The center of mass of three particles of masses 1 kg, 2 kg and 3 kg at (3, 3, 3) with reference to a fixed coordinate system. Where should a fourth particle of mass 4 kg should be placed, so that the center of mass of the system of all particles shifts to a point (1, 1, 1)?

Question

Dr. Abubacker · Accepted Answer

(b) Centre of Mass of a solid body is given by
$x_{CM} = \frac{\sum_{i = 1}^{n}{\mathrm{\Delta}m_{i}x_{i}}}{\sum_{j = 1}^{n}{\mathrm{\Delta}Mj}}$
$y_{CM} = \frac{\sum_{i = 1}^{n}{\mathrm{\Delta}m_{i}y_{i}}}{\sum_{j = 1}^{n}{\mathrm{\Delta}Mj}}$
$z_{CM} = \frac{\sum_{i = 1}^{n}{\mathrm{\Delta}m_{i}z_{i}}}{\sum_{j = 1}^{n}{\mathrm{\Delta}Mj}}$
$1 	imes x_{1} + 2 	imes x_{2} + 3 	imes x_{3} = (1 + 2 + 3)3$ …(i)
and $x_{1} = x_{2} = x_{3} = 3$
$x_{CM} = y_{CM} = z_{CM} = 1$ (given)
$1(1 + 2 + 3 + 4) = 1x_{1} + 2x_{2} + 3x_{3} + 4x_{4}$ …(ii)
Solving Eqs. (i) and (ii), we get
$4x_{4} = 10 - 18$
$x_{4} = - 2$
Similarly, $y_{4} = - 2,\ z_{4} = - 2$
The fourth particle must be placed at the point (-2, -2, -2).

Q. The center of mass of three particles of masses 1 kg, 2 kg and 3 kg at (3, 3, 3) with reference to a fixed coordinate system. Where should a fourth particle of mass 4 kg should be placed, so that the center of mass of the system of all particles shifts to a point (1, 1, 1)?

Explanation

Reviewed by Dr. Abubacker

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