Finding weight using Archimedes' principle

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Discussion Overview

The discussion revolves around calculating the weight of a rectangular block of solid carbon (graphite) floating at the interface of two immiscible liquids, specifically lubricating oil and water, using Archimedes' principle. Participants explore the implications of buoyancy and the relationship between the densities of the block and the liquids involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents a calculation approach to find the density of the lubricating oil based on the volume of the block immersed in each liquid.
  • Another participant questions the omission of the water's effect in the initial calculations.
  • A participant expresses uncertainty about their calculations when attempting to include water, noting that it leads to an illogical density for the oil.
  • There is a discussion about the buoyant forces acting on the block, with a focus on the relationship between the masses of the displaced liquids.
  • One participant proposes an equation that incorporates both the oil and water densities in relation to the block's volume.
  • A later reply confirms the proposed equation as correct, indicating a positive reception to the revised approach.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the need to consider both liquids in the calculations, but there is no consensus on the initial approach or the implications of the calculations regarding the densities involved.

Contextual Notes

The discussion includes assumptions about the relationship between the masses of the displaced liquids and the block, as well as the dependence on the specific densities of the materials involved. There are unresolved aspects regarding the calculations and the effects of the water on the overall buoyancy.

Who May Find This Useful

This discussion may be of interest to those studying fluid mechanics, buoyancy, or related fields in physics and engineering, particularly in understanding the application of Archimedes' principle in multi-liquid systems.

kari82
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A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids. The bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. Of the total block volume, 54.2% is immersed in the oil and the balance in the water. In a separate experiment, and empty flask is weighed, 35.3 cm^3 of the lubricating oil is poured into the flask, and the flask is reweighed. If the scale reading was 124.8g in the first weighting, what would it be in the second weighting?

mblock=m of oil displaced

density graphite=2.26g/cm^3

I'm not sure if I am right but I assumed that volume of oil=0.542volume of block

ρ(block) x V(block)=ρ(oil) x (0.542Vb)

ρ(oil)=ρ(b) x V(b)/0.542Vb=2.26/0.542= 4.17 g/cm^3

Then we can calculate the mass of the poured oil

m(oil)=ρ x V(poured) = 4.17 x 35.3 = 147.2g

m(total)=m(o) + m(f) = 147.2g +124.8g = 272g

I feel like I am missing something. Can someone please tell me if my approach is not correct? Thank you for your help!
 
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How do you justify not taking into account the effect of the water?
 
I cant.. And that was the main reason why I'm pretty sure I'm missing something. When I tried to include water into my calculations I was getting that the density of the oil is less than 1, which it does not make sense.

What I assumed was that the m(block)=m(oil displaced) and the m(block)=m(water displaced); therefore, the m(oil displaced)=m(water displaced)... Is that correct?
 
kari82 said:
What I assumed was that the m(block)=m(oil displaced) and the m(block)=m(water displaced); therefore, the m(oil displaced)=m(water displaced)... Is that correct?
No. What's the total buoyant force acting on the floating block?
 
mf(oil)g + mf(water)g ?
 
kari82 said:
mf(oil)g + mf(water)g ?
Yes. (Assuming that by mf(oil) you mean the mass of the displaced oil, etc.)

Express everything in terms of density and volume.
 
Thank you so much! So my correct equation will be:

ρ(block)V(block)=ρ(oil)0.542Vb + ρ(water)0.458Vb

Thank you again! :)
 
Looks good to me. ;-)
 

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