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Step-by-Step Solution
Step 1: Set up a coordinate system
Consider the rectangular sheet ABCD lying in the XY-plane with corner A at the origin
O(0,0), point B at (a, 0), point C at (a, b), and point D at (0, b). Hence, the sheet
has dimensions length = a and breadth = b.
Step 2: Determine the mass and center of mass of the full sheet
Let the total mass of the full sheet ABCD be $M$. Because the sheet is uniform,
its center of mass is at the geometric center:
$X_{\text{sheet}} = \dfrac{a}{2}, \quad Y_{\text{sheet}} = \dfrac{b}{2}.$
Step 3: Identify the removed portion and its center of mass
In the figure, the shaded part HBGO is cut off. Let its mass be $m$. Because the
sheet is uniform, $m$ is proportional to the area of the region HBGO. From the
problem's geometry (as given or derived), the center of mass of the removed portion
is located at some point
$ \bigl(x_{\text{cut}}, y_{\text{cut}}\bigr). $
In the provided solution, this coordinate works out to be
$ \bigl(\tfrac{3a}{4}, \tfrac{3b}{4}\bigr). $
Step 4: Apply the subtraction formula for the center of mass
When a portion of mass $m$ is removed from a bigger mass $M$ whose center of mass
is known, the new center of mass $(X_{cm},\, Y_{cm})$ of the remaining shape
(mass $M - m$) can be found using:
$
X_{cm} \;=\; \dfrac{M\,X_{\text{sheet}} \;-\; m\,X_{\text{cut}}}{M - m},
\quad
Y_{cm} \;=\; \dfrac{M\,Y_{\text{sheet}} \;-\; m\,Y_{\text{cut}}}{M - m}.
$
Step 5: Substitute known values
From the given/derived steps in the problem:
- $M\,X_{\text{sheet}} = M \times \dfrac{a}{2}.$
- $m\,X_{\text{cut}} = m \times \dfrac{3a}{4}.$
- $M\,Y_{\text{sheet}} = M \times \dfrac{b}{2}.$
- $m\,Y_{\text{cut}} = m \times \dfrac{3b}{4}.$
Sometimes, for convenience, one takes $M = 4m$ (depending on area subdivisions),
but the main idea is to keep track of each term consistently.
Step 6: Calculate $X_{cm}$
$
X_{cm}
= \dfrac{M \left(\tfrac{a}{2}\right) \;-\; m \left(\tfrac{3a}{4}\right)}{M - m}.
$
Using the ratio $M = 4m$ (often used when the cut-off portion is one-fourth of the sheet),
the numerator becomes
$
4m \left(\tfrac{a}{2}\right) \;-\; m \left(\tfrac{3a}{4}\right)
\;=\; 2ma \;-\; \tfrac{3ma}{4}
\;=\; \tfrac{5ma}{4}.
$
The denominator $M - m = 4m - m = 3m$, thus
$
X_{cm}
= \dfrac{\tfrac{5ma}{4}}{3m}
= \dfrac{5a}{12}.
$
Step 7: Calculate $Y_{cm}$
By a similar process for the vertical coordinate,
$
Y_{cm}
= \dfrac{M \left(\tfrac{b}{2}\right) - m \left(\tfrac{3b}{4}\right)}{M - m}
= \dfrac{\tfrac{5mb}{4}}{3m}
= \dfrac{5b}{12}.
$
Step 8: State the final coordinates of the remaining sheet
Therefore, the center of mass of the sheet after the shaded portion is removed
is
$
\left(\dfrac{5a}{12},\; \dfrac{5b}{12}\right).
$
Final Answer
$ \displaystyle \left( \frac{5a}{12}, \frac{5b}{12} \right). $
