Question

Three uniform bricks each of length $100\text{cm}$ and mass $m$ are arranged as shown in figure. The distance of centre of mass of the system of bricks from the wall is

• A. $91\text{cm}$
• B. $100\text{cm}$
• C. $75\text{cm}$
• D. $125\text{cm}$

Answer 1

The correct option is A $91\text{cm}$

Taking origin at the lower end of wall.
Centre of mass of brick $A$
$=(x_1,y_1)=(50,0)\text{cm}$
Centre of mass of brick $B$
$=(x_2,y_2)=(100,0)\text{cm}$
Centre of mass of brick $C$
$=(x_3,y_3)=(123,0)\text{cm}$
Let $x_{cm}$ be the distance of COM of system from the wall
Then, $x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{m\left(50\right)+m\left(100\right)+m\left(123\right)}{3m}$
$=\frac{273}{3}=91\text{cm}$