## Question

###### A system consists of the following masses in the $x y$ -plane: $4.0 \mathrm{~kg}$ at coordinates $(x=0, y=5.0 \mathrm{~m})$, $7.0 \mathrm{~kg}$ at $(3.0 \mathrm{~m}, 8.0 \mathrm{~m})$, and $5.0 \mathrm{~kg}$ at $(-3.0 \mathrm{~m},-6.0 \mathrm{~m})$. Find the position of its center of mass. $$ \begin{array}{l} x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(4.0 \mathrm{~kg})+(3.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-3.0 \mathrm{~m})(5.0 \mathrm{~kg})}{(4.0+7.0+5.0) \mathrm{kg}}=0.38 \ma

A system consists of the following masses in the $x y$ -plane: $4.0 \mathrm{~kg}$ at coordinates $(x=0, y=5.0 \mathrm{~m})$, $7.0 \mathrm{~kg}$ at $(3.0 \mathrm{~m}, 8.0 \mathrm{~m})$, and $5.0 \mathrm{~kg}$ at $(-3.0 \mathrm{~m},-6.0 \mathrm{~m})$. Find the position of its center of mass. $$ \begin{array}{l} x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(4.0 \mathrm{~kg})+(3.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-3.0 \mathrm{~m})(5.0 \mathrm{~kg})}{(4.0+7.0+5.0) \mathrm{kg}}=0.38 \mathrm{~m} \\ y_{\mathrm{cm}}=\frac{\sum y_{i} m_{i}}{\sum m_{i}}=\frac{(5.0 \mathrm{~m})(4.0 \mathrm{~kg})+(8.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-6.0 \mathrm{~m})(5.0 \mathrm{~kg})}{16 \mathrm{~kg}}=2.9 \mathrm{~m} \end{array} $$ and $z_{\mathrm{cm}}=0$. These distances are, of course, measured from the origin $(0,0,0,$, .