| Topic: |
Science > Physics |
| User: |
"Don1" |
| Date: |
05 Sep 2005 06:29:16 AM |
| Object: |
Weight varies with the rate of free fall |
Weight varies because the rate of free fall is not constant: They both
vary so that the ratio of a body's weight, divided by the rate at which
it will free fall is a constant: Anyplace; anytime.
Don
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| User: "Clemens W" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 08:15:24 AM |
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Don, producer of logical fallacies, wrote:
Weight varies because the rate of free fall is not constant
No. Wrong direction. You reversed cause and effect.
Your turn,
A. Friend
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| User: "Herman Trivilino" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 02:56:00 PM |
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"Clemens W" <a.friend@bigfoot.com> wrote ...
Weight varies because the rate of free fall is not constant
No. Wrong direction. You reversed cause and effect.
Not neccessarily, he didn't. The weight force is, by one (officially
sanctioned)definition, the net force required to make an object accelerate
at a rate equal to the LOCAL free fall acceleration. Therfore, if the local
free fall acceleration varies the weight force necessarily varies.
Some people adopt other definitions of the weight force, but I've never seen
an author of an introductory physics textbook do it in a way that I would
consider rational. If the weight force is defined, instead, as the
gravitational force then the ratio of the weight force to the mass gives a
value higher than the value used by those same authors!
Then, we end up getting in a discussion of "true weight" versus "apparent
weight". Such authors will then claim that what they had been calling the
weight all along was really the apparent weight. Apparently, the word takes
on different meanings depending on their present need!
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| User: "Gib Bogle" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 03:45:46 PM |
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Herman Trivilino wrote:
"Clemens W" <a.friend@bigfoot.com> wrote ...
Weight varies because the rate of free fall is not constant
No. Wrong direction. You reversed cause and effect.
Not neccessarily, he didn't. The weight force is, by one (officially
sanctioned)definition, the net force required to make an object accelerate
at a rate equal to the LOCAL free fall acceleration. Therfore, if the local
free fall acceleration varies the weight force necessarily varies.
The "rate of free fall" is a velocity, not an acceleration.
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| User: "tadchem" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 04:22:43 PM |
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"Gib Bogle" <bogle@ihug.too.much.spam.co.nz> wrote in message
news:dfialb$344$2@lust.ihug.co.nz...
<snip repost>
The "rate of free fall" is a velocity, not an acceleration.
In 'free fall' the velocity is always changing, the acceleration is not:
http://scienceworld.wolfram.com/physics/FreeFall.html
These equations only apply in a region in which the gravitation is
approximately constant - i.e. one's measurements are not precise enough to
reveal the variability.
Physicists often consider what happens in regions in which gravitation
varies significantly from place to place, such as in the fields of study
called 'celestial mechanics' and 'cosmology'. For this reason, the term
'free fall' is rarely used by physicists. It is mostly used by people who
have only a passing acquaintance with physics.
Tom Davidson
Richmond, VA
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| User: "Don1" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 04:17:26 PM |
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Gib Bogle wrote:
Herman Trivilino wrote:
"Clemens W" <a.friend@bigfoot.com> wrote ...
Weight varies because the rate of free fall is not constant
No. Wrong direction. You reversed cause and effect.
Not neccessarily, he didn't. The weight force is, by one (officially
sanctioned)definition, the net force required to make an object accelerate
at a rate equal to the LOCAL free fall acceleration. Therfore, if the local
free fall acceleration varies the weight force necessarily varies.
The "rate of free fall" is a velocity, not an acceleration.
The rate of free fall is the rate of _change_ in velocity:
Mathematically (vt-vi)/t = 2s/t^2 = g/2; at Earth's surface it is about
16.1'/sec^2.
Don
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| User: "Herman Trivilino" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 08:40:42 PM |
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"Don1" <dcshead@charter.net> wrote ...
The "rate of free fall" is a velocity, not an acceleration.
The rate of free fall is the rate of _change_ in velocity:
Mathematically (vt-vi)/t = 2s/t^2 = g/2; at Earth's surface it is about
16.1'/sec^2.
The velocity of a freely falling body changes by about 32.2 ft/s each
second. That's written as 32.2 ft/sē, NOT 16.1 ft/sē.
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| User: "Don1" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 10:01:49 PM |
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Herman Trivilino wrote:
"Don1" <dcshead@charter.net> wrote ...
The "rate of free fall" is a velocity, not an acceleration.
The rate of free fall is the rate of _change_ in velocity:
Mathematically (vt-vi)/t = 2s/t^2 = g/2; at Earth's surface it is about
16.1'/sec^2.
The velocity of a freely falling body changes by about 32.2 ft/s each
second. That's written as 32.2 ft/s2, NOT 16.1 ft/s2.
Yeah Herman: g=32.2'/sec^2; g/2=16.1'/sec^2.
Don
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| User: "Clemens W" |
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| Title: Re: Weight varies with the rate of free fall |
06 Sep 2005 05:55:02 AM |
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Don maxed out the bogometer again:
Mathematically (vt-vi)/t = 2s/t^2 = g/2;
No. Bogus. Your math is abominable.
DON'S MATHEMATICAL BLUNDERS, ERRORS and FALLACIES
Date Don Correct
05/09 (vt-vi)/t=2s/t^2=g/2 (vt-vi)/t=2s/t^2=g
27/08 ft/(vt-vi)=ft^2/s ft/(vt-vi)=ft^2/2s
23/08 a/2-16'/sec^2 g/2=16ft/sec^2
23/08 a-(vt-vi)/t a=(vt-vi)/t
12/06 g=(vt-vi)/t^2 g=(vt-vi)/t
12/06 (m)=ft/s/t m=Ft^2/2s
07/06 (m)=wa/fg m=F/a or m=w/g
28/05 1 slug = 32 ft sec^2/32 ft 1 slug = 32 lbf sec^2/32 ft
27/05 s=2(vt-vi)/t s=1/2(vt-vi)*t
21/05 1 slug = 1 lbf s^2/foot; _Not_ 1 slug = 1 lbf / (1 ft/s^2)
Don didn't even realize both equations are the same.
THE FOUR BASIC EQUATIONS DON WILL CONTINUOUSLY ABUSE...
BUT NEVER LEARN!
1. v = a * t
2. s = 0.5 * a * t^2
3. F = dp/dt = m*dv/dt = m*a
4. 1/(x/y) = y/x
Go back to school, Don!
A. Friend
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| User: "gwh" |
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| Title: Re: Weight varies with the rate of free fall |
06 Sep 2005 05:43:28 PM |
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I worked in the field of celestial mechanics for years, and to my best
recollection, it was universally accepted that a freely falling mass
has no weight. Maybe you folks (except Uncle Al, who, in my opinion,
has it right) define weight differently these days. If so, I can only
say, ain't semantics wonnerful?
wh
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| User: "Herman Trivilino" |
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| Title: Re: Weight varies with the rate of free fall |
06 Sep 2005 06:06:16 PM |
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"gwh" <ghughes@cei.net> wrote ...
I worked in the field of celestial mechanics for years, and to my best
recollection, it was universally accepted that a freely falling mass
has no weight. Maybe you folks (except Uncle Al, who, in my opinion,
has it right) define weight differently these days. If so, I can only
say, ain't semantics wonnerful?
Definitions vary, of course. I'm a proponent of the following definition of
the weight force. It's simply the net force required to make an object
accelerate at a rate equal to the LOCAL free fall acceleration.
So, for example, in a freely falling frame of reference you test the
acceleration of an object. It has zero acceleration in that LOCAL frame of
reference. The net force needed to make an object accelerate at this rate
is, of course, zero. Thus the weight force is, by definition, zero.
This is a condition that can be (very nearly) experienced in an orbiting
space station, and is referred to as weightlessness.
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| User: "tadchem" |
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| Title: Re: Weight varies with the rate of free fall |
06 Sep 2005 07:37:48 PM |
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"Herman Trivilino" <physhead@kingwoodREMOVECAPScable.com> wrote in message
news:1126048127_12695@spool6-east.superfeed.net...
<snip>
Definitions vary, of course. I'm a proponent of the following definition
of
the weight force.
I have always been a proponent of effective communication.
You must learn the definitions understood by the person you are trying to
communicate with and then use those definitions.
Once you get the lines of communication open and well-established you may
choose to quibble over semantic niceties. In my experience, however, unless
those common definitions are down in black and white, a lot of pointless and
misunderstood arguments over apple/oranges will ensue.
'Buenos dias' won't get you very far in Reykjavik.
Tom Davidson
Richmond, VA
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| User: "Dik T. Winter" |
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| Title: Re: Weight varies with the rate of free fall |
06 Sep 2005 09:28:06 PM |
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In article <GcqdnZ2dnZ34GRy0nZ2dnZWqg96dnZ2dRVn-yJ2dnZ0@comcast.com> "tadchem" <tadchemNOSPAM@comcast.net> writes:
....
'Buenos dias' won't get you very far in Reykjavik.
You may be surprised.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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| User: "Nick" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 05:49:05 PM |
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Don weight has no rate; not for gravity at least but for its
equivalent of accelerated motion it does.
Nothing moves in gravity's weight therefore there
is no rate to be applied!
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| User: "Uncle Al" |
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| Title: Re: Weight varies with the rate of free fall |
05 Sep 2005 06:42:41 PM |
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Don1 wrote:
Weight varies because the rate of free fall is not constant: They both
vary so that the ratio of a body's weight, divided by the rate at which
it will free fall is a constant: Anyplace; anytime.
Hey Dumb Donny *****, free fall is Minkowski space - no weight at
all for anything. Idiot.
Hey Dumb Donny *****, free fall is acceleration.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "G=EMC^2 Glazier" |
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| Title: Re: Weight varies with the rate of free fall |
06 Sep 2005 06:05:54 PM |
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Hi Uncle Al You tell em right. Add that the gravitational force creates
the rate of fall. Moon's gravity force 6 times weaker than
Earth's means if you drop a ball 3 feet up to match the speed it hits
the Earth's surface,you would have to drop it from a height of 18 feet
up on the Moon. Tricky thing happens when you let the ball go 18 feet
up it appears to start falling in slow motion. Best to keep in mind your
"Mass" stays the same on the Moon. Bert
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