Chapter 5, Problem 5.137RP

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

Fig. P5.137

To determine

The centroid of the plane shown.

Answer to Problem 5.137RP

The centroid of the plane area $\left(\overline{X},\overline{Y}\right)$ is $\left(5.67\text{in, 5}\text{.17}\text{in}\right)$.

Explanation of Solution

Refer Figure 1.

The plane is considered as two separate sections as in figure 1. Section 1 is a perpendicular triangle and section 2 is a rectangle.

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.

$A_2=lw$ (II)

Here, $A_2$ is the area of section 2, $l$ is the length of the rectangle and $w$ is the width of the rectangle.

Write an expression to calculate the area of the plane.

$A=A_1+A_2$ (III)

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

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

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

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 two sections in the plane. Rewrite equation (IV) according to the plane.

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

Here, $\overline{x_1}$ is the x component of the centroid of the triangle and $\overline{x_2}$ is the x component of the centroid of the rectangle.

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

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

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 (VI) according to the plane.

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

Here, $\overline{y_1}$ is the y component of the centroid of the triangle and $\overline{y_1}$ is the y component of the centroid of the rectangle.

Conclusion:

Substitute $12\text{in}$ for $b$, and $6\text{in}$ for $h$ in equation (I) to find $A_1$.

$\begin{array}{c}{A}_1=\frac{1}{2}\left(12\text{in}\right)\left(6\text{in}\right)\\ =36{\text{in}}^2\end{array}$

Substitute $6\text{in}$ for $l$ and, $3\text{in}$ for $w$ in equation (II) to find $A_2$.

$\begin{array}{c}{A}_2=\left(6\text{in}\right)\left(3\text{in}\right)\\ =18{\text{in}}^2\end{array}$

Substitute $36{\text{in}}^2$ for $A_1$, and $18{\text{in}}^2$ for $A_2$ in equation (III) to find $A$.

$\begin{array}{c}A=36{\text{in}}^2+18{\text{in}}^2\\ =54{\text{in}}^2\end{array}$

Substitute $4\text{in}$ for $\overline{x_1}$, $36{\text{in}}^2$ for $A_1$, $18{\text{in}}^2$ for $A_2$, $9\text{in}$ for $\overline{x_2}$, and $54{\text{in}}^2$ for $A$ in equation (V) to find $\overline{X}$.

$\begin{array}{c}\overline{X}=\frac{\left(4\text{in}\right)\left(36{\text{in}}^2\right)+\left(9\text{in}\right)\left(18{\text{in}}^2\right)}{54{\text{in}}^2}\\ =\frac{144{\text{in}}^3+162{\text{in}}^3}{54{\text{in}}^2}\\ =\frac{306{\text{in}}^3}{54{\text{in}}^2}\\ =5.67\text{in}\end{array}$

Substitute $4\text{in}$ for $\overline{y_1}$, $36{\text{in}}^2$ for $A_1$, $18{\text{in}}^2$ for $A_2$, $7.5\text{in}$ for $\overline{y_2}$, and $54{\text{in}}^2$ for $A$ in equation (VII) to find $\overline{Y}$.

$\begin{array}{c}\overline{X}=\frac{\left(4\text{in}\right)\left(36{\text{in}}^2\right)+\left(7.5\text{in}\right)\left(18{\text{in}}^2\right)}{54{\text{in}}^2}\\ =\frac{144{\text{in}}^3+135{\text{in}}^3}{54{\text{in}}^2}\\ =\frac{279{\text{in}}^3}{54{\text{in}}^2}\\ =5.17\text{in}\end{array}$$\overline{Y}=\frac{\overline{y_1}{A}_1+\overline{y_2}{A}_2}{A}$

Thus, the centroid of the plane area $\left(\overline{X},\overline{Y}\right)$ is $\left(5.67\text{in, 5}\text{.17}\text{in}\right)$.