Center of gravity 2

The center of mass will remain in place (Newton's First Law, applied to the entire system). The location of the center of mass is $\langle x \rangle = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}.$ Since the total mass of the system remains unchanged, we conclude that the numerator remains unchanged. This is the total "moment" of the system, $M = m_1x_1 + m_2x_2 + m_3x_3.$ Now two things happen: effectively, the mass difference of 30 kg is moved over a distance of 3 m relative to the rest of the system , and the system itself moves over an unknown distance $d$ . Thus the change in moment is $\Delta M = (85 - 55)\cdot 3 + (85 + 55 + 78)d.$ Equate this to zero and we find $d = -\frac{30\cdot 3}{85 + 55 + 78} = -0.416\ \text{m}.$

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A more "naive" approach would be to conclude that the center of mass shifts from $\langle x\rangle = \frac{85x_1 + 55x_2 + 78x_3}{85+55+78}$ to $\langle x'\rangle = \frac{85(x_1-3+d) + 55(x_2+3+d) + 78(x_3+d)}{85+55+78};$ equate the numerators, work out the brackets, cancel equal terms, and you end up with the same equation $(85 - 55)\cdot 3 + (85 + 55 + 78)d = 0.$

Initial center of mass is considering 85 kg person as origin.

$\frac{(0)(85) + (1.5)(78) + (3)(55)}{85 + 55 + 78} ..... (1)$

Final center of mass using same point

$\frac{(0)(55) + (1.5)(78) + (3)(85)}{85 + 55 + 78} ...... (2)$

Change in position = (2) - (1)

$\boxed{Change \approx 41.6cm}$