Amount of Force that Water Exerts on a Dam

[kawaiixsarang]

Homework Statement

Calculate the total force the Oahe Dam, placed on a structure by the reservoir behind it (in pounds and Newtons).

Oahe Dam: height=75meters, width=2835meters

Homework Equations

Pressure = Force / Area
Force = Pressure x Area
Hydrostatic Pressure: Atmospheric Pressure + (Density x Gravity x Height)

Density of Water= 1000kg/m^3
Gravity= 9.81 m/s^2

The Attempt at a Solution

- I can use Calculus on this, I know that... but I'm not in calculus yet, so I'm attempting to solve algebraically at least.

- I understand that pressure increases as the depth increases... so pressure isn't the same in all levels (right?)

What I tried doing is calculating pressure using (density x gravity x height) in every height...
For example,
(1000*9.8*1m) + (1000*9.8*2m) + (1000*9.8*3m)... so forth until I reach 75m.
Then I found force by using F=P*A...
the area would be (75m * 2835m)
& I get a total of 5.94 x 10^12 N as the force
BUTTTTT. I think the force is too high... =/

I should also consider atmospheric pressure, but I don't know where to include that (if I even should!)

PLEASE HELP MEEEE!

 

[Haths]

That is certianly one way to go about doing the problem. However...

Because 75m isn't very big in magnitude, and it's close to the Earths surface, then the force on a 'cube of water' due to gravity. Can be considered;

F=mg

Because our relationship for g is linear. Then we can suitably approximate it over the entire height of the dam.

So half of 75m is 37.5m

The presure at this depth is Rgh. So we find that the pressure is 3.68x10⁵ and is the average value for the pressure over the entire dams area. Hence with the area in total being 2.13x10⁵m² the force is approximatly 7.82x10¹⁰ note I left out the atmospheric pressure. If we include that 101.3kPa is 1 Atmos.

Then the pressure at average depth is Rgh + Atmos. or 4.69x10⁵, however remember that this same pressure is on the other side of the dam pushing back as well against the weight of water. Hence we don't really need to include it in the calculation.

The trouble with your calculation is that you are summing up sections of water that you have already calculated.

i.e. (1000*9.8*1m) + (1000*9.8*2m) + (1000*9.8*3m)...

The second term includes the first term, and the third includes the second and first. Which makes it inaccurate.

If you want to do it that way, it's an expansion of the Trapezium Rule and you should be doing each 'slice' seperatly. However I am quite confident that doing that is overcomplicating the problem. But I could be wrong.

Hope that helps,
Haths

 

[mr.survive]

U used d formula P = density x g x h very right n haths has correctly pointed out ur mistake of mixing up sections of previous layers in next terms..now d solution of ur problem s simple.. D approach s to find Pressure acting at the bottom layer n then using F=P.A.. - Now u r adding each LAYERS' pressure n that's only right when u have only one variable or otherwise u need calculus to sum-up d values(known as Integration).. Luckily u can Ignore change in Gravity as h <<<< R...so u get only one variable n that s height..so now simply do P=density x g x h(=75m) n then add to it d atmospheric pressure n then do F=P.A..well done..u did it :-)

 

[tiny-tim]

Welcome to PF!

Hi kawaiixsarang! Welcome to PF!

(1000*9.8*1m) + (1000*9.8*2m) + (1000*9.8*3m)... so forth until I reach 75m.

Yes, that's basically the correct method …

and we can tidy it up as (1000*9.8)*(1 + 2 + … + 74) …

now … what is the formula for sum in the second bracket?

I should also consider atmospheric pressure, but I don't know where to include that (if I even should!)

Atmospheric pressure will be the same on both sides of the dam (well, except for 75 feet of air, whose weight is negligible), so you can ignore it.

 

[kawaiixsarang]

and we can tidy it up as (1000*9.8)*(1 + 2 + … + 74) …

So I used the formula: n(n+1)/2 ; where n is the height of the dam.
When I sum up all the numbers on the right bracket I get a total of 2850.
And I multiplied that by 1000*9.8 and I get 2.793 x 10^7...
& I found F=PxA but it doesn't workkkk because I still get the same answer from the beginning 5.94 x 10^12 :/

What am I doing wrong??

 

[kawaiixsarang]

U used d formula P = density x g x h very right n haths has correctly pointed out ur mistake of mixing up sections of previous layers in next terms..now d solution of ur problem s simple.. D approach s to find Pressure acting at the bottom layer n then using F=P.A.. - Now u r adding each LAYERS' pressure n that's only right when u have only one variable or otherwise u need calculus to sum-up d values(known as Integration).. Luckily u can Ignore change in Gravity as h <<<< R...so u get only one variable n that s height..so now simply do P=density x g x h(=75m) n then add to it d atmospheric pressure n then do F=P.A..well done..u did it :-)

thank you.

i tried doing your suggestion, so I got:

1000 * 9.8 * 75 = 7.35 x 10^5 as pressure
Then I added it on to 101,300 N/m^3 (atmospheric pressure)
and I did F=PxA like you said...
which gave me 1.778 x 10^11

but according to Haths... it should be 7.8 x 10^10

What am I doing wrong? =/

 

[davieddy]

What I tried doing is calculating pressure using (density x gravity x height) in every height...
For example,
(1000*9.8*1m) + (1000*9.8*2m) + (1000*9.8*3m)... so forth until I reach 75m.
Then I found force by using F=P*A...
the area would be (75m * 2835m)
& I get a total of 5.94 x 10^12 N as the force
BUTTTTT. I think the force is too high... =/

Yes, that would give an answer ~75 times too high.

What you have done is add up the forces exerted on each
square metre in a vertical strip 1m wide.
To get the average pressure (force per square metre) you need to divide
by 75.
Or simply observe that there are 2835 of these metre wide vertical
strips in the wall, so multiply your sum by 2835 to get the total force
on the dam.

David

 

[davieddy]

thank you.

i tried doing your suggestion, so I got:

1000 * 9.8 * 75 = 7.35 x 10^5 as pressure
Then I added it on to 101,300 N/m^3 (atmospheric pressure)
and I did F=PxA like you said...
which gave me 1.778 x 10^11

but according to Haths... it should be 7.8 x 10^10

What am I doing wrong? =/

As tiny Tim said, the contribution that atmospheric pressure makes
to the water pressure is canceled out by the atmospheric pressure
on the outside of the wall.


1000*9.8*75 gives the pressure at the bottom.
What you want is the average pressure, which is half of this.

Now multiply by area and you get 7.8 x 10^10N

 

[kawaiixsarang]

Thank you very much for your help! =)

I'm assuming all of you tried giving me different ways to solve this problem??

 

[tiny-tim]

Hi kawaiixsarang! Thanks for the PM!

Easier to say (1 + 2 + … + 74) = 1/2 74*75 …

remember the general formula (1 + … + n) = 1/2 n(n+1)?

So the total is very nearly 1/2 75² …

and if you subdivided it even more, it would "tend to" exactly 1/2 75²

However, I've just noticed another more intuitive way of doing it (without caclulus) …

turn the dam on it side!

then the force on each part of the dam can be represented by a weight of water of height h …

that gives you a triangle of water of height h and width h, and the total force is the total weight fo the triangle, which is … ? ​

 

[kawaiixsarang]

Thank youuu! =)

I understand now. When I solve the entire problem I get out: 7.813e10 =)