CAIE A-Level Further Math 3.2.4 Centre of mass

Practise finding centres of mass for composite laminae and solids by replacing parts with equivalent particles and taking moments.

[Maximum number: 4]

Figure for Question 3.2.4—Centre of mass question 1 — CAIE A-Level Further Math AS

A uniform square lamina ABCD has sides of length 10 cm. The point E is on BC with $E C=7.5 \mathrm{~cm}$ , and the point F is on DC with $C F=x \mathrm{~cm}$ . The triangle EFC is removed from ABCD (see diagram). The centre of mass of the resulting shape ABEFD is a distance $\bar{x} \mathrm{~cm}$ from CB and a distance $\bar{y} \mathrm{~cm}$ from CD.

Show that $\bar{x}=\frac{400-x^{2}}{80-3 x}$ and find a corresponding expression for $\bar{y}$ .

The shape ABEFD is in equilibrium in a vertical plane with the edge DF resting on a smooth horizontal surface.

\cline { 2 - 4 } \multicolumn{1}{c|}{} & Area &
Centre of mass

from BC

Centre of mass

from DC

Square

100

5

Triangle

1 / 2 x.15 / 2

$\frac{1}{3} x$ $\frac{5}{2}$

Shape ABEFD

$100-\frac{15}{4} x$ $\bar{x}$ $\bar{y}$

\end{tabular}

Take moments about BC:
$\left(100-\frac{15}{4} x\right) \sigma \cdot \bar{x}=500 \sigma-\frac{15}{4} x \sigma \cdot \frac{1}{3} x$

(M1 for all terms present)
\end{tabular}

M1

$\bar{x}=\frac{400-x^{2}}{80-3 x}$ AG

A1

Take moments about DC:
$\left(100-\frac{15}{4} x\right) \cdot \bar{y}=100 \times 5-\frac{15}{4} x \cdot \frac{5}{2}$

M1

$\bar{y}=\frac{800-15 x}{160-6 x}$

A1

M1 for all terms present
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3.2.4—Centre of mass question 1