Four bricks of length $\mathbf{L}$ , identical and uniform, are stacked on a table in two ways, as shown in Fig. 12-83 (compare with Problem 63). We seek to maximize the overhang distance h in both arrangements. Find the optimum distances${\mathbf{a}}_{\mathbf{1}}$ ,${\mathbf{a}}_{\mathbf{2}}$ ,${\mathbf{b}}_{\mathbf{1}}$ , and${\mathbf{b}}_{\mathbf{2}}$ , and calculate hfor the two arrangements.

The length of the brick is L

You draw the free body diagram. The system is at equilibrium. For such a system, the vector sum of the forces acting on it is zero. The vector sum of the external torques acting on the beam at one point is zero. You can use the center of the mass concept. It is the point where the whole mass of the body can be supposed to be concentrated and the relative position of the distributed mass sums to zero.

$\begin{array}{c}{\mathbf{x}}_{\mathbf{cm}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}\\ \mathbf{\Sigma }{\overrightarrow{\mathbf{\tau }}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\end{array}$

You consider the origin of the x-axis at the edge of the table. At the edge of the table$\mathrm{x}=0$then the total center of mass of the block must be zero.

For figure (a):

In this figure, two bricks directly put one another hence total mass is $2\text{\hspace{0.17em}m}$ for these two bricks whose center is above the left edge of the bottom brick. There is a single brick at the upper right of mass $\mathrm{m}$ whose center is over the right edge of the bottom brick. The mass of the bottom brick is $\mathrm{m}$. Then the total center of mass is

$\begin{array}{c}{\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ 0=\frac{\left(2\mathrm{m}\right)({\mathrm{a}}_2-\mathrm{L})+{\mathrm{ma}}_2+\mathrm{m}\left({\mathrm{a}}_2-\frac{\mathrm{L}}{2}\right)}{2\mathrm{m}+\mathrm{m}+\mathrm{m}}\\ 0=\frac{\left(2\mathrm{m}\right)({\mathrm{a}}_2-\mathrm{L})+{\mathrm{ma}}_2+\mathrm{m}\left({\mathrm{a}}_2-\frac{\mathrm{L}}{2}\right)}{2\mathrm{m}+\mathrm{m}+\mathrm{m}}\\ 0=\frac{4{\mathrm{ma}}_2-3\mathrm{m}\left(\frac{\mathrm{L}}{2}\right)}{4\mathrm{m}}\\ {\mathrm{a}}_2=\frac{5\mathrm{L}}{8}\end{array}$

From the figure,

$\begin{array}{l}{\mathrm{a}}_1=\frac{\mathrm{L}}{2}\\ \mathrm{h}={\mathrm{a}}_1+{\mathrm{a}}_2\\ \mathrm{h}=\frac{\mathrm{L}}{2}+\frac{5\mathrm{L}}{8}\\ \mathrm{h}=\frac{9\mathrm{L}}{8}\end{array}$

For figure (b):

These four bricks where the center of mass of the top one and center of mass of the bottom one has the same value as

${\mathrm{x}}_{\mathrm{cm}}={\mathrm{b}}_2-\frac{\mathrm{L}}{2}$

The middle layer consists of two layers. The top brick is exerting downward forces on the middle bricks.

${\mathrm{F}}_{\mathrm{t}}=\frac{\mathrm{mg}}{2}$

This is the static equilibrium condition. Hence, the vector sum of the external torques acting on the ladder at one point is zero.

$\begin{array}{c}\mathrm{\Sigma }{\overrightarrow{\mathrm{\tau }}}_{\mathrm{net}}=0\\ \mathrm{mg}\left({\mathrm{b}}_1-\frac{\mathrm{L}}{2}\right)=\frac{\mathrm{mg}}{2}(\mathrm{L}-{\mathrm{b}}_1)\\ \left({\mathrm{b}}_1-\frac{\mathrm{L}}{2}\right)=\frac{1}{2}(\mathrm{L}-{\mathrm{b}}_1)\\ {\mathrm{b}}_1=\frac{2\mathrm{L}}{3}\end{array}$

From the figure,

$\begin{array}{c}{\mathrm{b}}_2=\frac{\mathrm{L}}{2}\\ \mathrm{h}={\mathrm{b}}_1+{\mathrm{b}}_2\\ \mathrm{h}=\frac{2\mathrm{L}}{3}+\frac{\mathrm{L}}{2}\\ \mathrm{h}=\frac{7\mathrm{L}}{6}\end{array}$