Solving Force on Water Tank Bottom on Mars

[12boone]

Homework Statement

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 145 kPa , and the depth of the water will be 14.2m . The pressure of the air in the building outside the tank will be 92.0 kPA. Find the net downward force on the tank's flat bottom, of area 1.80 m², exerted by the water and air inside the tank and the air outside the tank.

Homework Equations

P=F/A and P=pgh

The Attempt at a Solution

I plugged in P=1000(3.71)(14.2m)
Then I used that and converted it to kPA to plug in all the pressures combined to find F in the second equation. (52.7+92.0+145)=F/1.80m^2. I got 522. The answer was wrong. Help?

 

[alphysicist]

Hi 12boone,

The Attempt at a Solution

I plugged in P=1000(3.71)(14.2m)
Then I used that and converted it to kPA to plug in all the pressures combined to find F in the second equation. (52.7+92.0+145)=F/1.80m^2. I got 522. The answer was wrong. Help?

You did not mention what the units of your answer of 522 were, so there might be a problem there.

But the more important thing is that you are not taking into account that the force from the inside pressure pushes down, and the force from the outside pressure pushes up. Do you see what needs to be changed in your force equation? What do you get?

 

[thomate1]

I would like to commend you for your attempt at solving this problem. However, there are a few things that need to be addressed in your solution.

Firstly, the equation P=pgh is used to calculate the pressure of a fluid at a certain depth, where p is the density of the fluid, g is the acceleration due to gravity, and h is the depth. This equation is not applicable in this scenario as we are given the pressures at the surface and not the depth.

Secondly, when calculating the net downward force on the tank's bottom, we need to take into account the pressure of the air both inside and outside the tank. So the correct equation would be P=F/A, where P is the net pressure (difference between the internal and external pressures), F is the net downward force, and A is the area of the tank's bottom.

To solve for F, we can rearrange the equation to F=P*A. Plugging in the given values, we get F=(145-92)*(1.80m^2)=99.9 kN.

Therefore, the net downward force on the tank's bottom is 99.9 kN. This force is exerted by the combined weight of the water and air inside the tank, as well as the air outside the tank. It is important to consider all these factors when designing a pressurized water tank for a future colony on Mars.