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Hydrostatic problem  Impossible integral 
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#1
Dec2612, 01:43 PM

P: 248

Just studying hidrostatic over the internet and I saw the following problem:
A UTube filled with water, initially at rest in a horizontal table, has A1=40cm^{2}, A2 = 20cm^{2}, A3=30cm^{2}, h1 = 80cm, h2 = 20cm and L = 100cm (below piture). It is pressed in A1 by a constant vertical force F of 4N. Find the position x of the surface A1 in function of time (initial position is x=0). Given: g = 10m/s^{2}, density of water is 1kg/L How can I solve this? I tried some stuff but it didn't worked. A_{1}.dx_{1} = A_{2}.dx_{2} = A_{3}.dx_{3 }(where dx is the infinitesimal variation of position for a infinitesimal variation of time). By energy conservation: F.dx_{1} = Δ(mechanic energy) Potential Energy > Now imagine that the dx_{1}.A_{1} volume of water has just raised and become the dx_{2}.A_{2} volume. It raised dx_{1}.(A_{1}+A_{2})/(2A_{2}) Ep = dx_{1}.(A_{1}+A_{2})/(2A_{2}).dm.g But dm = A1.dx_{1}.ρ Ep = dx_{1}^{2}.(A_{1}+A_{2})/(2A_{2}).A_{1}.ρ.g Kinetic Enegrgy > Ek = ∫(dm.v.dv) dm = A_{1}.v_{1}.dt/ρ (v_{1} is the instantaneum velocity for the surface A_{1}) v = A_{1}.v_{1}/A (A is the area of the surface considered) dv = A_{1}.dv_{1}/A Ek = ∫[(A_{1}^{3}v1^{2}.dt.dv_{1})/(ρ.A^{2})] But I don't know how to solve this. I don't know even if it is right. Can anyone help me? 


#2
Dec2612, 02:08 PM

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Remember, water is incompressible.
Apply Pascal's law to the two columns after application of the force. 


#3
Dec2612, 03:48 PM

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But the corners bother me. Not sure how to use h_{2}. 


#4
Dec2612, 04:34 PM

P: 248

Hydrostatic problem  Impossible integral
I considered the process to be very slow (as the force is 4N). So I think the column of water tend to move orderly. I don't know how to use the h2 too! Actually this problem was from this university: http://www.ita.br/
It was a test applied on the school. You could use computers and you had to plot the graphic position xtime. The other exercises in the same test resulted in integrals that had to be seriesaproximated, so I think you have to do some assumptions. But I don't know what! []'s Joao 


#5
Dec2612, 04:41 PM

P: 248

How can I solve this? 


#6
Dec2612, 09:25 PM

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Working in terms of energy, you don't need dx's in the PE expression. Just consider that the total advancement of the force at time t is x. Abbreviating h1 to h:
PE = x^{2}.(A_{1}+A_{2})/(2A_{2}).A_{1}.ρ.g 2*KE = (hx)A_{1}ρv^{2} + Lρv^{2}A_{1}^{2}/A_{3} + (h+xA_{1}/A_{2})ρv^{2}A_{1}^{2}/A_{2} PE+KE = Fx v^{2}{h(1+A_{1}/A_{2})+LA_{1}/A_{3}+x((A_{1}/A_{2})^{2}1)} = 2Fx/(ρA_{1})  x^{2}.g(A_{1}+A_{2})/A_{2} 


#7
Dec2712, 10:38 PM

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v=dx/dt This results in a first order ordinary differential equation for calculating x as a function of t. After taking the square root of both sides of the equation, it probably would be best to substitute x=y^{2} and solve for y. 


#8
Dec2812, 09:19 PM

P: 248

v^{2} (3.73333 + 3 x) = 2 x  30 x^{2} http://www.wolframalpha.com/input/?i...+30+x[t]^2 What this means? 


#9
Dec2812, 11:52 PM

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As you can see, plugging in x=0 gives v=0, which is true. You would get the same trying to determine a trajectory under gravity using the energy equation y'^{2} = 2gy. This stops any numerical attempt to produce a plot from there  things never move. You can validly get around it by plugging in a very small nonzero starting value.



#10
Dec2912, 12:00 AM

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Wolframalpha won't be able to solve the equation until you work it into the form that I indicated. 


#11
Dec2912, 01:04 AM

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y'^{2} = 2gy If you take the square root of both sides, you get: [tex]y'=\sqrt{2gy}[/tex] Next, substitute y = z^{2} to get [tex]2z\frac{dz}{dt}=z\sqrt{2g}[/tex] or [tex]\frac{dz}{dt}=\frac{\sqrt{2g}}{2}[/tex] This can now easily be integrated with respect to t. The same type of approach applies to the problem at hand. 


#12
Dec2912, 02:44 PM

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#13
Dec2912, 03:48 PM

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#14
Dec2912, 06:04 PM

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Anyway, for the numbers given in the problem, I think it's safe to neglect the term in red. Then to a good approximation, the solution is SHM with a small amplitude. [Edit: I'm getting an amplitude of 10/3 cm and a period of about 2.2 s] 


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