Question

Three uniform-density spheres are positioned as follows:

• A$\mathbf{3}\mathbf{}\mathit{k}\mathit{g}$ sphere is centered at $<10,20,-5>\mathbf{}\mathit{m}$.
• A $\mathbf{5}\mathbf{}\mathit{k}\mathit{g}$sphere is centered at $<4,-15,8>\mathbf{}\mathit{m}$.
• A $\mathbf{6}\mathbf{}\mathit{k}\mathit{g}$sphere is centered at $<-7,10,9>\mathbf{}\mathit{m}$.

What is the location of the center of mass of this three-sphere system?

Short answer

The location of the center of mass of this three-sphere system is

Step-by-step solution

Identification of given data

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

Concept of the location of the center of mass of this three-sphere system

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.

Determination of the location of the center of mass of this three-sphere 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,

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

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

Finding y coordinates,

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

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

Finding z coordinates

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

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

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