Buoyancy calculating unknown density

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    Buoyancy Density
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Homework Help Overview

The problem involves a cube of wood floating in alcohol, with the goal of determining the density of both the wood and the alcohol based on the conditions of buoyancy and submersion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of density using mass and volume, with attempts to apply buoyancy principles. Questions arise regarding the relationship between the submerged volume of the wood and the density of the alcohol.

Discussion Status

Multiple approaches to the problem are being explored, with some participants questioning the assumptions made about pressure and buoyancy. There is no explicit consensus on the method to proceed, and guidance has been offered to reconsider basic principles.

Contextual Notes

Participants express uncertainty about the application of known equations and the implications of the block being half submerged. There is a mention of differing instructional methods and the potential for confusion regarding the concepts of buoyancy and density.

masterburn
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Homework Statement



A cube of wood, of mass 0.50 kg and side length 10.80 cm, is floating in a beaker containing alcohol, as shown.
a) What is the density of the wood?
b) Ifthe wood is exactly half submerged, find the density ofthe alcohol

Homework Equations



not sure how to go about to solve density of alcohol

The Attempt at a Solution



A) p=m/v = 0.5/ (0.1080^3) = 396.92 kg/m^3

B) density of alcohol = height of box/ height of water * known p
= 0.05/ height * 369.92

not sure how to do B at all ? am i even close
 
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Draw the diagram. When encountering occasional exercises of this type, I prefer to go back to basics and derive the equations in situ.

Imagine there is an ant (kitted out in scuba gear) walking around the bottom of the beaker. He doesn't feel any extra pressure on his body when he is in the shadow of the floating block compared to when he is out from under it. At any given depth, the pressure is the same everywhere at that level. (If it wasn't, then some fluid would rearrange itself until the pressure was equalised.)

So when the ant is under the floating block, pressing down on his body is a height of alcohol + 10.8cm of wood. When he is not directly under the block, pressing down on him is a certain height of alcohol. Formalize this into volumes, densities and pressures.

Can you continue from there?

(Acting above both of these is atmospheric pressure, but because it acts uniformly over both the block and the alcohol, it need not enter into the computations for density here.)
 
no i cannot continue from there not sure how ? since half the block is submerged and, the volume displaced is the amount from the submerged block therefore density unknown = h of block/ h of liquid * density known but since the h is equal wouldn't it cancel out and the density of the liquid = density of block ?
 
Notice that I didn't mention anything about volume displaced because I was not using that approach.
 
I'm not sure what way to approach ? it was the only way I was taught ? the only thing i know following your logic is density avg > density fluid since it only half sunk, so the free diagram is mg going down and fb going up then it would be Fb>mg, how so i have buoyancy force of FB but i think it wrong since it doesn't take into account of the block being half sunk
 

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