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 140 kPa, and the depth of the water will be 13.8 m. The pressure of the air in the building outside the tank will be 90.0 kPa.
Find the net downward force on the tank's flat bottom, of area 2.35 m^2, exerted by the water and air inside the tank and the air outside the tank.
Express your answer numerically in Newtons, to three significant figures.
Hydrostatic Pressure: p = p_0 + (rho)gd
Mass: m = vd
The Attempt at a Solution
Based on a force diagram for the pressure at a depth (d) in a liquid from my text:
Upward force = pA
Downward force = mg + (p_0)(A)
Net force = 0
To solve for m:
m = vd = (2.35m^2)(13.8m)(1000 kg/m^3) = 3.24 x 10^4 kg
Given pressure at the surface:
p_0 = 140 kPa + 90 kPa = 230 kPa = 2.3 x 10^5 Pa
mg = (3.24 x 10^4 kg)(3.71 m/s^2) = 1.20 x 10^5 N
(p_0)(A) = (2.3 x 10^5 Pa)(2.35m^2) = 5.41 x 10^4 N
1.20 x 10^5 N + 5.41 x 10^4 N = 6.61 x 10^5 N which is incorrect.
I also solved for p, which is probably unnecessary:
p = p_0 + (rho)gd
p = 2.3 x 10^5 Pa + (1000 kg/m^3)(3.71 m/s^2)(13.8m)
p = 2.82198 x 10^5 Pa