# My First Post: A Contradiction in Simple Free Fall?

Hello,

Let us imagine a solid in a free fall on planet Earth and let us neglect the effect of air resistance on the motion of the solid. According to Newton's Second Law of Motion, since the only force being applied on the solid is its Weight, then its acceleration must be its Weight divided by its Mass, that being approximately 9.81 m/s/s downwards. Logically, all the particles that constitute this solid must have the same magnitude of acceleration: 9.81 m/s/s downwards.

However, if we study any of those particles individually and apply Newton's Second Law of Motion, we would end up with a different result. The forces acting on a particle are its Weight and the intermolecular forces exerted by the other particles of the solid from different directions. No matter what the magnitudes of these intermolecular forces are, this particle must acquire an acceleration different from 9.81 m/s/s downwards.

How can this be?

Weam Abou Hamdan
Sunday, July 8, 2018

• russ_watters

russ_watters
Mentor
I'm not quite following what you are saying, but intermolecular forces are inside the object, so it is unrelated to the bulk movement of the object.

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I'm not quite following what you are saying, but "inter..." of course means "internal", so it is unrelated to the bulk movement of the object.
Indeed, internal forces do not affect the net acceleration of the solid as a whole. However, in the point of view of an individual particle, such internal forces are external, meaning they affect the net acceleration of the particle itself. What I am saying is that this is strange since the particles of a solid and the solid itself should logically acquire the same acceleration. The calculation does not show that.

Bystander
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I'm not sure how this is related to my post.

russ_watters
Mentor
Indeed, internal forces do not affect the net acceleration of the solid as a whole. However, in the point of view of an individual particle, such internal forces are external, meaning they affect the net acceleration of the particle itself.
Maybe try drawing a free body diagram and labeling the forces. Then maybe you will see the net of the internal forces on an object is zero. It must be, since the particles aren't moving (much) with respect to each other.

Maybe try drawing a free body diagram and labeling the forces. Then maybe you will see the net of the internal forces on an object is zero. It must be, since the particles aren't moving (much) with respect to each other.
When an object is falling, it acquires an acceleration of 9.81 m/s/s downwards since the only force exerted on the object is its Weight. Air resistance is neglected. The particles that make up this object are falling as well with an acceleration of 9.81 m/s/s. However, an individual particle experiences both Weight and other intermolecular forces. We take into consideration these intermolecular forces when studying individual particles since they are external forces in this case. On the other hand, we do not take these intermolecular forces into consideration when we are studying the object as a whole since they are internal forces in this case.

Nugatory
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The total force on each molecule is by definition the sum all the forces acting on it, which is to say the sum of the gravitational force on that molecule all the intermolecular forces. If the object is in free fall then the intermolecular forces on each molecule will add to zero, so each individual molecule is subject to a gravitational force that accelerates it downwards at 9.81 m/sec. When this happens to all the molecules at once the object as a whole accelerates downwards at that rate.

• russ_watters
The total force on each molecule is by definition the sum all the forces acting on it, which is to say the sum of the gravitational force on that molecule all the intermolecular forces. If the object is in free fall then the intermolecular forces on each molecule will add to zero, so each individual molecule is subject to a gravitational force that accelerates it downwards at 9.81 m/sec. When this happens to all the molecules at once the object as a whole accelerates downwards at that rate.
Why would the sum of all the intermolecular forces exerted on a particle be zero when an object is in free fall? For example, if we take a particle on the highest point of the solid and study it when the object is at rest, we will find that the sum of the intermolecular forces on that particle would cancel the particle's Weight, meaning the sum of the intermolecular forces on that particle cannot be zero. Why would that change when an object is falling?

Dale
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No matter what the magnitudes of these intermolecular forces are, this particle must acquire an acceleration different from 9.81 m/s/s downwards.
Unless the intermolecular forces sum to 0. Which they do.

Why would that change when an object is falling?
When it is falling the internal stress and strain go to zero.

• Bystander
Unless the intermolecular forces sum to 0. Which they do.
The sum of the intermolecular forces on a particle cannot be zero. For example, imagine a ball hanging from a spring. A particle on the lowest point of the ball is at equilibrium since it is not accelerating in any way. This particle experiences Weight and the intermolecular forces (it does not experience any force from the spring since it is too from the point of contact). If the sum of the intermolecular force was truly zero, the particle would simply fall. It does not.

• PeroK
https://en.wikipedia.org/wiki/Galileo's_Leaning_Tower_of_Pisa_experiment
I think Galileo figured it out pretty well. Look at the thought experiment he carried out.
His thought experiment involving the light object and relatively heavy objected connected by a string only demonstrates the idea that objects of different masses fall with the same acceleration. My argument in this thread is different, however.

Dale
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The sum of the intermolecular forces on a particle cannot be zero.
Sure it can. And in the case of free fall (rigid non rotating motion) it is zero.

For example, imagine a ball hanging from a spring. A particle on the lowest point of the ball is at equilibrium since it is not accelerating in any way. This particle experiences Weight and the intermolecular forces (it does not experience any force from the spring since it is too from the point of contact). If the sum of the intermolecular force was truly zero, the particle would simply fall. It does not.
Correct. Such a particle is not in free fall. There is a measurably non-zero stress and strain in the ball, and thus a non-zero intermolecular force in that case.

In free fall the stress and strain go to zero as does the intermolecular force.

• nasu
There is a measurably non-zero stress and strain in the ball, and thus a non-zero intermolecular force in that case.

In free fall the stress and strain go to zero as does the intermolecular force.

Is it possible you can give me an explanation in terms of mathematics? What exactly do you mean by stress and strain?

Nugatory
Mentor
Why would the sum of all the intermolecular forces exerted on a particle be zero when an object is in free fall? For example, if we take a particle on the highest point of the solid and study it when the object is at rest, we will find that the sum of the intermolecular forces on that particle would cancel the particle's Weight, meaning the sum of the intermolecular forces on that particle cannot be zero. Why would that change when an object is falling?
The intermolecular forces are different in the two situations.

Consider a molecule at the top of the object. Gravity is pulling it down, but when the object sitting at rest on the ground the molecules below it cannot fall because the molecules below it are in the way. Thus the object compresses very slightly as it is squeezed between the weight of the molecules at the top and the immobile ground below; this slight compression increases the intermolecular force on the top molecule just enough to cancel the gravitational force on it, leaving a net force and acceleration of zero.

When the object is in free fall, there's nothing pressing upwards from below, so the object doesn't compress and the intermolecular force on the topmost molecule is slightly less.

This situation is easier to understand if you start with a simpler problem: two weights, one above the other with a coil spring in between. Compare the tension in the spring and the net force on the the top weight when the apparatus is sitting on the ground and when it is in free fall. The intermolecular forces in a sold actually behave very much like an ideal spring; they resist more as the molecules are forced together or pulled apart.

• nasu, PeroK and Bystander
Thus the object compresses very slightly as it is squeezed between the weight of the molecules at the top and the immobile ground below; this slight compression increases the intermolecular force on the top molecule just enough to cancel the gravitational force on it, leaving a net force and acceleration of zero.
Is it possible you can give me an explanation in terms of mathematics?

Bystander
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Your profile states "age sixteen:" what level "in terms of mathematics" do you require?

His thought experiment involving the light object and relatively heavy objected connected by a string only demonstrates the idea that objects of different masses fall with the same acceleration. My argument in this thread is different, however.
The string is good analogy of an inter molecular force. And you can apply Galileos reasoning to real inter molecular forces for example those between the particles of the string. On a very small scale you could consider a falling diatomic molecule for example H2.

Your profile states "age sixteen:" what level "in terms of mathematics" do you require?
I wouldn't mind the mathematics despite its difficulty. I'll manage. All I need is mathematical proof.

The string is good analogy of an inter molecular force. And you can apply Galileos reasoning to real inter molecular forces for example those between the particles of the string. On a very small scale you could consider a falling diatomic molecule for example H2.
Why would the force exerted by the string in Galileo's experiment disappear during the free fall?

PeroK
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Is it possible you can give me an explanation in terms of mathematics? What exactly do you mean by stress and strain?

You may be interested in a video on YouTube by Veritasium: search for dropping a slinky in slow motion.

• Merlin3189 and Weam Abou Hamdan
You may be interested in a video on YouTube by Veritasium: search for dropping a slinky in slow motion.
It really is interesting. Thank you for the suggestion!

Dale
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What exactly do you mean by stress and strain?
Stress is the measure of the internal forces within a material, this is the intermolecular forces you are referring to. Strain is the amount that a material deforms in response to stress. These quantities are measurable, so we can measure when they are zero and when they are not.

Is it possible you can give me an explanation in terms of mathematics?
Sure. This is all based on Hooke’s law which states that for elastic materials the stress is proportional to the strain. So the force is given by f=-kx where k is the spring stiffness and x is the deformation from equilibrium.

When it is holding up a ball of mass m at rest then the force must be equal to mg and therefore the deformation is -mg/k. When it is in free fall then the deformation is 0 so the force is 0.

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CWatters
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If you consider a mass made of just two molecules, one above the other, then the inter molecular forces on each molecule are the same but act in opposite directions. So they sum to zero.

PeroK
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It really is interesting. Thank you for the suggestion!

I think the motion of the slinky answers your question. If you think of the slinky as a number of coils. Each coil has gravity and the weight of all the coils below pulling it down and the coil above pulling it up. Except the top coil which has an external force pulling it up.

When you release the slinky the internal forces between the coils stay the same and only the top coil falls. But, as it falls it releases the tension between it and the second coil, which then has no force holding it up and so on.

In a way all dropped objects act like this. But for a more rigid object the internal forces adjust much more quickly so you don't see any change in shape.

In fact, the idealised definition of a rigid body is that the internal forces are transmitted and adjust to the external force instantaneously throughout the body.

The slinky is in a way the opposite of a rigid body.

• Likith D and Weam Abou Hamdan
Hello,

Let us imagine a solid in a free fall on planet Earth and let us neglect the effect of air resistance on the motion of the solid. According to Newton's Second Law of Motion, since the only force being applied on the solid is its Weight, then its acceleration must be its Weight divided by its Mass, that being approximately 9.81 m/s/s downwards. Logically, all the particles that constitute this solid must have the same magnitude of acceleration: 9.81 m/s/s downwards.

However, if we study any of those particles individually and apply Newton's Second Law of Motion, we would end up with a different result. The forces acting on a particle are its Weight and the intermolecular forces exerted by the other particles of the solid from different directions. No matter what the magnitudes of these intermolecular forces are, this particle must acquire an acceleration different from 9.81 m/s/s downwards.

How can this be?

Weam Abou Hamdan
Sunday, July 8, 2018

Hey there. Your logic is valid, but you stopped one step shorter from the answer. Whathever internal forces exist in the solid, the net total (vector sum) is zero. The body behaves as a single total mass located at the body’s center of mass. You may check any classical dynamics textbook for a chapter on systems of particles. Any of them will show this behavior.

For short, any internal force doesn’t accelerate the particles because they are balanced by an equal and opposing force (Newton’s 3rd law) inside the object itself. When you apply Newton’s laws, you must apply all of them, not only the second one.

Thence, the net force inside the object when in free fall is zero not because of the nature of the external forces, but because of it being a rigid body.

If you consider a body which is not perfectly rigid (i.e., one in which its particles may move in respect to one another, under some sort of restoring force), this won’t apply. For whatever dislocation of a particle of the body that takes it away from its equilibrium position, internal forces will appear, so the object may (or may not) be restored to its equilibrium. This will change the dynamics of motion, though also not in the way you described it.

The intermolecular forces cancel each other out since there are other particles around the one you are focusing on. Thus their vector sum is zero and the only unbalanced force acting is the weight

Classically, label the particles in the object 1,2,3...N, so there are N particles. The force on particle i is mi g + Fij where mi is the mass of the particle, g is the acceleration due to gravity, and Fij is the intermolecular force of the j-th particle on the i-th particle. The total force on the object is ∑i mi g + ∑ijFij. If we suppose that the object is not rotating, and if it were not subject to gravity (i.e. motionless or in free fall), and we ignore the thermal motion of the particles, then the total force on a particle would be zero. ∑jFij = 0 so ∑ijFij=0 and the total force on the object is just ∑i mi g = m g where m=∑i mi.