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Re: [Phys-l] g...



But there is no "k" needed. In what my 1940 text calls the static systim,
1 pound of force measure the deflection of a standard spring. 1 lb accelerates a certain mass with an acceleration of 1 ft/s^2; we call that amount of mass 1 slug. 1 slug has a weight of ~32 ft/s^2. G has a value of about3.5x10^{-8}#(ft/slug)^2. F=ma in ordinary English units, no "k" needed.

I can make the arbitrariness more apparent by distinguishing between inertial and gravitational quantities. We choos a system of units, m-k-s,
where G=1. We do this by distinguishing gravitational from inertial mass. In an obvious notation, m_g=m_i/sqrt(G); recall that sqrt(G)~.8x10^{-5}.
Then the gravitational force between two masses, m and M is just
F=(m_gM_g)/r^_{2). All kinematics goes on as before, except that whenever dealing with gravitation, we must use gravitational masses. The "g" ambiguity is now clarified, there are two values of g. There is a g_i, with the usual value, for calculating how objects accelerate in free fall.
There is g_g =g_isqrt(G) for calculating the forces on gravitational masses. The numbers that plug into the student's calculators are, of course, the same as always.
You don't want to do it that way? It's a free country, so it's your (arbitrary) choice.
Regards,
Jack





On Mon, 20 Nov 2006, Ludwik Kowalski wrote:

On Nov 20, 2006, at 12:19 PM, Jack Uretsky wrote:

. . . Defining a system of units where a constant is equal to unit,
does
not make it non-arbitrary. The "arbitrary constants" are buried in the
choice of units. There are consistent sets of English units, which
used
to be taught, where the arbitrary constant in N2 is also unity.
Ludwik's
argument is, I think, circular, because the SI (and many other systems
of
units) aredefined so that there is no "k" in N2. The K_g (along with
K-e
and K-m) is incorporated into the Newtonian constant G.

The SI and CGS define their force units in terms of units of mass and
acceleration. One dyne, for example, was assigned to a net force
producing the acceleration of 1 cm/s^2 when m=1 gram. Likewise, F=1 N
is defined in terms of 1 m/s^2 and 1 kg. This "artificial trick" makes
k=1 dimensionless. Was a minor computational convenience responsible
for conceptual difficulties of students?

I was thinking about a situation in which units of m, a and F are
defined independently of each other, for example, 1 lb is a force
needed to deform a standard spring by a specified distance. (Our
kilogram is still defined in terms of an arbitrary-chosen standard
reference.) Then the N2 law would have to be F=k*m*a, as I described in
the message posted last night. I was not thinking about lb defined in
terms ft/s2, etc. Shortcuts can be costly; taking a shortcut one might
be lost. There is a Polish proverb about this.

Ludwik Kowalski
Let the perfect not be the enemy of the good.
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