enter image description here

I've read 4 different books and yet nobody explains why forces $F_1$ ($=p_1A_1$) and $F_2$ ($=p_2A_2$) point in different directions. Shouldn't $F_2$ point in the same direction as $v_2$?

Since we're assuming that parts of fluid between $a$ and $b$ have the same kinetic and potential energies (same holds for $c$ and $d$), why do all proofs state that the change in work: $W_2 - W_1$ is equal to the change in energy $E_2 - E_1$? Work is equal to the change in kinetic energy, so $W_2 = W_1 = 0$ (because we assumed that fluid between each pair of points has the same energy).

Then there's the problem of signs, how do we determine which sign to choose and how do potential energies come into the equation?


2 Answers 2


It's the definition of pressure. The pressure force is the force the stuff (fluid) external to the fluid in blue is exerting on the fluid in blue. It's like tension in a string, except with the sign changed.

In a string under tension the string outside the length you are interested in is pulling at both ends; in a rod or fluid under compression the outside is pushing at both ends.

  • $\begingroup$ What about the pressure from the left side for the surface $A_2$?The entire fluid is the same. $\endgroup$ Commented Aug 25, 2019 at 13:20
  • $\begingroup$ @user3711671: That force is the force exerted by the blue highlighted fluid on the fluid outside the highlighted regions. We dont care about that. We only want the force on the bit of fluid (the blue regions between the red arrows) whose motion we are examining. $\endgroup$
    – mike stone
    Commented Aug 25, 2019 at 13:26

@mike stone Does a great job at addressing your first point. To address you second point, it is true that the net work changes the kinetic energy, i.e. $W_{net}=\Delta K$. However, we are interested just in the work done by the external forces acting on the fluid. This means that $W_\text {ext}=\Delta E$. This is the work done by your forces on either end of the fluid segment.

Your third question is somewhat unclear to me, and this question runs dangerously close to being too broad by asking multiple questions, so I'll just leave it at this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.