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Phy 122

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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Column Height vs Pressure

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I don't understand a concept which seems to be absolutely fundatmental and essential to the work we are doing.

I can't grasp the concept of how to determine the pressure required to support a column of water.

I would calculate pressure by using the following formula

Pressure = density * gravity * height.

So to move a column of water 3 cm or .03 meters would require how much pressure?

Pressure = 1000kg/m^3 *9.8m/s^2 * .03m

Pressure = 294 kg m^2/m^3s^2

Pressure = 294 N/m^2 or 294 Pa

???First off, I'm not sure if that is correct, but I'm also not sure how we account for say the size of the tube we are raising the water in. It would seem to me that the size of the tube should somehow come into play, but I don't know how???

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Thanks for your help.

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You are correct in your calculation, and you are also correct that this is very fundamental to a lot of the work we're doing.

Check also the very first problem in the Introductory Problem Set.

Quick synopsis:

You are right that the pressure is density * g * height. I often use the expression rho g `dy, which is the difference in pressure due to a height difference of `dy. Textbooks often use rho g h for the pressure in a fluid at depth h.

The reason is pretty straightforward. If A is the cross-sectional area of the tube, then the volume of length `dy of the tube is cross-sectional area * height = A `dy and the mass of fluid in the tube is density * volume = rho A `dy.

The weight of the water in the tube would therefore be

weight = mass * g = rho A `dy * g

The pressure required to support this column is the force required per unit area. The force required is the weight of the water in the tube, the force is exerted on the base of the water column which has area A so the pressure is

pressure = force / area = rho A `dy * g / A = rho `dy * g.

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This applies in general to a fluid at any depth, since any horizontal cross-sectional area A in the fluid must support the fluid directly above it.

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