Question

Three uniform bricks each of length $60\text{cm}$ and mass $m$ are arranged as shown in the figure. Considering only the length of the bricks, find the distance of centre of mass of the system of bricks from the wall.

• A. $45\text{cm}$
• B. $55\text{cm}$
• C. $65\text{cm}$
• D. $35\text{cm}$

Answer 1

The correct option is B $55\text{cm}$

The $\text{COM}$ of the bricks will lie at the mid point of their lengths.

Since the brick $B$ is displaced by $30\text{cm}$ towards right and brick $C$ is displaced by $45\text{cm}$ towards right with respect to the wall, so their $\text{CM}$ will also be displaced by the same distance.

Centre of mass of brick $\left(A\right)\Rightarrow (x_1,y_1)=(30,0)$
Centre of mass of brick $\left(B\right)\Rightarrow (x_2,y_2)=(60,0)$
Centre of mass, of brick $\left(C\right)\Rightarrow (x_3,y_3)=(75,0)$

Let $x_{\text{COM}}$ be the distance of the centre of mass of the system from the wall, so applying the formula :

$x_{\text{COM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{m\left(30\right)+m\left(60\right)+m\left(75\right)}{3m}$
$\therefore x_{\text{COM}}=\frac{165}{3}=55\text{cm}$