Question

Four homogeneous brick, each of length $L=4\text{m}$, are arranged as shown in the figure. Each brick is displaced with respect to one in contact by $\frac{L}{8}.$ Find the $x-$ coordinate of centre of mass relative to origin $O$.

• A. $2.5\text{m}$
• B. $1.75\text{m}$
• C. $2\text{m}$
• D. $2.75\text{m}$

Answer 1

The correct option is D $2.75\text{m}$

length of each brick, $L=4\text{m}$, Let the mass of each brick be $m\text{kg}$.

$\Rightarrow$Initially all four bricks have their $\text{COM}$ at $\frac{L}{2}=2\text{m}$ from the $x-$axis.


After displacing the bricks, coordinates relative to origin $O$ :
$x-$ coordinate of $\text{COM}$ of brick $\left(1\right)\Rightarrow x_1=\frac{L}{2}$
$x-$ coordinate of $\text{COM}$ of brick $\left(2\right)\Rightarrow x_2=\frac{L}{2}+\frac{L}{8}=\frac{5}{8}L$
$x-$ coordinate of $\text{COM}$ of brick $\left(3\right)\Rightarrow x_3=\frac{L}{2}+\frac{L}{8}+\frac{L}{8}=\frac{6}{8}L$
$x-$ coordinate of $\text{COM}$ of brick $\left(4\right)\Rightarrow x_4=\frac{L}{2}+\frac{L}{8}+\frac{L}{8}+\frac{L}{8}+\frac{7}{8}L$
Replacing the bricks by point masses placed at their $\text{COM}$ , the $x-$ coordinate of $\text{COM}$ of system:

$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
$x_{\text{CM}}=\frac{m\left(\frac{L}{2}\right)+m\left(\frac{5L}{8}\right)+m\left(\frac{6L}{8}\right)+m\left(\frac{7L}{8}\right)}{m+m+m+m}$
$\Rightarrow x_{\text{CM}}=\frac{22}{32}L$
$\therefore x_{\text{CM}}=\frac{22}{32}\times 4=\frac{22}{8}=2.75\text{m}$