Re: Newton's Second Law of Motion is not even wrong.


slavek krepelka wrote:
> Dear PD,
>
> The force exerted by a mass which has mass value of 1 kg is weight of
> that mass, when acted upon by earth gravity. The equivalent value unit
> of 1 g is used as the unit of acceleration. When a body with mass of 1
> kg is accelerated so, that the force it exerts on whatever is pushing,
> or pulling it is equivalent to force it would weight on earth.
>
> Get it into your thick head. You are mixing weight and mass. Mass is
> always constant, as opposed to weight, which is subject to differences
> in the strength of gravitational field (lets say moon having 1/6 of the
> attractive force of earth), or the value of the rate of acceleration,
> therefore relative.
>
> That is why an astronaut in a space shuttle in a stable orbit weights
> nothing. The weight caused by earth gravitation is directionally
> countered by the weight caused by the centrifugal force, i.e.
> acceleration, while his mass remains unchanged at lets say 75 kg.
>
> Dip your nose into some elementary physics textbook or at least Webster
> (that is a standard American dictionary available even in some
> supermarkets) and lay off.
>
> Got it? Slavek

Well, since you doubt my understanding of elementary physics, let me
offer some feedback.

When looking at the 2nd law, F_net = m*a, do not confuse the left-hand
side with the right hand side. The left-hand side is for forces, the
right-hand side is for the effects of those forces. The weight is the
weight; it is a force, and it is not affected by acceleration.

Let's take an example of you riding down an elevator that is
accelerating downwards at 0.8 m/s^2. How does this acceleration arise?
It must be the *result* of forces acting on you. By inspection, there
are two forces acting on you: the force of gravity (downwards) and the
contact force of the floor of the elevator on your feet (upwards).
Let's blindly guess that your mass is 70 kg.

The right-hand side (taking up to be positive) is then (70 kg)(-0.8
m/s^2).

The cause of that acceleration is the sum of the forces, which is, on
the left-hand side
-(70 kg)(9.8 m/s^2) + F_floor = F_net.

You'll note in this sum that the first term, the weight, is just what
you would have expected for your weight if you were standing still on
the ground. We'll see that we do not need it to decrease. Furthermore,
since the 9.8 comes from G*M(earth)/R(earth)^2, we do not expect that
number to decrease just because you're going down in an elevator.
Moreover, your mass does not change simply because you're going down in
an elevator. Therefore, your weight, (70 kg)(9.8 m/s^2), is not
expected to change just because you're going down in an elevator.

Equating these two, we have
-(70 kg)(9.8 m/s^2) + F_floor = (70 kg)(-0.8 m/s^2)

Now we can do some algebra to find out what the contact force of the
floor on your feet is:
F_floor = (70 kg)(9.8 m/s^2) - (70 kg)(0.8 m/s^2) = (70 kg)(9.0 m/s^2)
= 630 N.

This contact force is somewhat less than your weight, 686 N. This is
illuminating, because it reveals that what we sense when the elevator
accelerates downward is not the drop in our weight, but the drop in the
contact force between the floor and our feet. In equilibrium, we're
accustomed to feeling our weight and that contact force being equal --
but this isn't an equilibrium situation.

Now, Slavek, this analysis is *exactly* the way that it would be
presented in an elementary physics text, and it is *entirely*
consistent with Newton's 2nd law. However, your poor understanding of
*apparent* weight has gotten in the way of your grip on the basic
principles here.

PD

>
> PD wrote:
> >
> > slavek krepelka wrote:
> > > PD wrote:
> > > >
> > > > slavek krepelka wrote:
> > > > > I have decided to show to whom ever may be interested:
> > > > >
> > > > > 1) that this law as stated by "F=ma" is down right wrong and incorrectly
> > > > > justified
> > > > > 2) where has the originator failed to correctly comprehend and interpret
> > > > > the standard experimental evidence
> > > > > 3) the correct relationship between force, mass and acceleration easily
> > > > > derived from static pressure in liquids and expressed by F=1/2ma^2.
> > > > >
> > > > > http://www3.sympatico.ca/slavek.krepelka/ttf2/force1.htm
> > > > >
> > > > > My kind regards, Slavek
> > > >
> > > > Your criticism is based on erroneous assumptions and incorrect
> > > > understanding of the term "weightlessness".
> > > >
> > > > You say for example:
> > > > ================
> > > > "Everybody" knows that the acceleration in the free fall is a balancing
> > > > act between inertia of a mass and gravitation. If there is no
> > > > acceleration of a body toward ground, the body weights the full
> > > > equivalent of its mass, i.e. its full weight, in our case 1 kg. When
> > > > the body is accelerating at the free fall rate toward ground, it is
> > > > weightless. When a body is hindered in its free fall by a force smaller
> > > > than the gravitational force and opposing the gravitational force, its
> > > > weight is in some proportion to the strength of the opposing force.
> > > > =================
> > > > This analysis is in fact incorrect, and it is not true that "everybody"
> > > > knows this. Indeed, I would say that only onebody know it.
> > > >
> > > > PD
> > >
> > > It appears that you have never taken a ride in a fast elevator. It might
> > > help your understanding of effects of acceleration on weight to take a
> > > once in a lifetime trip from your village and visit some larger urban
> > > center. This will give you and unprecedented opportunity to test ride a
> > > fast elevator in some high-rise building and directly observe what
> > > happens to weight with accelerating as well as decelerating of the
> > > elevator, up as well as down. You'll be truly amazed. Then talk to me
> > > about weight again if you still feel like it.
> > >
> >
> > On the contrary, Slavek. I've already discussed with you what goes on
> > in a fast elevator, and pointed out that what changes is not your
> > weight but the contact force between your feet and the floor. You
> > commented that this view of weight was peculiar, even though this is
> > precisely what is taught in textbooks and what is consistent with
> > Newton's 2nd law.
> >
> > PD

.

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