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Physics C‑class Mid‑importance

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Incorrect Physics

The article claims that the variations in the acceleration due to the gravity of the earth are due to so-called centrifugal forces. This is a poor explanation. I believe a better one would be:

One reason involves the difference in the distance from the centre of the Earth between the two positions due to the equatorial bulge – this leads to a variation in the gravitational field strength. The equator is further away from the centre of the Earth than the poles leading to a difference of about 0.05 m s–2. The second reason is due to the rotation of the Earth. The person on the equator experiences a centripetal acceleration. Given that the scales read the normal (or reaction) force N, in this case N = m(g–ac). Therefore there is a slight reduction (of the order of the first effect) —Preceding unsigned comment added by 132.181.7.1 (talk • contribs)

Further problems. The article states that weight is not the force on the object, but the reaction force ( from Newton's third law). However, the wikipedia article on weight does not define force this way! There is more than one way to define weight, and this article has chosen one, whereas the wikipedia article states that the other way is the usual way.

Another problem. There is massive confusion here about whether the force is a scalar or a vector. The force and the acceleration are both vectors. The direction might not always be stated, but that does not alter the fact that they are vectors. Very confusing. — Preceding unsigned comment added by 24.250.167.131 (talk) 00:09, 21 January 2014 (UTC)

While we're about it, the article quite correctly states that G is not a force but an acceleration. Could we please give some though to this having higher prominence? Flanker235 (talk) 11:24, 20 June 2014 (UTC)

g divided by time

Shouldn't this article include the difference of g force divided by time?

There are great difference of experiencing 10 g's for a millisecond and for ten seconds. A person have experienced and survived over 46 g's in a certain amount of time, but it is lethal to experience for example 25 g's over a minute. This difference may not be understandable in this article.

Something definitely needed here, particularly in reference to G onset. There is a prevailing thought, particularly among motorsport enthusiasts, that huge deceleration is survivable because of the short timeframe, as in the Kenny Bräck example. Were this true, we would not need crush zones in cars. The shorter the time the loading is experienced for, the more damaging the impact. Flanker235 (talk) 09:21, 10 June 2014 (UTC)

Roller coasters

I wondered about the paragraph on amusement rides, where it is said that they usually don't pull over 3 g with some listed exceptions. However, according to "rcdb.com" and other coaster-related sources, almost every looping coaster on the world pulls about 4-5 g on entering the loop (e.g. the Vekoma Boomerang which is found in many parks around the world is said to pull 5.2 g on its first inversion).—Preceding unsigned comment added by 141.203.254.65 (talk • contribs)

In traffic

At some typical force level examples for starting and stopping at red lights in traffic. Jidanni (talk) 02:03, 7 December 2013 (UTC)

Please include a section for non-scientists

just a few sentences that the non-science majors can use. — Preceding unsigned comment added by 70.189.173.196 (talk) 22:23, 20 August 2014 (UTC)

I've simplified the lede a bit. It's been trying not to say the obvious, which is that all the forces we're talking about, are mechanical contact forces between objects. They propagate through the object as mechanical stresses. These forces then cause what we call "g-force acceleration." This is why g-forces always (without exception) produce mechanical stress. This is because of their origin from a contact-point. Electromagnetic and gravitational forces, which act on all parts of an object at once, don't produce any g-force (g-force acceleration), and don't produce any mechanical stress. No matter how much you tow an object with gravity, it always experiences zero-g, which means zero g-force. Presumably the same would be true of a uniformly-charged object or uniformly magnetic object uniformly levitated or moved by a magnetic or electric field. Objects floating by buoyancy experience surface contact forces (as in a water bed). A frog floating in water would feel a g-force just as a person does in a water-bed. But a frog levitated in a strong magnetic field should feel something very close to zero-g, and no g-force acceleration. SBHarris 00:43, 21 August 2014 (UTC)

No, we are not talking about mechanical forces between objects here; those are impact (mechanics) or mechanical shock. An authoritative resource is MIL-STD-810 sect 516.6,para 2.1.1.1 which talks about shock being short duration where its frequency is near the natural frequency of the object. Thus a high speed fighter jet making a sharp turn produces g-forces. Also a racing car produces g-forces during a quick start. Also a roller-coaster produces g-forces. An automobile hitting a brick wall produces a shock. A dropped package produces a shock. A dropped glass of water produces an impact or shock. Shocks are also measured in g-s which is part of the confusion. g-forces relate to the perceived effects of longer accelerations. Let's work on the opening paragraph to clarify this. Pkgx (talk) 22:37, 6 February 2015 (UTC)

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