Question

You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 125 kPa , and the depth of the water will be 14.2 m . The pressure of the air in the building outside the tank will be 87.0 kPa .

Find the net downward force on the tank's flat bottom, of area 2.05 m^2 , exerted by the water and air inside the tank and the air outside the tank.

Answer 1

Pfluid = ρgh
Pfluid is pressure in Pa or N/m²
ρ is the density of the fluid in kg/m³
density of water at 20C = 0.998 g/cm³ = 998 kg/m³
density of seawater = 1.025 g/cm³ = 1025 kg/m³
g is the acceleration of gravity 3.71 m/s²
h is the height of the fluid above the object=14.2 m

Pfluid = ρgh = (998)(3.71)(14.2)

= 52576 Pa = 52.57kPa

Add to that 125 kPa to get 177.57 kPa or kN/m²
and subtract the outside air pressure to get 90.57 kPa or kN/m²

90.57 kN/m² x 2.05 m² = 185.66 kN