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: Anyone who's had an apple may know that pieces of an apple stick together, when picking up one piece a second piece may also come with the first piece. The same idea is tried on a golden apple. Consider two uniform hemispheres with radius $r=4 \mathrm{~cm}$ made of gold of density $\rho_{g}=19300 \mathrm{~kg} \mathrm{~m}^{-3}$. The top half is nailed to a support and the space between is filled with water.
<img_4253>

Given that the surface tension of water is $\gamma=0.072 \mathrm{~N} \mathrm{~m}^{-1}$ and that the contact angle between gold and water is $\theta=10^{\circ}$, what is the maximum distance between the two hemispheres so that the bottom half doesn't fall? Answer in millimeters.

Solution:
Let $h$ be the difference in height. There are 3 forces on the bottom hemisphere. The force from gravity, which has magnitude $\frac{2}{3} \pi \rho_{g} g r^{3}$, the force from the surface tension, and the force from the pressure difference at the top and bottom. The pressure difference is given by the young-laplace equation,
$$
\Delta P=\gamma\left(-\frac{1}{r}+\frac{2 \cos \theta}{h}\right) \approx \frac{2 \gamma \cos \theta}{h}
$$

The radii are found by some simple geometry. It is likely that $r$ will be much larger than the height, so we can neglect the $1 / r$ term. Now the force from surface tension is $2 \pi r \gamma \sin \theta$, since we take the vertical component. So we can now set the net force to 0 ,

$$
\frac{2}{3} \pi \rho_{g} g r^{3}=\pi r^{2} \Delta P+2 \pi r \gamma \sin \theta \Longrightarrow \frac{2}{3} \rho_{g} g r^{2}=(2 r \gamma \cos \theta) \frac{1}{h}+2 \gamma \sin \theta
$$

This is simple to solve,

$$
h=\frac{2 r \gamma \cos \theta}{2 \rho_{g} g r^{2} / 3-2 \gamma \sin \theta}=2.81 \times 10^{-5} \mathrm{~m}=0.0281 \mathrm{~mm}
$$

This is very small so our approximation from earlier is justified.

Final answer: 0.0281 mm