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seven homogeneous bricks, each of length L are arranged as shown in the figure. Each brick is displaced w.r.t the one in contact by L/10. Find the x-coordinate of the centre of mass relative to origin(figure in HCV- Q-3 pg 159 Centre of mass)

seven homogeneous bricks, each of length L are arranged as shown in the figure. Each brick is displaced w.r.t the one in contact by L/10. Find the x-coordinate of the centre of mass relative to origin(figure in HCV- Q-3 pg 159 Centre of mass)

Grade:upto college level

2 Answers

Nirmal Singh.
askIITians Faculty 44 Points
7 years ago

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Thanks & Best Regards,
Nirmal Singh
Askiitians faculty
Aman
39 Points
4 years ago
Let mass of brick be M and length be L 
Each are in contact with each other by distance L/10.Moreover they are placed on one over other .
Let the bricks be named as A,B,C,D,E,F,G 
They are placed that first and last bricks have same com 
Second and sixth have same com
Third and fifth have same com and
Fourth has different  com
 Therefore ,
Com of A,G=L/2
COM of B,F=L/2+L/10.    (AS 2 BRICKS HAVE BEEN DISPLACED BY DISTANCE L/10)
COM OF C,E =L/2+2L/10
COM OF D=L/2+3L/10
NOW COM OF X COORDINATE IS:-
X=(M1X1+M2X2+M3X3+M4X4+M5X5+M6X6+M7X7)/M1+M2+M3+M4+M5+M6+M7
={M*L/2+M*L/2+M*(L/2+L/10)+M*(L/2+L/10)+M*(L/2+2L/10)+M*(L/2+2L/10)+M*(L/2+3L/10)}/M+M+M+M+M+M+M
=2*ML/2+2*M(L/2+L/10)+2*M(L/2+2L/10)+M*(L/2+3L/10)/7M
=[ML+2M{(5L+L)/10}+2M{(5L+2L)/10}+M*{(5L+3L)/10}]/7M
=(10ML+12ML+14ML+8M)/10*7M
=44ML/70M
=44L/70 
OR
=22L/35
 

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