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Confusion in law of liquid pressure.

Pressure increases with increase in depth from its free surface. So, does it mean that $h_0$ increases or $h$ increases (height of the body). Not sure which height to consider here.

If it is $h_0$, then can we say that as we go down, the force exerted by the box increases on the liquid. If yes, then why is that? What is the derivation for it.

One point is that from the formula density of fluid gravity. Here, $h$ increases but this is the height of box, not height from the free surface.

Then, what is the right answer for this?

  • $\begingroup$ "So , does it mean that ℎ0 increases or h increases(Height of the body". I'm not sure what you mean by this. $\endgroup$ – Bob D Jan 19 at 14:39
  • $\begingroup$ h0 is the height from the free surface of liquid till the end surface of box. $\endgroup$ – srijan Sri Jan 19 at 14:41
  • $\begingroup$ @BobD not sure if that’s what the law means. If yes , then why ? $\endgroup$ – srijan Sri Jan 19 at 14:42

In an incompressible fluid in a gravitational field, the pressure at specific point in the fluid depends on the gravitational field strength, $g$ (some call this acceleration due to gravity, unfortunately), the density of the the fluid, $\rho$, and the depth (in the direction of the grav. field $\vec{g}$) of the point, $D$, plus whatever the pressure is at the top of the fluid.

To emphasize the concepts: The pressure in the fluid is a scaler value at a point and has the same value for a specific depth, no matter the lateral (horizontal) location. Force acting on some area, caused by this pressure, is the product of pressure and area and is directed perpendicular to area being analyzed.

For your diagram, the absolute pressure at a depth $h_o$ from the top of the liquid (at the bottom of the block is $$Pb=P_{atm}+\rho g h_o .$$ In many cases the atmospheric pressure can be ignored because pressure difference is often the quantity which drives the behavior of the system. Also notice that the pressure across the bottom of the block is the same at every point in the fluid at the same depth as the bottom. In similar fashion, the pressure across the top of the block and at every point at the same depth as the top is $$P_t=P_{atm}+\rho g (h_o-h) .$$

So the you see that $P_{atm}$ disappears and the presssure difference between the bottom and top of the block is $\Delta P = \rho g h$.

Buoyant force is a different conversation. While the result is rather simple, the proof does involve this pressure difference, which is independent of the actual depth of the block.

Summary - Pressure in the fluid depends on $h_o$. Pressure difference across a block of height $h$ depends on $h$, but not on $h_o$.


In $h \rho g$

h is the height or rather depth of that point from free surface. To find pressure at lower face of cube you consider height $h_°$ .

When you are taking the block down the depth of the point on lower surface increases that is which it is said that pressure increases with depth

the force exerted by box increases with depth

Yes you are right as we see that pressure is increasing so force by liquid on the side of block increases . Using newton's law we can say that equal force will be exerted by the block so it does increase

  • $\begingroup$ h is not height of body in formula but h from free surface till the end surface of box right .(As you say) $\endgroup$ – srijan Sri Jan 19 at 14:43
  • $\begingroup$ But in formula we write h and not h0. Pls check the edit $\endgroup$ – srijan Sri Jan 19 at 14:46
  • $\begingroup$ That is different it is buoyant force $\endgroup$ – Anusha Jan 19 at 14:48
  • $\begingroup$ This thing you are talking is due to hydrostatic pressure $\endgroup$ – Anusha Jan 19 at 14:53
  • 1
    $\begingroup$ Yes we do @Srijansingh7 $\endgroup$ – Anusha Jan 19 at 16:25

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