91Ó°ÊÓ

(1)

Where${\stackrel{\mathbf{→}}{\mathbf{r}}}_{\mathbf{cm}}$ is the position vector of the center of mass of a system of particles.

${\mathbf{m}}_{\mathbf{i}}$is the mass of the ith particle. ${\stackrel{\mathbf{→}}{\mathbf{r}}}_{\mathbf{i}}$ position vectors of ith particle.

Here we have the mass of both blocks is$\mathrm{m}$.

Now, the length of both blocks is $\mathrm{x}=\mathrm{L}$

(a)

From equation (1), we can write,

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$ (2)

Now, here mass of both blocks is the same. So, ${\mathrm{m}}_1={\mathrm{m}}_2=\mathrm{m}$

the center of the top block must be over the bottom block.

So, here${\mathrm{x}}_1=0$and ${\mathrm{x}}_2=\frac{\mathrm{L}}{2}$

So, equation (2) becomes,

Now, this is the center of the combination.

While the first block has a center of mass $\frac{\mathrm{L}}{2}$, so the maximum overhang possible is$\frac{\mathrm{L}}{4}+\frac{\mathrm{L}}{2}=\frac{3\mathrm{L}}{4}$

Hence, the maximum overhang possible is $\frac{3\mathrm{L}}{4}$.

(b)

We find the center of the mass of the third block with respect to the combination of the other two blocks is

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_{12}{\mathrm{x}}_{12}+{\mathrm{m}}_3{\mathrm{x}}_3}{{\mathrm{m}}_{12}+{\mathrm{m}}_3}$ (3)

Now, here mass of both blocks is the same. So, ${\mathrm{m}}_{12}=2\mathrm{m},\text{ â¶Ä‰}{\mathrm{m}}_3=\mathrm{m}$

the center of the top block must be over the bottom block.

So, here ${\mathrm{x}}_{12}=0$and ${\mathrm{x}}_3=\frac{\mathrm{L}}{2}$

So, equation (3) becomes,

$\begin{array}{c}{\mathrm{x}}_{\mathrm{cm}}=\frac{\mathrm{m}\left(0\right)+\mathrm{m}\left(\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{2\mathrm{m}+\mathrm{m}}\\ =\frac{\mathrm{L}}{6}\end{array}$

Now, this is the center of the combination of three blocks

While the combination of the first and second block has a center of mass $\frac{3\mathrm{L}}{4}$ (from step 3). So the maximum overhang possible is

$\frac{3\mathrm{L}}{4}+\frac{\mathrm{L}}{6}=\frac{11\mathrm{L}}{12}$

Hence, the maximum overhang possible is $\frac{11\mathrm{L}}{12}$.

for the center of the mass of the four blocks with respect to the combination of the other three blocks is

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_{123}{\mathrm{x}}_{123}+{\mathrm{m}}_4{\mathrm{x}}_4}{{\mathrm{m}}_{123}+{\mathrm{m}}_4}$ (4)

Now, here mass of both blocks is the same. So, ${\mathrm{m}}_{123}=3\mathrm{m},\text{ â¶Ä‰}{\mathrm{m}}_4=\mathrm{m}$

the center of the top block must be over the bottom block.

So, here${\mathrm{x}}_{123}=0$and ${\mathrm{x}}_4=\frac{\mathrm{L}}{2}$

So, equation (4) becomes,

Now, this is the center of the combination of three blocks

While the combination of the first three blocks has a center of mass $\frac{11\mathrm{L}}{12}$ So the maximum overhang possible is $\frac{11\mathrm{L}}{12}+\frac{\mathrm{L}}{8}=\frac{25\mathrm{L}}{24}$

Hence, the maximum overhang possible is $\frac{25\mathrm{L}}{24}$.

(c)

Here in step 4, we found that at four blocks, the maximum overhang is larger than the length of one block as,$\frac{25\mathrm{L}}{24}>\mathrm{L}$.

So, at four blocks the uppermost block is not directly over the table, so it is possible to make a stack of blocks.