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Force of Falling Bodies

  1. Jan 3, 2009 #1
    I posed this question to my teacher: I have a 2 kg object and it sits at rest on the floor. The force is mass * acceleration, so it's about 20 N. Now I drop the ball from, say, 20 feet above the ground. Same mass, same acceleration, so it has the same force, 20 N; but we know the object hits much harder. If it has the same force, why does it hit harder? Am I making the assumption that force and 'how hard it hits' are equivalent statements when they really aren't? She told me it had to do with a thing called 'momentum', and we would go over it next semester (I don't want to wait that long for an explanation!). If someone could set me straight on this, I'd really appreciate it. Thanks.
  2. jcsd
  3. Jan 3, 2009 #2

    Doc Al

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    The force you are talking about is the object's weight, the earth's gravitational pull on it. Why should that force be directly related to the contact force between the object and the ground upon collision? The object's weight acts even while the object is in mid-air and not even touching the floor.

    The contact force (not the object's weight) and "how hard it hits" are equivalent statements. And that contact force depends on the momentum of the object and also the nature of the colliding bodies. (Dropping a 2 kg steel ball onto concrete will give you a different impact force than dropping a 2 kg pillow onto a mattress.) The more momentum the object has and the quicker the object is stopped, the greater the impact force.
  4. Jan 3, 2009 #3
    Pupil...a good question, not easy to answer in a one liner...These "simple" questions often raise profound underlying questions...and require sound understanding to fully appreciate.

    There is a fundamental difference between the stationary and the falling ball in your example even though both feel a force mg: With a stationary ball, energy remains constant over time; but with the falling ball, the force of acceleration (mg) steadily increases the velocity and hence kinetic energy of the falling ball and when stopped this kinetic energy must be dissipated (otherwise the ball keeps moving). Work must be done to stop the ball.

    In terms of momentum: (from the above post)
    This quote becomes : FT= MV, a basic relationship you'll likely come across next semester. To stop something with momentum MV, you need to apply a force (F) over a time (T). For a given MV, you'll need a huge force F to stop the ball in a small time T...as noted above, if the fall were instead to a mattress (where T is extended), a smaller F is needed to more gradually slow the ball over a greater T.
  5. Jan 3, 2009 #4
    I am going to try and answer this question a la Einstein to see if I have understood the theory correctly and pose another related question.

    As the object falls it enters slower time 'zones' as the force of gravity increases closer to the earths surface and as the time slows the object sheds rest-energy (which is the same energy the object has when it sits on the floor), due to it entering slower time zones, it must gain energy of movement in order to maintain it's total energy (energy principal). When the object hits the ground it sheds it's energy of movement and re-gains it's total rest energy (?energy of movement becomes rest energy?). The more energy of movement the object has when it hits the ground the ‘harder it hits’.

    If the object gains movement energy as it enters slower time zones what is how does it begin entering slower time zones without energy of movement?
  6. Jan 3, 2009 #5

    Vanadium 50

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    Doc Al and Naty1 gave good replies.

    Electerr, I don't think that's the right way to go about this. While in principle, I suppose one could do the full GR calculation, it would just be the long way around. (One can go from Los Angeles to Pasadena via New York too...) Now, trying to do this a la GR without actually using the equations of GR is like trying to do the above in the dark without a map.

    Best to stick with Newton. There's a reason we teach it.
  7. Jan 3, 2009 #6
    Sorry... maybe I shouldn't post sub-topics on other peoples topics. I know Newton is easier in this case, I am just trying to see if I understand Einstein well enough to answer the question in another way and to get an answer about part of the theory I don't understand.
  8. Jan 3, 2009 #7
    I appreciate the feedback, guys!

    At the moment of impact, wouldn't all the forces sum up to equal ma? (excluding air resistance) At the moment of impact I have Fg and Fc (contact) working on the object, so Fg + Fc = ma. So if you know the acceleration of the object (how quickly it stops, that is), you can solve for Fc, and it will be some larger number? Or am I way off here?

    I can see your explanation if you use the concept of 'energy'. Perhaps I should stop asking these questions until we learn about energy and momentum?

    On a side note, did you get the equation FT = MV from F = MA, but replacing A by (V/T) and moving the T to the other side?

    I really appreciate the help guys; hope you can bear with me a little longer until I understand. Thanks.
  9. Jan 3, 2009 #8

    Doc Al

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    You are right on target.

    It doesn't hurt to ask. (It doesn't hurt to peek ahead in your book, either! :wink:)

    That's a perfectly fine way to derive what is known as the impulse-momentum theorem: FΔt = Δ(mV) Over a given time interval, the average (net) force F equals the mass time the average acceleration: F = mA = mΔv/Δt, thus FΔt = mΔv.
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