Terminal velocity and g-forces

Parthalan

Hi,

I need a little help with my understanding of g-forces. From what I can gather, this "force" is the acceleration experienced by the object expressed as a multiple of g (the usual acceleration due to gravity).

What has confused me, though, is the Wikipedia page, which says that "A sky-diver in a stable free-fall experiences 1g (full weight) after reaching terminal velocity". At terminal velocity, isn't the net force zero, and the velocity constant (therefore no acceleration)? Wouldn't that mean that the g-force is 0g? The article also states that an object in free fall (which is falling at 1g relative to the earth) experiences 0g, or weightlessness. Similarly, what is the g-force of an object which is at rest (relative to the earth, I guess)?

It's a very basic physics course, so the stuff about inertial reference frames goes a bit beyond our level of comprehension. Can someone please help me get this straight? I *think* I'm considering this relative to the earth, when they're considering it as relative to something else. Do we need to consider forces aside from the regular gravitational acceleration somehow?

Thanks

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LURCH

Hint; as yuo sit at your 'puter, you experience 1g of acceleration. If yu were sitting in a room that is in deep space (no nearby objects), and that room were accelerating in the direction of the cieling, yuo would exprience the same g-force. Think about where that g-force would be experienced by you, and what object in the room would be exerting that force against you. That should start your brain in the direction Wiki was talking about.

rcgldr

Homework Helper
What has confused me, though, is the Wikipedia page, which says that "A sky-diver in a stable free-fall experiences 1g (full weight) after reaching terminal velocity". At terminal velocity, isn't the net force zero, and the velocity constant (therefore no acceleration)? Wouldn't that mean that the g-force is 0g?
If the net force is zero, it simply means there's no acceleration. This doesn't mean that there aren't opposing forces that a person can feel. If you're standing on the ground, gravity is pulling down on you throughout your body, and the ground is pushing up against your feet. In the case of a sky-diver at terminal velocity, gravity is pulling down through the sky-divers body, and aerodynamic drag is pushing+pulling upwards on the skydiver. These are opposing forces that can be felt via compression or tension.

In the case of true free-fall, such as bouncing on a trampoline, or in an amusement park ride or in an aircraft creating a "zero g" environment, then there is downwards acceleration due to gravity, but there is no feeling of force since all components of the object (or person) are being accelerated at the same rate (so there's no compression or tension to "feel").

D H

Staff Emeritus
The term "g force" almost always refers to apparent weight divided by actual weight, not total force divided by actual weight. Actual weight is simply the force acting on an object due to gravity alone. Apparent weight is the total force acting on the object less the actual weight. An object with a constant velocity (e.g., a skydiver who has reached terminal velocity or a person standing on the surface of the earth) has zero total force acting on it. The apparent weight of the object is not zero: It is equal in magnitude to the actual weight. The g-force is thus one in the case of a non-accelerating object.

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