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

  1. Jul 8, 2018 #1
    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
     
  2. jcsd
  3. Jul 8, 2018 #2

    russ_watters

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    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.
     
    Last edited: Jul 8, 2018
  4. Jul 8, 2018 #3
    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.
     
  5. Jul 8, 2018 #4

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  6. Jul 8, 2018 #5
  7. Jul 8, 2018 #6

    russ_watters

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    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.
     
  8. Jul 8, 2018 #7
    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.
     
  9. Jul 8, 2018 #8

    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.
     
  10. Jul 8, 2018 #9
    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?
     
  11. Jul 8, 2018 #10

    Dale

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    Unless the intermolecular forces sum to 0. Which they do.

    When it is falling the internal stress and strain go to zero.
     
  12. Jul 8, 2018 #11
    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.
     
  13. Jul 8, 2018 #12
  14. Jul 8, 2018 #13
    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.
     
  15. Jul 8, 2018 #14

    Dale

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    Sure it can. And in the case of free fall (rigid non rotating motion) it is zero.

    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.
     
  16. Jul 8, 2018 #15
    Is it possible you can give me an explanation in terms of mathematics? What exactly do you mean by stress and strain?
     
  17. Jul 8, 2018 #16

    Nugatory

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    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.
     
  18. Jul 8, 2018 #17
    Is it possible you can give me an explanation in terms of mathematics?
     
  19. Jul 8, 2018 #18

    Bystander

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    Your profile states "age sixteen:" what level "in terms of mathematics" do you require?
     
  20. Jul 8, 2018 #19
    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.
     
  21. Jul 8, 2018 #20
    I wouldn't mind the mathematics despite its difficulty. I'll manage. All I need is mathematical proof.
     
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