Deriving density formulae from first principles

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

Can someone please help derive the relations below from first principles?

Also does someone please know what happens when $ρ_{object} = p_{fluid}$?

Many thanks!

 

[haruspex]

Homework Statement: Please see below
Relevant Equations: Please see below

Can someone please help derive the relations below from first principles?
View attachment 327173
Also does someone please know what happens when $ρ_{object} = p_{fluid}$?

Many thanks!

First, state those first principles.

 

[MatinSAR]

Homework Statement: Please see below
Relevant Equations: Please see below

Can someone please help derive the relations below from first principles?

Note that these relations are not correct in general.

Also does someone please know what happens when $ρ_{object} = p_{fluid}$?

You need to know about Archimedes' principle and Buoyant force to understand these.

 

[ChiralSuperfields]

First, state those first principles.

Thank you for your reply @haruspex! Sorry state which frist princpels?

Many thanks!

 

[ChiralSuperfields]

Note that these relations are not correct in general.

You need to know about Archimedes' principle and Buoyant force to understand these.

Thank you for your reply @MartinSAR!

Sorry, how are these relations not correct in general?

Many thanks!

 

[MatinSAR]

Thank you for your reply @haruspex! Sorry state which frist princpels?

Many thanks!

I guess:

 

[kuruman]

The thread
https://www.physicsforums.com/threads/finding-downward-force-on-immersed-object.1050821/
that you initiated not too long ago has everything that you need. Please study what has been said.

 

[MatinSAR]

Sorry, how are these relations not correct in general?

Consider a sphere that has cavity.
https://www.physicsforums.com/threads/density-of-a-sphere-that-has-a-cavity.1052083/#post-6880289
In this thread,it was stated that $\rho_{iron}>\rho_{water}$ but the sphere floats on water and it doesn't sink.

 

[haruspex]

state which frist princpels?

You cannot hope to derive something from first principles if you cannot say what those principles are. So the place to start is an attempt to state them.
Does the question specify it should be from first principles, or is that a wording you added?

 

[haruspex]

Consider a sphere that has cavity.
https://www.physicsforums.com/threads/density-of-a-sphere-that-has-a-cavity.1052083/#post-6880289
In this thread,it was stated that $\rho_{iron}>\rho_{water}$ but the sphere floats on water and it doesn't sink.

Sure, but to fix that (and composite body cases generally) you just have to define the density of an object as its average density. A bit trickier is non-convex bodies.

 

[haruspex]

Homework Statement: Please see below
Relevant Equations: Please see below

does someone please know what happens when ρobject=pfluid?

In the real world, there is no such thing as equality of continuous variables.
A more interesting question is that of stability. Different substances have different elastic moduli. A balloon may rise if placed at one depth in a fluid but sink if placed at a greater one.

 

[kuruman]

Also does someone please know what happens when $ρ_{object} = p_{fluid}$?

Fill a glass with water. Imagine a 1-cm cube of water in the middle of the glass. Does it float to the surface, does it sink to the bottom or does it stay where it is?

 

[ChiralSuperfields]

Thank you for your replies @MatinSAR, @kuruman, and @haruspex!

I think I understand now.

 

[kuruman]

Thank you for your replies @MatinSAR, @kuruman, and @haruspex!

I think I understand now.

You are welcome. Might you be interested in verifying your understanding by showing us how you derived the relations posted in #1 from first principles?

 

[ChiralSuperfields]

You are welcome. Might you be interested in verifying your understanding by showing us how you derived the relations posted in #1 from first principles?

Thank you for your reply @kuruman! That is very kind to offer. Yes I am interested, here it is:

For an object immersed in water the net force is

$\vec {F_{net}} = B\hat j - mg\hat j$
$F_{net} \hat j = B\hat j - mg\hat j$
$F_{net} \hat j = (B - mg)\hat j$ (Factoring out j-hat unit vector to and then cancelling them)
$F_{net} = B - mg$

$F_{net} = ρ_{fluid}V_{fluid}g - ρ_{object}V_{object}g$ (From Archimedes principle that states that the buoyancy force is equal to the weight of the displaced water)

Note: I let $V_{fluid}$ represent the volume of fluid displaced

Now we take special cases of our expression,
$F_{net} = ρ_{fluid}V_{fluid}g - ρ_{object}V_{object}g$

The first case we will consider $ρ_{object} > ρ_{fluid}$, and consider the object displacing the maximum amount of water to get the max $B$, therefore $V_{fluid} = V_{object}$

$F_{net} = ρ_{fluid}V_{object}g - ρ_{object}V_{object}g$
$F_{net} = V_{object}g(ρ_{fluid} - ρ_{object})$

And since $ρ_{object} > ρ_{fluid}$ then $ρ_{fluid} - ρ_{object} < 0$ so $F_{net} < 0$ which means that the object has a net force downward and will sink

For the second case we will consider $ρ_{object} < ρ_{fluid}$, and consider the object displacing the maximum amount of water to get the max $B$, therefore $V_{fluid} = V_{object}$

$F_{net} = ρ_{fluid}V_{object}g - ρ_{object}V_{object}g$
$F_{net} = V_{object}g(ρ_{fluid} - ρ_{object})$

And since $ρ_{object} < ρ_{fluid}$ then $ρ_{fluid} - ρ_{object} > 0$ so $F_{net} > 0$ which means that the object has a net force upwards and will float.

For the third case we will is $ρ_{object} = ρ_{fluid}$, and consider the object displacing the maximum amount of water to get the max $B$, therefore $V_{fluid} = V_{object}$

$F_{net} = ρ_{fluid}V_{object}g - ρ_{object}V_{object}g$
$F_{net} = V_{object}g(ρ_{fluid} - ρ_{object})$

And since $ρ_{object} = ρ_{fluid}$ then $ρ_{fluid} - ρ_{object} = 0$ so $F_{net} = 0$ which means that the object is in vertical equilibrium so if placed gently then it the object will float. However, if put into the water with an initial velocity it will remain sinking (Assuming no fluid resistance) in accordance with Newton First Law.

I think the main think I learnt from this was to consider the most extreme case where $B = B_{max}$ since this is the maximum amount of upwards force from the fluid that the object can get.

Is this please all correct?Many thanks!

 

[haruspex]

Thank you for your reply @kuruman! That is very kind to offer. Yes I am interested, here it is:

For an object immersed in water the net force is

$\vec {F_{net}} = B\hat j - mg\hat j$
$F_{net} \hat j = B\hat j - mg\hat j$
$F_{net} \hat j = (B - mg)\hat j$ (Factoring out j-hat unit vector to and then cancelling them)
$F_{net} = B - mg$

$F_{net} = ρ_{fluid}V_{fluid}g - ρ_{object}V_{object}g$ (From Archimedes principle that states that the buoyancy force is equal to the weight of the displaced water)

Note: I let $V_{fluid}$ represent the volume of fluid displaced

Now we take special cases of our expression,
$F_{net} = ρ_{fluid}V_{fluid}g - ρ_{object}V_{object}g$

The first case we will consider $ρ_{object} > ρ_{fluid}$, and consider the object displacing the maximum amount of water to get the max $B$, therefore $V_{fluid} = V_{object}$

$F_{net} = ρ_{fluid}V_{object}g - ρ_{object}V_{object}g$
$F_{net} = V_{object}g(ρ_{fluid} - ρ_{object})$

And since $ρ_{object} > ρ_{fluid}$ then $ρ_{fluid} - ρ_{object} < 0$ so $F_{net} < 0$ which means that the object has a net force downward and will sink

For the second case we will consider $ρ_{object} < ρ_{fluid}$, and consider the object displacing the maximum amount of water to get the max $B$, therefore $V_{fluid} = V_{object}$

$F_{net} = ρ_{fluid}V_{object}g - ρ_{object}V_{object}g$
$F_{net} = V_{object}g(ρ_{fluid} - ρ_{object})$

And since $ρ_{object} < ρ_{fluid}$ then $ρ_{fluid} - ρ_{object} > 0$ so $F_{net} > 0$ which means that the object has a net force upwards and will float.

For the third case we will is $ρ_{object} = ρ_{fluid}$, and consider the object displacing the maximum amount of water to get the max $B$, therefore $V_{fluid} = V_{object}$

$F_{net} = ρ_{fluid}V_{object}g - ρ_{object}V_{object}g$
$F_{net} = V_{object}g(ρ_{fluid} - ρ_{object})$

And since $ρ_{object} = ρ_{fluid}$ then $ρ_{fluid} - ρ_{object} = 0$ so $F_{net} = 0$ which means that the object is in vertical equilibrium so if placed gently then it the object will float. However, if put into the water with an initial velocity it will remain sinking (Assuming no fluid resistance) in accordance with Newton First Law.

I think the main think I learnt from this was to consider the most extreme case where $B = B_{max}$ since this is the maximum amount of upwards force from the fluid that the object can get.

Is this please all correct?Many thanks!

Pretty good. A couple of quibbles…

1. You don't consider what happens when a rising body reaches the surface. The problem statement says "and floats". Note that if there were no losses then, having surfaced, a body less dense than the fluid medium would rise so far that it would then descend again, bobbing up and down between the same two extremes for ever!

2. It is debatable whether Archimedes' principle constitutes a "first principle"; certainly not in the same way that Newton's laws of forces, accelerations and gravity are. Indeed, Archimedes' principle as generally stated omits a caveat: that the fluid can reach the whole surface of the body below the fluid level. Consider a suction pad stuck to the bottom of the tank.
You can avoid that difficulty by following the line of reasoning which, I have always assumed, Archimedes followed. Consider replacing the body with exactly that fluid which would fit in the portion of the body volume that is below the fluid level. What forces act on it? Would it be in equilibrium? What must be the buoyant force on it from the rest of the fluid?
Note what this line of reasoning leads to in the suction pad case.

 

[kuruman]

I was more interested in seeing what you consider a "first principle", something that @haruspex tried to get you to enunciate in post #2. Usually, the first principle comes first in a derivation and $\vec {F_{net}} = B\hat j - mg\hat j$ is not it.

Here, first principles are Newton's laws. Consider the second law, $\vec F_{net}=m\vec a$ and draw a FBD for a cube of volume $V$ of water floating in water. This cube's acceleration is zero, therefore the net force on it is zero. There are two entities exerting forces on the cube, the Earth that exerts force $mg=\rho_{H_2O}Vg$ down and the fluid that exerts buoyant force $B$ up. Since the acceleration of the cube is zero, the force exerted by the Earth must be equal to the net force $B$ exerted by the fluid. We conclude that the net force exerted by the fluid on volume $V$ is $B=\rho_{H_2O}Vg$ up. From Newton's third law we conclude that the net force exerted by volume $V$ on the surrounding water is $B'=\rho_{H_2O}Vg$ down.

Now imagine replacing the cube of water with a solid cube of unknown material having density $\rho_?$ and the same volume $V$. The surrounding water and the Earth are still the only entities exerting forces on the cube. The force exerted by the Earth on the cube is $mg=\rho_{?}Vg$ down. We have seen above that volume $V$ exerts force $B'=\rho_{H_2O}Vg$ down on the surrounding water which means that the surrounding water exerts force $B=\rho_{H_2O}Vg$ up on the lead of volume $V$. Thus the net force on the lead cube is $$F_{net}=-\rho_{?}Vg+\rho_{H_2O}Vg=\left(\rho_{H_2O}-\rho_?\right)Vg.$$This equation says that

• If $\rho_?$ is greater than $\rho_{H_2O}$, the net force will be negative and the object will sink.
• If $\rho_?$ is less than $\rho_{H_2O}$, the net force will be positive and the object will float.

It is the same equation that you arrived at except its derivation from the "first principles" of Newton's second and third laws is made transparent.

 

[ChiralSuperfields]

Pretty good. A couple of quibbles…

1. You don't consider what happens when a rising body reaches the surface. The problem statement says "and floats". Note that if there were no losses then, having surfaced, a body less dense than the fluid medium would rise so far that it would then descend again, bobbing up and down between the same two extremes for ever!

2. It is debatable whether Archimedes' principle constitutes a "first principle"; certainly not in the same way that Newton's laws of forces, accelerations and gravity are. Indeed, Archimedes' principle as generally stated omits a caveat: that the fluid can reach the whole surface of the body below the fluid level. Consider a suction pad stuck to the bottom of the tank.
You can avoid that difficulty by following the line of reasoning which, I have always assumed, Archimedes followed. Consider replacing the body with exactly that fluid which would fit in the portion of the body volume that is below the fluid level. What forces act on it? Would it be in equilibrium? What must be the buoyant force on it from the rest of the fluid?
Note what this line of reasoning leads to in the suction pad case.

I was more interested in seeing what you consider a "first principle", something that @haruspex tried to get you to enunciate in post #2. Usually, the first principle comes first in a derivation and $\vec {F_{net}} = B\hat j - mg\hat j$ is not it.

Here, first principles are Newton's laws. Consider the second law, $\vec F_{net}=m\vec a$ and draw a FBD for a cube of volume $V$ of water floating in water. This cube's acceleration is zero, therefore the net force on it is zero. There are two entities exerting forces on the cube, the Earth that exerts force $mg=\rho_{H_2O}Vg$ down and the fluid that exerts buoyant force $B$ up. Since the acceleration of the cube is zero, the force exerted by the Earth must be equal to the net force $B$ exerted by the fluid. We conclude that the net force exerted by the fluid on volume $V$ is $B=\rho_{H_2O}Vg$ up. From Newton's third law we conclude that the net force exerted by volume $V$ on the surrounding water is $B'=\rho_{H_2O}Vg$ down.

Now imagine replacing the cube of water with a solid cube of unknown material having density $\rho_?$ and the same volume $V$. The surrounding water and the Earth are still the only entities exerting forces on the cube. The force exerted by the Earth on the cube is $mg=\rho_{?}Vg$ down. We have seen above that volume $V$ exerts force $B'=\rho_{H_2O}Vg$ down on the surrounding water which means that the surrounding water exerts force $B=\rho_{H_2O}Vg$ up on the lead of volume $V$. Thus the net force on the lead cube is $$F_{net}=-\rho_{?}Vg+\rho_{H_2O}Vg=\left(\rho_{H_2O}-\rho_?\right)Vg.$$This equation says that

• If $\rho_?$ is greater than $\rho_{H_2O}$, the net force will be negative and the object will sink.
• If $\rho_?$ is less than $\rho_{H_2O}$, the net force will be positive and the object will float.

It is the same equation that you arrived at except its derivation from the "first principles" of Newton's second and third laws is made transparent.

Thank you for your replies @haruspex and @kuruman ! That is very helpful and I will look into it more.

Many thanks!