Question

Seven homogenous 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-ordinate of the centre of mass relative to the origin O shown.

• A. $\frac{22L}{5}$
• B. $\frac{2L}{35}$
• C. $\frac{22L}{35}$
• D. $\frac{12L}{35}$

Answer 1

The correct option is C $\frac{22L}{35}$

Let the mass of each brick be m units. Hence the total mass of the arrangement = 7m units.

Since the bricks are homogeneous and rectangular, the CoM of each brick will be at its center and mass of each brick will be assumed to be concentrated at this CoM.

x - Coordinate of the center of 1st and 7th brick = L/2 units

x - Coordinate of the center of 2nd and 6th brick = L/2+L/10 =6L/10 = 3L/5 units

x - Coordinate of the center of 3rd and 5th brick = L/2+L/10+L/10 =7L/10 units

x - Coordinate of the center of 4th brick = L/2+L/10+L/10+L/10 =8L/10 =4L/5 units

Hence x - Coordinate of the center of mass of the arrangement, X = (1/7m)*(2m*L/2+2m*3L/5+2m*7L/10+m*4L/5) = (1+6/5+14/10+ 4/5)L/7 =(10+12+14+8)L/70 =44L/70 =22L/35 units.

hence option (c) is correct.