According to this law, a 1 ton block of metal falling in air will exert a force of 10N on every single molecule in the air.

Let's simplify and take 1 single such molecule.

The 1 ton block exerts 10N on the molecule (action)

The molecule exerts 10N on the 1 ton block (reaction)

These 10N forces act on different bodies. I understand that.

However, we are not considering the molecule. We are considering the block which has an upward force on it (10N reaction from molecule), same as the downward force on it (10N due to gravity).

Why does the block move/fall ? It should stay still in air, because the 2 forces on it (upwards from air molecule, downwards from gravity) cancel each other out.

  • $\begingroup$ For how long does the block feel that upward force? And even if the up and down forces somehow cancelled permanently, why wouldn't the block continue to fall? Doesn't a body in motion stay in motion, absent some net force? $\endgroup$
    – WillO
    Commented Aug 28, 2022 at 17:53
  • $\begingroup$ Why do you say that the block will exert 10 N on every single molecule in the air? $\endgroup$
    – Sandejo
    Commented Aug 28, 2022 at 18:00
  • $\begingroup$ I do not follow where the 10 N comes from? The gravitational pull (the weight) in the block is around 10 000 N. Would you mind clarifying where the 10 N comes from, or is it just a typo? $\endgroup$
    – Steeven
    Commented Aug 28, 2022 at 20:00
  • $\begingroup$ javadesigner you should be able to comment here, feel free to ask for clarification or details. $\endgroup$
    – Kuhlambo
    Commented Aug 29, 2022 at 9:05
  • $\begingroup$ Hi - sorry, the 10N should have been 10KN. Was a typo. $\endgroup$ Commented Aug 31, 2022 at 2:47

3 Answers 3


Three main considerations here:

  1. the block interacts with a huge number of particles. The resultant on the block is teh sum of every single "small" interaction with the air molecules. Only a singular small interaction acts on each single molecule;

  2. during the initial fall of the block, there is no reason why drag should be equilibrated by gravity, since they are two independent types of forces. During this transient, the body accelerate downwards with acceleration $a = g - \frac{1}{2} \dfrac{\rho v^2 S c_D}{m}$;

  3. This is the main flaw: no net force means no acceleration, doesn't mean no velocity! a After the transient, equilibrium is reached at the velocity for which the drag equilibrates the weight of the block, i.e. acceleration is zero, $a = 0$ (and not velocity!), when $g - \frac{1}{2} \dfrac{\rho v_{e}^2 S c_D}{m} = 0$ i.e.

    $v_e = \sqrt{ \dfrac{2 m g}{\rho S c_D}}$.


While the block is falling, its acceleration is due to the gravitational force provided by the Earth, which is, in modulus, $F = mg = 9.81 \cdot 10^3 \text{N}$, about $10\text{kN}$. Newton's 3rd law applied to this force tells you that the block is pulling the Earth "up" with the very same force.

I assume that when you say

"10 N due to gravity"

you actually mean $10\text{kN}$.

This is not the force acting between the block and any of the molecules of the surrounding air.

During fall (but also at rest, for what matters) molecules collide with the block: during collision, the block will exert a force on each molecule which is identical, and opposite in direction, to the force that each molecule exerts on the block. However, these forces have nothing to do with the gravitational force acting on the block: it is another force, ultimately due to the electrostatic interaction between the molecule and the block during collision, which is quite complex.

Considering a single molecule, you might in principle calculate how large is this force by considering the initial (just before collision) momentum and kinetic energy of the block and of the molecule and by applying the conservation laws for these two quantities (to be fair, energy won't be exactly conserved). The point is that, once you have the final momenta, you could compute the force, but you would need to know how long the collision lasted - and you should assume that the force is constant!

Finally, as other answers already pointed out, when you put together many molecules, the forces due to collisions in one direction are on average balanced out by those in the opposite direction if the block is at rest. If it is moving, collision "from below" will overcome those "from above", and you get a very naive but intuitive model of viscous drag.

  • $\begingroup$ Alessio Perinelli- thanks for the explanation! $\endgroup$ Commented Aug 31, 2022 at 2:51

I think you may be confusing a motion carried by momentum, and acted upon by no force (Newton's 1st law, the principle of inertia) and falling, which is an accelerated motion, accelerated by the force of gravity, and is something very different.

If there where no gravity and the block was just coasting because inertia is keeping its momentum constant, you could be right to demand that the bock stops after pushing away the molecule, like a billiard ball that has transferred all is momentum to another ball. But we would have to assume the full momentum is transfer, which for a one ton block and a molecule is a bad assumption, because a tiny part of the blocks momentum is enough to accelerate any molecule out of the way.

But there is gravity, so the block constantly is feeling a force, it's being accelerated, it's momentum is growing (assuming we are far from terminal velocity).

Even if it pushes away a molecule and feels a counter force for one nano second, it's still being pulled down by gravity constantly, allowing it to push away all the air molecules until it finally hits the ground.

  • $\begingroup$ Moment of inertia ( en.wikipedia.org/wiki/Moment_of_inertia ) is a different thing and has no direct relation with Newton's 1st law. $\endgroup$ Commented Aug 28, 2022 at 22:21
  • $\begingroup$ @GiorgioP thanks for the hint. Newton's First law is also known as the principle of inertia, but I think i confused some English and German vocabulary, or something, who knows. I hope it's better now. $\endgroup$
    – Kuhlambo
    Commented Aug 29, 2022 at 7:06
  • $\begingroup$ @Kuhlambo - thanks for the explanation - very helpful! $\endgroup$ Commented Aug 31, 2022 at 2:51

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