Question

A uniform piece of sheet steel is shaped as in the figure below. (Both axes are marked in increments of 2). Compute the x and y coordinates of the center of mass of the piece.

Answer 1

"increments of 8" means the major divisions are 0,8,16,24 ? 

x axis, calculate the moment arms from 0 
3x4, 2x12, 1x20 
from an arbitrary C 
3(c-4) + 2(c-12) + (c-20) = 0 
3c - 12 + 2c -24 + c - 20 = 0 
6c = 56 
c = 9.33 

y axis 
3x3, 1x12, 2x20 
3(c-4) + 1(c-12) +2 (c-20) = 0 
3c - 12 + c - 12 + 2c - 40 = 0 
6c = 64 
c = 10.67 

so center is x = 9.33, y = 10.67 

Answer 2

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:

$(\bar{X},\bar{Y})=(2\frac{1}{3},2\frac{2}{3})$