What is the Maximum Weight of a Concrete Block Underwater?

AI Thread Summary
The discussion focuses on calculating the maximum weight of a concrete block that can be carried underwater, given that the density of concrete is 2200 kg/m³ and the density of water is 1000 kg/m³. Two approaches to the problem are presented, both leading to the conclusion that the maximum weight is approximately 45.83 kg. The calculations utilize Archimedes' Principle to account for buoyancy effects. The user also seeks clarification on the formula used, specifically regarding the appearance of two instances of the density of concrete. Overall, the thread emphasizes the application of physics principles to solve the problem accurately.
Delhi
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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
 
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I'd say it's correct.

In a more "elegant" form:

m' = m\frac{\rho_{conc}}{\rho_{conc}-\rho_{water}}
 
Quinzio said:
I'd say it's correct.

In a more "elegant" form:

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