Maximum weight underwater

  • Thread starter Delhi
  • Start date
  • #1
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Hello, I would like to ask if my calculations are correct and which approach I should use if the answer is correct (I apologize if my calculations look ugly, just joined this forum).

Homework Statement


On land, the maximum weight of a concrete block you can carry is 25kg. How massive block could you carry underwater, if the density of concrete is 2200kg/m³?

pwater = 1000 kg/m³
pconcrete = 2200 kg/m³
m1 = 25 kg
m2 = ?

G = mg
g = 9,81 m/s²


Homework Equations


Archimedes' Principle

Fapplied - G + Fbuoyancy = 0
G - Fbuoyancy


The Attempt at a Solution


Approach 1:

Fapplied - G + Fbuoyancy = 0 ->
m1g - m2g + m2pwaterg / pconcrete = 0

And I come to this:

m2 = -m1 / ( (pwater/pconcrete) - 1 )

m2 = 45,8333..... kg


Approach 2:

On land I need (F= mg) 245,25 N to carry the concrete block

G - Fbouyancy = 245,25 N
m2g - m2g(pwater / pconcrete) = 245,25N
m2g(1 - pwater / pconcrete) = 245,25 N
m2 = 245,25N / g(1 - pwater / pconcrete)

m2 = 45,8333... kg
 

Answers and Replies

  • #2
558
1
I'd say it's correct.

In a more "elegant" form:

[tex]m' = m\frac{\rho_{conc}}{\rho_{conc}-\rho_{water}}[/tex]
 
  • #3
2
0
I'd say it's correct.

In a more "elegant" form:

[tex]m' = m\frac{\rho_{conc}}{\rho_{conc}-\rho_{water}}[/tex]
Yes that looks elegant. This might sound like a stupid question but from which formula do you get two [tex]\rho_{conc}[/tex] ?
 

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