Question

Four uniform solid cubes of edges 10cm, 20cm, 30cm,

40cm are kept on the ground touching each other in

order. Locate the center of mass of their system.​

Answer 1

Explanation:

Again, if the weight of the 30cm cube passes through C' then OC'=45 cm. Again, if the weight of the 40cm cube passes through D' then OD'=80 cm. ⇒ d= 65 cm. Therefore, the center of mass of the system passes through 65 cm distance from the edge of the 10 cm cube.

Answer 2

Answer: (65 cm , 17.7 cm)

Explanation: Let the four cubes are denoted as 1, 2, 3, 4 .

The y- coordinate of the cubes are:

y1​=10/2​ = 5 ; y2 ​= 20/2​ = 10;  y3 ​= 30/2 ​= 15;  y4 ​= 40/2 ​= 20 (all are in cm)

The x-coordnate of the cubes are:

x1​ = 10/2​ = 5;  x2 ​= 10 + 20/2​ = 20;   x3​ = 10 + 20 + 30/2 ​= 45;

x4​ = 10 + 20 + 30 + 40​/2 = 80 (all are in cm)

The masses of the cubes are:

m1​ = ρ (10)^3 = 1000 ρ = M

m2​ = ρ (20)^3 = 8000 ρ = 8 M

m3​ = ρ (30)^3 = 27000 ρ = 27 M

m4​ = ρ (40)^3 = 64000 ρ = 64 M

Where ρ is the density of the material of the cubes.

So the Y coordinate of the center of mass :

$Y_{CM} = \frac{m1 y1 + m2y2 + m3y3 + m4 y4}{m1 + m2 + m3 + m4 }\\\\ = \frac{M\times5 + 8M\times10 + 27M \times 15 + 64M \times 20​ }{ M+8M+27M+64M}\\\\= 17.7 ~ cm$

and similarly,  the X coordinate of the centre of mass :

$X_{CM} = \frac{m1 x1 + m2x2 + m3x3 + m4 x4}{m1 + m2 + m3 + m4 }\\\\\\= 65 ~ cm$

So, the center of mass of the system is at the point (65 cm, 17.7 cm).