- #1

- 1

- 0

## Homework Statement

A reservoir filled with water to a height h has a hole with area A on height h1 with h1 < h.

What is the force on the reservoir? I neglected the fact, that the height of the reservoir itself changes.

## The Attempt at a Solution

I started with Bernoulli at the top and at the hole with p_a being the atmospheric pressure:

[tex]p_a + \rho g h =c [/tex] ...

[tex] p_a + \rho g h_1 +\frac{1}{2} \rho v_1 ^2 =c[/tex].

With this I get:

[tex]v_1 ^2 = 2g(h-h_1)[/tex]

Now I want the Force, with I being the impuls:

[tex]F=\frac{dI}{dt}=\frac{dV}{dt}\rho v_1=Av_1\rho v_1=A\rho v_1 ^2[/tex]

and therefore I get:

[tex] F=2gA\rho(h-h_1)[/tex]

I now would want to know, if this solution is correct and furthermore, why I can't just assume that the force is due to the lack of pressure on one side and therefore just the pressure on hight h1 times area A on the opposite side, but rather get two times this force with my solution.

Thanks in advance,

gaugie

edit: the post should actually be in the "introductory physics" forum, but was accidentally postet here. I don't know, if it is appropriate here, otherwise please move the post.

Last edited: