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This is NOT a homework question. The question here is just an example.

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In this question, after both the liquids are mixed and a homogenous mixture is created, I found the effective density of the mixture and found that the pressure at the base increases by 1/2(rho)gh. This was calculated by taking a smaller cylinder with area a inside the system and as the fluid inside this imaginary cylinder is in equilibrium, I took the pressure at the bottom into the area of that smaller cylinder as the weight of the fluid.

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The normal reaction on the container by the ground remains the same obviously as there is no change in the mass of the system.

Now what I think is that the pressure on the base × base area = normal force

As the base area hasn't changed but the pressure has, the normal force also changes? Which is clearly wrong.

I have clearly misunderstood something here and would like to know if the pressure at the bottom of a container and normal force are related and how.

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2 Answers 2

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Now what I think is that the pressure on the base × base area = normal force

If you mean the internal pressure in the vessel, then no. The vessel has rigid walls and the pressure inside is not necessarily related to the pressure outside.

Imagine a gas cylinder with a movable piston. We can depress the piston to increase the pressure inside without adding mass. But the normal force from the ground is unaltered.

If the fluid pressure on the bottom of the vessel increases, but the normal force does not, then some other force must be changing so that the bottom does not accelerate. This force comes from the walls of the cylinder.

The extra area of the bottom cylinder that is not under the upper cylinder is under internal pressure. Before the mixing the fluid is pushing up on this area with pressure $2\rho g H$, but after the mixing the pressure is $3 \rho g H$. This extra pressure creates an upward force that counteracts the force from the increased downward pressure on the base.

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  • $\begingroup$ Thank you for your gas cylinder example it really helped. Could you elaborate on how the "force from the walls of the cylinder" acts on the bottom? And regarding the pressure after mixing, I got the pressure acting on the base as 9/2(rho)gh (see comment on above answer). I did not understand where the pressure of 3(rho)gh acts, you said "pushing up on this area"? $\endgroup$
    – Jennie
    Commented Oct 8, 2021 at 3:53
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    $\begingroup$ $3\rho g H$ is the pressure at H above the base. The fluid is pushing up on that part of the vessel and down on the bottom of the vessel. The vessel is being pushed apart by the fluid, but the walls hold it together (assuming they are strong enough). $\endgroup$
    – BowlOfRed
    Commented Oct 8, 2021 at 7:07
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    $\begingroup$ It may help to calculate the downward force from pressure on the bottom of the vessel (with area = $2 \pi R$) and the upward force from pressure on the "shelf" of the vessel (with area = $1 \pi R$). $\endgroup$
    – BowlOfRed
    Commented Oct 8, 2021 at 7:10
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To get

the pressure at the base increases by 1/2(rho)gh

presumably you've found that the new density is $\frac{3\rho}{2}$ and done this:

Original pressure is $$2H\rho g + H (2\rho) g = 4H\rho g\tag 1$$

and the new pressure is

$$2H\frac{3\rho}{2} g + H\frac{3\rho}{2} g = \frac{9}{2}H\rho g\tag 2$$

But you should divide the contribution of the top cylinder, to the pressure at the base, by 2 due to the increase of area, as the force due to the weight of the top cylinder spreads evenly throughout, (first term) and change equations 1) and 2) to 3) and 4)

$$H\rho g + H (2\rho) g = 3H\rho g\tag 3$$

$$H\frac{3\rho}{2} g + H\frac{3\rho}{2} g = 3H\rho g\tag 4$$

so A) and B) seem ok.

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  • $\begingroup$ I understood your point regarding the contribution of weight of the top cylinder to the pressure. But as I have now explained, (see edited question) I took an imaginary cylinder with area a and equated the weight and the pressure at the base multiplied by AREA OF THE IMAGINARY CYLINDER. Why is this wrong? Also since we are dealing with pressures shouldn't the areas not affect the pressure and only affect the forces due to that pressure? $\endgroup$
    – Jennie
    Commented Oct 8, 2021 at 2:48
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    $\begingroup$ @ Jennie Yes, that is where equations 1) and 2) are from. There is an upwards force on the upper side of the bottom cylinder (the part not under the upper cylinder) and so even though pressure throughout the bottom cylinder is increased due to the pressure from the top one, the total resultant force at the base isn't Pressure x Big Area, but Pressure x Small Area. The resultant pressure (at the base), d,ue to the top cylinder, is weight of top cylinder divided by Big Area. I see you've got a similar answer whilst I was asleep! Hope it all makes sense to you now $\endgroup$ Commented Oct 8, 2021 at 7:20

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