Force by streaming out water

  • Thread starter gaugie
  • Start date
  • #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:

Answers and Replies

  • #2
1,384
0
Yes your solution is correct.
The force due to pressure difference is ρgA(h-h1).
The additional force comes due to the impulse of water jet streaming out of the reservoir.
 

Related Threads on Force by streaming out water

Replies
2
Views
3K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
2K
Replies
4
Views
4K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
5K
Replies
1
Views
3K
Replies
3
Views
1K
R
Replies
1
Views
7K
Replies
3
Views
2K
Top