Question

Let \({{\mathop{\rm v}\nolimits} _1},...,{v_k}\) be points in \({\mathbb{R}^3}\) and suppose that for \(j = 1,...,k\) an object with mass \({m_j}\) is located at point \({{\mathop{\rm v}\nolimits} _j}\). Physicists call such objects point masses. The total mass of the system of point masses is \(m = {m_1} + ... + {m_k}\).

The center of gravity (or center of mass) of the system is

\(\bar v = \frac{1}{m}\left[ {{m_1}{v_1} + ... + {m_k}{v_k}} \right]\). Compute the center of gravity of the system consisting of the following point masses (see the figure):

Point	Mass
\({{\mathop{\rm v}\nolimits} _1} = \left( {5, - 4,3} \right)\)	2 g
\({{\mathop{\rm v}\nolimits} _2} = \left( {5 = 4,3, - 2} \right)\)	5 g
\({{\mathop{\rm v}\nolimits} _3} = \left( { - 4, - 3, - 1} \right)\)	2 g
\({{\mathop{\rm v}\nolimits} _4} = \left( { - 9,8,6} \right)\)	1 g

Short answer

The center of gravity of the point masses is \(\bar v = \left[ {\begin{array}{*{20}{c}}{1.3}\\{0.9}\\0\end{array}} \right]\).

Step-by-step solution

Determine the total mass of the system

The total mass of the system of point masses is:

\(\begin{aligned}{c}m &= {m_1} + {m_2} + ... + {m_k}\\ &= 2 + 5 + 2 + 1\\ &= 10\end{aligned}\)

Write the given point as a vector

Write the given point as a vector.

\(\left\{ {{v_1},{v_2},{v_3},{v_4}} \right\} = \left\{ {\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}4\\3\\{ - 2}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 3}\\{ - 1}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 9}\\8\\6\end{array}} \right]} \right\}\)

Determine the center of gravity of the system

The sum of two vectors\({\mathop{\rm u}\nolimits} \)and\({\mathop{\rm v}\nolimits} \)is thevector addition \({\mathop{\rm u}\nolimits} + v\), which is obtained by adding the corresponding entries of \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \). The scalar multiple of vector \({\mathop{\rm u}\nolimits} \) by real number \(c\) is the vector \(c{\mathop{\rm u}\nolimits} \) obtained by multiplying each entry in \({\mathop{\rm u}\nolimits} \) by \(c\).

It is given that the center of gravity of the system is \(\bar v = \frac{1}{m}\left[ {{m_1}{v_1} + ... + {m_k}{v_k}} \right]\).

Use scalar multiple and vector addition to determine the center of gravity for the given point masses.

\(\begin{aligned}{c}\bar v &= \frac{1}{{10}}\left( {2{v_1} + 5{v_2} + 2{v_3} + {v_4}} \right)\\ &= \frac{1}{{10}}\left( {2\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\3\end{array}} \right] + 5\left[ {\begin{array}{*{20}{c}}4\\3\\{ - 2}\end{array}} \right] + 2\left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 3}\\{ - 1}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 9}\\8\\6\end{array}} \right]} \right)\\ &= \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}{10 + 20 - 8 - 9}\\{ - 8 + 15 - 6 + 8}\\{6 - 10 - 2 + 6}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{1.3}\\{0.9}\\0\end{array}} \right]\end{aligned}\)

Thus, the center of gravity of the point masses is \(\bar v = \left[ {\begin{array}{*{20}{c}}{1.3}\\{0.9}\\0\end{array}} \right]\).