# Questions about the formula P = rho*g*h

by jay.yoon314
Tags: formula, rhogh
 P: 22 Hello, I have two questions about the formula P = (rho)*g*h. The reasons for asking them are to clarify some points that are pertinent to a better comprehension of fluid statics, and also whether the ideas I am exploring in improving reverse osmosis water treatment, energetically speaking, are based on sound physics. The first question is as follows: Consider a "tube" of rather unusual shape that is described as follows: the lower half is cylindrical as well as the upper half, but the upper half's radius is less than that of the lower half, say by an order of magnitude. The cylinder as a whole can thus be described in a "piecewise" union of two cylinders of differing radii, with the interior volume being continuous throughout except for, of course, at the very top of the "upper cylinder" where the "tube" is open to the environment. Because the upper half has a radius that's 10 times less than that of the lower half, the cross sectional area of the upper half is 100 times less than that of the lower half. Since their respective heights are the same, as they are both halves of the whole height, the volume ratio of the upper half to the lower half is 1:100. If the entire interior of the "tube" is filled with a liquid, and that liquid has a uniform density everywhere throughout the tube, then the weight and mass ratios of the upper half to the lower half will also be 1:100, as this is what their volume ratios are as well. Now if the tube was filled so that the bottom half "filled up" all the way, but the upper half was completely free of any liquid, would it be the case that the pressure at any point in the liquid at the bottom of the tube would be only half of the pressure that would be exerted at that same point on the bottom if instead, as described in the previous paragraph, the tube were filled "all the way up" so that the top half was completely filled with liquid as well? The second question is as follows: Can you increase the pressure exerted at the bottom of a uniform/normal cylindrical tube filled halfway with water by floating a piece of wood (or any object that floats in water) on top of it? The fact that the wood is floating statically means that its weight, exerted downwards, and the buoyant force, exerted upwards by the displaced water on the wood, are equal. By Newton's Third Law, the reaction force of the buoyant force is exerted downward by the wood on the water, and is equal in magnitude to the buoyant force. Therefore, the gravitational force exerted on the wood and the reaction force of the buoyant force are equal both in direction and magnitude, and the former is exerted on the wood and the latter is exerted on the water. Call these forces Fgrav and Freaction, respectively. In other words, does the formula P = rho*g*h generalize to P = rho*g*h + Fgrav = rho*g*h + Freaction? (the latter equality being, of course, since Fgrav and Freaction are equal in both direction and magnitude?...assuming that this is indeed true...) Many thanks. These are not homework questions, I promise. It might look like one with the numbers removed and rephrased "conceptually," but please give me the benefit of the doubt. The reason I'm asking them is because I'm wondering whether reverse osmosis systems that are used to desalinate water can be designed to be more energetically efficient, specifically by obtaining the external pressure necessary to drive reverse osmosis in a process that doesn't require "artificial" pressurization/pumps, but rather uses the weight of the water to be desalinated itself as the "external force." My idea is only relevant to low-salinity water (water softening, improving the taste of already "fresh" water), as I've found that actual seawater exerts an enormous osmotic pressure that requires a water column hundreds of meters high. Thank you, Jay Yoon
HW Helper
P: 6,931
 Quote by jay.yoon314 Consider a "tube" of rather unusual shape
The shape doesn't matter, the pressure increases with depth (and density increases a tiny amount as pressure increases).

 Can you increase the pressure exerted at the bottom of a uniform/normal cylindrical tube filled halfway with water by floating a piece of wood (or any object that floats in water) on top of it?
Yes. The increase in pressure would depend on how much the top surface of the fluid was raised due to the piece of wood being added to the fluid.
 HW Helper P: 4,716 The pressure is due to what's above you. In your scenario, you should add atmospheric because that is pushing down on the surface of the fluid (open to the atmosphere). When you add the block of wood, it decreases the depth of water immediately under it, and succeeds in raising the water level everywhere else. Because pressure in a fluid equalizes along the horizontal plane, thanks to gravity, the new height of water tells you the pressure everywhere along the bottom of the tube, including directly in the shadow of that block of wood (where pressure is actually due to: the reduced depth of water + the density of the wood + atmospheric pressure).

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