CoG center

Find the position of the center of gravity of a system of four mass points having masses, m₁, m₂ = 2 m1, m₃ = 3 m1, and m₄ = 4 m₁, if they lie at the vertices of an isosceles tetrahedron. (in all cases, between adjacent material points, the distance is a=1. )

Final Answer:

x =  -0.05
y =  -0.0816
z =  0.7348

Step-by-step explanation:

$$\begin{gathered} m_1=1 \\ {m}_2=2 \\ {m}_3=3 \\ {m}_4=4 \\ a=1 \\ {h}_{33}=\sqrt{3}/2\cdot a \\ h=a\cdot \sqrt{6}/3=1\cdot \sqrt{6}/3\doteq 0.8165 \\ {h}_1=\frac{2}{3}\cdot h=\frac{2}{3}\cdot 0.8165\doteq 0.5443 \\ {h}_2=h-h_1=0.8165-0.5443\doteq 0.2722 \\ A=(0,0,0) \\ B=(0,h_1,h)=(0,0.54433105395182,0.81649658092773) \\ C=(a/2,-h_2,h)=(0.5,-0.27216552697591,0.81649658092773) \\ D=(-a/2,-h_2,h)=(-0.5,-0.27216552697591,0.81649658092773) \\ m=m_1+m_2+m_3+m_4=1+2+3+4=10 \\ x=\frac{m_1\cdot A_x+m_2\cdot B_x+m_3\cdot C_x+m_4\cdot D_x}{m}=\frac{1\cdot 0+2\cdot 0+3\cdot 0.5+4\cdot (-0.5)}{10}=-\frac{1}{20}=-0.05 \end{gathered}$$

$y=\frac{m_1\cdot A_y+m_2\cdot B_y+m_3\cdot C_y+m_4\cdot D_y}{m}=\frac{1\cdot 0+2\cdot 0.5443+3\cdot (-0.2722)+4\cdot (-0.2722)}{10}=-0.0816$

$z=\frac{m_1\cdot A_z+m_2\cdot B_z+m_3\cdot C_z+m_4\cdot D_z}{m}=\frac{1\cdot 0+2\cdot 0.8165+3\cdot 0.8165+4\cdot 0.8165}{10}=0.7348$

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