Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: For his art project, Weishaupt cut out $N=20$ wooden equilateral triangular blocks with a side length of $\ell=10 \mathrm{~cm}$ and a thickness of $t=2 \mathrm{~cm}$, each with the same mass and uniform density. He wishes to stack one on top of the other overhanging the edge of his table. In centimeters, what is the maximum overhang? Round to the nearest centimeter. A side view is shown below. Assume that all triangles are parallel to each other.

Note: This diagram is not to scale.

<img_4256>

Solution:
Let us consider $N=1$ equilateral triangles. From inspection, we need to place the triangle such that the center of mass lies at the edge of the table. The maximum overhang in this case is $(1-f) h$ where $h=\frac{\ell \sqrt{3}}{2}$ is the height of the triangle and $f h=\frac{h}{3}$ is the location of the center of mass.
If we wish to place a second triangle on top, we want to maximize the center of mass to be as far right as possible without the top block toppling. Placing the second block such that its center of mass is at the tip of the first triangle accomplishes this. However, the center of mass of the two triangles combined is now past the edge. Their center of mass is:

$$
x_{\mathrm{cm}}=\frac{f h+h}{2}=\frac{f+1}{2} h
$$

Thus the maximum overhang of the first block is now:

$$
h-\frac{f+1}{2} h=\frac{1-f}{2} h
$$

Now, we will place a third block such that it has the maximum overhang with respect to the top block and then shift the entire setup so that the center of mass of the system lies at the edge of the table. Following the same procedures, we find that the maximum overhang of the first block is:

$$
\frac{1-f}{3} h
$$

The overhang of the top two blocks are $(1-f) h$ and $\frac{1-f}{2} h$, unchanged from earlier. You can show via induction that the maximum overhang of the $n^{\text {th }}$ block (counting from the top downwards) is:

$$
\frac{1-f}{n} h
$$

so if there are 20 such blocks, then the total overhang (summing over all the blocks) is:

$$
\sum_{k=0}^{20} \frac{1-f}{k} h=(1-f) h H_{20}=\frac{2}{3} \frac{\ell \sqrt{3}}{2} H_{20}=\frac{\ell \sqrt{3}}{3} H_{20}=20.77 \mathrm{~cm} \approx 21 \mathrm{~cm}
$$

where $H_{N}$ is the $N^{\text {th }}$ harmonic number.

Final answer: 21 cm