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I don't follow the part of the solution, which I have underlined in green. Which equation would I get this from (if any)?

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usainlightning
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    Isn't just that if you assume that your fluid (or anything in the cylinder) is incompressible, meaning the density $\rho$ is constant, then $F=ma=\rho V a$, si if $a$ is fixed, $F$ increase porpotionnally to $V$ ? – yago Apr 21 '14 at 23:09
  • The equation comes from the fact that F = pA. The fluid must be incompressible in order to consider only the inlet and outlet pressures otherwise you would need to know the variation of pressure along the tube. @YannHamdaoui already stated the relation between force and volume but I don't see why acceleration should be constant. – Shibli Jun 04 '14 at 05:43
  • @YannHamdaoui Why have you let the acceleration be constant? – usainlightning Jul 09 '14 at 16:42
  • Note that $\rho$ need not be constant, in general. It is if you assume the fluid is incompressible. However, I assume this is a "pipe-flow" problem from some and the fluid is water, which explains the approximation. – user_of_math Jul 09 '14 at 18:01
  • @Yann Hamdaoui: Does incompressibility always imply that the density is constant? If I remember correctly, incompressibility means that the material derivative of the density is constant. Constant pressure means rather homogenity of the fluid, doesn't it? – Quickbeam2k1 Jul 11 '14 at 05:38
  • @Quickbeam2k1 I did a few googling about the vocabulary and it is a bit confusing. Moreover I speak about an incompressible fluid, which is often used in French and means that density is constant (it is closer to [homogeneous] incompressible material in English) and not incompressible flow (which is what you speak about). See for example http://en.wikipedia.org/wiki/Incompressible_flow#Difference_between_incompressible_flow_and_material. I don't really know which terminology is standard... – yago Jul 12 '14 at 08:34
  • @usainlightning actually what I wanted to say is between my previous statement and Jeb's answer. As he said, the acceleration need not te be constant to say that $F$ increases linearly when $V$ increases. But he is not totally right, because $a$ can't be anything in my opinion : the good assumption is to say that $a$ is independant of the volume. Then you can indeed see that $\frac{\partial F}{\partial V} = \text{ constant }$. – yago Jul 12 '14 at 08:40

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In part $d$ you show that $F \sim p dA$, where $p$ is some pressure and $dA$ is some small surface area. We know that $ F = m a$, but it's hard to talk about mass in this setting so let's multiply by $V/V$, Volume over Volume to get things in terms of density. Which is much easier to talk about for fluids. Using the fact that $\rho = m /V$. We obtain

$$ F = \rho V a$$

We assume the density of the liquid is constant. Why do we do this? The technical answer is because molecules don't shrink, if we have $m$ grams of water(for example) in a box of volume $V$, the water will equally distribute all the molecules due to dipole-dipole bonding or van der Waals-forces for different molecules. Thus we say the density is constant for nice fluids. Most things you work with are nice, so we can assume the above, hence the density is constant. Since density is now constant, we now see

$$ F \sim V $$

It doesn't matter what $a$ is (it doesn't have to be constant), we see any change in $V$ linearly changes $F$.

Jeb
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