Buoyancy - block of concrete is being raised from a lake

[nasu]

Never mind.

 

[nasu]

F + ρgV = mg

Yes, this is right. Just keep in mind that this "F" is not the same as in the first part.
You should use a different label.

Now solve this to find this new "F".
And find the ratio between this and the first one.

 

[KiNGGeexD]

So force out of the water assuming uniform lifting

F= mg

 

[KiNGGeexD]

Or rather

F(2)= mg

 

[nasu]

I have no idea what are you trying to do. And you notations do not help.

You had this equation
F + ρgV = mg

Let call this F' to avoid confusion. This is the pulling force when the block is in the air.
So the equilibrium of forces is expressed as
F'+Fb=W

where Fb is buoyant force and W is the weight.
Indeed we can write W=mg and Fb=ρgV.
Now solve this equation for F'.
Find F'=...
where ... is an expression, not a number.

 

[KiNGGeexD]

Isn't

F' + ρgV = mg

The force whilst in water? Otherwise why do we have the buoyancy force:(?

 

[nasu]

Isn't

F' + ρgV = mg

The force whilst in water? Otherwise why do we have the buoyancy force:(?

F' is that force. What you wrote above is an equation , a relationship between 3 forces, not the expression of a force.

So what will be F'? have you ever solve a linear equation? Do you understand what does it mean to solle the above equation to find F'?

 

[KiNGGeexD]

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]

So i would have 2mg = ρgV

 

[nasu]

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...

You forgot that the density of concrete is φ and not ρ.
And F is not equal to F' obviously.
after you put the right densities in the expressions, you have to calculate the ratio
F'/F.

Good luck.

 

[KiNGGeexD]

The ratio would surely just be the ratio of the two densities as the other terms are all constant?

 

[KiNGGeexD]

Ah but there would be a mass term on the denominator

ΦgV / ρgV -mG

 

[KiNGGeexD]

From the I can obtain

Φ / ρ- mg

And I have all of these variables at my disposal So the ratio of the forces would be the same as the ratio of the weights?

 

[KiNGGeexD]

This would yield

20/47 so roughly 42% or 0.42 of the weight must be in airW= mg so 50*9.8. = 490

So weight that must be in air 205.8 N

I know it doesn't ask for this latter part?

 

[NascentOxygen]

Ah but there would be a mass term on the denominator

ΦgV / ρgV -mG

Should there be parentheses in this expression?

Can you explain in words what this expression is.

 

[KiNGGeexD]

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 :)

 

[NascentOxygen]

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 :)

There were two questions in my last post. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

 

[KiNGGeexD]

No there shouldn't be parenthesis

 

[KiNGGeexD]

Or was it rhetorical?

 

[BvU]

Yes there should be parentheses. In the correct expression
Why don't you re-read post 1 and 38, and possibly also 40, instead of galloping off in the fog ?

You are asked for F'/F. You have F', you have F. How hard is it to write F'/F correctly ?

 

[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]