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This question already has an answer here:

The base area and water depth is supposed to be the same for all three containers.

3 containers

The pressure in a static fluid arises from the weight of the fluid and is given by the expression $P=\rho gh$.

The expression tells us that the fluid pressure at the bottom of all three containers is the same because the depth is the same.

If the pressure is the same then the normal force exerted by the bottom of the container on the fluid is the same since $F = P\times A$. But the normal force is also equal to the weight of the fluid which is different for all three containers. How is this possible? While I was trying to find an answer to this question on the internet, I found a couple of forums that said the reaction forces from the sides of the containers are responsible for this, but I can't see exactly how.

If anyone could show the math of how that would work out, that would be helpful.

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marked as duplicate by sammy gerbil, Yashas, John Rennie, Jon Custer, David Hammen Mar 8 '17 at 23:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See, in second example, there is more amount of water, but some force gets lessened due to a vertical component of force on wall. In third case, there is no net vertical force on walls. in first case, there is an upward force on the wall, which compensates for the less amount of water. You can calculate this explicitly. $\endgroup$ – samjoe Mar 6 '17 at 7:17
  • $\begingroup$ I was hoping for a explicit calculation. I tried to do it, but it didn't add up $\endgroup$ – E7_82_8E Mar 6 '17 at 17:24
  • $\begingroup$ @samjoe how would you go about calculating the magnitude of the downward component of the force with which the walls push against the fluid? $\endgroup$ – E7_82_8E Mar 6 '17 at 22:00
  • $\begingroup$ @sammygerbil I specifically ask for a quantitative explanation(ie. the math involved) in the post, so I guess it's sort of different? $\endgroup$ – E7_82_8E Mar 7 '17 at 23:51
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We know that pressure at height $h$ below a liquid surface is $\rho g h$. This pressure is exerted in every direction. We use this fact to calculate the force that liquid exerts on wall.

Consider an infinitesimal piece $dz$ at depth $z$ below surface of a liquid with volume density $\rho$. The container wall makes angle $\theta$ with base. The height of water surface is $h$ and radius of circular area at top is $r_o$. Let the radius of circular slice, at depth $z$ be $r$. All the rest quantities are marked in the figure.

From trigonometry, we get, $r = r_o+ z\cot(\theta)$

enter image description here

Now lets make a free body diagram of the slice of liquid. The forces acting on the slice are:

  • Downward force $mg=V(z)\rho g$ due to liquid above the slice, where $V(z)$ is volume of liquid above the slice.

  • Force $F$ due to wall: Now, consider the interface of liquid and container. The liquid exerts a pressure $\rho g h$ in all directions at depth $z$. This leads to a force on the wall due to liquid (normal to surface). But from Newton's third law, we have wall exerting an equal and opposite force on the liquid.
    This force is: $dF=P(z)\ dA$, where $dA$ is the curved area of slice in contact with container wall. Thus $dA = 2\pi r\csc(\theta) dz$

The key here is in the fact that the surface itself is slanted. Thus the second force has components in both horizontal and vertical directions. And as you can see, the horizontal components cancel, whereas vertical components add up. This compensates for the lesser amount of water.

NOTE: Sorry I didn't find time to do the calculations. I'll add a calculation if i get time, basically we have to integrate from $z=0$ to $h$

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  • $\begingroup$ Thank you, also found this article that goes through the whole calculation $\endgroup$ – E7_82_8E Mar 8 '17 at 0:09
  • $\begingroup$ @E7_82_8E Thanks a lot for the link, their integral looks much cleaner as they took a cone. I was taking a frustum and the integral looked a bit complicated, so I gave up :) $\endgroup$ – samjoe Mar 8 '17 at 6:20

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