- #36
KiNGGeexD
- 317
- 1
Isn't
F' + ρgV = mg
The force whilst in water? Otherwise why do we have the buoyancy force:(?
F' + ρgV = mg
The force whilst in water? Otherwise why do we have the buoyancy force:(?
KiNGGeexD said:Isn't
F' + ρgV = mg
The force whilst in water? Otherwise why do we have the buoyancy force:(?
You forgot that the density of concrete is φ and not ρ.KiNGGeexD said:Ok, I have my force in air
F-G=0
F= G where G= mg = ρgV
I'm assuming the density here would be that of concrete!
Then my second equation (in water)
F' + ρgV = mg
F'= mg- ρgV
Where i say density is that of water as buoyancy force is the weight of water displaced? So now I have my two pulling forces...
KiNGGeexD said:Ah but there would be a mass term on the denominator
ΦgV / ρgV -mG
There were two questions in my last post. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gifKiNGGeexD said:It is the ratio between the force in the air and the force when in the water.
Where φ and ρ are both densities but given different symbols to avoid confusion and because I can't subscript on my phone :)
KiNGGeexD said:(ΦgV)/ (ρgV -mg)
You won't get the same answer. Try it.KiNGGeexD said:That's what I done in #44, my answer was 0.42
Or as a fraction
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