Question

Your task in a science contest is to stack four identical uniform bricks, each of length L, so that the top brick is as far to the right as possible without the stack falling over. Is it possible, as Figure P12.61 shows, to stack the bricks such that no part of the top brick is over the table? Answer this question by determining the maximum possible value of d.

Short answer

The maximum possible value of d = 4/5 L

Step-by-step solution

Given information

4 bricks with length of L each is stacked as shown in given figure.

Explanation

First find the center of mass using

$x_{CoM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}\dots \dots \dots \dots \dots \text{(1)}$

From the figure we see each brick is shifted d/4 rightwards, from the previous one,

so

$$\begin{gathered} x_1=-\frac{L}{2}+\frac{d}{4} \\ {x}_2=-\frac{L}{2}+2\frac{d}{4}=-\frac{L}{2}+\frac{d}{2} \\ {x}_3=-\frac{L}{2}+\frac{3d}{4} \\ {x}_4=-\frac{L}{2}+4\frac{d}{4}=-\frac{l}{2}+d \end{gathered}$$

Use these in equation (1) we get

$$\begin{gathered} x_{CoM}=\frac{m\left(x_1+x_2+x_3+x_4\right)}{4m} \\ =\frac{1}{4}\left(-2L+\frac{5}{2}d\right) \end{gathered}$$

For the stack to be stable x_Com=0
This means

$$\begin{gathered} -2L+\frac{5}{2}d=0 \\ d=4/5\mathrm{L} \end{gathered}$$