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The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is _______.
$$\text{Total Mass } (M) = 2 + 3 + 15 = 20\text{ kg}$$
$$X_{\text{cm}} = \frac{2 \times (-5\sqrt{3}) + 3 \times (5\sqrt{3}) + 15 \times 0}{20} = \frac{\sqrt{3}}{4}\text{ cm}$$
$$Y_{\text{cm}} = \frac{2 \times \left(-\frac{5}{2}\right) + 3 \times \left(-\frac{5}{2}\right) + 15 \times \left(\frac{5}{2}\right)}{20} = \frac{25}{20} = 1.25\text{ cm}$$
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