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Does the Force of gravity act at an object every second?

  1. Jun 23, 2010 #1
    If something has a mass of 1 kg and is at some height at rest, and I drop it, it will experience of force due to gravity of about 10 N. So does that mean that every second that passes by the objects feels a force 10 N pulling it downwards? So lets say that object falls for 20 seconds, would that mean the object experienced a total force pulling it downwards of 200 N, or would that object have only experienced a force of 10 N downwards no matter how much time passes by?

    I also have another question that might build upon the right answer to this one.
     
  2. jcsd
  3. Jun 23, 2010 #2

    Pengwuino

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    It will only feel a force of 10N for the entire trip. If it did have 200N, then by F=ma, either the acceleration would have had to change (but we know it's always 9.8 m/s^2 near the surface of Earth) or the mass would have to change, both of which don't change.
     
  4. Jun 23, 2010 #3

    russ_watters

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    No, force is not typically multiplied by time. 20N is 20N.

    If you have $10 in your pocket for 20 seconds, does that mean you have $200?
     
  5. Jun 23, 2010 #4
    Thanks for your answers. My other question is if some object that weighs 10 N is resting at a surface, the object put a downward force of 10 N on the surface, and the surface puts an upward force of 10 N to balance it out. So where does the upward force from the surface come from? Does it get deflected upwards by the downward force from object? Or does it come from inside the surface?
     
  6. Jun 25, 2010 #5
    Hi sulemanma2 :smile: I think that a good way to think about this is to think of a piece of wood that is laid across two blocks. If you were to stand on the center of the piece of wood, it is not so hard to imagine that it would begin to bend or bow or even break depending on your weight and/or the length of the board between the blocks. Let's imagine that it does not break and only bends. Clearly the board is "pushing" back at you. If you cannot imagine this, think of what would happen if you suddenly jumped off of the board; surely it would "snap back" into it's original flat state; hence, there was some force pushing back in the direction of where you were standing.

    So the answer to your question is twofold: the force that pushes back comes from internal forces (i.e. the wood "trying not to break") and it acts via the surface of the wood where your feet are interacting with the board.
     
  7. Jun 26, 2010 #6

    DaveC426913

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    I had never thought about it that way until just this moment.

    We were always taught that Newton's Law applies to standing on the ground in that you push down on the Earth and it pushes back, but I never really grokked what it meant for the Earth to "push back". (How can a passive, solid object psuh back?)

    But your beam of wood shows it perfectly. The beam of wood really does push back; it yields and resists, because of its own internal structure (this becomes more obvious of you turn the wood vertically and simply push on it - it pushes back). And, of course, so does the Earth.

    Moreover, it then raises the question: well, what if the ground or wood did not push back with equal and opposite force?

    To which the answer is: then it is not ground or wood; it is something unresisting, like, say water. And then you push down on the water, it does not push back, so you sink!

    Thank you Samurai, a childhood mystery I can finally put to bed.
     
  8. Jun 26, 2010 #7

    russ_watters

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    Yes, water does push back with an equal and opposite force. What you're missing is that you won't be pushing on the water with a force equal to your weight.

    Forces always come in equal and opposite pairs.
     
  9. Jun 26, 2010 #8
    Hmmm... Much confusion has arisen due to this statement Russ :smile: You are right; I mean, Newton's third is Newton's third and makes no reservations as to what kind of object is experiencing a given force. But let's talk about this some more. If I were suspended such that my feet are just about to touch the water and then I was released from rest, what exactly happens? (Of course I sink, but in terms of the physics.)

    At the instant I touch the water, there are two forces acting on me: My weight and something else pushing back. What exactly is that something else? It is my natural instinct to say that it must be a buoyant force, but then I immediately reject that thought since at the instant my feet touch, there is no water displaced. So I feel like it must a normal force like that of the wood in my previous example. But then the water "breaks" as a result of something. That something is probably surface tension and is maybe analogous to the wood "bending moment."

    But I would love to get a real description of what is going on here.

    ~Casey
     
  10. Jun 26, 2010 #9

    DaveC426913

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    Correct me if I'm wrong but forces being balanced results in an object that is not accelerating. The fact that the object is accelerating (you are sinking into the water) is the indicator that the forces are not balanced.

    Forces will only come into balance again once you reach bouyant equilibrium, or you touch bottom.
     
  11. Jun 26, 2010 #10
    This is what confused me with Russ' post. Clearly there is an acceleration and thus an imbalance of forces. I thought that maybe by going through my thought experiment I could isolate the root of my confusion.
     
  12. Jun 26, 2010 #11

    D H

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    Dave, you and Russ are talking at cross purposes here. You are talking about the forces acting on one object. If the net force on some object is not zero it is necessarily undergoing an acceleration. That's Newton's second law. Russ is talking about Newton's third law. Newton's third law addresses the nature of the forces between two objects. If object A applies a force to object B, then object B is applying an equal but opposite force to object A.
     
  13. Jun 26, 2010 #12
    Cross purposes, maybe. But they are certainly related. What I am asserting, is that like the wood, the water exerts a reaction force that is equal and opposite. However, unlike the wood, the water allows an acceleration to occur. What I (and presumably Dave) was hoping to ascertain was exactly wherein lies the difference between the two.

    Presumably it is in the tendency of the "stuff" in question (wood or water) to stay together. In water I believe this is the property surface tension and in the wood, tensile strength.
     
  14. Jun 26, 2010 #13
    If a force, F, is exerted on a body. It is equally correct to say that the force -F was exerted on the forcer. It is not correct to say that there is a net force of 0 though.
     
  15. Jun 26, 2010 #14
    If the reaction - force pair is the only pair, then what say you?
     
  16. Jun 26, 2010 #15

    D H

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    Simple: Both objects are accelerating.

    Newton's third law is widely misunderstood. If object A is exerting a force on object B, object B is exerting an equal but opposite force on object A. Those equal-but-opposite forces necessarily act on the two different bodies. That the Earth is pulling you downwards via gravity and pushing you upwards due to the normal force is not an example of Newton's third law. Those are not equal but opposite forces, for one thing. More importantly, both forces (gravitation and the normal force) are acting on one body (you). Finally, the nature of the forces are very different. The third law reaction to the Earth pulling you downwards gravitationally is that you are pulling the Earth upwards gravitationally. The third law reaction to the Earth pushing you upwards via the normal force is you pushing the Earth downwards via the normal force.
     
  17. Jun 26, 2010 #16
    Splendid! I hate to say this, but I knew that! I feel like such a chump right now :redface:
     
  18. Jun 26, 2010 #17

    russ_watters

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    Perhaps, but the way Dave's post is worded implies that he thinks that you can apply (for example) 100N to the water while it applies 10N back at you.
     
  19. Jun 26, 2010 #18

    russ_watters

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    If you are suspended a few mm over the water and released, your weight is pulling you down (which is a force pair between you and the Earth), but there is no other external force acting on you: your inertia and acceleration provide the reaction force via f=ma.

    When you hit the water, the water provides very little resistance, so if you weigh 600N, perhaps 550 goes toward your acceleration and you apply 50N to the water.

    That 50N is applied directly to the water under your feet, accelerating it down, but also sideways around your feet due to friction and viscosity. It's the equivalent of aerodynamic drag, but in the water.
     
  20. Jun 28, 2010 #19
    Does that mean if an object is falling through the air with a weight of 10N, then the object is pulling the earth up with a force of 10N?
     
  21. Jun 28, 2010 #20

    Doc Al

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    Yes.
     
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