Chapter 5, Problem 5.138RP

5.137 and 5.138 Locate the centroid of the plane area shown.

Fig. P5.137

Fig. P5.138

To determine

The centroid of the plane shown.

Answer to Problem 5.138RP

The centroid of the plane area $\left(\overline{X},\overline{Y}\right)$ is $\left(92.0\text{mm, 23}\text{.3}\text{mm}\right)$.

Explanation of Solution

Refer Figures 1 and 2.

Figure 1

Figure 2

The plane is considered as three separate sections as in figure 1. Section 1 is a perpendicular triangle, section 2 is a square and section 3 is a quarter of a circle.

Write an expression to calculate the area of section 1.

$A_1=\frac{1}{2}bh$ (I)

Here, $A_1$ is the area of section 1, $b$ is the base of the triangle and $h$ is the height of the triangle.

Write an expression to calculate the area of section 2.

$A_2=a^2$ (II)

Here, $A_2$ is the area of section 2, $a$ is one side of the square.

Write an expression to calculate the area of section 3.

$A_3=-\frac{1}{4}\left(\pi r^2\right)$ (III)

Here, $A_3$ is the area of section 3 and $r$ is the radius of the circle.

Write an expression to calculate the area of the plane.

$A=A_1+A_2+A_3$ (IV)

Here, $A$ is the area of the plane.

Write an expression to calculate the x component of the centroid of the plane.

$\overline{X}=\frac{{\displaystyle \sum _1^n\left({\overline{x}}_i{A}_i\right)}}{A}$ (V)

Here, $\overline{X}$ is the x component of the centroid of the plane, $A_i$ is the area of each section and ${\overline{x}}_i$ is the centroid of each section.

There are three sections in the plane. Rewrite equation (V) according to the plane.

$\overline{X}=\frac{\overline{x_1}{A}_1+\overline{x_2}{A}_2+\overline{x_3}{A}_3}{A}$ (VI)

Here, $\overline{x_1}$ is the x component of the centroid of section 1, $\overline{x_2}$ is the x component of the centroid of section 2 and $\overline{x_3}$ is the x component of section 3.

Write an expression to calculate the y component of the centroid of the plane.

$\overline{Y}=\frac{{\displaystyle \sum _1^n\left({\overline{y}}_i{A}_i\right)}}{A}$ (VII)

Here, $\overline{Y}$ is the y component of the centroid of the plane and ${\overline{y}}_i$ is the centroid of each section.

There are two sections in the plane. Rewrite equation (VII) according to the plane.

$\overline{Y}=\frac{\overline{y_1}{A}_1+\overline{y_2}{A}_2+\overline{y_3}{A}_3}{A}$ (VIII)

Here, $\overline{y_1}$ is the y component of the centroid of section 1, $\overline{y_2}$ is the y component of the centroid of section 2 and $\overline{y_3}$ is the y component of section 3.

Conclusion:

Substitute $120\text{mm}$ for $b$, and $\text{75}\text{mm}$ for $h$ in equation (I) to find $A_1$.

$\begin{array}{c}{A}_1=\frac{1}{2}\left(120\text{mm}\right)\left(75\text{mm}\right)\\ =4500{\text{mm}}^2\end{array}$

Substitute $75\text{mm}$ for $a$ in equation (II) to find $A_2$.

$\begin{array}{c}{A}_2={\left(75\text{mm}\right)}^2\\ =5625{\text{mm}}^2\end{array}$

Substitute $75\text{mm}$ for $r$ in equation (III) to find $A_3$.

$\begin{array}{c}{A}_3=-\frac{1}{4}\pi {\left(75\text{mm}\right)}^2\\ =-4417.9{\text{mm}}^2\end{array}$

Substitute $4500{\text{mm}}^2$ for $A_1$, $5625{\text{mm}}^2$ for $A_2$, and $-4417.9{\text{mm}}^2$ for $A_3$ in equation (IV) to find $A$.

$\begin{array}{c}A=\left(4500{\text{mm}}^2\right)+\left(5625{\text{mm}}^2\right)+\left(-4417.9{\text{mm}}^2\right)\\ =5707.1{\text{mm}}^2\end{array}$

Substitute $80\text{mm}$ for $\overline{x_1}$, $4500{\text{mm}}^2$ for $A_1$, $5625{\text{mm}}^2$ for $A_2$, $157.5\text{mm}$ for $\overline{x_2}$, $163.169\text{mm}$ for $\overline{x_3}$, $-4417.9{\text{mm}}^2$ for $A_3$ and $5707.1{\text{mm}}^2$ for $A$ in equation (VI) to find $\overline{X}$.

$\begin{array}{c}\overline{X}=\frac{\left(80\text{mm}\right)\left(4500{\text{mm}}^2\right)+\left(157.5\text{mm}\right)\left(5625{\text{mm}}^2\right)+\left(163.169\text{mm}\right)\left(-4417.9{\text{mm}}^2\right)}{5707.1{\text{mm}}^2}\\ =\frac{360000{\text{mm}}^3+885940{\text{mm}}^3-720860{\text{mm}}^3}{5707.1{\text{mm}}^2}\\ =\frac{525080{\text{mm}}^3}{5707.1{\text{mm}}^2}\\ =92.0\text{mm}\end{array}$

Substitute $25\text{mm}$ for $\overline{y_1}$, $4500{\text{mm}}^2$ for $A_1$, $5625{\text{mm}}^2$ for $A_2$, $37.5\text{mm}$ for $\overline{y_2}$, $43.169{\text{mm}}^2$ for $\overline{y_3}$, $-4417.9{\text{mm}}^2$ for $A_3$ and $5707.1{\text{mm}}^2$ for $A$ in equation (VIII) to find $\overline{Y}$.

$\begin{array}{c}\overline{X}=\frac{\left(25\text{mm}\right)\left(4500{\text{mm}}^2\right)+\left(37.5\text{mm}\right)\left(5625{\text{mm}}^2\right)+\left(43.169\text{mm}\right)\left(-4417.9{\text{mm}}^2\right)}{5707.1{\text{mm}}^2}\\ =\frac{\text{112500}{\text{mm}}^3+210940{\text{mm}}^3-190716{\text{mm}}^3}{5707.1{\text{mm}}^2}\\ =\frac{132724{\text{mm}}^3}{5707.1{\text{mm}}^2}\\ =23.3\text{mm}\end{array}$

Thus, the centroid of the plane area $\left(\overline{X},\overline{Y}\right)$ is $\left(92.0\text{mm, 23}\text{.3}\text{mm}\right)$.