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 $\frac{\text{L}}{10}.$ Find the x-coordinate of the centre of mass relative to the origin shown. 
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Let ‘O’ (0, 0) be the origin of the system. Each brick is mass ‘M’ & length ‘L’. Each brick is displaced w.r.t. one in contact by $\frac{\text{L}}{10}$ $\therefore$ The X-coordinate of the centre of mass.
$\overline{\text{X}}_{\text{cm}}=\frac{\text{m}\Big(\frac{\text{L}}{2}\Big)+\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{L}}{10}\Big)+\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{2L}}{10}\Big)+\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{3L}}{10}\Big)+\\\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{3L}}{10}-\frac{\text{L}}{10}\Big)+\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{L}}{10}\Big)+\text{M}\Big(\frac{\text{L}}{2}\Big)}{\text{7m}}$ $=\frac{\frac{\text{L}}{2}+\frac{\text{L}}{2}+\frac{\text{L}}{10}+\frac{\text{L}}{2}+\frac{\text{L}}{5}+\frac{\text{L}}{2}+\frac{\text{3L}}{10}+\frac{\text{L}}{2}+\frac{\text{L}}{5}+\frac{\text{L}}{2}+\frac{\text{L}}{10}+\frac{\text{L}}{2}}{7}$ $=\frac{\frac{\text{7L}}{2}+\frac{\text{5L}}{10}+\frac{\text{2L}}{5}}{7}$ $=\frac{\text{35L}+\text{5L}+\text{4L}}{10\times7}=\frac{\text{44L}}{70}=\frac{11}{35}\text{L}$
$\overline{\text{X}}_{\text{cm}}=\frac{\text{m}\Big(\frac{\text{L}}{2}\Big)+\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{L}}{10}\Big)+\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{2L}}{10}\Big)+\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{3L}}{10}\Big)+\\\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{3L}}{10}-\frac{\text{L}}{10}\Big)+\text{m}\Big(\frac{\text{L}}{2}+\frac{\text{L}}{10}\Big)+\text{M}\Big(\frac{\text{L}}{2}\Big)}{\text{7m}}$ $=\frac{\frac{\text{L}}{2}+\frac{\text{L}}{2}+\frac{\text{L}}{10}+\frac{\text{L}}{2}+\frac{\text{L}}{5}+\frac{\text{L}}{2}+\frac{\text{3L}}{10}+\frac{\text{L}}{2}+\frac{\text{L}}{5}+\frac{\text{L}}{2}+\frac{\text{L}}{10}+\frac{\text{L}}{2}}{7}$ $=\frac{\frac{\text{7L}}{2}+\frac{\text{5L}}{10}+\frac{\text{2L}}{5}}{7}$ $=\frac{\text{35L}+\text{5L}+\text{4L}}{10\times7}=\frac{\text{44L}}{70}=\frac{11}{35}\text{L}$
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