \(AO\) is \(3a\), the length of \(OC\) is \(6a\) and the length of \(BD\) is \(16a\) (see diagram).
Find the distance of the centre of mass of the lamina from \(DB\). [3]
ABCD can be split in other ways, for example ADC and ABC.
Taking moments about DB:
72a2x = 24a2 ×−a + 48a2 ×2a
OR
Taking moments about A:
72a2x = 24a2×2a+48a2 ×5a
OR
Taking moments about G:
Answer
Marks
Guidance
24a 2( x+a )= 48a2 ×( 2a−x )
M1
Moments equation with masses in correct ratio.
x =a
A1
CWO
Alternative method for question 1
6a−3a
ADC: distance of centre of mass from BD = =a
3
6a−3a
ABC: distance of centre of mass from BD = =a
Answer
Marks
Guidance
3
B1
One calculation.
Second calculation or statement about symmetry
M1
x =a
A1
3
Answer
Marks
Area
Centre of mass
from DB
Answer
Marks
Guidance
ABD
24a2
– a
BCD
48a2
2a
Combined
72a2
x
Question
Answer
Marks
Question 1:
1 | Centre of mass
Area
from DB
ABD 24a2 – a
BCD 48a2 2a
Combined 72a2 x | B1 | All distances correct.
ABCD can be split in other ways, for example ADC and ABC.
Taking moments about DB:
72a2x = 24a2 ×−a + 48a2 ×2a
OR
Taking moments about A:
72a2x = 24a2×2a+48a2 ×5a
OR
Taking moments about G:
24a 2( x+a )= 48a2 ×( 2a−x ) | M1 | Moments equation with masses in correct ratio.
x =a | A1 | CWO
Alternative method for question 1
6a−3a
ADC: distance of centre of mass from BD = =a
3
6a−3a
ABC: distance of centre of mass from BD = =a
3 | B1 | One calculation.
Second calculation or statement about symmetry | M1
x =a | A1
3
Area | Centre of mass
from DB
ABD | 24a2 | – a
BCD | 48a2 | 2a
Combined | 72a2 | x
Question | Answer | Marks | Guidance
\includegraphics{figure_1}
A uniform lamina $ABCD$ consists of two isosceles triangles $ABD$ and $BCD$. The diagonals of $ABCD$ meet at the point $O$. The length of $AO$ is $3a$, the length of $OC$ is $6a$ and the length of $BD$ is $16a$ (see diagram).
Find the distance of the centre of mass of the lamina from $DB$. [3]
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q1 [3]}}