Chapter 6, Problem 14P

To determine

To Find: The distance of the centroid from the bottom of the square.

Answer to Problem 14P

The centroid of the shaded region from the base of is $3.92\text{cm}$ .

Explanation of Solution

Given:

The area of the triangle is $30{\text{cm}}^2$ and is cut from the corner of the square with the side $10\text{cm}$ .

The centroid of the remaining region is 4 cm right side of the square.

Calculation:

Consider the required diagram is shown in Figure 1

  

Figure 1

Consider the area of the square is $A=10\times 10$ that is $100{\text{cm}}^2$ .

Calculate the area of the triangle is,

  $A_T=\frac{1}{2}ab$

Consider the base of the triangle is $a$ and the height of the triangle is $b$ from the data so the area of the triangle is $30{\text{cm}}^3$ then,

  $\begin{array}{l}30{\text{cm}}^3=\frac{1}{2}ab\\ ab=60\end{array}$

Consider the centroid of the shaded region is,

  $x=\frac{A_0{x}_1-A_T{x}_2}{A_s-A_T}$

Then,

  $\begin{array}{l}6=\frac{100\left(5\right)-30\left(\frac{a}{3}\right)}{100-30}\\ a=8\text{cm}\end{array}$

Then, the value of $b$ is,

  $\begin{array}{l}\left(8\text{cm}\right)b=60\\ b=7.5\text{cm}\end{array}$

Also,

  $\begin{array}{c}y=\frac{A_S{y}_1-A_T{y}_2}{A_s-A_T}\\ =\frac{100\left(5\right)-30\left(10-\frac{b}{3}\right)}{100-30}\\ =\frac{100\left(5\right)-30\left(10-\frac{7.5}{3}\right)}{100-30}\\ =3.92\text{cm}\end{array}$

Thus, the centroid of the shaded region from the base of is $3.92\text{cm}$ .