Chapter 12, Problem 12.7P

To determine

The centre of gravity for the three object system.

Answer to Problem 12.7P

The centre of gravity for the three object system is $\left(2.54\text{m,}4.75\text{m}\right)$.

Explanation of Solution

Given info: The mass of the rod is $m_1=6.00\text{kg}$, the mass of the right triangle is $m_2=3.00\text{kg}$, the mass of the square is $m_3=5.00\text{kg}$.

The centre of gravity of the rod is,

$\left(x_1,y_1\right)=\left(\frac{x_{\text{r}1}+x_{\text{r}2}}{2},\frac{y_{\text{r}1}+y_{\text{r}2}}{2}\right)$

Here,

$x_{\text{r}1}$ is the first $x$-coordinate of the rod.

$x_{\text{r}2}$ is the second $x$-coordinate of the rod

$y_{\text{r}1}$ is the first $y$-coordinate of the rod.

$y_{\text{r}2}$ is the second $y$-coordinate of the rod.

Substitute $2\text{m}$ for $x_{\text{r}1}$, $9\text{m}$ for $x_{\text{r}2}$, $7\text{m}$ for $y_{\text{r}1}$ and $7\text{m}$ for $y_{\text{r}2}$ in the above equation.

$\begin{array}{c}\left(x_1,y_1\right)=\left(\frac{2\text{m}+9\text{m}}{2},\frac{7\text{m}+7\text{m}}{2}\right)\\ =\left(5.5\text{m},7\text{m}\right)\end{array}$

The centre of the gravity of the right angled triangle is at distance two third of the length of the base and one third of the height of the triangle from the base.

The centre of the gravity of the right angled triangle is,

$\left(x_2,y_2\right)=\left(x_{\text{t}1}+\frac{2}{3}\left(x_{\text{t}2}-x_{\text{t}1}\right),y_{\text{t}1}+\frac{1}{3}\left(y_{\text{t}2}-y_{\text{t}1}\right)\right)$

Here,

$x_{\text{t}1}$ is the first $x$-coordinate of the base of the right angled triangle.

$x_{\text{t}2}$ is the second $x$-coordinate of base of the right angled triangle.

$y_{\text{t}1}$ is the first $y$-coordinate of the right angled triangle.

$y_{\text{t}2}$ is the second $y$-coordinate of the right angled triangle.

Substitute $8\text{m}$ for $x_{\text{t}2}$, $4\text{m}$ for $x_{\text{t}1}$, $5\text{m}$ for $y_{\text{t}2}$ and $1\text{m}$ for $y_{\text{t}1}$ in the above equation.

$\begin{array}{c}\left(x_2,y_2\right)=\left(4\text{m}+\frac{2}{3}\left(8\text{m}-4\text{m}\right)+,1\text{m}+\frac{1}{3}\left(5\text{m}-1\text{m}\right)\right)\\ =\left(6.67\text{m},2.33\text{m}\right)\end{array}$

The centre of gravity of the square is,

$\left(x_3,y_3\right)=\left(\frac{x_{\text{s}1}+x_{\text{s}2}}{2},\frac{y_{\text{s}1}+y_{\text{s}2}}{2}\right)$

Here,

$x_{\text{s}1}$ is the first $x$-coordinate of the square.

$x_{\text{s}2}$ is the second $x$-coordinate of the square.

$y_{\text{s}1}$ is the first $y$-coordinate of the square.

$y_{\text{s}2}$ is the second $y$-coordinate of the square.

Substitute $-5\text{m}$ for $x_{\text{s}1}$, $-2\text{m}$ for $x_{\text{s}2}$, $5\text{m}$ for $y_{\text{s}1}$, $2\text{m}$ for $y_{\text{s}2}$ in the above equation.

$\begin{array}{c}\left(x_3,y_3\right)=\left(\frac{-5\text{m}-2\text{m}}{2},\frac{5\text{m}+2\text{m}}{2}\right)\\ =\left(-3.5\text{m},3.5\text{m}\right)\end{array}$

Formula to calculate center of gravity of three masses is,

$\left(x_{\text{cm}},y_{\text{cm}}\right)=\left(\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3},\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}\right)$

Here,

$x_{\text{cm}}$ is the $x$ coordinate of the centre of the gravity of complete system.

$y_{\text{cm}}$ is the $y$ coordinate of the centre of the gravity of complete system.

$x_1$ is the positions of centre of gravity of the mass $m_1$.

$x_2$ is the positions of centre of gravity of the  mass $m_2$.

$x_3$ is the positions of centre of gravity of the mass $m_3$.

$y_1$ is the position of the centre of gravity of the  mass $m_1$.

$y_2$ is the position of the centre of gravity of the mass $m_2$.

$y_3$ is the position of the centre of gravity of the mass $m_3$.

Substitute $6.00\text{kg}$ for $m_1$, $3.00\text{kg}$ for $m_2$, $5.00\text{kg}$ for $m_3$, $5.5\text{m}$ for $x_1$, $6.67\text{m}$ for $x_2$, $-3.5\text{m}$ for $x_3$, $7\text{m}$ for $y_1$, $2.33\text{m}$ for $y_2$, $3.5\text{m}$ for $y_3$ in the above equation.

$\left(x_{\text{cm}},y_{\text{cm}}\right)=\left(\begin{array}{l}\frac{\left(6.00\text{kg}\right)\left(5.5\text{m}\right)+\left(3.00\text{kg}\right)\left(6.67\text{m}\right)+\left(5.00\text{kg}\right)\left(-3.5\text{m}\right)}{6.00\text{kg}+3.00\text{kg}+5.00\text{kg}},\\ \frac{\left(6.00\text{kg}\right)\left(7\text{m}\right)+\left(3.00\text{kg}\right)\left(2.33\text{m}\right)+\left(5.00\text{kg}\right)\left(3.5\text{m}\right)}{6.00\text{kg}+3.00\text{kg}+5.00\text{kg}}\end{array}\right)$

Simplify the above equation.

$\begin{array}{c}\left(x_{\text{cm}},y_{\text{cm}}\right)=\left(\frac{35.51\text{kgm}}{6.00\text{kg}+3.00\text{kg}+5.00\text{kg}},\frac{66.49\text{kgm}}{6.00\text{kg}+3.00\text{kg}+5.00\text{kg}}\right)\\ =\left(\frac{35.51\text{kgm}}{14.00\text{kg}},\frac{66.49\text{kgm}}{14.00\text{kg}}\right)\\ =\left(2.536\text{m,}4.749\text{m}\right)\\ \approx \left(2.54\text{m,}4.75\text{m}\right)\end{array}$

Conclusion:

Therefore, the centre of gravity for the three object system is $\left(2.54\text{m,}4.75\text{m}\right)$.