If pressure equals force divided by area Why does area disappear in Pressure equals Rho x G x h x A all divided by A?

Answer 1

Ok, lets start with the basics:

let x = some distance [units = m]

let the area of some shape be a = x*x [units = m^2]

let the volume of some object be v = x*a [units = m^2]

we can see from above that the volume of an object contains information on its area.

So..

P [N/m^2]

rho [units = kg/m^3]

G [units = m/s^2]

h [units = m]

P = rho * G * h

N/m^2 = kg/m^3 * m/s^2 * m

N = kg*m/s^2 therefore:

N/m^2 = N/m^3 * m

canceling out m on top and bottom:

N/m^2 = N/m^2

The extra (* A / A) is not required. It reduces to 1. It is redundant to included this, as described above, because information on the objects area is already included within the volume.

Thank you for expanding the formulae from basic proofs. Actually this improves d

Thank you for showing the proofs from which this formulae derives. Actually this set of proofs improve the question. Personally I do and did realise how the formulae is derived from these basic proofs. What i was trying to impart to any enquiring mind is how the equation arrives at its final result and in nature takes no account of the millions of pounds of distributed atmospheric pressure, pressing down on the whole area of say the Atlantic Ocean, and because, and only because, of this equation has no effect on pressure at any depth. AS IF TO SAY THAT SOME `EXISTANCE` Who or Whatever has decreed " THIS IS HOW IT IS BECAUSE I SAY IT IS" Personally I wonder!

Lets work through an example here, involving water in the ocean. We will look at depths of 1m, 10m, 100m and 1,000m

rho sea water = ~1027kg/m^3

P = rho * G * H

1m

P = 1027kg/m^3 * 9.81m/s^2 * 1m = 10.07kPa_gage

10m

P = 1027kg/m^3 * 9.81m/s^2 * 10m = 100.7kPa_gage

100m

P = 1027kg/m^3 * 9.81m/s^2 * 100m = 1,007kPa_gage

1000m

P = 1027kg/m^3 * 9.81m/s^2 * 1000m = 10,007kPa_gage

The important part here is "gage". It means that all pressure is relative to the atmosphere. 1 ATM = 14.7psi or 101kPa. So if we wanted to find the pressure relative to a vacuum, "absolute", we simply add 101kPa to the number. After about 100m, atmospheric pressure only add 10% to the overall pressure. After 1000m, atmospheric pressure only adds 1% to the overall pressure. This is not to say that the atmospheric pressure is not great. 14.7 psi on your hand would be about the same as trying to hold up 235lb! (Its canceled on front and back so you don't notice). It most cases, water for example is about 1,000 times denser than air, pressures are indicated as "gage".