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Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. This principle is derived from pascal's law.

What I find unintuitive about the principle is that for objects with a higher density than water, shouldn't the pressure exerted against it by water also be larger?

My reasoning is that since the weight of the object is larger, the contact force exerted by the water below it should also be greater.

When deriving Pascal's law, the weight of the water body was taken into account. In this case, shouldn't the weight of the object be taken into account too?

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for objects with a higher density than water, shouldn't the pressure exerted against it by water also be larger?

No. Pressure is a function of state. For water at a given temperature and a given density, there is a pressure which can be found by the equation of state for water. Whether the water is touching a lump of lead or a wooden block is irrelevant.

weight of the object is larger, the contact force exerted by the water below it should also be greater

They have nothing to do with each other. One is a force exerted by literally the entire Earth due to gravity. The other is the force exerted by the water due to contact. They're completely separate phenomena.

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  • $\begingroup$ According to pascals law, shouldn't the weight of the water above a certain point directly affect its pressure? $\endgroup$
    – Luo Zeyuan
    Sep 16, 2018 at 9:41
  • $\begingroup$ Directly affect its pressure? No. Just the temperature and density is all you need. Pascal's law makes assumptions, specifically that everything is at equilibrium. In that case, it turns out that the gauge pressure is related to the weight of the water above, but taking it as a fundamental principle is a mistake. When you put a dense object in the water, it isn't at equilibrium. The assumptions that went into deriving Pascal's law don't hold. You can't apply it there. $\endgroup$
    – Mark Eichenlaub
    Sep 16, 2018 at 17:08
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My reasoning is that since the weight of the object is >larger, the contact force exerted by the water below it >should also be greater.

You are right - that's why the object is sinking, by pushing the water under it to the sides, where the pressure is lower.

Pascal's law, which states that the pressure at the sides should be the same as under the object, is not applicable here because it is not a static situation - the object is sinking.

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