Question

Consider the following mass distribution where the x and y coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 3.6 kg at (0.0, 4.1) m, and 4.0 kg at (2.9, 0.0) m. Where should a fourth object of 7.7 kg be placed so the center of gravity of the four-object arrangement will be at (0.0, 0.0) m?

Answer 1

Answer:

The fourth mass should be located at (-1.506 m, -1.917 m).

Explanation:

Given that each mass can be treated as puntual objects, the location of center of gravity can be determined by using weighted averages. That is:

$\bar x = \frac{x_{1}\cdot m_{1} + x_{2}\cdot m_{2} + x_{3} \cdot m_{3} + x_{4}\cdot m_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

$\bar y = \frac{y_{1}\cdot m_{1} + y_{2}\cdot m_{2} + y_{3} \cdot m_{3} + y_{4}\cdot m_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

Where:

$\bar x$, $\bar y$ - Horizontal and vertical component of the location of the center of gravity, measured in meters.

$x_{1}, x_{2}, x_{3}, x_{4}$ - Horizontal components of the location of first, second, third and fourth masses, measured in meters.  

$y_{1}, y_{2}, y_{3}, y_{4}$ - Vertical components of the location of first, second, third and fourth masses, measured in meters.

$m_{1}, m_{2}, m_{3}, m_{4}$ - Masses of first, second, third and fourth masses, measured in kilograms.  

If $m_{1} = 5\,kg$, $m_{2} = 3.6\,kg$, $m_{3} = 4\,kg$, $m_{4} = 7.7\,kg$, $\bar x = 0\,m$, $\bar y = 0\,m$, $x_{1} = 0\,m$, $x_{2} = 0\,m$, $x_{3} = 2.9\,m$, $y_{1} = 0\,m$, $y_{2} = 4.1\,m$ and $y_{3} = 0.0\,m$, then:

$0\,m = \frac{(0\,m)\cdot (5\,kg)+(0\,m)\cdot (3.6\,kg)+(2.9\,m)\cdot (4\,kg)+x_{4}\cdot (7.7\,kg)}{5\,kg + 3.6\,kg + 4\,kg + 7.7\,kg}$

$0\,m = \frac{(0\,m)\cdot (5\,kg)+(4.1\,m)\cdot (3.6\,kg)+(0\,m)\cdot (4\,kg)+y_{4}\cdot (7.7\,kg)}{5\,kg + 3.6\,kg + 4\,kg + 7.7\,kg}$

Both expression are simplified hereafter:

$\frac{4}{7}\,m + \frac{11}{29}\cdot x_{4} = 0\,m$

$\frac{738}{1015}\,m + \frac{11}{29}\cdot y_{4} = 0\,m$

The solution of this system of equation is $x_{4} = -\frac{116}{77}\,m$ ($x_{4} \approx -1.506\,m$) and $y_{4} = - \frac{738}{385}\,m$ ($y_{4}\approx -1.917\,m$).

The fourth mass should be located at (-1.506 m, -1.917 m).