Question

Please help The position of masses 4kg, 6kg, 7kg, 10kg, 2kg, and 12kg are (-1,1), (4,2), (-3,-2), (5,-4), (-2,4) and (3,-5) respectively. Determine the position of the center of mass of this system?​

Answer 1

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: $x_{cm}$ = $\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$, where m represents the masses and x represents the position.

In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.

x-values:

$x_{cm}$ = $\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ = $\frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12}$ = $\frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41}$ = $\frac{81}{41}$ = 1.9756

y-values:

$y_{cm}$ = $\frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ = $\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}$ = $\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}$ = $\frac{-90}{41}$ = -2.1951

center of mass:

(1.9756, -2.1951)