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Pressure: A Water Tank on Mars

  • Thread starter mantillab
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  • #1
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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 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.

Homework Equations



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

Downward forces:
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

Any suggestions?
 

Answers and Replies

  • #2
Dick
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As I read the question, p_0 is just 140kPa. The 90kPa is the pressure in the air outside the tank, and acts UPWARD on the tank bottom.
 
  • #3
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Thanks for your reply. If the p_0 is just 140kPa, then the downward pressure would be 4.49 x 10^5 N? (Incorrect answer according to MP).

I was assuming that the tank is on the floor/ground of the building, since the question states: "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." But maybe this is a false assumption?

According to the hints from the problem, these are the steps to solve:
1. Find the net force. (Equal to zero.)
2. Convert kPa to Pa. (I believe I did this.)
3. Use density of water to determine mass. (Solved for mass.)
4. Find the force exerted on the tank's bottom by the air outside the tank. (This is the snag.)
5. Find the force exerted on the tank's bottom by the water.
 
Last edited:
  • #4
HallsofIvy
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Thanks for your reply. If the p_0 is just 140kPa, then the downward pressure would be 4.49 x 10^5 N? (Incorrect answer according to MP).[./quote]
You mean force, not pressure, right?

I was assuming that the tank is on the floor/ground of the building, since the question states: "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." But maybe this is a false assumption?
Reading what you quote would make me come to the opposite conclusion. If the air pressure outside the tank is exerting any pressure on the bottom of the tank, then the tank can't be sitting on the floor.

According to the hints from the problem, these are the steps to solve:
1. Find the net force. (Equal to zero.)
The problem asks you to find the net force! Surely your first step is not to assume the answer is 0?

2. Convert kPa to Pa. (I believe I did this.)
3. Use density of water to determine mass. (Solved for mass.)
4. Find the force exerted on the tank's bottom by the air outside the tank. (This is the snag.)
Surely not- just multiply the pressure by the area. And remember that it is exerted upward.

5. Find the force exerted on the tank's bottom by the water.
Which is just the weight of the water. I don't see anywhere you say "find the force exerted on the tank's bottom by the pressurized air in the tank".
And, of course, there should be one more step: find the net force (which is NOT 0).
 
  • #5
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I guess I assumed that the net force was zero because of what my text has told me about hydrostatic pressure (static equilibrium with Fnet = 0).

If I attribute the pressure outside the tank to the upward forces...

Upward forces:
pA = (9 x 10^4 Pa)(2.35m^2) = 2.12 x 10^5 N

Downward forces:
mg = (3.24 x 10^4 kg)(3.71 m/s^2) = 1.20 x 10^5 N

(p_0)(A) = (1.4 x 10^5 Pa)(2.35m^2) = 3.29 x 10^5 N

Total: 1.20 x 10^5 N + 3.29 x 10^5 N = 4.69 x 10^5 N

Net: 4.69 x 10^5 N - 2.12 x 10^5 N = 2.57 x 10^5 N (incorrect)

It seems like I'm missing another downward force?
 
  • #6
alphysicist
Homework Helper
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Hi mantillab,

I guess I assumed that the net force was zero because of what my text has told me about hydrostatic pressure (static equilibrium with Fnet = 0).

If I attribute the pressure outside the tank to the upward forces...

Upward forces:
pA = (9 x 10^4 Pa)(2.35m^2) = 2.12 x 10^5 N

Downward forces:
mg = (3.24 x 10^4 kg)(3.71 m/s^2) = 1.20 x 10^5 N

(p_0)(A) = (1.4 x 10^5 Pa)(2.35m^2) = 3.29 x 10^5 N

Total: 1.20 x 10^5 N + 3.29 x 10^5 N = 4.69 x 10^5 N
There's a little arithmetic error in the above statement.
 
  • #7
Dick
Science Advisor
Homework Helper
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1.20+3.29 isn't equal to 4.69.
 
  • #8
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Ah, that's what I get for not wearing my glasses. Thanks everyone!
 

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