Answer with Explanation:
We are given that
[tex]m_1=5,m_2=4,m_3=3,m_4=6[/tex]
[tex]P_1=(-4,3),P_2=(0,5),P_3=(3,2),P_4=(1,-2)[/tex]
We have to find Mx ,My and center of mass of the system.
[tex]M_x=\sum_{i=1}^{i=n}m_iy_i[/tex]
[tex]M_x=5(3)+4(5)+3(2)+6(-2)[/tex]
[tex]M_x=29[/tex]
[tex]M_y=\sum_{i=1}^{i=n}m_ix_i[/tex]
[tex]M_y=5(-4)+4(0)+3(3)+6(1)[/tex]
[tex]M_y=-5[/tex]
Total mass,m=m1+m2+m3+m4=5+4+3+6=18
[tex]\bar x=\frac{M_y}{m}=\frac{-5}{18}[/tex]
[tex]\bar{y}=\frac{M_x}{m}=\frac{29}{18}[/tex]
Center of mass=(-5/18,29/18)
The center of mass of the system at the given points (-0.278, 1.61).
The given parameters;
The total mass is calculated as follows;
m = 5 + 4 + 3 + 6 = 18
The sum of the masses on the on x and y-direction;
[tex]M_x = \Sigma M_i y_i\\\\M_x = 5(3) + 4(5) + 3(2) + 6(-2) = 29\\\\M_y = \Sigma M_ix_i \\\\M_y = 5(-4)+ 4(0) + 3(3) + 6(1) = -5[/tex]
The center mass is calculated as follows;
[tex]M_y = \frac{29}{18} = 1.61 \\\\M_x = \frac{-5}{18} = -0.278 \\\\(M_x, M_y) = (-0.278, \ 1.61)[/tex]
Thus, the center of mass of the system at the given points (-0.278, 1.61).
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