The distance from the top of a block floating in the water to water

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Homework Help Overview

The problem involves a wooden block floating in fresh water, with specific dimensions and density provided. The goal is to determine the distance from the top of the block to the water surface.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations related to the weight of the block and the buoyant force, questioning unit consistency and the interpretation of the depth variable.

Discussion Status

Some participants have offered guidance on unit conversion and the distinction between immersed and un-immersed height. There is ongoing exploration of the calculations, with some confusion regarding the correct interpretation of the results.

Contextual Notes

Participants note the importance of consistent units and the need to clarify the height of the block in centimeters versus meters. There is mention of a potential misunderstanding regarding the final answer's validity based on unit discrepancies.

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Homework Statement


A 8 cm × 8 cm × 8 cm wood block with a density of 653.49 kg/m3 floats in water.
What is the distance from the top of the block to the water if the water is fresh?


Homework Equations





The Attempt at a Solution


weight= density*volume*g
weight=653.49*(512)*9.8
weight=3278951.424 Newton

weight=densitywater*g* 8 cm* 8 cm*depth
3278951.424=1000*9.81*8*8*d
3278951.424=627200*d
d=5.22792 cm...which is wrong

I have also tried subtracting 8 from 5.22792 and I got 2.77208 cm. I tried to submit it into my homework website and it is telling me I am wrong. I just don't understand what I am doing wrong.
 
Last edited:
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Yeah well, your units of length are not consistent throughout the calculation. In some quantities, you use centimetres and in other quantities you use metres.
 
Also, the "d" you are calculating is the height of the immersed part. You want the height of the un-immersed part.
 
So I would need to change the 8 cm to 0.08 m, right?

If I do that, then my answer would be;

weight= density*volume*g
weight=653.49*(5.14*10^-4)*9.8
weight= 3.2788 Newton

weight=densitywater*g* 8 cm* 8 cm*depth
3.2788=1000*9.8*0.08*0.08*d
3.2788=62.72*d
d=0.0522768 m?

And the height of the un-immersed part is 7.9477 m?
 
The last part is def. wrong. Remember, the height of the box is 8 cm, not 8 m. So you need to subtract your final answer from 0.08, not from 8. Assuming you did the calculations right (which I haven't checked), this means that more than half the box is immersed.
 
cepheid said:
The last part is def. wrong. Remember, the height of the box is 8 cm, not 8 m. So you need to subtract your final answer from 0.08, not from 8. Assuming you did the calculations right (which I haven't checked), this means that more than half the box is immersed.

I tried the answer 7.9477 m and it told me I was right. I don't understand how though because the 8 is in cm and the 0.0522768 is in m. I will have to talk to my professor about that. Thank you for your help.
 

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