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Suppose that we want to obtain liquid pressure in the below figure which is filled with water:

enter image description here

I don't know why I get to two different answers with two different ways:

  1. Get the cube and cylinder a system, we can write the below equation in order to calculate pressure: $$P=\frac{Mg}{A_1}=\frac{\rho (V1+V2)g}{A_1}=\rho gh_1+\rho gh_2\frac{A_2}{A_1}$$

  2. Suppose we have a column of water from bottom to top and write that equation for that: $$P=\frac{M_c g}{A_c}=\rho g(h_1+h_2)$$

Why the results are different?

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closed as off-topic by John Rennie, Jon Custer, sammy gerbil, Kyle Kanos, Bill N Jun 11 '18 at 2:54

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    $\begingroup$ You're obviously doing different things. In (1) you add the masses of a cube and a cylinder, and then divide by $A_1$, while in (2) you're dividing the mass of two cubes by $A_1$. $\endgroup$ – FGSUZ Jun 5 '18 at 9:24
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Your second formula is correct.

In the first one, you forgot to add the reaction force from the top wall of the cube, which acts vertically downward on water. The force on the top wall of the cube is due to water pressure (from the water column of height $h_2$), equal to $\rho gh_2 (A_1-A_2)=\rho gh_2A_1-\rho gV_2$. Therefore the total downward reaction force on the bottom of the cube is $(\rho gh_2A_1-\rho gV_2)+\rho g (V_1+V_2)=\rho gh_2A_1+\rho gV_1=\rho g(h_2+h_1)A_1$, which when divided by $A_1$ gives the pressure at the bottom of the cube.

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  • $\begingroup$ And don't forget to add $P_{atm}$ if the system is on Earth. $\endgroup$ – FGSUZ Jun 5 '18 at 9:26
  • $\begingroup$ @FGSUZ Absolutely right. $\endgroup$ – Deep Jun 5 '18 at 9:40
  • $\begingroup$ Isn't that internal force? if the cube and cylinder have very small mass these forces will be internal, right? so my equation become true $\endgroup$ – Panda Jun 6 '18 at 10:23
  • $\begingroup$ @Panda If you want to apply the formula P=F/A at the bottom of the cube, then you must calculate all contributions to force F on the bottom of the cube. One contribution is from weight of water (which you have considered) and the other is from any external force that is vertically acting on the water body. If you draw a free-body diagram of only the body of water (excluding container), accounting for all the vertical forces acting on it, the force balance will be clear to you. Mass of the container is irrelevant; only requirement is that the container be rigid to counter water pressure. $\endgroup$ – Deep Jun 6 '18 at 10:25
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According to your question:

Suppose that we want to obtain liquid pressure in the below figure which is filled with water:

Answer: Especially for a standing column of water as we have here, the cross-sectional area of the interconnecting part of the shapes is irrelevant. Area is relevant when calculating force ($F = PA$). Only the water density, gravitational acceleration, and total height of the water column are relevant here. Ignoring atmospheric pressure, your second solution equation is correct. The (hydrostatic) pressure at the bottom of the water column is:

$P = ρg(h_1+h_2)$

..and your first solution equation does not apply.

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