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Scenario description

Lets assume both containers have a capacity of 300 litres.

One is a vertical tube as shown in pic 1

Other one is more or less a V shaped containers as shown in pic 2

Both are at ground level

Question

No doubt that whole weight of 300 litres will act on the base of the tube (pic 1). But will the weight of the whole 300 litres act on the base of the container as shown in pic 2?

I.e if you keep both the containers on a weighing scale, for sure, it will show a weight of approx. 300 kg each.

But what I'm asking is whether the weight of whole 300 litres acts on the base of container 2 (since the container's walls are slanting, as opposed to the walls of the tube in pic 1).

If the whole weight will not act on the base of the container, can you state the reasoning and also the principle, so that I can further get to know about it?

Pic 1

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Pic 2

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            \________/
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  • $\begingroup$ If you are talking of the weight of the entire body, then it will always act perpendicularly down from the center of gravity. Otherwise a fluid (in your case a liquid) always exerts equal pressure in all directions at a given depth. This pressure varies with depth and is never affected by the shape of a container. $\endgroup$ – rahulgarg12342 Jul 15 '14 at 7:09
  • $\begingroup$ Do both containers contain 300L AND have the same base area? Or are the base areas different??? $\endgroup$ – Bryson S. Jul 15 '14 at 7:36
  • $\begingroup$ @rahulgarg12342 1) Yes, initially i had 300 litres and same base area but now after seeing all your replies i have a different doubt. the question is, " So does it mean the determinants for pressure are base area and height, and not volume of liquid??" $\endgroup$ – user52187 Jul 15 '14 at 11:05
  • $\begingroup$ @BrysonS. " So does it mean the determinants for pressure are base area and height, and not volume of liquid??" $\endgroup$ – user52187 Jul 15 '14 at 11:05
  • $\begingroup$ @user52187 The determinant for pressure is depth, and depth depends on how the volume is distributed. $\endgroup$ – Bryson S. Jul 15 '14 at 11:58
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No, the entire weight will not directly rest on the base of the slanted container (although it does indirectly). There are a number of ways to approach this, but the easiest way is to observe that the total force acting on the bottom of the container is equal to the sum of the hydrostatic pressure force (the pressure at the bottom of the container multiplied by its area) and the shear force around the edge of the base (draw a free-body diagram to convince yourself of this). The hydrostatic pressure depends only on the height of the column of fluid above the given location ($P=\rho gh$), and the shear force is equal to the weight of the fluid outside of the base (supported by the walls). Since both vessels contain the same volume, but the slanted container has a wider cross section above the base, the total height of the body of fluid will be less for the container with slanted walls than the other container. Thus, the hydrostatic pressure at the base will be less. But remember, any reduction in the pressure force will be compensated by an increase in the shear force around the edge of the base. The vertical walls support no vertical force, and so they exert no shear force on the base. The slanted walls do bear some of the weight, and so they transfer this force to the base via shear. In both cases, the sum of the hydrostatic pressure force plus the shear force (if there be any) is equal to the weight of the fluid, and this is what the scale measures.

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    $\begingroup$ We have to be careful here because the force measured by holding up the base is not just the base pressure times the area. It is that force plus the vertical shear from the slanted walls around the edge of the base. In either case, the full weight of the fluid must be borne. $\endgroup$ – Bryson S. Jul 15 '14 at 12:04
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As Rahulgarg mentioned, the pressure does not depend on the shape but on the depth. However, the direction of the force caused by pressure can be approximated as being normal to the surface, hence the total force on the sides will depend on the shape. For a fluid at rest like the one I think you are assuming, the pressure at any depth will be $p=p_0+\rho g h$, where $p_0$ is the atmospheric pressure at the surface, $\rho$ is the fluid density, $g$ the acceleration of gravity and $h$ the distance from the liquid's surface to the depth you want to know $p$. In your specific example the pressure at the bottom of case 2 will be smaller than in case one because of the different heights of the vessels.

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    $\begingroup$ The normal force against the floor (between the floor and the glass), as measured by a scale, will be the same in both cases, but not the pressure (force per unit area of the liquid against the glass) at the bottom of the liquid. $\endgroup$ – user16007 Jul 15 '14 at 7:28
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    $\begingroup$ ...And this is because the slanted walls resist some of the pressure force, which is in part balanced by the resistance to deformation of the vessel, and in part balanced by the reaction of the floor. $\endgroup$ – Joce Jul 15 '14 at 8:21
  • $\begingroup$ @Joce Understood julian fernandez, Still have one doubt. doesn't pressure indirectly mean weight, i.e for example if pressure is 1.5 bar (1 + 0.5), then it means 1.5 kg/sq.cm. So if i have a base area of 200 sq.cm, with 1.5 bar pressure, does it mean that the water above is exerting WEIGHT of 300 kg on base area?? i.e 1.5 kg/sq.cm * 200 sq.cm = 300 kgs. Is it right? $\endgroup$ – user52187 Jul 15 '14 at 11:47

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