Buoyancy - block of concrete is being raised from a lake

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Never mind.
 
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KiNGGeexD said:
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.
 
So force out of the water assuming uniform lifting

F= mg
 
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.
 
Isn't

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:(?

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'?
 
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:
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.
 
The ratio would surely just be the ratio of the two densities as the other terms are all constant?
 
Ah but there would be a mass term on the denominator

ΦgV / ρgV -mG
 
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?
 
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?
 
KiNGGeexD said:
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.
 
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:
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
 
Last edited by a moderator:
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 ?
 
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 said:
(ΦgV)/ (ρgV -mg)

Substitute for the data values in this expression, and see whether the answer looks right to you.
 
That's what I done in #44, my answer was 0.42

Or as a fraction

20/47
 
KiNGGeexD said:
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.
 
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
 
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?
 
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