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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 =$$ \cfrac{(m.\cfrac{L}{2}) + (m . \cfrac{6L}{10}) + (m . \cfrac{7L}{10}) + (m . \cfrac{8L}{10}) + (m. \cfrac{7L}{10}) + (m. \cfrac{6L}{10}) + (m. \cfrac{L}{2})}{m+m+m+m+m+m+m}$$
$$ = \cfrac{44L} { 7 \times 10}$$ 
$$ = \cfrac{22L}{35}$$

979864_1075039_ans_3dd56701bed64ada81384eecd845707b.png

Physics

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