# If $F=ma$, how can we experience both gravity and a normal force even though we are not accelerating?

As I sit in my chair, I experience a gravitational force pushing me into the chair and I'm also experiencing the normal force of the chair pushing back at me so I don't fall. According to Newton's Laws, $F=ma$ and I understand that gravitational acceleration near Earth is $-9.8\: \mathrm{m/s^2}$ so the normal force is $9.8\: \mathrm{m/s^2}$ times my mass.

What I don't understand is that if acceleration is change in velocity and my velocity is not changing (thus acceleration is zero), how is there a force?

• There is no net force. The downward force on you from gravity is balanced by the upward normal force from the Earth. No net force means no net acceleration. Feb 5 '14 at 6:37
• Say some one to remove the chair from the back, you will get to know whether you are accelerated or not! Feb 5 '14 at 10:34
• Posiible duplicate of: physics.stackexchange.com/q/97802 Feb 5 '14 at 11:11

You just need to be careful about the distinction between certain individual interactions (forces), and the net force on your body.

Newton's Second Law demands that the net force on your body is your mass times your acceleration. Your acceleration is zero when you're sitting still on Earth because the net force on your body is zero; the gravitational force pulling you downward balances the normal force of the ground pushing you upward. This does not mean that you can't feel the the normal force itself.

You will be able to feel any such contact force, even if the total force on your body is zero.

• In determining the net force on a body, don't we need to look at the individual forces and see what the sum of the vectors is? And are the individual forces still calculated with F=ma? Feb 5 '14 at 8:04
• @nonex Yes. In determining the net force we compute the vector sum of the individual forces. The individual forces are not calculated with $\mathbf F = m\mathbf a$. This is actually a crucial and important point. The force due to gravity, for example, is determined by Newton's Law of Gravitation. Near the surface of the Earth, it is approximately true that this force is $mg$ where $g$ is the acceleration due to gravity near the Earth's surface, but that fact is determined by making an approximation in the Law of Gravitation, not by using Newton's Second Law. Feb 5 '14 at 8:08
• If I stand on earth, I think we need to consider to cases, one that is discussed in the above answer, i.e gravitational force on me is equal to normal force on me thus my acceleration is zero. In the other case, I am exerting force on very very small mass of the earth, thus that very very small mass of earth exerts force on me. Here I am accelerated with 'a' which when multiplied with my mass will be equal to mass of the earth on which I exerted force multiplied by acceleration to which it was accelerated by me. Here though I was accelerated, I didn't move because my mass is sufficient...... Feb 5 '14 at 11:42
• .....to over come it. Some times we exert force on very huge mass, here it is accelerated but not sufficient to move. I think in the answer, acceleration of you is considered to be zero considering the first case, but taking into account second case you are actually accelerated. I didn't post this as answer because I was not sure. If anything wrong here I will be happy to learn. Feb 5 '14 at 11:50
• So is gravity a special case when it comes to calculating force on an object? And say if we're floating weightless in space, then any force we feel actually IS a product of mass and acceleration, in other words the velocity is actually changing and there's an actual acceleration? Feb 5 '14 at 17:06

Josh's answer is of course correct, but let me take a different perspective. According to general relativity you are accelerating and that's why you feel a force.

The maths behind this is described in twistor59's excellent answer to What is the weight equation through general relativity?. When you're at rest on the Earth's surface your acceleration is approximately:

$$a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$

where $r$ is the radius of the Earth and $M$ is the mass of the Earth.

Let me justify this outrageous claim. One way we could tell whether we are accelerating is whether we feel a force. When you're floating along in space then if you feel no force you can be sure you're not accelerating. Well, this is true for Newtonian physics but not for General Relativity. An astronaut in the International Space Station is weightless and feels no force, but from our perspective here on the Earch that astronaut is accelerating towards the Earth. We know they must be accelerating because they are travelling in a circular orbit round the Earth, and circular motion implies an acceleration towards the centre of the circle.

The point is that in GR acceleration is relative, and the acceleration measured depends on the observer. So if I'm trying to determine your acceleration when you're standing on the Earth's surface the answer I get depends on the observer I choose. Well the obvious observer to choose is one that feels no force, because feeling no force is the obvious reference point for zero acceleration.

So to determine your acceleration I compare your motion with that of a freely falling observer who is (momentarily) next to you. For example if you're standing at the edge of a cliff that observer would be falling off the cliff next to you. You're going to say the observer is accelerating downwards (presumably to an imminent and messy death!) but the observer is going to say that you are accelerating upwards. What's more, the observer is going to say that because they are feeling no force but you do feel a force, they can be sure it's you who is doing the accelerating.

This is how the equation above is derived. It compares the four-acceleration of someone at a fixed distance from a planet with the four-acceleration of a freely falling observer. It's actually just the acceleration calculated from Newton's law of gravity with a (usually) small relativistic correction.

At this point you're going to say that all this is very interesting (at least I hope so) but isn't it a bit silly to claim you're accelerating when you're obviously standing still? Well, the claim does neatly explain why you feel a force, and what's more it calculates exactly what that force is.

• I am wrestling with this question at the moment. Let's say, 'Fine', I accept your ludicrous GR notion that the acceleration depends on the observer. My free-falling observer claims that the guy standing on the cliff is accelerating at 9.8 m s-2 in a direction away from the center of the Earth. So when an objective observer measures both the falling observer and the static observed on the cliff, how come the distance for the observed object is static (i.e. the distance from the center of the Earth to the guy on the cliff doesn't change even though he's accelerating away from it)?
– user32023
Jul 2 '16 at 23:15
• Jul 3 '16 at 12:05

Technically you are correct in that you shouldn't be feeling a force (under the purview of classical mechanics) if you're not under acceleration (i.e. net force of a free body diagram on you is shown to be zero)

And that would be true if you were a rigid body. But you're not!

In reality, your skin that's in contact with the chair you're sitting on is suffering deformation.

Let's assess a case of an elastic sheet under load on a surface

          ↓ weight of load
--------------------------
↑ F
↓ F
--------------------------
↑ normal reaction


where F is the force exerted by the compressed elastic on both the surfaces.

Value of F can be calculated by Hooke's law which states that $$stress = k * strain$$ where k is the elastic modulus of the material.

There has been some research on the mechanical properties of skin, but I'm not sure if it can be approximated to an ideal elastic material.

Nonetheless, we can agree that skin does under go some deformation, and there are nerves in your body that feel this and send the signal to your brain.

Effectively, your receptors aren't even telling you that you're feeling a normal force or a gravitational force, that's your own inference. All it tells you is that an external pressure has deformed your skin.

And this is why you feel these forces even though you aren't under acceleration while an ideal rigid body may not.