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

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

 

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.

 

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.

 

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 m_i g + F_ij where m_i is the mass of the particle, g is the acceleration due to gravity, and F_ij is the intermolecular force of the j-th particle on the i-th particle. The total force on the object is ∑_i m_i g + ∑_ijF_ij. 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. ∑_jF_ij = 0 so ∑_ijF_ij=0 and the total force on the object is just ∑_i m_i g = m g where m=∑_i m_i.