A uniform rectangular block of length 20.0 cm is placed so that its centre of ma

Question

A uniform rectangular block of length 20.0 cm is placed so that its centre of mass is a distance of 6.17 cm away from the edge of the table. Since its centre of mass is still over the table (i.e. not sticking out past the edge), the block is stable. An identical block is placed on top of that block. How far from the edge of the table can the centre of mass of the top block be before the blocks become unstable? Take the edge of the table to be the origin of your coordinate system, with negative values representing positions that are over the table, and positive values representing positions that are past the edge.

Explanation / Answer

When the two blocks are placed over one another then their centre of mass will be at
Since both the blocks are identical therefore their masses are same
x = (mx1 + mx2) / (m + m)

Now taking edge of the table as origin with negative values representing positions that are over the table, and positive values representing positions that are past the edge

We get

x = m(-6.17) + m(x0) / 2m

Now Block will become unstable when the combined centre of mass of the block is at the edge of the table
or at x = 0
therefore
m(-6.17) + m(x0) / 2m = 0

We get
x0 = 6.17 (Answer)

Hence 6.17 cm from the edge of the table can the centre of mass of the top block be before the combined block becomes unstable

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