Hydrostatic Pressure PGH

  • Thread starter brett351
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  • #1
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My understanding is that the pressure below the surface of water is pgh.

p - density of water, g - accel of gravity, h - dist below surface

And that this relationship holds regardless of the shape of the container of water. (also, I'm neglecting the atmospheric pressure at the surface)

If I have a cylindrical beaker of water of height H and area A and put it on a scale, I can calc the reading of the scale two ways...

1) pressure at bottom times area is pgHA.

2) density times volume times g is also pgHA.

Both ways give the same reading. Now if the shape of the beaker is an hourglass which has an area at top and bottom of A (same as top and bottom of cylindrical beaker), the two ways don't yield the same result.

Method 1) yields the same result for both shapes but method 2) yields a smaller result for the hourglass. What wrong with my logic? Thanks, Brett.
 

Answers and Replies

  • #2
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I dont understand how you are measuring things. The two pressures are equal, but your method for measuring them is wrong.
 
  • #3
Doc Al
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My understanding is that the pressure below the surface of water is pgh.

p - density of water, g - accel of gravity, h - dist below surface

And that this relationship holds regardless of the shape of the container of water. (also, I'm neglecting the atmospheric pressure at the surface)
OK.

If I have a cylindrical beaker of water of height H and area A and put it on a scale, I can calc the reading of the scale two ways...

1) pressure at bottom times area is pgHA.

2) density times volume times g is also pgHA.
The scale just measures the weight of the beaker of water. Method 1 gives the water pressure*area on the inside bottom of the beaker--this only equals the weight in the special case where the walls are vertical.

Both ways give the same reading. Now if the shape of the beaker is an hourglass which has an area at top and bottom of A (same as top and bottom of cylindrical beaker), the two ways don't yield the same result.
Because the pressure*area on the bottom of the hourglass beaker does not equal the net force on the beaker due to the water. The water also pushes up on other areas of the beaker surface. If you added up the net force that the water exerts on the entire beaker, that would equal the weight of the water.
 
  • #4
russ_watters
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HA is the volume of a cylinder....
 
  • #5
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Thanks Doc Al. Now I see that I forgot to take into account the vertical components of the pressures on the side of the hourglass, Brett.
 

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