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This is a theoretical question.

Assume I have a $1000 \;\text{kg}$ box and just 1 molecule (say Oxygen) in a 100-meter-tall hollow vertical cylinder. The cylinder walls are much wider than the box and the box does not touch the walls of the cylinder. The box falls from the top of the cylinder under gravity.

  1. Does the box exert $10000 \;\text{N}$ force on the $\text{O}_2$ molecule?
  2. Does the $\text{O}_2$ molecule exert $10000 \;\text{N}$ force on the box?

Edit: to clarify, the molecule is stationary, situated midway vertically (ignore any effect of gravity on the molecule itself). The box falls from the top of the tube (under earth's gravity) and hits the molecule when it reaches halfway down. There is vacuum in the chamber - only the falling box and the molecule exist.

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  • $\begingroup$ Where is the molecule in the cylinder when the box is released from the top of the cylinder? By "size" difference due you mean difference in mass? Finally, when you say "hollow" do you mean there is a vacuum in the cylinder? $\endgroup$
    – Bob D
    Jun 18, 2021 at 15:06
  • $\begingroup$ I do not understand, where the molecule and where the box is. When the box falls, it is weightless, so when it is falling there will be no force between box and oxygen. If it stands, it would exert $10.000 \mathrm{N}$ to a ground, but it cannot stand on a single molecule and you say the box does not touch the cylinder (so I assume it does not touch the cylinder bottom). So to what exact setup and moment does your question about the third law refer to? $\endgroup$
    – Koschi
    Jun 18, 2021 at 15:15
  • $\begingroup$ Newton's third law applies regardless of the "size" (which I think you mean mass) of the object. However, it is not clear to me how the molecule and box interact. Assuming by "hollow" you mean there is a vacuum in the cylinder, then they can't interact if each is somewhere between the top and bottom of the cylinder (not touching the cylinder) since both will have the identical acceleration downward of $g$. $\endgroup$
    – Bob D
    Jun 18, 2021 at 15:21

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Yes. The three laws are fundamental to the analysis of Statics and Dynamics problems, regardless of size.

Let's say the molecule is in contact with the box at rest, before being dropped. When dropped, if the box were to accelerate the molecule and maintain the accumulated speed as the box falls, what force do you think is being applied to the molecule?

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Does the box exert 10000 N force on the O$_2$ molecule?

No. Anything that did exert such a force would be sufficient to hold the acceleration at zero (I'm assuming we're doing this experiment near the earth's surface). The single molecule will not exert such a force and the box will accelerate downward.

Does the O$_2$ molecule exert 10000 N force on the box?

No. The force each exerts on the other is identical in magnitude.

Just because the box can exert a particular force on the ground doesn't mean it can exert that force on some other object. One way to think about it is to imagine throwing a basketball. You can probably exert 100N on that box with your arm. Now try to exert 100N of force on a feather. You won't be able to do it. The interaction with the low-mass feather causes acceleration rather than large forces to develop.

The same happens with the box and the molecule. If they interact, most of the result will be the molecule accelerating away, not large forces or the box accelerating.

The forces between the two objects will be but small and identical in magnitude

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Earth attracts both the box and the molecule. By Newton's Third Law, both the box and the molecule attract the Earth with the same magnitude of force. Similarly the box attracts the molecule and the molecule attracts the box with the same magnitude of force.

However, the magnitude of acceleration is dependent on the mass of the body concerned. For example, in a two body system, the body with higher mass will have smaller acceleration.

The forces in the question are incorrect. Actual forces can be found using Newton's Law of Universal Gravitation.

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    $\begingroup$ @shaihorowitz Don't get jumpy. I never said acceleration due to gravity. $\endgroup$
    – Alpha Delta
    Jun 18, 2021 at 16:44
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We generally prefer to have simpler equations. Due to this reason we make certain approximations in physics. The value of $G$ is too small, so while dealing with bodies with small mass(when I say small I mean masses that are not comparable to the mass of celestial bodies) gravitational force of attraction and other forces (like electric force, etc) are neglected.

The force exerted by box on molecule = force exerted by molecule on box = $\frac{Gm_{box}m_{molecule}}{r^2}$. The net acceleration of box turns out to be $g−\frac{Gm_{molecule}}{r^2}$, since $$m_{molecule}<<1$$ hence $a_{box}≈g$.... The result that we use is a simplified one.

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The force between the box and the molecule will be tiny, not 10000N, although it will be reciprocal so Newton's third law will be obeyed.

You have made the example rather self-contradictory, since you have said that the box is falling as a result of gravity, yet we should ignore gravity as far as the molecule is concerned. However, for the sake of illustrating the principles involved, let us assume that gravity does not act on the molecule...

To make the scenario less ambiguous again, let us suppose the box is neutrally charged and composed of an inert substance that does not interact chemically with the O2 molecule, and that the molecule is motionless relative to the cylinder.

Given that, the approaching box will impart an electromagnetic force that will accelerate the molecule. The force will be electromagnetic- some type of Van Der Waals force, the exact form of which will depend on the shape of the surface of the box in the local area that comes closest to contact with the molecule.

The effect will be to accelerate the molecule to keep it at some equilibrium distance away from the box as the box falls. On average, therefore, the force exerted by the box on the molecule will be mg, where m is the mass of the molecule.

The principle would be clearer if the box were falling with a weightless string attached that went over a frictionless pulley and along a frictionless horizontal surface upon which the string lassoed the O2 molecule. As the box fell, it would drag the molecule along the horizontal surface. The molecule would move in pace with the box, accelerating at 1g.

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