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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?​

Sagot :

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: [tex]x_{cm}[/tex] = [tex]\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}}[/tex], 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:

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

y-values:

[tex]y_{cm}[/tex] = [tex]\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}}[/tex] = [tex]\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}[/tex] = [tex]\frac{-90}{41}[/tex] = -2.1951

center of mass:

(1.9756, -2.1951)