PD wrote:
> 
> 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

Unit "kg" is a unit of innertial mass in physics. Using it for weight is
folklore.