Question:
Three particles of masses $50 \mathrm{~g}, 100 \mathrm{~g}$ and $150 \mathrm{~g}$ are placed at the vertices of an equilateral triangle of side $1 \mathrm{~m}$ (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be :
Correct Option: , 3
Solution:
(3) $x_{\mathrm{cm}}=\frac{50 \times 0+100 \times 1+150 \times 0.5}{50+100+150}$
$=\frac{7}{12} \mathrm{~m}$
$y_{\mathrm{cm}}=\frac{50 \times 0+100 \times 0+150 \times \frac{\sqrt{3}}{2}}{50+100+150}$
$=\frac{\sqrt{3}}{4} \mathrm{~m}$
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