Answer:
\((\bar{X},\bar{Y})=(2\frac{1}{3},2\frac{2}{3})\)
Explanation:
Since the given sheet is of steel hence it has a homogeneous density, each piece of square measurement according to the graph will have equal mass.
For X coordinate of center of mass:
\(\bar{X}=\frac{m_1.x_1+m_2.x_2+m_3.x_3+m_4.x_4+m_5.x_5+m_6.x_6}{m_1+m_2+m_3+m_4+m_5+m_6}\)
where:
\(x_1,x_2,x_3,x_4,x_5,x_6\) are the respective geometric abscissa of square pieces.
Since,
Respective masses of the square pieces
\(m_1=m_2=m_3=m_4=m_5=m_6=m\,(let)\)
So,
\(\bar{X}=m\times \frac{(x_1+x_2+x_3+x_4+x_5+x_6)}{6m}\)
\(\bar{X}= \frac{(3+1+1+1+3+5)}{6}\)
\(\bar{X}=2\frac{1}{3}\)
For Y coordinate of center of mass:
\(\bar{Y}=\frac{m_1.y_1+m_2.y_2+m_3.y_3+m_4.y_4+m_5.y_5+m_6.y_6}{m_1+m_2+m_3+m_4+m_5+m_6}\)
where:
\(y_1,y_2,y_3,y_4,y_5,y_6\) are the respective geometric ordinates of square pieces.
Since,
Respective masses of the square pieces
\(m_1=m_2=m_3=m_4=m_5=m_6=m\,(let)\)
So,
\(\bar{Y}=m\times \frac{(y_1+y_2+y_3+y_4+y_5+y_6)}{6m}\)
\(\bar{Y}= \frac{(5+5+3+1+1+1)}{6}\)
\(\bar{Y}=2\frac{2}{3}\)
∴Center of mass of the given figure is:
\((\bar{X},\bar{Y})=(2\frac{1}{3},2\frac{2}{3})\)
