Question

the 11 boxes all have mass 1kg and side 1m
what is $y_{cm}$?
a. $y_{cm}=\frac{0.0(5)+1(6)+2(0)}{11}$
b. $y_{cm}=\frac{0.5(2)+1.5(0)+2.5(0)}{11}$
c. $y_{cm}=\frac{0.5(5)+1.5(6)+2.5(0)}{11}$
d. $y_{cm}=\frac{0.5(3)+1.5(3)+2.5(0)}{11}$
e. $y_{cm}=\frac{0.0(3)+1(3)+2(0)}{11}$

Explanation:

Step1: Calculate the \( y \)-coordinate of the center of mass for each layer

• For the bottom - layer (\( z = 0.5m \)): Count the number of boxes. There are \( 5 \) boxes in the bottom - layer.
• For the middle - layer (\( z = 1.5m \)): Count the number of boxes. There are \( 6 \) boxes in the middle - layer.
• For the top - layer (\( z = 2.5m \)): Count the number of boxes. There are \( 0 \) boxes in the top - layer.

Step2: Use the formula for the \( y \)-coordinate of the center of mass \( y_{cm}=\frac{\sum_{i = 1}^{n}m_iy_i}{\sum_{i = 1}^{n}m_i} \)

Since \( m_i = 1kg \) for all \( i \), \( \sum_{i = 1}^{n}m_i=11kg \), and \( y_{cm}=\frac{y_1m_1 + y_2m_2+\cdots+y_{11}m_{11}}{m_1 + m_2+\cdots+m_{11}}=\frac{y_1 + y_2+\cdots+y_{11}}{11} \) (because \( m_i = 1 \)).
If \( y_1=y_2=\cdots=y_5 = 0.5 \) (for the bottom - layer), \( y_6=y_7=\cdots=y_{11}=1.5 \) (for the middle - layer) (assuming the side - length of the cube is \( 1m \), the \( y \)-coordinate of the center of mass of a cube in the bottom - layer is \( 0.5m \) and in the middle - layer is \( 1.5m \)), then \( y_{cm}=\frac{0.5\times5 + 1.5\times6+2.5\times0}{11} \)

Answer:

C. \( y_{cm}=\frac{0.5(5)+1.5(6)+2.5(0)}{11} \)