Question

Seven homogeneous bricks, each of length L, are arranged as shown in figure. Each brick is displaced with respect to the one in contact by $L/10$. Find the x coordinate of the center of mass relative to the origin O.

Answer 1

x coordinate of center of mass of 1 = $\frac{L}{2}$

x coordinate of center of mass of 2 = $\frac{L}{2}+\frac{L}{10}$

x coordinate of center of mass of 3 = $\frac{L}{2}+\frac{L}{10}+\frac{L}{10}$

x coordinate of center of mass of 4 = $\frac{L}{2}+\frac{L}{10}+\frac{L}{10}+\frac{L}{10}$

x coordinate of center of mass of 5 = $\frac{L}{2}+\frac{L}{10}+\frac{L}{10}$

x coordinate of center of mass of 6 = $\frac{L}{2}+\frac{L}{10}$

x coordinate of center of mass of 7 = $\frac{L}{2}$

$\therefore$ x coordinate of combined system =$\frac{(m.\frac{L}{2})+(m.\frac{6L}{10})+(m.\frac{7L}{10})+(m.\frac{8L}{10})+(m.\frac{7L}{10})+(m.\frac{6L}{10})+(m.\frac{L}{2})}{m+m+m+m+m+m+m}$

$=\frac{44L}{7\times 10}$

$=\frac{22L}{35}$