# Fluid Mechanics Problem

1. Jan 19, 2007

### kingwinner

I am having great trouble understanding this problem and the way to solve it, could anyone explain in detail? Any help is appreciated!

1) An aquarium tank is 80cm long, 30cm wide, and 35cm deep and is filled to the top. What is the force of the water on the bottom of the tank, with dimensions 80cmx30cm, AND the force on the front window with dimensions 80cmx35cm, respectively?

For the first part, is the question actually asking for the normal force on the tank's bottom exerted by the water, or is it asking for the total weight of the water? Are these 2 values equal? If so, why?
By the way, I tried to use Force = pressure x area
=(p_o + rho*g*d) x area
=(101300+1000*9.8*0.35) x 0.8m x 0.3m
=25135.2N <----but this is not what I got when I calculate the weight of the water, how come their values are different? I don't see why...

Also, can someone help me on the second part? I know that this needs integration and I know that for a liquid, the pressure of WATER at a depth d is p=p_o+(rho)(g)(d), but in this case, what is the value for p_o? p_o=p_atm=101300Pa because there is air above the water, right?

(Note: This is not a homework question anymore. I did it for practice and to understand the major concepts of fluid mechanics!)

2. Jan 19, 2007

### cesiumfrog

in what way does atmospheric pressure contribute to the force on the fishtank?

3. Jan 20, 2007

### andrevdh

For the force of the water on the bottom you should not include the atmospheric pressure on top of the water (it is balanced by the atmospheric pressure on the other side of the glass anyway). Same for the second question - do not include $$p_o$$.

4. Jan 20, 2007

### JJ420

for the first part of your problem draw a freebody diagram.....hint-you should have 2 forces on it(Wieght and Normal)

for the second part of your problem use
F= (SpecificWieght)*(DepthToCentroid)*(Area)

then you will have to find where the force is acting on the wall
Depth= (DepthtoCentroid)+ I/(DepthtoCentroid * Area)
I Being the moment of inertia

5. Jan 21, 2007

### kingwinner

Hi,

Can you please explain how the atmospheric pressure is balanced on the top?

6. Jan 22, 2007

### andrevdh

If there were no water in the tank the atmospheric pressure would act on both sides of the bottom of the tank (The atmosperic pressure acts in all directions at each point a liquid. This is due to the fact that the atoms/molecules will collide with a test surface held in any orientation at such a point). The atmospheric pressure therefore puts no additional strain on the bottom. This is also true when there is water in the tank, but the question seems to ask specifically for the force caused only by the water in it.

7. Jan 23, 2007

### kingwinner

Hi,

But the pressure of the water at depth d is p=p_atm+(rho)(g)(d), am I right? This is the formula I see everywhere in my textbook. It never says p=(rho)(g)(d)

8. Jan 23, 2007

### andrevdh

Yes. That is correct. It means that the pressure at a depth d in a liquid has two contibuting factors:
1. The pressure on top of its surface (p_atm)
2. The pressure caused by the weight of the liquid from the surface up to that depth (rho x g x d).

If one would like to calculate the pressure caused by the water only (which in this case is important if one wants to know if the glass will break or not) one would need to consider only the (rho x g x d) factor because the same pressure that is acting on top of the surface of the water is also acting onthe outside of the bottom, therefore causing no additional strain on the bottom.

9. Jan 23, 2007

### kingwinner

So the atmospheric pressure on the sides are all balanced. How about the top and bottom? At the bottom of the tank, will it have atmospheric pressure? (I would assume it is resting on a flat surface, will there be air under it?)

10. Jan 24, 2007

### andrevdh

To "get rid" of atmospheric pressure is not easy - one need a "perfect fit" between the two surfaces - like with a sucker. The two surfaces effectively stick together when atmospheric pressure is eliminated between them. This is a result of an unbalanced pressure now acting on the outer surfaces of the two surfaces and fusing them together. One sometimes get this result if two glass panes are put on top of each other with a film of water in between. Normally atmospheric pressure will leak inbetween and into any space.

The problem does not mention a closed tank so the atmospheric pressure is acting only on the water surface on top, but this pressure is transferred to the inside of the bottom surface by the water.

11. Jan 25, 2007

### kingwinner

Then for the first part of the question:
"What is the force of the water on the BOTTOM of the tank, with dimensions 80cmx30cm"
How come I need to ignore atmospheric pressure for this part too? If I kept the atmospheric pressure p_o=101300 Pa, then I got the wrong answer...why?

12. Jan 26, 2007

### andrevdh

Well since you calculation is correct (25135 N) when you include the air pressure on top of the my assumption is that the question is only interested in the additional pressure caused by the weight of the water in which case one would not include the air pressure on top of the surface. The weight of the water is given by the last term

$$\rho g h$$

since the weight can be calculated with

$$W = mg = \rho V g = (\rho g h) A$$

13. Jan 29, 2007

### kingwinner

So the primary reason to neglect atmospheric pressure is because the atmospheric pressure inside is balanced by that of the outside.

Now let's say we have another question:
"What is the force of the water on the BOTTOM of the tank, provided that there is no air below the tank?"

Then, there can't be any atmospheric below the tank to balance...will the following calculation be correct now in this case?
Force = pressure x area
=(p_o + rho*g*d) x area
=(101300+1000*9.8*0.35) x 0.8m x 0.3m

14. Jan 29, 2007

### andrevdh

Yes.

When solving problems it is always a good idea to get clarity on what exactly is required to solve. Sometimes it is not clear and one need to use a bit of inferrent logic.

In this case my guess was that the problem was requiring one so solve for the additional force caused by the water in the tank since this could cause it to break if the glass cannot take the strain.