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jaumzaum
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Hydrostatic problem - Impossible integral!
Just studying hidrostatic over the internet and I saw the following problem:
A U-Tube filled with water, initially at rest in a horizontal table, has A1=40cm2, A2 = 20cm2, A3=30cm2, 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/s2, density of water is 1kg/L
http://img191.imageshack.us/img191/2825/78925268.png
How can I solve this? I tried some stuff but it didn't worked.
A1.dx1 = A2.dx2 = A3.dx3 (where dx is the infinitesimal variation of position for a infinitesimal variation of time).
By energy conservation:
F.dx1 = Δ(mechanic energy)
Potential Energy -> Now imagine that the dx1.A1 volume of water has just raised and become the dx2.A2 volume. It raised dx1.(A1+A2)/(2A2)
Ep = dx1.(A1+A2)/(2A2).dm.g
But dm = A1.dx1.ρ
Ep = dx12.(A1+A2)/(2A2).A1.ρ.g
Kinetic Enegrgy -> Ek = ∫(dm.v.dv)
dm = A1.v1.dt/ρ (v1 is the instantaneum velocity for the surface A1)
v = A1.v1/A (A is the area of the surface considered)
dv = A1.dv1/A
Ek = ∫[(A13v12.dt.dv1)/(ρ.A2)]
But I don't know how to solve this. I don't know even if it is right.
Can anyone help me?
Just studying hidrostatic over the internet and I saw the following problem:
A U-Tube filled with water, initially at rest in a horizontal table, has A1=40cm2, A2 = 20cm2, A3=30cm2, 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/s2, density of water is 1kg/L
http://img191.imageshack.us/img191/2825/78925268.png
How can I solve this? I tried some stuff but it didn't worked.
A1.dx1 = A2.dx2 = A3.dx3 (where dx is the infinitesimal variation of position for a infinitesimal variation of time).
By energy conservation:
F.dx1 = Δ(mechanic energy)
Potential Energy -> Now imagine that the dx1.A1 volume of water has just raised and become the dx2.A2 volume. It raised dx1.(A1+A2)/(2A2)
Ep = dx1.(A1+A2)/(2A2).dm.g
But dm = A1.dx1.ρ
Ep = dx12.(A1+A2)/(2A2).A1.ρ.g
Kinetic Enegrgy -> Ek = ∫(dm.v.dv)
dm = A1.v1.dt/ρ (v1 is the instantaneum velocity for the surface A1)
v = A1.v1/A (A is the area of the surface considered)
dv = A1.dv1/A
Ek = ∫[(A13v12.dt.dv1)/(ρ.A2)]
But I don't know how to solve this. I don't know even if it is right.
Can anyone help me?
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