Question

a flat (2 - dimensional) object consists of a 4 squares arranged as shown. each square has a uniform mass density, with the densities of the upper - right and lower - left squares are equal to each other and one - fourth the density of the upper - left square. the density of the lower - right square is ( 3.65\frac{kg}{m^{2}} ). find the density of the upper - left square, given that the center of mass of the entire object is at the point indicated. provide at least 1 decimal place

Explanation:

Step1: Set up variables

Let the density of the upper - left square be \(
ho_{ul}\), the density of the upper - right and lower - left squares be \(
ho_{ur}=
ho_{ll}=\frac{1}{4}
ho_{ul}\), and the density of the lower - right square be \(
ho_{lr} = 3.65\ \frac{kg}{m^{2}}\). Assume the side length of each square is \(a\), so the area of each square \(A=a^{2}\).
The coordinates of the centers of the four squares:

• Upper - left square \((x_{ul},y_{ul})=(-\frac{3a}{4}, \frac{3a}{4})\)
• Upper - right square \((x_{ur},y_{ur})=(\frac{3a}{4}, \frac{3a}{4})\)
• Lower - left square \((x_{ll},y_{ll})=(-\frac{3a}{4}, -\frac{3a}{4})\)
• Lower - right square \((x_{lr},y_{lr})=(\frac{3a}{4}, -\frac{3a}{4})\)

The formula for the \(x\) - coordinate of the center of mass \(x_{cm}=\frac{m_{ul}x_{ul}+m_{ur}x_{ur}+m_{ll}x_{ll}+m_{lr}x_{lr}}{m_{ul}+m_{ur}+m_{ll}+m_{lr}}\), and the formula for the \(y\) - coordinate of the center of mass \(y_{cm}=\frac{m_{ul}y_{ul}+m_{ur}y_{ur}+m_{ll}y_{ll}+m_{lr}y_{lr}}{m_{ul}+m_{ur}+m_{ll}+m_{lr}}\). Since \(x_{cm} = 0\) (symmetry in \(x\) - direction for center of mass), we focus on \(y\) - coordinate.

Mass \(m=
ho A\). So \(m_{ul}=
ho_{ul}A\), \(m_{ur}=\frac{1}{4}
ho_{ul}A\), \(m_{ll}=\frac{1}{4}
ho_{ul}A\), \(m_{lr}=
ho_{lr}A\)

\(y_{cm}=\frac{
ho_{ul}A\times\frac{3a}{4}+\frac{1}{4}
ho_{ul}A\times\frac{3a}{4}+\frac{1}{4}
ho_{ul}A\times(-\frac{3a}{4})+
ho_{lr}A\times(-\frac{3a}{4})}{
ho_{ul}A+\frac{1}{4}
ho_{ul}A+\frac{1}{4}
ho_{ul}A+
ho_{lr}A}\)

Since \(y_{cm} = 0\) (given center of mass at the indicated point), the numerator is \(0\)

$$
LATEXBLOCK0
$$

Answer:

\(3.65\ \frac{kg}{m^{2}}\)