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- Thread starter abdo799
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Your comment about a trapezium implies that there would be no water pressure on the angled sides? Of course that is not the case.

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SteamKing

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Any floating body can be overloaded, even the rectangular one.

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Correct me if i am wrong ( i am new to the world of physics) , but regarding the angled sides, doesn't pascal's second law apply here? if it does the force should be perpendicular to the surface , no upwards. If it doesn't apply , can you tell me please to calculate the water pressure on those angled sides ? PS: i am not arguing if the thing will float or sink, i know it will float , just trying to know why , thanks

Your comment about a trapezium implies that there would be no water pressure on the angled sides? Of course that is not the case.

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Chestermiller

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Mathematically, ##\vec F = - \oint_S P(S)\,\hat n(S)\,dS##, where ##P(S)## is the pressure at some point on the surface and ##\hat n(S)## is the unit outward normal to the surface at that point. This is a mess in general. However, the divergence theorem tells us how to rewrite this surface integral as a volume integral. This volume integral is much more tractable than is the surface integral and it is this volume integral that leads to Archimedes' principle.

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Chestermiller

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Yes. I had this approach in mind. In my response, I just wanted to explain how the pressure on the slanted sides of the body comes into play. This is a point of confusion for many new initiates.Mathematically, ##\vec F = - \oint_S P(S)\,\hat n(S)\,dS##, where ##P(S)## is the pressure at some point on the surface and ##\hat n(S)## is the unit outward normal to the surface at that point. This is a mess in general. However, the divergence theorem tells us how to rewrite this surface integral as a volume integral. This volume integral is much more tractable than is the surface integral and it is this volume integral that leads to Archimedes' principle.

Chet

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SteamKing

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It's wider at the top than the bottom.

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Thanks guys , you really helped me