# Ideal fluid hitting a wall

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Hi,

Im seeking for help in the following problem.

A flat vertical board is travelling in water which is to be considered as ideal. One of its ends is in water, the other one is outside the water. Its velocity is v with respect to its normal. What is the velocity of the water stream directed up the board?

I made the following attempts to solve the problem...

Go to the frame of reference of the board. In this reference frame water is simultaneously approaching the board from one direction with velocity v.
This water must leave that area somehow so it has to move up or down with some other velocity. Let this velocity be: u.
This is only possible if a steady layer of water is created in front of the board.
Let the thickness of it be: d.
Now if I write the continuity equation for that layer i.e. the total amount of fluid entering=leaving then I get an equation between d, u and v.
At this point I stuck. I dont know what else should I use. Conservation of momentum tells me that the velocity downwards= the velocity upwards=u. But It doesent contribute to the solution.
Then Im not sure if I could/should use energy conservation or not. Maybe Bernoullis equation?
I know that the solution is u=v regardless of any dimensions.
If someone could help me with that I would really appreciate it :D

Then the b part is: what is the velocity u if the board is making an angle with the horizontal, but I think I could solve that as myself if I have the solution for part a.

Thanks,

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haruspex
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not sure if I could/should use energy conservation or not.
In the reference frame of the board, I would say so. And I suspect it will lead to the given answer.

In the reference frame of the board, I would say so. And I suspect it will lead to the given answer.
Yes, however in th part b the answer is u=v/cosα where α is the angle with the vertical, u is the velocity of the water upwards. So there the velocity changes hence applying the energy concervation isnt as straightforward.

haruspex
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Remember that the given velocity v is in the direction normal to the board. It is not clear from your post whether this also applies in part b, but I do not see how else it can lead to the given answer.

Yes the velocity vector is perpendicular to the board but I dont see how could the velocity change if I consider energy conservation

haruspex
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Yes the velocity vector is perpendicular to the board but I dont see how could the velocity change if I consider energy conservation
As before, use the reference frame of the board to find the new velocity of the water, but remember to convert back to the "lab" frame after.
It might help to start by considering the extreme case.

• Istvan01
As before, use the reference frame of the board to find the new velocity of the water, but remember to convert back to the "lab" frame after.
It might help to start by considering the extreme case.
Is the velocity unchanged due to energy conservation? If so I cant figure out the solution. I also dont know why would it change. Maybe I mess something up with the geometry.

haruspex
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Is the velocity unchanged due to energy conservation?
For part a you used Bernoulli's equation, which is indeed energy conservation.
Do the same for part b and post your working.

• Istvan01
For part a you used Bernoulli's equation, which is indeed energy conservation.
Do the same for part b and post your working.
Energy concervation tells me that the velocity is constant since there is no pressure drop or change in height. So after collision water will move upwards with velocity v. I still don understand why its not the solution.
I was trying to take into account that new layers of water hit the board each time but that still did not change the result. (I considered a time Δt while a width of v*Δt/cosα water surface hit the board. I wrote down an equation if either part of that layer might get higher than u*Δt=v*Δt but I figured that thats not the case)

I also dont really understand what you meant by converting back to lab frame. Isnt the rising velocity is the same is both reference frame?

Thank you for help me out!

haruspex
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I also dont really understand what you meant by converting back to lab frame.
Hmm,. what I had in mind doesn't work, but try this.

With the board at some angle, but still moving normal to its own plane, it's a bit hard to figure out what is going on in its own reference frame. The individual molecules of water are still approaching the board at speed v, but the body of water, viewed as a whole, is moving "sideways" down the board.

It occurs to me that the water is unaffected by the sideways aspect of the relative movement. So answer this: for a given molecule of water, how fast is the board, as a surface, approaching it?

• Istvan01
So answer this: for a given molecule of water, how fast is the board, as a surface, approaching it?
If the molecule is stationary then with velocity v. If it has a "sideways" velocity then the square root of v2 and v2 where v is the sideways velocity. But Im not sure what you mean by this.

haruspex
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If the molecule is stationary then with velocity v. If it has a "sideways" velocity then the square root of v2 and v2 where v is the sideways velocity. But Im not sure what you mean by this.
That's because I expressed it poorly. Let me try again.

Consider the plane of the board as disembodied, so not a solud object, just an infinite plane. Movement within its plane is meaningless.
Consider the surface of the water in the same way. How fast does the board plane advance along the water plane?

• Istvan01
That's because I expressed it poorly. Let me try again.

Consider the plane of the board as disembodied, so not a solud object, just an infinite plane. Movement within its plane is meaningless.
Consider the surface of the water in the same way. How fast does the board plane advance along the water plane?
If I understand correctly than with velocity v. I still not undertand what you mean (not because you expressed it badly, I just dont see the connection). If you could explain in more details how it might lead to the solution I should fully understand it. Sorry if I dont see something obvious.

haruspex
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If I understand correctly than with velocity v. I still not undertand what you mean (not because you expressed it badly, I just dont see the connection). If you could explain in more details how it might lead to the solution I should fully understand it. Sorry if I dont see something obvious.
Consider an insect on the surface of the water. The board is moving up out of the water at speed v normal to its surface. But the insect does not see that. It sees where the board is emerging from the water, and that point is getting closer. How fast is it approaching the insect?

• Istvan01
Consider an insect on the surface of the water. The board is moving up out of the water at speed v normal to its surface. But the insect does not see that. It sees where the board is emerging from the water, and that point is getting closer. How fast is it approaching the insect?
Its v/cosα if Im not mistaken.

haruspex
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Its v/cosα if Im not mistaken.
Right!
So here's the thing... the effect on the water is the same whether the board moves normal to itself at speed v or horizontally at speed v sec(α). The question is, which of these views is appropriate for determining what the water will do?

• Istvan01
Right!
So here's the thing... the effect on the water is the same whether the board moves normal to itself at speed v or horizontally at speed v sec(α). The question is, which of these views is appropriate for determining what the water will do?
I guess the v*sec(α) and then due to energy conservation I get the result. I think I got it now though its not easy to understand.
Thanks for the help!

haruspex