If I understand correctly, I am “pushing” down onto the Earth with a force (a.k.a. weight). Likewise, the Earth is pushing up on me with a force of the same magnitude but opposite in direction.
Let’s talk about the first force mentioned above. If I am exerting a downward force onto the Earth, then this force must be equal to mass x acceleration, or
F = ma
where F is the force I am exerting onto the Earth (N), m is my mass (kg), and a is my acceleration toward the Earth (ms[sup]-2[/sup]).
Here is my question: how “real” is the acceleration in the above formula? Am I really accelerating toward the Earth at 9.8 ms[sup]-2[/sup]?
On the one hand, the acceleration in the above formula can’t be zero, else it would mean F = 0. We know F can’t be zero, since a bathroom scale indeed measures this force.
On the other hand, it’s pretty obvious I am not accelerating toward the Earth – I am at rest.
So what am I missing here? Am I – or am I not – accelerating toward the Earth?
You’re not. There is a gravitational force between you and the earth proportional to the product of your mass and the Earth’s and inversely proportional to the square of the distance between you and the Earth’s center of masses. F = GMm/d[sup]2[/sup] If you were not touching the Earth, you would be accelerating towards it with an acceleration proportional to that force and inversely proportional to your mass (F/m = a). The Earth would be accelerating towards you as well, but with its much bigger mass, the acceleration would be much smaller.
But there is also another force from the earth pushing on you (and you pushing on it). This is really an electromagnetic force of the electrons on the edges of your body pushing the electrons on the edge of the earth. This force is exactly the same magnitude as the other It’s what keeps you from falling into the earth and vice versa.
You had it right at the beginning. There are multiple forces acting on you. Gravity is pulling you downward. The force pushing you up is called the normal force. (That’s what a scale measures, how much normal force the scale applies to the bottom of your feet.)
So, you add up all the individual forces to get a combined net force. In this case, it’s zero (the normal force is clever like that). Put that net force in F=ma and you get:
0=ma
And so your acceleration is zero, as you figured it would be.
I understand the net force on me is zero, and thus my net acceleration must also be zero. But I am wondering about each component force, i.e. the force I am exerting on the Earth, and the force the Earth is exerting on me. These component forces are real, aren’t they? If they are real, does it mean the acceleration associated with each component force is also real?
Clearly not, since you can’t accelerate in multiple directions at once. Only net force causes net acceleration.
It’s not just “gravity” vs “normal force” that is being netted. The gravitational fields of every individual atom of the earth is pulling you in a slightly different direction, but the horizontal components of those pulls all average to zero to make a net pull of “down”. Likewise the effects of the electromagnetic fields of the individual molecules under your feet all net to the “normal force”.
I suppose if you wanted to associate an acceleration with each force that was acting on you, calculate the accelerations, and add them up to get a net acceleration, the math would still work. In the case you describe, they’d add up to 0.
But, I’ve never seen it done that way, and I can’t see any benefit to it. You wouldn’t experience these multiple component accelerations on your body in any way. They wouldn’t be measurable as distinct effects in any way. You can measure the speed of an object, and observe how that speed changes over time. Acceleration is a real, observable property, not just an accounting trick.
So, if you think of Physics as an elaborate framework for doing math, then you can probably get away with having each force produce its own acceleration. If you think of Physics as corresponding to the real world, then you do it the way I described; multiple forces acting on you, add them up, and there’s one acceleration corresponding to the sum of the forces.
I know exactly what you’re talking about. It doesn’t change my answer. A solid body (that wants to stay solid) experiences only one acceleration corresponding to the sum of the forces acting on it.
ETA: I suppose the closest thing that might apply is something like the forces that create tides. Some parts of the Earth are closer to the Moon than others, and thus the Moon’s gravity exerts a stronger pull on those parts. And the result is to slightly distort the Earth (which we observe as high and low tides). But, that analysis doesn’t consider the Earth as a single solid body, but as a collection of lots of smaller parts (down to the molecular level, if you want to take it that far). Again the answer is the same, each of those parts has only one acceleration associated with it.
IANAP, but I believe the two forces add to zero, but the two accelerations do not; my acceleration is much higher than the Earth’s acceleration (in my diagram, I believe a[sub]me[/sub] >> a[sub]Earth[/sub].)
It has always been my understanding that F is always equal to ma, regardless of whether it is a net force or a component force. If this is true, then I would think each acceleration associated with each component force must also be a very real acceleration.
There are two forces acting on you, gravity (pulling you toward the Earth) and the normal force (pushing you away from the Earth). Add those forces together, the sum is zero. Your acceleration is zero.
There are two forces acting on the Earth, gravity (pulling it toward you) and the normal force (pushing it away from you). Add those forces together, the sum is zero. The Earth’s acceleration is zero.
Now, if you want to remove the normal force, we can do that. Pretend you’ve just jumped out of an airplane. Ignore air resistance for the moment. Gravity is pulling you downward, and there’s nothing pushing you up, so you accelerate downward.
F is the same in both equations. So, assuming the Earth is much more massive than you are, you’ll accelerate toward the Earth a lot faster than the Earth will accelerate toward you.
Acceleration is, by definition, the rate of change of the velocity of an object. If something is moving at a constant velocity (even if it’s a constant 0) then its acceleration is 0. End of story.
Instead of starting with F and m as known quantities, and trying to figure out what a equals, let’s try something else. Your body has a certain mass that can be measured. Similarly, your acceleration can be measured. If you’re not moving, and a second later you’re still not moving, your acceleration is 0. Now, with m and a known, calculate F.
It’s easy to remove the normal force and show that small letter g = 9.8 m/s^2 is an acceleration and not “just” a convenient constant. Remove whatever supports you, such as the floor, and you’ll accelerate at g. The normal force doesn’t have the “same” reality in most situations, it’s simply the result of the floor or ground not deforming any further after you’ve placed your weight on it.
And so it’s a whole lot harder to remove g while keeping the normal force, but let me give you these two scenarios.
a) You’re standing on a floor. The Earth works on you with a force G, the floor with the opposite force N. The magnitude of N = mg. Is that g a real acceleration? It’s not possible to isolate it from the situation.
b) You’re standing on a rocket platform with thrust exactly balancing the force of gravity on you and the platform. This will be indistinguishable to you from scenario a, but here we can separate N from G. We turn the whole platform 90 degrees and you’re now accelerating horizontally at g.
Your total acceleration though is the vector sum of g horizontally and g vertically, just as it was when one of them was -g vertically and g vertically. Which accelerations do you consider real?
Forces due to fields (electric,magnetic, gravitational , etc) do not always produce acceleration. If they are producing zero acceleration, they are producing pressure… If they are producing their full acceleration, they are producing no pressure. The truth is that all force such as “a man pushes on a chair with 1000 Newtons of force” is really creating a pressure… if he pulls, its tension , not pressure ? well thats a negative pressure… Well its also useful to consider force as it is confusing to consider pressure… what if its in freefall ? no pressure, so whats gravity doing ? acting on each part of the object just the same as any other, so no internal pressure created…
Real forces, as evidenced by increased pressure at various places, may cancel and result in zero acceleration … Just consider your butt after sitting on a chair too long, if pressure was zero due to zero acceleration ?
So you can see, that where you say “F=ma”, its really saying “if there is acceleration, the net external force is given by that”… Its not talking about ALL forces (eg not internal or cancelling forces, as evidenced by pressure ), just the net external force… as in “a body remains at rest, or at constant speed, unless acted upon by a net external force”…
If you accept that F=m A, then there is no distinction at all between saying only the net force (of zero) has an acceleration associated with it, and saying each of the forces have their own acceleration and the accelerations sum to zero. I don’t mean, gee, look, the numbers work out the same. I’m saying the supposed distinction is at its deepest level of meaning an empty distinction, and you can go wherever the logic can take you from either of these two ways of thinking about it.
However, it’s not true that F = m A. That is an approximation that, on human scales, works very precisely. It is more true that f = dt(m V). That is, force equals the time derivative of the product of mass and velocity. It’s the rate of change of momentum. This is how Newton stated it in the first place, and he even pointed out that we don’t know whether m is independent of the other things, though it appears to be in experiments. Einstein and others demonstrated that m is specifically not independent of the other thing.
A more insightful description of the situation of standing on Earth would be that the space around you is accelerating downward, and to keep the distance between you and the ground you’re standing on constant, you have to have some upward force (which comes from the electrostatics et cetera). In other words, the forces are NOT in balance.
Einstein described an experiment involving being in a box, an elevator, and being unable to tell the difference between “the box is accelerating upward” and “there’s a gravitational attractor below the box” because there’s no distinction between these either.
That’s where you’re going wrong. What you said about F=ma just isn’t true. The F on the left hand side means “resultant of ALL the forces on the body”.
Other people have mentioned things about pressure, and about special relativistic corrections to Newton’s Laws, that have nothing whatsoever to do with the problems you’re having, and serve only to confuse the issue.
There’s another important mistake you’re making - you’re putting forces on different bodies into the same instance of F=ma. Don’t do that, it’s simply wrong.
If you are considering the motion of someone standing on the Earth, the forces that go on the LHS of Newton’s 2nd are, (1) the gravitational attraction from the Earth on the someone, and (2) the normal contact force from the Earth’s surface on that someone.
That someone’s influence on the Earth DOES NOT APPEAR in Newton’s second law if we are applying it to find the acceleration of the person.
Whenever you apply Newton’s 2nd Law you must keep these two principles in mind:
What body are you applying it to? You cannot apply it sensibly until you have made a decision on this. Once you have, ‘a’ is the acceleration of that body, m is the mass of that body and F is the resultant force on that body.
What forces are acting on the body? ALL of them must be combined on the left hand side. You can’t pick and choose which ones you want to put it, except in so far as you can perhaps say some are negligible for a particular calculation (e.g. air resistance for a lead ball dropped a couple of meters).
To be clear, this is what I meant by your ‘second mistake’. Read again what you’ve written about each of F, m and a. Your ‘F’ simply isn’t what should go into the equation, given that you have decided (from your m and a) that you are talking about your own motion.
F above ISN’T ‘the force I am exerting onto the Earth’. It’s the resultant of two forces FROM the Earth ON you.
Sorry for the triple post, but…
Your diagram is confused, and hence confusing. You’ve drawn two forces but they are not acting on the same body. Now you’re quite right that IF the only force acting on YOU was the earth’s gravity, THEN you would be accelerating downwards at 9.8 ms^-2. But that isn’t the only force acting on you. The earth’s surface is pushing you up. Net force on you: zero. Net acceleration of you:zero.
Your diagram ought to show four forces, and it ought to be clear about which body is experiencing which force:
The Earth’s gravity acting on you: arrow downwards, centred on you.
Your gravity acting on the Earth: arrow upwards, centred on the Earth.
The normal contact force of the Earth’s surface on you: arrow upwards, centred on you
The normal contact force from you, on the Earth’s surface: arrow downwards, centred on the Earth.
Forces (1) and (2) are equal and opposite, by Newton’s Third Law, but since each acts on a different object you DON’T put them both in N2. Ditto for the pair of forces (3) and (4).
The pair of forces that DOES go into N2 is (1)+(3) if we’re considering your (lack of) motion, and (2)+(4) if we’re considering the Earth’s.
Right, and one can see this by imagining being in a space ship that is accelerating with an acceleration of g. According to the narrative in the OP, to an outside observer this is now “real” acceleration because one can observe the velocity of the space ship increasing. But the astronaut inside perceives himself to be at rest with respect to the ship, just exactly as he would perceive it if the ship was on the surface of the earth. He couldn’t tell the difference, and Einstein said there was no difference.
But I have a question. If you measured the acceleration at the “down” side of the ship, then measured it at the “up” side, ISTM that you could tell you were in a gravity well if the “up” side acceleration was slightly lower. Indeed, if you were close to a very dense object that was small relative to its mass, the gravitational difference would be quite large and would exert significant tidal force. But if you were being accelerated by rocket thrust, wouldn’t the acceleration be uniform throughout?
I think I know the answer to my own question. The gravitational gradient is merely an uninteresting technicality. You’ll measure different gradients for different objects depending on their particular size and mass, but it doesn’t change the fact that gravity is the same fundamental force everywhere. The acceleration inside a space ship being powered by the thrust of a rocket engine is just a special case of exactly the same force where the gradient just happens to be zero, like the gravity well of an object of infinite size. IOW, you can tell the difference between rocket thrust and being in the gravity well of any real-world physical object, but it doesn’t change anything.