Buoyancy - block of concrete is being raised from a lake

[KiNGGeexD]

Thanks for you help

 

[KiNGGeexD]

I didn't realize I had the two mixed up, so the expression flips for it to be F'/F

I got confused as to which one I used for which, I'm assuming F' is the force when it's in water

 

[KiNGGeexD]

Is #44 not correct then?

 

[KiNGGeexD]

(ΦgV)/ (ρgV -mg)

 

[NascentOxygen]

(ΦgV)/ (ρgV -mg)

Substitute for the data values in this expression, and see whether the answer looks right to you.

 

[KiNGGeexD]

That's what I done in #44, my answer was 0.42

Or as a fraction

20/47

 

[NascentOxygen]

That's what I done in #44, my answer was 0.42

Or as a fraction

20/47

You won't get the same answer. Try it.

 

[KiNGGeexD]

Ok I mixed up the φ and ρ values
If I then put φ in as the density if concrete my answer is

48/19 which is 2.52

 

[KiNGGeexD]

Ok I had another look at the problemF1= gV(φ -ρ) the force to raise it to the surface?

And F2= gVφ
The force or raise it in air?

 

[KiNGGeexD]

 

[KiNGGeexD]

Ok I recalculated and got 0.39583Although if

F'/F is correct according to my expression I get a negative value asF' = mg - ρgV

And F = φgV

 

[BvU]

Horror. Look at what you are doing a little longer. That way you can make do with one post instead of six, and we can handle the entire thread in three or four posts instead of over sixty! Slap, Slap, Slap

You are asked for F₁ / F₂. I don't know why you change notation so often, except perhaps you like to torture yourself.

On your sketch, F₁ clearly points the same way as $\rho_w\, gV$. If you want to write a balance, write F₁ + $\rho_w\, gV$ = mg . Is that clear enough ? Read it over if it is not, don't quick reply.

In the next step, you can write F₂ = mg = $\rho_c\, gV$. Perhaps it is wise to use the same expression for the same thing. Why not change mg in the first to $\rho_c\, gV$ as well ? Less confusing, for you too.

In the ratio, you happily divide the numerator by V, but only the first term of the denominator. Aaaargh!

Then you swap to numbers, the last resort. A bit too early. To make up, you slap a dimension of Newtons on the quotient. No good. A ratio is a ratio and a force is a force. A ratio of forces is only a ratio and not a force.

Now, please please, write F₁ / F₂. Divide out properly and get something extremely simple.

 

[BvU]

No you don't get something negative. But I'll wait a few posts until you see the light...

 

[KiNGGeexD]

Oh ok soF1 + ρgV = mgAnd this is the force in water? I'm just trying to get this as clear as I can sorry F2= mg = ρgV And this is the force in air and I want the ratio of

F1/F2

However in my first expression I can write mg as ρgV but that would be the density of concrete the same as the density in the second expression

So I would have

ρ(concrete)gV- ρ(water)gV = F1And F2= ρ(concrete)gVSo my expression becomes

ρ(concrete)-ρ(water) divided by ρ(concrete)

Sorry for the shoddy notation here

 

[KiNGGeexD]

And when I plug my numbers in I get

7/12Or0.583

 

[BvU]

No need to apologize, and yes: you make me feel really happy by picking up the right thing!

In the end all you have is $$\rho_{\rm concrete} - \rho_{\rm water} \over \rho_{\rm concrete}$$ or indeed 14/24

Well done !

 

[KiNGGeexD]

I really don't see what was confusing me there! It was that pesky mg term I didn't think to change that as well! Thanks I won't forget this kind of problem! All part of learning I suppose! A lot more trivial than I though! Thanks so much for your help and for sticking with my irritating disregard for thought!

Cheers

 

[BvU]

Being stubborn can be a good quality too, especially if you are right and the others are not.

 

[KiNGGeexD]

Shame in this case I was horribly wrong!