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 140 kPa , and the depth of the water will be 13.6 m . The pressure of the air in the building outside the tank will be 85.0 kPa . Find the net downward force on the tank's flat bottom, of area 2.30 m2 , 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.

Answer 1

gIVEN
   acceleration due to gravity g = 3.71 m/s2
  
   depth of water h = 13.6 m

   pressure at the surface of the water 140 kPa,

   The pressure of the air in the building outside the tank will be 85.0 kPa

force on 2.30 m2 area is F = ?

we know that the pressure P = F/A ==> F = P*A

  
now for pressure , pressure on surface of water is P_e = 140000 Pa - 85000 Pa = 55000 Pa
  

   pressure of the wate is Pw = rho*g*h = 998 kg/m3*3.71 m/s2*13.6m = 50355.088 Pa

now total pressure is Pt = P_e + PW = 55000+50355.088 Pa = 105355.088 Pa

and the force on area of 2.30 m2 is , F = 105355.088*2.30 = 242316.70240 N = 242.317 kN