Question

A 3.0-kg mass is located at (0.0 m, 8.0 m), and a 1.0-kg mass is located at (12 m, 0.0 m) where (x, y) represents x and y coordinates respectively on x-y plane. You want to add a 4.0-kg mass so that the center of mass (or center of gravity) of the three-mass system will be at the origin. What should be the coordinates of the 3.0-kg mass

Answer 1

Answer:

The coordinates of the 4.0-kg mass must be $(x,y) = (-3\,m,-6\,m)$.

Explanation:

We have a system formed by three particles with known masses and locations with respect to origin. The location of the center of mass is calculated by means of weighted means. That is:

$\bar x = \frac{\Sigma\limits_{i=1}^{n}\,x_{i}\cdot m_{i}}{\Sigma\limits_{i=1}^{n}m_{i}}$ (Eq. 1)

$\bar y = \frac{\Sigma\limits_{i=1}^{n}\,y_{i}\cdot m_{i}}{\Sigma\limits_{i=1}^{n}m_{i}}$ (Eq. 2)

Where:

$\bar x$, $\bar y$ - Coordinates of the center of mass of the system, measured in meters.

$x_{i}$, $y_{i}$ - Coordinates of the i-th particle, measured in meters.

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

If we know that $m_{1} = 3\,kg$, $m_{2} = 1\,kg$, $m_{3} = 4\,kg$, $x_{1} = 0\,m$, $x_{2} = 12\,m$, $\bar x = 0\,m$, $y_{1} = 8\,m$, $y_{2} = 0\,m$ and $\bar y = 0\,m$, the location of the 4.0 kg mass is:

$\frac{(3\,kg)\cdot (0\,m)+(1\,kg)\cdot (12\,m)+(4\,kg)\cdot x_{3}}{3\,kg+1\,kg+4\,kg} = 0\,m$

$12+4\cdot x_{3} = 0$

$x_{3} = -3\,m$

$\frac{(3\,kg)\cdot (8\,m)+(1\,kg)\cdot (0\,m)+(4\,kg)\cdot y_{3}}{3\,kg+1\,kg+4\,kg}= 0\,m$

$24+4\cdot y_{3} = 0$

$y_{3} = -6\,m$

The coordinates of the 4.0-kg mass must be $(x,y) = (-3\,m,-6\,m)$.

Answer 2

The coordinates of the 4kg mass so that the center of mass of the three-mass system will be at the origin is (-3,-6)

Finding the coordinates:

The coordinates of the 3kg mass are (0.0 m, 8.0 m)

and the coordinates for the 1kg mass are  (12 m, 0.0 m)

Let the coordinates of the 4kg mass be (x,y)

The final coordinates of the center of mass of the system are (0,0),

so the equation for x-coordinate is given by:

$0=\frac{3\times0+1\times12+4\times x}{3+1+4}$

12 + 4x = 0

x = -3

the equation for y-coordinate is given by:

$0=\frac{3\times8+1\times0+4\times y}{3+1+4}$

24 + 4y = 0

y = -6

So, the coordinates of the 4kg mass are (-3,-6)

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