Buoyancy - block of concrete is being raised from a lake

[KiNGGeexD]

Buoyancy -- block of concrete is being raised from a lake

Question;

A 50kg block of concrete is being raised from a lake. What fraction of its weight in air is required to lift it while submerged?

My attempt,

I'm pretty sure this question pertains to Archimedes principle

I calculated the weight of the block (force due to gravity) and noted that the buoyancy force is equal to the weight of fluid displaced

Buoyancy = ρgV = ρgAy

And this is where it hit the buffers I'm not so sure how to proceed?

I think that it would be something like, calculate the buoyancy force and the force which would be needed to lift the block (in air) and then subtract the two as the buoyancy force would aid the elevation and then take some kind of ratio of that mass required and the given mass?Any help would be great I think I have given enough information:)

Thanks for any help in advanced:)

Regards

 

[BvU]

Buoyancy = ρgV = ρgAy is fine. Stick to V, because a "block" doesn't necessarily mean a rectangular block.
Something similar can be written down for the mass in terms of V and for the weight in air (check if you can ignore the buoyancy in air).

Since they want a fraction as a numerical answer, you will need the values of $\rho$ for concrete, water (and perhaps also for air).

Oh, and "submerged" can be taken to mean "fully submerged" I suppose.

 

[KiNGGeexD]

Ok I will give it a g

Thanks a bunch I really appreciate it

 

[KiNGGeexD]

So when I write mass in terms of ρV my volume is constant so I can cancel it and then go from there?

 

[KiNGGeexD]

Do I start with the initial conditionmg = ρgV where density is that of concrete? Just so I know I'm on the correct path?

 

[nasu]

This is not initial condition. It is true all the time for the given block.
But it is one of the equations you need, yes.

 

[KiNGGeexD]

So I would I then add the force applied to lifting the block

 

[nasu]

OK. What will be this force when the block is in air?
What will it be when it is submerged?

 

[KiNGGeexD]

mg would be still acting downwards and force of pulling would be

F= ma?

 

[KiNGGeexD]

When submerged the buoyancy aids the pulling force, and the only force acting down would be mg? Disregarding viscous forces and drag?

 

[nasu]

mg would be still acting downwards and force of pulling would be

F= ma?

What is "a" here?
In Newton's law, F must be the net force. The sum of all the forces acting on the body.
Here you can assume uniform lifting, with no acceleration.

 

[KiNGGeexD]

Ahh ok that helps a lot! I usually don't assume such things unless the problem states it explicitly, I'm the typical student who finds it hard to think on their own:(

So I am omitting acceleration as it is zero in this case?

ρVg = m(0)+ ρgV

Where the first rho is concrete density and the second is water?

 

[nasu]

What is m(0)?

 

[KiNGGeexD]

That's what i don't get if the acceleration is zero the

F= ma would be zero?

 

[nasu]

Yes, the net force will be zero. This means that the forces acting on the body are balanced.
For the case in air, you have two forces: the weight of the block and the force pulling up.
You said that the weight is ρgV. What is the force pulling up, taking into account that the two have to be balanced?

 

[KiNGGeexD]

So
ρgV- mg =0

ρgV= mg

 

[nasu]

No, ρVg and mg are two expressions of the same force.
You have two forces, gravity, G and the pulling force, F.
Newton's law applied to this situation is
F-G=0.

Now you can express G=mg=ρVg
so going back to the first equation,
F-ρVg=0 and
F=ρVg

This is the pulling force in the first case.

Now do the same thing for the case you have buoyant force too. Now you have three forces.
Again, find the pulling force, let say F'.

 

[KiNGGeexD]

Buoyancy force is ρgV also though so would I just add that force

 

[KiNGGeexD]

F= ρgV for the case in air

So buoyancy (when submerged)

mg= ρgV

Where rho would be that of water in this case where as it would be concrete in the first case

 

[nasu]

Just write the equilibrium of the forces first. Then you will know if (and where) you add it.

 

[KiNGGeexD]

So in water mg= ρgV + F

Where F is the force discussed previously?

 

[nasu]

Yes, if the ρ is density of water. You should use different symbols for different quantities.
Using ρ for both concrete and water is bad practice.
And the pulling force is different now so you should not use F which was the pulling force when in air.
Call it F' for example.

Now calculate this F' and compare with F found previously.

 

[KiNGGeexD]

mg =gV(ρ+φ)

Where phi is the density of concrete

 

[nasu]

Now, you don't listen.
The "F" in here is not the same as for the air. It is not ρVg or φVg.
It is a new force, F'. This is what you are supposed to find.
Maybe a drawing showing the forces may help you understand.

 

[NascentOxygen]

Where phi is the density of concrete

What value are you going to use as the density of concrete?

 

[KiNGGeexD]

What value are you going to use as the density of concrete?

2400 kg m^-3

 

[nasu]

You are not there yet.
Did you find an expression for the pulling force in the second case?

 

[KiNGGeexD]

F + ρgV = mg

 

[NascentOxygen]

2400 kg m^-3

So what volume of water will the concrete rock displace?

 

[KiNGGeexD]

It's the same as the volume of the block!

 

[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 ?