  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.

Solution

x coordinate of center of mass of 1 = $$\cfrac {L}{2}$$x coordinate of center of mass of 2 = $$\cfrac {L}{2} + \cfrac{L}{10}$$x coordinate of center of mass of 3 = $$\cfrac {L}{2} + \cfrac{L}{10} + \cfrac{L}{10}$$x coordinate of center of mass of 4 = $$\cfrac {L}{2} + \cfrac{L}{10} + \cfrac{L}{10} + \cfrac{L}{10}$$x coordinate of center of mass of 5 = $$\cfrac {L}{2} + \cfrac{L}{10} + \cfrac{L}{10}$$x coordinate of center of mass of 6 = $$\cfrac {L}{2} + \cfrac{L}{10}$$x coordinate of center of mass of 7 = $$\cfrac {L}{2}$$$$\therefore$$ x coordinate of combined system =$$\cfrac{(m.\cfrac{L}{2}) + (m . \cfrac{6L}{10}) + (m . \cfrac{7L}{10}) + (m . \cfrac{8L}{10}) + (m. \cfrac{7L}{10}) + (m. \cfrac{6L}{10}) + (m. \cfrac{L}{2})}{m+m+m+m+m+m+m}$$$$= \cfrac{44L} { 7 \times 10}$$ $$= \cfrac{22L}{35}$$ Physics

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