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Balloon in train

1. Homework Statement
As show in the figure below, balloon A (containing gas of density ##ρ_A##) and balloon B (containing gas of density ##ρ_B##) are each suspended by string from the ceiling of a train at rest. Balloon C (containing gas of density ##ρ_C## is attached to the train floor by string, and floats above the floor. The density of the air is ##ρ_o##, where ##ρ_C<ρ_o<ρ_B<ρ_A##. The balloon's mass, the string's mass at the movement of air inside the train are negligible.

The train begins moving with uniform accleration in a horzizontal direction (to the right in the figure). A, B, S come to rest with respect to the train and form, respectively, angles ##\theta_A;\theta_B;\theta_C## with the vertical. Choose the best represents the condition inside the train at this time.
3229684257_652254187_574_574.jpg

2. Homework Equations
F=-ma
3. The Attempt at a Solution

$$tan \theta_A=\frac{g\rho_A V-g\rho_oV}{V\rho_Aa}$$
$$tan \theta_B=\frac{g\rho_B V-g\rho_oV}{V\rho_Ba}$$
Because ##\dfrac{\rho_A-\rho_o}{\rho _A}>\dfrac{\rho_B-\rho_o}{\rho _B}##
and I choose 1. But the answer is 4. Where is my wrong?
 
Last edited:

jbriggs444

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If you are sitting in a chair in the train car, you can consider the train's forward acceleration as equivalent to a rearward pull of a gravity-like force. Together with gravity, the net effect is as if the direction of gravity had tilted.

What does this mean for the direction of buoyancy?
 
If you are sitting in a chair in the train car, you can consider the train's forward acceleration as equivalent to a rearward pull of a gravity-like force. Together with gravity, the net effect is as if the direction of gravity had tilted.

What does this mean for the direction of buoyancy?
the g' inside the train is equal ##g'=\sqrt{g^2+a^2}## but they are effected by Achimede force.
The 4 answer is right. But I dont understand why the balloon C is equilibrium.
 

CWatters

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the g' inside the train is equal g′=√g2+a2g′=g2+a2g'=\sqrt{g^2+a^2} but they are effected by Achimede force.
The 4 answer is right. But I dont understand why the balloon C is equilibrium.
The buoyancy force acts in the same direction as the net acceleration.

You think C should be moving?
 
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I think for most people it seems hard to believe that the helium balloon (or whatever gas it is using) would move in the direction that it does under acceleration of the vehicle. And probably the reason it is so counterintuitive is that everything we have seen that moves under acceleration of a vehicle moves in the opposite direction - including the direction our body tends to move, or something sitting loosely on the dashboard or on the floor.

I had wondered about this previously and had the opportunity to perform an experiment one time with a helium balloon in my car. It had a small weight attached to the string and I set it beside me as I was driving. It almost seemed a bit comical when the balloon leaned into the direction of the turn or leaned backward under braking and forward under acceleration - almost as if it was bracing itself for the normal movement that it was expecting.

An apple falls from a tree because it displaces the air that is less dense. Under acceleration, the air in the vehicle moves in the opposite direction of the acceleration because it displaces the less dense helium.
 
The buoyancy force acts in the same direction as the net acceleration.

You think C should be moving?
yes because ##\vec{P}+\vec{F}+\vec {F_a}## isnt equal ##\vec{T}##
 
I think for most people it seems hard to believe that the helium balloon (or whatever gas it is using) would move in the direction that it does under acceleration of the vehicle. And probably the reason it is so counterintuitive is that everything we have seen that moves under acceleration of a vehicle moves in the opposite direction - including the direction our body tends to move, or something sitting loosely on the dashboard or on the floor.

I had wondered about this previously and had the opportunity to perform an experiment one time with a helium balloon in my car. It had a small weight attached to the string and I set it beside me as I was driving. It almost seemed a bit comical when the balloon leaned into the direction of the turn or leaned backward under braking and forward under acceleration - almost as if it was bracing itself for the normal movement that it was expecting.

An apple falls from a tree because it displaces the air that is less dense. Under acceleration, the air in the vehicle moves in the opposite direction of the acceleration because it displaces the less dense helium.
Can you use mathematics equations to answer this question?
 

jbriggs444

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yes because ##\vec{P}+\vec{F}+\vec {F_a}## isnt equal ##\vec{T}##
Can you define ##\vec{P}##, ##\vec{F}##, ##\vec{F_a}## and ##\vec{T}## so that we know what physical inequality you are asserting? None of those variable names have been mentioned in the thread up to this point.
 
upload_2016-11-7_22-37-33.png

So Balloon C can't be equilibrium
 

jbriggs444

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As I read that drawing...

##\vec{T}## is the tension in the string pulling down and left.
##\vec{F}## is the leftward pseudo-force resulting from the use of the accelerating frame.
##\vec{F_a}## is the upward?! force arising from the pressure gradient in the air
##\vec{P}## is the downward force from gravity.

Which direction is the pressure gradient in the air?
 
Oh, I don't notice that pressure gradient in the air. But how 3 angels equal?
 
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jbriggs444

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Oh, I don't notice that pressure gradient in the air. But how 3 angels are equal?
Realize that the vector sum of the downward force of gravity and the leftward pseudo-force from the train's acceleration produce a result that is exactly like a motionless room that is tilted (and subject to slightly increased gravity).

The air is subject to the same net gravity as everything else. Its pressure gradient lines up accordingly.

Edit: I assume that you understand that "buoyancy", "the Archimedes force" and "the force arising from the pressure gradient" are all descriptions of the exact same thing.
 
Ok, now I find the gradien of air
$$ [p (x)- p (x + dx)] S = ma = a \rho_0 dx$$
So ## \Delta p = -a \rho_0 \Delta x \Rightarrow \frac{\Delta p}{\Delta x}=-a\rho_0=const##
So 3 balloon are effected by equal force
The force by gradien of C balloon is larger than the force F=ma of its
Realize that the vector sum of the downward force of gravity and the leftward pseudo-force from the train's acceleration produce a result that is exactly like a motionless room that is tilted (and subject to slightly increased gravity).
I think it wrong.
The gravity in train is:
$$g'=\sqrt{g_0^2+a^2}=\sqrt{(g-\frac{\rho_og}{\rho_i})^2+a^2}$$
So g' is denpendent with ##\frac{\rho_o}{\rho_i}##
Why 3 angles are equal?
 
Last edited:
SO I have the figure
upload_2016-11-8_1-39-51.png
 

jbriggs444

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Define your variables before you put them into formulas.
Explain what physical principle motivates the formulas that you write down.
A verbal description would make it much easier to decipher the chicken scratchings above.
 
Sorry.
Because English isn't my national language that showing my idea in science language is very difficult. But I can understand your idea. And I tried showing my idea
Following your 10#
#13. Firstly I find the gradien pressure force effect to 3 balloon And prove that C balloon doesn't move.
Secondly I show that g' (is virtual gravity acceleration) is dependent with ##\dfrac{\rho_0}{\rho_i}## (i=A,B,C) so the 3 angles can't equal.
But I don't inderstand why angle A is equal angle B.
Thanks your helping
 

CWatters

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Have you considered what happens if (instead of accelerating the container) you simply rotate the container so that the net acceleration (just due to gravity) points in a different direction...
Rotated.jpg
 
Have you considered what happens if (instead of accelerating the container) you simply rotate the container so that the net acceleration (just due to gravity) points in a different direction...
View attachment 108603
But the train moves horizontally with acceleration ##\vec{a}## and 3 balloon is effected by gravity and Achimede force
 

CWatters

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But the train moves horizontally with acceleration ⃗aa→\vec{a} and 3 balloon is effected by gravity and Achimede force
Correct.

Gravity and the horizontal acceleration add together (vector add) to make a net acceleration.

The "Archimedes force" (buoyancy force) always acts in the opposite direction to the net acceleration.

Rotated1.jpg
 

jbriggs444

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One confusion that I see in #13 may be the idea that g' is a force. It is an acceleration, not a force. The density of the object being accelerated is irrelevant.

You seem resistant to the notion that we can lump ordinary gravity and the pseudo-force resulting from the acceleration of the car into a single force and call it gravity. So let us discard that approach. Instead, we will adopt an inertial frame of reference in which the car is accelerating rightward.

Let us examine the direction of the pressure gradient in the air in the car. If the air in the car is to assume a steady state condition, this pressure gradient must be such that any small parcel of air is in equilibrium. The parcel will accelerate at the same rate as the rest of the car and its contents.

That parcel of air is subject to two forces. It is subject to a force from the pressure gradient in the air. And it is subject to a downward force from gravity.

Let us denote the downward force from gravity by ##\vec{F_g}##. This might be read as "Force of gravity".

Let us denote the force from the pressure gradient by ##\vec{F_a}##. This might be read as "Force from Archimedes".

We are given the rightward acceleration of the car and its contents as ##\vec{a}##.

Suppose that the parcel of air has mass ##m##. Newton's second law ##F=ma## can be written for the parcel as:

$$m\vec{g} + \vec{F_a} = m \vec{a}$$

Re-arranging:

$$\vec{F_a} = m\vec{a} - m\vec{g}$$

$$\frac{\vec{F_a}}{m} = \vec{a} - \vec{g}$$

Now let us expand the ##\vec{F_a}## term. If the mass of the small parcel is ##m## and its density is ##\rho_0## then its volume is ##\frac{m}{\rho_0}##. The buoyant force on the parcel is the pressure gradient (call it ##\vec{\Delta P}##) multiplied by the volume. So...

$$\frac{\vec{\Delta P}\frac{m}{\rho_0}}{m} = \frac{\vec{\Delta P}}{\rho_0} = \vec{a} - \vec{g}$$

Multiplying through by ##\rho_0##

$$\vec{\Delta P} = \rho_0 ( \vec{a} - \vec{g} )$$

All we care about is the direction. That direction is the direction of the vector sum of leftward a plus upward g.
 
I understanded, thanks for all helping
 

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