Finding magnitude of hydrostatic force per unit of circumference

Nov 2019
21
0
London
Hello all

I am trying to find the hydrostatic force of a cylindrical tank of water. I was hoping someone could help point me in the right direction.

I have a cylindrical water tank. The dimensions are as follows:-


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I need to find the magnitude of the hydro static force per unit of circumference.

I know the answer which is 354000 i just dont know how to get to it.

I currently do the following:-

Force = Density * Gravity * Distance from Surface to centriod of tank * Area
Density = 1000
Gravity = 9.8
Distance from surface to centriod of tank = 4.25
Area = 28.274

Force = 1000 * 9.8 * 4.25 * 28.274 = 1177612 Newtons

Circumference = 2 * Pie * Radius = 18.8496

I dont really know where to from here?

I thought i could just divide the Force by the Circumference but i have been told that this is incorrect.

Its the "per unit of circumference" that's really throwing me.

Can anyone help?

Thank you.
 
Oct 2017
610
312
Glasgow
It's a technicality. The formula you need to use is the same as the walls of a cuboid tank, except instead of using the tank width, use tank circumference.
 
Jun 2016
1,247
592
England
That is a bit of a cryptic answer Benit!
I had a look at this post, but was unable to make sense of it.
I came to the conclusion that I was fundamentally misunderstanding the problem
could you give us another clue?
 
Oct 2017
610
312
Glasgow
In the other thread, I posted this:

In general, for a submerged object, the pressure as a function of height is:

\(\displaystyle p(z) = \rho g z\)

Exploiting symmetry in the horizontal plane, we integrate over height only:

\(\displaystyle F = w \int_{0}^{h} \rho g z dz\)
\(\displaystyle = \rho g w \left[\frac{z^2}{2}\right]_{0}^{h}\)
\(\displaystyle = \rho g w \frac{h^2}{2}\)

where w is the width of the tank.
If you swap \(\displaystyle w\) for circumference, you have the solution for a cylindrical tank instead.

If you want force per unit length:

\(\displaystyle \frac{F}{w} = \rho g \frac{h^2}{2}\)
 
Last edited:
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Jun 2016
1,247
592
England
Ah, so the basic equation used by the OP was wrong (no h squared)
His equation looked believable, and I just assumed it was correct.
 
Oct 2017
610
312
Glasgow
If the surface area \(\displaystyle A = h C\), where C is circumference, then

\(\displaystyle \frac{F}{C} = \rho g \frac{h^2}{2} = \rho g \frac{h A}{2}\)

and the "distance to centroid" is \(\displaystyle \frac{h}{2}\), so the formula is correct.
 
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Nov 2019
21
0
London
If the surface area \(\displaystyle A = h C\), where C is circumference, then

\(\displaystyle \frac{F}{C} = \rho g \frac{h^2}{2} = \rho g \frac{h A}{2}\)

and the "distance to centroid" is \(\displaystyle \frac{h}{2}\), so the formula is correct.

Thank you benit13