Question

the 11 boxes all have mass 1kg and side 1m
what is x - coordinate of the position of the centre of mass, ( x_{cm} )?
(origin is bottom left corner, labelled \o\
a. ( x_{cm}=\frac{0.0(1)+1(2)+2(2)}{11} )
b. ( x_{cm}=\frac{0.5(1)+1.5(2)+2.5(2)}{11} )
c. ( x_{cm}=\frac{0.5(2)+1.5(5)+2.5(4)}{11} )
d. ( x_{cm}=\frac{0.5(1)+1.5(2)+2.5(4)}{11} )
e. ( x_{cm}=\frac{0.0(2)+1(5)+2(4)}{11} )

Explanation:

Step1: Calculate the \(x\) - coordinate of the center of mass of a single cube

For a cube of side \(a = 1m\), the \(x\) - coordinate of its center of mass is \(x_i=\frac{1}{2}+n\), where \(n\) is the number of unit lengths from the origin along the \(x\) - axis. The formula for the \(x\) - coordinate of the center of mass of a system of particles is \(x_{cm}=\frac{\sum_{i = 1}^{N}m_ix_i}{\sum_{i = 1}^{N}m_i}\). Here, \(m_i = 1kg\) for all \(i\), and \(\sum_{i=1}^{N}m_i=11kg\).

• In the \(x = 0.5m\) column (first column from the left), there is \(1\) cube.
• In the \(x = 1.5m\) column (second column from the left), there are \(2\) cubes.
• In the \(x = 2.5m\) column (third column from the left), there are \(4\) cubes.

Step2: Substitute into the center - of - mass formula

Substitute \(m_1 = 1,x_1=0.5\); \(m_2 = 2,x_2 = 1.5\); \(m_3=4,x_3 = 2.5\) into \(x_{cm}=\frac{\sum_{i = 1}^{N}m_ix_i}{\sum_{i = 1}^{N}m_i}\). We get \(x_{cm}=\frac{0.5\times1+1.5\times2 + 2.5\times4}{11}\)

Answer:

D. \(x_{cm}=\frac{0.5(1)+1.5(2)+2.5(4)}{11}\)