Q.Four uniform solid cubes of edges 10 cm, 20 cm, 30 cm and 40 cm are kept on the ground, touching each other in order (side by side, in a row, smallest to largest). Locate the centre of mass of their system.

[!TLDR] Placing the cubes in a row (10, 20, 30, 40 cm edges) and computing the mass-weighted average of both their horizontal centre positions and their (different) heights gives the system's c.m. at approximatel …

For a uniform solid cube of edge L, mass is proportional to $L^3$, and its own centre of mass is at height $L/2$ above the ground and at horizontal distance $L/2$ from its own near face. Placing the four cubes (edges 10, 20, 30, 40 cm) side by side touching each other in increasing order, with the very first cube's outer face at x=0: cube 1 (edge 10) occupies x=[0,10], its centre at $x_1=5$ cm, height $y_1=5$ cm, mass $\propto 10^3=1000$; cube 2 (edge 20) occupies x=[10,30], centre $x_2=20$ cm, height $y_2=10$ cm, mass $\propto 20^3=8000$; cube 3 (edge 30) occupies x=[30,60], centre $x_3=45$ cm, height $y_3=15$ cm, mass $\propto 30^3=27000$; cube 4 (edge 40) occupies x=[60,100], centre $x_4=80$ cm, height $y_4=20$ cm, mass $\propto 40^3=64000$. Total mass (in units of edge^3) $=1000+8000+27000+64000=100000$. Horizontal c.m.: $x_c=\frac{1000(5)+8000(20)+27000(45)+64000(80)}{100000}=\frac{5000+160000+1215000+5120000}{100000}=\frac{6500000}{100000}=65$ cm. V …

Treat each cube's own centre of mass as a point mass proportional to (edge length)^3, positioned at its geometric centre (both horizontally along the row and vertically at half its o …

Using edge length (or edge^2, i.e. face area) instead of edge^3 (volume, which is what mass is proportional to for uni …