# Confusing explanation - Archimedes' principle (AS Physcis)

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#1

How is the weight of the water displaced 8.83 x 103N, when the upthrust is not the same. The upthrust is 1000 x 1 x 9.81 = 9810N

Did I calculate the upthrust wrong or?
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1 month ago
#2
(Original post by GogetaORvegito?)
The upthrust is 1000 x 1 x 9.81 = 9810N
This calculation would be correct if the volume of displaced water is 1m3. But that's not the case as not all of the ice cube is submerged when floating. You're not given the volume of water displaced initially - the point of this problem is to calculate the volume of water displaced - so you can't calculate the upthrust using pVg.

Instead, you have to use the fact that, when the ice cube is floating, the upthrust will just be equal to its weight.
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#3
(Original post by nzy)
This calculation would be correct if the volume of displaced water is 1m3. But that's not the case as not all of the ice cube is submerged when floating. You're not given the volume of water displaced initially - the point of this problem is to calculate the volume of water displaced - so you can't calculate the upthrust using pVg.

Instead, you have to use the fact that, when the ice cube is floating, the upthrust will just be equal to its weight.
Oh ok then just one of those kinda things eh. However, I thought you could apply the equation for upthrust to anything regardless of how much is submerged? They used a floating object to derive it in a previous paragraph. I mean isn't the principle itself evident of this: "The upthrust 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"
Last edited by GogetaORvegito?; 1 month ago
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1 month ago
#4
(Original post by GogetaORvegito?)
However, I thought you could apply the equation for upthrust to anything regardless of how much is submerged? They used a floating object to derive it in a previous paragraph. I mean isn't the principle itself evident of this: "The upthrust 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"
The principle still applies, of course, but that's not the reason you can't use B = ρVg here - you aren't given V or B (the upthrust) at the beginning of the problem. V is the volume of fluid displaced, not the total volume of the immersed object, so you can't assume it's also 1.0m3 - again, V is value which the problem is trying to reach.

In order to find V, you first need to use Newton's First Law to deduce B. At this point you'd be able to calculate V using that very principle which you've quoted, B = ρVg, since you no longer have 2 unknowns.
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#5
(Original post by nzy)
The principle still applies, of course, but that's not the reason you can't use B = ρVg here - you aren't given V or B (the upthrust) at the beginning of the problem. V is the volume of fluid displaced, not the total volume of the immersed object, so you can't assume it's also 1.0m3 - again, V is value which the problem is trying to reach.

In order to find V, you first need to use Newton's First Law to deduce B. At this point you'd be able to calculate V using that very principle which you've quoted, B = ρVg, since you no longer have 2 unknowns.
Oh I see. But I just found out something using this information.
So now we can form the equation ρVg = 8.83 x 103N
Substituting the known values we get 1000 x 0.9 x 9.81 = 8829.
IT WAS A TYPO IN THE BOOK
Theres no reason they would times a number by 103 it was just 10^3 which matches my answer. They just rounded it as well.
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#6
nzy Although now i'm confused on how they found the Volume of the fluid displaced anyway.
They rearranged the upthrust equation to solve for V: 8.83 x 10^3 / (1000 x 9.81 ) = V

My question is, why did they use the weight of the whole ice cube. I thought it was only the weight of the fluid displaced that would be referenced here
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1 month ago
#7
(Original post by GogetaORvegito?)
My question is, why did they use the weight of the whole ice cube. I thought it was only the weight of the fluid displaced that would be referenced here
They did use the weight of the fluid displaced.
When the ice cube is floating, because of Newton's 1st Law, the weight of ice cube = upthrust from water, which, as you know, also = weight of water displaced.
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#8
(Original post by nzy)
They did use the weight of the fluid displaced.
When the ice cube is floating, because of Newton's 1st Law, the weight of ice cube = upthrust from water, which, as you know, also = weight of water displaced.
So in other words the actual weight of the ice cube is decreased because of the upthrust. Which is also the magnitude of the fluid that was displaced. This just gets confusing fast since this all is kinda like a loop of an explanation. Thanks for clearing things up for me! 0
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