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According to the Newton's third law when a body is placed on the floor the force of gravity acts downwards: the reaction force also acts so that it cancels its effect, so that the net force is zero. But why should we apply the force equal to its weight to displace it upwards?

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  • $\begingroup$ Welcome to Physics SE. This is very basic question, it is not clear for me why it is downvoted. $\endgroup$ – jaromrax Apr 25 '17 at 16:26
  • $\begingroup$ Try to apply the law : You have a free space without any forces and a body. Once you start to apply a Force on the body, the body will have to accelerate. If you - at some moment apply the same amount of Force in the opposite direction, the body will not stop - but stop to accelerate. Now your turn.... $\endgroup$ – jaromrax Apr 25 '17 at 16:34
  • $\begingroup$ @jaromrax Apparently it was downvoted exactly because it is a very basic question. $\endgroup$ – gented Apr 25 '17 at 21:21
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Abstract: It's an approximation.

Imagine what would happen if the force experienced by the body from the floor (the one you called reaction force) was not equal to the weight of the body. By Newton's Second Law: $$\vec{F}=m\vec{a}$$ Where $\vec{F}$ is the vector sum of all forces.

There are two possible scenarios. Imagine if the reaction force was greater than the weight force. Then their sum would add up to an upwards force. But by Newton's Second Law, you would have an upwards acceleration. Does anything resting on your floor look like it's accelerating upwards?

Now imagine if, instead, the reaction force was smaller. Then you would have a downwards net force, and again a downwards acceleration. That would mean the body would start sinking into the floor.

The latter is the case when the weight is just too much for the floor to handle. The object "sinks" into the floor because the floor itself breaks down. This is a particularly violent case, but it could happen. You see, the magnitude of the reaction force can "counterbalance" every weight up to a certain point. From that point on, the floor can't bear the downwards force: it's going to crumble under it.

Now that you've got all this set up, we can think about what it means to have motion on a body. Now, you can't go from rest ($\vec{v} = 0$) to anything at the snap of a finger. You need to go smoothly from a value to another, and you do that only when the variation with respect to time is a finite number. If you know calculus, I'm talking about finite derivatives. What does this mean? A body cannot, in principle, be able to go from a value of its velocity to another without some form of acceleration. And we already know that acceleration means force. When in any exercise you read "apply a force equal to its weight to displace", you're actually reading "the net force is zero". This means there is no acceleration. Read carefully: no acceleration does not imply no velocity. In fact, no acceleration implies constant velocity.

Often in exercises we simplify our model by saying "the object rests on the floor. (read: $\vec{v} = 0$) We apply a force equal to its weight to lift it up with a constant velocity." There has been a really really tiny moment where the lifting force was greater, so that velocity could change from rest to whatever velocity the exercise states; an instant later we diminish the magnitude of the lifting force, so that the net force is again zero. We just ignore that tiny moment.

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  • $\begingroup$ The last three paragraphs are unnecessary. Also, why would you say it is an approximation? It isn't. $\endgroup$ – gented Apr 25 '17 at 21:28
  • $\begingroup$ It is, since we're not taking into account that tiny fraction of time where the lifting force is greater in magnitude. Saying that "it starts from rest and then moves with constant velocity" surely is an approximation, for velocity - in Newtonian mechanics - only relies on continuous functions. Also, the user is asking why do we need a perfectly counterbalancing force to move the body upwards, which is what I explain in those paragraphs. I believe you should read the question again. $\endgroup$ – Niki Di Giano Apr 25 '17 at 23:44
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    $\begingroup$ I do not understand why you are assuming anything on the initial conditions: they can be whatever they are, but this is outside the scope of the question. Nowhere in the question is mentioned anything about the velocity or the motion, but only on the equilibrium condition. $\endgroup$ – gented Apr 26 '17 at 8:20
  • $\begingroup$ I'm just trying to target the problem. OP is probably puzzled by why you apply the same exact force on a body at rest and on a body that "displaces upward". It surely looks like a thought out of an exercise. I know I al assuming a lot: I'm taking a guess at the problem. It is a problem I've been asking myself in my early days in Physics. The one I wrote is the answer I got back then. Also, please read the question again. It clearly states "displace upwards". $\endgroup$ – Niki Di Giano Apr 26 '17 at 8:28
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The answer is physics!

Okay, that's not very helpful, but it is accurate. And your question is a very reasonable one. What's so magical about these reaction forces that let them perfectly equal the force of gravity? The answer is that they aren't perfect, but the imperfections are so small that you don't care about them!

In reality, the key force here is the electrostatic force. Electrons in the bottom of the body push away from electrons in the floor. Well, really the electrons push away from every other electron, but the ones in the bottom of the body and top of the floor are interesting to us here. If we assume an object is "above the floor," it's going to start accelerating downward under the force of gravity. As it gets closer to the floor, the repulsion between the electrons in the body and the floor increase. This effect happens to be very pronounced on small scales (picometers and nanometers). On larger scales than that, you don't see much of the repulsion, for various reasons.

So what actually happens is that the object moves into this small distance from the floor where it starts being worth our time to pay attention to these effects. The body will approach the floor until this repulsive force exactly counteracts gravity.

If the body's heavier, it will theoretically get "closer" to the floor than a light object, but for the kind of scales you are looking at, the difference is unimportant. The difference of 0.0000000001m just isn't a big deal for a lot of problems, so we can ignore it. There's also going to be some fancy slowing effects as the object approaches its equilibrium and starts to get noticeably affected by the electrostatics, but realistically this happens so ungodly fast that you don't worry about the 0.0000000000001 seconds it takes to stabilize.

You and I really don't want to have to think about all of these electrostatic forces, because they'd be a computational nightmare and it wouldn't really add anything. So instead, we lump them all together as a "normal force," which is the name we give for a reactionary force caused by one hard object pushing against another. We note that, for the scales you and I care about, it's effective to just say "the normal force is exactly equal to the force of the object pushing down on it." In reality, there's another 3 physics courses worth of stuff going on, but you don't gain any fidelity by knowing it!

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If we have a thousand newton dumbbell on the surface of the earth, then the earth should supply a thousand newton upward force for support so that the object stays in equilibrium:

$$\sum F_y=1000-1000=0$$

Now let's say you pull the object up with a force of $F_2$ then the new sum of the forces at the instant you start pulling is,

$$\sum F=F_2=ma$$

The acceleration is $a=\frac{F_2}{m}$ right? So the object accelerates up and so your able to lift it right?

Only there is a catch, when the object accelerates up: it eventually looses contact of the support force from the ground so immediately after it looses contact of the ground the actual $\sum F$ is:

$$\sum F=F_2-1000=ma$$

From which we see that in order for the the object to keep accelerating up we need $a>0$. This is only possible if

$$F_2-1000>0$$

Which implies,

$$F_2>1000$$

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