A water tank on mars helpp pressure related

[biggy135]

Description: 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 150 kPa, and the depth of the water will be 14.2 meters. The pressure of the air in the building outside the tank will be 88.0 kPa.

Question:

Find the net downward force on the tank's flat bottom, of area 2.30 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.

So I'm new here and I cannot firgure this problem out because my professor did not even talk on this. But i am sure someone here can probably solve this easily. Thanks so much

 

[biggy135]

Any help anyone? I have tried using the formula for pressure at different heights in a liquid: P = pgh

But i really can't firgure this out? Any help at all??

 

[AngeloG]

*edit for Moderator*

Let's start out with the given information and equations.

Given
g = 3.71 m/s

Pressure Outside = 88.0 kPa = 88,000 Pascals = 88,000 N / m^2
Pressure Inside [on top of surface] = 150 kPa = 150,000 Pascals = 150,000 N / m^2
Pressure Inside [on the bottom] = ? kPa = ? Pascals = ? N / m^2

Density of Water: 1x10^3 Kg / m^3

-> Cylinder Dimensions
Bottom Area: 2.30 m^2
Height (y2 - y1 basically): 14.2 m

Thought
You know you want to get Newtons as a final answer. How would you do this mathematically first of all using what's given.

Link this idea with equations dealing with pressures [remember, the fluid is not moving so Bernoulli's equation doesn't matter].

Using your thoughts on pressure; what can you use [in terms of equations] to find various information.

A basic idea is Pressure = Force / Area. Switching back and forth [using algebra] we can find the Force, the Area or the Pressure.

 

[Gokul43201]

AngeloG and biggy: Welcome to PF, both of you. And please read the posting guidelines for Homework Help.

https://www.physicsforums.com/showthread.php?t=5374

We try to guide people to the answers (rather than feed it to them). This way, not only are they more likely to learn how to think, but they'll also remember the better.

 

[biggy135]

*edit for Moderator*

Let's start out with the given information and equations.

Given
g = 3.71 m/s

Pressure Outside = 88.0 kPa = 88,000 Pascals = 88,000 N / m^2
Pressure Inside [on top of surface] = 150 kPa = 150,000 Pascals = 150,000 N / m^2
Pressure Inside [on the bottom] = ? kPa = ? Pascals = ? N / m^2

Density of Water: 1x10^3 Kg / m^3

-> Cylinder Dimensions
Bottom Area: 2.30 m^2
Height (y2 - y1 basically): 14.2 m

Thought
You know you want to get Newtons as a final answer. How would you do this mathematically first of all using what's given.

Link this idea with equations dealing with pressures [remember, the fluid is not moving so Bernoulli's equation doesn't matter].

Using your thoughts on pressure; what can you use [in terms of equations] to find various information.

A basic idea is Pressure = Force / Area. Switching back and forth [using algebra] we can find the Force, the Area or the Pressure.

Well, i did what you said but i still can't get the right answer. This is so frustrating...

Since the i know the pressure of one...150,000 N/m^2 than divided by 6.60 as the area of flat service gives me the force of that force.

I calculated the pressure of the other by taking the density * gravity * height which gave me 52,682. Which i assume is not kPa but might be? I than take that and divide it by the same area since i believe this is in N/m^2. I than the two forces since the problem states for the net downward force.

But that answer does not work? Ahhh...I might hit my computer haha.

 

[biggy135]

Hint 4. Find the force exerted on the tank's bottom by the air outside the tank

Write an expression for the force exerted on the tank's bottom by the air outside the tank.
Hint 1. Pressure and force

The pressure in a gas is defined as the normal force exerted by the gas on a surface in contact with it. If the force is the same at all points of a finite plane surface with area , then the pressure is uniform and given by

than...

Write an expression for the force exerted on the tank's bottom by the water in the tank. Keep in mind that the water tank is located on Mars, so weights depend on the acceleration due to gravity on that particular planet.

So each of these wanted the answers in Newtons. Each time i tried what you said were correct it didn't say it was correct. This is under the hint section which tries to guide you while you have to get the right answer for it to help.

AHHH lol...

Ive honestly worked on this problem for 4-5 hours. I feel so stupid.

The question asks " Find the net downward force on the tank's flat bottom, of area 2.30 , exerted by the water and air inside the tank and the air outside the tank." So I think you are right about adding the three downward forces. But it doenst seem to work! If it asks for Newtons and i have it in SI units...like 150,000 N/m^2 than in N's, itd be 150. Correct? I am so confused


SOO...

THis is soooo frustrating. So to find the pressure of the air outside on the bottom of the tank should be this:

Pressure 88 kPa so 88,000 N/m^2 which is 88 Netwons?
But to find the FORCE it is Pressure = Force/Area so it would be pressure x area to solve for force.

So 88,000 N/m^2 x 6.60 is 13333.33 which is not the right answer? What am i doing wrong!

 

[Gokul43201]

Well, i did what you said but i still can't get the right answer. This is so frustrating...

Since the i know the pressure of one...150,000 N/m^2 than divided by 6.60 as the area of flat service gives me the force of that force.

1. Where does this 6.60 number come from?
2. Force = pressure times area (not pressure divided by area)

 

[biggy135]

right. I got it wrong above but if you look at the end of what i wrote...i got it right. 6.60 is the area because the area of the bottom of the tank is 2.30 m^2.

 

[Gokul43201]

6.60 is the area because the area of the bottom of the tank is 2.30 m^2.

If the area is 2.3 m^2, then how can it also be 6.6 m^2?

 

[biggy135]

but i never used it as 6.6m^2 did i? I only used 6.6meters...no squared

 

[Gokul43201]

What is 6.6m and how did you get this number from 2.3m^2?

What are the units of area?

 

[biggy135]

Ok so i got the answer on the downward force by the air on the outside air on teh bottom of the tank. I took 88000 x 2.30 = 2.02 x 10^5 N

But i still can't firgure out what the pressure would be on the bottom of the tank by the water in the tank...

I took density * gravity * height to get = 52682 N/m^2

Then to find the force of that it should be...52682 N/m^2 * 2.30 = 1.21 * 10^5 N/m^2 (121168.6 N/m^2) so it could also be 121.2 in Newtons. But that isn't right for some reason! :(

 

[Gokul43201]

Ok so i got the answer on the downward force by the air on the outside air on teh bottom of the tank. I took 88000 x 2.30 = 2.02 x 10^5 N

This is actually an upward force. The air below the tank is pushing up on the floor of the tank.

But i still can't firgure out what the pressure would be on the bottom of the tank by the water in the tank...

I took density * gravity * height to get = 52682 N/m^2

This is correct.

Then to find the force of that it should be...52682 N/m^2 * 2.30 = 1.21 * 10^5 N/m^2 (121168.6 N/m^2)

What are the units of force? It is not N/m^2. You'll see the error if you write down the units for all quantities in the equation above (include the units for the area as well)

so it could also be 121.2 in Newtons.

You need to learn about units and conversions before you can proceed. I strongly suggest you spend a few hours brushing up on the units/dimensional analysis.

1.21 * 10^5 N/m2 can not be the same thing as 121.2 N. You are comparing quantities with different units...which is meaningless.

 

[biggy135]

sorry. I was confusing myself...

So ..."Write an expression for the force exerted on the tank's bottom by the water in the tank. Keep in mind that the water tank is located on Mars, so weights depend on the acceleration due to gravity on that particular planet."

Therefore i took the pressure at the surface of the water...150,000 N/m^2 + (1000)(3.71)(14.2) = 202700 * 2.30 = 466210 N

Which was the correct answer for that part. I than take that and subtact it by 2.02 x 10^5 N. And than i got the correct answer of 264210 N (2.64 x 10^5 N). Thanks everyone!