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So if I have any big object with a mass let's say 100 kg on the floor, why can't I with a force of 1 N not push that object upwards? If we add all the forces of the object (see picture below)

enter image description here

What's wrong in my thinking, is the gravitational force somehow getting bigger to match the difference in forces? All help is appreciated (and I'm sorry if I posted this in the wrong place)

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  • $\begingroup$ I'm not clever enough to answer it myself, but I personally think and answer that accounts for the work done in compressing the floor (and the box) by the effects of gravity and the two masses ('earth' being the other) would give the OP a deeper (yet still incomplete!) understanding... Imagining everything as completely rigid objects can only get you so far before intuition and observation lead to quandaries like this. Yes/no ? $\endgroup$ – Lamar Latrell Jun 2 '17 at 23:28
  • $\begingroup$ All the normal force does is act to determine that your object isn't going to drill through the ground. It's not constant and will change based on outside forces, i.e. will decrease if you try to lift up the object. $\endgroup$ – Goldname Jun 19 '17 at 16:25
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The gravitational force would not be getting bigger, but the upward normal force would be getting smaller. You can try this out by putting an object on a scale, and then pull up on it a bit. The object won't be lifted, but the reading on the scale will go down.

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    $\begingroup$ So the "space" which the normal force is shrunken by is taken up by the force i push up? $\endgroup$ – Iram Haque Jun 2 '17 at 17:09
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    $\begingroup$ Yes, exactly. It might help to think of the normal force as a reaction. It only pushes up as much as it needs to. Imagine the object starts 1 m above the table, with your 1 N force and gravity. Clearly, it would accelerate downward toward the table. When it reaches the table and stops, we know the forces are in equilibrium (because there is no acceleration). Then, starting with gravity and your 1 N force, we find that the normal force MUST be (W - 1 N), which is 1 N less than when the object is only under the influence of gravity. $\endgroup$ – Pawr Jun 2 '17 at 17:16
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    $\begingroup$ @KevinWells No, the normal force is by definition the force of the table or whatever solid object is pushing back as a response to a force acting on some other object. It's technically provided by the electromagnetic forces between the atoms. The force you provide from your hand is not part of the normal force; it's something else. (I want to say it's a frictional force, since you're applying pressure on the sides and not in the direction of the gravity.) $\endgroup$ – jpmc26 Jun 2 '17 at 22:22
  • $\begingroup$ @jpmc26 If you're calling the normal force electromagnetic there are only at most four types of forces - fewer at ridiculously high temperatures. $\endgroup$ – wizzwizz4 Jun 3 '17 at 16:45
  • $\begingroup$ @wizzwizz4 That's not what I'm saying. I'm only saying that's what the normal force in the classical model traces down to in our more modern model. I'm well aware of the fact we now say there are 4 fundamental forces. $\endgroup$ – jpmc26 Jun 3 '17 at 16:54
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Your equations assume the reaction force of the floor with the box is constant. However, what should happen is $R=Mg-F$ where R is the reaction force, Mg is the weight of the box and F is the force you exert, I.e. R decreases as you exert an upwards force but the box still remains in contact with the floor. In order to lift the box of the floor, your force F has to excceed Mg such that $R=0$.

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Stop using counterforces to calculate problems.

This is a warning I cannot stress enough. If you want to calculate problems, choose the one object you are interested in and draw which forces are active on this object. In your case this means

either:

  • the body with 100 kg mass. Gravitation forces the body down. Now you can calculate if the body crushes thrcough the floor or how compressed the body will be if you know its elastic modulus, what force is necessary to overcome friction and so on.

or

  • the floor. It is compressed by a body of 100 kg mass. So you can calculate the pressure if you know how much surface area the body has or you can calculate the bending of the floor caused by the mass and so on.

Never both.

As you already found out, you only get garbage solutions out of it because you hide which problem you are actually solving: Are you trying to solve the movement of the body or the movement of the floor? The lifting force you are using does not act on both body and floor and therefore you get confused by nonsensical answers. What you can do is solving each problem from every available viewpoint (body/floor) to see how the forces interact, but never mix forces with counterforces for solutions.

ADDENDUM:
I see that the problem is still not clear.

Let's imagine the 100 kg body is a human. Let's take two of them, one of him is put on the ISS, the other one stays on earth. Both are staying still.

If the drawing would be fine, then it is no difference between both bodies: Both have balanced forces, the one on the ISS having 0 N, the other one having 100 daN (~ 1 kg) in opposite directions, each canceling the other out.

So they should feel no difference if the drawing is right? But there is a difference, on Earth we feel pulled down. Our bones are compressed. The floor on which we are standing is bent and is depressed around our feet.

If the forces are really canceled out, then the originator of the question should be right: Applying 1 daN upward should move the body upward. But it does not: Mysteriously the "counter force" is now decreased by 1 daN. What causes exactly the strange "counter force"? Hm, the force is caused by the compressed molecules under our feet. Why are the molecules are compressed? Because the weight of our body...

The real problem is that the drawing shows the forces as two arrows, two separate entities when in fact for Earth only one is really active, gravitation. This single active (!) force cause all the effects we are observing, all things like bone compression and floor depression are simply reactive forces in response and do not exist independently from the main force. For this reason we should not include "counter forces" in our drawings and calculations as separate entities. It causes the following well observed paradoxons:

  • Unmoveable or freely moveable objects: If there is always a counterforce to a force, things could never move. Or, if counterforce is always equal to a force, things are always balanced and could be moved with ease (see above).

  • Counterforce chaining: Human does not move, human pushes floor down, floor pushes human up. Floor does not move, floor pushes human up, so floor itself must push down. House does not move, floor pushes down, so house must push up. And so on. It is not exactly wrong, but completely confounding and more or less superfluous.

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    $\begingroup$ this seems like high school physics, so how he incoperated the forces in his drawing was fine. He is doing exactly what you said: choose one object and draw which forces are active on this object (which are the normal and gravitational force, they are both active on the object described). The object is experiencing both the normal and gravititational force, if it would only experience one of them like you describe then the object would accelerate up or down. He didn't ever draw forces from the floor, he just didn't understand that normal force is a counter force of gravitational force $\endgroup$ – Thomas W Jun 3 '17 at 10:30
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There are two things wrong with your thinking. You are assuming that the normal force provided by the ground is "always there" and that it does not change.
Ignoring the "how" for now, let's just take the case where the mass is not touching the ground. Clearly, the normal force from the ground goes to zero and disappears!
At what point does this case happens? when the force pulling the mass up = mg.
What happens prior to that? As the force pulling up increases, the normal force from the ground changes from fn = mg to fn = 0. This shows that the mass will not move until the lifting force is larger than mg.

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So let's have a feel of this scene! Actually the floor is applying the force on the object to make it still(to balance its weight). Now, when you're trying to move it in vertical direction then you have to the "work"(not that which has a form of energy) what floor is doing and some extra force because you made the block to leave the touch. I hope you got the " feel" !

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