The perks of being a sea snail

The radius of the snail is $r = 3$ $cm = 0.03$ $m$ . So, the area of the snail is $A = \pi r^2 = \pi \times 0.0009$ $m^2$ .

Now, the volume of the sea water that vertically above the snail $V = A \times h = \pi \times 0.0009\times 3$ $m^3$ $= \pi \times 0.0027$ $m^3$

We know, $m = ρ\times V$ and according to Newton's $2$ nd law $F = ma = mg$ where, $a = g$

So, $F = mg =( ρ\times V) \times g = (1000 \times \pi \times 0.0027 \times g)$ $N$
$= 2.7\times \pi \times g$ $N$ $= 2.7 \times 3.14 \times 9.8$ $N$
$= \boxed {83.084}$ $N$

Considering the all the forces acting on the snail, a) Buoyant force of water on the snail i.e.,
$F_{B} = \rho g V = 0.55417 N$ , where $V= \frac {2}{3} \pi r^{3}$ is the volume of the displaced water and $r = .03m$ , \row is density of water ( $1000kg/m^3$ ) b) Weight exerted by the column of water above it

i.e., $F_ {W} = \rho g V’= 83.1265 N$ where V’ is the volume of water above the snail which is given by $V’=\pi r^{2} h$ where h is the height of the column of water above it( $h=3m$ ).

So, the net vertical force acting one the snail is given by,

$F = F_{W} - F_{B}$

substituting the values we get, $F = 83.1265 - .55417 = 82.572 \approx 82.57 N$

Thus, net vertical force on the snail is $82.57 N$ .

Imagine you have a huge cilinder arround the hemisphere, with a base with area of $\pi \cdot (0.03)^2$ and therefore a volume of $3 \cdot \pi \cdot (0.03)^2$ . The volume of the hemisphere is $\frac {4}{6} \cdot \pi \cdot (0.03)^3$ . Subtracting the volume of the hemisphere of the volume of the cilinder gives us the amount of volume ocupied by water above the hemisphere. That volume times the water density gives us the waight of the water above the hemisphere. Multiplying that weigh by the absolute of the acceleration of gravity, since it wants only the force's magnitude, we get the final result: $(3 \cdot \pi \cdot (0.03)^2 - \frac {4}{6} \cdot \pi \cdot (0.03)^3) \cdot 1000 \cdot 9,8$

I would think you just find the volume of water on top of the snail:

Take a cylindrical column right above the snail:

V = πr²h

However, you need to subtract the volume of the snail (the hemisphere):

V = ½ * 4π/3 * r³

So the total volume of water above the snail is:

V = πr²h - 2π/3 * r³ = πr² * (h - 2r/3)

Now the total mass of the column of water is simply this volume time the density of water:

m = ρV = ρπr² * (h - 2r/3)

The total force due to this mass is simply mg (assuming constant gravity which is a perfectly valid assumption):

F = mg = gρV = gρπr² * (h - 2r/3) --> now plug in

g = 9.8 m/s², ρ = 1000 kg/m³, r = 0.03 m, and h = 3 m -->

F = 9.8 * 1000 * π * (0.03)² * (3 - 2*0.03/3) ~ 82.57 N

Force acts on the sea slug is equal to the weight of water above it.

Hence we have the volume of water column is $V$ = $hS$ - $\frac{1}{2}$ $\frac{4}{3}$ $\pi$ $r^3$ = $8.426\times10^{-3}$ ( $m^3$ ) since sea slug is a hemisphere.

So the weight of water is $P$ = $\rho$ $Vg$ = $82.576$ (N)

The area that the snail takes up on the bottom of the ocean is $\pi (0.03 m)^2 \approx 2.827*10^{-3} m^2$ and the volume of the snail is $\frac{1}{2}\frac{4}{3}\pi(0.03 m)^3 \approx 5.655*10^{-5} m^3$ . Therefore, the mass of the water column above the snail is $1000 \frac{kg}{m^3} *(2.827*10^{-3}*3 m^3 -5.655*10^{-5} m^3) \approx 8.426$ $kg$ . Therefore, the force of gravity on the water column, which is the vertical force on the snail is $8.426 kg *9.81 \frac{m}{s^2} \approx \fbox {82.576 N}$ .

If the force is $F$ and the pressure is $P$ and the area we need to consider is $A$ , we have that $F = PA$ . Since $A$ is the cross-sectional area, it is $\pi r^2$ , where $r$ is the radius. Thus: $F = P \cdot \pi r^2.$ But we know that $P = \rho g h$ , where $\rho$ is the density, $g$ is the acceleration of gravity, and $h$ is the depth. So: $F = \rho g h \cdot \pi r^2.$ Using $\rho = 1000 \, \text{kg/m}^3$ , $g = 9.81 \, \text{m/s}^2$ , $h = 3 \, \text{m}$ , and $r = 0.03 \, \text{m}$ , we get $F = \boxed {82.6 \, \text{N}}$ .

A hemisphere has a volume of $\frac{1}{2} \cdot \frac{4}{3}\pi r^3$ and it touches the ground at an area of $\pi r^2$ , so its average height is

$\frac{\frac{1}{2} \cdot \frac{4}{3}\pi r^3}{\pi r^2} = \frac{2r}{3} = 2\text{cm.}$

Therefore, the amount of water that's pressing on the snail is

$h \cdot \pi r^2 = (300-2) \cdot \pi \cdot 3^2 \approx 8426\text{cm}^3 = 0.008426\text{m}^3.$

This water has a mass of $0.008426\text{m}^3 \cdot 1000 \text{kg/m}^3 = 8.426$ kg. Therefore, the vertical force of the water is $m \cdot g = 8.426 \cdot 9.8 = \boxed{82.6}$ N.