This is the derivation for gauge pressure inside a fluid. A cylinder has been considered with base area A and height h inside a fluid. The derivation goes along the lines :

The net horizontal forces on the liquid are zero and the vertical forces must balance out the weight of the liquid. Here I don't understand why buoyancy hasn't been considered. But let's put it aside for a second. The difference between the forces exerted on the bottom and top surface must be equal to the wight of the cylinder.

$$(P_{top}-P_{bot})A = mg$$

And,density $\rho = \frac{m}{Ah}$. Therefore,

$$P_{top}-P_{bot} = \rho gh$$

They further say that since area hasn't occurred in the expression, the area/shape of the object doesn't matter (As long as the density doesn't change, right?). And the pressure is same at the same depth at all horizontal points. But if we hadn't replaced $m$ for $\rho$ in the expression, height wouldn't have been in the expression at all! So does height not matter in the first expression, $P_{top} - P_{bot} = \frac{mg}{A}$?


1 Answer 1


Think of that pressure difference, times the area, as providing the force to balance the weight mg of the fluid.

It just has to hold it up by matching the weight. That’s sufficient, whether the liquid is a tall sample of less dense liquid or a short one that’s more dense. What matters is the resulting weight.

You could also write it in terms of height and density too, of course, instead of just weight.

  • $\begingroup$ I forgot to add, they further assumed the top point to be exposed to the atmosphere to calculate gauge pressure. So now the expression was, P(b) = P(atm) + dgh and height here represents depth and then they said that depth matters where as area doesn't. So same question again, what if I didn't replace mass with density, height wouldn't occurred. $\endgroup$
    – user662650
    Commented Dec 4, 2019 at 6:44

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