Calculating the Net Downward Force on a Cylindrical Water Tank on Mars

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SUMMARY

The net downward force on a cylindrical water tank on Mars is calculated using the pressure exerted by the water and the air inside and outside the tank. Given a pressure of 110 kPa at the water's surface and an external air pressure of 91 kPa, the downward force from the water is 3.93x10^5 N, while the upward force from the air is 2.18x10^5 N. To find the net force, one must consider the volume of water, its mass, and the gravitational acceleration on Mars (3.71 m/s²). The correct approach involves summing the forces from the water and the air inside the tank.

PREREQUISITES
  • Understanding of fluid mechanics principles, specifically pressure calculations.
  • Knowledge of gravitational effects on Mars, specifically 3.71 m/s².
  • Familiarity with pressure units, particularly kilopascals (kPa).
  • Basic algebra for force calculations using the equation p = F/A.
NEXT STEPS
  • Calculate the volume of water in the tank to determine its mass.
  • Learn how to convert pressure measurements from kPa to Newtons using area.
  • Explore the concept of buoyancy and its effects on submerged objects.
  • Study the impact of atmospheric pressure variations on fluid dynamics in extraterrestrial environments.
USEFUL FOR

Students in physics or engineering courses, particularly those focusing on fluid mechanics and gravitational effects in extraterrestrial environments, as well as professionals involved in aerospace engineering and planetary science.

MissKaylaPaige
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Homework Statement



I have a cylindrical water tank of mars, where acc. due to gravity= 3.71

The pressure at the surface of the water is 110 kPa, and the depth of the water is 14.5 m

The pressure of the air outside the tank is 91 kPa

Find the net downward force on the tanks flat bottom of area 2.4 m^2, exerted by the water and air inside the tank and the air outside the tank



Homework Equations



p= F/A



The Attempt at a Solution



alright, I found the downward force to be 3.93x10^5 N
upward force to be 2.18x10^5

I thought the sum of these forces would give me the net forces but I must've been mistaken because the answer was wrong.
 
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You've got the water pushing down, so you need to find the volume of water and figure out the mass(I guess you assume 1 mL= 1g) and mg(but not the normal g!)will be the downward force from the water alone

Also the air is pushing down on it, as is the outside air, both can be found using your equation

Don't forget if you use kPa and m^2 you get your answer in kN
 
I know the force from the outside air is 2.18x10^5 N, and the force of the water inside is 3.93x10^5 N, so if I find the force of the air inside the tank as well, the sum of them will give me what I'm looking for?
 

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