Balloon in Sealed Box: Does a Suspended Rock Affect Weight? | Forum Discussion

[carbonick]

Hello all. I'm new to the forum, but I have a feeling I'll be hanging around a lot.
I have a question for all of you. If you have a rock suspended by a helium balloon inside a sealed box (the balloon is not touching the top of the box), would the box weigh the same as it would if the rock was just sitting in the box?
I think it would, because the air in the box has to hold up the balloon/rock, but I'd like some way to prove this. Is there an equation or Free Body Diagram that would illustrate this? Would the weight only stay the same if the balloon was inflated in the sealed box (it's a big box)? Could someone help me understand this?

Everyone says I'm an idiot for thinking the weight would be the same, and maybe I am. Anyone care to prove it?

 

[Q_Goest]

You're correct in assuming the box would weigh more. The most simple way of looking at this is to draw your control volume around the box and it's obvious the box plus all the contents have some mass which then will weigh something. This is true regardless of if the helium balloon is pushing on the ceiling of the box, or the rock is resting on the floor. Everything inside the control volume contributes to the mass of the CV an equal amount.

To look at it from the perspective of the balloon inside the box is a bit more difficult. There's a pressure gradient in the air between the top and bottom of the balloon. The air pressure on the bottom of the balloon is greater than the air pressure on the top of the balloon, so the net force upward is the air pressure pushing up times the area it operates on minus the air pressure pushing down times the area it operates on. Since the balloon is actually spherical, you'd need to integrate over the surface, but that only confuses the issue.

By putting the balloon inside the box, you increase the density of the gas that's already inside by the amount equal to the additional volume displaced by what you put inside. This creates a higher pressure of air pushing on the bottom of the box, along with a higher pressure of air pushing on the top of the box. The air pressure pushing down on the floor of the box minus the air pressure pushing up on the ceiling of the box, times the area it is pushing against will equal an additional force, which will be equal to the force needed to suspend the balloon given in your example.

 

[carbonick]

Thanks.

Does anyone else want to chime in on this? This should be simple for most of you, but more info would really help me out.

 

[Nenad]

is the box is sealed, the rock and the baloon are an internal part of the box, meaning that they contribute to the weight and centre of mass of the box. It doesn't matter what the baloon or rock do inside of the box, it is internal change and it will not change the weight or mass of the box.

Regards,

Nenad

 

[Q_Goest]

Just a thought for clarification: you mentioned a "sealed box" which to me means there is the same mass of air in the box before and after you put the balloon and rock in. If the box is open, then when you put the balloon/rock in, air will have to be displaced. In that case the box weighs the same since the mass of air displaced is equal to the mass of the balloon/rock.

 

[Davorak]

Just a thought for clarification: you mentioned a "sealed box" which to me means there is the same mass of air in the box before and after you put the balloon and rock in. If the box is open, then when you put the balloon/rock in, air will have to be displaced. In that case the box weighs the same since the mass of air displaced is equal to the mass of the balloon/rock.

Q_Goest, the box will still increase in weight when the box is opened and then resealed. The surrounding air may also provided an upward force from buoyancy but this is a separate issue.

The force which supports the balloon is ultimately tided to the box since the box is air tight. The balloon on a microscopic level is supported by the air molecules and the air molecules are supported by the box.

Take the following situation:

The box is hermetically sealed with a balloon supporting a rock trapped inside. Now this box will have a very small pressure gradient from the bottom to the top do to the influence of gravity.

The balloon rock system has been set so that in equilibrium the balloon rests in the middle and the bottom.

If the box is lifted very quickly the balloon and rock will stay relatively still and hence will be closer to the bottom of the box.

The balloon is now out of equilibrium and starts moving to the middle of the box again.(It still moves to the center of the box since the force of gravity is the same and hence the pressure gradient.) A force is required to move the balloon in the upward direction. Now:
<br /> \int \vec{F}\cdot d\vec{l}<br />
The force points in the same direction of the path vector hence work is being performed. This work must be performed by what ever is holding the box up.

What ever provides the work to move the balloon is also required to maintain a force to counteract the weight of the balloon.

Make sense?

I have also heard of a similar problem involving a Cargo plane and pigeons.

 

[Q_Goest]

Davorak,
Everything you said is perfectly correct. But note there are two possible cases, 1) the box has the same mass of air in it before and after and 2) the balloon/rock displace a mass of air equal to the mass of the balloon/rock, thus the mass of air is less after the balloon/rock is put in (by an amount equal to the mass of the balloon/rock.

Here's case 1. The box is sealed, and the balloon is then added such that the mass of air in the box does not change. In this case, the pressure inside the box will rise after the balloon is added, since this balloon takes up space. In this case the box will have more mass and will weigh more.

Here's case 2. The box is at atmospheric pressure before anything is added, and it is still at atmospheric pressure after the balloon and rock is put inside. In this case, I'm claiming the box will weigh the same. Per the original post, the balloon/rock is floating which means the mass of air displaced by the balloon/rock will be equal to the mass of air they displace. I believe you're suggesting the box will weigh more. Is that really what you're suggesting?

Regarding pigeons inside an aircraft, you're correct that regardless of if they are flying or sitting on a perch inside the airplane, the aircraft will weigh more. But in this case, the pigeons are pushing air down to suspend themselves. If you were underneath the birds, you'd feel a flow of air down on you. In the case of the balloon/rock, this system is "floating" and standing beneath you'd not feel any air blowing on you. A flying bird is different from a floating balloon.

 

[Davorak]

If the bottom of the box is magnetic and there is a big heavy magnetic inside the box floating without touching the sides of the box will the box weigh more. My answer is yes.

In this case the force(weight) of the magnetic is transferred to the box through the magnetic field.

In the case of the balloon it is transferred to through air.(Molecular collisions).

So yes I am staying the box will weigh more. The point of my example however is to suggest that the balloon acts like a milk in a milk carton will. When you move the milk carton the milk will move but delayed since it is not directly attacked to the sides of the carton.

If the box is weighed in a vacuum the scale should show all of the mass present in the box whether it is floating or not.

 

[Q_Goest]

Question: What has more mass, the floating balloon/rock, or the air it displaces?
Answer: Neither, the mass is the same. If you disagree, please state why.

Question: For an open container, what happens to the mass of air when a balloon/rock is placed inside?
Answer: It leaves the box. (Note that for a 'sealed' container, we are assuming the air can not leave the box when the balloon/rock is added.)

Question: If the mass of air that leaves the open container is equal to the mass of the balloon/rock, does the container have more mass inside before or after the balloon/rock is added?

 

[Davorak]

Right well I was stupid. Your right I did not think things through.

Then it just matters if you put the balloon at the top of the box, the bottom, or let it come to equilibrium before closing the box, but this will change the answer by very little.

As you said:
Mass of balloon/rock = mass of displaced air.