91Ó°ÊÓ

• A $3kg$sphere is centered at .
• A $5kg$sphere is centered at .
• A $6kg$sphere is centered at.

The location of the center of mass of the system is determined by considering the average positions of all the objects acting in the system.

The following is the formula for finding the center of mass of this three-sphere system,

$$\begin{gathered} M{X}_{CM}=m_1{x}_1+m_2{x}_2+m_3{x}_3 \\ M{Y}_{CM}=m_1{y}_1+m_2{y}_2+m_3{y}_3 \\ M{Z}_{CM}=m_1{z}_1+m_2{z}_2+m_3{z}_3 \end{gathered}$$

Here,

$m_1=3kg,m_2=5kg,m_3=6kg$

Substitute these values in above expression,

Finding x coordinates,

$$\begin{gathered} M{X}_{CM}=m_1{x}_1+m_2{x}_2+m_3{x}_3 \\ =\left(3×10\right)+\left(5×4\right)+\left(6×-7\right) \\ M{X}_{CM}=8 \end{gathered}$$

$$\begin{gathered} X_{CM}=\frac{8}{M} \\ =\frac{8}{m_1+m_2+m_3} \\ =\frac{8}{3+5+6} \\ {X}_{Cm}=0.57m \end{gathered}$$

Finding y coordinates,

$$\begin{gathered} M{Y}_{CM}=m_1{y}_1+m_2{y}_2+m_3{y}_3 \\ =\left(3×20\right)+\left(5×-15\right)+\left(6×10\right) \\ M{y}_{CM}=45 \end{gathered}$$

$$\begin{gathered} y_{CM}=\frac{45}{M} \\ =\frac{45}{m_1+m_2+m_3} \\ =\frac{45}{3+5+6} \\ {y}_{CM}=3.21m \end{gathered}$$

Finding z coordinates

$$\begin{gathered} M{z}_{CM}=m_1{z}_1+m_2{z}_2+m_3{z}_3 \\ =\left(3×-5\right)+\left(5×8\right)+\left(6×9\right) \\ M{z}_{CM}=79 \end{gathered}$$

$$\begin{gathered} z_{CM}=\frac{79}{M} \\ =\frac{79}{m_1+m_2+m_3} \\ =\frac{79}{3+5+6} \\ {z}_{CM}=5.64m \end{gathered}$$

Hence, the location of the center of mass of this three-sphere system is role="math" localid="1653931917585" $\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{57}\mathbf{,}\mathbf{3}\mathbf{.}\mathbf{21}\mathbf{,}\mathbf{5}\mathbf{.}\mathbf{64}\mathbf{>}\mathbf{}\mathit{m}$.