In article "Parse Tree" <parsetree at hotmail.com> writes:
<
<"Cary Kittrell" <cary at afone.as.arizona.edu> wrote in message
<news:ahk14h$ib0$1 at oasis.ccit.arizona.edu...
<> In article "John Knight" <johnknight at usa.com> 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. The
<> 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 of
<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.
<
<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 spring.
-- cary
