Center of gravity

The mass points are distributed in space as specified by coordinates and weight. Find the center of gravity of the mass points system:

A₁ [-17; 10; 9] m₁ = 23 kg
A₂ [-16; -19; 0] m₂ = 31 kg
A₃ [4; -14; -10] m₃ = 94 kg
A₄ [-7; -14; 1] m₄ = 85 kg
A₅ [17; -9; -12] m₅ = 59 kg

Final Answer:

x₀ =  -0.3527
y₀ =  -11.6301
z₀ =  -4.6438

Step-by-step explanation:

$$\begin{gathered} x_1=-17;y_1=10;z_1=9 \\ {m}_1=23 \\ {x}_2=-16;y_2=-19;z_2=0 \\ {m}_2=31 \\ {x}_3=4;y_3=-14;z_3=-10 \\ {m}_3=94 \\ {x}_4=-7;y_4=-14;z_4=1 \\ {m}_4=85 \\ {x}_5=17;y_5=-9;z_5=-12 \\ {m}_5=59 \\ m=m_1+m_2+m_3+m_4+m_5=23+31+94+85+59=292 \\ {M}_x=m_1\cdot x_1+m_2\cdot x_2+m_3\cdot x_3+m_4\cdot x_4+m_5\cdot x_5=23\cdot (-17)+31\cdot (-16)+94\cdot 4+85\cdot (-7)+59\cdot 17=-103 \\ {M}_y=m_1\cdot y_1+m_2\cdot y_2+m_3\cdot y_3+m_4\cdot y_4+m_5\cdot y_5=23\cdot 10+31\cdot (-19)+94\cdot (-14)+85\cdot (-14)+59\cdot (-9)=-3396 \\ {M}_z=m_1\cdot z_1+m_2\cdot z_2+m_3\cdot z_3+m_4\cdot z_4+m_5\cdot z_5=23\cdot 9+31\cdot 0+94\cdot (-10)+85\cdot 1+59\cdot (-12)=-1356 \\ {x}_0=\frac{M_x}{m}=\frac{(-103)}{292}=-0.3527 \end{gathered}$$

$y_0=\frac{M_y}{m}=\frac{(-3396)}{292}=-11.6301$

$z_0=\frac{M_z}{m}=\frac{(-1356)}{292}=-4.6438$

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