Q.
Three bricks each of length $L$ and mass $M$ are arranged as shown from the wall. The distance of the centre of mass of the system from the wall is
Solution:
$ {{x}_{1}}=\frac{L}{2} $
$ {{x}_{2}}=L $
$ {{x}_{3}}=L+\frac{L}{4} $
$ {{x}_{3}}=\frac{5L}{4} $
$ \therefore $ $ {{x}_{CM}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{2}}{{x}_{3}}}{3m} $
$ =\frac{\frac{mL}{2}+mL+\frac{m5L}{4}}{3m} $
$ =\frac{11L}{12} $
