Question

Four uniform solid cubes of edges 10cm, 20cm, 30cm and 40 cm are kept on the ground, touching each other in order.
Locate centre of mass of their system.​

Answer 1

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put (0,0){\framebox(10,10){}}\put (10,0){\framebox(20,20){}}\put (30,0){\framebox(30,30){}}\put (60,0){\framebox(40,40){}}\put(37.5,12.5){\circle*{1}}\put(36.9,10.5){\tiny(x,\ y)}\put(-5,-2.5){\tiny(0,0)}\put(4.4,6){\tiny(5,5)}\put(19.4,11){\tiny(20,10)}\put(44.4,16){\tiny(45,15)}\put(79.4,21){\tiny(80,20)}\put(0,5){\vector (1,0){5}}\put(10,10){\vector (1,0){10}}\put(30,15){\vector (1,0){15}}\put(60,20){\vector (1,0){20}}\put(5,0){\vector (0,1){5}}\put(20,0){\vector (0,1){10}}\put(45,0){\vector (0,1){15}}\put(80,0){\vector (0,1){20}}\end{picture}$

Here the four solid cubes are of uniform mass.

We have the expression for the center of mass of a system of particles,

$\overline {a}=\dfrac {\displaystyle\sum_{i=1}^nm_ia_i}{\displaystyle\sum_{i=1}^nm_i}$

But if the system consists of particles of uniform mass,

$\overline {a}=\dfrac {1}{n}\displaystyle\sum_{i=1}^na_i$

Let (x, y) be the position of center of mass of the system. Then, from the fig.,

$x=\dfrac {5+20+45+80}{4}\\\\\\x=37.5$

And,

$y=\dfrac {5+10+15+20}{4}\\\\\\y=12.5$

Hence the center of mass is at the point $\underline {\underline {(37.5,\ 12.5)}}.$