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I was doing some circular motion questions, and in every example to calculate a person's apparent weight/force that they experience, you find the normal force. For example when you're at the top of a Ferris wheel, you experience less weight, as $F_N = mv^2 / r - F_g$. Or, if the earth is spinning fast enough, you'll feel weightless when FN is $0$. It's not because the net force is 0, but because the normal force is $0$.

Another example is when a pilot does a loop. At the top of the loop, the force they experience is FN, and not Fg + FN. This particular example especially confuses me, as I feel like the pilot should definitely be experiencing both the normal force from the seat as well as gravity. Why is the force that we feel the normal force, and not the net force? I can't wrap my head around it. I found this thread, but it's not a very intuitive answer for me.

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The force of gravity acts throughout our body and doesn't stretch or compress us.

When we stand on the ground, the normal force from the ground acts at one place, our feet and so we feel it as it tends to try to compress us.

In this example the net force is zero, but we experience the force on our feet.

If you were in a lift moving downwards that suddenly stopped, you'd feel heavier. The force of gravity has been constant, it's the force at our feet that has changed and increased.

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    $\begingroup$ Gravity does compress us, just a tiny bit. In my opinion this is much linked to physiology. An overweight person feels it, than is quite difficult to say from where the feel comes of. What is certain is that the normal force is applied to the feet or whatever the contact point. $\endgroup$
    – Alchimista
    Nov 15 '21 at 11:42
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    $\begingroup$ @Alchimista Yes, the normal force is applied to the feet. Gravity may seem to compress us, but only due to it causing the normal force when we stand on something. On it's own gravity isn't felt. In space for example, orbiting earth, an overweight person wouldn't feel the effects of gravity and might like the sensation of being freed from the things that normally press on them, their feet or back, for example, whilst laying down. $\endgroup$ Nov 15 '21 at 11:50
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    $\begingroup$ yes clear. It seems becoming semantic. But it is not that the ground lifts to compress us towards an immobile head. That is also linked to the above comment. I can apply weight to the head, or to the com. $\endgroup$
    – Alchimista
    Nov 15 '21 at 11:55
  • $\begingroup$ By the way, I suddenly started to feel force more than 5 minutes ago :)) because focusing on my sensation :)) indeed most of it comes from the feet.,but we knew it. $\endgroup$
    – Alchimista
    Nov 15 '21 at 12:02
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    $\begingroup$ As others have pointed out, I think there is a semantic issue. In physics we sometime say that an object feels a force if it is subject to the force. In your problems, feel means feel. We are subject to the force of gravity; it pulls us down. But it's the normal force that is causing the skin on our feet to compress sending nerve signals to the brain which we interpret as feel. $\endgroup$
    – garyp
    Nov 15 '21 at 12:37
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Why is the force that we feel the normal force, and not the net force?

This is part of the key insight that led to the development of general relativity (GR). Einstein called it his “happiest thought”. Today it is known as the equivalence principle.

For a little background, we are all introduced to physics with Newton’s laws. What is covered only much later is that they only apply in inertial reference frames. In accelerated, rotating, or otherwise non-inertial frames, we have to introduce so-called fictitious forces. Fictitious forces cannot be directly measured or detected by any means. They can only be inferred by their effects in motion in the non-inertial frame, and they can be made to go away simply by transforming to an inertial frame.

It turns out that gravity is a special case. Uniform gravity is a fictitious force, only tidal gravity is a real force. Uniform gravity, like any fictitious force, is completely unmeasurable. Any time you think you are measuring uniform gravity, you are actually measuring something else, usually the normal force as you discovered. So in problems like this, where the distances are too small for tidal gravity, gravity is a fictitious force in the sense that it has no measurable effect. It causes no measurable anything, it only is used to explain the motion of objects.

For example, an object on a scale registers a 100 N force. So there is a measurable 100 N real force pointing upward. If we were in an inertial frame, like a space station in deep space, then they would be accelerating as they go by us, like if the scale were on the floor of a passing rocket. But if they are observed not to be accelerating, like with us on the ground observing the scale also on the ground, then we infer the presence of a fictitious force pointing down. This fictitious force has no directly measurable effect and is only inferred from the fact that the object is not accelerating.

This is the equivalence principle. Only tidal gravity is real (it is spacetime curvature). Locally gravity is a fictitious force that has no measurable effects at distances too small for tidal gravity to matter

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  • $\begingroup$ I can imagine a scale exerting a normal force on an object, but under Newton's third law, this is entirely equivalent to the object exerting a downward force on the scale. Since those two forces are just two sides of the same coin, I'm really having a hard time seeing how this does not count as "measuring" the force due to gravity. Compare and contrast holding a spring-scale in one's hand and spinning it around (to measure the fictitious centrifugal force, which is just the action to oppose the reaction of tension in the spring). $\endgroup$
    – Kevin
    Nov 16 '21 at 18:06
  • $\begingroup$ @Kevin the downward force which is the 3rd law pair for the normal force is also a normal force. It is not a gravitational force, even if though points downward. If you measure that force you are still not measuring gravity. $\endgroup$
    – Dale
    Nov 16 '21 at 18:50
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When we "experience" a force like gravity, it actually is a very distributed process. It stretches and distorts every part of the body just a little. We have strain detectors build into our body to detect this.

However, these are only differential measurements. It's a difference in forces that causes us to stretch like this. Think of a spring. If you hold a spring such that the forces are identical on both ends, it won't stretch.

If we are in 'free fall,' the forces on all of cells is the same, so there's no stretch, no feeling of force. We need some cells to have different forces on them than others before we sense weight.

This could happen any number of ways, but in the situations you are thinking of, the differential in force is that there is some normal force pushing on us somewhere. This is what creates the differential squishing and straining that our body eventually detects.

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At the top or bottom of a loop, the net force (which determines the acceleration) does include gravity. The normal force may change.

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