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

Click For Summary
SUMMARY

The problem involves calculating the distance from the top of an 8 cm × 8 cm × 8 cm wood block with a density of 653.49 kg/m³ to the water surface when floating. The correct approach requires consistent unit usage, converting dimensions to meters. The final calculation shows that the height of the un-immersed part of the block is approximately 7.9477 cm, derived from the correct application of buoyancy principles and density equations.

PREREQUISITES
  • Understanding of buoyancy principles
  • Knowledge of density calculations
  • Familiarity with unit conversion (cm to m)
  • Ability to apply Newton's laws in fluid mechanics
NEXT STEPS
  • Review Archimedes' principle and its applications in fluid mechanics
  • Study unit conversion techniques for physics problems
  • Learn about the relationship between density, volume, and weight
  • Explore common mistakes in buoyancy calculations and how to avoid them
USEFUL FOR

Students studying physics, particularly those focusing on fluid mechanics and buoyancy, as well as educators looking for examples of common calculation errors in homework problems.

PoPrOcKsRoCk
Messages
8
Reaction score
0

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:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 21 ·
Replies
21
Views
21K