brain sizes: Einstein's and women's

Parse Tree parsetree at
Tue Jul 23 22:35:21 EST 2002

"Cary Kittrell" <cary at> wrote in message
news:ahkjf2$rvp$1 at
> In article  "Parse Tree" <parsetree at> writes:
> <
> <"Cary Kittrell" <cary at> wrote in message
> <news:ahkcm8$og6$1 at
> <> In article  "Parse Tree" <parsetree at> writes:
> <> <
> <> <"Cary Kittrell" <cary at> wrote in message
> <> <news:ahk14h$ib0$1 at
> <> <> In article  "John Knight" <johnknight at> writes:
> <> <>
> <> <> <Sqr means square-root of the equation in the parenthesis ().
> <> <> <So, the resulting velocity would be the same, as the same time is
> <spent
> <> <on
> <> <> <the fall, and the tension would be zero.
> <> <>
> <> <> The "resulting velocity would be the same" if both masses were
> <> <experiencing
> <> <> the same acceleration the instant of release, but they were not.
> <> <> bottom mass was experiencing -2mg downards due to gravity and +2mg
> <upwards
> <> <> due to the tension in the spring.  The upper mass is experiencing a
> <> <> now unopposed -mg downwards due to gravity and a -2mg downwards due
> <> <> to the same spring tension.  You figure it out.
> <> <
> <> <But they really are experiencing the same acceleration at the instant
> <> <release.
> <>
> <> That is right as far is acceleration due to gravity is concerned,
> <> as is implied in my statement above.  But each body is experiencing
> <> additional forces due to the spring, so they will no be subject
> <> to the same accelerations.  If you mentally switch off gravity,
> <> the two bodies will move towards one another with an acceleration
> <> proportional to the 2mg tension in the spring.  If you now switch
> <> gravity back on, the whole system will accelerate downards at
> <> 1 g, but this acts equally on the whole system, so you're back
> <> to considering things in the frame of referrence of the system
> <> itself -- as Jet implied.
> <
> <This is not true.  If you switch off gravity, then each sphere will stay
> <rest.  Firstly, you're assuming tension again, and secondly, the tension
> <assume exists only because of gravity.
> <
> This suddenly grows more interesting.  I've been reading "spring"; it
> in fact says "string".

Yes, I noticed that.  But it's true even with a real spring.

> <Regardless, you can simulate this using two balls and a string.  Just put
> <them on a table and attach them with some string.  Then pull on them and
> <release.  They don't move together with a force proprotionational to how
> <much you pulled them apart.
> <
> If you assume an infinitely strong string, then you are correct. Otherwise
> they will indeed move, unless you've stretched the string inelastically.
> However, I'm being picky, and you're on to the intention of the question,
> I think.

They will move, but they won't move the same amount that they are pulled on.
Even a spring won't, unless it is perfectly elastic.

> <> <Also, you're assuming the value of the unknown.
> <> <
> <>
> <> Um, beg pardon?  Assuming the value of what unknown?  If you mean
> <> the spring tension, I simply said that the /initial/ spring
> <> tension is 2mg, because the lower mass is being pulled downwards
> <> by a force of 2mg due to gravity.  Since it isn't moving initially,
> <> there must be an equal and opposite force: 2mg of tension in the
> <
> <The initial spring tension is unknown.  You're assuming that the bottom
> <sphere is suspended from the top one.  It simply says that it's suspended
> <rest.  Which could simply mean that the system is suspended at rest.  Who
> <knows?  Actually, I find many of these questions to be very imprecise.
> <
> <Regardless, the acceleration of the system is g.  And the acceleration of
> <all of the parts are g.  Thus the string's tension should be 0.
> Assuming an infinitely strong string -- one whose relaxation is zero --
> you are correct.

Yes.  There are too many assumptions in these questions though.  I can see
why they're difficult.  There was another question about probability which
didn't even seem to specify if the two values involved were independent or

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