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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?

sammy gerbil
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nonex
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3 Answers3

14

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.

joshphysics
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8

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.

Community
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John Rennie
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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.

Peeyush Kushwaha
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