Archimedes Principle: Calculating Relative Density for a Block of Oak and Lead

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Homework Help Overview

The discussion revolves around calculating the relative density of a block of oak using Archimedes' principle. The original poster presents a scenario involving the weights of oak and lead in different environments, aiming to find the relative density of the oak based on given weights in air and water.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations related to the forces acting on the oak and lead, questioning the signs used in the equations. There are inquiries about how to determine the buoyant force acting on the oak when submerged in water.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's calculations and questioning the interpretation of negative forces. Some guidance is offered regarding the manipulation of equations and understanding the implications of buoyancy.

Contextual Notes

Participants are navigating through the implications of negative forces in the context of buoyancy and are addressing potential errors in the original calculations. There is a focus on understanding the relationships between the weights of the objects in different states.

Nanu Nana
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. Homework Statement
A block oak weighs 90 N in air. A piece of lead weighs 130 N in water. The two weigh together 100N in water. Calculate the relative density of the wood.

Homework Equations


F=mxg
F(archimedes)= density x Volume x g

The Attempt at a Solution


F (g)oak = 90 N
F= mxg
m= 90N/9.81 m/s2
m= 9.17 gram
Oak in water = 130-100 N =30 N
Farchimedes oak = 90N-30N =60N
60N =1000kg/m3 x V x 9.81 m/s2
V=60N/9810
V=0.00611
Now that I have calculated V I can find density
rho = 9.17kg/ 0.00611 m3
density = 1499
Relative density is 1499 /1000 kg/m3 = 1.499
But the correct answer is 0.75
 
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Check your signs. In water, oak+lead=100N, while lead =130N.
 
How to do find N Force of oak in water ??
 
Nanu Nana said:
How to do find N Force of oak in water ??
Your method was ok but you made a mistake with the signs. Manipulate the two equations I wrote above.
 
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100+130?
 
Nanu Nana said:
100+130?
Please, don't make wild guesses.
If Ow+Lw=100 and Lw=130, what is Ow?
 
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Sorry /D its -30 so 90-(-30) = 120
 
Nanu Nana said:
Sorry /D its -30 so 90-(-30) = 120
Right. Does this get you to the right answer?
(How do you interpret the weight of -30N?)
 
haruspex said:
Right. Does this get you to the right answer?
(How do you interpret the weight of -30N?)
Yes I just calculated it. Negative force means opposite direction . But i don't know in this context
 
  • #10
Nanu Nana said:
Yes I just calculated it. Negative force means opposite direction . But i don't know in this context
If the wood were immersed in the water without being bound to the lead, what would happen?
 
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  • #11
It floats ofcourse
 
  • #12
Nanu Nana said:
It floats ofcourse
Right, hence the negative net force.
 
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  • #13
haruspex said:
Right, hence the negative net force.
Oh I see . thank you very much :)
 

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