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Sagot :
Given,
Three masses
m₁=2 kg
m₂=3 kg
m₃=1 kg
The coordinate of mass 1 is (x₁,y₁)=(0,0)
The coordinate of mass 2 is (x₂,y₂)=(-1,0)
The coordinate of mass 3 is (x₃,y₃)=(1,1)
The x-coordinate of the center of mass is given by,
[tex]x=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}[/tex]On substituting the known values,
[tex]\begin{gathered} x=\frac{2\times0+3\times-1+1\times1}{2+3+1} \\ =-\frac{1}{3} \end{gathered}[/tex]And the y-coordinate of the center of mass is given by,
[tex]y=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}[/tex]On substituting the known values,
[tex]\begin{gathered} y=\frac{2\times0+3\times0+1\times1}{2+3+1} \\ =\frac{1}{6} \end{gathered}[/tex]Thus the center of mass is at (-1/3, 1/6)
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