I with a static fluids question

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Homework Help Overview

The discussion revolves around a static fluids problem involving a wooden block floating on water and the addition of a lead block. The original poster seeks assistance in determining the mass of the lead block when the wooden block becomes fully submerged.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Archimedes' Principle and question the original poster's equations and calculations. There is an exploration of the relationship between the weight of the displaced water and the weights of the blocks.

Discussion Status

Some participants have provided guidance on correcting the original poster's approach and calculations. There is an acknowledgment of errors in arithmetic, and the discussion reflects a collaborative effort to identify and rectify misunderstandings without reaching a definitive conclusion.

Contextual Notes

The original poster expresses uncertainty about their calculations and seeks clarification on specific steps, indicating a learning process that involves checking assumptions and verifying results.

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[SOLVED] i need help with a static fluids question

Homework Statement


A block of wood of 1.8 kg mass floats on water with 64% of its volume submerged. A lead block is placed on the wood and the wood is then fully submerged. Find the mass of the lead block.



Homework Equations



Fb = pvg

The Attempt at a Solution



(1000)(.64v) = Pblock (.36v)

Pblock = 1777.78

p = m/v...m/p = v

1.8/1777.78 = .001 m^3 = total volume

i have no clue what to do next, or even if what i have done so far is correct. please anyone who can help me?
 
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Use Archimedes' Principle.

Your eqn under attempt is wrong.

Weight of block = Fb_up = wt of water displaced = (64/100)*vol of block*density of water
=> 1.8*9.8 = 1000*0.64*V. You get V

Applying the same idea, can you do the 2nd part. Remember, the whole block plus the lead is being supported by the wt of the water displaced by block alone.
 
Shooting star said:
Use Archimedes' Principle.

Your eqn under attempt is wrong.

Weight of block = Fb_up = wt of water displaced = (64/100)*vol of block*density of water
=> 1.8*9.8 = 1000*0.64*V. You get V

Applying the same idea, can you do the 2nd part. Remember, the whole block plus the lead is being supported by the wt of the water displaced by block alone.

ok i did what you said. i obtained a v of .02756... and plugged that into Fb = 1000*(.02756...)*9.8, which gave me a Fb of 270.1125. I set this equal to mg and divided by 9.8 to get a final answer of 27.56. I then subtracted the 1.8 kg of wood and for some reason I'm wrong. Can you find my error? thanks soo much
 
I'm getting the mass of lead block as 1.01 kg.

I'm a bit tired of calculating, so if that's the correct answer, then you check your calc. Otherwise, we'll have to identify the source of error.
 
Last edited:
Shooting star said:
I'm getting the mass of lead block as 1.01 kg.

I'm a bit tired of calculating, so if that's the correct answer, then you check your calc. Otherwise, we'll have to identify the source of error.

haha I've learned to never blame the calculator, blame the user ;]. it was my own careless arithmetic error that caused me to get the wrong answer. when i got down to 27.5625 i subtracted 1.8 kg instead of subtracting (1.8*9.8)N and so i got an answer that was too large.

thanks SO much for the help, i really appreciate it!
 

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